{"title": "100 doors", "language": "Python 2.5+", "task": "There are 100 doors in a row that are all initially closed. \n\nYou make 100 passes by the doors. \n\nThe first time through, visit every door and  ''toggle''  the door  (if the door is closed,  open it;   if it is open,  close it). \n\nThe second time, only visit every 2nd door   (door #2, #4, #6, ...),   and toggle it.  \n\nThe third time, visit every 3rd door   (door #3, #6, #9, ...), etc,   until you only visit the 100th door.\n\n\n;Task:\nAnswer the question:   what state are the doors in after the last pass?   Which are open, which are closed?\n\n\n'''Alternate:'''  \nAs noted in this page's   discussion page,   the only doors that remain open are those whose numbers are perfect squares.\n\nOpening only those doors is an   optimization   that may also be expressed; \nhowever, as should be obvious, this defeats the intent of comparing implementations across programming languages.\n\n", "solution": "doors = [False] * 100\nfor i in range(100):\n   for j in range(i, 100, i+1):\n       doors[j] = not doors[j]\n   print(\"Door %d:\" % (i+1), 'open' if doors[i] else 'close')\n"}
{"title": "100 doors", "language": "Python 3.x", "task": "There are 100 doors in a row that are all initially closed. \n\nYou make 100 passes by the doors. \n\nThe first time through, visit every door and  ''toggle''  the door  (if the door is closed,  open it;   if it is open,  close it). \n\nThe second time, only visit every 2nd door   (door #2, #4, #6, ...),   and toggle it.  \n\nThe third time, visit every 3rd door   (door #3, #6, #9, ...), etc,   until you only visit the 100th door.\n\n\n;Task:\nAnswer the question:   what state are the doors in after the last pass?   Which are open, which are closed?\n\n\n'''Alternate:'''  \nAs noted in this page's   discussion page,   the only doors that remain open are those whose numbers are perfect squares.\n\nOpening only those doors is an   optimization   that may also be expressed; \nhowever, as should be obvious, this defeats the intent of comparing implementations across programming languages.\n\n", "solution": "for i in range(1, 101):\n    if i**0.5 % 1:\n        state='closed'\n    else:\n        state='open'\n    print(\"Door {}:{}\".format(i, state))\n"}
{"title": "100 prisoners", "language": "Python", "task": "The Problem:\n* 100 prisoners are individually numbered 1 to 100\n* A room having a cupboard of 100 opaque drawers numbered 1 to 100, that cannot be seen from outside.\n* Cards numbered 1 to 100 are placed randomly, one to a drawer, and the drawers all closed; at the start.\n* Prisoners start outside the room\n:* They can decide some strategy before any enter the room.\n:* Prisoners enter the room one by one, can open a drawer, inspect the card number in the drawer, then close the drawer.\n:* A prisoner can open no more than 50 drawers.\n:* A prisoner tries to find his own number.\n:* A prisoner finding his own number is then held apart from the others.\n* If '''all''' 100 prisoners find their own numbers then they will all be pardoned. If ''any'' don't then ''all'' sentences stand. \n\n\n;The task:\n# Simulate several thousand instances of the game where the prisoners randomly open drawers\n# Simulate several thousand instances of the game where the prisoners use the optimal strategy mentioned in the Wikipedia article, of:\n:* First opening the drawer whose outside number is his prisoner number.\n:* If the card within has his number then he succeeds otherwise he opens the drawer with the same number as that of the revealed card. (until he opens his maximum).\n\n\nShow and compare the computed probabilities of success for the two strategies, here, on this page.\n\n\n;References:\n# The unbelievable solution to the 100 prisoner puzzle standupmaths (Video).\n# [[wp:100 prisoners problem]]\n# 100 Prisoners Escape Puzzle DataGenetics.\n# Random permutation statistics#One hundred prisoners on Wikipedia.\n\n", "solution": "import random\n\ndef play_random(n):\n    # using 0-99 instead of ranges 1-100\n    pardoned = 0\n    in_drawer = list(range(100))\n    sampler = list(range(100))\n    for _round in range(n):\n        random.shuffle(in_drawer)\n        found = False\n        for prisoner in range(100):\n            found = False\n            for reveal in random.sample(sampler, 50):\n                card = in_drawer[reveal]\n                if card == prisoner:\n                    found = True\n                    break\n            if not found:\n                break\n        if found:\n            pardoned += 1\n    return pardoned / n * 100   # %\n\ndef play_optimal(n):\n    # using 0-99 instead of ranges 1-100\n    pardoned = 0\n    in_drawer = list(range(100))\n    for _round in range(n):\n        random.shuffle(in_drawer)\n        for prisoner in range(100):\n            reveal = prisoner\n            found = False\n            for go in range(50):\n                card = in_drawer[reveal]\n                if card == prisoner:\n                    found = True\n                    break\n                reveal = card\n            if not found:\n                break\n        if found:\n            pardoned += 1\n    return pardoned / n * 100   # %\n\nif __name__ == '__main__':\n    n = 100_000\n    print(\" Simulation count:\", n)\n    print(f\" Random play wins: {play_random(n):4.1f}% of simulations\")\n    print(f\"Optimal play wins: {play_optimal(n):4.1f}% of simulations\")"}
{"title": "100 prisoners", "language": "Python 3.7", "task": "The Problem:\n* 100 prisoners are individually numbered 1 to 100\n* A room having a cupboard of 100 opaque drawers numbered 1 to 100, that cannot be seen from outside.\n* Cards numbered 1 to 100 are placed randomly, one to a drawer, and the drawers all closed; at the start.\n* Prisoners start outside the room\n:* They can decide some strategy before any enter the room.\n:* Prisoners enter the room one by one, can open a drawer, inspect the card number in the drawer, then close the drawer.\n:* A prisoner can open no more than 50 drawers.\n:* A prisoner tries to find his own number.\n:* A prisoner finding his own number is then held apart from the others.\n* If '''all''' 100 prisoners find their own numbers then they will all be pardoned. If ''any'' don't then ''all'' sentences stand. \n\n\n;The task:\n# Simulate several thousand instances of the game where the prisoners randomly open drawers\n# Simulate several thousand instances of the game where the prisoners use the optimal strategy mentioned in the Wikipedia article, of:\n:* First opening the drawer whose outside number is his prisoner number.\n:* If the card within has his number then he succeeds otherwise he opens the drawer with the same number as that of the revealed card. (until he opens his maximum).\n\n\nShow and compare the computed probabilities of success for the two strategies, here, on this page.\n\n\n;References:\n# The unbelievable solution to the 100 prisoner puzzle standupmaths (Video).\n# [[wp:100 prisoners problem]]\n# 100 Prisoners Escape Puzzle DataGenetics.\n# Random permutation statistics#One hundred prisoners on Wikipedia.\n\n", "solution": "'''100 Prisoners'''\n\nfrom random import randint, sample\n\n\n# allChainedPathsAreShort :: Int -> IO (0|1)\ndef allChainedPathsAreShort(n):\n    '''1 if none of the index-chasing cycles in a shuffled\n       sample of [1..n] cards are longer than half the\n       sample size. Otherwise, 0.\n    '''\n    limit = n // 2\n    xs = range(1, 1 + n)\n    shuffled = sample(xs, k=n)\n\n    # A cycle of boxes, drawn from a shuffled\n    # sample, which includes the given target.\n    def cycleIncluding(target):\n        boxChain = [target]\n        v = shuffled[target - 1]\n        while v != target:\n            boxChain.append(v)\n            v = shuffled[v - 1]\n        return boxChain\n\n    # Nothing if the target list is empty, or if the cycle which contains the\n    # first target is larger than half the sample size.\n    # Otherwise, just a cycle of enchained boxes containing the first target\n    # in the list, tupled with the residue of any remaining targets which\n    # fall outside that cycle.\n    def boxCycle(targets):\n        if targets:\n            boxChain = cycleIncluding(targets[0])\n            return Just((\n                difference(targets[1:])(boxChain),\n                boxChain\n            )) if limit >= len(boxChain) else Nothing()\n        else:\n            return Nothing()\n\n    # No cycles longer than half of total box count ?\n    return int(n == sum(map(len, unfoldr(boxCycle)(xs))))\n\n\n# randomTrialResult :: RandomIO (0|1) -> Int -> (0|1)\ndef randomTrialResult(coin):\n    '''1 if every one of the prisoners finds their ticket\n       in an arbitrary half of the sample. Otherwise 0.\n    '''\n    return lambda n: int(all(\n        coin(x) for x in range(1, 1 + n)\n    ))\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Two sampling techniques constrasted with 100 drawers\n       and 100 prisoners, over 100,000 trial runs.\n    '''\n    halfOfDrawers = randomRInt(0)(1)\n\n    def optimalDrawerSampling(x):\n        return allChainedPathsAreShort(x)\n\n    def randomDrawerSampling(x):\n        return randomTrialResult(halfOfDrawers)(x)\n\n    # kSamplesWithNBoxes :: Int -> Int -> String\n    def kSamplesWithNBoxes(k):\n        tests = range(1, 1 + k)\n        return lambda n: '\\n\\n' + fTable(\n            str(k) + ' tests of optimal vs random drawer-sampling ' +\n            'with ' + str(n) + ' boxes: \\n'\n        )(fName)(lambda r: '{:.2%}'.format(r))(\n            lambda f: sum(f(n) for x in tests) / k\n        )([\n            optimalDrawerSampling,\n            randomDrawerSampling,\n        ])\n\n    print(kSamplesWithNBoxes(10000)(10))\n\n    print(kSamplesWithNBoxes(10000)(100))\n\n    print(kSamplesWithNBoxes(100000)(100))\n\n\n# ------------------------DISPLAY--------------------------\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# fname :: (a -> b) -> String\ndef fName(f):\n    '''Name bound to the given function.'''\n    return f.__name__\n\n\n# ------------------------GENERIC -------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# difference :: Eq a => [a] -> [a] -> [a]\ndef difference(xs):\n    '''All elements of xs, except any also found in ys.'''\n    return lambda ys: list(set(xs) - set(ys))\n\n\n# randomRInt :: Int -> Int -> IO () -> Int\ndef randomRInt(m):\n    '''The return value of randomRInt is itself\n       a function. The returned function, whenever\n       called, yields a a new pseudo-random integer\n       in the range [m..n].\n    '''\n    return lambda n: lambda _: randint(m, n)\n\n\n# unfoldr(lambda x: Just((x, x - 1)) if 0 != x else Nothing())(10)\n# -> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]\n# unfoldr :: (b -> Maybe (a, b)) -> b -> [a]\ndef unfoldr(f):\n    '''Dual to reduce or foldr.\n       Where catamorphism reduces a list to a summary value,\n       the anamorphic unfoldr builds a list from a seed value.\n       As long as f returns Just(a, b), a is prepended to the list,\n       and the residual b is used as the argument for the next\n       application of f.\n       When f returns Nothing, the completed list is returned.\n    '''\n    def go(v):\n        xr = v, v\n        xs = []\n        while True:\n            mb = f(xr[0])\n            if mb.get('Nothing'):\n                return xs\n            else:\n                xr = mb.get('Just')\n                xs.append(xr[1])\n        return xs\n    return lambda x: go(x)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "15 puzzle game", "language": "Python 3.X", "task": "Implement the Fifteen Puzzle Game.\n\n\nThe   '''15-puzzle'''   is also known as: \n:::*   '''Fifteen Puzzle'''\n:::*   '''Gem Puzzle'''\n:::*   '''Boss Puzzle'''\n:::*   '''Game of Fifteen'''\n:::*   '''Mystic Square'''\n:::*   '''14-15 Puzzle'''\n:::*   and some others.\n\n\n;Related Tasks:\n:*   15 Puzzle Solver\n:*   [[16 Puzzle Game]]\n\n", "solution": "''' Python 3.6.5 code using Tkinter graphical user interface.''' \n\nfrom tkinter import *\nfrom tkinter import messagebox\nimport random\n\n# ************************************************\n\nclass Board:\n    def __init__(self, playable=True):\n        while True:\n            # list of text for game squares:\n            self.lot = [str(i) for i in range(1,16)] + ['']\n            if not playable:\n                break\n            # list of text for game squares randomized:\n            random.shuffle(self.lot)\n            if self.is_solvable():\n                break\n            \n        # game board is 2D array of game squares:\n        self.bd = []\n        i = 0\n        for r in range(4):\n            row = []\n            for c in range(4):\n                row.append(Square(r,c,self.lot[i]))\n                i += 1\n            self.bd.append(row)\n            \n    # How to check if an instance of a 15 puzzle\n    # is solvable is explained here:\n    # https://www.geeksforgeeks.org/check-instance-15-puzzle-solvable/\n    # I only coded for the case where N is even.\n    def is_solvable(self):\n        inv = self.get_inversions()\n        odd = self.is_odd_row()\n        if inv % 2 == 0 and odd:\n            return True\n        if inv % 2 == 1 and not odd:\n            return True\n        return False\n        \n    def get_inversions(self):\n        cnt = 0\n        for i, x in enumerate(self.lot[:-1]):\n            if x != '':\n                for y in self.lot[i+1:]:\n                    if y != '' and int(x) > int(y):\n                        cnt += 1\n        return cnt\n\n    # returns True if open square is in odd row from bottom:\n    def is_odd_row(self):\n        idx = self.lot.index('')\n        return idx in [4,5,6,7,12,13,14,15]           \n\n    # returns name, text, and button object at row & col:\n    def get_item(self, r, c):\n        return self.bd[r][c].get()\n\n    def get_square(self, r, c):\n        return self.bd[r][c]\n\n    def game_won(self):\n        goal = [str(i) for i in range(1,16)] + ['']\n        i = 0\n        for r in range(4):\n            for c in range(4):\n                nm, txt, btn = self.get_item(r,c)\n                if txt != goal[i]:\n                    return False\n                i += 1\n        return True\n                \n# ************************************************\n\nclass Square:       # ['btn00', '0', None]\n    def __init__(self, row, col, txt):\n        self.row = row\n        self.col = col\n        self.name = 'btn' + str(row) + str(col)\n        self.txt = txt\n        self.btn = None\n        \n    def get(self):\n            return [self.name, self.txt, self.btn]\n\n    def set_btn(self, btn):\n        self.btn = btn\n\n    def set_txt(self, txt):\n        self.txt = txt\n\n# ************************************************\n\nclass Game:\n    def __init__(self, gw):\n        self.window = gw\n\n        # game data:\n        self.bd = None\n        self.playable = False\n\n        # top frame:\n        self.top_fr = Frame(gw,\n                            width=600,\n                            height=100,\n                            bg='light green')\n        self.top_fr.pack(fill=X)\n\n        self.hdg = Label(self.top_fr,\n                         text='  15 PUZZLE GAME  ',\n                         font='arial 22 bold',\n                         fg='Navy Blue',\n                         bg='white')\n        self.hdg.place(relx=0.5, rely=0.4,\n                       anchor=CENTER)\n\n        self.dir = Label(self.top_fr,\n                 text=\"(Click 'New Game' to begin)\",\n                 font='arial 12 ',\n                 fg='Navy Blue',\n                 bg='light green')\n        self.dir.place(relx=0.5, rely=0.8,\n                       anchor=CENTER)\n\n        self.play_btn = Button(self.top_fr,\n                               text='New \\nGame',\n                               bd=5,\n                               bg='PaleGreen4',\n                               fg='White',\n                               font='times 12 bold',\n                               command=self.new_game)\n        self.play_btn.place(relx=0.92, rely=0.5,\n                       anchor=E)\n\n        # bottom frame:\n        self.btm_fr = Frame(gw,\n                            width=600,\n                            height=500,\n                            bg='light steel blue')\n        self.btm_fr.pack(fill=X)\n\n        # board frame:\n        self.bd_fr = Frame(self.btm_fr,\n                           width=400+2,\n                           height=400+2,\n                           relief='solid',\n                           bd=1,\n                           bg='lemon chiffon')\n        self.bd_fr.place(relx=0.5, rely=0.5,\n                         anchor=CENTER)\n\n        self.play_game()\n\n# ************************************************\n\n    def new_game(self):\n        self.playable = True\n        self.dir.config(text='(Click on a square to move it)')\n        self.play_game()\n\n    def play_game(self):\n        # place squares on board:\n        if self.playable:\n            btn_state = 'normal'\n        else:\n            btn_state = 'disable'\n        self.bd = Board(self.playable)               \n        objh = 100  # widget height\n        objw = 100  # widget width\n        objx = 0    # x-position of widget in frame\n        objy = 0    # y-position of widget in frame\n\n        for r in range(4):\n            for c in range(4):\n                nm, txt, btn = self.bd.get_item(r,c)\n                bg_color = 'RosyBrown1'\n                if txt == '':\n                    bg_color = 'White'           \n                game_btn = Button(self.bd_fr,\n                                  text=txt,\n                                  relief='solid',\n                                  bd=1,\n                                  bg=bg_color,\n                                  font='times 12 bold',\n                                  state=btn_state,\n                                  command=lambda x=nm: self.clicked(x))\n                game_btn.place(x=objx, y=objy,\n                               height=objh, width=objw)\n                \n                sq = self.bd.get_square(r,c)\n                sq.set_btn(game_btn)\n                \n                objx = objx + objw\n            objx = 0\n            objy = objy + objh\n\n    # processing when a square is clicked:\n    def clicked(self, nm):\n        r, c = int(nm[3]), int(nm[4])\n        nm_fr, txt_fr, btn_fr = self.bd.get_item(r,c)\n        \n        # cannot 'move' open square to itself:\n        if not txt_fr:\n            messagebox.showerror(\n                'Error Message',\n                'Please select \"square\" to be moved')\n            return\n\n        # 'move' square to open square if 'adjacent' to it:            \n        adjs = [(r-1,c), (r, c-1), (r, c+1), (r+1, c)]\n        for x, y in adjs:\n            if 0 <= x <= 3 and 0 <= y <= 3:\n                nm_to, txt_to, btn_to = self.bd.get_item(x,y)\n                if not txt_to:\n                    sq = self.bd.get_square(x,y)\n                    sq.set_txt(txt_fr)\n                    sq = self.bd.get_square(r,c)\n                    sq.set_txt(txt_to)\n                    btn_to.config(text=txt_fr,\n                                  bg='RosyBrown1')\n                    btn_fr.config(text=txt_to,\n                                  bg='White')\n                    # check if game is won:              \n                    if self.bd.game_won():\n                        ans = messagebox.askquestion(\n                            'You won!!!   Play again?')\n                        if ans == 'no':\n                            self.window.destroy()\n                        else:\n                            self.new_game()\n                    return\n                \n        # cannot move 'non-adjacent' square to open square:\n        messagebox.showerror(\n            'Error Message',\n            'Illigal move, Try again')\n        return\n\n# ************************************************\n\nroot = Tk()\nroot.title('15 Puzzle Game')\nroot.geometry('600x600+100+50')\nroot.resizable(False, False)\ng = Game(root)\nroot.mainloop()\n"}
{"title": "21 game", "language": "Python 2.X and 3.X", "task": "'''21''' is a two player game, the game is played by choosing \na number ('''1''', '''2''', or '''3''')  to be added to the ''running total''.\n\nThe game is won by the player whose chosen number causes the ''running total''\nto reach ''exactly'' '''21'''.\n\nThe ''running total'' starts at zero. \nOne player will be the computer.\n \nPlayers alternate supplying a number to be added to the ''running total''. \n\n\n;Task:\nWrite a computer program that will:\n::* do the prompting (or provide a button menu), \n::* check for errors and display appropriate error messages, \n::* do the additions (add a chosen number to the ''running total''), \n::* display the ''running total'', \n::* provide a mechanism for the player to quit/exit/halt/stop/close the program,\n::* issue a notification when there is a winner, and\n::* determine who goes first (maybe a random or user choice, or can be specified when the game begins). \n\n", "solution": "from random import randint\ndef start():\n\tgame_count=0\n\tprint(\"Enter q to quit at any time.\\nThe computer will choose first.\\nRunning total is now {}\".format(game_count))\n\troundno=1\n\twhile game_count<21:\n\t\tprint(\"\\nROUND {}: \\n\".format(roundno))\n\t\tt = select_count(game_count)\n\t\tgame_count = game_count+t\n\t\tprint(\"Running total is now {}\\n\".format(game_count))\n\t\tif game_count>=21:\n\t\t\tprint(\"So, commiserations, the computer has won!\")\n\t\t\treturn 0\n\t\tt = request_count()\n\t\tif not t:\n\t\t\tprint('OK,quitting the game')\n\t\t\treturn -1\n\t\tgame_count = game_count+t\n\t\tprint(\"Running total is now {}\\n\".format(game_count))\n\t\tif game_count>=21:\n\t\t\tprint(\"So, congratulations, you've won!\")\n\t\t\treturn 1\n\t\troundno+=1\n\ndef select_count(game_count):\n\t'''selects a random number if the game_count is less than 18. otherwise chooses the winning number'''\n\tif game_count<18:\n\t\tt= randint(1,3)\n\telse:\n\t\tt = 21-game_count\n\tprint(\"The computer chooses {}\".format(t))\n\treturn t\n\ndef request_count():\n\t'''request user input between 1,2 and 3. It will continue till either quit(q) or one of those numbers is requested.'''\n\tt=\"\"\n\twhile True:\n\t\ttry:\n\t\t\tt = raw_input('Your choice 1 to 3 :')\n\t\t\tif int(t) in [1,2,3]:\n\t\t\t\treturn int(t)\n\t\t\telse:\n\t\t\t\tprint(\"Out of range, try again\")\n\t\texcept:\n\t\t\tif t==\"q\":\n\t\t\t\treturn None\n\t\t\telse:\n\t\t\t\tprint(\"Invalid Entry, try again\")\n\nc=0\nm=0\nr=True\nwhile r:\n\to = start()\n\tif o==-1:\n\t\tbreak\n\telse:\n\t\tc+=1 if o==0 else 0\n\t\tm+=1 if o==1 else 0\n\tprint(\"Computer wins {0} game, human wins {1} games\".format(c,m))\n\tt = raw_input(\"Another game?(press y to continue):\")\n\tr = (t==\"y\")"}
{"title": "24 game", "language": "Python", "task": "The 24 Game tests one's mental arithmetic. \n\n\n;Task\nWrite a program that displays four digits, each from 1 --> 9 (inclusive) with repetitions allowed.\n\nThe program should prompt for the player to enter an arithmetic expression using ''just'' those, and ''all'' of those four digits, used exactly ''once'' each. The program should ''check'' then evaluate the expression. \n\nThe goal is for the player to enter an expression that (numerically) evaluates to '''24'''.\n*  Only the following operators/functions are allowed: multiplication, division, addition, subtraction\n*  Division should use floating point or rational arithmetic, etc, to preserve remainders.\n*  Brackets are allowed, if using an infix expression evaluator.\n*  Forming multiple digit numbers from the supplied digits is ''disallowed''. (So an answer of 12+12 when given 1, 2, 2, and 1 is wrong).\n*  The order of the digits when given does not have to be preserved.\n\n\n;Notes\n* The type of expression evaluator used is not mandated. An RPN evaluator is equally acceptable for example.\n* The task is not for the program to generate the expression, or test whether an expression is even possible.\n\n\n;Related tasks\n* [[24 game/Solve]]\n\n\n;Reference\n* The 24 Game on h2g2.\n\n", "solution": "'''\n The 24 Game\n\n Given any four digits in the range 1 to 9, which may have repetitions,\n Using just the +, -, *, and / operators; and the possible use of\n brackets, (), show how to make an answer of 24.\n\n An answer of \"q\" will quit the game.\n An answer of \"!\" will generate a new set of four digits.\n Otherwise you are repeatedly asked for an expression until it evaluates to 24\n\n Note: you cannot form multiple digit numbers from the supplied digits,\n so an answer of 12+12 when given 1, 2, 2, and 1 would not be allowed.\n\n'''\n\nfrom __future__ import division, print_function\nimport random, ast, re\nimport sys\n\nif sys.version_info[0] < 3: input = raw_input\n\ndef choose4():\n    'four random digits >0 as characters'\n    return [str(random.randint(1,9)) for i in range(4)]\n\ndef welcome(digits):\n    print (__doc__)\n    print (\"Your four digits: \" + ' '.join(digits))\n\ndef check(answer, digits):\n    allowed = set('() +-*/\\t'+''.join(digits))\n    ok = all(ch in allowed for ch in answer) and \\\n         all(digits.count(dig) == answer.count(dig) for dig in set(digits)) \\\n         and not re.search('\\d\\d', answer)\n    if ok:\n        try:\n            ast.parse(answer)\n        except:\n            ok = False\n    return ok\n\ndef main():    \n    digits = choose4()\n    welcome(digits)\n    trial = 0\n    answer = ''\n    chk = ans = False\n    while not (chk and ans == 24):\n        trial +=1\n        answer = input(\"Expression %i: \" % trial)\n        chk = check(answer, digits)\n        if answer.lower() == 'q':\n            break\n        if answer == '!':\n            digits = choose4()\n            print (\"New digits:\", ' '.join(digits))\n            continue\n        if not chk:\n            print (\"The input '%s' was wonky!\" % answer)\n        else:\n            ans = eval(answer)\n            print (\" = \", ans)\n            if ans == 24:\n                print (\"Thats right!\")\n    print (\"Thank you and goodbye\")   \n\nif __name__ == '__main__': main() "}
{"title": "4-rings or 4-squares puzzle", "language": "Python", "task": "Replace       '''a, b, c, d, e, f,'''   and\n  '''g '''       with the decimal\ndigits   LOW   --->   HIGH\nsuch that the sum of the letters inside of each of the four large squares add up to\nthe same sum.\n\n\n            +--------------+      +--------------+\n            |              |      |              |\n            |      a       |      |      e       |\n            |              |      |              |\n            |          +---+------+---+      +---+---------+\n            |          |   |      |   |      |   |         |\n            |          | b |      | d |      | f |         |\n            |          |   |      |   |      |   |         |\n            |          |   |      |   |      |   |         |\n            +----------+---+      +---+------+---+         |\n                       |       c      |      |      g      |\n                       |              |      |             |\n                       |              |      |             |\n                       +--------------+      +-------------+\n\n\nShow all output here.\n\n\n:*   Show all solutions for each letter being unique with\n         LOW=1     HIGH=7\n:*   Show all solutions for each letter being unique with\n         LOW=3     HIGH=9\n:*   Show only the   ''number''   of solutions when each letter can be non-unique\n         LOW=0     HIGH=9\n\n\n;Related task:\n* [[Solve the no connection puzzle]]\n\n", "solution": "def foursquares(lo,hi,unique,show):\n\n    def acd_iter():\n        \"\"\"\n        Iterates through all the possible valid values of \n        a, c, and d.\n        \n        a = c + d\n        \"\"\"\n        for c in range(lo,hi+1):\n            for d in range(lo,hi+1):\n                if (not unique) or (c <> d):\n                    a = c + d\n                    if a >= lo and a <= hi:\n                        if (not unique) or (c <> 0 and d <> 0):\n                            yield (a,c,d)\n                            \n    def ge_iter():\n        \"\"\"\n        Iterates through all the possible valid values of \n        g and e.\n        \n        g = d + e\n        \"\"\"\n        for e in range(lo,hi+1):\n            if (not unique) or (e not in (a,c,d)):\n                g = d + e\n                if g >= lo and g <= hi:\n                    if (not unique) or (g not in (a,c,d,e)):\n                        yield (g,e)\n                        \n    def bf_iter():\n        \"\"\"\n        Iterates through all the possible valid values of \n        b and f.\n        \n        b = e + f - c\n        \"\"\"\n        for f in range(lo,hi+1):\n            if (not unique) or (f not in (a,c,d,g,e)):\n                b = e + f - c\n                if b >= lo and b <= hi:\n                    if (not unique) or (b not in (a,c,d,g,e,f)):\n                        yield (b,f)\n\n    solutions = 0                    \n    acd_itr = acd_iter()              \n    for acd in acd_itr:\n        a,c,d = acd\n        ge_itr = ge_iter()\n        for ge in ge_itr:\n            g,e = ge\n            bf_itr = bf_iter()\n            for bf in bf_itr:\n                b,f = bf\n                solutions += 1\n                if show:\n                    print str((a,b,c,d,e,f,g))[1:-1]\n    if unique:\n        uorn = \"unique\"\n    else:\n        uorn = \"non-unique\"\n               \n    print str(solutions)+\" \"+uorn+\" solutions in \"+str(lo)+\" to \"+str(hi)\n    print"}
{"title": "4-rings or 4-squares puzzle", "language": "Python 3.7", "task": "Replace       '''a, b, c, d, e, f,'''   and\n  '''g '''       with the decimal\ndigits   LOW   --->   HIGH\nsuch that the sum of the letters inside of each of the four large squares add up to\nthe same sum.\n\n\n            +--------------+      +--------------+\n            |              |      |              |\n            |      a       |      |      e       |\n            |              |      |              |\n            |          +---+------+---+      +---+---------+\n            |          |   |      |   |      |   |         |\n            |          | b |      | d |      | f |         |\n            |          |   |      |   |      |   |         |\n            |          |   |      |   |      |   |         |\n            +----------+---+      +---+------+---+         |\n                       |       c      |      |      g      |\n                       |              |      |             |\n                       |              |      |             |\n                       +--------------+      +-------------+\n\n\nShow all output here.\n\n\n:*   Show all solutions for each letter being unique with\n         LOW=1     HIGH=7\n:*   Show all solutions for each letter being unique with\n         LOW=3     HIGH=9\n:*   Show only the   ''number''   of solutions when each letter can be non-unique\n         LOW=0     HIGH=9\n\n\n;Related task:\n* [[Solve the no connection puzzle]]\n\n", "solution": "'''4-rings or 4-squares puzzle'''\n\nfrom itertools import chain\n\n\n# rings :: noRepeatedDigits -> DigitList -> Lists of solutions\n# rings :: Bool -> [Int] -> [[Int]]\ndef rings(uniq):\n    '''Sets of unique or non-unique integer values\n       (drawn from the `digits` argument)\n       for each of the seven names [a..g] such that:\n       (a + b) == (b + c + d) == (d + e + f) == (f + g)\n    '''\n    def go(digits):\n        ns = sorted(digits, reverse=True)\n        h = ns[0]\n\n        # CENTRAL DIGIT :: d\n        def central(d):\n            xs = list(filter(lambda x: h >= (d + x), ns))\n\n            # LEFT NEIGHBOUR AND LEFTMOST :: c and a\n            def left(c):\n                a = c + d\n                if a > h:\n                    return []\n                else:\n                    # RIGHT NEIGHBOUR AND RIGHTMOST :: e and g\n                    def right(e):\n                        g = d + e\n                        if ((g > h) or (uniq and (g == c))):\n                            return []\n                        else:\n                            agDelta = a - g\n                            bfs = difference(ns)(\n                                [d, c, e, g, a]\n                            ) if uniq else ns\n\n                            # MID LEFT AND RIGHT :: b and f\n                            def midLeftRight(b):\n                                f = b + agDelta\n                                return [[a, b, c, d, e, f, g]] if (\n                                    (f in bfs) and (\n                                        (not uniq) or (\n                                            f not in [a, b, c, d, e, g]\n                                        )\n                                    )\n                                ) else []\n\n    # CANDIDATE DIGITS BOUND TO POSITIONS [a .. g] --------\n\n                            return concatMap(midLeftRight)(bfs)\n\n                    return concatMap(right)(\n                        difference(xs)([d, c, a]) if uniq else ns\n                    )\n\n            return concatMap(left)(\n                delete(d)(xs) if uniq else ns\n            )\n\n        return concatMap(central)(ns)\n\n    return lambda digits: go(digits) if digits else []\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Testing unique digits [1..7], [3..9] and unrestricted digits'''\n\n    print(main.__doc__ + ':\\n')\n    print(unlines(map(\n        lambda tpl: '\\nrings' + repr(tpl) + ':\\n\\n' + unlines(\n            map(repr, uncurry(rings)(*tpl))\n        ), [\n            (True, enumFromTo(1)(7)),\n            (True, enumFromTo(3)(9))\n        ]\n    )))\n    tpl = (False, enumFromTo(0)(9))\n    print(\n        '\\n\\nlen(rings' + repr(tpl) + '):\\n\\n' +\n        str(len(uncurry(rings)(*tpl)))\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).\n    '''\n    return lambda xs: list(\n        chain.from_iterable(map(f, xs))\n    )\n\n\n# delete :: Eq a => a -> [a] -> [a]\ndef delete(x):\n    '''xs with the first of any instances of x removed.'''\n    def go(xs):\n        xs.remove(x)\n        return xs\n    return lambda xs: go(list(xs)) if (\n        x in xs\n    ) else list(xs)\n\n\n#  difference :: Eq a => [a] -> [a] -> [a]\ndef difference(xs):\n    '''All elements of ys except any also found in xs'''\n    def go(ys):\n        s = set(ys)\n        return [x for x in xs if x not in s]\n    return lambda ys: go(ys)\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    '''A function over a pair of arguments,\n       derived from a vanilla or curried function.\n    '''\n    return lambda x, y: f(x)(y)\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string formed by the intercalation\n       of a list of strings with the newline character.\n    '''\n    return '\\n'.join(xs)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "99 bottles of beer", "language": "Python 3.6+", "task": "Display the complete lyrics for the song:     '''99 Bottles of Beer on the Wall'''. \n\n\n;The beer song:\nThe lyrics follow this form:\n\n\n:::              99 bottles of beer on the wall\n:::              99 bottles of beer\n:::              Take one down, pass it around\n:::              98 bottles of beer on the wall\n    \n:::              98 bottles of beer on the wall\n:::              98 bottles of beer\n:::              Take one down, pass it around\n:::              97 bottles of beer on the wall\n\n\n... and so on, until reaching   '''0'''     (zero).\n\nGrammatical support for   ''1 bottle of beer''   is optional. \n\nAs with any puzzle, try to do it in as creative/concise/comical a way \nas possible (simple, obvious solutions allowed, too).\n\n\n\n\n;See also: \n*   http://99-bottles-of-beer.net/\n*   [[:Category:99_Bottles_of_Beer]]\n*   [[:Category:Programming language families]]\n*   Wikipedia 99 bottles of beer\n\n", "solution": "\"\"\"\nExcercise of style. An overkill for the task :-D\n\n1. OOP, with abstract class and implementation with much common magic methods\n2. you can customize:\n    a. the initial number\n    b. the name of the item and its plural\n    c. the string to display when there's no more items\n    d. the normal action\n    e. the final action\n    f. the template used, for foreign languages\n3. strofas of the song are created with multiprocessing\n4. when you launch it as a script, you can specify an optional parameter for \n   the number of initial items\n\"\"\"\n\nfrom string import Template\nfrom abc import ABC, abstractmethod\nfrom multiprocessing.pool import Pool as ProcPool\nfrom functools import partial\nimport sys\n\nclass Song(ABC):\n    @abstractmethod\n    def sing(self):\n        \"\"\"\n        it must return the song as a text-like object\n        \"\"\"\n        \n        pass\n\nclass MuchItemsSomewhere(Song):\n    eq_attrs = (\n        \"initial_number\", \n        \"zero_items\", \n        \"action1\", \n        \"action2\", \n        \"item\", \n        \"items\", \n        \"strofa_tpl\"\n    )\n    \n    hash_attrs = eq_attrs\n    repr_attrs = eq_attrs\n    \n    __slots__ = eq_attrs + (\"_repr\", \"_hash\")\n    \n    def __init__(\n        self, \n        items = \"bottles of beer\", \n        item = \"bottle of beer\",\n        where = \"on the wall\",\n        initial_number = None,\n        zero_items = \"No more\",\n        action1 = \"Take one down, pass it around\",\n        action2 = \"Go to the store, buy some more\",\n        template = None,\n    ):\n        initial_number_true = 99 if initial_number is None else initial_number\n        \n        try:\n            is_initial_number_int = (initial_number_true % 1) == 0\n        except Exception:\n            is_initial_number_int = False\n        \n        if not is_initial_number_int:\n            raise ValueError(\"`initial_number` parameter must be None or a int-like object\")\n        \n        if initial_number_true < 0:\n            raise ValueError(\"`initial_number` parameter must be >=0\")\n        \n        \n        true_tpl = template or \"\"\"\\\n$i $items1 $where\n$i $items1\n$action\n$j $items2 $where\"\"\"\n        \n        strofa_tpl_tmp = Template(true_tpl)\n        strofa_tpl = Template(strofa_tpl_tmp.safe_substitute(where=where))\n        \n        self.zero_items = zero_items\n        self.action1 = action1\n        self.action2 = action2\n        self.initial_number = initial_number_true\n        self.item = item\n        self.items = items\n        self.strofa_tpl = strofa_tpl\n        self._hash = None\n        self._repr = None\n    \n    def strofa(self, number):\n        zero_items = self.zero_items\n        item = self.item\n        items = self.items\n        \n        if number == 0:\n            i = zero_items\n            action = self.action2\n            j = self.initial_number\n        else:\n            i = number\n            action = self.action1\n            j = i - 1\n        \n        if i == 1:\n            items1 = item\n            j = zero_items\n        else:\n            items1 = items\n        \n        if j == 1:\n            items2 = item\n        else:\n            items2 = items\n        \n        return self.strofa_tpl.substitute(\n            i = i, \n            j = j, \n            action = action, \n            items1 = items1, \n            items2 = items2\n        )\n    \n    def sing(self):\n        with ProcPool() as proc_pool:\n            strofa = self.strofa\n            initial_number = self.initial_number\n            args = range(initial_number, -1, -1)\n            return \"\\n\\n\".join(proc_pool.map(strofa, args))\n    \n    def __copy__(self, *args, **kwargs):\n        return self\n    \n    def __deepcopy__(self, *args, **kwargs):\n        return self\n    \n    def __eq__(self, other, *args, **kwargs):\n        if self is other:\n            return True\n        \n        getmyattr = partial(getattr, self)\n        getotherattr = partial(getattr, other)\n        eq_attrs = self.eq_attrs\n        \n        for attr in eq_attrs:\n            val = getmyattr(attr)\n            \n            try:\n                val2 = getotherattr(attr)\n            except Exception:\n                return False\n            \n            if attr == \"strofa_tpl\":\n                val_true = val.safe_substitute()\n                val2_true = val.safe_substitute()\n            else:\n                val_true = val\n                val2_true = val\n            \n            if val_true != val2_true:\n                return False\n        \n        return True\n    \n    def __hash__(self, *args, **kwargs):\n        _hash = self._hash\n        \n        if _hash is None:\n            getmyattr = partial(getattr, self)\n            attrs = self.hash_attrs\n            hash_true = self._hash = hash(tuple(map(getmyattr, attrs)))\n        else:\n            hash_true = _hash\n        \n        return hash_true\n    \n    def __repr__(self, *args, **kwargs):\n        _repr = self._repr\n        \n        if _repr is None:\n            repr_attrs = self.repr_attrs\n            getmyattr = partial(getattr, self)\n            \n            attrs = []\n            \n            for attr in repr_attrs:\n                val = getmyattr(attr)\n                \n                if attr == \"strofa_tpl\":\n                    val_true = val.safe_substitute()\n                else:\n                    val_true = val\n                \n                attrs.append(f\"{attr}={repr(val_true)}\")\n            \n            repr_true = self._repr = f\"{self.__class__.__name__}({', '.join(attrs)})\"\n        else:\n            repr_true = _repr\n        \n        return repr_true\n\ndef muchBeersOnTheWall(num):\n    song = MuchItemsSomewhere(initial_number=num)\n    \n    return song.sing()\n\ndef balladOfProgrammer(num):\n    \"\"\"\n    Prints\n    \"99 Subtle Bugs in Production\"\n    or\n    \"The Ballad of Programmer\"\n    \"\"\"\n    \n    song = MuchItemsSomewhere(\n        initial_number = num,\n        items = \"subtle bugs\",\n        item = \"subtle bug\",\n        where = \"in Production\",\n        action1 = \"Debug and catch, commit a patch\",\n        action2 = \"Release the fixes, wait for some tickets\",\n        zero_items = \"Zarro\",\n    )\n\n    return song.sing()\n\ndef main(num):\n    print(f\"### {num} Bottles of Beers on the Wall ###\")\n    print()\n    print(muchBeersOnTheWall(num))\n    print()\n    print()\n    print('### \"The Ballad of Programmer\", by Marco Sulla')\n    print()\n    print(balladOfProgrammer(num))\n\nif __name__ == \"__main__\":\n    # Ok, argparse is **really** too much\n    argv = sys.argv\n    \n    if len(argv) == 1:\n        num = None\n    elif len(argv) == 2:\n        try:\n            num = int(argv[1])\n        except Exception:\n            raise ValueError(\n                f\"{__file__} parameter must be an integer, or can be omitted\"\n            )\n    else:\n        raise RuntimeError(f\"{__file__} takes one parameter at max\")\n    \n    main(num)\n\n__all__ = (Song.__name__, MuchItemsSomewhere.__name__, muchBeersOnTheWall.__name__, balladOfProgrammer.__name__)\n"}
{"title": "99 bottles of beer", "language": "Python 3", "task": "Display the complete lyrics for the song:     '''99 Bottles of Beer on the Wall'''. \n\n\n;The beer song:\nThe lyrics follow this form:\n\n\n:::              99 bottles of beer on the wall\n:::              99 bottles of beer\n:::              Take one down, pass it around\n:::              98 bottles of beer on the wall\n    \n:::              98 bottles of beer on the wall\n:::              98 bottles of beer\n:::              Take one down, pass it around\n:::              97 bottles of beer on the wall\n\n\n... and so on, until reaching   '''0'''     (zero).\n\nGrammatical support for   ''1 bottle of beer''   is optional. \n\nAs with any puzzle, try to do it in as creative/concise/comical a way \nas possible (simple, obvious solutions allowed, too).\n\n\n\n\n;See also: \n*   http://99-bottles-of-beer.net/\n*   [[:Category:99_Bottles_of_Beer]]\n*   [[:Category:Programming language families]]\n*   Wikipedia 99 bottles of beer\n\n", "solution": "\"\"\"\n    99 Bottles of Beer on the Wall made functional\n    \n    Main function accepts a number of parameters, so you can specify a name of \n    the drink, its container and other things. English only.\n\"\"\"\n\nfrom functools import partial\nfrom typing import Callable\n\n\ndef regular_plural(noun: str) -> str:\n    \"\"\"English rule to get the plural form of a word\"\"\"\n    if noun[-1] == \"s\":\n        return noun + \"es\"\n    \n    return noun + \"s\"\n\n\ndef beer_song(\n    *,\n    location: str = 'on the wall',\n    distribution: str = 'Take one down, pass it around',\n    solution: str = 'Better go to the store to buy some more!',\n    container: str = 'bottle',\n    plurifier: Callable[[str], str] = regular_plural,\n    liquid: str = \"beer\",\n    initial_count: int = 99,\n) -> str:\n    \"\"\"\n    Return the lyrics of the beer song\n    :param location: initial location of the drink\n    :param distribution: specifies the process of its distribution\n    :param solution: what happens when we run out of drinks\n    :param container: bottle/barrel/flask or other containers\n    :param plurifier: function converting a word to its plural form\n    :param liquid: the name of the drink in the given container\n    :param initial_count: how many containers available initially\n    \"\"\"\n    \n    verse = partial(\n        get_verse,\n        initial_count = initial_count, \n        location = location,\n        distribution = distribution,\n        solution = solution,\n        container = container,\n        plurifier = plurifier,\n        liquid = liquid,\n    )\n    \n    verses = map(verse, range(initial_count, -1, -1))\n    return '\\n\\n'.join(verses)\n\n\ndef get_verse(\n    count: int,\n    *,\n    initial_count: str,\n    location: str,\n    distribution: str,\n    solution: str,\n    container: str,\n    plurifier: Callable[[str], str],\n    liquid: str,\n) -> str:\n    \"\"\"Returns the verse for the given amount of drinks\"\"\"\n    \n    asset = partial(\n        get_asset,\n        container = container,\n        plurifier = plurifier,\n        liquid = liquid,\n    )\n    \n    current_asset = asset(count)\n    next_number = count - 1 if count else initial_count\n    next_asset = asset(next_number)\n    action = distribution if count else solution\n    \n    inventory = partial(\n        get_inventory,\n        location = location,\n    )\n    \n    return '\\n'.join((\n        inventory(current_asset),\n        current_asset,\n        action,\n        inventory(next_asset),\n    ))\n\n\ndef get_inventory(\n    asset: str,\n    *,\n    location: str,\n) -> str:\n    \"\"\"\n    Used to return the first or the fourth line of the verse\n\n    >>> get_inventory(\"10 bottles of beer\", location=\"on the wall\")\n    \"10 bottles of beer on the wall\"\n    \"\"\"\n    return ' '.join((asset, location))\n\n\ndef get_asset(\n    count: int,\n    *,\n    container: str,\n    plurifier: Callable[[str], str],\n    liquid: str,\n) -> str:\n    \"\"\"\n    Quantified asset\n    \n    >>> get_asset(0, container=\"jar\", plurifier=regular_plural, liquid='milk')\n    \"No more jars of milk\"\n    \"\"\"\n    \n    containers = plurifier(container) if count != 1 else container\n    spelled_out_quantity = str(count) if count else \"No more\" \n    return ' '.join((spelled_out_quantity, containers, \"of\", liquid))\n\n\nif __name__ == '__main__':\n    print(beer_song())\n"}
{"title": "99 bottles of beer", "language": "Python 3.7+", "task": "Display the complete lyrics for the song:     '''99 Bottles of Beer on the Wall'''. \n\n\n;The beer song:\nThe lyrics follow this form:\n\n\n:::              99 bottles of beer on the wall\n:::              99 bottles of beer\n:::              Take one down, pass it around\n:::              98 bottles of beer on the wall\n    \n:::              98 bottles of beer on the wall\n:::              98 bottles of beer\n:::              Take one down, pass it around\n:::              97 bottles of beer on the wall\n\n\n... and so on, until reaching   '''0'''     (zero).\n\nGrammatical support for   ''1 bottle of beer''   is optional. \n\nAs with any puzzle, try to do it in as creative/concise/comical a way \nas possible (simple, obvious solutions allowed, too).\n\n\n\n\n;See also: \n*   http://99-bottles-of-beer.net/\n*   [[:Category:99_Bottles_of_Beer]]\n*   [[:Category:Programming language families]]\n*   Wikipedia 99 bottles of beer\n\n", "solution": "'''99 Units of Disposable Asset'''\n\n\nfrom itertools import chain\n\n\n# main :: IO ()\ndef main():\n    '''Modalised asset dispersal procedure.'''\n\n    # localisation :: (String, String, String)\n    localisation = (\n        'on the wall',\n        'Take one down, pass it around',\n        'Better go to the store to buy some more'\n    )\n\n    print(unlines(map(\n        incantation(localisation),\n        enumFromThenTo(99)(98)(0)\n    )))\n\n\n# incantation :: (String, String, String) -> Int -> String\ndef incantation(localisation):\n    '''Versification of asset disposal\n       and inventory update.'''\n\n    location, distribution, solution = localisation\n\n    def inventory(n):\n        return unwords([asset(n), location])\n    return lambda n: solution if 0 == n else (\n        unlines([\n            inventory(n),\n            asset(n),\n            distribution,\n            inventory(pred(n))\n        ])\n    )\n\n\n# asset :: Int -> String\ndef asset(n):\n    '''Quantified asset.'''\n    def suffix(n):\n        return [] if 1 == n else 's'\n    return unwords([\n        str(n),\n        concat(reversed(concat(cons(suffix(n))([\"elttob\"]))))\n    ])\n\n\n# GENERIC -------------------------------------------------\n\n# concat :: [[a]] -> [a]\n# concat :: [String] -> String\ndef concat(xxs):\n    '''The concatenation of all the elements in a list.'''\n    xs = list(chain.from_iterable(xxs))\n    unit = '' if isinstance(xs, str) else []\n    return unit if not xs else (\n        ''.join(xs) if isinstance(xs[0], str) else xs\n    )\n\n\n# cons :: a -> [a] -> [a]\ndef cons(x):\n    '''Construction of a list from x as head,\n       and xs as tail.'''\n    return lambda xs: [x] + xs if (\n        isinstance(xs, list)\n    ) else chain([x], xs)\n\n\n# enumFromThenTo :: Int -> Int -> Int -> [Int]\ndef enumFromThenTo(m):\n    '''Integer values enumerated from m to n\n       with a step defined by nxt-m.'''\n    def go(nxt, n):\n        d = nxt - m\n        return list(range(m, d + n, d))\n    return lambda nxt: lambda n: (\n        go(nxt, n)\n    )\n\n\n# pred ::  Enum a => a -> a\ndef pred(x):\n    '''The predecessor of a value. For numeric types, (- 1).'''\n    return x - 1 if isinstance(x, int) else (\n        chr(ord(x) - 1)\n    )\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.'''\n    return '\\n'.join(xs)\n\n\n# unwords :: [String] -> String\ndef unwords(xs):\n    '''A space-separated string derived from\n       a list of words.'''\n    return ' '.join(xs)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "9 billion names of God the integer", "language": "Python", "task": "This task is a variation of the short story by Arthur C. Clarke.\n \n(Solvers should be aware of the consequences of completing this task.)\n\nIn detail, to specify what is meant by a   \"name\":\n:The integer 1 has 1 name       \"1\".\n:The integer 2 has 2 names   \"1+1\",   and   \"2\".\n:The integer 3 has 3 names   \"1+1+1\",   \"2+1\",   and   \"3\".\n:The integer 4 has 5 names   \"1+1+1+1\",   \"2+1+1\",   \"2+2\",   \"3+1\",   \"4\".\n:The integer 5 has 7 names   \"1+1+1+1+1\",   \"2+1+1+1\",   \"2+2+1\",   \"3+1+1\",   \"3+2\",   \"4+1\",   \"5\".\n\n\n;Task\nDisplay the first 25 rows of a number triangle which begins:\n\n                                      1\n                                    1   1\n                                  1   1   1 \n                                1   2   1   1\n                              1   2   2   1   1\n                            1   3   3   2   1   1\n\n\nWhere row   n   corresponds to integer   n,   and each column   C   in row   m   from left to right corresponds to the number of names beginning with   C.\n\nA function   G(n)   should return the sum of the   n-th   row. \n\nDemonstrate this function by displaying:   G(23),   G(123),   G(1234),   and   G(12345).  \n\nOptionally note that the sum of the   n-th   row   P(n)   is the      integer partition function. \n\nDemonstrate this is equivalent to   G(n)   by displaying:   P(23),   P(123),   P(1234),   and   P(12345).\n\n\n;Extra credit\n\nIf your environment is able, plot   P(n)   against   n   for   n=1\\ldots 999.\n\n;Related tasks\n* [[Partition function P]]\n\n", "solution": "cache = [[1]]\ndef cumu(n):\n    for l in range(len(cache), n+1):\n        r = [0]\n        for x in range(1, l+1):\n            r.append(r[-1] + cache[l-x][min(x, l-x)])\n        cache.append(r)\n    return cache[n]\n\ndef row(n):\n    r = cumu(n)\n    return [r[i+1] - r[i] for i in range(n)]\n\nprint \"rows:\"\nfor x in range(1, 11): print \"%2d:\"%x, row(x)\n\n\nprint \"\\nsums:\"\nfor x in [23, 123, 1234, 12345]: print x, cumu(x)[-1]"}
{"title": "9 billion names of God the integer", "language": "Python from C", "task": "This task is a variation of the short story by Arthur C. Clarke.\n \n(Solvers should be aware of the consequences of completing this task.)\n\nIn detail, to specify what is meant by a   \"name\":\n:The integer 1 has 1 name       \"1\".\n:The integer 2 has 2 names   \"1+1\",   and   \"2\".\n:The integer 3 has 3 names   \"1+1+1\",   \"2+1\",   and   \"3\".\n:The integer 4 has 5 names   \"1+1+1+1\",   \"2+1+1\",   \"2+2\",   \"3+1\",   \"4\".\n:The integer 5 has 7 names   \"1+1+1+1+1\",   \"2+1+1+1\",   \"2+2+1\",   \"3+1+1\",   \"3+2\",   \"4+1\",   \"5\".\n\n\n;Task\nDisplay the first 25 rows of a number triangle which begins:\n\n                                      1\n                                    1   1\n                                  1   1   1 \n                                1   2   1   1\n                              1   2   2   1   1\n                            1   3   3   2   1   1\n\n\nWhere row   n   corresponds to integer   n,   and each column   C   in row   m   from left to right corresponds to the number of names beginning with   C.\n\nA function   G(n)   should return the sum of the   n-th   row. \n\nDemonstrate this function by displaying:   G(23),   G(123),   G(1234),   and   G(12345).  \n\nOptionally note that the sum of the   n-th   row   P(n)   is the      integer partition function. \n\nDemonstrate this is equivalent to   G(n)   by displaying:   P(23),   P(123),   P(1234),   and   P(12345).\n\n\n;Extra credit\n\nIf your environment is able, plot   P(n)   against   n   for   n=1\\ldots 999.\n\n;Related tasks\n* [[Partition function P]]\n\n", "solution": "def partitions(n):\n    partitions.p.append(0)\n\n    for k in xrange(1, n + 1):\n        d = n - k * (3 * k - 1) // 2\n        if d < 0:\n            break\n\n        if k & 1:\n            partitions.p[n] += partitions.p[d]\n        else:\n            partitions.p[n] -= partitions.p[d]\n\n        d -= k\n        if d < 0:\n            break\n\n        if k & 1:\n            partitions.p[n] += partitions.p[d]\n        else:\n            partitions.p[n] -= partitions.p[d]\n\n    return partitions.p[-1]\n\npartitions.p = [1]\n\ndef main():\n    ns = set([23, 123, 1234, 12345])\n    max_ns = max(ns)\n\n    for i in xrange(1, max_ns + 1):\n        if i > max_ns:\n            break\n        p = partitions(i)\n        if i in ns:\n            print \"%6d: %s\" % (i, p)\n\nmain()"}
{"title": "A+B", "language": "Python", "task": "'''A+B'''   --- a classic problem in programming contests,   it's given so contestants can gain familiarity with the online judging system being used.\n\n\n;Task:\nGiven two integers,   '''A''' and '''B'''. \n\nTheir sum needs to be calculated.\n\n\n;Input data:\nTwo integers are written in the input stream, separated by space(s):\n: (-1000 \\le A,B \\le +1000)\n\n\n;Output data:\nThe required output is one integer:   the sum of '''A''' and '''B'''.\n\n\n;Example:\n::{|class=\"standard\"\n ! input  \n ! output  \n |-\n | 2 2 \n | 4 \n |-\n | 3 2 \n | 5 \n |}\n\n", "solution": "a = int(input(\"First number: \"))\nb = int(input(\"Second number: \"))\nprint(\"Result:\", a+b)"}
{"title": "ABC problem", "language": "Python", "task": "You are given a collection of ABC blocks   (maybe like the ones you had when you were a kid).  \n\nThere are twenty blocks with two letters on each block. \n\nA complete alphabet is guaranteed amongst all sides of the blocks. \n\nThe sample collection of blocks: \n  (B O)\n  (X K)\n  (D Q)\n  (C P)\n  (N A)\n  (G T)\n  (R E)\n  (T G)\n  (Q D)\n  (F S)\n  (J W)\n  (H U)\n  (V I)\n  (A N)\n  (O B)\n  (E R)\n  (F S)\n  (L Y)\n  (P C)\n  (Z M)\n\n\n;Task:\nWrite a function that takes a string (word) and determines whether the word can be spelled with the given collection of blocks. \n\n\nThe rules are simple:\n::#   Once a letter on a block is used that block cannot be used again\n::#   The function should be case-insensitive\n::#   Show the output on this page for the following 7 words in the following example\n\n\n;Example:\n    >>> can_make_word(\"A\")\n    True\n    >>> can_make_word(\"BARK\")\n    True\n    >>> can_make_word(\"BOOK\")\n    False\n    >>> can_make_word(\"TREAT\")\n    True\n    >>> can_make_word(\"COMMON\")\n    False\n    >>> can_make_word(\"SQUAD\")\n    True\n    >>> can_make_word(\"CONFUSE\")\n    True\n\n", "solution": "'''\nNote that this code is broken, e.g., it won't work when \nblocks = [(\"A\", \"B\"), (\"A\",\"C\")] and the word is \"AB\", where the answer\nshould be True, but the code returns False.\n'''\nblocks = [(\"B\", \"O\"),\n          (\"X\", \"K\"),\n          (\"D\", \"Q\"),\n          (\"C\", \"P\"),\n          (\"N\", \"A\"),\n          (\"G\", \"T\"),\n          (\"R\", \"E\"),\n          (\"T\", \"G\"),\n          (\"Q\", \"D\"),\n          (\"F\", \"S\"),\n          (\"J\", \"W\"),\n          (\"H\", \"U\"),\n          (\"V\", \"I\"),\n          (\"A\", \"N\"),\n          (\"O\", \"B\"),\n          (\"E\", \"R\"),\n          (\"F\", \"S\"),\n          (\"L\", \"Y\"),\n          (\"P\", \"C\"),\n          (\"Z\", \"M\")]\n\n\ndef can_make_word(word, block_collection=blocks):\n    \"\"\"\n    Return True if `word` can be made from the blocks in `block_collection`.\n\n    >>> can_make_word(\"\")\n    False\n    >>> can_make_word(\"a\")\n    True\n    >>> can_make_word(\"bark\")\n    True\n    >>> can_make_word(\"book\")\n    False\n    >>> can_make_word(\"treat\")\n    True\n    >>> can_make_word(\"common\")\n    False\n    >>> can_make_word(\"squad\")\n    True\n    >>> can_make_word(\"coNFused\")\n    True\n    \"\"\"\n    if not word:\n        return False\n\n    blocks_remaining = block_collection[:]\n    for char in word.upper():\n        for block in blocks_remaining:\n            if char in block:\n                blocks_remaining.remove(block)\n                break\n        else:\n            return False\n    return True\n\n\nif __name__ == \"__main__\":\n    import doctest\n    doctest.testmod()\n    print(\", \".join(\"'%s': %s\" % (w, can_make_word(w)) for w in\n                    [\"\", \"a\", \"baRk\", \"booK\", \"treat\", \n                     \"COMMON\", \"squad\", \"Confused\"]))\n"}
{"title": "ASCII art diagram converter", "language": "Python", "task": "Given the RFC 1035 message diagram from Section 4.1.1 (Header section format) as a string:\nhttp://www.ietf.org/rfc/rfc1035.txt\n\n +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n |                      ID                       |\n +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n |QR|   Opcode  |AA|TC|RD|RA|   Z    |   RCODE   |\n +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n |                    QDCOUNT                    |\n +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n |                    ANCOUNT                    |\n +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n |                    NSCOUNT                    |\n +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n |                    ARCOUNT                    |\n +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n\nWhere (every column of the table is 1 bit):\n\n ID is 16 bits\n QR = Query (0) or Response (1)\n Opcode = Four bits defining kind of query:\n   0:    a standard query (QUERY)\n   1:    an inverse query (IQUERY)\n   2:    a server status request (STATUS)\n   3-15: reserved for future use\n AA = Authoritative Answer bit\n TC = Truncation bit\n RD = Recursion Desired bit\n RA = Recursion Available bit\n Z = Reserved\n RCODE = Response code\n QC = Question Count\n ANC = Answer Count\n AUC = Authority Count\n ADC = Additional Count\n\nWrite a function, member function, class or template that accepts a similar multi-line string as input to define a data structure or something else able to decode or store a header with that specified bit structure.\n\nIf your language has macros, introspection, code generation, or powerful enough templates, then accept such string at compile-time to define the header data structure statically.\n\nSuch \"Header\" function or template should accept a table with 8, 16, 32 or 64 columns, and any number of rows. For simplicity the only allowed symbols to define the table are + - | (plus, minus, pipe), and whitespace. Lines of the input string composed just of whitespace should be ignored. Leading and trailing whitespace in the input string should be ignored, as well as  before and after each table row. The box for each bit of the diagram takes four chars \"+--+\". The code should perform a little of validation of the input string, but for brevity a full validation is not required.\n\nBonus: perform a thoroughly validation of the input string.\n\n", "solution": "\"\"\"\nhttp://rosettacode.org/wiki/ASCII_art_diagram_converter\n\nPython example based off Go example:\n\nhttp://rosettacode.org/wiki/ASCII_art_diagram_converter#Go\n\n\"\"\"\n\ndef validate(diagram):\n\n    # trim empty lines\n    \n    rawlines = diagram.splitlines()\n    lines = []\n    for line in rawlines:\n        if line != '':\n            lines.append(line)\n            \n    # validate non-empty lines\n            \n    if len(lines) == 0:\n        print('diagram has no non-empty lines!')\n        return None\n        \n    width = len(lines[0])\n    cols = (width - 1) // 3\n    \n    if cols not in [8, 16, 32, 64]: \n        print('number of columns should be 8, 16, 32 or 64')\n        return None\n        \n    if len(lines)%2 == 0:\n        print('number of non-empty lines should be odd')\n        return None\n    \n    if lines[0] != (('+--' * cols)+'+'):\n            print('incorrect header line')\n            return None\n\n    for i in range(len(lines)):\n        line=lines[i]\n        if i == 0:\n            continue\n        elif i%2 == 0:\n            if line != lines[0]:\n                print('incorrect separator line')\n                return None\n        elif len(line) != width:\n            print('inconsistent line widths')\n            return None\n        elif line[0] != '|' or line[width-1] != '|':\n            print(\"non-separator lines must begin and end with '|'\")    \n            return None\n    \n    return lines\n\n\"\"\"\n\nresults is list of lists like:\n\n[[name, bits, start, end],...\n\n\"\"\"\n\ndef decode(lines):\n    print(\"Name     Bits  Start  End\")\n    print(\"=======  ====  =====  ===\")\n    \n    startbit = 0\n    \n    results = []\n    \n    for line in lines:\n        infield=False\n        for c in line:\n            if not infield and c == '|':\n                infield = True\n                spaces = 0\n                name = ''\n            elif infield:\n                if c == ' ':\n                    spaces += 1\n                elif c != '|':\n                    name += c\n                else:\n                    bits = (spaces + len(name) + 1) // 3\n                    endbit = startbit + bits - 1\n                    print('{0:7}    {1:2d}     {2:2d}   {3:2d}'.format(name, bits, startbit, endbit))\n                    reslist = [name, bits, startbit, endbit]\n                    results.append(reslist)\n                    spaces = 0\n                    name = ''\n                    startbit += bits\n                    \n    return results\n                        \ndef unpack(results, hex):\n    print(\"\\nTest string in hex:\")\n    print(hex)\n    print(\"\\nTest string in binary:\")\n    bin = f'{int(hex, 16):0>{4*len(hex)}b}'\n    print(bin)\n    print(\"\\nUnpacked:\\n\")\n    print(\"Name     Size  Bit pattern\")\n    print(\"=======  ====  ================\")\n    for r in results:\n        name = r[0]\n        size = r[1]\n        startbit = r[2]\n        endbit = r[3]\n        bitpattern = bin[startbit:endbit+1]\n        print('{0:7}    {1:2d}  {2:16}'.format(name, size, bitpattern))\n\n\ndiagram = \"\"\"\n+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n|                      ID                       |\n+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n|QR|   Opcode  |AA|TC|RD|RA|   Z    |   RCODE   |\n+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n|                    QDCOUNT                    |\n+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n|                    ANCOUNT                    |\n+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n|                    NSCOUNT                    |\n+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n|                    ARCOUNT                    |\n+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\n\n\"\"\"\n\nlines = validate(diagram)\n\nif lines == None:\n    print(\"No lines returned\")\nelse:\n    print(\" \")\n    print(\"Diagram after trimming whitespace and removal of blank lines:\")\n    print(\" \")\n    for line in lines:\n        print(line)\n        \n    print(\" \")\n    print(\"Decoded:\")\n    print(\" \")\n\n    results = decode(lines)    \n    \n    # test string\n    \n    hex = \"78477bbf5496e12e1bf169a4\" \n    \n    unpack(results, hex)\n"}
{"title": "AVL tree", "language": "Python", "task": "{{wikipedia|AVL tree}}\n\n\n\nIn computer science, an '''AVL tree''' is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; at no time do they differ by more than one because rebalancing is done ensure this is the case. Lookup, insertion, and deletion all take O(log ''n'') time in both the average and worst cases, where n is the number of nodes in the tree prior to the operation. Insertions and deletions may require the tree to be rebalanced by one or more tree rotations. Note the tree of nodes comprise a set, so duplicate node keys are not allowed.\n\nAVL trees are often compared with red-black trees because they support the same set of operations and because red-black trees also take O(log ''n'') time for the basic operations.  Because AVL trees are more rigidly balanced, they are faster than red-black trees for lookup-intensive applications. Similar to red-black trees, AVL trees are height-balanced, but in general not weight-balanced nor m-balanced; that is, sibling nodes can have hugely differing numbers of descendants.\n\n\n;Task:\nImplement an AVL tree in the language of choice, and provide at least basic operations.\n\n;Related task\n[[Red_black_tree_sort]]\n\n", "solution": "# Module: calculus.py\n\nimport enum\n\nclass entry_not_found(Exception):\n   \"\"\"Raised when an entry is not found in a collection\"\"\"\n   pass\n\nclass entry_already_exists(Exception):\n   \"\"\"Raised when an entry already exists in a collection\"\"\"\n   pass\n\nclass state(enum.Enum):\n   header = 0\n   left_high = 1\n   right_high = 2\n   balanced = 3\n\nclass direction(enum.Enum):\n   from_left = 0\n   from_right = 1\n\nfrom abc import ABC, abstractmethod\n\nclass comparer(ABC):\n\n    @abstractmethod\n    def compare(self,t):\n        pass\n\nclass node(comparer):\n \n    def __init__(self):\n        self.parent = None\n        self.left = self\n        self.right = self\n        self.balance = state.header\n\n    def compare(self,t):\n        if self.key < t:\n             return -1\n        elif t < self.key:\n             return 1\n        else:\n             return 0\n\n    def is_header(self):\n        return self.balance == state.header\n\n    def length(self):\n        if self != None:\n           if self.left != None:\n              left = self.left.length()\n           else:\n              left = 0\n           if self.right != None:   \n              right = self.right.length()\n           else:\n              right = 0\n              \n           return left + right + 1\n        else:\n           return 0\n    \n    def rotate_left(self):\n         _parent = self.parent\n         x = self.right\n         self.parent = x\n         x.parent = _parent\n         if x.left is not None:\n             x.left.parent = self\n         self.right = x.left\n         x.left = self\n         return x\n    \n \n    def rotate_right(self):\n        _parent = self.parent\n        x = self.left\n        self.parent = x\n        x.parent = _parent;\n        if x.right is not None:\n            x.right.parent = self\n        self.left = x.right\n        x.right = self\n        return x\n\n    def balance_left(self):\n       \n       _left = self.left\n\n       if _left is None:\n          return self;\n       \n       if _left.balance == state.left_high:\n                self.balance = state.balanced\n                _left.balance = state.balanced\n                self = self.rotate_right()\n       elif _left.balance == state.right_high: \n                subright = _left.right\n                if subright.balance == state.balanced:\n                        self.balance = state.balanced\n                        _left.balance = state.balanced\n                elif subright.balance == state.right_high:\n                        self.balance = state.balanced\n                        _left.balance = state.left_high\n                elif subright.balance == left_high:\n                        root.balance = state.right_high\n                        _left.balance = state.balanced\n                subright.balance = state.balanced\n                _left = _left.rotate_left()\n                self.left = _left\n                self = self.rotate_right()\n       elif _left.balance == state.balanced:\n               self.balance = state.left_high\n               _left.balance = state.right_high\n               self = self.rotate_right()\n       return self;\n   \n    def balance_right(self):\n\n       _right = self.right\n\n       if _right is None:\n          return self;\n       \n       if _right.balance == state.right_high:\n                self.balance = state.balanced\n                _right.balance = state.balanced\n                self = self.rotate_left()\n       elif _right.balance == state.left_high:\n                subleft = _right.left;\n                if subleft.balance == state.balanced:\n                        self.balance = state.balanced\n                        _right.balance = state.balanced\n                elif subleft.balance == state.left_high:\n                        self.balance = state.balanced\n                        _right.balance = state.right_high\n                elif subleft.balance == state.right_high:\n                        self.balance = state.left_high\n                        _right.balance = state.balanced\n                subleft.balance = state.balanced\n                _right = _right.rotate_right()\n                self.right = _right\n                self = self.rotate_left()\n       elif _right.balance == state.balanced:\n                self.balance = state.right_high\n                _right.balance = state.left_high\n                self = self.rotate_left()\n       return self\n\n\n    def balance_tree(self, direct):\n        taller = True\n        while taller:\n            _parent = self.parent;\n            if _parent.left == self:\n                next_from =  direction.from_left\n            else:\n                next_from = direction.from_right;\n\n            if direct == direction.from_left:\n                if self.balance == state.left_high:\n                        if _parent.is_header():\n                            _parent.parent = _parent.parent.balance_left()\n                        elif _parent.left == self:\n                            _parent.left = _parent.left.balance_left()\n                        else:\n                            _parent.right = _parent.right.balance_left()\n                        taller = False\n \n                elif self.balance == state.balanced:\n                        self.balance = state.left_high\n                        taller = True\n      \n                elif self.balance == state.right_high:\n                        self.balance = state.balanced\n                        taller = False\n            else:\n              if self.balance == state.left_high:\n                        self.balance = state.balanced\n                        taller = False\n  \n              elif self.balance ==  state.balanced:\n                        self.balance = state.right_high\n                        taller = True\n  \n              elif self.balance ==  state.right_high:\n                        if _parent.is_header():\n                            _parent.parent = _parent.parent.balance_right()\n                        elif _parent.left == self:\n                            _parent.left = _parent.left.balance_right()\n                        else:\n                            _parent.right = _parent.right.balance_right()\n                        taller = False\n  \n            if taller:\n                if _parent.is_header():\n                    taller = False\n                else:\n                    self = _parent\n                    direct = next_from\n\n    def balance_tree_remove(self, _from):\n      \n        if self.is_header():\n            return;\n\n        shorter = True;\n\n        while shorter:\n            _parent = self.parent;\n            if _parent.left == self:\n                next_from = direction.from_left\n            else:\n                next_from = direction.from_right\n\n            if _from == direction.from_left:\n                if self.balance == state.left_high:\n                        shorter = True\n \n                elif self.balance == state.balanced:\n                        self.balance = state.right_high;\n                        shorter = False\n  \n                elif self.balance == state.right_high:\n                        if self.right is not None:\n                            if self.right.balance == state.balanced:\n                                shorter = False\n                            else:\n                                shorter = True\n                        else:\n                            shorter = False;\n\n                        if _parent.is_header():\n                            _parent.parent = _parent.parent.balance_right()\n                        elif _parent.left == self:\n                            _parent.left = _parent.left.balance_right();\n                        else:\n                            _parent.right = _parent.right.balance_right()\n            \n            else:\n                if self.balance == state.right_high:\n                        self.balance = state.balanced\n                        shorter = True\n  \n                elif self.balance == state.balanced:\n                        self.balance = state.left_high\n                        shorter = False\n                 \n                elif self.balance == state.left_high:\n\n                        if self.left is not None:\n                            if self.left.balance == state.balanced:\n                                shorter = False\n                            else:\n                                shorter = True\n                        else:\n                           short = False;\n\n                        if _parent.is_header():\n                            _parent.parent = _parent.parent.balance_left();\n                        elif _parent.left == self:\n                            _parent.left = _parent.left.balance_left();\n                        else:\n                            _parent.right = _parent.right.balance_left();\n \n            if shorter:\n               if _parent.is_header():\n                    shorter = False\n               else: \n                    _from = next_from\n                    self = _parent\n\n    def previous(self):\n        if self.is_header():\n            return self.right\n\n        if self.left is not None:\n            y = self.left\n            while y.right is not None:\n                y = y.right\n            return y\n         \n        else: \n            y = self.parent;\n            if y.is_header():\n                return y\n\n            x = self\n            while x == y.left:\n                x = y\n                y = y.parent\n\n            return y\n        \n    def next(self):\n        if self.is_header():\n            return self.left\n\n        if self.right is not None:\n            y = self.right\n            while y.left is not None:\n                y = y.left\n            return y;\n         \n        else:\n            y = self.parent\n            if y.is_header():\n                return y\n\n            x = self;         \n            while x == y.right:\n                x = y\n                y = y.parent;\n                \n            return y\n\n    def swap_nodes(a, b):\n       \n        if b == a.left:\n            if b.left is not None:\n                b.left.parent = a\n\n            if b.right is not None:\n                b.right.parent = a\n\n            if a.right is not None:\n                a.right.parent = b\n\n            if not a.parent.is_header():\n                if a.parent.left == a:\n                    a.parent.left = b\n                else:\n                    a.parent.right = b;\n            else:\n                a.parent.parent = b\n\n            b.parent = a.parent\n            a.parent = b\n\n            a.left = b.left\n            b.left = a\n\n            temp = a.right\n            a.right = b.right\n            b.right = temp\n        elif b == a.right:\n            if b.right is not None:\n                b.right.parent = a\n                \n            if b.left is not None:\n               b.left.parent = a\n\n            if a.left is not None:\n               a.left.parent = b\n\n            if not a.parent.is_header(): \n                if a.parent.left == a:\n                    a.parent.left = b\n                else:\n                    a.parent.right = b\n            else:\n               a.parent.parent = b\n\n            b.parent = a.parent\n            a.parent = b\n\n            a.right = b.right\n            b.right = a\n\n            temp = a.left\n            a.left = b.left\n            b.left = temp\n        elif a == b.left:\n            if a.left is not None:\n                a.left.parent = b\n                \n            if a.right is not None:\n                a.right.parent = b\n\n            if b.right is not None:\n                b.right.parent = a\n\n            if not parent.is_header(): \n                if b.parent.left == b:\n                    b.parent.left = a\n                else:\n                    b.parent.right = a\n            else:\n                b.parent.parent = a\n\n            a.parent = b.parent\n            b.parent = a\n\n            b.left = a.left\n            a.left = b\n\n            temp = a.right\n            a.right = b.right\n            b.right = temp\n        elif a == b.right:\n            if a.right is not None:\n                a.right.parent = b\n            if a.left is not None:\n               a.left.parent = b\n\n            if b.left is not None:\n               b.left.parent = a\n\n            if not b.parent.is_header():\n                if b.parent.left == b:\n                    b.parent.left = a\n                else:\n                    b.parent.right = a\n            else:\n                b.parent.parent = a\n\n            a.parent = b.parent\n            b.parent = a\n\n            b.right = a.right\n            a.right = b\n\n            temp = a.left\n            a.left = b.left\n            b.left = temp\n        else:\n            if a.parent == b.parent:\n                temp = a.parent.left\n                a.parent.left = a.parent.right\n                a.parent.right = temp\n            else:\n                if not a.parent.is_header():\n                    if a.parent.left == a:\n                        a.parent.left = b\n                    else:\n                        a.parent.right = b\n                else:\n                    a.parent.parent = b\n\n                if not b.parent.is_header():\n                    if b.parent.left == b:\n                        b.parent.left = a\n                    else:\n                        b.parent.right = a\n                else:\n                    b.parent.parent = a\n            \n            if b.left is not None:\n                b.left.parent = a\n                \n            if b.right is not None:\n                b.right.parent = a\n\n            if a.left is not None:\n                a.left.parent = b\n                \n            if a.right is not None:\n                a.right.parent = b\n\n            temp1 = a.left\n            a.left = b.left\n            b.left = temp1\n\n            temp2 = a.right\n            a.right = b.right\n            b.right = temp2\n\n            temp3 = a.parent\n            a.parent = b.parent\n            b.parent = temp3\n        \n        balance = a.balance\n        a.balance = b.balance\n        b.balance = balance\n    \nclass parent_node(node):\n\n    def __init__(self, parent):\n        self.parent = parent\n        self.left = None\n        self.right = None\n        self.balance = state.balanced\n\nclass set_node(node):\n\n    def __init__(self, parent, key):\n        self.parent = parent\n        self.left = None\n        self.right = None\n        self.balance = state.balanced\n        self.key = key\n\nclass ordered_set:\n    \n    def __init__(self):\n        self.header = node()\n\n    def __iter__(self):\n        self.node = self.header\n        return self\n    \n    def __next__(self):\n        self.node = self.node.next()\n        if self.node.is_header():\n            raise StopIteration\n        return self.node.key\n\n    def __delitem__(self, key):\n          self.remove(key)\n\n    def __lt__(self, other):\n        first1 = self.header.left\n        last1 = self.header\n        first2 = other.header.left\n        last2 = other.header\n\n        while (first1 != last1) and (first2 != last2):\n           l =  first1.key < first2.key\n           if not l: \n              first1 = first1.next();\n              first2 = first2.next();\n           else:\n              return True;\n  \n        a = self.__len__()\n        b = other.__len__()\n        return a < b\n\n    def __hash__(self):\n        h = 0\n        for i in self:\n            h = h + i.__hash__()\n        return h    \n\n    def __eq__(self, other):\n       if self < other:\n          return False\n       if other < self:\n          return False\n       return True\n     \n    def __ne__(self, other):\n       if self < other:\n          return True\n       if other < self:\n          return True\n       return False\n\n    def __len__(self):\n        return self.header.parent.length()\n\n    def __getitem__(self, key):\n          return self.contains(key)\n\n    def __str__(self):\n       l = self.header.right\n       s = \"{\"\n       i = self.header.left\n       h = self.header\n       while i != h:\n           s = s + i.key.__str__()\n           if i != l:\n               s = s + \",\"\n           i = i.next()\n\n       s = s + \"}\"\n       return s\n\n    def __or__(self, other):\n       r = ordered_set()\n       \n       first1 = self.header.left\n       last1 = self.header\n       first2 = other.header.left\n       last2 = other.header\n       \n       while first1 != last1 and first2 != last2:\n          les = first1.key < first2.key\n          graater = first2.key < first1.key\n\n          if les:\n             r.add(first1.key)\n             first1 = first1.next()\n          elif graater:\n             r.add(first2.key)\n             first2 = first2.next()\n          else:\n             r.add(first1.key)\n             first1 = first1.next()\n             first2 = first2.next()\n             \n       while first1 != last1:\n          r.add(first1.key)\n          first1 = first1.next()\n                        \n       while first2 != last2:\n          r.add(first2.key)\n          first2 = first2.next()\n\n       return r\n\n    def __and__(self, other):\n       r = ordered_set()\n       \n       first1 = self.header.left\n       last1 = self.header\n       first2 = other.header.left\n       last2 = other.header\n       \n       while first1 != last1 and first2 != last2:\n          les = first1.key < first2.key\n          graater = first2.key < first1.key\n\n          if les:\n             first1 = first1.next()\n          elif graater:\n             first2 = first2.next()\n          else:\n             r.add(first1.key)\n             first1 = first1.next()\n             first2 = first2.next()\n  \n       return r\n\n    def __xor__(self, other):\n       r = ordered_set()\n       \n       first1 = self.header.left\n       last1 = self.header\n       first2 = other.header.left\n       last2 = other.header\n       \n       while first1 != last1 and first2 != last2:\n          les = first1.key < first2.key\n          graater = first2.key < first1.key\n\n          if les:\n             r.add(first1.key)\n             first1 = first1.next()\n          elif graater:\n             r.add(first2.key)\n             first2 = first2.next()\n          else:\n             first1 = first1.next()\n             first2 = first2.next()\n             \n       while first1 != last1:\n          r.add(first1.key)\n          first1 = first1.next()\n                        \n       while first2 != last2:\n          r.add(first2.key)\n          first2 = first2.next()\n\n       return r\n\n\n    def __sub__(self, other):\n       r = ordered_set()\n       \n       first1 = self.header.left\n       last1 = self.header\n       first2 = other.header.left\n       last2 = other.header\n       \n       while first1 != last1 and first2 != last2:\n          les = first1.key < first2.key\n          graater = first2.key < first1.key\n\n          if les:\n             r.add(first1.key)\n             first1 = first1.next()\n          elif graater:\n             r.add(first2.key)\n             first2 = first2.next()\n          else:\n             first1 = first1.next()\n             first2 = first2.next()\n             \n       while first1 != last1:\n          r.add(first1.key)\n          first1 = first1.next()\n\n       return r\n \n    def __lshift__(self, data):\n       self.add(data)\n       return self\n\n    def __rshift__(self, data):\n       self.remove(data)\n       return self\n\n    def is_subset(self, other):\n       first1 = self.header.left\n       last1 = self.header\n       first2 = other.header.left\n       last2 = other.header\n\n       is_subet = True\n\n       while first1 != last1 and first2 != last2:\n          if first1.key < first2.key:\n              is_subset = False\n              break\n          elif first2.key < first1.key:\n             first2 = first2.next()\n          else:\n             first1 = first1.next()\n             first2 = first2.next()\n \n          if is_subet:\n             if first1 != last1:\n                is_subet = False\n \n       return is_subet\n\n    def is_superset(self,other):\n       return other.is_subset(self)\n  \n    def add(self, data):\n            if self.header.parent is None:\n                self.header.parent = set_node(self.header,data)\n                self.header.left = self.header.parent\n                self.header.right = self.header.parent\n            else:\n                \n                root = self.header.parent\n\n                while True:\n                    c = root.compare(data)\n                    if c >= 0:\n                        if root.left is not None:\n                            root = root.left\n                        else:\n                            new_node = set_node(root,data)\n                            root.left = new_node\n                            \n                            if self.header.left == root:\n                                 self.header.left = new_node\n                            root.balance_tree(direction.from_left)\n                            return\n                        \n                    else:\n                        if root.right is not None:\n                            root = root.right\n                        else:\n                            new_node = set_node(root, data)\n                            root.right = new_node\n                            if self.header.right == root:\n                                  self.header.right = new_node\n                            root.balance_tree(direction.from_right)\n                            return\n                    \n    def remove(self,data):\n        root = self.header.parent;\n\n        while True:\n            if root is None:\n                raise entry_not_found(\"Entry not found in collection\")\n                \n            c  = root.compare(data)\n\n            if c < 0:\n               root = root.left;\n\n            elif c > 0:\n               root = root.right;\n\n            else:\n                 \n                 if root.left is not None:\n                     if root.right is not None: \n                         replace = root.left\n                         while replace.right is not None:\n                             replace = replace.right\n                         root.swap_nodes(replace)\n                         \n                 _parent = root.parent\n\n                 if _parent.left == root:\n                     _from = direction.from_left\n                 else:\n                     _from = direction.from_right\n\n                 if self.header.left == root:\n                                \n                     n = root.next();\n                 \n                     if n.is_header():\n                         self.header.left = self.header\n                         self.header.right = self.header\n                     else:\n                        self.header.left = n\n                 elif self.header.right == root: \n\n                     p = root.previous();\n\n                     if p.is_header():\n                          self.header.left = self.header\n                          self.header.right = self.header\n                     else:\n                          self.header.right = p\n\n                 if root.left is None:\n                     if _parent == self.header:\n                         self.header.parent = root.right\n                     elif _parent.left == root:\n                         _parent.left = root.right\n                     else:\n                         _parent.right = root.right\n\n                     if root.right is not None:\n                          root.right.parent = _parent\n                            \n                 else:\n                     if _parent == self.header:\n                          self.header.parent = root.left\n                     elif _parent.left == root:\n                         _parent.left = root.left\n                     else:\n                         _parent.right = root.left\n\n                     if root.left is not None:\n                         root.left.parent = _parent;\n\n\n                 _parent.balance_tree_remove(_from)\n                 return   \n\n    def contains(self,data):\n        root = self.header.parent;\n\n        while True:\n            if root == None:\n                return False\n\n            c  = root.compare(data);\n\n            if c > 0:\n               root = root.left;\n\n            elif c < 0:\n               root = root.right;\n\n            else:\n           \n                 return True  \n\n   \n    def find(self,data):\n        root = self.header.parent;\n\n        while True:\n            if root == None:\n                raise entry_not_found(\"An entry is not found in a collection\")\n\n            c  = root.compare(data);\n\n            if c > 0:\n               root = root.left;\n\n            elif c < 0:\n               root = root.right;\n\n            else:\n           \n                 return root.key;  \n            \nclass key_value(comparer):\n\n    def __init__(self, key, value):\n        self.key = key\n        self.value = value\n\n    def compare(self,kv):\n        if self.key < kv.key:\n             return -1\n        elif kv.key < self.key:\n             return 1\n        else:\n             return 0\n\n    def __lt__(self, other):\n        return self.key < other.key\n\n    def __str__(self):\n        return '(' + self.key.__str__() + ',' + self.value.__str__() + ')'\n\n    def __eq__(self, other):\n       return self.key == other.key\n\n    def __hash__(self):\n        return hash(self.key)\n \n\nclass dictionary:\n\n    def __init__(self):\n        self.set = ordered_set()\n        return None\n\n    def __lt__(self, other):\n       if self.keys() < other.keys():\n          return true\n\n       if other.keys() < self.keys():\n          return false\n         \n       first1 = self.set.header.left\n       last1 = self.set.header\n       first2 = other.set.header.left\n       last2 = other.set.header\n\n       while (first1 != last1) and (first2 != last2):\n          l =  first1.key.value < first2.key.value\n          if not l: \n             first1 = first1.next();\n             first2 = first2.next();\n          else:\n             return True;\n  \n       a = self.__len__()\n       b = other.__len__()\n       return a < b\n\n\n    def add(self, key, value):\n       try:\n           self.set.remove(key_value(key,None))\n       except entry_not_found:\n            pass  \n       self.set.add(key_value(key,value))\n       return\n\n    def remove(self, key):\n       self.set.remove(key_value(key,None))\n       return\n\n    def clear(self):\n       self.set.header = node()\n\n    def sort(self):\n    \n      sort_bag = bag()\n      for e in self:\n        sort_bag.add(e.value)\n      keys_set = self.keys()\n      self.clear()\n      i = sort_bag.__iter__()\n      i = sort_bag.__next__()\n      try:\n        for e in keys_set:\n          self.add(e,i)\n          i = sort_bag.__next__()\n      except:\n         return        \n\n    def keys(self):\n         keys_set = ordered_set()\n         for e in self:\n             keys_set.add(e.key)\n         return keys_set  \n   \n    def __len__(self):\n        return self.set.header.parent.length()\n\n    def __str__(self):\n       l = self.set.header.right;\n       s = \"{\"\n       i = self.set.header.left;\n       h = self.set.header;\n       while i != h:\n           s = s + \"(\"\n           s = s + i.key.key.__str__()\n           s = s + \",\"\n           s = s + i.key.value.__str__()\n           s = s + \")\"\n           if i != l:\n               s = s + \",\"\n           i = i.next()\n\n       s = s + \"}\"\n       return s;\n\n    def __iter__(self):\n       \n        self.set.node = self.set.header\n        return self\n    \n    def __next__(self):\n        self.set.node = self.set.node.next()\n        if self.set.node.is_header():\n            raise StopIteration\n        return key_value(self.set.node.key.key,self.set.node.key.value)\n\n    def __getitem__(self, key):\n          kv = self.set.find(key_value(key,None))\n          return kv.value\n\n    def __setitem__(self, key, value):\n          self.add(key,value)\n          return\n\n    def __delitem__(self, key):\n          self.set.remove(key_value(key,None))\n\n\nclass array:\n\n    def __init__(self):\n        self.dictionary = dictionary()\n        return None\n      \n    def __len__(self):\n        return self.dictionary.__len__()\n\n    def push(self, value):\n       k = self.dictionary.set.header.right\n       if k == self.dictionary.set.header:\n           self.dictionary.add(0,value)\n       else:\n           self.dictionary.add(k.key.key+1,value)\n       return\n\n    def pop(self):\n       if self.dictionary.set.header.parent != None:\n          data = self.dictionary.set.header.right.key.value\n          self.remove(self.dictionary.set.header.right.key.key)\n          return data\n\n    def add(self, key, value):\n       try:\n          self.dictionary.remove(key)\n       except entry_not_found:\n          pass\n       self.dictionary.add(key,value)          \n       return\n\n    def remove(self, key):\n       self.dictionary.remove(key)\n       return\n\n    def sort(self):\n       self.dictionary.sort()\n\n    def clear(self):\n      self.dictionary.header = node();\n      \n\n    def __iter__(self):\n        self.dictionary.node = self.dictionary.set.header\n        return self\n    \n    def __next__(self):\n        self.dictionary.node = self.dictionary.node.next()\n        if self.dictionary.node.is_header():\n            raise StopIteration\n        return self.dictionary.node.key.value\n\n    def __getitem__(self, key):\n          kv = self.dictionary.set.find(key_value(key,None))\n          return kv.value\n\n    def __setitem__(self, key, value):\n          self.add(key,value)\n          return\n\n    def __delitem__(self, key):\n          self.dictionary.remove(key)\n\n    def __lshift__(self, data):\n         self.push(data)\n         return self\n\n    def __lt__(self, other):\n       return self.dictionary < other.dictionary\n \n    def __str__(self):\n       l = self.dictionary.set.header.right;\n       s = \"{\"\n       i = self.dictionary.set.header.left;\n       h = self.dictionary.set.header;\n       while i != h:\n           s = s + i.key.value.__str__()\n           if i != l:\n               s = s + \",\"\n           i = i.next()\n\n       s = s + \"}\"\n       return s;\n          \n\nclass bag:\n    \n    def __init__(self):\n        self.header = node()\n      \n    def __iter__(self):\n        self.node = self.header\n        return self\n\n    def __delitem__(self, key):\n          self.remove(key)\n    \n    def __next__(self):\n        self.node = self.node.next()\n        if self.node.is_header():\n            raise StopIteration\n        return self.node.key\n\n    def __str__(self):\n       l = self.header.right;\n       s = \"(\"\n       i = self.header.left;\n       h = self.header;\n       while i != h:\n           s = s + i.key.__str__()\n           if i != l:\n               s = s + \",\"\n           i = i.next()\n\n       s = s + \")\"\n       return s;\n\n    def __len__(self):\n        return self.header.parent.length()\n\n    def __lshift__(self, data):\n       self.add(data)\n       return self\n\n    def add(self, data):\n            if self.header.parent is None:\n                self.header.parent = set_node(self.header,data)\n                self.header.left = self.header.parent\n                self.header.right = self.header.parent\n            else:\n                \n                root = self.header.parent\n\n                while True:\n                    c = root.compare(data)\n                    if c >= 0:\n                        if root.left is not None:\n                            root = root.left\n                        else:\n                            new_node = set_node(root,data)\n                            root.left = new_node\n                            \n                            if self.header.left == root:\n                                 self.header.left = new_node\n\n                            root.balance_tree(direction.from_left)\n                            return\n                        \n                    else:\n                        if root.right is not None:\n                            root = root.right\n                        else:\n                            new_node = set_node(root, data)\n                            root.right = new_node\n\n                            if self.header.right == root:\n                                  self.header.right = new_node\n\n                            root.balance_tree(direction.from_right)\n                            return\n \n    def remove_first(self,data):\n       \n        root = self.header.parent;\n\n        while True:\n            if root is None:\n                return False;\n\n            c  = root.compare(data);\n\n            if c > 0:\n               root = root.left;\n\n            elif c < 0:\n               root = root.right;\n\n            else:\n                 \n                 if root.left is not None:\n                     if root.right is not None: \n                         replace = root.left;\n                         while replace.right is not None:\n                             replace = replace.right;\n                         root.swap_nodes(replace);\n                         \n                 _parent = root.parent\n\n                 if _parent.left == root:\n                     _from = direction.from_left\n                 else:\n                     _from = direction.from_right\n\n                 if self.header.left == root:\n                                \n                     n = root.next();\n                 \n                     if n.is_header():\n                         self.header.left = self.header\n                         self.header.right = self.header\n                     else:\n                        self.header.left = n;\n                 elif self.header.right == root: \n\n                     p = root.previous();\n\n                     if p.is_header():\n                          self.header.left = self.header\n                          self.header.right = self.header\n                     else:\n                          self.header.right = p\n\n                 if root.left is None:\n                     if _parent == self.header:\n                         self.header.parent = root.right\n                     elif _parent.left == root:\n                         _parent.left = root.right\n                     else:\n                         _parent.right = root.right\n\n                     if root.right is not None:\n                          root.right.parent = _parent\n                            \n                 else:\n                     if _parent == self.header:\n                          self.header.parent = root.left\n                     elif _parent.left == root:\n                         _parent.left = root.left\n                     else:\n                         _parent.right = root.left\n\n                     if root.left is not None:\n                         root.left.parent = _parent;\n\n\n                 _parent.balance_tree_remove(_from)\n                 return True;\n\n    def remove(self,data):\n       success = self.remove_first(data)\n       while success:\n          success = self.remove_first(data)\n\n    def remove_node(self, root):\n       \n        if root.left != None and root.right != None:\n            replace = root.left\n            while replace.right != None:\n               replace = replace.right\n            root.swap_nodes(replace)\n\n        parent = root.parent;\n\n        if parent.left == root:\n           next_from = direction.from_left\n        else:\n           next_from = direction.from_right\n\n        if self.header.left == root:\n            n = root.next()\n\n            if n.is_header():\n                self.header.left = self.header;\n                self.header.right = self.header\n            else:\n                self.header.left = n\n        elif self.header.right == root:\n             p = root.previous()\n\n             if p.is_header(): \n                root.header.left = root.header\n                root.header.right = header\n             else:\n                self.header.right = p\n\n        if root.left == None:\n            if parent == self.header:\n                self.header.parent = root.right\n            elif parent.left == root:\n                parent.left = root.right\n            else:\n                parent.right = root.right\n\n            if root.right != None:\n               root.right.parent = parent\n        else:\n            if parent == self.header:\n                self.header.parent = root.left\n            elif parent.left == root:\n                parent.left = root.left\n            else:\n                parent.right = root.left\n\n            if root.left != None:\n               root.left.parent = parent;\n\n        parent.balance_tree_remove(next_from)\n    \n    def remove_at(self, data, ophset):\n \n            p = self.search(data);\n\n            if p == None:\n                return\n            else:\n                lower = p\n                after = after(data)\n \n            s = 0\n            while True:\n                if ophset == s:\n                    remove_node(lower);\n                    return;\n                lower = lower.next_node()\n                if after == lower:\n                   break\n                s = s+1\n            \n            return\n\n    def search(self, key):\n        s = before(key)\n        s.next()\n        if s.is_header():\n           return None\n        c = s.compare(s.key)\n        if c != 0:\n           return None\n        return s\n    \n  \n    def before(self, data):\n        y = self.header;\n        x = self.header.parent;\n\n        while x != None:\n            if x.compare(data) >= 0:\n                x = x.left;\n            else:\n                y = x;\n                x = x.right;\n        return y\n    \n    def after(self, data):\n        y = self.header;\n        x = self.header.parent;\n\n        while x != None:\n            if x.compare(data) > 0:\n                y = x\n                x = x.left\n            else:\n                x = x.right\n\n        return y;\n    \n \n    def find(self,data):\n        root = self.header.parent;\n\n        results = array()\n        \n        while True:\n            if root is None:\n                break;\n\n            p = self.before(data)\n            p = p.next()\n            if not p.is_header():\n               i = p\n               l = self.after(data)\n               while i != l:\n                  results.push(i.key)\n                  i = i.next()\n         \n               return results\n            else:\n               break;\n            \n        return results\n    \nclass bag_dictionary:\n\n    def __init__(self):\n        self.bag = bag()\n        return None\n\n    def add(self, key, value):\n       self.bag.add(key_value(key,value))\n       return\n\n    def remove(self, key):\n       self.bag.remove(key_value(key,None))\n       return\n\n    def remove_at(self, key, index):\n       self.bag.remove_at(key_value(key,None), index)\n       return\n\n    def clear(self):\n       self.bag.header = node()\n\n    def __len__(self):\n        return self.bag.header.parent.length()\n\n    def __str__(self):\n       l = self.bag.header.right;\n       s = \"{\"\n       i = self.bag.header.left;\n       h = self.bag.header;\n       while i != h:\n           s = s + \"(\"\n           s = s + i.key.key.__str__()\n           s = s + \",\"\n           s = s + i.key.value.__str__()\n           s = s + \")\"\n           if i != l:\n               s = s + \",\"\n           i = i.next()\n\n       s = s + \"}\"\n       return s;\n\n    def __iter__(self):\n       \n        self.bag.node = self.bag.header\n        return self\n    \n    def __next__(self):\n        self.bag.node = self.bag.node.next()\n        if self.bag.node.is_header():\n            raise StopIteration\n        return key_value(self.bag.node.key.key,self.bag.node.key.value)\n\n    def __getitem__(self, key):\n          kv_array = self.bag.find(key_value(key,None))\n          return kv_array\n\n    def __setitem__(self, key, value):\n          self.add(key,value)\n          return\n\n    def __delitem__(self, key):\n          self.bag.remove(key_value(key,None))\n\nclass unordered_set:\n\n    def __init__(self):\n        self.bag_dictionary = bag_dictionary()\n\n    def __len__(self):\n        return self.bag_dictionary.__len__()\n\n    def __hash__(self):\n        h = 0\n        for i in self:\n            h = h + i.__hash__()\n        return h    \n\n    def __eq__(self, other):\n        for t in self:\n           if not other.contains(t):\n              return False\n        for u in other:\n           if self.contains(u):\n              return False\n        return true;\n\n    def __ne__(self, other):\n        return not self == other\n      \n    def __or__(self, other):\n       r = unordered_set()\n       \n       for t in self:\n          r.add(t);\n          \n       for u in other:\n          if not self.contains(u):\n             r.add(u);\n\n       return r\n\n    def __and__(self, other):\n       r = unordered_set()\n   \n       for t in self:\n          if other.contains(t):\n              r.add(t)\n              \n       for u in other:\n              if self.contains(u) and not r.contains(u):\n                  r.add(u);\n  \n       return r\n\n    def __xor__(self, other):\n       r = unordered_set()\n       \n       for t in self:\n          if not other.contains(t):\n             r.add(t)\n             \n       for u in other:\n          if not self.contains(u) and not r.contains(u):\n             r.add(u)\n             \n       return r\n\n\n    def __sub__(self, other):\n       r = ordered_set()\n       \n       for t in self:\n          if not other.contains(t):\n             r.add(t);\n             \n       return r\n \n    def __lshift__(self, data):\n       self.add(data)\n       return self\n\n    def __rshift__(self, data):\n       self.remove(data)\n       return self\n\n    def __getitem__(self, key):\n          return self.contains(key)\n\n    def is_subset(self, other):\n\n       is_subet = True\n\n       for t in self:\n          if not other.contains(t):\n             subset = False\n             break\n            \n       return is_subet\n\n    def is_superset(self,other):\n       return other.is_subset(self)\n\n\n    def add(self, value):\n       if not self.contains(value):\n           self.bag_dictionary.add(hash(value),value)\n       else:\n          raise entry_already_exists(\"Entry already exists in the unordered set\")\n\n    def contains(self, data):\n            if self.bag_dictionary.bag.header.parent == None:\n                return False;\n            else:\n                index = hash(data);\n\n                _search = self.bag_dictionary.bag.header.parent;\n\n                search_index =  _search.key.key;\n\n                if index < search_index:\n                   _search = _search.left\n\n                elif index > search_index:\n                   _search = _search.right\n\n                if _search == None:\n                    return False\n\n                while _search != None:\n                    search_index =  _search.key.key;\n\n                    if index < search_index:\n                       _search = _search.left\n\n                    elif index > search_index:\n                       _search = _search.right\n\n                    else:\n                       break\n\n                if _search == None:\n                   return False\n\n                return self.contains_node(data, _search)\n \n    def contains_node(self,data,_node):\n       \n        previous = _node.previous()\n        save = _node\n\n        while not previous.is_header() and previous.key.key == _node.key.key:\n            save = previous;\n            previous = previous.previous()\n      \n        c = _node.key.value\n        _node = save\n        if c == data:\n           return True\n\n        next = _node.next()\n        while not next.is_header() and next.key.key == _node.key.key:\n            _node = next\n            c = _node.key.value\n            if c == data:\n               return True;\n            next = _node.next()\n \n        return False;\n      \n    def find(self,data,_node):\n       \n        previous = _node.previous()\n        save = _node\n\n        while not previous.is_header() and previous.key.key == _node.key.key:\n            save = previous;\n            previous = previous.previous();\n \n        _node = save;\n        c = _node.key.value\n        if c == data:\n           return _node\n\n        next = _node.next()\n        while not next.is_header() and next.key.key == _node.key.key:\n            _node = next\n            c = _node.data.value\n            if c == data:\n               return _node\n            next = _node.next()\n \n        return None\n    \n    def search(self, data):\n        if self.bag_dictionary.bag.header.parent == None:\n            return None\n        else:\n            index = hash(data)\n\n            _search = self.bag_dictionary.bag.header.parent\n\n            c = _search.key.key\n\n            if index < c:\n               _search = _search.left;\n\n            elif index > c:\n               _search = _search.right;\n\n            while _search != None:\n\n                if index != c:\n                   break\n               \n                c = _search.key.key\n\n                if index < c:\n                   _search = _search.left;\n\n                elif index > c:\n                   _search = _search.right;\n\n                else:\n                   break\n\n            if _search == None:\n               return None\n\n            return self.find(data, _search)\n\n    def remove(self,data):\n       found = self.search(data);\n       if found != None:\n          self.bag_dictionary.bag.remove_node(found);\n       else:\n          raise entry_not_found(\"Entry not found in the unordered set\")\n \n    def clear(self):\n       self.bag_dictionary.bag.header = node()\n\n    def __str__(self):\n       l = self.bag_dictionary.bag.header.right;\n       s = \"{\"\n       i = self.bag_dictionary.bag.header.left;\n       h = self.bag_dictionary.bag.header;\n       while i != h:\n           s = s + i.key.value.__str__()\n           if i != l:\n               s = s + \",\"\n           i = i.next()\n\n       s = s + \"}\"\n       return s;\n\n    def __iter__(self):\n       \n        self.bag_dictionary.bag.node = self.bag_dictionary.bag.header\n        return self\n    \n    def __next__(self):\n        self.bag_dictionary.bag.node = self.bag_dictionary.bag.node.next()\n        if self.bag_dictionary.bag.node.is_header():\n            raise StopIteration\n        return self.bag_dictionary.bag.node.key.value\n\n\nclass map:\n\n    def __init__(self):\n        self.set = unordered_set()\n        return None\n\n    def __len__(self):\n        return self.set.__len__()\n\n    def add(self, key, value):\n       try:\n           self.set.remove(key_value(key,None))\n       except entry_not_found:\n            pass  \n       self.set.add(key_value(key,value))\n       return\n\n    def remove(self, key):\n       self.set.remove(key_value(key,None))\n       return\n\n    def clear(self):\n       self.set.clear()\n\n    def __str__(self):\n       l = self.set.bag_dictionary.bag.header.right;\n       s = \"{\"\n       i = self.set.bag_dictionary.bag.header.left;\n       h = self.set.bag_dictionary.bag.header;\n       while i != h:\n           s = s + \"(\"\n           s = s + i.key.value.key.__str__()\n           s = s + \",\"\n           s = s + i.key.value.value.__str__()\n           s = s + \")\"\n           if i != l:\n               s = s + \",\"\n           i = i.next()\n\n       s = s + \"}\"\n       return s;\n\n    def __iter__(self):\n       \n        self.set.node = self.set.bag_dictionary.bag.header\n        return self\n    \n    def __next__(self):\n        self.set.node = self.set.node.next()\n        if self.set.node.is_header():\n            raise StopIteration\n        return key_value(self.set.node.key.key,self.set.node.key.value)\n\n    def __getitem__(self, key):\n          kv = self.set.find(key_value(key,None))\n          return kv.value\n\n    def __setitem__(self, key, value):\n          self.add(key,value)\n          return\n\n    def __delitem__(self, key):\n          self.remove(key)\n"}
{"title": "Abbreviations, automatic", "language": "Python from Kotlin", "task": "The use of   abbreviations    (also sometimes called synonyms, nicknames, AKAs, or aliases)    can be an\neasy way to add flexibility when specifying or using commands, sub-commands, options, etc.\n\n\n\nIt would make a list of words easier to maintain   (as words are added, changed, and/or deleted)   if\nthe minimum abbreviation length of that list could be automatically (programmatically) determined.\n\n\nFor this task, use the list (below) of the days-of-the-week names that are expressed in about a hundred languages   (note that there is a blank line in the list).\n\nSunday Monday Tuesday Wednesday Thursday Friday Saturday\nSondag Maandag Dinsdag Woensdag Donderdag Vrydag Saterdag\nE_djele E_hene E_marte E_merkure E_enjte E_premte E_shtune\nEhud Segno Maksegno Erob Hamus Arbe Kedame\nAl_Ahad Al_Ithinin Al_Tholatha'a Al_Arbia'a Al_Kamis Al_Gomia'a Al_Sabit\nGuiragui Yergou_shapti Yerek_shapti Tchorek_shapti Hink_shapti Ourpat Shapat\ndomingu llunes martes miercoles xueves vienres sabadu\nBazar_gUnU Birinci_gUn Ckinci_gUn UcUncU_gUn DOrdUncU_gUn Bes,inci_gUn Altonco_gUn\nIgande Astelehen Astearte Asteazken Ostegun Ostiral Larunbat\nRobi_bar Shom_bar Mongal_bar Budhh_bar BRihashpati_bar Shukro_bar Shoni_bar\nNedjelja Ponedeljak Utorak Srijeda Cxetvrtak Petak Subota\nDisul Dilun Dimeurzh Dimerc'her Diriaou Digwener Disadorn\nnedelia ponedelnik vtornik sriada chetvartak petak sabota\nsing_kei_yaht sing_kei_yat sing_kei_yee sing_kei_saam sing_kei_sie sing_kei_ng sing_kei_luk\nDiumenge Dilluns Dimarts Dimecres Dijous Divendres Dissabte\nDzeenkk-eh Dzeehn_kk-ehreh Dzeehn_kk-ehreh_nah_kay_dzeeneh Tah_neesee_dzeehn_neh Deehn_ghee_dzee-neh Tl-oowey_tts-el_dehlee Dzeentt-ahzee\ndy_Sul dy_Lun dy_Meurth dy_Mergher dy_You dy_Gwener dy_Sadorn\nDimanch Lendi Madi Mekredi Jedi Vandredi Samdi\nnedjelja ponedjeljak utorak srijeda cxetvrtak petak subota\nnede^le ponde^li utery str^eda c^tvrtek patek sobota\nSondee Mondee Tiisiday Walansedee TOOsedee Feraadee Satadee\ns0ndag mandag tirsdag onsdag torsdag fredag l0rdag\nzondag maandag dinsdag woensdag donderdag vrijdag zaterdag\nDiman^co Lundo Mardo Merkredo ^Jaudo Vendredo Sabato\npUhapaev esmaspaev teisipaev kolmapaev neljapaev reede laupaev\n\nDiu_prima Diu_sequima Diu_tritima Diu_quartima Diu_quintima Diu_sextima Diu_sabbata\nsunnudagur manadagur tysdaguy mikudagur hosdagur friggjadagur leygardagur\nYek_Sham'beh Do_Sham'beh Seh_Sham'beh Cha'har_Sham'beh Panj_Sham'beh Jom'eh Sham'beh\nsunnuntai maanantai tiistai keskiviiko torsktai perjantai lauantai\ndimanche lundi mardi mercredi jeudi vendredi samedi\nSnein Moandei Tiisdei Woansdei Tonersdei Freed Sneon\nDomingo Segunda_feira Martes Mercores Joves Venres Sabado\nk'vira orshabati samshabati otkhshabati khutshabati p'arask'evi shabati\nSonntag Montag Dienstag Mittwoch Donnerstag Freitag Samstag\nKiriaki' Defte'ra Tri'ti Teta'rti Pe'mpti Paraskebi' Sa'bato\nravivaar somvaar mangalvaar budhvaar guruvaar shukravaar shanivaar\npopule po`akahi po`alua po`akolu po`aha po`alima po`aono\nYom_rishon Yom_sheni Yom_shlishi Yom_revi'i Yom_chamishi Yom_shishi Shabat\nravivara somavar mangalavar budhavara brahaspativar shukravara shanivar\nvasarnap hetfo kedd szerda csutortok pentek szombat\nSunnudagur Manudagur +ridjudagur Midvikudagar Fimmtudagur FOstudagur Laugardagur\nsundio lundio mardio merkurdio jovdio venerdio saturdio\nMinggu Senin Selasa Rabu Kamis Jumat Sabtu\nDominica Lunedi Martedi Mercuridi Jovedi Venerdi Sabbato\nDe_Domhnaigh De_Luain De_Mairt De_Ceadaoin De_ardaoin De_hAoine De_Sathairn\ndomenica lunedi martedi mercoledi giovedi venerdi sabato\nNichiyou_bi Getzuyou_bi Kayou_bi Suiyou_bi Mokuyou_bi Kin'you_bi Doyou_bi\nIl-yo-il Wol-yo-il Hwa-yo-il Su-yo-il Mok-yo-il Kum-yo-il To-yo-il\nDies_Dominica Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Saturni\nsve-tdien pirmdien otrdien tresvdien ceturtdien piektdien sestdien\nSekmadienis Pirmadienis Antradienis Trec^iadienis Ketvirtadienis Penktadienis S^es^tadienis\nWangu Kazooba Walumbe Mukasa Kiwanuka Nnagawonye Wamunyi\nxing-_qi-_ri xing-_qi-_yi-. xing-_qi-_er xing-_qi-_san-. xing-_qi-_si xing-_qi-_wuv. xing-_qi-_liu\nJedoonee Jelune Jemayrt Jecrean Jardaim Jeheiney Jesam\nJabot Manre Juje Wonje Taije Balaire Jarere\ngeminrongo minomishi martes mierkoles misheushi bernashi mishabaro\nAhad Isnin Selasa Rabu Khamis Jumaat Sabtu\nsphndag mandag tirsdag onsdag torsdag fredag lphrdag\nlo_dimenge lo_diluns lo_dimarc lo_dimercres lo_dijous lo_divendres lo_dissabte\ndjadomingo djaluna djamars djarason djaweps djabierna djasabra\nNiedziela Poniedzial/ek Wtorek S,roda Czwartek Pia,tek Sobota\nDomingo segunda-feire terca-feire quarta-feire quinta-feire sexta-feira sabado\nDomingo Lunes martes Miercoles Jueves Viernes Sabado\nDuminica Luni Mart'i Miercuri Joi Vineri Sambata\nvoskresenie ponedelnik vtornik sreda chetverg pyatnitsa subbota\nSunday Di-luain Di-mairt Di-ciadain Di-ardaoin Di-haoine Di-sathurne\nnedjelja ponedjeljak utorak sreda cxetvrtak petak subota\nSontaha Mmantaha Labobedi Laboraro Labone Labohlano Moqebelo\nIridha- Sandhudha- Anga.haruwa-dha- Badha-dha- Brahaspa.thindha- Sikura-dha- Sena.sura-dha-\nnedel^a pondelok utorok streda s^tvrtok piatok sobota\nNedelja Ponedeljek Torek Sreda Cxetrtek Petek Sobota\ndomingo lunes martes miercoles jueves viernes sabado\nsonde mundey tude-wroko dride-wroko fode-wroko freyda Saturday\nJumapili Jumatatu Jumanne Jumatano Alhamisi Ijumaa Jumamosi\nsondag mandag tisdag onsdag torsdag fredag lordag\nLinggo Lunes Martes Miyerkoles Huwebes Biyernes Sabado\nLe-pai-jit Pai-it Pai-ji Pai-san Pai-si Pai-gO. Pai-lak\nwan-ar-tit wan-tjan wan-ang-kaan wan-phoet wan-pha-ru-hat-sa-boh-die wan-sook wan-sao\nTshipi Mosupologo Labobedi Laboraro Labone Labotlhano Matlhatso\nPazar Pazartesi Sali Car,samba Per,sembe Cuma Cumartesi\nnedilya ponedilok vivtorok sereda chetver pyatnytsya subota\nChu?_Nha.t Thu*_Hai Thu*_Ba Thu*_Tu* Thu*_Na'm Thu*_Sau Thu*_Ba?y\ndydd_Sul dyds_Llun dydd_Mawrth dyds_Mercher dydd_Iau dydd_Gwener dyds_Sadwrn\nDibeer Altine Talaata Allarba Al_xebes Aljuma Gaaw\niCawa uMvulo uLwesibini uLwesithathu uLuwesine uLwesihlanu uMgqibelo\nzuntik montik dinstik mitvokh donershtik fraytik shabes\niSonto uMsombuluko uLwesibili uLwesithathu uLwesine uLwesihlanu uMgqibelo\nDies_Dominica Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Saturni\nBazar_gUnU Bazar_aertaesi Caers,aenbae_axs,amo Caers,aenbae_gUnU CUmae_axs,amo CUmae_gUnU CUmae_Senbae\nSun Moon Mars Mercury Jove Venus Saturn\nzondag maandag dinsdag woensdag donderdag vrijdag zaterdag\nKoseEraa GyoOraa BenEraa Kuoraa YOwaaraa FeEraa Memenaa\nSonntag Montag Dienstag Mittwoch Donnerstag Freitag Sonnabend\nDomingo Luns Terza_feira Corta_feira Xoves Venres Sabado\nDies_Solis Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Sabbatum\nxing-_qi-_tian xing-_qi-_yi-. xing-_qi-_er xing-_qi-_san-. xing-_qi-_si xing-_qi-_wuv. xing-_qi-_liu\ndjadomingu djaluna djamars djarason djaweps djabierne djasabra\nKillachau Atichau Quoyllurchau Illapachau Chaskachau Kuychichau Intichau\n\n''Caveat:   The list (above) most surely contains errors (or, at the least, differences) of what the actual (or true) names for the days-of-the-week.''\n\n\nTo make this Rosetta Code task page as small as possible, if processing the complete list, read the days-of-the-week from a file (that is created from the above list).\n\n\nNotes concerning the above list of words\n::*   each line has a list of days-of-the-week for a language, separated by at least one blank\n::*   the words on each line happen to be in order, from Sunday --> Saturday\n::*   most lines have words in mixed case and some have all manner of accented words and other characters\n::*   some words were translated to the nearest character that was available to ''code page''   '''437'''\n::*   the characters in the words are not restricted except that they may not have imbedded blanks\n::*   for this example, the use of an underscore ('''_''') was used to indicate a blank in a word\n\n\n;Task:\n::*   The list of words   (days of the week)   needn't be verified/validated.\n::*   Write a function to find the (numeric) minimum length abbreviation for each line that would make abbreviations unique.\n::*   A blank line   (or a null line)   should return a null string.\n::*   Process and show the output for at least the first '''five''' lines of the file.\n::*   Show all output here.\n\n\n", "solution": "def shortest_abbreviation_length(line, list_size):\n    words = line.split()\n    word_count = len(words)\n    # Can't give true answer with unexpected number of entries\n    if word_count != list_size:\n        raise ValueError(f'Not enough entries, expected {list_size} found {word_count}')\n\n    # Find the small slice length that gives list_size unique values\n    abbreviation_length = 1\n    abbreviations = set()\n    while(True):\n        abbreviations = {word[:abbreviation_length] for word in words}\n        if len(abbreviations) == list_size:\n            return abbreviation_length\n        abbreviation_length += 1\n        abbreviations.clear()\n\ndef automatic_abbreviations(filename, words_per_line):\n    with open(filename) as file:\n        for line in file:\n            line = line.rstrip()\n            if len(line) > 0:\n                length = shortest_abbreviation_length(line, words_per_line)\n                print(f'{length:2} {line}')\n            else:\n                print()\n\nautomatic_abbreviations('daysOfWeek.txt', 7)"}
{"title": "Abbreviations, automatic", "language": "Python 3", "task": "The use of   abbreviations    (also sometimes called synonyms, nicknames, AKAs, or aliases)    can be an\neasy way to add flexibility when specifying or using commands, sub-commands, options, etc.\n\n\n\nIt would make a list of words easier to maintain   (as words are added, changed, and/or deleted)   if\nthe minimum abbreviation length of that list could be automatically (programmatically) determined.\n\n\nFor this task, use the list (below) of the days-of-the-week names that are expressed in about a hundred languages   (note that there is a blank line in the list).\n\nSunday Monday Tuesday Wednesday Thursday Friday Saturday\nSondag Maandag Dinsdag Woensdag Donderdag Vrydag Saterdag\nE_djele E_hene E_marte E_merkure E_enjte E_premte E_shtune\nEhud Segno Maksegno Erob Hamus Arbe Kedame\nAl_Ahad Al_Ithinin Al_Tholatha'a Al_Arbia'a Al_Kamis Al_Gomia'a Al_Sabit\nGuiragui Yergou_shapti Yerek_shapti Tchorek_shapti Hink_shapti Ourpat Shapat\ndomingu llunes martes miercoles xueves vienres sabadu\nBazar_gUnU Birinci_gUn Ckinci_gUn UcUncU_gUn DOrdUncU_gUn Bes,inci_gUn Altonco_gUn\nIgande Astelehen Astearte Asteazken Ostegun Ostiral Larunbat\nRobi_bar Shom_bar Mongal_bar Budhh_bar BRihashpati_bar Shukro_bar Shoni_bar\nNedjelja Ponedeljak Utorak Srijeda Cxetvrtak Petak Subota\nDisul Dilun Dimeurzh Dimerc'her Diriaou Digwener Disadorn\nnedelia ponedelnik vtornik sriada chetvartak petak sabota\nsing_kei_yaht sing_kei_yat sing_kei_yee sing_kei_saam sing_kei_sie sing_kei_ng sing_kei_luk\nDiumenge Dilluns Dimarts Dimecres Dijous Divendres Dissabte\nDzeenkk-eh Dzeehn_kk-ehreh Dzeehn_kk-ehreh_nah_kay_dzeeneh Tah_neesee_dzeehn_neh Deehn_ghee_dzee-neh Tl-oowey_tts-el_dehlee Dzeentt-ahzee\ndy_Sul dy_Lun dy_Meurth dy_Mergher dy_You dy_Gwener dy_Sadorn\nDimanch Lendi Madi Mekredi Jedi Vandredi Samdi\nnedjelja ponedjeljak utorak srijeda cxetvrtak petak subota\nnede^le ponde^li utery str^eda c^tvrtek patek sobota\nSondee Mondee Tiisiday Walansedee TOOsedee Feraadee Satadee\ns0ndag mandag tirsdag onsdag torsdag fredag l0rdag\nzondag maandag dinsdag woensdag donderdag vrijdag zaterdag\nDiman^co Lundo Mardo Merkredo ^Jaudo Vendredo Sabato\npUhapaev esmaspaev teisipaev kolmapaev neljapaev reede laupaev\n\nDiu_prima Diu_sequima Diu_tritima Diu_quartima Diu_quintima Diu_sextima Diu_sabbata\nsunnudagur manadagur tysdaguy mikudagur hosdagur friggjadagur leygardagur\nYek_Sham'beh Do_Sham'beh Seh_Sham'beh Cha'har_Sham'beh Panj_Sham'beh Jom'eh Sham'beh\nsunnuntai maanantai tiistai keskiviiko torsktai perjantai lauantai\ndimanche lundi mardi mercredi jeudi vendredi samedi\nSnein Moandei Tiisdei Woansdei Tonersdei Freed Sneon\nDomingo Segunda_feira Martes Mercores Joves Venres Sabado\nk'vira orshabati samshabati otkhshabati khutshabati p'arask'evi shabati\nSonntag Montag Dienstag Mittwoch Donnerstag Freitag Samstag\nKiriaki' Defte'ra Tri'ti Teta'rti Pe'mpti Paraskebi' Sa'bato\nravivaar somvaar mangalvaar budhvaar guruvaar shukravaar shanivaar\npopule po`akahi po`alua po`akolu po`aha po`alima po`aono\nYom_rishon Yom_sheni Yom_shlishi Yom_revi'i Yom_chamishi Yom_shishi Shabat\nravivara somavar mangalavar budhavara brahaspativar shukravara shanivar\nvasarnap hetfo kedd szerda csutortok pentek szombat\nSunnudagur Manudagur +ridjudagur Midvikudagar Fimmtudagur FOstudagur Laugardagur\nsundio lundio mardio merkurdio jovdio venerdio saturdio\nMinggu Senin Selasa Rabu Kamis Jumat Sabtu\nDominica Lunedi Martedi Mercuridi Jovedi Venerdi Sabbato\nDe_Domhnaigh De_Luain De_Mairt De_Ceadaoin De_ardaoin De_hAoine De_Sathairn\ndomenica lunedi martedi mercoledi giovedi venerdi sabato\nNichiyou_bi Getzuyou_bi Kayou_bi Suiyou_bi Mokuyou_bi Kin'you_bi Doyou_bi\nIl-yo-il Wol-yo-il Hwa-yo-il Su-yo-il Mok-yo-il Kum-yo-il To-yo-il\nDies_Dominica Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Saturni\nsve-tdien pirmdien otrdien tresvdien ceturtdien piektdien sestdien\nSekmadienis Pirmadienis Antradienis Trec^iadienis Ketvirtadienis Penktadienis S^es^tadienis\nWangu Kazooba Walumbe Mukasa Kiwanuka Nnagawonye Wamunyi\nxing-_qi-_ri xing-_qi-_yi-. xing-_qi-_er xing-_qi-_san-. xing-_qi-_si xing-_qi-_wuv. xing-_qi-_liu\nJedoonee Jelune Jemayrt Jecrean Jardaim Jeheiney Jesam\nJabot Manre Juje Wonje Taije Balaire Jarere\ngeminrongo minomishi martes mierkoles misheushi bernashi mishabaro\nAhad Isnin Selasa Rabu Khamis Jumaat Sabtu\nsphndag mandag tirsdag onsdag torsdag fredag lphrdag\nlo_dimenge lo_diluns lo_dimarc lo_dimercres lo_dijous lo_divendres lo_dissabte\ndjadomingo djaluna djamars djarason djaweps djabierna djasabra\nNiedziela Poniedzial/ek Wtorek S,roda Czwartek Pia,tek Sobota\nDomingo segunda-feire terca-feire quarta-feire quinta-feire sexta-feira sabado\nDomingo Lunes martes Miercoles Jueves Viernes Sabado\nDuminica Luni Mart'i Miercuri Joi Vineri Sambata\nvoskresenie ponedelnik vtornik sreda chetverg pyatnitsa subbota\nSunday Di-luain Di-mairt Di-ciadain Di-ardaoin Di-haoine Di-sathurne\nnedjelja ponedjeljak utorak sreda cxetvrtak petak subota\nSontaha Mmantaha Labobedi Laboraro Labone Labohlano Moqebelo\nIridha- Sandhudha- Anga.haruwa-dha- Badha-dha- Brahaspa.thindha- Sikura-dha- Sena.sura-dha-\nnedel^a pondelok utorok streda s^tvrtok piatok sobota\nNedelja Ponedeljek Torek Sreda Cxetrtek Petek Sobota\ndomingo lunes martes miercoles jueves viernes sabado\nsonde mundey tude-wroko dride-wroko fode-wroko freyda Saturday\nJumapili Jumatatu Jumanne Jumatano Alhamisi Ijumaa Jumamosi\nsondag mandag tisdag onsdag torsdag fredag lordag\nLinggo Lunes Martes Miyerkoles Huwebes Biyernes Sabado\nLe-pai-jit Pai-it Pai-ji Pai-san Pai-si Pai-gO. Pai-lak\nwan-ar-tit wan-tjan wan-ang-kaan wan-phoet wan-pha-ru-hat-sa-boh-die wan-sook wan-sao\nTshipi Mosupologo Labobedi Laboraro Labone Labotlhano Matlhatso\nPazar Pazartesi Sali Car,samba Per,sembe Cuma Cumartesi\nnedilya ponedilok vivtorok sereda chetver pyatnytsya subota\nChu?_Nha.t Thu*_Hai Thu*_Ba Thu*_Tu* Thu*_Na'm Thu*_Sau Thu*_Ba?y\ndydd_Sul dyds_Llun dydd_Mawrth dyds_Mercher dydd_Iau dydd_Gwener dyds_Sadwrn\nDibeer Altine Talaata Allarba Al_xebes Aljuma Gaaw\niCawa uMvulo uLwesibini uLwesithathu uLuwesine uLwesihlanu uMgqibelo\nzuntik montik dinstik mitvokh donershtik fraytik shabes\niSonto uMsombuluko uLwesibili uLwesithathu uLwesine uLwesihlanu uMgqibelo\nDies_Dominica Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Saturni\nBazar_gUnU Bazar_aertaesi Caers,aenbae_axs,amo Caers,aenbae_gUnU CUmae_axs,amo CUmae_gUnU CUmae_Senbae\nSun Moon Mars Mercury Jove Venus Saturn\nzondag maandag dinsdag woensdag donderdag vrijdag zaterdag\nKoseEraa GyoOraa BenEraa Kuoraa YOwaaraa FeEraa Memenaa\nSonntag Montag Dienstag Mittwoch Donnerstag Freitag Sonnabend\nDomingo Luns Terza_feira Corta_feira Xoves Venres Sabado\nDies_Solis Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Sabbatum\nxing-_qi-_tian xing-_qi-_yi-. xing-_qi-_er xing-_qi-_san-. xing-_qi-_si xing-_qi-_wuv. xing-_qi-_liu\ndjadomingu djaluna djamars djarason djaweps djabierne djasabra\nKillachau Atichau Quoyllurchau Illapachau Chaskachau Kuychichau Intichau\n\n''Caveat:   The list (above) most surely contains errors (or, at the least, differences) of what the actual (or true) names for the days-of-the-week.''\n\n\nTo make this Rosetta Code task page as small as possible, if processing the complete list, read the days-of-the-week from a file (that is created from the above list).\n\n\nNotes concerning the above list of words\n::*   each line has a list of days-of-the-week for a language, separated by at least one blank\n::*   the words on each line happen to be in order, from Sunday --> Saturday\n::*   most lines have words in mixed case and some have all manner of accented words and other characters\n::*   some words were translated to the nearest character that was available to ''code page''   '''437'''\n::*   the characters in the words are not restricted except that they may not have imbedded blanks\n::*   for this example, the use of an underscore ('''_''') was used to indicate a blank in a word\n\n\n;Task:\n::*   The list of words   (days of the week)   needn't be verified/validated.\n::*   Write a function to find the (numeric) minimum length abbreviation for each line that would make abbreviations unique.\n::*   A blank line   (or a null line)   should return a null string.\n::*   Process and show the output for at least the first '''five''' lines of the file.\n::*   Show all output here.\n\n\n", "solution": "'''Automatic abbreviations'''\n\nfrom itertools import (accumulate, chain)\nfrom os.path import expanduser\n\n\n# abbrevLen :: [String] -> Int\ndef abbrevLen(xs):\n    '''The minimum length of prefix required to obtain\n       a unique abbreviation for each string in xs.'''\n    n = len(xs)\n\n    return next(\n        len(a[0]) for a in map(\n            compose(nub)(map_(concat)),\n            transpose(list(map(inits, xs)))\n        ) if n == len(a)\n    )\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''Test'''\n\n    xs = map_(strip)(\n        lines(readFile('weekDayNames.txt'))\n    )\n    for i, n in enumerate(map(compose(abbrevLen)(words), xs)):\n        print(n, '  ', xs[i])\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# concat :: [String] -> String\ndef concat(xs):\n    '''The concatenation of a list of strings.'''\n    return ''.join(xs)\n\n\n# inits :: [a] -> [[a]]\ndef inits(xs):\n    '''all initial segments of xs, shortest first.'''\n    return list(scanl(lambda a, x: a + [x])(\n        []\n    )(list(xs)))\n\n\n# lines :: String -> [String]\ndef lines(s):\n    '''A list of strings,\n       (containing no newline characters)\n       derived from a single new-line delimited string.'''\n    return s.splitlines()\n\n\n# map :: (a -> b) -> [a] -> [b]\ndef map_(f):\n    '''The list obtained by applying f\n       to each element of xs.'''\n    return lambda xs: list(map(f, xs))\n\n\n# nub :: [a] -> [a]\ndef nub(xs):\n    '''A list containing the same elements as xs,\n       without duplicates, in the order of their\n       first occurrence.'''\n    return list(dict.fromkeys(xs))\n\n\n# readFile :: FilePath -> IO String\ndef readFile(fp):\n    '''The contents of any file at the path\n       derived by expanding any ~ in fp.'''\n    with open(expanduser(fp), 'r', encoding='utf-8') as f:\n        return f.read()\n\n\n# scanl :: (b -> a -> b) -> b -> [a] -> [b]\ndef scanl(f):\n    '''scanl is like reduce, but returns a succession of\n       intermediate values, building from the left.'''\n    return lambda a: lambda xs: (\n        accumulate(chain([a], xs), f)\n    )\n\n\n# strip :: String -> String\ndef strip(s):\n    '''A copy of s without any leading or trailling\n       white space.'''\n    return s.strip()\n\n\n# transpose :: Matrix a -> Matrix a\ndef transpose(m):\n    '''The rows and columns of the argument transposed.\n       (The matrix containers and rows can be lists or tuples).'''\n    if m:\n        inner = type(m[0])\n        z = zip(*m)\n        return (type(m))(\n            map(inner, z) if tuple != inner else z\n        )\n    else:\n        return m\n\n\n# words :: String -> [String]\ndef words(s):\n    '''A list of words delimited by characters\n       representing white space.'''\n    return s.split()\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Abbreviations, easy", "language": "Python 3.6", "task": "This task is an easier (to code) variant of the Rosetta Code task:    [[Abbreviations, simple]].\n\n\nFor this task, the following   ''command table''   will be used:\n    Add ALTer  BAckup Bottom  CAppend Change SCHANGE  CInsert CLAst COMPress COpy\n    COUnt COVerlay CURsor DELete CDelete Down DUPlicate Xedit EXPand EXTract Find\n    NFind NFINDUp NFUp CFind FINdup FUp FOrward GET Help HEXType Input POWerinput\n    Join SPlit SPLTJOIN  LOAD  Locate CLocate  LOWercase UPPercase  LPrefix MACRO\n    MErge MODify MOve MSG Next Overlay PARSE PREServe PURge PUT PUTD  Query  QUIT\n    READ  RECover REFRESH RENum REPeat  Replace CReplace  RESet  RESTore  RGTLEFT\n    RIght LEft  SAVE  SET SHift SI  SORT  SOS  STAck STATus  TOP TRAnsfer Type Up\n\n\n\n\nNotes concerning the above   ''command table'':\n::*   it can be thought of as one long literal string   (with blanks at end-of-lines)\n::*   it may have superfluous blanks\n::*   it may be in any case (lower/upper/mixed)\n::*   the order of the words in the   ''command table''   must be preserved as shown\n::*   the user input(s) may be in any case (upper/lower/mixed)\n::*   commands will be restricted to the Latin alphabet   (A --> Z,   a --> z)\n::*   A valid abbreviation is a word that has:\n:::*   at least the minimum length of the number of capital letters of the word in the ''command table''\n:::*   compares equal (regardless of case) to the leading characters of the word in the ''command table''\n:::*   a length not longer than the word in the ''command table''\n::::*   '''ALT''',   '''aLt''',   '''ALTE''',   and   '''ALTER'''   are all abbreviations of   '''ALTer'''\n::::*   '''AL''',   '''ALF''',   '''ALTERS''',   '''TER''',   and   '''A'''    aren't valid abbreviations of   '''ALTer'''\n::::*   The number of capital letters in   '''ALTer'''   indicates that any abbreviation for   '''ALTer'''   must be at least three letters\n::::*   Any word longer than five characters can't be an abbreviation for   '''ALTer'''\n::::*   '''o''',    '''ov''',    '''oVe''',    '''over''',    '''overL''',    '''overla'''   are all acceptable abbreviations for   '''Overlay'''\n::*   if there isn't any lowercase letters in the word in the ''command table'',   then there isn't an abbreviation permitted\n\n\n\n;Task:\n::*   The command table needn't be verified/validated.\n::*   Write a function to validate if the user \"words\"   (given as input)   are valid   (in the ''command table'').\n::*   If the word   is   valid,   then return the full uppercase version of that \"word\".\n::*   If the word        isn't     valid,   then return the lowercase string:    '''*error*'''        (7 characters).\n::*   A blank input   (or a null input)   should return a null string.\n::*   Show all output here.\n\n\n;An example test case to be used for this task:\nFor a user string of:\n  riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin\nthe computer program should return the string:\n  RIGHT REPEAT *error* PUT MOVE RESTORE *error* *error* *error* POWERINPUT\n\n\n", "solution": "command_table_text = \\\n\"\"\"Add ALTer  BAckup Bottom  CAppend Change SCHANGE  CInsert CLAst COMPress COpy\nCOUnt COVerlay CURsor DELete CDelete Down DUPlicate Xedit EXPand EXTract Find\nNFind NFINDUp NFUp CFind FINdup FUp FOrward GET Help HEXType Input POWerinput\nJoin SPlit SPLTJOIN  LOAD  Locate CLocate  LOWercase UPPercase  LPrefix MACRO\nMErge MODify MOve MSG Next Overlay PARSE PREServe PURge PUT PUTD  Query  QUIT\nREAD  RECover REFRESH RENum REPeat  Replace CReplace  RESet  RESTore  RGTLEFT\nRIght LEft  SAVE  SET SHift SI  SORT  SOS  STAck STATus  TOP TRAnsfer Type Up\"\"\"\n\nuser_words = \"riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin\"\n\ndef find_abbreviations_length(command_table_text):\n    \"\"\" find the minimal abbreviation length for each word by counting capital letters.\n        a word that does not have capital letters gets it's full length as the minimum.\n    \"\"\"\n    command_table = dict()\n    for word in command_table_text.split():\n        abbr_len = sum(1 for c in word if c.isupper())\n        if abbr_len == 0:\n            abbr_len = len(word)\n        command_table[word] = abbr_len\n    return command_table\n\ndef find_abbreviations(command_table):\n    \"\"\" for each command insert all possible abbreviations\"\"\"\n    abbreviations = dict()\n    for command, min_abbr_len in command_table.items():\n        for l in range(min_abbr_len, len(command)+1):\n            abbr = command[:l].lower()\n            abbreviations[abbr] = command.upper()\n    return abbreviations\n\ndef parse_user_string(user_string, abbreviations):\n    user_words = [word.lower() for word in user_string.split()]\n    commands = [abbreviations.get(user_word, \"*error*\") for user_word in user_words]\n    return \" \".join(commands)\n\ncommand_table = find_abbreviations_length(command_table_text)\nabbreviations_table = find_abbreviations(command_table)\n\nfull_words = parse_user_string(user_words, abbreviations_table)\n\nprint(\"user words:\", user_words)\nprint(\"full words:\", full_words)"}
{"title": "Abbreviations, simple", "language": "Python 3.6", "task": "The use of   abbreviations    (also sometimes called synonyms, nicknames, AKAs, or aliases)    can be an \neasy way to add flexibility when specifying or using commands, sub-commands, options, etc.\n\n\n\nFor this task, the following   ''command table''   will be used:\n    add 1  alter 3  backup 2  bottom 1  Cappend 2  change 1  Schange  Cinsert 2  Clast 3\n    compress 4 copy 2 count 3 Coverlay 3 cursor 3  delete 3 Cdelete 2  down 1  duplicate\n    3 xEdit 1 expand 3 extract 3  find 1 Nfind 2 Nfindup 6 NfUP 3 Cfind 2 findUP 3 fUP 2\n    forward 2  get  help 1 hexType 4  input 1 powerInput 3  join 1 split 2 spltJOIN load\n    locate 1 Clocate 2 lowerCase 3 upperCase 3 Lprefix 2  macro  merge 2 modify 3 move 2\n    msg  next 1 overlay 1 parse preserve 4 purge 3 put putD query 1 quit  read recover 3\n    refresh renum 3 repeat 3 replace 1 Creplace 2 reset 3 restore 4 rgtLEFT right 2 left\n    2  save  set  shift 2  si  sort  sos  stack 3 status 4 top  transfer 3  type 1  up 1\n\n\n\n\nNotes concerning the above   ''command table'':\n::*   it can be thought of as one long literal string   (with blanks at end-of-lines)\n::*   it may have superfluous blanks\n::*   it may be in any case (lower/upper/mixed)\n::*   the order of the words in the   ''command table''   must be preserved as shown\n::*   the user input(s) may be in any case (upper/lower/mixed)\n::*   commands will be restricted to the Latin alphabet   (A --> Z,   a --> z)\n::*   a command is followed by an optional number, which indicates the minimum abbreviation\n::*   A valid abbreviation is a word that has:\n:::*   at least the minimum length of the word's minimum number in the ''command table''\n:::*   compares equal (regardless of case) to the leading characters of the word in the ''command table''\n:::*   a length not longer than the word in the ''command table''\n::::*   '''ALT''',   '''aLt''',   '''ALTE''',   and   '''ALTER'''   are all abbreviations of   '''ALTER 3'''\n::::*   '''AL''',   '''ALF''',   '''ALTERS''',   '''TER''',   and   '''A'''    aren't valid abbreviations of   '''ALTER 3'''\n::::*   The   '''3'''   indicates that any abbreviation for   '''ALTER'''   must be at least three characters\n::::*   Any word longer than five characters can't be an abbreviation for   '''ALTER'''\n::::*   '''o''',    '''ov''',    '''oVe''',    '''over''',    '''overL''',    '''overla'''   are all acceptable abbreviations for   '''overlay 1'''\n::*   if there isn't a number after the command,   then there isn't an abbreviation permitted\n\n\n\n;Task:\n::*   The command table needn't be verified/validated.\n::*   Write a function to validate if the user \"words\"   (given as input)   are valid   (in the ''command table'').\n::*   If the word   is   valid,   then return the full uppercase version of that \"word\".\n::*   If the word        isn't     valid,   then return the lowercase string:    '''*error*'''        (7 characters).\n::*   A blank input   (or a null input)   should return a null string.\n::*   Show all output here.\n\n\n;An example test case to be used for this task:\nFor a user string of:\n  riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin\nthe computer program should return the string:\n  RIGHT REPEAT *error* PUT MOVE RESTORE *error* *error* *error* POWERINPUT\n\n\n", "solution": "command_table_text = \"\"\"add 1  alter 3  backup 2  bottom 1  Cappend 2  change 1  Schange  Cinsert 2  Clast 3\n   compress 4 copy 2 count 3 Coverlay 3 cursor 3  delete 3 Cdelete 2  down 1  duplicate\n   3 xEdit 1 expand 3 extract 3  find 1 Nfind 2 Nfindup 6 NfUP 3 Cfind 2 findUP 3 fUP 2\n   forward 2  get  help 1 hexType 4  input 1 powerInput 3  join 1 split 2 spltJOIN load\n   locate 1 Clocate 2 lowerCase 3 upperCase 3 Lprefix 2  macro  merge 2 modify 3 move 2\n   msg  next 1 overlay 1 parse preserve 4 purge 3 put putD query 1 quit  read recover 3\n   refresh renum 3 repeat 3 replace 1 Creplace 2 reset 3 restore 4 rgtLEFT right 2 left\n   2  save  set  shift 2  si  sort  sos  stack 3 status 4 top  transfer 3  type 1  up 1\"\"\"\n\nuser_words = \"riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin\"\n\n\ndef find_abbreviations_length(command_table_text):\n    \"\"\" find the minimal abbreviation length for each word.\n        a word that does not have minimum abbreviation length specified\n        gets it's full lengths as the minimum.\n    \"\"\"\n    command_table = dict()\n    input_iter = iter(command_table_text.split())\n\n    word = None\n    try:\n        while True:\n            if word is None:\n                word = next(input_iter)\n            abbr_len = next(input_iter, len(word))\n            try:\n                command_table[word] = int(abbr_len)\n                word = None\n            except ValueError:\n                command_table[word] = len(word)\n                word = abbr_len\n    except StopIteration:\n        pass\n    return command_table\n\n\ndef find_abbreviations(command_table):\n    \"\"\" for each command insert all possible abbreviations\"\"\"\n    abbreviations = dict()\n    for command, min_abbr_len in command_table.items():\n        for l in range(min_abbr_len, len(command)+1):\n            abbr = command[:l].lower()\n            abbreviations[abbr] = command.upper()\n    return abbreviations\n\n\ndef parse_user_string(user_string, abbreviations):\n    user_words = [word.lower() for word in user_string.split()]\n    commands = [abbreviations.get(user_word, \"*error*\") for user_word in user_words]\n    return \" \".join(commands)\n\n\ncommand_table = find_abbreviations_length(command_table_text)\nabbreviations_table = find_abbreviations(command_table)\n\nfull_words = parse_user_string(user_words, abbreviations_table)\n\nprint(\"user words:\", user_words)\nprint(\"full words:\", full_words)\n"}
{"title": "Abbreviations, simple", "language": "Python 3.7", "task": "The use of   abbreviations    (also sometimes called synonyms, nicknames, AKAs, or aliases)    can be an \neasy way to add flexibility when specifying or using commands, sub-commands, options, etc.\n\n\n\nFor this task, the following   ''command table''   will be used:\n    add 1  alter 3  backup 2  bottom 1  Cappend 2  change 1  Schange  Cinsert 2  Clast 3\n    compress 4 copy 2 count 3 Coverlay 3 cursor 3  delete 3 Cdelete 2  down 1  duplicate\n    3 xEdit 1 expand 3 extract 3  find 1 Nfind 2 Nfindup 6 NfUP 3 Cfind 2 findUP 3 fUP 2\n    forward 2  get  help 1 hexType 4  input 1 powerInput 3  join 1 split 2 spltJOIN load\n    locate 1 Clocate 2 lowerCase 3 upperCase 3 Lprefix 2  macro  merge 2 modify 3 move 2\n    msg  next 1 overlay 1 parse preserve 4 purge 3 put putD query 1 quit  read recover 3\n    refresh renum 3 repeat 3 replace 1 Creplace 2 reset 3 restore 4 rgtLEFT right 2 left\n    2  save  set  shift 2  si  sort  sos  stack 3 status 4 top  transfer 3  type 1  up 1\n\n\n\n\nNotes concerning the above   ''command table'':\n::*   it can be thought of as one long literal string   (with blanks at end-of-lines)\n::*   it may have superfluous blanks\n::*   it may be in any case (lower/upper/mixed)\n::*   the order of the words in the   ''command table''   must be preserved as shown\n::*   the user input(s) may be in any case (upper/lower/mixed)\n::*   commands will be restricted to the Latin alphabet   (A --> Z,   a --> z)\n::*   a command is followed by an optional number, which indicates the minimum abbreviation\n::*   A valid abbreviation is a word that has:\n:::*   at least the minimum length of the word's minimum number in the ''command table''\n:::*   compares equal (regardless of case) to the leading characters of the word in the ''command table''\n:::*   a length not longer than the word in the ''command table''\n::::*   '''ALT''',   '''aLt''',   '''ALTE''',   and   '''ALTER'''   are all abbreviations of   '''ALTER 3'''\n::::*   '''AL''',   '''ALF''',   '''ALTERS''',   '''TER''',   and   '''A'''    aren't valid abbreviations of   '''ALTER 3'''\n::::*   The   '''3'''   indicates that any abbreviation for   '''ALTER'''   must be at least three characters\n::::*   Any word longer than five characters can't be an abbreviation for   '''ALTER'''\n::::*   '''o''',    '''ov''',    '''oVe''',    '''over''',    '''overL''',    '''overla'''   are all acceptable abbreviations for   '''overlay 1'''\n::*   if there isn't a number after the command,   then there isn't an abbreviation permitted\n\n\n\n;Task:\n::*   The command table needn't be verified/validated.\n::*   Write a function to validate if the user \"words\"   (given as input)   are valid   (in the ''command table'').\n::*   If the word   is   valid,   then return the full uppercase version of that \"word\".\n::*   If the word        isn't     valid,   then return the lowercase string:    '''*error*'''        (7 characters).\n::*   A blank input   (or a null input)   should return a null string.\n::*   Show all output here.\n\n\n;An example test case to be used for this task:\nFor a user string of:\n  riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin\nthe computer program should return the string:\n  RIGHT REPEAT *error* PUT MOVE RESTORE *error* *error* *error* POWERINPUT\n\n\n", "solution": "'''Simple abbreviations'''\n\nfrom functools import reduce\nimport re\n\n\n# withExpansions :: [(String, Int)] -> String -> String\ndef withExpansions(table):\n    '''A string in which all words are either\n       expanded (if they match abbreviations in\n       a supplied table), or replaced with an\n       '*error*' string.\n    '''\n    return lambda s: unwords(map(\n        expanded(table), words(s)\n    ))\n\n\n# expanded :: [(String, Int)] -> String -> String\ndef expanded(table):\n    '''An abbreviation (or error string) for\n       a given word, based on a table of full\n       strings and minimum abbreviation lengths.\n    '''\n    def expansion(k):\n        u = k.upper()\n        lng = len(k)\n\n        def p(wn):\n            w, n = wn\n            return w.startswith(u) and lng >= n\n        return find(p)(table) if k else Just(('', 0))\n\n    return lambda s: maybe('*error*')(fst)(expansion(s))\n\n\n# cmdsFromString :: String -> [(String, Int)]\ndef cmdsFromString(s):\n    '''A simple rule-base consisting of a\n       list of tuples [(\n          upper case expansion string,\n          minimum character count of abbreviation\n       )],\n       obtained by a parse of single input string.\n    '''\n    def go(a, x):\n        xs, n = a\n        return (xs, int(x)) if x.isdigit() else (\n            ([(x.upper(), n)] + xs, 0)\n        )\n    return fst(reduce(\n        go,\n        reversed(re.split(r'\\s+', s)),\n        ([], 0)\n    ))\n\n\n# TEST -------------------------------------------------\ndef main():\n    '''Tests of abbreviation expansions'''\n\n    # table :: [(String, Int)]\n    table = cmdsFromString(\n        '''add 1  alter 3  backup 2  bottom 1  Cappend 2  change 1\n            Schange Cinsert 2  Clast 3 compress 4 copy 2 count 3 Coverlay 3\n            cursor 3  delete 3 Cdelete 2  down 1  duplicate 3 xEdit 1 expand 3\n            extract 3  find 1 Nfind 2 Nfindup 6 NfUP 3 Cfind 2 findUP 3 fUP 2\n            forward 2  get  help 1 hexType 4 input 1 powerInput 3  join 1\n            split 2 spltJOIN load locate 1 Clocate 2 lowerCase 3 upperCase 3\n            Lprefix 2  macro  merge 2 modify 3 move 2 msg  next 1 overlay 1\n            parse preserve 4 purge 3 put putD query 1 quit read recover 3\n            refresh renum 3 repeat 3 replace 1 Creplace 2 reset 3 restore 4\n            rgtLEFT right 2 left 2  save  set  shift 2  si  sort  sos stack 3\n            status 4 top  transfer 3  type 1  up 1'''\n    )\n\n    # tests :: [String]\n    tests = [\n        'riG   rePEAT copies  put mo   rest    types   fup.    6      poweRin',\n        ''\n    ]\n\n    print(\n        fTable(__doc__ + ':\\n')(lambda s: \"'\" + s + \"'\")(\n            lambda s: \"\\n\\t'\" + s + \"'\"\n        )(withExpansions(table))(tests)\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# find :: (a -> Bool) -> [a] -> Maybe a\ndef find(p):\n    '''Just the first element in the list that matches p,\n       or Nothing if no elements match.'''\n    def go(xs):\n        for x in xs:\n            if p(x):\n                return Just(x)\n        return Nothing()\n    return lambda xs: go(xs)\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# fTable :: String -> (a -> String)\n#                  -> (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function ->\n                 fx display function ->\n          f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + '\\n'.join([\n            xShow(x).rjust(w, ' ') + (' -> ') + fxShow(f(x))\n            for x in xs\n        ])\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if m is Nothing,\n       or the application of f to x,\n       where m is Just(x).'''\n    return lambda f: lambda m: v if m.get('Nothing') else (\n        f(m.get('Just'))\n    )\n\n\n# unwords :: [String] -> String\ndef unwords(xs):\n    '''A space-separated string derived from\n       a list of words.'''\n    return ' '.join(xs)\n\n\n# words :: String -> [String]\ndef words(s):\n    '''A list of words delimited by characters\n       representing white space.'''\n    return re.split(r'\\s+', s)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Abelian sandpile model", "language": "Python", "task": "{{wikipedia|Abelian sandpile model}}\nImplement the '''Abelian sandpile model''' also known as '''Bak-Tang-Wiesenfeld model'''. Its history, mathematical definition and properties can be found under its wikipedia article.\n\nThe task requires the creation of a 2D grid of arbitrary size on which \"piles of sand\" can be placed. Any \"pile\" that has 4 or more sand particles on it ''collapses'', resulting in '''four particles being subtracted from the pile''' and '''distributed among its neighbors.'''\n\nIt is recommended to display the output in some kind of image format, as terminal emulators are usually too small to display images larger than a few dozen characters tall. As an example of how to accomplish this, see the Bitmap/Write a PPM file task.\nExamples up to 2^30, wow!\njavascript running on web\n'''Examples:'''\n\n\n0 0 0 0 0    0 0 0 0 0\n0 0 0 0 0    0 0 1 0 0\n0 0 4 0 0 -> 0 1 0 1 0\n0 0 0 0 0    0 0 1 0 0\n0 0 0 0 0    0 0 0 0 0\n\n0 0 0 0 0    0 0 0 0 0\n0 0 0 0 0    0 0 1 0 0\n0 0 6 0 0 -> 0 1 2 1 0\n0 0 0 0 0    0 0 1 0 0\n0 0 0 0 0    0 0 0 0 0\n\n0  0 0  0  0    0 0 1 0 0\n0  0 0  0  0    0 2 1 2 0\n0  0 16 0  0 -> 1 1 0 1 1\n0  0 0  0  0    0 2 1 2 0\n0  0 0  0  0    0 0 1 0 0\n\n", "solution": "import numpy as np\nimport matplotlib.pyplot as plt\n\n\ndef iterate(grid):\n    changed = False\n    for ii, arr in enumerate(grid):\n        for jj, val in enumerate(arr):\n            if val > 3:\n                grid[ii, jj] -= 4\n                if ii > 0:\n                    grid[ii - 1, jj] += 1\n                if ii < len(grid)-1:\n                    grid[ii + 1, jj] += 1\n                if jj > 0:\n                    grid[ii, jj - 1] += 1\n                if jj < len(grid)-1:\n                    grid[ii, jj + 1] += 1\n                changed = True\n    return grid, changed\n\n\ndef simulate(grid):\n    while True:\n        grid, changed = iterate(grid)\n        if not changed:\n            return grid\n\n\nif __name__ == '__main__':\n    start_grid = np.zeros((10, 10))\n    start_grid[4:5, 4:5] = 64\n    final_grid = simulate(start_grid.copy())\n    plt.figure()\n    plt.gray()\n    plt.imshow(start_grid)\n    plt.figure()\n    plt.gray()\n    plt.imshow(final_grid)"}
{"title": "Abelian sandpile model/Identity", "language": "Python", "task": "Our sandpiles are based on a 3 by 3 rectangular grid giving nine areas that \ncontain a number from 0 to 3 inclusive. (The numbers are said to represent \ngrains of sand in each area of the sandpile).\n\nE.g. s1 =\n    \n    1 2 0\n    2 1 1\n    0 1 3\n\n\nand s2 =\n\n    2 1 3\n    1 0 1\n    0 1 0\n \n\nAddition on sandpiles is done by adding numbers in corresponding grid areas,\nso for the above:\n\n              1 2 0     2 1 3     3 3 3\n    s1 + s2 = 2 1 1  +  1 0 1  =  3 1 2\n              0 1 3     0 1 0     0 2 3\n\n\nIf the addition would result in more than 3 \"grains of sand\" in any area then \nthose areas cause the whole sandpile to become \"unstable\" and the sandpile \nareas are \"toppled\" in an \"avalanche\" until the \"stable\" result is obtained.\n\nAny unstable area (with a number >= 4), is \"toppled\" by loosing one grain of \nsand to each of its four horizontal or vertical neighbours. Grains are lost \nat the edge of the grid, but otherwise increase the number in neighbouring \ncells by one, whilst decreasing the count in the toppled cell by four in each \ntoppling.\n\nA toppling may give an adjacent area more than four grains of sand leading to\na chain of topplings called an \"avalanche\".\nE.g.\n    \n    4 3 3     0 4 3     1 0 4     1 1 0     2 1 0\n    3 1 2 ==> 4 1 2 ==> 4 2 2 ==> 4 2 3 ==> 0 3 3\n    0 2 3     0 2 3     0 2 3     0 2 3     1 2 3\n\n\nThe final result is the stable sandpile on the right. \n\n'''Note:''' The order in which cells are toppled does not affect the final result.\n\n;Task:\n* Create a class or datastructure and functions to represent and operate on sandpiles. \n* Confirm the result of the avalanche of topplings shown above\n* Confirm that s1 + s2 == s2 + s1  # Show the stable results\n* If s3 is the sandpile with number 3 in every grid area, and s3_id is the following sandpile:\n\n    2 1 2  \n    1 0 1  \n    2 1 2\n\n\n* Show that s3 + s3_id == s3\n* Show that s3_id + s3_id == s3_id\n\n\nShow confirming output here, with your examples.\n\n\n;References:\n*    https://www.youtube.com/watch?v=1MtEUErz7Gg\n*    https://en.wikipedia.org/wiki/Abelian_sandpile_model\n\n", "solution": "'''Abelian Sandpile \u2013 Identity'''\n\nfrom operator import add, eq\n\n\n# -------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Tests of cascades and additions'''\n    s0 = [[4, 3, 3], [3, 1, 2], [0, 2, 3]]\n    s1 = [[1, 2, 0], [2, 1, 1], [0, 1, 3]]\n    s2 = [[2, 1, 3], [1, 0, 1], [0, 1, 0]]\n    s3 = [[3, 3, 3], [3, 3, 3], [3, 3, 3]]\n    s3_id = [[2, 1, 2], [1, 0, 1], [2, 1, 2]]\n\n    series = list(cascadeSeries(s0))\n    for expr in [\n            'Cascade:',\n            showSandPiles(\n                [(' ', series[0])] + [\n                    (':', xs) for xs in series[1:]\n                ]\n            ),\n            '',\n            f's1 + s2 == s2 + s1 -> {addSand(s1)(s2) == addSand(s2)(s1)}',\n            showSandPiles([\n                (' ', s1),\n                ('+', s2),\n                ('=', addSand(s1)(s2))\n            ]),\n            '',\n            showSandPiles([\n                (' ', s2),\n                ('+', s1),\n                ('=', addSand(s2)(s1))\n            ]),\n            '',\n            f's3 + s3_id == s3 -> {addSand(s3)(s3_id) == s3}',\n            showSandPiles([\n                (' ', s3),\n                ('+', s3_id),\n                ('=', addSand(s3)(s3_id))\n            ]),\n            '',\n            f's3_id + s3_id == s3_id -> {addSand(s3_id)(s3_id) == s3_id}',\n            showSandPiles([\n                (' ', s3_id),\n                ('+', s3_id),\n                ('=', addSand(s3_id)(s3_id))\n            ]),\n\n    ]:\n        print(expr)\n\n\n# ----------------------- SANDPILES ------------------------\n\n# addSand :: [[Int]] -> [[Int]] -> [[Int]]\ndef addSand(xs):\n    '''The stabilised sum of two sandpiles.\n    '''\n    def go(ys):\n        return cascadeSeries(\n            chunksOf(len(xs))(\n                map(\n                    add,\n                    concat(xs),\n                    concat(ys)\n                )\n            )\n        )[-1]\n    return go\n\n\n# cascadeSeries :: [[Int]] -> [[[Int]]]\ndef cascadeSeries(rows):\n    '''The sequence of states from a given\n       sand pile to a stable condition.\n    '''\n    xs = list(rows)\n    w = len(xs)\n    return [\n        list(chunksOf(w)(x)) for x\n        in convergence(eq)(\n            iterate(nextState(w))(\n                concat(xs)\n            )\n        )\n    ]\n\n\n# convergence :: (a -> a -> Bool) -> [a] -> [a]\ndef convergence(p):\n    '''All items of xs to the point where the binary\n       p returns True over two successive values.\n    '''\n    def go(xs):\n        def conv(prev, ys):\n            y = next(ys)\n            return [prev] + (\n                [] if p(prev, y) else conv(y, ys)\n            )\n        return conv(next(xs), xs)\n    return go\n\n\n# nextState Int -> Int -> [Int] -> [Int]\ndef nextState(w):\n    '''The next state of a (potentially unstable)\n       flattened sand-pile matrix of row length w.\n    '''\n    def go(xs):\n        def tumble(i):\n            neighbours = indexNeighbours(w)(i)\n            return [\n                1 + k if j in neighbours else (\n                    k - (1 + w) if j == i else k\n                ) for (j, k) in enumerate(xs)\n            ]\n        return maybe(xs)(tumble)(\n            findIndex(lambda x: w < x)(xs)\n        )\n    return go\n\n\n# indexNeighbours :: Int -> Int -> [Int]\ndef indexNeighbours(w):\n    '''Indices vertically and horizontally adjoining the\n       given index in a flattened matrix of dimension w.\n    '''\n    def go(i):\n        lastCol = w - 1\n        iSqr = (w * w)\n        col = i % w\n        return [\n            j for j in [i - w, i + w]\n            if -1 < j < iSqr\n        ] + ([i - 1] if 0 != col else []) + (\n            [1 + i] if lastCol != col else []\n        )\n    return go\n\n\n# ------------------------ DISPLAY -------------------------\n\n# showSandPiles :: [(String, [[Int]])] -> String\ndef showSandPiles(pairs):\n    '''Indented multi-line representation\n       of a sequence of matrices, delimited\n       by preceding operators or indents.\n    '''\n    return '\\n'.join([\n        ' '.join([' '.join(map(str, seq)) for seq in tpl])\n        for tpl in zip(*[\n            zip(\n                *[list(str(pfx).center(len(rows)))]\n                + list(zip(*rows))\n            )\n            for (pfx, rows) in pairs\n        ])\n    ])\n\n\n# ------------------------ GENERIC -------------------------\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n, subdividing the\n       contents of xs. Where the length of xs is not evenly\n       divible, the final list will be shorter than n.\n    '''\n    def go(xs):\n        ys = list(xs)\n        return (\n            ys[i:n + i] for i in range(0, len(ys), n)\n        ) if 0 < n else None\n    return go\n\n\n# concat :: [[a]] -> [a]\ndef concat(xs):\n    '''The concatenation of all\n       elements in a list.\n    '''\n    return [x for lst in xs for x in lst]\n\n\n# findIndex :: (a -> Bool) -> [a] -> Maybe Int\ndef findIndex(p):\n    '''Just the first index at which an\n       element in xs matches p,\n       or Nothing if no elements match.\n    '''\n    def go(xs):\n        return next(\n            (i for (i, x) in enumerate(xs) if p(x)),\n            None\n        )\n    return go\n\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated\n       applications of f to x.\n    '''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return go\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if x is None,\n       or the application of f to x.\n    '''\n    def go(f):\n        def g(x):\n            return v if None is x else f(x)\n        return g\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Abundant odd numbers", "language": "Python from Visual Basic .NET", "task": "An Abundant number is a number '''n''' for which the   ''sum of divisors''   '''s(n) > 2n''',\nor,   equivalently,   the   ''sum of proper divisors''   (or aliquot sum)       '''s(n) > n'''.\n\n\n;E.G.:\n'''12'''   is abundant, it has the proper divisors     '''1,2,3,4 & 6'''     which sum to   '''16'''   ( > '''12''' or '''n'''); \n       or alternately,   has the sigma sum of   '''1,2,3,4,6 & 12'''   which sum to   '''28'''   ( > '''24''' or '''2n''').\n\n\nAbundant numbers are common, though '''even''' abundant numbers seem to be much more common than '''odd''' abundant numbers. \n\nTo make things more interesting, this task is specifically about finding   ''odd abundant numbers''.\n\n\n;Task\n*Find and display here: at least the first 25 abundant odd numbers and either their proper divisor sum or sigma sum.\n*Find and display here: the one thousandth abundant odd number and either its proper divisor sum or sigma sum.\n*Find and display here: the first abundant odd number greater than one billion (109) and either its proper divisor sum or sigma sum.\n\n\n;References:\n:*   OEIS:A005231: Odd abundant numbers (odd numbers n whose sum of divisors exceeds 2n)\n:*   American Journal of Mathematics, Vol. 35, No. 4 (Oct., 1913), pp. 413-422 - Finiteness of the Odd Perfect and Primitive Abundant Numbers with n Distinct Prime Factors (LE Dickson)\n\n", "solution": "#!/usr/bin/python\n# Abundant odd numbers - Python\n\noddNumber  = 1\naCount  = 0\ndSum  = 0\n \nfrom math import sqrt\n \ndef divisorSum(n):\n    sum = 1\n    i = int(sqrt(n)+1)\n \n    for d in range (2, i):\n        if n % d == 0:\n            sum += d\n            otherD = n // d\n            if otherD != d:\n                sum += otherD\n    return sum\n \nprint (\"The first 25 abundant odd numbers:\")\nwhile aCount  < 25:\n    dSum  = divisorSum(oddNumber )\n    if dSum  > oddNumber :\n        aCount  += 1\n        print(\"{0:5} proper divisor sum: {1}\". format(oddNumber ,dSum ))\n    oddNumber  += 2\n \nwhile aCount  < 1000:\n    dSum  = divisorSum(oddNumber )\n    if dSum  > oddNumber :\n        aCount  += 1\n    oddNumber  += 2\nprint (\"\\n1000th abundant odd number:\")\nprint (\"    \",(oddNumber - 2),\" proper divisor sum: \",dSum)\n \noddNumber  = 1000000001\nfound  = False\nwhile not found :\n    dSum  = divisorSum(oddNumber )\n    if dSum  > oddNumber :\n        found  = True\n        print (\"\\nFirst abundant odd number > 1 000 000 000:\")\n        print (\"    \",oddNumber,\" proper divisor sum: \",dSum)\n    oddNumber  += 2"}
{"title": "Accumulator factory", "language": "Python 2.x/3.x", "task": "{{requires|Mutable State}}\n\nA problem posed by Paul Graham is that of creating a function that takes a single (numeric) argument and which returns another function that is an accumulator. The returned accumulator function in turn also takes a single numeric argument, and returns the sum of all the numeric values passed in so far to that accumulator (including the initial value passed when the accumulator was created).\n\n\n;Rules:\nThe detailed rules are at http://paulgraham.com/accgensub.html and are reproduced here for simplicity (with additions in ''small italic text'').\n:Before you submit an example, make sure the function\n\n:# Takes a number n and returns a function (lets call it g), that takes a number i, and returns n incremented by the accumulation of i from every call of function g(i).Although these exact function and parameter names need not be used\n:# Works for any numeric type-- i.e. can take both ints and floats and returns functions that can take both ints and floats. (It is not enough simply to convert all input to floats. An accumulator that has only seen integers must return integers.) ''(i.e., if the language doesn't allow for numeric polymorphism, you have to use overloading or something like that)''\n:# Generates functions that return the sum of every number ever passed to them, not just the most recent. ''(This requires a piece of state to hold the accumulated value, which in turn means that pure functional languages can't be used for this task.)''\n:# Returns a real function, meaning something that you can use wherever you could use a function you had defined in the ordinary way in the text of your program. ''(Follow your language's conventions here.)''\n:# Doesn't store the accumulated value or the returned functions in a way that could cause them to be inadvertently modified by other code. ''(No global variables or other such things.)''\n: E.g. if after the example, you added the following code (in a made-up language) ''where the factory function is called foo'':\n:: x = foo(1); \nx(5); \nfoo(3);\nprint x(2.3);\n: It should print 8.3. ''(There is no need to print the form of the accumulator function returned by foo(3); it's not part of the task at all.)''\n\n\n;Task:\nCreate a function that implements the described rules. \n\n\nIt need not handle any special error cases not described above. The simplest way to implement the task as described is typically to use a closure, providing the language supports them.\n\nWhere it is not possible to hold exactly to the constraints above, describe the deviations.\n\n", "solution": ">>> def accumulator(sum):\n  def f(n):\n    f.sum += n\n    return f.sum\n  f.sum = sum\n  return f\n\n>>> x = accumulator(1)\n>>> x(5)\n6\n>>> x(2.3)\n8.3000000000000007\n>>> x = accumulator(1)\n>>> x(5)\n6\n>>> x(2.3)\n8.3000000000000007\n>>> x2 = accumulator(3)\n>>> x2(5)\n8\n>>> x2(3.3)\n11.300000000000001\n>>> x(0)\n8.3000000000000007\n>>> x2(0)\n11.300000000000001"}
{"title": "Accumulator factory", "language": "Python 3.x", "task": "{{requires|Mutable State}}\n\nA problem posed by Paul Graham is that of creating a function that takes a single (numeric) argument and which returns another function that is an accumulator. The returned accumulator function in turn also takes a single numeric argument, and returns the sum of all the numeric values passed in so far to that accumulator (including the initial value passed when the accumulator was created).\n\n\n;Rules:\nThe detailed rules are at http://paulgraham.com/accgensub.html and are reproduced here for simplicity (with additions in ''small italic text'').\n:Before you submit an example, make sure the function\n\n:# Takes a number n and returns a function (lets call it g), that takes a number i, and returns n incremented by the accumulation of i from every call of function g(i).Although these exact function and parameter names need not be used\n:# Works for any numeric type-- i.e. can take both ints and floats and returns functions that can take both ints and floats. (It is not enough simply to convert all input to floats. An accumulator that has only seen integers must return integers.) ''(i.e., if the language doesn't allow for numeric polymorphism, you have to use overloading or something like that)''\n:# Generates functions that return the sum of every number ever passed to them, not just the most recent. ''(This requires a piece of state to hold the accumulated value, which in turn means that pure functional languages can't be used for this task.)''\n:# Returns a real function, meaning something that you can use wherever you could use a function you had defined in the ordinary way in the text of your program. ''(Follow your language's conventions here.)''\n:# Doesn't store the accumulated value or the returned functions in a way that could cause them to be inadvertently modified by other code. ''(No global variables or other such things.)''\n: E.g. if after the example, you added the following code (in a made-up language) ''where the factory function is called foo'':\n:: x = foo(1); \nx(5); \nfoo(3);\nprint x(2.3);\n: It should print 8.3. ''(There is no need to print the form of the accumulator function returned by foo(3); it's not part of the task at all.)''\n\n\n;Task:\nCreate a function that implements the described rules. \n\n\nIt need not handle any special error cases not described above. The simplest way to implement the task as described is typically to use a closure, providing the language supports them.\n\nWhere it is not possible to hold exactly to the constraints above, describe the deviations.\n\n", "solution": "def accumulator(sum):\n  def f(n):\n    nonlocal sum\n    sum += n\n    return sum\n  return f\n\nx = accumulator(1)\nx(5)\nprint(accumulator(3))\nprint(x(2.3))"}
{"title": "Accumulator factory", "language": "Python 2.5+", "task": "{{requires|Mutable State}}\n\nA problem posed by Paul Graham is that of creating a function that takes a single (numeric) argument and which returns another function that is an accumulator. The returned accumulator function in turn also takes a single numeric argument, and returns the sum of all the numeric values passed in so far to that accumulator (including the initial value passed when the accumulator was created).\n\n\n;Rules:\nThe detailed rules are at http://paulgraham.com/accgensub.html and are reproduced here for simplicity (with additions in ''small italic text'').\n:Before you submit an example, make sure the function\n\n:# Takes a number n and returns a function (lets call it g), that takes a number i, and returns n incremented by the accumulation of i from every call of function g(i).Although these exact function and parameter names need not be used\n:# Works for any numeric type-- i.e. can take both ints and floats and returns functions that can take both ints and floats. (It is not enough simply to convert all input to floats. An accumulator that has only seen integers must return integers.) ''(i.e., if the language doesn't allow for numeric polymorphism, you have to use overloading or something like that)''\n:# Generates functions that return the sum of every number ever passed to them, not just the most recent. ''(This requires a piece of state to hold the accumulated value, which in turn means that pure functional languages can't be used for this task.)''\n:# Returns a real function, meaning something that you can use wherever you could use a function you had defined in the ordinary way in the text of your program. ''(Follow your language's conventions here.)''\n:# Doesn't store the accumulated value or the returned functions in a way that could cause them to be inadvertently modified by other code. ''(No global variables or other such things.)''\n: E.g. if after the example, you added the following code (in a made-up language) ''where the factory function is called foo'':\n:: x = foo(1); \nx(5); \nfoo(3);\nprint x(2.3);\n: It should print 8.3. ''(There is no need to print the form of the accumulator function returned by foo(3); it's not part of the task at all.)''\n\n\n;Task:\nCreate a function that implements the described rules. \n\n\nIt need not handle any special error cases not described above. The simplest way to implement the task as described is typically to use a closure, providing the language supports them.\n\nWhere it is not possible to hold exactly to the constraints above, describe the deviations.\n\n", "solution": "def accumulator(sum):\n  while True:\n    sum += yield sum\n\nx = accumulator(1)\nx.send(None)\nx.send(5)\nprint(accumulator(3))\nprint(x.send(2.3))"}
{"title": "Achilles numbers", "language": "Python", "task": "{{Wikipedia|Achilles number}}\n\n\nAn '''Achilles number''' is a number that is powerful but imperfect. ''Named after Achilles, a hero of the Trojan war, who was also powerful but imperfect.''\n\n\nA positive integer '''n''' is a powerful number if, for every prime factor '''p''' of '''n''', '''p2''' is also a divisor.\n\nIn other words, every prime factor appears at least squared in the factorization.\n\nAll '''Achilles numbers''' are powerful. However, not all powerful numbers are '''Achilles numbers''': only those that cannot be represented as '''mk''', where '''m''' and '''k''' are positive integers greater than '''1'''.\n\n\nA '''''strong'' Achilles number''' is an '''Achilles number''' whose '''Euler totient ()''' is also an '''Achilles number'''.\n\n\n\n;E.G.\n\n'''108''' is a powerful number. Its prime factorization is '''22 x 33''', and thus its prime factors are '''2''' and '''3'''. Both '''22 = 4''' and  '''32 = 9''' are divisors of '''108'''. However, '''108''' cannot be represented as '''mk''', where '''m''' and '''k''' are positive integers greater than '''1''', so '''108''' is an '''Achilles number'''.\n\n'''360''' is ''not'' an '''Achilles number''' because it is not powerful. One of its prime factors is '''5''' but '''360''' is not divisible by '''52 = 25'''.\n\nFinally, '''784''' is ''not'' an '''Achilles number'''. It is a powerful number, because not only are '''2''' and '''7''' its only prime factors, but also '''22 = 4''' and '''72 = 49''' are divisors of it. Nonetheless, it is a perfect power; its square root is an even integer, so it is ''not'' an '''Achilles number'''.\n\n\n'''500 = 22 x 53''' is a '''''strong'' Achilles number''' as its Euler totient, '''(500)''', is '''200 = 23 x 52''' which is ''also'' an '''Achilles number'''.\n\n\n\n;Task\n\n* Find and show the first 50 '''Achilles numbers'''.\n* Find and show at least the first 20 '''''strong'' Achilles numbers'''.\n* For at least 2 through 5, show the count of '''Achilles numbers''' with that many digits.\n\n\n;See also\n\n;* Wikipedia: Achilles number\n;* OEIS:A052486 - Achilles numbers - powerful but imperfect numbers\n;* OEIS:A194085 - Strong Achilles numbers: Achilles numbers m such that phi(m) is also an Achilles number\n;* Related task: Powerful numbers\n;* Related task: Totient function\n\n\n", "solution": "from math import gcd\nfrom sympy import factorint\n \ndef is_Achilles(n):\n    p = factorint(n).values()\n    return all(i > 1 for i in p) and gcd(*p) == 1\n\ndef is_strong_Achilles(n):\n    return is_Achilles(n) and is_Achilles(totient(n))\n \ndef test_strong_Achilles(nachilles, nstrongachilles):\n    # task 1\n    print('First', nachilles, 'Achilles numbers:')\n    n, found = 0, 0\n    while found < nachilles:\n        if is_Achilles(n):\n            found += 1\n            print(f'{n: 8,}', end='\\n' if found % 10 == 0 else '')\n        n += 1\n\n    # task 2\n    print('\\nFirst', nstrongachilles, 'strong Achilles numbers:')\n    n, found = 0, 0\n    while found < nstrongachilles:\n        if is_strong_Achilles(n):\n            found += 1\n            print(f'{n: 9,}', end='\\n' if found % 10 == 0 else '')\n        n += 1\n\n    # task 3\n    print('\\nCount of Achilles numbers for various intervals:')\n    intervals = [[10, 99], [100, 999], [1000, 9999], [10000, 99999], [100000, 999999]]\n    for interval in intervals:\n        print(f'{interval}:', sum(is_Achilles(i) for i in range(*interval)))\n\n\ntest_strong_Achilles(50, 100)\n"}
{"title": "Aliquot sequence classifications", "language": "Python", "task": "An aliquot sequence of a positive integer K is defined recursively as the first member\nbeing K and subsequent members being the sum of the [[Proper divisors]] of the previous term.\n\n:* If the terms eventually reach 0 then the series for K is said to '''terminate'''.\n\n:There are several classifications for non termination:\n:* If the second term is K then all future terms are also K and so the sequence repeats from the first term with period 1 and K is called '''perfect'''.\n:* If the third term ''would'' be repeating K then the sequence repeats with period 2 and K is called '''amicable'''.\n:* If the Nth term ''would'' be repeating K for the first time, with N > 3 then the sequence repeats with period N - 1 and K is called '''sociable'''.\n:Perfect, amicable and sociable numbers eventually repeat the original number K; there are other repetitions...\n:* Some K have a sequence that eventually forms a periodic repetition of period 1 but of a number other than K, for example 95 which forms the sequence 95, 25, 6, 6, 6, ... such K are called '''aspiring'''.\n:* K that have a sequence that eventually forms a periodic repetition of period >= 2 but of a number other than K, for example 562 which forms the sequence 562, 284, 220, 284, 220, ... such K are called '''cyclic'''.\n\n:And finally:\n:* Some K form aliquot sequences that are not known to be either terminating or periodic; these K are to be called '''non-terminating'''. For the purposes of this task, K is to be classed as non-terminating if it has not been otherwise classed after generating '''16''' terms or if any term of the sequence is greater than 2**47 = 140,737,488,355,328. \n\n\n;Task:\n# Create routine(s) to generate the aliquot sequence of a positive integer enough to classify it according to the classifications given above.\n# Use it to display the classification and sequences of the numbers one to ten inclusive.\n# Use it to show the classification and sequences of the following integers, in order:\n:: 11, 12, 28, 496, 220, 1184,  12496, 1264460, 790, 909, 562, 1064, 1488, and optionally 15355717786080.\n\nShow all output on this page.\n\n\n;Related tasks:\n*   [[Abundant, deficient and perfect number classifications]]. (Classifications from only the first two members of the whole sequence).\n*   [[Proper divisors]]\n*   [[Amicable pairs]]\n\n", "solution": "from proper_divisors import proper_divs\nfrom functools import lru_cache\n\n\n@lru_cache()\ndef pdsum(n): \n    return sum(proper_divs(n))\n    \n    \ndef aliquot(n, maxlen=16, maxterm=2**47):\n    if n == 0:\n        return 'terminating', [0]\n    s, slen, new = [n], 1, n\n    while slen <= maxlen and new < maxterm:\n        new = pdsum(s[-1])\n        if new in s:\n            if s[0] == new:\n                if slen == 1:\n                    return 'perfect', s\n                elif slen == 2:\n                    return 'amicable', s\n                else:\n                    return 'sociable of length %i' % slen, s\n            elif s[-1] == new:\n                return 'aspiring', s\n            else:\n                return 'cyclic back to %i' % new, s\n        elif new == 0:\n            return 'terminating', s + [0]\n        else:\n            s.append(new)\n            slen += 1\n    else:\n        return 'non-terminating', s\n                \nif __name__ == '__main__':\n    for n in range(1, 11): \n        print('%s: %r' % aliquot(n))\n    print()\n    for n in [11, 12, 28, 496, 220, 1184,  12496, 1264460, 790, 909, 562, 1064, 1488, 15355717786080]: \n        print('%s: %r' % aliquot(n))"}
{"title": "Almkvist-Giullera formula for pi", "language": "Python", "task": "The Almkvist-Giullera formula for calculating   1/p2   is based on the Calabi-Yau\ndifferential equations of order 4 and 5,   which were originally used to describe certain manifolds\nin string theory. \n\n\nThe formula is:\n::::: 1/p2 = (25/3)  0  ((6n)! / (n!6))(532n2 + 126n + 9) / 10002n+1\n\n\nThis formula can be used to calculate the constant   p-2,   and thus to calculate   p.\n\nNote that, because the product of all terms but the power of 1000 can be calculated as an integer,\nthe terms in the series can be separated into a large integer term:\n\n:::::  (25) (6n)! (532n2 + 126n + 9) / (3(n!)6)        (***)\n\nmultiplied by a negative integer power of 10:\n\n:::::  10-(6n + 3) \n\n\n;Task:\n:* Print the integer portions (the starred formula, which is without the power of 1000 divisor) of the first 10 terms of the series.\n:* Use the complete formula to calculate and print p to 70 decimal digits of precision.\n\n\n;Reference:\n:*  Gert Almkvist and Jesus Guillera, Ramanujan-like series for 1/p2 and string theory, Experimental Mathematics, 21 (2012), page 2, formula 1.\n\n", "solution": "import mpmath as mp\n\nwith mp.workdps(72):\n\n    def integer_term(n):\n        p = 532 * n * n + 126 * n + 9\n        return (p * 2**5 * mp.factorial(6 * n)) / (3 * mp.factorial(n)**6)\n\n    def exponent_term(n):\n        return -(mp.mpf(\"6.0\") * n + 3)\n\n    def nthterm(n):\n        return integer_term(n) * mp.mpf(\"10.0\")**exponent_term(n)\n\n\n    for n in range(10):\n        print(\"Term \", n, '  ', int(integer_term(n)))\n\n\n    def almkvist_guillera(floatprecision):\n        summed, nextadd = mp.mpf('0.0'), mp.mpf('0.0')\n        for n in range(100000000):\n            nextadd = summed + nthterm(n)\n            if abs(nextadd - summed) < 10.0**(-floatprecision):\n                break\n\n            summed = nextadd\n\n        return nextadd\n\n\n    print('\\n\u03c0 to 70 digits is ', end='')\n    mp.nprint(mp.mpf(1.0 / mp.sqrt(almkvist_guillera(70))), 71)\n    print('mpmath \u03c0 is       ', end='')\n    mp.nprint(mp.pi, 71)\n"}
{"title": "Amb", "language": "Python", "task": "Define and give an example of the Amb operator.\n\nThe Amb operator (short for \"ambiguous\") expresses nondeterminism. This doesn't refer to randomness (as in \"nondeterministic universe\") but is closely related to the term as it is used in automata theory (\"non-deterministic finite automaton\").\n\nThe Amb operator takes a variable number of expressions (or values if that's simpler in the language) and yields a correct one which will satisfy a constraint in some future computation, thereby avoiding failure.\n\nProblems whose solution the Amb operator naturally expresses can be approached with other tools, such as explicit nested iterations over data sets, or with pattern matching. By contrast, the Amb operator appears integrated into the language. Invocations of Amb are not wrapped in any visible loops or other search patterns; they appear to be independent. \n\nEssentially Amb(x, y, z) splits the computation into three possible futures: a future in which the value x is yielded, a future in which the value y is yielded and a future in which the value z is yielded. The future which leads to a successful subsequent computation is chosen. The other \"parallel universes\" somehow go away.   Amb called with no arguments fails.\n\nFor simplicity, one of the domain values usable with Amb may denote failure, if that is convenient. For instance, it is convenient if a Boolean false denotes failure, so that Amb(false) fails, and thus constraints can be expressed using Boolean expressions like Amb(x * y == 8) which unless x and y add to four.\n\nA pseudo-code program which satisfies this constraint might look like:\n\nlet x = Amb(1, 2, 3)\nlet y = Amb(7, 6, 4, 5)\nAmb(x * y = 8)\nprint x, y\n\nThe output is 2 4 because Amb(1, 2, 3) correctly chooses the future in which x has value 2, Amb(7, 6, 4, 5) chooses 4 and consequently Amb(x * y = 8) produces a success.\n\nAlternatively, failure could be represented using strictly Amb():\n\nunless x * y = 8 do Amb()\n\nOr else Amb could take the form of two operators or functions: one for producing values and one for enforcing constraints:\n\nlet x = Ambsel(1, 2, 3)\nlet y = Ambsel(4, 5, 6)\nAmbassert(x * y = 8)\nprint x, y\n\nwhere Ambassert behaves like Amb() if the Boolean expression is false, otherwise it allows the future computation to take place, without yielding any value.\n\nThe task is to somehow implement Amb, and demonstrate it with a program which chooses one word from each of the following four sets of character strings to generate a four-word sentence:\n\n#\"the\" \"that\" \"a\"\n#\"frog\" \"elephant\" \"thing\"\n#\"walked\" \"treaded\" \"grows\"\n#\"slowly\" \"quickly\"\n\nThe constraint to be satisfied is that the last character of each word (other than the last) is the same as the first character of its successor.\n\nThe only successful sentence is \"that thing grows slowly\"; other combinations do not satisfy the constraint and thus fail.\n\nThe goal of this task isn't to simply process the four lists of words with explicit, deterministic program flow such as nested iteration, to trivially demonstrate the correct output. The goal is to implement the Amb operator, or a facsimile thereof that is possible within the language limitations.\n\n", "solution": "import itertools as _itertools\n \nclass Amb(object):\n    def __init__(self):\n        self._names2values   = {}       # set of values for each global name\n        self._func           = None     # Boolean constraint function\n        self._valueiterator  = None     # itertools.product of names values\n        self._funcargnames   = None     # Constraint parameter names\n \n    def __call__(self, arg=None):\n        if hasattr(arg, '__code__'):                \n            ##\n            ## Called with a constraint function. \n            ##\n            globls = arg.__globals__ if hasattr(arg, '__globals__') else arg.func_globals\n            # Names used in constraint\n            argv = arg.__code__.co_varnames[:arg.__code__.co_argcount]\n            for name in argv:\n                if name not in self._names2values:\n                    assert name in globls, \\\n                           \"Global name %s not found in function globals\" % name\n                    self._names2values[name] = globls[name]\n            # Gather the range of values of all names used in the constraint\n            valuesets = [self._names2values[name] for name in argv]\n            self._valueiterator = _itertools.product(*valuesets)\n            self._func = arg\n            self._funcargnames = argv\n            return self\n        elif arg is not None:\n            ##\n            ## Assume called with an iterable set of values\n            ##\n            arg = frozenset(arg)\n            return arg\n        else:\n            ##\n            ## blank call tries to return next solution\n            ##\n            return self._nextinsearch()\n \n    def _nextinsearch(self):\n        arg = self._func\n        globls = arg.__globals__\n        argv = self._funcargnames\n        found = False\n        for values in self._valueiterator:\n            if arg(*values):\n                # Set globals.\n                found = True\n                for n, v in zip(argv, values):\n                    globls[n] = v\n                break\n        if not found: raise StopIteration\n        return values\n \n    def __iter__(self):\n        return self\n \n    def __next__(self):\n        return self()\n    next = __next__ # Python 2\n \nif __name__ == '__main__':\n    if True:\n        amb = Amb()\n \n        print(\"\\nSmall Pythagorean triples problem:\")\n        x = amb(range(1,11))\n        y = amb(range(1,11))\n        z = amb(range(1,11))\n \n        for _dummy in amb( lambda x, y, z: x*x + y*y == z*z ):\n            print ('%s %s %s' % (x, y, z))\n \n \n    if True:\n        amb = Amb()\n \n        print(\"\\nRosetta Code Amb problem:\")\n        w1 = amb([\"the\", \"that\", \"a\"])\n        w2 = amb([\"frog\", \"elephant\", \"thing\"])\n        w3 = amb([\"walked\", \"treaded\", \"grows\"])\n        w4 = amb([\"slowly\", \"quickly\"])\n \n        for _dummy in amb( lambda w1, w2, w3, w4: \\\n                             w1[-1] == w2[0] and \\\n                             w2[-1] == w3[0] and \\\n                             w3[-1] == w4[0] ):\n            print ('%s %s %s %s' % (w1, w2, w3, w4))\n \n    if True:\n        amb = Amb()\n \n        print(\"\\nAmb problem from \"\n            \"http://www.randomhacks.net/articles/2005/10/11/amb-operator:\")\n        x = amb([1, 2, 3])\n        y = amb([4, 5, 6])\n \n        for _dummy in amb( lambda x, y: x * y != 8 ):\n            print ('%s %s' % (x, y))"}
{"title": "Amb", "language": "Python from Haskell", "task": "Define and give an example of the Amb operator.\n\nThe Amb operator (short for \"ambiguous\") expresses nondeterminism. This doesn't refer to randomness (as in \"nondeterministic universe\") but is closely related to the term as it is used in automata theory (\"non-deterministic finite automaton\").\n\nThe Amb operator takes a variable number of expressions (or values if that's simpler in the language) and yields a correct one which will satisfy a constraint in some future computation, thereby avoiding failure.\n\nProblems whose solution the Amb operator naturally expresses can be approached with other tools, such as explicit nested iterations over data sets, or with pattern matching. By contrast, the Amb operator appears integrated into the language. Invocations of Amb are not wrapped in any visible loops or other search patterns; they appear to be independent. \n\nEssentially Amb(x, y, z) splits the computation into three possible futures: a future in which the value x is yielded, a future in which the value y is yielded and a future in which the value z is yielded. The future which leads to a successful subsequent computation is chosen. The other \"parallel universes\" somehow go away.   Amb called with no arguments fails.\n\nFor simplicity, one of the domain values usable with Amb may denote failure, if that is convenient. For instance, it is convenient if a Boolean false denotes failure, so that Amb(false) fails, and thus constraints can be expressed using Boolean expressions like Amb(x * y == 8) which unless x and y add to four.\n\nA pseudo-code program which satisfies this constraint might look like:\n\nlet x = Amb(1, 2, 3)\nlet y = Amb(7, 6, 4, 5)\nAmb(x * y = 8)\nprint x, y\n\nThe output is 2 4 because Amb(1, 2, 3) correctly chooses the future in which x has value 2, Amb(7, 6, 4, 5) chooses 4 and consequently Amb(x * y = 8) produces a success.\n\nAlternatively, failure could be represented using strictly Amb():\n\nunless x * y = 8 do Amb()\n\nOr else Amb could take the form of two operators or functions: one for producing values and one for enforcing constraints:\n\nlet x = Ambsel(1, 2, 3)\nlet y = Ambsel(4, 5, 6)\nAmbassert(x * y = 8)\nprint x, y\n\nwhere Ambassert behaves like Amb() if the Boolean expression is false, otherwise it allows the future computation to take place, without yielding any value.\n\nThe task is to somehow implement Amb, and demonstrate it with a program which chooses one word from each of the following four sets of character strings to generate a four-word sentence:\n\n#\"the\" \"that\" \"a\"\n#\"frog\" \"elephant\" \"thing\"\n#\"walked\" \"treaded\" \"grows\"\n#\"slowly\" \"quickly\"\n\nThe constraint to be satisfied is that the last character of each word (other than the last) is the same as the first character of its successor.\n\nThe only successful sentence is \"that thing grows slowly\"; other combinations do not satisfy the constraint and thus fail.\n\nThe goal of this task isn't to simply process the four lists of words with explicit, deterministic program flow such as nested iteration, to trivially demonstrate the correct output. The goal is to implement the Amb operator, or a facsimile thereof that is possible within the language limitations.\n\n", "solution": "# joins :: String -> String -> Bool\ndef joins(a, b):\n    return a[-1] == b[0]\n\n\nprint (\n    [\n        ' '.join([w1, w2, w3, w4])\n        for w1 in ['the', 'that', 'a']\n        for w2 in ['frog', 'elephant', 'thing']\n        for w3 in ['walked', 'treaded', 'grows']\n        for w4 in ['slowly', 'quickly']\n        if joins(w1, w2) and joins(w2, w3) and joins(w3, w4)\n    ]\n)"}
{"title": "Anagrams/Deranged anagrams", "language": "Python", "task": "Two or more words are said to be anagrams if they have the same characters, but in a different order. \n\nBy analogy with derangements we define a  ''deranged anagram'' as two words with the same characters, but in which the same character does ''not'' appear in the same position in both words.\n\n;Task\n\nUse the word list at unixdict to find and display the longest deranged anagram. \n\n\n;Related\n* [[Permutations/Derangements]]\n* Best shuffle\n\n{{Related tasks/Word plays}}\n\n\n\n", "solution": "import urllib.request\nfrom collections import defaultdict\nfrom itertools import combinations\n\ndef getwords(url='http://www.puzzlers.org/pub/wordlists/unixdict.txt'):\n    return list(set(urllib.request.urlopen(url).read().decode().split()))\n\ndef find_anagrams(words):\n    anagram = defaultdict(list) # map sorted chars to anagrams\n    for word in words:\n        anagram[tuple(sorted(word))].append( word )\n    return dict((key, words) for key, words in anagram.items()\n                if len(words) > 1)\n\ndef is_deranged(words):\n    'returns pairs of words that have no character in the same position'\n    return [ (word1, word2)\n             for word1,word2 in combinations(words, 2)\n             if all(ch1 != ch2 for ch1, ch2 in zip(word1, word2)) ]\n\ndef largest_deranged_ana(anagrams):\n    ordered_anagrams = sorted(anagrams.items(),\n                              key=lambda x:(-len(x[0]), x[0]))\n    for _, words in ordered_anagrams:\n        deranged_pairs = is_deranged(words)\n        if deranged_pairs:\n            return deranged_pairs\n    return []\n\nif __name__ == '__main__':\n    words = getwords('http://www.puzzlers.org/pub/wordlists/unixdict.txt')\n    print(\"Word count:\", len(words))\n\n    anagrams = find_anagrams(words)\n    print(\"Anagram count:\", len(anagrams),\"\\n\")\n\n    print(\"Longest anagrams with no characters in the same position:\")\n    print('  ' + '\\n  '.join(', '.join(pairs)\n                             for pairs in largest_deranged_ana(anagrams)))"}
{"title": "Angle difference between two bearings", "language": "Python from C++", "task": "Finding the angle between two bearings is often confusing.[https://stackoverflow.com/questions/16180595/find-the-angle-between-two-bearings]\n\n\n;Task:\nFind the angle which is the result of the subtraction '''b2 - b1''', where '''b1''' and '''b2''' are the bearings. \n\nInput bearings are expressed in the range   '''-180'''   to   '''+180'''   degrees. \nThe  result  is also expressed in the range   '''-180'''   to   '''+180'''   degrees. \n\n\nCompute the angle for the following pairs:\n*  20 degrees ('''b1''') and 45 degrees ('''b2''')\n* -45 and  45\n* -85 and  90\n* -95 and  90\n* -45 and 125\n* -45 and 145\n*  29.4803 and  -88.6381\n* -78.3251 and -159.036\n\n\n;Optional extra:  \nAllow the input bearings to be any (finite) value. \n\n\n;Test cases:\n*  -70099.74233810938   and   29840.67437876723\n* -165313.6666297357    and   33693.9894517456\n*    1174.8380510598456 and -154146.66490124757\n*   60175.77306795546   and   42213.07192354373\n\n", "solution": "from __future__ import print_function\n \ndef getDifference(b1, b2):\n\tr = (b2 - b1) % 360.0\n\t# Python modulus has same sign as divisor, which is positive here,\n\t# so no need to consider negative case\n\tif r >= 180.0:\n\t\tr -= 360.0\n\treturn r\n \nif __name__ == \"__main__\":\n\tprint (\"Input in -180 to +180 range\")\n\tprint (getDifference(20.0, 45.0))\n\tprint (getDifference(-45.0, 45.0))\n\tprint (getDifference(-85.0, 90.0))\n\tprint (getDifference(-95.0, 90.0))\n\tprint (getDifference(-45.0, 125.0))\n\tprint (getDifference(-45.0, 145.0))\n\tprint (getDifference(-45.0, 125.0))\n\tprint (getDifference(-45.0, 145.0))\n\tprint (getDifference(29.4803, -88.6381))\n\tprint (getDifference(-78.3251, -159.036))\n \n\tprint (\"Input in wider range\")\n\tprint (getDifference(-70099.74233810938, 29840.67437876723))\n\tprint (getDifference(-165313.6666297357, 33693.9894517456))\n\tprint (getDifference(1174.8380510598456, -154146.66490124757))\n\tprint (getDifference(60175.77306795546, 42213.07192354373))"}
{"title": "Angle difference between two bearings", "language": "Python 3.7", "task": "Finding the angle between two bearings is often confusing.[https://stackoverflow.com/questions/16180595/find-the-angle-between-two-bearings]\n\n\n;Task:\nFind the angle which is the result of the subtraction '''b2 - b1''', where '''b1''' and '''b2''' are the bearings. \n\nInput bearings are expressed in the range   '''-180'''   to   '''+180'''   degrees. \nThe  result  is also expressed in the range   '''-180'''   to   '''+180'''   degrees. \n\n\nCompute the angle for the following pairs:\n*  20 degrees ('''b1''') and 45 degrees ('''b2''')\n* -45 and  45\n* -85 and  90\n* -95 and  90\n* -45 and 125\n* -45 and 145\n*  29.4803 and  -88.6381\n* -78.3251 and -159.036\n\n\n;Optional extra:  \nAllow the input bearings to be any (finite) value. \n\n\n;Test cases:\n*  -70099.74233810938   and   29840.67437876723\n* -165313.6666297357    and   33693.9894517456\n*    1174.8380510598456 and -154146.66490124757\n*   60175.77306795546   and   42213.07192354373\n\n", "solution": "'''Difference between two bearings'''\n\nfrom math import (acos, cos, pi, sin)\n\n\n# bearingDelta :: Radians -> Radians -> Radians\ndef bearingDelta(ar):\n    '''Difference between two bearings,\n       expressed in radians.'''\n    def go(br):\n        [(ax, ay), (bx, by)] = [\n            (sin(x), cos(x)) for x in [ar, br]\n        ]\n        # cross-product > 0 ?\n        sign = +1 if 0 < ((ay * bx) - (by * ax)) else -1\n        # sign * dot-product\n        return sign * acos((ax * bx) + (ay * by))\n    return lambda br: go(br)\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test and display'''\n\n    # showMap :: Degrees -> Degrees -> String\n    def showMap(da, db):\n        return unwords(\n            str(x).rjust(n) for n, x in\n            [\n                (22, str(da) + ' +'),\n                (24, str(db) + '  -> '),\n                (7, round(\n                    degrees(\n                        bearingDelta\n                        (radians(da))\n                        (radians(db))\n                    ), 2)\n                 )\n            ]\n        )\n\n    print(__doc__ + ':')\n    print(\n        unlines(showMap(a, b) for a, b in [\n            (20, 45),\n            (-45, 45),\n            (-85, 90),\n            (-95, 90),\n            (-45, 125),\n            (-45, 145),\n            (-70099.74233810938, 29840.67437876723),\n            (-165313.6666297357, 33693.9894517456),\n            (1174.8380510598456, -154146.66490124757),\n            (60175.77306795546, 42213.07192354373)\n        ]))\n\n\n# GENERIC ----------------------------------------------\n\n\n# radians :: Float x => Degrees x -> Radians x\ndef radians(x):\n    '''Radians derived from degrees.'''\n    return pi * x / 180\n\n\n# degrees :: Float x => Radians x -> Degrees x\ndef degrees(x):\n    '''Degrees derived from radians.'''\n    return 180 * x / pi\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single newline-delimited string derived\n       from a list of strings.'''\n    return '\\n'.join(xs)\n\n\n# unwords :: [String] -> String\ndef unwords(xs):\n    '''A space-separated string derived from\n       a list of words.'''\n    return ' '.join(xs)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Angle difference between two bearings", "language": "Python 3.6", "task": "Finding the angle between two bearings is often confusing.[https://stackoverflow.com/questions/16180595/find-the-angle-between-two-bearings]\n\n\n;Task:\nFind the angle which is the result of the subtraction '''b2 - b1''', where '''b1''' and '''b2''' are the bearings. \n\nInput bearings are expressed in the range   '''-180'''   to   '''+180'''   degrees. \nThe  result  is also expressed in the range   '''-180'''   to   '''+180'''   degrees. \n\n\nCompute the angle for the following pairs:\n*  20 degrees ('''b1''') and 45 degrees ('''b2''')\n* -45 and  45\n* -85 and  90\n* -95 and  90\n* -45 and 125\n* -45 and 145\n*  29.4803 and  -88.6381\n* -78.3251 and -159.036\n\n\n;Optional extra:  \nAllow the input bearings to be any (finite) value. \n\n\n;Test cases:\n*  -70099.74233810938   and   29840.67437876723\n* -165313.6666297357    and   33693.9894517456\n*    1174.8380510598456 and -154146.66490124757\n*   60175.77306795546   and   42213.07192354373\n\n", "solution": "\"\"\"\nDifference between two bearings\n\"\"\"\n\ndef  delta_bearing(b1 , b2):\n\treturn ((b2-b1+540)%360)-180\n\ndataSet = [[20, 45], [-45, 45], [-85, 90], [-95, 90], [-45, 125], [-45, 145], \\\n\t[29.4803, -88.6381], [-78.3251, -159.036], \\\n\t[-70099.74233810938, 29840.67437876723], \\\n\t[-165313.6666297357, 33693.9894517456], \\\n\t[1174.8380510598456, -154146.66490124757], \\\n\t[60175.77306795546, 42213.07192354373]]\n\nprint('.{:-^19}.{:-^19}.{:-^9}.' .format(\" b1 \", \" b2 \", \" \u0394 b \" ))\nfor \u0394 in dataSet: \n\tprint('|{: > 19}|{: > 19}|{: > 9.4f}|' .format(\u0394[0], \u0394[1],delta_bearing(\u0394[0],\u0394[1])))"}
{"title": "Anti-primes", "language": "Python", "task": "The anti-primes \n(or highly composite numbers, sequence A002182 in the OEIS) \nare the natural numbers with more factors than any smaller than itself.\n\n\n;Task:\nGenerate and show here, the first twenty anti-primes.\n\n\n;Related tasks:\n:*   [[Factors of an integer]]\n:*   [[Sieve of Eratosthenes]]\n\n", "solution": "from itertools import chain, count, cycle, islice, accumulate\n \ndef factors(n):\n    def prime_powers(n):\n        for c in accumulate(chain([2, 1, 2], cycle([2,4]))):\n            if c*c > n: break\n            if n%c: continue\n            d,p = (), c\n            while not n%c:\n                n,p,d = n//c, p*c, d+(p,)\n            yield d\n        if n > 1: yield n,\n \n    r = [1]\n    for e in prime_powers(n):\n        r += [a*b for a in r for b in e]\n    return r\n    \ndef antiprimes():\n    mx = 0\n    yield 1\n    for c in count(2,2):\n        if c >= 58: break\n        ln = len(factors(c))\n        if ln > mx:\n            yield c\n            mx = ln\n    for c in count(60,30):\n        ln = len(factors(c))\n        if ln > mx:\n            yield c\n            mx = ln  \n\nif __name__ == '__main__':\n    print(*islice(antiprimes(), 40)))"}
{"title": "Apply a digital filter (direct form II transposed)", "language": "Python", "task": "Digital filters are used to apply a mathematical operation to a sampled signal. One of the common formulations is the \"direct form II transposed\" which can represent both infinite impulse response (IIR) and finite impulse response (FIR) filters, as well as being more numerically stable than other forms. [https://ccrma.stanford.edu/~jos/fp/Transposed_Direct_Forms.html]\n\n;Task:\n\nFilter a signal using an order 3 low-pass Butterworth filter. The coefficients for the filter are a=[1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17] and b = [0.16666667, 0.5, 0.5, 0.16666667]\n\nThe signal that needs filtering is the following vector: [-0.917843918645, 0.141984778794, 1.20536903482, 0.190286794412, -0.662370894973, -1.00700480494, -0.404707073677 ,0.800482325044, 0.743500089861, 1.01090520172, 0.741527555207, 0.277841675195, 0.400833448236, -0.2085993586, -0.172842103641, -0.134316096293, 0.0259303398477, 0.490105989562, 0.549391221511, 0.9047198589]\n\n;See also:\n[Wikipedia on Butterworth filters]\n\n\n", "solution": "#!/bin/python\nfrom __future__ import print_function\nfrom scipy import signal\nimport matplotlib.pyplot as plt\n\nif __name__==\"__main__\":\n\tsig = [-0.917843918645,0.141984778794,1.20536903482,0.190286794412,-0.662370894973,-1.00700480494,\n\t\t-0.404707073677,0.800482325044,0.743500089861,1.01090520172,0.741527555207,\n\t\t0.277841675195,0.400833448236,-0.2085993586,-0.172842103641,-0.134316096293,\n\t\t0.0259303398477,0.490105989562,0.549391221511,0.9047198589]\n\n\t#Create an order 3 lowpass butterworth filter\n\t#Generated using b, a = signal.butter(3, 0.5)\n\ta = [1.00000000, -2.77555756e-16, 3.33333333e-01, -1.85037171e-17]\n\tb = [0.16666667, 0.5, 0.5, 0.16666667]\n\n\t#Apply the filter to signal\n\tfilt = signal.lfilter(b, a, sig)\n\tprint (filt)\n\n\tplt.plot(sig, 'b')\n\tplt.plot(filt, 'r--')\n\tplt.show()"}
{"title": "Approximate equality", "language": "Python", "task": "Sometimes, when testing whether the solution to a task (for example, here on Rosetta Code) is correct, the\ndifference in floating point calculations between different language implementations becomes significant. \n\nFor example, a difference between '''32''' bit and '''64''' bit floating point calculations may appear by \nabout the 8th significant digit in base 10 arithmetic.\n\n\n;Task:\nCreate a function which returns true if two floating point numbers are approximately equal. \n\n\nThe function should allow for differences in the magnitude of numbers, so that, for example, \n'''100000000000000.01'''   may be approximately equal to   '''100000000000000.011''', \neven though   '''100.01'''   is not approximately equal to   '''100.011'''.\n\nIf the language has such a feature in its standard library, this may be used instead of a custom function.\n\nShow the function results with comparisons on the following pairs of values:\n:#     100000000000000.01,                        100000000000000.011       (note: should return ''true'')\n:#     100.01,                                    100.011                                                     (note: should return ''false'')\n:#     10000000000000.001 / 10000.0,   1000000000.0000001000\n:#     0.001,                                     0.0010000001\n:#     0.000000000000000000000101,                0.0\n:#      sqrt(2) * sqrt(2),                    2.0\n:#     -sqrt(2) * sqrt(2),                        -2.0\n:#     3.14159265358979323846,                    3.14159265358979324\n\nAnswers should be true for the first example and false in the second, so that just rounding the numbers to a fixed number of decimals should not be enough. Otherwise answers may vary and still be correct. See the Python code for one type of solution.\n\n\n__TOC__\n\n", "solution": "from numpy import sqrt\nfrom math import isclose\n\ntestvalues = [[100000000000000.01,           100000000000000.011],\n              [100.01,                       100.011],\n              [10000000000000.001 / 10000.0, 1000000000.0000001000],\n              [0.001,                        0.0010000001],\n              [0.000000000000000000000101,   0.0],\n              [sqrt(2) * sqrt(2),            2.0],\n              [-sqrt(2) * sqrt(2),          -2.0],\n              [3.14159265358979323846,       3.14159265358979324]]\n\nfor (x, y) in testvalues:\n    maybenot = \"is\" if isclose(x, y) else \"is NOT\"\n    print(x, maybenot, \"approximately equal to \", y)\n\n"}
{"title": "Archimedean spiral", "language": "Python", "task": "The Archimedean spiral is a spiral named after the Greek mathematician Archimedes. \n\n\nAn Archimedean spiral can be described by the equation:\n\n:\\, r=a+b\\theta\n\nwith real numbers ''a'' and ''b''.\n\n\n;Task\nDraw an Archimedean spiral.\n\n", "solution": "from turtle import *\nfrom math import *\ncolor(\"blue\")\ndown()\nfor i in range(200):\n    t = i / 20 * pi\n    x = (1 + 5 * t) * cos(t)\n    y = (1 + 5 * t) * sin(t)\n    goto(x, y)\nup()\ndone()"}
{"title": "Arena storage pool", "language": "Python", "task": "Dynamically allocated objects take their memory from a [[heap]]. \n\nThe memory for an object is provided by an '''allocator''' which maintains the storage pool used for the [[heap]].\n\nOften a call to allocator is denoted as\nP := new T\nwhere   '''T'''   is the type of an allocated object,   and   '''P'''   is a [[reference]] to the object.\n\nThe storage pool chosen by the allocator can be determined by either:\n* the object type   '''T'''\n* the type of pointer   '''P'''\n\n\nIn the former case objects can be allocated only in one storage pool. \n\nIn the latter case objects of the type can be allocated in any storage pool or on the [[stack]].\n\n\n;Task:\nThe task is to show how allocators and user-defined storage pools are supported by the language. \n\nIn particular:\n# define an arena storage pool.   An arena is a pool in which objects are allocated individually, but freed by groups.\n# allocate some objects (e.g., integers) in the pool.\n\n\nExplain what controls the storage pool choice in the language.\n\n", "solution": "In Python:\n* Everything is an object.\n* Objects are dynamically allocated.\n* Unused objects are garbage collected.\n\nWhere objects appear from, or disappear to, is treated as an implementation detail.\n\nStatements, such as assignments, class and function definitions, and import statements can create objects and assign names to them which can be seen as assigning a reference to objects. Objects can also be referred to from other objects e.g. in collections such as lists.<br>\nWhen names go out of scope, or objects are explicitly destroyed, references to objects are diminished. Python's implementation keeps track of references to objects and marks objects that have no remaining references so that they become candidates for '[[wp:Garbage collection (computer science)|garbage collection]]' at a later time.\n\n"}
{"title": "Arithmetic-geometric mean", "language": "Python", "task": "{{wikipedia|Arithmetic-geometric mean}}\n\n\n;Task:\nWrite a function to compute the arithmetic-geometric mean of two numbers.\n\n\nThe arithmetic-geometric mean of two numbers can be (usefully) denoted as \\mathrm{agm}(a,g), and is equal to the limit of the sequence:\n: a_0 = a; \\qquad g_0 = g\n: a_{n+1} = \\tfrac{1}{2}(a_n + g_n); \\quad g_{n+1} = \\sqrt{a_n g_n}.\nSince the limit of a_n-g_n tends (rapidly) to zero with iterations, this is an efficient method.\n\nDemonstrate the function by calculating:\n:\\mathrm{agm}(1,1/\\sqrt{2})\n \n\n;Also see:\n*   mathworld.wolfram.com/Arithmetic-Geometric Mean\n\n", "solution": "from math import sqrt\n\ndef agm(a0, g0, tolerance=1e-10):\n    \"\"\"\n    Calculating the arithmetic-geometric mean of two numbers a0, g0.\n\n    tolerance     the tolerance for the converged \n                  value of the arithmetic-geometric mean\n                  (default value = 1e-10)\n    \"\"\"\n    an, gn = (a0 + g0) / 2.0, sqrt(a0 * g0)\n    while abs(an - gn) > tolerance:\n        an, gn = (an + gn) / 2.0, sqrt(an * gn)\n    return an\n\nprint agm(1, 1 / sqrt(2))"}
{"title": "Arithmetic-geometric mean/Calculate Pi", "language": "Python from Ruby", "task": "Almkvist Berndt 1988 begins with an investigation of why the agm is such an efficient algorithm, and proves that it converges quadratically. This is an efficient method to calculate \\pi.\n\nWith the same notations used in [[Arithmetic-geometric mean]], we can summarize the paper by writing:\n\n\\pi =\n\\frac{4\\; \\mathrm{agm}(1, 1/\\sqrt{2})^2}\n{1 - \\sum\\limits_{n=1}^{\\infty} 2^{n+1}(a_n^2-g_n^2)}\n\n\nThis allows you to make the approximation, for any large   '''N''':\n\n\\pi \\approx\n\\frac{4\\; a_N^2}\n{1 - \\sum\\limits_{k=1}^N 2^{k+1}(a_k^2-g_k^2)}\n\n\nThe purpose of this task is to demonstrate how to use this approximation in order to compute a large number of decimals of \\pi.\n\n", "solution": "from decimal import *\n\nD = Decimal\ngetcontext().prec = 100\na = n = D(1)\ng, z, half = 1 / D(2).sqrt(), D(0.25), D(0.5)\nfor i in range(18):\n    x = [(a + g) * half, (a * g).sqrt()]\n    var = x[0] - a\n    z -= var * var * n\n    n += n\n    a, g = x    \nprint(a * a / z)"}
{"title": "Arithmetic derivative", "language": "Python", "task": "The '''arithmetic derivative''' of an integer (more specifically, the\n'''Lagarias arithmetic derivative''') is a function defined for integers, based on prime\nfactorization, by analogy with the product rule for the derivative of a function that is\nused in mathematical analysis. Accordingly, for natural numbers n, the arithmetic \nderivative D(n) is defined as follows:\n\n;*D(0) = D(1) = 0.\n;*D(p) = 1 for any prime p.\n;*D(mn) = D(m)n + mD(n) for any m,n  N. (Leibniz rule for derivatives).\n\nAdditionally, for negative integers the arithmetic derivative may be defined as -D(-n) (n < 0).\n\n; Examples\n\nD(2) = 1 and D(3) = 1 (both are prime) so if mn = 2 * 3, D(6) = (1)(3) + (1)(2) = 5.\n\nD(9) = D(3)(3) + D(3)(3) = 6\n\nD(27) = D(3)*9 + D(9)*3 = 9 + 18 = 27\n\nD(30) = D(5)(6) + D(6)(5) = 6 + 5 * 5 = 31.\n\n; Task\n\nFind and show the arithmetic derivatives for -99 through 100.\n\n; Stretch task\n\nFind (the arithmetic derivative of 10^m) then divided by 7, where m is from 1 to 20.\n\n; See also\n\n;* OEIS:A003415 - a(n) = n' = arithmetic derivative of n.\n;*Wikipedia: Arithmetic Derivative\n\n\n", "solution": "from sympy.ntheory import factorint\n\ndef D(n):\n    if n < 0:\n        return -D(-n)\n    elif n < 2:\n        return 0\n    else:\n        fdict = factorint(n)\n        if len(fdict) == 1 and 1 in fdict: # is prime\n            return 1\n        return sum([n * e // p for p, e in fdict.items()])\n\nfor n in range(-99, 101):\n    print('{:5}'.format(D(n)), end='\\n' if n % 10 == 0 else '')\n\nprint()\nfor m in range(1, 21):\n    print('(D for 10**{}) divided by 7 is {}'.format(m, D(10 ** m) // 7))\n\n"}
{"title": "Arithmetic evaluation", "language": "Python", "task": "Create a program which parses and evaluates arithmetic expressions.\n;Requirements:\n* An abstract-syntax tree (AST) for the expression must be created from parsing the input. \n* The AST must be used in evaluation, also, so the input may not be directly evaluated (e.g. by calling eval or a similar language feature.) \n* The expression will be a string or list of symbols like \"(1+3)*7\". \n* The four symbols + - * / must be supported as binary operators with conventional precedence rules. \n* Precedence-control parentheses must also be supported.\n\n\n;Note:\nFor those who don't remember, mathematical precedence is as follows:\n* Parentheses\n* Multiplication/Division (left to right)\n* Addition/Subtraction (left to right)\n\n\n;C.f: \n* [[24 game Player]].\n* [[Parsing/RPN calculator algorithm]].\n* [[Parsing/RPN to infix conversion]].\n\n", "solution": "import operator\n\nclass AstNode(object):\n   def __init__( self, opr, left, right ):\n      self.opr = opr\n      self.l = left\n      self.r = right\n\n   def eval(self):\n      return self.opr(self.l.eval(), self.r.eval())\n\nclass LeafNode(object):\n   def __init__( self, valStrg ):\n      self.v = int(valStrg)\n\n   def eval(self):\n      return self.v\n\nclass Yaccer(object):\n   def __init__(self):\n      self.operstak = []\n      self.nodestak =[]\n      self.__dict__.update(self.state1)\n\n   def v1( self, valStrg ):\n      # Value String\n      self.nodestak.append( LeafNode(valStrg))\n      self.__dict__.update(self.state2)\n      #print 'push', valStrg\n\n   def o2( self, operchar ):\n      # Operator character or open paren in state1\n      def openParen(a,b):\n         return 0\t\t# function should not be called\n\n      opDict= { '+': ( operator.add, 2, 2 ),\n         '-': (operator.sub, 2, 2 ),\n         '*': (operator.mul, 3, 3 ),\n         '/': (operator.div, 3, 3 ),\n         '^': ( pow,         4, 5 ),  # right associative exponentiation for grins\n         '(': ( openParen,   0, 8 )\n         }\n      operPrecidence = opDict[operchar][2]\n      self.redeuce(operPrecidence)\n\n      self.operstak.append(opDict[operchar])\n      self.__dict__.update(self.state1)\n      # print 'pushop', operchar\n\n   def syntaxErr(self, char ):\n      # Open Parenthesis \n      print 'parse error - near operator \"%s\"' %char\n\n   def pc2( self,operchar ):\n      # Close Parenthesis\n      # reduce node until matching open paren found \n      self.redeuce( 1 )\n      if len(self.operstak)>0:\n         self.operstak.pop()\t\t# pop off open parenthesis\n      else:\n         print 'Error - no open parenthesis matches close parens.'\n      self.__dict__.update(self.state2)\n\n   def end(self):\n      self.redeuce(0)\n      return self.nodestak.pop()\n\n   def redeuce(self, precidence):\n      while len(self.operstak)>0:\n         tailOper = self.operstak[-1]\n         if tailOper[1] < precidence: break\n\n         tailOper = self.operstak.pop()\n         vrgt = self.nodestak.pop()\n         vlft= self.nodestak.pop()\n         self.nodestak.append( AstNode(tailOper[0], vlft, vrgt))\n         # print 'reduce'\n\n   state1 = { 'v': v1, 'o':syntaxErr, 'po':o2, 'pc':syntaxErr }\n   state2 = { 'v': syntaxErr, 'o':o2, 'po':syntaxErr, 'pc':pc2 }\n\n\ndef Lex( exprssn, p ):\n   bgn = None\n   cp = -1\n   for c in exprssn:\n      cp += 1\n      if c in '+-/*^()':         # throw in exponentiation (^)for grins\n         if bgn is not None:\n            p.v(p, exprssn[bgn:cp])\n            bgn = None\n         if c=='(': p.po(p, c)\n         elif c==')':p.pc(p, c)\n         else: p.o(p, c)\n      elif c in ' \\t':\n         if bgn is not None:\n            p.v(p, exprssn[bgn:cp])\n            bgn = None\n      elif c in '0123456789':\n         if bgn is None:\n            bgn = cp\n      else:\n         print 'Invalid character in expression'\n         if bgn is not None:\n            p.v(p, exprssn[bgn:cp])\n            bgn = None\n         \n   if bgn is not None:\n      p.v(p, exprssn[bgn:cp+1])\n      bgn = None\n   return p.end()\n\n\nexpr = raw_input(\"Expression:\")\nastTree = Lex( expr, Yaccer())\nprint expr, '=',astTree.eval()"}
{"title": "Arithmetic numbers", "language": "Python", "task": "Definition\nA positive integer '''n''' is an arithmetic number if the average of its positive divisors is also an integer.\n\nClearly all odd primes '''p''' must be arithmetic numbers because their only divisors are '''1''' and '''p''' whose sum is even and hence their average must be an integer. However, the prime number '''2''' is not an arithmetic number because the average of its divisors is 1.5.\n\n;Example\n30 is an arithmetic number because its 7 divisors are: [1, 2, 3, 5, 6, 10, 15, 30], their sum is 72 and average 9 which is an integer. \n\n;Task\nCalculate and show here:\n\n1. The first 100 arithmetic numbers.\n\n2. The '''x'''th arithmetic number where '''x''' = 1,000 and '''x''' = 10,000.\n\n3. How many of the first '''x''' arithmetic numbers are composite.\n\nNote that, technically, the arithmetic number '''1''' is neither prime nor composite.\n\n;Stretch\nCarry out the same exercise in 2. and 3. above for '''x''' = 100,000 and '''x''' = 1,000,000.\n\n;References\n\n* Wikipedia: Arithmetic number\n* OEIS:A003601 - Numbers n such that the average of the divisors of n is an integer\n\n", "solution": "def factors(n: int):\n    f = set([1, n])\n    i = 2\n    while True:\n        j = n // i\n        if j < i:\n            break\n        if i * j == n:\n            f.add(i)\n            f.add(j)\n        i += 1\n    return f\n\narithmetic_count = 0\ncomposite_count = 0\nn = 1\nwhile arithmetic_count <= 1000000:\n    f = factors(n)\n    if (sum(f)/len(f)).is_integer():\n        arithmetic_count += 1\n        if len(f) > 2:\n            composite_count += 1\n        if arithmetic_count <= 100:\n            print(f'{n:3d} ', end='')\n            if arithmetic_count % 10 == 0:\n                print()\n        if arithmetic_count in (1000, 10000, 100000, 1000000):\n            print(f'\\n{arithmetic_count}th arithmetic number is {n}')\n            print(f'Number of composite arithmetic numbers <= {n}: {composite_count}')\n    n += 1"}
{"title": "Ascending primes", "language": "Python", "task": "Generate and show all primes with strictly ascending decimal digits.\n\nAside: Try solving without peeking at existing solutions. I had a weird idea for generating\na prime sieve faster, which needless to say didn't pan out. The solution may be p(r)etty trivial\nbut generating them quickly is at least mildly interesting.\nTip: filtering all 7,027,260 primes below 123,456,789 probably won't kill you, but there is\nat least one significantly better and much faster way, needing a mere 511 odd/prime tests.\n\n\n;See also\n;* OEIS:A052015 - Primes with distinct digits in ascending order\n\n\n;Related:\n*[[Primes with digits in nondecreasing order]] (infinite series allowing duplicate digits, whereas this isn't and doesn't)\n*[[Pandigital prime]] (whereas this is the smallest, with gaps in the used digits being permitted)\n\n\n", "solution": "from sympy import isprime\n\ndef ascending(x=0):\n    for y in range(x*10 + (x%10) + 1, x*10 + 10):\n        yield from ascending(y)\n        yield(y)\n\nprint(sorted(x for x in ascending() if isprime(x)))"}
{"title": "Ascending primes", "language": "Python from Pascal", "task": "Generate and show all primes with strictly ascending decimal digits.\n\nAside: Try solving without peeking at existing solutions. I had a weird idea for generating\na prime sieve faster, which needless to say didn't pan out. The solution may be p(r)etty trivial\nbut generating them quickly is at least mildly interesting.\nTip: filtering all 7,027,260 primes below 123,456,789 probably won't kill you, but there is\nat least one significantly better and much faster way, needing a mere 511 odd/prime tests.\n\n\n;See also\n;* OEIS:A052015 - Primes with distinct digits in ascending order\n\n\n;Related:\n*[[Primes with digits in nondecreasing order]] (infinite series allowing duplicate digits, whereas this isn't and doesn't)\n*[[Pandigital prime]] (whereas this is the smallest, with gaps in the used digits being permitted)\n\n\n", "solution": "def isprime(n):\n    if n == 2: return True\n    if n == 1 or n % 2 == 0: return False\n    root1 = int(n**0.5) + 1;\n    for k in range(3, root1, 2):\n        if n % k == 0: return False\n    return True\n\nqueue = [k for k in range(1, 10)]\nprimes = []\n\nwhile queue:\n    n = queue.pop(0)\n    if isprime(n):\n        primes.append(n)\n    queue.extend(n * 10 + k for k in range(n % 10 + 1, 10))\n\nprint(primes)"}
{"title": "Associative array/Merging", "language": "Python", "task": "Define two associative arrays, where one represents the following \"base\" data:\n\n::::: {| class=\"wikitable\"\n|+\n| '''Key''' || '''Value'''\n|-\n| \"name\" || \"Rocket Skates\"\n|-\n| \"price\" || 12.75\n|-\n| \"color\" || \"yellow\"\n|}\n\nAnd the other represents \"update\" data:\n\n::::: {| class=\"wikitable\"\n|+\n| '''Key''' || '''Value'''\n|-\n| \"price\" || 15.25\n|-\n| \"color\" || \"red\"\n|-\n| \"year\" || 1974\n|}\n\nMerge these into a new associative array that contains every key found in either of the source ones.  Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table.  If possible, do this in a way that does not mutate the original two associative arrays.  Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:\n\n::::: {| class=\"wikitable\"\n|+\n| '''Key''' || '''Value'''\n|-\n| \"name\" || \"Rocket Skates\"\n|-\n| \"price\" || 15.25\n|-\n| \"color\" || \"red\"\n|-\n| \"year\" || 1974\n|}\n\n", "solution": "As of Python 3.5, this can be solved with the dictionary unpacking operator.\n"}
{"title": "Associative array/Merging", "language": "Python 3.5+", "task": "Define two associative arrays, where one represents the following \"base\" data:\n\n::::: {| class=\"wikitable\"\n|+\n| '''Key''' || '''Value'''\n|-\n| \"name\" || \"Rocket Skates\"\n|-\n| \"price\" || 12.75\n|-\n| \"color\" || \"yellow\"\n|}\n\nAnd the other represents \"update\" data:\n\n::::: {| class=\"wikitable\"\n|+\n| '''Key''' || '''Value'''\n|-\n| \"price\" || 15.25\n|-\n| \"color\" || \"red\"\n|-\n| \"year\" || 1974\n|}\n\nMerge these into a new associative array that contains every key found in either of the source ones.  Each key should map to the value in the second (update) table if that exists, or else to the value in the first (base) table.  If possible, do this in a way that does not mutate the original two associative arrays.  Obviously this should be done in a way that would work for any data, not just the specific data given here, but in this example the result should be:\n\n::::: {| class=\"wikitable\"\n|+\n| '''Key''' || '''Value'''\n|-\n| \"name\" || \"Rocket Skates\"\n|-\n| \"price\" || 15.25\n|-\n| \"color\" || \"red\"\n|-\n| \"year\" || 1974\n|}\n\n", "solution": "base = {\"name\":\"Rocket Skates\", \"price\":12.75, \"color\":\"yellow\"}\nupdate = {\"price\":15.25, \"color\":\"red\", \"year\":1974}\n\nresult = {**base, **update}\n\nprint(result)"}
{"title": "Attractive numbers", "language": "Python 2.7.12", "task": "A number is an   ''attractive number''   if the number of its prime factors (whether distinct or not) is also prime.\n\n\n;Example:\nThe number   '''20''',   whose prime decomposition is   '''2 x 2 x 5''',   is an   ''attractive number''   because the number of its prime factors   ('''3''')   is also prime.\n\n\n;Task:\nShow sequence items up to   '''120'''.\n\n\n;Reference:\n:*   The OEIS entry:   A063989: Numbers with a prime number of prime divisors.\n\n", "solution": "from sympy import sieve # library for primes\n\ndef get_pfct(n): \n\ti = 2; factors = []\n\twhile i * i <= n:\n\t\tif n % i:\n\t\t\ti += 1\n\t\telse:\n\t\t\tn //= i\n\t\t\tfactors.append(i)\n\tif n > 1:\n\t\tfactors.append(n)\n\treturn len(factors) \n\nsieve.extend(110) # first 110 primes...\nprimes=sieve._list\n\npool=[]\n\nfor each in xrange(0,121):\n\tpool.append(get_pfct(each))\n\nfor i,each in enumerate(pool):\n\tif each in primes:\n\t\tprint i,"}
{"title": "Attractive numbers", "language": "Python 3.7", "task": "A number is an   ''attractive number''   if the number of its prime factors (whether distinct or not) is also prime.\n\n\n;Example:\nThe number   '''20''',   whose prime decomposition is   '''2 x 2 x 5''',   is an   ''attractive number''   because the number of its prime factors   ('''3''')   is also prime.\n\n\n;Task:\nShow sequence items up to   '''120'''.\n\n\n;Reference:\n:*   The OEIS entry:   A063989: Numbers with a prime number of prime divisors.\n\n", "solution": "'''Attractive numbers'''\n\nfrom itertools import chain, count, takewhile\nfrom functools import reduce\n\n\n# attractiveNumbers :: () -> [Int]\ndef attractiveNumbers():\n    '''A non-finite stream of attractive numbers.\n       (OEIS A063989)\n    '''\n    return filter(\n        compose(\n            isPrime,\n            len,\n            primeDecomposition\n        ),\n        count(1)\n    )\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''Attractive numbers drawn from the range [1..120]'''\n    for row in chunksOf(15)(list(\n            takewhile(\n                lambda x: 120 >= x,\n                attractiveNumbers()\n            )\n    )):\n        print(' '.join(map(\n            compose(justifyRight(3)(' '), str),\n            row\n        )))\n\n\n# GENERAL FUNCTIONS ---------------------------------------\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n, subdividing the\n       contents of xs. Where the length of xs is not evenly\n       divible, the final list will be shorter than n.\n    '''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    return lambda x: reduce(\n        lambda a, f: f(a),\n        fs[::-1], x\n    )\n\n\n# We only need light implementations\n# of prime functions here:\n\n# primeDecomposition :: Int -> [Int]\ndef primeDecomposition(n):\n    '''List of integers representing the\n       prime decomposition of n.\n    '''\n    def go(n, p):\n        return [p] + go(n // p, p) if (\n            0 == n % p\n        ) else []\n    return list(chain.from_iterable(map(\n        lambda p: go(n, p) if isPrime(p) else [],\n        range(2, 1 + n)\n    )))\n\n\n# isPrime :: Int -> Bool\ndef isPrime(n):\n    '''True if n is prime.'''\n    if n in (2, 3):\n        return True\n    if 2 > n or 0 == n % 2:\n        return False\n    if 9 > n:\n        return True\n    if 0 == n % 3:\n        return False\n\n    return not any(map(\n        lambda x: 0 == n % x or 0 == n % (2 + x),\n        range(5, 1 + int(n ** 0.5), 6)\n    ))\n\n\n# justifyRight :: Int -> Char -> String -> String\ndef justifyRight(n):\n    '''A string padded at left to length n,\n       using the padding character c.\n    '''\n    return lambda c: lambda s: s.rjust(n, c)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Average loop length", "language": "Python from C", "task": "Let f be a uniformly-randomly chosen mapping from the numbers 1..N to the numbers 1..N (note: not necessarily a permutation of 1..N; the mapping could produce a number in more than one way or not at all). At some point, the sequence 1, f(1), f(f(1))... will contain a repetition, a number that occurring for the second time in the sequence.\n\n\n;Task:\nWrite a program or a script that estimates, for each N, the average length until the first such repetition.\n\nAlso calculate this expected length using an analytical formula, and optionally compare the simulated result with the theoretical one.\n\n\nThis problem comes from the end of Donald Knuth's Christmas tree lecture 2011.\n\nExample of expected output:\n\n N    average    analytical    (error)\n===  =========  ============  =========\n  1     1.0000        1.0000  (  0.00%)\n  2     1.4992        1.5000  (  0.05%)\n  3     1.8784        1.8889  (  0.56%)\n  4     2.2316        2.2188  (  0.58%)\n  5     2.4982        2.5104  (  0.49%)\n  6     2.7897        2.7747  (  0.54%)\n  7     3.0153        3.0181  (  0.09%)\n  8     3.2429        3.2450  (  0.07%)\n  9     3.4536        3.4583  (  0.14%)\n 10     3.6649        3.6602  (  0.13%)\n 11     3.8091        3.8524  (  1.12%)\n 12     3.9986        4.0361  (  0.93%)\n 13     4.2074        4.2123  (  0.12%)\n 14     4.3711        4.3820  (  0.25%)\n 15     4.5275        4.5458  (  0.40%)\n 16     4.6755        4.7043  (  0.61%)\n 17     4.8877        4.8579  (  0.61%)\n 18     4.9951        5.0071  (  0.24%)\n 19     5.1312        5.1522  (  0.41%)\n 20     5.2699        5.2936  (  0.45%)\n\n", "solution": "from __future__ import division # Only necessary for Python 2.X\nfrom math import factorial\nfrom random import randrange\n\nMAX_N = 20\nTIMES = 1000000\n\ndef analytical(n):\n\treturn sum(factorial(n) / pow(n, i) / factorial(n -i) for i in range(1, n+1))\n\ndef test(n, times):\n    count = 0\n    for i in range(times):\n        x, bits = 1, 0\n        while not (bits & x):\n            count += 1\n            bits |= x\n            x = 1 << randrange(n)\n    return count / times\n\nif __name__ == '__main__':\n    print(\" n\\tavg\\texp.\\tdiff\\n-------------------------------\")\n    for n in range(1, MAX_N+1):\n        avg = test(n, TIMES)\n        theory = analytical(n)\n        diff = (avg / theory - 1) * 100\n        print(\"%2d %8.4f %8.4f %6.3f%%\" % (n, avg, theory, diff))"}
{"title": "Averages/Mean angle", "language": "Python 2.6+", "task": "{{Related tasks/Statistical measures}}\n\n\n", "solution": ">>> from cmath import rect, phase\n>>> from math import radians, degrees\n>>> def mean_angle(deg):\n...     return degrees(phase(sum(rect(1, radians(d)) for d in deg)/len(deg)))\n... \n>>> for angles in [[350, 10], [90, 180, 270, 360], [10, 20, 30]]:\n...     print('The mean angle of', angles, 'is:', round(mean_angle(angles), 12), 'degrees')\n...     \nThe mean angle of [350, 10] is: -0.0 degrees\nThe mean angle of [90, 180, 270, 360] is: -90.0 degrees\nThe mean angle of [10, 20, 30] is: 20.0 degrees\n>>> "}
{"title": "Averages/Pythagorean means", "language": "Python 3", "task": "{{Related tasks/Statistical measures}}\n\n\n", "solution": "from operator import mul\nfrom functools import reduce\n\n\ndef amean(num):\n    return sum(num) / len(num)\n\n\ndef gmean(num):\n    return reduce(mul, num, 1)**(1 / len(num))\n\n\ndef hmean(num):\n    return len(num) / sum(1 / n for n in num)\n\n\nnumbers = range(1, 11)  # 1..10\na, g, h = amean(numbers), gmean(numbers), hmean(numbers)\nprint(a, g, h)\nassert a >= g >= h"}
{"title": "Averages/Root mean square", "language": "Python 3", "task": "Task\n\nCompute the   Root mean square   of the numbers 1..10.\n\n\nThe   ''root mean square''   is also known by its initials RMS (or rms), and as the '''quadratic mean'''.\n\nThe RMS is calculated as the mean of the squares of the numbers, square-rooted:\n\n\n::: x_{\\mathrm{rms}} = \\sqrt {{{x_1}^2 + {x_2}^2 + \\cdots + {x_n}^2} \\over n}. \n\n\n;See also\n\n{{Related tasks/Statistical measures}}\n\n\n", "solution": "from functools import (reduce)\nfrom math import (sqrt)\n\n\n# rootMeanSquare :: [Num] -> Float\ndef rootMeanSquare(xs):\n    return sqrt(reduce(lambda a, x: a + x * x, xs, 0) / len(xs))\n\n\nprint(\n    rootMeanSquare([1, 2, 3, 4, 5, 6, 7, 8, 9, 10])\n)"}
{"title": "Babbage problem", "language": "Python", "task": "Charles Babbage\nCharles Babbage's analytical engine.\n\nCharles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:\n{{quote\n | What is the smallest positive integer whose square ends in the digits 269,696?\n | Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.\n}}\nHe thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.\n\n\n;Task\n\nThe task is to find out if Babbage had the right answer -- and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand. \nAs Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.\n\nFor these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].\n\n\n;Motivation\n\nThe aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.\n\n", "solution": "# Lines that start by # are a comments:\n# they will be ignored by the machine\n\nn=0 # n is a variable and its value is 0\n\n# we will increase its value by one until\n# its square ends in 269,696\n\nwhile n**2 % 1000000 != 269696:\n\n    # n**2 -> n squared\n    # %    -> 'modulo' or remainder after division\n    # !=   -> not equal to\n    \n    n += 1 # += -> increase by a certain number\n\nprint(n) # prints n\n"}
{"title": "Babbage problem", "language": "Python 3.7", "task": "Charles Babbage\nCharles Babbage's analytical engine.\n\nCharles Babbage, looking ahead to the sorts of problems his Analytical Engine would be able to solve, gave this example:\n{{quote\n | What is the smallest positive integer whose square ends in the digits 269,696?\n | Babbage, letter to Lord Bowden, 1837; see Hollingdale and Tootill, Electronic Computers, second edition, 1970, p. 125.\n}}\nHe thought the answer might be 99,736, whose square is 9,947,269,696; but he couldn't be certain.\n\n\n;Task\n\nThe task is to find out if Babbage had the right answer -- and to do so, as far as your language allows it, in code that Babbage himself would have been able to read and understand. \nAs Babbage evidently solved the task with pencil and paper, a similar efficient solution is preferred.\n\nFor these purposes, Charles Babbage may be taken to be an intelligent person, familiar with mathematics and with the idea of a computer; he has written the first drafts of simple computer programmes in tabular form. [Babbage Archive Series L].\n\n\n;Motivation\n\nThe aim of the task is to write a program that is sufficiently clear and well-documented for such a person to be able to read it and be confident that it does indeed solve the specified problem.\n\n", "solution": "'''Babbage problem'''\n\nfrom math import (floor, sqrt)\nfrom itertools import (islice)\n\n\n# squaresWithSuffix :: Int -> Gen [Int]\ndef squaresWithSuffix(n):\n    '''Non finite stream of squares with a given suffix.'''\n    stem = 10 ** len(str(n))\n    i = 0\n    while True:\n        i = until(lambda x: isPerfectSquare(n + (stem * x)))(\n            succ\n        )(i)\n        yield n + (stem * i)\n        i = succ(i)\n\n\n# isPerfectSquare :: Int -> Bool\ndef isPerfectSquare(n):\n    '''True if n is a perfect square.'''\n    r = sqrt(n)\n    return r == floor(r)\n\n\n# TEST ----------------------------------------------------\n\n# main :: IO ()\ndef main():\n    '''Smallest positive integers whose squares end in the digits 269,696'''\n    print(\n        fTable(main.__doc__ + ':\\n')(\n            lambda n: str(int(sqrt(n))) + '^2'\n        )(repr)(identity)(\n            take(10)(squaresWithSuffix(269696))\n        )\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# identity :: a -> a\ndef identity(x):\n    '''The identity function.'''\n    return x\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value.\n       For numeric types, (1 +).\n    '''\n    return 1 + x if isinstance(x, int) else (\n        chr(1 + ord(x))\n    )\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, (list, tuple))\n        else list(islice(xs, n))\n    )\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f, x):\n        v = x\n        while not p(v):\n            v = f(v)\n        return v\n    return lambda f: lambda x: go(f, x)\n\n\n# FORMATTING ----------------------------------------------\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Babylonian spiral", "language": "Python", "task": "The '''Babylonian spiral''' is a sequence of points in the plane that are created so as to\ncontinuously minimally increase in vector length and minimally bend in vector direction,\nwhile always moving from point to point on strictly integral coordinates. Of the two criteria\nof length and angle, the length has priority.\n\n; Examples\n\nP(1) and P(2) are defined to be at (x = 0, y = 0) and (x = 0, y = 1). The first vector is\nfrom P(1) to P(2). It is vertical and of length 1. Note that the square of that length is 1.\n\nNext in sequence is the vector from P(2) to P(3). This should be the smallest distance to a\npoint with integral (x, y) which is longer than the last vector (that is, 1). It should also bend clockwise\nmore than zero radians, but otherwise to the least degree.\n\nThe point chosen for P(3) that fits criteria is (x = 1, y = 2). Note the length of the vector\nfrom P(2) to P(3) is 2, which squared is 2. The lengths of the vectors thus determined can be given by a sorted\nlist of possible sums of two integer squares, including 0 as a square.\n\n; Task\n\nFind and show the first 40 (x, y) coordinates of the Babylonian spiral.\n\n; Stretch task\n\nShow in your program how to calculate and plot the first 10000 points in the sequence. Your result\nshould look similar to the graph shown at the OEIS: [[File:From_oies_A@97346_A297347_plot2a.png]]\n\n; See also\n\n;* OEIS:A256111 - squared distance to the origin of the n-th vertex on a Babylonian Spiral.\n;* OEIS:A297346 - List of successive x-coordinates in the Babylonian Spiral.\n;* OEIS:A297347 - List of successive y-coordinates in the Babylonian Spiral.\n\n\n\n", "solution": "\"\"\" Rosetta Code task rosettacode.org/wiki/Babylonian_spiral \"\"\"\n\nfrom itertools import accumulate\nfrom math import isqrt, atan2, tau\nfrom matplotlib.pyplot import axis, plot, show\n\n\nsquare_cache = []\n\ndef babylonian_spiral(nsteps):\n    \"\"\"\n    Get the points for each step along a Babylonia spiral of `nsteps` steps.\n    Origin is at (0, 0) with first step one unit in the positive direction along\n    the vertical (y) axis. The other points are selected to have integer x and y\n    coordinates, progressively concatenating the next longest vector with integer\n    x and y coordinates on the grid. The direction change of the  new vector is\n    chosen to be nonzero and clockwise in a direction that minimizes the change\n    in direction from the previous vector.\n    \n    See also: oeis.org/A256111, oeis.org/A297346, oeis.org/A297347\n    \"\"\"\n    if len(square_cache) <= nsteps:\n        square_cache.extend([x * x for x in range(len(square_cache), nsteps)])\n    xydeltas = [(0, 0), (0, 1)]\n    \u03b4squared = 1\n    for _ in range(nsteps - 2):\n        x, y = xydeltas[-1]\n        \u03b8 = atan2(y, x)\n        candidates = []\n        while not candidates:\n            \u03b4squared += 1\n            for i, a in enumerate(square_cache):\n                if a > \u03b4squared // 2:\n                    break\n                for j in range(isqrt(\u03b4squared) + 1, 0, -1):\n                    b = square_cache[j]\n                    if a + b < \u03b4squared:\n                        break\n                    if a + b == \u03b4squared:\n                        candidates.extend([(i, j), (-i, j), (i, -j), (-i, -j), (j, i), (-j, i),\n                           (j, -i), (-j, -i)])\n\n        p = min(candidates, key=lambda d: (\u03b8 - atan2(d[1], d[0])) % tau)\n        xydeltas.append(p)\n\n    return list(accumulate(xydeltas, lambda a, b: (a[0] + b[0], a[1] + b[1])))\n\n\npoints10000 = babylonian_spiral(10000)\nprint(\"The first 40 Babylonian spiral points are:\")\nfor i, p in enumerate(points10000[:40]):\n     print(str(p).ljust(10), end = '\\n' if (i + 1) % 10 == 0 else '')\n\n# stretch portion of task\nplot(*zip(*points10000), color=\"navy\", linewidth=0.2)\naxis('scaled')\nshow()\n"}
{"title": "Balanced brackets", "language": "Python", "task": "'''Task''': \n* Generate a string with   '''N'''   opening brackets   '''['''   and with   '''N'''   closing brackets   ''']''',   in some arbitrary order. \n* Determine whether the generated string is ''balanced''; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.\n\n\n\n;Examples:\n    (empty)      OK\n    []           OK   \n    [][]         OK   \n    [[][]]       OK \n    ][         NOT OK\n    ][][       NOT OK\n    []][[]     NOT OK\n\n", "solution": ">>> def gen(N):\n...     txt = ['[', ']'] * N\n...     random.shuffle( txt )\n...     return ''.join(txt)\n... \n>>> def balanced(txt):\n...     braced = 0\n...     for ch in txt:\n...         if ch == '[': braced += 1\n...         if ch == ']':\n...             braced -= 1\n...             if braced < 0: return False\n...     return braced == 0\n... \n>>> for txt in (gen(N) for N in range(10)):\n...     print (\"%-22r is%s balanced\" % (txt, '' if balanced(txt) else ' not'))\n... \n''                     is balanced\n'[]'                   is balanced\n'[][]'                 is balanced\n'][[[]]'               is not balanced\n'[]][[][]'             is not balanced\n'[][[][]]]['           is not balanced\n'][]][][[]][['         is not balanced\n'[[]]]]][]][[[['       is not balanced\n'[[[[]][]]][[][]]'     is balanced\n'][[][[]]][]]][[[[]'   is not balanced"}
{"title": "Balanced brackets", "language": "Python 3.2", "task": "'''Task''': \n* Generate a string with   '''N'''   opening brackets   '''['''   and with   '''N'''   closing brackets   ''']''',   in some arbitrary order. \n* Determine whether the generated string is ''balanced''; that is, whether it consists entirely of pairs of opening/closing brackets (in that order), none of which mis-nest.\n\n\n\n;Examples:\n    (empty)      OK\n    []           OK   \n    [][]         OK   \n    [[][]]       OK \n    ][         NOT OK\n    ][][       NOT OK\n    []][[]     NOT OK\n\n", "solution": ">>> from itertools import accumulate\n>>> from random import shuffle\n>>> def gen(n):\n...     txt = list('[]' * n)\n...     shuffle(txt)\n...     return ''.join(txt)\n...\n>>> def balanced(txt):\n...     brackets = ({'[': 1, ']': -1}.get(ch, 0) for ch in txt)\n...     return all(x>=0 for x in accumulate(brackets))\n...\n>>> for txt in (gen(N) for N in range(10)):\n...     print (\"%-22r is%s balanced\" % (txt, '' if balanced(txt) else ' not'))\n...\n''                     is balanced\n']['                   is not balanced\n'[]]['                 is not balanced\n']][[[]'               is not balanced\n'][[][][]'             is not balanced\n'[[[][][]]]'           is balanced\n'][[[][][]][]'         is not balanced\n'][]][][[]][[]['       is not balanced\n'][[]]][][[]][[[]'     is not balanced\n'][[][[]]]][[[]][]['   is not balanced"}
{"title": "Balanced ternary", "language": "Python from Common Lisp", "task": "Balanced ternary is a way of representing numbers.  Unlike the prevailing binary representation, a balanced ternary integer is in base 3, and each digit can have the values 1, 0, or -1.  \n\n\n;Examples:\nDecimal 11 = 32 + 31 - 30, thus it can be written as \"++-\"\n\nDecimal  6 = 32 - 31 + 0 x 30, thus it can be written as \"+-0\"\n\n\n;Task:\nImplement balanced ternary representation of integers with the following:\n# Support arbitrarily large integers, both positive and negative;\n# Provide ways to convert to and from text strings, using digits '+', '-' and '0' (unless you are already using strings to represent balanced ternary; but see requirement 5).\n# Provide ways to convert to and from native integer type (unless, improbably, your platform's native integer type ''is'' balanced ternary).  If your native integers can't support arbitrary length, overflows during conversion must be indicated.\n# Provide ways to perform addition, negation and multiplication directly on balanced ternary integers; do ''not'' convert to native integers first.\n# Make your implementation efficient, with a reasonable definition of \"efficient\" (and with a reasonable definition of \"reasonable\").\n\n\n'''Test case''' With balanced ternaries ''a'' from string \"+-0++0+\", ''b'' from native integer -436, ''c'' \"+-++-\":\n* write out ''a'', ''b'' and ''c'' in decimal notation;\n* calculate ''a'' x (''b'' - ''c''), write out the result in both ternary and decimal notations.\n\n\n'''Note:''' The pages floating point balanced ternary.\n\n", "solution": "class BalancedTernary:\n    # Represented as a list of 0, 1 or -1s, with least significant digit first.\n\n    str2dig = {'+': 1, '-': -1, '0': 0} # immutable\n    dig2str = {1: '+', -1: '-', 0: '0'} # immutable\n    table = ((0, -1), (1, -1), (-1, 0), (0, 0), (1, 0), (-1, 1), (0, 1)) # immutable\n\n    def __init__(self, inp):\n        if isinstance(inp, str):\n            self.digits = [BalancedTernary.str2dig[c] for c in reversed(inp)]\n        elif isinstance(inp, int):\n            self.digits = self._int2ternary(inp)\n        elif isinstance(inp, BalancedTernary):\n            self.digits = list(inp.digits)\n        elif isinstance(inp, list):\n            if all(d in (0, 1, -1) for d in inp):\n                self.digits = list(inp)\n            else:\n                raise ValueError(\"BalancedTernary: Wrong input digits.\")\n        else:\n            raise TypeError(\"BalancedTernary: Wrong constructor input.\")\n\n    @staticmethod\n    def _int2ternary(n):\n        if n == 0: return []\n        if (n % 3) == 0: return [0] + BalancedTernary._int2ternary(n // 3)\n        if (n % 3) == 1: return [1] + BalancedTernary._int2ternary(n // 3)\n        if (n % 3) == 2: return [-1] + BalancedTernary._int2ternary((n + 1) // 3)\n\n    def to_int(self):\n        return reduce(lambda y,x: x + 3 * y, reversed(self.digits), 0)\n\n    def __repr__(self):\n        if not self.digits: return \"0\"\n        return \"\".join(BalancedTernary.dig2str[d] for d in reversed(self.digits))\n\n    @staticmethod\n    def _neg(digs):\n        return [-d for d in digs]\n\n    def __neg__(self):\n        return BalancedTernary(BalancedTernary._neg(self.digits))\n\n    @staticmethod\n    def _add(a, b, c=0):\n        if not (a and b):\n            if c == 0:\n                return a or b\n            else:\n                return BalancedTernary._add([c], a or b)\n        else:\n            (d, c) = BalancedTernary.table[3 + (a[0] if a else 0) + (b[0] if b else 0) + c]\n            res = BalancedTernary._add(a[1:], b[1:], c)\n            # trim leading zeros\n            if res or d != 0:\n                return [d] + res\n            else:\n                return res\n\n    def __add__(self, b):\n        return BalancedTernary(BalancedTernary._add(self.digits, b.digits))\n\n    def __sub__(self, b):\n        return self + (-b)\n\n    @staticmethod\n    def _mul(a, b):\n        if not (a and b):\n            return []\n        else:\n            if   a[0] == -1: x = BalancedTernary._neg(b)\n            elif a[0] ==  0: x = []\n            elif a[0] ==  1: x = b\n            else: assert False\n            y = [0] + BalancedTernary._mul(a[1:], b)\n            return BalancedTernary._add(x, y)\n\n    def __mul__(self, b):\n        return BalancedTernary(BalancedTernary._mul(self.digits, b.digits))\n\n\ndef main():\n    a = BalancedTernary(\"+-0++0+\")\n    print \"a:\", a.to_int(), a\n\n    b = BalancedTernary(-436)\n    print \"b:\", b.to_int(), b\n\n    c = BalancedTernary(\"+-++-\")\n    print \"c:\", c.to_int(), c\n\n    r = a * (b - c)\n    print \"a * (b - c):\", r.to_int(), r\n\nmain()"}
{"title": "Barnsley fern", "language": "Python", "task": "A Barnsley fern is a fractal named after British mathematician Michael Barnsley and can be created using an iterated function system (IFS).\n\n\n;Task:\nCreate this fractal fern, using the following transformations:\n* f1   (chosen 1% of the time)\n         xn + 1 = 0\n         yn + 1 = 0.16 yn\n\n* f2   (chosen 85% of the time)\n         xn + 1 = 0.85 xn + 0.04 yn\n         yn + 1 = -0.04 xn + 0.85 yn + 1.6\n\n* f3   (chosen 7% of the time)\n         xn + 1 = 0.2 xn - 0.26 yn\n         yn + 1 = 0.23 xn + 0.22 yn + 1.6\n\n* f4   (chosen 7% of the time)\n         xn + 1 = -0.15 xn + 0.28 yn\n         yn + 1 = 0.26 xn + 0.24 yn + 0.44.\n\nStarting position: x = 0, y = 0\n\n", "solution": "import random\nfrom PIL import Image\n\n\nclass BarnsleyFern(object):\n    def __init__(self, img_width, img_height, paint_color=(0, 150, 0),\n                 bg_color=(255, 255, 255)):\n        self.img_width, self.img_height = img_width, img_height\n        self.paint_color = paint_color\n        self.x, self.y = 0, 0\n        self.age = 0\n\n        self.fern = Image.new('RGB', (img_width, img_height), bg_color)\n        self.pix = self.fern.load()\n        self.pix[self.scale(0, 0)] = paint_color\n\n    def scale(self, x, y):\n        h = (x + 2.182)*(self.img_width - 1)/4.8378\n        k = (9.9983 - y)*(self.img_height - 1)/9.9983\n        return h, k\n\n    def transform(self, x, y):\n        rand = random.uniform(0, 100)\n        if rand < 1:\n            return 0, 0.16*y\n        elif 1 <= rand < 86:\n            return 0.85*x + 0.04*y, -0.04*x + 0.85*y + 1.6\n        elif 86 <= rand < 93:\n            return 0.2*x - 0.26*y, 0.23*x + 0.22*y + 1.6\n        else:\n            return -0.15*x + 0.28*y, 0.26*x + 0.24*y + 0.44\n\n    def iterate(self, iterations):\n        for _ in range(iterations):\n            self.x, self.y = self.transform(self.x, self.y)\n            self.pix[self.scale(self.x, self.y)] = self.paint_color\n        self.age += iterations\n\nfern = BarnsleyFern(500, 500)\nfern.iterate(1000000)\nfern.fern.show()\n"}
{"title": "Base64 decode data", "language": "Python", "task": "See [[Base64 encode data]]. \n\nNow write a program that takes the output of the [[Base64 encode data]] task as input and regenerate the original file. \n\nWhen working on the VBA implementation I found several 'solutions' on the net, including one from the software maker himself, that showed output with incorrect padding. Obviously with incorrect padding in the output you can not decode correctly to the original file again.\n\n", "solution": "import base64\ndata = 'VG8gZXJyIGlzIGh1bWFuLCBidXQgdG8gcmVhbGx5IGZvdWwgdGhpbmdzIHVwIHlvdSBuZWVkIGEgY29tcHV0ZXIuCiAgICAtLSBQYXVsIFIuIEVocmxpY2g='\nprint(base64.b64decode(data).decode('utf-8'))\n"}
{"title": "Bell numbers", "language": "Python 2.7", "task": "Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly '''n''' elements. Each element of the sequence '''Bn''' is the number of partitions of a set of size '''n''' where order of the elements and order of the partitions are non-significant. E.G.: '''{a b}''' is the same as '''{b a}''' and '''{a} {b}''' is the same as '''{b} {a}'''.\n\n\n;So:\n\n:'''B0 = 1''' trivially. There is only one way to partition a set with zero elements. '''{ }'''\n\n:'''B1 = 1''' There is only one way to partition a set with one element. '''{a}'''\n\n:'''B2 = 2''' Two elements may be partitioned in two ways. '''{a} {b}, {a b}'''\n\n:'''B3 = 5''' Three elements may be partitioned in five ways '''{a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}'''\n\n: and so on.\n\n\nA simple way to find the Bell numbers is construct a '''Bell triangle''', also known as an '''Aitken's array''' or '''Peirce triangle''', and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.\n\n\n;Task:\n\nWrite a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the '''first 15''' and (if your language supports big Integers) '''50th''' elements of the sequence. \n\nIf you ''do'' use the Bell triangle method to generate the numbers, also show the '''first ten rows''' of the Bell triangle.\n\n\n;See also:\n\n:* '''OEIS:A000110 Bell or exponential numbers'''\n:* '''OEIS:A011971 Aitken's array'''\n\n", "solution": "def bellTriangle(n):\n    tri = [None] * n\n    for i in xrange(n):\n        tri[i] = [0] * i\n    tri[1][0] = 1\n    for i in xrange(2, n):\n        tri[i][0] = tri[i - 1][i - 2]\n        for j in xrange(1, i):\n            tri[i][j] = tri[i][j - 1] + tri[i - 1][j - 1]\n    return tri\n\ndef main():\n    bt = bellTriangle(51)\n    print \"First fifteen and fiftieth Bell numbers:\"\n    for i in xrange(1, 16):\n        print \"%2d: %d\" % (i, bt[i][0])\n    print \"50:\", bt[50][0]\n    print\n    print \"The first ten rows of Bell's triangle:\"\n    for i in xrange(1, 11):\n        print bt[i]\n\nmain()"}
{"title": "Bell numbers", "language": "Python 3.7", "task": "Bell or exponential numbers are enumerations of the number of different ways to partition a set that has exactly '''n''' elements. Each element of the sequence '''Bn''' is the number of partitions of a set of size '''n''' where order of the elements and order of the partitions are non-significant. E.G.: '''{a b}''' is the same as '''{b a}''' and '''{a} {b}''' is the same as '''{b} {a}'''.\n\n\n;So:\n\n:'''B0 = 1''' trivially. There is only one way to partition a set with zero elements. '''{ }'''\n\n:'''B1 = 1''' There is only one way to partition a set with one element. '''{a}'''\n\n:'''B2 = 2''' Two elements may be partitioned in two ways. '''{a} {b}, {a b}'''\n\n:'''B3 = 5''' Three elements may be partitioned in five ways '''{a} {b} {c}, {a b} {c}, {a} {b c}, {a c} {b}, {a b c}'''\n\n: and so on.\n\n\nA simple way to find the Bell numbers is construct a '''Bell triangle''', also known as an '''Aitken's array''' or '''Peirce triangle''', and read off the numbers in the first column of each row. There are other generating algorithms though, and you are free to choose the best / most appropriate for your case.\n\n\n;Task:\n\nWrite a routine (function, generator, whatever) to generate the Bell number sequence and call the routine to show here, on this page at least the '''first 15''' and (if your language supports big Integers) '''50th''' elements of the sequence. \n\nIf you ''do'' use the Bell triangle method to generate the numbers, also show the '''first ten rows''' of the Bell triangle.\n\n\n;See also:\n\n:* '''OEIS:A000110 Bell or exponential numbers'''\n:* '''OEIS:A011971 Aitken's array'''\n\n", "solution": "'''Bell numbers'''\n\nfrom itertools import accumulate, chain, islice\nfrom operator import add, itemgetter\nfrom functools import reduce\n\n\n# bellNumbers :: [Int]\ndef bellNumbers():\n    '''Bell or exponential numbers.\n       A000110\n    '''\n    return map(itemgetter(0), bellTriangle())\n\n\n# bellTriangle :: [[Int]]\ndef bellTriangle():\n    '''Bell triangle.'''\n    return map(\n        itemgetter(1),\n        iterate(\n            compose(\n                bimap(last)(identity),\n                list, uncurry(scanl(add))\n            )\n        )((1, [1]))\n    )\n\n\n# ------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Tests'''\n    showIndex = compose(repr, succ, itemgetter(0))\n    showValue = compose(repr, itemgetter(1))\n    print(\n        fTable(\n            'First fifteen Bell numbers:'\n        )(showIndex)(showValue)(identity)(list(\n            enumerate(take(15)(bellNumbers()))\n        ))\n    )\n\n    print('\\nFiftieth Bell number:')\n    bells = bellNumbers()\n    drop(49)(bells)\n    print(\n        next(bells)\n    )\n\n    print(\n        fTable(\n            \"\\nFirst 10 rows of Bell's triangle:\"\n        )(showIndex)(showValue)(identity)(list(\n            enumerate(take(10)(bellTriangle()))\n        ))\n    )\n\n\n# ------------------------ GENERIC ------------------------\n\n# bimap :: (a -> b) -> (c -> d) -> (a, c) -> (b, d)\ndef bimap(f):\n    '''Tuple instance of bimap.\n       A tuple of the application of f and g to the\n       first and second values respectively.\n    '''\n    def go(g):\n        def gox(x):\n            return (f(x), g(x))\n        return gox\n    return go\n\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    def go(f, g):\n        def fg(x):\n            return f(g(x))\n        return fg\n    return reduce(go, fs, identity)\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.\n    '''\n    def go(xs):\n        if isinstance(xs, (list, tuple, str)):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return go\n\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def gox(xShow):\n        def gofx(fxShow):\n            def gof(f):\n                def goxs(xs):\n                    ys = [xShow(x) for x in xs]\n                    w = max(map(len, ys))\n\n                    def arrowed(x, y):\n                        return y.rjust(w, ' ') + ' -> ' + fxShow(f(x))\n                    return s + '\\n' + '\\n'.join(\n                        map(arrowed, xs, ys)\n                    )\n                return goxs\n            return gof\n        return gofx\n    return gox\n\n\n# identity :: a -> a\ndef identity(x):\n    '''The identity function.'''\n    return x\n\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated\n       applications of f to x.\n    '''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return go\n\n\n# last :: [a] -> a\ndef last(xs):\n    '''The last element of a non-empty list.'''\n    return xs[-1]\n\n\n# scanl :: (b -> a -> b) -> b -> [a] -> [b]\ndef scanl(f):\n    '''scanl is like reduce, but returns a succession of\n       intermediate values, building from the left.\n    '''\n    def go(a):\n        def g(xs):\n            return accumulate(chain([a], xs), f)\n        return g\n    return go\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value.\n       For numeric types, (1 +).\n    '''\n    return 1 + x\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    def go(xs):\n        return (\n            xs[0:n]\n            if isinstance(xs, (list, tuple))\n            else list(islice(xs, n))\n        )\n    return go\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    '''A function over a tuple,\n       derived from a curried function.\n    '''\n    def go(tpl):\n        return f(tpl[0])(tpl[1])\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Benford's law", "language": "Python", "task": "{{Wikipedia|Benford's_law}}\n\n\n'''Benford's law''', also called the '''first-digit law''', refers to the frequency distribution of digits in many (but not all) real-life sources of data. \n\nIn this distribution, the number 1 occurs as the first digit about 30% of the time, while larger numbers occur in that position less frequently: 9 as the first digit less than 5% of the time. This distribution of first digits is the same as the widths of gridlines on a logarithmic scale. \n\nBenford's law also concerns the expected distribution for digits beyond the first, which approach a uniform distribution.\n\nThis result has been found to apply to a wide variety of data sets, including electricity bills, street addresses, stock prices, population numbers, death rates, lengths of rivers, physical and mathematical constants, and processes described by power laws (which are very common in nature). It tends to be most accurate when values are distributed across multiple orders of magnitude.\n\nA set of numbers is said to satisfy Benford's law if the leading digit d  (d \\in \\{1, \\ldots, 9\\}) occurs with probability\n\n:::: P(d) = \\log_{10}(d+1)-\\log_{10}(d) = \\log_{10}\\left(1+\\frac{1}{d}\\right)\n\nFor this task, write (a) routine(s) to calculate the distribution of first significant (non-zero) digits in a collection of numbers, then display the actual vs. expected distribution in the way most convenient for your language (table / graph / histogram / whatever).\n\nUse the first 1000 numbers from the Fibonacci sequence as your data set. No need to show how the Fibonacci numbers are obtained. \n\nYou can generate them or load them from a file; whichever is easiest. \n\nDisplay your actual vs expected distribution.\n\n\n''For extra credit:'' Show the distribution for one other set of numbers from a page on Wikipedia. State which Wikipedia page it can be obtained from and what the set enumerates. Again, no need to display the actual list of numbers or the code to load them.\n\n\n;See also:\n* numberphile.com.\n* A starting page on Wolfram Mathworld is {{Wolfram|Benfords|Law}}.\n\n", "solution": "from __future__ import division\nfrom itertools import islice, count\nfrom collections import Counter\nfrom math import log10\nfrom random import randint\n\nexpected = [log10(1+1/d) for d in range(1,10)]\n\ndef fib():\n    a,b = 1,1\n    while True:\n        yield a\n        a,b = b,a+b\n\n# powers of 3 as a test sequence\ndef power_of_threes():\n    return (3**k for k in count(0))\n\ndef heads(s):\n    for a in s: yield int(str(a)[0])\n\ndef show_dist(title, s):\n    c = Counter(s)\n    size = sum(c.values())\n    res = [c[d]/size for d in range(1,10)]\n\n    print(\"\\n%s Benfords deviation\" % title)\n    for r, e in zip(res, expected):\n        print(\"%5.1f%% %5.1f%%  %5.1f%%\" % (r*100., e*100., abs(r - e)*100.))\n\ndef rand1000():\n    while True: yield randint(1,9999)\n\nif __name__ == '__main__':\n    show_dist(\"fibbed\", islice(heads(fib()), 1000))\n    show_dist(\"threes\", islice(heads(power_of_threes()), 1000))\n\n    # just to show that not all kind-of-random sets behave like that\n    show_dist(\"random\", islice(heads(rand1000()), 10000))"}
{"title": "Best shuffle", "language": "Python", "task": "Shuffle the characters of a string in such a way that as many of the character values are in a different position as possible. \n\nA shuffle that produces a randomized result among the best choices is to be preferred. A deterministic approach that produces the same sequence every time is acceptable as an alternative.\n\nDisplay the result as follows: \n\n original string, shuffled string, (score) \n\nThe score gives the number of positions whose character value did ''not'' change. \n\n\n;Example:\n tree, eetr, (0)\n\n\n;Test cases:\n abracadabra\n seesaw\n elk\n grrrrrr\n up\n a\n\n\n;Related tasks\n*   [[Anagrams/Deranged anagrams]]\n*   [[Permutations/Derangements]]\n\n\n\n", "solution": "#!/usr/bin/env python\n\ndef best_shuffle(s):\n    # Count the supply of characters.\n    from collections import defaultdict\n    count = defaultdict(int)\n    for c in s:\n        count[c] += 1\n\n    # Shuffle the characters.\n    r = []\n    for x in s:\n        # Find the best character to replace x.\n        best = None\n        rankb = -2\n        for c, rankc in count.items():\n            # Prefer characters with more supply.\n            # (Save characters with less supply.)\n            # Avoid identical characters.\n            if c == x: rankc = -1\n            if rankc > rankb:\n                best = c\n                rankb = rankc\n\n        # Add character to list. Remove it from supply.\n        r.append(best)\n        count[best] -= 1\n        if count[best] >= 0: del count[best]\n\n    # If the final letter became stuck (as \"ababcd\" became \"bacabd\",\n    # and the final \"d\" became stuck), then fix it.\n    i = len(s) - 1\n    if r[i] == s[i]:\n        for j in range(i):\n            if r[i] != s[j] and r[j] != s[i]:\n                r[i], r[j] = r[j], r[i]\n                break\n\n    # Convert list to string. PEP 8, \"Style Guide for Python Code\",\n    # suggests that ''.join() is faster than + when concatenating\n    # many strings. See http://www.python.org/dev/peps/pep-0008/\n    r = ''.join(r)\n\n    score = sum(x == y for x, y in zip(r, s))\n\n    return (r, score)\n\nfor s in \"abracadabra\", \"seesaw\", \"elk\", \"grrrrrr\", \"up\", \"a\":\n    shuffled, score = best_shuffle(s)\n    print(\"%s, %s, (%d)\" % (s, shuffled, score))"}
{"title": "Bin given limits", "language": "Python", "task": "You are given a list of n ascending, unique numbers which are to form limits\nfor n+1 bins which count how many of a large set of input numbers fall in the\nrange of each bin.\n\n(Assuming zero-based indexing)\n\n    bin[0] counts how many inputs are < limit[0]\n    bin[1] counts how many inputs are >= limit[0] and < limit[1]\n    ..''\n    bin[n-1] counts how many inputs are >= limit[n-2] and < limit[n-1]\n    bin[n] counts how many inputs are >= limit[n-1]\n\n;Task:\nThe task is to create a function that given the ascending limits and a stream/\nlist of numbers, will return the bins; together with another function that\ngiven the same list of limits and the binning will ''print the limit of each bin\ntogether with the count of items that fell in the range''.\n\nAssume the numbers to bin are too large to practically sort.\n\n;Task examples:\nPart 1: Bin using the following limits the given input data\n\n    limits  = [23, 37, 43, 53, 67, 83]\n    data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,\n            16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]\n\nPart 2: Bin using the following limits the given input data\n\n    limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]\n    data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,\n            416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,\n            655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,\n            346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,\n            345,110,720,917,313,845,426,  9,457,628,410,723,354,895,881,953,677,137,397, 97,\n            854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157,  5,316,395,\n            787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,\n            698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,\n            605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,\n            466, 23,707,467, 33,670,921,180,991,396,160,436,717,918,  8,374,101,684,727,749]\n\nShow output here, on this page.\n", "solution": "from bisect import bisect_right\n\ndef bin_it(limits: list, data: list) -> list:\n    \"Bin data according to (ascending) limits.\"\n    bins = [0] * (len(limits) + 1)      # adds under/over range bins too\n    for d in data:\n        bins[bisect_right(limits, d)] += 1\n    return bins\n\ndef bin_print(limits: list, bins: list) -> list:\n    print(f\"          < {limits[0]:3} := {bins[0]:3}\")\n    for lo, hi, count in zip(limits, limits[1:], bins[1:]):\n        print(f\">= {lo:3} .. < {hi:3} := {count:3}\")\n    print(f\">= {limits[-1]:3}          := {bins[-1]:3}\")\n\n\nif __name__ == \"__main__\":\n    print(\"RC FIRST EXAMPLE\\n\")\n    limits  = [23, 37, 43, 53, 67, 83]\n    data = [95,21,94,12,99,4,70,75,83,93,52,80,57,5,53,86,65,17,92,83,71,61,54,58,47,\n            16, 8, 9,32,84,7,87,46,19,30,37,96,6,98,40,79,97,45,64,60,29,49,36,43,55]\n    bins = bin_it(limits, data)\n    bin_print(limits, bins)\n\n    print(\"\\nRC SECOND EXAMPLE\\n\")\n    limits = [14, 18, 249, 312, 389, 392, 513, 591, 634, 720]\n    data = [445,814,519,697,700,130,255,889,481,122,932, 77,323,525,570,219,367,523,442,933,\n            416,589,930,373,202,253,775, 47,731,685,293,126,133,450,545,100,741,583,763,306,\n            655,267,248,477,549,238, 62,678, 98,534,622,907,406,714,184,391,913, 42,560,247,\n            346,860, 56,138,546, 38,985,948, 58,213,799,319,390,634,458,945,733,507,916,123,\n            345,110,720,917,313,845,426,  9,457,628,410,723,354,895,881,953,677,137,397, 97,\n            854,740, 83,216,421, 94,517,479,292,963,376,981,480, 39,257,272,157,  5,316,395,\n            787,942,456,242,759,898,576, 67,298,425,894,435,831,241,989,614,987,770,384,692,\n            698,765,331,487,251,600,879,342,982,527,736,795,585, 40, 54,901,408,359,577,237,\n            605,847,353,968,832,205,838,427,876,959,686,646,835,127,621,892,443,198,988,791,\n            466, 23,707,467, 33,670,921,180,991,396,160,436,717,918,  8,374,101,684,727,749]\n    bins = bin_it(limits, data)\n    bin_print(limits, bins)"}
{"title": "Bioinformatics/Global alignment", "language": "Python from Go", "task": "Global alignment is designed to search for highly similar regions in two or more DNA sequences, where the\nsequences appear in the same order and orientation, fitting the sequences in as pieces in a puzzle.\n\nCurrent DNA sequencers find the sequence for multiple small segments of DNA which have mostly randomly formed by splitting a much larger DNA molecule into shorter segments. When re-assembling such segments of DNA sequences into a larger sequence to form, for example, the DNA coding for the relevant gene, the overlaps between multiple shorter sequences are commonly used to decide how the longer sequence is to be assembled. For example, \"AAGATGGA\", GGAGCGCATC\", and \"ATCGCAATAAGGA\" can be assembled into the sequence \"AAGATGGAGCGCATCGCAATAAGGA\" by noting that \"GGA\" is at the tail of the first string and head of the second string and \"ATC\" likewise is at the tail\nof the second and head of the third string.\n\nWhen looking for the best global alignment in the output strings produced by DNA sequences, there are\ntypically a large number of such overlaps among a large number of sequences. In such a case, the ordering\nthat results in the shortest common superstring is generrally preferred.\n\nFinding such a supersequence is an NP-hard problem, and many algorithms have been proposed to\nshorten calculations, especially when many very long sequences are matched.\n\nThe shortest common superstring as used in bioinfomatics here differs from the string task\n[[Shortest_common_supersequence]]. In that task, a supersequence\nmay have other characters interposed as long as the characters of each subsequence appear in order,\nso that (abcbdab, abdcaba) -> abdcabdab.  In this task, (abcbdab, abdcaba) -> abcbdabdcaba.\n\n\n;Task:\n:*   Given N non-identical strings of characters A, C, G, and T representing N DNA sequences, find the shortest DNA sequence containing all N sequences.\n\n:*    Handle cases where two sequences are identical or one sequence is entirely contained in another.\n\n:*   Print the resulting sequence along with its size (its base count) and a count of each base in the sequence.\n\n:*   Find the shortest common superstring for the following four examples:\n:   \n (\"TA\", \"AAG\", \"TA\", \"GAA\", \"TA\")\n\n (\"CATTAGGG\", \"ATTAG\", \"GGG\", \"TA\")\n\n (\"AAGAUGGA\", \"GGAGCGCAUC\", \"AUCGCAAUAAGGA\")\n\n (\"ATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTAT\",\n \"GGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGT\",\n \"CTATGTTCTTATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\",\n \"TGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATC\",\n \"AACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\",\n \"GCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTC\",\n \"CGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTTCGATTCTGCTTATAACACTATGTTCT\",\n \"TGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATC\",\n \"CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATGCTCGTGC\",\n \"GATGGAGCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTTCGATT\",\n \"TTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATC\",\n \"CTATGTTCTTATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\",\n \"TCTCTTAAACTCCTGCTAAATGCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGA\")\n\n;Related tasks:\n:* Bioinformatics base count.\n:* Bioinformatics sequence mutation.\n\n", "solution": "import os\n\nfrom collections import Counter\nfrom functools import reduce\nfrom itertools import permutations\n\nBASES = (\"A\", \"C\", \"G\", \"T\")\n\n\ndef deduplicate(sequences):\n    \"\"\"Return the set of sequences with those that are a substring\n    of others removed too.\"\"\"\n    sequences = set(sequences)\n    duplicates = set()\n\n    for s, t in permutations(sequences, 2):\n        if s != t and s in t:\n            duplicates.add(s)\n\n    return sequences - duplicates\n\n\ndef smash(s, t):\n    \"\"\"Return `s` concatenated with `t`. The longest suffix of `s`\n    that matches a prefix of `t` will be removed.\"\"\"\n    for i in range(len(s)):\n        if t.startswith(s[i:]):\n            return s[:i] + t\n    return s + t\n\n\ndef shortest_superstring(sequences):\n    \"\"\"Return the shortest superstring covering all sequences. If\n    there are multiple shortest superstrings, an arbitrary\n    superstring is returned.\"\"\"\n    sequences = deduplicate(sequences)\n    shortest = \"\".join(sequences)\n\n    for perm in permutations(sequences):\n        superstring = reduce(smash, perm)\n        if len(superstring) < len(shortest):\n            shortest = superstring\n\n    return shortest\n\n\ndef shortest_superstrings(sequences):\n    \"\"\"Return a list of all shortest superstrings that cover\n    `sequences`.\"\"\"\n    sequences = deduplicate(sequences)\n\n    shortest = set([\"\".join(sequences)])\n    shortest_length = sum(len(s) for s in sequences)\n\n    for perm in permutations(sequences):\n        superstring = reduce(smash, perm)\n        superstring_length = len(superstring)\n        if superstring_length < shortest_length:\n            shortest.clear()\n            shortest.add(superstring)\n            shortest_length = superstring_length\n        elif superstring_length == shortest_length:\n            shortest.add(superstring)\n\n    return shortest\n\n\ndef print_report(sequence):\n    \"\"\"Writes a report to stdout for the given DNA sequence.\"\"\"\n    buf = [f\"Nucleotide counts for {sequence}:\\n\"]\n\n    counts = Counter(sequence)\n    for base in BASES:\n        buf.append(f\"{base:>10}{counts.get(base, 0):>12}\")\n\n    other = sum(v for k, v in counts.items() if k not in BASES)\n    buf.append(f\"{'Other':>10}{other:>12}\")\n\n    buf.append(\" \" * 5 + \"_\" * 17)\n    buf.append(f\"{'Total length':>17}{sum(counts.values()):>5}\")\n\n    print(os.linesep.join(buf), \"\\n\")\n\n\nif __name__ == \"__main__\":\n    test_cases = [\n        (\"TA\", \"AAG\", \"TA\", \"GAA\", \"TA\"),\n        (\"CATTAGGG\", \"ATTAG\", \"GGG\", \"TA\"),\n        (\"AAGAUGGA\", \"GGAGCGCAUC\", \"AUCGCAAUAAGGA\"),\n        (\n            \"ATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTAT\",\n            \"GGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGT\",\n            \"CTATGTTCTTATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\",\n            \"TGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATC\",\n            \"AACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\",\n            \"GCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTC\",\n            \"CGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTTCGATTCTGCTTATAACACTATGTTCT\",\n            \"TGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATC\",\n            \"CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATGCTCGTGC\",\n            \"GATGGAGCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTTCGATT\",\n            \"TTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGATGGAGCGCATC\",\n            \"CTATGTTCTTATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\",\n            \"TCTCTTAAACTCCTGCTAAATGCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTGAGGACAAAGGTCAAGA\",\n        ),\n    ]\n\n    for case in test_cases:\n        for superstring in shortest_superstrings(case):\n            print_report(superstring)\n\n    # or ..\n    #\n    # for case in test_cases:\n    #     print_report(shortest_superstring(case))\n    #\n    # .. if you don't want all possible shortest superstrings.\n"}
{"title": "Bioinformatics/Sequence mutation", "language": "Python", "task": "Given a string of characters A, C, G, and T representing a DNA sequence write a routine to mutate the sequence, (string) by:\n# Choosing a random base position in the sequence.\n# Mutate the sequence by  doing one of either:\n## '''S'''wap the base at that position by changing it to one of A, C, G, or T. (which has a chance of swapping the base for the same base)\n## '''D'''elete the chosen base at the position.\n## '''I'''nsert another base randomly chosen from A,C, G, or T into the sequence at that position.\n# Randomly generate a test DNA sequence of at least 200 bases\n# \"Pretty print\" the sequence and a count of its size, and the count of each base in the sequence\n# Mutate the sequence ten times.\n# \"Pretty print\" the sequence after all mutations, and a count of its size, and the count of each base in the sequence.\n\n;Extra credit:\n* Give more information on the individual mutations applied.\n* Allow mutations to be weighted and/or chosen.\n", "solution": "import random\nfrom collections import Counter\n\ndef basecount(dna):\n    return sorted(Counter(dna).items())\n\ndef seq_split(dna, n=50):\n    return [dna[i: i+n] for i in range(0, len(dna), n)]\n\ndef seq_pp(dna, n=50):\n    for i, part in enumerate(seq_split(dna, n)):\n        print(f\"{i*n:>5}: {part}\")\n    print(\"\\n  BASECOUNT:\")\n    tot = 0\n    for base, count in basecount(dna):\n        print(f\"    {base:>3}: {count}\")\n        tot += count\n    base, count = 'TOT', tot\n    print(f\"    {base:>3}= {count}\")\n\ndef seq_mutate(dna, count=1, kinds=\"IDSSSS\", choice=\"ATCG\" ):\n    mutation = []\n    k2txt = dict(I='Insert', D='Delete', S='Substitute')\n    for _ in range(count):\n        kind = random.choice(kinds)\n        index = random.randint(0, len(dna))\n        if kind == 'I':    # Insert\n            dna = dna[:index] + random.choice(choice) + dna[index:]\n        elif kind == 'D' and dna:  # Delete\n            dna = dna[:index] + dna[index+1:]\n        elif kind == 'S' and dna:  # Substitute\n            dna = dna[:index] + random.choice(choice) + dna[index+1:]\n        mutation.append((k2txt[kind], index))\n    return dna, mutation\n\nif __name__ == '__main__':\n    length = 250\n    print(\"SEQUENCE:\")\n    sequence = ''.join(random.choices('ACGT', weights=(1, 0.8, .9, 1.1), k=length))\n    seq_pp(sequence)\n    print(\"\\n\\nMUTATIONS:\")\n    mseq, m = seq_mutate(sequence, 10)\n    for kind, index in m:\n        print(f\" {kind:>10} @{index}\")\n    print()\n    seq_pp(mseq)"}
{"title": "Bioinformatics/base count", "language": "Python", "task": "Given this string representing ordered DNA bases:\n\nCGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG\nCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG\nAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT\nGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\nCGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG\nTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\nTTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT\nCGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG\nTCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC\nGACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT\n\n\n\n;Task:\n:*   \"Pretty print\" the sequence followed by a summary of the counts of each of the bases:   ('''A''', '''C''', '''G''', and '''T''')   in the sequence \n:*   print the total count of each base in the string.\n\n\n\n", "solution": "from collections import Counter\n\ndef basecount(dna):\n    return sorted(Counter(dna).items())\n\ndef seq_split(dna, n=50):\n    return [dna[i: i+n] for i in range(0, len(dna), n)]\n\ndef seq_pp(dna, n=50):\n    for i, part in enumerate(seq_split(dna, n)):\n        print(f\"{i*n:>5}: {part}\")\n    print(\"\\n  BASECOUNT:\")\n    tot = 0\n    for base, count in basecount(dna):\n        print(f\"    {base:>3}: {count}\")\n        tot += count\n    base, count = 'TOT', tot\n    print(f\"    {base:>3}= {count}\")\n    \nif __name__ == '__main__':\n    print(\"SEQUENCE:\")\n    sequence = '''\\\nCGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG\\\nCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG\\\nAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT\\\nGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\\\nCGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG\\\nTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\\\nTTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT\\\nCGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG\\\nTCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC\\\nGACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT'''\n    seq_pp(sequence)\n"}
{"title": "Bioinformatics/base count", "language": "Python  3.10.5 ", "task": "Given this string representing ordered DNA bases:\n\nCGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG\nCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG\nAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT\nGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\nCGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG\nTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\nTTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT\nCGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG\nTCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC\nGACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT\n\n\n\n;Task:\n:*   \"Pretty print\" the sequence followed by a summary of the counts of each of the bases:   ('''A''', '''C''', '''G''', and '''T''')   in the sequence \n:*   print the total count of each base in the string.\n\n\n\n", "solution": "\"\"\"\n\tPython 3.10.5 (main, Jun  6 2022, 18:49:26) [GCC 12.1.0] on linux\n\n\tCreated on Wed 2022/08/17 11:19:31\n\t\n\"\"\"\n\n\ndef main ():\n\n\tdef DispCount () :\n\n\t    return f'\\n\\nBases :\\n\\n' + f''.join ( [ f'{i} =\\t{D [ i ]:4d}\\n' for i in  sorted ( BoI ) ] )\n\n\n\tS =\t'CGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATGCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG' \\\n\t\t'AGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGATGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT' \\\n\t\t'CGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGGTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA' \\\n\t\t'TTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTATCGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG' \\\n\t\t'TCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGACGACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT'\n\n\tAll   = set( S ) \n\t\n\tBoI   = set ( [ \"A\",\"C\",\"G\",\"T\" ] )\n\t\n\tother = All - BoI\n\t\n\tD     = { k : S.count ( k ) for k in All }\n\t\n\tprint ( 'Sequence:\\n\\n')\n\n\tprint ( ''.join ( [ f'{k:4d} : {S [ k: k + 50 ]}\\n' for k in range ( 0, len ( S ), 50 ) ] ) )\n\n\tprint ( f'{DispCount ()} \\n------------')\n\n\tprint ( '' if ( other == set () ) else f'Other\\t{sum ( [ D [ k ] for k in sorted ( other ) ] ):4d}\\n\\n' )\n\n\tprint ( f'\u03a3 = \\t {sum ( [ D [ k ] for k in sorted ( All ) ] ) } \\n============\\n')\n\t\n\tpass\n\n\ndef test ():\n\n\tpass\n\n\n## START\n\nLIVE = True\n\nif ( __name__ == '__main__' ) :\n\n\tmain () if LIVE else test ()\n\n\n"}
{"title": "Bioinformatics/base count", "language": "Python 3.7", "task": "Given this string representing ordered DNA bases:\n\nCGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG\nCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG\nAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT\nGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\nCGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG\nTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\nTTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT\nCGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG\nTCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC\nGACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT\n\n\n\n;Task:\n:*   \"Pretty print\" the sequence followed by a summary of the counts of each of the bases:   ('''A''', '''C''', '''G''', and '''T''')   in the sequence \n:*   print the total count of each base in the string.\n\n\n\n", "solution": "'''Bioinformatics \u2013 base count'''\n\nfrom itertools import count\nfrom functools import reduce\n\n\n# genBankFormatWithBaseCounts :: String -> String\ndef genBankFormatWithBaseCounts(sequence):\n    '''DNA Sequence displayed in a subset of the GenBank format.\n       See example at foot of:\n       https://www.genomatix.de/online_help/help/sequence_formats.html\n    '''\n    ks, totals = zip(*baseCounts(sequence))\n    ns = list(map(str, totals))\n    w = 2 + max(map(len, ns))\n\n    return '\\n'.join([\n        'DEFINITION  len=' + str(sum(totals)),\n        'BASE COUNT  ' + ''.join(\n            n.rjust(w) + ' ' + k.lower() for (k, n)\n            in zip(ks, ns)\n        ),\n        'ORIGIN'\n    ] + [\n        str(i).rjust(9) + ' ' + k for i, k\n        in zip(\n            count(1, 60),\n            [\n                ' '.join(row) for row in\n                chunksOf(6)(chunksOf(10)(sequence))\n            ]\n        )\n    ] + ['//'])\n\n\n# baseCounts :: String -> Zip [(String, Int)]\ndef baseCounts(baseString):\n    '''Sums for each base type in the given sequence string, with\n       a fifth sum for any characters not drawn from {A, C, G, T}.'''\n    bases = {\n        'A': 0,\n        'C': 1,\n        'G': 2,\n        'T': 3\n    }\n    return zip(\n        list(bases.keys()) + ['Other'],\n        foldl(\n            lambda a: compose(\n                nthArrow(succ)(a),\n                flip(curry(bases.get))(4)\n            )\n        )((0, 0, 0, 0, 0))(baseString)\n    )\n\n\n# -------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Base counts and sequence displayed in GenBank format\n    '''\n    print(\n        genBankFormatWithBaseCounts('''\\\nCGTAAAAAATTACAACGTCCTTTGGCTATCTCTTAAACTCCTGCTAAATG\\\nCTCGTGCTTTCCAATTATGTAAGCGTTCCGAGACGGGGTGGTCGATTCTG\\\nAGGACAAAGGTCAAGATGGAGCGCATCGAACGCAATAAGGATCATTTGAT\\\nGGGACGTTTCGTCGACAAAGTCTTGTTTCGAGAGTAACGGCTACCGTCTT\\\nCGATTCTGCTTATAACACTATGTTCTTATGAAATGGATGTTCTGAGTTGG\\\nTCAGTCCCAATGTGCGGGGTTTCTTTTAGTACGTCGGGAGTGGTATTATA\\\nTTTAATTTTTCTATATAGCGATCTGTATTTAAGCAATTCATTTAGGTTAT\\\nCGCCGCGATGCTCGGTTCGGACCGCCAAGCATCTGGCTCCACTGCTAGTG\\\nTCCTAAATTTGAATGGCAAACACAAATAAGATTTAGCAATTCGTGTAGAC\\\nGACCGGGGACTTGCATGATGGGAGCAGCTTTGTTAAACTACGAACGTAAT''')\n    )\n\n\n# ------------------------ GENERIC -------------------------\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n, subdividing the\n       contents of xs. Where the length of xs is not evenly\n       divible, the final list will be shorter than n.\n    '''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    def go(f, g):\n        def fg(x):\n            return f(g(x))\n        return fg\n    return reduce(go, fs, lambda x: x)\n\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    '''A curried function derived\n       from an uncurried function.\n    '''\n    return lambda x: lambda y: f(x, y)\n\n\n# flip :: (a -> b -> c) -> b -> a -> c\ndef flip(f):\n    '''The (curried or uncurried) function f with its\n       arguments reversed.\n    '''\n    return lambda a: lambda b: f(b)(a)\n\n\n# foldl :: (a -> b -> a) -> a -> [b] -> a\ndef foldl(f):\n    '''Left to right reduction of a list,\n       using the binary operator f, and\n       starting with an initial value a.\n    '''\n    def go(acc, xs):\n        return reduce(lambda a, x: f(a)(x), xs, acc)\n    return lambda acc: lambda xs: go(acc, xs)\n\n\n# nthArrow :: (a -> b) -> Tuple -> Int -> Tuple\ndef nthArrow(f):\n    '''A simple function lifted to one which applies\n       to a tuple, transforming only its nth value.\n    '''\n    def go(v, n):\n        return v if n > len(v) else [\n            x if n != i else f(x)\n            for i, x in enumerate(v)\n        ]\n    return lambda tpl: lambda n: tuple(go(tpl, n))\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value.\n       For numeric types, (1 +).\n    '''\n    return 1 + x\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Biorhythms", "language": "Python", "task": "For a while in the late 70s, the pseudoscience of biorhythms was popular enough to rival astrology, with kiosks in malls that would give you your weekly printout. It was also a popular entry in \"Things to Do with your Pocket Calculator\" lists. You can read up on the history at Wikipedia, but the main takeaway is that unlike astrology, the math behind biorhythms is dead simple.\n\nIt's based on the number of days since your birth. The premise is that three cycles of unspecified provenance govern certain aspects of everyone's lives - specifically, how they're feeling physically, emotionally, and mentally. The best part is that not only do these cycles somehow have the same respective lengths for all humans of any age, gender, weight, genetic background, etc, but those lengths are an exact number of days. And the pattern is in each case a perfect sine curve. Absolutely miraculous!\n\nTo compute your biorhythmic profile for a given day, the first thing you need is the number of days between that day and your birth, so the answers in [[Days between dates]] are probably a good starting point. (Strictly speaking, the biorhythms start at 0 at the moment of your birth, so if you know time of day you can narrow things down further, but in general these operate at whole-day granularity.) Then take the residue of that day count modulo each of the the cycle lengths to calculate where the day falls on each of the three sinusoidal journeys. \n\nThe three cycles and their lengths are as follows:\n:::{| class=\"wikitable\"\n! Cycle \n! Length\n|- \n|Physical \n| 23 days\n|- \n|Emotional \n| 28 days\n|- \n|Mental   \n| 33 days\n|}\n\nThe first half of each cycle is in \"plus\" territory, with a peak at the quarter-way point; the second half in \"minus\" territory, with a valley at the three-quarters mark. You can calculate a specific value between -1 and +1 for the ''k''th day of an ''n''-day cycle by computing '''sin( 2p''k'' / ''n'' )'''. The days where a cycle crosses the axis in either direction are called \"critical\" days, although with a cycle value of 0 they're also said to be the most neutral, which seems contradictory.\n\nThe task: write a subroutine, function, or program that will, given a birthdate and a target date, output the three biorhythmic values for the day. You may optionally include a text description of the position and the trend (e.g. \"up and rising\", \"peak\", \"up but falling\", \"critical\", \"down and falling\", \"valley\", \"down but rising\"), an indication of the date on which the next notable event (peak, valley, or crossing) falls, or even a graph of the cycles around the target date. Demonstrate the functionality for dates of your choice.\n\nExample run of my Raku implementation:\nraku br.raku 1943-03-09 1972-07-11\n", "solution": "\"\"\"\n\nPython implementation of\n\nhttp://rosettacode.org/wiki/Biorhythms\n\n\"\"\"\n\nfrom datetime import date, timedelta\nfrom math import floor, sin, pi\n\ndef biorhythms(birthdate,targetdate):\n    \"\"\"\n    Print out biorhythm data for targetdate assuming you were\n    born on birthdate.\n    \n    birthdate and targetdata are strings in this format:\n    \n    YYYY-MM-DD e.g. 1964-12-26\n    \"\"\"\n    \n    # print dates\n    \n    print(\"Born: \"+birthdate+\" Target: \"+targetdate)    \n    \n    # convert to date types - Python 3.7 or later\n    \n    birthdate = date.fromisoformat(birthdate)\n    targetdate = date.fromisoformat(targetdate)\n    \n    # days between\n    \n    days = (targetdate - birthdate).days\n    \n    print(\"Day: \"+str(days))\n    \n    # cycle logic - mostly from Julia example\n    \n    cycle_labels = [\"Physical\", \"Emotional\", \"Mental\"]\n    cycle_lengths = [23, 28, 33]\n    quadrants = [(\"up and rising\", \"peak\"), (\"up but falling\", \"transition\"),\n                   (\"down and falling\", \"valley\"), (\"down but rising\", \"transition\")]\n    \n    for i in range(3):\n        label = cycle_labels[i]\n        length = cycle_lengths[i]\n        position = days % length\n        quadrant = int(floor((4 * position) / length))\n        percentage = int(round(100 * sin(2 * pi * position / length),0))\n        transition_date = targetdate + timedelta(days=floor((quadrant + 1)/4 * length) - position)\n        trend, next = quadrants[quadrant]\n        \n        if percentage > 95:\n            description = \"peak\"\n        elif percentage < -95:\n             description = \"valley\"\n        elif abs(percentage) < 5:\n             description = \"critical transition\"\n        else:\n             description = str(percentage)+\"% (\"+trend+\", next \"+next+\" \"+str(transition_date)+\")\"\n        print(label+\" day \"+str(position)+\": \"+description)\n    \n    \nbiorhythms(\"1943-03-09\",\"1972-07-11\")\n"}
{"title": "Bitcoin/address validation", "language": "Python", "task": "Write a program that takes a bitcoin address as argument, \nand checks whether or not this address is valid.\n\nA bitcoin address uses a base58 encoding, which uses an alphabet of the characters 0 .. 9, A ..Z, a .. z, but without the four characters:\n:::*   0   zero\n:::*   O   uppercase oh\n:::*   I   uppercase eye\n:::*   l   lowercase ell\n\n\nWith this encoding, a bitcoin address encodes 25 bytes:\n* the first byte is the version number, which will be zero for this task ;\n* the next twenty bytes are a [[RIPEMD-160]] digest, but you don't have to know that for this task:  you can consider them a pure arbitrary data ;\n* the last four bytes are a checksum check.  They are the first four bytes of a double [[SHA-256]] digest of the previous 21 bytes.\n\n\nTo check the bitcoin address, you must read the first twenty-one bytes, compute the checksum, and check that it corresponds to the last four bytes.\n\nThe program can either return a boolean value or throw an exception when not valid.\n\nYou can use a digest library for [[SHA-256]].\n\n\n;Example of a bitcoin address:\n\n 1AGNa15ZQXAZUgFiqJ2i7Z2DPU2J6hW62i\n\n\nIt doesn't belong to anyone and is part of the test suite of the bitcoin software.   \nYou can change a few characters in this string and check that it'll fail the test.\n\n", "solution": "from hashlib import sha256\n\ndigits58 = '123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'\n\ndef decode_base58(bc, length):\n    n = 0\n    for char in bc:\n        n = n * 58 + digits58.index(char)\n    return n.to_bytes(length, 'big')\ndef check_bc(bc):\n    try:\n        bcbytes = decode_base58(bc, 25)\n        return bcbytes[-4:] == sha256(sha256(bcbytes[:-4]).digest()).digest()[:4]\n    except Exception:\n        return False\n\nprint(check_bc('1AGNa15ZQXAZUgFiqJ3i7Z2DPU2J6hW62i'))\nprint(check_bc(\"17NdbrSGoUotzeGCcMMCqnFkEvLymoou9j\"))"}
{"title": "Bitcoin/public point to address", "language": "Python", "task": "elliptic curve public point into a short ASCII string.  The purpose of this task is to perform such a conversion.\n\nThe encoding steps are:\n* take the X and Y coordinates of the given public point, and concatenate them in order to have a 64 byte-longed string ;\n* add one byte prefix equal to 4 (it is a convention for this way of encoding a public point) ;\n* compute the [[SHA-256]] of this string ;\n* compute the [[RIPEMD-160]] of this SHA-256 digest ;\n* compute the checksum of the concatenation of the version number digit (a single zero byte) and this RIPEMD-160 digest, as described in [[bitcoin/address validation]] ;\n* Base-58 encode (see below) the concatenation of the version number (zero in this case), the ripemd digest and the checksum\n\nThe base-58 encoding is based on an alphabet of alphanumeric characters (numbers, upper case and lower case, in that order) but without the four characters 0, O, l and I.\n\nHere is an example public point:\nX = 0x50863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B2352\nY = 0x2CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6\n\nThe corresponding address should be:\n16UwLL9Risc3QfPqBUvKofHmBQ7wMtjvM\n\nNb.  The leading '1' is not significant as 1 is zero in base-58.  It is however often added to the bitcoin address for various reasons.  There can actually be several of them.  You can ignore this and output an address without the leading 1.\n\n''Extra credit:'' add a verification procedure about the public point, making sure it belongs to the secp256k1 elliptic curve\n", "solution": "#!/usr/bin/env python3\n\nimport binascii\nimport functools\nimport hashlib\n\ndigits58 = b'123456789ABCDEFGHJKLMNPQRSTUVWXYZabcdefghijkmnopqrstuvwxyz'\n\ndef b58(n):\n    return b58(n//58) + digits58[n%58:n%58+1] if n else b''\n\ndef public_point_to_address(x, y):\n    c = b'\\x04' + binascii.unhexlify(x) + binascii.unhexlify(y)\n    r = hashlib.new('ripemd160')\n    r.update(hashlib.sha256(c).digest())\n    c = b'\\x00' + r.digest()\n    d = hashlib.sha256(hashlib.sha256(c).digest()).digest()\n    return b58(functools.reduce(lambda n, b: n<<8|b, c + d[:4]))\n\nif __name__ == '__main__':\n    print(public_point_to_address(\n        b'50863AD64A87AE8A2FE83C1AF1A8403CB53F53E486D8511DAD8A04887E5B2352',\n        b'2CD470243453A299FA9E77237716103ABC11A1DF38855ED6F2EE187E9C582BA6'))"}
{"title": "Bitwise IO", "language": "Python", "task": "The aim of this task is to write functions (or create a class if yourlanguage is Object Oriented and you prefer) for reading and writing sequences of\nbits, most significant bit first. While the output of a asciiprint \"STRING\" is the ASCII byte sequence\n\"S\", \"T\", \"R\", \"I\", \"N\", \"G\", the output of a \"print\" of the bits sequence\n0101011101010 (13 bits) must be 0101011101010; real I/O is performed always\n''quantized'' by byte (avoiding endianness issues and relying on underlying\nbuffering for performance), therefore you must obtain as output the bytes\n0101 0111 0101 0'''000''' (bold bits are padding bits), i.e. in hexadecimal 57 50.\n\nAs test, you can implement a '''rough''' (e.g. don't care about error handling or\nother issues) compression/decompression program for ASCII sequences\nof bytes, i.e. bytes for which the most significant bit is always unused, so that you can write\nseven bits instead of eight (each 8 bytes of input, we write 7 bytes of output).\n\nThese bit oriented I/O functions can be used to implement compressors and\ndecompressors; e.g. Dynamic and Static Huffman encodings use variable length\nbits sequences, while LZW (see [[LZW compression]]) use fixed or variable ''words''\nnine (or more) bits long.\n\n* Limits in the maximum number of bits that can be written/read in a single read/write operation are allowed.\n* Errors handling is not mandatory\n\n", "solution": "class BitWriter(object):\n    def __init__(self, f):\n        self.accumulator = 0\n        self.bcount = 0\n        self.out = f\n\n    def __enter__(self):\n        return self\n\n    def __exit__(self, exc_type, exc_val, exc_tb):\n        self.flush()\n\n    def __del__(self):\n        try:\n            self.flush()\n        except ValueError:   # I/O operation on closed file.\n            pass\n\n    def _writebit(self, bit):\n        if self.bcount == 8:\n            self.flush()\n        if bit > 0:\n            self.accumulator |= 1 << 7-self.bcount\n        self.bcount += 1\n\n    def writebits(self, bits, n):\n        while n > 0:\n            self._writebit(bits & 1 << n-1)\n            n -= 1\n\n    def flush(self):\n        self.out.write(bytearray([self.accumulator]))\n        self.accumulator = 0\n        self.bcount = 0\n\n\nclass BitReader(object):\n    def __init__(self, f):\n        self.input = f\n        self.accumulator = 0\n        self.bcount = 0\n        self.read = 0\n\n    def __enter__(self):\n        return self\n\n    def __exit__(self, exc_type, exc_val, exc_tb):\n        pass\n\n    def _readbit(self):\n        if not self.bcount:\n            a = self.input.read(1)\n            if a:\n                self.accumulator = ord(a)\n            self.bcount = 8\n            self.read = len(a)\n        rv = (self.accumulator & (1 << self.bcount-1)) >> self.bcount-1\n        self.bcount -= 1\n        return rv\n\n    def readbits(self, n):\n        v = 0\n        while n > 0:\n            v = (v << 1) | self._readbit()\n            n -= 1\n        return v\n\nif __name__ == '__main__':\n    import os\n    import sys\n    # Determine this module's name from it's file name and import it.\n    module_name = os.path.splitext(os.path.basename(__file__))[0]\n    bitio = __import__(module_name)\n\n    with open('bitio_test.dat', 'wb') as outfile:\n        with bitio.BitWriter(outfile) as writer:\n            chars = '12345abcde'\n            for ch in chars:\n                writer.writebits(ord(ch), 7)\n\n    with open('bitio_test.dat', 'rb') as infile:\n        with bitio.BitReader(infile) as reader:\n            chars = []\n            while True:\n                x = reader.readbits(7)\n                if not reader.read:  # End-of-file?\n                    break\n                chars.append(chr(x))\n            print(''.join(chars))\n"}
{"title": "Box the compass", "language": "Python", "task": "Avast me hearties!\nThere be many a land lubber that knows naught of the pirate ways and gives direction by degree!\nThey know not how to box the compass! \n\n\n;Task description:\n# Create a function that takes a heading in degrees and returns the correct 32-point compass heading.\n# Use the function to print and display a table of Index, Compass point, and Degree; rather like the corresponding columns from, the first table of the wikipedia article, but use only the following 33 headings as input:\n:[0.0, 16.87, 16.88, 33.75, 50.62, 50.63, 67.5, 84.37, 84.38, 101.25, 118.12, 118.13, 135.0, 151.87, 151.88, 168.75, 185.62, 185.63, 202.5, 219.37, 219.38, 236.25, 253.12, 253.13, 270.0, 286.87, 286.88, 303.75, 320.62, 320.63, 337.5, 354.37, 354.38]. (They should give the same order of points but are spread throughout the ranges of acceptance).\n\n\n;Notes;\n* The headings and indices can be calculated from this pseudocode:\nfor i in 0..32 inclusive:\n    heading = i * 11.25\n    case i %3:\n      if 1: heading += 5.62; break\n      if 2: heading -= 5.62; break\n    end\n    index = ( i mod 32) + 1\n* The column of indices can be thought of as an enumeration of the thirty two cardinal points (see talk page)..\n\n", "solution": "majors   = 'north east south west'.split()\nmajors   *= 2 # no need for modulo later\nquarter1 = 'N,N by E,N-NE,NE by N,NE,NE by E,E-NE,E by N'.split(',')\nquarter2 = [p.replace('NE','EN') for p in quarter1]\n\ndef degrees2compasspoint(d):\n    d = (d % 360) + 360/64\n    majorindex, minor = divmod(d, 90.)\n    majorindex = int(majorindex)\n    minorindex  = int( (minor*4) // 45 )\n    p1, p2 = majors[majorindex: majorindex+2]\n    if p1 in {'north', 'south'}:\n        q = quarter1\n    else:\n        q = quarter2\n    return q[minorindex].replace('N', p1).replace('E', p2).capitalize()\n\nif __name__ == '__main__':\n    for i in range(33):\n        d = i * 11.25\n        m = i % 3\n        if   m == 1: d += 5.62\n        elif m == 2: d -= 5.62\n        n = i % 32 + 1\n        print( '%2i %-18s %7.2f\u00b0' % (n, degrees2compasspoint(d), d) )"}
{"title": "Boyer-Moore string search", "language": "Python", "task": "This algorithm is designed for pattern searching on certain types of devices which are backtracking-unfriendly such as Tape drives and Hard disks.\n\nIt works by first caching a segment of data from storage and match it against the pattern from the highest position backward to the lowest. If the matching fails, it will cache next segment of data and move the start point forward to skip the portion of data which will definitely fail to match. This continues until we successfully match the pattern or the pointer exceeds the data length.\n\nFollow this link for more information about this algorithm.\n\nTo properly test this algorithm, it would be good to search for a string which contains repeated subsequences, such as alfalfa and the text being searched should not only include a match but that match should be preceded by words which partially match, such as alfredo, behalf, calfskin, halfhearted, malfunction or severalfold.\n", "solution": "\"\"\"\nThis version uses ASCII for case-sensitive matching. For Unicode you may want to match in UTF-8\nbytes instead of creating a 0x10FFFF-sized table.\n\"\"\"\n\nfrom typing import List\n\nALPHABET_SIZE = 256\n\ndef alphabet_index(c: str) -> int:\n    \"\"\"Return the index of the given ASCII character. \"\"\"\n    val = ord(c)\n    assert 0 <= val < ALPHABET_SIZE\n    return val\n\ndef match_length(S: str, idx1: int, idx2: int) -> int:\n    \"\"\"Return the length of the match of the substrings of S beginning at idx1 and idx2.\"\"\"\n    if idx1 == idx2:\n        return len(S) - idx1\n    match_count = 0\n    while idx1 < len(S) and idx2 < len(S) and S[idx1] == S[idx2]:\n        match_count += 1\n        idx1 += 1\n        idx2 += 1\n    return match_count\n\ndef fundamental_preprocess(S: str) -> List[int]:\n    \"\"\"Return Z, the Fundamental Preprocessing of S.\n\n    Z[i] is the length of the substring beginning at i which is also a prefix of S.\n    This pre-processing is done in O(n) time, where n is the length of S.\n    \"\"\"\n    if len(S) == 0:  # Handles case of empty string\n        return []\n    if len(S) == 1:  # Handles case of single-character string\n        return [1]\n    z = [0 for x in S]\n    z[0] = len(S)\n    z[1] = match_length(S, 0, 1)\n    for i in range(2, 1 + z[1]):  # Optimization from exercise 1-5\n        z[i] = z[1] - i + 1\n    # Defines lower and upper limits of z-box\n    l = 0\n    r = 0\n    for i in range(2 + z[1], len(S)):\n        if i <= r:  # i falls within existing z-box\n            k = i - l\n            b = z[k]\n            a = r - i + 1\n            if b < a:  # b ends within existing z-box\n                z[i] = b\n            else:  # b ends at or after the end of the z-box, we need to do an explicit match to the right of the z-box\n                z[i] = a + match_length(S, a, r + 1)\n                l = i\n                r = i + z[i] - 1\n        else:  # i does not reside within existing z-box\n            z[i] = match_length(S, 0, i)\n            if z[i] > 0:\n                l = i\n                r = i + z[i] - 1\n    return z\n\ndef bad_character_table(S: str) -> List[List[int]]:\n    \"\"\"\n    Generates R for S, which is an array indexed by the position of some character c in the\n    ASCII table. At that index in R is an array of length |S|+1, specifying for each\n    index i in S (plus the index after S) the next location of character c encountered when\n    traversing S from right to left starting at i. This is used for a constant-time lookup\n    for the bad character rule in the Boyer-Moore string search algorithm, although it has\n    a much larger size than non-constant-time solutions.\n    \"\"\"\n    if len(S) == 0:\n        return [[] for a in range(ALPHABET_SIZE)]\n    R = [[-1] for a in range(ALPHABET_SIZE)]\n    alpha = [-1 for a in range(ALPHABET_SIZE)]\n    for i, c in enumerate(S):\n        alpha[alphabet_index(c)] = i\n        for j, a in enumerate(alpha):\n            R[j].append(a)\n    return R\n\ndef good_suffix_table(S: str) -> List[int]:\n    \"\"\"\n    Generates L for S, an array used in the implementation of the strong good suffix rule.\n    L[i] = k, the largest position in S such that S[i:] (the suffix of S starting at i) matches\n    a suffix of S[:k] (a substring in S ending at k). Used in Boyer-Moore, L gives an amount to\n    shift P relative to T such that no instances of P in T are skipped and a suffix of P[:L[i]]\n    matches the substring of T matched by a suffix of P in the previous match attempt.\n    Specifically, if the mismatch took place at position i-1 in P, the shift magnitude is given\n    by the equation len(P) - L[i]. In the case that L[i] = -1, the full shift table is used.\n    Since only proper suffixes matter, L[0] = -1.\n    \"\"\"\n    L = [-1 for c in S]\n    N = fundamental_preprocess(S[::-1])  # S[::-1] reverses S\n    N.reverse()\n    for j in range(0, len(S) - 1):\n        i = len(S) - N[j]\n        if i != len(S):\n            L[i] = j\n    return L\n\ndef full_shift_table(S: str) -> List[int]:\n    \"\"\"\n    Generates F for S, an array used in a special case of the good suffix rule in the Boyer-Moore\n    string search algorithm. F[i] is the length of the longest suffix of S[i:] that is also a\n    prefix of S. In the cases it is used, the shift magnitude of the pattern P relative to the\n    text T is len(P) - F[i] for a mismatch occurring at i-1.\n    \"\"\"\n    F = [0 for c in S]\n    Z = fundamental_preprocess(S)\n    longest = 0\n    for i, zv in enumerate(reversed(Z)):\n        longest = max(zv, longest) if zv == i + 1 else longest\n        F[-i - 1] = longest\n    return F\n\ndef string_search(P, T) -> List[int]:\n    \"\"\"\n    Implementation of the Boyer-Moore string search algorithm. This finds all occurrences of P\n    in T, and incorporates numerous ways of pre-processing the pattern to determine the optimal\n    amount to shift the string and skip comparisons. In practice it runs in O(m) (and even\n    sublinear) time, where m is the length of T. This implementation performs a case-sensitive\n    search on ASCII characters. P must be ASCII as well.\n    \"\"\"\n    if len(P) == 0 or len(T) == 0 or len(T) < len(P):\n        return []\n\n    matches = []\n\n    # Preprocessing\n    R = bad_character_table(P)\n    L = good_suffix_table(P)\n    F = full_shift_table(P)\n\n    k = len(P) - 1      # Represents alignment of end of P relative to T\n    previous_k = -1     # Represents alignment in previous phase (Galil's rule)\n    while k < len(T):\n        i = len(P) - 1  # Character to compare in P\n        h = k           # Character to compare in T\n        while i >= 0 and h > previous_k and P[i] == T[h]:  # Matches starting from end of P\n            i -= 1\n            h -= 1\n        if i == -1 or h == previous_k:  # Match has been found (Galil's rule)\n            matches.append(k - len(P) + 1)\n            k += len(P) - F[1] if len(P) > 1 else 1\n        else:  # No match, shift by max of bad character and good suffix rules\n            char_shift = i - R[alphabet_index(T[h])][i]\n            if i + 1 == len(P):  # Mismatch happened on first attempt\n                suffix_shift = 1\n            elif L[i + 1] == -1:  # Matched suffix does not appear anywhere in P\n                suffix_shift = len(P) - F[i + 1]\n            else:               # Matched suffix appears in P\n                suffix_shift = len(P) - 1 - L[i + 1]\n            shift = max(char_shift, suffix_shift)\n            previous_k = k if shift >= i + 1 else previous_k  # Galil's rule\n            k += shift\n    return matches\n\nTEXT1 = 'InhisbookseriesTheArtofComputerProgrammingpublishedbyAddisonWesleyDKnuthusesanimaginarycomputertheMIXanditsassociatedmachinecodeandassemblylanguagestoillustratetheconceptsandalgorithmsastheyarepresented'\nTEXT2 = \"Nearby farms grew a half acre of alfalfa on the dairy's behalf, with bales of all that alfalfa exchanged for milk.\"\nPAT1, PAT2, PAT3 = 'put', 'and', 'alfalfa'\n\nprint(\"Found\", PAT1, \"at:\", string_search(PAT1, TEXT1))\nprint(\"Found\", PAT2, \"at:\", string_search(PAT2, TEXT1))\nprint(\"Found\", PAT3, \"at:\", string_search(PAT3, TEXT2))\n"}
{"title": "Brazilian numbers", "language": "Python", "task": "Brazilian numbers are so called as they were first formally presented at the 1994 math Olympiad ''Olimpiada Iberoamericana de Matematica'' in Fortaleza, Brazil.\n\nBrazilian numbers are defined as:\n\nThe set of positive integer numbers where each number '''N''' has at least one natural number '''B''' where '''1 < B < N-1''' where the representation of '''N''' in '''base B''' has all equal digits.\n\n\n;E.G.:\n\n:* '''1, 2 & 3''' can not be Brazilian; there is no base '''B''' that satisfies the condition '''1 < B < N-1'''.\n:* '''4''' is not Brazilian; '''4''' in '''base 2''' is '''100'''. The digits are not all the same.\n:* '''5''' is not Brazilian; '''5''' in '''base 2''' is '''101''', in '''base 3''' is '''12'''. There is no representation where the digits are the same.\n:* '''6''' is not Brazilian; '''6''' in '''base 2''' is '''110''', in '''base 3''' is '''20''', in '''base 4''' is '''12'''. There is no representation where the digits are the same.\n:* '''7''' ''is'' Brazilian; '''7''' in '''base 2''' is '''111'''. There is at least one representation where the digits are all the same.\n:* '''8''' ''is'' Brazilian; '''8''' in '''base 3''' is '''22'''. There is at least one representation where the digits are all the same.\n:* ''and so on...''\n\n\nAll even integers '''2P >= 8''' are Brazilian because '''2P = 2(P-1) + 2''', which is '''22''' in '''base P-1''' when '''P-1 > 2'''. That becomes true when '''P >= 4'''.\nMore common: for all all integers '''R''' and '''S''', where '''R > 1''' and also '''S-1 > R''', then '''R*S''' is Brazilian because '''R*S = R(S-1) + R''', which is '''RR''' in '''base S-1'''\nThe only problematic numbers are squares of primes, where R = S. Only 11^2 is brazilian to base 3.   \nAll prime integers, that are brazilian, can only have the digit '''1'''. Otherwise one could factor out the digit, therefore it cannot be a prime number. Mostly in form of '''111''' to base Integer(sqrt(prime number)). Must be an odd count of '''1''' to stay odd like primes >  2\n\n;Task:\n\nWrite a routine (function, whatever) to determine if a number is Brazilian and use the routine to show here, on this page;\n\n:* the first '''20''' Brazilian numbers;\n:* the first '''20 odd''' Brazilian numbers;\n:* the first '''20 prime''' Brazilian numbers;\n\n\n;See also:\n\n:* '''OEIS:A125134 - Brazilian numbers'''\n:* '''OEIS:A257521 - Odd Brazilian numbers'''\n:* '''OEIS:A085104 - Prime Brazilian numbers'''\n\n", "solution": "'''Brazilian numbers'''\n\nfrom itertools import count, islice\n\n\n# isBrazil :: Int -> Bool\ndef isBrazil(n):\n    '''True if n is a Brazilian number,\n       in the sense of OEIS:A125134.\n    '''\n    return 7 <= n and (\n        0 == n % 2 or any(\n            map(monoDigit(n), range(2, n - 1))\n        )\n    )\n\n\n# monoDigit :: Int -> Int -> Bool\ndef monoDigit(n):\n    '''True if all the digits of n,\n       in the given base, are the same.\n    '''\n    def go(base):\n        def g(b, n):\n            (q, d) = divmod(n, b)\n\n            def p(qr):\n                return d != qr[1] or 0 == qr[0]\n\n            def f(qr):\n                return divmod(qr[0], b)\n            return d == until(p)(f)(\n                (q, d)\n            )[1]\n        return g(base, n)\n    return go\n\n\n# -------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''First 20 members each of:\n        OEIS:A125134\n        OEIS:A257521\n        OEIS:A085104\n    '''\n    for kxs in ([\n            (' ', count(1)),\n            (' odd ', count(1, 2)),\n            (' prime ', primes())\n    ]):\n        print(\n            'First 20' + kxs[0] + 'Brazilians:\\n' +\n            showList(take(20)(filter(isBrazil, kxs[1]))) + '\\n'\n        )\n\n\n# ------------------- GENERIC FUNCTIONS --------------------\n\n# primes :: [Int]\ndef primes():\n    ''' Non finite sequence of prime numbers.\n    '''\n    n = 2\n    dct = {}\n    while True:\n        if n in dct:\n            for p in dct[n]:\n                dct.setdefault(n + p, []).append(p)\n            del dct[n]\n        else:\n            yield n\n            dct[n * n] = [n]\n        n = 1 + n\n\n\n# showList :: [a] -> String\ndef showList(xs):\n    '''Stringification of a list.'''\n    return '[' + ','.join(str(x) for x in xs) + ']'\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    def go(xs):\n        return (\n            xs[0:n]\n            if isinstance(xs, (list, tuple))\n            else list(islice(xs, n))\n        )\n    return go\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f):\n        def g(x):\n            v = x\n            while not p(v):\n                v = f(v)\n            return v\n        return g\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Break OO privacy", "language": "Python", "task": "Show how to access private or protected members of a class in an object-oriented language from outside an instance of the class, without calling non-private or non-protected members of the class as a proxy.  \nThe intent is to show how a debugger, serializer, or other meta-programming tool might access information that is barred by normal access methods to the object  but can nevertheless be accessed from within the language by some provided escape hatch or reflection mechanism. \nThe intent is specifically not to demonstrate heroic measures such as peeking and poking raw memory.\n\nNote that cheating on your type system is almost universally regarded \nas unidiomatic at best, and poor programming practice at worst.  \nNonetheless, if your language intentionally maintains a double-standard for OO privacy, here's where you can show it off.\n", "solution": ">>> class MyClassName:\n\t__private = 123\n\tnon_private = __private * 2\n\n\t\n>>> mine = MyClassName()\n>>> mine.non_private\n246\n>>> mine.__private\nTraceback (most recent call last):\n  File \"<pyshell#23>\", line 1, in <module>\n    mine.__private\nAttributeError: 'MyClassName' object has no attribute '__private'\n>>> mine._MyClassName__private\n123\n>>> "}
{"title": "Brilliant numbers", "language": "Python", "task": "'''Brilliant numbers''' are a subset of semiprime numbers. Specifically, they are numbers that are the product of exactly two prime numbers '''that both have the same number of digits when expressed in base 10'''.\n\n''Brilliant numbers are useful in cryptography and when testing prime factoring algorithms.''\n \n\n\n;E.G.\n* '''3 x 3''' (9) is a brilliant number. \n* '''2 x 7''' (14) is a brilliant number.\n* '''113 x 691''' (78083) is a brilliant number.\n* '''2 x 31''' (62) is semiprime, but is '''not''' a brilliant number (different number of digits in the two factors).\n\n\n\n;Task\n* Find and display the first 100 brilliant numbers.\n* For the orders of magnitude 1 through 6, find and show the first brilliant number greater than or equal to the order of magnitude, and, its position in the series (or the count of brilliant numbers up to that point).\n\n\n;Stretch\n* Continue for larger orders of magnitude.\n\n\n;See also\n;* Numbers Aplenty - Brilliant numbers\n;* OEIS:A078972 - Brilliant numbers: semiprimes whose prime factors have the same number of decimal digits\n", "solution": "from primesieve.numpy import primes\nfrom math import isqrt\nimport numpy as np\n\nmax_order = 9\nblocks = [primes(10**n, 10**(n + 1)) for n in range(max_order)]\n\ndef smallest_brilliant(lb):\n    pos = 1\n    root = isqrt(lb)\n\n    for blk in blocks:\n        n = len(blk)\n        if blk[-1]*blk[-1] < lb:\n            pos += n*(n + 1)//2\n            continue\n\n        i = np.searchsorted(blk, root, 'left')\n        i += blk[i]*blk[i] < lb\n\n        if not i:\n            return blk[0]*blk[0], pos\n\n        p = blk[:i + 1]\n        q = (lb - 1)//p\n        idx = np.searchsorted(blk, q, 'right')\n\n        sel = idx < n\n        p, idx = p[sel], idx[sel]\n        q = blk[idx]\n\n        sel = q >= p\n        p, q, idx = p[sel], q[sel], idx[sel]\n\n        pos += np.sum(idx - np.arange(len(idx)))\n        return np.min(p*q), pos\n\nres = []\np = 0\nfor i in range(100):\n    p, _ = smallest_brilliant(p + 1)\n    res.append(p)\n\nprint(f'first 100 are {res}')\n\nfor i in range(max_order*2):\n    thresh = 10**i\n    p, pos = smallest_brilliant(thresh)\n    print(f'Above 10^{i:2d}: {p:20d} at #{pos}')"}
{"title": "Burrows\u2013Wheeler transform", "language": "Python", "task": "{{Wikipedia|Burrows-Wheeler_transform}}\n\n\nThe Burrows-Wheeler transform (BWT, also called block-sorting compression) rearranges a character string into runs of similar characters. \n\nThis is useful for compression, since it tends to be easy to compress a string that has runs of repeated characters by techniques such as move-to-front transform and run-length encoding. \n\nMore importantly, the transformation is reversible, without needing to store any additional data. \n\nThe BWT is thus a \"free\" method of improving the efficiency of text compression algorithms, costing only some extra computation.\n\n\nSource: Burrows-Wheeler transform\n\n", "solution": "def bwt(s):\n    \"\"\"Apply Burrows-Wheeler transform to input string.\"\"\"\n    assert \"\\002\" not in s and \"\\003\" not in s, \"Input string cannot contain STX and ETX characters\"\n    s = \"\\002\" + s + \"\\003\"  # Add start and end of text marker\n    table = sorted(s[i:] + s[:i] for i in range(len(s)))  # Table of rotations of string\n    last_column = [row[-1:] for row in table]  # Last characters of each row\n    return \"\".join(last_column)  # Convert list of characters into string\n\n\ndef ibwt(r):\n    \"\"\"Apply inverse Burrows-Wheeler transform.\"\"\"\n    table = [\"\"] * len(r)  # Make empty table\n    for i in range(len(r)):\n        table = sorted(r[i] + table[i] for i in range(len(r)))  # Add a column of r\n    s = [row for row in table if row.endswith(\"\\003\")][0]  # Find the correct row (ending in ETX)\n    return s.rstrip(\"\\003\").strip(\"\\002\")  # Get rid of start and end markers\n"}
{"title": "CSV data manipulation", "language": "Python", "task": "CSV spreadsheet files are suitable for storing tabular data in a relatively portable way. \n\nThe CSV format is flexible but somewhat ill-defined.  \n\nFor present purposes, authors may assume that the data fields contain no commas, backslashes, or quotation marks.\n\n\n;Task:\nRead a CSV file, change some values and save the changes back to a file.\n\nFor this task we will use the following CSV file:\n\n C1,C2,C3,C4,C5\n 1,5,9,13,17\n 2,6,10,14,18\n 3,7,11,15,19\n 4,8,12,16,20\n\nSuggestions\n\n Show how to add a column, headed 'SUM', of the sums of the rows.\n If possible, illustrate the use of built-in or standard functions, methods, or libraries, that handle generic CSV files.\n\n\n", "solution": "import fileinput\n\nchangerow, changecolumn, changevalue = 2, 4, '\"Spam\"'\n\nwith fileinput.input('csv_data_manipulation.csv', inplace=True) as f:\n    for line in f:\n        if fileinput.filelineno() == changerow:\n            fields = line.rstrip().split(',')\n            fields[changecolumn-1] = changevalue\n            line = ','.join(fields) + '\\n'\n        print(line, end='')"}
{"title": "CSV to HTML translation", "language": "Python", "task": "Consider a simplified CSV format where all rows are separated by a newline \nand all columns are separated by commas. \n\nNo commas are allowed as field data, but the data may contain \nother characters and character sequences that would \nnormally be   ''escaped''   when converted to HTML\n\n\n;Task:\nCreate a function that takes a string representation of the CSV data\nand returns a text string of an HTML table representing the CSV data. \n\nUse the following  data as the CSV text to convert, and show your output.\n: Character,Speech\n: The multitude,The messiah! Show us the messiah!\n: Brians mother,Now you listen here! He's not the messiah; he's a very naughty boy! Now go away!\n: The multitude,Who are you?\n: Brians mother,I'm his mother; that's who!\n: The multitude,Behold his mother! Behold his mother!\n\n\n;Extra credit:\n''Optionally'' allow special formatting for the first row of the table as if it is the tables header row \n(via  preferably; CSS if you must).\n\n", "solution": "from csv import DictReader\nfrom xml.etree import ElementTree as ET\n\ndef csv2html_robust(txt, header=True, attr=None):\n    # Use DictReader because, despite what the docs say, reader() doesn't\n    # return an object with .fieldnames\n    # (DictReader expects an iterable that returns lines, so split on \\n)\n    reader = DictReader(txt.split('\\n'))\n\n    table = ET.Element(\"TABLE\", **attr.get('TABLE', {}))\n    thead_tr = ET.SubElement(\n        ET.SubElement(table, \"THEAD\", **attr.get('THEAD', {})),\n        \"TR\")\n    tbody = ET.SubElement(table, \"TBODY\", **attr.get('TBODY', {}))\n\n    if header:\n        for name in reader.fieldnames:\n            ET.SubElement(thead_tr, \"TD\").text = name\n\n    for row in reader:\n        tr_elem = ET.SubElement(tbody, \"TR\", **attr.get('TR', {}))\n\n        # Use reader.fieldnames to query `row` in the correct order.\n        # (`row` isn't an OrderedDict prior to Python 3.6)\n        for field in reader.fieldnames:\n            td_elem = ET.SubElement(tr_elem, \"TD\", **attr.get('TD', {}))\n            td_elem.text = row[field]\n\n    return ET.tostring(table, method='html')\n\nhtmltxt = csv2html_robust(csvtxt, True, {\n    'TABLE': {'border': \"1\", 'summary': \"csv2html extra program output\"},\n    'THEAD': {'bgcolor': \"yellow\"},\n    'TBODY': {'bgcolor': \"orange\"}\n})\n\nprint(htmltxt.decode('utf8'))"}
{"title": "Calculating the value of e", "language": "Python", "task": "Calculate the value of   ''e''.\n\n\n(''e''   is also known as   ''Euler's number''   and   ''Napier's constant''.)\n\n\nSee details: Calculating the value of e\n\n", "solution": "import math\n#Implementation of Brother's formula\ne0 = 0\ne = 2\nn = 0\nfact = 1\nwhile(e-e0 > 1e-15):\n\te0 = e\n\tn += 1\n\tfact *= 2*n*(2*n+1)\n\te += (2.*n+2)/fact\n\nprint \"Computed e = \"+str(e)\nprint \"Real e = \"+str(math.e)\nprint \"Error = \"+str(math.e-e)\nprint \"Number of iterations = \"+str(n)"}
{"title": "Calculating the value of e", "language": "Python 3.7", "task": "Calculate the value of   ''e''.\n\n\n(''e''   is also known as   ''Euler's number''   and   ''Napier's constant''.)\n\n\nSee details: Calculating the value of e\n\n", "solution": "'''Calculating an approximate value for e'''\n\nfrom itertools import (accumulate, chain)\nfrom functools import (reduce)\nfrom operator import (mul)\n\n\n# eApprox :: () -> Float\ndef eApprox():\n    '''Approximation to the value of e.'''\n    return reduce(\n        lambda a, x: a + 1 / x,\n        scanl(mul)(1)(\n            range(1, 18)\n        ),\n        0\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test'''\n\n    print(\n        eApprox()\n    )\n\n\n# GENERIC ABSTRACTIONS ------------------------------------\n\n# scanl is like reduce, but returns a succession of\n# intermediate values, building from the left.\n# See, for example, under `scan` in the Lists chapter of\n# the language-independent Bird & Wadler 1988.\n\n# scanl :: (b -> a -> b) -> b -> [a] -> [b]\ndef scanl(f):\n    '''scanl is like reduce, but returns a succession of\n       intermediate values, building from the left.'''\n    return lambda a: lambda xs: (\n        accumulate(chain([a], xs), f)\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Calkin-Wilf sequence", "language": "Python", "task": "The '''Calkin-Wilf sequence''' contains every nonnegative rational number exactly once. \n\nIt can be calculated recursively as follows:\n\n        {{math|a1}}   =  {{math|1}} \n {{math|an+1}}  =  {{math|1/(2an+1-an)}} for n > 1 \n\n\n;Task part 1:\n* Show on this page terms 1 through 20 of the Calkin-Wilf sequence.\n\nTo avoid floating point error, you may want to use a rational number data type.\n\n\nIt is also possible, given a non-negative rational number, to determine where it appears in the sequence without calculating the sequence. The procedure is to get the continued fraction representation of the rational and use it as the run-length encoding of the binary representation of the term number, beginning from the end of the continued fraction. \nIt only works if the number of terms in the continued fraction is odd- use either of the two equivalent representations to achieve this:\n\n        {{math|[a0; a1, a2, ..., an]}}  =  {{math|[a0; a1, a2 ,..., an-1, 1]}} \n\n\n;Example:\nThe fraction   '''9/4'''   has odd continued fraction representation      {{math|2; 3, 1}},     giving a binary representation of   '''100011''', \nwhich means   '''9/4'''   appears as the   '''35th'''   term of the sequence.\n\n\n;Task part 2:\n* Find the position of the number   '''83116''''''/''''''51639'''   in the Calkin-Wilf sequence.\n\n\n;See also:\n* Wikipedia entry:  Calkin-Wilf tree\n* [[Continued fraction]]\n* [[Continued fraction/Arithmetic/Construct from rational number]]\n\n", "solution": "from fractions import Fraction\nfrom math import floor\nfrom itertools import islice, groupby\n\n\ndef cw():\n    a = Fraction(1)\n    while True:\n        yield a\n        a = 1 / (2 * floor(a) + 1 - a)\n\ndef r2cf(rational):\n    num, den = rational.numerator, rational.denominator\n    while den:\n        num, (digit, den) = den, divmod(num, den)\n        yield digit\n\ndef get_term_num(rational):\n    ans, dig, pwr = 0, 1, 0\n    for n in r2cf(rational):\n        for _ in range(n):\n            ans |= dig << pwr\n            pwr += 1\n        dig ^= 1\n    return ans\n\n          \nif __name__ == '__main__':\n    print('TERMS 1..20: ', ', '.join(str(x) for x in islice(cw(), 20)))\n    x = Fraction(83116, 51639)\n    print(f\"\\n{x} is the {get_term_num(x):_}'th term.\")"}
{"title": "Call a function", "language": "Python", "task": "Demonstrate the different syntax and semantics provided for calling a function. \n\n\nThis may include:\n:*   Calling a function that requires no arguments\n:*   Calling a function with a fixed number of arguments\n:*   Calling a function with optional arguments\n:*   Calling a function with a variable number of arguments\n:*   Calling a function with named arguments\n:*   Using a function in statement context\n:*   Using a function in first-class context within an expression\n:*   Obtaining the return value of a function\n:*   Distinguishing built-in functions and user-defined functions\n:*   Distinguishing subroutines and functions\n;*   Stating whether arguments are passed by value or by reference\n;*   Is partial application possible and how\n\n\nThis task is ''not'' about defining functions.\n\n", "solution": "def no_args():\n    pass\n# call\nno_args()\n\ndef fixed_args(x, y):\n    print('x=%r, y=%r' % (x, y))\n# call\nfixed_args(1, 2)        # x=1, y=2\n\n## Can also called them using the parameter names, in either order:\nfixed_args(y=2, x=1)\n\n## Can also \"apply\" fixed_args() to a sequence:\nmyargs=(1,2) # tuple\nfixed_args(*myargs)\n\ndef opt_args(x=1):\n    print(x)\n# calls\nopt_args()              # 1\nopt_args(3.141)         # 3.141\n\ndef var_args(*v):\n    print(v)\n# calls\t\nvar_args(1, 2, 3)       # (1, 2, 3)\nvar_args(1, (2,3))      # (1, (2, 3))\nvar_args()              # ()\n\n## Named arguments\nfixed_args(y=2, x=1)    # x=1, y=2\n\n## As a statement\nif 1:\n    no_args()\n\n## First-class within an expression\nassert no_args() is None\n\ndef return_something():\n    return 1\nx = return_something()\n\ndef is_builtin(x):\n\tprint(x.__name__ in dir(__builtins__))\n# calls\nis_builtin(pow)         # True\nis_builtin(is_builtin)  # False\n\n# Very liberal function definition\n\ndef takes_anything(*args, **kwargs):\n    for each in args:\n        print(each)\n    for key, value in sorted(kwargs.items()):\n        print(\"%s:%s\" % (key, value))\n    # Passing those to another, wrapped, function:\n    wrapped_fn(*args, **kwargs)\n    # (Function being wrapped can have any parameter list\n    # ... that doesn't have to match this prototype)\n\n## A subroutine is merely a function that has no explicit\n## return statement and will return None.\n\n## Python uses \"Call by Object Reference\".\n## See, for example, http://www.python-course.eu/passing_arguments.php\n\n## For partial function application see:\n##   http://rosettacode.org/wiki/Partial_function_application#Python"}
{"title": "Canonicalize CIDR", "language": "Python from Perl", "task": "Implement a function or program that, given a range of IPv4 addresses in CIDR notation (dotted-decimal/network-bits), will return/output the same range in canonical form. \n\nThat is, the IP address portion of the output CIDR block must not contain any set (1) bits in the host part of the address.\n\n\n;Example:\nGiven   '''87.70.141.1/22''',    your code should output    '''87.70.140.0/22'''\n\n\n;Explanation:\nAn Internet Protocol version 4 address is a 32-bit value, conventionally represented as a number in base 256 using dotted-decimal notation, where each base-256 digit is given in decimal and the digits are separated by periods. Logically, this 32-bit value represents two components: the leftmost (most-significant) bits determine the network portion of the address, while the rightmost (least-significant) bits determine the host portion. Classless Internet Domain Routing block notation indicates where the boundary between these two components is for a given address by adding a slash followed by the number of bits in the network portion.\n\nIn general, CIDR blocks stand in for the entire set of IP addresses sharing the same network component, so it's common to see access control lists that specify individual IP addresses using /32 to indicate that only the one address is included. Software accepting this notation as input often expects it to be entered in canonical form, in which the host bits are all zeroes. But network admins sometimes skip this step and just enter the address of a specific host on the subnet with the network size, resulting in a non-canonical entry.\n\nThe example address, 87.70.141.1/22, represents binary 0101011101000110100011 / 0100000001, with the / indicating the network/host division. To canonicalize, clear all the bits to the right of the / and convert back to dotted decimal: 0101011101000110100011 / 0000000000 - 87.70.140.0.\n\n\n;More examples for testing\n\n        36.18.154.103/12    -  36.16.0.0/12\n        62.62.197.11/29     -  62.62.197.8/29\n        67.137.119.181/4    -  64.0.0.0/4\n        161.214.74.21/24    -  161.214.74.0/24\n        184.232.176.184/18  -  184.232.128.0/18\n\n\n", "solution": "#!/usr/bin/env python\n# canonicalize a CIDR block specification:\n# make sure none of the host bits are set\n\nimport sys\nfrom socket import inet_aton, inet_ntoa\nfrom struct import pack, unpack\n\nargs = sys.argv[1:]\nif len(args) == 0:\n    args = sys.stdin.readlines()\n\nfor cidr in args:\n   # IP in dotted-decimal / bits in network part\n   dotted, size_str = cidr.split('/')\n   size = int(size_str)\n\n   numeric = unpack('!I', inet_aton(dotted))[0]  # IP as an integer\n   binary = f'{numeric:#034b}'                   # then as a padded binary string\n   prefix = binary[:size + 2]                    # just the network part\n                                                 #   (34 and +2 are to account\n                                                 #    for leading '0b')\n\n   canon_binary = prefix + '0' * (32 - size)     # replace host part with all zeroes\n   canon_numeric = int(canon_binary, 2)          # convert back to integer\n   canon_dotted = inet_ntoa(pack('!I',\n                            (canon_numeric)))    # and then to dotted-decimal\n   print(f'{canon_dotted}/{size}')               # output result"}
{"title": "Cantor set", "language": "Python", "task": "Draw a Cantor set.\n\n\nSee details at this Wikipedia webpage:   Cantor set\n\n", "solution": "WIDTH = 81\nHEIGHT = 5\n\nlines=[]\ndef cantor(start, len, index):\n    seg = len / 3\n    if seg == 0:\n        return None\n    for it in xrange(HEIGHT-index):\n        i = index + it\n        for jt in xrange(seg):\n            j = start + seg + jt\n            pos = i * WIDTH + j\n            lines[pos] = ' '\n    cantor(start,           seg, index + 1)\n    cantor(start + seg * 2, seg, index + 1)\n    return None\n\nlines = ['*'] * (WIDTH*HEIGHT)\ncantor(0, WIDTH, 1)\n\nfor i in xrange(HEIGHT):\n    beg = WIDTH * i\n    print ''.join(lines[beg : beg+WIDTH])"}
{"title": "Cantor set", "language": "Python 3.7", "task": "Draw a Cantor set.\n\n\nSee details at this Wikipedia webpage:   Cantor set\n\n", "solution": "'''A Cantor set generator, and two different\n   representations of its output.\n'''\n\nfrom itertools import (islice, chain)\nfrom fractions import Fraction\nfrom functools import (reduce)\n\n\n# ----------------------- CANTOR SET -----------------------\n\n# cantor :: Generator [[(Fraction, Fraction)]]\ndef cantor():\n    '''A non-finite stream of successive Cantor\n       partitions of the line, in the form of\n       lists of fraction pairs.\n    '''\n    def go(xy):\n        (x, y) = xy\n        third = Fraction(y - x, 3)\n        return [(x, x + third), (y - third, y)]\n\n    return iterate(\n        concatMap(go)\n    )(\n        [(0, 1)]\n    )\n\n\n# fractionLists :: [(Fraction, Fraction)] -> String\ndef fractionLists(xs):\n    '''A fraction pair representation of a\n       Cantor-partitioned line.\n    '''\n    def go(xy):\n        return ', '.join(map(showRatio, xy))\n    return ' '.join('(' + go(x) + ')' for x in xs)\n\n\n# intervalBars :: [(Fraction, Fraction)] -> String\ndef intervalBars(w):\n    '''A block diagram representation of a\n       Cantor-partitioned line.\n    '''\n    def go(xs):\n        def show(a, tpl):\n            [x, y] = [int(w * r) for r in tpl]\n            return (\n                y,\n                (' ' * (x - a)) + ('\u2588' * (y - x))\n            )\n        return mapAccumL(show)(0)(xs)\n    return lambda xs: ''.join(go(xs)[1])\n\n\n# -------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Testing the generation of successive\n       Cantor subdivisions of the line, and\n       displaying them both as lines of fraction\n       pairs and as graphic interval bars.\n    '''\n    xs = list(islice(cantor(), 4))\n    w = max(xy[1].denominator for xy in xs[-1])\n    print(\n        '\\n'.join(map(fractionLists, xs)),\n        '\\n'\n    )\n    print(\n        '\\n'.join(map(intervalBars(w), xs))\n    )\n\n\n# ------------------------ GENERIC -------------------------\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).'''\n    return lambda xs: list(\n        chain.from_iterable(map(f, xs))\n    )\n\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated\n       applications of f to x.\n    '''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return go\n\n\n# mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])\ndef mapAccumL(f):\n    '''A tuple of an accumulation and a list derived by a\n       combined map and fold,\n       with accumulation from left to right.\n    '''\n    def go(a, x):\n        tpl = f(a[0], x)\n        return (tpl[0], a[1] + [tpl[1]])\n    return lambda acc: lambda xs: (\n        reduce(go, xs, (acc, []))\n    )\n\n\n# showRatio :: Ratio -> String\ndef showRatio(r):\n    '''String representation of the ratio r.'''\n    d = r.denominator\n    return str(r.numerator) + (\n        '/' + str(d) if 1 != d else ''\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Cartesian product of two or more lists", "language": "Python", "task": "Show one or more idiomatic ways of generating the Cartesian product of two arbitrary lists in your language.\n\nDemonstrate that your function/method correctly returns:\n::{1, 2} x {3, 4} = {(1, 3), (1, 4), (2, 3), (2, 4)}\n\nand, in contrast:\n::{3, 4} x {1, 2} = {(3, 1), (3, 2), (4, 1), (4, 2)}\n\nAlso demonstrate, using your function/method, that the product of an empty list with any other list is empty.\n:: {1, 2} x {} = {}\n:: {} x {1, 2} = {}\n\nFor extra credit, show or write a function returning the n-ary product of an arbitrary number of lists, each of arbitrary length. Your function might, for example, accept a single argument which is itself a list of lists, and return the n-ary product of those lists.\n\nUse your n-ary Cartesian product function to show the following products:\n:: {1776, 1789} x {7, 12} x {4, 14, 23} x {0, 1}\n:: {1, 2, 3} x {30} x {500, 100}\n:: {1, 2, 3} x {} x {500, 100}\n\n\n", "solution": "import itertools\n\ndef cp(lsts):\n    return list(itertools.product(*lsts))\n\nif __name__ == '__main__':\n    from pprint import pprint as pp\n    \n    for lists in [[[1,2],[3,4]], [[3,4],[1,2]], [[], [1, 2]], [[1, 2], []],\n                  ((1776, 1789),  (7, 12), (4, 14, 23), (0, 1)),\n                  ((1, 2, 3), (30,), (500, 100)),\n                  ((1, 2, 3), (), (500, 100))]:\n        print(lists, '=>')\n        pp(cp(lists), indent=2)\n"}
{"title": "Casting out nines", "language": "Python", "task": "Task   (in three parts):\n\n\n;Part 1\nWrite a procedure (say \\mathit{co9}(x)) which implements Casting Out Nines as described by returning the checksum for x. Demonstrate the procedure using the examples given there, or others you may consider lucky.\n\nNote that this function does nothing more than calculate the least positive residue, modulo 9. Many of the solutions omit Part 1 for this reason. Many languages have a modulo operator, of which this is a trivial application.\n\nWith that understanding, solutions to Part 1, if given, are encouraged to follow the naive pencil-and-paper or mental arithmetic of repeated digit addition understood to be \"casting out nines\", or some approach other than just reducing modulo 9 using a built-in operator. Solutions for part 2 and 3 are not required to make use of the function presented in part 1.\n\n;Part 2\nNotwithstanding past Intel microcode errors, checking computer calculations like this would not be sensible. To find a computer use for your procedure:\n: Consider the statement \"318682 is 101558 + 217124 and squared is 101558217124\" (see: [[Kaprekar numbers#Casting Out Nines (fast)]]).\n: note that 318682 has the same checksum as (101558 + 217124);\n: note that 101558217124 has the same checksum as (101558 + 217124) because for a Kaprekar they are made up of the same digits (sometimes with extra zeroes);\n: note that this implies that for Kaprekar numbers the checksum of k equals the checksum of k^2.\n\nDemonstrate that your procedure can be used to generate or filter a range of numbers with the property \\mathit{co9}(k) = \\mathit{co9}(k^2) and show that this subset is a small proportion of the range and contains all the Kaprekar in the range.\n\n;Part 3\nConsidering this MathWorld page, produce a efficient algorithm based on the more mathematical treatment of Casting Out Nines, and realizing:\n: \\mathit{co9}(x) is the residual of x mod 9;\n: the procedure can be extended to bases other than 9.\n\nDemonstrate your algorithm by generating or filtering a range of numbers with the property k%(\\mathit{Base}-1) == (k^2)%(\\mathit{Base}-1) and show that this subset is a small proportion of the range and contains all the Kaprekar in the range.\n\n;related tasks\n* [[First perfect square in base N with N unique digits]]\n* [[Kaprekar numbers]]\n\n", "solution": "# Casting out Nines\n#\n# Nigel Galloway: June 27th., 2012,\n#\ndef CastOut(Base=10, Start=1, End=999999):\n  ran = [y for y in range(Base-1) if y%(Base-1) == (y*y)%(Base-1)]\n  x,y = divmod(Start, Base-1)\n  while True:\n    for n in ran:\n      k = (Base-1)*x + n\n      if k < Start:\n        continue\n      if k > End:\n        return\n      yield k\n    x += 1\n\nfor V in CastOut(Base=16,Start=1,End=255):\n  print(V, end=' ')"}
{"title": "Catalan numbers/Pascal's triangle", "language": "Python from C++", "task": "Print out the first   '''15'''   Catalan numbers by extracting them from Pascal's triangle.\n\n\n;See:\n*   Catalan Numbers and the Pascal Triangle.      This method enables calculation of Catalan Numbers using only addition and subtraction.\n\n*   Catalan's Triangle for a Number Triangle that generates Catalan Numbers using only addition.\n*   Sequence A000108 on OEIS has a lot of information on Catalan Numbers.\n\n;Related Tasks:\n[[Pascal's triangle]]\n\n", "solution": ">>> n = 15\n>>> t = [0] * (n + 2)\n>>> t[1] = 1\n>>> for i in range(1, n + 1):\n\tfor j in range(i, 1, -1): t[j] += t[j - 1]\n\tt[i + 1] = t[i]\n\tfor j in range(i + 1, 1, -1): t[j] += t[j - 1]\n\tprint(t[i+1] - t[i], end=' ')\n\n\t\n1 2 5 14 42 132 429 1430 4862 16796 58786 208012 742900 2674440 9694845 \n>>> "}
{"title": "Catalan numbers/Pascal's triangle", "language": "Python 2.7", "task": "Print out the first   '''15'''   Catalan numbers by extracting them from Pascal's triangle.\n\n\n;See:\n*   Catalan Numbers and the Pascal Triangle.      This method enables calculation of Catalan Numbers using only addition and subtraction.\n\n*   Catalan's Triangle for a Number Triangle that generates Catalan Numbers using only addition.\n*   Sequence A000108 on OEIS has a lot of information on Catalan Numbers.\n\n;Related Tasks:\n[[Pascal's triangle]]\n\n", "solution": "def catalan_number(n):\n    nm = dm = 1\n    for k in range(2, n+1):\n      nm, dm = ( nm*(n+k), dm*k )\n    return nm/dm\n \nprint [catalan_number(n) for n in range(1, 16)]\n \n[1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845]"}
{"title": "Catalan numbers/Pascal's triangle", "language": "Python 3.7", "task": "Print out the first   '''15'''   Catalan numbers by extracting them from Pascal's triangle.\n\n\n;See:\n*   Catalan Numbers and the Pascal Triangle.      This method enables calculation of Catalan Numbers using only addition and subtraction.\n\n*   Catalan's Triangle for a Number Triangle that generates Catalan Numbers using only addition.\n*   Sequence A000108 on OEIS has a lot of information on Catalan Numbers.\n\n;Related Tasks:\n[[Pascal's triangle]]\n\n", "solution": "'''Catalan numbers from Pascal's triangle'''\n\nfrom itertools import (islice)\nfrom operator import (add)\n\n\n# nCatalans :: Int -> [Int]\ndef nCatalans(n):\n    '''The first n Catalan numbers,\n       derived from Pascal's triangle.'''\n\n    # diff :: [Int] -> Int\n    def diff(xs):\n        '''Difference between the first two items in the list,\n           if its length is more than one.\n           Otherwise, the first (only) item in the list.'''\n        return (\n            xs[0] - (xs[1] if 1 < len(xs) else 0)\n        ) if xs else None\n    return list(map(\n        compose(diff)(uncurry(drop)),\n        enumerate(map(fst, take(n)(\n            everyOther(\n                pascalTriangle()\n            )\n        )))\n    ))\n\n\n# pascalTriangle :: Gen [[Int]]\ndef pascalTriangle():\n    '''A non-finite stream of\n       Pascal's triangle rows.'''\n    return iterate(nextPascal)([1])\n\n\n# nextPascal :: [Int] -> [Int]\ndef nextPascal(xs):\n    '''A row of Pascal's triangle\n       derived from a preceding row.'''\n    return zipWith(add)([0] + xs)(xs + [0])\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''First 16 Catalan numbers.'''\n\n    print(\n        nCatalans(16)\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.'''\n    def go(xs):\n        if isinstance(xs, list):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return lambda xs: go(xs)\n\n\n# everyOther :: Gen [a] -> Gen [a]\ndef everyOther(g):\n    '''Every other item of a generator stream.'''\n    while True:\n        yield take(1)(g)\n        take(1)(g)      # Consumed, not yielded.\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First component of a pair.'''\n    return tpl[0]\n\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated applications of f to x.'''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return lambda x: go(x)\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    '''A function over a tuple\n       derived from a curried function.'''\n    return lambda xy: f(xy[0])(\n        xy[1]\n    )\n\n\n# zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]\ndef zipWith(f):\n    '''A list constructed by zipping with a\n       custom function, rather than with the\n       default tuple constructor.'''\n    return lambda xs: lambda ys: (\n        list(map(f, xs, ys))\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Catamorphism", "language": "Python", "task": "''Reduce'' is a function or method that is used to take the values in an array or a list and apply a function to successive members of the list to produce (or reduce them to), a single value. \n\n\n;Task:\nShow how ''reduce'' (or ''foldl'' or ''foldr'' etc), work (or would be implemented) in your language.\n\n\n;See also:\n* Wikipedia article:   Fold\n* Wikipedia article:   Catamorphism\n\n", "solution": ">>> # Python 2.X\n>>> from operator import add\n>>> listoflists = [['the', 'cat'], ['sat', 'on'], ['the', 'mat']]\n>>> help(reduce)\nHelp on built-in function reduce in module __builtin__:\n\nreduce(...)\n    reduce(function, sequence[, initial]) -> value\n    \n    Apply a function of two arguments cumulatively to the items of a sequence,\n    from left to right, so as to reduce the sequence to a single value.\n    For example, reduce(lambda x, y: x+y, [1, 2, 3, 4, 5]) calculates\n    ((((1+2)+3)+4)+5).  If initial is present, it is placed before the items\n    of the sequence in the calculation, and serves as a default when the\n    sequence is empty.\n\n>>> reduce(add, listoflists, [])\n['the', 'cat', 'sat', 'on', 'the', 'mat']\n>>> "}
{"title": "Chaocipher", "language": "Python", "task": "Description:\nThe Chaocipher was invented by J.F.Byrne in 1918 and, although simple by modern cryptographic standards, does not appear to have been broken until the algorithm was finally disclosed by his family in 2010.\n\nThe algorithm is described in this paper by M.Rubin in 2010 and there is a C# implementation here.\n\n\n;Task:\nCode the algorithm in your language and to test that it works with the plaintext 'WELLDONEISBETTERTHANWELLSAID' used in the paper itself.\n\n", "solution": "# Python3 implementation of Chaocipher \n# left wheel = ciphertext wheel\n# right wheel = plaintext wheel\n\ndef main():\n    # letters only! makealpha(key) helps generate lalpha/ralpha. \n    lalpha = \"HXUCZVAMDSLKPEFJRIGTWOBNYQ\"\n    ralpha = \"PTLNBQDEOYSFAVZKGJRIHWXUMC\"\n    msg = \"WELLDONEISBETTERTHANWELLSAID\"\n\n    print(\"L:\", lalpha)\n    print(\"R:\", ralpha)\n    print(\"I:\", msg)\n    print(\"O:\", do_chao(msg, lalpha, ralpha, 1, 0), \"\\n\")\n    \n    do_chao(msg, lalpha, ralpha, 1, 1)\n\ndef do_chao(msg, lalpha, ralpha, en=1, show=0):\n    msg = correct_case(msg)\n    out = \"\"    \n    if show:\n        print(\"=\"*54)        \n        print(10*\" \" + \"left:\" + 21*\" \" + \"right: \")\n        print(\"=\"*54)        \n        print(lalpha, ralpha, \"\\n\")\n    for L in msg:\n        if en:\n            lalpha, ralpha = rotate_wheels(lalpha, ralpha, L)\n            out += lalpha[0]\n        else:\n            ralpha, lalpha = rotate_wheels(ralpha, lalpha, L)\n            out += ralpha[0]\n        lalpha, ralpha = scramble_wheels(lalpha, ralpha)\n        if show:\n            print(lalpha, ralpha)            \n    return out\n    \ndef makealpha(key=\"\"):\n    alpha = \"ABCDEFGHIJKLMNOPQRSTUVWXYZ\"\n    z = set()\n    key = [x.upper() for x in (key + alpha[::-1])\n           if not (x.upper() in z or z.add(x.upper()))]\n    return \"\".join(key)\n\ndef correct_case(string):\n    return \"\".join([s.upper() for s in string if s.isalpha()])\n\ndef permu(alp, num):\n    alp = alp[:num], alp[num:]\n    return \"\".join(alp[::-1])\n\ndef rotate_wheels(lalph, ralph, key):\n    newin = ralph.index(key)\n    return permu(lalph, newin), permu(ralph, newin)    \n\ndef scramble_wheels(lalph, ralph):\n    # LEFT = cipher wheel \n    # Cycle second[1] through nadir[14] forward\n    lalph = list(lalph)\n    lalph = \"\".join([*lalph[0],\n                    *lalph[2:14],\n                    lalph[1],\n                    *lalph[14:]])\n    \n    # RIGHT = plain wheel                    \n    # Send the zenith[0] character to the end[25],\n    # cycle third[2] through nadir[14] characters forward\n    ralph = list(ralph)\n    ralph = \"\".join([*ralph[1:3],\n                     *ralph[4:15],\n                     ralph[3],\n                     *ralph[15:],\n                     ralph[0]])\n    return lalph, ralph\n\nmain()"}
{"title": "Chaocipher", "language": "Python 3.7", "task": "Description:\nThe Chaocipher was invented by J.F.Byrne in 1918 and, although simple by modern cryptographic standards, does not appear to have been broken until the algorithm was finally disclosed by his family in 2010.\n\nThe algorithm is described in this paper by M.Rubin in 2010 and there is a C# implementation here.\n\n\n;Task:\nCode the algorithm in your language and to test that it works with the plaintext 'WELLDONEISBETTERTHANWELLSAID' used in the paper itself.\n\n", "solution": "'''Chaocipher'''\n\nfrom itertools import chain, cycle, islice\n\n\n# chao :: String -> String -> Bool -> String -> String\ndef chao(l):\n    '''Chaocipher encoding or decoding for the given\n       left and right 'wheels'.\n       A ciphertext is returned if the boolean flag\n       is True, and a plaintext if the flag is False.\n    '''\n    def go(l, r, plain, xxs):\n        if xxs:\n            (src, dst) = (l, r) if plain else (r, l)\n            (x, xs) = (xxs[0], xxs[1:])\n\n            def chaoProcess(n):\n                return [dst[n]] + go(\n                    shifted(1)(14)(rotated(n, l)),\n                    compose(shifted(2)(14))(shifted(0)(26))(\n                        rotated(n, r)\n                    ),\n                    plain,\n                    xs\n                )\n\n            return maybe('')(chaoProcess)(\n                elemIndex(x)(src)\n            )\n        else:\n            return []\n    return lambda r: lambda plain: lambda xxs: concat(go(\n        l, r, plain, xxs\n    ))\n\n\n# rotated :: Int -> [a] -> [a]\ndef rotated(z, s):\n    '''Rotation of string s by z characters.'''\n    return take(len(s))(\n        drop(z)(\n            cycle(s)\n        )\n    )\n\n\n# shifted :: Int -> Int -> [a] -> [a]\ndef shifted(src):\n    '''The string s with a set of its characters cyclically\n       shifted from a source index to a destination index.\n    '''\n    def go(dst, s):\n        (a, b) = splitAt(dst)(s)\n        (x, y) = splitAt(src)(a)\n        return concat([x, rotated(1, y), b])\n    return lambda dst: lambda s: go(dst, s)\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Print the plain text, followed by\n       a corresponding cipher text,\n       and a decode of that cipher text.\n    '''\n    chaoWheels = chao(\n        \"HXUCZVAMDSLKPEFJRIGTWOBNYQ\"\n    )(\n        \"PTLNBQDEOYSFAVZKGJRIHWXUMC\"\n    )\n    plainText = \"WELLDONEISBETTERTHANWELLSAID\"\n    cipherText = chaoWheels(False)(plainText)\n\n    print(plainText)\n    print(cipherText)\n    print(\n        chaoWheels(True)(cipherText)\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# concat :: [[a]] -> [a]\n# concat :: [String] -> String\ndef concat(xs):\n    '''The concatenation of all the elements\n       in a list or iterable.\n    '''\n    def f(ys):\n        zs = list(chain(*ys))\n        return ''.join(zs) if isinstance(ys[0], str) else zs\n\n    return (\n        f(xs) if isinstance(xs, list) else (\n            chain.from_iterable(xs)\n        )\n    ) if xs else []\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.\n    '''\n    def go(xs):\n        if isinstance(xs, (list, tuple, str)):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return lambda xs: go(xs)\n\n\n# elemIndex :: Eq a => a -> [a] -> Maybe Int\ndef elemIndex(x):\n    '''Just the index of the first element in xs\n       which is equal to x,\n       or Nothing if there is no such element.\n    '''\n    def go(xs):\n        try:\n            return Just(xs.index(x))\n        except ValueError:\n            return Nothing()\n    return lambda xs: go(xs)\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if m is Nothing,\n       or the application of f to x,\n       where m is Just(x).\n    '''\n    return lambda f: lambda m: v if None is m or m.get('Nothing') else (\n        f(m.get('Just'))\n    )\n\n\n# splitAt :: Int -> [a] -> ([a], [a])\ndef splitAt(n):\n    '''A tuple pairing the prefix of length n\n       with the rest of xs.\n    '''\n    return lambda xs: (xs[0:n], xs[n:])\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, (list, tuple))\n        else list(islice(xs, n))\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Chaos game", "language": "Python", "task": "The Chaos Game is a method of generating the attractor of an iterated function system (IFS). \n\nOne of the best-known and simplest examples creates a fractal, using a polygon and an initial point selected at random.\n\n\n;Task\nPlay the Chaos Game using the corners of an equilateral triangle as the reference points.   Add a starting point at random (preferably inside the triangle).   Then add the next point halfway between the starting point and one of the reference points.   This reference point is chosen at random.\n\nAfter a sufficient number of iterations, the image of a Sierpinski Triangle should emerge.\n\n\n;See also\n* The Game of Chaos\n\n", "solution": "import argparse\nimport random\nimport shapely.geometry as geometry\nimport numpy as np\nimport matplotlib.pyplot as plt\nimport matplotlib.animation as animation\n\n\ndef main(args):\n    # Styles\n    plt.style.use(\"ggplot\")\n\n    # Creating figure\n    fig = plt.figure()\n    line, = plt.plot([], [], \".\")\n\n    # Limit axes\n    plt.xlim(0, 1)\n    plt.ylim(0, 1)\n\n    # Titles\n    title = \"Chaos Game\"\n    plt.title(title)\n    fig.canvas.set_window_title(title)\n\n    # Getting data\n    data = get_data(args.frames)\n\n    # Creating animation\n    line_ani = animation.FuncAnimation(\n        fig=fig,\n        func=update_line,\n        frames=args.frames,\n        fargs=(data, line),\n        interval=args.interval,\n        repeat=False\n    )\n\n    # To save the animation install ffmpeg and uncomment\n    # line_ani.save(\"chaos_game.gif\")\n\n    plt.show()\n\n\ndef get_data(n):\n    \"\"\"\n    Get data to plot\n    \"\"\"\n    leg = 1\n    triangle = get_triangle(leg)\n    cur_point = gen_point_within_poly(triangle)\n    data = []\n    for _ in range(n):\n        data.append((cur_point.x, cur_point.y))\n        cur_point = next_point(triangle, cur_point)\n    return data\n\n\ndef get_triangle(n):\n    \"\"\"\n    Create right triangle\n    \"\"\"\n    ax = ay = 0.0\n    a = ax, ay\n\n    bx = 0.5  *  n\n    by = 0.75 * (n ** 2)\n    b = bx, by\n\n    cx = n\n    cy = 0.0\n    c = cx, cy\n\n    triangle = geometry.Polygon([a, b, c])\n    return triangle\n\n\ndef gen_point_within_poly(poly):\n    \"\"\"\n    Generate random point inside given polygon\n    \"\"\"\n    minx, miny, maxx, maxy = poly.bounds\n    while True:\n        x = random.uniform(minx, maxx)\n        y = random.uniform(miny, maxy)\n        point = geometry.Point(x, y)\n        if point.within(poly):\n            return point\n\n\ndef next_point(poly, point):\n    \"\"\"\n    Generate next point according to chaos game rules\n    \"\"\"\n    vertices = poly.boundary.coords[:-1]  # Last point is the same as the first one\n    random_vertex = geometry.Point(random.choice(vertices))\n    line = geometry.linestring.LineString([point, random_vertex])\n    return line.centroid\n\n\ndef update_line(num, data, line):\n    \"\"\"\n    Update line with new points\n    \"\"\"\n    new_data = zip(*data[:num]) or [(), ()]\n    line.set_data(new_data)\n    return line,\n\n\nif __name__ == \"__main__\":\n    arg_parser = argparse.ArgumentParser(description=\"Chaos Game by Suenweek (c) 2017\")\n    arg_parser.add_argument(\"-f\", dest=\"frames\", type=int, default=1000)\n    arg_parser.add_argument(\"-i\", dest=\"interval\", type=int, default=10)\n\n    main(arg_parser.parse_args())\n\n"}
{"title": "Check Machin-like formulas", "language": "Python", "task": "Machin-like formulas   are useful for efficiently computing numerical approximations for \\pi\n\n\n;Task:\nVerify the following Machin-like formulas are correct by calculating the value of '''tan'''   (''right hand side)'' for each equation using exact arithmetic and showing they equal '''1''':\n\n: {\\pi\\over4} = \\arctan{1\\over2} + \\arctan{1\\over3} \n: {\\pi\\over4} = 2 \\arctan{1\\over3} + \\arctan{1\\over7}\n: {\\pi\\over4} = 4 \\arctan{1\\over5} - \\arctan{1\\over239}\n: {\\pi\\over4} = 5 \\arctan{1\\over7} + 2 \\arctan{3\\over79}\n: {\\pi\\over4} = 5 \\arctan{29\\over278} + 7 \\arctan{3\\over79}\n: {\\pi\\over4} = \\arctan{1\\over2} + \\arctan{1\\over5} + \\arctan{1\\over8} \n: {\\pi\\over4} = 4 \\arctan{1\\over5} - \\arctan{1\\over70} + \\arctan{1\\over99} \n: {\\pi\\over4} = 5 \\arctan{1\\over7} + 4 \\arctan{1\\over53} + 2 \\arctan{1\\over4443}\n: {\\pi\\over4} = 6 \\arctan{1\\over8} + 2 \\arctan{1\\over57} + \\arctan{1\\over239}\n: {\\pi\\over4} = 8 \\arctan{1\\over10} - \\arctan{1\\over239} - 4 \\arctan{1\\over515}\n: {\\pi\\over4} = 12 \\arctan{1\\over18} + 8 \\arctan{1\\over57} - 5 \\arctan{1\\over239}\n: {\\pi\\over4} = 16 \\arctan{1\\over21} + 3 \\arctan{1\\over239} + 4 \\arctan{3\\over1042}\n: {\\pi\\over4} = 22 \\arctan{1\\over28} + 2 \\arctan{1\\over443} - 5 \\arctan{1\\over1393} - 10 \\arctan{1\\over11018}\n: {\\pi\\over4} = 22 \\arctan{1\\over38} + 17 \\arctan{7\\over601} + 10 \\arctan{7\\over8149}\n: {\\pi\\over4} = 44 \\arctan{1\\over57} + 7 \\arctan{1\\over239} - 12 \\arctan{1\\over682} + 24 \\arctan{1\\over12943}\n: {\\pi\\over4} = 88 \\arctan{1\\over172} + 51 \\arctan{1\\over239} + 32 \\arctan{1\\over682} + 44 \\arctan{1\\over5357} + 68 \\arctan{1\\over12943}\n\nand confirm that the following formula is ''incorrect'' by showing   '''tan'''   (''right hand side)''   is ''not''   '''1''':\n\n: {\\pi\\over4} = 88 \\arctan{1\\over172} + 51 \\arctan{1\\over239} + 32 \\arctan{1\\over682} + 44 \\arctan{1\\over5357} + 68 \\arctan{1\\over12944}\n\nThese identities are useful in calculating the values:\n: \\tan(a + b) = {\\tan(a) + \\tan(b) \\over 1 - \\tan(a) \\tan(b)}\n\n: \\tan\\left(\\arctan{a \\over b}\\right) = {a \\over b}\n\n: \\tan(-a) = -\\tan(a)\n\nYou can store the equations in any convenient data structure, but for extra credit parse them from human-readable text input.\n\nNote: to formally prove the formula correct, it would have to be shown that ''{-3 pi \\over 4} < right hand side < {5 pi \\over 4}'' due to ''\\tan()'' periodicity.\n\n\n", "solution": "import re\nfrom fractions import Fraction\nfrom pprint import pprint as pp\n\n\nequationtext = '''\\\n  pi/4 = arctan(1/2) + arctan(1/3) \n  pi/4 = 2*arctan(1/3) + arctan(1/7)\n  pi/4 = 4*arctan(1/5) - arctan(1/239)\n  pi/4 = 5*arctan(1/7) + 2*arctan(3/79)\n  pi/4 = 5*arctan(29/278) + 7*arctan(3/79)\n  pi/4 = arctan(1/2) + arctan(1/5) + arctan(1/8) \n  pi/4 = 4*arctan(1/5) - arctan(1/70) + arctan(1/99) \n  pi/4 = 5*arctan(1/7) + 4*arctan(1/53) + 2*arctan(1/4443)\n  pi/4 = 6*arctan(1/8) + 2*arctan(1/57) + arctan(1/239)\n  pi/4 = 8*arctan(1/10) - arctan(1/239) - 4*arctan(1/515)\n  pi/4 = 12*arctan(1/18) + 8*arctan(1/57) - 5*arctan(1/239)\n  pi/4 = 16*arctan(1/21) + 3*arctan(1/239) + 4*arctan(3/1042)\n  pi/4 = 22*arctan(1/28) + 2*arctan(1/443) - 5*arctan(1/1393) - 10*arctan(1/11018)\n  pi/4 = 22*arctan(1/38) + 17*arctan(7/601) + 10*arctan(7/8149)\n  pi/4 = 44*arctan(1/57) + 7*arctan(1/239) - 12*arctan(1/682) + 24*arctan(1/12943)\n  pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12943)\n  pi/4 = 88*arctan(1/172) + 51*arctan(1/239) + 32*arctan(1/682) + 44*arctan(1/5357) + 68*arctan(1/12944)\n'''\n\ndef parse_eqn(equationtext=equationtext):\n    eqn_re = re.compile(r\"\"\"(?mx)\n    (?P<lhs> ^ \\s* pi/4 \\s* = \\s*)?             # LHS of equation\n    (?:                                         # RHS\n        \\s* (?P<sign> [+-])? \\s* \n        (?: (?P<mult> \\d+) \\s* \\*)? \n        \\s* arctan\\( (?P<numer> \\d+) / (?P<denom> \\d+)\n    )\"\"\")\n\n    found = eqn_re.findall(equationtext)\n    machins, part = [], []\n    for lhs, sign, mult, numer, denom in eqn_re.findall(equationtext):\n        if lhs and part:\n            machins.append(part)\n            part = []\n        part.append( ( (-1 if sign == '-' else 1) * ( int(mult) if mult else 1),\n                       Fraction(int(numer), (int(denom) if denom else 1)) ) )\n    machins.append(part)\n    return machins\n\n\ndef tans(xs):\n    xslen = len(xs)\n    if xslen == 1:\n        return tanEval(*xs[0])\n    aa, bb = xs[:xslen//2], xs[xslen//2:]\n    a, b = tans(aa), tans(bb)\n    return (a + b) / (1 - a * b)\n\ndef tanEval(coef, f):\n    if coef == 1:\n        return f\n    if coef < 0:\n        return -tanEval(-coef, f)\n    ca = coef // 2\n    cb = coef - ca\n    a, b = tanEval(ca, f), tanEval(cb, f)\n    return (a + b) / (1 - a * b)\n\n\nif __name__ == '__main__':\n    machins = parse_eqn()\n    #pp(machins, width=160)\n    for machin, eqn in zip(machins, equationtext.split('\\n')):\n        ans = tans(machin)\n        print('%5s: %s' % ( ('OK' if ans == 1 else 'ERROR'), eqn))"}
{"title": "Check input device is a terminal", "language": "Python", "task": "Demonstrate how to check whether the input device is a terminal or not.\n\n\n;Related task:\n*   [[Check output device is a terminal]]\n\n", "solution": "from sys import stdin\nif stdin.isatty():\n    print(\"Input comes from tty.\")\nelse:\n    print(\"Input doesn't come from tty.\")"}
{"title": "Check output device is a terminal", "language": "Python", "task": "Demonstrate how to check whether the output device is a terminal or not.\n\n\n;Related task:\n*   [[Check input device is a terminal]]\n\n", "solution": "from sys import stdout\nif stdout.isatty():\n    print 'The output device is a teletype. Or something like a teletype.'\nelse:\n    print 'The output device isn\\'t like a teletype.'"}
{"title": "Cheryl's birthday", "language": "Python 3", "task": "Albert and Bernard just became friends with Cheryl, and they want to know when her birthday is.\n\nCheryl gave them a list of ten possible dates:\n      May 15,     May 16,     May 19\n      June 17,    June 18\n      July 14,    July 16\n      August 14,  August 15,  August 17\n\nCheryl then tells Albert the   ''month''   of birth,   and Bernard the   ''day''   (of the month)   of birth.\n  1)  Albert:   I don't know when Cheryl's birthday is, but I know that Bernard does not know too.\n  2)  Bernard:  At first I don't know when Cheryl's birthday is, but I know now.\n  3)  Albert:   Then I also know when Cheryl's birthday is.\n\n\n;Task\nWrite a computer program to deduce, by successive elimination, Cheryl's birthday.\n\n\n;Related task:\n* [[Sum and Product Puzzle]]\n\n\n;References\n* Wikipedia article of the same name.\n* Tuple Relational Calculus\n\n", "solution": "'''Cheryl's Birthday'''\n\nfrom itertools import groupby\nfrom re import split\n\n\n# main :: IO ()\ndef main():\n    '''Derivation of the date.'''\n\n    month, day = 0, 1\n    print(\n        # (3 :: A \"Then I also know\")\n        # (A's month contains only one remaining day)\n        uniquePairing(month)(\n            # (2 :: B \"I know now\")\n            # (B's day is paired with only one remaining month)\n            uniquePairing(day)(\n                # (1 :: A \"I know that Bernard does not know\")\n                # (A's month is not among those with unique days)\n                monthsWithUniqueDays(False)([\n                    # 0 :: Cheryl's list:\n                    tuple(x.split()) for x in\n                    split(\n                        ', ',\n                        'May 15, May 16, May 19, ' +\n                        'June 17, June 18, ' +\n                        'July 14, July 16, ' +\n                        'Aug 14, Aug 15, Aug 17'\n                    )\n                ])\n            )\n        )\n    )\n\n\n# ------------------- QUERY FUNCTIONS --------------------\n\n# monthsWithUniqueDays :: Bool -> [(Month, Day)] -> [(Month, Day)]\ndef monthsWithUniqueDays(blnInclude):\n    '''The subset of months with (or without) unique days.\n    '''\n    def go(xs):\n        month, day = 0, 1\n        months = [fst(x) for x in uniquePairing(day)(xs)]\n        return [\n            md for md in xs\n            if blnInclude or not (md[month] in months)\n        ]\n    return go\n\n\n# uniquePairing :: DatePart -> [(Month, Day)] -> [(Month, Day)]\ndef uniquePairing(i):\n    '''Subset of months (or days) with a unique intersection.\n    '''\n    def go(xs):\n        def inner(md):\n            dct = md[i]\n            uniques = [\n                k for k in dct.keys()\n                if 1 == len(dct[k])\n            ]\n            return [tpl for tpl in xs if tpl[i] in uniques]\n        return inner\n    return ap(bindPairs)(go)\n\n\n# bindPairs :: [(Month, Day)] ->\n# ((Dict String [String], Dict String [String])\n# -> [(Month, Day)]) -> [(Month, Day)]\ndef bindPairs(xs):\n    '''List monad injection operator for lists\n       of (Month, Day) pairs.\n    '''\n    return lambda f: f(\n        (\n            dictFromPairs(xs),\n            dictFromPairs(\n                [(b, a) for (a, b) in xs]\n            )\n        )\n    )\n\n\n# dictFromPairs :: [(Month, Day)] -> Dict Text [Text]\ndef dictFromPairs(xs):\n    '''A dictionary derived from a list of\n       month day pairs.\n    '''\n    return {\n        k: [snd(x) for x in m] for k, m in groupby(\n            sorted(xs, key=fst), key=fst\n        )\n    }\n\n\n# ----------------------- GENERIC ------------------------\n\n# ap :: (a -> b -> c) -> (a -> b) -> a -> c\ndef ap(f):\n    '''Applicative instance for functions.\n    '''\n    def go(g):\n        def fxgx(x):\n            return f(x)(\n                g(x)\n            )\n        return fxgx\n    return go\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First component of a pair.\n    '''\n    return tpl[0]\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second component of a pair.\n    '''\n    return tpl[1]\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Chinese remainder theorem", "language": "Python", "task": "Suppose   n_1,   n_2,   \\ldots,   n_k   are positive [[integer]]s that are pairwise co-prime.   \n\nThen, for any given sequence of integers   a_1,   a_2,   \\dots,   a_k,   there exists an integer   x   solving the following system of simultaneous congruences:\n\n::: \\begin{align}\n  x &\\equiv a_1 \\pmod{n_1} \\\\\n  x &\\equiv a_2 \\pmod{n_2} \\\\\n    &{}\\  \\  \\vdots \\\\\n  x &\\equiv a_k \\pmod{n_k}\n\\end{align}\n\nFurthermore, all solutions   x   of this system are congruent modulo the product,    N=n_1n_2\\ldots n_k.\n\n\n;Task:\nWrite a program to solve a system of linear congruences by applying the   Chinese Remainder Theorem. \n\nIf the system of equations cannot be solved, your program must somehow indicate this. \n\n(It may throw an exception or return a special false value.) \n\nSince there are infinitely many solutions, the program should return the unique solution   s   where   0 \\leq s \\leq n_1n_2\\ldots n_k.\n\n\n''Show the functionality of this program'' by printing the result such that the   n's   are   [3,5,7]   and the   a's   are   [2,3,2].\n\n\n'''Algorithm''':   The following algorithm only applies if the   n_i's   are pairwise co-prime. \n\nSuppose, as above, that a solution is required for the system of congruences:\n\n::: x \\equiv a_i \\pmod{n_i} \\quad\\mathrm{for}\\; i = 1, \\ldots, k\n\nAgain, to begin, the product   N = n_1n_2 \\ldots n_k   is defined. \n\nThen a solution   x   can be found as follows:\n\nFor each   i,   the integers   n_i   and   N/n_i   are co-prime. \n\nUsing the   Extended Euclidean algorithm,   we can find integers   r_i   and   s_i   such that   r_i n_i + s_i N/n_i = 1. \n\nThen, one solution to the system of simultaneous congruences is:\n\n::: x = \\sum_{i=1}^k a_i s_i N/n_i\n\nand the minimal solution,\n\n::: x \\pmod{N}.\n\n", "solution": "# Python 2.7\ndef chinese_remainder(n, a):\n    sum = 0\n    prod = reduce(lambda a, b: a*b, n)\n\n    for n_i, a_i in zip(n, a):\n        p = prod / n_i\n        sum += a_i * mul_inv(p, n_i) * p\n    return sum % prod\n\n\ndef mul_inv(a, b):\n    b0 = b\n    x0, x1 = 0, 1\n    if b == 1: return 1\n    while a > 1:\n        q = a / b\n        a, b = b, a%b\n        x0, x1 = x1 - q * x0, x0\n    if x1 < 0: x1 += b0\n    return x1\n\nif __name__ == '__main__':\n    n = [3, 5, 7]\n    a = [2, 3, 2]\n    print chinese_remainder(n, a)"}
{"title": "Chinese remainder theorem", "language": "Python 3.7", "task": "Suppose   n_1,   n_2,   \\ldots,   n_k   are positive [[integer]]s that are pairwise co-prime.   \n\nThen, for any given sequence of integers   a_1,   a_2,   \\dots,   a_k,   there exists an integer   x   solving the following system of simultaneous congruences:\n\n::: \\begin{align}\n  x &\\equiv a_1 \\pmod{n_1} \\\\\n  x &\\equiv a_2 \\pmod{n_2} \\\\\n    &{}\\  \\  \\vdots \\\\\n  x &\\equiv a_k \\pmod{n_k}\n\\end{align}\n\nFurthermore, all solutions   x   of this system are congruent modulo the product,    N=n_1n_2\\ldots n_k.\n\n\n;Task:\nWrite a program to solve a system of linear congruences by applying the   Chinese Remainder Theorem. \n\nIf the system of equations cannot be solved, your program must somehow indicate this. \n\n(It may throw an exception or return a special false value.) \n\nSince there are infinitely many solutions, the program should return the unique solution   s   where   0 \\leq s \\leq n_1n_2\\ldots n_k.\n\n\n''Show the functionality of this program'' by printing the result such that the   n's   are   [3,5,7]   and the   a's   are   [2,3,2].\n\n\n'''Algorithm''':   The following algorithm only applies if the   n_i's   are pairwise co-prime. \n\nSuppose, as above, that a solution is required for the system of congruences:\n\n::: x \\equiv a_i \\pmod{n_i} \\quad\\mathrm{for}\\; i = 1, \\ldots, k\n\nAgain, to begin, the product   N = n_1n_2 \\ldots n_k   is defined. \n\nThen a solution   x   can be found as follows:\n\nFor each   i,   the integers   n_i   and   N/n_i   are co-prime. \n\nUsing the   Extended Euclidean algorithm,   we can find integers   r_i   and   s_i   such that   r_i n_i + s_i N/n_i = 1. \n\nThen, one solution to the system of simultaneous congruences is:\n\n::: x = \\sum_{i=1}^k a_i s_i N/n_i\n\nand the minimal solution,\n\n::: x \\pmod{N}.\n\n", "solution": "'''Chinese remainder theorem'''\n\nfrom operator import (add, mul)\nfrom functools import reduce\n\n\n# cnRemainder :: [Int] -> [Int] -> Either String Int\ndef cnRemainder(ms):\n    '''Chinese remainder theorem.\n       (moduli, residues) -> Either explanation or solution\n    '''\n    def go(ms, rs):\n        mp = numericProduct(ms)\n        cms = [(mp // x) for x in ms]\n\n        def possibleSoln(invs):\n            return Right(\n                sum(map(\n                    mul,\n                    cms, map(mul, rs, invs)\n                )) % mp\n            )\n        return bindLR(\n            zipWithEither(modMultInv)(cms)(ms)\n        )(possibleSoln)\n\n    return lambda rs: go(ms, rs)\n\n\n# modMultInv :: Int -> Int -> Either String Int\ndef modMultInv(a, b):\n    '''Modular multiplicative inverse.'''\n    x, y = eGcd(a, b)\n    return Right(x) if 1 == (a * x + b * y) else (\n        Left('no modular inverse for ' + str(a) + ' and ' + str(b))\n    )\n\n\n# egcd :: Int -> Int -> (Int, Int)\ndef eGcd(a, b):\n    '''Extended greatest common divisor.'''\n    def go(a, b):\n        if 0 == b:\n            return (1, 0)\n        else:\n            q, r = divmod(a, b)\n            (s, t) = go(b, r)\n            return (t, s - q * t)\n    return go(a, b)\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Tests of soluble and insoluble cases.'''\n\n    print(\n        fTable(\n            __doc__ + ':\\n\\n         (moduli, residues) -> ' + (\n                'Either solution or explanation\\n'\n            )\n        )(repr)(\n            either(compose(quoted(\"'\"))(curry(add)('No solution: ')))(\n                compose(quoted(' '))(repr)\n            )\n        )(uncurry(cnRemainder))([\n            ([10, 4, 12], [11, 12, 13]),\n            ([11, 12, 13], [10, 4, 12]),\n            ([10, 4, 9], [11, 22, 19]),\n            ([3, 5, 7], [2, 3, 2]),\n            ([2, 3, 2], [3, 5, 7])\n        ])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# Left :: a -> Either a b\ndef Left(x):\n    '''Constructor for an empty Either (option type) value\n       with an associated string.'''\n    return {'type': 'Either', 'Right': None, 'Left': x}\n\n\n# Right :: b -> Either a b\ndef Right(x):\n    '''Constructor for a populated Either (option type) value'''\n    return {'type': 'Either', 'Left': None, 'Right': x}\n# any :: (a -> Bool) -> [a] -> Bool\n\n\ndef any_(p):\n    '''True if p(x) holds for at least\n       one item in xs.'''\n    def go(xs):\n        for x in xs:\n            if p(x):\n                return True\n        return False\n    return lambda xs: go(xs)\n\n\n# bindLR (>>=) :: Either a -> (a -> Either b) -> Either b\ndef bindLR(m):\n    '''Either monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.'''\n    return lambda mf: (\n        mf(m.get('Right')) if None is m.get('Left') else m\n    )\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    '''A curried function derived\n       from an uncurried function.'''\n    return lambda a: lambda b: f(a, b)\n\n\n# either :: (a -> c) -> (b -> c) -> Either a b -> c\ndef either(fl):\n    '''The application of fl to e if e is a Left value,\n       or the application of fr to e if e is a Right value.'''\n    return lambda fr: lambda e: fl(e['Left']) if (\n        None is e['Right']\n    ) else fr(e['Right'])\n\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function ->\n                 fx display function ->\n          f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + '\\n'.join([\n            xShow(x).rjust(w, ' ') + (' -> ') + fxShow(f(x))\n            for x in xs\n        ])\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# numericProduct :: [Num] -> Num\ndef numericProduct(xs):\n    '''The arithmetic product of all numbers in xs.'''\n    return reduce(mul, xs, 1)\n\n\n# partitionEithers :: [Either a b] -> ([a],[b])\ndef partitionEithers(lrs):\n    '''A list of Either values partitioned into a tuple\n       of two lists, with all Left elements extracted\n       into the first list, and Right elements\n       extracted into the second list.\n    '''\n    def go(a, x):\n        ls, rs = a\n        r = x.get('Right')\n        return (ls + [x.get('Left')], rs) if None is r else (\n            ls, rs + [r]\n        )\n    return reduce(go, lrs, ([], []))\n\n\n# quoted :: Char -> String -> String\ndef quoted(c):\n    '''A string flanked on both sides\n       by a specified quote character.\n    '''\n    return lambda s: c + s + c\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    '''A function over a tuple,\n       derived from a curried function.'''\n    return lambda xy: f(xy[0])(xy[1])\n\n\n# zipWithEither :: (a -> b -> Either String  c)\n#            -> [a] -> [b] -> Either String [c]\ndef zipWithEither(f):\n    '''Either a list of results if f succeeds with every pair\n       in the zip of xs and ys, or an explanatory string\n       if any application of f returns no result.\n    '''\n    def go(xs, ys):\n        ls, rs = partitionEithers(map(f, xs, ys))\n        return Left(ls[0]) if ls else Right(rs)\n    return lambda xs: lambda ys: go(xs, ys)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Chinese zodiac", "language": "Python from Ruby", "task": "Determine the Chinese zodiac sign and related associations for a given year.\nTraditionally, the Chinese have counted years using two lists of labels, one of length 10 (the \"celestial stems\") and one of length 12 (the \"terrestrial branches\"). The labels do not really have any meaning outside their positions in the two lists; they're simply a traditional enumeration device, used much as Westerners use letters and numbers. They were historically used for months and days as well as years, and the stems are still sometimes used for school grades.\n\nYears cycle through both lists concurrently, so that both stem and branch advance each year; if we used Roman letters for the stems and numbers for the branches, consecutive years would be labeled A1, B2, C3, etc. Since the two lists are different lengths, they cycle back to their beginning at different points: after J10 we get A11, and then after B12 we get C1. The result is a repeating 60-year pattern within which each pair of names occurs only once.\n\nMapping the branches to twelve traditional animal deities results in the well-known \"Chinese zodiac\", assigning each year to a given animal. For example, Sunday, January 22, 2023 CE (in the common Gregorian calendar) began the lunisolar Year of the Rabbit.\n\nThe celestial stems do not have a one-to-one mapping like that of the branches to animals; however, the five pairs of consecutive stems are each associated with one of the five traditional Chinese elements (Wood, Fire, Earth, Metal, and Water). Further, one of the two years within each element is assigned to yin, the other to yang.\n\nThus, the Chinese year beginning in 2023 CE is also the yin year of Water. Since 12 is an even number, the association between animals and yin/yang doesn't change; consecutive Years of the Rabbit will cycle through the five elements, but will always be yin.\n\n;Task: Create a subroutine or program that will return or output the animal, yin/yang association, and element for the lunisolar year that begins in a given CE year.\n\nYou may optionally provide more information in the form of the year's numerical position within the 60-year cycle and/or its actual Chinese stem-branch name (in Han characters or Pinyin transliteration).\n\n;Requisite information:\n* The animal cycle runs in this order: Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig.\n* The element cycle runs in this order: Wood, Fire, Earth, Metal, Water. \n* Each element gets two consecutive years; a yang followed by a yin.\n* The current 60-year cycle began in 1984; any multiple of 60 years from that point may be used to reckon from.\n\nThus, year 1 of a cycle is the year of the Wood Rat (yang), year 2 the Wood Ox (yin), and year 3 the Fire Tiger (yang). The year 2023 - which, as already noted, is the year of the Water Rabbit (yin) - is the 40th year of the current cycle. \n\n;Information for optional task:\n* The ten celestial stems are '''Jia ''' ''jia'', '''Yi ''' ''yi'', '''Bing ''' ''bing'', '''Ding ''' ''ding'', '''Wu ''' ''wu'', '''Ji ''' ''ji'', '''Geng ''' ''geng'', '''Xin ''' ''xin'', '''Ren ''' ''ren'', and '''Gui ''' ''gui''. With the ASCII version of Pinyin tones, the names are written \"jia3\", \"yi3\", \"bing3\", \"ding1\", \"wu4\", \"ji3\", \"geng1\", \"xin1\", \"ren2\", and \"gui3\".\n* The twelve terrestrial branches are '''Zi ''' ''zi'', '''Chou ''' ''chou'', '''Yin ''' ''yin'', '''Mao ''' ''mao'', '''Chen ''' ''chen'', '''Si ''' ''si'', '''Wu ''' ''wu'', '''Wei ''' ''wei'', '''Shen ''' ''shen'', '''You ''' ''you'', '''Xu ''' ''xu'', '''Hai ''' ''hai''. In ASCII Pinyin, those are \"zi3\", \"chou3\", \"yin2\", \"mao3\", \"chen2\", \"si4\", \"wu3\", \"wei4\", \"shen1\", \"you3\", \"xu1\", and \"hai4\".\n\nTherefore 1984 was '''Jia Zi ''' (''jia-zi'', or jia3-zi3).  2023 is '''Gui Mao ''' (''gui-mao'' or gui3-mao3).\n", "solution": "# coding: utf-8\n\nfrom __future__ import print_function\nfrom datetime import datetime\n\npinyin = {\n  '\u7532': 'ji\u0103',\n  '\u4e59': 'y\u012d',\n  '\u4e19': 'b\u012dng',\n  '\u4e01': 'd\u012bng',\n  '\u620a': 'w\u00f9',\n  '\u5df1': 'j\u012d',\n  '\u5e9a': 'g\u0113ng',\n  '\u8f9b': 'x\u012bn',\n  '\u58ec': 'r\u00e9n',\n  '\u7678': 'g\u016di',\n\n  '\u5b50': 'z\u012d',\n  '\u4e11': 'ch\u014fu',\n  '\u5bc5': 'y\u00edn',\n  '\u536f': 'm\u0103o',\n  '\u8fb0': 'ch\u00e9n',\n  '\u5df3': 's\u00ec',\n  '\u5348': 'w\u016d',\n  '\u672a': 'w\u00e8i',\n  '\u7533': 'sh\u0113n',\n  '\u9149': 'y\u014fu',\n  '\u620c': 'x\u016b',\n  '\u4ea5': 'h\u00e0i'\n}\n\nanimals = ['Rat', 'Ox', 'Tiger', 'Rabbit', 'Dragon', 'Snake',\n           'Horse', 'Goat', 'Monkey', 'Rooster', 'Dog', 'Pig']\nelements = ['Wood', 'Fire', 'Earth', 'Metal', 'Water']\n\ncelestial = ['\u7532', '\u4e59', '\u4e19', '\u4e01', '\u620a', '\u5df1', '\u5e9a', '\u8f9b', '\u58ec', '\u7678']\nterrestrial = ['\u5b50', '\u4e11', '\u5bc5', '\u536f', '\u8fb0', '\u5df3', '\u5348', '\u672a', '\u7533', '\u9149', '\u620c', '\u4ea5']\naspects = ['yang', 'yin']\n\n\ndef calculate(year):\n    BASE = 4\n    year = int(year)\n    cycle_year = year - BASE\n    stem_number = cycle_year % 10\n    stem_han = celestial[stem_number]\n    stem_pinyin = pinyin[stem_han]\n    element_number = stem_number // 2\n    element = elements[element_number]\n    branch_number = cycle_year % 12\n    branch_han = terrestrial[branch_number]\n    branch_pinyin = pinyin[branch_han]\n    animal = animals[branch_number]\n    aspect_number = cycle_year % 2\n    aspect = aspects[aspect_number]\n    index = cycle_year % 60 + 1\n    print(\"{}: {}{} ({}-{}, {} {}; {} - year {} of the cycle)\"\n          .format(year, stem_han, branch_han,\n                  stem_pinyin, branch_pinyin, element, animal, aspect, index))\n\n\ncurrent_year = datetime.now().year\nyears = [1935, 1938, 1968, 1972, 1976, current_year]\nfor year in years:\n    calculate(year)"}
{"title": "Chinese zodiac", "language": "Python 3.7", "task": "Determine the Chinese zodiac sign and related associations for a given year.\nTraditionally, the Chinese have counted years using two lists of labels, one of length 10 (the \"celestial stems\") and one of length 12 (the \"terrestrial branches\"). The labels do not really have any meaning outside their positions in the two lists; they're simply a traditional enumeration device, used much as Westerners use letters and numbers. They were historically used for months and days as well as years, and the stems are still sometimes used for school grades.\n\nYears cycle through both lists concurrently, so that both stem and branch advance each year; if we used Roman letters for the stems and numbers for the branches, consecutive years would be labeled A1, B2, C3, etc. Since the two lists are different lengths, they cycle back to their beginning at different points: after J10 we get A11, and then after B12 we get C1. The result is a repeating 60-year pattern within which each pair of names occurs only once.\n\nMapping the branches to twelve traditional animal deities results in the well-known \"Chinese zodiac\", assigning each year to a given animal. For example, Sunday, January 22, 2023 CE (in the common Gregorian calendar) began the lunisolar Year of the Rabbit.\n\nThe celestial stems do not have a one-to-one mapping like that of the branches to animals; however, the five pairs of consecutive stems are each associated with one of the five traditional Chinese elements (Wood, Fire, Earth, Metal, and Water). Further, one of the two years within each element is assigned to yin, the other to yang.\n\nThus, the Chinese year beginning in 2023 CE is also the yin year of Water. Since 12 is an even number, the association between animals and yin/yang doesn't change; consecutive Years of the Rabbit will cycle through the five elements, but will always be yin.\n\n;Task: Create a subroutine or program that will return or output the animal, yin/yang association, and element for the lunisolar year that begins in a given CE year.\n\nYou may optionally provide more information in the form of the year's numerical position within the 60-year cycle and/or its actual Chinese stem-branch name (in Han characters or Pinyin transliteration).\n\n;Requisite information:\n* The animal cycle runs in this order: Rat, Ox, Tiger, Rabbit, Dragon, Snake, Horse, Goat, Monkey, Rooster, Dog, Pig.\n* The element cycle runs in this order: Wood, Fire, Earth, Metal, Water. \n* Each element gets two consecutive years; a yang followed by a yin.\n* The current 60-year cycle began in 1984; any multiple of 60 years from that point may be used to reckon from.\n\nThus, year 1 of a cycle is the year of the Wood Rat (yang), year 2 the Wood Ox (yin), and year 3 the Fire Tiger (yang). The year 2023 - which, as already noted, is the year of the Water Rabbit (yin) - is the 40th year of the current cycle. \n\n;Information for optional task:\n* The ten celestial stems are '''Jia ''' ''jia'', '''Yi ''' ''yi'', '''Bing ''' ''bing'', '''Ding ''' ''ding'', '''Wu ''' ''wu'', '''Ji ''' ''ji'', '''Geng ''' ''geng'', '''Xin ''' ''xin'', '''Ren ''' ''ren'', and '''Gui ''' ''gui''. With the ASCII version of Pinyin tones, the names are written \"jia3\", \"yi3\", \"bing3\", \"ding1\", \"wu4\", \"ji3\", \"geng1\", \"xin1\", \"ren2\", and \"gui3\".\n* The twelve terrestrial branches are '''Zi ''' ''zi'', '''Chou ''' ''chou'', '''Yin ''' ''yin'', '''Mao ''' ''mao'', '''Chen ''' ''chen'', '''Si ''' ''si'', '''Wu ''' ''wu'', '''Wei ''' ''wei'', '''Shen ''' ''shen'', '''You ''' ''you'', '''Xu ''' ''xu'', '''Hai ''' ''hai''. In ASCII Pinyin, those are \"zi3\", \"chou3\", \"yin2\", \"mao3\", \"chen2\", \"si4\", \"wu3\", \"wei4\", \"shen1\", \"you3\", \"xu1\", and \"hai4\".\n\nTherefore 1984 was '''Jia Zi ''' (''jia-zi'', or jia3-zi3).  2023 is '''Gui Mao ''' (''gui-mao'' or gui3-mao3).\n", "solution": "'''Chinese zodiac'''\n\nfrom functools import (reduce)\nfrom datetime import datetime\n\n\n# TRADITIONAL STRINGS -------------------------------------\n\n# zodiacNames :: Dict\ndef zodiacNames():\n    '''\u5929\u5e72 tiangan \u2013 10 heavenly stems\n       \u5730\u652f dizhi \u2013 12 terrestrial branches\n       \u4e94\u884c wuxing \u2013 5 elements\n       \u751f\u8096 shengxiao \u2013 12 symbolic animals\n       \u9634\u9633 yinyang - dark and light\n    '''\n    return dict(\n        zip(\n            ['tian', 'di', 'wu', 'sx', 'yy'],\n            map(\n                lambda tpl: list(\n                    zip(* [tpl[0]] + list(\n                        map(\n                            lambda x: x.split(),\n                            tpl[1:])\n                    ))\n                ),\n                [\n                    # \u5929\u5e72 tiangan \u2013 10 heavenly stems\n                    ('\u7532\u4e59\u4e19\u4e01\u620a\u5df1\u5e9a\u8f9b\u58ec\u7678',\n                     'ji\u0103 y\u012d b\u012dng d\u012bng w\u00f9 j\u012d g\u0113ng x\u012bn r\u00e9n g\u016di'),\n\n                    # \u5730\u652f dizhi \u2013 12 terrestrial branches\n                    ('\u5b50\u4e11\u5bc5\u536f\u8fb0\u5df3\u5348\u672a\u7533\u9149\u620c\u4ea5',\n                     'z\u012d ch\u014fu y\u00edn m\u0103o ch\u00e9n s\u00ec w\u016d w\u00e8i sh\u0113n y\u014fu x\u016b h\u00e0i'),\n\n                    # \u4e94\u884c wuxing \u2013 5 elements\n                    ('\u6728\u706b\u571f\u91d1\u6c34',\n                     'm\u00f9 hu\u01d2 t\u01d4 j\u012bn shu\u01d0',\n                     'wood fire earth metal water'),\n\n                    # \u5341\u4e8c\u751f\u8096 shengxiao \u2013 12 symbolic animals\n                    ('\u9f20\u725b\u864e\u5154\u9f8d\u86c7\u99ac\u7f8a\u7334\u9e21\u72d7\u8c6c',\n                     'sh\u01d4 ni\u00fa h\u01d4 t\u00f9 l\u00f3ng sh\u00e9 m\u01ce y\u00e1ng h\u00f3u j\u012b g\u01d2u zh\u016b',\n                     'rat ox tiger rabbit dragon snake horse goat ' +\n                     'monkey rooster dog pig'\n                     ),\n\n                    # \u9634\u9633 yinyang\n                    ('\u9633\u9634', 'y\u00e1ng y\u012bn')\n                ]\n            )))\n\n\n# zodiacYear :: Dict -> [[String]]\ndef zodiacYear(dct):\n    '''A string of strings containing the\n       Chinese zodiac tokens for a given year.\n    '''\n    def tokens(y):\n        iYear = y - 4\n        iStem = iYear % 10\n        iBranch = iYear % 12\n        (hStem, pStem) = dct['tian'][iStem]\n        (hBranch, pBranch) = dct['di'][iBranch]\n        yy = iYear % 2\n        return [\n            [str(y), '', ''],\n            [\n                hStem + hBranch,\n                pStem + pBranch,\n                str((iYear % 60) + 1) + '/60'\n            ],\n            list(dct['wu'][iStem // 2]),\n            list(dct['sx'][iBranch]),\n            list(dct['yy'][int(yy)]) + ['dark' if yy else 'light']\n        ]\n    return lambda year: tokens(year)\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Writing out wiki tables displaying Chinese zodiac\n       details for a given list of years.\n    '''\n    print('\\n'.join(\n        list(map(\n            zodiacTable(zodiacNames()),\n            [\n                1935, 1938, 1949,\n                1968, 1972, 1976,\n                datetime.now().year\n            ]\n        ))\n    ))\n\n\n# WIKI TABLES  --------------------------------------------\n\n# zodiacTable :: Dict -> Int -> String\ndef zodiacTable(tokens):\n    '''A wiki table displaying Chinese zodiac\n       details for a a given year.\n    '''\n    return lambda y: wikiTable({\n        'class': 'wikitable',\n        'colwidth': '70px'\n    })(transpose(zodiacYear(tokens)(y)))\n\n\n# wikiTable :: Dict -> [[a]] -> String\ndef wikiTable(opts):\n    '''List of lists rendered as a wiki table string.'''\n    def colWidth():\n        return 'width:' + opts['colwidth'] + '; ' if (\n            'colwidth' in opts\n        ) else ''\n\n    def cellStyle():\n        return opts['cell'] if 'cell' in opts else ''\n\n    return lambda rows: '{| ' + reduce(\n        lambda a, k: (\n            a + k + '=\"' + opts[k] + '\" ' if k in opts else a\n        ),\n        ['class', 'style'],\n        ''\n    ) + '\\n' + '\\n|-\\n'.join(\n        '\\n'.join(\n            ('|' if (0 != i and ('cell' not in opts)) else (\n                '|style=\"' + colWidth() + cellStyle() + '\"|'\n            )) + (\n                str(x) or ' '\n            ) for x in row\n        ) for i, row in enumerate(rows)\n    ) + '\\n|}\\n\\n'\n\n\n# GENERIC -------------------------------------------------\n\n# transpose :: Matrix a -> Matrix a\ndef transpose(m):\n    '''The rows and columns of the argument transposed.\n       (The matrix containers and rows can be lists or tuples).'''\n    if m:\n        inner = type(m[0])\n        z = zip(*m)\n        return (type(m))(\n            map(inner, z) if tuple != inner else z\n        )\n    else:\n        return m\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Church numerals", "language": "Python 3.7", "task": "In the Church encoding of natural numbers, the number N is encoded by a function that applies its first argument N times to its second argument.\n\n* '''Church zero''' always returns the identity function, regardless of its first argument. In other words, the first argument is not applied to the second argument at all.\n* '''Church one''' applies its first argument f just once to its second argument x, yielding '''f(x)'''\n* '''Church two''' applies its first argument f twice to its second argument x, yielding '''f(f(x))'''\n* and each successive Church numeral applies its first argument one additional time to its second argument, '''f(f(f(x)))''', '''f(f(f(f(x))))''' ...  The Church numeral 4, for example, returns a quadruple composition of the function supplied as its first argument.\n\n\nArithmetic operations on natural numbers can be similarly represented as functions on Church numerals.\n\nIn your language define:\n\n* Church Zero,\n* a Church successor function (a function on a Church numeral which returns the next Church numeral in the series),\n* functions for Addition, Multiplication and Exponentiation over Church numerals,\n* a function to convert integers to corresponding Church numerals,\n* and a function to convert Church numerals to corresponding integers.\n\n\nYou should:\n\n* Derive Church numerals three and four in terms of Church zero and a Church successor function.\n* use Church numeral arithmetic to obtain the the sum and the product of Church 3 and Church 4,\n* similarly obtain 4^3 and 3^4 in terms of Church numerals, using a Church numeral exponentiation function,\n* convert each result back to an integer, and return it or print it to the console.\n\n\n", "solution": "'''Church numerals'''\n\nfrom itertools import repeat\nfrom functools import reduce\n\n\n# ----- CHURCH ENCODINGS OF NUMERALS AND OPERATIONS ------\n\ndef churchZero():\n    '''The identity function.\n       No applications of any supplied f\n       to its argument.\n    '''\n    return lambda f: identity\n\n\ndef churchSucc(cn):\n    '''The successor of a given\n       Church numeral. One additional\n       application of f. Equivalent to\n       the arithmetic addition of one.\n    '''\n    return lambda f: compose(f)(cn(f))\n\n\ndef churchAdd(m):\n    '''The arithmetic sum of two Church numerals.'''\n    return lambda n: lambda f: compose(m(f))(n(f))\n\n\ndef churchMult(m):\n    '''The arithmetic product of two Church numerals.'''\n    return lambda n: compose(m)(n)\n\n\ndef churchExp(m):\n    '''Exponentiation of Church numerals. m^n'''\n    return lambda n: n(m)\n\n\ndef churchFromInt(n):\n    '''The Church numeral equivalent of\n       a given integer.\n    '''\n    return lambda f: (\n        foldl\n        (compose)\n        (identity)\n        (replicate(n)(f))\n    )\n\n\n# OR, alternatively:\ndef churchFromInt_(n):\n    '''The Church numeral equivalent of a given\n       integer, by explicit recursion.\n    '''\n    if 0 == n:\n        return churchZero()\n    else:\n        return churchSucc(churchFromInt(n - 1))\n\n\ndef intFromChurch(cn):\n    '''The integer equivalent of a\n       given Church numeral.\n    '''\n    return cn(succ)(0)\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    'Tests'\n\n    cThree = churchFromInt(3)\n    cFour = churchFromInt(4)\n\n    print(list(map(intFromChurch, [\n        churchAdd(cThree)(cFour),\n        churchMult(cThree)(cFour),\n        churchExp(cFour)(cThree),\n        churchExp(cThree)(cFour),\n    ])))\n\n\n# ------------------ GENERIC FUNCTIONS -------------------\n\n# compose (flip (.)) :: (a -> b) -> (b -> c) -> a -> c\ndef compose(f):\n    '''A left to right composition of two\n       functions f and g'''\n    return lambda g: lambda x: g(f(x))\n\n\n# foldl :: (a -> b -> a) -> a -> [b] -> a\ndef foldl(f):\n    '''Left to right reduction of a list,\n       using the binary operator f, and\n       starting with an initial value a.\n    '''\n    def go(acc, xs):\n        return reduce(lambda a, x: f(a)(x), xs, acc)\n    return lambda acc: lambda xs: go(acc, xs)\n\n\n# identity :: a -> a\ndef identity(x):\n    '''The identity function.'''\n    return x\n\n\n# replicate :: Int -> a -> [a]\ndef replicate(n):\n    '''A list of length n in which every\n       element has the value x.\n    '''\n    return lambda x: repeat(x, n)\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value.\n       For numeric types, (1 +).\n    '''\n    return 1 + x if isinstance(x, int) else (\n        chr(1 + ord(x))\n    )\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Circles of given radius through two points", "language": "Python", "task": "2 circles with a given radius through 2 points in 2D space.\n\nGiven two points on a plane and a radius, usually two circles of given radius can be drawn through the points. \n;Exceptions:\n# r==0.0 should be treated as never describing circles (except in the case where the points are coincident).\n# If the points are coincident then an infinite number of circles with the point on their circumference can be drawn, unless r==0.0 as well which then collapses the circles to a point.\n# If the points form a diameter then return two identical circles ''or'' return a single circle, according to which is the most natural mechanism for the implementation language.\n# If the points are too far apart then no circles can be drawn.\n\n\n;Task detail:\n* Write a function/subroutine/method/... that takes two points and a radius and returns the two circles through those points, ''or some indication of special cases where two, possibly equal, circles cannot be returned''.\n* Show here the output for the following inputs:\n\n      p1                p2           r\n0.1234, 0.9876    0.8765, 0.2345    2.0\n0.0000, 2.0000    0.0000, 0.0000    1.0\n0.1234, 0.9876    0.1234, 0.9876    2.0\n0.1234, 0.9876    0.8765, 0.2345    0.5\n0.1234, 0.9876    0.1234, 0.9876    0.0\n\n\n\n;Related task:\n*   [[Total circles area]].\n\n\n;See also:\n*   Finding the Center of a Circle from 2 Points and Radius from Math forum @ Drexel\n\n", "solution": "from collections import namedtuple\nfrom math import sqrt\n\nPt = namedtuple('Pt', 'x, y')\nCircle = Cir = namedtuple('Circle', 'x, y, r')\n\ndef circles_from_p1p2r(p1, p2, r):\n    'Following explanation at http://mathforum.org/library/drmath/view/53027.html'\n    if r == 0.0:\n        raise ValueError('radius of zero')\n    (x1, y1), (x2, y2) = p1, p2\n    if p1 == p2:\n        raise ValueError('coincident points gives infinite number of Circles')\n    # delta x, delta y between points\n    dx, dy = x2 - x1, y2 - y1\n    # dist between points\n    q = sqrt(dx**2 + dy**2)\n    if q > 2.0*r:\n        raise ValueError('separation of points > diameter')\n    # halfway point\n    x3, y3 = (x1+x2)/2, (y1+y2)/2\n    # distance along the mirror line\n    d = sqrt(r**2-(q/2)**2)\n    # One answer\n    c1 = Cir(x = x3 - d*dy/q,\n             y = y3 + d*dx/q,\n             r = abs(r))\n    # The other answer\n    c2 = Cir(x = x3 + d*dy/q,\n             y = y3 - d*dx/q,\n             r = abs(r))\n    return c1, c2\n\nif __name__ == '__main__':\n    for p1, p2, r in [(Pt(0.1234, 0.9876), Pt(0.8765, 0.2345), 2.0),\n                      (Pt(0.0000, 2.0000), Pt(0.0000, 0.0000), 1.0),\n                      (Pt(0.1234, 0.9876), Pt(0.1234, 0.9876), 2.0),\n                      (Pt(0.1234, 0.9876), Pt(0.8765, 0.2345), 0.5),\n                      (Pt(0.1234, 0.9876), Pt(0.1234, 0.9876), 0.0)]:\n        print('Through points:\\n  %r,\\n  %r\\n  and radius %f\\nYou can construct the following circles:'\n              % (p1, p2, r))\n        try:\n            print('  %r\\n  %r\\n' % circles_from_p1p2r(p1, p2, r))\n        except ValueError as v:\n            print('  ERROR: %s\\n' % (v.args[0],))"}
{"title": "Cistercian numerals", "language": "Python", "task": "Cistercian numerals were used across Europe by Cistercian monks during the Late Medieval Period as an alternative to Roman numerals. They were used to represent base 10 integers from '''0''' to '''9999'''.\n\n;How they work\nAll Cistercian numerals begin with a vertical line segment, which by itself represents the number '''0'''. Then, glyphs representing the digits '''1''' through '''9''' are optionally added to the four quadrants surrounding the vertical line segment. These glyphs are drawn with vertical and horizontal symmetry about the initial line segment. Each quadrant corresponds to a digit place in the number:\n\n:* The '''upper-right''' quadrant represents the '''ones''' place.\n:* The '''upper-left''' quadrant represents the '''tens''' place.\n:* The '''lower-right''' quadrant represents the '''hundreds''' place.\n:* The '''lower-left''' quadrant represents the '''thousands''' place.\n\nPlease consult the following image for examples of Cistercian numerals showing each glyph: [https://upload.wikimedia.org/wikipedia/commons/6/67/Cistercian_digits_%28vertical%29.svg]\n\n;Task\n\n:* Write a function/procedure/routine to display any given Cistercian numeral. This could be done by drawing to the display, creating an image, or even as text (as long as it is a reasonable facsimile).\n\n:* Use the routine to show the following Cistercian numerals:\n\n::* 0\n::* 1\n::* 20\n::* 300\n::* 4000\n::* 5555\n::* 6789\n::* And a number of your choice!\n\n;Notes\nDue to the inability to upload images to Rosetta Code as of this task's creation, showing output here on this page is not required. However, it is welcomed -- especially for text output.\n\n;See also\n\n:* '''Numberphile - The Forgotten Number System'''\n:* '''dcode.fr - Online Cistercian numeral converter'''\n", "solution": "# -*- coding: utf-8 -*-\n\"\"\"\nSome UTF-8 chars used:\n    \n\u203e\t8254\t203E\t&oline;\tOVERLINE\n\u2503\t9475\t2503\t \tBOX DRAWINGS HEAVY VERTICAL\n\u2571\t9585\t2571\t \tBOX DRAWINGS LIGHT DIAGONAL UPPER RIGHT TO LOWER LEFT\n\u2572\t9586\t2572\t \tBOX DRAWINGS LIGHT DIAGONAL UPPER LEFT TO LOWER RIGHT\n\u25f8\t9720\t25F8\t \tUPPER LEFT TRIANGLE\n\u25f9\t9721\t25F9\t \tUPPER RIGHT TRIANGLE\n\u25fa\t9722\t25FA\t \tLOWER LEFT TRIANGLE\n\u25fb\t9723\t25FB\t \tWHITE MEDIUM SQUARE\n\u25ff\t9727\t25FF\t \tLOWER RIGHT TRIANGLE\n\n\"\"\"\n\n#%% digit sections\n\ndef _init():\n    \"digit sections for forming numbers\"\n    digi_bits = \"\"\"\n#0  1   2  3  4  5  6   7   8   9\n#\n .  \u203e   _  \u2572  \u2571  \u25f8  .|  \u203e|  _|  \u25fb\n#\n .  \u203e   _  \u2571  \u2572  \u25f9  |.  |\u203e  |_  \u25fb\n#\n .  _  \u203e   \u2571  \u2572  \u25fa  .|  _|  \u203e|  \u25fb\n#\n .  _  \u203e   \u2572  \u2571  \u25ff  |.  |_  |\u203e  \u25fb\n \n\"\"\".strip()\n\n    lines = [[d.replace('.', ' ') for d in ln.strip().split()]\n             for ln in digi_bits.strip().split('\\n')\n             if '#' not in ln]\n    formats = '<2 >2 <2 >2'.split()\n    digits = [[f\"{dig:{f}}\" for dig in line]\n              for f, line in zip(formats, lines)]\n\n    return digits\n\n_digits = _init()\n\n\n#%% int to 3-line strings\ndef _to_digits(n):\n    assert 0 <= n < 10_000 and int(n) == n\n    \n    return [int(digit) for digit in f\"{int(n):04}\"][::-1]\n\ndef num_to_lines(n):\n    global _digits\n    d = _to_digits(n)\n    lines = [\n        ''.join((_digits[1][d[1]], '\u2503',  _digits[0][d[0]])),\n        ''.join((_digits[0][   0], '\u2503',  _digits[0][   0])),\n        ''.join((_digits[3][d[3]], '\u2503',  _digits[2][d[2]])),\n        ]\n    \n    return lines\n\ndef cjoin(c1, c2, spaces='   '):\n    return [spaces.join(by_row) for by_row in zip(c1, c2)]\n\n#%% main\nif __name__ == '__main__':\n    #n = 6666\n    #print(f\"Arabic {n} to Cistercian:\\n\")\n    #print('\\n'.join(num_to_lines(n)))\n    \n    for pow10 in range(4):    \n        step = 10 ** pow10\n        print(f'\\nArabic {step}-to-{9*step} by {step} in Cistercian:\\n')\n        lines = num_to_lines(step)\n        for n in range(step*2, step*10, step):\n            lines = cjoin(lines, num_to_lines(n))\n        print('\\n'.join(lines))\n    \n\n    numbers = [0, 5555, 6789, 6666]\n    print(f'\\nArabic {str(numbers)[1:-1]} in Cistercian:\\n')\n    lines = num_to_lines(numbers[0])\n    for n in numbers[1:]:\n        lines = cjoin(lines, num_to_lines(n))\n    print('\\n'.join(lines))"}
{"title": "Closures/Value capture", "language": "Python", "task": "Create a list of ten functions, in the simplest manner possible   (anonymous functions are encouraged),   such that the function at index   '' i ''    (you may choose to start    '' i ''    from either    '''0'''    or    '''1'''),     when run, should return the square of the index,   that is,    '' i '' 2. \n\nDisplay the result of running any but the last function, to demonstrate that the function indeed remembers its value.\n\n\n;Goal:\nDemonstrate how to create a series of independent closures based on the same template but maintain separate copies of the variable closed over. \n\nIn imperative languages, one would generally use a loop with a mutable counter variable. \n\nFor each function to maintain the correct number, it has to capture the ''value'' of the variable at the time it was created, rather than just a reference to the variable, which would have a different value by the time the function was run.\n\nSee also: [[Multiple distinct objects]]\n", "solution": "funcs = []\nfor i in range(10):\n    funcs.append((lambda i: lambda: i * i)(i))\nprint funcs[3]() # prints 9"}
{"title": "Colorful numbers", "language": "Python", "task": "A '''colorful number''' is a non-negative base 10 integer where the product of every sub group of consecutive digits is unique.\n\n\n;E.G.\n\n24753 is a colorful number. 2, 4, 7, 5, 3, (2x4)8, (4x7)28, (7x5)35, (5x3)15, (2x4x7)56, (4x7x5)140, (7x5x3)105, (2x4x7x5)280, (4x7x5x3)420, (2x4x7x5x3)840\n\nEvery product is unique.\n\n\n2346 is '''not''' a colorful number. 2, 3, 4, '''6''', (2x3)'''6''', (3x4)12, (4x6)24, (2x3x4)48, (3x4x6)72, (2x3x4x6)144\n\nThe product '''6''' is repeated.\n\n\nSingle digit numbers '''are''' considered to be colorful. A colorful number larger than 9 cannot contain a repeated digit, the digit 0 or the digit 1. As a consequence, there is a firm upper limit for colorful numbers; no colorful number can have more than 8 digits.\n\n\n;Task\n\n* Write a routine (subroutine, function, procedure, whatever it may be called in your language) to test if a number is a colorful number or not.\n* Use that routine to find all of the colorful numbers less than 100.\n* Use that routine to find the largest possible colorful number.\n\n\n;Stretch\n\n* Find and display the count of colorful numbers in each order of magnitude.\n* Find and show the total count of '''all''' colorful numbers.\n\n\n''Colorful numbers have no real number theory application. They are more a recreational math puzzle than a useful tool.''\n\n", "solution": "from math import prod\n\nlargest = [0]\n\ndef iscolorful(n):\n    if 0 <= n < 10:\n        return True\n    dig = [int(c) for c in str(n)]\n    if 1 in dig or 0 in dig or len(dig) > len(set(dig)):\n        return False\n    products = list(set(dig))\n    for i in range(len(dig)):\n        for j in range(i+2, len(dig)+1):\n            p = prod(dig[i:j])\n            if p in products:\n                return False\n            products.append(p)\n\n    largest[0] = max(n, largest[0])\n    return True\n\nprint('Colorful numbers for 1:25, 26:50, 51:75, and 76:100:')\nfor i in range(1, 101, 25):\n    for j in range(25):\n        if iscolorful(i + j):\n            print(f'{i + j: 5,}', end='')\n    print()\n\ncsum = 0\nfor i in range(8):\n    j = 0 if i == 0 else 10**i\n    k = 10**(i+1) - 1\n    n = sum(iscolorful(x) for x in range(j, k+1))\n    csum += n\n    print(f'The count of colorful numbers between {j} and {k} is {n}.')\n\nprint(f'The largest possible colorful number is {largest[0]}.')\nprint(f'The total number of colorful numbers is {csum}.')\n"}
{"title": "Colour bars/Display", "language": "Python", "task": "Display a series of vertical color bars across the width of the display. \n\nThe color bars should either use:\n:::*   the system palette,   or \n:::*   the sequence of colors: \n::::::*   black\n::::::*   red\n::::::*   green\n::::::*   blue\n::::::*   magenta\n::::::*   cyan\n::::::*   yellow\n::::::*   white\n\n", "solution": "#!/usr/bin/env python\n#vertical coloured stripes in window in Python 2.7.1\n\nfrom livewires import *\n\nhoriz=640; vert=480\nbegin_graphics(width=horiz,height=vert,title=\"v_stripes\",background=Colour.black)\nNameColors=[\"black\",\"red\",\"green\",\"dark_blue\",\"purple\",\"blue\",\"yellow\",\"white\"]\nstepik=horiz/len(NameColors)\n\nfor index,each in enumerate(NameColors):\n\tExcStrng=\"set_colour(Colour.\"+each+\")\"\n\texec ExcStrng\n\tbox(index*stepik,0,(index+1)*stepik,vert,filled=1)\n\nwhile keys_pressed() != ['x']: # press x key to terminate program\n\tpass\n\nend_graphics()\n"}
{"title": "Colour pinstripe/Printer", "language": "Python", "task": "The task is to create 1 point wide colour vertical pinstripes with a sufficient number of pinstripes to span the entire width of the colour graphics printer. The pinstripes should alternate between each individual cartridge ink and ink pair and black and white pinstripes should be included. A typical pinstripe sequence woud be black, red, green, blue, magenta, cyan, yellow, white.\n\nAfter the first inch of printing, we switch to a wider 2 pixel wide vertical pinstripe pattern. and to 3 point wide vertical for the next inch, and then 4 point wide, etc. This trend continues for the entire length of the page (or for 12 inches of run length in the case of a printer using continuous roll stationery). After printing the test pattern the page is ejected (or the test pattern is rolled clear of the printer enclosure, in the case of continuous roll printers).\n\nNote that it is an acceptable solution to use the smallest marks that the language provides, rather than working at native printer resolution, where this is not achievable from within the language.\n\nOptionally, on systems where the printer resolution cannot be determined, it is permissible to prompt the user for printer resolution, and to calculate point size based on user input, enabling fractional point sizes to be used.\n\n", "solution": "from turtle import *\nfrom PIL import Image\nimport time\nimport subprocess\n\n\"\"\"\n\nOnly works on Windows. Assumes that you have Ghostscript\ninstalled and in your path.\n\nhttps://www.ghostscript.com/download/gsdnld.html\n\nHard coded to 100 pixels per inch.\n\n\"\"\"\n\ncolors = [\"black\", \"red\", \"green\", \"blue\", \"magenta\", \"cyan\", \"yellow\", \"white\"]\n\nscreen = getscreen()\n\n# width and height in pixels\n# aspect ratio for 11 by 8.5 paper\n\ninch_width = 11.0\ninch_height = 8.5\n\npixels_per_inch = 100\n\npix_width = int(inch_width*pixels_per_inch)\npix_height = int(inch_height*pixels_per_inch)\n\nscreen.setup (width=pix_width, height=pix_height, startx=0, starty=0)\n\nscreen.screensize(pix_width,pix_height)\n\n# center is 0,0\n\n# get coordinates of the edges\n\nleft_edge = -screen.window_width()//2\n\nright_edge = screen.window_width()//2\n\nbottom_edge = -screen.window_height()//2\n\ntop_edge = screen.window_height()//2\n\n# draw quickly\n\nscreen.delay(0)\nscreen.tracer(5)\n\nfor inch in range(int(inch_width)-1):\n    line_width = inch + 1\n    pensize(line_width)\n    colornum = 0\n\n    min_x = left_edge + (inch * pixels_per_inch)\n    max_x = left_edge + ((inch+1) * pixels_per_inch)\n    \n    for y in range(bottom_edge,top_edge,line_width):\n        penup()\n        pencolor(colors[colornum])\n        colornum = (colornum + 1) % len(colors)\n        setposition(min_x,y)\n        pendown()\n        setposition(max_x,y)\n         \nscreen.getcanvas().postscript(file=\"striped.eps\")\n\n# convert to jpeg\n# won't work without Ghostscript.\n\nim = Image.open(\"striped.eps\")\nim.save(\"striped.jpg\")\n\n# Got idea from http://rosettacode.org/wiki/Colour_pinstripe/Printer#Go\n    \nsubprocess.run([\"mspaint\", \"/pt\", \"striped.jpg\"])\n"}
{"title": "Comma quibbling", "language": "Python", "task": "Comma quibbling is a task originally set by Eric Lippert in his blog.\n\n\n;Task:\n\nWrite a function to generate a string output which is the concatenation of input words from a list/sequence where:\n# An input of no words produces the output string of just the two brace characters \"{}\".\n# An input of just one word, e.g. [\"ABC\"], produces the output string of the word inside the two braces, e.g. \"{ABC}\".\n# An input of two words, e.g. [\"ABC\", \"DEF\"], produces the output string of the two words inside the two braces with the words separated by the string \" and \", e.g. \"{ABC and DEF}\".\n# An input of three or more words, e.g. [\"ABC\", \"DEF\", \"G\", \"H\"], produces the output string of all but the last word separated by \", \" with the last word separated by \" and \" and all within braces; e.g. \"{ABC, DEF, G and H}\".\n\n\nTest your function with the following series of inputs showing your output here on this page:\n* []                       # (No input words).\n* [\"ABC\"]\n* [\"ABC\", \"DEF\"]\n* [\"ABC\", \"DEF\", \"G\", \"H\"]\n\n\nNote: Assume words are non-empty strings of uppercase characters for this task.\n\n", "solution": ">>> def strcat(sequence):\n    return '{%s}' % ', '.join(sequence)[::-1].replace(',', 'dna ', 1)[::-1]\n\n>>> for seq in ([], [\"ABC\"], [\"ABC\", \"DEF\"], [\"ABC\", \"DEF\", \"G\", \"H\"]):\n    print('Input: %-24r -> Output: %r' % (seq, strcat(seq)))\n\n\t\nInput: []                       -> Output: '{}'\nInput: ['ABC']                  -> Output: '{ABC}'\nInput: ['ABC', 'DEF']           -> Output: '{ABC and DEF}'\nInput: ['ABC', 'DEF', 'G', 'H'] -> Output: '{ABC, DEF, G and H}'\n>>> "}
{"title": "Command-line arguments", "language": "Python", "task": "{{selection|Short Circuit|Console Program Basics}} Retrieve the list of command-line arguments given to the program. For programs that only print the arguments when run directly, see [[Scripted main]].\nSee also [[Program name]].\n\nFor parsing command line arguments intelligently, see [[Parsing command-line arguments]].\n\nExample command line:\n\n myprogram -c \"alpha beta\" -h \"gamma\"\n\n", "solution": "import sys\nprogram_name = sys.argv[0]\narguments = sys.argv[1:]\ncount = len(arguments)"}
{"title": "Compare a list of strings", "language": "Python", "task": "Given a   list   of arbitrarily many strings, show how to:\n\n*   test if they are all lexically '''equal'''\n*   test if every string is lexically '''less than''' the one after it  ''(i.e. whether the list is in strict ascending order)''\n\n\nEach of those two tests should result in a single true or false value, which could be used as the condition of an    if    statement or similar. \n\nIf the input list has less than two elements, the tests should always return true.\n\nThere is ''no'' need to provide a complete program and output.\n \nAssume that the strings are already stored in an array/list/sequence/tuple variable (whatever is most idiomatic) with the name   strings,   and just show the expressions for performing those two tests on it (plus of course any includes and custom functions etc. that it needs),   with as little distractions as possible.\n\nTry to write your solution in a way that does not modify the original list,   but if it does then please add a note to make that clear to readers.\n\nIf you need further guidance/clarification,   see [[#Perl]] and [[#Python]] for solutions that use implicit short-circuiting loops,   and [[#Raku]] for a solution that gets away with simply using a built-in language feature. \n\n", "solution": "from operator import (eq, lt)\n\n\nxs = [\"alpha\", \"beta\", \"gamma\", \"delta\", \"epsilon\", \"zeta\",\n      \"eta\", \"theta\", \"iota\", \"kappa\", \"lambda\", \"mu\"]\n\nys = [\"alpha\", \"beta\", \"gamma\", \"delta\", \"epsilon\", \"zeta\",\n      \"eta\", \"theta\", \"iota\", \"kappa\", \"lambda\", \"mu\"]\n\naz = sorted(xs)\n\nprint (\n    all(map(eq, xs, ys)),\n\n    all(map(lt, xs, xs[1:])),\n\n    all(map(lt, az, az[1:]))\n)"}
{"title": "Compiler/AST interpreter", "language": "Python", "task": "The C and Python versions can be considered reference implementations.\n\n;Related Tasks\n\n* Lexical Analyzer task\n* Syntax Analyzer task\n* Code Generator task\n* Virtual Machine Interpreter task\n\n\n__TOC__\n\n", "solution": "from __future__ import print_function\nimport sys, shlex, operator\n\nnd_Ident, nd_String, nd_Integer, nd_Sequence, nd_If, nd_Prtc, nd_Prts, nd_Prti, nd_While, \\\nnd_Assign, nd_Negate, nd_Not, nd_Mul, nd_Div, nd_Mod, nd_Add, nd_Sub, nd_Lss, nd_Leq,     \\\nnd_Gtr, nd_Geq, nd_Eql, nd_Neq, nd_And, nd_Or = range(25)\n\nall_syms = {\n    \"Identifier\"  : nd_Ident,    \"String\"      : nd_String,\n    \"Integer\"     : nd_Integer,  \"Sequence\"    : nd_Sequence,\n    \"If\"          : nd_If,       \"Prtc\"        : nd_Prtc,\n    \"Prts\"        : nd_Prts,     \"Prti\"        : nd_Prti,\n    \"While\"       : nd_While,    \"Assign\"      : nd_Assign,\n    \"Negate\"      : nd_Negate,   \"Not\"         : nd_Not,\n    \"Multiply\"    : nd_Mul,      \"Divide\"      : nd_Div,\n    \"Mod\"         : nd_Mod,      \"Add\"         : nd_Add,\n    \"Subtract\"    : nd_Sub,      \"Less\"        : nd_Lss,\n    \"LessEqual\"   : nd_Leq,      \"Greater\"     : nd_Gtr,\n    \"GreaterEqual\": nd_Geq,      \"Equal\"       : nd_Eql,\n    \"NotEqual\"    : nd_Neq,      \"And\"         : nd_And,\n    \"Or\"          : nd_Or}\n\ninput_file  = None\nglobals     = {}\n\n#*** show error and exit\ndef error(msg):\n    print(\"%s\" % (msg))\n    exit(1)\n\nclass Node:\n    def __init__(self, node_type, left = None, right = None, value = None):\n        self.node_type  = node_type\n        self.left  = left\n        self.right = right\n        self.value = value\n\n#***\ndef make_node(oper, left, right = None):\n    return Node(oper, left, right)\n\n#***\ndef make_leaf(oper, n):\n    return Node(oper, value = n)\n\n#***\ndef fetch_var(var_name):\n    n = globals.get(var_name, None)\n    if n == None:\n        globals[var_name] = n = 0\n    return n\n\n#***\ndef interp(x):\n    global globals\n\n    if x == None: return None\n    elif x.node_type == nd_Integer: return int(x.value)\n    elif x.node_type == nd_Ident:   return fetch_var(x.value)\n    elif x.node_type == nd_String:  return x.value\n\n    elif x.node_type == nd_Assign:\n                    globals[x.left.value] = interp(x.right)\n                    return None\n    elif x.node_type == nd_Add:     return interp(x.left) +   interp(x.right)\n    elif x.node_type == nd_Sub:     return interp(x.left) -   interp(x.right)\n    elif x.node_type == nd_Mul:     return interp(x.left) *   interp(x.right)\n    # use C like division semantics\n    # another way: abs(x) / abs(y) * cmp(x, 0) * cmp(y, 0)\n    elif x.node_type == nd_Div:     return int(float(interp(x.left)) / interp(x.right))\n    elif x.node_type == nd_Mod:     return int(float(interp(x.left)) % interp(x.right))\n    elif x.node_type == nd_Lss:     return interp(x.left) <   interp(x.right)\n    elif x.node_type == nd_Gtr:     return interp(x.left) >   interp(x.right)\n    elif x.node_type == nd_Leq:     return interp(x.left) <=  interp(x.right)\n    elif x.node_type == nd_Geq:     return interp(x.left) >=  interp(x.right)\n    elif x.node_type == nd_Eql:     return interp(x.left) ==  interp(x.right)\n    elif x.node_type == nd_Neq:     return interp(x.left) !=  interp(x.right)\n    elif x.node_type == nd_And:     return interp(x.left) and interp(x.right)\n    elif x.node_type == nd_Or:      return interp(x.left) or  interp(x.right)\n    elif x.node_type == nd_Negate:  return -interp(x.left)\n    elif x.node_type == nd_Not:     return not interp(x.left)\n\n    elif x.node_type ==  nd_If:\n                    if (interp(x.left)):\n                        interp(x.right.left)\n                    else:\n                        interp(x.right.right)\n                    return None\n\n    elif x.node_type == nd_While:\n                    while (interp(x.left)):\n                        interp(x.right)\n                    return None\n\n    elif x.node_type == nd_Prtc:\n                    print(\"%c\" % (interp(x.left)), end='')\n                    return None\n\n    elif x.node_type == nd_Prti:\n                    print(\"%d\" % (interp(x.left)), end='')\n                    return None\n\n    elif x.node_type == nd_Prts:\n                    print(interp(x.left), end='')\n                    return None\n\n    elif x.node_type == nd_Sequence:\n                    interp(x.left)\n                    interp(x.right)\n                    return None\n    else:\n        error(\"error in code generator - found %d, expecting operator\" % (x.node_type))\n\ndef str_trans(srce):\n    dest = \"\"\n    i = 0\n    srce = srce[1:-1]\n    while i < len(srce):\n        if srce[i] == '\\\\' and i + 1 < len(srce):\n            if srce[i + 1] == 'n':\n                dest += '\\n'\n                i += 2\n            elif srce[i + 1] == '\\\\':\n                dest += '\\\\'\n                i += 2\n        else:\n            dest += srce[i]\n            i += 1\n\n    return dest\n\ndef load_ast():\n    line = input_file.readline()\n    line_list = shlex.split(line, False, False)\n\n    text = line_list[0]\n\n    value = None\n    if len(line_list) > 1:\n        value = line_list[1]\n        if value.isdigit():\n            value = int(value)\n\n    if text == \";\":\n        return None\n    node_type = all_syms[text]\n    if value != None:\n        if node_type == nd_String:\n            value = str_trans(value)\n\n        return make_leaf(node_type, value)\n    left = load_ast()\n    right = load_ast()\n    return make_node(node_type, left, right)\n\n#*** main driver\ninput_file = sys.stdin\nif len(sys.argv) > 1:\n    try:\n        input_file = open(sys.argv[1], \"r\", 4096)\n    except IOError as e:\n        error(0, 0, \"Can't open %s\" % sys.argv[1])\n\nn = load_ast()\ninterp(n)"}
{"title": "Compiler/code generator", "language": "Python", "task": "The C and Python versions can be considered reference implementations.\n\n;Related Tasks\n\n* Lexical Analyzer task\n* Syntax Analyzer task\n* Virtual Machine Interpreter task\n* AST Interpreter task\n\n\n__TOC__\n\n", "solution": "from __future__ import print_function\nimport sys, struct, shlex, operator\n\nnd_Ident, nd_String, nd_Integer, nd_Sequence, nd_If, nd_Prtc, nd_Prts, nd_Prti, nd_While, \\\nnd_Assign, nd_Negate, nd_Not, nd_Mul, nd_Div, nd_Mod, nd_Add, nd_Sub, nd_Lss, nd_Leq,     \\\nnd_Gtr, nd_Geq, nd_Eql, nd_Neq, nd_And, nd_Or = range(25)\n\nall_syms = {\n    \"Identifier\"  : nd_Ident,    \"String\"      : nd_String,\n    \"Integer\"     : nd_Integer,  \"Sequence\"    : nd_Sequence,\n    \"If\"          : nd_If,       \"Prtc\"        : nd_Prtc,\n    \"Prts\"        : nd_Prts,     \"Prti\"        : nd_Prti,\n    \"While\"       : nd_While,    \"Assign\"      : nd_Assign,\n    \"Negate\"      : nd_Negate,   \"Not\"         : nd_Not,\n    \"Multiply\"    : nd_Mul,      \"Divide\"      : nd_Div,\n    \"Mod\"         : nd_Mod,      \"Add\"         : nd_Add,\n    \"Subtract\"    : nd_Sub,      \"Less\"        : nd_Lss,\n    \"LessEqual\"   : nd_Leq,      \"Greater\"     : nd_Gtr,\n    \"GreaterEqual\": nd_Geq,      \"Equal\"       : nd_Eql,\n    \"NotEqual\"    : nd_Neq,      \"And\"         : nd_And,\n    \"Or\"          : nd_Or}\n\nFETCH, STORE, PUSH, ADD, SUB, MUL, DIV, MOD, LT, GT, LE, GE, EQ, NE, AND, OR, NEG, NOT, \\\nJMP, JZ, PRTC, PRTS, PRTI, HALT = range(24)\n\noperators = {nd_Lss: LT, nd_Gtr: GT, nd_Leq: LE, nd_Geq: GE, nd_Eql: EQ, nd_Neq: NE,\n    nd_And: AND, nd_Or: OR, nd_Sub: SUB, nd_Add: ADD, nd_Div: DIV, nd_Mul: MUL, nd_Mod: MOD}\n\nunary_operators = {nd_Negate: NEG, nd_Not: NOT}\n\ninput_file  = None\ncode        = bytearray()\nstring_pool = {}\nglobals     = {}\nstring_n    = 0\nglobals_n   = 0\nword_size   = 4\n\n#*** show error and exit\ndef error(msg):\n    print(\"%s\" % (msg))\n    exit(1)\n\ndef int_to_bytes(val):\n    return struct.pack(\"<i\", val)\n\ndef bytes_to_int(bstr):\n    return struct.unpack(\"<i\", bstr)\n\nclass Node:\n    def __init__(self, node_type, left = None, right = None, value = None):\n        self.node_type  = node_type\n        self.left  = left\n        self.right = right\n        self.value = value\n\n#***\ndef make_node(oper, left, right = None):\n    return Node(oper, left, right)\n\n#***\ndef make_leaf(oper, n):\n    return Node(oper, value = n)\n\n#***\ndef emit_byte(x):\n    code.append(x)\n\n#***\ndef emit_word(x):\n    s = int_to_bytes(x)\n    for x in s:\n        code.append(x)\n\ndef emit_word_at(at, n):\n    code[at:at+word_size] = int_to_bytes(n)\n\ndef hole():\n    t = len(code)\n    emit_word(0)\n    return t\n\n#***\ndef fetch_var_offset(name):\n    global globals_n\n\n    n = globals.get(name, None)\n    if n == None:\n        globals[name] = globals_n\n        n = globals_n\n        globals_n += 1\n    return n\n\n#***\ndef fetch_string_offset(the_string):\n    global string_n\n\n    n = string_pool.get(the_string, None)\n    if n == None:\n        string_pool[the_string] = string_n\n        n = string_n\n        string_n += 1\n    return n\n\n#***\ndef code_gen(x):\n    if x == None: return\n    elif x.node_type == nd_Ident:\n        emit_byte(FETCH)\n        n = fetch_var_offset(x.value)\n        emit_word(n)\n    elif x.node_type == nd_Integer:\n        emit_byte(PUSH)\n        emit_word(x.value)\n    elif x.node_type == nd_String:\n        emit_byte(PUSH)\n        n = fetch_string_offset(x.value)\n        emit_word(n)\n    elif x.node_type == nd_Assign:\n        n = fetch_var_offset(x.left.value)\n        code_gen(x.right)\n        emit_byte(STORE)\n        emit_word(n)\n    elif x.node_type == nd_If:\n        code_gen(x.left)              # expr\n        emit_byte(JZ)                 # if false, jump\n        p1 = hole()                   # make room for jump dest\n        code_gen(x.right.left)        # if true statements\n        if (x.right.right != None):\n            emit_byte(JMP)            # jump over else statements\n            p2 = hole()\n        emit_word_at(p1, len(code) - p1)\n        if (x.right.right != None):\n            code_gen(x.right.right)   # else statements\n            emit_word_at(p2, len(code) - p2)\n    elif x.node_type == nd_While:\n        p1 = len(code)\n        code_gen(x.left)\n        emit_byte(JZ)\n        p2 = hole()\n        code_gen(x.right)\n        emit_byte(JMP)                       # jump back to the top\n        emit_word(p1 - len(code))\n        emit_word_at(p2, len(code) - p2)\n    elif x.node_type == nd_Sequence:\n        code_gen(x.left)\n        code_gen(x.right)\n    elif x.node_type == nd_Prtc:\n        code_gen(x.left)\n        emit_byte(PRTC)\n    elif x.node_type == nd_Prti:\n        code_gen(x.left)\n        emit_byte(PRTI)\n    elif x.node_type == nd_Prts:\n        code_gen(x.left)\n        emit_byte(PRTS)\n    elif x.node_type in operators:\n        code_gen(x.left)\n        code_gen(x.right)\n        emit_byte(operators[x.node_type])\n    elif x.node_type in unary_operators:\n        code_gen(x.left)\n        emit_byte(unary_operators[x.node_type])\n    else:\n        error(\"error in code generator - found %d, expecting operator\" % (x.node_type))\n\n#***\ndef code_finish():\n    emit_byte(HALT)\n\n#***\ndef list_code():\n    print(\"Datasize: %d Strings: %d\" % (len(globals), len(string_pool)))\n\n    for k in sorted(string_pool, key=string_pool.get):\n        print(k)\n\n    pc = 0\n    while pc < len(code):\n        print(\"%4d \" % (pc), end='')\n        op = code[pc]\n        pc += 1\n        if op == FETCH:\n            x = bytes_to_int(code[pc:pc+word_size])[0]\n            print(\"fetch [%d]\" % (x));\n            pc += word_size\n        elif op == STORE:\n            x = bytes_to_int(code[pc:pc+word_size])[0]\n            print(\"store [%d]\" % (x));\n            pc += word_size\n        elif op == PUSH:\n            x = bytes_to_int(code[pc:pc+word_size])[0]\n            print(\"push  %d\" % (x));\n            pc += word_size\n        elif op == ADD:   print(\"add\")\n        elif op == SUB:   print(\"sub\")\n        elif op == MUL:   print(\"mul\")\n        elif op == DIV:   print(\"div\")\n        elif op == MOD:   print(\"mod\")\n        elif op == LT:    print(\"lt\")\n        elif op == GT:    print(\"gt\")\n        elif op == LE:    print(\"le\")\n        elif op == GE:    print(\"ge\")\n        elif op == EQ:    print(\"eq\")\n        elif op == NE:    print(\"ne\")\n        elif op == AND:   print(\"and\")\n        elif op == OR:    print(\"or\")\n        elif op == NEG:   print(\"neg\")\n        elif op == NOT:   print(\"not\")\n        elif op == JMP:\n            x = bytes_to_int(code[pc:pc+word_size])[0]\n            print(\"jmp    (%d) %d\" % (x, pc + x));\n            pc += word_size\n        elif op == JZ:\n            x = bytes_to_int(code[pc:pc+word_size])[0]\n            print(\"jz     (%d) %d\" % (x, pc + x));\n            pc += word_size\n        elif op == PRTC:  print(\"prtc\")\n        elif op == PRTI:  print(\"prti\")\n        elif op == PRTS:  print(\"prts\")\n        elif op == HALT:  print(\"halt\")\n        else: error(\"list_code: Unknown opcode %d\", (op));\n\ndef load_ast():\n    line = input_file.readline()\n    line_list = shlex.split(line, False, False)\n\n    text = line_list[0]\n    if text == \";\":\n        return None\n    node_type = all_syms[text]\n\n    if len(line_list) > 1:\n        value = line_list[1]\n        if value.isdigit():\n            value = int(value)\n        return make_leaf(node_type, value)\n\n    left = load_ast()\n    right = load_ast()\n    return make_node(node_type, left, right)\n\n#*** main driver\ninput_file = sys.stdin\nif len(sys.argv) > 1:\n    try:\n        input_file = open(sys.argv[1], \"r\", 4096)\n    except IOError as e:\n        error(\"Can't open %s\" % sys.argv[1])\n\nn = load_ast()\ncode_gen(n)\ncode_finish()\nlist_code()"}
{"title": "Compiler/lexical analyzer", "language": "Python", "task": "The C and Python versions can be considered reference implementations.\n\n\n;Related Tasks\n* Syntax Analyzer task\n* Code Generator task\n* Virtual Machine Interpreter task\n* AST Interpreter task\n\n\n", "solution": "from __future__ import print_function\nimport sys\n\n# following two must remain in the same order\n\ntk_EOI, tk_Mul, tk_Div, tk_Mod, tk_Add, tk_Sub, tk_Negate, tk_Not, tk_Lss, tk_Leq, tk_Gtr, \\\ntk_Geq, tk_Eq, tk_Neq, tk_Assign, tk_And, tk_Or, tk_If, tk_Else, tk_While, tk_Print,       \\\ntk_Putc, tk_Lparen, tk_Rparen, tk_Lbrace, tk_Rbrace, tk_Semi, tk_Comma, tk_Ident,          \\\ntk_Integer, tk_String = range(31)\n\nall_syms = [\"End_of_input\", \"Op_multiply\", \"Op_divide\", \"Op_mod\", \"Op_add\", \"Op_subtract\",\n    \"Op_negate\", \"Op_not\", \"Op_less\", \"Op_lessequal\", \"Op_greater\", \"Op_greaterequal\",\n    \"Op_equal\", \"Op_notequal\", \"Op_assign\", \"Op_and\", \"Op_or\", \"Keyword_if\",\n    \"Keyword_else\", \"Keyword_while\", \"Keyword_print\", \"Keyword_putc\", \"LeftParen\",\n    \"RightParen\", \"LeftBrace\", \"RightBrace\", \"Semicolon\", \"Comma\", \"Identifier\",\n    \"Integer\", \"String\"]\n\n# single character only symbols\nsymbols = { '{': tk_Lbrace, '}': tk_Rbrace, '(': tk_Lparen, ')': tk_Rparen, '+': tk_Add, '-': tk_Sub,\n    '*': tk_Mul, '%': tk_Mod, ';': tk_Semi, ',': tk_Comma }\n\nkey_words = {'if': tk_If, 'else': tk_Else, 'print': tk_Print, 'putc': tk_Putc, 'while': tk_While}\n\nthe_ch = \" \"    # dummy first char - but it must be a space\nthe_col = 0\nthe_line = 1\ninput_file = None\n\n#*** show error and exit\ndef error(line, col, msg):\n    print(line, col, msg)\n    exit(1)\n\n#*** get the next character from the input\ndef next_ch():\n    global the_ch, the_col, the_line\n\n    the_ch = input_file.read(1)\n    the_col += 1\n    if the_ch == '\\n':\n        the_line += 1\n        the_col = 0\n    return the_ch\n\n#*** 'x' - character constants\ndef char_lit(err_line, err_col):\n    n = ord(next_ch())              # skip opening quote\n    if the_ch == '\\'':\n        error(err_line, err_col, \"empty character constant\")\n    elif the_ch == '\\\\':\n        next_ch()\n        if the_ch == 'n':\n            n = 10\n        elif the_ch == '\\\\':\n            n = ord('\\\\')\n        else:\n            error(err_line, err_col, \"unknown escape sequence \\\\%c\" % (the_ch))\n    if next_ch() != '\\'':\n        error(err_line, err_col, \"multi-character constant\")\n    next_ch()\n    return tk_Integer, err_line, err_col, n\n\n#*** process divide or comments\ndef div_or_cmt(err_line, err_col):\n    if next_ch() != '*':\n        return tk_Div, err_line, err_col\n\n    # comment found\n    next_ch()\n    while True:\n        if the_ch == '*':\n            if next_ch() == '/':\n                next_ch()\n                return gettok()\n        elif len(the_ch) == 0:\n            error(err_line, err_col, \"EOF in comment\")\n        else:\n            next_ch()\n\n#*** \"string\"\ndef string_lit(start, err_line, err_col):\n    global the_ch\n    text = \"\"\n\n    while next_ch() != start:\n        if len(the_ch) == 0:\n            error(err_line, err_col, \"EOF while scanning string literal\")\n        if the_ch == '\\n':\n            error(err_line, err_col, \"EOL while scanning string literal\")\n        if the_ch == '\\\\':\n            next_ch()\n            if the_ch != 'n':\n                error(err_line, err_col, \"escape sequence unknown \\\\%c\" % the_ch)\n            the_ch = '\\n'\n        text += the_ch\n\n    next_ch()\n    return tk_String, err_line, err_col, text\n\n#*** handle identifiers and integers\ndef ident_or_int(err_line, err_col):\n    is_number = True\n    text = \"\"\n\n    while the_ch.isalnum() or the_ch == '_':\n        text += the_ch\n        if not the_ch.isdigit():\n            is_number = False\n        next_ch()\n\n    if len(text) == 0:\n        error(err_line, err_col, \"ident_or_int: unrecognized character: (%d) '%c'\" % (ord(the_ch), the_ch))\n\n    if text[0].isdigit():\n        if not is_number:\n            error(err_line, err_col, \"invalid number: %s\" % (text))\n        n = int(text)\n        return tk_Integer, err_line, err_col, n\n\n    if text in key_words:\n        return key_words[text], err_line, err_col\n\n    return tk_Ident, err_line, err_col, text\n\n#*** look ahead for '>=', etc.\ndef follow(expect, ifyes, ifno, err_line, err_col):\n    if next_ch() == expect:\n        next_ch()\n        return ifyes, err_line, err_col\n\n    if ifno == tk_EOI:\n        error(err_line, err_col, \"follow: unrecognized character: (%d) '%c'\" % (ord(the_ch), the_ch))\n\n    return ifno, err_line, err_col\n\n#*** return the next token type\ndef gettok():\n    while the_ch.isspace():\n        next_ch()\n\n    err_line = the_line\n    err_col  = the_col\n\n    if len(the_ch) == 0:    return tk_EOI, err_line, err_col\n    elif the_ch == '/':     return div_or_cmt(err_line, err_col)\n    elif the_ch == '\\'':    return char_lit(err_line, err_col)\n    elif the_ch == '<':     return follow('=', tk_Leq, tk_Lss,    err_line, err_col)\n    elif the_ch == '>':     return follow('=', tk_Geq, tk_Gtr,    err_line, err_col)\n    elif the_ch == '=':     return follow('=', tk_Eq,  tk_Assign, err_line, err_col)\n    elif the_ch == '!':     return follow('=', tk_Neq, tk_Not,    err_line, err_col)\n    elif the_ch == '&':     return follow('&', tk_And, tk_EOI,    err_line, err_col)\n    elif the_ch == '|':     return follow('|', tk_Or,  tk_EOI,    err_line, err_col)\n    elif the_ch == '\"':     return string_lit(the_ch, err_line, err_col)\n    elif the_ch in symbols:\n        sym = symbols[the_ch]\n        next_ch()\n        return sym, err_line, err_col\n    else: return ident_or_int(err_line, err_col)\n\n#*** main driver\ninput_file = sys.stdin\nif len(sys.argv) > 1:\n    try:\n        input_file = open(sys.argv[1], \"r\", 4096)\n    except IOError as e:\n        error(0, 0, \"Can't open %s\" % sys.argv[1])\n\nwhile True:\n    t = gettok()\n    tok  = t[0]\n    line = t[1]\n    col  = t[2]\n\n    print(\"%5d  %5d   %-14s\" % (line, col, all_syms[tok]), end='')\n\n    if tok == tk_Integer:  print(\"   %5d\" % (t[3]))\n    elif tok == tk_Ident:  print(\"  %s\" %   (t[3]))\n    elif tok == tk_String: print('  \"%s\"' % (t[3]))\n    else:                  print(\"\")\n\n    if tok == tk_EOI:\n        break"}
{"title": "Compiler/syntax analyzer", "language": "Python", "task": "The C and Python versions can be considered reference implementations.\n\n;Related Tasks\n\n* Lexical Analyzer task\n* Code Generator task\n* Virtual Machine Interpreter task\n* AST Interpreter task\n\n\n__TOC__\n\n", "solution": "from __future__ import print_function\nimport sys, shlex, operator\n\ntk_EOI, tk_Mul, tk_Div, tk_Mod, tk_Add, tk_Sub, tk_Negate, tk_Not, tk_Lss, tk_Leq, tk_Gtr, \\\ntk_Geq, tk_Eql, tk_Neq, tk_Assign, tk_And, tk_Or, tk_If, tk_Else, tk_While, tk_Print,      \\\ntk_Putc, tk_Lparen, tk_Rparen, tk_Lbrace, tk_Rbrace, tk_Semi, tk_Comma, tk_Ident,          \\\ntk_Integer, tk_String = range(31)\n\nnd_Ident, nd_String, nd_Integer, nd_Sequence, nd_If, nd_Prtc, nd_Prts, nd_Prti, nd_While, \\\nnd_Assign, nd_Negate, nd_Not, nd_Mul, nd_Div, nd_Mod, nd_Add, nd_Sub, nd_Lss, nd_Leq,     \\\nnd_Gtr, nd_Geq, nd_Eql, nd_Neq, nd_And, nd_Or = range(25)\n\n# must have same order as above\nTokens = [\n    [\"EOI\"             , False, False, False, -1, -1        ],\n    [\"*\"               , False, True,  False, 13, nd_Mul    ],\n    [\"/\"               , False, True,  False, 13, nd_Div    ],\n    [\"%\"               , False, True,  False, 13, nd_Mod    ],\n    [\"+\"               , False, True,  False, 12, nd_Add    ],\n    [\"-\"               , False, True,  False, 12, nd_Sub    ],\n    [\"-\"               , False, False, True,  14, nd_Negate ],\n    [\"!\"               , False, False, True,  14, nd_Not    ],\n    [\"<\"               , False, True,  False, 10, nd_Lss    ],\n    [\"<=\"              , False, True,  False, 10, nd_Leq    ],\n    [\">\"               , False, True,  False, 10, nd_Gtr    ],\n    [\">=\"              , False, True,  False, 10, nd_Geq    ],\n    [\"==\"              , False, True,  False,  9, nd_Eql    ],\n    [\"!=\"              , False, True,  False,  9, nd_Neq    ],\n    [\"=\"               , False, False, False, -1, nd_Assign ],\n    [\"&&\"              , False, True,  False,  5, nd_And    ],\n    [\"||\"              , False, True,  False,  4, nd_Or     ],\n    [\"if\"              , False, False, False, -1, nd_If     ],\n    [\"else\"            , False, False, False, -1, -1        ],\n    [\"while\"           , False, False, False, -1, nd_While  ],\n    [\"print\"           , False, False, False, -1, -1        ],\n    [\"putc\"            , False, False, False, -1, -1        ],\n    [\"(\"               , False, False, False, -1, -1        ],\n    [\")\"               , False, False, False, -1, -1        ],\n    [\"{\"               , False, False, False, -1, -1        ],\n    [\"}\"               , False, False, False, -1, -1        ],\n    [\";\"               , False, False, False, -1, -1        ],\n    [\",\"               , False, False, False, -1, -1        ],\n    [\"Ident\"           , False, False, False, -1, nd_Ident  ],\n    [\"Integer literal\" , False, False, False, -1, nd_Integer],\n    [\"String literal\"  , False, False, False, -1, nd_String ]\n    ]\n\nall_syms = {\"End_of_input\"   : tk_EOI,     \"Op_multiply\"    : tk_Mul,\n            \"Op_divide\"      : tk_Div,     \"Op_mod\"         : tk_Mod,\n            \"Op_add\"         : tk_Add,     \"Op_subtract\"    : tk_Sub,\n            \"Op_negate\"      : tk_Negate,  \"Op_not\"         : tk_Not,\n            \"Op_less\"        : tk_Lss,     \"Op_lessequal\"   : tk_Leq,\n            \"Op_greater\"     : tk_Gtr,     \"Op_greaterequal\": tk_Geq,\n            \"Op_equal\"       : tk_Eql,     \"Op_notequal\"    : tk_Neq,\n            \"Op_assign\"      : tk_Assign,  \"Op_and\"         : tk_And,\n            \"Op_or\"          : tk_Or,      \"Keyword_if\"     : tk_If,\n            \"Keyword_else\"   : tk_Else,    \"Keyword_while\"  : tk_While,\n            \"Keyword_print\"  : tk_Print,   \"Keyword_putc\"   : tk_Putc,\n            \"LeftParen\"      : tk_Lparen,  \"RightParen\"     : tk_Rparen,\n            \"LeftBrace\"      : tk_Lbrace,  \"RightBrace\"     : tk_Rbrace,\n            \"Semicolon\"      : tk_Semi,    \"Comma\"          : tk_Comma,\n            \"Identifier\"     : tk_Ident,   \"Integer\"        : tk_Integer,\n            \"String\"         : tk_String}\n\nDisplay_nodes = [\"Identifier\", \"String\", \"Integer\", \"Sequence\", \"If\", \"Prtc\", \"Prts\",\n    \"Prti\", \"While\", \"Assign\", \"Negate\", \"Not\", \"Multiply\", \"Divide\", \"Mod\", \"Add\",\n    \"Subtract\", \"Less\", \"LessEqual\", \"Greater\", \"GreaterEqual\", \"Equal\", \"NotEqual\",\n    \"And\", \"Or\"]\n\nTK_NAME         = 0\nTK_RIGHT_ASSOC  = 1\nTK_IS_BINARY    = 2\nTK_IS_UNARY     = 3\nTK_PRECEDENCE   = 4\nTK_NODE         = 5\n\ninput_file = None\nerr_line   = None\nerr_col    = None\ntok        = None\ntok_text   = None\n\n#*** show error and exit\ndef error(msg):\n    print(\"(%d, %d) %s\" % (int(err_line), int(err_col), msg))\n    exit(1)\n\n#***\ndef gettok():\n    global err_line, err_col, tok, tok_text, tok_other\n    line = input_file.readline()\n    if len(line) == 0:\n        error(\"empty line\")\n\n    line_list = shlex.split(line, False, False)\n    # line col Ident var_name\n    # 0    1   2     3\n\n    err_line = line_list[0]\n    err_col  = line_list[1]\n    tok_text = line_list[2]\n\n    tok = all_syms.get(tok_text)\n    if tok == None:\n        error(\"Unknown token %s\" % (tok_text))\n\n    tok_other = None\n    if tok in [tk_Integer, tk_Ident, tk_String]:\n        tok_other = line_list[3]\n\nclass Node:\n    def __init__(self, node_type, left = None, right = None, value = None):\n        self.node_type  = node_type\n        self.left  = left\n        self.right = right\n        self.value = value\n\n#***\ndef make_node(oper, left, right = None):\n    return Node(oper, left, right)\n\n#***\ndef make_leaf(oper, n):\n    return Node(oper, value = n)\n\n#***\ndef expect(msg, s):\n    if tok == s:\n        gettok()\n        return\n    error(\"%s: Expecting '%s', found '%s'\" % (msg, Tokens[s][TK_NAME], Tokens[tok][TK_NAME]))\n\n#***\ndef expr(p):\n    x = None\n\n    if tok == tk_Lparen:\n        x = paren_expr()\n    elif tok in [tk_Sub, tk_Add]:\n        op = (tk_Negate if tok == tk_Sub else tk_Add)\n        gettok()\n        node = expr(Tokens[tk_Negate][TK_PRECEDENCE])\n        x = (make_node(nd_Negate, node) if op == tk_Negate else node)\n    elif tok == tk_Not:\n        gettok()\n        x = make_node(nd_Not, expr(Tokens[tk_Not][TK_PRECEDENCE]))\n    elif tok == tk_Ident:\n        x = make_leaf(nd_Ident, tok_other)\n        gettok()\n    elif tok == tk_Integer:\n        x = make_leaf(nd_Integer, tok_other)\n        gettok()\n    else:\n        error(\"Expecting a primary, found: %s\" % (Tokens[tok][TK_NAME]))\n\n    while Tokens[tok][TK_IS_BINARY] and Tokens[tok][TK_PRECEDENCE] >= p:\n        op = tok\n        gettok()\n        q = Tokens[op][TK_PRECEDENCE]\n        if not Tokens[op][TK_RIGHT_ASSOC]:\n            q += 1\n\n        node = expr(q)\n        x = make_node(Tokens[op][TK_NODE], x, node)\n\n    return x\n\n#***\ndef paren_expr():\n    expect(\"paren_expr\", tk_Lparen)\n    node = expr(0)\n    expect(\"paren_expr\", tk_Rparen)\n    return node\n\n#***\ndef stmt():\n    t = None\n\n    if tok == tk_If:\n        gettok()\n        e = paren_expr()\n        s = stmt()\n        s2 = None\n        if tok == tk_Else:\n            gettok()\n            s2 = stmt()\n        t = make_node(nd_If, e, make_node(nd_If, s, s2))\n    elif tok == tk_Putc:\n        gettok()\n        e = paren_expr()\n        t = make_node(nd_Prtc, e)\n        expect(\"Putc\", tk_Semi)\n    elif tok == tk_Print:\n        gettok()\n        expect(\"Print\", tk_Lparen)\n        while True:\n            if tok == tk_String:\n                e = make_node(nd_Prts, make_leaf(nd_String, tok_other))\n                gettok()\n            else:\n                e = make_node(nd_Prti, expr(0))\n\n            t = make_node(nd_Sequence, t, e)\n            if tok != tk_Comma:\n                break\n            gettok()\n        expect(\"Print\", tk_Rparen)\n        expect(\"Print\", tk_Semi)\n    elif tok == tk_Semi:\n        gettok()\n    elif tok == tk_Ident:\n        v = make_leaf(nd_Ident, tok_other)\n        gettok()\n        expect(\"assign\", tk_Assign)\n        e = expr(0)\n        t = make_node(nd_Assign, v, e)\n        expect(\"assign\", tk_Semi)\n    elif tok == tk_While:\n        gettok()\n        e = paren_expr()\n        s = stmt()\n        t = make_node(nd_While, e, s)\n    elif tok == tk_Lbrace:\n        gettok()\n        while tok != tk_Rbrace and tok != tk_EOI:\n            t = make_node(nd_Sequence, t, stmt())\n        expect(\"Lbrace\", tk_Rbrace)\n    elif tok == tk_EOI:\n        pass\n    else:\n        error(\"Expecting start of statement, found: %s\" % (Tokens[tok][TK_NAME]))\n\n    return t\n\n#***\ndef parse():\n    t = None\n    gettok()\n    while True:\n        t = make_node(nd_Sequence, t, stmt())\n        if tok == tk_EOI or t == None:\n            break\n    return t\n\ndef prt_ast(t):\n    if t == None:\n        print(\";\")\n    else:\n        print(\"%-14s\" % (Display_nodes[t.node_type]), end='')\n        if t.node_type in [nd_Ident, nd_Integer]:\n            print(\"%s\" % (t.value))\n        elif t.node_type == nd_String:\n            print(\"%s\" %(t.value))\n        else:\n            print(\"\")\n            prt_ast(t.left)\n            prt_ast(t.right)\n\n#*** main driver\ninput_file = sys.stdin\nif len(sys.argv) > 1:\n    try:\n        input_file = open(sys.argv[1], \"r\", 4096)\n    except IOError as e:\n        error(0, 0, \"Can't open %s\" % sys.argv[1])\nt = parse()\nprt_ast(t)"}
{"title": "Compiler/virtual machine interpreter", "language": "Python", "task": "The C and Python versions can be considered reference implementations.\n\n;Related Tasks\n\n* Lexical Analyzer task\n* Syntax Analyzer task\n* Code Generator task\n* AST Interpreter task\n\n\n__TOC__\n\n", "solution": "from __future__ import print_function\nimport sys, struct\n\nFETCH, STORE, PUSH, ADD, SUB, MUL, DIV, MOD, LT, GT, LE, GE, EQ, NE, AND, OR, NEG, NOT, \\\nJMP, JZ, PRTC, PRTS, PRTI, HALT = range(24)\n\ncode_map = {\n    \"fetch\": FETCH,\n    \"store\": STORE,\n    \"push\":  PUSH,\n    \"add\":   ADD,\n    \"sub\":   SUB,\n    \"mul\":   MUL,\n    \"div\":   DIV,\n    \"mod\":   MOD,\n    \"lt\":    LT,\n    \"gt\":    GT,\n    \"le\":    LE,\n    \"ge\":    GE,\n    \"eq\":    EQ,\n    \"ne\":    NE,\n    \"and\":   AND,\n    \"or\":    OR,\n    \"not\":   NOT,\n    \"neg\":   NEG,\n    \"jmp\":   JMP,\n    \"jz\":    JZ,\n    \"prtc\":  PRTC,\n    \"prts\":  PRTS,\n    \"prti\":  PRTI,\n    \"halt\":  HALT\n}\n\ninput_file  = None\ncode        = bytearray()\nstring_pool = []\nword_size   = 4\n\n#*** show error and exit\ndef error(msg):\n    print(\"%s\" % (msg))\n    exit(1)\n\ndef int_to_bytes(val):\n    return struct.pack(\"<i\", val)\n\ndef bytes_to_int(bstr):\n    return struct.unpack(\"<i\", bstr)\n\n#***\ndef emit_byte(x):\n    code.append(x)\n\n#***\ndef emit_word(x):\n    s = int_to_bytes(x)\n    for x in s:\n        code.append(x)\n\n#***\ndef run_vm(data_size):\n    stack = [0 for i in range(data_size + 1)]\n    pc = 0\n    while True:\n        op = code[pc]\n        pc += 1\n\n        if op == FETCH:\n            stack.append(stack[bytes_to_int(code[pc:pc+word_size])[0]]);\n            pc += word_size\n        elif op == STORE:\n            stack[bytes_to_int(code[pc:pc+word_size])[0]] = stack.pop();\n            pc += word_size\n        elif op == PUSH:\n            stack.append(bytes_to_int(code[pc:pc+word_size])[0]);\n            pc += word_size\n        elif op == ADD:   stack[-2] += stack[-1]; stack.pop()\n        elif op == SUB:   stack[-2] -= stack[-1]; stack.pop()\n        elif op == MUL:   stack[-2] *= stack[-1]; stack.pop()\n        # use C like division semantics\n        elif op == DIV:   stack[-2] = int(float(stack[-2]) / stack[-1]); stack.pop()\n        elif op == MOD:   stack[-2] = int(float(stack[-2]) % stack[-1]); stack.pop()\n        elif op == LT:    stack[-2] = stack[-2] <  stack[-1]; stack.pop()\n        elif op == GT:    stack[-2] = stack[-2] >  stack[-1]; stack.pop()\n        elif op == LE:    stack[-2] = stack[-2] <= stack[-1]; stack.pop()\n        elif op == GE:    stack[-2] = stack[-2] >= stack[-1]; stack.pop()\n        elif op == EQ:    stack[-2] = stack[-2] == stack[-1]; stack.pop()\n        elif op == NE:    stack[-2] = stack[-2] != stack[-1]; stack.pop()\n        elif op == AND:   stack[-2] = stack[-2] and stack[-1]; stack.pop()\n        elif op == OR:    stack[-2] = stack[-2] or  stack[-1]; stack.pop()\n        elif op == NEG:   stack[-1] = -stack[-1]\n        elif op == NOT:   stack[-1] = not stack[-1]\n        elif op == JMP:   pc += bytes_to_int(code[pc:pc+word_size])[0]\n        elif op == JZ:\n            if stack.pop():\n                pc += word_size\n            else:\n                pc += bytes_to_int(code[pc:pc+word_size])[0]\n        elif op == PRTC:  print(\"%c\" % (stack[-1]), end=''); stack.pop()\n        elif op == PRTS:  print(\"%s\" % (string_pool[stack[-1]]), end=''); stack.pop()\n        elif op == PRTI:  print(\"%d\" % (stack[-1]), end=''); stack.pop()\n        elif op == HALT:  break\n\ndef str_trans(srce):\n    dest = \"\"\n    i = 0\n    while i < len(srce):\n        if srce[i] == '\\\\' and i + 1 < len(srce):\n            if srce[i + 1] == 'n':\n                dest += '\\n'\n                i += 2\n            elif srce[i + 1] == '\\\\':\n                dest += '\\\\'\n                i += 2\n        else:\n            dest += srce[i]\n            i += 1\n\n    return dest\n\n#***\ndef load_code():\n    global string_pool\n\n    line = input_file.readline()\n    if len(line) == 0:\n        error(\"empty line\")\n\n    line_list = line.split()\n    data_size = int(line_list[1])\n    n_strings = int(line_list[3])\n\n    for i in range(n_strings):\n        string_pool.append(str_trans(input_file.readline().strip('\"\\n')))\n\n    while True:\n        line = input_file.readline()\n        if len(line) == 0:\n            break\n        line_list = line.split()\n        offset = int(line_list[0])\n        instr  = line_list[1]\n        opcode = code_map.get(instr)\n        if opcode == None:\n            error(\"Unknown instruction %s at %d\" % (instr, offset))\n        emit_byte(opcode)\n        if opcode in [JMP, JZ]:\n            p = int(line_list[3])\n            emit_word(p - (offset + 1))\n        elif opcode == PUSH:\n            value = int(line_list[2])\n            emit_word(value)\n        elif opcode in [FETCH, STORE]:\n            value = int(line_list[2].strip('[]'))\n            emit_word(value)\n\n    return data_size\n\n#*** main driver\ninput_file = sys.stdin\nif len(sys.argv) > 1:\n    try:\n        input_file = open(sys.argv[1], \"r\", 4096)\n    except IOError as e:\n        error(0, 0, \"Can't open %s\" % sys.argv[1])\n\ndata_size = load_code()\nrun_vm(data_size)"}
{"title": "Composite numbers k with no single digit factors whose factors are all substrings of k", "language": "Python", "task": "Find the composite numbers '''k''' in base 10, that have no single digit prime factors and whose prime factors are all a substring of '''k'''.\n\n\n;Task\n\n* Find and show here, on this page, the first ten elements of the sequence.\n\n\n\n;Stretch\n\n* Find and show the next ten elements.\n\n\n\n\n", "solution": "from sympy import isprime, factorint\n\ndef contains_its_prime_factors_all_over_7(n):\n    if n < 10 or isprime(n):\n        return False\n    strn = str(n)\n    pfacs = factorint(n).keys()\n    return all(f > 9 and str(f) in strn for f in pfacs)\n\nfound = 0\nfor n in range(1_000_000_000):\n    if contains_its_prime_factors_all_over_7(n):\n        found += 1\n        print(f'{n: 12,}', end = '\\n' if found % 10 == 0 else '')\n        if found == 20:\n            break\n"}
{"title": "Conjugate transpose", "language": "Python", "task": "Suppose that a conjugate transpose of M is a matrix M^H containing the [[complex conjugate]]s of the [[matrix transposition]] of M.\n\n::: (M^H)_{ji} = \\overline{M_{ij}}\n\n\nThis means that   row j, column i of the conjugate transpose equals the \ncomplex conjugate of row i, column j of the original matrix.\n\n\nIn the next list, M must also be a square matrix.\n\n* A Hermitian matrix equals its own conjugate transpose: M^H = M.\n* A multiplication with its conjugate transpose: M^HM = MM^H.\n* A iff M^HM = I_n and iff MM^H = I_n, where I_n is the identity matrix.\n\n\n\n;Task:\nGiven some matrix of complex numbers, find its conjugate transpose. \n\nAlso determine if the matrix is a:\n::* Hermitian matrix,\n::* normal matrix,  or \n::* unitary matrix.\n\n\n;See also:\n* MathWorld entry: conjugate transpose\n* MathWorld entry: Hermitian matrix\n* MathWorld entry: normal matrix\n* MathWorld entry: unitary matrix\n\n", "solution": "def conjugate_transpose(m):\n    return tuple(tuple(n.conjugate() for n in row) for row in zip(*m))\n\ndef mmul( ma, mb):\n    return tuple(tuple(sum( ea*eb for ea,eb in zip(a,b)) for b in zip(*mb)) for a in ma)\n\ndef mi(size):\n    'Complex Identity matrix'\n    sz = range(size)\n    m = [[0 + 0j for i in sz] for j in sz]\n    for i in range(size):\n        m[i][i] = 1 + 0j\n    return tuple(tuple(row) for row in m)\n\ndef __allsame(vector):\n    first, rest = vector[0], vector[1:]\n    return all(i == first for i in rest)\n\ndef __allnearsame(vector, eps=1e-14):\n    first, rest = vector[0], vector[1:]\n    return all(abs(first.real - i.real) < eps and abs(first.imag - i.imag) < eps\n               for i in rest)\n\ndef isequal(matrices, eps=1e-14):\n    'Check any number of matrices for equality within eps'\n    x = [len(m) for m in matrices]\n    if not __allsame(x): return False\n    y = [len(m[0]) for m in matrices]\n    if not __allsame(y): return False\n    for s in range(x[0]):\n        for t in range(y[0]):\n            if not __allnearsame([m[s][t] for m in matrices], eps): return False\n    return True\n    \n\ndef ishermitian(m, ct):\n    return isequal([m, ct])\n\ndef isnormal(m, ct):\n    return isequal([mmul(m, ct), mmul(ct, m)])\n\ndef isunitary(m, ct):\n    mct, ctm = mmul(m, ct), mmul(ct, m)\n    mctx, mcty, cmx, ctmy = len(mct), len(mct[0]), len(ctm), len(ctm[0])\n    ident = mi(mctx)\n    return isequal([mct, ctm, ident])\n\ndef printm(comment, m):\n    print(comment)\n    fields = [['%g%+gj' % (f.real, f.imag) for f in row] for row in m]\n    width = max(max(len(f) for f in row) for row in fields)\n    lines = (', '.join('%*s' % (width, f) for f in row) for row in fields)\n    print('\\n'.join(lines))\n\nif __name__ == '__main__':\n    for matrix in [\n            ((( 3.000+0.000j), (+2.000+1.000j)), \n            (( 2.000-1.000j), (+1.000+0.000j))),\n\n            ((( 1.000+0.000j), (+1.000+0.000j), (+0.000+0.000j)), \n            (( 0.000+0.000j), (+1.000+0.000j), (+1.000+0.000j)), \n            (( 1.000+0.000j), (+0.000+0.000j), (+1.000+0.000j))),\n\n            ((( 2**0.5/2+0.000j), (+2**0.5/2+0.000j), (+0.000+0.000j)), \n            (( 0.000+2**0.5/2j), (+0.000-2**0.5/2j), (+0.000+0.000j)), \n            (( 0.000+0.000j), (+0.000+0.000j), (+0.000+1.000j)))]:\n        printm('\\nMatrix:', matrix)\n        ct = conjugate_transpose(matrix)\n        printm('Its conjugate transpose:', ct)\n        print('Hermitian? %s.' % ishermitian(matrix, ct))\n        print('Normal?    %s.' % isnormal(matrix, ct))\n        print('Unitary?   %s.' % isunitary(matrix, ct))"}
{"title": "Continued fraction", "language": "Python 2.6+ and 3.x", "task": "A number may be represented as a continued fraction (see Mathworld for more information) as follows:\n:a_0 + \\cfrac{b_1}{a_1 + \\cfrac{b_2}{a_2 + \\cfrac{b_3}{a_3 + \\ddots}}}\n\nThe task is to write a program which generates such a number and prints a real representation of it. The code should be tested by calculating and printing the square root of 2, Napier's Constant, and Pi, using the following coefficients:\n\nFor the square root of 2, use a_0 = 1 then a_N = 2. b_N is always 1.\n\n:\\sqrt{2} = 1 + \\cfrac{1}{2 + \\cfrac{1}{2 + \\cfrac{1}{2 + \\ddots}}}\n\nFor Napier's Constant, use a_0 = 2, then a_N = N. b_1 = 1 then b_N = N-1.\n\n:e = 2 + \\cfrac{1}{1 + \\cfrac{1}{2 + \\cfrac{2}{3 + \\cfrac{3}{4 + \\ddots}}}}\n\nFor Pi, use a_0 = 3 then a_N = 6. b_N = (2N-1)^2.\n\n:\\pi = 3 + \\cfrac{1}{6 + \\cfrac{9}{6 + \\cfrac{25}{6 + \\ddots}}}\n\n\n;See also:\n:*   [[Continued fraction/Arithmetic]] for tasks that do arithmetic over continued fractions.\n\n", "solution": "from fractions import Fraction\nimport itertools\ntry: zip = itertools.izip\nexcept: pass\n \n# The Continued Fraction\ndef CF(a, b, t):\n  terms = list(itertools.islice(zip(a, b), t))\n  z = Fraction(1,1)\n  for a, b in reversed(terms):\n    z = a + b / z\n  return z\n \n# Approximates a fraction to a string\ndef pRes(x, d):\n  q, x = divmod(x, 1)\n  res = str(q)\n  res += \".\"\n  for i in range(d):\n    x *= 10\n    q, x = divmod(x, 1)\n    res += str(q)\n  return res\n \n# Test the Continued Fraction for sqrt2\ndef sqrt2_a():\n  yield 1\n  for x in itertools.repeat(2):\n    yield x\n \ndef sqrt2_b():\n  for x in itertools.repeat(1):\n    yield x\n \ncf = CF(sqrt2_a(), sqrt2_b(), 950)\nprint(pRes(cf, 200))\n#1.41421356237309504880168872420969807856967187537694807317667973799073247846210703885038753432764157273501384623091229702492483605585073721264412149709993583141322266592750559275579995050115278206057147\n \n \n# Test the Continued Fraction for Napier's Constant\ndef Napier_a():\n  yield 2\n  for x in itertools.count(1):\n    yield x\n \ndef Napier_b():\n  yield 1\n  for x in itertools.count(1):\n    yield x\n \ncf = CF(Napier_a(), Napier_b(), 950)\nprint(pRes(cf, 200))\n#2.71828182845904523536028747135266249775724709369995957496696762772407663035354759457138217852516642742746639193200305992181741359662904357290033429526059563073813232862794349076323382988075319525101901\n \n# Test the Continued Fraction for Pi\ndef Pi_a():\n  yield 3\n  for x in itertools.repeat(6):\n    yield x\n \ndef Pi_b():\n  for x in itertools.count(1,2):\n    yield x*x\n \ncf = CF(Pi_a(), Pi_b(), 950)\nprint(pRes(cf, 10))\n#3.1415926532"}
{"title": "Continued fraction", "language": "Python from D", "task": "A number may be represented as a continued fraction (see Mathworld for more information) as follows:\n:a_0 + \\cfrac{b_1}{a_1 + \\cfrac{b_2}{a_2 + \\cfrac{b_3}{a_3 + \\ddots}}}\n\nThe task is to write a program which generates such a number and prints a real representation of it. The code should be tested by calculating and printing the square root of 2, Napier's Constant, and Pi, using the following coefficients:\n\nFor the square root of 2, use a_0 = 1 then a_N = 2. b_N is always 1.\n\n:\\sqrt{2} = 1 + \\cfrac{1}{2 + \\cfrac{1}{2 + \\cfrac{1}{2 + \\ddots}}}\n\nFor Napier's Constant, use a_0 = 2, then a_N = N. b_1 = 1 then b_N = N-1.\n\n:e = 2 + \\cfrac{1}{1 + \\cfrac{1}{2 + \\cfrac{2}{3 + \\cfrac{3}{4 + \\ddots}}}}\n\nFor Pi, use a_0 = 3 then a_N = 6. b_N = (2N-1)^2.\n\n:\\pi = 3 + \\cfrac{1}{6 + \\cfrac{9}{6 + \\cfrac{25}{6 + \\ddots}}}\n\n\n;See also:\n:*   [[Continued fraction/Arithmetic]] for tasks that do arithmetic over continued fractions.\n\n", "solution": "from decimal import Decimal, getcontext\n\ndef calc(fun, n):\n    temp = Decimal(\"0.0\")\n\n    for ni in xrange(n+1, 0, -1):\n        (a, b) = fun(ni)\n        temp = Decimal(b) / (a + temp)\n\n    return fun(0)[0] + temp\n\ndef fsqrt2(n):\n    return (2 if n > 0 else 1, 1)\n\ndef fnapier(n):\n    return (n if n > 0 else 2, (n - 1) if n > 1 else 1)\n\ndef fpi(n):\n    return (6 if n > 0 else 3, (2 * n - 1) ** 2)\n\ngetcontext().prec = 50\nprint calc(fsqrt2, 200)\nprint calc(fnapier, 200)\nprint calc(fpi, 200)"}
{"title": "Continued fraction/Arithmetic/Construct from rational number", "language": "Python from Ruby", "task": "To understand this task in context please see [[Continued fraction arithmetic]]\nThe purpose of this task is to write a function \\mathit{r2cf}(\\mathrm{int} N_1, \\mathrm{int} N_2), or \\mathit{r2cf}(\\mathrm{Fraction} N), which will output a continued fraction assuming:\n:N_1 is the numerator\n:N_2 is the denominator\n  \nThe function should output its results one digit at a time each time it is called, in a manner sometimes described as lazy evaluation.\n\nTo achieve this it must determine: the integer part; and remainder part, of N_1 divided by N_2. It then sets N_1 to N_2 and N_2 to the determined remainder part. It then outputs the determined integer part. It does this until \\mathrm{abs}(N_2) is zero.\n\nDemonstrate the function by outputing the continued fraction for:\n: 1/2\n: 3\n: 23/8\n: 13/11\n: 22/7\n: -151/77\n\\sqrt 2 should approach [1; 2, 2, 2, 2, \\ldots] try ever closer rational approximations until boredom gets the better of you:\n: 14142,10000\n: 141421,100000\n: 1414214,1000000\n: 14142136,10000000\n\nTry :\n: 31,10\n: 314,100\n: 3142,1000\n: 31428,10000\n: 314285,100000\n: 3142857,1000000\n: 31428571,10000000\n: 314285714,100000000\n\nObserve how this rational number behaves differently to \\sqrt 2 and convince yourself that, in the same way as 3.7 may be represented as 3.70 when an extra decimal place is required, [3;7] may be represented as [3;7,\\infty] when an extra term is required.\n\n", "solution": "def r2cf(n1,n2):\n  while n2:\n    n1, (t1, n2) = n2, divmod(n1, n2)\n    yield t1\n\nprint(list(r2cf(1,2)))    # => [0, 2]\nprint(list(r2cf(3,1)))    # => [3]\nprint(list(r2cf(23,8)))    # => [2, 1, 7]\nprint(list(r2cf(13,11)))    # => [1, 5, 2]\nprint(list(r2cf(22,7)))    # => [3, 7]\nprint(list(r2cf(14142,10000)))    # => [1, 2, 2, 2, 2, 2, 1, 1, 29]\nprint(list(r2cf(141421,100000)))    # => [1, 2, 2, 2, 2, 2, 2, 3, 1, 1, 3, 1, 7, 2]\nprint(list(r2cf(1414214,1000000)))    # => [1, 2, 2, 2, 2, 2, 2, 2, 3, 6, 1, 2, 1, 12]\nprint(list(r2cf(14142136,10000000)))    # => [1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 6, 1, 2, 4, 1, 1, 2]"}
{"title": "Continued fraction/Arithmetic/G(matrix ng, continued fraction n)", "language": "Python from Ruby", "task": "This task investigates mathmatical operations that can be performed on a single continued fraction. This requires only a baby version of NG:\n: \\begin{bmatrix}\n  a_1 & a \\\\\n  b_1 & b \n\\end{bmatrix}\nI may perform perform the following operations:\n:Input the next term of N1\n:Output a term of the continued fraction resulting from the operation.\n\nI output a term if the integer parts of \\frac{a}{b} and \\frac{a_1}{b_1} are equal. Otherwise I input a term from N. If I need a term from N but N has no more terms I inject \\infty.\n\nWhen I input a term t my internal state: \\begin{bmatrix}\n  a_1 & a \\\\\n  b_1 & b \n\\end{bmatrix} is transposed thus \\begin{bmatrix} a + a_1 * t & a_1  \\\\\n  b + b_1 * t & b_1 \n\\end{bmatrix}\n\nWhen I output a term t my internal state: \\begin{bmatrix}\n  a_1 & a \\\\\n  b_1 & b \n\\end{bmatrix} is transposed thus \\begin{bmatrix} b_1 & b  \\\\\n  a_1 - b_1 * t & a - b * t \n\\end{bmatrix}\n\nWhen I need a term t but there are no more my internal state: \\begin{bmatrix}\n  a_1 & a \\\\\n  b_1 & b \n\\end{bmatrix} is transposed thus \\begin{bmatrix} a_1 & a_1 \\\\\n  b_1 & b_1 \n\\end{bmatrix}\n\nI am done when b1 and b are zero.\n\nDemonstrate your solution by calculating:\n:[1;5,2] + 1/2\n:[3;7] + 1/2   \n:[3;7] divided by 4\nUsing a generator for \\sqrt{2} (e.g., from [[Continued fraction]]) calculate \\frac{1}{\\sqrt{2}}. You are now at the starting line for using Continued Fractions to implement [[Arithmetic-geometric mean]] without ulps and epsilons.\n\nThe first step in implementing [[Arithmetic-geometric mean]] is to calculate \\frac{1 + \\frac{1}{\\sqrt{2}}}{2} do this now to cross the starting line and begin the race.\n\n", "solution": "class NG:\n  def __init__(self, a1, a, b1, b):\n    self.a1, self.a, self.b1, self.b = a1, a, b1, b\n\n  def ingress(self, n):\n    self.a, self.a1 = self.a1, self.a + self.a1 * n\n    self.b, self.b1 = self.b1, self.b + self.b1 * n\n\n  @property\n  def needterm(self):\n    return (self.b == 0 or self.b1 == 0) or not self.a//self.b == self.a1//self.b1\n\n  @property\n  def egress(self):\n    n = self.a // self.b\n    self.a,  self.b  = self.b,  self.a  - self.b  * n\n    self.a1, self.b1 = self.b1, self.a1 - self.b1 * n\n    return n\n\n  @property\n  def egress_done(self):\n    if self.needterm: self.a, self.b = self.a1, self.b1\n    return self.egress\n\n  @property\n  def done(self):\n    return self.b == 0 and self.b1 == 0"}
{"title": "Convert decimal number to rational", "language": "Python 2.6+", "task": "The task is to write a program to transform a decimal number into a fraction in lowest terms.\n\nIt is not always possible to do this exactly. For instance, while rational numbers can be converted to decimal representation, some of them need an infinite number of digits to be represented exactly in decimal form. Namely, repeating decimals such as 1/3 = 0.333...\n\nBecause of this, the following fractions cannot be obtained (reliably) unless the language has some way of representing repeating decimals:\n* 67 / 74 = 0.9(054) = 0.9054054...\n* 14 / 27 = 0.(518)  = 0.518518...\n\nAcceptable output:\n\n* 0.9054054 - 4527027 / 5000000\n* 0.518518  - 259259 / 500000\n\nFinite decimals are of course no problem:\n\n* 0.75      - 3 / 4 \n\n", "solution": ">>> from fractions import Fraction\n>>> for d in (0.9054054, 0.518518, 0.75): print(d, Fraction.from_float(d).limit_denominator(100))\n\n0.9054054 67/74\n0.518518 14/27\n0.75 3/4\n>>> for d in '0.9054054 0.518518 0.75'.split(): print(d, Fraction(d))\n\n0.9054054 4527027/5000000\n0.518518 259259/500000\n0.75 3/4\n>>> "}
{"title": "Convert decimal number to rational", "language": "Python 3.7", "task": "The task is to write a program to transform a decimal number into a fraction in lowest terms.\n\nIt is not always possible to do this exactly. For instance, while rational numbers can be converted to decimal representation, some of them need an infinite number of digits to be represented exactly in decimal form. Namely, repeating decimals such as 1/3 = 0.333...\n\nBecause of this, the following fractions cannot be obtained (reliably) unless the language has some way of representing repeating decimals:\n* 67 / 74 = 0.9(054) = 0.9054054...\n* 14 / 27 = 0.(518)  = 0.518518...\n\nAcceptable output:\n\n* 0.9054054 - 4527027 / 5000000\n* 0.518518  - 259259 / 500000\n\nFinite decimals are of course no problem:\n\n* 0.75      - 3 / 4 \n\n", "solution": "'''Approximate rationals from decimals'''\n\nfrom math import (floor, gcd)\nimport sys\n\n\n# approxRatio :: Float -> Float -> Ratio\ndef approxRatio(epsilon):\n    '''The simplest rational approximation to\n       n within the margin given by epsilon.\n    '''\n    def gcde(e, x, y):\n        def _gcd(a, b):\n            return a if b < e else _gcd(b, a % b)\n        return _gcd(abs(x), abs(y))\n    return lambda n: (lambda c=(\n        gcde(epsilon if 0 < epsilon else (0.0001), 1, n)\n    ): ratio(floor(n / c))(floor(1 / c)))()\n\n\n# main :: IO ()\ndef main():\n    '''Conversions at different levels of precision.'''\n\n    xs = [0.9054054, 0.518518, 0.75]\n    print(\n        fTable(__doc__ + ' (epsilon of 1/10000):\\n')(str)(\n            lambda r: showRatio(r) + ' -> ' + repr(fromRatio(r))\n        )(\n            approxRatio(1 / 10000)\n        )(xs)\n    )\n    print('\\n')\n\n    e = minBound(float)\n    print(\n        fTable(__doc__ + ' (epsilon of ' + repr(e) + '):\\n')(str)(\n            lambda r: showRatio(r) + ' -> ' + repr(fromRatio(r))\n        )(\n            approxRatio(e)\n        )(xs)\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# fromRatio :: Ratio Int -> Float\ndef fromRatio(r):\n    '''A floating point value derived from a\n       a rational value.\n    '''\n    return r.get('numerator') / r.get('denominator')\n\n\n# minBound :: Bounded Type -> a\ndef minBound(t):\n    '''Minimum value for a bounded type.'''\n    maxsize = sys.maxsize\n    float_infomin = sys.float_info.min\n    return {\n        int: (-maxsize - 1),\n        float: float_infomin,\n        bool: False,\n        str: chr(0)\n    }[t]\n\n\n# ratio :: Int -> Int -> Ratio Int\ndef ratio(n):\n    '''Rational value constructed\n       from a numerator and a denominator.\n    '''\n    def go(n, d):\n        g = gcd(n, d)\n        return {\n            'type': 'Ratio',\n            'numerator': n // g, 'denominator': d // g\n        }\n    return lambda d: go(n * signum(d), abs(d))\n\n\n# showRatio :: Ratio -> String\ndef showRatio(r):\n    '''String representation of the ratio r.'''\n    d = r.get('denominator')\n    return str(r.get('numerator')) + (\n        ' / ' + str(d) if 1 != d else ''\n    )\n\n\n# signum :: Num -> Num\ndef signum(n):\n    '''The sign of n.'''\n    return -1 if 0 > n else (1 if 0 < n else 0)\n\n\n# DISPLAY -------------------------------------------------\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Convert seconds to compound duration", "language": "Python", "task": "Write a function or program which:\n*   takes a positive integer representing a duration in seconds as input (e.g., 100), and\n*   returns a string which shows the same duration decomposed into:\n:::*   weeks,\n:::*   days, \n:::*   hours, \n:::*   minutes,   and \n:::*   seconds.\n\nThis is detailed below (e.g., \"2 hr, 59 sec\").\n\n\nDemonstrate that it passes the following three test-cases:\n\n'''''Test Cases'''''\n\n:::::{| class=\"wikitable\"\n|-\n! input number\n! output string\n|-\n| 7259\n| 2 hr, 59 sec\n|-\n| 86400\n| 1 d\n|-\n| 6000000\n| 9 wk, 6 d, 10 hr, 40 min\n|}\n\n'''''Details'''''\n\nThe following five units should be used:\n:::::{| class=\"wikitable\"\n|-\n! unit\n! suffix used in output\n! conversion\n|-\n| week\n| wk\n| 1 week = 7 days\n|-\n| day\n| d\n| 1 day = 24 hours\n|-\n| hour\n| hr\n| 1 hour = 60 minutes\n|-\n| minute\n| min\n| 1 minute = 60 seconds\n|-\n| second\n| sec\n| \n|}\n\nHowever, '''only''' include quantities with non-zero values in the output (e.g., return \"1 d\" and not \"0 wk, 1 d, 0 hr, 0 min, 0 sec\").\n\nGive larger units precedence over smaller ones as much as possible (e.g., return 2 min, 10 sec and not 1 min, 70 sec or 130 sec)\n\nMimic the formatting shown in the test-cases (quantities sorted from largest unit to smallest and separated by comma+space; value and unit of each quantity separated by space).\n\n\n", "solution": "'''Compound duration'''\n\nfrom functools import reduce\nfrom itertools import chain\n\n\n# compoundDurationFromUnits :: [Num] -> [String] -> Num -> [(Num, String)]\ndef compoundDurationFromUnits(qs):\n    '''A list of compound string representions of a number n of time units,\n       in terms of the multiples given in qs, and the labels given in ks.\n    '''\n    return lambda ks: lambda n: list(\n        chain.from_iterable(map(\n            lambda v, k: [(v, k)] if 0 < v else [],\n            mapAccumR(\n                lambda a, x: divmod(a, x) if 0 < x else (1, a)\n            )(n)(qs)[1],\n            ks\n        ))\n    )\n\n\n# --------------------------TEST---------------------------\n# main :: IO ()\ndef main():\n    '''Tests of various durations, with a\n       particular set of units and labels.\n    '''\n\n    print(\n        fTable('Compound durations from numbers of seconds:\\n')(str)(\n            quoted(\"'\")\n        )(\n            lambda n: ', '.join([\n                str(v) + ' ' + k for v, k in\n                compoundDurationFromUnits([0, 7, 24, 60, 60])(\n                    ['wk', 'd', 'hr', 'min', 'sec']\n                )(n)\n            ])\n        )([7259, 86400, 6000000])\n    )\n\n\n# -------------------------GENERIC-------------------------\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# mapAccumR :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])\ndef mapAccumR(f):\n    '''A tuple of an accumulation and a list derived by a combined\n       map and fold, with accumulation from right to left.\n    '''\n    def go(a, x):\n        acc, y = f(a[0], x)\n        return (acc, [y] + a[1])\n    return lambda acc: lambda xs: (\n        reduce(go, reversed(xs), (acc, []))\n    )\n\n\n# quoted :: Char -> String -> String\ndef quoted(c):\n    '''A string flanked on both sides\n       by a specified quote character.\n    '''\n    return lambda s: c + s + c\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Copy stdin to stdout", "language": "Python", "task": "Create an executable file that copies stdin to stdout, or else a script that does so through the invocation of an interpreter at the command line.\n\n", "solution": "python -c 'import sys; sys.stdout.write(sys.stdin.read())'\n\n"}
{"title": "Count the coins", "language": "Python from Go", "task": "There are four types of common coins in   US   currency: \n:::#   quarters   (25 cents)\n:::#   dimes   (10 cents)\n:::#   nickels   (5 cents),   and \n:::#   pennies   (1 cent)  \n\n\nThere are six ways to make change for 15 cents:\n:::#   A dime and a nickel \n:::#   A dime and 5 pennies\n:::#   3 nickels\n:::#   2 nickels and 5 pennies\n:::#   A nickel and 10 pennies\n:::#   15 pennies\n\n\n;Task:\nHow many ways are there to make change for a dollar using these common coins?     (1 dollar = 100 cents).\n\n\n;Optional:\nLess common are dollar coins (100 cents);   and very rare are half dollars (50 cents).   With the addition of these two coins, how many ways are there to make change for $1000? \n\n(Note:   the answer is larger than   232).\n\n\n;References:\n* an algorithm from the book ''Structure and Interpretation of Computer Programs''.\n* an article in the algorithmist.\n* Change-making problem on Wikipedia.\n\n", "solution": "def changes(amount, coins):\n    ways = [0] * (amount + 1)\n    ways[0] = 1\n    for coin in coins:\n        for j in xrange(coin, amount + 1):\n            ways[j] += ways[j - coin]\n    return ways[amount]\n\nprint changes(100, [1, 5, 10, 25])\nprint changes(100000, [1, 5, 10, 25, 50, 100])"}
{"title": "Count the coins", "language": "Python from C", "task": "There are four types of common coins in   US   currency: \n:::#   quarters   (25 cents)\n:::#   dimes   (10 cents)\n:::#   nickels   (5 cents),   and \n:::#   pennies   (1 cent)  \n\n\nThere are six ways to make change for 15 cents:\n:::#   A dime and a nickel \n:::#   A dime and 5 pennies\n:::#   3 nickels\n:::#   2 nickels and 5 pennies\n:::#   A nickel and 10 pennies\n:::#   15 pennies\n\n\n;Task:\nHow many ways are there to make change for a dollar using these common coins?     (1 dollar = 100 cents).\n\n\n;Optional:\nLess common are dollar coins (100 cents);   and very rare are half dollars (50 cents).   With the addition of these two coins, how many ways are there to make change for $1000? \n\n(Note:   the answer is larger than   232).\n\n\n;References:\n* an algorithm from the book ''Structure and Interpretation of Computer Programs''.\n* an article in the algorithmist.\n* Change-making problem on Wikipedia.\n\n", "solution": "try:\n    import psyco\n    psyco.full()\nexcept ImportError:\n    pass\n\ndef count_changes(amount_cents, coins):\n    n = len(coins)\n    # max([]) instead of max() for Psyco\n    cycle = max([c+1 for c in coins if c <= amount_cents]) * n\n    table = [0] * cycle\n    for i in xrange(n):\n        table[i] = 1\n\n    pos = n\n    for s in xrange(1, amount_cents + 1):\n        for i in xrange(n):\n            if i == 0 and pos >= cycle:\n                pos = 0\n            if coins[i] <= s:\n                q = pos - coins[i] * n\n                table[pos]= table[q] if (q >= 0) else table[q + cycle]\n            if i:\n                table[pos] += table[pos - 1]\n            pos += 1\n    return table[pos - 1]\n\ndef main():\n    us_coins = [100, 50, 25, 10, 5, 1]\n    eu_coins = [200, 100, 50, 20, 10, 5, 2, 1]\n\n    for coins in (us_coins, eu_coins):\n        print count_changes(     100, coins[2:])\n        print count_changes(  100000, coins)\n        print count_changes( 1000000, coins)\n        print count_changes(10000000, coins), \"\\n\"\n\nmain()"}
{"title": "Create an HTML table", "language": "Python", "task": "Create an HTML table. \n* The table body should have at least three rows of three columns.\n* Each of these three columns should be labelled \"X\", \"Y\", and \"Z\". \n* An extra column should be added at either the extreme left or the extreme right of the table that has no heading, but is filled with sequential row numbers. \n* The rows of the \"X\", \"Y\", and \"Z\" columns should be filled with random or sequential integers having 4 digits or less. \n* The numbers should be aligned in the same fashion for all columns.\n\n", "solution": "import random\n\ndef rand9999():\n    return random.randint(1000, 9999)\n\ndef tag(attr='', **kwargs):\n    for tag, txt in kwargs.items():\n        return '<{tag}{attr}>{txt}</{tag}>'.format(**locals())\n\nif __name__ == '__main__':\n    header = tag(tr=''.join(tag(th=txt) for txt in ',X,Y,Z'.split(','))) + '\\n'\n    rows = '\\n'.join(tag(tr=tag(' style=\"font-weight: bold;\"', td=i)\n                                    + ''.join(tag(td=rand9999())\n                                              for j in range(3)))\n                     for i in range(1, 6))\n    table = tag(table='\\n' + header + rows + '\\n')\n    print(table)"}
{"title": "Create an HTML table", "language": "Python 3.6", "task": "Create an HTML table. \n* The table body should have at least three rows of three columns.\n* Each of these three columns should be labelled \"X\", \"Y\", and \"Z\". \n* An extra column should be added at either the extreme left or the extreme right of the table that has no heading, but is filled with sequential row numbers. \n* The rows of the \"X\", \"Y\", and \"Z\" columns should be filled with random or sequential integers having 4 digits or less. \n* The numbers should be aligned in the same fashion for all columns.\n\n", "solution": "from functools import (reduce)\nimport itertools\nimport random\n\n\n# HTML RENDERING ----------------------------------------\n\n# treeHTML :: tree\n#      {tag :: String, text :: String, kvs :: Dict}\n#      -> HTML String\ndef treeHTML(tree):\n    return foldTree(\n        lambda x: lambda xs: (\n            f\"<{x['tag'] + attribString(x)}>\" + (\n                str(x['text']) if 'text' in x else '\\n'\n            ) + ''.join(xs) + f\"</{x['tag']}>\\n\"\n        )\n    )(tree)\n\n\n# attribString :: Dict -> String\ndef attribString(dct):\n    kvs = dct['kvs'] if 'kvs' in dct else None\n    return ' ' + reduce(\n        lambda a, k: a + k + '=\"' + kvs[k] + '\" ',\n        kvs.keys(), ''\n    ).strip() if kvs else ''\n\n\n# HTML TABLE FROM GENERATED DATA ------------------------\n\n\ndef main():\n    # Number of columns and rows to generate.\n    n = 3\n\n    # Table details -------------------------------------\n    strCaption = 'Table generated with Python'\n    colNames = take(n)(enumFrom('A'))\n    dataRows = map(\n        lambda x: (x, map(\n            lambda _: random.randint(100, 9999),\n            colNames\n        )), take(n)(enumFrom(1)))\n    tableStyle = {\n        'style': \"width:25%; border:2px solid silver;\"\n    }\n    trStyle = {\n        'style': \"border:1px solid silver;text-align:right;\"\n    }\n\n    # TREE STRUCTURE OF TABLE ---------------------------\n    tableTree = Node({'tag': 'table', 'kvs': tableStyle})([\n        Node({\n            'tag': 'caption',\n            'text': strCaption\n        })([]),\n\n        # HEADER ROW --------------------------------\n        (Node({'tag': 'tr'})(\n            Node({\n                'tag': 'th',\n                'kvs': {'style': 'text-align:right;'},\n                'text': k\n            })([]) for k in ([''] + colNames)\n        ))\n    ] +\n        # DATA ROWS ---------------------------------\n        list(Node({'tag': 'tr', 'kvs': trStyle})(\n            [Node({'tag': 'th', 'text': tpl[0]})([])] +\n            list(Node(\n                {'tag': 'td', 'text': str(v)})([]) for v in tpl[1]\n            )\n        ) for tpl in dataRows)\n    )\n\n    print(\n        treeHTML(tableTree)\n        # dataRows\n    )\n\n\n# GENERIC -----------------------------------------------\n\n# Node :: a -> [Tree a] -> Tree a\ndef Node(v):\n    return lambda xs: {'type': 'Node', 'root': v, 'nest': xs}\n\n\n# enumFrom :: Enum a => a -> [a]\ndef enumFrom(x):\n    return itertools.count(x) if type(x) is int else (\n        map(chr, itertools.count(ord(x)))\n    )\n\n\n# foldTree :: (a -> [b] -> b) -> Tree a -> b\ndef foldTree(f):\n    def go(node):\n        return f(node['root'])(\n            list(map(go, node['nest']))\n        )\n    return lambda tree: go(tree)\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(itertools.islice(xs, n))\n    )\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Cullen and Woodall numbers", "language": "Python", "task": "A Cullen number is a number of the form '''n x 2n + 1''' where '''n''' is a natural number.\n\nA Woodall number is very similar. It is a number of the form '''n x 2n - 1''' where '''n''' is a natural number.\n\nSo for each '''n''' the associated Cullen number and Woodall number differ by 2.\n\n''Woodall numbers are sometimes referred to as Riesel numbers or Cullen numbers of the second kind.''\n\n\n'''Cullen primes''' are Cullen numbers that are prime. Similarly, '''Woodall primes''' are Woodall numbers that are prime.\n\nIt is common to list the Cullen and Woodall primes by the value of '''n''' rather than the full evaluated expression. They tend to get very large very quickly. For example, the third Cullen prime, '''n''' == 4713, has 1423 digits when evaluated.\n\n\n;Task\n\n* Write procedures to find Cullen numbers and Woodall numbers. \n\n* Use those procedures to find and show here, on this page the first 20 of each.\n\n\n;Stretch\n\n* Find and show the first '''5''' '''Cullen primes''' in terms of '''n'''.\n\n* Find and show the first '''12''' '''Woodall primes''' in terms of '''n'''.\n\n\n;See also\n\n* OEIS:A002064 - Cullen numbers: a(n) = n*2^n + 1\n\n* OEIS:A003261 - Woodall (or Riesel) numbers: n*2^n - 1\n\n* OEIS:A005849 - Indices of prime Cullen numbers: numbers k such that k*2^k + 1 is prime\n\n* OEIS:A002234 - Numbers k such that the Woodall number k*2^k - 1 is prime\n\n\n", "solution": "print(\"working...\")\nprint(\"First 20 Cullen numbers:\")\n\nfor n in range(1,21):\n    num = n*pow(2,n)+1\n    print(str(num),end= \" \")\n\nprint()\nprint(\"First 20 Woodall numbers:\")\n\nfor n in range(1,21):\n    num = n*pow(2,n)-1\n    print(str(num),end=\" \")\n\nprint()\nprint(\"done...\")\n"}
{"title": "Cullen and Woodall numbers", "language": "Python from Quackery", "task": "A Cullen number is a number of the form '''n x 2n + 1''' where '''n''' is a natural number.\n\nA Woodall number is very similar. It is a number of the form '''n x 2n - 1''' where '''n''' is a natural number.\n\nSo for each '''n''' the associated Cullen number and Woodall number differ by 2.\n\n''Woodall numbers are sometimes referred to as Riesel numbers or Cullen numbers of the second kind.''\n\n\n'''Cullen primes''' are Cullen numbers that are prime. Similarly, '''Woodall primes''' are Woodall numbers that are prime.\n\nIt is common to list the Cullen and Woodall primes by the value of '''n''' rather than the full evaluated expression. They tend to get very large very quickly. For example, the third Cullen prime, '''n''' == 4713, has 1423 digits when evaluated.\n\n\n;Task\n\n* Write procedures to find Cullen numbers and Woodall numbers. \n\n* Use those procedures to find and show here, on this page the first 20 of each.\n\n\n;Stretch\n\n* Find and show the first '''5''' '''Cullen primes''' in terms of '''n'''.\n\n* Find and show the first '''12''' '''Woodall primes''' in terms of '''n'''.\n\n\n;See also\n\n* OEIS:A002064 - Cullen numbers: a(n) = n*2^n + 1\n\n* OEIS:A003261 - Woodall (or Riesel) numbers: n*2^n - 1\n\n* OEIS:A005849 - Indices of prime Cullen numbers: numbers k such that k*2^k + 1 is prime\n\n* OEIS:A002234 - Numbers k such that the Woodall number k*2^k - 1 is prime\n\n\n", "solution": "def cullen(n): return((n<<n)+1)\n\t\ndef woodall(n): return((n<<n)-1)\n\nprint(\"First 20 Cullen numbers:\")\nfor i in range(1,21):\n\tprint(cullen(i),end=\" \")\nprint()\nprint()\nprint(\"First 20 Woodall numbers:\")\nfor i in range(1,21): \n\tprint(woodall(i),end=\" \")\nprint()"}
{"title": "Currency", "language": "Python", "task": "Show how to represent currency in a simple example, using a data type that represent exact values of dollars and cents.  \n\n\n;Note:\nThe '''IEEE 754''' binary floating point representations of numbers like   '''2.86'''   and   '''.0765'''   are not exact.\n\nFor this example, data will be two items with prices in dollars and cents, a quantity for each, and a tax rate. \n \nUse the values:\n::* 4000000000000000 hamburgers at $5.50 each       (four quadrillion burgers)\n::* 2 milkshakes at $2.86 each, and \n::* a tax rate of 7.65%.  \n\n\n(That number of hamburgers is a 4 with 15 zeros after it.    The number is contrived to exclude naive task solutions using 64 bit floating point types.)  \n\nCompute and output (show results on this page):\n::* the total price before tax\n::* the tax\n::* the total with tax  \n\n\nThe tax value must be computed by rounding to the nearest whole cent and this exact value must be added to the total price before tax.  \n\nThe output must show dollars and cents with a decimal point. \n \nThe three results displayed should be:\n::* 22000000000000005.72\n::* 1683000000000000.44 \n::* 23683000000000006.16  \n\n\nDollar signs and thousands separators are optional.\n\n", "solution": "from decimal import Decimal as D\nfrom collections import namedtuple\n\nItem = namedtuple('Item', 'price, quant')\n\nitems = dict( hamburger=Item(D('5.50'), D('4000000000000000')),\n              milkshake=Item(D('2.86'), D('2')) )\ntax_rate = D('0.0765')\n\nfmt = \"%-10s %8s %18s %22s\"\nprint(fmt % tuple('Item Price Quantity Extension'.upper().split()))\n\ntotal_before_tax = 0\nfor item, (price, quant) in sorted(items.items()):\n    ext = price * quant\n    print(fmt % (item, price, quant, ext))\n    total_before_tax += ext\nprint(fmt % ('', '', '', '--------------------'))\nprint(fmt % ('', '', 'subtotal', total_before_tax))\n\ntax = (tax_rate * total_before_tax).quantize(D('0.00'))\nprint(fmt % ('', '', 'Tax', tax))\n\ntotal = total_before_tax + tax\nprint(fmt % ('', '', '', '--------------------'))\nprint(fmt % ('', '', 'Total', total))"}
{"title": "Currying", "language": "Python", "task": "{{Wikipedia|Currying}}\n\n\n;Task:\nCreate a simple demonstrative example of Currying in a specific language.  \n\nAdd any historic details as to how the feature made its way into the language.\n\n\n\n", "solution": "# AUTOMATIC CURRYING AND UNCURRYING OF EXISTING FUNCTIONS\n\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    return lambda a: lambda b: f(a, b)\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    return lambda x, y: f(x)(y)\n\n\n# EXAMPLES --------------------------------------\n\n# A plain uncurried function with 2 arguments,\n\n# justifyLeft :: Int -> String -> String\ndef justifyLeft(n, s):\n    return (s + (n * ' '))[:n]\n\n\n# and a similar, but manually curried, function.\n\n# justifyRight :: Int -> String -> String\ndef justifyRight(n):\n    return lambda s: (\n        ((n * ' ') + s)[-n:]\n    )\n\n\n# CURRYING and UNCURRYING at run-time:\n\ndef main():\n    for s in [\n        'Manually curried using a lambda:',\n        '\\n'.join(map(\n            justifyRight(5),\n            ['1', '9', '10', '99', '100', '1000']\n        )),\n\n        '\\nAutomatically uncurried:',\n        uncurry(justifyRight)(5, '10000'),\n\n        '\\nAutomatically curried',\n        '\\n'.join(map(\n            curry(justifyLeft)(10),\n            ['1', '9', '10', '99', '100', '1000']\n        ))\n    ]:\n        print (s)\n\n\nmain()"}
{"title": "Curzon numbers", "language": "Python", "task": "A '''Curzon number''' is defined to be a positive integer '''n''' for which '''2n + 1''' is evenly divisible by  '''2 x n + 1'''.\n\n'''Generalized Curzon numbers''' are those where the positive integer '''n''', using a base integer '''k''', satisfy the condition that '''kn + 1''' is evenly divisible by  '''k x n + 1'''.\n\n''Base here does not imply the radix of the counting system; rather the integer the equation is based on. All calculations should be done in base 10.'' \n\nGeneralized Curzon numbers only exist for even base integers. \n\n\n;Task \n\n* Find and show the first '''50 Generalized Curzon numbers''' for even base integers from '''2''' through '''10'''.\n\n\n;Stretch\n\n* Find and show the '''one thousandth'''.\n\n\n;See also\n\n;* Numbers Aplenty - Curzon numbers\n;* OEIS:A224486 - Numbers k such that 2*k+1 divides 2^k+1 (Curzon numbers)\n\n''and even though it is not specifically mentioned that they are Curzon numbers:''\n\n;* OEIS:A230076 - (A007521(n)-1)/4 (Generalized Curzon numbers with a base 4)\n\n", "solution": "def is_Curzon(n, k):\n    r = k * n\n    return pow(k, n, r + 1) == r\n\nfor k in [2, 4, 6, 8, 10]:\n    n, curzons = 1, []\n    while len(curzons) < 1000:\n        if is_Curzon(n, k):\n            curzons.append(n)\n        n += 1\n    print(f'Curzon numbers with k = {k}:')\n    for i, c in enumerate(curzons[:50]):\n        print(f'{c: 5,}', end='\\n' if (i + 1) % 25 == 0 else '')\n    print(f'    Thousandth Curzon with k = {k}: {curzons[999]}.\\n')"}
{"title": "Cut a rectangle", "language": "Python from D", "task": "A given rectangle is made from ''m'' x ''n'' squares. If ''m'' and ''n'' are not both odd, then it is possible to cut a path through the rectangle along the square edges such that the rectangle splits into two connected pieces with the same shape (after rotating one of the pieces by 180deg). All such paths for 2 x 2 and 4 x 3 rectangles are shown below.\n\n[[file:rect-cut.svg]]\n\nWrite a program that calculates the number of different ways to cut an ''m'' x ''n'' rectangle.  Optionally, show each of the cuts.\n\nPossibly related task: [[Maze generation]] for depth-first search.\n\n", "solution": "try:\n    import psyco\nexcept ImportError:\n    pass\nelse:\n    psyco.full()\n\nw, h = 0, 0\ncount = 0\nvis = []\n\ndef cwalk(y, x, d):\n    global vis, count, w, h\n    if not y or y == h or not x or x == w:\n        count += 1\n        return\n\n    vis[y][x] = vis[h - y][w - x] = 1\n\n    if x and not vis[y][x - 1]:\n        cwalk(y, x - 1, d | 1)\n    if (d & 1) and x < w and not vis[y][x+1]:\n        cwalk(y, x + 1, d|1)\n    if y and not vis[y - 1][x]:\n        cwalk(y - 1, x, d | 2)\n    if (d & 2) and y < h and not vis[y + 1][x]:\n        cwalk(y + 1, x, d | 2)\n\n    vis[y][x] = vis[h - y][w - x] = 0\n\ndef count_only(x, y):\n    global vis, count, w, h\n    count = 0\n    w = x\n    h = y\n\n    if (h * w) & 1:\n        return count\n    if h & 1:\n        w, h = h, w\n\n    vis = [[0] * (w + 1) for _ in xrange(h + 1)]\n    vis[h // 2][w // 2] = 1\n\n    if w & 1:\n        vis[h // 2][w // 2 + 1] = 1\n\n    res = 0\n    if w > 1:\n        cwalk(h // 2, w // 2 - 1, 1)\n        res = 2 * count - 1\n        count = 0\n        if w != h:\n            cwalk(h // 2 + 1, w // 2, 3 if (w & 1) else 2)\n\n        res += 2 * count - (not (w & 1))\n    else:\n        res = 1\n\n    if w == h:\n        res = 2 * res + 2\n    return res\n\ndef main():\n    for y in xrange(1, 10):\n        for x in xrange(1, y + 1):\n            if not (x & 1) or not (y & 1):\n                print \"%d x %d: %d\" % (y, x, count_only(x, y))\n\nmain()"}
{"title": "Cyclotomic polynomial", "language": "Python", "task": "The nth Cyclotomic polynomial, for any positive integer n, is the unique irreducible polynomial of largest degree with integer coefficients that is a divisor of x^n - 1, and is not a divisor of x^k - 1 for any k < n.\n\n\n;Task:\n*   Find and print the first 30 cyclotomic polynomials.\n*   Find and print the order of the first 10 cyclotomic polynomials that have n or -n as a coefficient.\n\n\n;See also\n*   Wikipedia article, Cyclotomic polynomial, showing ways to calculate them.\n*   The sequence A013594 with the smallest order of cyclotomic polynomial containing n or -n as a coefficient.\n\n", "solution": "from itertools import count, chain\nfrom collections import deque\n\ndef primes(_cache=[2, 3]):\n    yield from _cache\n    for n in count(_cache[-1]+2, 2):\n        if isprime(n):\n            _cache.append(n)\n            yield n\n\ndef isprime(n):\n    for p in primes():\n        if n%p == 0:\n            return False\n        if p*p > n:\n            return True\n\ndef factors(n):\n    for p in primes():\n    # prime factoring is such a non-issue for small numbers that, for\n    # this example, we might even just say\n    # for p in count(2):\n        if p*p > n:\n            if n > 1:\n                yield(n, 1, 1)\n            break\n\n        if n%p == 0:\n            cnt = 0\n            while True:\n                n, cnt = n//p, cnt+1\n                if n%p != 0: break\n            yield p, cnt, n\n# ^^ not the most sophisticated prime number routines, because no need\n\n# Returns (list1, list2) representing the division between\n# two polinomials. A list p of integers means the product\n#   (x^p[0] - 1) * (x^p[1] - 1) * ...\ndef cyclotomic(n):\n    def poly_div(num, den):\n        return (num[0] + den[1], num[1] + den[0])\n\n    def elevate(poly, n): # replace poly p(x) with p(x**n)\n        powerup = lambda p, n: [a*n for a in p]\n        return poly if n == 1 else (powerup(poly[0], n), powerup(poly[1], n))\n\n\n    if n == 0:\n        return ([], [])\n    if n == 1:\n        return ([1], [])\n\n    p, m, r = next(factors(n))\n    poly = cyclotomic(r)\n    return elevate(poly_div(elevate(poly, p), poly), p**(m-1))\n\ndef to_text(poly):\n    def getx(c, e):\n        if e == 0:\n            return '1'\n        elif e == 1:\n            return 'x'\n        return 'x' + (''.join('\u2070\u00b9\u00b2\u00b3\u2074\u2075\u2076\u2077\u2078\u2079'[i] for i in map(int, str(e))))\n\n    parts = []\n    for (c,e) in (poly):\n        if c < 0:\n            coef = ' - ' if c == -1 else f' - {-c} '\n        else:\n            coef = (parts and ' + ' or '') if c == 1 else f' + {c}'\n        parts.append(coef + getx(c,e))\n    return ''.join(parts)\n\ndef terms(poly):\n    # convert above representation of division to (coef, power) pairs\n\n    def merge(a, b):\n        # a, b should be deques. They may change during the course.\n        while a or b:\n            l = a[0] if a else (0, -1) # sentinel value\n            r = b[0] if b else (0, -1)\n            if l[1] > r[1]:\n                a.popleft()\n            elif l[1] < r[1]:\n                b.popleft()\n                l = r\n            else:\n                a.popleft()\n                b.popleft()\n                l = (l[0] + r[0], l[1])\n            yield l\n\n    def mul(poly, p): # p means polynomial x^p - 1\n        poly = list(poly)\n        return merge(deque((c, e+p) for c,e in poly),\n                     deque((-c, e) for c,e in poly))\n\n    def div(poly, p): # p means polynomial x^p - 1\n        q = deque()\n        for c,e in merge(deque(poly), q):\n            if c:\n                q.append((c, e - p))\n                yield (c, e - p)\n            if e == p: break\n\n    p = [(1, 0)]  # 1*x^0, i.e. 1\n\n    for x in poly[0]: # numerator\n        p = mul(p, x)\n    for x in sorted(poly[1], reverse=True): # denominator\n        p = div(p, x)\n    return p\n\nfor n in chain(range(11), [2]):\n    print(f'{n}: {to_text(terms(cyclotomic(n)))}')\n\nwant = 1\nfor n in count():\n    c = [c for c,_ in terms(cyclotomic(n))]\n    while want in c or -want in c:\n        print(f'C[{want}]: {n}')\n        want += 1"}
{"title": "Damm algorithm", "language": "Python", "task": "The '''Damm''' algorithm is a checksum algorithm which detects all single digit errors and adjacent transposition errors. \n\n\nThe algorithm is named after H. Michael Damm.\n\n\n;Task:\nVerify the checksum, stored as last digit of an input.\n\n\n\n", "solution": "def damm(num: int) -> bool:\n    row = 0\n    for digit in str(num):\n        row = _matrix[row][int(digit)] \n    return row == 0\n\n_matrix = (\n    (0, 3, 1, 7, 5, 9, 8, 6, 4, 2),\n    (7, 0, 9, 2, 1, 5, 4, 8, 6, 3),\n    (4, 2, 0, 6, 8, 7, 1, 3, 5, 9),\n    (1, 7, 5, 0, 9, 8, 3, 4, 2, 6),\n    (6, 1, 2, 3, 0, 4, 5, 9, 7, 8),\n    (3, 6, 7, 4, 2, 0, 9, 5, 8, 1),\n    (5, 8, 6, 9, 7, 2, 0, 1, 3, 4),\n    (8, 9, 4, 5, 3, 6, 2, 0, 1, 7),\n    (9, 4, 3, 8, 6, 1, 7, 2, 0, 5),\n    (2, 5, 8, 1, 4, 3, 6, 7, 9, 0)\n)\n\nif __name__ == '__main__':\n    for test in [5724, 5727, 112946]:\n        print(f'{test}\\t Validates as: {damm(test)}')"}
{"title": "De Bruijn sequences", "language": "Python", "task": "{{DISPLAYTITLE:de Bruijn sequences}}\nThe sequences are named after the Dutch mathematician   Nicolaas Govert de Bruijn.\n\n\nA note on Dutch capitalization:   Nicolaas' last name is   '''de Bruijn''',   the   '''de'''   isn't normally capitalized\nunless it's the first word in a sentence.   Rosetta Code (more or less by default or by fiat) requires the first word in the task name to be\ncapitalized.\n\n\nIn combinatorial mathematics,   a   '''de Bruijn sequence'''   of order    ''n''    on\na    size-''k''    alphabet (computer science)    ''A''    is a cyclic sequence in which every\npossible    length-''n''    string (computer science, formal theory)   on    ''A''    occurs\nexactly once as a contiguous substring.\n\n\nSuch a sequence is denoted by    ''B''(''k'', ''n'')    and has\nlength   ''k''''n'',   which is also the number of distinct substrings of\nlength   ''n''   on   ''A'';    \nde Bruijn sequences are therefore optimally short.\n\n\nThere are:\n                          (k!)k(n-1)   /   kn\ndistinct de Bruijn sequences    ''B''(''k'', ''n''). \n\n\n;Task:\nFor this Rosetta Code task,   a   '''de Bruijn'''   sequence is to be generated that can be used to shorten a brute-force attack on\na   PIN-like   code lock that does not have an \"enter\"\nkey and accepts the last    ''n''    digits entered.\n\n\nNote:   automated teller machines (ATMs)   used to work like\nthis,   but their software has been updated to not allow a brute-force attack.\n\n\n;Example:\nA   digital door lock   with a 4-digit code would\nhave ''B'' (10, 4) solutions,   with a length of   '''10,000'''   (digits).\n\nTherefore, only at most     '''10,000 + 3'''     (as the solutions are cyclic or ''wrap-around'')   presses are needed to\nopen the lock.\n\nTrying all 4-digit codes separately would require   '''4 x 10,000'''   or   '''40,000'''   presses.\n\n\n;Task requirements:\n:*   Generate a de Bruijn sequence for a 4-digit (decimal) PIN code.\n:::*   Show the length of the generated de Bruijn sequence.\n:::*   (There are many possible de Bruijn sequences that solve this task,   one solution is shown on the ''discussion'' page).\n:::*   Show the first and last   '''130'''   digits of the de Bruijn sequence.\n:*   Verify that all four-digit (decimal)   '''1,000'''   PIN codes are contained within the de Bruijn sequence.\n:::*   0000, 0001, 0002, 0003,   ...   9996, 9997, 9998, 9999   (note the leading zeros).\n:*   Reverse the de Bruijn sequence.\n:*   Again, perform the (above) verification test.\n:*   Replace the 4,444th digit with a period (.) in the original de Bruijn sequence.\n:::*   Perform the verification test (again).   There should be four PIN codes missing.\n\n\n(The last requirement is to ensure that the verification tests performs correctly.   The verification processes should list\nany and all missing PIN codes.)\n\nShow all output here, on this page.\n\n\n\n;References:\n:*   Wikipedia           entry:   de Bruijn sequence.\n:*   MathWorld           entry:   de Bruijn sequence.\n:*   An  OEIS  entry:   A166315 lexicographically earliest binary de Bruijn sequences, B(2,n)     --- Not B(10,4),   but possibly relevant.\n\n", "solution": "# from https://en.wikipedia.org/wiki/De_Bruijn_sequence\n\ndef de_bruijn(k, n):\n    \"\"\"\n    de Bruijn sequence for alphabet k\n    and subsequences of length n.\n    \"\"\"\n    try:\n        # let's see if k can be cast to an integer;\n        # if so, make our alphabet a list\n        _ = int(k)\n        alphabet = list(map(str, range(k)))\n\n    except (ValueError, TypeError):\n        alphabet = k\n        k = len(k)\n\n    a = [0] * k * n\n    sequence = []\n\n    def db(t, p):\n        if t > n:\n            if n % p == 0:\n                sequence.extend(a[1:p + 1])\n        else:\n            a[t] = a[t - p]\n            db(t + 1, p)\n            for j in range(a[t - p] + 1, k):\n                a[t] = j\n                db(t + 1, t)\n    db(1, 1)\n    return \"\".join(alphabet[i] for i in sequence)\n    \ndef validate(db):\n    \"\"\"\n    \n    Check that all 10,000 combinations of 0-9 are present in \n    De Bruijn string db.\n    \n    Validating the reversed deBruijn sequence:\n      No errors found\n    \n    Validating the overlaid deBruijn sequence:\n      4 errors found:\n        PIN number 1459 missing\n        PIN number 4591 missing\n        PIN number 5814 missing\n        PIN number 8145 missing\n    \n    \"\"\"\n    \n    dbwithwrap = db+db[0:3]\n    \n    digits = '0123456789'\n    \n    errorstrings = []\n    \n    for d1 in digits:\n        for d2 in digits:\n            for d3 in digits:\n                for d4 in digits:\n                    teststring = d1+d2+d3+d4\n                    if teststring not in dbwithwrap:\n                        errorstrings.append(teststring)\n                        \n    if len(errorstrings) > 0:\n        print(\"  \"+str(len(errorstrings))+\" errors found:\")\n        for e in errorstrings:\n            print(\"  PIN number \"+e+\"  missing\")\n    else:\n        print(\"  No errors found\")\n\ndb = de_bruijn(10, 4)\n\nprint(\" \")\nprint(\"The length of the de Bruijn sequence is \", str(len(db)))\nprint(\" \")\nprint(\"The first 130 digits of the de Bruijn sequence are: \"+db[0:130])\nprint(\" \")\nprint(\"The last 130 digits of the de Bruijn sequence are: \"+db[-130:])\nprint(\" \")\nprint(\"Validating the deBruijn sequence:\")\nvalidate(db)\ndbreversed = db[::-1]\nprint(\" \")\nprint(\"Validating the reversed deBruijn sequence:\")\nvalidate(dbreversed)\ndboverlaid = db[0:4443]+'.'+db[4444:]\nprint(\" \")\nprint(\"Validating the overlaid deBruijn sequence:\")\nvalidate(dboverlaid)\n"}
{"title": "Deceptive numbers", "language": "Python", "task": "Repunits are numbers that consist entirely of repetitions of the digit one (unity).  The notation '''Rn''' symbolizes the repunit made up of '''n''' ones.\n\nEvery prime '''p''' larger than 5, evenly divides the repunit '''Rp-1'''.\n\n\n;E.G.\n\nThe repunit '''R6''' is evenly divisible by '''7'''.\n\n111111 / 7 = 15873\n\nThe repunit '''R42''' is evenly divisible by '''43'''.\n\n111111111111111111111111111111111111111111 / 43 = 2583979328165374677002583979328165374677\n\nAnd so on.\n\n\nThere are composite numbers that also have this same property. They are often referred to as ''deceptive non-primes'' or ''deceptive numbers''.\n\n\nThe repunit '''R90''' is evenly divisible by the composite number '''91''' (=7*13).\n\n111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111 / 91 = 1221001221001221001221001221001221001221001221001221001221001221001221001221001221001221\n\n\n;Task \n\n* Find and show at least the first '''10 deceptive numbers'''; composite numbers '''n''' that evenly divide the repunit '''Rn-1'''\n\n\n;See also\n\n;* Numbers Aplenty - Deceptive numbers\n;* OEIS:A000864 - Deceptive nonprimes: composite numbers k that divide the repunit R_{k-1}\n\n", "solution": "from itertools import count, islice\nfrom math import isqrt\n\ndef is_deceptive(n):\n    if n & 1 and n % 3 and n % 5 and pow(10, n - 1, n) == 1:\n        for d in range(7, isqrt(n) + 1, 6):\n            if not (n % d and n % (d + 4)): return True\n    return False\n\nprint(*islice(filter(is_deceptive, count()), 100))"}
{"title": "Deming's funnel", "language": "Python from Racket", "task": "W Edwards Deming was an American statistician and management guru who used physical demonstrations to illuminate his teachings.  In one demonstration Deming repeatedly dropped marbles through a funnel at a target, marking where they landed, and observing the resulting pattern.  He applied a sequence of \"rules\" to try to improve performance.  In each case the experiment begins with the funnel positioned directly over the target.\n\n* '''Rule 1''': The funnel remains directly above the target.\n* '''Rule 2''': Adjust the funnel position by shifting the target to compensate after each drop.  E.g. If the last drop missed 1 cm east, move the funnel 1 cm to the west of its current position.\n* '''Rule 3''': As rule 2, but first move the funnel back over the target, before making the adjustment.  E.g. If the funnel is 2 cm north, and the marble lands 3 cm north, move the funnel 3 cm south of the target.\n* '''Rule 4''': The funnel is moved directly over the last place a marble landed.\n\nApply the four rules to the set of 50 pseudorandom displacements provided (e.g in the Racket solution) for the dxs and dys.  '''Output''': calculate the mean and standard-deviations of the resulting x and y values for each rule.  \n\nNote that rules 2, 3, and 4 give successively worse results.  Trying to deterministically compensate for a random process is counter-productive, but -- according to Deming -- quite a popular pastime: see the Further Information, below for examples.\n\n'''Stretch goal 1''': Generate fresh pseudorandom data.  The radial displacement of the drop from the funnel position is given by a Gaussian distribution (standard deviation is 1.0) and the angle of displacement is uniformly distributed.\n\n'''Stretch goal 2''': Show scatter plots of all four results.\n\n\n;Further information:\n* Further explanation and interpretation\n* Video demonstration of the funnel experiment at the Mayo Clinic.\n\n", "solution": "import math \n\ndxs = [-0.533, 0.27, 0.859, -0.043, -0.205, -0.127, -0.071, 0.275, 1.251,\n       -0.231, -0.401, 0.269, 0.491, 0.951, 1.15, 0.001, -0.382, 0.161, 0.915,\n       2.08, -2.337, 0.034, -0.126, 0.014, 0.709, 0.129, -1.093, -0.483, -1.193, \n       0.02, -0.051, 0.047, -0.095, 0.695, 0.34, -0.182, 0.287, 0.213, -0.423,\n       -0.021, -0.134, 1.798, 0.021, -1.099, -0.361, 1.636, -1.134, 1.315, 0.201, \n       0.034, 0.097, -0.17, 0.054, -0.553, -0.024, -0.181, -0.7, -0.361, -0.789,\n       0.279, -0.174, -0.009, -0.323, -0.658, 0.348, -0.528, 0.881, 0.021, -0.853,\n       0.157, 0.648, 1.774, -1.043, 0.051, 0.021, 0.247, -0.31, 0.171, 0.0, 0.106,\n       0.024, -0.386, 0.962, 0.765, -0.125, -0.289, 0.521, 0.017, 0.281, -0.749,\n       -0.149, -2.436, -0.909, 0.394, -0.113, -0.598, 0.443, -0.521, -0.799, \n       0.087]\n\ndys = [0.136, 0.717, 0.459, -0.225, 1.392, 0.385, 0.121, -0.395, 0.49, -0.682,\n       -0.065, 0.242, -0.288, 0.658, 0.459, 0.0, 0.426, 0.205, -0.765, -2.188, \n       -0.742, -0.01, 0.089, 0.208, 0.585, 0.633, -0.444, -0.351, -1.087, 0.199,\n       0.701, 0.096, -0.025, -0.868, 1.051, 0.157, 0.216, 0.162, 0.249, -0.007, \n       0.009, 0.508, -0.79, 0.723, 0.881, -0.508, 0.393, -0.226, 0.71, 0.038, \n       -0.217, 0.831, 0.48, 0.407, 0.447, -0.295, 1.126, 0.38, 0.549, -0.445, \n       -0.046, 0.428, -0.074, 0.217, -0.822, 0.491, 1.347, -0.141, 1.23, -0.044, \n       0.079, 0.219, 0.698, 0.275, 0.056, 0.031, 0.421, 0.064, 0.721, 0.104, \n       -0.729, 0.65, -1.103, 0.154, -1.72, 0.051, -0.385, 0.477, 1.537, -0.901, \n       0.939, -0.411, 0.341, -0.411, 0.106, 0.224, -0.947, -1.424, -0.542, -1.032]\n\ndef funnel(dxs, rule):\n    x, rxs = 0, []\n    for dx in dxs:\n        rxs.append(x + dx)\n        x = rule(x, dx)\n    return rxs\n\ndef mean(xs): return sum(xs) / len(xs)\n\ndef stddev(xs):\n    m = mean(xs)\n    return math.sqrt(sum((x-m)**2 for x in xs) / len(xs))\n\ndef experiment(label, rule):\n    rxs, rys = funnel(dxs, rule), funnel(dys, rule)\n    print label\n    print 'Mean x, y    : %.4f, %.4f' % (mean(rxs), mean(rys))\n    print 'Std dev x, y : %.4f, %.4f' % (stddev(rxs), stddev(rys))\n    print\n\n\nexperiment('Rule 1:', lambda z, dz: 0)\nexperiment('Rule 2:', lambda z, dz: -dz)\nexperiment('Rule 3:', lambda z, dz: -(z+dz))\nexperiment('Rule 4:', lambda z, dz: z+dz)"}
{"title": "Department numbers", "language": "Python", "task": "There is a highly organized city that has decided to assign a number to each of their departments:\n::*   police department\n::*   sanitation department\n::*   fire department \n\n\nEach department can have a number between   '''1'''   and   '''7'''   (inclusive).\n\nThe three department numbers are to be unique (different from each other) and must add up to   '''12'''.\n\nThe Chief of the Police doesn't like odd numbers and wants to have an even number for his department.\n\n\n;Task:\nWrite a computer program which outputs all valid combinations.\n\n\nPossible output   (for the 1st and 14th solutions):\n  \n  --police--  --sanitation--  --fire-- \n      2             3            7 \n      6             5            1\n\n", "solution": "from itertools import permutations\n \ndef solve():\n    c, p, f, s = \"\\\\,Police,Fire,Sanitation\".split(',')\n    print(f\"{c:>3}  {p:^6} {f:^4} {s:^10}\")\n    c = 1\n    for p, f, s in permutations(range(1, 8), r=3):\n        if p + s + f == 12 and p % 2 == 0:\n            print(f\"{c:>3}: {p:^6} {f:^4} {s:^10}\")\n            c += 1\n \nif __name__ == '__main__':\n    solve()"}
{"title": "Department numbers", "language": "Python 3", "task": "There is a highly organized city that has decided to assign a number to each of their departments:\n::*   police department\n::*   sanitation department\n::*   fire department \n\n\nEach department can have a number between   '''1'''   and   '''7'''   (inclusive).\n\nThe three department numbers are to be unique (different from each other) and must add up to   '''12'''.\n\nThe Chief of the Police doesn't like odd numbers and wants to have an even number for his department.\n\n\n;Task:\nWrite a computer program which outputs all valid combinations.\n\n\nPossible output   (for the 1st and 14th solutions):\n  \n  --police--  --sanitation--  --fire-- \n      2             3            7 \n      6             5            1\n\n", "solution": "'''Department numbers'''\n\nfrom itertools import (chain)\nfrom operator import (ne)\n\n\n# options :: Int -> Int -> Int -> [(Int, Int, Int)]\ndef options(lo, hi, total):\n    '''Eligible integer triples.'''\n    ds = enumFromTo(lo)(hi)\n    return bind(filter(even, ds))(\n        lambda x: bind(filter(curry(ne)(x), ds))(\n            lambda y: bind([total - (x + y)])(\n                lambda z: [(x, y, z)] if (\n                    z != y and lo <= z <= hi\n                ) else []\n            )\n        )\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test'''\n\n    xs = options(1, 7, 12)\n    print(('Police', 'Sanitation', 'Fire'))\n    for tpl in xs:\n        print(tpl)\n    print('\\nNo. of options: ' + str(len(xs)))\n\n\n# GENERIC ABSTRACTIONS ------------------------------------\n\n# bind (>>=) :: [a] -> (a -> [b]) -> [b]\ndef bind(xs):\n    '''List monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.'''\n    return lambda f: list(\n        chain.from_iterable(\n            map(f, xs)\n        )\n    )\n\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    '''A curried function derived\n       from an uncurried function.'''\n    return lambda a: lambda b: f(a, b)\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# even :: Int -> Bool\ndef even(x):\n    '''True if x is an integer\n       multiple of two.'''\n    return 0 == x % 2\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Department numbers", "language": "Python 3.7", "task": "There is a highly organized city that has decided to assign a number to each of their departments:\n::*   police department\n::*   sanitation department\n::*   fire department \n\n\nEach department can have a number between   '''1'''   and   '''7'''   (inclusive).\n\nThe three department numbers are to be unique (different from each other) and must add up to   '''12'''.\n\nThe Chief of the Police doesn't like odd numbers and wants to have an even number for his department.\n\n\n;Task:\nWrite a computer program which outputs all valid combinations.\n\n\nPossible output   (for the 1st and 14th solutions):\n  \n  --police--  --sanitation--  --fire-- \n      2             3            7 \n      6             5            1\n\n", "solution": "'''Department numbers'''\n\nfrom operator import ne\n\n\n# options :: Int -> Int -> Int -> [(Int, Int, Int)]\ndef options(lo, hi, total):\n    '''Eligible triples.'''\n    ds = enumFromTo(lo)(hi)\n    return [\n        (x, y, z)\n        for x in filter(even, ds)\n        for y in filter(curry(ne)(x), ds)\n        for z in [total - (x + y)]\n        if y != z and lo <= z <= hi\n    ]\n\n\n# Or with less tightly-constrained generation,\n# and more winnowing work downstream:\n\n# options2 :: Int -> Int -> Int -> [(Int, Int, Int)]\ndef options2(lo, hi, total):\n    '''Eligible triples.'''\n    ds = enumFromTo(lo)(hi)\n    return [\n        (x, y, z)\n        for x in ds\n        for y in ds\n        for z in [total - (x + y)]\n        if even(x) and y not in [x, z] and lo <= z <= hi\n    ]\n\n\n# GENERIC -------------------------------------------------\n\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    '''A curried function derived\n       from an uncurried function.'''\n    return lambda a: lambda b: f(a, b)\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# even :: Int -> Bool\ndef even(x):\n    '''True if x is an integer\n       multiple of two.'''\n    return 0 == x % 2\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.'''\n    return '\\n'.join(xs)\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test'''\n\n    xs = options(1, 7, 12)\n    print(('Police', 'Sanitation', 'Fire'))\n    print(unlines(map(str, xs)))\n    print('\\nNo. of options: ' + str(len(xs)))\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Descending primes", "language": "Python", "task": "Generate and show all primes with strictly descending decimal digits.\n\n;See also\n;* OEIS:A052014 - Primes with distinct digits in descending order\n\n;Related:\n*[[Ascending primes]]\n\n\n", "solution": "from sympy import isprime\n\ndef descending(xs=range(10)):\n    for x in xs:\n        yield x\n        yield from descending(x*10 + d for d in range(x%10))\n\nfor i, p in enumerate(sorted(filter(isprime, descending()))):\n    print(f'{p:9d}', end=' ' if (1 + i)%8 else '\\n')\n\nprint()"}
{"title": "Detect division by zero", "language": "Python", "task": "Write a function to detect a   ''divide by zero error''   without checking if the denominator is zero.\n\n", "solution": "def div_check(x, y):\n  try:\n    x / y\n  except ZeroDivisionError:\n    return True\n  else:\n    return False"}
{"title": "Determinant and permanent", "language": "Python", "task": "permanent of the matrix.\n\nThe determinant is given by\n:: \\det(A) = \\sum_\\sigma\\sgn(\\sigma)\\prod_{i=1}^n M_{i,\\sigma_i}\nwhile the permanent is given by\n::  \\operatorname{perm}(A)=\\sum_\\sigma\\prod_{i=1}^n M_{i,\\sigma_i}\nIn both cases the sum is over the permutations \\sigma of the permutations of 1, 2, ..., ''n''. (A permutation's sign is 1 if there are an even number of inversions and -1 otherwise; see parity of a permutation.)\n\nMore efficient algorithms for the determinant are known: [[LU decomposition]], see for example [[wp:LU decomposition#Computing the determinant]]. Efficient methods for calculating the permanent are not known.\n\n\n;Related task:\n* [[Permutations by swapping]]\n\n", "solution": "from itertools import permutations\nfrom operator import mul\nfrom math import fsum\nfrom spermutations import spermutations\n\ndef prod(lst):\n    return reduce(mul, lst, 1)\n\ndef perm(a):\n    n = len(a)\n    r = range(n)\n    s = permutations(r)\n    return fsum(prod(a[i][sigma[i]] for i in r) for sigma in s)\n\ndef det(a):\n    n = len(a)\n    r = range(n)\n    s = spermutations(n)\n    return fsum(sign * prod(a[i][sigma[i]] for i in r)\n                for sigma, sign in s)\n\nif __name__ == '__main__':\n    from pprint import pprint as pp\n\n    for a in ( \n            [\n             [1, 2], \n             [3, 4]], \n\n            [\n             [1, 2, 3, 4],\n             [4, 5, 6, 7],\n             [7, 8, 9, 10],\n             [10, 11, 12, 13]],        \n\n            [\n             [ 0,  1,  2,  3,  4],\n             [ 5,  6,  7,  8,  9],\n             [10, 11, 12, 13, 14],\n             [15, 16, 17, 18, 19],\n             [20, 21, 22, 23, 24]],\n        ):\n        print('')\n        pp(a)\n        print('Perm: %s Det: %s' % (perm(a), det(a)))"}
{"title": "Determine if a string has all the same characters", "language": "Python", "task": "Given a character string   (which may be empty, or have a length of zero characters):\n::*   create a function/procedure/routine to:\n::::*    determine if all the characters in the string are the same\n::::*    indicate if or which character is different from the previous character\n::*   display each string and its length   (as the strings are being examined)\n::*   a zero-length (empty) string shall be considered as all the same character(s)\n::*   process the strings from left-to-right\n::*   if       all the same character,   display a message saying such\n::*   if                  not all the same character,   then:\n::::*    display a message saying such\n::::*    display what character is different\n::::*    only the 1st different character need be displayed\n::::*    display where the different character is in the string\n::::*    the above messages can be part of a single message\n::::*    display the hexadecimal value of the different character\n\n\nUse (at least) these seven test values   (strings):\n:::*   a string of length   0   (an empty string)\n:::*   a string of length   3   which contains three blanks\n:::*   a string of length   1   which contains:   '''2'''\n:::*   a string of length   3   which contains:   '''333'''\n:::*   a string of length   3   which contains:   '''.55'''\n:::*   a string of length   6   which contains:   '''tttTTT'''\n:::*   a string of length   9   with a blank in the middle:   '''4444   444k'''\n\n\n\nShow all output here on this page.\n\n\n", "solution": "===Functional===\n\nWhat we are testing here is the cardinality of the set of characters from which a string is drawn, so the first thought might well be to use '''set'''. \n\nOn the other hand, '''itertools.groupby''' has the advantage of yielding richer information (the list of groups is ordered), for less work.\n\n"}
{"title": "Determine if a string has all the same characters", "language": "Python 3.7", "task": "Given a character string   (which may be empty, or have a length of zero characters):\n::*   create a function/procedure/routine to:\n::::*    determine if all the characters in the string are the same\n::::*    indicate if or which character is different from the previous character\n::*   display each string and its length   (as the strings are being examined)\n::*   a zero-length (empty) string shall be considered as all the same character(s)\n::*   process the strings from left-to-right\n::*   if       all the same character,   display a message saying such\n::*   if                  not all the same character,   then:\n::::*    display a message saying such\n::::*    display what character is different\n::::*    only the 1st different character need be displayed\n::::*    display where the different character is in the string\n::::*    the above messages can be part of a single message\n::::*    display the hexadecimal value of the different character\n\n\nUse (at least) these seven test values   (strings):\n:::*   a string of length   0   (an empty string)\n:::*   a string of length   3   which contains three blanks\n:::*   a string of length   1   which contains:   '''2'''\n:::*   a string of length   3   which contains:   '''333'''\n:::*   a string of length   3   which contains:   '''.55'''\n:::*   a string of length   6   which contains:   '''tttTTT'''\n:::*   a string of length   9   with a blank in the middle:   '''4444   444k'''\n\n\n\nShow all output here on this page.\n\n\n", "solution": "'''Determine if a string has all the same characters'''\n\nfrom itertools import groupby\n\n\n# firstDifferingCharLR :: String -> Either String Dict\ndef firstDifferingCharLR(s):\n    '''Either a message reporting that no character changes were\n       seen, or a dictionary with details of the  first character\n       (if any) that differs from that at the head of the string.\n    '''\n    def details(xs):\n        c = xs[1][0]\n        return {\n            'char': repr(c),\n            'hex': hex(ord(c)),\n            'index': s.index(c),\n            'total': len(s)\n        }\n    xs = list(groupby(s))\n    return Right(details(xs)) if 1 < len(xs) else (\n        Left('Total length ' + str(len(s)) + ' - No character changes.')\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test of 7 strings'''\n\n    print(fTable('First, if any, points of difference:\\n')(repr)(\n        either(identity)(\n            lambda dct: dct['char'] + ' (' + dct['hex'] +\n            ') at character ' + str(1 + dct['index']) +\n            ' of ' + str(dct['total']) + '.'\n        )\n    )(firstDifferingCharLR)([\n        '',\n        '   ',\n        '2',\n        '333',\n        '.55',\n        'tttTTT',\n        '4444 444'\n    ]))\n\n\n# GENERIC -------------------------------------------------\n\n# either :: (a -> c) -> (b -> c) -> Either a b -> c\ndef either(fl):\n    '''The application of fl to e if e is a Left value,\n       or the application of fr to e if e is a Right value.\n    '''\n    return lambda fr: lambda e: fl(e['Left']) if (\n        None is e['Right']\n    ) else fr(e['Right'])\n\n\n# identity :: a -> a\ndef identity(x):\n    '''The identity function.'''\n    return x\n\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# Left :: a -> Either a b\ndef Left(x):\n    '''Constructor for an empty Either (option type) value\n       with an associated string.\n    '''\n    return {'type': 'Either', 'Right': None, 'Left': x}\n\n\n# Right :: b -> Either a b\ndef Right(x):\n    '''Constructor for a populated Either (option type) value'''\n    return {'type': 'Either', 'Left': None, 'Right': x}\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Determine if a string has all unique characters", "language": "Python", "task": "Given a character string   (which may be empty, or have a length of zero characters):\n::*   create a function/procedure/routine to:\n::::*    determine if all the characters in the string are unique\n::::*    indicate if or which character is duplicated and where\n::*   display each string and its length   (as the strings are being examined)\n::*   a zero-length (empty) string shall be considered as unique\n::*   process the strings from left-to-right\n::*   if       unique,   display a message saying such\n::*   if                  not unique,   then:\n::::*    display a message saying such\n::::*    display what character is duplicated\n::::*    only the 1st non-unique character need be displayed\n::::*    display where \"both\" duplicated characters are in the string\n::::*    the above messages can be part of a single message\n::::*    display the hexadecimal value of the duplicated character\n\n\n\nUse (at least) these five test values   (strings):\n:::*   a string of length     0   (an empty string)\n:::*   a string of length     1   which is a single period   ('''.''')\n:::*   a string of length     6   which contains:   '''abcABC'''\n:::*   a string of length     7   which contains a blank in the middle:   '''XYZ  ZYX'''\n:::*   a string of length         36   which   ''doesn't''   contain the letter \"oh\":\n::::::::  '''1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ'''\n\n\n\nShow all output here on this page.\n\n\n", "solution": "'''Determine if a string has all unique characters'''\n\nfrom itertools import groupby\n\n\n# duplicatedCharIndices :: String -> Maybe (Char, [Int])\ndef duplicatedCharIndices(s):\n    '''Just the first duplicated character, and\n       the indices of its occurrence, or\n       Nothing if there are no duplications.\n    '''\n    def go(xs):\n        if 1 < len(xs):\n            duplicates = list(filter(lambda kv: 1 < len(kv[1]), [\n                (k, list(v)) for k, v in groupby(\n                    sorted(xs, key=swap),\n                    key=snd\n                )\n            ]))\n            return Just(second(fmap(fst))(\n                sorted(\n                    duplicates,\n                    key=lambda kv: kv[1][0]\n                )[0]\n            )) if duplicates else Nothing()\n        else:\n            return Nothing()\n    return go(list(enumerate(s)))\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test over various strings.'''\n\n    def showSample(s):\n        return repr(s) + ' (' + str(len(s)) + ')'\n\n    def showDuplicate(cix):\n        c, ix = cix\n        return repr(c) + (\n            ' (' + hex(ord(c)) + ') at ' + repr(ix)\n        )\n\n    print(\n        fTable('First duplicated character, if any:')(\n            showSample\n        )(maybe('None')(showDuplicate))(duplicatedCharIndices)([\n            '', '.', 'abcABC', 'XYZ ZYX',\n            '1234567890ABCDEFGHIJKLMN0PQRSTUVWXYZ'\n        ])\n    )\n\n\n# FORMATTING ----------------------------------------------\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# fmap :: (a -> b) -> [a] -> [b]\ndef fmap(f):\n    '''fmap over a list.\n       f lifted to a function over a list.\n    '''\n    return lambda xs: [f(x) for x in xs]\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# head :: [a] -> a\ndef head(xs):\n    '''The first element of a non-empty list.'''\n    return xs[0] if isinstance(xs, list) else next(xs)\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if m is Nothing,\n       or the application of f to x,\n       where m is Just(x).\n    '''\n    return lambda f: lambda m: v if (\n        None is m or m.get('Nothing')\n    ) else f(m.get('Just'))\n\n\n# second :: (a -> b) -> ((c, a) -> (c, b))\ndef second(f):\n    '''A simple function lifted to a function over a tuple,\n       with f applied only to the second of two values.\n    '''\n    return lambda xy: (xy[0], f(xy[1]))\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second member of a pair.'''\n    return tpl[1]\n\n\n# swap :: (a, b) -> (b, a)\ndef swap(tpl):\n    '''The swapped components of a pair.'''\n    return (tpl[1], tpl[0])\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Determine if a string is collapsible", "language": "Python", "task": "Determine if a character string is   ''collapsible''.\n\nAnd if so,   collapse the string   (by removing   ''immediately repeated''   characters).\n\n\n\nIf a character string has   ''immediately repeated''   character(s),   the repeated characters are to be\ndeleted (removed),   but not the primary (1st) character(s).\n\n\nAn   ''immediately repeated''   character is any character that is   immediately   followed by an\nidentical character (or characters).   Another word choice could've been   ''duplicated character'',   but that\nmight have ruled out   (to some readers)   triplicated characters   ***   or more.\n\n\n{This Rosetta Code task was inspired by a newly introduced   (as of around November 2019)   '''PL/I'''   BIF:   '''collapse'''.}\n\n\n;Examples:\nIn the following character string:\n\n\n  The better the 4-wheel drive, the further you'll be from help when ya get stuck! \n\n\nOnly the 2nd   '''t''',   '''e''', and   '''l'''   are repeated characters,   indicated\nby underscores (above),   even though they (those characters) appear elsewhere in the character string.\n\n\n\nSo, after ''collapsing'' the string, the result would be:\n\n  The beter the 4-whel drive, the further you'l be from help when ya get stuck! \n\n\n\n\nAnother example:\nIn the following character string:\n\n  headmistressship \n\n\nThe \"collapsed\" string would be:\n\n  headmistreship \n\n\n\n;Task:\nWrite a subroutine/function/procedure/routine***   to\nlocate   ''repeated''   characters and   ''collapse''   (delete)   them from the character\nstring.   The character string can be processed from either direction.\n\n\nShow all output here, on this page:\n:*   the   original string and its length\n:*   the       resultant string and its length\n:*   the above strings should be \"bracketed\" with   '''<<<'''   and   '''>>>'''   (to delineate blanks)\n;*   <<<<<<Guillemets may be used instead for \"bracketing\" for the more artistic programmers,   shown used here>>>>>>\n\n\n\n\nUse (at least) the following five strings,   all strings are length seventy-two (characters, including blanks),   except\nthe 1st string:\n\n  string\n  number\n         ++\n    1    |+-----------------------------------------------------------------------+   <######  a null string  (length zero)\n    2    |\"If I were two-faced, would I be wearing this one?\" --- Abraham Lincoln |\n    3    |..1111111111111111111111111111111111111111111111111111111111111117777888|\n    4    |I never give 'em hell, I just tell the truth, and they think it's hell. |\n    5    |                                                    --- Harry S Truman  |   <######  has many repeated blanks\n         +------------------------------------------------------------------------+\n\n\n", "solution": "from itertools import groupby\n\ndef collapser(txt):\n    return ''.join(item for item, grp in groupby(txt))\n\nif __name__ == '__main__':\n    strings = [\n            \"\",\n            '\"If I were two-faced, would I be wearing this one?\" --- Abraham Lincoln ',\n            \"..1111111111111111111111111111111111111111111111111111111111111117777888\",\n            \"I never give 'em hell, I just tell the truth, and they think it's hell. \",\n            \"                                                   ---  Harry S Truman  \",\n            \"The better the 4-wheel drive, the further you'll be from help when ya get stuck!\",\n            \"headmistressship\",\n            \"aardvark\",\n            \"\ud83d\ude0d\ud83d\ude00\ud83d\ude4c\ud83d\udc83\ud83d\ude0d\ud83d\ude0d\ud83d\ude0d\ud83d\ude4c\",\n            ]\n    for txt in strings:\n        this = \"Original\"\n        print(f\"\\n{this:14} Size: {len(txt)} \u00ab\u00ab\u00ab{txt}\u00bb\u00bb\u00bb\" )\n        this = \"Collapsed\"\n        sqz = collapser(txt)\n        print(f\"{this:>14} Size: {len(sqz)} \u00ab\u00ab\u00ab{sqz}\u00bb\u00bb\u00bb\" )"}
{"title": "Determine if a string is squeezable", "language": "Python", "task": "Determine if a character string is   ''squeezable''.\n\nAnd if so,   squeeze the string   (by removing any number of\na   ''specified''   ''immediately repeated''   character).\n\n\nThis task is very similar to the task     '''Determine if a character string is collapsible'''     except\nthat only a specified character is   ''squeezed''   instead of any character that is ''immediately repeated''.\n\n\nIf a character string has a specified   ''immediately repeated''   character(s),   the repeated characters are to be\ndeleted (removed),   but not the primary (1st) character(s).\n\n\nA specified   ''immediately repeated''   character is any specified character that is   immediately  \nfollowed by an identical character (or characters).   Another word choice could've been   ''duplicated\ncharacter'',   but that might have ruled out   (to some readers)   triplicated characters   ***   or more.\n\n\n{This Rosetta Code task was inspired by a newly introduced   (as of around\nNovember 2019)   '''PL/I'''   BIF:   '''squeeze'''.}\n\n\n;Examples:\nIn the following character string with a specified   ''immediately repeated''   character of   '''e''':\n\n\n  The better the 4-wheel drive, the further you'll be from help when ya get stuck! \n\n\nOnly the 2nd   '''e'''   is an specified repeated character,   indicated by an underscore\n(above),   even though they (the characters) appear elsewhere in the character string.\n\n\n\nSo, after ''squeezing'' the string, the result would be:\n\n  The better the 4-whel drive, the further you'll be from help when ya get stuck! \n\n\n\n\nAnother example:\nIn the following character string,   using a specified immediately repeated character   '''s''':\n\n  headmistressship \n\n\nThe \"squeezed\" string would be:\n\n  headmistreship \n\n\n\n;Task:\nWrite a subroutine/function/procedure/routine***   to locate a   ''specified immediately repeated''   character\nand   ''squeeze''   (delete)   them from the character string.   The\ncharacter string can be processed from either direction.\n\n\nShow all output here, on this page:\n:*   the   specified repeated character   (to be searched for and possibly ''squeezed''):\n:*   the   original string and its length\n:*   the       resultant string and its length\n:*   the above strings should be \"bracketed\" with   '''<<<'''   and   '''>>>'''   (to delineate blanks)\n;*   <<<<<<Guillemets may be used instead for \"bracketing\" for the more artistic programmers,   shown used here>>>>>>\n\n\n\nUse (at least) the following five strings,   all strings are length seventy-two (characters, including blanks),   except\nthe 1st string:\n\n                                                                                   immediately\n  string                                                                            repeated\n  number                                                                            character\n                                                                                      ( |   a blank,  a minus,  a seven,  a period)\n         ++\n    1    |+-----------------------------------------------------------------------+    ' '    <######  a null string  (length zero)\n    2    |\"If I were two-faced, would I be wearing this one?\" --- Abraham Lincoln |    '-'\n    3    |..1111111111111111111111111111111111111111111111111111111111111117777888|    '7'\n    4    |I never give 'em hell, I just tell the truth, and they think it's hell. |    '.'\n    5    |                                                    --- Harry S Truman  |  (below)  <######  has many repeated blanks\n         +------------------------------------------------------------------------+     |\n                                                                                        |\n                                                                                        |\n         For the 5th string  (Truman's signature line),  use each of these  specified immediately  repeated characters:\n                                   *  a blank\n                                   *  a minus\n                                   *  a lowercase  '''r'''\n\n\nNote:   there should be seven results shown,   one each for the 1st four strings,   and three results for\nthe 5th string.\n\n\n", "solution": "from itertools import groupby\n\ndef squeezer(s, txt):\n    return ''.join(item if item == s else ''.join(grp)\n                   for item, grp in groupby(txt))\n\nif __name__ == '__main__':\n    strings = [\n            \"\",\n            '\"If I were two-faced, would I be wearing this one?\" --- Abraham Lincoln ',\n            \"..1111111111111111111111111111111111111111111111111111111111111117777888\",\n            \"I never give 'em hell, I just tell the truth, and they think it's hell. \",\n            \"                                                   ---  Harry S Truman  \",\n            \"The better the 4-wheel drive, the further you'll be from help when ya get stuck!\",\n            \"headmistressship\",\n            \"aardvark\",\n            \"\ud83d\ude0d\ud83d\ude00\ud83d\ude4c\ud83d\udc83\ud83d\ude0d\ud83d\ude0d\ud83d\ude0d\ud83d\ude4c\",\n            ]\n    squeezers = ' ,-,7,., -r,e,s,a,\ud83d\ude0d'.split(',')\n    for txt, chars in zip(strings, squeezers):\n        this = \"Original\"\n        print(f\"\\n{this:14} Size: {len(txt)} \u00ab\u00ab\u00ab{txt}\u00bb\u00bb\u00bb\" )\n        for ch in chars:\n            this = f\"Squeezer '{ch}'\"\n            sqz = squeezer(ch, txt)\n            print(f\"{this:>14} Size: {len(sqz)} \u00ab\u00ab\u00ab{sqz}\u00bb\u00bb\u00bb\" )"}
{"title": "Dice game probabilities", "language": "Python", "task": "Two players have a set of dice each. The first player has nine dice with four faces each, with numbers one to four. The second player has six normal dice with six faces each, each face has the usual numbers from one to six.\n\nThey roll their dice and sum the totals of the faces. The player with the highest total wins (it's a draw if the totals are the same). What's the probability of the first player beating the second player?\n\nLater the two players use a different set of dice each. Now the first player has five dice with ten faces each, and the second player has six dice with seven faces each. Now what's the probability of the first player beating the second player?\n\nThis task was adapted from the Project Euler Problem n.205:\nhttps://projecteuler.net/problem=205\n\n", "solution": "from itertools import product\n\ndef gen_dict(n_faces, n_dice):\n    counts = [0] * ((n_faces + 1) * n_dice)\n    for t in product(range(1, n_faces + 1), repeat=n_dice):\n        counts[sum(t)] += 1\n    return counts, n_faces ** n_dice\n\ndef beating_probability(n_sides1, n_dice1, n_sides2, n_dice2):\n    c1, p1 = gen_dict(n_sides1, n_dice1)\n    c2, p2 = gen_dict(n_sides2, n_dice2)\n    p12 = float(p1 * p2)\n\n    return sum(p[1] * q[1] / p12\n               for p, q in product(enumerate(c1), enumerate(c2))\n               if p[0] > q[0])\n\nprint beating_probability(4, 9, 6, 6)\nprint beating_probability(10, 5, 7, 6)"}
{"title": "Digital root", "language": "Python", "task": "The digital root, X, of a number, n, is calculated:\n: find X as the sum of the digits of n\n: find a new X by summing the digits of X, repeating until X has only one digit.\n\nThe additive persistence is the number of summations required to obtain the single digit. \n\nThe task is to calculate the additive persistence and the digital root of a number, e.g.:\n:627615 has additive persistence 2 and digital root of 9;\n:39390 has additive persistence 2 and digital root of 6;\n:588225 has additive persistence 2 and digital root of 3;\n:393900588225 has additive persistence 2 and digital root of 9;\n\nThe digital root may be calculated in bases other than 10.\n\n\n;See:\n* [[Casting out nines]] for this wiki's use of this procedure.\n* [[Digital root/Multiplicative digital root]]\n* [[Sum digits of an integer]]\n* Digital root sequence on OEIS\n* Additive persistence sequence on OEIS\n* [[Iterated digits squaring]]\n\n", "solution": "def digital_root (n):\n    ap = 0\n    n = abs(int(n))\n    while n >= 10:\n        n = sum(int(digit) for digit in str(n))\n        ap += 1\n    return ap, n\n\nif __name__ == '__main__':\n    for n in [627615, 39390, 588225, 393900588225, 55]:\n        persistance, root = digital_root(n)\n        print(\"%12i has additive persistance %2i and digital root %i.\" \n              % (n, persistance, root))"}
{"title": "Disarium numbers", "language": "Python", "task": "A '''Disarium number''' is an integer where the sum of each digit raised to the power of its position in the number, is equal to the number.\n\n\n;E.G.\n\n'''135''' is a '''Disarium number''':\n\n 11 + 32 + 53 == 1 + 9 + 125 == 135\n\nThere are a finite number of '''Disarium numbers'''.\n\n\n;Task\n\n* Find and display the first 18 '''Disarium numbers'''.\n\n\n;Stretch\n\n* Find and display all 20 '''Disarium numbers'''.\n\n\n;See also\n\n;* Geeks for Geeks - Disarium numbers\n;* OEIS:A032799 - Numbers n such that n equals the sum of its digits raised to the consecutive powers (1,2,3,...)\n;* Related task: Narcissistic decimal number\n;* Related task: Own digits power sum  ''Which seems to be the same task as Narcissistic decimal number...''\n\n", "solution": "#!/usr/bin/python\n\ndef isDisarium(n):\n    digitos = len(str(n))\n    suma = 0\n    x = n\n    while x != 0:\n        suma += (x % 10) ** digitos\n        digitos -= 1\n        x //= 10\n    if suma == n:\n        return True\n    else:\n        return False\n\nif __name__ == '__main__':\n    limite = 19\n    cont = 0\n    n = 0\n    print(\"The first\",limite,\"Disarium numbers are:\")\n    while cont < limite:\n        if isDisarium(n):\n            print(n, end = \" \")\n            cont += 1\n        n += 1"}
{"title": "Display a linear combination", "language": "Python", "task": "Display a finite linear combination in an infinite vector basis (e_1, e_2,\\ldots).\n\nWrite a function that, when given a finite list of scalars (\\alpha^1,\\alpha^2,\\ldots), creates a string representing the linear combination \\sum_i\\alpha^i e_i in an explicit format often used in mathematics, that is:\n\n:\\alpha^{i_1}e_{i_1}\\pm|\\alpha^{i_2}|e_{i_2}\\pm|\\alpha^{i_3}|e_{i_3}\\pm\\ldots\n\nwhere \\alpha^{i_k}\\neq 0\n\nThe output must comply to the following rules:\n*   don't show null terms, unless the whole combination is null. \n:::::::  '''e(1)'''     is fine,     '''e(1) + 0*e(3)'''     or     '''e(1) + 0'''     is wrong.\n*   don't show scalars when they are equal to one or minus one. \n:::::::  '''e(3)'''     is fine,     '''1*e(3)'''     is wrong.\n*   don't prefix by a minus sign if it follows a preceding term.   Instead you use subtraction. \n:::::::  '''e(4) - e(5)'''     is fine,     '''e(4) + -e(5)'''     is wrong.\n\n\nShow here output for the following lists of scalars:\n\n 1)    1,  2,  3\n 2)    0,  1,  2,  3\n 3)    1,  0,  3,  4\n 4)    1,  2,  0\n 5)    0,  0,  0\n 6)    0\n 7)    1,  1,  1\n 8)   -1, -1, -1\n 9)   -1, -2,  0, -3\n10)   -1\n\n\n", "solution": "def linear(x):\n    return ' + '.join(['{}e({})'.format('-' if v == -1 else '' if v == 1 else str(v) + '*', i + 1)\n        for i, v in enumerate(x) if v] or ['0']).replace(' + -', ' - ')\n\nlist(map(lambda x: print(linear(x)), [[1, 2, 3], [0, 1, 2, 3], [1, 0, 3, 4], [1, 2, 0],\n        [0, 0, 0], [0], [1, 1, 1], [-1, -1, -1], [-1, -2, 0, 3], [-1]]))\n"}
{"title": "Display an outline as a nested table", "language": "Python", "task": "{| class=\"wikitable\" style=\"text-align: center;\"\n|-\n| style=\"background: #ffffe6; \" colspan=7 | Display an outline as a nested table.\n|-\n| style=\"background: #ffebd2; \" colspan=3 | Parse the outline to a tree,\n| style=\"background: #f0fff0; \" colspan=2 | count the leaves descending from each node,\n| style=\"background: #e6ffff; \" colspan=2 | and write out a table with 'colspan' values\n|-\n| style=\"background: #ffebd2; \" | measuring the indent of each line,\n| style=\"background: #ffebd2; \" | translating the indentation to a nested structure,\n| style=\"background: #ffebd2; \" | and padding the tree to even depth.\n| style=\"background: #f0fff0; \" | defining the width of a leaf as 1,\n| style=\"background: #f0fff0; \" | and the width of a parent node as a sum.\n| style=\"background: #e6ffff; \" | either as a wiki table,\n| style=\"background: #e6ffff; \" | or as HTML.\n|-\n|  | \n|  | \n|  | \n|  | \n| style=\"background: #f0fff0; \" | (The sum of the widths of its children)\n|  | \n|  | \n|}\n\nThe graphic representation of outlines is a staple of mind-mapping and the planning of papers, reports, and speeches.\n\n;Task:\nGiven a outline with at least 3 levels of indentation, for example:\n\nDisplay an outline as a nested table.\n    Parse the outline to a tree,\n        measuring the indent of each line,\n        translating the indentation to a nested structure,\n        and padding the tree to even depth.\n    count the leaves descending from each node,\n        defining the width of a leaf as 1,\n        and the width of a parent node as a sum.\n            (The sum of the widths of its children)\n    and write out a table with 'colspan' values\n        either as a wiki table,\n        or as HTML.\n\nwrite a program in your language which translates your outline into a nested table, with WikiTable or HTML colspan values attached (where needed) to parent nodes in the nested table.\n\nThe WikiTable at the top of this page was generated from the indented outline shown above, producing the following markup string:\n\n{| class=\"wikitable\" style=\"text-align: center;\"\n|-\n| style=\"background: #ffffe6; \" colspan=7 | Display an outline as a nested table.\n|-\n| style=\"background: #ffebd2; \" colspan=3 | Parse the outline to a tree,\n| style=\"background: #f0fff0; \" colspan=2 | count the leaves descending from each node,\n| style=\"background: #e6ffff; \" colspan=2 | and write out a table with 'colspan' values\n|-\n| style=\"background: #ffebd2; \" | measuring the indent of each line,\n| style=\"background: #ffebd2; \" | translating the indentation to a nested structure,\n| style=\"background: #ffebd2; \" | and padding the tree to even depth.\n| style=\"background: #f0fff0; \" | defining the width of a leaf as 1,\n| style=\"background: #f0fff0; \" | and the width of a parent node as a sum.\n| style=\"background: #e6ffff; \" | either as a wiki table,\n| style=\"background: #e6ffff; \" | or as HTML.\n|-\n|  | \n|  | \n|  | \n|  | \n| style=\"background: #f0fff0; \" | (The sum of the widths of its children)\n|  | \n|  | \n|}\n\n;Extra credit:\nUse background color to distinguish the main stages of your outline, so that the subtree of each node at level two is consistently colored, and the edges between adjacent subtrees are immediately revealed.\n\n\n;Output:\nDisplay your nested table on this page.\n\n", "solution": "\"\"\"Display an outline as a nested table. Requires Python >=3.6.\"\"\"\n\nimport itertools\nimport re\nimport sys\n\nfrom collections import deque\nfrom typing import NamedTuple\n\n\nRE_OUTLINE = re.compile(r\"^((?: |\\t)*)(.+)$\", re.M)\n\nCOLORS = itertools.cycle(\n    [\n        \"#ffffe6\",\n        \"#ffebd2\",\n        \"#f0fff0\",\n        \"#e6ffff\",\n        \"#ffeeff\",\n    ]\n)\n\n\nclass Node:\n    def __init__(self, indent, value, parent, children=None):\n        self.indent = indent\n        self.value = value\n        self.parent = parent\n        self.children = children or []\n\n        self.color = None\n\n    def depth(self):\n        if self.parent:\n            return self.parent.depth() + 1\n        return -1\n\n    def height(self):\n        \"\"\"Height of the subtree rooted at this node.\"\"\"\n        if not self.children:\n            return 0\n        return max(child.height() for child in self.children) + 1\n\n    def colspan(self):\n        if self.leaf:\n            return 1\n        return sum(child.colspan() for child in self.children)\n\n    @property\n    def leaf(self):\n        return not bool(self.children)\n\n    def __iter__(self):\n        # Level order tree traversal.\n        q = deque()\n        q.append(self)\n        while q:\n            node = q.popleft()\n            yield node\n            q.extend(node.children)\n\n\nclass Token(NamedTuple):\n    indent: int\n    value: str\n\n\ndef tokenize(outline):\n    \"\"\"Generate ``Token``s from the given outline.\"\"\"\n    for match in RE_OUTLINE.finditer(outline):\n        indent, value = match.groups()\n        yield Token(len(indent), value)\n\n\ndef parse(outline):\n    \"\"\"Return the given outline as a tree of ``Node``s.\"\"\"\n    # Split the outline into lines and count the level of indentation.\n    tokens = list(tokenize(outline))\n\n    # Parse the tokens into a tree of nodes.\n    temp_root = Node(-1, \"\", None)\n    _parse(tokens, 0, temp_root)\n\n    # Pad the tree so that all branches have the same depth.\n    root = temp_root.children[0]\n    pad_tree(root, root.height())\n\n    return root\n\n\ndef _parse(tokens, index, node):\n    \"\"\"Recursively build a tree of nodes.\n\n    Args:\n        tokens (list): A collection of ``Token``s.\n        index (int): Index of the current token.\n        node (Node): Potential parent or sibling node.\n    \"\"\"\n    # Base case. No more lines.\n    if index >= len(tokens):\n        return\n\n    token = tokens[index]\n\n    if token.indent == node.indent:\n        # A sibling of node\n        current = Node(token.indent, token.value, node.parent)\n        node.parent.children.append(current)\n        _parse(tokens, index + 1, current)\n\n    elif token.indent > node.indent:\n        # A child of node\n        current = Node(token.indent, token.value, node)\n        node.children.append(current)\n        _parse(tokens, index + 1, current)\n\n    elif token.indent < node.indent:\n        # Try the node's parent until we find a sibling.\n        _parse(tokens, index, node.parent)\n\n\ndef pad_tree(node, height):\n    \"\"\"Pad the tree with blank nodes so all branches have the same depth.\"\"\"\n    if node.leaf and node.depth() < height:\n        pad_node = Node(node.indent + 1, \"\", node)\n        node.children.append(pad_node)\n\n    for child in node.children:\n        pad_tree(child, height)\n\n\ndef color_tree(node):\n    \"\"\"Walk the tree and color each node as we go.\"\"\"\n    if not node.value:\n        node.color = \"#F9F9F9\"\n    elif node.depth() <= 1:\n        node.color = next(COLORS)\n    else:\n        node.color = node.parent.color\n\n    for child in node.children:\n        color_tree(child)\n\n\ndef table_data(node):\n    \"\"\"Return an HTML table data element for the given node.\"\"\"\n    indent = \"    \"\n\n    if node.colspan() > 1:\n        colspan = f'colspan=\"{node.colspan()}\"'\n    else:\n        colspan = \"\"\n\n    if node.color:\n        style = f'style=\"background-color: {node.color};\"'\n    else:\n        style = \"\"\n\n    attrs = \" \".join([colspan, style])\n    return f\"{indent}<td{attrs}>{node.value}</td>\"\n\n\ndef html_table(tree):\n    \"\"\"Return the tree as an HTML table.\"\"\"\n    # Number of columns in the table.\n    table_cols = tree.colspan()\n\n    # Running count of columns in the current row.\n    row_cols = 0\n\n    # HTML buffer\n    buf = [\"<table style='text-align: center;'>\"]\n\n    # Breadth first iteration.\n    for node in tree:\n        if row_cols == 0:\n            buf.append(\"  <tr>\")\n\n        buf.append(table_data(node))\n        row_cols += node.colspan()\n\n        if row_cols == table_cols:\n            buf.append(\"  </tr>\")\n            row_cols = 0\n\n    buf.append(\"</table>\")\n    return \"\\n\".join(buf)\n\n\ndef wiki_table_data(node):\n    \"\"\"Return an wiki table data string for the given node.\"\"\"\n    if not node.value:\n        return \"|  |\"\n\n    if node.colspan() > 1:\n        colspan = f\"colspan={node.colspan()}\"\n    else:\n        colspan = \"\"\n\n    if node.color:\n        style = f'style=\"background: {node.color};\"'\n    else:\n        style = \"\"\n\n    attrs = \" \".join([colspan, style])\n    return f\"| {attrs} | {node.value}\"\n\n\ndef wiki_table(tree):\n    \"\"\"Return the tree as a wiki table.\"\"\"\n    # Number of columns in the table.\n    table_cols = tree.colspan()\n\n    # Running count of columns in the current row.\n    row_cols = 0\n\n    # HTML buffer\n    buf = ['{| class=\"wikitable\" style=\"text-align: center;\"']\n\n    for node in tree:\n        if row_cols == 0:\n            buf.append(\"|-\")\n\n        buf.append(wiki_table_data(node))\n        row_cols += node.colspan()\n\n        if row_cols == table_cols:\n            row_cols = 0\n\n    buf.append(\"|}\")\n    return \"\\n\".join(buf)\n\n\ndef example(table_format=\"wiki\"):\n    \"\"\"Write an example table to stdout in either HTML or Wiki format.\"\"\"\n\n    outline = (\n        \"Display an outline as a nested table.\\n\"\n        \"    Parse the outline to a tree,\\n\"\n        \"        measuring the indent of each line,\\n\"\n        \"        translating the indentation to a nested structure,\\n\"\n        \"        and padding the tree to even depth.\\n\"\n        \"    count the leaves descending from each node,\\n\"\n        \"        defining the width of a leaf as 1,\\n\"\n        \"        and the width of a parent node as a sum.\\n\"\n        \"            (The sum of the widths of its children)\\n\"\n        \"    and write out a table with 'colspan' values\\n\"\n        \"        either as a wiki table,\\n\"\n        \"        or as HTML.\"\n    )\n\n    tree = parse(outline)\n    color_tree(tree)\n\n    if table_format == \"wiki\":\n        print(wiki_table(tree))\n    else:\n        print(html_table(tree))\n\n\nif __name__ == \"__main__\":\n    args = sys.argv[1:]\n\n    if len(args) == 1:\n        table_format = args[0]\n    else:\n        table_format = \"wiki\"\n\n    example(table_format)"}
{"title": "Distance and Bearing", "language": "Python", "task": "It is very important in aviation to have knowledge of the nearby airports at any time in flight. \n;Task:\nDetermine the distance and bearing from an Airplane to the 20 nearest Airports whenever requested.\nUse the non-commercial data from openflights.org airports.dat as reference.\n\n\nA request comes from an airplane at position  ( latitude, longitude ): ( '''51.514669, 2.198581''' ).\n\n\nYour report should contain the following information from table airports.dat (column shown in brackets):\n\nName(2), Country(4), ICAO(6), Distance and Bearing calculated from Latitude(7) and Longitude(8).  \n\n\nDistance is measured in nautical miles (NM). Resolution is 0.1 NM.\n\nBearing is measured in degrees (deg). 0deg = 360deg = north then clockwise 90deg = east, 180deg = south, 270deg = west. Resolution is 1deg.\n  \n\n;See:\n:*   openflights.org/data:  Airport, airline and route data\n:*   Movable Type Scripts:  Calculate distance, bearing and more between Latitude/Longitude points\n\n", "solution": "''' Rosetta Code task Distance_and_Bearing '''\n\nfrom math import radians, degrees, sin, cos, asin, atan2, sqrt\nfrom pandas import read_csv\n\n\nEARTH_RADIUS_KM = 6372.8\nTASK_CONVERT_NM =  0.0094174\nAIRPORT_DATA_FILE = 'airports.dat.txt'\n\nQUERY_LATITUDE, QUERY_LONGITUDE = 51.514669, 2.198581\n\n\ndef haversine(lat1, lon1, lat2, lon2):\n    '''\n    Given two latitude, longitude pairs in degrees for two points on the Earth,\n    get distance (nautical miles) and initial direction of travel (degrees)\n    for travel from lat1, lon1 to lat2, lon2\n    '''\n    rlat1, rlon1, rlat2, rlon2 = [radians(x) for x in [lat1, lon1, lat2, lon2]]\n    dlat = rlat2 - rlat1\n    dlon = rlon2 - rlon1\n    arc = sin(dlat / 2) ** 2 + cos(rlat1) * cos(rlat2) * sin(dlon / 2) ** 2\n    clen = 2.0 * degrees(asin(sqrt(arc)))\n    theta = atan2(sin(dlon) * cos(rlat2),\n                  cos(rlat1) * sin(rlat2) - sin(rlat1) * cos(rlat2) * cos(dlon))\n    theta = (degrees(theta) + 360) % 360\n    return EARTH_RADIUS_KM * clen * TASK_CONVERT_NM, theta\n\n\ndef find_nearest_airports(latitude, longitude, wanted=20, csv=AIRPORT_DATA_FILE):\n    ''' Given latitude and longitude, find `wanted` closest airports in database file csv. '''\n    airports = read_csv(csv, header=None, usecols=[1, 3, 5, 6, 7], names=[\n                        'Name', 'Country', 'ICAO', 'Latitude', 'Longitude'])\n    airports['Distance'] = 0.0\n    airports['Bearing'] = 0\n    for (idx, row) in enumerate(airports.itertuples()):\n        distance, bearing = haversine(\n            latitude, longitude, row.Latitude, row.Longitude)\n        airports.at[idx, 'Distance'] = round(distance, ndigits=1)\n        airports.at[idx, 'Bearing'] = int(round(bearing))\n\n    airports.sort_values(by=['Distance'], ignore_index=True, inplace=True)\n    return airports.loc[0:wanted-1, ['Name', 'Country', 'ICAO', 'Distance', 'Bearing']]\n\n\nprint(find_nearest_airports(QUERY_LATITUDE, QUERY_LONGITUDE))\n"}
{"title": "Diversity prediction theorem", "language": "Python 3.7", "task": "The   ''wisdom of the crowd''   is the collective opinion of a group of individuals rather than that of a single expert.\n\nWisdom-of-the-crowds research routinely attributes the superiority of crowd averages over individual judgments to the elimination of individual noise,   an explanation that assumes independence of the individual judgments from each other. \n\nThus the crowd tends to make its best decisions if it is made up of diverse opinions and ideologies.\n\n\nScott E. Page introduced the diversity prediction theorem: \n: ''The squared error of the collective prediction equals the average squared error minus the predictive diversity''. \n\n\nTherefore,   when the diversity in a group is large,   the error of the crowd is small.\n\n\n;Definitions:\n::*   Average Individual Error:   Average of the individual squared errors\n::*   Collective Error:           Squared error of the collective prediction\n::*   Prediction Diversity:       Average squared distance from the individual predictions to the collective prediction\n::*   Diversity Prediction Theorem:   ''Given a crowd of predictive models'',     then\n::::::   Collective Error   =   Average Individual Error   -   Prediction Diversity\n\n;Task:\nFor a given   true   value and a number of number of estimates (from a crowd),   show   (here on this page):\n:::*   the true value   and   the crowd estimates\n:::*   the average error\n:::*   the crowd error\n:::*   the prediction diversity\n\n\nUse   (at least)   these two examples:\n:::*   a true value of   '''49'''   with crowd estimates of:   ''' 48   47   51'''\n:::*   a true value of   '''49'''   with crowd estimates of:   ''' 48   47   51   42'''\n\n\n\n;Also see:\n:*   Wikipedia entry:   Wisdom of the crowd\n:*   University of Michigan: PDF paper         (exists on a web archive,   the ''Wayback Machine'').\n\n", "solution": "'''Diversity prediction theorem'''\n\nfrom itertools import chain\nfrom functools import reduce\n\n\n#  diversityValues :: Num a => a -> [a] ->\n#  { mean-Error :: a, crowd-error :: a, diversity :: a }\ndef diversityValues(x):\n    '''The mean error, crowd error and\n       diversity, for a given observation x\n       and a non-empty list of predictions ps.\n    '''\n    def go(ps):\n        mp = mean(ps)\n        return {\n            'mean-error': meanErrorSquared(x)(ps),\n            'crowd-error': pow(x - mp, 2),\n            'diversity': meanErrorSquared(mp)(ps)\n        }\n    return go\n\n\n# meanErrorSquared :: Num -> [Num] -> Num\ndef meanErrorSquared(x):\n    '''The mean of the squared differences\n       between the observed value x and\n       a non-empty list of predictions ps.\n    '''\n    def go(ps):\n        return mean([\n            pow(p - x, 2) for p in ps\n        ])\n    return go\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Observed value: 49,\n       prediction lists: various.\n    '''\n\n    print(unlines(map(\n        showDiversityValues(49),\n        [\n            [48, 47, 51],\n            [48, 47, 51, 42],\n            [50, '?', 50, {}, 50],  # Non-numeric values.\n            []                      # Missing predictions.\n        ]\n    )))\n    print(unlines(map(\n        showDiversityValues('49'),  # String in place of number.\n        [\n            [50, 50, 50],\n            [40, 35, 40],\n        ]\n    )))\n\n\n# ---------------------- FORMATTING ----------------------\n\n# showDiversityValues :: Num -> [Num] -> Either String String\ndef showDiversityValues(x):\n    '''Formatted string representation\n       of diversity values for a given\n       observation x and a non-empty\n       list of predictions p.\n    '''\n    def go(ps):\n        def showDict(dct):\n            w = 4 + max(map(len, dct.keys()))\n\n            def showKV(a, kv):\n                k, v = kv\n                return a + k.rjust(w, ' ') + (\n                    ' : ' + showPrecision(3)(v) + '\\n'\n                )\n            return 'Predictions: ' + showList(ps) + ' ->\\n' + (\n                reduce(showKV, dct.items(), '')\n            )\n\n        def showProblem(e):\n            return (\n                unlines(map(indented(1), e)) if (\n                    isinstance(e, list)\n                ) else indented(1)(repr(e))\n            ) + '\\n'\n\n        return 'Observation:  ' + repr(x) + '\\n' + (\n            either(showProblem)(showDict)(\n                bindLR(numLR(x))(\n                    lambda n: bindLR(numsLR(ps))(\n                        compose(Right, diversityValues(n))\n                    )\n                )\n            )\n        )\n    return go\n\n\n# ------------------ GENERIC FUNCTIONS -------------------\n\n# Left :: a -> Either a b\ndef Left(x):\n    '''Constructor for an empty Either (option type) value\n       with an associated string.\n    '''\n    return {'type': 'Either', 'Right': None, 'Left': x}\n\n\n# Right :: b -> Either a b\ndef Right(x):\n    '''Constructor for a populated Either (option type) value'''\n    return {'type': 'Either', 'Left': None, 'Right': x}\n\n\n# bindLR (>>=) :: Either a -> (a -> Either b) -> Either b\ndef bindLR(m):\n    '''Either monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.\n    '''\n    def go(mf):\n        return (\n            mf(m.get('Right')) if None is m.get('Left') else m\n        )\n    return go\n\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    def go(f, g):\n        def fg(x):\n            return f(g(x))\n        return fg\n    return reduce(go, fs, identity)\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).\n    '''\n    def go(xs):\n        return chain.from_iterable(map(f, xs))\n    return go\n\n\n# either :: (a -> c) -> (b -> c) -> Either a b -> c\ndef either(fl):\n    '''The application of fl to e if e is a Left value,\n       or the application of fr to e if e is a Right value.\n    '''\n    return lambda fr: lambda e: fl(e['Left']) if (\n        None is e['Right']\n    ) else fr(e['Right'])\n\n\n# identity :: a -> a\ndef identity(x):\n    '''The identity function.'''\n    return x\n\n\n# indented :: Int -> String -> String\ndef indented(n):\n    '''String indented by n multiples\n       of four spaces.\n    '''\n    return lambda s: (4 * ' ' * n) + s\n\n# mean :: [Num] -> Float\ndef mean(xs):\n    '''Arithmetic mean of a list\n       of numeric values.\n    '''\n    return sum(xs) / float(len(xs))\n\n\n# numLR :: a -> Either String Num\ndef numLR(x):\n    '''Either Right x if x is a float or int,\n       or a Left explanatory message.'''\n    return Right(x) if (\n        isinstance(x, (float, int))\n    ) else Left(\n        'Expected number, saw: ' + (\n            str(type(x)) + ' ' + repr(x)\n        )\n    )\n\n\n# numsLR :: [a] -> Either String [Num]\ndef numsLR(xs):\n    '''Either Right xs if all xs are float or int,\n       or a Left explanatory message.'''\n    def go(ns):\n        ls, rs = partitionEithers(map(numLR, ns))\n        return Left(ls) if ls else Right(rs)\n    return bindLR(\n        Right(xs) if (\n            bool(xs) and isinstance(xs, list)\n        ) else Left(\n            'Expected a non-empty list, saw: ' + (\n                str(type(xs)) + ' ' + repr(xs)\n            )\n        )\n    )(go)\n\n\n# partitionEithers :: [Either a b] -> ([a],[b])\ndef partitionEithers(lrs):\n    '''A list of Either values partitioned into a tuple\n       of two lists, with all Left elements extracted\n       into the first list, and Right elements\n       extracted into the second list.\n    '''\n    def go(a, x):\n        ls, rs = a\n        r = x.get('Right')\n        return (ls + [x.get('Left')], rs) if None is r else (\n            ls, rs + [r]\n        )\n    return reduce(go, lrs, ([], []))\n\n\n# showList :: [a] -> String\ndef showList(xs):\n    '''Compact string representation of a list'''\n    return '[' + ','.join(str(x) for x in xs) + ']'\n\n\n# showPrecision :: Int -> Float -> String\ndef showPrecision(n):\n    '''A string showing a floating point number\n       at a given degree of precision.'''\n    def go(x):\n        return str(round(x, n))\n    return go\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.'''\n    return '\\n'.join(xs)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Doomsday rule", "language": "Python", "task": " About the task\nJohn Conway (1937-2020), was a mathematician who also invented several mathematically\noriented computer pastimes, such as the famous Game of Life cellular automaton program.\nDr. Conway invented a simple algorithm for finding the day of the week, given any date.\nThe algorithm was based on calculating the distance of a given date from certain\n\"anchor days\" which follow a pattern for the day of the week upon which they fall.\n\n; Algorithm\nThe formula is calculated assuming that Sunday is 0, Monday 1, and so forth with Saturday 7, and\n\n    doomsday = (Tuesday(or 2) + 5(y mod 4) + 4(y mod 100) + 6(y mod 400)) % 7\n\nwhich, for 2021, is 0 (Sunday).\n\nTo calculate the day of the week, we then count days from a close doomsday,\nwith these as charted here by month, then add the doomsday for the year,\nthen get the remainder after dividing by 7. This should give us the number\ncorresponding to the day of the week for that date.\n\n    Month\t        Doomsday Dates for Month\n    --------------------------------------------\n    January (common years)   3, 10, 17, 24, 31\n    January (leap years)     4, 11, 18, 25\n    February (common years)  7, 14, 21, 28\n    February (leap years)    1, 8, 15, 22, 29\n    March                    7, 14, 21, 28\n    April                    4, 11, 18, 25\n    May                      2, 9, 16, 23, 30\n    June                     6, 13, 20, 27\n    July                     4, 11, 18, 25\n    August                   1, 8, 15, 22, 29\n    September                5, 12, 19, 26\n    October                  3, 10, 17, 24, 31\n    November                 7, 14, 21, 28\n    December                 5, 12, 19, 26\n\n; Task\n\nGiven the following dates:\n\n*   1800-01-06 (January 6, 1800)\n*   1875-03-29 (March 29, 1875)\n*   1915-12-07 (December 7, 1915)\n*   1970-12-23 (December 23, 1970)\n*   2043-05-14 (May 14, 2043)\n*   2077-02-12 (February 12, 2077)\n*   2101-04-02 (April 2, 2101)\n\n\nUse Conway's Doomsday rule to calculate the day of the week for each date.\n\n\n; see also\n*   Doomsday rule\n*   Tomorrow is the Day After Doomsday (p.28)\n\n\n\n\n", "solution": "from datetime import date\nfrom calendar import isleap\n\ndef weekday(d):\n    days = [\"Sunday\", \"Monday\", \"Tuesday\", \"Wednesday\", \"Thursday\",\n            \"Friday\", \"Saturday\"]\n    dooms = [\n        [3, 7, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5],\n        [4, 1, 7, 4, 2, 6, 4, 1, 5, 3, 7, 5]\n    ]\n    \n    c = d.year // 100\n    r = d.year % 100\n    s = r // 12\n    t = r % 12\n    c_anchor = (5 * (c % 4) + 2) % 7\n    doomsday = (s + t + (t // 4) + c_anchor) % 7\n    anchorday = dooms[isleap(d.year)][d.month - 1]\n    weekday = (doomsday + d.day - anchorday + 7) % 7\n    return days[weekday]\n\ndates = [date(*x) for x in\n    [(1800, 1, 6), (1875, 3, 29), (1915, 12, 7), (1970, 12, 23),\n     (2043, 5, 14), (2077, 2, 12), (2101, 4, 2)]\n]\n\nfor d in dates:\n    tense = \"was\" if d < date.today() else \"is\" if d == date.today() else \"will be\"\n    print(\"{} {} a {}\".format(d.strftime(\"%B %d, %Y\"), tense, weekday(d)))"}
{"title": "Dot product", "language": "Python", "task": "Create a function/use an in-built function, to compute the   '''dot product''',   also known as the   '''scalar product'''   of two vectors. \n\nIf possible, make the vectors of arbitrary length.\n\n\nAs an example, compute the dot product of the vectors:\n::::     [1,  3, -5]       and \n::::     [4,      -2, -1]  \n\nIf implementing the dot product of two vectors directly:\n:::*   each vector must be the same length\n:::*   multiply corresponding terms from each vector\n:::*   sum the products   (to produce the answer)\n\n\n;Related task:\n*   [[Vector products]]\n\n", "solution": "def dotp(a,b):\n    assert len(a) == len(b), 'Vector sizes must match'\n    return sum(aterm * bterm for aterm,bterm in zip(a, b))\n\nif __name__ == '__main__':\n    a, b = [1, 3, -5], [4, -2, -1]\n    assert dotp(a,b) == 3"}
{"title": "Dot product", "language": "Python 3.7", "task": "Create a function/use an in-built function, to compute the   '''dot product''',   also known as the   '''scalar product'''   of two vectors. \n\nIf possible, make the vectors of arbitrary length.\n\n\nAs an example, compute the dot product of the vectors:\n::::     [1,  3, -5]       and \n::::     [4,      -2, -1]  \n\nIf implementing the dot product of two vectors directly:\n:::*   each vector must be the same length\n:::*   multiply corresponding terms from each vector\n:::*   sum the products   (to produce the answer)\n\n\n;Related task:\n*   [[Vector products]]\n\n", "solution": "'''Dot product'''\n\nfrom operator import (mul)\n\n\n# dotProduct :: Num a => [a] -> [a] -> Either String a\ndef dotProduct(xs):\n    '''Either the dot product of xs and ys,\n       or a string reporting unmatched vector sizes.\n    '''\n    return lambda ys: Left('vector sizes differ') if (\n        len(xs) != len(ys)\n    ) else Right(sum(map(mul, xs, ys)))\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Dot product of other vectors with [1, 3, -5]'''\n\n    print(\n        fTable(main.__doc__ + ':\\n')(str)(str)(\n            compose(\n                either(append('Undefined :: '))(str)\n            )(dotProduct([1, 3, -5]))\n        )([[4, -2, -1, 8], [4, -2], [4, 2, -1], [4, -2, -1]])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# Left :: a -> Either a b\ndef Left(x):\n    '''Constructor for an empty Either (option type) value\n       with an associated string.\n    '''\n    return {'type': 'Either', 'Right': None, 'Left': x}\n\n\n# Right :: b -> Either a b\ndef Right(x):\n    '''Constructor for a populated Either (option type) value'''\n    return {'type': 'Either', 'Left': None, 'Right': x}\n\n\n# append (++) :: [a] -> [a] -> [a]\n# append (++) :: String -> String -> String\ndef append(xs):\n    '''Two lists or strings combined into one.'''\n    return lambda ys: xs + ys\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# either :: (a -> c) -> (b -> c) -> Either a b -> c\ndef either(fl):\n    '''The application of fl to e if e is a Left value,\n       or the application of fr to e if e is a Right value.\n    '''\n    return lambda fr: lambda e: fl(e['Left']) if (\n        None is e['Right']\n    ) else fr(e['Right'])\n\n\n# FORMATTING ----------------------------------------------\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Draw a clock", "language": "Python", "task": "Draw a clock.  \n\n\nMore specific:\n# Draw a time keeping device. It can be a stopwatch, hourglass, sundial, a mouth counting \"one thousand and one\", anything. Only showing the seconds is required, e.g.: a watch with just a second hand will suffice. However, it must clearly change every second, and the change must cycle every so often (one minute, 30 seconds, etc.) It must be ''drawn''; printing a string of numbers to your terminal doesn't qualify. Both text-based and graphical drawing are OK.\n# The clock is unlikely to be used to control space flights, so it needs not be hyper-accurate, but it should be usable, meaning if one can read the seconds off the clock, it must agree with the system clock.\n# A clock is rarely (never?) a major application: don't be a CPU hog and poll the system timer every microsecond, use a proper timer/signal/event from your system or language instead. For a bad example, many OpenGL programs update the frame-buffer in a busy loop even if no redraw is needed, which is very undesirable for this task.\n# A clock is rarely (never?) a major application: try to keep your code simple and to the point. Don't write something too elaborate or convoluted, instead do whatever is natural, concise and clear in your language.\n\n\n;Key points\n* animate simple object\n* timed event \n* polling system resources \n* code clarity\n\n", "solution": "[http://www.thinkgeek.com/gadgets/watches/6a17/ Think Geek Binary Clock]\n"}
{"title": "Draw a clock", "language": "Python 2.6+, 3.0+", "task": "Draw a clock.  \n\n\nMore specific:\n# Draw a time keeping device. It can be a stopwatch, hourglass, sundial, a mouth counting \"one thousand and one\", anything. Only showing the seconds is required, e.g.: a watch with just a second hand will suffice. However, it must clearly change every second, and the change must cycle every so often (one minute, 30 seconds, etc.) It must be ''drawn''; printing a string of numbers to your terminal doesn't qualify. Both text-based and graphical drawing are OK.\n# The clock is unlikely to be used to control space flights, so it needs not be hyper-accurate, but it should be usable, meaning if one can read the seconds off the clock, it must agree with the system clock.\n# A clock is rarely (never?) a major application: don't be a CPU hog and poll the system timer every microsecond, use a proper timer/signal/event from your system or language instead. For a bad example, many OpenGL programs update the frame-buffer in a busy loop even if no redraw is needed, which is very undesirable for this task.\n# A clock is rarely (never?) a major application: try to keep your code simple and to the point. Don't write something too elaborate or convoluted, instead do whatever is natural, concise and clear in your language.\n\n\n;Key points\n* animate simple object\n* timed event \n* polling system resources \n* code clarity\n\n", "solution": "import time\n\ndef chunks(l, n=5):\n    return [l[i:i+n] for i in range(0, len(l), n)]\n\ndef binary(n, digits=8):\n    n=int(n)\n    return '{0:0{1}b}'.format(n, digits)\n\ndef secs(n):\n    n=int(n)\n    h='x' * n\n    return \"|\".join(chunks(h))\n\ndef bin_bit(h):\n    h=h.replace(\"1\",\"x\")\n    h=h.replace(\"0\",\" \")\n    return \"|\".join(list(h))\n\n\nx=str(time.ctime()).split()\ny=x[3].split(\":\")\n\ns=y[-1]\ny=map(binary,y[:-1])\n\nprint bin_bit(y[0])\nprint\nprint bin_bit(y[1])\nprint\nprint secs(s)"}
{"title": "Draw a rotating cube", "language": "Python 2.7.9", "task": "Task\nDraw a rotating cube. \n\nIt should be oriented with one vertex pointing straight up, and its opposite vertex on the main diagonal (the one farthest away) straight down. It can be solid or wire-frame, and you can use ASCII art if your language doesn't have graphical capabilities. Perspective is optional.\n\n\n;Related tasks\n* Draw a cuboid\n* write language name in 3D ASCII\n\n", "solution": "from visual import *\nscene.title = \"VPython: Draw a rotating cube\"\n\nscene.range = 2\nscene.autocenter = True\n\nprint \"Drag with right mousebutton to rotate view.\"\nprint \"Drag up+down with middle mousebutton to zoom.\"\n\ndeg45 = math.radians(45.0)  # 0.785398163397\n\ncube = box()    # using defaults, see http://www.vpython.org/contents/docs/defaults.html \ncube.rotate( angle=deg45, axis=(1,0,0) )\ncube.rotate( angle=deg45, axis=(0,0,1) )\n\nwhile True:                 # Animation-loop\n    rate(50)\n    cube.rotate( angle=0.005, axis=(0,1,0) )\n"}
{"title": "Draw a sphere", "language": "Python from C", "task": "{{requires|Graphics}}\n\n\n;Task:\nDraw a sphere. \n\nThe sphere can be represented graphically, or in ASCII art, depending on the language capabilities. \n\nEither static or rotational projection is acceptable for this task.\n\n\n;Related tasks:\n* draw a cuboid\n* draw a rotating cube\n* write language name in 3D ASCII\n* draw a Deathstar\n\n", "solution": "import math\n\nshades = ('.',':','!','*','o','e','&','#','%','@')\n\ndef normalize(v):\n\tlen = math.sqrt(v[0]**2 + v[1]**2 + v[2]**2)\n\treturn (v[0]/len, v[1]/len, v[2]/len)\n\ndef dot(x,y):\n\td = x[0]*y[0] + x[1]*y[1] + x[2]*y[2]\n\treturn -d if d < 0 else 0\n\ndef draw_sphere(r, k, ambient, light):\n\tfor i in range(int(math.floor(-r)),int(math.ceil(r)+1)):\n\t\tx = i + 0.5\n\t\tline = ''\n\n\t\tfor j in range(int(math.floor(-2*r)),int(math.ceil(2*r)+1)):\n\t\t\ty = j/2 + 0.5\n\t\t\tif x*x + y*y <= r*r:\n\t\t\t\tvec = normalize((x,y,math.sqrt(r*r - x*x - y*y)))\n\t\t\t\tb = dot(light,vec)**k + ambient\n\t\t\t\tintensity = int((1-b)*(len(shades)-1))\n\t\t\t\tline += shades[intensity] if 0 <= intensity < len(shades) else shades[0]\n\t\t\telse:\n\t\t\t\tline += ' '\n\n\t\tprint(line)\n\nlight = normalize((30,30,-50))\ndraw_sphere(20,4,0.1, light)\ndraw_sphere(10,2,0.4, light)"}
{"title": "Draw a sphere", "language": "Python 2.7.5", "task": "{{requires|Graphics}}\n\n\n;Task:\nDraw a sphere. \n\nThe sphere can be represented graphically, or in ASCII art, depending on the language capabilities. \n\nEither static or rotational projection is acceptable for this task.\n\n\n;Related tasks:\n* draw a cuboid\n* draw a rotating cube\n* write language name in 3D ASCII\n* draw a Deathstar\n\n", "solution": "from __future__ import print_function, division\nfrom visual import *\n\ntitle = \"VPython: Draw a sphere\"\nscene.title = title\nprint( \"%s\\n\" % title )\n\nprint( 'Drag with right mousebutton to rotate view'  )\nprint( 'Drag up+down with middle mousebutton to zoom')\n\nscene.autocenter = True\n\n# uncomment any (or all) of those variants:\nS1 = sphere(pos=(0.0, 0.0, 0.0), radius=1.0, color=color.blue)\n#S2 = sphere(pos=(2.0, 0.0, 0.0), radius=1.0, material=materials.earth)\n#S3 = sphere(pos=(0.0, 2.0, 0.0), radius=1.0, material=materials.BlueMarble)\n#S4 = sphere(pos=(0.0, 0.0, 2.0), radius=1.0,\n#            color=color.orange, material=materials.marble)\n\nwhile True:                 # Animation-loop\n    rate(100)\n    pass                    # no animation in this demo\n"}
{"title": "Dutch national flag problem", "language": "Python", "task": "The Dutch national flag is composed of three coloured bands in the order:\n::*   red     (top)\n::*   then white,   and\n::*   lastly blue   (at the bottom). \n\n\nThe problem posed by Edsger Dijkstra is:\n:Given a number of red, blue and white balls in random order, arrange them in the order of the colours in the Dutch national flag.\nWhen the problem was first posed, Dijkstra then went on to successively refine a solution, minimising the number of swaps and the number of times the colour of a ball needed to determined and restricting the balls to end in an array, ...\n\n\n;Task\n# Generate a randomized order of balls ''ensuring that they are not in the order of the Dutch national flag''.\n# Sort the balls in a way idiomatic to your language.\n# Check the sorted balls ''are'' in the order of the Dutch national flag.\n\n\n;C.f.:\n* Dutch national flag problem\n* Probabilistic analysis of algorithms for the Dutch national flag problem by Wei-Mei Chen. (pdf)\n\n", "solution": "import random\n\ncolours_in_order = 'Red White Blue'.split()\n\ndef dutch_flag_sort(items, order=colours_in_order):\n    'return sort of items using the given order'\n    reverse_index = dict((x,i) for i,x in enumerate(order))\n    return sorted(items, key=lambda x: reverse_index[x])\n\ndef dutch_flag_check(items, order=colours_in_order):\n    'Return True if each item of items is in the given order'\n    reverse_index = dict((x,i) for i,x in enumerate(order))\n    order_of_items = [reverse_index[item] for item in items]\n    return all(x <= y for x, y in zip(order_of_items, order_of_items[1:]))\n\ndef random_balls(mx=5):\n    'Select from 1 to mx balls of each colour, randomly'\n    balls = sum([[colour] * random.randint(1, mx)\n                 for colour in colours_in_order], [])\n    random.shuffle(balls)\n    return balls\n\ndef main():\n    # Ensure we start unsorted\n    while True:\n        balls = random_balls()\n        if not dutch_flag_check(balls):\n            break\n    print(\"Original Ball order:\", balls)\n    sorted_balls = dutch_flag_sort(balls)\n    print(\"Sorted Ball Order:\", sorted_balls)\n    assert dutch_flag_check(sorted_balls), 'Whoops. Not sorted!'\n\nif __name__ == '__main__':\n    main()"}
{"title": "EKG sequence convergence", "language": "Python", "task": "The sequence is from the natural numbers and is defined by:\n* a(1) = 1; \n* a(2) = Start = 2;\n* for n > 2, a(n) shares at least one prime factor with a(n-1) and is the ''smallest'' such natural number ''not already used''.\n\nThe sequence is called the EKG sequence (after its visual similarity to an electrocardiogram when graphed).\n\nVariants of the sequence can be generated starting 1, N where N is any natural number larger than one. For the purposes of this task let us call: \n* The sequence described above , starting 1, 2, ... the EKG(2) sequence;\n* the sequence starting 1, 3, ... the EKG(3) sequence; \n* ... the sequence starting 1, N, ... the EKG(N) sequence.\n\n\n;Convergence\nIf an algorithm that keeps track of the minimum amount of numbers and their corresponding prime factors used to generate the next term is used, then this may be known as the generators essential '''state'''. Two EKG generators with differing starts can converge to produce the same sequence after initial differences.\nEKG(N1) and EKG(N2) are said to to have converged at and after generation a(c) if state_of(EKG(N1).a(c)) == state_of(EKG(N2).a(c)).\n\n\n;Task:\n# Calculate and show here the first 10 members of EKG(2).\n# Calculate and show here the first 10 members of EKG(5).\n# Calculate and show here the first 10 members of EKG(7).\n# Calculate and show here the first 10 members of EKG(9).\n# Calculate and show here the first 10 members of EKG(10).\n# Calculate and show here at which term EKG(5) and EKG(7) converge   ('''stretch goal''').\n\n;Related Tasks:\n# [[Greatest common divisor]]\n# [[Sieve of Eratosthenes]]\n\n;Reference:\n* The EKG Sequence and the Tree of Numbers. (Video).\n\n", "solution": "from itertools import count, islice, takewhile\nfrom math import gcd\n\ndef EKG_gen(start=2):\n    \"\"\"\\\n    Generate the next term of the EKG together with the minimum cache of \n    numbers left in its production; (the \"state\" of the generator).\n    Using math.gcd\n    \"\"\"\n    c = count(start + 1)\n    last, so_far = start, list(range(2, start))\n    yield 1, []\n    yield last, []\n    while True:\n        for index, sf in enumerate(so_far):\n            if gcd(last, sf) > 1:\n                last = so_far.pop(index)\n                yield last, so_far[::]\n                break\n        else:\n            so_far.append(next(c))\n\ndef find_convergence(ekgs=(5,7)):\n    \"Returns the convergence point or zero if not found within the limit\"\n    ekg = [EKG_gen(n) for n in ekgs]\n    for e in ekg:\n        next(e)    # skip initial 1 in each sequence\n    return 2 + len(list(takewhile(lambda state: not all(state[0] == s for  s in state[1:]),\n                                  zip(*ekg))))\n\nif __name__ == '__main__':\n    for start in 2, 5, 7, 9, 10:\n        print(f\"EKG({start}):\", str([n[0] for n in islice(EKG_gen(start), 10)])[1: -1])\n    print(f\"\\nEKG(5) and EKG(7) converge at term {find_convergence(ekgs=(5,7))}!\")"}
{"title": "Earliest difference between prime gaps", "language": "Python from Julia, Wren", "task": "When calculating prime numbers > 2, the difference between adjacent primes is always an even number. This difference, also referred to as the gap, varies in an random pattern; at least, no pattern has ever been discovered, and it is strongly conjectured that no pattern exists. However, it is also conjectured that between some two adjacent primes will be a gap corresponding to every positive even integer.\n\n\n\n{|class=\"wikitable\"\n!gap!!minimalstartingprime!!endingprime\n|-\n|2||3||5\n|-\n|4||7||11\n|-\n|6||23||29\n|-\n|8||89||97\n|-\n|10||139||149\n|-\n|12||199||211\n|-\n|14||113||127\n|-\n|16||1831||1847\n|-\n|18||523||541\n|-\n|20||887||907\n|-\n|22||1129||1151\n|-\n|24||1669||1693\n|-\n|26||2477||2503\n|-\n|28||2971||2999\n|-\n|30||4297||4327\n|}\n\nThis task involves locating the minimal primes corresponding to those gaps.\n\nThough every gap value exists, they don't seem to come in any particular order. For example, this table shows the gaps and minimum starting value primes for 2 through 30:\n\n\nFor the purposes of this task, considering only primes greater than 2, consider prime gaps that differ by exactly two to be adjacent.\n\n\n\n;Task\n\nFor each order of magnitude '''m''' from '''101''' through '''106''':\n\n* Find the first two sets of minimum adjacent prime gaps where the absolute value of the difference between the prime gap start values is greater than '''m'''.\n\n\n;E.G.\n\nFor an '''m''' of '''101'''; \n\nThe start value of gap 2 is 3, the start value of gap 4 is 7, the difference is 7 - 3 or 4. 4 < '''101''' so keep going.\n\nThe start value of gap 4 is 7, the start value of gap 6 is 23, the difference is 23 - 7, or 16. 16 > '''101''' so this the earliest adjacent gap difference > '''101'''.\n\n\n;Stretch goal\n\n* Do the same for '''107''' and '''108''' (and higher?) orders of magnitude\n\nNote: the earliest value found for each order of magnitude may not be unique, in fact, ''is'' not unique; also, with the gaps in ascending order, the minimal starting values are not strictly ascending.\n\n", "solution": "\"\"\" https://rosettacode.org/wiki/Earliest_difference_between_prime_gaps \"\"\"\n\nfrom primesieve import primes\n\nLIMIT = 10**9\npri = primes(LIMIT * 5)\ngapstarts = {}\nfor i in range(1, len(pri)):\n    if pri[i] - pri[i - 1] not in gapstarts:\n        gapstarts[pri[i] - pri[i - 1]] = pri[i - 1]\n\nPM, GAP1, = 10, 2\nwhile True:\n    while GAP1 not in gapstarts:\n        GAP1 += 2\n    start1 = gapstarts[GAP1]\n    GAP2 = GAP1 + 2\n    if GAP2 not in gapstarts:\n        GAP1 = GAP2 + 2\n        continue\n    start2 = gapstarts[GAP2]\n    diff = abs(start2 - start1)\n    if diff > PM:\n        print(f\"Earliest difference >{PM: ,} between adjacent prime gap starting primes:\")\n        print(f\"Gap {GAP1} starts at{start1: ,}, gap {GAP2} starts at{start2: ,}, difference is{diff: ,}.\\n\")\n        if PM == LIMIT:\n            break\n        PM *= 10\n    else:\n        GAP1 = GAP2\n"}
{"title": "Eban numbers", "language": "Python", "task": "Definition:\nAn   '''eban'''   number is a number that has no letter    '''e'''    in it when the number is spelled in English.\n\nOr more literally,   spelled numbers that contain the letter   '''e'''   are banned.\n\n\nThe American version of spelling numbers will be used here   (as opposed to the British).\n\n'''2,000,000,000'''   is two billion,   ''not''   two milliard.\n\n\nOnly numbers less than   '''one sextillion'''   ('''1021''')   will be considered in/for this task.\n\nThis will allow optimizations to be used.\n\n\n\n;Task:\n:::*   show all eban numbers   <=   '''1,000'''   (in a horizontal format),                   and a count\n:::*   show all eban numbers between   '''1,000'''   and   '''4,000'''   (inclusive),   and a count\n:::*   show a count of all eban numbers up and including            '''10,000'''\n:::*   show a count of all eban numbers up and including                '''100,000'''\n:::*   show a count of all eban numbers up and including                   '''1,000,000'''\n:::*   show a count of all eban numbers up and including                      '''10,000,000'''\n:::*   show all output here.\n\n\n;See also:\n:*   The MathWorld entry:    eban numbers.\n:*   The OEIS      entry:   A6933,                eban numbers.\n:*   [[Number names]].\n\n", "solution": "# Use inflect\n\n\"\"\"\n\n  show all eban numbers <= 1,000 (in a horizontal format), and a count\n  show all eban numbers between 1,000 and 4,000 (inclusive), and a count\n  show a count of all eban numbers up and including 10,000\n  show a count of all eban numbers up and including 100,000\n  show a count of all eban numbers up and including 1,000,000\n  show a count of all eban numbers up and including 10,000,000\n  \n\"\"\"\n\nimport inflect\nimport time\n\nbefore = time.perf_counter()\n\np = inflect.engine()\n\n# eban numbers <= 1000\n\nprint(' ')\nprint('eban numbers up to and including 1000:')\nprint(' ')\n\ncount = 0\n\nfor i in range(1,1001):\n    if not 'e' in p.number_to_words(i):\n        print(str(i)+' ',end='')\n        count += 1\n        \nprint(' ')\nprint(' ')\nprint('count = '+str(count))\nprint(' ')\n\n# eban numbers 1000 to 4000\n\nprint(' ')\nprint('eban numbers between 1000 and 4000 (inclusive):')\nprint(' ')\n\ncount = 0\n\nfor i in range(1000,4001):\n    if not 'e' in p.number_to_words(i):\n        print(str(i)+' ',end='')\n        count += 1\n        \nprint(' ')\nprint(' ')\nprint('count = '+str(count))\nprint(' ')\n\n# eban numbers up to 10000\n\nprint(' ')\nprint('eban numbers up to and including 10000:')\nprint(' ')\n\ncount = 0\n\nfor i in range(1,10001):\n    if not 'e' in p.number_to_words(i):\n        count += 1\n        \nprint(' ')\nprint('count = '+str(count))\nprint(' ')\n\n# eban numbers up to 100000\n\nprint(' ')\nprint('eban numbers up to and including 100000:')\nprint(' ')\n\ncount = 0\n\nfor i in range(1,100001):\n    if not 'e' in p.number_to_words(i):\n        count += 1\n        \nprint(' ')\nprint('count = '+str(count))\nprint(' ')\n\n# eban numbers up to 1000000\n\nprint(' ')\nprint('eban numbers up to and including 1000000:')\nprint(' ')\n\ncount = 0\n\nfor i in range(1,1000001):\n    if not 'e' in p.number_to_words(i):\n        count += 1\n        \nprint(' ')\nprint('count = '+str(count))\nprint(' ')\n\n# eban numbers up to 10000000\n\nprint(' ')\nprint('eban numbers up to and including 10000000:')\nprint(' ')\n\ncount = 0\n\nfor i in range(1,10000001):\n    if not 'e' in p.number_to_words(i):\n        count += 1\n        \nprint(' ')\nprint('count = '+str(count))\nprint(' ')\n\nafter = time.perf_counter()\n\nprint(\" \")\nprint(\"Run time in seconds: \"+str(after - before))\n"}
{"title": "Eertree", "language": "Python", "task": "An '''eertree''' is a data structure designed for efficient processing of certain palindrome tasks, for instance counting the number of sub-palindromes in an input string. \n\nThe data structure has commonalities to both ''tries'' and ''suffix trees''.\n  See links below. \n\n\n;Task:\nConstruct an eertree for the string \"eertree\", then output all sub-palindromes by traversing the tree.\n\n\n;See also:\n*   Wikipedia entry:   trie.\n*   Wikipedia entry:   suffix tree \n*   Cornell University Library, Computer Science, Data Structures and Algorithms ---> EERTREE: An Efficient Data Structure for Processing Palindromes in Strings.\n\n", "solution": "#!/bin/python\nfrom __future__ import print_function\n\nclass Node(object):\n\tdef __init__(self):\n\t\tself.edges = {} # edges (or forward links)\n\t\tself.link = None # suffix link (backward links)\n\t\tself.len = 0 # the length of the node\n\nclass Eertree(object):\n\tdef __init__(self):\n\t\tself.nodes = []\n\t\t# two initial root nodes\n\t\tself.rto = Node() #odd length root node, or node -1\n\t\tself.rte = Node() #even length root node, or node 0\n\n\t\t# Initialize empty tree\n\t\tself.rto.link = self.rte.link = self.rto;\n\t\tself.rto.len = -1\n\t\tself.rte.len = 0\n\t\tself.S = [0] # accumulated input string, T=S[1..i]\n\t\tself.maxSufT = self.rte # maximum suffix of tree T\n\n\tdef get_max_suffix_pal(self, startNode, a):\n\t\t# We traverse the suffix-palindromes of T in the order of decreasing length.\n\t\t# For each palindrome we read its length k and compare T[i-k] against a\n\t\t# until we get an equality or arrive at the -1 node.\n\t\tu = startNode\n\t\ti = len(self.S)\n\t\tk = u.len\n\t\twhile id(u) != id(self.rto) and self.S[i - k - 1] != a:\n\t\t\tassert id(u) != id(u.link) #Prevent infinte loop\n\t\t\tu = u.link\n\t\t\tk = u.len\n\n\t\treturn u\n\t\n\tdef add(self, a):\n\n\t\t# We need to find the maximum suffix-palindrome P of Ta\n\t\t# Start by finding maximum suffix-palindrome Q of T.\n\t\t# To do this, we traverse the suffix-palindromes of T\n\t\t# in the order of decreasing length, starting with maxSuf(T)\n\t\tQ = self.get_max_suffix_pal(self.maxSufT, a)\n\n\t\t# We check Q to see whether it has an outgoing edge labeled by a.\n\t\tcreateANewNode = not a in Q.edges\n\n\t\tif createANewNode:\n\t\t\t# We create the node P of length Q+2\n\t\t\tP = Node()\n\t\t\tself.nodes.append(P)\n\t\t\tP.len = Q.len + 2\n\t\t\tif P.len == 1:\n\t\t\t\t# if P = a, create the suffix link (P,0)\n\t\t\t\tP.link = self.rte\n\t\t\telse:\n\t\t\t\t# It remains to create the suffix link from P if |P|>1. Just\n\t\t\t\t# continue traversing suffix-palindromes of T starting with the suffix \n\t\t\t\t# link of Q.\n\t\t\t\tP.link = self.get_max_suffix_pal(Q.link, a).edges[a]\n\n\t\t\t# create the edge (Q,P)\n\t\t\tQ.edges[a] = P\n\n\t\t#P becomes the new maxSufT\n\t\tself.maxSufT = Q.edges[a]\n\n\t\t#Store accumulated input string\n\t\tself.S.append(a)\n\n\t\treturn createANewNode\n\t\n\tdef get_sub_palindromes(self, nd, nodesToHere, charsToHere, result):\n\t\t#Each node represents a palindrome, which can be reconstructed\n\t\t#by the path from the root node to each non-root node.\n\n\t\t#Traverse all edges, since they represent other palindromes\n\t\tfor lnkName in nd.edges:\n\t\t\tnd2 = nd.edges[lnkName] #The lnkName is the character used for this edge\n\t\t\tself.get_sub_palindromes(nd2, nodesToHere+[nd2], charsToHere+[lnkName], result)\n\n\t\t#Reconstruct based on charsToHere characters.\n\t\tif id(nd) != id(self.rto) and id(nd) != id(self.rte): #Don't print for root nodes\n\t\t\ttmp = \"\".join(charsToHere)\n\t\t\tif id(nodesToHere[0]) == id(self.rte): #Even string\n\t\t\t\tassembled = tmp[::-1] + tmp\n\t\t\telse: #Odd string\n\t\t\t\tassembled = tmp[::-1] + tmp[1:]\n\t\t\tresult.append(assembled)\n\nif __name__==\"__main__\":\n\tst = \"eertree\"\n\tprint (\"Processing string\", st)\n\teertree = Eertree()\n\tfor ch in st:\n\t\teertree.add(ch)\n\n\tprint (\"Number of sub-palindromes:\", len(eertree.nodes))\n\n\t#Traverse tree to find sub-palindromes\n\tresult = []\n\teertree.get_sub_palindromes(eertree.rto, [eertree.rto], [], result) #Odd length words\n\teertree.get_sub_palindromes(eertree.rte, [eertree.rte], [], result) #Even length words\n\tprint (\"Sub-palindromes:\", result)"}
{"title": "Egyptian division", "language": "Python", "task": "Egyptian division is a method of dividing integers using addition and \ndoubling that is similar to the algorithm of [[Ethiopian multiplication]]\n\n'''Algorithm:'''\n\nGiven two numbers where the '''dividend''' is to be divided by the '''divisor''':\n\n# Start the construction of a table of two columns: '''powers_of_2''', and '''doublings'''; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.\n# Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.\n# Continue with successive i'th rows of 2^i and 2^i * divisor.\n# Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.\n# We now assemble two separate sums that both start as zero, called here '''answer''' and '''accumulator'''\n# Consider each row of the table, in the ''reverse'' order of its construction.\n# If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.\n# When the first row has been considered as above, then the integer division of dividend by divisor is given by answer. (And the remainder is given by the absolute value of accumulator - dividend).\n\n\n'''Example: 580 / 34'''\n\n''' Table creation: '''\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n|-\n| 1\n| 34\n|-\n| 2\n| 68\n|-\n| 4\n| 136\n|-\n| 8\n| 272\n|-\n| 16\n| 544\n|}\n\n''' Initialization of sums: '''\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| 16\n| 544\n|\n\n|\n\n|-\n|\n|\n| 0\n| 0\n|}\n\n''' Considering table rows, bottom-up: '''\n\nWhen a row is considered it is shown crossed out if it is not accumulated, or '''bold''' if the row causes summations.\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n| 16\n| 544\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n| 16\n| 544\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n| 16\n| 544\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n| 16\n| 544\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| '''1'''\n| '''34'''\n| 17\n| 578\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n;Answer:\nSo 580 divided by 34 using the Egyptian method is '''17''' remainder (578 - 580) or '''2'''.\n\n\n;Task:\nThe task is to create a function that does Egyptian division. The function should\nclosely follow the description above in using a list/array of powers of two, and\nanother of doublings.\n\n* Functions should be clear interpretations of the algorithm.\n* Use the function to divide 580 by 34 and show the answer '''here, on this page'''.\n\n\n;Related tasks:\n:*   Egyptian fractions\n\n\n;References:\n:*   Egyptian Number System\n\n", "solution": "from itertools import product\n\ndef egyptian_divmod(dividend, divisor):\n    assert divisor != 0\n    pwrs, dbls = [1], [divisor]\n    while dbls[-1] <= dividend:\n        pwrs.append(pwrs[-1] * 2)\n        dbls.append(pwrs[-1] * divisor)\n    ans, accum = 0, 0\n    for pwr, dbl in zip(pwrs[-2::-1], dbls[-2::-1]):\n        if accum + dbl <= dividend:\n            accum += dbl\n            ans += pwr\n    return ans, abs(accum - dividend)\n\nif __name__ == \"__main__\":\n    # Test it gives the same results as the divmod built-in\n    for i, j in product(range(13), range(1, 13)):\n            assert egyptian_divmod(i, j) == divmod(i, j)\n    # Mandated result\n    i, j = 580, 34\n    print(f'{i} divided by {j} using the Egyption method is %i remainder %i'\n          % egyptian_divmod(i, j))"}
{"title": "Egyptian division", "language": "Python from Haskell", "task": "Egyptian division is a method of dividing integers using addition and \ndoubling that is similar to the algorithm of [[Ethiopian multiplication]]\n\n'''Algorithm:'''\n\nGiven two numbers where the '''dividend''' is to be divided by the '''divisor''':\n\n# Start the construction of a table of two columns: '''powers_of_2''', and '''doublings'''; by a first row of a 1 (i.e. 2^0) in the first column and 1 times the divisor in the first row second column.\n# Create the second row with columns of 2 (i.e 2^1), and 2 * divisor in order.\n# Continue with successive i'th rows of 2^i and 2^i * divisor.\n# Stop adding rows, and keep only those rows, where 2^i * divisor is less than or equal to the dividend.\n# We now assemble two separate sums that both start as zero, called here '''answer''' and '''accumulator'''\n# Consider each row of the table, in the ''reverse'' order of its construction.\n# If the current value of the accumulator added to the doublings cell would be less than or equal to the dividend then add it to the accumulator, as well as adding the powers_of_2 cell value to the answer.\n# When the first row has been considered as above, then the integer division of dividend by divisor is given by answer. (And the remainder is given by the absolute value of accumulator - dividend).\n\n\n'''Example: 580 / 34'''\n\n''' Table creation: '''\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n|-\n| 1\n| 34\n|-\n| 2\n| 68\n|-\n| 4\n| 136\n|-\n| 8\n| 272\n|-\n| 16\n| 544\n|}\n\n''' Initialization of sums: '''\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| 16\n| 544\n|\n\n|\n\n|-\n|\n|\n| 0\n| 0\n|}\n\n''' Considering table rows, bottom-up: '''\n\nWhen a row is considered it is shown crossed out if it is not accumulated, or '''bold''' if the row causes summations.\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n| 16\n| 544\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n| 16\n| 544\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n| 16\n| 544\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| 1\n| 34\n|\n\n|\n\n|-\n| 2\n| 68\n| 16\n| 544\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n::: {| class=\"wikitable\"\n! powers_of_2\n! doublings\n! answer\n! accumulator\n|-\n| '''1'''\n| '''34'''\n| 17\n| 578\n|-\n| 2\n| 68\n|\n\n|\n\n|-\n| 4\n| 136\n|\n\n|\n\n|-\n| 8\n| 272\n|\n\n|\n\n|-\n| '''16'''\n| '''544'''\n|\n\n|\n\n|}\n\n;Answer:\nSo 580 divided by 34 using the Egyptian method is '''17''' remainder (578 - 580) or '''2'''.\n\n\n;Task:\nThe task is to create a function that does Egyptian division. The function should\nclosely follow the description above in using a list/array of powers of two, and\nanother of doublings.\n\n* Functions should be clear interpretations of the algorithm.\n* Use the function to divide 580 by 34 and show the answer '''here, on this page'''.\n\n\n;Related tasks:\n:*   Egyptian fractions\n\n\n;References:\n:*   Egyptian Number System\n\n", "solution": "'''Quotient and remainder of division by the Rhind papyrus method.'''\n\nfrom functools import reduce\n\n\n# eqyptianQuotRem :: Int -> Int -> (Int, Int)\ndef eqyptianQuotRem(m):\n    '''Quotient and remainder derived by the Eqyptian method.'''\n\n    def expansion(xi):\n        '''Doubled value, and next power of two - both by self addition.'''\n        x, i = xi\n        return Nothing() if x > m else Just(\n            ((x + x, i + i), xi)\n        )\n\n    def collapse(qr, ix):\n        '''Addition of a power of two to the quotient,\n           and subtraction of a paired value from the remainder.'''\n        i, x = ix\n        q, r = qr\n        return (q + i, r - x) if x < r else qr\n\n    return lambda n: reduce(\n        collapse,\n        unfoldl(expansion)(\n            (1, n)\n        ),\n        (0, m)\n    )\n\n\n# ------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Test'''\n    print(\n        eqyptianQuotRem(580)(34)\n    )\n\n\n# ------------------- GENERIC FUNCTIONS -------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# unfoldl(lambda x: Just(((x - 1), x)) if 0 != x else Nothing())(10)\n# -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n# unfoldl :: (b -> Maybe (b, a)) -> b -> [a]\ndef unfoldl(f):\n    '''Dual to reduce or foldl.\n       Where these reduce a list to a summary value, unfoldl\n       builds a list from a seed value.\n       Where f returns Just(a, b), a is appended to the list,\n       and the residual b is used as the argument for the next\n       application of f.\n       When f returns Nothing, the completed list is returned.\n    '''\n    def go(v):\n        x, r = v, v\n        xs = []\n        while True:\n            mb = f(x)\n            if mb.get('Nothing'):\n                return xs\n            else:\n                x, r = mb.get('Just')\n                xs.insert(0, r)\n        return xs\n    return go\n\n\n# MAIN ----------------------------------------------------\nif __name__ == '__main__':\n    main()"}
{"title": "Elementary cellular automaton", "language": "Python", "task": "An '''cellular automaton where there are two possible states (labeled 0 and 1) and the rule to determine the state of a cell in the next generation depends only on the current state of the cell and its two immediate neighbors.  Those three values can be encoded with three bits.\n\nThe rules of evolution are then encoded with eight bits indicating the outcome of each of the eight possibilities 111, 110, 101, 100, 011, 010, 001 and 000 in this order.  Thus for instance the rule 13 means that a state is updated to 1 only in the cases 011, 010 and 000, since 13 in binary is 0b00001101.\n\n\n;Task:\nCreate a subroutine, program or function that allows to create and visualize the evolution of any of the 256 possible elementary cellular automaton of arbitrary space length and for any given initial state.  You can demonstrate your solution with any automaton of your choice.\n\nThe space state should ''wrap'':  this means that the left-most cell should be considered as the right neighbor of the right-most cell, and reciprocally.\n\nThis task is basically a generalization of [[one-dimensional cellular automata]].\n\n\n;See also\n* Cellular automata (natureofcode.com)\n\n", "solution": "def eca(cells, rule):\n    lencells = len(cells)\n    c = \"0\" + cells + \"0\"    # Zero pad the ends\n    rulebits = '{0:08b}'.format(rule)\n    neighbours2next = {'{0:03b}'.format(n):rulebits[::-1][n] for n in range(8)}\n    yield c[1:-1]\n    while True:\n        c = ''.join(['0',\n                     ''.join(neighbours2next[c[i-1:i+2]]\n                             for i in range(1,lencells+1)),\n                     '0'])\n        yield c[1:-1]\n\nif __name__ == '__main__':\n    lines, start, rules = 50, '0000000001000000000', (90, 30, 122)\n    zipped = [range(lines)] + [eca(start, rule) for rule in rules]\n    print('\\n   Rules: %r' % (rules,))\n    for data in zip(*zipped):\n        i = data[0]\n        cells = data[1:]\n        print('%2i: %s' % (i, '    '.join(cells).replace('0', '.').replace('1', '#')))"}
{"title": "Elementary cellular automaton/Infinite length", "language": "Python", "task": "The purpose of this task is to create a version of an [[Elementary cellular automaton]] whose number of cells is only limited by the memory size of the computer.\n\nTo be precise, consider the state of the automaton to be made of an infinite number of cells, but with a bounded support.  In other words, to describe the state of the automaton, you need a finite number of adjacent cells, along with their individual state, and you then consider that the individual state of each of all other cells is the negation of the closest individual cell among the previously defined finite number of cells.\n\nExamples:\n\n\n1        ->   ..., 0, 0,      1,      0, 0, ...\n0, 1     ->   ..., 1, 1,   0, 1,      0, 0, ...\n1, 0, 1  ->   ..., 0, 0,   1, 0, 1,   0, 0, ...\n\n\nMore complex methods can be imagined, provided it is possible to somehow encode the infinite sections.  But for this task we will stick to this simple version.\n\n", "solution": "def _notcell(c):\n    return '0' if c == '1' else '1'\n\ndef eca_infinite(cells, rule):\n    lencells = len(cells)\n    rulebits = '{0:08b}'.format(rule)\n    neighbours2next = {'{0:03b}'.format(n):rulebits[::-1][n] for n in range(8)}\n    c = cells\n    while True:\n        yield c\n        c = _notcell(c[0])*2 + c + _notcell(c[-1])*2    # Extend and pad the ends\n\n        c = ''.join(neighbours2next[c[i-1:i+2]] for i in range(1,len(c) - 1))\n        #yield c[1:-1]\n\nif __name__ == '__main__':\n    lines = 25\n    for rule in (90, 30):\n        print('\\nRule: %i' % rule)\n        for i, c in zip(range(lines), eca_infinite('1', rule)):\n            print('%2i: %s%s' % (i, ' '*(lines - i), c.replace('0', '.').replace('1', '#')))"}
{"title": "Elementary cellular automaton/Random number generator", "language": "Python", "task": "Mathematica software for its default random number generator.\n\nSteven Wolfram's recommendation for random number generation from rule 30 consists in extracting successive bits in a fixed position in the array of cells, as the automaton changes state.\n\nThe purpose of this task is to demonstrate this.  With the code written in the most significant.\n\nYou can pick which ever length you want for the initial array but it should be visible in the code so that your output can be reproduced with an other language.\n\nFor extra-credits, you will make this algorithm run as fast as possible in your language, for instance with an extensive use of bitwise logic.\n\n;Reference:\n* Cellular automata: Is Rule 30 random? (PDF).\n\n\n", "solution": "from elementary_cellular_automaton import eca, eca_wrap\n\ndef rule30bytes(lencells=100):\n    cells = '1' + '0' * (lencells - 1)\n    gen = eca(cells, 30)\n    while True:\n        yield int(''.join(next(gen)[0] for i in range(8)), 2)\n\nif __name__ == '__main__':\n    print([b for i,b in zip(range(10), rule30bytes())])"}
{"title": "Elliptic curve arithmetic", "language": "Python from C", "task": "digital signatures. \n\nThe purpose of this task is to implement a simplified (without modular arithmetic) version of the elliptic curve arithmetic which is required by the   elliptic curve DSA   protocol. \n\nIn a nutshell, an elliptic curve is a bi-dimensional curve defined by the following relation between the '''x''' and '''y''' coordinates of any point on the curve:\n\n::::   y^2 = x^3 + a x + b\n\n'''a''' and '''b''' are arbitrary parameters that define the specific curve which is used.  \n\nFor this particular task, we'll use the following parameters:\n\n::::    a=0,   b=7 \n\nThe most interesting thing about elliptic curves is the fact that it is possible to define a   group   structure on it.  \n\nTo do so we define an   internal composition   rule with an additive notation '''+''',   such that for any three distinct points '''P''', '''Q''' and '''R''' on the curve, whenever these points are aligned, we have:\n\n::::    P + Q + R = 0  \n\nHere   '''0'''   (zero)   is the ''infinity point'',   for which the '''x''' and '''y''' values are not defined.   It's basically the same kind of point which defines the horizon in   projective geometry.  \n\nWe'll also assume here that this infinity point is unique and defines the   neutral element   of the addition.\n\nThis was not the definition of the addition, but only its desired property.    For a more accurate definition, we proceed as such:\n\nGiven any three aligned points '''P''', '''Q''' and '''R''',   we define the sum   '''S = P + Q'''   as the point (possibly the infinity point) such that   '''S''', '''R'''   and the infinity point are aligned.\n\nConsidering the symmetry of the curve around the x-axis, it's easy to convince oneself that two points '''S''' and '''R''' can be aligned with the infinity point if and only if '''S''' and '''R''' are symmetric of one another towards the x-axis   (because in that case there is no other candidate than the infinity point to complete the alignment triplet).\n\n'''S''' is thus defined as the symmetric of '''R''' towards the '''x''' axis.\n\nThe task consists in defining the addition which, for any two points of the curve, returns the sum of these two points.   You will pick two random points on the curve, compute their sum and show that the symmetric of the sum is aligned with the two initial points. \n\nYou will use the '''a''' and '''b''' parameters of secp256k1, i.e.  respectively zero and seven.\n\n''Hint'':   You might need to define a \"doubling\" function, that returns '''P+P''' for any given point '''P'''.\n\n''Extra credit'':   define the full elliptic curve arithmetic (still not modular, though) by defining a \"multiply\" function that returns, \nfor any point '''P''' and integer '''n''',   the point '''P + P + ... + P'''     ('''n''' times).\n\n", "solution": "#!/usr/bin/env python3\n\nclass Point:\n    b = 7\n    def __init__(self, x=float('inf'), y=float('inf')):\n        self.x = x\n        self.y = y\n\n    def copy(self):\n        return Point(self.x, self.y)\n\n    def is_zero(self):\n        return self.x > 1e20 or self.x < -1e20\n\n    def neg(self):\n        return Point(self.x, -self.y)\n\n    def dbl(self):\n        if self.is_zero():\n            return self.copy()\n        try:\n            L = (3 * self.x * self.x) / (2 * self.y)\n        except ZeroDivisionError:\n            return Point()\n        x = L * L - 2 * self.x\n        return Point(x, L * (self.x - x) - self.y)\n\n    def add(self, q):\n        if self.x == q.x and self.y == q.y:\n            return self.dbl()\n        if self.is_zero():\n            return q.copy()\n        if q.is_zero():\n            return self.copy()\n        try:\n            L = (q.y - self.y) / (q.x - self.x)\n        except ZeroDivisionError:\n            return Point()\n        x = L * L - self.x - q.x\n        return Point(x, L * (self.x - x) - self.y)\n\n    def mul(self, n):\n        p = self.copy()\n        r = Point()\n        i = 1\n        while i <= n:\n            if i&n:\n                r = r.add(p)\n            p = p.dbl()\n            i <<= 1\n        return r\n\n    def __str__(self):\n        return \"({:.3f}, {:.3f})\".format(self.x, self.y)\n\ndef show(s, p):\n    print(s, \"Zero\" if p.is_zero() else p)\n\ndef from_y(y):\n    n = y * y - Point.b\n    x = n**(1./3) if n>=0 else -((-n)**(1./3))\n    return Point(x, y)\n\n# demonstrate\na = from_y(1)\nb = from_y(2)\nshow(\"a =\", a)\nshow(\"b =\", b)\nc = a.add(b)\nshow(\"c = a + b =\", c)\nd = c.neg()\nshow(\"d = -c =\", d)\nshow(\"c + d =\", c.add(d))\nshow(\"a + b + d =\", a.add(b.add(d)))\nshow(\"a * 12345 =\", a.mul(12345))"}
{"title": "Empty directory", "language": "Python 2.x", "task": "Starting with a path to some directory, determine whether the directory is empty.\n\nAn empty directory contains no files nor subdirectories. \nWith [[Unix]] or [[Windows]] systems, every directory contains an entry for \".\" and almost every directory contains \"..\" (except for a root directory); an empty directory contains no other entries.\n\n", "solution": "import os;\nif os.listdir(raw_input(\"directory\")):\n    print \"not empty\"\nelse:\n    print \"empty\"\n"}
{"title": "Empty string", "language": "Python", "task": "Languages may have features for dealing specifically with empty strings \n(those containing no characters).\n\n\n;Task:\n::*   Demonstrate how to assign an empty string to a variable.\n::*   Demonstrate how to check that a string is empty.\n::*   Demonstrate how to check that a string is not empty.\n\n\n\n", "solution": "s = ''\n# or:\ns = str()\n\nif not s or s == '':\n   print(\"String is empty\")\n\nif len(s) == 0:\n    print(\"String is empty\")\nelse:\n    print(\"String not empty\")\n\n\n# boolean test function for python2 and python3\n# test for regular (non-unicode) strings\n# unicode strings\n# None \ndef emptystring(s):\n   if isinstance(s, (''.__class__ , u''.__class__) ):\n      if len(s) == 0: \n         return True\n      else \n         return False\n\n   elif s is None:\n        return True\n"}
{"title": "Entropy/Narcissist", "language": "Python 3.4", "task": "Write a computer program that computes and shows its own   [[entropy]].\n\n\n;Related Tasks: \n:*   [[Fibonacci_word]]\n:*   [[Entropy]]\n\n", "solution": "import math\nfrom collections import Counter\n\ndef entropy(s):\n    p, lns = Counter(s), float(len(s))\n    return -sum( count/lns * math.log(count/lns, 2) for count in p.values())\n\nwith open(__file__) as f:\n    b=f.read()\n    \nprint(entropy(b))"}
{"title": "Equilibrium index", "language": "Python", "task": "An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices. \n\n\nFor example, in a sequence   A:\n\n:::::   A_0 = -7\n:::::   A_1 =  1\n:::::   A_2 =  5\n:::::   A_3 =  2\n:::::   A_4 = -4\n:::::   A_5 =  3\n:::::   A_6 =  0\n\n3   is an equilibrium index, because:\n\n:::::   A_0 + A_1 + A_2 = A_4 + A_5 + A_6\n\n6   is also an equilibrium index, because:\n\n:::::   A_0 + A_1 + A_2 + A_3 + A_4 + A_5 = 0\n\n(sum of zero elements is zero) \n\n7   is not an equilibrium index, because it is not a valid index of sequence A.\n\n\n;Task;\nWrite a function that, given a sequence, returns its equilibrium indices (if any).\n\nAssume that the sequence may be very long.\n\n", "solution": "f = (eqindex2Pass, eqindexMultiPass, eqindex1Pass)\nd = ([-7, 1, 5, 2, -4, 3, 0],\n     [2, 4, 6],\n     [2, 9, 2],\n     [1, -1, 1, -1, 1, -1, 1])\n\nfor data in d:\n    print(\"d = %r\" % data)\n    for func in f:\n        print(\"  %16s(d) -> %r\" % (func.__name__, list(func(data))))"}
{"title": "Esthetic numbers", "language": "Python 3.9.5", "task": "An '''esthetic number''' is a positive integer where every adjacent digit differs from its neighbour by 1.\n\n\n;E.G.\n\n:* '''12''' is an esthetic number. One and two differ by 1.\n\n:* '''5654''' is an esthetic number. Each digit is exactly 1 away from its neighbour.\n\n:* '''890''' is '''not''' an esthetic number. Nine and zero differ by 9.\n\n\nThese examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers ''are'' included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.\n\nEsthetic numbers are also sometimes referred to as stepping numbers.\n\n\n;Task\n\n:* Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.\n\n:* Use that routine to find esthetic numbers in bases '''2''' through '''16''' and display, here on this page, the esthectic numbers from index '''(base x 4)''' through index '''(base x 6)''', inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)\n\n:* Find and display, here on this page, the '''base 10''' esthetic numbers with a magnitude between '''1000''' and '''9999'''.\n\n:* Stretch: Find and display, here on this page, the '''base 10''' esthetic numbers with a magnitude between '''1.0e8''' and '''1.3e8'''.\n\n\n;Related task:\n*   numbers with equal rises and falls\n\n\n;See also:\n:;*OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1\n:;*Numbers Aplenty - Esthetic numbers\n:;*Geeks for Geeks - Stepping numbers\n\n", "solution": "from collections import deque\nfrom itertools import dropwhile, islice, takewhile\nfrom textwrap import wrap\nfrom typing import Iterable, Iterator\n\n\nDigits = str  # Alias for the return type of to_digits()\n\n\ndef esthetic_nums(base: int) -> Iterator[int]:\n    \"\"\"Generate the esthetic number sequence for a given base\n\n    >>> list(islice(esthetic_nums(base=10), 20))\n    [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 21, 23, 32, 34, 43, 45, 54, 56, 65]\n    \"\"\"\n    queue: deque[tuple[int, int]] = deque()\n    queue.extendleft((d, d) for d in range(1, base))\n    while True:\n        num, lsd = queue.pop()\n        yield num\n        new_lsds = (d for d in (lsd - 1, lsd + 1) if 0 <= d < base)\n        num *= base  # Shift num left one digit\n        queue.extendleft((num + d, d) for d in new_lsds)\n\n\ndef to_digits(num: int, base: int) -> Digits:\n    \"\"\"Return a representation of an integer as digits in a given base\n\n    >>> to_digits(0x3def84f0ce, base=16)\n    '3def84f0ce'\n    \"\"\"\n    digits: list[str] = []\n    while num:\n        num, d = divmod(num, base)\n        digits.append(\"0123456789abcdef\"[d])\n    return \"\".join(reversed(digits)) if digits else \"0\"\n\n\ndef pprint_it(it: Iterable[str], indent: int = 4, width: int = 80) -> None:\n    \"\"\"Pretty print an iterable which returns strings\n\n    >>> pprint_it(map(str, range(20)), indent=0, width=40)\n    0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,\n    12, 13, 14, 15, 16, 17, 18, 19\n    <BLANKLINE>\n    \"\"\"\n    joined = \", \".join(it)\n    lines = wrap(joined, width=width - indent)\n    for line in lines:\n        print(f\"{indent*' '}{line}\")\n    print()\n\n\ndef task_2() -> None:\n    nums: Iterator[int]\n    for base in range(2, 16 + 1):\n        start, stop = 4 * base, 6 * base\n        nums = esthetic_nums(base)\n        nums = islice(nums, start - 1, stop)  # start and stop are 1-based indices\n        print(\n            f\"Base-{base} esthetic numbers from \"\n            f\"index {start} through index {stop} inclusive:\\n\"\n        )\n        pprint_it(to_digits(num, base) for num in nums)\n\n\ndef task_3(lower: int, upper: int, base: int = 10) -> None:\n    nums: Iterator[int] = esthetic_nums(base)\n    nums = dropwhile(lambda num: num < lower, nums)\n    nums = takewhile(lambda num: num <= upper, nums)\n    print(\n        f\"Base-{base} esthetic numbers with \"\n        f\"magnitude between {lower:,} and {upper:,}:\\n\"\n    )\n    pprint_it(to_digits(num, base) for num in nums)\n\n\nif __name__ == \"__main__\":\n    print(\"======\\nTask 2\\n======\\n\")\n    task_2()\n\n    print(\"======\\nTask 3\\n======\\n\")\n    task_3(1_000, 9_999)\n\n    print(\"======\\nTask 4\\n======\\n\")\n    task_3(100_000_000, 130_000_000)\n\n"}
{"title": "Esthetic numbers", "language": "Python from Haskell", "task": "An '''esthetic number''' is a positive integer where every adjacent digit differs from its neighbour by 1.\n\n\n;E.G.\n\n:* '''12''' is an esthetic number. One and two differ by 1.\n\n:* '''5654''' is an esthetic number. Each digit is exactly 1 away from its neighbour.\n\n:* '''890''' is '''not''' an esthetic number. Nine and zero differ by 9.\n\n\nThese examples are nominally in base 10 but the concept extends easily to numbers in other bases. Traditionally, single digit numbers ''are'' included in esthetic numbers; zero may or may not be. For our purposes, for this task, do not include zero (0) as an esthetic number. Do not include numbers with leading zeros.\n\nEsthetic numbers are also sometimes referred to as stepping numbers.\n\n\n;Task\n\n:* Write a routine (function, procedure, whatever) to find esthetic numbers in a given base.\n\n:* Use that routine to find esthetic numbers in bases '''2''' through '''16''' and display, here on this page, the esthectic numbers from index '''(base x 4)''' through index '''(base x 6)''', inclusive. (E.G. for base 2: 8th through 12th, for base 6: 24th through 36th, etc.)\n\n:* Find and display, here on this page, the '''base 10''' esthetic numbers with a magnitude between '''1000''' and '''9999'''.\n\n:* Stretch: Find and display, here on this page, the '''base 10''' esthetic numbers with a magnitude between '''1.0e8''' and '''1.3e8'''.\n\n\n;Related task:\n*   numbers with equal rises and falls\n\n\n;See also:\n:;*OEIS A033075 - Positive numbers n such that all pairs of consecutive decimal digits differ by 1\n:;*Numbers Aplenty - Esthetic numbers\n:;*Geeks for Geeks - Stepping numbers\n\n", "solution": "'''Esthetic numbers'''\n\nfrom functools import reduce\nfrom itertools import (\n    accumulate, chain, count, dropwhile,\n    islice, product, takewhile\n)\nfrom operator import add\nfrom string import digits, ascii_lowercase\nfrom textwrap import wrap\n\n\n# estheticNumbersInBase :: Int -> [Int]\ndef estheticNumbersInBase(b):\n    '''Infinite stream of numbers which are\n       esthetic in a given base.\n    '''\n    return concatMap(\n        compose(\n            lambda deltas: concatMap(\n                lambda headDigit: concatMap(\n                    compose(\n                        fromBaseDigits(b),\n                        scanl(add)(headDigit)\n                    )\n                )(deltas)\n            )(range(1, b)),\n            replicateList([-1, 1])\n        )\n    )(count(0))\n\n\n# ------------------------ TESTS -------------------------\ndef main():\n    '''Specified tests'''\n    def samples(b):\n        i, j = b * 4, b * 6\n        return '\\n'.join([\n            f'Esthetics [{i}..{j}] for base {b}:',\n            unlines(wrap(\n                unwords([\n                    showInBase(b)(n) for n in compose(\n                        drop(i - 1), take(j)\n                    )(\n                        estheticNumbersInBase(b)\n                    )\n                ]), 60\n            ))\n        ])\n\n    def takeInRange(a, b):\n        return compose(\n            dropWhile(lambda x: x < a),\n            takeWhile(lambda x: x <= b)\n        )\n\n    print(\n        '\\n\\n'.join([\n            samples(b) for b in range(2, 1 + 16)\n        ])\n    )\n    for (lo, hi) in [(1000, 9999), (100_000_000, 130_000_000)]:\n        print(f'\\nBase 10 Esthetics in range [{lo}..{hi}]:')\n        print(\n            unlines(wrap(\n                unwords(\n                    str(x) for x in takeInRange(lo, hi)(\n                        estheticNumbersInBase(10)\n                    )\n                ), 60\n            ))\n        )\n\n\n# ------------------- BASES AND DIGITS -------------------\n\n# fromBaseDigits :: Int -> [Int] -> [Int]\ndef fromBaseDigits(b):\n    '''An empty list if any digits are out of range for\n       the base. Otherwise a list containing an integer.\n    '''\n    def go(digitList):\n        maybeNum = reduce(\n            lambda r, d: None if r is None or (\n                0 > d or d >= b\n            ) else r * b + d,\n            digitList, 0\n        )\n        return [] if None is maybeNum else [maybeNum]\n    return go\n\n\n# toBaseDigits :: Int -> Int -> [Int]\ndef toBaseDigits(b):\n    '''A list of the digits of n in base b.\n    '''\n    def f(x):\n        return None if 0 == x else (\n            divmod(x, b)[::-1]\n        )\n    return lambda n: list(reversed(unfoldr(f)(n)))\n\n\n# showInBase :: Int -> Int -> String\ndef showInBase(b):\n    '''String representation of n in base b.\n    '''\n    charSet = digits + ascii_lowercase\n    return lambda n: ''.join([\n        charSet[i] for i in toBaseDigits(b)(n)\n    ])\n\n\n# ----------------------- GENERIC ------------------------\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    def go(f, g):\n        def fg(x):\n            return f(g(x))\n        return fg\n    return reduce(go, fs, lambda x: x)\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been\n       mapped.\n       The list monad can be derived by using a function f\n       which wraps its output in a list, (using an empty\n       list to represent computational failure).\n    '''\n    def go(xs):\n        return chain.from_iterable(map(f, xs))\n    return go\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.\n    '''\n    def go(xs):\n        if isinstance(xs, (list, tuple, str)):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return go\n\n\n# dropWhile :: (a -> Bool) -> [a] -> [a]\n# dropWhile :: (Char -> Bool) -> String -> String\ndef dropWhile(p):\n    '''The suffix remainining after takeWhile p xs.\n    '''\n    return lambda xs: list(\n        dropwhile(p, xs)\n    )\n\n\n# replicateList :: [a] -> Int -> [[a]]\ndef replicateList(xs):\n    '''All distinct lists of length n that\n       consist of elements drawn from xs.\n    '''\n    def rep(n):\n        def go(x):\n            return [[]] if 1 > x else [\n                ([a] + b) for (a, b) in product(\n                    xs, go(x - 1)\n                )\n            ]\n        return go(n)\n    return rep\n\n\n# scanl :: (b -> a -> b) -> b -> [a] -> [b]\ndef scanl(f):\n    '''scanl is like reduce, but defines a succession of\n       intermediate values, building from the left.\n    '''\n    def go(a):\n        def g(xs):\n            return accumulate(chain([a], xs), f)\n        return g\n    return go\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    def go(xs):\n        return list(islice(xs, n))\n    return go\n\n\n# takeWhile :: (a -> Bool) -> [a] -> [a]\n# takeWhile :: (Char -> Bool) -> String -> String\ndef takeWhile(p):\n    '''The longest (possibly empty) prefix of xs\n       in which all elements satisfy p.\n    '''\n    return lambda xs: list(\n        takewhile(p, xs)\n    )\n\n\n# unfoldr :: (b -> Maybe (a, b)) -> b -> [a]\ndef unfoldr(f):\n    '''Dual to reduce or foldr.\n       Where catamorphism reduces a list to a summary value,\n       the anamorphic unfoldr builds a list from a seed value.\n       As long as f returns (a, b) a is prepended to the list,\n       and the residual b is used as the argument for the next\n       application of f.\n       When f returns None, the completed list is returned.\n    '''\n    def go(v):\n        xr = v, v\n        xs = []\n        while True:\n            xr = f(xr[1])\n            if None is not xr:\n                xs.append(xr[0])\n            else:\n                return xs\n    return go\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string formed by the intercalation\n       of a list of strings with the newline character.\n    '''\n    return '\\n'.join(xs)\n\n\n# unwords :: [String] -> String\ndef unwords(xs):\n    '''A space-separated string derived\n       from a list of words.\n    '''\n    return ' '.join(xs)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Euclid-Mullin sequence", "language": "Python", "task": "Definition\nThe Euclid-Mullin sequence is an infinite sequence of distinct prime numbers, in which each element is the least prime factor of one plus the product of all earlier elements. \n\nThe first element is usually assumed to be 2. So the second element is : (2) + 1 = 3 and the third element is : (2 x 3) + 1 = 7 as this is prime.\n\nAlthough intermingled with smaller elements, the sequence can produce very large elements quite quickly and only the first 51 have been computed at the time of writing.\n\n;Task\nCompute and show here the first '''16''' elements of the sequence or, if your language does not support arbitrary precision arithmetic, as many as you can.\n\n;Stretch goal\nCompute the next '''11''' elements of the sequence.\n\n;Reference\nOEIS sequence A000945\n\n", "solution": "\"\"\" Rosetta code task: Euclid-Mullin_sequence \"\"\"\n\nfrom primePy import primes\n\ndef euclid_mullin():\n    \"\"\" generate Euclid-Mullin sequence \"\"\"\n    total = 1\n    while True:\n        next_iter = primes.factor(total + 1)\n        total *= next_iter\n        yield next_iter\n\nGEN = euclid_mullin()\nprint('First 16 Euclid-Mullin numbers:', ', '.join(str(next(GEN)) for _ in range(16)))\n"}
{"title": "Euler's identity", "language": "Python", "task": "{{Wikipedia|Euler's_identity}}\n\n\nIn mathematics, ''Euler's identity'' is the equality:\n\n                ei\\pi + 1 = 0\n\nwhere\n\n    e is Euler's number, the base of natural logarithms,\n    ''i'' is the imaginary unit, which satisfies ''i''2 = -1, and\n    \\pi is pi, the ratio of the circumference of a circle to its diameter.\n\nEuler's identity is often cited as an example of deep mathematical beauty. Three of the basic arithmetic operations occur exactly once each: addition, multiplication, and exponentiation. The identity also links five fundamental mathematical constants:\n\n    The number 0.\n    The number 1.\n    The number \\pi (\\pi = 3.14159+),\n    The number e (e = 2.71828+), which occurs widely in mathematical analysis.\n    The number ''i'', the imaginary unit of the complex numbers.\n\n;Task\nShow in your language that Euler's identity is true. As much as possible and practical, mimic the Euler's identity equation. \n\nMost languages are limited to IEEE 754 floating point calculations so will have some error in the calculation. \n\nIf that is the case, or there is some other limitation, show \nthat ei\\pi + 1 is ''approximately'' equal to zero and \nshow the amount of error in the calculation.\n\nIf your language is capable of symbolic calculations, show \nthat ei\\pi + 1 is ''exactly'' equal to zero for bonus kudos points.\n\n", "solution": ">>> import math\n>>> math.e ** (math.pi * 1j) + 1\n1.2246467991473532e-16j"}
{"title": "Euler's sum of powers conjecture", "language": "Python", "task": "There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.\n\nThis conjecture is called Euler's sum of powers conjecture and can be stated as such:\n\n:At least k positive kth powers are required to sum to a kth power, except for the trivial case of one kth power: yk = yk.\n\nIn 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.\n\nThe task consists in writing a program to search for an integer solution of x_0^5 + x_1^5 + x_2^5 + x_3^5 = y^5 where all x_i and y are distinct integers between 0 and 250 (exclusive). Show an answer here.\n\nRelated tasks are:\n* [[Pythagorean quadruples]]. \n* [[Pythagorean triples]].\n\n\n\n", "solution": "def eulers_sum_of_powers():\n    max_n = 250\n    pow_5 = [n**5 for n in range(max_n)]\n    pow5_to_n = {n**5: n for n in range(max_n)}\n    for x0 in range(1, max_n):\n        for x1 in range(1, x0):\n            for x2 in range(1, x1):\n                for x3 in range(1, x2):\n                    pow_5_sum = sum(pow_5[i] for i in (x0, x1, x2, x3))\n                    if pow_5_sum in pow5_to_n:\n                        y = pow5_to_n[pow_5_sum]\n                        return (x0, x1, x2, x3, y)\n\nprint(\"%i**5 + %i**5 + %i**5 + %i**5 == %i**5\" % eulers_sum_of_powers())"}
{"title": "Euler's sum of powers conjecture", "language": "Python 2.6+", "task": "There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.\n\nThis conjecture is called Euler's sum of powers conjecture and can be stated as such:\n\n:At least k positive kth powers are required to sum to a kth power, except for the trivial case of one kth power: yk = yk.\n\nIn 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.\n\nThe task consists in writing a program to search for an integer solution of x_0^5 + x_1^5 + x_2^5 + x_3^5 = y^5 where all x_i and y are distinct integers between 0 and 250 (exclusive). Show an answer here.\n\nRelated tasks are:\n* [[Pythagorean quadruples]]. \n* [[Pythagorean triples]].\n\n\n\n", "solution": "from itertools import combinations\n\ndef eulers_sum_of_powers():\n    max_n = 250\n    pow_5 = [n**5 for n in range(max_n)]\n    pow5_to_n = {n**5: n for n in range(max_n)}\n    for x0, x1, x2, x3 in combinations(range(1, max_n), 4):\n        pow_5_sum = sum(pow_5[i] for i in (x0, x1, x2, x3))\n        if pow_5_sum in pow5_to_n:\n            y = pow5_to_n[pow_5_sum]\n            return (x0, x1, x2, x3, y)\n\nprint(\"%i**5 + %i**5 + %i**5 + %i**5 == %i**5\" % eulers_sum_of_powers())"}
{"title": "Euler's sum of powers conjecture", "language": "Python 3.7", "task": "There is a conjecture in mathematics that held for over two hundred years before it was disproved by the finding of a counterexample in 1966 by Lander and Parkin.\n\nThis conjecture is called Euler's sum of powers conjecture and can be stated as such:\n\n:At least k positive kth powers are required to sum to a kth power, except for the trivial case of one kth power: yk = yk.\n\nIn 1966, Leon J. Lander and Thomas R. Parkin used a brute-force search on a CDC 6600 computer restricting numbers to those less than 250.\n\nThe task consists in writing a program to search for an integer solution of x_0^5 + x_1^5 + x_2^5 + x_3^5 = y^5 where all x_i and y are distinct integers between 0 and 250 (exclusive). Show an answer here.\n\nRelated tasks are:\n* [[Pythagorean quadruples]]. \n* [[Pythagorean triples]].\n\n\n\n", "solution": "'''Euler's sum of powers conjecture'''\n\nfrom itertools import (chain, takewhile)\n\n\n# main :: IO ()\ndef main():\n    '''Search for counter-example'''\n\n    xs = enumFromTo(1)(249)\n\n    powerMap = {x**5: x for x in xs}\n    sumMap = {\n        x**5 + y**5: (x, y)\n        for x in xs[1:]\n        for y in xs if x > y\n    }\n\n    # isExample :: (Int, Int) -> Bool\n    def isExample(ps):\n        p, s = ps\n        return p - s in sumMap\n\n    # display :: (Int, Int) -> String\n    def display(ps):\n        p, s = ps\n        a, b = sumMap[p - s]\n        c, d = sumMap[s]\n        return '^5 + '.join([str(n) for n in [a, b, c, d]]) + (\n            '^5 = ' + str(powerMap[p]) + '^5'\n        )\n\n    print(__doc__ + ' \u2013 counter-example:\\n')\n    print(\n        maybe('No counter-example found.')(display)(\n            find(isExample)(\n                bind(powerMap.keys())(\n                    lambda p: bind(\n                        takewhile(\n                            lambda x: p > x,\n                            sumMap.keys()\n                        )\n                    )(lambda s: [(p, s)])\n                )\n            )\n        )\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: () -> Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# bind (>>=) :: [a] -> (a -> [b]) -> [b]\ndef bind(xs):\n    '''List monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.\n    '''\n    def go(f):\n        return chain.from_iterable(map(f, xs))\n    return go\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: range(m, 1 + n)\n\n\n# find :: (a -> Bool) -> [a] -> Maybe a\ndef find(p):\n    '''Just the first element in the list that matches p,\n       or Nothing if no elements match.\n    '''\n    def go(xs):\n        try:\n            return Just(next(x for x in xs if p(x)))\n        except StopIteration:\n            return Nothing()\n    return go\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if m is Nothing,\n       or the application of f to x,\n       where m is Just(x).\n    '''\n    return lambda f: lambda m: v if (\n        None is m or m.get('Nothing')\n    ) else f(m.get('Just'))\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Even or odd", "language": "Python", "task": "Test whether an integer is even or odd.\n\nThere is more than one way to solve this task:\n\n* Use the even and odd predicates, if the language provides them.\n* Check the least significant digit. With binary integers, ''i iff ''i'' is even, or equals 1 iff ''i'' is odd.\n* Divide ''i'' by 2. The remainder equals 0 iff ''i'' is even. The remainder equals +1 or -1 iff ''i'' is odd.\n* Use modular congruences:\n** ''i''  0 (mod 2) iff ''i'' is even.\n** ''i''  1 (mod 2) iff ''i'' is odd.\n\n", "solution": ">>> def is_odd(i): return bool(i & 1)\n\n>>> def is_even(i): return not is_odd(i)\n\n>>> [(j, is_odd(j)) for j in range(10)]\n[(0, False), (1, True), (2, False), (3, True), (4, False), (5, True), (6, False), (7, True), (8, False), (9, True)]\n>>> [(j, is_even(j)) for j in range(10)]\n[(0, True), (1, False), (2, True), (3, False), (4, True), (5, False), (6, True), (7, False), (8, True), (9, False)]\n>>> "}
{"title": "Evolutionary algorithm", "language": "Python", "task": "Starting with:\n* The target string: \"METHINKS IT IS LIKE A WEASEL\".\n* An array of random characters chosen from the set of upper-case letters together with the space, and of the same length as the target string. (Call it the parent).\n* A fitness function that computes the 'closeness' of its argument to the target string.\n* A mutate function that given a string and a mutation rate returns a copy of the string, with some characters probably mutated.\n* While the parent is not yet the target:\n:* copy the parent C times, each time allowing some random probability that another character might be substituted using mutate.\n:* Assess the fitness of the parent and all the copies to the target and make the most fit string the new parent, discarding the others.\n:* repeat until the parent converges, (hopefully), to the target.\n\n\n;See also:\n*   Wikipedia entry:   Weasel algorithm.\n*   Wikipedia entry:   Evolutionary algorithm.\n\n\nNote: to aid comparison, try and ensure the variables and functions mentioned in the task description appear in solutions\n\nA cursory examination of a few of the solutions reveals that the instructions have not been followed rigorously in some solutions. Specifically,\n* While the parent is not yet the target:\n:* copy the parent C times, each time allowing some random probability that another character might be substituted using mutate.\n\nNote that some of the the solutions given retain characters in the mutated string that are ''correct'' in the target string. However, the instruction above does not state to retain any of the characters while performing the mutation. Although some may believe to do so is implied from the use of \"converges\"\n\n (:* repeat until the parent converges, (hopefully), to the target.\n\nStrictly speaking, the new parent should be selected from the new pool of mutations, and then the new parent used to generate the next set of mutations with parent characters getting retained only by ''not'' being mutated. It then becomes possible that the new set of mutations has no member that is fitter than the parent!\n\nAs illustration of this error, the code for 8th has the following remark.\n\n Create a new string based on the TOS, '''changing randomly any characters which\n don't already match the target''':\n\n''NOTE:'' this has been changed, the 8th version is completely random now\n\nClearly, this algo will be applying the mutation function only to the parent characters that don't match to the target characters!\n\nTo ensure that the new parent is never less fit than the prior parent, both the parent and all of the latest mutations are subjected to the fitness test to select the next parent.\n\n", "solution": "from string import letters\nfrom random import choice, random\n \ntarget  = list(\"METHINKS IT IS LIKE A WEASEL\")\ncharset = letters + ' '\nparent  = [choice(charset) for _ in range(len(target))]\nminmutaterate  = .09\nC = range(100)\n \nperfectfitness = float(len(target))\n    \ndef fitness(trial):\n    'Sum of matching chars by position'\n    return sum(t==h for t,h in zip(trial, target))\n \ndef mutaterate():\n    'Less mutation the closer the fit of the parent'\n    return 1-((perfectfitness - fitness(parent)) / perfectfitness * (1 - minmutaterate))\n \ndef mutate(parent, rate):\n    return [(ch if random() <= rate else choice(charset)) for ch in parent]\n \ndef que():\n    '(from the favourite saying of Manuel in Fawlty Towers)'\n    print (\"#%-4i, fitness: %4.1f%%, '%s'\" %\n           (iterations, fitness(parent)*100./perfectfitness, ''.join(parent)))\n\ndef mate(a, b):\n    place = 0\n    if choice(xrange(10)) < 7:\n        place = choice(xrange(len(target)))\n    else:\n        return a, b\n    \n    return a, b, a[:place] + b[place:], b[:place] + a[place:]\n\niterations = 0\ncenter = len(C)/2\nwhile parent != target:\n    rate = mutaterate()\n    iterations += 1\n    if iterations % 100 == 0: que()\n    copies = [ mutate(parent, rate) for _ in C ]  + [parent]\n    parent1 = max(copies[:center], key=fitness)\n    parent2 = max(copies[center:], key=fitness)\n    parent = max(mate(parent1, parent2), key=fitness)\nque()"}
{"title": "Executable library", "language": "Python", "task": "The general idea behind an executable library is to create a library \nthat when used as a library does one thing; \nbut has the ability to be run directly via command line. \nThus the API comes with a CLI in the very same source code file.\n\n'''Task detail'''\n\n* Create a library/module/dll/shared object/... for a programming language that contains a function/method called hailstone that is a function taking a positive integer and returns the [[Hailstone sequence]] for that number.\n\n* The library, when executed directly should satisfy the remaining requirements of the [[Hailstone sequence]] task:\n:: 2.  Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1\n:: 3.  Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.\n\n* Create a second executable to calculate the following:\n** Use the library's hailstone function, in the standard manner, (or document how this use deviates from standard use of a library), together with extra code in this executable, to find the hailstone length returned most often for 1 <= n < 100,000.\n\n* Explain any extra setup/run steps needed to complete the task.\n\n'''Notes:''' \n* It is assumed that for a language that overwhelmingly ships in a compiled form, such as C, the library must also be an executable and the compiled user of that library is to do so without changing the compiled library. I.e. the compile tool-chain is assumed ''not'' to be present in the runtime environment.\n* Interpreters are present in the runtime environment.\n\n", "solution": "from collections import Counter\n\ndef function_length_frequency(func, hrange):\n    return Counter(len(func(n)) for n in hrange).most_common()\n\nif __name__ == '__main__':\n    from executable_hailstone_library import hailstone\n\n    upto = 100000\n    hlen, freq = function_length_frequency(hailstone, range(1, upto))[0]\n    print(\"The length of hailstone sequence that is most common for\\n\"\n          \"hailstone(n) where 1<=n<%i, is %i. It occurs %i times.\"\n          % (upto, hlen, freq))"}
{"title": "Execute Computer/Zero", "language": "Python", "task": "Computer/zero Assembly}}\n\n;Task:\nCreate a [[Computer/zero Assembly emulator. You may consider this webpage as a reference implementation. Output the results of the sample programs \"2+2\" and \"7*8\" found there.\n\n:*   The virtual machine \"bytecode\" needs to be able to modify itself (or at least act as though it can) while the virtual machine is running, to be consistent with the reference implementation.\n:*   For output, it is sufficient to have the implementation of the STP opcode return the accumulator to your actual language, and then use your standard printing routines to output it.\n\n;Bonus Points: Run all 5 sample programs at the aforementioned website and output their results.\n\n\n\n", "solution": "\"\"\"Computer/zero Assembly emulator. Requires Python >= 3.7\"\"\"\n\nimport re\n\nfrom typing import Dict\nfrom typing import Iterable\nfrom typing import List\nfrom typing import NamedTuple\nfrom typing import Optional\nfrom typing import Tuple\n\n\nNOP = 0b000\nLDA = 0b001\nSTA = 0b010\nADD = 0b011\nSUB = 0b100\nBRZ = 0b101\nJMP = 0b110\nSTP = 0b111\n\nOPCODES = {\n    \"NOP\": NOP,\n    \"LDA\": LDA,\n    \"STA\": STA,\n    \"ADD\": ADD,\n    \"SUB\": SUB,\n    \"BRZ\": BRZ,\n    \"JMP\": JMP,\n    \"STP\": STP,\n}\n\nRE_INSTRUCTION = re.compile(\n    r\"\\s*\"\n    r\"(?:(?P<label>\\w+):)?\"\n    r\"\\s*\"\n    rf\"(?P<opcode>{'|'.join(OPCODES)})?\"\n    r\"\\s*\"\n    r\"(?P<argument>\\w+)?\"\n    r\"\\s*\"\n    r\"(?:;(?P<comment>[\\w\\s]+))?\"\n)\n\n\nclass AssemblySyntaxError(Exception):\n    pass\n\n\nclass Instruction(NamedTuple):\n    label: Optional[str]\n    opcode: Optional[str]\n    argument: Optional[str]\n    comment: Optional[str]\n\n\ndef parse(assembly: str) -> Tuple[List[Instruction], Dict[str, int]]:\n    instructions: List[Instruction] = []\n    labels: Dict[str, int] = {}\n    linenum: int = 0\n\n    for line in assembly.split(\"\\n\"):\n        match = RE_INSTRUCTION.match(line)\n\n        if not match:\n            raise AssemblySyntaxError(f\"{line}: {linenum}\")\n\n        instructions.append(Instruction(**match.groupdict()))\n        label = match.group(1)\n        if label:\n            labels[label] = linenum\n\n        linenum += 1\n\n    return instructions, labels\n\n\ndef compile(instructions: List[Instruction], labels: Dict[str, int]) -> Iterable[int]:\n    for instruction in instructions:\n        if instruction.argument is None:\n            argument = 0\n        elif instruction.argument.isnumeric():\n            argument = int(instruction.argument)\n        else:\n            argument = labels[instruction.argument]\n\n        if instruction.opcode:\n            yield OPCODES[instruction.opcode] << 5 | argument\n        else:\n            yield argument\n\n\ndef run(bytecode: bytes) -> int:\n    accumulator = 0\n    program_counter = 0\n    memory = list(bytecode)[:32] + [0 for _ in range(32 - len(bytecode))]\n\n    while program_counter < 32:\n        operation = memory[program_counter] >> 5\n        argument = memory[program_counter] & 0b11111\n        program_counter += 1\n\n        if operation == NOP:\n            continue\n        elif operation == LDA:\n            accumulator = memory[argument]\n        elif operation == STA:\n            memory[argument] = accumulator\n        elif operation == ADD:\n            accumulator = (accumulator + memory[argument]) % 256\n        elif operation == SUB:\n            accumulator = (accumulator - memory[argument]) % 256\n        elif operation == BRZ:\n            if accumulator == 0:\n                program_counter = argument\n        elif operation == JMP:\n            program_counter = argument\n        elif operation == STP:\n            break\n        else:\n            raise Exception(f\"error: {operation} {argument}\")\n\n    return accumulator\n\n\nSAMPLES = [\n    \"\"\"\\\n            LDA   x\n            ADD   y       ; accumulator = x + y\n            STP\n    x:            2\n    y:            2\n    \"\"\",\n    \"\"\"\\\n    loop:   LDA   prodt\n            ADD   x\n            STA   prodt\n            LDA   y\n            SUB   one\n            STA   y\n            BRZ   done\n            JMP   loop\n    done:   LDA   prodt   ; to display it\n            STP\n    x:            8\n    y:            7\n    prodt:        0\n    one:          1\n    \"\"\",\n    \"\"\"\\\n    loop:   LDA   n\n            STA   temp\n            ADD   m\n            STA   n\n            LDA   temp\n            STA   m\n            LDA   count\n            SUB   one\n            BRZ   done\n            STA   count\n            JMP   loop\n    done:   LDA   n       ; to display it\n            STP\n    m:            1\n    n:            1\n    temp:         0\n    count:        8       ; valid range: 1-11\n    one:          1\n    \"\"\",\n    \"\"\"\\\n    start:  LDA   load\n            ADD   car     ; head of list\n            STA   ldcar\n            ADD   one\n            STA   ldcdr   ; next CONS cell\n    ldcar:  NOP\n            STA   value\n    ldcdr:  NOP\n            BRZ   done    ; 0 stands for NIL\n            STA   car\n            JMP   start\n    done:   LDA   value   ; CAR of last CONS\n            STP\n    load:   LDA   0\n    value:        0\n    car:          28\n    one:          1\n                        ; order of CONS cells\n                        ; in memory\n                        ; does not matter\n                6\n                0       ; 0 stands for NIL\n                2       ; (CADR ls)\n                26      ; (CDDR ls) -- etc.\n                5\n                20\n                3\n                30\n                1       ; value of (CAR ls)\n                22      ; points to (CDR ls)\n                4\n                24\n    \"\"\",\n    \"\"\"\\\n    p:            0       ; NOP in first round\n    c:            0\n    start:  STP           ; wait for p's move\n    pmove:  NOP\n            LDA   pmove\n            SUB   cmove\n            BRZ   same\n            LDA   pmove\n            STA   cmove   ; tit for tat\n            BRZ   cdeft\n            LDA   c       ; p defected, c did not\n            ADD   three\n            STA   c\n            JMP   start\n    cdeft:  LDA   p\n            ADD   three\n            STA   p\n            JMP   start\n    same:   LDA   pmove\n            STA   cmove   ; tit for tat\n            LDA   p\n            ADD   one\n            ADD   pmove\n            STA   p\n            LDA   c\n            ADD   one\n            ADD   pmove\n            STA   c\n            JMP   start\n    cmove:        0       ; co-operate initially\n    one:          1\n    three:        3\n    \"\"\",\n    \"\"\"\\\n    LDA  3\n    SUB  4\n    STP  0\n         0\n         255\n    \"\"\",\n    \"\"\"\\\n    LDA  3\n    SUB  4\n    STP  0\n         0\n         1\n    \"\"\",\n    \"\"\"\\\n    LDA  3\n    ADD  4\n    STP  0\n         1\n         255\n    \"\"\",\n]\n\n\ndef main() -> None:\n    for sample in SAMPLES:\n        instructions, labels = parse(sample)\n        bytecode = bytes(compile(instructions, labels))\n        result = run(bytecode)\n        print(result)\n\n\nif __name__ == \"__main__\":\n    main()\n"}
{"title": "Execute SNUSP", "language": "Python from Go", "task": "SNUSP}}'''RCSNUSP''' is a set of [[SNUSP compilers and interpreters written for Rosetta Code in a variety of languages. Below are links to each of the versions of RCSNUSP.\n\nAn implementation need only properly implement the Core SNUSP instructions ('$', '\\', '/', '+', '-', '<', '>', ',', '.', '!', and '?'). Modular SNUSP ('#', '@') and Bloated SNUSP (':', ';', '%', and '&') are also allowed, but not required. Any extra characters that you implement should be noted in the description of your implementation. Any cell size is allowed, EOF support is optional, as is whether you have bounded or unbounded memory.\n\n", "solution": "#!/usr/bin/env python3\n\nHW = r'''\n/++++!/===========?\\>++.>+.+++++++..+++\\\n\\+++\\ | /+>+++++++>/ /++++++++++<<.++>./\n$+++/ | \\+++++++++>\\ \\+++++.>.+++.-----\\\n      \\==-<<<<+>+++/ /=.>.+>.--------.-/'''\n\ndef snusp(store, code):\n    ds = bytearray(store)  # data store\n    dp = 0                 # data pointer\n    cs = code.splitlines() # 2 dimensional code store\n    ipr, ipc = 0, 0        # instruction pointers in row and column\n    for r, row in enumerate(cs):\n        try:\n            ipc = row.index('$')\n            ipr = r\n            break\n        except ValueError:\n            pass\n    rt, dn, lt, up = range(4)\n    id = rt  # instruction direction.  starting direction is always rt\n    def step():\n        nonlocal ipr, ipc\n        if id&1:\n            ipr += 1 - (id&2)\n        else:\n            ipc += 1 - (id&2)\n    while ipr >= 0 and ipr < len(cs) and ipc >= 0 and ipc < len(cs[ipr]):\n        op = cs[ipr][ipc]\n        if op == '>':\n            dp += 1\n        elif op == '<':\n            dp -= 1\n        elif op == '+':\n            ds[dp] += 1\n        elif op == '-':\n            ds[dp] -= 1\n        elif op == '.':\n            print(chr(ds[dp]), end='')\n        elif op == ',':\n            ds[dp] = input()\n        elif op == '/':\n            id = ~id\n        elif op == '\\\\':\n            id ^= 1\n        elif op == '!':\n            step()\n        elif op == '?':\n            if not ds[dp]:\n                step()\n        step()\n\nif __name__ == '__main__':\n    snusp(5, HW)"}
{"title": "Exponentiation order", "language": "Python", "task": "This task will demonstrate the order of exponentiation   ('''xy''')    when there are multiple exponents.\n\n(Many programming languages,   especially those with extended-precision integer arithmetic,   usually support one of  **, ^, |  or some such for exponentiation.)\n\n\n;Task requirements\nShow the result of a language's evaluation of multiple exponentiation (either as an integer or floating point).\n\nIf your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.\n\n\nUsing whatever operator or syntax your language supports (if any), show the results in three lines (with identification):\n  \n::::*   5**3**2 \n::::*   (5**3)**2\n::::*   5**(3**2)\n\n\nIf there are other methods (or formats) of multiple exponentiations, show them as well. \n\n\n;See also:\n* MathWorld entry:   exponentiation\n\n\n;Related tasks:\n*   exponentiation operator\n*   arbitrary-precision integers (included)\n*   [[Exponentiation with infix operators in (or operating on) the base]]\n\n", "solution": ">>> 5**3**2\n1953125\n>>> (5**3)**2\n15625\n>>> 5**(3**2)\n1953125\n>>> # The following is not normally done\n>>> try: from functools import reduce # Py3K\nexcept: pass\n\n>>> reduce(pow, (5, 3, 2))\n15625\n>>> "}
{"title": "Exponentiation with infix operators in (or operating on) the base", "language": "Python", "task": "(Many programming languages,   especially those with extended-precision integer arithmetic,   usually\nsupport one of  **, ^, |  or some such for exponentiation.)\n\n\nSome languages treat/honor infix operators when performing exponentiation   (raising\nnumbers to some power by the language's exponentiation operator,   if the computer\nprogramming language has one).\n\n\nOther programming languages may make use of the   '''POW'''   or some other BIF\n  ('''B'''uilt-'''I'''n '''F'''function),   or some other library service.\n\nIf your language's exponentiation operator is not one of the usual ones, please comment on how to recognize it.\n\n\nThis task will deal with the case where there is some form of an   ''infix operator''   operating\nin   (or operating on)   the base.\n\n\n;Example:\nA negative five raised to the 3rd power could be specified as:\n    -5  ** 3          or as\n   -(5) ** 3          or as\n   (-5) ** 3          or as something else\n\n\n(Not all computer programming languages have an exponential operator and/or support these syntax expression(s).\n\n\n;Task:\n:*   compute and display exponentiation with a possible infix operator, whether specified and/or implied/inferred.\n:*   Raise the following numbers   (integer or real):\n:::*   -5     and\n:::*   +5\n:*   to the following powers:\n:::*   2nd     and\n:::*   3rd\n:*   using the following expressions   (if applicable in your language):\n:::*   -x**p\n:::*   -(x)**p\n:::*   (-x)**p\n:::*   -(x**p)\n:*   Show here (on this page) the four (or more) types of symbolic expressions for each number and power.\n\n\nTry to present the results in the same format/manner as the other programming entries to make any differences apparent.\n\n\nThe variables may be of any type(s) that is/are applicable in your language.\n\n\n;Related tasks:\n*   [[Exponentiation order]]\n*   [[Exponentiation operator]]\n*   [[Arbitrary-precision integers (included)]]\n*   [[Parsing/RPN to infix conversion]]\n*   [[Operator precedence]]\n\n;References:\n* Wikipedia: Order of operations in Programming languages\n\n", "solution": "from itertools import product\n\nxx = '-5 +5'.split()\npp = '2 3'.split()\ntexts = '-x**p -(x)**p (-x)**p -(x**p)'.split()\n\nprint('Integer variable exponentiation')\nfor x, p in product(xx, pp):\n    print(f'  x,p = {x:2},{p}; ', end=' ')\n    x, p = int(x), int(p)\n    print('; '.join(f\"{t} =={eval(t):4}\" for t in texts))\n\nprint('\\nBonus integer literal exponentiation')\nX, P = 'xp'\nxx.insert(0, ' 5')\ntexts.insert(0, 'x**p')\nfor x, p in product(xx, pp):\n    texts2 = [t.replace(X, x).replace(P, p) for t in texts]\n    print(' ', '; '.join(f\"{t2} =={eval(t2):4}\" for t2 in texts2))"}
{"title": "Extend your language", "language": "Python", "task": "{{Control Structures}}Some programming languages allow you to extend the language. While this can be done to a certain degree in most languages (e.g. by using macros), other languages go much further. Most notably in the Forth and Lisp families, programming per se is done by extending the language without any formal distinction between built-in and user-defined elements.\n\nIf your language supports it, show how to introduce a new flow control mechanism. A practical and useful example is a four-way branch:\n\nOccasionally, code must be written that depends on ''two'' conditions, resulting in up to four branches (depending on whether both, only the first, only the second, or none of the conditions are \"true\"). In a C-like language this could look like the following:\n\n   if (condition1isTrue) {\n      if (condition2isTrue)\n         bothConditionsAreTrue();\n      else\n         firstConditionIsTrue();\n   }\n   else if (condition2isTrue)\n      secondConditionIsTrue();\n   else\n      noConditionIsTrue();\n\nBesides being rather cluttered, the statement(s) for 'condition2isTrue' must be written down twice. If 'condition2isTrue' were a lengthy and involved expression, it would be quite unreadable, and the code generated by the compiler might be unnecessarily large.\n\nThis can be improved by introducing a new keyword '''if2'''. It is similar to '''if''', but takes two conditional statements instead of one, and up to three 'else' statements. One proposal (in pseudo-C syntax) might be:\n\n   if2 (condition1isTrue) (condition2isTrue)\n      bothConditionsAreTrue();\n   else1\n      firstConditionIsTrue();\n   else2\n      secondConditionIsTrue();\n   else\n      noConditionIsTrue();\n\nPick the syntax which suits your language. The keywords 'else1' and 'else2' are just examples. The new conditional expression should look, nest and behave analogously to the language's built-in 'if' statement.\n", "solution": "a, b = 1, 0\n\nif (c1 := a == 1) and (c2 := b == 3):\n  print('a = 1 and b = 3')\nelif c1:\n  print('a = 1 and b <> 3')\nelif c2:\n  print('a <> 1 and b = 3')\nelse:\n  print('a <> 1 and b <> 3')"}
{"title": "Extreme floating point values", "language": "Python", "task": "The IEEE floating point specification defines certain 'extreme' floating point values such as minus zero, -0.0, a value distinct from plus zero; not a number, NaN; and plus and minus infinity.\n\nThe task is to use expressions involving other 'normal' floating point values in your language to calculate these, (and maybe other), extreme floating point values in your language and assign them to variables. \n\nPrint the values of these variables if possible; and show some arithmetic with these values and variables. \n\nIf your language can directly enter these extreme floating point values then show it.\n\n\n;See also:\n*   What Every Computer Scientist Should Know About Floating-Point Arithmetic\n\n\n;Related tasks:\n*   [[Infinity]]\n*   [[Detect division by zero]]\n*   [[Literals/Floating point]]\n\n", "solution": ">>> # Extreme values from expressions\n>>> inf = 1e234 * 1e234\n>>> _inf = 1e234 * -1e234\n>>> _zero = 1 / _inf\n>>> nan = inf + _inf\n>>> inf, _inf, _zero, nan\n(inf, -inf, -0.0, nan)\n>>> # Print\n>>> for value in (inf, _inf, _zero, nan): print (value)\n\ninf\n-inf\n-0.0\nnan\n>>> # Extreme values from other means\n>>> float('nan')\nnan\n>>> float('inf')\ninf\n>>> float('-inf')\n-inf\n>>> -0.\n-0.0\n>>> # Some arithmetic\n>>> nan == nan\nFalse\n>>> nan is nan\nTrue\n>>> 0. == -0.\nTrue\n>>> 0. is -0.\nFalse\n>>> inf + _inf\nnan\n>>> 0.0 * nan\nnan\n>>> nan * 0.0\nnan\n>>> 0.0 * inf\nnan\n>>> inf * 0.0\nnan"}
{"title": "FASTA format", "language": "Python", "task": "In FASTA.  \n\nA FASTA file can contain several strings, each identified by a name marked by a > (greater than) character at the beginning of the line.\n\n\n;Task:\nWrite a program that reads a FASTA file such as:\n\n>Rosetta_Example_1\nTHERECANBENOSPACE\n>Rosetta_Example_2\nTHERECANBESEVERAL\nLINESBUTTHEYALLMUST\nBECONCATENATED\n\n", "solution": "import io\n\nFASTA='''\\\n>Rosetta_Example_1\nTHERECANBENOSPACE\n>Rosetta_Example_2\nTHERECANBESEVERAL\nLINESBUTTHEYALLMUST\nBECONCATENATED'''\n\ninfile = io.StringIO(FASTA)\n\ndef fasta_parse(infile):\n    key = ''\n    for line in infile:\n        if line.startswith('>'):\n            if key:\n                yield key, val\n            key, val = line[1:].rstrip().split()[0], ''\n        elif key:\n            val += line.rstrip()\n    if key:\n        yield key, val\n\nprint('\\n'.join('%s: %s' % keyval for keyval in fasta_parse(infile)))"}
{"title": "Faces from a mesh", "language": "Python", "task": "A mesh defining a surface has uniquely numbered vertices, and named, \nsimple-polygonal faces described usually by an ordered list of edge numbers \ngoing around the face,  \n\n\nFor example:\nExternal image of two faces\nRough textual version without edges:\n\n\n      1         \n                        17\n 7   A\n             B\n\n       11                     \n                  23\n\n\n\n* A is the triangle (1, 11, 7), or equally (7, 11, 1), going anti-clockwise, or\nany of all the rotations of those ordered vertices.\n      1         \n                        \n 7   A\n            \n\n       11\n\n* B is the four-sided face (1, 17, 23, 11), or equally (23, 17, 1, 11) or any \nof their rotations.\n1         \n                  17\n\n       B\n\n 11                     \n            23\n\nLet's call the above the '''perimeter format''' as it traces around the perimeter.\n\n;A second format:\nA separate algorithm returns polygonal faces consisting of a face name and an unordered \nset of edge definitions for each face.\n* A single edge is described by the vertex numbers at its two ends, always in \nascending order.\n* All edges for the face are given, but in an undefined order.\n\nFor example face A could be described by the edges (1, 11), (7, 11), and (1, 7)\n(The  order of each vertex number in an edge is ascending, but the order in \nwhich the edges are stated is arbitrary).\n\nSimilarly face B could be described by the edges (11, 23), (1, 17), (17, 23),\nand (1, 11) in arbitrary order of the edges. \n\nLet's call this second format the '''edge format'''.\n\n\n\n;Task:\n'''1.''' Write a routine to check if two perimeter formatted faces have the same perimeter.  Use it to check if the following pairs of perimeters are the same:\n Q: (8, 1, 3)\n R: (1, 3, 8)\n\n U: (18, 8, 14, 10, 12, 17, 19)\n V: (8, 14, 10, 12, 17, 19, 18)\n\n'''2.''' Write a routine and use it to transform the following faces from edge to perimeter format.\n E: {(1, 11), (7, 11), (1, 7)}\n F: {(11, 23), (1, 17), (17, 23), (1, 11)}\n G: {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)}\n H: {(1, 3), (9, 11), (3, 11), (1, 11)}\n\n\nShow your output here.\n\n", "solution": "def perim_equal(p1, p2):\n    # Cheap tests first\n    if len(p1) != len(p2) or set(p1) != set(p2):\n        return False\n    if any(p2 == (p1[n:] + p1[:n]) for n in range(len(p1))):\n        return True\n    p2 = p2[::-1] # not inplace\n    return any(p2 == (p1[n:] + p1[:n]) for n in range(len(p1)))\n\ndef edge_to_periphery(e):\n    edges = sorted(e)\n    p = list(edges.pop(0)) if edges else []\n    last = p[-1] if p else None\n    while edges:\n        for n, (i, j) in enumerate(edges):\n            if i == last:\n                p.append(j)\n                last = j\n                edges.pop(n)\n                break\n            elif j == last:\n                p.append(i)\n                last = i\n                edges.pop(n)\n                break\n        else:\n            #raise ValueError(f'Invalid edge format: {e}')\n            return \">>>Error! Invalid edge format<<<\"\n    return p[:-1]\n\nif __name__ == '__main__':\n    print('Perimeter format equality checks:')\n    for eq_check in [\n            { 'Q': (8, 1, 3),\n              'R': (1, 3, 8)},\n            { 'U': (18, 8, 14, 10, 12, 17, 19),\n              'V': (8, 14, 10, 12, 17, 19, 18)} ]:\n        (n1, p1), (n2, p2) = eq_check.items()\n        eq = '==' if perim_equal(p1, p2) else '!='\n        print(' ', n1, eq, n2)\n\n    print('\\nEdge to perimeter format translations:')\n    edge_d = {\n     'E': {(1, 11), (7, 11), (1, 7)},\n     'F': {(11, 23), (1, 17), (17, 23), (1, 11)},\n     'G': {(8, 14), (17, 19), (10, 12), (10, 14), (12, 17), (8, 18), (18, 19)},\n     'H': {(1, 3), (9, 11), (3, 11), (1, 11)}\n            }\n    for name, edges in edge_d.items():\n        print(f\"  {name}: {edges}\\n     -> {edge_to_periphery(edges)}\")"}
{"title": "Factorial base numbers indexing permutations of a collection", "language": "Python", "task": "You need a random arrangement of a deck of cards, you are sick of lame ways of doing this. This task is a super-cool way of doing this using factorial base numbers.\nThe first 25 factorial base numbers in increasing order are: 0.0.0, 0.0.1, 0.1.0, 0.1.1, 0.2.0, 0.2.1, 1.0.0, 1.0.1, 1.1.0, 1.1.1,1.2.0, 1.2.1, 2.0.0, 2.0.1, 2.1.0, 2.1.1, 2.2.0, 2.2.1, 3.0.0, 3.0.1, 3.1.0, 3.1.1, 3.2.0, 3.2.1, 1.0.0.0\nObserve that the least significant digit is base 2 the next base 3, in general an n-digit factorial base number has digits n..1 in base n+1..2.\n\nI want to produce a 1 to 1 mapping between these numbers and permutations:-\n        0.0.0 -> 0123\n        0.0.1 -> 0132\n        0.1.0 -> 0213\n        0.1.1 -> 0231\n        0.2.0 -> 0312\n        0.2.1 -> 0321\n        1.0.0 -> 1023\n        1.0.1 -> 1032\n        1.1.0 -> 1203\n        1.1.1 -> 1230\n        1.2.0 -> 1302\n        1.2.1 -> 1320\n        2.0.0 -> 2013\n        2.0.1 -> 2031\n        2.1.0 -> 2103\n        2.1.1 -> 2130\n        2.2.0 -> 2301\n        2.2.1 -> 2310\n        3.0.0 -> 3012\n        3.0.1 -> 3021\n        3.1.0 -> 3102\n        3.1.1 -> 3120\n        3.2.0 -> 3201\n        3.2.1 -> 3210\n\nThe following psudo-code will do this:\nStarting with m=0 and O, an array of elements to be permutated, for each digit g starting with the most significant digit in the factorial base number.\n\nIf g is greater than zero, rotate the elements from m to m+g in O (see example)\nIncrement m and repeat the first step using the next most significant digit until the factorial base number is exhausted.\nFor example: using the factorial base number 2.0.1 and O = 0 1 2 3 where place 0 in both is the most significant (left-most) digit/element.\n\nStep 1: m=0 g=2; Rotate places 0 through 2. 0 1 2 3 becomes 2 0 1 3\nStep 2: m=1 g=0; No action.\nStep 3: m=2 g=1; Rotate places 2 through 3. 2 0 1 3 becomes 2 0 3 1\n\nLet me work 2.0.1 and 0123\n     step 1 n=0 g=2 O=2013\n     step 2 n=1 g=0 so no action\n     step 3 n=2 g=1 O=2031\n\nThe task:\n\n  First use your function to recreate the above table.\n  Secondly use your function to generate all permutaions of 11 digits, perhaps count them don't display them, compare this method with\n     methods in rc's permutations task.\n  Thirdly here following are two ramdom 51 digit factorial base numbers I prepared earlier:\n    39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0\n    51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1\n    use your function to crate the corresponding permutation of the following shoe of cards:\n       AKQJ1098765432AKQJ1098765432AKQJ1098765432AKQJ1098765432\n  Finally create your own 51 digit factorial base number and produce the corresponding permutation of the above shoe\n\n", "solution": "\"\"\"\n\nhttp://rosettacode.org/wiki/Factorial_base_numbers_indexing_permutations_of_a_collection\n\nhttps://en.wikipedia.org/wiki/Factorial_number_system\n\n\"\"\"\n\nimport math\n\ndef apply_perm(omega,fbn):\n    \"\"\"\n    \n    omega contains a list which will be permuted (scrambled)\n    based on fbm.\n    \n    fbm is a list which represents a factorial base number.\n    \n    This function just translates the pseudo code in the \n    Rosetta Code task.\n    \n    \"\"\"\n    for m in range(len(fbn)):\n        g = fbn[m]\n        if g > 0:\n            # do rotation\n            # save last number\n            new_first = omega[m+g]\n            # move numbers right\n            omega[m+1:m+g+1] = omega[m:m+g]\n            # put last number first\n            omega[m] = new_first\n            \n    return omega\n    \ndef int_to_fbn(i):\n    \"\"\"\n    \n    convert integer i to factorial based number\n    \n    \"\"\"\n    current = i\n    divisor = 2\n    new_fbn = []\n    while current > 0:\n        remainder = current % divisor\n        current = current // divisor\n        new_fbn.append(remainder)\n        divisor += 1\n    \n    return list(reversed(new_fbn))\n    \ndef leading_zeros(l,n):\n   \"\"\"\n   \n   If list l has less than n elements returns l with enough 0 elements\n   in front of the list to make it length n.\n   \n   \"\"\"\n   if len(l) < n:\n       return(([0] * (n - len(l))) + l)\n   else:\n       return l\n\ndef get_fbn(n):\n    \"\"\"\n    \n    Return the n! + 1 first Factorial Based Numbers starting with zero.\n            \n    \"\"\"\n    max = math.factorial(n)\n    \n    for i in range(max):\n        # from Wikipedia article\n        current = i\n        divisor = 1\n        new_fbn = int_to_fbn(i)\n        yield leading_zeros(new_fbn,n-1)\n        \ndef print_write(f, line):\n    \"\"\"\n\n    prints to console and\n    output file f\n\n    \"\"\"\n    print(line)\n    f.write(str(line)+'\\n')     \n    \ndef dot_format(l):\n    \"\"\"\n    Take a list l that is a factorial based number\n    and returns it in dot format.\n    \n    i.e. [0, 2, 1] becomes 0.2.1\n    \"\"\"\n    # empty list\n    if len(l) < 1:\n        return \"\"\n    # start with just first element no dot\n    dot_string = str(l[0])\n    # add rest if any with dots\n    for e in l[1:]:\n        dot_string += \".\"+str(e)\n        \n    return dot_string\n    \ndef str_format(l):\n    \"\"\"\n    Take a list l and returns a string\n    of those elements converted to strings.\n    \"\"\"\n    if len(l) < 1:\n        return \"\"\n        \n    new_string = \"\"\n        \n    for e in l:\n        new_string += str(e)\n    \n    return new_string \n    \nwith open(\"output.html\", \"w\", encoding=\"utf-8\") as f:\n    f.write(\"\\n\")\n    \n    # first print list\n        \n    omega=[0,1,2,3]\n    \n    four_list = get_fbn(4)\n    \n    for l in four_list:\n        print_write(f,dot_format(l)+' -> '+str_format(apply_perm(omega[:],l)))\n        \n    print_write(f,\" \")\n    \n    # now generate this output:\n    #\n    # Permutations generated = 39916800\n    # compared to 11! which  = 39916800\n    \n        \n    num_permutations = 0\n    \n    for p in get_fbn(11):\n        num_permutations += 1\n        if num_permutations % 1000000 == 0:\n            print_write(f,\"permutations so far = \"+str(num_permutations))\n    \n    print_write(f,\" \")\n    print_write(f,\"Permutations generated = \"+str(num_permutations))\n    print_write(f,\"compared to 11! which  = \"+str(math.factorial(11)))\n    \n    print_write(f,\" \")\n    \n       \n    \n    \"\"\"\n    \n    u\"\\u2660\" - spade\n    \n    u\"\\u2665\" - heart\n    \n    u\"\\u2666\" - diamond\n    \n    u\"\\u2663\" - club\n        \n    \"\"\"\n    \n    shoe = []\n    \n    for suit in [u\"\\u2660\",u\"\\u2665\",u\"\\u2666\",u\"\\u2663\"]:\n        for value in ['A','K','Q','J','10','9','8','7','6','5','4','3','2']:\n            shoe.append(value+suit)\n                    \n    print_write(f,str_format(shoe))\n    \n    p1 = [39,49,7,47,29,30,2,12,10,3,29,37,33,17,12,31,29,34,17,25,2,4,25,4,1,14,20,6,21,18,1,1,1,4,0,5,15,12,4,3,10,10,9,1,6,5,5,3,0,0,0]\n    \n    p2 = [51,48,16,22,3,0,19,34,29,1,36,30,12,32,12,29,30,26,14,21,8,12,1,3,10,4,7,17,6,21,8,12,15,15,13,15,7,3,12,11,9,5,5,6,6,3,4,0,3,2,1]\n    \n    print_write(f,\" \")\n    print_write(f,dot_format(p1))\n    print_write(f,\" \")\n    print_write(f,str_format(apply_perm(shoe[:],p1)))\n    \n    print_write(f,\" \")\n    print_write(f,dot_format(p2))\n    print_write(f,\" \")\n    print_write(f,str_format(apply_perm(shoe[:],p2)))\n\n    # generate random 51 digit factorial based number\n    \n    import random\n    \n    max = math.factorial(52)\n    \n    random_int = random.randint(0, max-1)\n\n    myperm = leading_zeros(int_to_fbn(random_int),51)\n    \n    print(len(myperm))\n    \n    print_write(f,\" \")\n    print_write(f,dot_format(myperm))\n    print_write(f,\" \")\n    print_write(f,str_format(apply_perm(shoe[:],myperm)))\n\n    f.write(\"\\n\")\n\n"}
{"title": "Factorions", "language": "Python from C", "task": "Definition:\nA factorion is a natural number that equals the sum of the factorials of its digits. \n\n\n;Example: \n'''145'''   is a factorion in base '''10''' because:\n\n 1! + 4! + 5!   =   1 + 24 + 120   =   145 \n\n\n\nIt can be shown (see talk page) that no factorion in base '''10''' can exceed   '''1,499,999'''.\n\n\n;Task:\nWrite a program in your language to demonstrate, by calculating and printing out the factorions, that:\n:*   There are   '''3'''   factorions in base   '''9'''\n:*   There are   '''4'''   factorions in base '''10'''\n:*   There are   '''5'''   factorions in base '''11''' \n:*   There are   '''2'''   factorions in base '''12'''     (up to the same upper bound as for base '''10''')\n\n\n;See also:\n:* '''Wikipedia article'''\n:* '''OEIS:A014080 - Factorions in base 10'''\n:* '''OEIS:A193163 - Factorions in base n'''\n\n", "solution": "fact = [1] # cache factorials from 0 to 11\nfor n in range(1, 12):\n    fact.append(fact[n-1] * n)\n\nfor b in range(9, 12+1):\n    print(f\"The factorions for base {b} are:\")\n    for i in range(1, 1500000):\n        fact_sum = 0\n        j = i\n        while j > 0:\n            d = j % b\n            fact_sum += fact[d]\n            j = j//b\n        if fact_sum == i:\n            print(i, end=\" \")\n    print(\"\\n\")\n"}
{"title": "Fairshare between two and more", "language": "Python", "task": "The [[Thue-Morse]] sequence is a sequence of ones and zeros that if two people\ntake turns in the given order, the first persons turn for every '0' in the\nsequence, the second for every '1'; then this is shown to give a fairer, more\nequitable sharing of resources. (Football penalty shoot-outs for example, might\nnot favour the team that goes first as much if the penalty takers take turns\naccording to the Thue-Morse sequence and took 2^n penalties)\n\nThe Thue-Morse sequence of ones-and-zeroes can be generated by:\n:''\"When counting in binary, the digit sum modulo 2 is the Thue-Morse sequence\"''\n\n\n;Sharing fairly between two or more:\nUse this method:\n:''When counting base '''b''', the digit sum modulo '''b''' is the Thue-Morse sequence of fairer sharing between '''b''' people.''\n\n\n;Task\nCounting from zero;   using a function/method/routine to express an integer count in base '''b''',\nsum the digits modulo '''b''' to produce the next member of the Thue-Morse fairshare series for '''b''' people.\n\n\nShow the first 25 terms of the fairshare sequence:\n:*   For two people:\n:*   For three people\n:*   For five people\n:*   For eleven people\n\n\n;Related tasks: \n:*   [[Non-decimal radices/Convert]]\n:*   [[Thue-Morse]]\n\n\n;See also:\n:*   A010060, A053838, A053840: The On-Line Encyclopedia of Integer Sequences(r) (OEIS(r))\n\n", "solution": "from itertools import count, islice\n\ndef _basechange_int(num, b):\n    \"\"\"\n    Return list of ints representing positive num in base b\n\n    >>> b = 3\n    >>> print(b, [_basechange_int(num, b) for num in range(11)])\n    3 [[0], [1], [2], [1, 0], [1, 1], [1, 2], [2, 0], [2, 1], [2, 2], [1, 0, 0], [1, 0, 1]]\n    >>>\n    \"\"\"\n    if num == 0:\n        return [0]\n    result = []\n    while num != 0:\n        num, d = divmod(num, b)\n        result.append(d)\n    return result[::-1]\n\ndef fairshare(b=2):\n    for i in count():\n        yield sum(_basechange_int(i, b)) % b\n\nif __name__ == '__main__':\n    for b in (2, 3, 5, 11):\n        print(f\"{b:>2}: {str(list(islice(fairshare(b), 25)))[1:-1]}\")"}
{"title": "Farey sequence", "language": "Python", "task": "The   Farey sequence   ''' ''F''n'''   of order   '''n'''   is the sequence of completely reduced fractions between   '''0'''   and   '''1'''   which, when in lowest terms, have denominators less than or equal to   '''n''',   arranged in order of increasing size.\n\nThe   ''Farey sequence''   is sometimes incorrectly called a   ''Farey series''. \n\n\nEach Farey sequence:\n:::*   starts with the value   '''0'''   (zero),   denoted by the fraction      \\frac{0}{1} \n:::*   ends with the value   '''1'''   (unity),   denoted by the fraction    \\frac{1}{1}.\n\n\nThe Farey sequences of orders   '''1'''   to   '''5'''   are:\n\n:::: {\\bf\\it{F}}_1 = \\frac{0}{1}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_2 = \\frac{0}{1}, \\frac{1}{2}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_3 = \\frac{0}{1}, \\frac{1}{3}, \\frac{1}{2}, \\frac{2}{3}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_4 = \\frac{0}{1}, \\frac{1}{4}, \\frac{1}{3}, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{4}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_5 = \\frac{0}{1}, \\frac{1}{5}, \\frac{1}{4}, \\frac{1}{3}, \\frac{2}{5}, \\frac{1}{2}, \\frac{3}{5}, \\frac{2}{3}, \\frac{3}{4}, \\frac{4}{5}, \\frac{1}{1}\n\n;Task\n*   Compute and show the Farey sequence for orders   '''1'''   through   '''11'''   (inclusive).\n*   Compute and display the   ''number''   of fractions in the Farey sequence for order   '''100'''   through   '''1,000'''   (inclusive)   by hundreds.\n*   Show the fractions as    '''n/d'''    (using the solidus [or slash] to separate the numerator from the denominator).   \n\n\nThe length   (the number of fractions)   of a Farey sequence asymptotically approaches:\n\n:::::::::::  3 x n2   /   \\pi2 \n\n;See also:\n*    OEIS sequence           A006842 numerators of Farey series of order 1, 2, *** \n*    OEIS sequence           A006843 denominators of Farey series of order 1, 2, *** \n*    OEIS sequence           A005728 number of fractions in Farey series of order n \n*   MathWorld entry          Farey sequence\n*   Wikipedia   entry   Farey sequence\n\n", "solution": "from fractions import Fraction\n\n\nclass Fr(Fraction):\n    def __repr__(self):\n        return '(%s/%s)' % (self.numerator, self.denominator)\n\n\ndef farey(n, length=False):\n    if not length:\n        return [Fr(0, 1)] + sorted({Fr(m, k) for k in range(1, n+1) for m in range(1, k+1)})\n    else:\n        #return 1         +    len({Fr(m, k) for k in range(1, n+1) for m in range(1, k+1)})\n        return  (n*(n+3))//2 - sum(farey(n//k, True) for k in range(2, n+1))\n        \nif __name__ == '__main__':\n    print('Farey sequence for order 1 through 11 (inclusive):')\n    for n in range(1, 12): \n        print(farey(n))\n    print('Number of fractions in the Farey sequence for order 100 through 1,000 (inclusive) by hundreds:')\n    print([farey(i, length=True) for i in range(100, 1001, 100)])"}
{"title": "Farey sequence", "language": "Python 3.7", "task": "The   Farey sequence   ''' ''F''n'''   of order   '''n'''   is the sequence of completely reduced fractions between   '''0'''   and   '''1'''   which, when in lowest terms, have denominators less than or equal to   '''n''',   arranged in order of increasing size.\n\nThe   ''Farey sequence''   is sometimes incorrectly called a   ''Farey series''. \n\n\nEach Farey sequence:\n:::*   starts with the value   '''0'''   (zero),   denoted by the fraction      \\frac{0}{1} \n:::*   ends with the value   '''1'''   (unity),   denoted by the fraction    \\frac{1}{1}.\n\n\nThe Farey sequences of orders   '''1'''   to   '''5'''   are:\n\n:::: {\\bf\\it{F}}_1 = \\frac{0}{1}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_2 = \\frac{0}{1}, \\frac{1}{2}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_3 = \\frac{0}{1}, \\frac{1}{3}, \\frac{1}{2}, \\frac{2}{3}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_4 = \\frac{0}{1}, \\frac{1}{4}, \\frac{1}{3}, \\frac{1}{2}, \\frac{2}{3}, \\frac{3}{4}, \\frac{1}{1}\n:\n:::: {\\bf\\it{F}}_5 = \\frac{0}{1}, \\frac{1}{5}, \\frac{1}{4}, \\frac{1}{3}, \\frac{2}{5}, \\frac{1}{2}, \\frac{3}{5}, \\frac{2}{3}, \\frac{3}{4}, \\frac{4}{5}, \\frac{1}{1}\n\n;Task\n*   Compute and show the Farey sequence for orders   '''1'''   through   '''11'''   (inclusive).\n*   Compute and display the   ''number''   of fractions in the Farey sequence for order   '''100'''   through   '''1,000'''   (inclusive)   by hundreds.\n*   Show the fractions as    '''n/d'''    (using the solidus [or slash] to separate the numerator from the denominator).   \n\n\nThe length   (the number of fractions)   of a Farey sequence asymptotically approaches:\n\n:::::::::::  3 x n2   /   \\pi2 \n\n;See also:\n*    OEIS sequence           A006842 numerators of Farey series of order 1, 2, *** \n*    OEIS sequence           A006843 denominators of Farey series of order 1, 2, *** \n*    OEIS sequence           A005728 number of fractions in Farey series of order n \n*   MathWorld entry          Farey sequence\n*   Wikipedia   entry   Farey sequence\n\n", "solution": "'''Farey sequence'''\n\nfrom itertools import (chain, count, islice)\nfrom math import gcd\n\n\n# farey :: Int -> [Ratio Int]\ndef farey(n):\n    '''Farey sequence of order n.'''\n    return sorted(\n        nubBy(on(eq)(fromRatio))(\n            bind(enumFromTo(1)(n))(\n                lambda k: bind(enumFromTo(0)(k))(\n                    lambda m: [ratio(m)(k)]\n                )\n            )\n        ),\n        key=fromRatio\n    ) + [ratio(1)(1)]\n\n\n# fareyLength :: Int -> Int\ndef fareyLength(n):\n    '''Number of terms in a Farey sequence\n       of order n.'''\n    def go(x):\n        return (x * (x + 3)) // 2 - sum(\n            go(x // k) for k in enumFromTo(2)(x)\n        )\n    return go(n)\n\n\n# showFarey :: [Ratio Int] -> String\ndef showFarey(xs):\n    '''Stringification of a Farey sequence.'''\n    return '(' + ', '.join(map(showRatio, xs)) + ')'\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Tests'''\n\n    print(\n        fTable(\n            'Farey sequence for orders 1-11 (inclusive):\\n'\n        )(str)(showFarey)(\n            farey\n        )(enumFromTo(1)(11))\n    )\n    print(\n        fTable(\n            '\\n\\nNumber of fractions in the Farey sequence ' +\n            'for order 100 through 1,000 (inclusive) by hundreds:\\n'\n        )(str)(str)(\n            fareyLength\n        )(enumFromThenTo(100)(200)(1000))\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# bind(>>=) :: [a] -> (a -> [b]) -> [b]\ndef bind(xs):\n    '''List monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.'''\n    return lambda f: list(\n        chain.from_iterable(\n            map(f, xs)\n        )\n    )\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# enumFromThenTo :: Int -> Int -> Int -> [Int]\ndef enumFromThenTo(m):\n    '''Integer values enumerated from m to n\n       with a step defined by nxt-m.\n    '''\n    def go(nxt, n):\n        d = nxt - m\n        return islice(count(0), m, d + n, d)\n    return lambda nxt: lambda n: (\n        list(go(nxt, n))\n    )\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# eq (==) :: Eq a => a -> a -> Bool\ndef eq(a):\n    '''Simple equality of a and b.'''\n    return lambda b: a == b\n\n\n# fromRatio :: Ratio Int -> Float\ndef fromRatio(r):\n    '''A floating point value derived from a\n       a rational value.\n    '''\n    return r.get('numerator') / r.get('denominator')\n\n\n# nubBy :: (a -> a -> Bool) -> [a] -> [a]\ndef nubBy(p):\n    '''A sublist of xs from which all duplicates,\n       (as defined by the equality predicate p)\n       are excluded.\n    '''\n    def go(xs):\n        if not xs:\n            return []\n        x = xs[0]\n        return [x] + go(\n            list(filter(\n                lambda y: not p(x)(y),\n                xs[1:]\n            ))\n        )\n    return lambda xs: go(xs)\n\n\n# on :: (b -> b -> c) -> (a -> b) -> a -> a -> c\ndef on(f):\n    '''A function returning the value of applying\n      the binary f to g(a) g(b)\n    '''\n    return lambda g: lambda a: lambda b: f(g(a))(g(b))\n\n\n# ratio :: Int -> Int -> Ratio Int\ndef ratio(n):\n    '''Rational value constructed\n       from a numerator and a denominator.\n    '''\n    def go(n, d):\n        g = gcd(n, d)\n        return {\n            'type': 'Ratio',\n            'numerator': n // g, 'denominator': d // g\n        }\n    return lambda d: go(n * signum(d), abs(d))\n\n\n# showRatio :: Ratio -> String\ndef showRatio(r):\n    '''String representation of the ratio r.'''\n    d = r.get('denominator')\n    return str(r.get('numerator')) + (\n        '/' + str(d) if 1 != d else ''\n    )\n\n\n# signum :: Num -> Num\ndef signum(n):\n    '''The sign of n.'''\n    return -1 if 0 > n else (1 if 0 < n else 0)\n\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.\n    '''\n    return '\\n'.join(xs)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Fast Fourier transform", "language": "Python", "task": "Calculate the   FFT   (Fast Fourier Transform)   of an input sequence.\n\nThe most general case allows for complex numbers at the input \nand results in a sequence of equal length, again of complex numbers. \nIf you need to restrict yourself to real numbers, the output should \nbe the magnitude   (i.e.:   sqrt(re2 + im2))   of the complex result. \n\nThe classic version is the recursive Cooley-Tukey FFT. Wikipedia has pseudo-code for that. \nFurther optimizations are possible but not required.\n\n", "solution": "from cmath import exp, pi\n\ndef fft(x):\n    N = len(x)\n    if N <= 1: return x\n    even = fft(x[0::2])\n    odd =  fft(x[1::2])\n    T= [exp(-2j*pi*k/N)*odd[k] for k in range(N//2)]\n    return [even[k] + T[k] for k in range(N//2)] + \\\n           [even[k] - T[k] for k in range(N//2)]\n\nprint( ' '.join(\"%5.3f\" % abs(f) \n                for f in fft([1.0, 1.0, 1.0, 1.0, 0.0, 0.0, 0.0, 0.0])) )"}
{"title": "Feigenbaum constant calculation", "language": "Python from D", "task": "Calculate the Feigenbaum constant. \n\n\n;See:\n:*   Details in the Wikipedia article:   Feigenbaum constant.\n\n", "solution": "max_it = 13\nmax_it_j = 10\na1 = 1.0\na2 = 0.0\nd1 = 3.2\na = 0.0\n\nprint \" i       d\"\nfor i in range(2, max_it + 1):\n    a = a1 + (a1 - a2) / d1\n    for j in range(1, max_it_j + 1):\n        x = 0.0\n        y = 0.0\n        for k in range(1, (1 << i) + 1):\n            y = 1.0 - 2.0 * y * x\n            x = a - x * x\n        a = a - x / y\n    d = (a1 - a2) / (a - a1)\n    print(\"{0:2d}    {1:.8f}\".format(i, d))\n    d1 = d\n    a2 = a1\n    a1 = a"}
{"title": "Fibonacci n-step number sequences", "language": "Python", "task": "These number series are an expansion of the ordinary [[Fibonacci sequence]] where:\n# For n = 2 we have the Fibonacci sequence; with initial values [1, 1] and F_k^2 = F_{k-1}^2 + F_{k-2}^2\n# For n = 3 we have the tribonacci sequence; with initial values [1, 1, 2] and F_k^3 = F_{k-1}^3 + F_{k-2}^3 + F_{k-3}^3\n# For n = 4 we have the tetranacci sequence; with initial values [1, 1, 2, 4] and F_k^4 = F_{k-1}^4 + F_{k-2}^4 + F_{k-3}^4 + F_{k-4}^4...\n# For general n>2 we have the Fibonacci n-step sequence - F_k^n; with initial values of the first n values of the (n-1)'th Fibonacci n-step sequence F_k^{n-1}; and k'th value of this n'th sequence being F_k^n = \\sum_{i=1}^{(n)} {F_{k-i}^{(n)}}\n\nFor small values of n, Greek numeric prefixes are sometimes used to individually name each series.\n\n:::: {| style=\"text-align: left;\" border=\"4\" cellpadding=\"2\" cellspacing=\"2\"\n|+ Fibonacci n-step sequences\n|- style=\"background-color: rgb(255, 204, 255);\"\n! n !! Series name !! Values\n|-\n|  2 ||  fibonacci || 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...\n|-\n|  3 || tribonacci || 1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...\n|-\n|  4 || tetranacci || 1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...\n|-\n|  5 || pentanacci || 1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...\n|-\n|  6 ||  hexanacci || 1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...\n|-\n|  7 || heptanacci || 1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...\n|-\n|  8 ||  octonacci || 1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...\n|-\n|  9 ||  nonanacci || 1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...\n|-\n| 10 ||  decanacci || 1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...\n|}\n\nAllied sequences can be generated where the initial values are changed:\n: '''The Lucas series''' sums the two preceding values like the fibonacci series for n=2 but uses [2, 1] as its initial values.\n\n\n\n;Task:\n# Write a function to generate Fibonacci n-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.\n# Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.\n\n\n;Related tasks:\n*   [[Fibonacci sequence]]\n*   Wolfram Mathworld\n*   [[Hofstadter Q sequence]]\n*   [[Leonardo numbers]]\n\n\n;Also see:\n*   Lucas Numbers - Numberphile (Video)\n*   Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)\n*   Wikipedia, Lucas number\n*   MathWorld, Fibonacci Number\n*   Some identities for r-Fibonacci numbers\n*   OEIS Fibonacci numbers\n*   OEIS Lucas numbers\n\n", "solution": ">>> class Fiblike():\n\tdef __init__(self, start):\n\t\tself.addnum = len(start)\n\t\tself.memo = start[:]\n\tdef __call__(self, n):\n\t\ttry:\n\t\t\treturn self.memo[n]\n\t\texcept IndexError:\n\t\t\tans = sum(self(i) for i in range(n-self.addnum, n))\n\t\t\tself.memo.append(ans)\n\t\t\treturn ans\n\n\t\t\n>>> fibo = Fiblike([1,1])\n>>> [fibo(i) for i in range(10)]\n[1, 1, 2, 3, 5, 8, 13, 21, 34, 55]\n>>> lucas = Fiblike([2,1])\n>>> [lucas(i) for i in range(10)]\n[2, 1, 3, 4, 7, 11, 18, 29, 47, 76]\n>>> for n, name in zip(range(2,11), 'fibo tribo tetra penta hexa hepta octo nona deca'.split()) :\n\tfibber = Fiblike([1] + [2**i for i in range(n-1)])\n\tprint('n=%2i, %5snacci -> %s ...' % (n, name, ' '.join(str(fibber(i)) for i in range(15))))\n\n\t\nn= 2,  fibonacci -> 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...\nn= 3, tribonacci -> 1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...\nn= 4, tetranacci -> 1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...\nn= 5, pentanacci -> 1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...\nn= 6,  hexanacci -> 1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...\nn= 7, heptanacci -> 1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...\nn= 8,  octonacci -> 1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...\nn= 9,  nonanacci -> 1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...\nn=10,  decanacci -> 1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...\n>>> "}
{"title": "Fibonacci n-step number sequences", "language": "Python 3.7", "task": "These number series are an expansion of the ordinary [[Fibonacci sequence]] where:\n# For n = 2 we have the Fibonacci sequence; with initial values [1, 1] and F_k^2 = F_{k-1}^2 + F_{k-2}^2\n# For n = 3 we have the tribonacci sequence; with initial values [1, 1, 2] and F_k^3 = F_{k-1}^3 + F_{k-2}^3 + F_{k-3}^3\n# For n = 4 we have the tetranacci sequence; with initial values [1, 1, 2, 4] and F_k^4 = F_{k-1}^4 + F_{k-2}^4 + F_{k-3}^4 + F_{k-4}^4...\n# For general n>2 we have the Fibonacci n-step sequence - F_k^n; with initial values of the first n values of the (n-1)'th Fibonacci n-step sequence F_k^{n-1}; and k'th value of this n'th sequence being F_k^n = \\sum_{i=1}^{(n)} {F_{k-i}^{(n)}}\n\nFor small values of n, Greek numeric prefixes are sometimes used to individually name each series.\n\n:::: {| style=\"text-align: left;\" border=\"4\" cellpadding=\"2\" cellspacing=\"2\"\n|+ Fibonacci n-step sequences\n|- style=\"background-color: rgb(255, 204, 255);\"\n! n !! Series name !! Values\n|-\n|  2 ||  fibonacci || 1 1 2 3 5 8 13 21 34 55 89 144 233 377 610 ...\n|-\n|  3 || tribonacci || 1 1 2 4 7 13 24 44 81 149 274 504 927 1705 3136 ...\n|-\n|  4 || tetranacci || 1 1 2 4 8 15 29 56 108 208 401 773 1490 2872 5536 ...\n|-\n|  5 || pentanacci || 1 1 2 4 8 16 31 61 120 236 464 912 1793 3525 6930 ...\n|-\n|  6 ||  hexanacci || 1 1 2 4 8 16 32 63 125 248 492 976 1936 3840 7617 ...\n|-\n|  7 || heptanacci || 1 1 2 4 8 16 32 64 127 253 504 1004 2000 3984 7936 ...\n|-\n|  8 ||  octonacci || 1 1 2 4 8 16 32 64 128 255 509 1016 2028 4048 8080 ...\n|-\n|  9 ||  nonanacci || 1 1 2 4 8 16 32 64 128 256 511 1021 2040 4076 8144 ...\n|-\n| 10 ||  decanacci || 1 1 2 4 8 16 32 64 128 256 512 1023 2045 4088 8172 ...\n|}\n\nAllied sequences can be generated where the initial values are changed:\n: '''The Lucas series''' sums the two preceding values like the fibonacci series for n=2 but uses [2, 1] as its initial values.\n\n\n\n;Task:\n# Write a function to generate Fibonacci n-step number sequences given its initial values and assuming the number of initial values determines how many previous values are summed to make the next number of the series.\n# Use this to print and show here at least the first ten members of the Fibo/tribo/tetra-nacci and Lucas sequences.\n\n\n;Related tasks:\n*   [[Fibonacci sequence]]\n*   Wolfram Mathworld\n*   [[Hofstadter Q sequence]]\n*   [[Leonardo numbers]]\n\n\n;Also see:\n*   Lucas Numbers - Numberphile (Video)\n*   Tribonacci Numbers (and the Rauzy Fractal) - Numberphile (Video)\n*   Wikipedia, Lucas number\n*   MathWorld, Fibonacci Number\n*   Some identities for r-Fibonacci numbers\n*   OEIS Fibonacci numbers\n*   OEIS Lucas numbers\n\n", "solution": "'''Fibonacci n-step number sequences'''\n\nfrom itertools import chain, count, islice\n\n\n# A000032 :: () -> [Int]\ndef A000032():\n    '''Non finite sequence of Lucas numbers.\n    '''\n    return unfoldr(recurrence(2))([2, 1])\n\n\n# nStepFibonacci :: Int -> [Int]\ndef nStepFibonacci(n):\n    '''Non-finite series of N-step Fibonacci numbers,\n       defined by a recurrence relation.\n    '''\n    return unfoldr(recurrence(n))(\n        take(n)(\n            chain(\n                [1],\n                (2 ** i for i in count(0))\n            )\n        )\n    )\n\n\n# recurrence :: Int -> [Int] -> Int\ndef recurrence(n):\n    '''Recurrence relation in Fibonacci and related series.\n    '''\n    def go(xs):\n        h, *t = xs\n        return h, t + [sum(take(n)(xs))]\n    return go\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''First 15 terms each n-step Fibonacci(n) series\n       where n is drawn from [2..8]\n    '''\n    labels = \"fibo tribo tetra penta hexa hepta octo nona deca\"\n    table = list(\n        chain(\n            [['lucas:'] + [\n                str(x) for x in take(15)(A000032())]\n             ],\n            map(\n                lambda k, n: list(\n                    chain(\n                        [k + 'nacci:'],\n                        (\n                            str(x) for x\n                            in take(15)(nStepFibonacci(n))\n                        )\n                    )\n                ),\n                labels.split(),\n                count(2)\n            )\n        )\n    )\n    print('Recurrence relation series:\\n')\n    print(\n        spacedTable(table)\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    def go(xs):\n        return (\n            xs[0:n]\n            if isinstance(xs, (list, tuple))\n            else list(islice(xs, n))\n        )\n    return go\n\n\n# unfoldr :: (b -> Maybe (a, b)) -> b -> [a]\ndef unfoldr(f):\n    '''Generic anamorphism.\n       A lazy (generator) list unfolded from a seed value by\n       repeated application of f until no residue remains.\n       Dual to fold/reduce.\n       f returns either None, or just (value, residue).\n       For a strict output value, wrap in list().\n    '''\n    def go(x):\n        valueResidue = f(x)\n        while None is not valueResidue:\n            yield valueResidue[0]\n            valueResidue = f(valueResidue[1])\n    return go\n\n\n# ---------------------- FORMATTING ----------------------\n\n# spacedTable :: [[String]] -> String\ndef spacedTable(rows):\n    columnWidths = [\n        max([len(x) for x in col])\n        for col in zip(*rows)\n    ]\n    return '\\n'.join([\n        ' '.join(\n            map(\n                lambda x, w: x.rjust(w, ' '),\n                row, columnWidths\n            )\n        )\n        for row in rows\n    ])\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Fibonacci word", "language": "Python", "task": "The   Fibonacci Word   may be created in a manner analogous to the   Fibonacci Sequence    as described here:\n\n     Define   F_Word1   as   '''1'''\n     Define   F_Word2   as   '''0'''\n     Form     F_Word3   as   F_Word2     concatenated with   F_Word1    i.e.:   '''01'''\n     Form     F_Wordn   as   F_Wordn-1   concatenated with   F_wordn-2\n\n\n;Task:\nPerform the above steps for     n = 37.\n\nYou may display the first few but not the larger values of   n.\n{Doing so will get the task's author into trouble with them what be (again!).} \n\nInstead, create a table for   F_Words   '''1'''   to   '''37'''   which shows:\n::*   The number of characters in the word\n::*   The word's [[Entropy]]\n\n\n;Related tasks:  \n*   Fibonacci word/fractal\n*   [[Entropy]]\n*   [[Entropy/Narcissist]]\n\n", "solution": ">>> import math\n>>> from collections import Counter\n>>> \n>>> def entropy(s):\n...     p, lns = Counter(s), float(len(s))\n...     return -sum( count/lns * math.log(count/lns, 2) for count in p.values())\n... \n>>> \n>>> def fibword(nmax=37):\n...     fwords = ['1', '0']\n...     print('%-3s %10s %-10s %s' % tuple('N Length Entropy Fibword'.split()))\n...     def pr(n, fwords):\n...         while len(fwords) < n:\n...             fwords += [''.join(fwords[-2:][::-1])]\n...         v = fwords[n-1]\n...         print('%3i %10i %10.7g %s' % (n, len(v), entropy(v), v if len(v) < 20 else '<too long>'))\n...     for n in range(1, nmax+1): pr(n, fwords)\n... \n>>> fibword()\nN       Length Entropy    Fibword\n  1          1         -0 1\n  2          1         -0 0\n  3          2          1 01\n  4          3  0.9182958 010\n  5          5  0.9709506 01001\n  6          8   0.954434 01001010\n  7         13  0.9612366 0100101001001\n  8         21  0.9587119 <too long>\n  9         34  0.9596869 <too long>\n 10         55   0.959316 <too long>\n 11         89  0.9594579 <too long>\n 12        144  0.9594038 <too long>\n 13        233  0.9594244 <too long>\n 14        377  0.9594165 <too long>\n 15        610  0.9594196 <too long>\n 16        987  0.9594184 <too long>\n 17       1597  0.9594188 <too long>\n 18       2584  0.9594187 <too long>\n 19       4181  0.9594187 <too long>\n 20       6765  0.9594187 <too long>\n 21      10946  0.9594187 <too long>\n 22      17711  0.9594187 <too long>\n 23      28657  0.9594187 <too long>\n 24      46368  0.9594187 <too long>\n 25      75025  0.9594187 <too long>\n 26     121393  0.9594187 <too long>\n 27     196418  0.9594187 <too long>\n 28     317811  0.9594187 <too long>\n 29     514229  0.9594187 <too long>\n 30     832040  0.9594187 <too long>\n 31    1346269  0.9594187 <too long>\n 32    2178309  0.9594187 <too long>\n 33    3524578  0.9594187 <too long>\n 34    5702887  0.9594187 <too long>\n 35    9227465  0.9594187 <too long>\n 36   14930352  0.9594187 <too long>\n 37   24157817  0.9594187 <too long>\n>>> "}
{"title": "File extension is in extensions list", "language": "Python", "task": "Checking if a file is in a certain category of file formats with known extensions (e.g. archive files, or image files) is a common problem in practice, and may be approached differently from extracting and outputting an arbitrary extension ''(see e.g. FileNameExtensionFilter in Java)''.\n\nIt also requires less assumptions about the format of an extension, because the calling code can decide what extensions are valid.\n\nFor these reasons, this task exists in addition to the [[Extract file extension]] task.\n\n\n;Related tasks: \n* [[Extract file extension]]\n* [[String matching]]\n\n", "solution": "'''Check for a specific set of file extensions'''\n\n\n# extensionFound :: [Extension] -> FileName -> Maybe Extension\ndef extensionFound(xs):\n    '''Nothing if no matching extension is found,\n       or Just the extension (drawn from xs, and\n       a suffix of the filename, immediately following\n       a dot character).\n    '''\n    return lambda fn: find(fn.lower().endswith)(\n        ['.' + x.lower() for x in xs]\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Check filenames for a particular set of extensions.'''\n\n    # checkExtension :: FileName -> Maybe Extension\n    def checkExtension(fn):\n        return extensionFound([\n            'zip', 'rar', '7z', 'gz', 'archive', 'A##', 'tar.bz2'\n        ])(fn)\n\n    print(\n        fTable(__doc__ + ':\\n')(str)(str)(\n            compose(fromMaybe('n/a'))(checkExtension)\n        )([\n            'MyData.a##',\n            'MyData.tar.Gz',\n            'MyData.gzip',\n            'MyData.7z.backup',\n            'MyData...',\n            'MyData',\n            'MyData_v1.0.tar.bz2',\n            'MyData_v1.0.bz2'\n        ])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# find :: (a -> Bool) -> [a] -> Maybe a\ndef find(p):\n    '''Just the first element in the list that matches p,\n       or Nothing if no elements match.\n    '''\n    def go(xs):\n        for x in xs:\n            if p(x):\n                return Just(x)\n        return Nothing()\n    return lambda xs: go(xs)\n\n\n# fromMaybe :: a -> Maybe a -> a\ndef fromMaybe(x):\n    '''The default value x if mb is Nothing,\n       or the value contained in mb.\n    '''\n    return lambda mb: x if (\n        mb.get('Nothing')\n    ) else mb.get('Just')\n\n\n# DISPLAY -------------------------------------------------\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "File size distribution", "language": "Python", "task": "Beginning from the current directory, or optionally from a directory specified as a command-line argument, determine how many files there are of various sizes in a directory hierarchy.\n\n\nMy suggestion is to sort by logarithmn of file size, since a few bytes here or there, or even a factor of two or three, may not be that significant.\n\nDon't forget that empty files may exist, to serve as a marker.\n\n\nIs your file system predominantly devoted to a large number of smaller files, or a smaller number of huge files?\n\n", "solution": "import sys, os\nfrom collections import Counter\n\ndef dodir(path):\n    global h\n\n    for name in os.listdir(path):\n        p = os.path.join(path, name)\n\n        if os.path.islink(p):\n            pass\n        elif os.path.isfile(p):\n            h[os.stat(p).st_size] += 1\n        elif os.path.isdir(p):\n            dodir(p)\n        else:\n            pass\n\ndef main(arg):\n    global h\n    h = Counter()\n    for dir in arg:\n        dodir(dir)\n\n    s = n = 0\n    for k, v in sorted(h.items()):\n        print(\"Size %d -> %d file(s)\" % (k, v))\n        n += v\n        s += k * v\n    print(\"Total %d bytes for %d files\" % (s, n))\n\nmain(sys.argv[1:])"}
{"title": "Find Chess960 starting position identifier", "language": "Python 3.10.5 2022-06-28", "task": "Starting Position Identifier number (\"SP-ID\"), and generate the corresponding position. \n\n;Task:\nThis task is to go the other way: given a starting array of pieces (provided in any form that suits your implementation, whether string or list or array, of letters or Unicode chess symbols or enum values, etc.), derive its unique SP-ID. For example, given the starting array '''QNRBBNKR''' (or '''''' or ''''''), which we assume is given as seen from White's side of the board from left to right, your (sub)program should return 105; given the starting lineup of standard chess, it should return 518.\n\nYou may assume the input is a valid Chess960 position; detecting invalid input (including illegal characters or starting arrays with the bishops on the same color square or the king not between the two rooks) is optional.\n\n;Algorithm:\n\nThe derivation is the inverse of the algorithm given at Wikipedia, and goes like this (we'll use the standard chess setup as an example).\n\n1. Ignoring the Queen and Bishops, find the positions of the Knights within the remaining five spaces (in the standard array they're in the second and fourth positions), and then find the index number of that combination. There's a table at the above Wikipedia article, but it's just the possible positions sorted left to right and numbered 0 to 9: 0='''NN---''', 1='''N-N--''', 2='''N--N-''', 3='''N---N''', 4='''-NN--''', etc; our pair is combination number 5. Call this number N. '''N=5'''\n\n2. Still ignoring the Bishops, find the position of the Queen in the remaining 6 spaces; number them 0..5 from left to right and call the index of the Queen's position Q. In our example, '''Q=2'''.\n\n3. Finally, find the positions of the two bishops within their respective sets of four like-colored squares. It's important to note here that the board in chess is placed such that the leftmost position on the home row is on a dark square and the rightmost a light. So if we number the squares of each color 0..3 from left to right, the dark bishop in the standard position is on square 1 ('''D=1'''), and the light bishop is on square 2 ('''L=2''').\n\n4. Then the position number is given by '''4(4(6N + Q)+D)+L''', which reduces to '''96N + 16Q + 4D + L'''. In our example, that's 96x5 + 16x2 + 4x1 + 2 = 480 + 32 + 4 + 2 = 518.\n\nNote that an earlier iteration of this page contained an incorrect description of the algorithm which would give the same SP-ID for both of the following two positions.\n\n    RQNBBKRN = 601\n    RNQBBKRN = 617\n\n", "solution": "# optional, but task function depends on it as written\ndef validate_position(candidate: str):\n    assert (\n        len(candidate) == 8\n    ), f\"candidate position has invalide len = {len(candidate)}\"\n\n    valid_pieces = {\"R\": 2, \"N\": 2, \"B\": 2, \"Q\": 1, \"K\": 1}\n    assert {\n        piece for piece in candidate\n    } == valid_pieces.keys(), f\"candidate position contains invalid pieces\"\n    for piece_type in valid_pieces.keys():\n        assert (\n            candidate.count(piece_type) == valid_pieces[piece_type]\n        ), f\"piece type '{piece_type}' has invalid count\"\n\n    bishops_pos = [index for index, \n                   value in enumerate(candidate) if value == \"B\"]\n    assert (\n        bishops_pos[0] % 2 != bishops_pos[1] % 2\n    ), f\"candidate position has both bishops in the same color\"\n\n    assert [piece for piece in candidate if piece in \"RK\"] == [\n        \"R\",\n        \"K\",\n        \"R\",\n    ], \"candidate position has K outside of RR\"\n\n\ndef calc_position(start_pos: str):\n    try:\n        validate_position(start_pos)\n    except AssertionError:\n        raise AssertionError\n    # step 1\n    subset_step1 = [piece for piece in start_pos if piece not in \"QB\"]\n    nights_positions = [\n        index for index, value in enumerate(subset_step1) if value == \"N\"\n    ]\n    nights_table = {\n        (0, 1): 0,\n        (0, 2): 1,\n        (0, 3): 2,\n        (0, 4): 3,\n        (1, 2): 4,\n        (1, 3): 5,\n        (1, 4): 6,\n        (2, 3): 7,\n        (2, 4): 8,\n        (3, 4): 9,\n    }\n    N = nights_table.get(tuple(nights_positions))\n\n    # step 2\n    subset_step2 = [piece for piece in start_pos if piece != \"B\"]\n    Q = subset_step2.index(\"Q\")\n\n    # step 3\n    dark_squares = [\n        piece for index, piece in enumerate(start_pos) if index in range(0, 9, 2)\n    ]\n    light_squares = [\n        piece for index, piece in enumerate(start_pos) if index in range(1, 9, 2)\n    ]\n    D = dark_squares.index(\"B\")\n    L = light_squares.index(\"B\")\n\n    return 4 * (4 * (6*N + Q) + D) + L\n\nif __name__ == '__main__':\n    for example in [\"QNRBBNKR\", \"RNBQKBNR\", \"RQNBBKRN\", \"RNQBBKRN\"]:\n        print(f'Position: {example}; Chess960 PID= {calc_position(example)}')"}
{"title": "Find duplicate files", "language": "Python", "task": "In a large directory structure it is easy to inadvertently leave unnecessary copies of files around, which can use considerable disk space and create confusion. \n\n\n;Task:\nCreate a program which, given a minimum size and a folder/directory, will find all files of at least ''size'' bytes with duplicate contents under the directory and output or show the sets of duplicate files in order of decreasing size.\n\nThe program may be command-line or graphical, and duplicate content may be determined by direct comparison or by calculating a hash of the data. \n\nSpecify which filesystems or operating systems your program works with if it has any filesystem- or OS-specific requirements. \n\nIdentify hard links (filenames referencing the same content) in the output if applicable for the filesystem. \n\nFor extra points, detect when whole directory sub-trees are identical, or optionally remove or link identical files.\n\n", "solution": "from __future__ import print_function\nimport os\nimport hashlib\nimport datetime\n\ndef FindDuplicateFiles(pth, minSize = 0, hashName = \"md5\"):\n    knownFiles = {}\n\n    #Analyse files\n    for root, dirs, files in os.walk(pth):\n        for fina in files:\n            fullFina = os.path.join(root, fina)\n            isSymLink = os.path.islink(fullFina)\n            if isSymLink:\n                continue # Skip symlinks\n            si = os.path.getsize(fullFina)\n            if si < minSize:\n                continue\n            if si not in knownFiles:\n                knownFiles[si] = {}\n            h = hashlib.new(hashName)\n            h.update(open(fullFina, \"rb\").read())\n            hashed = h.digest()\n            if hashed in knownFiles[si]:\n                fileRec = knownFiles[si][hashed]\n                fileRec.append(fullFina)\n            else:\n                knownFiles[si][hashed] = [fullFina]\n\n    #Print result\n    sizeList = list(knownFiles.keys())\n    sizeList.sort(reverse=True)\n    for si in sizeList:\n        filesAtThisSize = knownFiles[si]\n        for hashVal in filesAtThisSize:\n            if len(filesAtThisSize[hashVal]) < 2:\n                continue\n            fullFinaLi = filesAtThisSize[hashVal]\n            print (\"=======Duplicate=======\")\n            for fullFina in fullFinaLi:\n                st = os.stat(fullFina)\n                isHardLink = st.st_nlink > 1 \n                infoStr = []\n                if isHardLink:\n                    infoStr.append(\"(Hard linked)\")\n                fmtModTime = datetime.datetime.utcfromtimestamp(st.st_mtime).strftime('%Y-%m-%dT%H:%M:%SZ')\n                print (fmtModTime, si, os.path.relpath(fullFina, pth), \" \".join(infoStr))\n\nif __name__==\"__main__\":\n\n    FindDuplicateFiles('/home/tim/Dropbox', 1024*1024)\n"}
{"title": "Find if a point is within a triangle", "language": "Python", "task": "Find if a point is within a triangle.\n\n\n;Task:\n    \n::*   Assume points are on a plane defined by (x, y) real number coordinates.\n\n::*   Given a point P(x, y) and a triangle formed by points A, B, and C, determine if P is within triangle ABC. \n\n::*   You may use any algorithm.  \n\n::*   Bonus: explain why the algorithm you chose works.\n\n\n;Related tasks:\n*   [[Determine_if_two_triangles_overlap]]\n\n\n;Also see:\n:* Discussion of several methods. [[http://totologic.blogspot.com/2014/01/accurate-point-in-triangle-test.html]]\n:* Determine if a point is in a polygon [[https://en.wikipedia.org/wiki/Point_in_polygon]]\n:* Triangle based coordinate systems [[https://en.wikipedia.org/wiki/Barycentric_coordinate_system]]\n:* Wolfram entry [[https://mathworld.wolfram.com/TriangleInterior.html]]\n\n", "solution": "\"\"\" find if point is in a triangle \"\"\"\n\nfrom sympy.geometry import Point, Triangle\n\ndef sign(pt1, pt2, pt3):\n    \"\"\" which side of plane cut by line (pt2, pt3) is pt1 on? \"\"\"\n    return (pt1.x - pt3.x) * (pt2.y - pt3.y) - (pt2.x - pt3.x) * (pt1.y - pt3.y)\n\n\ndef iswithin(point, pt1, pt2, pt3):\n    \"\"\" \n    Determine if point is within triangle formed by points p1, p2, p3.\n    If so, the point will be on the same side of each of the half planes\n    defined by vectors p1p2, p2p3, and p3p1. zval is positive if outside,\n    negative if inside such a plane. All should be positive or all negative\n    if point is within the triangle.\n    \"\"\"\n    zval1 = sign(point, pt1, pt2)\n    zval2 = sign(point, pt2, pt3)\n    zval3 = sign(point, pt3, pt1)\n    notanyneg = zval1 >= 0 and zval2 >= 0 and zval3 >= 0\n    notanypos = zval1 <= 0 and zval2 <= 0 and zval3 <= 0\n    return notanyneg or notanypos\n\nif __name__ == \"__main__\":\n    POINTS = [Point(0, 0)]\n    TRI = Triangle(Point(1.5, 2.4), Point(5.1, -3.1), Point(-3.8, 0.5))\n    for pnt in POINTS:\n        a, b, c = TRI.vertices\n        isornot = \"is\" if iswithin(pnt, a, b, c) else \"is not\"\n        print(\"Point\", pnt, isornot, \"within the triangle\", TRI)\n"}
{"title": "Find limit of recursion", "language": "Python", "task": "{{selection|Short Circuit|Console Program Basics}}\n\n \n\n\n;Task:\nFind the limit of recursion.\n\n", "solution": "def recurseDeeper(counter):\n    try:\n        print(counter)\n        recurseDeeper(counter + 1)\n    except RecursionError:\n        print(\"RecursionError at depth\", counter)\n        recurseDeeper(counter + 1)"}
{"title": "Find palindromic numbers in both binary and ternary bases", "language": "Python", "task": "*   Find and show (in decimal) the first six numbers (non-negative integers) that are   palindromes   in   ''both'':\n:::*   base 2\n:::*   base 3\n*   Display   '''0'''   (zero) as the first number found, even though some other definitions ignore it.\n*   Optionally, show the decimal number found in its binary and ternary form.\n*   Show all output here.\n\n\nIt's permissible to assume the first two numbers and simply list them.\n\n\n;See also\n*   Sequence A60792,   numbers that are palindromic in bases 2 and 3 on ''The On-Line Encyclopedia of Integer Sequences''.\n\n", "solution": "from itertools import islice\n\ndigits = \"0123456789abcdefghijklmnopqrstuvwxyz\"\n\ndef baseN(num,b):\n  if num == 0: return \"0\"\n  result = \"\"\n  while num != 0:\n    num, d = divmod(num, b)\n    result += digits[d]\n  return result[::-1] # reverse\n\ndef pal2(num):\n    if num == 0 or num == 1: return True\n    based = bin(num)[2:]\n    return based == based[::-1]\n\ndef pal_23():\n    yield 0\n    yield 1\n    n = 1\n    while True:\n        n += 1\n        b = baseN(n, 3)\n        revb = b[::-1]\n        #if len(b) > 12: break\n        for trial in ('{0}{1}'.format(b, revb), '{0}0{1}'.format(b, revb),\n                      '{0}1{1}'.format(b, revb), '{0}2{1}'.format(b, revb)):\n            t = int(trial, 3)\n            if pal2(t):\n                yield t\n\nfor pal23 in islice(pal_23(), 6):\n    print(pal23, baseN(pal23, 3), baseN(pal23, 2))"}
{"title": "Find palindromic numbers in both binary and ternary bases", "language": "Python 3.7", "task": "*   Find and show (in decimal) the first six numbers (non-negative integers) that are   palindromes   in   ''both'':\n:::*   base 2\n:::*   base 3\n*   Display   '''0'''   (zero) as the first number found, even though some other definitions ignore it.\n*   Optionally, show the decimal number found in its binary and ternary form.\n*   Show all output here.\n\n\nIt's permissible to assume the first two numbers and simply list them.\n\n\n;See also\n*   Sequence A60792,   numbers that are palindromic in bases 2 and 3 on ''The On-Line Encyclopedia of Integer Sequences''.\n\n", "solution": "'''Numbers with palindromic digit strings in both binary and ternary'''\n\nfrom itertools import (islice)\n\n\n# palinBoth :: Generator [Int]\ndef palinBoth():\n    '''Non finite stream of dually palindromic integers.'''\n    yield 0, '0', '0'\n    ibt = 1, '1', '1'\n\n    yield ibt\n    while True:\n        ibt = until(isBoth)(psucc)(psucc(ibt))\n        yield int(ibt[2], 3), ibt[1], ibt[2]\n\n\n# isBoth :: (Int, String, String) -> Bool\ndef isBoth(ibt):\n    '''True if the binary string is palindromic (as\n       the ternary string is already known to be).\n    '''\n    b = ibt[1]\n    return b == b[::-1]\n\n\n# psucc :: (Int, String, String) -> (Int, String, String)\ndef psucc(ibt):\n    '''The next triple of index, binary\n       and (palindromic) ternary string\n    '''\n    d = 1 + ibt[0]\n    s = showBase3(d)\n    pal = s + '1' + s[::-1]\n    return d, bin(int(pal, 3))[2:], pal\n\n\n# showBase3 :: Int -> String\ndef showBase3(n):\n    '''Ternary digit string for integer n.'''\n    return showIntAtBase(3)(\n        lambda i: '012'[i]\n    )(n)('')\n\n\n# ------------------------- TEST -------------------------\ndef main():\n    '''Integers with palindromic digits in\n       both binary and ternary bases.\n    '''\n\n    xs = take(6)(palinBoth())\n    d, b, t = xs[-1]\n    bw = len(b)\n    tw = len(t)\n\n    print(\n        fTable(\n            label('rjust')(('Decimal', len(str(d)))) +\n            ''.join(map(\n                label('center'),\n                [('Binary', bw), ('Ternary', tw)]\n            )) + '\\n'\n        )(compose(str)(fst))(\n            lambda p: p[1].center(bw, ' ') +\n            '    ' + p[2].center(tw, ' ')\n        )(identity)(xs)\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# identity :: a -> a\ndef identity(x):\n    '''The identity function.'''\n    return x\n\n\n# showIntAtBase :: Int -> (Int -> String) -> Int -> String -> String\ndef showIntAtBase(base):\n    '''String representation of an integer in a given base,\n       using a supplied function for the string representation\n       of digits.\n    '''\n    def wrap(toChr, n, rs):\n        def go(nd, r):\n            n, d = nd\n            r_ = toChr(d) + r\n            return go(divmod(n, base), r_) if 0 != n else r_\n        return 'unsupported base' if 1 >= base else (\n            'negative number' if 0 > n else (\n                go(divmod(n, base), rs))\n        )\n    return lambda toChr: lambda n: lambda rs: (\n        wrap(toChr, n, rs)\n    )\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    def go(xs):\n        return (\n            xs[0:n]\n            if isinstance(xs, (list, tuple))\n            else list(islice(xs, n))\n        )\n    return go\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f):\n        def g(x):\n            v = x\n            while not p(v):\n                v = f(v)\n            return v\n        return g\n    return go\n\n\n# ---------------------- FORMATTING ----------------------\n\n# label :: Method String -> (String, Int)\ndef label(k):\n    '''Stringification, using the named justification\n       method (ljust|centre|rjust) of the label,\n       and the specified amount of white space.\n    '''\n    def go(sw):\n        s, w = sw\n        return getattr(s, k)(w, ' ') + '    '\n    return go\n\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Find the intersection of a line with a plane", "language": "Python", "task": "Finding the intersection of an infinite ray with a plane in 3D is an important topic in collision detection.\n\n\n;Task:\nFind the point of intersection for the infinite ray with direction   (0, -1, -1)   passing through position   (0, 0, 10)   with the infinite plane with a normal vector of   (0, 0, 1)   and which passes through [0, 0, 5].\n\n", "solution": "#!/bin/python\nfrom __future__ import print_function\nimport numpy as np\n\ndef LinePlaneCollision(planeNormal, planePoint, rayDirection, rayPoint, epsilon=1e-6):\n\n\tndotu = planeNormal.dot(rayDirection)\n\tif abs(ndotu) < epsilon:\n\t\traise RuntimeError(\"no intersection or line is within plane\")\n\n\tw = rayPoint - planePoint\n\tsi = -planeNormal.dot(w) / ndotu\n\tPsi = w + si * rayDirection + planePoint\n\treturn Psi\n\n\nif __name__==\"__main__\":\n\t#Define plane\n\tplaneNormal = np.array([0, 0, 1])\n\tplanePoint = np.array([0, 0, 5]) #Any point on the plane\n\n\t#Define ray\n\trayDirection = np.array([0, -1, -1])\n\trayPoint = np.array([0, 0, 10]) #Any point along the ray\n\n\tPsi = LinePlaneCollision(planeNormal, planePoint, rayDirection, rayPoint)\n\tprint (\"intersection at\", Psi)"}
{"title": "Find the intersection of two lines", "language": "Python", "task": "Finding the intersection of two lines that are in the same plane is an important topic in collision detection.[http://mathworld.wolfram.com/Line-LineIntersection.html]\n\n;Task:\nFind the point of intersection of two lines in 2D. \n\n\nThe 1st     line passes though    (4,0)    and    (6,10) . \nThe 2nd line passes though    (0,3)    and    (10,7) .\n\n", "solution": "def line_intersect(Ax1, Ay1, Ax2, Ay2, Bx1, By1, Bx2, By2):\n    \"\"\" returns a (x, y) tuple or None if there is no intersection \"\"\"\n    d = (By2 - By1) * (Ax2 - Ax1) - (Bx2 - Bx1) * (Ay2 - Ay1)\n    if d:\n        uA = ((Bx2 - Bx1) * (Ay1 - By1) - (By2 - By1) * (Ax1 - Bx1)) / d\n        uB = ((Ax2 - Ax1) * (Ay1 - By1) - (Ay2 - Ay1) * (Ax1 - Bx1)) / d\n    else:\n        return\n    if not(0 <= uA <= 1 and 0 <= uB <= 1):\n        return\n    x = Ax1 + uA * (Ax2 - Ax1)\n    y = Ay1 + uA * (Ay2 - Ay1)\n    \n    return x, y\n\nif __name__ == '__main__':\n    (a, b), (c, d) = (4, 0), (6, 10)  # try (4, 0), (6, 4)\n    (e, f), (g, h) = (0, 3), (10, 7)  # for non intersecting test\n    pt = line_intersect(a, b, c, d, e, f, g, h)\n    print(pt)"}
{"title": "Find the intersection of two lines", "language": "Python 3.7", "task": "Finding the intersection of two lines that are in the same plane is an important topic in collision detection.[http://mathworld.wolfram.com/Line-LineIntersection.html]\n\n;Task:\nFind the point of intersection of two lines in 2D. \n\n\nThe 1st     line passes though    (4,0)    and    (6,10) . \nThe 2nd line passes though    (0,3)    and    (10,7) .\n\n", "solution": "'''The intersection of two lines.'''\n\nfrom itertools import product\n\n\n# intersection :: Line -> Line -> Either String Point\ndef intersection(ab):\n    '''Either the point at which the lines ab and pq\n       intersect, or a message string indicating that\n       they are parallel and have no intersection.'''\n    def delta(f):\n        return lambda x: f(fst(x)) - f(snd(x))\n\n    def prodDiff(abcd):\n        [a, b, c, d] = abcd\n        return (a * d) - (b * c)\n\n    def go(pq):\n        [abDX, pqDX, abDY, pqDY] = apList(\n            [delta(fst), delta(snd)]\n        )([ab, pq])\n        determinant = prodDiff([abDX, abDY, pqDX, pqDY])\n\n        def point():\n            [abD, pqD] = map(\n                lambda xy: prodDiff(\n                    apList([fst, snd])([fst(xy), snd(xy)])\n                ), [ab, pq]\n            )\n            return apList(\n                [lambda abpq: prodDiff(\n                    [abD, fst(abpq), pqD, snd(abpq)]) / determinant]\n            )(\n                [(abDX, pqDX), (abDY, pqDY)]\n            )\n        return Right(point()) if 0 != determinant else Left(\n            '( Parallel lines - no intersection )'\n        )\n\n    return lambda pq: bindLR(go(pq))(\n        lambda xs: Right((fst(xs), snd(xs)))\n    )\n\n\n# --------------------------TEST---------------------------\n# main :: IO()\ndef main():\n    '''Test'''\n\n    # Left(message - no intersection) or Right(point)\n    # lrPoint :: Either String Point\n    lrPoint = intersection(\n        ((4.0, 0.0), (6.0, 10.0))\n    )(\n        ((0.0, 3.0), (10.0, 7.0))\n    )\n    print(\n        lrPoint['Left'] or lrPoint['Right']\n    )\n\n\n# --------------------GENERIC FUNCTIONS--------------------\n\n# Left :: a -> Either a b\ndef Left(x):\n    '''Constructor for an empty Either (option type) value\n       with an associated string.'''\n    return {'type': 'Either', 'Right': None, 'Left': x}\n\n\n# Right :: b -> Either a b\ndef Right(x):\n    '''Constructor for a populated Either (option type) value'''\n    return {'type': 'Either', 'Left': None, 'Right': x}\n\n\n# apList (<*>) :: [(a -> b)] -> [a] -> [b]\ndef apList(fs):\n    '''The application of each of a list of functions,\n       to each of a list of values.\n    '''\n    def go(fx):\n        f, x = fx\n        return f(x)\n    return lambda xs: [\n        go(x) for x\n        in product(fs, xs)\n    ]\n\n\n# bindLR (>>=) :: Either a -> (a -> Either b) -> Either b\ndef bindLR(m):\n    '''Either monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.'''\n    return lambda mf: (\n        mf(m.get('Right')) if None is m.get('Left') else m\n    )\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second member of a pair.'''\n    return tpl[1]\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Find the last Sunday of each month", "language": "Python", "task": "Write a program or a script that returns the last Sundays of each month of a given year. The year may be given through any simple input method in your language (command line, std in, etc).\n\nExample of an expected output:\n\n./last_sundays 2013\n2013-01-27\n2013-02-24\n2013-03-31\n2013-04-28\n2013-05-26\n2013-06-30\n2013-07-28\n2013-08-25\n2013-09-29\n2013-10-27\n2013-11-24\n2013-12-29\n\n;Related tasks\n* [[Day of the week]]\n* [[Five weekends]]\n* [[Last Friday of each month]]\n\n", "solution": "import sys\nimport calendar\n\nyear = 2013\nif len(sys.argv) > 1:\n    try:\n        year = int(sys.argv[-1])\n    except ValueError:\n        pass\n\nfor month in range(1, 13):\n    last_sunday = max(week[-1] for week in calendar.monthcalendar(year, month))\n    print('{}-{}-{:2}'.format(year, calendar.month_abbr[month], last_sunday))\n"}
{"title": "Find the missing permutation", "language": "Python", "task": "ABCD\n                    CABD\n                    ACDB\n                    DACB\n                    BCDA\n                    ACBD\n                    ADCB\n                    CDAB\n                    DABC\n                    BCAD\n                    CADB\n                    CDBA\n                    CBAD\n                    ABDC\n                    ADBC\n                    BDCA\n                    DCBA\n                    BACD\n                    BADC\n                    BDAC\n                    CBDA\n                    DBCA\n                    DCAB\n\n\nListed above are   all-but-one   of the permutations of the symbols   '''A''',   '''B''',   '''C''',   and   '''D''',   ''except''   for one permutation that's   ''not''   listed.  \n\n\n;Task:\nFind that missing permutation.\n\n\n;Methods:\n* Obvious method: \n         enumerate all permutations of   '''A''',  '''B''',  '''C''',  and  '''D''',  \n         and then look for the missing permutation. \n\n* alternate method:\n         Hint:  if all permutations were shown above,  how many \n         times would  '''A'''  appear in each position?     \n         What is the  ''parity''  of this number?\n\n* another alternate method:\n         Hint:  if you add up the letter values of each column, \n         does a missing letter   '''A''',  '''B''',  '''C''',  and  '''D'''   from each\n         column cause the total value for each column to be unique?\n\n\n;Related task:\n*   [[Permutations]])\n\n", "solution": "===Python: Calculate difference when compared to all permutations===\n"}
{"title": "Find the missing permutation", "language": "Python 2.6+", "task": "ABCD\n                    CABD\n                    ACDB\n                    DACB\n                    BCDA\n                    ACBD\n                    ADCB\n                    CDAB\n                    DABC\n                    BCAD\n                    CADB\n                    CDBA\n                    CBAD\n                    ABDC\n                    ADBC\n                    BDCA\n                    DCBA\n                    BACD\n                    BADC\n                    BDAC\n                    CBDA\n                    DBCA\n                    DCAB\n\n\nListed above are   all-but-one   of the permutations of the symbols   '''A''',   '''B''',   '''C''',   and   '''D''',   ''except''   for one permutation that's   ''not''   listed.  \n\n\n;Task:\nFind that missing permutation.\n\n\n;Methods:\n* Obvious method: \n         enumerate all permutations of   '''A''',  '''B''',  '''C''',  and  '''D''',  \n         and then look for the missing permutation. \n\n* alternate method:\n         Hint:  if all permutations were shown above,  how many \n         times would  '''A'''  appear in each position?     \n         What is the  ''parity''  of this number?\n\n* another alternate method:\n         Hint:  if you add up the letter values of each column, \n         does a missing letter   '''A''',  '''B''',  '''C''',  and  '''D'''   from each\n         column cause the total value for each column to be unique?\n\n\n;Related task:\n*   [[Permutations]])\n\n", "solution": "def missing_permutation(arr):\n  \"Find the missing permutation in an array of N! - 1 permutations.\"\n  \n  # We won't validate every precondition, but we do have some basic\n  # guards.\n  if len(arr) == 0: raise Exception(\"Need more data\")\n  if len(arr) == 1:\n      return [arr[0][1] + arr[0][0]]\n  \n  # Now we know that for each position in the string, elements should appear\n  # an even number of times (N-1 >= 2).  We can use a set to detect the element appearing\n  # an odd number of times.  Detect odd occurrences by toggling admission/expulsion\n  # to and from the set for each value encountered.  At the end of each pass one element\n  # will remain in the set.\n  missing_permutation = ''\n  for pos in range(len(arr[0])):\n      s = set()\n      for permutation in arr:\n          c = permutation[pos]\n          if c in s:\n            s.remove(c)\n          else:\n            s.add(c)\n      missing_permutation += list(s)[0]\n  return missing_permutation\n  \ngiven = '''ABCD CABD ACDB DACB BCDA ACBD ADCB CDAB DABC BCAD CADB CDBA\n           CBAD ABDC ADBC BDCA DCBA BACD BADC BDAC CBDA DBCA DCAB'''.split()\n           \nprint missing_permutation(given)\n"}
{"title": "Find the missing permutation", "language": "Python from JavaScript", "task": "ABCD\n                    CABD\n                    ACDB\n                    DACB\n                    BCDA\n                    ACBD\n                    ADCB\n                    CDAB\n                    DABC\n                    BCAD\n                    CADB\n                    CDBA\n                    CBAD\n                    ABDC\n                    ADBC\n                    BDCA\n                    DCBA\n                    BACD\n                    BADC\n                    BDAC\n                    CBDA\n                    DBCA\n                    DCAB\n\n\nListed above are   all-but-one   of the permutations of the symbols   '''A''',   '''B''',   '''C''',   and   '''D''',   ''except''   for one permutation that's   ''not''   listed.  \n\n\n;Task:\nFind that missing permutation.\n\n\n;Methods:\n* Obvious method: \n         enumerate all permutations of   '''A''',  '''B''',  '''C''',  and  '''D''',  \n         and then look for the missing permutation. \n\n* alternate method:\n         Hint:  if all permutations were shown above,  how many \n         times would  '''A'''  appear in each position?     \n         What is the  ''parity''  of this number?\n\n* another alternate method:\n         Hint:  if you add up the letter values of each column, \n         does a missing letter   '''A''',  '''B''',  '''C''',  and  '''D'''   from each\n         column cause the total value for each column to be unique?\n\n\n;Related task:\n*   [[Permutations]])\n\n", "solution": "'''Find the missing permutation'''\n\nfrom functools import reduce\nfrom operator import xor\n\n\nprint(''.join([\n    chr(i) for i in reduce(\n        lambda a, s: map(\n            xor,\n            a,\n            [ord(c) for c in list(s)]\n        ), [\n            'ABCD', 'CABD', 'ACDB', 'DACB',\n            'BCDA', 'ACBD', 'ADCB', 'CDAB',\n            'DABC', 'BCAD', 'CADB', 'CDBA',\n            'CBAD', 'ABDC', 'ADBC', 'BDCA',\n            'DCBA', 'BACD', 'BADC', 'BDAC',\n            'CBDA', 'DBCA', 'DCAB'\n        ],\n        [0, 0, 0, 0]\n    )\n]))"}
{"title": "First-class functions/Use numbers analogously", "language": "Python", "task": "In [[First-class functions]], a language is showing how its manipulation of functions is similar to its manipulation of other types.\n\nThis tasks aim is to compare and contrast a language's implementation of first class functions, with its normal handling of numbers.\n\n\nWrite a program to create an ordered collection of a mixture of literally typed and expressions producing a real number, together with another ordered collection of their multiplicative inverses. Try and use the following pseudo-code to generate the numbers for the ordered collections:\n   x  = 2.0\n   xi = 0.5\n   y  = 4.0\n   yi = 0.25\n   z  = x + y\n   zi = 1.0 / ( x + y )\n\nCreate a function ''multiplier'', that given two numbers as arguments returns a function that when called with one argument, returns the result of multiplying the two arguments to the call to multiplier that created it and the argument in the call:\n  new_function = multiplier(n1,n2)\n  # where new_function(m) returns the result of n1 * n2 * m\n\nApplying the multiplier of a number and its inverse from the two ordered collections of numbers in pairs, show that the result in each case is one.\n'''Compare and contrast the resultant program with the corresponding entry in [[First-class functions]].''' They should be close.\n\nTo paraphrase the task description: Do what was done before, but with numbers rather than functions\n\n", "solution": ">>> # Some built in functions and their inverses\n>>> from math import sin, cos, acos, asin\n>>> # Add a user defined function and its inverse\n>>> cube = lambda x: x * x * x\n>>> croot = lambda x: x ** (1/3.0)\n>>> # First class functions allow run-time creation of functions from functions\n>>> # return function compose(f,g)(x) == f(g(x))\n>>> compose = lambda f1, f2: ( lambda x: f1(f2(x)) )\n>>> # first class functions should be able to be members of collection types\n>>> funclist = [sin, cos, cube]\n>>> funclisti = [asin, acos, croot]\n>>> # Apply functions from lists as easily as integers\n>>> [compose(inversef, f)(.5) for f, inversef in zip(funclist, funclisti)]\n[0.5, 0.4999999999999999, 0.5]\n>>>"}
{"title": "First perfect square in base n with n unique digits", "language": "Python 3.7", "task": "Find the first perfect square in a given base '''N''' that has at least '''N''' digits and\nexactly '''N''' ''significant unique'' digits when expressed in base '''N'''.\n\nE.G. In base '''10''', the first perfect square with at least '''10''' unique digits is '''1026753849''' ('''320432''').\n\nYou may use analytical methods to reduce the search space, but the code must do a search. Do not use magic numbers or just feed the code the answer to verify it is correct.\n\n;Task\n\n* Find and display here, on this page, the first perfect square in base '''N''', with '''N''' significant unique digits when expressed in base '''N''', for each of base '''2''' through '''12'''. Display each number in the base '''N''' for which it was calculated.\n\n* (optional) Do the same for bases '''13''' through '''16'''.\n\n* (stretch goal) Continue on for bases '''17''' - '''??''' (Big Integer math)\n\n\n;See also:\n\n;* OEIS A260182: smallest square that is pandigital in base n.\n\n;Related task\n\n;* [[Casting out nines]]\n\n", "solution": "'''Perfect squares using every digit in a given base.'''\n\nfrom itertools import count, dropwhile, repeat\nfrom math import ceil, sqrt\nfrom time import time\n\n\n# allDigitSquare :: Int -> Int -> Int\ndef allDigitSquare(base, above):\n    '''The lowest perfect square which\n       requires all digits in the given base.\n    '''\n    bools = list(repeat(True, base))\n    return next(\n        dropwhile(\n            missingDigitsAtBase(base, bools),\n            count(\n                max(\n                    above,\n                    ceil(sqrt(int(\n                        '10' + '0123456789abcdef'[2:base],\n                        base\n                    )))\n                )\n            )\n        )\n    )\n\n\n# missingDigitsAtBase :: Int -> [Bool] -> Int -> Bool\ndef missingDigitsAtBase(base, bools):\n    '''Fusion of representing the square of integer N at a\n       given base with checking whether all digits of\n       that base contribute to N^2.\n       Clears the bool at a digit position to False when used.\n       True if any positions remain uncleared (unused).\n    '''\n    def go(x):\n        xs = bools.copy()\n        while x:\n            xs[x % base] = False\n            x //= base\n        return any(xs)\n    return lambda n: go(n * n)\n\n\n# digit :: Int -> Char\ndef digit(n):\n    '''Digit character for given integer.'''\n    return '0123456789abcdef'[n]\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Smallest perfect squares using all digits in bases 2-16'''\n\n    start = time()\n\n    print(main.__doc__ + ':\\n\\nBase      Root    Square')\n    q = 0\n    for b in enumFromTo(2)(16):\n        q = allDigitSquare(b, q)\n        print(\n            str(b).rjust(2, ' ') + ' -> ' +\n            showIntAtBase(b)(digit)(q)('').rjust(8, ' ') +\n            ' -> ' +\n            showIntAtBase(b)(digit)(q * q)('')\n        )\n\n    print(\n        '\\nc. ' + str(ceil(time() - start)) + ' seconds.'\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# showIntAtBase :: Int -> (Int -> String) -> Int -> \n# String -> String\ndef showIntAtBase(base):\n    '''String representation of an integer in a given base,\n       using a supplied function for the string representation\n       of digits.\n    '''\n    def wrap(toChr, n, rs):\n        def go(nd, r):\n            n, d = nd\n            r_ = toChr(d) + r\n            return go(divmod(n, base), r_) if 0 != n else r_\n        return 'unsupported base' if 1 >= base else (\n            'negative number' if 0 > n else (\n                go(divmod(n, base), rs))\n        )\n    return lambda toChr: lambda n: lambda rs: (\n        wrap(toChr, n, rs)\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()\n"}
{"title": "First power of 2 that has leading decimal digits of 12", "language": "Python", "task": "(This task is taken from a   ''Project Euler''   problem.)\n\n(All numbers herein are expressed in base ten.)\n\n\n'''27   =   128'''   and   '''7'''   is\nthe first power of   '''2'''   whose leading decimal digits are   '''12'''.\n\nThe next power of   '''2'''   whose leading decimal digits\nare   '''12'''   is   '''80''',\n'''280   =   1208925819614629174706176'''.\n\n\nDefine     '''  '' p''(''L,n'')'''      to be the ''' ''n''th'''-smallest\nvalue of   ''' ''j'' '''   such that the base ten representation\nof    '''2''j'''''    begins with the digits of   ''' ''L'' '''.\n\n     So   ''p''(12, 1) =  7    and\n          ''p''(12, 2) = 80\n\n\nYou are also given that:\n          ''p''(123, 45)   =   12710\n\n\n;Task:\n::*   find: \n:::::*     '''   ''p''(12,  1)        ''' \n:::::*     '''   ''p''(12,  2)        ''' \n:::::*     '''   ''p''(123, 45)       ''' \n:::::*     '''   ''p''(123, 12345)    ''' \n:::::*     '''   ''p''(123, 678910)   ''' \n::*   display the results here, on this page.\n\n", "solution": "from math import log, modf, floor\n\ndef p(l, n, pwr=2):\n    l = int(abs(l))\n    digitcount = floor(log(l, 10))\n    log10pwr = log(pwr, 10)\n    raised, found = -1, 0\n    while found < n:\n        raised += 1\n        firstdigits = floor(10**(modf(log10pwr * raised)[0] + digitcount))\n        if firstdigits == l:\n            found += 1\n    return raised\n\n\nif __name__ == '__main__':\n    for l, n in [(12, 1), (12, 2), (123, 45), (123, 12345), (123, 678910)]:\n        print(f\"p({l}, {n}) =\", p(l, n))"}
{"title": "Fivenum", "language": "Python from Perl", "task": "Many big data or scientific programs use boxplots to show distributions of data.   In addition, sometimes saving large arrays for boxplots can be impractical and use extreme amounts of RAM.   It can be useful to save large arrays as arrays with five numbers to save memory.\n\nFor example, the   '''R'''   programming language implements Tukey's five-number summary as the '''fivenum''' function.\n\n\n;Task:\nGiven an array of numbers, compute the five-number summary.\n\n\n;Note:  \nWhile these five numbers can be used to draw a boxplot,   statistical packages will typically need extra data. \n\nMoreover, while there is a consensus about the \"box\" of the boxplot,   there are variations among statistical packages for the whiskers.\n\n", "solution": "from __future__ import division\nimport math\nimport sys\n\ndef fivenum(array):\n    n = len(array)\n    if n == 0:\n        print(\"you entered an empty array.\")\n        sys.exit()\n    x = sorted(array)\n    \n    n4 = math.floor((n+3.0)/2.0)/2.0\n    d = [1, n4, (n+1)/2, n+1-n4, n]\n    sum_array = []\n    \n    for e in range(5):\n        floor = int(math.floor(d[e] - 1))\n        ceil = int(math.ceil(d[e] - 1))\n        sum_array.append(0.5 * (x[floor] + x[ceil]))\n    \n    return sum_array\n\nx = [0.14082834, 0.09748790, 1.73131507, 0.87636009, -1.95059594, 0.73438555, -0.03035726, 1.46675970,\n-0.74621349, -0.72588772, 0.63905160, 0.61501527, -0.98983780, -1.00447874, -0.62759469, 0.66206163,\n1.04312009, -0.10305385, 0.75775634, 0.32566578]\n\ny = fivenum(x)\nprint(y)"}
{"title": "Fixed length records", "language": "Python", "task": "Fixed length read/write\n\nBefore terminals, computers commonly used punch card readers or paper tape input.\n\nA common format before these devices were superseded by terminal technology was based on the Hollerith code, Hollerith code.\n\nThese input devices handled 80 columns per card and had a limited character set, encoded by punching holes in one or more rows of the card for each column.\n\nThese devices assumed/demanded a fixed line width of 80 characters, newlines were not required (and could not even be encoded in some systems).\n\n;Task:\nWrite a program to read 80 column fixed length records (no newline terminators (but newline characters allowed in the data)) and then write out the reverse of each line as fixed length 80 column records. \n\nSamples here use printable characters, but that is not a given with fixed length data.  Filenames used are sample.txt, infile.dat, outfile.dat.\n\n'''Note:''' There are no newlines, inputs and outputs are fixed at 80 columns, no more, no less, space padded.  Fixed length data is 8 bit complete.  NUL bytes of zero are allowed.\n\nThese fixed length formats are still in wide use on mainframes, with JCL and with COBOL (which commonly use EBCDIC encoding and not ASCII).  Most of the large players in day to day financial transactions know all about fixed length records and the expression ''logical record length''.\n\n;Sample data:\nTo create the sample input file, use an editor that supports fixed length records or use a conversion utility.  For instance, most GNU/Linux versions of '''dd''' support blocking and unblocking records with a conversion byte size.\n\nLine 1...1.........2.........3.........4.........5.........6.........7.........8\nLine 2\nLine 3\nLine 4\n\nLine 6\nLine 7\n     Indented line 8............................................................\nLine 9                                                                 RT MARGIN\nprompt$ dd if=sample.txt of=infile.dat cbs=80 conv=block will create a fixed length record file of 80 bytes given newline delimited text input.\n\nprompt$ dd if=infile.dat cbs=80 conv=unblock will display a file with 80 byte logical record lengths to standard out as standard text with newlines.\n\n\n;Bonus round:\nForth systems often include BLOCK words.  A block is 1024 bytes.  Source code is stored as 16 lines of 64 characters each (again, no newline character or sequence to mark the end of a line).\n\nWrite a program to convert a block file to text (using newlines). Trailing spaces should be excluded from the output.\n\nAlso demonstrate how to convert from a normal text file to block form. All lines either truncated or padded to 64 characters with no newline terminators. The last block filled to be exactly 1024 characters by adding blanks if needed. Assume a full range of 8 bit byte values for each character.\n\nThe COBOL example uses forth.txt and forth.blk filenames.\n\n", "solution": "infile = open('infile.dat', 'rb')\noutfile = open('outfile.dat', 'wb')\n\nwhile True:\n    onerecord = infile.read(80)\n    if len(onerecord) < 80:\n        break\n    onerecordreversed = bytes(reversed(onerecord))\n    outfile.write(onerecordreversed)\n\ninfile.close()\noutfile.close()\n"}
{"title": "Flatten a list", "language": "Python", "task": "Write a function to flatten the nesting in an arbitrary list of values. \n\nYour program should work on the equivalent of this list:\n   [[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]\nWhere the correct result would be the list:\n    [1, 2, 3, 4, 5, 6, 7, 8]\n\n;Related task:\n*   [[Tree traversal]]\n\n", "solution": ">>> def flatten(itr):\n>>>    for x in itr:\n>>>        try:\n>>>            yield from flatten(x)\n>>>        except:\n>>>            yield x\n\n>>> lst = [[1], 2, [[3,4], 5], [[[]]], [[[6]]], 7, 8, []]\n\n>>> list(flatten(lst))\n[1, 2, 3, 4, 5, 6, 7, 8]\n\n>>> tuple(flatten(lst))\n(1, 2, 3, 4, 5, 6, 7, 8)\n\n>>>for i in flatten(lst):\n>>>    print(i)\n1\n2\n3\n4\n5\n6\n7\n8"}
{"title": "Flatten a list", "language": "Python 3.7", "task": "Write a function to flatten the nesting in an arbitrary list of values. \n\nYour program should work on the equivalent of this list:\n   [[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]\nWhere the correct result would be the list:\n    [1, 2, 3, 4, 5, 6, 7, 8]\n\n;Related task:\n*   [[Tree traversal]]\n\n", "solution": "'''Flatten a list'''\n\nfrom functools import (reduce)\nfrom itertools import (chain)\n\n\ndef flatten(xs):\n    '''A flat list of atomic values derived\n       from a nested list.\n    '''\n    return reduce(\n        lambda a, x: a + list(until(every(notList))(\n            concatMap(pureList)\n        )([x])),\n        xs, []\n    )\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''From nested list to flattened list'''\n\n    print(main.__doc__ + ':\\n\\n')\n    xs = [[1], 2, [[3, 4], 5], [[[]]], [[[6]]], 7, 8, []]\n    print(\n        repr(xs) + ' -> ' + repr(flatten(xs))\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).\n    '''\n    return lambda xs: list(\n        chain.from_iterable(map(f, xs))\n    )\n\n\n# every :: (a -> Bool) -> [a] -> Bool\ndef every(p):\n    '''True if p(x) holds for every x in xs'''\n    def go(p, xs):\n        return all(map(p, xs))\n    return lambda xs: go(p, xs)\n\n\n# notList :: a -> Bool\ndef notList(x):\n    '''True if the value x is not a list.'''\n    return not isinstance(x, list)\n\n\n# pureList :: a -> [b]\ndef pureList(x):\n    '''x if x is a list, othewise [x]'''\n    return x if isinstance(x, list) else [x]\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.'''\n    def go(f, x):\n        v = x\n        while not p(v):\n            v = f(v)\n        return v\n    return lambda f: lambda x: go(f, x)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Flipping bits game", "language": "Python", "task": "The game:\nGiven an   '''NxN'''   square array of zeroes or ones in an initial configuration,   and a target configuration of zeroes and ones.\n\n\nThe game is to transform one to the other in as few moves as possible by inverting whole numbered rows or whole lettered \ncolumns at once   (as one move).\n\nIn an inversion.   any  '''1'''  becomes  '''0''',   and any  '''0'''  becomes  '''1'''  for that whole row or column.\n\n\n;Task:\nCreate a program to score for the Flipping bits game.\n# The game should create an original random target configuration and a starting configuration.\n# Ensure that the starting position is ''never'' the target position.\n# The target position must be guaranteed as reachable from the starting position.   (One possible way to do this is to generate the start position by legal flips from a random target position.   The flips will always be reversible back to the target from the given start position).\n# The number of moves taken so far should be shown.\n\n\nShow an example of a short game here, on this page, for a   '''3x3'''   array of bits.\n\n", "solution": "\"\"\"\nGiven a %i by %i sqare array of zeroes or ones in an initial\nconfiguration, and a target configuration of zeroes and ones\nThe task is to transform one to the other in as few moves as \npossible by inverting whole numbered rows or whole lettered \ncolumns at once.\nIn an inversion any 1 becomes 0 and any 0 becomes 1 for that\nwhole row or column.\n\n\"\"\"\n\nfrom random import randrange\nfrom copy import deepcopy\nfrom string import ascii_lowercase\n\n\ntry:    # 2to3 fix\n    input = raw_input\nexcept:\n    pass\n\nN = 3   # N x N Square arrray\n\nboard  = [[0]* N for i in range(N)]\n\ndef setbits(board, count=1):\n    for i in range(count):\n        board[randrange(N)][randrange(N)] ^= 1\n\ndef shuffle(board, count=1):\n    for i in range(count):\n        if randrange(0, 2):\n            fliprow(randrange(N))\n        else:\n            flipcol(randrange(N))\n\n\ndef pr(board, comment=''):\n    print(str(comment))\n    print('     ' + ' '.join(ascii_lowercase[i] for i in range(N)))\n    print('  ' + '\\n  '.join(' '.join(['%2s' % j] + [str(i) for i in line])\n                             for j, line in enumerate(board, 1)))\n\ndef init(board):\n    setbits(board, count=randrange(N)+1)\n    target = deepcopy(board)\n    while board == target:\n        shuffle(board, count=2 * N)\n    prompt = '  X, T, or 1-%i / %s-%s to flip: ' % (N, ascii_lowercase[0], \n                                                    ascii_lowercase[N-1])\n    return target, prompt\n\ndef fliprow(i):\n    board[i-1][:] = [x ^ 1 for x in board[i-1] ]\n    \ndef flipcol(i):\n    for row in board:\n        row[i] ^= 1\n\nif __name__ == '__main__':\n    print(__doc__ % (N, N))\n    target, prompt = init(board)\n    pr(target, 'Target configuration is:')\n    print('')\n    turns = 0\n    while board != target:\n        turns += 1\n        pr(board, '%i:' % turns)\n        ans = input(prompt).strip()\n        if (len(ans) == 1 \n            and ans in ascii_lowercase and ascii_lowercase.index(ans) < N):\n            flipcol(ascii_lowercase.index(ans))\n        elif ans and all(ch in '0123456789' for ch in ans) and 1 <= int(ans) <= N:\n            fliprow(int(ans))\n        elif ans == 'T':\n            pr(target, 'Target configuration is:')\n            turns -= 1\n        elif ans == 'X':\n            break\n        else:\n            print(\"  I don't understand %r... Try again. \"\n                  \"(X to exit or T to show target)\\n\" % ans[:9])\n            turns -= 1\n    else:\n        print('\\nWell done!\\nBye.')"}
{"title": "Floyd's triangle", "language": "Python", "task": "Floyd's triangle   lists the natural numbers in a right triangle aligned to the left where \n* the first row is   '''1'''     (unity)\n* successive rows start towards the left with the next number followed by successive naturals listing one more number than the line above.\n\n\nThe first few lines of a Floyd triangle looks like this:\n\n 1\n 2  3\n 4  5  6\n 7  8  9 10\n11 12 13 14 15\n\n\n\n;Task:\n:# Write a program to generate and display here the first   n   lines of a Floyd triangle. (Use   n=5   and   n=14   rows).\n:# Ensure that when displayed in a mono-space font, the numbers line up in vertical columns as shown and that only one space separates numbers of the last row.\n\n", "solution": ">>> def floyd(rowcount=5):\n\trows = [[1]]\n\twhile len(rows) < rowcount:\n\t\tn = rows[-1][-1] + 1\n\t\trows.append(list(range(n, n + len(rows[-1]) + 1)))\n\treturn rows\n\n>>> floyd()\n[[1], [2, 3], [4, 5, 6], [7, 8, 9, 10], [11, 12, 13, 14, 15]]\n>>> def pfloyd(rows=[[1], [2, 3], [4, 5, 6], [7, 8, 9, 10]]):\n\tcolspace = [len(str(n)) for n in rows[-1]]\n\tfor row in rows:\n\t\tprint( ' '.join('%*i' % space_n for space_n in zip(colspace, row)))\n\n\t\t\n>>> pfloyd()\n1\n2 3\n4 5 6\n7 8 9 10\n>>> pfloyd(floyd(5))\n 1\n 2  3\n 4  5  6\n 7  8  9 10\n11 12 13 14 15\n>>> pfloyd(floyd(14))\n 1\n 2  3\n 4  5  6\n 7  8  9 10\n11 12 13 14 15\n16 17 18 19 20 21\n22 23 24 25 26 27 28\n29 30 31 32 33 34 35 36\n37 38 39 40 41 42 43 44  45\n46 47 48 49 50 51 52 53  54  55\n56 57 58 59 60 61 62 63  64  65  66\n67 68 69 70 71 72 73 74  75  76  77  78\n79 80 81 82 83 84 85 86  87  88  89  90  91\n92 93 94 95 96 97 98 99 100 101 102 103 104 105\n>>> "}
{"title": "Floyd's triangle", "language": "Python 3", "task": "Floyd's triangle   lists the natural numbers in a right triangle aligned to the left where \n* the first row is   '''1'''     (unity)\n* successive rows start towards the left with the next number followed by successive naturals listing one more number than the line above.\n\n\nThe first few lines of a Floyd triangle looks like this:\n\n 1\n 2  3\n 4  5  6\n 7  8  9 10\n11 12 13 14 15\n\n\n\n;Task:\n:# Write a program to generate and display here the first   n   lines of a Floyd triangle. (Use   n=5   and   n=14   rows).\n:# Ensure that when displayed in a mono-space font, the numbers line up in vertical columns as shown and that only one space separates numbers of the last row.\n\n", "solution": "'''Floyd triangle in terms of iterate(f)(x)'''\n\nfrom itertools import islice\n\n\n# floyd :: Int -> [[Int]]\ndef floyd(n):\n    '''n rows of a Floyd triangle.'''\n    return take(n)(iterate(nextFloyd)([1]))\n\n\n# nextFloyd :: [Int] -> [Int]\ndef nextFloyd(xs):\n    '''A Floyd triangle row derived from\n       the preceding row.'''\n    n = succ(len(xs))\n    return [1] if n < 2 else (\n        enumFromTo(succ(n * pred(n) // 2))(\n            n * succ(n) // 2\n        )\n    )\n\n\n# showFloyd :: [[Int]] -> String\ndef showFloyd(xs):\n    '''A stringification of Floyd triangle rows.'''\n    return unlines(str(x) for x in xs)\n\n\n# main :: IO ()\ndef main():\n    '''Test'''\n    print(showFloyd(\n        floyd(5)\n    ))\n\n\n# GENERIC ABSTRACTIONS ------------------------------------\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated applications of f to x.'''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return lambda x: go(x)\n\n\n# pred ::  Enum a => a -> a\ndef pred(x):\n    '''The predecessor of a value. For numeric types, (- 1).'''\n    return x - 1 if isinstance(x, int) else (\n        chr(ord(x) - 1)\n    )\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value. For numeric types, (1 +).'''\n    return 1 + x if isinstance(x, int) else (\n        chr(1 + ord(x))\n    )\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.'''\n    return '\\n'.join(xs)\n\n\n# MAIN ----------------------------------------------------\nif __name__ == '__main__':\n    main()"}
{"title": "Four bit adder", "language": "Python", "task": "\"''Simulate''\" a four-bit adder.\n \nThis design can be realized using four 1-bit full adders. \nEach of these 1-bit full adders can be built with two gate. ; \n\nFinally a half adder can be made using an   ''xor''   gate and an   ''and''   gate. \n\nThe   ''xor''   gate can be made using two   ''not''s,   two   ''and''s   and one   ''or''.\n\n'''Not''',   '''or'''   and   '''and''',   the only allowed \"gates\" for the task, can be \"imitated\" by using the bitwise operators of your language. \n\nIf there is not a ''bit type'' in your language, to be sure that the   ''not''   does not \"invert\" all the other bits of the basic type   (e.g. a byte)   we are not interested in,   you can use an extra   ''nand''   (''and''   then   ''not'')   with the constant   '''1'''   on one input.\n\nInstead of optimizing and reducing the number of gates used for the final 4-bit adder,   build it in the most straightforward way,   ''connecting'' the other \"constructive blocks\",   in turn made of \"simpler\" and \"smaller\" ones.\n\n{|\n|+Schematics of the \"constructive blocks\"\n!(Xor gate with ANDs, ORs and NOTs)        \n!   (A half adder)                \n!          (A full adder)            \n!                (A 4-bit adder)                           \n|-\n|Xor gate done with ands, ors and nots\n|A half adder\n|A full adder\n|A 4-bit adder\n|}\n\n\n\nSolutions should try to be as descriptive as possible, making it as easy as possible to identify \"connections\" between higher-order \"blocks\". \n\nIt is not mandatory to replicate the syntax of higher-order blocks in the atomic \"gate\" blocks, i.e. basic \"gate\" operations can be performed as usual bitwise operations, or they can be \"wrapped\" in a ''block'' in order to expose the same syntax of higher-order blocks, at implementers' choice.\n\nTo test the implementation, show the sum of two four-bit numbers (in binary).\n\n\n", "solution": "def xor(a, b): return (a and not b) or (b and not a)\n\ndef ha(a, b): return xor(a, b), a and b     # sum, carry\n\ndef fa(a, b, ci):\n    s0, c0 = ha(ci, a)\n    s1, c1 = ha(s0, b)\n    return s1, c0 or c1     # sum, carry\n\ndef fa4(a, b):\n    width = 4\n    ci = [None] * width\n    co = [None] * width\n    s  = [None] * width\n    for i in range(width):\n        s[i], co[i] = fa(a[i], b[i], co[i-1] if i else 0)\n    return s, co[-1]\n\ndef int2bus(n, width=4):\n    return [int(c) for c in \"{0:0{1}b}\".format(n, width)[::-1]]\n\ndef bus2int(b):\n    return sum(1 << i for i, bit in enumerate(b) if bit)\n\ndef test_fa4():\n    width = 4\n    tot = [None] * (width + 1)\n    for a in range(2**width):\n        for b in range(2**width):\n            tot[:width], tot[width] = fa4(int2bus(a), int2bus(b))\n            assert a + b == bus2int(tot), \"totals don't match: %i + %i != %s\" % (a, b, tot)\n\n\nif __name__ == '__main__':\n   test_fa4()"}
{"title": "Four is magic", "language": "Python", "task": "Write a subroutine, function, whatever it may be called in your language, that takes an integer number and returns an English text sequence starting with the English cardinal representation of that integer, the word 'is' and then the English cardinal representation of the count of characters that made up the first word, followed by a comma.\n\nContinue the sequence by using the previous count word as the first word of the next phrase, append 'is' and the cardinal count of the letters in ''that'' word. \n\nContinue until you reach four. Since four has four characters, finish by adding the words 'four is magic' and a period. All integers will eventually wind up at four.\n\nFor instance, suppose your are given the integer '''3'''. Convert '''3''' to '''Three''', add ''' is ''', then the cardinal character count of three, or '''five''', with a comma to separate if from the next phrase. Continue the sequence '''five is four,''' (five has four letters), and finally, '''four is magic.''' \n\n      '''Three is five, five is four, four is magic.'''\n\nFor reference, here are outputs for 0 through 9.\n\n      Zero is four, four is magic.\n      One is three, three is five, five is four, four is magic.\n      Two is three, three is five, five is four, four is magic.\n      Three is five, five is four, four is magic.\n      Four is magic.\n      Five is four, four is magic.\n      Six is three, three is five, five is four, four is magic.\n      Seven is five, five is four, four is magic.\n      Eight is five, five is four, four is magic.\n      Nine is four, four is magic.\n\n\n;Some task guidelines:\n:* You may assume the input will only contain integer numbers.\n:* Cardinal numbers between 20 and 100 may use either hyphens or spaces as word separators but they must use a word separator. ('''23''' is '''twenty three''' or '''twenty-three''' not '''twentythree'''.)\n:* Cardinal number conversions should follow the  English short scale. (billion is 1e9, trillion is 1e12, etc.)\n:* Cardinal numbers should not include commas. ('''20140''' is '''twenty thousand one hundred forty''' not '''twenty thousand, one hundred forty'''.)\n:* When converted to a string, '''100''' should be '''one hundred''', not '''a hundred''' or '''hundred''', '''1000''' should be '''one thousand''', not '''a thousand''' or '''thousand'''.\n:* When converted to a string, there should be no '''and''' in the cardinal string. '''130''' should be '''one hundred thirty''' not '''one hundred and thirty'''.\n:* When counting characters, count ''all'' of the characters in the cardinal number including spaces and hyphens. '''One hundred fifty-one''' should be '''21''' not '''18'''.\n:* The output should follow the format \"N is K, K is M, M is ... four is magic.\" (unless the input is 4, in which case the output should simply be \"four is magic.\")\n:* The output can either be the return value from the function, or be displayed from within the function.\n:* You are encouraged, though not mandated to use proper sentence capitalization.\n:* You may optionally support negative numbers. '''-7''' is '''negative seven'''.\n:* Show the output here for a small representative sample of values, at least 5 but no more than 25. You are free to choose which which numbers to use for output demonstration.\n\n\nYou can choose to use a library, (module, external routine, whatever) to do the cardinal conversions as long as the code is easily and freely available to the public.  \n\nIf you roll your own, make the routine accept at minimum any integer from 0 up to 999999. If you use a pre-made library, support at least up to unsigned 64 bit integers. (or the largest integer supported in your language if it is less.)\n\nFour is magic is a popular code-golf task. '''This is not code golf.''' Write legible, idiomatic and well formatted code.\n\n\n;Related tasks:\n:*   [[Four is the number of_letters in the ...]]\n:*   [[Look-and-say sequence]]\n:*   [[Number names]]\n:*   [[Self-describing numbers]]\n:*   [[Summarize and say sequence]]\n:*   [[Spelling of ordinal numbers]]\n:*   [[De Bruijn sequences]]\n\n", "solution": "import random\nfrom collections import OrderedDict\n\nnumbers = {  # taken from https://en.wikipedia.org/wiki/Names_of_large_numbers#cite_ref-a_14-3\n    1: 'one',\n    2: 'two',\n    3: 'three',\n    4: 'four',\n    5: 'five',\n    6: 'six',\n    7: 'seven',\n    8: 'eight',\n    9: 'nine',\n    10: 'ten',\n    11: 'eleven',\n    12: 'twelve',\n    13: 'thirteen',\n    14: 'fourteen',\n    15: 'fifteen',\n    16: 'sixteen',\n    17: 'seventeen',\n    18: 'eighteen',\n    19: 'nineteen',\n    20: 'twenty',\n    30: 'thirty',\n    40: 'forty',\n    50: 'fifty',\n    60: 'sixty',\n    70: 'seventy',\n    80: 'eighty',\n    90: 'ninety',\n    100: 'hundred',\n    1000: 'thousand',\n    10 ** 6: 'million',\n    10 ** 9: 'billion',\n    10 ** 12: 'trillion',\n    10 ** 15: 'quadrillion',\n    10 ** 18: 'quintillion',\n    10 ** 21: 'sextillion',\n    10 ** 24: 'septillion',\n    10 ** 27: 'octillion',\n    10 ** 30: 'nonillion',\n    10 ** 33: 'decillion',\n    10 ** 36: 'undecillion',\n    10 ** 39: 'duodecillion',\n    10 ** 42: 'tredecillion',\n    10 ** 45: 'quattuordecillion',\n    10 ** 48: 'quinquadecillion',\n    10 ** 51: 'sedecillion',\n    10 ** 54: 'septendecillion',\n    10 ** 57: 'octodecillion',\n    10 ** 60: 'novendecillion',\n    10 ** 63: 'vigintillion',\n    10 ** 66: 'unvigintillion',\n    10 ** 69: 'duovigintillion',\n    10 ** 72: 'tresvigintillion',\n    10 ** 75: 'quattuorvigintillion',\n    10 ** 78: 'quinquavigintillion',\n    10 ** 81: 'sesvigintillion',\n    10 ** 84: 'septemvigintillion',\n    10 ** 87: 'octovigintillion',\n    10 ** 90: 'novemvigintillion',\n    10 ** 93: 'trigintillion',\n    10 ** 96: 'untrigintillion',\n    10 ** 99: 'duotrigintillion',\n    10 ** 102: 'trestrigintillion',\n    10 ** 105: 'quattuortrigintillion',\n    10 ** 108: 'quinquatrigintillion',\n    10 ** 111: 'sestrigintillion',\n    10 ** 114: 'septentrigintillion',\n    10 ** 117: 'octotrigintillion',\n    10 ** 120: 'noventrigintillion',\n    10 ** 123: 'quadragintillion',\n    10 ** 153: 'quinquagintillion',\n    10 ** 183: 'sexagintillion',\n    10 ** 213: 'septuagintillion',\n    10 ** 243: 'octogintillion',\n    10 ** 273: 'nonagintillion',\n    10 ** 303: 'centillion',\n    10 ** 306: 'uncentillion',\n    10 ** 309: 'duocentillion',\n    10 ** 312: 'trescentillion',\n    10 ** 333: 'decicentillion',\n    10 ** 336: 'undecicentillion',\n    10 ** 363: 'viginticentillion',\n    10 ** 366: 'unviginticentillion',\n    10 ** 393: 'trigintacentillion',\n    10 ** 423: 'quadragintacentillion',\n    10 ** 453: 'quinquagintacentillion',\n    10 ** 483: 'sexagintacentillion',\n    10 ** 513: 'septuagintacentillion',\n    10 ** 543: 'octogintacentillion',\n    10 ** 573: 'nonagintacentillion',\n    10 ** 603: 'ducentillion',\n    10 ** 903: 'trecentillion',\n    10 ** 1203: 'quadringentillion',\n    10 ** 1503: 'quingentillion',\n    10 ** 1803: 'sescentillion',\n    10 ** 2103: 'septingentillion',\n    10 ** 2403: 'octingentillion',\n    10 ** 2703: 'nongentillion',\n    10 ** 3003: 'millinillion'\n}\nnumbers = OrderedDict(sorted(numbers.items(), key=lambda t: t[0], reverse=True))\n\n\ndef string_representation(i: int) -> str:\n    \"\"\"\n    Return the english string representation of an integer\n    \"\"\"\n    if i == 0:\n        return 'zero'\n\n    words = ['negative'] if i < 0 else []\n    working_copy = abs(i)\n\n    for key, value in numbers.items():\n        if key <= working_copy:\n            times = int(working_copy / key)\n\n            if key >= 100:\n                words.append(string_representation(times))\n\n            words.append(value)\n            working_copy -= times * key\n\n        if working_copy == 0:\n            break\n\n    return ' '.join(words)\n\n\ndef next_phrase(i: int):\n    \"\"\"\n    Generate all the phrases\n    \"\"\"\n    while not i == 4:  # Generate phrases until four is reached\n        str_i = string_representation(i)\n        len_i = len(str_i)\n\n        yield str_i, 'is', string_representation(len_i)\n\n        i = len_i\n\n    # the last phrase\n    yield string_representation(i), 'is', 'magic'\n\n\ndef magic(i: int) -> str:\n    phrases = []\n\n    for phrase in next_phrase(i):\n        phrases.append(' '.join(phrase))\n\n    return f'{\", \".join(phrases)}.'.capitalize()\n\n\nif __name__ == '__main__':\n\n    for j in (random.randint(0, 10 ** 3) for i in range(5)):\n        print(j, ':\\n', magic(j), '\\n')\n\n    for j in (random.randint(-10 ** 24, 10 ** 24) for i in range(2)):\n        print(j, ':\\n', magic(j), '\\n')"}
{"title": "Four is the number of letters in the ...", "language": "Python", "task": "The      '''Four is ...'''      sequence is based on the counting of the number of\nletters in the words of the (never-ending) sentence:\n   Four is the number of letters in the first word of this sentence, two in the second,\n   three in the third, six in the fourth, two in the fifth, seven in the sixth, *** \n\n\n;Definitions and directives:\n:*   English is to be used in spelling numbers.\n:*   '''Letters'''   are defined as the upper- and lowercase letters in the Latin alphabet   ('''A-->Z'''   and   '''a-->z''').\n:*   Commas are not counted,   nor are hyphens (dashes or minus signs).\n:*   '''twenty-three'''   has eleven letters.\n:*   '''twenty-three'''   is considered one word   (which is hyphenated).\n:*   no   ''' ''and'' '''   words are to be used when spelling a (English) word for a number.\n:*   The American version of numbers will be used here in this task   (as opposed to the British version).\n          '''2,000,000,000'''   is two billion,   ''not''   two milliard.\n\n\n;Task:\n:*   Write a driver (invoking routine) and a function (subroutine/routine***) that returns the sequence (for any positive integer) of the number of letters in the first   '''N'''    words in the never-ending sentence.   For instance, the portion of the never-ending sentence shown above (2nd sentence of this task's preamble),   the sequence would be:\n          '''4  2  3  6  2  7'''\n:*   Only construct as much as is needed for the never-ending sentence.\n:*   Write a driver (invoking routine) to show the number of letters in the   Nth   word,   ''as well as''   showing the   Nth   word itself.\n:*   After each test case, show the total number of characters   (including blanks, commas, and punctuation)   of the sentence that was constructed.\n:*   Show all output here.\n\n\n;Test cases:\n  Display the first  201  numbers in the sequence   (and the total number of characters in the sentence).\n  Display the number of letters  (and the word itself)  of the       1,000th  word.\n  Display the number of letters  (and the word itself)  of the      10,000th  word.\n  Display the number of letters  (and the word itself)  of the     100,000th  word.\n  Display the number of letters  (and the word itself)  of the   1,000,000th  word.\n  Display the number of letters  (and the word itself)  of the  10,000,000th  word  (optional).\n\n\n;Related tasks:\n:*   [[Four is magic]]\n:*   [[Look-and-say sequence]]\n:*   [[Number names]]\n:*   [[Self-describing numbers]]\n:*   [[Self-referential sequence]]\n:*   [[Spelling of ordinal numbers]]\n\n\n;Also see:\n:*   See the OEIS sequence  A72425  \"Four is the number of letters...\".\n:*   See the OEIS sequence  A72424  \"Five's the number of letters...\"\n\n", "solution": "# Python implementation of Rosetta Code Task \n# http://rosettacode.org/wiki/Four_is_the_number_of_letters_in_the_...\n# Uses inflect\n# https://pypi.org/project/inflect/\n\nimport inflect\n\ndef count_letters(word):\n    \"\"\"\n    count letters ignore , or -, or space\n    \"\"\"\n    count = 0\n    for letter in word:\n        if letter != ',' and letter !='-' and letter !=' ':\n            count += 1\n            \n    return count\n    \ndef split_with_spaces(sentence):\n    \"\"\"\n    Takes string with partial sentence and returns\n    list of words with spaces included.\n    \n    Leading space is attached to first word.\n    Later spaces attached to prior word.\n    \"\"\"\n    sentence_list = []\n    curr_word = \"\"\n    for c in sentence:\n        if c == \" \" and curr_word != \"\":\n            # append space to end of non-empty words\n            # assumed no more than 1 consecutive space.\n            sentence_list.append(curr_word+\" \")\n            curr_word = \"\"\n        else:\n            curr_word += c\n    \n    # add trailing word that does not end with a space        \n    \n    if len(curr_word) > 0:\n        sentence_list.append(curr_word)\n    \n    return sentence_list\n    \ndef my_num_to_words(p, my_number):\n    \"\"\"\n    Front end to inflect's number_to_words\n    \n    Get's rid of ands and commas in large numbers.\n    \"\"\"\n    \n    number_string_list = p.number_to_words(my_number, wantlist=True, andword='')\n    \n    number_string = number_string_list[0]\n    \n    for i in range(1,len(number_string_list)):\n        number_string += \" \" + number_string_list[i]\n    \n    return number_string\n        \ndef build_sentence(p, max_words):\n    \"\"\"\n    \n    Builds at most max_words of the task following the pattern:\n    \n    Four is the number of letters in the first word of this sentence, two in the second,\n    three in the third, six in the fourth, two in the fifth, seven in the sixth,\n    \n    \"\"\"\n    \n    # start with first part of sentence up first comma as a list\n    \n    sentence_list = split_with_spaces(\"Four is the number of letters in the first word of this sentence,\")\n      \n    num_words = 13\n    \n    # which word number we are doing next\n    # two/second is first one in loop\n    \n    word_number = 2\n    \n    # loop until sentance is at least as long as needs be\n    \n    while num_words < max_words:\n        # Build something like\n        # ,two in the second\n        \n        # get second or whatever we are on\n        \n        ordinal_string = my_num_to_words(p, p.ordinal(word_number))\n        \n        # get two or whatever the length is of the word_number word\n        \n        word_number_string = my_num_to_words(p, count_letters(sentence_list[word_number - 1]))\n        \n        # sentence addition\n        \n        new_string = \" \"+word_number_string+\" in the \"+ordinal_string+\",\"\n\n        new_list = split_with_spaces(new_string)\n        \n        sentence_list += new_list\n\n        # add new word count\n        \n        num_words += len(new_list)\n        \n        # increment word number\n        \n        word_number += 1\n        \n    return sentence_list, num_words\n    \ndef word_and_counts(word_num):\n    \"\"\"\n    \n    Print's lines like this:\n    \n    Word     1000 is \"in\", with 2 letters.  Length of sentence so far: 6279\n\n    \"\"\"\n        \n    sentence_list, num_words = build_sentence(p, word_num)\n    \n    word_str = sentence_list[word_num - 1].strip(' ,')\n    \n    num_letters = len(word_str)\n    \n    num_characters = 0\n    \n    for word in sentence_list:\n       num_characters += len(word)\n       \n    print('Word {0:8d} is \"{1}\", with {2} letters.  Length of the sentence so far: {3}  '.format(word_num,word_str,num_letters,num_characters))\n   \n    \np = inflect.engine()\n\nsentence_list, num_words = build_sentence(p, 201)\n\nprint(\" \")\nprint(\"The lengths of the first 201 words are:\")\nprint(\" \")\n\nprint('{0:3d}:  '.format(1),end='')\n\ntotal_characters = 0\n\nfor word_index in range(201):\n\n    word_length = count_letters(sentence_list[word_index])\n    \n    total_characters += len(sentence_list[word_index])\n    \n    print('{0:2d}'.format(word_length),end='')\n    if (word_index+1) % 20 == 0:\n        # newline every 20\n        print(\" \")\n        print('{0:3d}:  '.format(word_index + 2),end='')\n    else:\n        print(\" \",end='')\n \nprint(\" \")\nprint(\" \")\nprint(\"Length of the sentence so far: \"+str(total_characters))\nprint(\" \")\n\n\"\"\"\n\nExpected output this part:\n\nWord     1000 is \"in\", with 2 letters.  Length of the sentence so far: 6279\nWord    10000 is \"in\", with 2 letters.  Length of the sentence so far: 64140\nWord   100000 is \"one\", with 3 letters.  Length of the sentence so far: 659474\nWord  1000000 is \"the\", with 3 letters.  Length of the sentence so far: 7113621\nWord 10000000 is \"thousand\", with 8 letters.  Length of the sentence so far: 70995756\n\n\"\"\"\n\nword_and_counts(1000)\nword_and_counts(10000)\nword_and_counts(100000)\nword_and_counts(1000000)\nword_and_counts(10000000)\n"}
{"title": "Functional coverage tree", "language": "Python", "task": "Functional coverage is a measure of how much a particular function of a system \nhas been verified as correct. It is used heavily in tracking the completeness \nof the verification of complex System on Chip (SoC) integrated circuits, where \nit can also be used to track how well the functional ''requirements'' of the \nsystem have been verified.\n\nThis task uses a sub-set of the calculations sometimes used in tracking \nfunctional coverage but uses a more familiar(?) scenario.\n\n;Task Description:\nThe head of the clean-up crews for \"The Men in a very dark shade of grey when \nviewed at night\" has been tasked with managing the cleansing of two properties\nafter an incident involving aliens.\n\nShe arranges the task hierarchically with a manager for the crews working on \neach house who return with a breakdown of how they will report on progress in \neach house.\n\nThe overall hierarchy of (sub)tasks is as follows, \ncleaning\n    house1\n        bedrooms\n        bathrooms\n            bathroom1\n            bathroom2\n            outside lavatory\n        attic\n        kitchen\n        living rooms\n            lounge\n            dining room\n            conservatory\n            playroom\n        basement\n        garage\n        garden\n    house2\n        upstairs\n            bedrooms\n                suite 1\n                suite 2\n                bedroom 3\n                bedroom 4\n            bathroom\n            toilet\n            attics\n        groundfloor\n            kitchen\n            living rooms\n                lounge\n                dining room\n                conservatory\n                playroom\n            wet room & toilet\n            garage\n            garden\n            hot tub suite\n        basement\n            cellars\n            wine cellar\n            cinema\n\nThe head of cleanup knows that her managers will report fractional completion of leaf tasks (tasks with no child tasks of their own), and she knows that she will want to modify the weight of values of completion as she sees fit.\n\nSome time into the cleaning, and some coverage reports have come in and she thinks see needs to weight the big house2 60-40 with respect to coverage from house1 She prefers a tabular view of her data where missing weights are assumed to be 1.0 and missing coverage 0.0.\nNAME_HIERARCHY                  |WEIGHT  |COVERAGE  |\ncleaning                        |        |          |\n    house1                      |40      |          |\n        bedrooms                |        |0.25      |\n        bathrooms               |        |          |\n            bathroom1           |        |0.5       |\n            bathroom2           |        |          |\n            outside_lavatory    |        |1         |\n        attic                   |        |0.75      |\n        kitchen                 |        |0.1       |\n        living_rooms            |        |          |\n            lounge              |        |          |\n            dining_room         |        |          |\n            conservatory        |        |          |\n            playroom            |        |1         |\n        basement                |        |          |\n        garage                  |        |          |\n        garden                  |        |0.8       |\n    house2                      |60      |          |\n        upstairs                |        |          |\n            bedrooms            |        |          |\n                suite_1         |        |          |\n                suite_2         |        |          |\n                bedroom_3       |        |          |\n                bedroom_4       |        |          |\n            bathroom            |        |          |\n            toilet              |        |          |\n            attics              |        |0.6       |\n        groundfloor             |        |          |\n            kitchen             |        |          |\n            living_rooms        |        |          |\n                lounge          |        |          |\n                dining_room     |        |          |\n                conservatory    |        |          |\n                playroom        |        |          |\n            wet_room_&_toilet   |        |          |\n            garage              |        |          |\n            garden              |        |0.9       |\n            hot_tub_suite       |        |1         |\n        basement                |        |          |\n            cellars             |        |1         |\n            wine_cellar         |        |1         |\n            cinema              |        |0.75      |\n\n;Calculation:\nThe coverage of a node in the tree is calculated as the weighted average of the coverage of its children evaluated bottom-upwards in the tree.\n\n'''The task is to''' calculate the overall coverage of the cleaning task and display the coverage at all levels of the hierarchy on this page, in a manner that visually shows the hierarchy, weights and coverage of all nodes.\n\n;Extra Credit:\nAfter calculating the coverage for all nodes, one can also calculate the additional/delta top level coverage that would occur if any (sub)task were to be fully covered from its current fractional coverage. This is done by multiplying the extra coverage that could be gained 1-coverage for any node, by the product of the `powers` of its parent nodes from the top down to the node.\nThe power of a direct child of any parent is given by the power of the parent multiplied by the weight of the child divided by the sum of the weights of all the direct children.\n\nThe pseudo code would be:\n\n    method delta_calculation(this, power):\n        sum_of_weights = sum(node.weight for node in children)\n        this.delta  = (1 - this.coverage) * power\n        for node in self.children:\n            node.delta_calculation(power * node.weight / sum_of_weights)\n        return this.delta\n\nFollowed by a call to:\n\n    top.delta_calculation(power=1)\n\n\n'''Note:''' to aid in getting the data into your program you might want to use an alternative, more functional description of the starting data given on the discussion page.\n\n", "solution": "from itertools import zip_longest\n\n\nfc2 = '''\\\ncleaning,,\n    house1,40,\n        bedrooms,,.25\n        bathrooms,,\n            bathroom1,,.5\n            bathroom2,,\n            outside_lavatory,,1\n        attic,,.75\n        kitchen,,.1\n        living_rooms,,\n            lounge,,\n            dining_room,,\n            conservatory,,\n            playroom,,1\n        basement,,\n        garage,,\n        garden,,.8\n    house2,60,\n        upstairs,,\n            bedrooms,,\n                suite_1,,\n                suite_2,,\n                bedroom_3,,\n                bedroom_4,,\n            bathroom,,\n            toilet,,\n            attics,,.6\n        groundfloor,,\n            kitchen,,\n            living_rooms,,\n                lounge,,\n                dining_room,,\n                conservatory,,\n                playroom,,\n            wet_room_&_toilet,,\n            garage,,\n            garden,,.9\n            hot_tub_suite,,1\n        basement,,\n            cellars,,1\n            wine_cellar,,1\n            cinema,,.75\n\n'''\n\nNAME, WT, COV = 0, 1, 2\n\ndef right_type(txt):\n    try:\n        return float(txt)\n    except ValueError:\n        return txt\n\ndef commas_to_list(the_list, lines, start_indent=0):\n    '''\n    Output format is a nest of lists and tuples\n    lists are for coverage leaves without children items in the list are name, weight, coverage\n    tuples are 2-tuples for nodes with children. The first element is a list representing the\n    name, weight, coverage of the node (some to be calculated); the second element is a list of\n    child elements which may be 2-tuples or lists as above.\n    \n    the_list is modified in-place\n    lines must be a generator of successive lines of input like fc2\n    '''\n    for n, line in lines:\n        indent = 0\n        while line.startswith(' ' * (4 * indent)):\n            indent += 1\n        indent -= 1\n        fields = [right_type(f) for f in line.strip().split(',')]\n        if indent == start_indent:\n            the_list.append(fields)\n        elif indent > start_indent:\n            lst = [fields]\n            sub = commas_to_list(lst, lines, indent)\n            the_list[-1] = (the_list[-1], lst)\n            if sub not in (None, ['']) :\n                the_list.append(sub)\n        else:\n            return fields if fields else None\n    return None\n\n\ndef pptreefields(lst, indent=0, widths=['%-32s', '%-8g', '%-10g']):\n    '''\n    Pretty prints the format described from function commas_to_list as a table with \n    names in the first column suitably indented and all columns having a fixed \n    minimum column width.\n    '''\n    lhs = ' ' * (4 * indent)\n    for item in lst:\n        if type(item) != tuple:\n            name, *rest = item\n            print(widths[0] % (lhs + name), end='|')\n            for width, item in zip_longest(widths[1:len(rest)], rest, fillvalue=widths[-1]):\n                if type(item) == str:\n                    width = width[:-1] + 's'\n                print(width % item, end='|')\n            print()\n        else:\n            item, children = item\n            name, *rest = item\n            print(widths[0] % (lhs + name), end='|')\n            for width, item in zip_longest(widths[1:len(rest)], rest, fillvalue=widths[-1]):\n                if type(item) == str:\n                    width = width[:-1] + 's'\n                print(width % item, end='|')\n            print()\n            pptreefields(children, indent+1)\n\n\ndef default_field(node_list):\n    node_list[WT] = node_list[WT] if node_list[WT] else 1.0\n    node_list[COV] = node_list[COV] if node_list[COV] else 0.0\n\ndef depth_first(tree, visitor=default_field):\n    for item in tree:\n        if type(item) == tuple:\n            item, children = item\n            depth_first(children, visitor)\n        visitor(item)\n            \n\ndef covercalc(tree):\n    '''\n    Depth first weighted average of coverage\n    '''\n    sum_covwt, sum_wt = 0, 0\n    for item in tree:\n        if type(item) == tuple:\n            item, children = item\n            item[COV] = covercalc(children)\n        sum_wt  += item[WT]\n        sum_covwt += item[COV] * item[WT]\n    cov = sum_covwt / sum_wt\n    return cov\n\nif __name__ == '__main__':        \n    lstc = []\n    commas_to_list(lstc, ((n, ln) for n, ln in enumerate(fc2.split('\\n'))))\n    #pp(lstc, width=1, indent=4, compact=1)\n    \n    #print('\\n\\nEXPANDED DEFAULTS\\n')\n    depth_first(lstc)\n    #pptreefields(['NAME_HIERARCHY WEIGHT COVERAGE'.split()] + lstc)\n    \n    print('\\n\\nTOP COVERAGE = %f\\n' % covercalc(lstc))\n    depth_first(lstc)\n    pptreefields(['NAME_HIERARCHY WEIGHT COVERAGE'.split()] + lstc)"}
{"title": "Functional coverage tree", "language": "Python 3.7", "task": "Functional coverage is a measure of how much a particular function of a system \nhas been verified as correct. It is used heavily in tracking the completeness \nof the verification of complex System on Chip (SoC) integrated circuits, where \nit can also be used to track how well the functional ''requirements'' of the \nsystem have been verified.\n\nThis task uses a sub-set of the calculations sometimes used in tracking \nfunctional coverage but uses a more familiar(?) scenario.\n\n;Task Description:\nThe head of the clean-up crews for \"The Men in a very dark shade of grey when \nviewed at night\" has been tasked with managing the cleansing of two properties\nafter an incident involving aliens.\n\nShe arranges the task hierarchically with a manager for the crews working on \neach house who return with a breakdown of how they will report on progress in \neach house.\n\nThe overall hierarchy of (sub)tasks is as follows, \ncleaning\n    house1\n        bedrooms\n        bathrooms\n            bathroom1\n            bathroom2\n            outside lavatory\n        attic\n        kitchen\n        living rooms\n            lounge\n            dining room\n            conservatory\n            playroom\n        basement\n        garage\n        garden\n    house2\n        upstairs\n            bedrooms\n                suite 1\n                suite 2\n                bedroom 3\n                bedroom 4\n            bathroom\n            toilet\n            attics\n        groundfloor\n            kitchen\n            living rooms\n                lounge\n                dining room\n                conservatory\n                playroom\n            wet room & toilet\n            garage\n            garden\n            hot tub suite\n        basement\n            cellars\n            wine cellar\n            cinema\n\nThe head of cleanup knows that her managers will report fractional completion of leaf tasks (tasks with no child tasks of their own), and she knows that she will want to modify the weight of values of completion as she sees fit.\n\nSome time into the cleaning, and some coverage reports have come in and she thinks see needs to weight the big house2 60-40 with respect to coverage from house1 She prefers a tabular view of her data where missing weights are assumed to be 1.0 and missing coverage 0.0.\nNAME_HIERARCHY                  |WEIGHT  |COVERAGE  |\ncleaning                        |        |          |\n    house1                      |40      |          |\n        bedrooms                |        |0.25      |\n        bathrooms               |        |          |\n            bathroom1           |        |0.5       |\n            bathroom2           |        |          |\n            outside_lavatory    |        |1         |\n        attic                   |        |0.75      |\n        kitchen                 |        |0.1       |\n        living_rooms            |        |          |\n            lounge              |        |          |\n            dining_room         |        |          |\n            conservatory        |        |          |\n            playroom            |        |1         |\n        basement                |        |          |\n        garage                  |        |          |\n        garden                  |        |0.8       |\n    house2                      |60      |          |\n        upstairs                |        |          |\n            bedrooms            |        |          |\n                suite_1         |        |          |\n                suite_2         |        |          |\n                bedroom_3       |        |          |\n                bedroom_4       |        |          |\n            bathroom            |        |          |\n            toilet              |        |          |\n            attics              |        |0.6       |\n        groundfloor             |        |          |\n            kitchen             |        |          |\n            living_rooms        |        |          |\n                lounge          |        |          |\n                dining_room     |        |          |\n                conservatory    |        |          |\n                playroom        |        |          |\n            wet_room_&_toilet   |        |          |\n            garage              |        |          |\n            garden              |        |0.9       |\n            hot_tub_suite       |        |1         |\n        basement                |        |          |\n            cellars             |        |1         |\n            wine_cellar         |        |1         |\n            cinema              |        |0.75      |\n\n;Calculation:\nThe coverage of a node in the tree is calculated as the weighted average of the coverage of its children evaluated bottom-upwards in the tree.\n\n'''The task is to''' calculate the overall coverage of the cleaning task and display the coverage at all levels of the hierarchy on this page, in a manner that visually shows the hierarchy, weights and coverage of all nodes.\n\n;Extra Credit:\nAfter calculating the coverage for all nodes, one can also calculate the additional/delta top level coverage that would occur if any (sub)task were to be fully covered from its current fractional coverage. This is done by multiplying the extra coverage that could be gained 1-coverage for any node, by the product of the `powers` of its parent nodes from the top down to the node.\nThe power of a direct child of any parent is given by the power of the parent multiplied by the weight of the child divided by the sum of the weights of all the direct children.\n\nThe pseudo code would be:\n\n    method delta_calculation(this, power):\n        sum_of_weights = sum(node.weight for node in children)\n        this.delta  = (1 - this.coverage) * power\n        for node in self.children:\n            node.delta_calculation(power * node.weight / sum_of_weights)\n        return this.delta\n\nFollowed by a call to:\n\n    top.delta_calculation(power=1)\n\n\n'''Note:''' to aid in getting the data into your program you might want to use an alternative, more functional description of the starting data given on the discussion page.\n\n", "solution": "'''Functional coverage tree'''\n\nfrom itertools import chain, product\nfrom functools import reduce\n\n\n# main :: IO ()\ndef main():\n    '''Tabular outline serialisation of a parse tree\n       decorated with computations of:\n       1. Weighted coverage of each tree node.\n       2. Each node's share of the total project's\n          remaining work.\n    '''\n    columnWidths = [31, 9, 9, 9]\n    delimiter = '|'\n\n    reportLines = REPORT.splitlines()\n    columnTitles = init(columnNames(delimiter)(reportLines[0]))\n\n    # ------ SERIALISATION OF DECORATED PARSE TREE -------\n    print(titleLine(delimiter)(columnWidths)(\n        columnTitles + ['share of residue']\n    ))\n    print(indentedLinesFromTree('    ', tabulation(columnWidths))(\n\n        # -------- TWO COMPUTATIONS BY TRAVERSAL ---------\n        withResidueShares(1.0)(\n            foldTree(weightedCoverage)(\n\n                # --- TREE FROM PARSE OF OUTLINE TEXT ----\n                fmapTree(\n                    recordFromKeysDefaultsDelimiterAndLine(\n                        columnTitles\n                    )(\n                        [str, float, float])([\n                            '?', 1.0, 0.0\n                        ])(delimiter)\n                )(\n                    forestFromIndentLevels(\n                        indentLevelsFromLines(\n                            reportLines[1:]\n                        )\n                    )[0]\n                )\n            )\n        )\n    ))\n\n\n# ---- WEIGHTED COVERAGE, AND SHARE OF TOTAL RESIDUE -----\n\n# weightedCoverage :: Tree Dict ->\n# [Tree Dict] -> Tree Dict\ndef weightedCoverage(x):\n    '''The weighted coverage of a tree node,\n       as a function of the weighted averages\n       of its children.\n    '''\n    def go(xs):\n        cws = [\n            (r['coverage'], r['weight']) for r\n            in [root(x) for x in xs]\n        ]\n        totalWeight = reduce(lambda a, x: a + x[1], cws, 0)\n        return Node(dict(\n            x, **{\n                'coverage': round(reduce(\n                    lambda a, cw: a + (cw[0] * cw[1]),\n                    cws, x['coverage']\n                ) / (totalWeight if 0 < totalWeight else 1), 5)\n            }\n        ))(xs)\n    return go\n\n\n# withResidueShares :: Float -> Tree Dict -> Tree Dict\ndef withResidueShares(shareOfTotal):\n    '''A Tree of dictionaries additionally decorated with each\n       node's proportion of the total project's outstanding work.\n    '''\n    def go(fraction, node):\n        [nodeRoot, nodeNest] = ap([root, nest])([node])\n        weights = [root(x)['weight'] for x in nodeNest]\n        siblingsTotal = sum(weights)\n        return Node(\n            insertDict('residual_share')(\n                round(fraction * (1 - nodeRoot['coverage']), 5)\n            )(nodeRoot)\n        )(\n            map(\n                go,\n                [fraction * (w / siblingsTotal) for w in weights],\n                nodeNest\n            )\n        )\n    return lambda tree: go(shareOfTotal, tree)\n\n\n# ------------------ OUTLINE TABULATION ------------------\n\n# tabulation :: [Int] -> String -> Dict -> String\ndef tabulation(columnWidths):\n    '''Indented string representation of a node\n       in a functional coverage tree.\n    '''\n    return lambda indent, dct: '| '.join(map(\n        lambda k, w: (\n            (indent if 10 < w else '') + str(dct.get(k, ''))\n        ).ljust(w, ' '),\n        dct.keys(),\n        columnWidths\n    ))\n\n\n# titleLine :: String -> [Int] -> [String] -> String\ndef titleLine(delimiter):\n    '''A string consisting of a spaced and delimited\n       series of upper-case column titles.\n    '''\n    return lambda columnWidths: lambda ks: (\n        delimiter + ' '\n    ).join(map(\n        lambda k, w: k.ljust(w, ' '),\n        [k.upper() for k in ks],\n        columnWidths\n    ))\n\n\n# ------------ GENERIC AND REUSABLE FUNCTIONS ------------\n\n# Node :: a -> [Tree a] -> Tree a\ndef Node(v):\n    '''Constructor for a Tree node which connects a\n       value of some kind to a list of zero or\n       more child trees.\n    '''\n    return lambda xs: {'type': 'Tree', 'root': v, 'nest': xs}\n\n\n# ap (<*>) :: [(a -> b)] -> [a] -> [b]\ndef ap(fs):\n    '''The application of each of a list of functions,\n       to each of a list of values.\n    '''\n    def go(xs):\n        return [\n            f(x) for (f, x)\n            in product(fs, xs)\n        ]\n    return go\n\n\n# columnNames :: String -> String -> [String]\ndef columnNames(delimiter):\n    '''A list of lower-case keys derived from\n       a header line and a delimiter character.\n    '''\n    return compose(\n        fmapList(compose(toLower, strip)),\n        splitOn(delimiter)\n    )\n\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    return lambda x: reduce(\n        lambda a, f: f(a),\n        fs[::-1], x\n    )\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).\n    '''\n    def go(xs):\n        return chain.from_iterable(map(f, xs))\n    return go\n\n\n# div :: Int -> Int -> Int\ndef div(x):\n    '''Integer division.'''\n    return lambda y: x // y\n\n\n# first :: (a -> b) -> ((a, c) -> (b, c))\ndef first(f):\n    '''A simple function lifted to a function over a tuple,\n       with f applied only the first of two values.\n    '''\n    return lambda xy: (f(xy[0]), xy[1])\n\n\n# flip :: (a -> b -> c) -> b -> a -> c\ndef flip(f):\n    '''The (curried or uncurried) function f with its\n       arguments reversed.\n    '''\n    return lambda a: lambda b: f(b)(a)\n\n\n# fmapList :: (a -> b) -> [a] -> [b]\ndef fmapList(f):\n    '''fmap over a list.\n       f lifted to a function over a list.\n    '''\n    return lambda xs: [f(x) for x in xs]\n\n\n# fmapTree :: (a -> b) -> Tree a -> Tree b\ndef fmapTree(f):\n    '''A new tree holding the results of\n       an application of f to each root in\n       the existing tree.\n    '''\n    def go(x):\n        return Node(\n            f(x['root'])\n        )([go(v) for v in x['nest']])\n    return go\n\n\n# foldTree :: (a -> [b] -> b) -> Tree a -> b\ndef foldTree(f):\n    '''The catamorphism on trees. A summary\n       value defined by a depth-first fold.\n    '''\n    def go(node):\n        return f(root(node))([\n            go(x) for x in nest(node)\n        ])\n    return go\n\n\n# forestFromIndentLevels :: [(Int, a)] -> [Tree a]\ndef forestFromIndentLevels(tuples):\n    '''A list of trees derived from a list of values paired\n       with integers giving their levels of indentation.\n    '''\n    def go(xs):\n        if xs:\n            intIndent, v = xs[0]\n            firstTreeLines, rest = span(\n                lambda x: intIndent < x[0]\n            )(xs[1:])\n            return [Node(v)(go(firstTreeLines))] + go(rest)\n        else:\n            return []\n    return go(tuples)\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# indentLevelsFromLines :: [String] -> [(Int, String)]\ndef indentLevelsFromLines(xs):\n    '''Each input line stripped of leading\n       white space, and tupled with a preceding integer\n       giving its level of indentation from 0 upwards.\n    '''\n    indentTextPairs = list(map(\n        compose(first(len), span(isSpace)),\n        xs\n    ))\n    indentUnit = min(concatMap(\n        lambda x: [x[0]] if x[0] else []\n    )(indentTextPairs))\n    return list(map(\n        first(flip(div)(indentUnit)),\n        indentTextPairs\n    ))\n\n\n# indentedLinesFromTree :: String -> (String -> a -> String) ->\n# [Tree a] -> String\ndef indentedLinesFromTree(strTab, f):\n    '''An indented line rendering of a tree, in which\n       the function f stringifies a root value.\n    '''\n    def go(indent):\n        return lambda node: [f(indent, node['root'])] + list(\n            concatMap(\n                go(strTab + indent)\n            )(node['nest'])\n        )\n    return lambda tree: '\\n'.join(go('')(tree))\n\n\n# init :: [a] -> [a]\ndef init(xs):\n    '''A list containing all the elements\n       of xs except the last.\n    '''\n    return xs[:-1]\n\n\n# insertDict :: String -> a -> Dict -> Dict\ndef insertDict(k):\n    '''A new dictionary updated with a (k, v) pair.'''\n    def go(v, dct):\n        return dict(dct, **{k: v})\n    return lambda v: lambda dct: go(v, dct)\n\n\n# isSpace :: Char -> Bool\n# isSpace :: String -> Bool\ndef isSpace(s):\n    '''True if s is not empty, and\n       contains only white space.\n    '''\n    return s.isspace()\n\n\n# lt (<) :: Ord a => a -> a -> Bool\ndef lt(x):\n    '''True if x < y.'''\n    return lambda y: (x < y)\n\n\n# nest :: Tree a -> [Tree a]\ndef nest(t):\n    '''Accessor function for children of tree node.'''\n    return t['nest'] if 'nest' in t else None\n\n\n# recordFromKeysDefaultsAndLine :: String ->\n# { name :: String, weight :: Float, completion :: Float }\ndef recordFromKeysDefaultsDelimiterAndLine(columnTitles):\n    '''A dictionary of key-value pairs, derived from a\n       delimited string, together with ordered lists of\n       key-names, types, default values, and a delimiter.\n    '''\n    return lambda ts: lambda vs: lambda delim: lambda s: dict(\n        map(\n            lambda k, t, v, x: (k, t(x) if x else v),\n            columnTitles, ts, vs,\n            map(strip, splitOn(delim)(s))\n        )\n    )\n\n\n# root :: Tree a -> a\ndef root(t):\n    '''Accessor function for data of tree node.'''\n    return t['root'] if 'root' in t else None\n\n\n# strip :: String -> String\ndef strip(s):\n    '''A copy of s without any leading or trailling\n       white space.\n    '''\n    return s.strip()\n\n\n# span :: (a -> Bool) -> [a] -> ([a], [a])\ndef span(p):\n    '''The longest (possibly empty) prefix of xs\n       that contains only elements satisfying p,\n       tupled with the remainder of xs.\n       span p xs is equivalent to (takeWhile p xs, dropWhile p xs).\n    '''\n    def match(ab):\n        b = ab[1]\n        return not b or not p(b[0])\n\n    def f(ab):\n        a, b = ab\n        return a + [b[0]], b[1:]\n\n    def go(xs):\n        return until(match)(f)(([], xs))\n    return go\n\n\n# splitOn :: String -> String -> [String]\ndef splitOn(pat):\n    '''A list of the strings delimited by\n       instances of a given pattern in s.\n    '''\n    return lambda xs: (\n        xs.split(pat) if isinstance(xs, str) else None\n    )\n\n\n# toLower :: String -> String\ndef toLower(s):\n    '''String in lower case.'''\n    return s.lower()\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f):\n        def g(x):\n            v = x\n            while not p(v):\n                v = f(v)\n            return v\n        return g\n    return go\n\n\n# MAIN ----------------------------------------------------\nif __name__ == '__main__':\n    REPORT = '''NAME_HIERARCHY                  |WEIGHT  |COVERAGE  |\n    cleaning                        |        |          |\n        house1                      |40      |          |\n            bedrooms                |        |0.25      |\n            bathrooms               |        |          |\n                bathroom1           |        |0.5       |\n                bathroom2           |        |          |\n                outside_lavatory    |        |1         |\n            attic                   |        |0.75      |\n            kitchen                 |        |0.1       |\n            living_rooms            |        |          |\n                lounge              |        |          |\n                dining_room         |        |          |\n                conservatory        |        |          |\n                playroom            |        |1         |\n            basement                |        |          |\n            garage                  |        |          |\n            garden                  |        |0.8       |\n        house2                      |60      |          |\n            upstairs                |        |          |\n                bedrooms            |        |          |\n                    suite_1         |        |          |\n                    suite_2         |        |          |\n                    bedroom_3       |        |          |\n                    bedroom_4       |        |          |\n                bathroom            |        |          |\n                toilet              |        |          |\n                attics              |        |0.6       |\n            groundfloor             |        |          |\n                kitchen             |        |          |\n                living_rooms        |        |          |\n                    lounge          |        |          |\n                    dining_room     |        |          |\n                    conservatory    |        |          |\n                    playroom        |        |          |\n                wet_room_&_toilet   |        |          |\n                garage              |        |          |\n                garden              |        |0.9       |\n                hot_tub_suite       |        |1         |\n            basement                |        |          |\n                cellars             |        |1         |\n                wine_cellar         |        |1         |\n                cinema              |        |0.75      |'''\n    main()"}
{"title": "Fusc sequence", "language": "Python", "task": "Definitions:\nThe   '''fusc'''   integer sequence is defined as:\n::*   fusc(0) = 0\n::*   fusc(1) = 1\n::*   for '''n'''>1,   the   '''n'''th   term is defined as:\n::::*   if   '''n'''   is       even;     fusc(n) = fusc(n/2)\n::::*   if   '''n'''   is   odd;     fusc(n) = fusc((n-1)/2)   +   fusc((n+1)/2)\n\n\nNote that MathWorld's definition starts with unity, not zero.   This task will be using the OEIS' version   (above).\n\n\n\n;An observation:\n:::::*   fusc(A) = fusc(B)\n\nwhere   '''A'''   is some non-negative integer expressed in binary,   and\nwhere   '''B'''   is the binary value of   '''A'''   reversed.\n\n\n\nFusc numbers are also known as:\n::*   fusc function   (named by Dijkstra, 1982)\n::*   Stern's Diatomic series   (although it starts with unity, not zero)\n::*   Stern-Brocot sequence   (although it starts with unity, not zero)\n\n\n\n;Task:\n::*   show the first   '''61'''   fusc numbers (starting at zero) in a horizontal format.\n::*   show the fusc number (and its index) whose length is greater than any previous fusc number length.\n::::*   (the length is the number of decimal digits when the fusc number is expressed in base ten.)\n::*   show all numbers with commas   (if appropriate).\n::*   show all output here.\n\n\n;Related task:\n::*   RosettaCode Stern-Brocot sequence\n\n\n\n;Also see:\n::*   the MathWorld entry:   Stern's Diatomic Series.\n::*   the OEIS      entry:   A2487.\n\n", "solution": "from collections import deque\nfrom itertools import islice, count\n\n\ndef fusc():\n    q = deque([1])\n    yield 0\n    yield 1\n\n    while True:\n        x = q.popleft()\n        q.append(x)\n        yield x\n\n        x += q[0]\n        q.append(x)\n        yield x\n\n\ndef longest_fusc():\n    sofar = 0\n    for i, f in zip(count(), fusc()):\n        if f >= sofar:\n            yield(i, f)\n            sofar = 10 * sofar or 10\n\n\nprint('First 61:')\nprint(list(islice(fusc(), 61)))\n\nprint('\\nLength records:')\nfor i, f in islice(longest_fusc(), 6):\n    print(f'fusc({i}) = {f}')\n"}
{"title": "Gapful numbers", "language": "Python", "task": "Numbers   (positive integers expressed in base ten)   that are (evenly) divisible by the number formed by the\nfirst and last digit are known as   '''gapful numbers'''.\n\n\n''Evenly divisible''   means divisible with   no   remainder.\n\n\nAll   one-   and two-digit   numbers have this property and are trivially excluded.   Only\nnumbers    >=  '''100'''   will be considered for this Rosetta Code task.\n\n\n;Example:\n'''187'''   is a   '''gapful'''   number because it is evenly divisible by the\nnumber   '''17'''   which is formed by the first and last decimal digits\nof    '''187'''. \n\n\nAbout   7.46%   of positive integers are   ''gapful''. \n\n\n;Task:\n:*   Generate and show all sets of numbers (below) on one line (horizontally) with a title,   here on this page\n:*   Show the first   '''30'''   gapful numbers\n:*   Show the first   '''15'''   gapful numbers    >=           '''1,000,000'''\n:*   Show the first   '''10'''   gapful numbers    >=                               '''1,000,000,000'''\n\n\n;Related tasks:\n:*   Harshad or Niven series.\n:*   palindromic gapful numbers.\n:*   largest number divisible by its digits.\n\n\n;Also see:\n:*   The OEIS entry:   A108343 gapful numbers.\n:*   numbersaplenty gapful numbers\n\n", "solution": "from itertools import islice, count\nfor start, n in [(100, 30), (1_000_000, 15), (1_000_000_000, 10)]:\n    print(f\"\\nFirst {n} gapful numbers from {start:_}\")\n    print(list(islice(( x for x in count(start) \n                        if (x % (int(str(x)[0]) * 10 + (x % 10)) == 0) )\n                      , n)))"}
{"title": "Gauss-Jordan matrix inversion", "language": "Python", "task": "Invert matrix   '''A'''   using Gauss-Jordan method.\n\n'''A'''   being an   '''n''' x '''n'''   matrix.\n\n", "solution": "import numpy as np\nfrom numpy.linalg import inv\na = np.array([[1., 2., 3.], [4., 1., 6.],[ 7., 8., 9.]])\nainv = inv(a)\n\nprint(a)\nprint(ainv)\n"}
{"title": "Gaussian elimination", "language": "Python", "task": "Solve   '''Ax=b'''   using Gaussian elimination then backwards substitution. \n\n'''A'''   being an   '''n''' by '''n'''   matrix. \n\nAlso,   '''x''' and '''b'''   are   '''n''' by '''1'''   vectors. \n\nTo improve accuracy, please use partial pivoting and scaling.\n\n\n;See also:\n:*   the Wikipedia entry:   Gaussian elimination\n\n", "solution": "# The 'gauss' function takes two matrices, 'a' and 'b', with 'a' square, and it return the determinant of 'a' and a matrix 'x' such that a*x = b.\n# If 'b' is the identity, then 'x' is the inverse of 'a'.\n\nimport copy\nfrom fractions import Fraction\n\ndef gauss(a, b):\n    a = copy.deepcopy(a)\n    b = copy.deepcopy(b)\n    n = len(a)\n    p = len(b[0])\n    det = 1\n    for i in range(n - 1):\n        k = i\n        for j in range(i + 1, n):\n            if abs(a[j][i]) > abs(a[k][i]):\n                k = j\n        if k != i:\n            a[i], a[k] = a[k], a[i]\n            b[i], b[k] = b[k], b[i]\n            det = -det\n            \n        for j in range(i + 1, n):\n            t = a[j][i]/a[i][i]\n            for k in range(i + 1, n):\n                a[j][k] -= t*a[i][k]\n            for k in range(p):\n                b[j][k] -= t*b[i][k]\n                \n    for i in range(n - 1, -1, -1):\n        for j in range(i + 1, n):\n            t = a[i][j]\n            for k in range(p):\n                b[i][k] -= t*b[j][k]\n        t = 1/a[i][i]\n        det *= a[i][i]\n        for j in range(p):\n            b[i][j] *= t\n    return det, b\n\ndef zeromat(p, q):\n    return [[0]*q for i in range(p)]\n\ndef matmul(a, b):\n    n, p = len(a), len(a[0])\n    p1, q = len(b), len(b[0])\n    if p != p1:\n        raise ValueError(\"Incompatible dimensions\")\n    c = zeromat(n, q)\n    for i in range(n):\n        for j in range(q):\n                c[i][j] = sum(a[i][k]*b[k][j] for k in range(p))\n    return c\n\n\ndef mapmat(f, a):\n    return [list(map(f, v)) for v in a]\n\ndef ratmat(a):\n    return mapmat(Fraction, a)\n\n# As an example, compute the determinant and inverse of 3x3 magic square\n\na = [[2, 9, 4], [7, 5, 3], [6, 1, 8]]\nb = [[1, 0, 0], [0, 1, 0], [0, 0, 1]]\ndet, c = gauss(a, b)\n\ndet\n-360.0\n\nc\n[[-0.10277777777777776, 0.18888888888888888, -0.019444444444444438],\n[0.10555555555555554, 0.02222222222222223, -0.061111111111111116],\n[0.0638888888888889, -0.14444444444444446, 0.14722222222222223]]\n\n# Check product\nmatmul(a, c)\n[[1.0, 0.0, 0.0], [5.551115123125783e-17, 1.0, 0.0],\n[1.1102230246251565e-16, -2.220446049250313e-16, 1.0]]\n\n# Same with fractions, so the result is exact\n\ndet, c = gauss(ratmat(a), ratmat(b))\n\ndet\nFraction(-360, 1)\n\nc\n[[Fraction(-37, 360), Fraction(17, 90), Fraction(-7, 360)],\n[Fraction(19, 180), Fraction(1, 45), Fraction(-11, 180)],\n[Fraction(23, 360), Fraction(-13, 90), Fraction(53, 360)]]\n\nmatmul(a, c)\n[[Fraction(1, 1), Fraction(0, 1), Fraction(0, 1)],\n[Fraction(0, 1), Fraction(1, 1), Fraction(0, 1)],\n[Fraction(0, 1), Fraction(0, 1), Fraction(1, 1)]]"}
{"title": "Generate Chess960 starting position", "language": "Python", "task": "Bobby Fischer.  Unlike other variants of the game, Chess960 does not require a different material, but instead relies on a random initial position, with a few constraints:\n\n* as in the standard chess game, all eight white pawns must be placed on the second rank.\n* White pieces must stand on the first rank as in the standard game, in random column order but with the two following constraints:\n** the bishops must be placed on opposite color squares (i.e. they must be an odd number of spaces apart or there must be an even number of spaces between them)\n** the King must be between two rooks (with any number of other pieces between them all)\n* Black pawns and pieces must be placed respectively on the seventh and eighth ranks, mirroring the white pawns and pieces, just as in the standard game.  (That is, their positions are not independently randomized.)\n\n\nWith those constraints there are '''960''' possible starting positions, thus the name of the variant.\n\n\n;Task:\nThe purpose of this task is to write a program that can randomly generate any one of the 960 Chess960 initial positions. You will show the result as the first rank displayed with Chess symbols in Unicode:  or with the letters '''K'''ing '''Q'''ueen '''R'''ook '''B'''ishop k'''N'''ight.\n\n", "solution": "from random import choice\n\ndef random960():\n    start = ['R', 'K', 'R']         # Subsequent order unchanged by insertions.\n    #\n    for piece in ['Q', 'N', 'N']:\n        start.insert(choice(range(len(start)+1)), piece)\n    #\n    bishpos = choice(range(len(start)+1))\n    start.insert(bishpos, 'B')\n    start.insert(choice(range(bishpos + 1, len(start) + 1, 2)), 'B')\n    return start\n    return ''.join(start).upper()\n\nprint(random960())"}
{"title": "Generate random chess position", "language": "Python", "task": "Generate a random chess position in FEN format.  \n\nThe position does not have to be realistic or even balanced,  but it must comply to the following rules:\n:* there is one and only one king of each color  (one black king and one white king);\n:* the kings must not be placed on adjacent squares;\n:* there can not be any pawn in the promotion square  (no white pawn in the eighth rank, and no black pawn in the first rank);\n:* including the kings, up to 32 pieces of either color can be placed. \n:* There is no requirement for material balance between sides. \n:* The picking of pieces does not have to comply to a regular chess set --- there can be five knights, twenty rooks, whatever ... as long as the total number of pieces do not exceed thirty-two.  \n:* it is white's turn.\n:* It's assumed that both sides have lost castling rights and that there is no possibility for   ''en passant''   (the FEN should thus end in w - - 0 1).\n\n\nNo requirement is made regarding the probability distribution of your method, but your program should be able to span a reasonably representative sample of all possible positions.  For instance, programs that would always generate positions with say five pieces on the board, or with kings on a corner, would not be considered truly random.\n\n", "solution": "import random\n\nboard = [[\" \" for x in range(8)] for y in range(8)]\npiece_list = [\"R\", \"N\", \"B\", \"Q\", \"P\"]\n\n\ndef place_kings(brd):\n\twhile True:\n\t\trank_white, file_white, rank_black, file_black = random.randint(0,7), random.randint(0,7), random.randint(0,7), random.randint(0,7)\n\t\tdiff_list = [abs(rank_white - rank_black),  abs(file_white - file_black)]\n\t\tif sum(diff_list) > 2 or set(diff_list) == set([0, 2]):\n\t\t\tbrd[rank_white][file_white], brd[rank_black][file_black] = \"K\", \"k\"\n\t\t\tbreak\n\ndef populate_board(brd, wp, bp):\n\tfor x in range(2):\n\t\tif x == 0:\n\t\t\tpiece_amount = wp\n\t\t\tpieces = piece_list\n\t\telse:\n\t\t\tpiece_amount = bp\n\t\t\tpieces = [s.lower() for s in piece_list]\n\t\twhile piece_amount != 0:\n\t\t\tpiece_rank, piece_file = random.randint(0, 7), random.randint(0, 7)\n\t\t\tpiece = random.choice(pieces)\n\t\t\tif brd[piece_rank][piece_file] == \" \" and pawn_on_promotion_square(piece, piece_rank) == False:\n\t\t\t\tbrd[piece_rank][piece_file] = piece\n\t\t\t\tpiece_amount -= 1\n\ndef fen_from_board(brd):\n\tfen = \"\"\n\tfor x in brd:\n\t\tn = 0\n\t\tfor y in x:\n\t\t\tif y == \" \":\n\t\t\t\tn += 1\n\t\t\telse:\n\t\t\t\tif n != 0:\n\t\t\t\t\tfen += str(n)\n\t\t\t\tfen += y\n\t\t\t\tn = 0\n\t\tif n != 0:\n\t\t\tfen += str(n)\n\t\tfen += \"/\" if fen.count(\"/\") < 7 else \"\"\n\tfen += \" w - - 0 1\\n\"\n\treturn fen\n\ndef pawn_on_promotion_square(pc, pr):\n\tif pc == \"P\" and pr == 0:\n\t\treturn True\n\telif pc == \"p\" and pr == 7:\n\t\treturn True\n\treturn False\n\n\ndef start():\n\tpiece_amount_white, piece_amount_black = random.randint(0, 15), random.randint(0, 15)\n\tplace_kings(board)\n\tpopulate_board(board, piece_amount_white, piece_amount_black)\n\tprint(fen_from_board(board))\n\tfor x in board:\n\t\tprint(x)\n\n#entry point\nstart()\n"}
{"title": "Generator/Exponential", "language": "Python", "task": "A generator is an executable entity (like a function or procedure) that contains code that yields a sequence of values, one at a time, so that each time you call the generator, the next value in the sequence is provided.\n \nGenerators are often built on top of coroutines or objects so that the internal state of the object is handled \"naturally\". \n\nGenerators are often used in situations where a sequence is potentially infinite, and where it is possible to construct the next value of the sequence with only minimal state.\n\n\n;Task:\n* Create a function that returns a generation of the m'th powers of the positive integers starting from zero, in order, and without obvious or simple upper limit. (Any upper limit to the generator should not be stated in the source but should be down to factors such as the languages natural integer size limit or computational time/size).\n* Use it to create a generator of:\n:::*   Squares.\n:::*   Cubes. \n* Create a new generator that filters all cubes from the generator of squares.\n* Drop the first 20 values from this last generator of filtered results, and then show the next 10 values.\n\n\nNote that this task ''requires'' the use of generators in the calculation of the result.\n\n\n;Also see:\n* Generator\n\n", "solution": "In Python, any function that contains a yield statement becomes a generator. The standard libraries itertools module provides the following functions used in the solution: [http://docs.python.org/library/itertools.html#itertools.count count], that will count up from zero; and [http://docs.python.org/library/itertools.html#itertools.islice islice], which will take a slice from an iterator/generator.\n\n"}
{"title": "Generator/Exponential", "language": "Python 2.6+ and 3.x", "task": "A generator is an executable entity (like a function or procedure) that contains code that yields a sequence of values, one at a time, so that each time you call the generator, the next value in the sequence is provided.\n \nGenerators are often built on top of coroutines or objects so that the internal state of the object is handled \"naturally\". \n\nGenerators are often used in situations where a sequence is potentially infinite, and where it is possible to construct the next value of the sequence with only minimal state.\n\n\n;Task:\n* Create a function that returns a generation of the m'th powers of the positive integers starting from zero, in order, and without obvious or simple upper limit. (Any upper limit to the generator should not be stated in the source but should be down to factors such as the languages natural integer size limit or computational time/size).\n* Use it to create a generator of:\n:::*   Squares.\n:::*   Cubes. \n* Create a new generator that filters all cubes from the generator of squares.\n* Drop the first 20 values from this last generator of filtered results, and then show the next 10 values.\n\n\nNote that this task ''requires'' the use of generators in the calculation of the result.\n\n\n;Also see:\n* Generator\n\n", "solution": "from itertools import islice, count\n\ndef powers(m):\n    for n in count():\n        yield n ** m\n    \ndef filtered(s1, s2):\n    v, f = next(s1), next(s2)\n    while True:\n        if v > f:\n            f = next(s2)\n            continue\n        elif v < f:\n            yield v\n        v = next(s1)\n\nsquares, cubes = powers(2), powers(3)\nf = filtered(squares, cubes)\nprint(list(islice(f, 20, 30)))"}
{"title": "Generator/Exponential", "language": "Python 3.7", "task": "A generator is an executable entity (like a function or procedure) that contains code that yields a sequence of values, one at a time, so that each time you call the generator, the next value in the sequence is provided.\n \nGenerators are often built on top of coroutines or objects so that the internal state of the object is handled \"naturally\". \n\nGenerators are often used in situations where a sequence is potentially infinite, and where it is possible to construct the next value of the sequence with only minimal state.\n\n\n;Task:\n* Create a function that returns a generation of the m'th powers of the positive integers starting from zero, in order, and without obvious or simple upper limit. (Any upper limit to the generator should not be stated in the source but should be down to factors such as the languages natural integer size limit or computational time/size).\n* Use it to create a generator of:\n:::*   Squares.\n:::*   Cubes. \n* Create a new generator that filters all cubes from the generator of squares.\n* Drop the first 20 values from this last generator of filtered results, and then show the next 10 values.\n\n\nNote that this task ''requires'' the use of generators in the calculation of the result.\n\n\n;Also see:\n* Generator\n\n", "solution": "'''Exponentials as generators'''\n\nfrom itertools import count, islice\n\n\n# powers :: Gen [Int]\ndef powers(n):\n    '''A non-finite succession of integers,\n       starting at zero,\n       raised to the nth power.'''\n\n    def f(x):\n        return pow(x, n)\n\n    return map(f, count(0))\n\n\n# main :: IO ()\ndef main():\n    '''Taking the difference between two derived generators.'''\n    print(\n        take(10)(\n            drop(20)(\n                differenceGen(powers(2))(\n                    powers(3)\n                )\n            )\n        )\n    )\n\n\n# GENERIC -------------------------------------------------\n\n\n# differenceGen :: Gen [a] -> Gen [a] -> Gen [a]\ndef differenceGen(ga):\n    '''All values of ga except any\n       already seen in gb.'''\n    def go(a, b):\n        stream = zip(a, b)\n        bs = set([])\n        while True:\n            xy = next(stream, None)\n            if None is not xy:\n                x, y = xy\n                bs.add(y)\n                if x not in bs:\n                    yield x\n            else:\n                return\n    return lambda gb: go(ga, gb)\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.'''\n    def go(xs):\n        if isinstance(xs, list):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return lambda xs: go(xs)\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Get system command output", "language": "Python", "task": "Task\nExecute a system command and get its output into the program. The output may be stored in any kind of collection (array, list, etc.). \n\n;Related task\n*  Execute a system command\n\n", "solution": ">>> import subprocess\n>>> returned_text = subprocess.check_output(\"dir\", shell=True, universal_newlines=True)\n>>> type(returned_text)\n<class 'str'>\n>>> print(returned_text)\n Volume in drive C is Windows\n Volume Serial Number is 44X7-73CE\n\n Directory of C:\\Python33\n\n04/07/2013  06:40    <DIR>          .\n04/07/2013  06:40    <DIR>          ..\n27/05/2013  07:10    <DIR>          DLLs\n27/05/2013  07:10    <DIR>          Doc\n27/05/2013  07:10    <DIR>          include\n27/05/2013  07:10    <DIR>          Lib\n27/05/2013  07:10    <DIR>          libs\n16/05/2013  00:15            33,326 LICENSE.txt\n15/05/2013  22:49           214,554 NEWS.txt\n16/05/2013  00:03            26,624 python.exe\n16/05/2013  00:03            27,136 pythonw.exe\n15/05/2013  22:49             6,701 README.txt\n27/05/2013  07:10    <DIR>          tcl\n27/05/2013  07:10    <DIR>          Tools\n16/05/2013  00:02            43,008 w9xpopen.exe\n               6 File(s)        351,349 bytes\n               9 Dir(s)  46,326,947,840 bytes free\n\n>>> # Ref: https://docs.python.org/3/library/subprocess.html"}
{"title": "Giuga numbers", "language": "Python from FreeBASIC", "task": "Definition\nA '''Giuga number''' is a composite number '''n''' which is such that each of its distinct prime factors\n'''f''' divide (n/f - 1) exactly.\n\nAll known Giuga numbers are even though it is not known for certain that there are no odd examples. \n\n;Example\n30 is a Giuga number because its distinct prime factors are 2, 3 and 5 and:\n* 30/2 - 1 = 14 is divisible by 2\n* 30/3 - 1 = 9 is divisible by 3\n* 30/5 - 1 = 5 is divisible by 5\n\n;Task\nDetermine and show here the first four Giuga numbers.\n\n;Stretch\nDetermine the fifth Giuga number and any more you have the patience for.\n\n;References\n\n* Wikipedia: Giuga number\n* OEIS:A007850 - Giuga numbers\n\n", "solution": "#!/usr/bin/python\n\nfrom math import sqrt\n\ndef isGiuga(m):\n    n = m\n    f = 2\n    l = sqrt(n)\n    while True:\n        if n % f == 0:\n            if ((m / f) - 1) % f != 0:\n                return False\n            n /= f\n            if f > n:\n                return True\n        else:\n            f += 1\n            if f > l:\n                return False\n\n\nif __name__ == '__main__':\n    n = 3\n    c = 0\n    print(\"The first 4 Giuga numbers are: \")\n    while c < 4:\n        if isGiuga(n):\n            c += 1\n            print(n)\n        n += 1"}
{"title": "Globally replace text in several files", "language": "Python", "task": "Replace every occurring instance of a piece of text in a group of text files with another one. \n\n\nFor this task we want to replace the text   \"'''Goodbye London!'''\"   with   \"'''Hello New York!'''\"   for a list of files.\n\n", "solution": "import fileinput\n\nfor line in fileinput.input(inplace=True):\n    print(line.replace('Goodbye London!', 'Hello New York!'), end='')\n"}
{"title": "Graph colouring", "language": "Python", "task": "A Graph is a collection of nodes\n(or vertices), connected by edges (or not).\nNodes directly connected by edges are called neighbours.\n\nIn our representation of graphs, nodes are numbered and edges are represented\nby the two node numbers connected by the edge separated by a dash.\nEdges define the nodes being connected. Only unconnected nodes ''need'' a separate\ndescription.\n\nFor example,\n0-1 1-2 2-0 3\nDescribes the following graph. Note that node 3 has no neighbours\n\n\n;Example graph:\n\n+---+\n| 3 |\n+---+\n\n  +-------------------+\n  |                   |\n+---+     +---+     +---+\n| 0 | --- | 1 | --- | 2 |\n+---+     +---+     +---+\n\n\nA useful internal datastructure for a graph and for later graph algorithms is\nas a mapping between each node and the set/list of its neighbours.\n\nIn the above example:\n0 maps-to 1 and 2\n1 maps to 2 and 0\n2 maps-to 1 and 0\n3 maps-to \n\n;Graph colouring task:\nColour the vertices of a given graph so that no edge is between verticies of\nthe same colour.\n\n* Integers may be used to denote different colours.\n* Algorithm should do better than just assigning each vertex a separate colour. The idea is to minimise the number of colours used, although no algorithm short of exhaustive search for the minimum is known at present, (and exhaustive search is '''not''' a requirement).\n* Show for each edge, the colours assigned on each vertex.\n* Show the total number of nodes, edges, and colours used for each graph.\n\n;Use the following graphs:\n;Ex1:\n\n        0-1 1-2 2-0 3\n\n\n+---+\n| 3 |\n+---+\n\n  +-------------------+\n  |                   |\n+---+     +---+     +---+\n| 0 | --- | 1 | --- | 2 |\n+---+     +---+     +---+\n\n;Ex2:\nThe wp articles left-side graph\n\n    1-6 1-7 1-8 2-5 2-7 2-8 3-5 3-6 3-8 4-5 4-6 4-7\n\n\n\n  +----------------------------------+\n  |                                  |\n  |                      +---+       |\n  |    +-----------------| 3 | ------+----+\n  |    |                 +---+       |    |\n  |    |                   |         |    |\n  |    |                   |         |    |\n  |    |                   |         |    |\n  |  +---+     +---+     +---+     +---+  |\n  |  | 8 | --- | 1 | --- | 6 | --- | 4 |  |\n  |  +---+     +---+     +---+     +---+  |\n  |    |         |                   |    |\n  |    |         |                   |    |\n  |    |         |                   |    |\n  |    |       +---+     +---+     +---+  |\n  +----+------ | 7 | --- | 2 | --- | 5 | -+\n       |       +---+     +---+     +---+\n       |                   |\n       +-------------------+\n\n;Ex3:\nThe wp articles right-side graph which is the same graph as Ex2, but with\ndifferent node orderings and namings.\n\n    1-4 1-6 1-8 3-2 3-6 3-8 5-2 5-4 5-8 7-2 7-4 7-6\n\n\n\n  +----------------------------------+\n  |                                  |\n  |                      +---+       |\n  |    +-----------------| 5 | ------+----+\n  |    |                 +---+       |    |\n  |    |                   |         |    |\n  |    |                   |         |    |\n  |    |                   |         |    |\n  |  +---+     +---+     +---+     +---+  |\n  |  | 8 | --- | 1 | --- | 4 | --- | 7 |  |\n  |  +---+     +---+     +---+     +---+  |\n  |    |         |                   |    |\n  |    |         |                   |    |\n  |    |         |                   |    |\n  |    |       +---+     +---+     +---+  |\n  +----+------ | 6 | --- | 3 | --- | 2 | -+\n       |       +---+     +---+     +---+\n       |                   |\n       +-------------------+\n\n;Ex4:\nThis is the same graph, node naming, and edge order as Ex2 except some of the edges x-y are flipped to y-x.\nThis might alter the node order used in the greedy algorithm leading to differing numbers of colours.\n\n    1-6 7-1 8-1 5-2 2-7 2-8 3-5 6-3 3-8 4-5 4-6 4-7\n\n\n\n                      +-------------------------------------------------+\n                      |                                                 |\n                      |                                                 |\n  +-------------------+---------+                                       |\n  |                   |         |                                       |\n+---+     +---+     +---+     +---+     +---+     +---+     +---+     +---+\n| 4 | --- | 5 | --- | 2 | --- | 7 | --- | 1 | --- | 6 | --- | 3 | --- | 8 |\n+---+     +---+     +---+     +---+     +---+     +---+     +---+     +---+\n  |         |                             |         |         |         |\n  +---------+-----------------------------+---------+         |         |\n            |                             |                   |         |\n            |                             |                   |         |\n            +-----------------------------+-------------------+         |\n                                          |                             |\n                                          |                             |\n                                          +-----------------------------+\n\n\n;References:\n* Greedy coloring Wikipedia.\n* Graph Coloring : Greedy Algorithm & Welsh Powell Algorithm by  Priyank Jain.\n\n\n", "solution": "import re\nfrom collections import defaultdict\nfrom itertools import count\n\n\nconnection_re = r\"\"\"\n    (?: (?P<N1>\\d+) - (?P<N2>\\d+) | (?P<N>\\d+) (?!\\s*-))\n    \"\"\"\n\nclass Graph:\n\n    def __init__(self, name, connections):\n        self.name = name\n        self.connections = connections\n        g = self.graph = defaultdict(list)  # maps vertex to direct connections\n\n        matches = re.finditer(connection_re, connections,\n                              re.MULTILINE | re.VERBOSE)\n        for match in matches:\n            n1, n2, n = match.groups()\n            if n:\n                g[n] += []\n            else:\n                g[n1].append(n2)    # Each the neighbour of the other\n                g[n2].append(n1)\n\n    def greedy_colour(self, order=None):\n        \"Greedy colourisation algo.\"\n        if order is None:\n            order = self.graph      # Choose something\n        colour = self.colour = {}\n        neighbours = self.graph\n        for node in order:\n            used_neighbour_colours = (colour[nbr] for nbr in neighbours[node]\n                                      if nbr in colour)\n            colour[node] = first_avail_int(used_neighbour_colours)\n        self.pp_colours()\n        return colour\n\n    def pp_colours(self):\n        print(f\"\\n{self.name}\")\n        c = self.colour\n        e = canonical_edges = set()\n        for n1, neighbours in sorted(self.graph.items()):\n            if neighbours:\n                for n2 in neighbours:\n                    edge = tuple(sorted([n1, n2]))\n                    if edge not in canonical_edges:\n                        print(f\"       {n1}-{n2}: Colour: {c[n1]}, {c[n2]}\")\n                        canonical_edges.add(edge)\n            else:\n                print(f\"         {n1}: Colour: {c[n1]}\")\n        lc = len(set(c.values()))\n        print(f\"    #Nodes: {len(c)}\\n    #Edges: {len(e)}\\n  #Colours: {lc}\")\n\n\ndef first_avail_int(data):\n    \"return lowest int 0... not in data\"\n    d = set(data)\n    for i in count():\n        if i not in d:\n            return i\n\n\nif __name__ == '__main__':\n    for name, connections in [\n            ('Ex1', \"0-1 1-2 2-0 3\"),\n            ('Ex2', \"1-6 1-7 1-8 2-5 2-7 2-8 3-5 3-6 3-8 4-5 4-6 4-7\"),\n            ('Ex3', \"1-4 1-6 1-8 3-2 3-6 3-8 5-2 5-4 5-8 7-2 7-4 7-6\"),\n            ('Ex4', \"1-6 7-1 8-1 5-2 2-7 2-8 3-5 6-3 3-8 4-5 4-6 4-7\"),\n            ]:\n        g = Graph(name, connections)\n        g.greedy_colour()"}
{"title": "Gray code", "language": "Python", "task": "Karnaugh maps in order from left to right or top to bottom.\nCreate functions to encode a number to and decode a number from Gray code. \n\nDisplay the normal binary representations, Gray code representations, and decoded Gray code values for all 5-bit binary numbers (0-31 inclusive, leading 0's not necessary).\n\nThere are many possible Gray codes. The following encodes what is called \"binary reflected Gray code.\"\n\nEncoding (MSB is bit 0, b is binary, g is Gray code):\n\nif b[i-1] = 1\n   g[i] = not b[i]\nelse\n   g[i] = b[i]\n\nOr:\n\ng = b xor (b logically right shifted 1 time)\n\nDecoding (MSB is bit 0, b is binary, g is Gray code):\n\nb[0] = g[0]\n\nfor other bits:\nb[i] = g[i] xor b[i-1]\n\n;Reference\n* Converting Between Gray and Binary Codes. It includes step-by-step animations.\n\n", "solution": "def gray_encode(n):\n    return n ^ n >> 1\n\ndef gray_decode(n):\n    m = n >> 1\n    while m:\n        n ^= m\n        m >>= 1\n    return n\n\n\nif __name__ == '__main__':\n    print(\"DEC,   BIN =>  GRAY => DEC\")\n    for i in range(32):\n        gray = gray_encode(i)\n        dec = gray_decode(gray)\n        print(f\" {i:>2d}, {i:>05b} => {gray:>05b} => {dec:>2d}\")"}
{"title": "Greatest subsequential sum", "language": "Python", "task": "Given a sequence of integers, find a continuous subsequence which maximizes the sum of its elements, that is, the elements of no other single subsequence add up to a value larger than this one. \n\n\nAn empty subsequence is considered to have the sum of   '''0''';   thus if all elements are negative, the result must be the empty sequence.\n\n", "solution": "f = maxsumit\nassert f([]) == []\nassert f([-1]) == []\nassert f([0])  == []\nassert f([1])       == [1]\nassert f([1, 0])    == [1]\nassert f([0, 1])    == [0, 1]   \nassert f([0, 1, 0]) == [0, 1]   \nassert f([2])         == [2]\nassert f([2, -1])     == [2]\nassert f([-1, 2])     == [2]\nassert f([-1, 2, -1]) == [2]\nassert f([2, -1, 3])         == [2, -1, 3]\nassert f([2, -1, 3, -1])     == [2, -1, 3] \nassert f([-1, 2, -1, 3])     == [2, -1, 3]\nassert f([-1, 2, -1, 3, -1]) == [2, -1, 3]\nassert f([-1, 1, 2, -5, -6]) == [1,2]"}
{"title": "Greatest subsequential sum", "language": "Python 3.7", "task": "Given a sequence of integers, find a continuous subsequence which maximizes the sum of its elements, that is, the elements of no other single subsequence add up to a value larger than this one. \n\n\nAn empty subsequence is considered to have the sum of   '''0''';   thus if all elements are negative, the result must be the empty sequence.\n\n", "solution": "'''Greatest subsequential sum'''\n\nfrom functools import (reduce)\n\n\n# maxSubseq :: [Int] -> [Int] -> (Int, [Int])\ndef maxSubseq(xs):\n    '''Subsequence of xs with the maximum sum'''\n    def go(ab, x):\n        (m1, m2) = ab[0]\n        hi = max((0, []), (m1 + x, m2 + [x]))\n        return (hi, max(ab[1], hi))\n    return reduce(go, xs, ((0, []), (0, [])))[1]\n\n\n# TEST -----------------------------------------------------------\nprint(\n    maxSubseq(\n        [-1, -2, 3, 5, 6, -2, -1, 4, -4, 2, -1]\n    )\n)"}
{"title": "Greedy algorithm for Egyptian fractions", "language": "Python", "task": "An   Egyptian fraction   is the sum of distinct unit fractions such as: \n\n::::   \\tfrac{1}{2} + \\tfrac{1}{3} + \\tfrac{1}{16} \\,(= \\tfrac{43}{48}) \n\nEach fraction in the expression has a numerator equal to   '''1'''   (unity)   and a denominator that is a positive integer,   and all the denominators are distinct   (i.e., no repetitions).  \n\nFibonacci's   Greedy algorithm for Egyptian fractions   expands the fraction     \\tfrac{x}{y}     to be represented by repeatedly performing the replacement\n\n::::   \\frac{x}{y} = \\frac{1}{\\lceil y/x\\rceil} + \\frac{(-y)\\!\\!\\!\\!\\mod x}{y\\lceil y/x\\rceil}  \n\n\n(simplifying the 2nd term in this replacement as necessary, and where     \\lceil x \\rceil     is the   ''ceiling''   function).\n\n\n\nFor this task,   Proper and improper fractions   must be able to be expressed.\n\n\nProper  fractions   are of the form    \\tfrac{a}{b}    where    a    and    b    are positive integers, such that    a < b,     and \n\nimproper fractions are of the form    \\tfrac{a}{b}    where    a    and    b    are positive integers, such that    ''a'' >= ''b''. \n\n\n(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.)\n\nFor improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [''n''].\n\n\n;Task requirements:\n*   show the Egyptian fractions for:  \\tfrac{43}{48}  and  \\tfrac{5}{121}  and  \\tfrac{2014}{59} \n*   for all proper fractions,    \\tfrac{a}{b}    where    a    and    b    are positive one-or two-digit (decimal) integers, find and show an Egyptian fraction that has:\n::*   the largest number of terms,\n::*   the largest denominator.\n*   for all one-, two-, and three-digit integers,   find and show (as above).     {extra credit}\n\n\n;Also see:\n*   Wolfram MathWorld(tm) entry: Egyptian fraction\n\n", "solution": "from fractions import Fraction\nfrom math import ceil\n\nclass Fr(Fraction):\n    def __repr__(self):\n        return '%s/%s' % (self.numerator, self.denominator)\n\ndef ef(fr):\n    ans = []\n    if fr >= 1:\n        if fr.denominator == 1:\n            return [[int(fr)], Fr(0, 1)]\n        intfr = int(fr)\n        ans, fr = [[intfr]], fr - intfr\n    x, y = fr.numerator, fr.denominator\n    while x != 1:\n        ans.append(Fr(1, ceil(1/fr)))\n        fr = Fr(-y % x, y* ceil(1/fr))\n        x, y = fr.numerator, fr.denominator\n    ans.append(fr)\n    return ans\n\nif __name__ == '__main__':\n    for fr in [Fr(43, 48), Fr(5, 121), Fr(2014, 59)]:\n        print('%r \u2500\u25ba %s' % (fr, ' '.join(str(x) for x in ef(fr))))\n    lenmax = denommax = (0, None) \n    for fr in set(Fr(a, b) for a in range(1,100) for b in range(1, 100)):\n        e = ef(fr)\n        #assert sum((f[0] if type(f) is list else f) for f in e) == fr, 'Whoops!'\n        elen, edenom = len(e), e[-1].denominator\n        if elen > lenmax[0]:\n            lenmax = (elen, fr, e)\n        if edenom > denommax[0]:\n            denommax = (edenom, fr, e)\n    print('Term max is %r with %i terms' % (lenmax[1], lenmax[0]))\n    dstr = str(denommax[0])\n    print('Denominator max is %r with %i digits %s...%s' %\n          (denommax[1], len(dstr), dstr[:5], dstr[-5:]))"}
{"title": "Greedy algorithm for Egyptian fractions", "language": "Python 3.7", "task": "An   Egyptian fraction   is the sum of distinct unit fractions such as: \n\n::::   \\tfrac{1}{2} + \\tfrac{1}{3} + \\tfrac{1}{16} \\,(= \\tfrac{43}{48}) \n\nEach fraction in the expression has a numerator equal to   '''1'''   (unity)   and a denominator that is a positive integer,   and all the denominators are distinct   (i.e., no repetitions).  \n\nFibonacci's   Greedy algorithm for Egyptian fractions   expands the fraction     \\tfrac{x}{y}     to be represented by repeatedly performing the replacement\n\n::::   \\frac{x}{y} = \\frac{1}{\\lceil y/x\\rceil} + \\frac{(-y)\\!\\!\\!\\!\\mod x}{y\\lceil y/x\\rceil}  \n\n\n(simplifying the 2nd term in this replacement as necessary, and where     \\lceil x \\rceil     is the   ''ceiling''   function).\n\n\n\nFor this task,   Proper and improper fractions   must be able to be expressed.\n\n\nProper  fractions   are of the form    \\tfrac{a}{b}    where    a    and    b    are positive integers, such that    a < b,     and \n\nimproper fractions are of the form    \\tfrac{a}{b}    where    a    and    b    are positive integers, such that    ''a'' >= ''b''. \n\n\n(See the REXX programming example to view one method of expressing the whole number part of an improper fraction.)\n\nFor improper fractions, the integer part of any improper fraction should be first isolated and shown preceding the Egyptian unit fractions, and be surrounded by square brackets [''n''].\n\n\n;Task requirements:\n*   show the Egyptian fractions for:  \\tfrac{43}{48}  and  \\tfrac{5}{121}  and  \\tfrac{2014}{59} \n*   for all proper fractions,    \\tfrac{a}{b}    where    a    and    b    are positive one-or two-digit (decimal) integers, find and show an Egyptian fraction that has:\n::*   the largest number of terms,\n::*   the largest denominator.\n*   for all one-, two-, and three-digit integers,   find and show (as above).     {extra credit}\n\n\n;Also see:\n*   Wolfram MathWorld(tm) entry: Egyptian fraction\n\n", "solution": "'''Egyptian fractions'''\n\nfrom fractions import Fraction\nfrom functools import reduce\nfrom operator import neg\n\n\n# eqyptianFraction :: Ratio Int -> Ratio Int\ndef eqyptianFraction(nd):\n    '''The rational number nd as a sum\n       of the series of unit fractions\n       obtained by application of the\n       greedy algorithm.'''\n    def go(x):\n        n, d = x.numerator, x.denominator\n        r = 1 + d // n if n else None\n        return Just((0, x) if 1 == n else (\n            (fr(n % d, d), fr(n // d, 1)) if n > d else (\n                fr(-d % n, d * r), fr(1, r)\n            )\n        )) if n else Nothing()\n    fr = Fraction\n    f = unfoldr(go)\n    return list(map(neg, f(-nd))) if 0 > nd else f(nd)\n\n\n# TESTS ---------------------------------------------------\n\n# maxEqyptianFraction :: Int -> (Ratio Int -> a)\n#                               -> (Ratio Int, a)\ndef maxEqyptianFraction(nDigits):\n    '''An Egyptian Fraction, representing a\n       proper fraction with numerators and\n       denominators of up to n digits each,\n       which returns a maximal value for the\n       supplied function f.'''\n\n    # maxVals :: ([Ratio Int], a) -> (Ratio Int, a)\n    #                               -> ([Ratio Int], a)\n    def maxima(xsv, ndfx):\n        xs, v = xsv\n        nd, fx = ndfx\n        return ([nd], fx) if fx > v else (\n            xs + [nd], v\n        ) if fx == v and nd not in xs else xsv\n\n    # go :: (Ratio Int -> a) -> ([Ratio Int], a)\n    def go(f):\n        iLast = int(nDigits * '9')\n        fs, mx = reduce(\n            maxima, [\n                (nd, f(eqyptianFraction(nd))) for nd in [\n                    Fraction(n, d)\n                    for n in enumFromTo(1)(iLast)\n                    for d in enumFromTo(1 + n)(iLast)\n                ]\n            ],\n            ([], 0)\n        )\n        return f.__name__ + ' -> [' + ', '.join(\n            map(str, fs)\n        ) + '] -> ' + str(mx)\n    return lambda f: go(f)\n\n\n# main :: IO ()\ndef main():\n    '''Tests'''\n\n    ef = eqyptianFraction\n    fr = Fraction\n\n    print('Three values as Eqyptian fractions:')\n    print('\\n'.join([\n        str(fr(*nd)) + ' -> ' + ' + '.join(map(str, ef(fr(*nd))))\n        for nd in [(43, 48), (5, 121), (2014, 59)]\n    ]))\n\n    # maxDenominator :: [Ratio Int] -> Int\n    def maxDenominator(ef):\n        return max(map(lambda nd: nd.denominator, ef))\n\n    # maxTermCount :: [Ratio Int] -> Int\n    def maxTermCount(ef):\n        return len(ef)\n\n    for i in [1, 2, 3]:\n        print(\n            '\\nMaxima for proper fractions with up to ' + (\n                str(i) + ' digit(s):'\n            )\n        )\n        for f in [maxTermCount, maxDenominator]:\n            print(maxEqyptianFraction(i)(f))\n\n\n# GENERIC -------------------------------------------------\n\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.'''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# unfoldr :: (b -> Maybe (b, a)) -> b -> [a]\ndef unfoldr(f):\n    '''Dual to reduce or foldr.\n       Where catamorphism reduces a list to a summary value,\n       the anamorphic unfoldr builds a list from a seed value.\n       As long as f returns Just(a, b), a is prepended to the list,\n       and the residual b is used as the argument for the next\n       application of f.\n       When f returns Nothing, the completed list is returned.'''\n    def go(xr):\n        mb = f(xr[0])\n        if mb.get('Nothing'):\n            return []\n        else:\n            y, r = mb.get('Just')\n            return [r] + go((y, r))\n\n    return lambda x: go((x, x))\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Hailstone sequence", "language": "Python", "task": "The Hailstone sequence of numbers can be generated from a starting positive integer,   n   by:\n*   If   n   is     '''1'''     then the sequence ends.\n*   If   n   is   '''even''' then the next   n   of the sequence    = n/2 \n*   If   n   is   '''odd'''    then the next   n   of the sequence    = (3 * n) + 1 \n\n\nThe (unproven) Collatz conjecture is that the hailstone sequence for any starting number always terminates.\n\n\nThis sequence was named by Lothar Collatz in 1937   (or possibly in 1939),   and is also known as (the):\n:::*   hailstone sequence,   hailstone numbers\n:::*   3x + 2 mapping,   3n + 1 problem\n:::*   Collatz sequence\n:::*   Hasse's algorithm\n:::*   Kakutani's problem\n:::*   Syracuse algorithm,   Syracuse problem\n:::*   Thwaites conjecture \n:::*   Ulam's problem\n\n\nThe hailstone sequence is also known as   ''hailstone numbers''   (because the values are usually subject to multiple descents and ascents like hailstones in a cloud).\n\n\n;Task:\n# Create a routine to generate the hailstone sequence for a number.\n# Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1\n# Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.    (But don't show the actual sequence!)\n\n\n;See also:\n*   xkcd (humourous).\n*   The Notorious Collatz conjecture Terence Tao, UCLA (Presentation, pdf).\n*   The Simplest Math Problem No One Can Solve Veritasium (video, sponsored).\n\n", "solution": "def hailstone(n):\n    seq = [n]\n    while n > 1:\n        n = 3 * n + 1 if n & 1 else n // 2\n        seq.append(n)\n    return seq\n\n\nif __name__ == '__main__':\n    h = hailstone(27)\n    assert (len(h) == 112\n            and h[:4] == [27, 82, 41, 124]\n            and h[-4:] == [8, 4, 2, 1])\n    max_length, n = max((len(hailstone(i)), i) for i in range(1, 100_000))\n    print(f\"Maximum length {max_length} was found for hailstone({n}) \"\n          f\"for numbers <100,000\")"}
{"title": "Hailstone sequence", "language": "Python 3.7", "task": "The Hailstone sequence of numbers can be generated from a starting positive integer,   n   by:\n*   If   n   is     '''1'''     then the sequence ends.\n*   If   n   is   '''even''' then the next   n   of the sequence    = n/2 \n*   If   n   is   '''odd'''    then the next   n   of the sequence    = (3 * n) + 1 \n\n\nThe (unproven) Collatz conjecture is that the hailstone sequence for any starting number always terminates.\n\n\nThis sequence was named by Lothar Collatz in 1937   (or possibly in 1939),   and is also known as (the):\n:::*   hailstone sequence,   hailstone numbers\n:::*   3x + 2 mapping,   3n + 1 problem\n:::*   Collatz sequence\n:::*   Hasse's algorithm\n:::*   Kakutani's problem\n:::*   Syracuse algorithm,   Syracuse problem\n:::*   Thwaites conjecture \n:::*   Ulam's problem\n\n\nThe hailstone sequence is also known as   ''hailstone numbers''   (because the values are usually subject to multiple descents and ascents like hailstones in a cloud).\n\n\n;Task:\n# Create a routine to generate the hailstone sequence for a number.\n# Use the routine to show that the hailstone sequence for the number 27 has 112 elements starting with 27, 82, 41, 124 and ending with 8, 4, 2, 1\n# Show the number less than 100,000 which has the longest hailstone sequence together with that sequence's length.    (But don't show the actual sequence!)\n\n\n;See also:\n*   xkcd (humourous).\n*   The Notorious Collatz conjecture Terence Tao, UCLA (Presentation, pdf).\n*   The Simplest Math Problem No One Can Solve Veritasium (video, sponsored).\n\n", "solution": "'''Hailstone sequences'''\n\nfrom itertools import (islice, takewhile)\n\n\n# hailstone :: Int -> [Int]\ndef hailstone(x):\n    '''Hailstone sequence starting with x.'''\n    def p(n):\n        return 1 != n\n    return list(takewhile(p, iterate(collatz)(x))) + [1]\n\n\n# collatz :: Int -> Int\ndef collatz(n):\n    '''Next integer in the hailstone sequence.'''\n    return 3 * n + 1 if 1 & n else n // 2\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Tests.'''\n\n    n = 27\n    xs = hailstone(n)\n    print(unlines([\n        f'The hailstone sequence for {n} has {len(xs)} elements,',\n        f'starting with {take(4)(xs)},',\n        f'and ending with {drop(len(xs) - 4)(xs)}.\\n'\n    ]))\n\n    (a, b) = (1, 99999)\n    (i, x) = max(\n        enumerate(\n            map(compose(len)(hailstone), enumFromTo(a)(b))\n        ),\n        key=snd\n    )\n    print(unlines([\n        f'The number in the range {a}..{b} '\n        f'which produces the longest sequence is {1 + i},',\n        f'generating a hailstone sequence of {x} integers.'\n    ]))\n\n\n# ----------------------- GENERIC ------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.\n    '''\n    def go(xs):\n        if isinstance(xs, (list, tuple, str)):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return go\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: range(m, 1 + n)\n\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated\n       applications of f to x.\n    '''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return go\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second component of a tuple.'''\n    return tpl[1]\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    def go(xs):\n        return (\n            xs[0:n]\n            if isinstance(xs, (list, tuple))\n            else list(islice(xs, n))\n        )\n    return go\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single newline-delimited string derived\n       from a list of strings.'''\n    return '\\n'.join(xs)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Harmonic series", "language": "Python", "task": "{{Wikipedia|Harmonic number}}\n\n\nIn mathematics, the '''n-th''' harmonic number is the sum of the reciprocals of the first '''n''' natural numbers: \n\n\n'''H''n'' = 1 + 1/2 + 1/3 + ... + 1/n'''\n\nThe series of harmonic numbers thus obtained is often loosely referred to as the harmonic series. \n\nHarmonic numbers are closely related to the Euler-Mascheroni constant.\n\nThe harmonic series is divergent, albeit quite slowly, and grows toward infinity.\n\n\n;Task\n* Write a function (routine, procedure, whatever it may be called in your language) to generate harmonic numbers.\n* Use that procedure to show the values of the first 20 harmonic numbers.\n* Find and show the position in the series of the first value greater than the integers 1 through 5\n\n\n;Stretch\n* Find and show the position in the series of the first value greater than the integers 6 through 10\n\n\n;Related\n* [[Egyptian fractions]]\n\n\n", "solution": "from  fractions import Fraction\n\ndef harmonic_series():\n    n, h = Fraction(1), Fraction(1)\n    while True:\n        yield h\n        h += 1 / (n + 1)\n        n += 1\n\nif __name__ == '__main__':\n    from itertools import islice\n    for n, d in (h.as_integer_ratio() for h in islice(harmonic_series(), 20)):\n        print(n, '/', d)"}
{"title": "Harshad or Niven series", "language": "Python", "task": "The Harshad or Niven numbers are positive integers >= 1 that are divisible by the sum of their digits. \n\nFor example,   '''42'''   is a Harshad number as   '''42'''   is divisible by   ('''4''' + '''2''')   without remainder.\n\nAssume that the series is defined as the numbers in increasing order.\n\n\n;Task:\nThe task is to create a function/method/procedure to generate successive members of the Harshad sequence. \n\nUse it to:\n::*   list the first '''20''' members of the sequence,   and\n::*   list the first Harshad number greater than '''1000'''.\n\n\nShow your output here.\n\n\n;Related task\n:*   Increasing gaps between consecutive Niven numbers\n\n\n;See also\n*    OEIS: A005349\n\n", "solution": ">>> import itertools\n>>> def harshad():\n\tfor n in itertools.count(1):\n\t\tif n % sum(int(ch) for ch in str(n)) == 0:\n\t\t\tyield n\n\n\t\t\n>>> list(itertools.islice(harshad(), 0, 20))\n[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42]\n>>> for n in harshad():\n\tif n > 1000:\n\t\tprint(n)\n\t\tbreak\n\n\t\n1002\n>>> "}
{"title": "Harshad or Niven series", "language": "Python 3.7", "task": "The Harshad or Niven numbers are positive integers >= 1 that are divisible by the sum of their digits. \n\nFor example,   '''42'''   is a Harshad number as   '''42'''   is divisible by   ('''4''' + '''2''')   without remainder.\n\nAssume that the series is defined as the numbers in increasing order.\n\n\n;Task:\nThe task is to create a function/method/procedure to generate successive members of the Harshad sequence. \n\nUse it to:\n::*   list the first '''20''' members of the sequence,   and\n::*   list the first Harshad number greater than '''1000'''.\n\n\nShow your output here.\n\n\n;Related task\n:*   Increasing gaps between consecutive Niven numbers\n\n\n;See also\n*    OEIS: A005349\n\n", "solution": "'''Harshad or Niven series'''\n\nfrom itertools import count, dropwhile, islice\n\n\n# harshads :: () -> [Int]\ndef harshads():\n    '''Harshad series'''\n    return (\n        x for x in count(1)\n        if 0 == x % digitSum(x)\n    )\n\n\n# digitSum :: Int -> Int\ndef digitSum(n):\n    '''Sum of the decimal digits of n.'''\n    def go(x):\n        return None if 0 == x else divmod(x, 10)\n    return sum(unfoldl(go)(n))\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''First 20, and first above 1000.'''\n\n    def firstTwenty(xs):\n        return take(20)(xs)\n\n    def firstAbove1000(xs):\n        return take(1)(\n            dropwhile(lambda x: 1000 >= x, xs)\n        )\n\n    print(\n        fTable(__doc__ + ':\\n')(\n            lambda x: x.__name__\n        )(showList)(lambda f: f(harshads()))([\n            firstTwenty,\n            firstAbove1000\n        ])\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, (list, tuple))\n        else list(islice(xs, n))\n    )\n\n\n# unfoldl :: (b -> Maybe (b, a)) -> b -> [a]\ndef unfoldl(f):\n    '''A lazy (generator) list unfolded from a seed value\n       by repeated application of f until no residue remains.\n       Dual to fold/reduce.\n       f returns either None or just (residue, value).\n       For a strict output list, wrap the result with list()\n    '''\n    def go(v):\n        residueValue = f(v)\n        while residueValue:\n            yield residueValue[1]\n            residueValue = f(residueValue[0])\n    return go\n\n\n# ----------------------- DISPLAY ------------------------\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function ->\n       fx display function -> f -> xs -> tabular string.\n    '''\n    def gox(xShow):\n        def gofx(fxShow):\n            def gof(f):\n                def goxs(xs):\n                    ys = [xShow(x) for x in xs]\n                    w = max(map(len, ys))\n\n                    def arrowed(x, y):\n                        return y.rjust(w, ' ') + ' -> ' + (\n                            fxShow(f(x))\n                        )\n                    return s + '\\n' + '\\n'.join(\n                        map(arrowed, xs, ys)\n                    )\n                return goxs\n            return gof\n        return gofx\n    return gox\n\n\n# showList :: [a] -> String\ndef showList(xs):\n    '''Stringification of a list.'''\n    return '[' + ','.join(repr(x) for x in xs) + ']'\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Hash join", "language": "Python", "task": "{| class=\"wikitable\"\n|-\n! Input\n! Output\n|-\n|\n\n{| style=\"border:none; border-collapse:collapse;\"\n|-\n| style=\"border:none\" | ''A'' = \n| style=\"border:none\" |\n\n{| class=\"wikitable\"\n|-\n! Age !! Name\n|-\n| 27 || Jonah\n|-\n| 18 || Alan\n|-\n| 28 || Glory\n|-\n| 18 || Popeye\n|-\n| 28 || Alan\n|}\n\n| style=\"border:none; padding-left:1.5em;\" rowspan=\"2\" |\n| style=\"border:none\" | ''B'' = \n| style=\"border:none\" |\n\n{| class=\"wikitable\"\n|-\n! Character !! Nemesis\n|-\n| Jonah || Whales\n|-\n| Jonah || Spiders\n|-\n| Alan || Ghosts\n|-\n| Alan || Zombies\n|-\n| Glory || Buffy\n|}\n\n|-\n| style=\"border:none\" | ''jA'' =\n| style=\"border:none\" | Name (i.e. column 1)\n\n| style=\"border:none\" | ''jB'' =\n| style=\"border:none\" | Character (i.e. column 0)\n|}\n\n|\n\n{| class=\"wikitable\" style=\"margin-left:1em\"\n|-\n! A.Age !! A.Name !! B.Character !! B.Nemesis\n|-\n| 27 || Jonah || Jonah || Whales\n|-\n| 27 || Jonah || Jonah || Spiders\n|-\n| 18 || Alan || Alan || Ghosts\n|-\n| 18 || Alan || Alan || Zombies\n|-\n| 28 || Glory || Glory || Buffy\n|-\n| 28 || Alan || Alan || Ghosts\n|-\n| 28 || Alan || Alan || Zombies\n|}\n\n|}\n\nThe order of the rows in the output table is not significant.\nIf you're using numerically indexed arrays to represent table rows (rather than referring to columns by name), you could represent the output rows in the form [[27, \"Jonah\"], [\"Jonah\", \"Whales\"]].\n\n\n", "solution": "from collections import defaultdict\n\ndef hashJoin(table1, index1, table2, index2):\n    h = defaultdict(list)\n    # hash phase\n    for s in table1:\n        h[s[index1]].append(s)\n    # join phase\n    return [(s, r) for r in table2 for s in h[r[index2]]]\n\ntable1 = [(27, \"Jonah\"),\n          (18, \"Alan\"),\n          (28, \"Glory\"),\n          (18, \"Popeye\"),\n          (28, \"Alan\")]\ntable2 = [(\"Jonah\", \"Whales\"),\n          (\"Jonah\", \"Spiders\"),\n          (\"Alan\", \"Ghosts\"),\n          (\"Alan\", \"Zombies\"),\n          (\"Glory\", \"Buffy\")]\n\nfor row in hashJoin(table1, 1, table2, 0):\n    print(row)"}
{"title": "Haversine formula", "language": "Python", "task": "{{Wikipedia}}\n\n\nThe '''haversine formula''' is an equation important in navigation, giving great-circle distances between two points on a sphere from their longitudes and latitudes. \n\nIt is a special case of a more general formula  in spherical trigonometry, the '''law of haversines''', relating the sides and angles of spherical \"triangles\".\n\n\n;Task:\nImplement a great-circle distance function, or use a library function, \nto show the great-circle distance between:\n* Nashville International Airport (BNA)   in Nashville, TN, USA,   which is: \n    '''N''' 36deg7.2',   '''W''' 86deg40.2'     (36.12,   -86.67)           -and-\n* Los Angeles International Airport (LAX)  in Los Angeles, CA, USA,   which is:\n    '''N''' 33deg56.4',  '''W''' 118deg24.0'    (33.94,  -118.40) \n\n\nUser Kaimbridge clarified on the Talk page:\n\n -- 6371.0 km is the authalic radius based on/extracted from surface area;\n -- 6372.8 km is an approximation of the radius of the average circumference\n    (i.e., the average great-elliptic or great-circle radius), where the\n     boundaries are the meridian (6367.45 km) and the equator (6378.14 km).\n\nUsing either of these values results, of course, in differing distances:\n\n 6371.0 km -> 2886.44444283798329974715782394574671655 km;\n 6372.8 km -> 2887.25995060711033944886005029688505340 km;\n (results extended for accuracy check:  Given that the radii are only\n  approximations anyways, .01'  1.0621333 km and .001\"  .00177 km,\n  practical precision required is certainly no greater than about\n  .0000001----i.e., .1 mm!)\n\nAs distances are segments of great circles/circumferences, it is\nrecommended that the latter value (r = 6372.8 km) be used (which\nmost of the given solutions have already adopted, anyways). \n\n\nMost of the examples below adopted Kaimbridge's recommended value of\n6372.8 km for the earth radius.  However, the derivation of this\nellipsoidal quadratic mean radius\nis wrong (the averaging over azimuth is biased).  When applying these\nexamples in real applications, it is better to use the\nmean earth radius,\n6371 km.  This value is recommended by the International Union of\nGeodesy and Geophysics and it minimizes the RMS relative error between the\ngreat circle and geodesic distance.\n\n", "solution": "from math import radians, sin, cos, sqrt, asin\n\n\ndef haversine(lat1, lon1, lat2, lon2):\n    R = 6372.8  # Earth radius in kilometers\n\n    dLat = radians(lat2 - lat1)\n    dLon = radians(lon2 - lon1)\n    lat1 = radians(lat1)\n    lat2 = radians(lat2)\n\n    a = sin(dLat / 2)**2 + cos(lat1) * cos(lat2) * sin(dLon / 2)**2\n    c = 2 * asin(sqrt(a))\n\n    return R * c\n\n>>> haversine(36.12, -86.67, 33.94, -118.40)\n2887.2599506071106\n>>> "}
{"title": "Hello world/Line printer", "language": "Python", "task": "Cause a line printer attached to the computer to print a line containing the message:    Hello World! \n\n\n;Note:\nA line printer is not the same as standard output. \n\nA   line printer   was an older-style printer which prints one line at a time to a continuous ream of paper. \n\nWith some systems, a line printer can be any device attached to an appropriate port (such as a parallel port).\n\n", "solution": "lp = open(\"/dev/lp0\",\"w\")\nlp.write(\"Hello World!\\n\")\nlp.close()"}
{"title": "Here document", "language": "Python", "task": "A   ''here document''    (or \"heredoc\")   is a way of specifying a text block, preserving the line breaks, indentation and other whitespace within the text. \n\nDepending on the language being used, a   ''here document''   is constructed using a command followed by \"<<\" (or some other symbol) followed by a token string. \n\nThe text block will then start on the next line, and will be followed by the chosen token at the beginning of the following line, which is used to mark the end of the text block.\n\n\n;Task:\nDemonstrate the use of   ''here documents''   within the language.\n\n;Related task:\n*   [[Documentation]]\n\n", "solution": "print(\"\"\"\\\nUsage: thingy [OPTIONS]\n     -h                        Display this usage message\n     -H hostname               Hostname to connect to\n\"\"\")"}
{"title": "Heronian triangles", "language": "Python", "task": "Hero's formula for the area of a triangle given the length of its three sides    ''a'',   ''b'',   and   ''c''   is given by:\n\n:::: A = \\sqrt{s(s-a)(s-b)(s-c)},\n\nwhere   ''s''   is half the perimeter of the triangle; that is,\n\n:::: s=\\frac{a+b+c}{2}.\n\n'''Heronian triangles'''\nare triangles whose sides ''and area'' are all integers.\n: An example is the triangle with sides   '''3, 4, 5'''   whose area is   '''6'''   (and whose perimeter is   '''12'''). \n\n\nNote that any triangle whose sides are all an integer multiple of   '''3, 4, 5''';   such as   '''6, 8, 10,'''   will also be a Heronian triangle.\n\nDefine a '''Primitive Heronian triangle''' as a Heronian triangle where the greatest common divisor \nof all three sides is   '''1'''   (unity). \n\nThis will exclude, for example, triangle   '''6, 8, 10.'''\n\n\n;Task:\n# Create a named function/method/procedure/... that implements Hero's formula.\n# Use the function to generate all the ''primitive'' Heronian triangles with sides <= 200.\n# Show the count of how many triangles are found.\n# Order the triangles by first increasing area, then by increasing perimeter, then by increasing maximum side lengths\n# Show the first ten ordered triangles in a table of sides, perimeter, and area.\n# Show a similar ordered table for those triangles with area = 210\n\n\nShow all output here.\n\n'''Note''': when generating triangles it may help to restrict a <= b <= c\n", "solution": "from __future__ import division, print_function\nfrom math import gcd, sqrt\n\n\ndef hero(a, b, c):\n    s = (a + b + c) / 2\n    a2 = s * (s - a) * (s - b) * (s - c)\n    return sqrt(a2) if a2 > 0 else 0\n\n\ndef is_heronian(a, b, c):\n    a = hero(a, b, c)\n    return a > 0 and a.is_integer()\n\n\ndef gcd3(x, y, z):\n    return gcd(gcd(x, y), z)\n\n\nif __name__ == '__main__':\n    MAXSIDE = 200\n\n    N = 1 + MAXSIDE\n    h = [(x, y, z)\n         for x in range(1, N)\n         for y in range(x, N)\n         for z in range(y, N) if (x + y > z) and\n         1 == gcd3(x, y, z) and\n         is_heronian(x, y, z)]\n\n    # By increasing area, perimeter, then sides\n    h.sort(key=lambda x: (hero(*x), sum(x), x[::-1]))\n\n    print(\n        'Primitive Heronian triangles with sides up to %i:' % MAXSIDE, len(h)\n    )\n    print('\\nFirst ten when ordered by increasing area, then perimeter,',\n          'then maximum sides:')\n    print('\\n'.join('  %14r perim: %3i area: %i'\n                    % (sides, sum(sides), hero(*sides)) for sides in h[:10]))\n    print('\\nAll with area 210 subject to the previous ordering:')\n    print('\\n'.join('  %14r perim: %3i area: %i'\n                    % (sides, sum(sides), hero(*sides)) for sides in h\n                    if hero(*sides) == 210))\n"}
{"title": "Hex words", "language": "Python", "task": "Definition\nFor the purposes of this task a '''hex word''' means a word which (in lower case form) consists entirely of the letters '''a, b, c, d, e''' and '''f'''.\n\n;Task\nUsing unixdict.txt, find all hex words with '''4''' letters or more.\n\nConvert each such word to its decimal equivalent and compute its base 10 digital root.\n\nDisplay all three in increasing order of digital root and show the total count of such words.\n\nKeeping only words which contain at least '''4''' distinct letters, display the same statistics but in decreasing order of decimal equivalent together with their total count.\n\n", "solution": "def digroot(n):\n    while n > 9:\n        n = sum([int(d) for d in str(n)])\n    return n\n\nwith open('unixdict.txt') as f:\n    lines = [w.strip() for w in f.readlines()]\n    words = [w for w in lines if len(w) >= 4 and all(c in 'abcdef' for c in w)]\n    results = [[w, int(w, 16)] for w in words]\n    for a in results:\n        a.append(digroot(a[1]))\n        \n    print(f\"Hex words in unixdict.txt:\\nRoot  Word      Base 10\\n\", \"-\"*22)\n    for a in sorted(results, key=lambda x:x[2]):\n        print(f\"{a[2]}     {a[0]:6}{a[1]:10}\")\n        \n    print(\"Total count of these words:\", len(results))\n    print(\"\\nHex words with > 3 distinct letters:\\nRoot  Word      Base 10\\n\", \"-\"*22)\n    results = [a for a in results if len(set(str(a[0]))) > 3]\n    for a in sorted(results, key=lambda x:x[2]):\n        print(f\"{a[2]}     {a[0]:6}{a[1]:10}\")\n        \n    print(\"Total count of those words:\", len(results))\n"}
{"title": "Hickerson series of almost integers", "language": "Python", "task": "The following function,   due to D. Hickerson,   is said to generate \"Almost integers\" by the \n\"Almost Integer\" page of Wolfram MathWorld,   (December 31 2013).   (See formula numbered   '''51'''.)\n\n\nThe function is:           h(n) = {\\operatorname{n}!\\over2(\\ln{2})^{n+1}}\n\n\nIt is said to produce \"almost integers\" for   '''n'''   between   '''1'''   and   '''17'''.  \nThe purpose of the task is to verify this assertion.\n\nAssume that an \"almost integer\" has '''either a nine or a zero as its first digit after the decimal point''' of its decimal string representation\n\n\n;Task:\nCalculate all values of the function checking and stating which are \"almost integers\".\n\nNote: Use extended/arbitrary precision numbers in your calculation if necessary to ensure you have adequate precision of results as for example:\n                h(18) = 3385534663256845326.39...\n\n", "solution": "from decimal import Decimal\nimport math\n\ndef h(n):\n    'Simple, reduced precision calculation'\n    return math.factorial(n) / (2 * math.log(2) ** (n + 1))\n    \ndef h2(n):\n    'Extended precision Hickerson function'\n    return Decimal(math.factorial(n)) / (2 * Decimal(2).ln() ** (n + 1))\n\nfor n in range(18):\n    x = h2(n)\n    norm = str(x.normalize())\n    almostinteger = (' Nearly integer' \n                     if 'E' not in norm and ('.0' in norm or '.9' in norm) \n                     else ' NOT nearly integer!')\n    print('n:%2i h:%s%s' % (n, norm, almostinteger))"}
{"title": "History variables", "language": "Python", "task": "''Storing the history of objects in a program is a common task. \nMaintaining the history of an object in a program has traditionally required programmers either to write specific code for handling the historical data, or to use a library which supports history logging.''\n\n''History variables are variables in a programming language which store not only their current value, but also the values they have contained in the past. Some existing languages do provide support for history variables. However these languages typically have many limits and restrictions on use of history variables.\n'' \n\n\"History Variables: \nThe Semantics, Formal Correctness, and Implementation of History Variables \nin an Imperative Programming Language\" by Mallon and Takaoka\n\nConcept also discussed on LtU and Patents.com.\n\n;Task: \nDemonstrate History variable support: \n* enable history variable support (if needed) \n* define a history variable\n* assign three values\n* non-destructively display the history\n* recall the three values. \n\nFor extra points, if the language of choice does not support history variables, \ndemonstrate how this might be implemented.\n\n", "solution": "import sys\n\nHIST = {}\n\ndef trace(frame, event, arg):\n    for name,val in frame.f_locals.items():\n        if name not in HIST:\n            HIST[name] = []\n        else:\n            if HIST[name][-1] is val:\n                continue\n        HIST[name].append(val)\n    return trace\n\ndef undo(name):\n    HIST[name].pop(-1)\n    return HIST[name][-1]\n\ndef main():\n    a = 10\n    a = 20\n\n    for i in range(5):\n        c = i\n\n    print \"c:\", c, \"-> undo x3 ->\",\n    c = undo('c')\n    c = undo('c')\n    c = undo('c')\n    print c\n    print 'HIST:', HIST\n\nsys.settrace(trace)\nmain()"}
{"title": "Hofstadter-Conway $10,000 sequence", "language": "Python", "task": "The definition of the sequence is colloquially described as:\n*   Starting with the list [1,1],\n*   Take the last number in the list so far: 1, I'll call it x.\n*   Count forward x places from the beginning of the list to find the first number to add (1)\n*   Count backward x places from the end of the list to find the second number to add (1)\n*   Add the two indexed numbers from the list and the result becomes the next number in the list (1+1)\n*   This would then produce [1,1,2] where 2 is the third element of the sequence.\n\nNote that indexing for the description above starts from alternately the left and right ends of the list and starts from an index of ''one''.\n\nA less wordy description of the sequence is:\n    a(1)=a(2)=1\n    a(n)=a(a(n-1))+a(n-a(n-1))\n\nThe sequence begins:\n    1, 1, 2, 2, 3, 4, 4, 4, 5, ...\n\nInteresting features of the sequence are that:\n*   a(n)/n   tends to   0.5   as   n   grows towards infinity.\n*   a(n)/n   where   n   is a power of   2   is   0.5\n*   For   n>4   the maximal value of   a(n)/n   between successive powers of 2 decreases.\n\na(n) / n   for   n   in   1..256 \n\n\nThe sequence is so named because John Conway offered a prize of $10,000 to the first person who could\nfind the first position,   p   in the sequence where \n    |a(n)/n| < 0.55  for all  n > p\nIt was later found that Hofstadter had also done prior work on the sequence.\n\nThe 'prize' was won quite quickly by Dr. Colin L. Mallows who proved the properties of the sequence and allowed him to find the value of   n   (which is much smaller than the 3,173,375,556 quoted in the NYT article).\n\n\n;Task:\n#   Create a routine to generate members of the Hofstadter-Conway $10,000 sequence.\n#   Use it to show the maxima of   a(n)/n   between successive powers of two up to   2**20\n#   As a stretch goal:   compute the value of   n   that would have won the prize and confirm it is true for   n   up to 2**20\n\n\n\n;Also see:\n*   Conways Challenge Sequence, Mallows' own account.\n*   Mathworld Article.\n\n", "solution": "from __future__ import division\n\ndef maxandmallows(nmaxpower2):\n    nmax = 2**nmaxpower2\n    mx = (0.5, 2)\n    mxpow2 = []\n    mallows = None\n\n    # Hofstadter-Conway sequence starts at hc[1],\n    # hc[0] is not part of the series.\n    hc = [None, 1, 1]\n\n    for n in range(2, nmax + 1):\n        ratio = hc[n] / n\n        if ratio > mx[0]:\n            mx = (ratio, n)\n        if ratio >= 0.55:\n            mallows = n\n        if ratio == 0.5:\n            print(\"In the region %7i < n <= %7i: max a(n)/n = %6.4f at  n = %i\" %\n\t\t  (n//2, n, mx[0], mx[1]))\n            mxpow2.append(mx[0])\n            mx = (ratio, n)\n        hc.append(hc[hc[n]] + hc[-hc[n]])\n\n    return hc, mallows if mxpow2 and mxpow2[-1] < 0.55 and n > 4 else None\n\nif __name__ == '__main__':\n    hc, mallows = maxandmallows(20)\n    if mallows:\n        print(\"\\nYou too might have won $1000 with the mallows number of %i\" % mallows)\n"}
{"title": "Hofstadter Figure-Figure sequences", "language": "Python", "task": "These two sequences of positive integers are defined as:\n:::: \\begin{align}\nR(1)&=1\\ ;\\ S(1)=2 \\\\\nR(n)&=R(n-1)+S(n-1), \\quad n>1.\n\\end{align}\n\nThe sequence S(n) is further defined as the sequence of positive integers '''''not''''' present in R(n).\n\nSequence R starts: \n    1, 3, 7, 12, 18, ...\nSequence S starts: \n    2, 4, 5, 6, 8, ...\n\n\n;Task:\n# Create two functions named '''ffr''' and '''ffs''' that when given '''n''' return '''R(n)''' or '''S(n)''' respectively.(Note that R(1) = 1 and S(1) = 2 to avoid off-by-one errors).\n# No maximum value for '''n''' should be assumed.\n# Calculate and show that the first ten values of '''R''' are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69\n# Calculate and show that the first 40 values of '''ffr''' plus the first 960 values of '''ffs''' include all the integers from 1 to 1000 exactly once.\n\n\n;References:\n* Sloane's A005228 and A030124.\n* Wolfram MathWorld\n* Wikipedia: Hofstadter Figure-Figure sequences.\n\n", "solution": "def ffr(n):\n    if n < 1 or type(n) != int: raise ValueError(\"n must be an int >= 1\")\n    try:\n        return ffr.r[n]\n    except IndexError:\n        r, s = ffr.r, ffs.s\n        ffr_n_1 = ffr(n-1)\n        lastr = r[-1]\n        # extend s up to, and one past, last r \n        s += list(range(s[-1] + 1, lastr))\n        if s[-1] < lastr: s += [lastr + 1]\n        # access s[n-1] temporarily extending s if necessary\n        len_s = len(s)\n        ffs_n_1 = s[n-1] if len_s > n else (n - len_s) + s[-1]\n        ans = ffr_n_1 + ffs_n_1\n        r.append(ans)\n        return ans\nffr.r = [None, 1]\n\ndef ffs(n):\n    if n < 1 or type(n) != int: raise ValueError(\"n must be an int >= 1\")\n    try:\n        return ffs.s[n]\n    except IndexError:\n        r, s = ffr.r, ffs.s\n        for i in range(len(r), n+2):\n            ffr(i)\n            if len(s) > n:\n                return s[n]\n        raise Exception(\"Whoops!\")\nffs.s = [None, 2]\n\nif __name__ == '__main__':\n    first10 = [ffr(i) for i in range(1,11)]\n    assert first10 == [1, 3, 7, 12, 18, 26, 35, 45, 56, 69], \"ffr() value error(s)\"\n    print(\"ffr(n) for n = [1..10] is\", first10)\n    #\n    bin = [None] + [0]*1000\n    for i in range(40, 0, -1):\n        bin[ffr(i)] += 1\n    for i in range(960, 0, -1):\n        bin[ffs(i)] += 1\n    if all(b == 1 for b in bin[1:1000]):\n        print(\"All Integers 1..1000 found OK\")\n    else:\n        print(\"All Integers 1..1000 NOT found only once: ERROR\")"}
{"title": "Hofstadter Figure-Figure sequences", "language": "Python from Haskell", "task": "These two sequences of positive integers are defined as:\n:::: \\begin{align}\nR(1)&=1\\ ;\\ S(1)=2 \\\\\nR(n)&=R(n-1)+S(n-1), \\quad n>1.\n\\end{align}\n\nThe sequence S(n) is further defined as the sequence of positive integers '''''not''''' present in R(n).\n\nSequence R starts: \n    1, 3, 7, 12, 18, ...\nSequence S starts: \n    2, 4, 5, 6, 8, ...\n\n\n;Task:\n# Create two functions named '''ffr''' and '''ffs''' that when given '''n''' return '''R(n)''' or '''S(n)''' respectively.(Note that R(1) = 1 and S(1) = 2 to avoid off-by-one errors).\n# No maximum value for '''n''' should be assumed.\n# Calculate and show that the first ten values of '''R''' are: 1, 3, 7, 12, 18, 26, 35, 45, 56, and 69\n# Calculate and show that the first 40 values of '''ffr''' plus the first 960 values of '''ffs''' include all the integers from 1 to 1000 exactly once.\n\n\n;References:\n* Sloane's A005228 and A030124.\n* Wolfram MathWorld\n* Wikipedia: Hofstadter Figure-Figure sequences.\n\n", "solution": "from itertools import islice\n\ndef R():\n\tn = 1\n\tyield n\n\tfor s in S():\n\t\tn += s\n\t\tyield n;\n\ndef S():\n\tyield 2\n\tyield 4\n\tu = 5\n\tfor r in R():\n\t\tif r <= u: continue;\n\t\tfor x in range(u, r): yield x\n\t\tu = r + 1\n\ndef lst(s, n): return list(islice(s(), n))\n\nprint \"R:\", lst(R, 10)\nprint \"S:\", lst(S, 10)\nprint sorted(lst(R, 40) + lst(S, 960)) == list(range(1,1001))\n\n# perf test case\n# print sum(lst(R, 10000000))"}
{"title": "Hofstadter Q sequence", "language": "Python", "task": "The Hofstadter Q sequence is defined as:\n:: \\begin{align}\nQ(1)&=Q(2)=1, \\\\\nQ(n)&=Q\\big(n-Q(n-1)\\big)+Q\\big(n-Q(n-2)\\big), \\quad n>2.\n\\end{align}\n\n\nIt is defined like the [[Fibonacci sequence]], but whereas the next term in the Fibonacci sequence is the sum of the previous two terms, in the Q sequence the previous two terms tell you how far to go back in the Q sequence to find the two numbers to sum to make the next term of the sequence.\n\n\n;Task:\n* Confirm and display that the first ten terms of the sequence are: 1, 1, 2, 3, 3, 4, 5, 5, 6, and 6 \n* Confirm and display that the 1000th term is:   502\n\n\n;Optional extra credit\n* Count and display how many times a member of the sequence is less than its preceding term for terms up to and including the 100,000th term.\n* Ensure that the extra credit solution   ''safely''   handles being initially asked for an '''n'''th term where   '''n'''   is large.\n(This point is to ensure that caching and/or recursion limits, if it is a concern, is correctly handled).\n\n", "solution": "from sys import getrecursionlimit\n\ndef q1(n):\n    if n < 1 or type(n) != int: raise ValueError(\"n must be an int >= 1\")\n    try:\n        return q.seq[n]\n    except IndexError:\n        len_q, rlimit = len(q.seq), getrecursionlimit()\n        if (n - len_q) > (rlimit // 5):\n            for i in range(len_q, n, rlimit // 5):\n                q(i)\n        ans = q(n - q(n - 1)) + q(n - q(n - 2))\n        q.seq.append(ans)\n        return ans\n\nif __name__ == '__main__':\n    tmp = q1(100000)\n    print(\"Q(i+1) < Q(i) for i [1..100000] is true %i times.\" %\n          sum(k1 < k0 for k0, k1 in zip(q.seq[1:], q.seq[2:])))"}
{"title": "Horner's rule for polynomial evaluation", "language": "Python", "task": "A fast scheme for evaluating a polynomial such as:\n: -19+7x-4x^2+6x^3\\,\nwhen\n: x=3\\;.\nis to arrange the computation as follows:\n: ((((0) x + 6) x + (-4)) x + 7) x + (-19)\\;\nAnd compute the result from the innermost brackets outwards as in this pseudocode:\n coefficients ''':=''' [-19, 7, -4, 6] ''# list coefficients of all x^0..x^n in order''\n x ''':=''' 3\n accumulator ''':=''' 0\n '''for''' i '''in''' ''length''(coefficients) '''downto''' 1 '''do'''\n     ''# Assumes 1-based indexing for arrays''\n     accumulator ''':=''' ( accumulator * x ) + coefficients[i]\n '''done'''\n ''# accumulator now has the answer''\n\n'''Task Description'''\n:Create a routine that takes a list of coefficients of a polynomial in order of increasing powers of x; together with a value of x to compute its value at, and return the value of the polynomial at that value using Horner's rule.\n\nCf. [[Formal power series]]\n\n", "solution": ">>> try: from functools import reduce\nexcept: pass\n\n>>> def horner(coeffs, x):\n\treturn reduce(lambda acc, c: acc * x + c, reversed(coeffs), 0)\n\n>>> horner( (-19, 7, -4, 6), 3)\n128"}
{"title": "ISBN13 check digit", "language": "Python", "task": "Validate the check digit of an ISBN-13 code:\n::*   Multiply every other digit by  '''3'''.\n::*   Add these numbers and the other digits.\n::*   Take the remainder of this number after division by  '''10'''.\n::*   If it is  '''0''',   the ISBN-13 check digit is correct.\n\n\nYou might use the following codes for testing:\n::::*   978-0596528126              (good)\n::::*   978-0596528120          (bad)\n::::*   978-1788399081              (good)\n::::*   978-1788399083          (bad)\n\n\nShow output here, on this page\n\n\n;See also:\n:*   for details:   13-digit ISBN method of validation.       (installs cookies.)\n\n", "solution": "def is_isbn13(n):\n    n = n.replace('-','').replace(' ', '')\n    if len(n) != 13:\n        return False\n    product = (sum(int(ch) for ch in n[::2]) \n               + sum(int(ch) * 3 for ch in n[1::2]))\n    return product % 10 == 0\n\nif __name__ == '__main__':\n    tests = '''\n978-1734314502\n978-1734314509\n978-1788399081\n978-1788399083'''.strip().split()\n    for t in tests:\n        print(f\"ISBN13 {t} validates {is_isbn13(t)}\")"}
{"title": "I before E except after C", "language": "Python", "task": "The phrase      \"I before E, except after C\"     is a \nwidely known mnemonic which is supposed to help when spelling English words.\n\n\n;Task:\nUsing the word list from   http://wiki.puzzlers.org/pub/wordlists/unixdict.txt, \ncheck if the two sub-clauses of the phrase are plausible individually:\n:::#   ''\"I before E when not preceded by C\"''\n:::#   ''\"E before I when preceded by C\"''\n\n\nIf both sub-phrases are plausible then the original phrase can be said to be plausible.\n\nSomething is plausible if the number of words having the feature is more than two times the number of words having the opposite feature (where feature is 'ie' or 'ei' preceded or not by 'c' as appropriate).\n \n\n;Stretch goal:\nAs a stretch goal use the entries from the table of Word Frequencies in Written and Spoken English: based on the British National Corpus, (selecting those rows with three space or tab separated words only), to see if the phrase is plausible when word frequencies are taken into account.\n\n\n''Show your output here as well as your program.''\n\n\n\n\n;cf.:\n* Schools to rethink 'i before e' - BBC news, 20 June 2009\n* I Before E Except After C - QI Series 8 Ep 14, (humorous)\n* Companion website for the book: \"Word Frequencies in Written and Spoken English: based on the British National Corpus\".\n\n", "solution": "import urllib.request\nimport re\n\nPLAUSIBILITY_RATIO = 2\n\ndef plausibility_check(comment, x, y):\n    print('\\n  Checking plausibility of: %s' % comment)\n    if x > PLAUSIBILITY_RATIO * y:\n        print('    PLAUSIBLE. As we have counts of %i vs %i, a ratio of %4.1f times'\n              % (x, y, x / y))\n    else:\n        if x > y:\n            print('    IMPLAUSIBLE. As although we have counts of %i vs %i, a ratio of %4.1f times does not make it plausible'\n                  % (x, y, x / y))\n        else:\n            print('    IMPLAUSIBLE, probably contra-indicated. As we have counts of %i vs %i, a ratio of %4.1f times'\n                  % (x, y, x / y))\n    return x > PLAUSIBILITY_RATIO * y\n\ndef simple_stats(url='http://wiki.puzzlers.org/pub/wordlists/unixdict.txt'):\n    words = urllib.request.urlopen(url).read().decode().lower().split()\n    cie = len({word for word in words if 'cie' in word})\n    cei = len({word for word in words if 'cei' in word})\n    not_c_ie = len({word for word in words if re.search(r'(^ie|[^c]ie)', word)})\n    not_c_ei = len({word for word in words if re.search(r'(^ei|[^c]ei)', word)})\n    return cei, cie, not_c_ie, not_c_ei\n\ndef print_result(cei, cie, not_c_ie, not_c_ei):\n    if ( plausibility_check('I before E when not preceded by C', not_c_ie, not_c_ei)\n         & plausibility_check('E before I when preceded by C', cei, cie) ):\n        print('\\nOVERALL IT IS PLAUSIBLE!')\n    else:\n        print('\\nOVERALL IT IS IMPLAUSIBLE!')\n    print('(To be plausible, one count must exceed another by %i times)' % PLAUSIBILITY_RATIO)\n\nprint('Checking plausibility of \"I before E except after C\":')\nprint_result(*simple_stats())"}
{"title": "Identity matrix", "language": "Python", "task": "Build an   identity matrix   of a size known at run-time. \n\n\nAn ''identity matrix'' is a square matrix of size '''''n'' x ''n''''', \nwhere the diagonal elements are all '''1'''s (ones), \nand all the other elements are all '''0'''s (zeroes).\n\n\nI_n = \\begin{bmatrix}\n  1      & 0      & 0      & \\cdots & 0      \\\\\n  0      & 1      & 0      & \\cdots & 0      \\\\\n  0      & 0      & 1      & \\cdots & 0      \\\\\n  \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n  0      & 0      & 0      & \\cdots & 1      \\\\\n\\end{bmatrix}\n\n\n;Related tasks:\n*   [[Spiral matrix]]\n*   [[Zig-zag matrix]] \n*   [[Ulam_spiral_(for_primes)]]\n\n", "solution": "def identity(size):\n    matrix = [[0]*size for i in range(size)]\n    #matrix = [[0] * size] * size    #Has a flaw. See http://stackoverflow.com/questions/240178/unexpected-feature-in-a-python-list-of-lists\n\n    for i in range(size):\n        matrix[i][i] = 1\n    \n    for rows in matrix:\n        for elements in rows:\n            print elements,\n        print \"\""}
{"title": "Identity matrix", "language": "Python 3.7", "task": "Build an   identity matrix   of a size known at run-time. \n\n\nAn ''identity matrix'' is a square matrix of size '''''n'' x ''n''''', \nwhere the diagonal elements are all '''1'''s (ones), \nand all the other elements are all '''0'''s (zeroes).\n\n\nI_n = \\begin{bmatrix}\n  1      & 0      & 0      & \\cdots & 0      \\\\\n  0      & 1      & 0      & \\cdots & 0      \\\\\n  0      & 0      & 1      & \\cdots & 0      \\\\\n  \\vdots & \\vdots & \\vdots & \\ddots & \\vdots \\\\\n  0      & 0      & 0      & \\cdots & 1      \\\\\n\\end{bmatrix}\n\n\n;Related tasks:\n*   [[Spiral matrix]]\n*   [[Zig-zag matrix]] \n*   [[Ulam_spiral_(for_primes)]]\n\n", "solution": "'''Identity matrices by maps and equivalent list comprehensions'''\n\nimport operator\n\n\n# idMatrix :: Int -> [[Int]]\ndef idMatrix(n):\n    '''Identity matrix of order n,\n       expressed as a nested map.\n    '''\n    eq = curry(operator.eq)\n    xs = range(0, n)\n    return list(map(\n        lambda x: list(map(\n            compose(int)(eq(x)),\n            xs\n        )),\n        xs\n    ))\n\n\n# idMatrix3 :: Int -> [[Int]]\ndef idMatrix2(n):\n    '''Identity matrix of order n,\n       expressed as a nested comprehension.\n    '''\n    xs = range(0, n)\n    return ([int(x == y) for x in xs] for y in xs)\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''\n        Identity matrix of dimension five,\n        by two different routes.\n    '''\n    for f in [idMatrix, idMatrix2]:\n        print(\n            '\\n' + f.__name__ + ':',\n            '\\n\\n' + '\\n'.join(map(str, f(5))),\n        )\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    '''A curried function derived\n       from an uncurried function.'''\n    return lambda a: lambda b: f(a, b)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Idiomatically determine all the characters that can be used for symbols", "language": "Python", "task": "Idiomatically determine all the characters that can be used for ''symbols''.\nThe word ''symbols'' is meant things like names of variables, procedures (i.e., named fragments of programs, functions, subroutines, routines), statement labels, events or conditions, and in general, anything a computer programmer can choose to ''name'', but not being restricted to this list. ''Identifiers'' might be another name for ''symbols''.\n\nThe method should find the characters regardless of the hardware architecture that is being used  (ASCII, EBCDIC, or other).\n\n;Task requirements\n\nDisplay the set of all the characters that can be used for symbols which can be used (allowed) by the computer program. \nYou may want to mention what hardware architecture is being used, and if applicable, the operating system.\n\nNote that most languages have additional restrictions on what characters can't be used for the first character of a variable or statement label, for instance.  These type of restrictions needn't be addressed here (but can be mentioned).\n\n;See also\n* Idiomatically determine all the lowercase and uppercase letters.\n\n", "solution": "See [[Idiomatically_determine_all_the_lowercase_and_uppercase_letters#Python|String class isidentifier]].\n\n"}
{"title": "Idiomatically determine all the lowercase and uppercase letters", "language": "Python", "task": "Idiomatically determine all the lowercase and uppercase letters   (of the Latin [English] alphabet)   being used currently by a computer programming language.\nThe method should find the letters regardless of the hardware architecture that is being used (ASCII, EBCDIC, or other).\n \n\n;Task requirements\n\nDisplay the set of all:\n::::::*   lowercase letters \n::::::*   uppercase letters\n\nthat can be used (allowed) by the computer program,\n\nwhere   ''letter''   is a member of the Latin (English) alphabet:     '''a''' --> '''z'''     and     '''A''' --> '''Z'''. \n\n\nYou may want to mention what hardware architecture is being used, and if applicable, the operating system.\n\n\n;See also\n* Idiomatically determine all the characters that can be used for symbols.\n\n", "solution": "classes = (str.isupper, str.islower, str.isalnum, str.isalpha, str.isdecimal,\n           str.isdigit, str.isidentifier, str.isnumeric, str.isprintable,\n           str.isspace, str.istitle)\n\nfor stringclass in classes:\n    chars = ''.join(chr(i) for i in range(0x10FFFF+1) if stringclass(chr(i)))\n    print('\\nString class %s has %i characters the first of which are:\\n  %r'\n          % (stringclass.__name__, len(chars), chars[:100]))"}
{"title": "Imaginary base numbers", "language": "Python from C++", "task": "Imaginary base numbers are a non-standard positional numeral system which uses an imaginary number as its radix. The most common is quater-imaginary with radix 2i. \n\n''The quater-imaginary numeral system was first proposed by Donald Knuth in 1955 as a submission for a high school science talent search.  [Ref.]''\n\nOther imaginary bases are possible too but are not as widely discussed and aren't specifically named.\n\n'''Task:''' Write a set of procedures (functions, subroutines, however they are referred to in your language) to convert base 10 numbers to an imaginary base and back. \n\nAt a minimum, support quater-imaginary (base 2i).\n\nFor extra kudos, support positive or negative bases 2i through 6i (or higher).\n\nAs a stretch goal, support converting non-integer numbers ( E.G. 227.65625+10.859375i ) to an imaginary base.\n\nSee Wikipedia: Quater-imaginary_base for more details. \n\nFor reference, here are some some decimal and complex numbers converted to quater-imaginary.\n\n\n\n\n\nBase 10\nBase 2i\n\n\n1\n 1\n\n\n2\n 2\n\n\n3\n 3\n\n\n4\n 10300\n\n\n5\n 10301\n\n\n6\n 10302\n\n\n7\n 10303\n\n\n8\n 10200\n\n\n9\n 10201\n\n\n10\n 10202\n\n\n11\n 10203\n\n\n12\n 10100\n\n\n13\n 10101\n\n\n14\n 10102\n\n\n15\n 10103\n\n\n16\n 10000\n\n\n\n\n\n\nBase 10\nBase 2i\n\n\n-1\n 103\n\n\n-2\n 102\n\n\n-3\n 101\n\n\n-4\n 100\n\n\n-5\n 203\n\n\n-6\n 202\n\n\n-7\n 201\n\n\n-8\n 200\n\n\n-9\n 303\n\n\n-10\n 302\n\n\n-11\n 301\n\n\n-12\n 300\n\n\n-13\n 1030003\n\n\n-14\n 1030002\n\n\n-15\n 1030001\n\n\n-16\n 1030000\n\n\n\n\n\n\nBase 10\nBase 2i\n\n\n1i\n10.2\n\n\n2i\n10.0\n\n\n3i\n20.2\n\n\n4i\n20.0\n\n\n5i\n30.2\n\n\n6i\n30.0\n\n\n7i\n103000.2\n\n\n8i\n103000.0\n\n\n9i\n103010.2\n\n\n10i\n103010.0\n\n\n11i\n103020.2\n\n\n12i\n103020.0\n\n\n13i\n103030.2\n\n\n14i\n103030.0\n\n\n15i\n102000.2\n\n\n16i\n102000.0\n\n\n\n\n\n\nBase 10\nBase 2i\n\n\n-1i\n0.2\n\n\n-2i\n1030.0\n\n\n-3i\n1030.2\n\n\n-4i\n1020.0\n\n\n-5i\n1020.2\n\n\n-6i\n1010.0\n\n\n-7i\n1010.2\n\n\n-8i\n1000.0\n\n\n-9i\n1000.2\n\n\n-10i\n2030.0\n\n\n-11i\n2030.2\n\n\n-12i\n2020.0\n\n\n-13i\n2020.2\n\n\n-14i\n2010.0\n\n\n-15i\n2010.2\n\n\n-16i\n2000.0\n\n\n\n\n\n", "solution": "import math\nimport re\n\ndef inv(c):\n    denom = c.real * c.real + c.imag * c.imag\n    return complex(c.real / denom, -c.imag / denom)\n\nclass QuaterImaginary:\n    twoI = complex(0, 2)\n    invTwoI = inv(twoI)\n\n    def __init__(self, str):\n        if not re.match(\"^[0123.]+$\", str) or str.count('.') > 1:\n            raise Exception('Invalid base 2i number')\n        self.b2i = str\n\n    def toComplex(self):\n        pointPos = self.b2i.find('.')\n        posLen = len(self.b2i) if (pointPos < 0) else pointPos\n        sum = complex(0, 0)\n        prod = complex(1, 0)\n        for j in xrange(0, posLen):\n            k = int(self.b2i[posLen - 1 - j])\n            if k > 0:\n                sum += prod * k\n            prod *= QuaterImaginary.twoI\n        if pointPos != -1:\n            prod = QuaterImaginary.invTwoI\n            for j in xrange(posLen + 1, len(self.b2i)):\n                k = int(self.b2i[j])\n                if k > 0:\n                    sum += prod * k\n                prod *= QuaterImaginary.invTwoI\n        return sum\n\n    def __str__(self):\n        return str(self.b2i)\n\ndef toQuaterImaginary(c):\n    if c.real == 0.0 and c.imag == 0.0:\n        return QuaterImaginary(\"0\")\n\n    re = int(c.real)\n    im = int(c.imag)\n    fi = -1\n    ss = \"\"\n    while re != 0:\n        re, rem = divmod(re, -4)\n        if rem < 0:\n            rem += 4\n            re += 1\n        ss += str(rem) + '0'\n    if im != 0:\n        f = c.imag / 2\n        im = int(math.ceil(f))\n        f = -4 * (f - im)\n        index = 1\n        while im != 0:\n            im, rem = divmod(im, -4)\n            if rem < 0:\n                rem += 4\n                im += 1\n            if index < len(ss):\n                ss[index] = str(rem)\n            else:\n                ss += '0' + str(rem)\n            index = index + 2\n        fi = int(f)\n    ss = ss[::-1]\n    if fi != -1:\n        ss += '.' + str(fi)\n    ss = ss.lstrip('0')\n    if ss[0] == '.':\n        ss = '0' + ss\n    return QuaterImaginary(ss)\n\nfor i in xrange(1,17):\n    c1 = complex(i, 0)\n    qi = toQuaterImaginary(c1)\n    c2 = qi.toComplex()\n    print \"{0:8} -> {1:>8} -> {2:8}     \".format(c1, qi, c2),\n\n    c1 = -c1\n    qi = toQuaterImaginary(c1)\n    c2 = qi.toComplex()\n    print \"{0:8} -> {1:>8} -> {2:8}\".format(c1, qi, c2)\nprint\n\nfor i in xrange(1,17):\n    c1 = complex(0, i)\n    qi = toQuaterImaginary(c1)\n    c2 = qi.toComplex()\n    print \"{0:8} -> {1:>8} -> {2:8}     \".format(c1, qi, c2),\n\n    c1 = -c1\n    qi = toQuaterImaginary(c1)\n    c2 = qi.toComplex()\n    print \"{0:8} -> {1:>8} -> {2:8}\".format(c1, qi, c2)\n\nprint \"done\"\n"}
{"title": "Increasing gaps between consecutive Niven numbers", "language": "Python", "task": "Note:   '''Niven'''   numbers are also called   '''Harshad'''        numbers.\n::::   They are also called                     '''multidigital'''   numbers.\n\n\n'''Niven''' numbers are positive integers which are evenly divisible by the sum of its\ndigits   (expressed in base ten).\n\n''Evenly divisible''   means   ''divisible with no remainder''.\n\n\n;Task:\n:*   find the gap (difference) of a Niven number from the previous Niven number\n:*   if the gap is   ''larger''   than the (highest) previous gap,   then:\n:::*   show the index (occurrence) of the gap      (the 1st gap is   '''1''')\n:::*   show the index of the Niven number that starts the gap     (1st Niven number is   '''1''',   33rd Niven number is   '''100''')\n:::*   show the Niven number that starts the gap\n:::*   show all numbers with comma separators where appropriate   (optional)\n:::*   I.E.:   the gap size of   '''60'''   starts at the   33,494th   Niven number which is Niven number   '''297,864'''\n:*   show all increasing gaps up to the   ten millionth   ('''10,000,000th''')   Niven number\n:*   (optional)   show all gaps up to whatever limit is feasible/practical/realistic/reasonable/sensible/viable on your computer\n:*   show all output here, on this page\n\n\n;Related task:\n:*   Harshad or Niven series.\n\n\n;Also see:\n:*   Journal of Integer Sequences, Vol. 6 (2004), Article 03.2.5, Large and Small Gaps Between Consecutive Niven Numbers.\n:*   (PDF) version of the (above) article by Doyon.\n\n", "solution": "\"\"\"\n\nPython implementation of\n\nhttp://rosettacode.org/wiki/Increasing_gaps_between_consecutive_Niven_numbers\n\n\"\"\"\n\n# based on C example\n\n# Returns the sum of the digits of n given the\n# sum of the digits of n - 1\ndef digit_sum(n, sum):\n    sum += 1\n    while n > 0 and n % 10 == 0:\n        sum -= 9\n        n /= 10\n    \n    return sum\n    \nprevious = 1\ngap = 0\nsum = 0\nniven_index = 0\ngap_index = 1\n \nprint(\"Gap index  Gap    Niven index    Niven number\")\n\nniven = 1\n\nwhile gap_index <= 22:\n    sum = digit_sum(niven, sum)\n    if niven % sum == 0:\n        if niven > previous + gap:\n            gap = niven - previous;\n            print('{0:9d} {1:4d}  {2:13d}     {3:11d}'.format(gap_index, gap, niven_index, previous))\n            gap_index += 1\n        previous = niven\n        niven_index += 1\n    niven += 1\n"}
{"title": "Index finite lists of positive integers", "language": "Python", "task": "It is known that the set of finite lists of positive integers is    countable.  \n\nThis means that there exists a subset of natural integers which can be mapped to the set of finite lists of positive integers.  \n\n\n;Task:\nImplement such a mapping:\n:*   write a function     ''rank''     which assigns an integer to any finite, arbitrarily long list of arbitrary large positive integers.\n:*   write a function          ''unrank''        which is the   ''rank''    inverse function.\n\n\nDemonstrate your solution by:\n:*   picking a random-length list of random positive integers\n:*   turn it into an integer,   and \n:*   get the list back.\n\n\nThere are many ways to do this.   Feel free to choose any one you like.\n\n\n;Extra credit:\nMake the   ''rank''   function as a    bijection   and show   ''unrank(n)''   for   '''n'''   varying from   '''0'''   to   '''10'''.\n\n", "solution": "def rank(x): return int('a'.join(map(str, [1] + x)), 11)\n\ndef unrank(n):\n\ts = ''\n\twhile n: s,n = \"0123456789a\"[n%11] + s, n//11\n\treturn map(int, s.split('a'))[1:]\n\nl = [1, 2, 3, 10, 100, 987654321]\nprint l\nn = rank(l)\nprint n\nl = unrank(n)\nprint l"}
{"title": "Integer overflow", "language": "Python", "task": "Some languages support one or more integer types of the underlying processor.\n\nThis integer types have fixed size;   usually   '''8'''-bit,   '''16'''-bit,   '''32'''-bit,   or   '''64'''-bit.\nThe integers supported by such a type can be   ''signed''   or   ''unsigned''.\n\nArithmetic for machine level integers can often be done by single CPU instructions.\nThis allows high performance and is the main reason to support machine level integers.\n\n\n;Definition:\nAn integer overflow happens when the result of a computation does not fit into the fixed size integer.\nThe result can be too small or too big to be representable in the fixed size integer.\n\n\n;Task:\nWhen a language has fixed size integer types, create a program that\ndoes arithmetic computations for the fixed size integers of the language.\n\nThese computations must be done such that the result would overflow.\n\nThe program should demonstrate what the following expressions do.\n\n\nFor 32-bit signed integers:\n::::: {|class=\"wikitable\"\n!Expression\n!Result that does not fit into a 32-bit signed integer\n|-\n| -(-2147483647-1)\n| 2147483648\n|-\n| 2000000000 + 2000000000\n| 4000000000\n|-\n| -2147483647 - 2147483647\n| -4294967294\n|-\n| 46341 * 46341\n| 2147488281\n|-\n| (-2147483647-1) / -1\n| 2147483648\n|}\n\nFor 64-bit signed integers:\n::: {|class=\"wikitable\"\n!Expression\n!Result that does not fit into a 64-bit signed integer\n|-\n| -(-9223372036854775807-1)\n| 9223372036854775808\n|-\n| 5000000000000000000+5000000000000000000\n| 10000000000000000000\n|-\n| -9223372036854775807 - 9223372036854775807\n| -18446744073709551614\n|-\n| 3037000500 * 3037000500\n| 9223372037000250000\n|-\n| (-9223372036854775807-1) / -1\n| 9223372036854775808\n|}\n\nFor 32-bit unsigned integers:\n::::: {|class=\"wikitable\"\n!Expression\n!Result that does not fit into a 32-bit unsigned integer\n|-\n| -4294967295\n| -4294967295\n|-\n| 3000000000 + 3000000000\n| 6000000000\n|-\n| 2147483647 - 4294967295\n| -2147483648\n|-\n| 65537 * 65537\n| 4295098369\n|}\n\nFor 64-bit unsigned integers:\n::: {|class=\"wikitable\"\n!Expression\n!Result that does not fit into a 64-bit unsigned integer\n|-\n| -18446744073709551615\n| -18446744073709551615\n|-\n| 10000000000000000000 + 10000000000000000000\n| 20000000000000000000\n|-\n| 9223372036854775807 - 18446744073709551615\n| -9223372036854775808\n|-\n| 4294967296 * 4294967296\n| 18446744073709551616\n|}\n\n\n;Notes:\n:*   When the integer overflow does trigger an exception show how the exception is caught.\n:*   When the integer overflow produces some value,   print it.\n:*   It should be explicitly noted when an integer overflow is not recognized,   the program continues with wrong results.\n:*   This should be done for signed and unsigned integers of various sizes supported by the computer programming language.\n:*   When a language has no fixed size integer type,   or when no integer overflow can occur for other reasons,   this should be noted.\n:*   It is okay to mention,   when a language supports unlimited precision integers,   but this task is NOT the place to demonstrate the   capabilities of unlimited precision integers.\n\n", "solution": "Python 3.4.1 (v3.4.1:c0e311e010fc, May 18 2014, 10:38:22) [MSC v.1600 32 bit (Intel)] on win32\nType \"copyright\", \"credits\" or \"license()\" for more information.\n>>> for calc in '''   -(-2147483647-1)\n   2000000000 + 2000000000\n   -2147483647 - 2147483647\n   46341 * 46341\n   (-2147483647-1) / -1'''.split('\\n'):\n\tans = eval(calc)\n\tprint('Expression: %r evaluates to %s of type %s'\n\t      % (calc.strip(), ans, type(ans)))\n\n\t\nExpression: '-(-2147483647-1)' evaluates to 2147483648 of type <class 'int'>\nExpression: '2000000000 + 2000000000' evaluates to 4000000000 of type <class 'int'>\nExpression: '-2147483647 - 2147483647' evaluates to -4294967294 of type <class 'int'>\nExpression: '46341 * 46341' evaluates to 2147488281 of type <class 'int'>\nExpression: '(-2147483647-1) / -1' evaluates to 2147483648.0 of type <class 'float'>\n>>> "}
{"title": "Integer sequence", "language": "Python", "task": "Create a program that, when run, would display all integers from   '''1'''   to    '''    '''    (or any relevant implementation limit),   in sequence   (i.e.   1, 2, 3, 4, etc)   if given enough time.\n\n\nAn example may not be able to reach arbitrarily-large numbers based on implementations limits.   For example, if integers are represented as a 32-bit unsigned value with 0 as the smallest representable value, the largest representable value would be 4,294,967,295.   Some languages support arbitrarily-large numbers as a built-in feature, while others make use of a module or library.\n\nIf appropriate, provide an example which reflect the language implementation's common built-in limits as well as an example which supports arbitrarily large numbers, and describe the nature of such limitations--or lack thereof.\n\n", "solution": "from itertools import count\n\nfor i in count(): \n    print(i)"}
{"title": "Intersecting number wheels", "language": "Python", "task": "A number wheel has:\n* A ''name'' which is an uppercase letter.\n* A set of ordered ''values'' which are either ''numbers'' or ''names''.\n\nA ''number'' is generated/yielded from a named wheel by:\n:1. Starting at the first value of the named wheel and advancing through subsequent values and wrapping around to the first value to form a \"wheel\":\n::1.a If the value is a number, yield it.\n::1.b If the value is a name, yield the next value from the named wheel\n::1.c Advance the position of this wheel.\n\nGiven the wheel\n: A: 1 2 3\nthe number 1 is first generated, then 2, then 3, 1, 2, 3, 1, ...\n\n'''Note:''' When more than one wheel is defined as a set of intersecting wheels then the \nfirst named wheel is assumed to be the one that values are generated from.\n\n;Examples:\nGiven the wheels:\n    A: 1 B 2\n    B: 3 4\nThe series of numbers generated starts:\n    1, 3, 2, 1, 4, 2, 1, 3, 2, 1, 4, 2, 1, 3, 2...\n\nThe intersections of number wheels can be more complex, (and might loop forever),\nand wheels may be multiply connected. \n\n'''Note:''' If a named wheel is referenced more than \nonce by one or many other wheels, then there is only one position of the wheel \nthat is advanced by each and all references to it.\n\nE.g.\n  A:  1 D D\n  D:  6 7 8\n  Generates:\n    1 6 7 1 8 6 1 7 8 1 6 7 1 8 6 1 7 8 1 6 ...    \n\n;Task:\nGenerate and show the first twenty terms of the sequence of numbers generated \nfrom these groups:\n\n    Intersecting Number Wheel group:\n      A:  1 2 3\n    \n    Intersecting Number Wheel group:\n      A:  1 B 2\n      B:  3 4\n    \n    Intersecting Number Wheel group:\n      A:  1 D D\n      D:  6 7 8\n    \n    Intersecting Number Wheel group:\n      A:  1 B C\n      B:  3 4\n      C:  5 B\n\nShow your output here, on this page.\n\n\n", "solution": "from itertools import islice\n\nclass INW():\n    \"\"\"\n    Intersecting Number Wheels\n    represented as a dict mapping\n    name to tuple of values.\n    \"\"\"\n\n    def __init__(self, **wheels):\n        self._wheels = wheels\n        self.isect = {name: self._wstate(name, wheel) \n                      for name, wheel in wheels.items()}\n    \n    def _wstate(self, name, wheel):\n        \"Wheel state holder\"\n        assert all(val in self._wheels for val in wheel if type(val) == str), \\\n               f\"ERROR: Interconnected wheel not found in {name}: {wheel}\"\n        pos = 0\n        ln = len(wheel)\n        while True:\n            nxt, pos = wheel[pos % ln], pos + 1\n            yield next(self.isect[nxt]) if type(nxt) == str else nxt\n                \n    def __iter__(self):\n        base_wheel_name = next(self.isect.__iter__())\n        yield from self.isect[base_wheel_name]\n        \n    def __repr__(self):\n        return f\"{self.__class__.__name__}({self._wheels})\"\n    \n    def __str__(self):\n        txt = \"Intersecting Number Wheel group:\"\n        for name, wheel in self._wheels.items():\n            txt += f\"\\n  {name+':':4}\" + ' '.join(str(v) for v in wheel)\n        return txt\n\ndef first(iter, n):\n    \"Pretty print first few terms\"\n    return ' '.join(f\"{nxt}\" for nxt in islice(iter, n))\n\nif __name__ == '__main__':\n    for group in[\n      {'A': (1, 2, 3)},\n      {'A': (1, 'B', 2),\n       'B': (3, 4)},\n      {'A': (1, 'D', 'D'),\n       'D': (6, 7, 8)},\n      {'A': (1, 'B', 'C'),\n       'B': (3, 4),\n       'C': (5, 'B')}, # 135143145...\n     ]:\n        w = INW(**group)\n        print(f\"{w}\\n  Generates:\\n    {first(w, 20)} ...\\n\")"}
{"title": "Intersecting number wheels", "language": "Python 3.7", "task": "A number wheel has:\n* A ''name'' which is an uppercase letter.\n* A set of ordered ''values'' which are either ''numbers'' or ''names''.\n\nA ''number'' is generated/yielded from a named wheel by:\n:1. Starting at the first value of the named wheel and advancing through subsequent values and wrapping around to the first value to form a \"wheel\":\n::1.a If the value is a number, yield it.\n::1.b If the value is a name, yield the next value from the named wheel\n::1.c Advance the position of this wheel.\n\nGiven the wheel\n: A: 1 2 3\nthe number 1 is first generated, then 2, then 3, 1, 2, 3, 1, ...\n\n'''Note:''' When more than one wheel is defined as a set of intersecting wheels then the \nfirst named wheel is assumed to be the one that values are generated from.\n\n;Examples:\nGiven the wheels:\n    A: 1 B 2\n    B: 3 4\nThe series of numbers generated starts:\n    1, 3, 2, 1, 4, 2, 1, 3, 2, 1, 4, 2, 1, 3, 2...\n\nThe intersections of number wheels can be more complex, (and might loop forever),\nand wheels may be multiply connected. \n\n'''Note:''' If a named wheel is referenced more than \nonce by one or many other wheels, then there is only one position of the wheel \nthat is advanced by each and all references to it.\n\nE.g.\n  A:  1 D D\n  D:  6 7 8\n  Generates:\n    1 6 7 1 8 6 1 7 8 1 6 7 1 8 6 1 7 8 1 6 ...    \n\n;Task:\nGenerate and show the first twenty terms of the sequence of numbers generated \nfrom these groups:\n\n    Intersecting Number Wheel group:\n      A:  1 2 3\n    \n    Intersecting Number Wheel group:\n      A:  1 B 2\n      B:  3 4\n    \n    Intersecting Number Wheel group:\n      A:  1 D D\n      D:  6 7 8\n    \n    Intersecting Number Wheel group:\n      A:  1 B C\n      B:  3 4\n      C:  5 B\n\nShow your output here, on this page.\n\n\n", "solution": "'''Intersecting number wheels'''\n\nfrom itertools import cycle, islice\nfrom functools import reduce\n\n\n# clockWorkTick :: Dict -> (Dict, Char)\ndef clockWorkTick(wheelMap):\n    '''The new state of the wheels, tupled with a\n       digit found by recursive descent from a single\n       click of the first wheel.'''\n    def click(wheels):\n        def go(wheelName):\n            wheel = wheels.get(wheelName, ['?'])\n            v = wheel[0]\n            return (Tuple if v.isdigit() or '?' == v else click)(\n                insertDict(wheelName)(leftRotate(wheel))(wheels)\n            )(v)\n        return go\n    return click(wheelMap)('A')\n\n\n# leftRotate :: [a] -> String\ndef leftRotate(xs):\n    ''' A string shifted cyclically towards\n        the left by one position.\n    '''\n    return ''.join(islice(cycle(xs), 1, 1 + len(xs)))\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''First twenty values from each set of test wheels.'''\n\n    wheelMaps = [dict(kvs) for kvs in [\n        [('A', \"123\")],\n        [('A', \"1B2\"), ('B', \"34\")],\n        [('A', \"1DD\"), ('D', \"678\")],\n        [('A', \"1BC\"), ('B', \"34\"), ('C', \"5B\")]\n    ]]\n    print('New state of wheel sets, after 20 clicks of each:\\n')\n    for wheels, series in [\n            mapAccumL(compose(const)(clockWorkTick))(\n                dct\n            )(' ' * 20) for dct in wheelMaps\n    ]:\n        print((wheels, ''.join(series)))\n\n    print('\\nInital states:')\n    for x in wheelMaps:\n        print(x)\n\n\n# ----------------------- GENERIC ------------------------\n\n# Tuple (,) :: a -> b -> (a, b)\ndef Tuple(x):\n    '''Constructor for a pair of values,\n       possibly of two different types.\n    '''\n    return lambda y: (\n        x + (y,)\n    ) if isinstance(x, tuple) else (x, y)\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# const :: a -> b -> a\ndef const(k):\n    '''The latter of two arguments,\n       with the first discarded.\n    '''\n    return lambda _: k\n\n\n# insertDict :: String -> a -> Dict -> Dict\ndef insertDict(k):\n    '''A new dictionary updated with a (k, v) pair.'''\n    def go(v, dct):\n        return dict(dct, **{k: v})\n    return lambda v: lambda dct: go(v, dct)\n\n\n# mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])\ndef mapAccumL(f):\n    '''A tuple of an accumulation and a map\n       with accumulation from left to right.\n    '''\n    def nxt(a, x):\n        tpl = f(a[0])(x)\n        return tpl[0], a[1] + [tpl[1]]\n\n    def go(acc):\n        def g(xs):\n            return reduce(nxt, xs, (acc, []))\n        return g\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Isqrt (integer square root) of X", "language": "Python 2.7", "task": "Sometimes a function is needed to find the integer square root of   '''X''',   where   '''X'''   can be a\nreal non-negative number.\n\nOften   '''X'''   is actually a non-negative integer.\n\nFor the purposes of this task,   '''X'''   can be an integer or a real number,   but if it\nsimplifies things in your computer programming language,   assume it's an integer.\n\n\nOne of the most common uses of   '''Isqrt'''   is in the division of an integer by all factors   (or\nprimes)   up to the   \n X     of that\ninteger,   either to find the factors of that integer,   or to determine primality.\n\n\nAn alternative method for finding the   '''Isqrt'''   of a number is to\ncalculate:        floor( sqrt(X) ) \n::*   where    '''sqrt'''    is the   square root   function for non-negative real numbers,   and\n::*   where   '''floor'''         is the   floor         function for real numbers.\n\n\nIf the hardware supports the computation of (real) square roots,   the above method might be a faster method for\nsmall numbers that don't have very many significant (decimal) digits.\n\nHowever, floating point arithmetic is limited in the number of   (binary or decimal)   digits that it can support.\n\n\n;Pseudo-code using quadratic residue:\nFor this task, the integer square root of a non-negative number will be computed using a version\nof   ''quadratic residue'',   which has the advantage that no   ''floating point''   calculations are\nused,   only integer arithmetic. \n\nFurthermore, the two divisions can be performed by bit shifting,   and the one multiplication can also be be performed by bit shifting or additions.\n\nThe disadvantage is the limitation of the size of the largest integer that a particular computer programming language can support.\n\n\nPseudo-code of a procedure for finding the integer square root of   '''X'''       (all variables are integers):\n          q <-- 1                                /*initialize  Q  to unity.  */\n                                   /*find a power of 4 that's greater than X.*/\n                   perform  while q <= x         /*perform while  Q <= X.    */\n                   q <-- q * 4                   /*multiply  Q  by  four.    */\n                   end  /*perform*/\n                                                 /*Q  is now greater than  X.*/\n          z <-- x                                /*set  Z  to the value of X.*/\n          r <-- 0                                /*initialize  R  to zero.   */\n                   perform  while q > 1          /*perform while  Q > unity. */\n                   q <-- q / 4                   /*integer divide by  four.  */\n                   t <-- z - r - q               /*compute value of  T.      */\n                   r <-- r / 2                   /*integer divide by  two.   */\n                   if t >= 0  then do            \n                                   z <-- t       /*set  Z  to value of  T.   */\n                                   r <-- r + q   /*compute new value of  R.  */\n                                   end\n                   end  /*perform*/\n                                                 /*R  is now the  Isqrt(X).  */\n \n          /* Sidenote: Also, Z is now the remainder after square root (i.e.  */\n          /*           R^2 + Z = X, so if Z = 0 then X is a perfect square). */\n\nAnother version for the (above)   1st   '''perform'''   is:\n                   perform  until q > X          /*perform until  Q > X.     */\n                   q <-- q * 4                   /*multiply  Q  by  four.    */\n                   end  /*perform*/\n\n\nInteger square roots of some values:\n Isqrt( 0)  is   0               Isqrt(60)  is  7                Isqrt( 99)  is   9\n Isqrt( 1)  is   1               Isqrt(61)  is  7                Isqrt(100)  is  10\n Isqrt( 2)  is   1               Isqrt(62)  is  7                Isqrt(102)  is  10\n Isqrt( 3)  is   1               Isqrt(63)  is  7\n Isqrt( 4)  is   2               Isqrt(64)  is  8                Isqet(120)  is  10\n Isqrt( 5)  is   2               Isqrt(65)  is  8                Isqrt(121)  is  11\n Isqrt( 6)  is   2               Isqrt(66)  is  8                Isqrt(122)  is  11\n Isqrt( 7)  is   2               Isqrt(67)  is  8\n Isqrt( 8)  is   2               Isqrt(68)  is  8                Isqrt(143)  is  11\n Isqrt( 9)  is   3               Isqrt(69)  is  8                Isqrt(144)  is  12\n Isqrt(10)  is   3               Isqrt(70)  is  8                Isqrt(145)  is  12\n\n\n;Task:\nCompute and show all output here   (on this page)   for:\n::*   the Isqrt of the     integers     from     '''0''' ---> '''65'''                (inclusive), shown in a       horizontal       format.\n::*   the Isqrt of the   odd powers               from          '''71''' ---> '''773'''   (inclusive), shown in a   vertical   format.\n::*   use commas in the displaying of larger numbers.\n\n\nYou can show more numbers for the 2nd requirement if the displays fits on one screen on Rosetta Code.\nIf your computer programming language only supports smaller integers,   show what you can.\n\n\n;Related tasks:\n:*   sequence of non-squares\n:*   integer roots\n:*   square root by hand\n\n", "solution": "def isqrt ( x ):\n    q = 1\n    while q <= x : \n        q *= 4\n    z,r = x,0\n    while q > 1 :\n        q  /= 4\n        t,r = z-r-q,r/2\n        if t >= 0 :\n            z,r = t,r+q\n    return r \n\nprint ' '.join( '%d'%isqrt( n ) for n in xrange( 66 ))\nprint '\\n'.join( '{0:114,} = isqrt( 7^{1:3} )'.format( isqrt( 7**n ),n ) for n in range( 1,204,2 ))"}
{"title": "Iterated digits squaring", "language": "Python from D", "task": "If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89:\n15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89\n7 -> 49 -> 97 -> 130 -> 10 -> 1\n\nAn example in Python:\n\n>>> step = lambda x: sum(int(d) ** 2 for d in str(x))\n>>> iterate = lambda x: x if x in [1, 89] else iterate(step(x))\n>>> [iterate(x) for x in xrange(1, 20)]\n[1, 89, 89, 89, 89, 89, 1, 89, 89, 1, 89, 89, 1, 89, 89, 89, 89, 89, 1]\n\n\n;Task:\n: Count how many number chains for integers 1 <= n < 100_000_000 end with a value 89.\nOr, for much less credit - (showing that your algorithm and/or language is slow):\n: Count how many number chains for integers 1 <= n < 1_000_000 end with a value 89.\n\nThis problem derives from the Project Euler problem 92.\n\nFor a quick algorithm for this task see the talk page\n\n\n;Related tasks:\n* [[Combinations with repetitions]]\n* [[Digital root]]\n* [[Digital root/Multiplicative digital root]]\n\n", "solution": "from math import ceil, log10, factorial\n\ndef next_step(x):\n    result = 0\n    while x > 0:\n        result += (x % 10) ** 2\n        x /= 10\n    return result\n\ndef check(number):\n    candidate = 0\n    for n in number:\n        candidate = candidate * 10 + n\n\n    while candidate != 89 and candidate != 1:\n        candidate = next_step(candidate)\n\n    if candidate == 89:\n        digits_count = [0] * 10\n        for d in number:\n            digits_count[d] += 1\n\n        result = factorial(len(number))\n        for c in digits_count:\n            result /= factorial(c)\n        return result\n\n    return 0\n\ndef main():\n    limit = 100000000\n    cache_size = int(ceil(log10(limit)))\n    assert 10 ** cache_size == limit\n\n    number = [0] * cache_size\n    result = 0\n    i = cache_size - 1\n\n    while True:\n        if i == 0 and number[i] == 9:\n            break\n        if i == cache_size - 1 and number[i] < 9:\n            number[i] += 1\n            result += check(number)\n        elif number[i] == 9:\n            i -= 1\n        else:\n            number[i] += 1\n            for j in xrange(i + 1, cache_size):\n                number[j] = number[i]\n            i = cache_size - 1\n            result += check(number)\n\n    print result\n\nmain()"}
{"title": "Iterated digits squaring", "language": "Python from Ruby", "task": "If you add the square of the digits of a Natural number (an integer bigger than zero), you always end with either 1 or 89:\n15 -> 26 -> 40 -> 16 -> 37 -> 58 -> 89\n7 -> 49 -> 97 -> 130 -> 10 -> 1\n\nAn example in Python:\n\n>>> step = lambda x: sum(int(d) ** 2 for d in str(x))\n>>> iterate = lambda x: x if x in [1, 89] else iterate(step(x))\n>>> [iterate(x) for x in xrange(1, 20)]\n[1, 89, 89, 89, 89, 89, 1, 89, 89, 1, 89, 89, 1, 89, 89, 89, 89, 89, 1]\n\n\n;Task:\n: Count how many number chains for integers 1 <= n < 100_000_000 end with a value 89.\nOr, for much less credit - (showing that your algorithm and/or language is slow):\n: Count how many number chains for integers 1 <= n < 1_000_000 end with a value 89.\n\nThis problem derives from the Project Euler problem 92.\n\nFor a quick algorithm for this task see the talk page\n\n\n;Related tasks:\n* [[Combinations with repetitions]]\n* [[Digital root]]\n* [[Digital root/Multiplicative digital root]]\n\n", "solution": "from itertools import combinations_with_replacement\nfrom array import array\nfrom time import clock\nD = 8\nF = [1, 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800, 479001600, 6227020800, 87178291200, 1307674368000, 20922789888000, 355687428096000]\ndef b(n):\n    yield 1\n    for g in range(1,n+1):\n        gn = g\n        res = 0\n        while gn > 0:\n            gn,rem = divmod(gn,10)\n            res += rem**2\n        if res==89:\n            yield 0\n        else:\n            yield res\nN = array('I',b(81*D))\nfor n in range(2,len(N)):\n    q = N[n]\n    while q>1:\n        q = N[q]\n    N[n] = q\n\nes = clock()\nz = 0\nfor n in combinations_with_replacement(range(10),D):\n    t = 0\n    for g in n:\n        t += g*g\n    if N[t] == 0:\n        continue\n    t = [0,0,0,0,0,0,0,0,0,0]\n    for g in n:\n        t[g] += 1\n    t1 = F[D]\n    for g in t:\n        t1 /= F[g]\n    z += t1\nee = clock() - es\nprint \"\\nD==\" + str(D) + \"\\n  \" + str(z) + \" numbers produce 1 and \" + str(10**D-z) + \" numbers produce 89\"\nprint \"Time ~= \" + str(ee) + \" secs\"\n"}
{"title": "JSON", "language": "Python 2.6+", "task": "Load a JSON string into a data structure.  \nAlso, create a new data structure and serialize it into JSON.\n\nUse objects and arrays (as appropriate for your language) \nand make sure your JSON is valid (https://jsonformatter.org).\n\n", "solution": ">>> import json\n>>> data = json.loads('{ \"foo\": 1, \"bar\": [10, \"apples\"] }')\n>>> sample = { \"blue\": [1,2], \"ocean\": \"water\" }\n>>> json_string = json.dumps(sample)\n>>> json_string\n'{\"blue\": [1, 2], \"ocean\": \"water\"}'\n>>> sample\n{'blue': [1, 2], 'ocean': 'water'}\n>>> data\n{'foo': 1, 'bar': [10, 'apples']}"}
{"title": "Jacobi symbol", "language": "Python", "task": "The '''Jacobi symbol''' is a multiplicative function that generalizes the Legendre symbol. Specifically, the Jacobi symbol (a | n) equals the product of the Legendre symbols (a | p_i)^(k_i), where n = p_1^(k_1)*p_2^(k_2)*...*p_i^(k_i) and the Legendre symbol (a | p) denotes the value of  a ^ ((p-1)/2) (mod p)\n* (a | p)     1      if a is a square (mod p)\n* (a | p)         -1      if a is not a square (mod p)\n* (a | p)     0      if a  0 \n\nIf n is prime, then the Jacobi symbol (a | n) equals the Legendre symbol (a | n).\n\n;Task:\nCalculate the Jacobi symbol (a | n).\n\n;Reference:\n* Wikipedia article on Jacobi symbol.\n\n", "solution": "def jacobi(a, n):\n    if n <= 0:\n        raise ValueError(\"'n' must be a positive integer.\")\n    if n % 2 == 0:\n        raise ValueError(\"'n' must be odd.\")\n    a %= n\n    result = 1\n    while a != 0:\n        while a % 2 == 0:\n            a /= 2\n            n_mod_8 = n % 8\n            if n_mod_8 in (3, 5):\n                result = -result\n        a, n = n, a\n        if a % 4 == 3 and n % 4 == 3:\n            result = -result\n        a %= n\n    if n == 1:\n        return result\n    else:\n        return 0"}
{"title": "Jacobsthal numbers", "language": "Python from Phix", "task": "'''Jacobsthal numbers''' are an integer sequence related to Fibonacci numbers. Similar to Fibonacci, where each term is the sum of the previous two terms, each term is the sum of the previous, plus twice the one before that. Traditionally the sequence starts with the given terms 0, 1.\n\n \n    J0 = 0\n    J1 = 1\n    Jn = Jn-1 + 2 x Jn-2\n\n\nTerms may be calculated directly using one of several possible formulas:\n \n    Jn = ( 2n - (-1)n ) / 3\n \n\n\n'''Jacobsthal-Lucas numbers''' are very similar. They have the same recurrence relationship, the only difference is an initial starting value '''J0 = 2''' rather than '''J0 = 0'''.\n\nTerms may be calculated directly using one of several possible formulas:\n    \n    JLn = 2n + (-1)n\n\n\n'''Jacobsthal oblong numbers''' is the sequence obtained from multiplying each '''Jacobsthal number''' '''Jn''' by its direct successor '''Jn+1'''.\n\n\n'''Jacobsthal primes''' are '''Jacobsthal numbers''' that are prime.\n\n\n\n;Task\n* Find and display the first 30 '''Jacobsthal numbers'''\n* Find and display the first 30 '''Jacobsthal-Lucas numbers'''\n* Find and display the first 20 '''Jacobsthal oblong numbers'''\n* Find and display at least the first 10 '''Jacobsthal primes'''\n\n\n\n;See also\n;* Wikipedia: Jacobsthal number\n;* Numbers Aplenty - Jacobsthal number\n;* OEIS:A001045 - Jacobsthal sequence (or Jacobsthal numbers)\n;* OEIS:A014551 - Jacobsthal-Lucas numbers.\n;* OEIS:A084175 - Jacobsthal oblong numbers\n;* OEIS:A049883 - Primes in the Jacobsthal sequence\n;* Related task: Fibonacci sequence\n;* Related task: Leonardo numbers\n\n\n\n\n", "solution": "#!/usr/bin/python\nfrom math import floor, pow\n\ndef isPrime(n):\n    for i in range(2, int(n**0.5) + 1):\n        if n % i == 0:\n            return False        \n    return True\n\ndef odd(n):\n    return n and 1 != 0\n    \ndef jacobsthal(n):\n    return floor((pow(2,n)+odd(n))/3)\n\ndef jacobsthal_lucas(n):\n    return int(pow(2,n)+pow(-1,n))\n\ndef jacobsthal_oblong(n):\n    return jacobsthal(n)*jacobsthal(n+1)\n\n\nif __name__ == '__main__':\n    print(\"First 30 Jacobsthal numbers:\")\n    for j in range(0, 30):\n        print(jacobsthal(j), end=\"  \")\n\n    print(\"\\n\\nFirst 30 Jacobsthal-Lucas numbers: \")\n    for j in range(0, 30):\n        print(jacobsthal_lucas(j), end = '\\t')\n\n    print(\"\\n\\nFirst 20 Jacobsthal oblong numbers: \")\n    for j in range(0, 20):\n        print(jacobsthal_oblong(j), end=\"  \")\n\n    print(\"\\n\\nFirst 10 Jacobsthal primes: \")\n    for j in range(3, 33):\n        if isPrime(jacobsthal(j)):\n            print(jacobsthal(j))"}
{"title": "Jaro similarity", "language": "Python 3", "task": "The Jaro distance is a measure of edit distance between two strings; its inverse, called the ''Jaro similarity'', is a measure of two strings' similarity: the higher the value, the more similar the strings are. The score is normalized such that   '''0'''   equates to no similarities and   '''1'''   is an exact match.\n\n\n;;Definition\n\nThe Jaro similarity   d_j   of two given strings   s_1   and   s_2   is\n\n: d_j = \\left\\{\n\n\\begin{array}{l l}\n  0 & \\text{if }m = 0\\\\\n  \\frac{1}{3}\\left(\\frac{m}{|s_1|} + \\frac{m}{|s_2|} + \\frac{m-t}{m}\\right) & \\text{otherwise} \\end{array} \\right.\n\nWhere:\n\n* m   is the number of ''matching characters'';\n* t   is half the number of ''transpositions''.\n\n\nTwo characters from   s_1   and   s_2   respectively, are considered ''matching'' only if they are the same and not farther apart than   \\left\\lfloor\\frac{\\max(|s_1|,|s_2|)}{2}\\right\\rfloor-1 characters.\n\nEach character of   s_1   is compared with all its matching characters in   s_2. Each difference in position is half a ''transposition''; that is, the number of transpositions is half the number of characters which are common to the two strings but occupy different positions in each one.\n\n\n;;Example\n\nGiven the strings   s_1   ''DWAYNE''   and   s_2   ''DUANE''   we find:\n\n* m = 4\n* |s_1| = 6\n* |s_2| = 5\n* t = 0\n\n\nWe find a Jaro score of:\n\n: d_j = \\frac{1}{3}\\left(\\frac{4}{6} + \\frac{4}{5} + \\frac{4-0}{4}\\right) = 0.822\n\n\n;Task\n\nImplement the Jaro algorithm and show the similarity scores for each of the following pairs:\n\n* (\"MARTHA\", \"MARHTA\")\n* (\"DIXON\", \"DICKSONX\")\n* (\"JELLYFISH\", \"SMELLYFISH\")\n\n\n; See also\n* Jaro-Winkler distance on Wikipedia.\n\n", "solution": "'''Jaro distance between two strings'''\n\nfrom functools import reduce\nimport itertools\n\n\n# --------------------- JARO FUNCTION ----------------------\n\n# jaro :: String -> String -> Float\ndef jaro(x):\n    '''The Jaro distance between two strings.'''\n    def go(s1, s2):\n        m, t = ap(compose(Tuple, len))(\n            transpositionSum\n        )(matches(s1, s2))\n        return 0 if 0 == m else (\n            (1 / 3) * ((m / len(s1)) + (\n                m / len(s2)\n            ) + ((m - t) / m))\n        )\n    return lambda y: go(x, y)\n\n\n# -------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Sample word pairs'''\n\n    print(\n        fTable('Jaro distances:\\n')(str)(\n            showPrecision(3)\n        )(\n            uncurry(jaro)\n        )([\n            (\"DWAYNE\", \"DUANE\"),\n            (\"MARTHA\", \"MARHTA\"),\n            (\"DIXON\", \"DICKSONX\"),\n            (\"JELLYFISH\", \"SMELLYFISH\")\n        ])\n    )\n\n\n# ----------------- JARO HELPER FUNCTIONS ------------------\n\n# transpositionSum :: [(Int, Char)] -> Int\ndef transpositionSum(xs):\n    '''A count of the transpositions in xs.'''\n    def f(a, xy):\n        x, y = xy\n        return 1 + a if fst(x) > fst(y) else a\n    return reduce(f, zip(xs, xs[1:]), 0)\n\n\n# matches :: String -> String -> [(Int, Char)]\ndef matches(s1, s2):\n    '''A list of (Index, Char) correspondences\n       between the two strings s1 and s2.'''\n\n    [(_, xs), (l2, ys)] = sorted(map(\n        ap(compose(Tuple, len))(list), [s1, s2]\n    ))\n    r = l2 // 2 - 1\n\n    # match :: (Int, (Char, Int)) -> (Int, Char)\n    def match(a, nc):\n        n, c = nc\n        offset = max(0, n - (1 + r))\n\n        def indexChar(x):\n            return a + [(offset + x, c)]\n\n        return maybe(a)(indexChar)(\n            elemIndex(c)(\n                drop(offset)(take(n + r)(ys))\n            )\n        )\n    return reduce(match, enumerate(xs), [])\n\n\n# ------------------- GENERIC FUNCTIONS --------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# Tuple (,) :: a -> b -> (a, b)\ndef Tuple(x):\n    '''Constructor for a pair of values,\n       possibly of two different types.\n    '''\n    def go(y):\n        return (\n            x + (y,)\n        ) if isinstance(x, tuple) else (x, y)\n    return go\n\n\n# ap :: (a -> b -> c) -> (a -> b) -> a -> c\ndef ap(f):\n    '''Applicative instance for functions.\n    '''\n    def go(g):\n        def fxgx(x):\n            return f(x)(\n                g(x)\n            )\n        return fxgx\n    return go\n\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    def go(f, g):\n        def fg(x):\n            return f(g(x))\n        return fg\n    return reduce(go, fs, lambda x: x)\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.\n    '''\n    def go(xs):\n        if isinstance(xs, (list, tuple, str)):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return go\n\n\n# elemIndex :: Eq a => a -> [a] -> Maybe Int\ndef elemIndex(x):\n    '''Just the index of the first element in xs\n       which is equal to x,\n       or Nothing if there is no such element.\n    '''\n    def go(xs):\n        try:\n            return Just(xs.index(x))\n        except ValueError:\n            return Nothing()\n    return go\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if m is Nothing,\n       or the application of f to x,\n       where m is Just(x).\n    '''\n    return lambda f: lambda m: v if (\n        None is m or m.get('Nothing')\n    ) else f(m.get('Just'))\n\n\n# showPrecision Int -> Float -> String\ndef showPrecision(n):\n    '''A string showing a floating point number\n       at a given degree of precision.'''\n    return lambda x: str(round(x, n))\n\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def gox(xShow):\n        def gofx(fxShow):\n            def gof(f):\n                def goxs(xs):\n                    ys = [xShow(x) for x in xs]\n                    w = max(map(len, ys))\n\n                    def arrowed(x, y):\n                        return y.rjust(w, ' ') + (\n                            ' -> ' + fxShow(f(x))\n                        )\n                    return s + '\\n' + '\\n'.join(\n                        map(arrowed, xs, ys)\n                    )\n                return goxs\n            return gof\n        return gofx\n    return gox\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    islice = itertools.islice\n\n    def go(xs):\n        return (\n            xs[0:n]\n            if isinstance(xs, (list, tuple))\n            else list(islice(xs, n))\n        )\n    return go\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    '''A function over a tuple derived from a curried function.'''\n    return lambda xy: f(xy[0])(\n        xy[1]\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Jewels and stones", "language": "Python", "task": "Create a function which takes two string parameters: 'stones' and 'jewels' and returns an integer.\n\nBoth strings can contain any number of upper or lower case letters. However, in the case of 'jewels', all letters must be distinct.\n\nThe function should count (and return) how many 'stones' are 'jewels' or, in other words, how many letters in 'stones' are also letters in 'jewels'.\n\n\nNote that:\n:# Only letters in the ISO basic Latin alphabet i.e. 'A to Z' or 'a to z' need be considered. \n:# A lower case letter is considered to be different from its upper case equivalent for this purpose, i.e., 'a' != 'A'.\n:# The parameters do not need to have exactly the same names.\n:# Validating the arguments is unnecessary.  \n\nSo, for example, if passed \"aAAbbbb\" for 'stones' and \"aA\" for 'jewels', the function should return 3.\n\nThis task was inspired by this problem.\n\n\n\n", "solution": "def countJewels(s, j):\n    return sum(x in j for x in s)\n\nprint countJewels(\"aAAbbbb\", \"aA\")\nprint countJewels(\"ZZ\", \"z\")"}
{"title": "Juggler sequence", "language": "Python", "task": "Background of the   '''juggler sequence''':\n\nJuggler sequences were publicized by an American mathematician and author   Clifford A. Pickover.   The name of the sequence gets it's name from the similarity of the rising and falling nature of the numbers in the sequences,   much like balls in the hands of a juggler.\n\n\n;Description\nA juggler sequence is an integer sequence that starts with a positive integer a[0], with each subsequent term in the sequence being defined by the recurrence relation: \n\n             a[k + 1]  =  floor(a[k] ^ 0.5)    if a[k] is even   ''' ''or'' '''  \n    a[k + 1]  =  floor(a[k] ^ 1.5)    if a[k] is odd                                   \n\nIf a juggler sequence reaches 1, then all subsequent terms are equal to 1. This is known to be the case for initial terms up to 1,000,000 but it is not known whether all juggler sequences after that will eventually reach 1.\n\n\n;Task:\nCompute and show here the following statistics for juggler sequences with an initial term of a[n] where n is between '''20''' and '''39''' inclusive:\n\n* l[n] - the number of terms needed to reach 1.\n* h[n] - the maximum value reached in that sequence.\n* i[n] - the index of the term (starting from 0) at which the maximum is (first) reached.\n\nIf your language supports ''big integers'' with an integer square root function, also compute and show here the same statistics for as many as you reasonably can of the following values for n:\n\n113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443, 275485, 1267909, 2264915, 5812827\n\nThose with fast languages and fast machines may also like to try their luck at n = 7110201.\n\nHowever, as h[n] for most of these numbers is thousands or millions of digits long, show instead of h[n]:\n\n* d[n] - the number of digits in h[n]\n\nThe results can be (partially) verified against the table here.\n \n\n;Related tasks:\n*   [[Hailstone sequence]]\n*   [[Yellowstone sequence]]\n*   [[Isqrt_(integer_square_root)_of_X]]\n\n\n;See also:\n* [[oeis:A007320]] Number of steps needed for Juggler sequence started at n to reach 1 \t\t\n* [[oeis:A094716]] Largest value in the Juggler sequence started at n\n\n", "solution": "from math import isqrt\n\ndef juggler(k, countdig=True, maxiters=1000):\n    m, maxj, maxjpos = k, k, 0\n    for i in range(1, maxiters):\n        m = isqrt(m) if m % 2 == 0 else isqrt(m * m * m)\n        if m >= maxj:\n            maxj, maxjpos  = m, i\n        if m == 1:\n            print(f\"{k: 9}{i: 6,}{maxjpos: 6}{len(str(maxj)) if countdig else maxj: 20,}{' digits' if countdig else ''}\")\n            return i\n\n    print(\"ERROR: Juggler series starting with $k did not converge in $maxiters iterations\")\n\n\nprint(\"       n    l(n)  i(n)       h(n) or d(n)\\n-------------------------------------------\")\nfor k in range(20, 40):\n    juggler(k, False)\n\nfor k in [113, 173, 193, 2183, 11229, 15065, 15845, 30817, 48443, 275485, 1267909]:\n    juggler(k)\n"}
{"title": "Julia set", "language": "Python from zkl", "task": "Task\nGenerate and draw a Julia set. \n\n\n;Related tasks\n*   Mandelbrot Set\n\n", "solution": "\"\"\"\nSolution from:\nhttps://codereview.stackexchange.com/questions/210271/generating-julia-set\n\"\"\"\nfrom functools import partial\nfrom numbers import Complex\nfrom typing import Callable\n\nimport matplotlib.pyplot as plt\nimport numpy as np\n\n\ndef douady_hubbard_polynomial(z: Complex,\n                              c: Complex) -> Complex:\n    \"\"\"\n    Monic and centered quadratic complex polynomial\n    https://en.wikipedia.org/wiki/Complex_quadratic_polynomial#Map\n    \"\"\"\n    return z ** 2 + c\n\n\ndef julia_set(mapping: Callable[[Complex], Complex],\n              *,\n              min_coordinate: Complex,\n              max_coordinate: Complex,\n              width: int,\n              height: int,\n              iterations_count: int = 256,\n              threshold: float = 2.) -> np.ndarray:\n    \"\"\"\n    As described in https://en.wikipedia.org/wiki/Julia_set\n    :param mapping: function defining Julia set\n    :param min_coordinate: bottom-left complex plane coordinate\n    :param max_coordinate: upper-right complex plane coordinate\n    :param height: pixels in vertical axis\n    :param width: pixels in horizontal axis\n    :param iterations_count: number of iterations\n    :param threshold: if the magnitude of z becomes greater\n    than the threshold we assume that it will diverge to infinity\n    :return: 2D pixels array of intensities\n    \"\"\"\n    im, re = np.ogrid[min_coordinate.imag: max_coordinate.imag: height * 1j,\n                      min_coordinate.real: max_coordinate.real: width * 1j]\n    z = (re + 1j * im).flatten()\n\n    live, = np.indices(z.shape)  # indexes of pixels that have not escaped\n    iterations = np.empty_like(z, dtype=int)\n\n    for i in range(iterations_count):\n        z_live = z[live] = mapping(z[live])\n        escaped = abs(z_live) > threshold\n        iterations[live[escaped]] = i\n        live = live[~escaped]\n        if live.size == 0:\n            break\n    else:\n        iterations[live] = iterations_count\n\n    return iterations.reshape((height, width))\n\n\nif __name__ == '__main__':\n    mapping = partial(douady_hubbard_polynomial,\n                      c=-0.7 + 0.27015j)  # type: Callable[[Complex], Complex]\n\n    image = julia_set(mapping,\n                      min_coordinate=-1.5 - 1j,\n                      max_coordinate=1.5 + 1j,\n                      width=800,\n                      height=600)\n    plt.axis('off')\n    plt.imshow(image,\n               cmap='nipy_spectral_r',\n               origin='lower')\n    plt.show()\n"}
{"title": "Jump anywhere", "language": "Python", "task": "Imperative programs like to jump around, but some languages restrict these jumps. Many structured languages restrict their [[conditional structures]] and [[loops]] to ''local jumps'' within a function. Some assembly languages limit certain jumps or branches to a small range.\n\nThis task is to demonstrate a local jump and a global jump and the various other types of jumps that the language supports. \nFor the purpose of this task, the jumps need not be used for a single purpose and you have the freedom to use these jumps for different purposes. \nYou may also defer to more specific tasks, like [[Exceptions]] or [[Generator]]. \nThis task provides a \"grab bag\" for several types of jumps. There are ''non-local jumps'' across function calls, or ''long jumps'' to anywhere within a program. Anywhere means not only to the tops of functions!\n\n* Some languages can ''go to'' any global label in a program.\n* Some languages can break multiple function calls, also known as ''unwinding the call stack''.\n* Some languages can save a ''continuation''. The program can later continue from the same place. So you can jump anywhere, but only if you have a previous visit there (to save the continuation).\n\nThese jumps are not all alike. \nA simple ''goto'' never touches the call stack. \nA continuation saves the call stack, so you can continue a function call after it ends.\n\n\n;Task:\nUse your language to demonstrate the various types of jumps that it supports. \n\nBecause the possibilities vary by language, this task is not specific. \nYou have the freedom to use these jumps for different purposes. \nYou may also defer to more specific tasks, like [[Exceptions]] or [[Generator]].\n\n", "solution": "# Example 2: Restarting a loop:\nfrom goto import goto, label\nlabel .start\nfor i in range(1, 4):\n    print i\n    if i == 2:\n        try:\n            output = message\n        except NameError:\n            print \"Oops - forgot to define 'message'!  Start again.\"\n            message = \"Hello world\"\n            goto .start\nprint output, \"\\n\"\n"}
{"title": "K-d tree", "language": "Python from D", "task": "{{wikipedia|K-d tree}}\n\nA k-d tree (short for ''k''-dimensional tree) is a space-partitioning data structure for organizing points in a k-dimensional space. k-d trees are a useful data structure for several applications, such as searches involving a multidimensional search key (e.g. range searches and nearest neighbor searches). \nk-d trees are a special case of binary space partitioning trees.\n\nk-d trees are not suitable, however, for efficiently finding the nearest neighbor in high dimensional spaces. As a general rule, if the dimensionality is ''k'', the number of points in the data, ''N'', should be ''N''  2''k''. \nOtherwise, when k-d trees are used with high-dimensional data, most of the points in the tree will be evaluated and the efficiency is no better than exhaustive search, and other methods such as approximate nearest-neighbor are used instead.\n\n'''Task:''' Construct a k-d tree and perform a nearest neighbor search for two example data sets:\n\n# The Wikipedia example data of [(2,3), (5,4), (9,6), (4,7), (8,1), (7,2)].\n# 1000 3-d points uniformly distributed in a 3-d cube.\n\nFor the Wikipedia example, find the nearest neighbor to point (9, 2)\nFor the random data, pick a random location and find the nearest neighbor.\n\nIn addition, instrument your code to count the number of nodes visited in the nearest neighbor search.  Count a node as visited if any field of it is accessed.\n\nOutput should show the point searched for, the point found, \nthe distance to the point, and the number of nodes visited.\n\nThere are variant algorithms for constructing the tree.  \nYou can use a simple median strategy or implement something more efficient.  \nVariants of the nearest neighbor search include nearest N neighbors, approximate nearest neighbor, and range searches.  \nYou do not have to implement these.  \nThe requirement for this task is specifically the nearest single neighbor.  \nAlso there are algorithms for inserting, deleting, and balancing k-d trees.  \nThese are also not required for the task.\n\n", "solution": "from random import seed, random\nfrom time import time\nfrom operator import itemgetter\nfrom collections import namedtuple\nfrom math import sqrt\nfrom copy import deepcopy\n\n\ndef sqd(p1, p2):\n    return sum((c1 - c2) ** 2 for c1, c2 in zip(p1, p2))\n\n\nclass KdNode(object):\n    __slots__ = (\"dom_elt\", \"split\", \"left\", \"right\")\n\n    def __init__(self, dom_elt, split, left, right):\n        self.dom_elt = dom_elt\n        self.split = split\n        self.left = left\n        self.right = right\n\n\nclass Orthotope(object):\n    __slots__ = (\"min\", \"max\")\n\n    def __init__(self, mi, ma):\n        self.min, self.max = mi, ma\n\n\nclass KdTree(object):\n    __slots__ = (\"n\", \"bounds\")\n\n    def __init__(self, pts, bounds):\n        def nk2(split, exset):\n            if not exset:\n                return None\n            exset.sort(key=itemgetter(split))\n            m = len(exset) // 2\n            d = exset[m]\n            while m + 1 < len(exset) and exset[m + 1][split] == d[split]:\n                m += 1\n            d = exset[m]\n\n\n            s2 = (split + 1) % len(d)  # cycle coordinates\n            return KdNode(d, split, nk2(s2, exset[:m]),\n                                    nk2(s2, exset[m + 1:]))\n        self.n = nk2(0, pts)\n        self.bounds = bounds\n\nT3 = namedtuple(\"T3\", \"nearest dist_sqd nodes_visited\")\n\n\ndef find_nearest(k, t, p):\n    def nn(kd, target, hr, max_dist_sqd):\n        if kd is None:\n            return T3([0.0] * k, float(\"inf\"), 0)\n\n        nodes_visited = 1\n        s = kd.split\n        pivot = kd.dom_elt\n        left_hr = deepcopy(hr)\n        right_hr = deepcopy(hr)\n        left_hr.max[s] = pivot[s]\n        right_hr.min[s] = pivot[s]\n\n        if target[s] <= pivot[s]:\n            nearer_kd, nearer_hr = kd.left, left_hr\n            further_kd, further_hr = kd.right, right_hr\n        else:\n            nearer_kd, nearer_hr = kd.right, right_hr\n            further_kd, further_hr = kd.left, left_hr\n\n        n1 = nn(nearer_kd, target, nearer_hr, max_dist_sqd)\n        nearest = n1.nearest\n        dist_sqd = n1.dist_sqd\n        nodes_visited += n1.nodes_visited\n\n        if dist_sqd < max_dist_sqd:\n            max_dist_sqd = dist_sqd\n        d = (pivot[s] - target[s]) ** 2\n        if d > max_dist_sqd:\n            return T3(nearest, dist_sqd, nodes_visited)\n        d = sqd(pivot, target)\n        if d < dist_sqd:\n            nearest = pivot\n            dist_sqd = d\n            max_dist_sqd = dist_sqd\n\n        n2 = nn(further_kd, target, further_hr, max_dist_sqd)\n        nodes_visited += n2.nodes_visited\n        if n2.dist_sqd < dist_sqd:\n            nearest = n2.nearest\n            dist_sqd = n2.dist_sqd\n\n        return T3(nearest, dist_sqd, nodes_visited)\n\n    return nn(t.n, p, t.bounds, float(\"inf\"))\n\n\ndef show_nearest(k, heading, kd, p):\n    print(heading + \":\")\n    print(\"Point:           \", p)\n    n = find_nearest(k, kd, p)\n    print(\"Nearest neighbor:\", n.nearest)\n    print(\"Distance:        \", sqrt(n.dist_sqd))\n    print(\"Nodes visited:   \", n.nodes_visited, \"\\n\")\n\n\ndef random_point(k):\n    return [random() for _ in range(k)]\n\n\ndef random_points(k, n):\n    return [random_point(k) for _ in range(n)]\n\nif __name__ == \"__main__\":\n    seed(1)\n    P = lambda *coords: list(coords)\n    kd1 = KdTree([P(2, 3), P(5, 4), P(9, 6), P(4, 7), P(8, 1), P(7, 2)],\n                  Orthotope(P(0, 0), P(10, 10)))\n    show_nearest(2, \"Wikipedia example data\", kd1, P(9, 2))\n\n    N = 400000\n    t0 = time()\n    kd2 = KdTree(random_points(3, N), Orthotope(P(0, 0, 0), P(1, 1, 1)))\n    t1 = time()\n    text = lambda *parts: \"\".join(map(str, parts))\n    show_nearest(2, text(\"k-d tree with \", N,\n                         \" random 3D points (generation time: \",\n                         t1-t0, \"s)\"),\n                 kd2, random_point(3))"}
{"title": "Kaprekar numbers", "language": "Python", "task": "A positive integer is a Kaprekar number if:\n* It is   '''1'''     (unity)\n* The decimal representation of its square may be split once into two parts consisting of positive integers which sum to the original number. \nNote that a split resulting in a part consisting purely of 0s is not valid, \nas 0 is not considered positive.\n\n\n;Example Kaprekar numbers:\n* 2223 is a Kaprekar number, as 2223 * 2223 = 4941729, 4941729 may be split to 494 and 1729, and 494 + 1729 = 2223.\n* The series of Kaprekar numbers is known as A006886, and begins as 1, 9, 45, 55, ....\n\n\n;Example process:\n10000 (1002) splitting from left to right:\n* The first split is [1, 0000], and is invalid; the 0000 element consists entirely of 0s, and 0 is not considered positive.\n* Slight optimization opportunity: When splitting from left to right, once the right part consists entirely of 0s, no further testing is needed; all further splits would also be invalid.\n\n\n;Task:\nGenerate and show all Kaprekar numbers less than 10,000. \n\n\n;Extra credit:\nOptionally, count (and report the count of) how many Kaprekar numbers are less than 1,000,000.\n\n\n;Extra extra credit:\nThe concept of Kaprekar numbers is not limited to base 10 (i.e. decimal numbers); \nif you can, show that Kaprekar numbers exist in other bases too.  \n\n\nFor this purpose, do the following:\n* Find all Kaprekar numbers for base 17 between 1 and 1,000,000 (one million);\n* Display each of them in base 10 representation;\n* Optionally, using base 17 representation (use letters 'a' to 'g' for digits 10(10) to 16(10)), display each of the numbers, its square, and where to split the square. \n \nFor example, 225(10) is \"d4\" in base 17, its square \"a52g\", and a5(17) + 2g(17) = d4(17), so the display would be something like:225   d4  a52g  a5 + 2g\n\n\n;Reference:\n* The Kaprekar Numbers by Douglas E. Iannucci (2000). PDF version\n\n\n;Related task:\n*   [[Casting out nines]]\n\n", "solution": "def encode(n, base):\n    result = \"\"\n    while n:\n        n, d = divmod(n, base)\n        if d < 10:\n            result += str(d)\n        else:\n            result += chr(d - 10 + ord(\"a\"))\n    return result[::-1]\ndef Kaprekar(n, base):\n    if n == '1':\n        return True\n    sq = encode((int(n, base)**2), base)\n    for i in range(1,len(sq)):\n        if (int(sq[:i], base) + int(sq[i:], base) == int(n, base)) and (int(sq[:i], base) > 0) and (int(sq[i:], base)>0):\n            return True\n    return False\ndef Find(m, n, base):\n    return [encode(i, base) for i in range(m,n+1) if Kaprekar(encode(i, base), base)]\n\nm = int(raw_input('Where to start?\\n'))\nn = int(raw_input('Where to stop?\\n'))\nbase = int(raw_input('Enter base:'))\nKNumbers = Find(m, n, base)\nfor i in KNumbers:\n    print i\nprint 'The number of Kaprekar Numbers found are',\nprint len(KNumbers)\nraw_input()"}
{"title": "Kernighans large earthquake problem", "language": "Python", "task": "Brian Kernighan, in a lecture at the University of Nottingham, described a problem on which this task is based.\n\n;Problem:\nYou are given a a data file of thousands of lines; each of three `whitespace` separated fields: a date, a one word name and the magnitude of the event.\n\nExample lines from the file would be lines like:\n8/27/1883    Krakatoa            8.8\n5/18/1980    MountStHelens       7.6\n3/13/2009    CostaRica           5.1\n\n;Task:\n* Create a program or script invocation to find all the events with magnitude greater than 6\n* Assuming an appropriate name e.g. \"data.txt\" for the file:\n:# Either: Show how your program is invoked to process a data file of that name.\n:# Or: Incorporate the file name into the program, (as it is assumed that the program is single use).\n\n", "solution": "from os.path import expanduser\nfrom functools import (reduce)\nfrom itertools import (chain)\n\n\n# largeQuakes :: Int -> [String] -> [(String, String, String)]\ndef largeQuakes(n):\n    def quake(threshold):\n        def go(x):\n            ws = x.split()\n            return [tuple(ws)] if threshold < float(ws[2]) else []\n        return lambda x: go(x)\n    return concatMap(quake(n))\n\n\n# main :: IO ()\ndef main():\n    print (\n        largeQuakes(6)(\n            open(expanduser('~/data.txt')).read().splitlines()\n        )\n    )\n\n\n# GENERIC ABSTRACTION -------------------------------------\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    return lambda xs: list(\n        chain.from_iterable(\n            map(f, xs)\n        )\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Keyboard input/Obtain a Y or N response", "language": "Python", "task": "Obtain a valid   '''Y'''   or   '''N'''   response from the [[input device::keyboard]]. \n\nThe keyboard should be flushed, so that any outstanding key-presses are removed, preventing any existing   '''Y'''   or   '''N'''   key-press from being evaluated. \n\nThe response should be obtained as soon as   '''Y'''   or   '''N'''   are pressed, and there should be no need to press an   enter   key.\n\n", "solution": "#!/usr/bin/env python\n# -*- coding: utf-8 -*-\nfrom curses import wrapper\n#\n#\ndef main(stdscr):\n  # const\n  #y = ord(\"y\")\n  #n = ord(\"n\")\n  while True:\n    # keyboard input interceptor|listener\n    #window.nodelay(yes)\n    # - If yes is 1, getch() will be non-blocking.\n    # return char code\n    #kb_Inpt = stdscr.getch()\n    # return string\n    kb_Inpt = stdscr.getkey()\n    #if kb_Inpt == (y or n):\n    if kb_Inpt.lower() == ('y' or 'n'):\n      break\n      return None\n  #\n  return None\n#\n#*** unit test ***#\nif __name__ == \"__main__\":\n  #\n  wrapper(main)"}
{"title": "Knight's tour", "language": "Python", "task": "Task\nProblem: you have a standard 8x8 chessboard, empty but for a single knight on some square.   Your task is to emit a series of legal knight moves that result in the knight visiting every square on the chessboard exactly once. Note that it is ''not'' a requirement that the tour be \"closed\"; that is, the knight need not end within a single move of its start position.\n\nInput and output may be textual or graphical, according to the conventions of the programming environment.  If textual, squares should be indicated in algebraic notation.  The output should indicate the order in which the knight visits the squares, starting with the initial position.  The form of the output may be a diagram of the board with the squares numbered according to visitation sequence, or a textual list of algebraic coordinates in order, or even an actual animation of the knight moving around the chessboard.\n\nInput: starting square\n\nOutput: move sequence\n\n\n;Related tasks\n* [[A* search algorithm]]\n* [[N-queens problem]]\n* [[Solve a Hidato puzzle]]\n* [[Solve a Holy Knight's tour]]\n* [[Solve a Hopido puzzle]]\n* [[Solve a Numbrix puzzle]]\n* [[Solve the no connection puzzle]]\n\n", "solution": "import copy\n\nboardsize=6\n_kmoves = ((2,1), (1,2), (-1,2), (-2,1), (-2,-1), (-1,-2), (1,-2), (2,-1)) \n\n\ndef chess2index(chess, boardsize=boardsize):\n    'Convert Algebraic chess notation to internal index format'\n    chess = chess.strip().lower()\n    x = ord(chess[0]) - ord('a')\n    y = boardsize - int(chess[1:])\n    return (x, y)\n    \ndef boardstring(board, boardsize=boardsize):\n    r = range(boardsize)\n    lines = ''\n    for y in r:\n        lines += '\\n' + ','.join('%2i' % board[(x,y)] if board[(x,y)] else '  '\n                                 for x in r)\n    return lines\n    \ndef knightmoves(board, P, boardsize=boardsize):\n    Px, Py = P\n    kmoves = set((Px+x, Py+y) for x,y in _kmoves)\n    kmoves = set( (x,y)\n                  for x,y in kmoves\n                  if 0 <= x < boardsize\n                     and 0 <= y < boardsize\n                     and not board[(x,y)] )\n    return kmoves\n\ndef accessibility(board, P, boardsize=boardsize):\n    access = []\n    brd = copy.deepcopy(board)\n    for pos in knightmoves(board, P, boardsize=boardsize):\n        brd[pos] = -1\n        access.append( (len(knightmoves(brd, pos, boardsize=boardsize)), pos) )\n        brd[pos] = 0\n    return access\n    \ndef knights_tour(start, boardsize=boardsize, _debug=False):\n    board = {(x,y):0 for x in range(boardsize) for y in range(boardsize)}\n    move = 1\n    P = chess2index(start, boardsize)\n    board[P] = move\n    move += 1\n    if _debug:\n        print(boardstring(board, boardsize=boardsize))\n    while move <= len(board):\n        P = min(accessibility(board, P, boardsize))[1]\n        board[P] = move\n        move += 1\n        if _debug:\n            print(boardstring(board, boardsize=boardsize))\n            input('\\n%2i next: ' % move)\n    return board\n\nif __name__ == '__main__':\n    while 1:\n        boardsize = int(input('\\nboardsize: '))\n        if boardsize < 5:\n            continue\n        start = input('Start position: ')\n        board = knights_tour(start, boardsize)\n        print(boardstring(board, boardsize=boardsize))"}
{"title": "Knuth's algorithm S", "language": "Python 3.x", "task": "This is a method of randomly sampling n items from a set of M items, with equal probability; where M >= n and M, the number of items is unknown until the end.\nThis means that the equal probability sampling should be maintained  for all successive items > n as they become available (although the content of successive samples can change).\n\n\n;The algorithm:\n:* Select the first n items as the sample as they become available;\n:* For the i-th item where i > n, have a random chance of n/i of keeping it.  If failing this chance, the sample remains the same.  If not, have it randomly (1/n) replace one of the previously selected n items of the sample.\n:* Repeat   2nd step   for any subsequent items.\n\n\n;The Task:\n:* Create a function s_of_n_creator that given n the maximum sample size, returns a function s_of_n that takes one parameter, item.\n:* Function  s_of_n when called with successive items returns an equi-weighted random sample of up to n of its items so far, each time it is called, calculated using Knuths Algorithm S.\n:* Test your functions by printing and showing the frequency of occurrences of the selected digits from 100,000 repetitions of:\n:::# Use the s_of_n_creator with n == 3 to generate an s_of_n.\n:::# call s_of_n with each of the digits 0 to 9 in order, keeping the returned three digits of its random sampling from its last call with argument item=9.\n\n\nNote: A class taking n and generating a callable instance/function might also be used.\n\n\n;Reference:\n* The Art of Computer Programming, Vol 2, 3.4.2 p.142\n\n\n;Related tasks: \n* [[One of n lines in a file]] \n* [[Accumulator factory]]\n\n", "solution": "from random import randrange\n\ndef s_of_n_creator(n):\n    sample, i = [], 0\n    def s_of_n(item):\n        nonlocal i\n\n        i += 1\n        if i <= n:\n            # Keep first n items\n            sample.append(item)\n        elif randrange(i) < n:\n            # Keep item\n            sample[randrange(n)] = item\n        return sample\n    return s_of_n\n\nif __name__ == '__main__':\n    bin = [0]* 10\n    items = range(10)\n    print(\"Single run samples for n = 3:\")\n    s_of_n = s_of_n_creator(3)\n    for item in items:\n        sample = s_of_n(item)\n        print(\"  Item: %i -> sample: %s\" % (item, sample))\n    #\n    for trial in range(100000):\n        s_of_n = s_of_n_creator(3)\n        for item in items:\n            sample = s_of_n(item)\n        for s in sample:\n            bin[s] += 1\n    print(\"\\nTest item frequencies for 100000 runs:\\n \",\n          '\\n  '.join(\"%i:%i\" % x for x in enumerate(bin)))"}
{"title": "Koch curve", "language": "Python", "task": "Draw a Koch curve. See details: Koch curve\n\n", "solution": "===Functional===\nGenerates SVG for a Koch snowflake. To view, save as a text file with the extension .svg, and open in a browser.\n\n"}
{"title": "Koch curve", "language": "Python from Haskell", "task": "Draw a Koch curve. See details: Koch curve\n\n", "solution": "'''Koch curve'''\n\nfrom math import cos, pi, sin\nfrom operator import add, sub\nfrom itertools import chain\n\n\n# kochSnowflake :: Int -> (Float, Float) -> (Float, Float) -> [(Float, Float)]\ndef kochSnowflake(n, a, b):\n    '''List of points on a Koch snowflake of order n, derived\n       from an equilateral triangle with base a b.\n    '''\n    points = [a, equilateralApex(a, b), b]\n    return chain.from_iterable(map(\n        kochCurve(n),\n        points,\n        points[1:] + [points[0]]\n    ))\n\n\n# kochCurve :: Int -> (Float, Float) -> (Float, Float)\n#                  -> [(Float, Float)]\ndef kochCurve(n):\n    '''List of points on a Koch curve of order n,\n       starting at point ab, and ending at point xy.\n    '''\n    def koch(n):\n        def goTuple(abxy):\n            ab, xy = abxy\n            if 0 == n:\n                return [xy]\n            else:\n                mp, mq = midThirdOfLine(ab, xy)\n                points = [\n                    ab,\n                    mp,\n                    equilateralApex(mp, mq),\n                    mq,\n                    xy\n                ]\n                return list(\n                    chain.from_iterable(map(\n                        koch(n - 1),\n                        zip(points, points[1:])\n                    ))\n                )\n        return goTuple\n\n    def go(ab, xy):\n        return [ab] + koch(n)((ab, xy))\n    return go\n\n\n# equilateralApex :: (Float, Float) -> (Float, Float) -> (Float, Float)\ndef equilateralApex(p, q):\n    '''Apex of triangle with base p q.\n    '''\n    return rotatedPoint(pi / 3)(p, q)\n\n\n# rotatedPoint :: Float -> (Float, Float) ->\n#                (Float, Float) -> (Float, Float)\ndef rotatedPoint(theta):\n    '''The point ab rotated theta radians\n        around the origin xy.\n    '''\n    def go(xy, ab):\n        ox, oy = xy\n        a, b = ab\n        dx, dy = rotatedVector(theta, (a - ox, oy - b))\n        return ox + dx, oy - dy\n    return go\n\n\n# rotatedVector :: Float -> (Float, Float) -> (Float, Float)\ndef rotatedVector(theta, xy):\n    '''The vector xy rotated by theta radians.\n    '''\n    x, y = xy\n    return (\n        x * cos(theta) - y * sin(theta),\n        x * sin(theta) + y * cos(theta)\n    )\n\n\n# midThirdOfLine :: (Float, Float) -> (Float, Float)\n#                -> ((Float, Float), (Float, Float))\ndef midThirdOfLine(ab, xy):\n    '''Second of three equal segments of\n       the line between ab and xy.\n    '''\n    vector = [x / 3 for x in map(sub, xy, ab)]\n\n    def f(p):\n        return tuple(map(add, vector, p))\n    p = f(ab)\n    return (p, f(p))\n\n\n# -------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''SVG for Koch snowflake of order 4.\n    '''\n    print(\n        svgFromPoints(1024)(\n            kochSnowflake(\n                4, (200, 600), (800, 600)\n            )\n        )\n    )\n\n\n# -------------------------- SVG ---------------------------\n\n# svgFromPoints :: Int -> [(Float, Float)] -> SVG String\ndef svgFromPoints(w):\n    '''Width of square canvas -> Point list -> SVG string.\n    '''\n    def go(xys):\n        xs = ' '.join(map(\n            lambda xy: str(round(xy[0], 2)) + ' ' + str(round(xy[1], 2)),\n            xys\n        ))\n        return '\\n'.join([\n            '<svg xmlns=\"http://www.w3.org/2000/svg\"',\n            f'width=\"512\" height=\"512\" viewBox=\"5 5 {w} {w}\">',\n            f'<path d=\"M{xs}\" ',\n            'stroke-width=\"2\" stroke=\"red\" fill=\"transparent\"/>',\n            '</svg>'\n        ])\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Kolakoski sequence", "language": "Python", "task": "The natural numbers, (excluding zero); with the property that:\n: ''if you form a new sequence from the counts of runs of the same number in the first sequence, this new sequence is the same as the first sequence''.\n\n;Example:\nThis is ''not'' a Kolakoski sequence:\n1,1,2,2,2,1,2,2,1,2,...\nIts sequence of run counts, (sometimes called a run length encoding, (RLE); but a true RLE also gives the character that each run encodes), is calculated like this:\n\n: Starting from the leftmost number of the sequence we have 2 ones, followed by 3 twos, then 1 ones, 2 twos, 1 one, ...\n\nThe above gives the RLE of:\n2, 3, 1, 2, 1, ...\n\nThe original sequence is different from its RLE in this case. '''It would be the same for a true Kolakoski sequence'''.\n\n;Creating a Kolakoski sequence:\n\nLets start with the two numbers (1, 2) that we will cycle through; i.e. they will be used in this order: 1,2,1,2,1,2,....\n\n# We start the sequence s with the first item from the cycle c: 1\n# An index, k, into the, (expanding), sequence will step, or index through each item of the sequence s from the first, at its own rate. \nWe will arrange that the k'th item of s states how many ''times'' the ''last'' item of sshould appear at the end of s.\n\nWe started s with 1 and therefore s[k] states that it should appear only the 1 time.\n\nIncrement k\nGet the next item from c and append it to the end of sequence s. s will then become: 1, 2\nk was moved to the second item in the list and s[k] states that it should appear two times, so append another of the last item to the sequence s: 1, 2,2\nIncrement k\nAppend the next item from the cycle to the list: 1, 2,2, 1\nk is now at the third item in the list that states that the last item should appear twice so add another copy of the last item to the sequence s: 1, 2,2, 1,1\nincrement k\n\n...\n\n'''Note''' that the RLE of 1, 2, 2, 1, 1, ... begins 1, 2, 2 which is the beginning of the original sequence. The generation algorithm ensures that this will always be the case.\n\n;Task:\n# Create a routine/proceedure/function/... that given an initial ordered list/array/tuple etc of the natural numbers (1, 2), returns the next number from the list when accessed in a cycle.\n# Create another routine that when given the initial ordered list (1, 2) and the minimum length of the sequence  to generate; uses the first routine and the algorithm above, to generate at least the requested first members of the kolakoski sequence.\n# Create a routine that when given a sequence, creates the run length encoding of that sequence (as defined above) and returns the result of checking if sequence starts with the exact members of its RLE. (But ''note'', due to sampling, do not compare the last member of the RLE).\n# Show, on this page, (compactly), the first 20 members of the sequence generated from (1, 2)\n# Check the sequence againt its RLE.\n# Show, on this page, the first 20 members of the sequence generated from (2, 1)\n# Check the sequence againt its RLE.\n# Show, on this page, the first 30 members of the Kolakoski sequence generated from (1, 3, 1, 2)\n# Check the sequence againt its RLE.\n# Show, on this page, the first 30 members of the Kolakoski sequence generated from (1, 3, 2, 1)\n# Check the sequence againt its RLE.\n(There are rules on generating Kolakoski sequences from this method that are broken by the last example)\n\n", "solution": "import itertools\n\ndef cycler(start_items):\n\treturn itertools.cycle(start_items).__next__\n\ndef _kolakoski_gen(start_items):\n    s, k = [], 0\n    c = cycler(start_items)\n    while True:\n        c_next = c()\n        s.append(c_next)\n        sk = s[k]\n        yield sk\n        if sk > 1:\n            s += [c_next] * (sk - 1)\n        k += 1\n\ndef kolakoski(start_items=(1, 2), length=20):\n    return list(itertools.islice(_kolakoski_gen(start_items), length))\n\ndef _run_len_encoding(truncated_series):\n    return [len(list(group)) for grouper, group in itertools.groupby(truncated_series)][:-1]\n\ndef is_series_eq_its_rle(series):\n    rle = _run_len_encoding(series)\n    return (series[:len(rle)] == rle) if rle else not series\n\nif __name__ == '__main__':\n    for start_items, length in [((1, 2), 20), ((2, 1), 20), \n                                ((1, 3, 1, 2), 30), ((1, 3, 2, 1), 30)]:\n        print(f'\\n## {length} members of the series generated from {start_items} is:')\n        s = kolakoski(start_items, length)\n        print(f'  {s}')\n        ans = 'YES' if is_series_eq_its_rle(s) else 'NO'\n        print(f'  Does it look like a Kolakoski sequence: {ans}')"}
{"title": "Kosaraju", "language": "Python", "task": "{{wikipedia|Graph}}\n\n\nKosaraju's algorithm (also known as the Kosaraju-Sharir algorithm) is a linear time algorithm to find the strongly connected components of a directed graph. Aho, Hopcroft and Ullman credit it to an unpublished paper from 1978 by S. Rao Kosaraju. The same algorithm was independently discovered by Micha Sharir and published by him in 1981. It makes use of the fact that the transpose graph (the same graph with the direction of every edge reversed) has exactly the same strongly connected components as the original graph.\n\nFor this task consider the directed graph with these connections:\n  0 -> 1\n  1 -> 2\n  2 -> 0\n  3 -> 1,  3 -> 2,  3 -> 4\n  4 -> 3,  4 -> 5\n  5 -> 2,  5 -> 6\n  6 -> 5\n  7 -> 4, 7  -> 6, 7 -> 7\n\nAnd report the kosaraju strongly connected component for each node.\n\n;References:\n* The article on Wikipedia.\n\n", "solution": "def kosaraju(g):\n    class nonlocal: pass\n\n    # 1. For each vertex u of the graph, mark u as unvisited. Let l be empty.\n    size = len(g)\n\n    vis = [False]*size # vertexes that have been visited\n    l = [0]*size\n    nonlocal.x = size\n    t = [[]]*size   # transpose graph\n\n    def visit(u):\n        if not vis[u]:\n            vis[u] = True\n            for v in g[u]:\n                visit(v)\n                t[v] = t[v] + [u]\n            nonlocal.x = nonlocal.x - 1\n            l[nonlocal.x] = u\n\n    # 2. For each vertex u of the graph do visit(u)\n    for u in range(len(g)):\n        visit(u)\n    c = [0]*size\n\n    def assign(u, root):\n        if vis[u]:\n            vis[u] = False\n            c[u] = root\n            for v in t[u]:\n                assign(v, root)\n\n    # 3: For each element u of l in order, do assign(u, u)\n    for u in l:\n        assign(u, u)\n\n    return c\n\ng = [[1], [2], [0], [1,2,4], [3,5], [2,6], [5], [4,6,7]]\nprint kosaraju(g)"}
{"title": "Lah numbers", "language": "Python", "task": "Lah numbers, sometimes referred to as ''Stirling numbers of the third kind'', are coefficients of polynomial expansions expressing rising factorials in terms of falling factorials.\n\nUnsigned Lah numbers count the number of ways a set of '''n''' elements can be partitioned into '''k''' non-empty linearly ordered subsets.\n\nLah numbers are closely related to Stirling numbers of the first & second kinds, and may be derived from them.\n\nLah numbers obey the identities and relations:\n   L(n, 0), L(0, k) = 0   # for n, k > 0\n   L(n, n) = 1\n   L(n, 1) = n!\n   L(n, k) =           ( n! * (n - 1)! ) / ( k! * (k - 1)! ) / (n - k)!      # For unsigned Lah numbers\n      ''or''\n   L(n, k) = (-1)**n * ( n! * (n - 1)! ) / ( k! * (k - 1)! ) / (n - k)!      # For   signed Lah numbers\n\n;Task:\n:* Write a routine (function, procedure, whatever) to find '''unsigned Lah numbers'''. There are several methods to generate unsigned Lah numbers. You are free to choose the most appropriate for your language. If your language has a built-in, or easily, publicly available library implementation, it is acceptable to use that.\n\n:* Using the routine, generate and show here, on this page, a table (or triangle) showing the unsigned Lah numbers, '''L(n, k)''', up to '''L(12, 12)'''. it is optional to show the row / column for n == 0 and k == 0. It is optional to show places where L(n, k) == 0 (when k > n).\n\n:* If your language supports large integers, find and show here, on this page, the maximum value of '''L(n, k)''' where '''n == 100'''.\n\n\n;See also:\n\n:* '''Wikipedia - Lah number'''\n:* '''OEIS:A105278 - Unsigned Lah numbers'''\n:* '''OEIS:A008297 - Signed Lah numbers'''\n\n\n;Related Tasks:\n\n:* '''Stirling numbers of the first kind'''\n:* '''Stirling numbers of the second kind'''\n:* '''Bell numbers'''\n\n\n", "solution": "from math import (comb,\n                  factorial)\n\n\ndef lah(n, k):\n    if k == 1:\n        return factorial(n)\n    if k == n:\n        return 1\n    if k > n:\n        return 0\n    if k < 1 or n < 1:\n        return 0\n    return comb(n, k) * factorial(n - 1) // factorial(k - 1)\n\n\ndef main():\n    print(\"Unsigned Lah numbers: L(n, k):\")\n    print(\"n/k \", end='\\t')\n    for i in range(13):\n        print(\"%11d\" % i, end='\\t')\n    print()\n    for row in range(13):\n        print(\"%-4d\" % row, end='\\t')\n        for i in range(row + 1):\n            l = lah(row, i)\n            print(\"%11d\" % l, end='\\t')\n        print()\n    print(\"\\nMaximum value from the L(100, *) row:\")\n    max_val = max(lah(100, a) for a in range(100))\n    print(max_val)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Largest int from concatenated ints", "language": "Python", "task": "Given a set of positive integers, write a function to order the integers in such a way that the concatenation of the numbers forms the largest possible integer and return this integer.\n\nUse the following two sets of integers as tests   and   show your program output here.\n\n:::::*   {1, 34, 3, 98, 9, 76, 45, 4}\n:::::*   {54, 546, 548, 60}\n\n\n;Possible algorithms:\n# A solution could be found by trying all combinations and return the best. \n# Another way to solve this is to note that in the best arrangement, for any two adjacent original integers '''X''' and '''Y''', the concatenation '''X''' followed by '''Y''' will be numerically greater than or equal to the concatenation '''Y''' followed by '''X.\n# Yet another way to solve this is to pad the integers to the same size by repeating the digits then sort using these repeated integers as a sort key.\n\n\n;See also:\n*   Algorithms: What is the most efficient way to arrange the given numbers to form the biggest number?\n*   Constructing the largest number possible by rearranging a list\n\n", "solution": "try:\n    cmp     # Python 2 OK or NameError in Python 3\n    def maxnum(x):\n        return ''.join(sorted((str(n) for n in x),\n                              cmp=lambda x,y:cmp(y+x, x+y)))\nexcept NameError:\n    # Python 3\n    from functools import cmp_to_key\n    def cmp(x, y):\n        return -1 if x<y else ( 0 if x==y else 1)\n    def maxnum(x):\n        return ''.join(sorted((str(n) for n in x),\n                              key=cmp_to_key(lambda x,y:cmp(y+x, x+y))))\n\nfor numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:\n    print('Numbers: %r\\n  Largest integer: %15s' % (numbers, maxnum(numbers)))"}
{"title": "Largest int from concatenated ints", "language": "Python 2.6+", "task": "Given a set of positive integers, write a function to order the integers in such a way that the concatenation of the numbers forms the largest possible integer and return this integer.\n\nUse the following two sets of integers as tests   and   show your program output here.\n\n:::::*   {1, 34, 3, 98, 9, 76, 45, 4}\n:::::*   {54, 546, 548, 60}\n\n\n;Possible algorithms:\n# A solution could be found by trying all combinations and return the best. \n# Another way to solve this is to note that in the best arrangement, for any two adjacent original integers '''X''' and '''Y''', the concatenation '''X''' followed by '''Y''' will be numerically greater than or equal to the concatenation '''Y''' followed by '''X.\n# Yet another way to solve this is to pad the integers to the same size by repeating the digits then sort using these repeated integers as a sort key.\n\n\n;See also:\n*   Algorithms: What is the most efficient way to arrange the given numbers to form the biggest number?\n*   Constructing the largest number possible by rearranging a list\n\n", "solution": "from fractions import Fraction\nfrom math import log10\n\ndef maxnum(x):\n    return ''.join(str(n) for n in sorted(x, reverse=True,\n                          key=lambda i: Fraction(i, 10**(int(log10(i))+1)-1)))\n\nfor numbers in [(1, 34, 3, 98, 9, 76, 45, 4), (54, 546, 548, 60)]:\n    print('Numbers: %r\\n  Largest integer: %15s' % (numbers, maxnum(numbers)))"}
{"title": "Largest number divisible by its digits", "language": "Python", "task": "Find the largest base 10 integer whose digits are all different,   and   is evenly divisible by each of its individual digits.\n\n\nThese numbers are also known as   '''Lynch-Bell numbers''',   numbers   '''n'''   such that the\n(base ten) digits are all different (and do not include zero)   and   '''n'''   is divisible by each of its individual digits. \n\n\n;Example:\n'''135'''   is evenly divisible by   '''1''',   '''3''',   and   '''5'''.\n\n\nNote that the digit zero (0) can not be in the number as integer division by zero is undefined. \n\nThe digits must all be unique so a base ten number will have at most '''9''' digits.\n\nFeel free to use analytics and clever algorithms to reduce the search space your example needs to visit, but it must do an actual search. (Don't just feed it the answer and verify it is correct.)\n\n\n;Stretch goal:\nDo the same thing for hexadecimal.\n\n\n;Related tasks:\n:*   gapful numbers.\n:*   palindromic gapful numbers. \n\n\n;Also see:\n:*   The OEIS sequence:   A115569: Lynch-Bell numbers. \n\n", "solution": "===base 10===\nUsing the insights presented in the preamble to the Raku (base 10) example:\n"}
{"title": "Largest number divisible by its digits", "language": "Python 3.7", "task": "Find the largest base 10 integer whose digits are all different,   and   is evenly divisible by each of its individual digits.\n\n\nThese numbers are also known as   '''Lynch-Bell numbers''',   numbers   '''n'''   such that the\n(base ten) digits are all different (and do not include zero)   and   '''n'''   is divisible by each of its individual digits. \n\n\n;Example:\n'''135'''   is evenly divisible by   '''1''',   '''3''',   and   '''5'''.\n\n\nNote that the digit zero (0) can not be in the number as integer division by zero is undefined. \n\nThe digits must all be unique so a base ten number will have at most '''9''' digits.\n\nFeel free to use analytics and clever algorithms to reduce the search space your example needs to visit, but it must do an actual search. (Don't just feed it the answer and verify it is correct.)\n\n\n;Stretch goal:\nDo the same thing for hexadecimal.\n\n\n;Related tasks:\n:*   gapful numbers.\n:*   palindromic gapful numbers. \n\n\n;Also see:\n:*   The OEIS sequence:   A115569: Lynch-Bell numbers. \n\n", "solution": "'''Largest number divisible by its hex digits'''\n\nfrom functools import (reduce)\nfrom math import (gcd)\n\n\n# main :: IO ()\ndef main():\n    '''First integer evenly divisible by each of its\n       hex digits, none of which appear more than once.\n    '''\n\n    # Least common multiple of digits [1..15]\n    # ( -> 360360 )\n    lcmDigits = foldl1(lcm)(\n        enumFromTo(1)(15)\n    )\n    allDigits = 0xfedcba987654321\n\n    # ( -> 1147797409030632360 )\n    upperLimit = allDigits - (allDigits % lcmDigits)\n\n    # Possible numbers\n    xs = enumFromThenToNext(upperLimit)(\n        upperLimit - lcmDigits\n    )(1)\n\n    print(\n        hex(\n            until(lambda x: 15 == len(set(showHex(x))))(\n                lambda _: next(xs)\n            )(next(xs))\n        )\n    )   # --> 0xfedcb59726a1348\n\n\n# ------------------ GENERIC FUNCTIONS -------------------\n\n# enumFromThenToNext :: Int -> Int -> Int -> Gen [Int]\ndef enumFromThenToNext(m):\n    '''Non-finite series of integer values enumerated\n       from m to n with a step size defined by nxt-m.\n    '''\n    def go(m, nxt):\n        d = nxt - m\n        v = m\n        while True:\n            yield v\n            v = d + v\n    return lambda nxt: lambda n: go(m, nxt)\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: range(m, 1 + n)\n\n\n# foldl1 :: (a -> a -> a) -> [a] -> a\ndef foldl1(f):\n    '''Left to right reduction of the\n       non-empty list xs, using the binary\n       operator f, with the head of xs\n       as the initial acccumulator value.\n    '''\n    return lambda xs: reduce(\n        lambda a, x: f(a)(x), xs[1:], xs[0]\n    ) if xs else None\n\n\n# lcm :: Int -> Int -> Int\ndef lcm(x):\n    '''The smallest positive integer divisible\n       without remainder by both x and y.\n    '''\n    return lambda y: (\n        0 if (0 == x or 0 == y) else abs(\n            y * (x // gcd(x, y))\n        )\n    )\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f, x):\n        v = x\n        while not p(v):\n            v = f(v)\n        return v\n    return lambda f: lambda x: go(f, x)\n\n\n# showHex :: Int -> String\ndef showHex(n):\n    '''Hexadecimal string representation\n       of an integer value.\n    '''\n    return hex(n)[2:]\n\n\n# MAIN --\nif __name__ == '__main__':\n    main()"}
{"title": "Largest proper divisor of n", "language": "Python", "task": "a(1) = 1; for n > 1, a(n) = '''largest''' proper divisor of n, where '''n < 101 '''.\n\n", "solution": "def lpd(n):\n    for i in range(n-1,0,-1):\n        if n%i==0: return i\n    return 1\n\nfor i in range(1,101):\n    print(\"{:3}\".format(lpd(i)), end=i%10==0 and '\\n' or '')"}
{"title": "Last Friday of each month", "language": "Python", "task": "Write a program or a script that returns the date of the last Fridays of each month of a given year. \n\nThe year may be given through any simple input method in your language (command line, std in, etc).\n\n\nExample of an expected output:\n./last_fridays 2012\n2012-01-27\n2012-02-24\n2012-03-30\n2012-04-27\n2012-05-25\n2012-06-29\n2012-07-27\n2012-08-31\n2012-09-28\n2012-10-26\n2012-11-30\n2012-12-28\n\n\n;Related tasks\n* [[Five weekends]]\n* [[Day of the week]]\n* [[Find the last Sunday of each month]]\n\n", "solution": "import calendar\n\ndef last_fridays(year):\n    for month in range(1, 13):\n        last_friday = max(week[calendar.FRIDAY]\n            for week in calendar.monthcalendar(year, month))\n        print('{:4d}-{:02d}-{:02d}'.format(year, month, last_friday))"}
{"title": "Last letter-first letter", "language": "Python", "task": "A certain children's game involves starting with a word in a particular category.   Each participant in turn says a word, but that word must begin with the final letter of the previous word.   Once a word has been given, it cannot be repeated.   If an opponent cannot give a word in the category, they fall out of the game. \n\n\nFor example, with   \"animals\"   as the category,\n\nChild 1: dog \nChild 2: goldfish\nChild 1: hippopotamus\nChild 2: snake\n...\n\n\n\n;Task:\nTake the following selection of 70 English Pokemon names   (extracted from   Wikipedia's list of Pokemon)   and generate the/a sequence with the highest possible number of Pokemon names where the subsequent name starts with the final letter of the preceding name. \n\nNo Pokemon name is to be repeated.\n\n\naudino bagon baltoy banette bidoof braviary bronzor carracosta charmeleon\ncresselia croagunk darmanitan deino emboar emolga exeggcute gabite\ngirafarig gulpin haxorus heatmor heatran ivysaur jellicent jumpluff kangaskhan\nkricketune landorus ledyba loudred lumineon lunatone machamp magnezone mamoswine\nnosepass petilil pidgeotto pikachu pinsir poliwrath poochyena porygon2\nporygonz registeel relicanth remoraid rufflet sableye scolipede scrafty seaking\nsealeo silcoon simisear snivy snorlax spoink starly tirtouga trapinch treecko\ntyrogue vigoroth vulpix wailord wartortle whismur wingull yamask\n\n\n\nExtra brownie points for dealing with the full list of   646   names.\n\n", "solution": "import psyco\n\nnsolutions = 0\n\ndef search(sequences, ord_minc, curr_word, current_path,\n           current_path_len, longest_path):\n    global nsolutions\n\n    current_path[current_path_len] = curr_word\n    current_path_len += 1\n\n    if current_path_len == len(longest_path):\n        nsolutions += 1\n    elif current_path_len > len(longest_path):\n        nsolutions = 1\n        longest_path[:] = current_path[:current_path_len]\n\n    # recursive search\n    last_char_index = ord(curr_word[-1]) - ord_minc\n    if last_char_index >= 0 and last_char_index < len(sequences):\n        for pair in sequences[last_char_index]:\n            if not pair[1]:\n                pair[1] = True\n                search(sequences, ord_minc, pair[0], current_path,\n                       current_path_len, longest_path)\n                pair[1] = False\n\n\ndef find_longest_chain(words):\n    ord_minc = ord(min(word[0] for word in words))\n    ord_maxc = ord(max(word[0] for word in words))\n    sequences = [[] for _ in xrange(ord_maxc - ord_minc + 1)]\n    for word in words:\n        sequences[ord(word[0]) - ord_minc].append([word, False])\n\n    current_path = [None] * len(words)\n    longest_path = []\n\n    # try each item as possible start\n    for seq in sequences:\n        for pair in seq:\n            pair[1] = True\n            search(sequences, ord_minc, pair[0],\n                   current_path, 0, longest_path)\n            pair[1] = False\n\n    return longest_path\n\n\ndef main():\n    global nsolutions\n\n    pokemon = \"\"\"audino bagon baltoy banette bidoof braviary\nbronzor carracosta charmeleon cresselia croagunk darmanitan deino\nemboar emolga exeggcute gabite girafarig gulpin haxorus heatmor\nheatran ivysaur jellicent jumpluff kangaskhan kricketune landorus\nledyba loudred lumineon lunatone machamp magnezone mamoswine nosepass\npetilil pidgeotto pikachu pinsir poliwrath poochyena porygon2\nporygonz registeel relicanth remoraid rufflet sableye scolipede\nscrafty seaking sealeo silcoon simisear snivy snorlax spoink starly\ntirtouga trapinch treecko tyrogue vigoroth vulpix wailord wartortle\nwhismur wingull yamask\"\"\".lower().split()\n\n    # remove duplicates\n    pokemon = sorted(set(pokemon))\n\n    sol = find_longest_chain(pokemon)\n    print \"Maximum path length:\", len(sol)\n    print \"Paths of that length:\", nsolutions\n    print \"Example path of that length:\"\n    for i in xrange(0, len(sol), 7):\n        print \" \", \" \".join(sol[i : i+7])\n\npsyco.full()\nmain()"}
{"title": "Latin Squares in reduced form", "language": "Python from D", "task": "A Latin Square is in its reduced form if the first row and first column contain items in their natural order. The order n is the number of items. For any given n there is a set of reduced Latin Squares whose size increases rapidly with n. g is a number which identifies a unique element within the set of reduced Latin Squares of order n. The objective of this task is to construct the set of all Latin Squares of a given order and to provide a means which given suitable values for g any element within the set may be obtained.\n\nFor a reduced Latin Square the first row is always 1 to n. The second row is all [[Permutations/Derangements]] of 1 to n starting with 2. The third row is all [[Permutations/Derangements]] of 1 to n starting with 3 which do not clash (do not have the same item in any column) with row 2. The fourth row is all [[Permutations/Derangements]] of 1 to n starting with 4 which do not clash with rows 2 or 3. Likewise continuing to the nth row.\n\nDemonstrate by:\n* displaying the four reduced Latin Squares of order 4.\n* for n = 1 to 6 (or more) produce the set of reduced Latin Squares; produce a table which shows the size of the set of reduced Latin Squares and compares this value times n! times (n-1)! with the values in OEIS A002860.\n\n", "solution": "def dList(n, start):\n    start -= 1 # use 0 basing\n    a = range(n)\n    a[start] = a[0]\n    a[0] = start\n    a[1:] = sorted(a[1:])\n    first = a[1]\n    # rescursive closure permutes a[1:]\n    r = []\n    def recurse(last):\n        if (last == first):\n            # bottom of recursion. you get here once for each permutation.\n            # test if permutation is deranged.\n            # yes, save a copy with 1 based indexing\n            for j,v in enumerate(a[1:]):\n                if j + 1 == v:\n                    return # no, ignore it\n            b = [x + 1 for x in a]\n            r.append(b)\n            return\n        for i in xrange(last, 0, -1):\n            a[i], a[last] = a[last], a[i]\n            recurse(last - 1)\n            a[i], a[last] = a[last], a[i]\n    recurse(n - 1)\n    return r\n\ndef printSquare(latin,n):\n    for row in latin:\n        print row\n    print\n\ndef reducedLatinSquares(n,echo):\n    if n <= 0:\n        if echo:\n            print []\n        return 0\n    elif n == 1:\n        if echo:\n            print [1]\n        return 1\n\n    rlatin = [None] * n\n    for i in xrange(n):\n        rlatin[i] = [None] * n\n    # first row\n    for j in xrange(0, n):\n        rlatin[0][j] = j + 1\n\n    class OuterScope:\n        count = 0\n    def recurse(i):\n        rows = dList(n, i)\n\n        for r in xrange(len(rows)):\n            rlatin[i - 1] = rows[r]\n            justContinue = False\n            k = 0\n            while not justContinue and k < i - 1:\n                for j in xrange(1, n):\n                    if rlatin[k][j] == rlatin[i - 1][j]:\n                        if r < len(rows) - 1:\n                            justContinue = True\n                            break\n                        if i > 2:\n                            return\n                k += 1\n            if not justContinue:\n                if i < n:\n                    recurse(i + 1)\n                else:\n                    OuterScope.count += 1\n                    if echo:\n                        printSquare(rlatin, n)\n\n    # remaining rows\n    recurse(2)\n    return OuterScope.count\n\ndef factorial(n):\n    if n == 0:\n        return 1\n    prod = 1\n    for i in xrange(2, n + 1):\n        prod *= i\n    return prod\n\nprint \"The four reduced latin squares of order 4 are:\\n\"\nreducedLatinSquares(4,True)\n\nprint \"The size of the set of reduced latin squares for the following orders\"\nprint \"and hence the total number of latin squares of these orders are:\\n\"\nfor n in xrange(1, 7):\n    size = reducedLatinSquares(n, False)\n    f = factorial(n - 1)\n    f *= f * n * size\n    print \"Order %d: Size %-4d x %d! x %d! => Total %d\" % (n, size, n, n - 1, f)"}
{"title": "Law of cosines - triples", "language": "Python", "task": "The Law of cosines states that for an angle g, (gamma) of any triangle, if the sides adjacent to the angle are A and B and the side opposite is C; then the lengths of the sides are related by this formula:\n           A2 + B2 - 2ABcos(g) = C2 \n\n;Specific angles:\nFor an angle of of   '''90o'''   this becomes the more familiar \"Pythagoras equation\":\n           A2 + B2  =  C2 \n\nFor an angle of   '''60o'''   this becomes the less familiar  equation:\n           A2 + B2 - AB  =  C2 \n\nAnd finally for an angle of   '''120o'''   this becomes the equation:\n           A2 + B2 + AB  =  C2 \n\n\n;Task:\n*   Find all integer solutions (in order) to the three specific cases, distinguishing between each angle being considered.\n*   Restrain all sides to the integers   '''1..13'''   inclusive.\n*   Show how many results there are for each of the three angles mentioned above.\n*   Display results on this page.\n\n\nNote: Triangles with the same length sides but different order are to be treated as the same. \n\n;Optional Extra credit:\n* How many 60deg integer triples are there for sides in the range 1..10_000 ''where the sides are not all of the same length''.\n\n\n;Related Task\n* [[Pythagorean triples]]\n\n\n;See also:\n* Visualising Pythagoras: ultimate proofs and crazy contortions Mathlogger Video\n\n", "solution": "N = 13\n\ndef method1(N=N):\n    squares = [x**2 for x in range(0, N+1)]\n    sqrset = set(squares)\n    tri90, tri60, tri120 = (set() for _ in range(3))\n    for a in range(1, N+1):\n        a2 = squares[a]\n        for b in range(1, a + 1):\n            b2 = squares[b]\n            c2 = a2 + b2\n            if c2 in sqrset:\n                tri90.add(tuple(sorted((a, b, int(c2**0.5)))))\n            ab = a * b\n            c2 -= ab\n            if c2 in sqrset:\n                tri60.add(tuple(sorted((a, b, int(c2**0.5)))))\n            c2 += 2 * ab\n            if c2 in sqrset:\n                tri120.add(tuple(sorted((a, b, int(c2**0.5)))))\n    return  sorted(tri90), sorted(tri60), sorted(tri120)\n#%%\nif __name__ == '__main__':\n    print(f'Integer triangular triples for sides 1..{N}:')\n    for angle, triples in zip([90, 60, 120], method1(N)):\n        print(f'  {angle:3}\u00b0 has {len(triples)} solutions:\\n    {triples}')\n    _, t60, _ = method1(10_000)\n    notsame = sum(1 for a, b, c in t60 if a != b or b != c)\n    print('Extra credit:', notsame)"}
{"title": "Least common multiple", "language": "Python", "task": "Compute the   least common multiple   (LCM)   of two integers.\n\nGiven   ''m''   and   ''n'',   the least common multiple is the smallest positive integer that has both   ''m''   and   ''n''   as factors. \n\n\n;Example:\nThe least common multiple of   '''12'''   and   '''18'''   is   '''36''',       because:\n:*   '''12'''   is a factor     ('''12''' x '''3''' = '''36'''),     and \n:*   '''18'''   is a factor     ('''18''' x '''2''' = '''36'''),     and \n:*   there is no positive integer less than   '''36'''   that has both factors. \n\n\nAs a special case,   if either   ''m''   or   ''n''   is zero,   then the least common multiple is zero.\n\n\nOne way to calculate the least common multiple is to iterate all the multiples of   ''m'',   until you find one that is also a multiple of   ''n''.\n\nIf you already have   ''gcd''   for [[greatest common divisor]],   then this formula calculates   ''lcm''.\n\n\n:::: \\operatorname{lcm}(m, n) = \\frac{|m \\times n|}{\\operatorname{gcd}(m, n)}\n\n\nOne can also find   ''lcm''   by merging the [[prime decomposition]]s of both   ''m''   and   ''n''.\n\n\n;Related task\n:*   greatest common divisor.\n\n\n;See also:\n*   MathWorld entry:   Least Common Multiple.\n*   Wikipedia entry:   Least common multiple.\n\n", "solution": "'''Least common multiple'''\n\nfrom inspect import signature\n\n\n# lcm :: Int -> Int -> Int\ndef lcm(x):\n    '''The smallest positive integer divisible\n       without remainder by both x and y.\n    '''\n    return lambda y: 0 if 0 in (x, y) else abs(\n        y * (x // gcd_(x)(y))\n    )\n\n\n# gcd_ :: Int -> Int -> Int\ndef gcd_(x):\n    '''The greatest common divisor in terms of\n       the divisibility preordering.\n    '''\n    def go(a, b):\n        return go(b, a % b) if 0 != b else a\n    return lambda y: go(abs(x), abs(y))\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Tests'''\n\n    print(\n        fTable(\n            __doc__ + 's of 60 and [12..20]:'\n        )(repr)(repr)(\n            lcm(60)\n        )(enumFromTo(12)(20))\n    )\n\n    pairs = [(0, 2), (2, 0), (-6, 14), (12, 18)]\n    print(\n        fTable(\n            '\\n\\n' + __doc__ + 's of ' + repr(pairs) + ':'\n        )(repr)(repr)(\n            uncurry(lcm)\n        )(pairs)\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    '''A function over a tuple, derived from\n       a vanilla or curried function.\n    '''\n    if 1 < len(signature(f).parameters):\n        return lambda xy: f(*xy)\n    else:\n        return lambda xy: f(xy[0])(xy[1])\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.\n    '''\n    return '\\n'.join(xs)\n\n\n# FORMATTING ----------------------------------------------\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Least common multiple", "language": "Python from Tcl", "task": "Compute the   least common multiple   (LCM)   of two integers.\n\nGiven   ''m''   and   ''n'',   the least common multiple is the smallest positive integer that has both   ''m''   and   ''n''   as factors. \n\n\n;Example:\nThe least common multiple of   '''12'''   and   '''18'''   is   '''36''',       because:\n:*   '''12'''   is a factor     ('''12''' x '''3''' = '''36'''),     and \n:*   '''18'''   is a factor     ('''18''' x '''2''' = '''36'''),     and \n:*   there is no positive integer less than   '''36'''   that has both factors. \n\n\nAs a special case,   if either   ''m''   or   ''n''   is zero,   then the least common multiple is zero.\n\n\nOne way to calculate the least common multiple is to iterate all the multiples of   ''m'',   until you find one that is also a multiple of   ''n''.\n\nIf you already have   ''gcd''   for [[greatest common divisor]],   then this formula calculates   ''lcm''.\n\n\n:::: \\operatorname{lcm}(m, n) = \\frac{|m \\times n|}{\\operatorname{gcd}(m, n)}\n\n\nOne can also find   ''lcm''   by merging the [[prime decomposition]]s of both   ''m''   and   ''n''.\n\n\n;Related task\n:*   greatest common divisor.\n\n\n;See also:\n*   MathWorld entry:   Least Common Multiple.\n*   Wikipedia entry:   Least common multiple.\n\n", "solution": ">>> def lcm(p,q):\n\tp, q = abs(p), abs(q)\n\tm = p * q\n\tif not m: return 0\n\twhile True:\n\t\tp %= q\n\t\tif not p: return m // q\n\t\tq %= p\n\t\tif not q: return m // p\n\n\t\t\n>>> lcm(-6, 14)\n42\n>>> lcm(12, 18)\n36\n>>> lcm(2, 0)\n0\n>>> "}
{"title": "Left factorials", "language": "Python", "task": "'''Left factorials''',   !n,   may refer to either   ''subfactorials''   or to   ''factorial sums''; \nthe same notation can be confusingly seen being used for the two different definitions.\n\nSometimes,   ''subfactorials''   (also known as ''derangements'')   may use any of the notations: \n:::::::*     !''n''`   \n:::::::*     !''n''  \n:::::::*     ''n''!  \n\n\n(It may not be visually obvious, but the last example uses an upside-down exclamation mark.)\n\n\nThis Rosetta Code task will be using this formula   (''factorial sums'')   for   '''left factorial''':\n\n:::::    !n = \\sum_{k=0}^{n-1} k! \n\n:::: where\n\n:::::   !0 = 0\n\n\n\n;Task\nDisplay the left factorials for:\n*   zero      through        ten     (inclusive)\n*   20   through   110          (inclusive)   by tens\n\n\nDisplay the length (in decimal digits) of the left factorials for:\n*   1,000   through   10,000   (inclusive), by thousands.\n\n\n;Also see:\n*   The OEIS entry: A003422 left factorials\n*   The MathWorld entry: left factorial\n*   The MathWorld entry: factorial sums\n*   The MathWorld entry: subfactorial\n\n\n;Related task:\n*   permutations/derangements (subfactorials)\n\n", "solution": "from itertools import islice\n\ndef lfact():\n    yield 0\n    fact, summ, n = 1, 0, 1 \n    while 1:\n        fact, summ, n = fact*n, summ + fact, n + 1\n        yield summ\n\nprint('first 11:\\n  %r' % [lf for i, lf in zip(range(11), lfact())])\nprint('20 through 110 (inclusive) by tens:')\nfor lf in islice(lfact(), 20, 111, 10):\n    print(lf)\nprint('Digits in 1,000 through 10,000 (inclusive) by thousands:\\n  %r' \n      % [len(str(lf)) for lf in islice(lfact(), 1000, 10001, 1000)] )"}
{"title": "Left factorials", "language": "Python from Haskell", "task": "'''Left factorials''',   !n,   may refer to either   ''subfactorials''   or to   ''factorial sums''; \nthe same notation can be confusingly seen being used for the two different definitions.\n\nSometimes,   ''subfactorials''   (also known as ''derangements'')   may use any of the notations: \n:::::::*     !''n''`   \n:::::::*     !''n''  \n:::::::*     ''n''!  \n\n\n(It may not be visually obvious, but the last example uses an upside-down exclamation mark.)\n\n\nThis Rosetta Code task will be using this formula   (''factorial sums'')   for   '''left factorial''':\n\n:::::    !n = \\sum_{k=0}^{n-1} k! \n\n:::: where\n\n:::::   !0 = 0\n\n\n\n;Task\nDisplay the left factorials for:\n*   zero      through        ten     (inclusive)\n*   20   through   110          (inclusive)   by tens\n\n\nDisplay the length (in decimal digits) of the left factorials for:\n*   1,000   through   10,000   (inclusive), by thousands.\n\n\n;Also see:\n*   The OEIS entry: A003422 left factorials\n*   The MathWorld entry: left factorial\n*   The MathWorld entry: factorial sums\n*   The MathWorld entry: subfactorial\n\n\n;Related task:\n*   permutations/derangements (subfactorials)\n\n", "solution": "'''Left factorials'''\n\nfrom itertools import (accumulate, chain, count, islice)\nfrom operator import (mul, add)\n\n\n# leftFact :: [Integer]\ndef leftFact():\n    '''Left factorial series defined in terms\n       of the factorial series.\n    '''\n    return accumulate(\n        chain([0], fact()), add\n    )\n\n\n# fact :: [Integer]\ndef fact():\n    '''The factorial series.\n    '''\n    return accumulate(\n        chain([1], count(1)), mul\n    )\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Tests'''\n    print(\n        'Terms 0 thru 10 inclusive:\\n  %r'\n        % take(11)(leftFact())\n    )\n\n    print('\\nTerms 20 thru 110 (inclusive) by tens:')\n    for x in takeFromThenTo(20)(30)(110)(leftFact()):\n        print(x)\n\n    print(\n        '\\n\\nDigit counts for terms 1k through 10k (inclusive) by k:\\n  %r'\n        % list(map(\n            compose(len)(str),\n            takeFromThenTo(1000)(2000)(10000)(\n                leftFact()\n            )\n        ))\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# scanl :: (b -> a -> b) -> b -> [a] -> [b]\ndef scanl(f):\n    '''scanl is like reduce, but defines a succession of\n       intermediate values, building from the left.\n    '''\n    def go(a):\n        def g(xs):\n            return accumulate(chain([a], xs), f)\n        return g\n    return go\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# takeFromThenTo :: Int -> Int -> Int -> [a] -> [a]\ndef takeFromThenTo(a):\n    '''Values drawn from a series betweens positions a and b\n       at intervals of size z'''\n    return lambda b: lambda z: lambda xs: islice(\n        xs, a, 1 + z, b - a\n    )\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Levenshtein distance", "language": "Python", "task": "{{Wikipedia}}\n\nIn information theory and computer science, the '''Levenshtein distance''' is a edit distance). The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character. \n\n\n;Example:\nThe Levenshtein distance between \"'''kitten'''\" and \"'''sitting'''\" is 3, since the following three edits change one into the other, and there isn't a way to do it with fewer than three edits:\n::#   '''k'''itten    '''s'''itten       (substitution of 'k' with 's')\n::#   sitt'''e'''n    sitt'''i'''n       (substitution of 'e' with 'i')\n::#   sittin          sittin'''g'''      (insert 'g' at the end).\n\n\n''The Levenshtein distance between   \"'''rosettacode'''\",   \"'''raisethysword'''\"   is   '''8'''.\n\n''The distance between two strings is same as that when both strings are reversed.''\n\n\n;Task:\nImplements a Levenshtein distance function, or uses a library function, to show the Levenshtein distance between   \"kitten\"   and   \"sitting\". \n\n\n;Related task:\n*   [[Longest common subsequence]]\n\n\n\n", "solution": "def levenshteinDistance(str1, str2):\n    m = len(str1)\n    n = len(str2)\n    d = [[i] for i in range(1, m + 1)]   # d matrix rows\n    d.insert(0, list(range(0, n + 1)))   # d matrix columns\n    for j in range(1, n + 1):\n        for i in range(1, m + 1):\n            if str1[i - 1] == str2[j - 1]:   # Python (string) is 0-based\n                substitutionCost = 0\n            else:\n                substitutionCost = 1\n            d[i].insert(j, min(d[i - 1][j] + 1,\n                               d[i][j - 1] + 1,\n                               d[i - 1][j - 1] + substitutionCost))\n    return d[-1][-1]\n\nprint(levenshteinDistance(\"kitten\",\"sitting\"))\nprint(levenshteinDistance(\"rosettacode\",\"raisethysword\"))"}
{"title": "Levenshtein distance", "language": "Python 3.7", "task": "{{Wikipedia}}\n\nIn information theory and computer science, the '''Levenshtein distance''' is a edit distance). The Levenshtein distance between two strings is defined as the minimum number of edits needed to transform one string into the other, with the allowable edit operations being insertion, deletion, or substitution of a single character. \n\n\n;Example:\nThe Levenshtein distance between \"'''kitten'''\" and \"'''sitting'''\" is 3, since the following three edits change one into the other, and there isn't a way to do it with fewer than three edits:\n::#   '''k'''itten    '''s'''itten       (substitution of 'k' with 's')\n::#   sitt'''e'''n    sitt'''i'''n       (substitution of 'e' with 'i')\n::#   sittin          sittin'''g'''      (insert 'g' at the end).\n\n\n''The Levenshtein distance between   \"'''rosettacode'''\",   \"'''raisethysword'''\"   is   '''8'''.\n\n''The distance between two strings is same as that when both strings are reversed.''\n\n\n;Task:\nImplements a Levenshtein distance function, or uses a library function, to show the Levenshtein distance between   \"kitten\"   and   \"sitting\". \n\n\n;Related task:\n*   [[Longest common subsequence]]\n\n\n\n", "solution": "'''Levenshtein distance'''\n\nfrom itertools import (accumulate, chain, islice)\nfrom functools import reduce\n\n\n# levenshtein :: String -> String -> Int\ndef levenshtein(sa):\n    '''Levenshtein distance between\n       two strings.'''\n    s1 = list(sa)\n\n    # go :: [Int] -> Char -> [Int]\n    def go(ns, c):\n        n, ns1 = ns[0], ns[1:]\n\n        # gap :: Int -> (Char, Int, Int) -> Int\n        def gap(z, c1xy):\n            c1, x, y = c1xy\n            return min(\n                succ(y),\n                succ(z),\n                succ(x) if c != c1 else x\n            )\n        return scanl(gap)(succ(n))(\n            zip(s1, ns, ns1)\n        )\n    return lambda sb: reduce(\n        go, list(sb), enumFromTo(0)(len(s1))\n    )[-1]\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Tests'''\n\n    pairs = [\n        ('rosettacode', 'raisethysword'),\n        ('kitten', 'sitting'),\n        ('saturday', 'sunday')\n    ]\n\n    print(\n        tabulated(\n            'Levenshtein minimum edit distances:\\n'\n        )(str)(str)(\n            uncurry(levenshtein)\n        )(concat(map(\n            list,\n            zip(pairs, map(swap, pairs))\n        )))\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# concat :: [[a]] -> [a]\n# concat :: [String] -> String\ndef concat(xxs):\n    '''The concatenation of all the elements in a list.'''\n    xs = list(chain.from_iterable(xxs))\n    unit = '' if isinstance(xs, str) else []\n    return unit if not xs else (\n        ''.join(xs) if isinstance(xs[0], str) else xs\n    )\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# scanl :: (b -> a -> b) -> b -> [a] -> [b]\ndef scanl(f):\n    '''scanl is like reduce, but returns a succession of\n       intermediate values, building from the left.'''\n    return lambda a: lambda xs: (\n        list(accumulate(chain([a], xs), f))\n    )\n\n\n# swap :: (a, b) -> (b, a)\ndef swap(tpl):\n    '''The swapped components of a pair.'''\n    return (tpl[1], tpl[0])\n\n\n# succ :: Int => Int -> Int\ndef succ(x):\n    '''The successor of a value.\n       For numeric types, (1 +).'''\n    return 1 + x\n\n\n# tabulated :: String -> (a -> String) ->\n#                        (b -> String) ->\n#                        (a -> b) -> [a] -> String\ndef tabulated(s):\n    '''Heading -> x display function ->\n                 fx display function ->\n                 f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + '\\n'.join([\n            xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x))\n            for x in xs\n\n        ])\n    return lambda xShow: lambda fxShow: (\n        lambda f: lambda xs: go(\n            xShow, fxShow, f, xs\n        )\n    )\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# uncurry :: (a -> b -> c) -> ((a, b) -> c)\ndef uncurry(f):\n    '''A function over a tuple\n       derived from a curried function.'''\n    return lambda xy: f(xy[0])(\n        xy[1]\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Levenshtein distance/Alignment", "language": "Python", "task": "The [[Levenshtein distance]] algorithm returns the number of atomic operations (insertion, deletion or edition) that must be performed on a string in order to obtain an other one, but it does not say anything about the actual operations used or their order.\n\nAn alignment is a notation used to describe the operations used to turn a string into an other.  At some point in the strings, the minus character ('-') is placed in order to signify that a character must be added at this very place.  For instance, an alignment between the words 'place' and 'palace' is:\n\n\nP-LACE\nPALACE\n\n\n\n;Task:\nWrite a function that shows the alignment of two strings for the corresponding levenshtein distance.  \n\nAs an example, use the words \"rosettacode\" and \"raisethysword\".\n\nYou can either implement an algorithm, or use a dedicated library (thus showing us how it is named in your language).\n\n", "solution": "from difflib import ndiff\n\ndef levenshtein(str1, str2):\n    result = \"\"\n    pos, removed = 0, 0\n    for x in ndiff(str1, str2):\n        if pos<len(str1) and str1[pos] == x[2]:\n          pos += 1\n          result += x[2]\n          if x[0] == \"-\":\n              removed += 1\n          continue\n        else:\n          if removed > 0:\n            removed -=1\n          else:\n            result += \"-\"\n    print(result)\n\nlevenshtein(\"place\",\"palace\")\nlevenshtein(\"rosettacode\",\"raisethysword\")"}
{"title": "List rooted trees", "language": "Python", "task": "You came back from grocery shopping.    After putting away all the goods, you are left with a pile of plastic bags, which you want to save for later use, so you take one bag and stuff all the others into it, and throw it under the sink.   In doing so, you realize that there are various ways of nesting the bags, with all bags viewed as identical.\n\nIf we use a matching pair of parentheses to represent a bag, the ways are:\n\nFor 1 bag, there's one way:\n  ()\t<- a bag\n\nfor 2 bags, there's one way:\n  (())\t<- one bag in another\n\nfor 3 bags, there are two:\n  ((())) <- 3 bags nested Russian doll style\n  (()()) <- 2 bags side by side, inside the third\n\nfor 4 bags, four:\n  (()()())\n  ((())())\n  ((()()))\n  (((())))\n\nNote that because all bags are identical, the two 4-bag strings ((())()) and (()(())) represent the same configuration.\n\nIt's easy to see that each configuration for ''n'' bags represents a ''n''-node rooted tree, where a bag is a tree node, and a bag with its content forms a subtree.  The outermost bag is the tree root.  Number of configurations for given ''n'' is given by OEIS A81.\n\n\n;Task: \nWrite a program that, when given ''n'', enumerates all ways of nesting ''n'' bags.   You can use the parentheses notation above, or any tree representation that's unambiguous and preferably intuitive.\n\nThis task asks for enumeration of trees only; for counting solutions without enumeration, that OEIS page lists various formulas, but that's not encouraged by this task, especially if implementing it would significantly increase code size.\n\nAs an example output, run 5 bags.   There should be 9 ways.\n\n", "solution": "def bags(n,cache={}):\n\tif not n: return [(0, \"\")]\n\n\tupto = sum([bags(x) for x in range(n-1, 0, -1)], [])\n\treturn [(c+1, '('+s+')') for c,s in bagchain((0, \"\"), n-1, upto)]\n\ndef bagchain(x, n, bb, start=0):\n\tif not n: return [x]\n\n\tout = []\n\tfor i in range(start, len(bb)):\n\t\tc,s = bb[i]\n\t\tif c <= n: out += bagchain((x[0] + c, x[1] + s), n-c, bb, i)\n\treturn out\n\n# Maybe this lessens eye strain. Maybe not.\ndef replace_brackets(s):\n\tdepth,out = 0,[]\n\tfor c in s:\n\t\tif c == '(':\n\t\t\tout.append(\"([{\"[depth%3])\n\t\t\tdepth += 1\n\t\telse:\n\t\t\tdepth -= 1\n\t\t\tout.append(\")]}\"[depth%3])\n\treturn \"\".join(out)\n\nfor x in bags(5): print(replace_brackets(x[1]))"}
{"title": "Long literals, with continuations", "language": "Python", "task": "This task is about writing a computer program that has long literals   (character\nliterals that may require specifying the words/tokens on more than one (source)\nline,   either with continuations or some other method, such as abutments or\nconcatenations   (or some other mechanisms).\n\n\nThe literal is to be in the form of a \"list\",    a literal that contains many\nwords (tokens) separated by a blank (space),   in this case   (so as to have a\ncommon list),   the (English) names of the chemical elements of the periodic table.\n\n\nThe list is to be in (ascending) order of the (chemical) element's atomic number:\n\n ''hydrogen helium lithium beryllium boron carbon nitrogen oxygen fluorine neon sodium aluminum silicon ...''\n\n... up to the last known (named) chemical element   (at this time).\n\n\nDo not include any of the   \"unnamed\"   chemical element names such as:\n\n ''ununennium unquadnilium triunhexium penthextrium penthexpentium septhexunium octenntrium ennennbium''\n\n\nTo make computer programming languages comparable,   the statement widths should be\nrestricted to less than   '''81'''   bytes (characters),   or less\nif a computer programming language has more restrictive limitations or standards.\n\nAlso mention what column the programming statements can start in if   ''not''  \nin column one.\n\n\nThe list   ''may''   have leading/embedded/trailing blanks during the\ndeclaration   (the actual program statements),   this is allow the list to be\nmore readable.   The \"final\" list shouldn't have any leading/trailing or superfluous\nblanks   (when stored in the program's \"memory\").\n\nThis list should be written with the idea in mind that the\nprogram   ''will''   be updated,   most likely someone other than the\noriginal author,   as there will be newer (discovered) elements of the periodic\ntable being added   (possibly in the near future).   These future updates\nshould be one of the primary concerns in writing these programs and it should be \"easy\"\nfor someone else to add chemical elements to the list   (within the computer\nprogram).\n\nAttention should be paid so as to not exceed the   ''clause length''   of\ncontinued or specified statements,   if there is such a restriction.   If the\nlimit is greater than (say) 4,000 bytes or so,   it needn't be mentioned here.\n\n\n;Task:\n:*   Write a computer program (by whatever name) to contain a list of the known elements.\n:*   The program should eventually contain a long literal of words   (the elements).\n:*   The literal should show how one could create a long list of blank-delineated words.\n:*   The \"final\" (stored) list should only have a single blank between elements.\n:*   Try to use the most idiomatic approach(es) in creating the final list.\n:*   Use continuation if possible, and/or show alternatives   (possibly using concatenation).\n:*   Use a program comment to explain what the continuation character is if it isn't obvious.\n:*   The program should contain a variable that has the date of the last update/revision.\n:*   The program, when run, should display with verbiage:\n:::*   The last update/revision date   (and should be unambiguous).\n:::*   The number of chemical elements in the list.\n:::*   The name of the highest (last) element name.\n\n\nShow all output here, on this page.\n\n\n\n", "solution": "\"\"\"Long string literal. Requires Python 3.6+ for f-strings.\"\"\"\n\nrevision = \"October 13th 2020\"\n\n# String literal continuation. Notice the trailing space at the end of each\n# line but the last, and the lack of commas. There is exactly one \"blank\"\n# between each item in the list.\nelements = (\n    \"hydrogen helium lithium beryllium boron carbon nitrogen oxygen fluorine \"\n    \"neon sodium magnesium aluminum silicon phosphorous sulfur chlorine argon \"\n    \"potassium calcium scandium titanium vanadium chromium manganese iron \"\n    \"cobalt nickel copper zinc gallium germanium arsenic selenium bromine \"\n    \"krypton rubidium strontium yttrium zirconium niobium molybdenum \"\n    \"technetium ruthenium rhodium palladium silver cadmium indium tin \"\n    \"antimony tellurium iodine xenon cesium barium lanthanum cerium \"\n    \"praseodymium neodymium promethium samarium europium gadolinium terbium \"\n    \"dysprosium holmium erbium thulium ytterbium lutetium hafnium tantalum \"\n    \"tungsten rhenium osmium iridium platinum gold mercury thallium lead \"\n    \"bismuth polonium astatine radon francium radium actinium thorium \"\n    \"protactinium uranium neptunium plutonium americium curium berkelium \"\n    \"californium einsteinium fermium mendelevium nobelium lawrencium \"\n    \"rutherfordium dubnium seaborgium bohrium hassium meitnerium darmstadtium \"\n    \"roentgenium copernicium nihonium flerovium moscovium livermorium \"\n    \"tennessine oganesson\"\n)\n\n\ndef report():\n    \"\"\"Write a summary report to stdout.\"\"\"\n    items = elements.split()\n\n    print(f\"Last revision date: {revision}\")\n    print(f\"Number of elements: {len(items)}\")\n    print(f\"Last element      : {items[-1]}\")\n\n\nif __name__ == \"__main__\":\n    report()\n"}
{"title": "Long year", "language": "Python 3.7", "task": "Most years have 52 weeks, some have 53, according to ISO8601.\n\n\n;Task:\nWrite a function which determines if a given year is long (53 weeks) or not, and demonstrate it.\n\n", "solution": "'''Long Year ?'''\n\nfrom datetime import date\n\n\n# longYear :: Year Int -> Bool\ndef longYear(y):\n    '''True if the ISO year y has 53 weeks.'''\n    return 52 < date(y, 12, 28).isocalendar()[1]\n\n\n# --------------------------TEST---------------------------\n# main :: IO ()\ndef main():\n    '''Longer (53 week) years in the range 2000-2100'''\n    for year in [\n            x for x in range(2000, 1 + 2100)\n            if longYear(x)\n    ]:\n        print(year)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Longest common subsequence", "language": "Python", "task": "'''Introduction'''\n\nDefine a ''subsequence'' to be any output string obtained by deleting zero or more symbols from an input string.\n\nThe '''Longest Common Subsequence''' ('''LCS''') is a subsequence of maximum length common to two or more strings.\n\nLet ''A''  ''A''[0]... ''A''[m - 1] and ''B''  ''B''[0]... ''B''[n - 1], m < n be strings drawn from an alphabet S of size s, containing every distinct symbol in A + B.\n\nAn ordered pair (i, j) will be referred to as a match if ''A''[i] = ''B''[j], where 0 <= i < m and 0 <= j < n.\n\nThe set of matches '''M''' defines a relation over matches: '''M'''[i, j] = (i, j)  '''M'''.\n\nDefine a ''non-strict'' product-order (<=) over ordered pairs, such that (i1, j1) <= (i2, j2) = i1 <= i2 and j1 <= j2.  We define (>=) similarly.\n\nWe say ordered pairs p1 and p2 are ''comparable'' if either p1 <= p2 or p1 >= p2 holds.  If i1 < i2 and j2 < j1 (or i2 < i1 and j1 < j2) then neither p1 <= p2 nor p1 >= p2 are possible, and we say p1 and p2 are ''incomparable''.\n\nDefine the ''strict'' product-order (<) over ordered pairs, such that (i1, j1) < (i2, j2) = i1 < i2 and j1 < j2.  We define (>) similarly.\n\nA chain '''C''' is a subset of '''M''' consisting of at least one element m; and where either m1 < m2 or m1 > m2 for every pair of distinct elements m1 and m2.  An antichain '''D''' is any subset of '''M''' in which every pair of distinct elements m1 and m2 are incomparable.\n\nA chain can be visualized as a strictly increasing curve that passes through matches (i, j) in the m*n coordinate space of '''M'''[i, j].\n\nEvery Common Sequence of length ''q'' corresponds to a chain of cardinality ''q'', over the set of matches '''M'''.  Thus, finding an LCS can be restated as the problem of finding a chain of maximum cardinality ''p''.\n\nAccording to [Dilworth 1950], this cardinality ''p'' equals the minimum number of disjoint antichains into which '''M''' can be decomposed.  Note that such a decomposition into the minimal number p of disjoint antichains may not be unique.\n\n'''Background'''\n\nWhere the number of symbols appearing in matches is small relative to the length of the input strings, reuse of the symbols increases; and the number of matches will tend towards O(''m*n'') quadratic growth.  This occurs, for example, in the Bioinformatics application of nucleotide and protein sequencing.\n\nThe divide-and-conquer approach of [Hirschberg 1975] limits the space required to O(''n'').  However, this approach requires O(''m*n'') time even in the best case.\n\nThis quadratic time dependency may become prohibitive, given very long input strings.  Thus, heuristics are often favored over optimal Dynamic Programming solutions.\n\nIn the application of comparing file revisions, records from the input files form a large symbol space; and the number of symbols approaches the length of the LCS.  In this case the number of matches reduces to linear, O(''n'') growth.\n\nA binary search optimization due to [Hunt and Szymanski 1977] can be applied to the basic Dynamic Programming approach, resulting in an expected performance of O(''n log m'').  Performance can degrade to O(''m*n log m'') time in the worst case, as the number of matches grows to O(''m*n'').\n\n'''Note'''\n\n[Rick 2000] describes a linear-space algorithm with a time bound of O(''n*s + p*min(m, n - p)'').\n\n'''Legend'''\n\n A, B are input strings of lengths m, n respectively\n p is the length of the LCS\n M is the set of matches (i, j) such that A[i] = B[j]\n r is the magnitude of M\n s is the magnitude of the alphabet S of distinct symbols in A + B\n\n'''References'''\n\n[Dilworth 1950] \"A decomposition theorem for partially ordered sets\"\nby Robert P. Dilworth, published January 1950,\nAnnals of Mathematics [Volume 51, Number 1, ''pp.'' 161-166]\n\n[Goeman and Clausen 2002] \"A New Practical Linear Space Algorithm for the Longest Common\nSubsequence Problem\" by Heiko Goeman and Michael Clausen,\npublished 2002, Kybernetika [Volume 38, Issue 1, ''pp.'' 45-66]\n\n[Hirschberg 1975] \"A linear space algorithm for computing maximal common subsequences\"\nby Daniel S. Hirschberg, published June 1975\nCommunications of the ACM [Volume 18, Number 6, ''pp.'' 341-343]\n\n[Hunt and McIlroy 1976] \"An Algorithm for Differential File Comparison\"\nby James W. Hunt and M. Douglas McIlroy, June 1976\nComputing Science Technical Report, Bell Laboratories 41\n\n[Hunt and Szymanski 1977] \"A Fast Algorithm for Computing Longest Common Subsequences\"\nby James W. Hunt and Thomas G. Szymanski, published May 1977\nCommunications of the ACM [Volume 20, Number 5, ''pp.'' 350-353]\n\n[Rick 2000] \"Simple and fast linear space computation of longest common subsequences\"\nby Claus Rick, received 17 March 2000, Information Processing Letters,\nElsevier Science [Volume 75, ''pp.'' 275-281]\n\n\n'''Examples'''\n\nThe sequences \"1234\" and \"1224533324\" have an LCS of \"1234\":\n '''1234'''\n '''12'''245'''3'''332'''4'''\n\nFor a string example, consider the sequences \"thisisatest\" and \"testing123testing\". An LCS would be \"tsitest\":\n '''t'''hi'''si'''sa'''test'''\n '''t'''e'''s'''t'''i'''ng123'''test'''ing\n\nIn this puzzle, your code only needs to deal with strings. Write a function which returns an LCS of two strings (case-sensitive). You don't need to show multiple LCS's.\n\nFor more information on this problem please see Wikipedia.\n\n\n", "solution": "def lcs(a, b):\n    # generate matrix of length of longest common subsequence for substrings of both words\n    lengths = [[0] * (len(b)+1) for _ in range(len(a)+1)]\n    for i, x in enumerate(a):\n        for j, y in enumerate(b):\n            if x == y:\n                lengths[i+1][j+1] = lengths[i][j] + 1\n            else:\n                lengths[i+1][j+1] = max(lengths[i+1][j], lengths[i][j+1])\n\n    # read a substring from the matrix\n    result = ''\n    j = len(b)\n    for i in range(1, len(a)+1):\n        if lengths[i][j] != lengths[i-1][j]:\n            result += a[i-1]\n\n    return result"}
{"title": "Longest common substring", "language": "Python", "task": "Write a function that returns the longest common substring of two strings. \n\nUse it within a program that demonstrates sample output from the function, which will consist of the longest common substring between \"thisisatest\" and \"testing123testing\". \n\nNote that substrings are consecutive characters within a string.   This distinguishes them from subsequences, which is any sequence of characters within a string, even if there are extraneous characters in between them. \n\nHence, the [[longest common subsequence]] between \"thisisatest\" and \"testing123testing\" is \"tsitest\", whereas the longest common sub''string'' is just \"test\".\n\n\n\n\n;References:\n*Generalize Suffix Tree\n*[[Ukkonen's Suffix Tree Construction]]\n\n", "solution": "s1 = \"thisisatest\"\ns2 = \"testing123testing\"\nlen1, len2 = len(s1), len(s2)\nir, jr = 0, -1\nfor i1 in range(len1):\n    i2 = s2.find(s1[i1])\n    while i2 >= 0:\n        j1, j2 = i1, i2\n        while j1 < len1 and j2 < len2 and s2[j2] == s1[j1]:\n            if j1-i1 >= jr-ir:\n                ir, jr = i1, j1\n            j1 += 1; j2 += 1\n        i2 = s2.find(s1[i1], i2+1)\nprint (s1[ir:jr+1])"}
{"title": "Longest common substring", "language": "Python from JavaScript", "task": "Write a function that returns the longest common substring of two strings. \n\nUse it within a program that demonstrates sample output from the function, which will consist of the longest common substring between \"thisisatest\" and \"testing123testing\". \n\nNote that substrings are consecutive characters within a string.   This distinguishes them from subsequences, which is any sequence of characters within a string, even if there are extraneous characters in between them. \n\nHence, the [[longest common subsequence]] between \"thisisatest\" and \"testing123testing\" is \"tsitest\", whereas the longest common sub''string'' is just \"test\".\n\n\n\n\n;References:\n*Generalize Suffix Tree\n*[[Ukkonen's Suffix Tree Construction]]\n\n", "solution": "'''Longest common substring'''\n\nfrom itertools import accumulate, chain\nfrom functools import reduce\n\n\n# longestCommon :: String -> String -> String\ndef longestCommon(s1):\n    '''The longest common substring of\n       two given strings.\n    '''\n    def go(s2):\n        return max(intersect(\n            *map(lambda s: map(\n                concat,\n                concatMap(tails)(\n                    compose(tail, list, inits)(s)\n                )\n            ), [s1, s2])\n        ), key=len)\n    return go\n\n\n# ------------------------- TEST -------------------------\ndef main():\n    '''Test'''\n    print(\n        longestCommon(\n            \"testing123testing\"\n        )(\n            \"thisisatest\"\n        )\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    def go(f, g):\n        def fg(x):\n            return f(g(x))\n        return fg\n    return reduce(go, fs, lambda x: x)\n\n\n# concat :: [String] -> String\ndef concat(xs):\n    '''The concatenation of all the elements\n       in a list or iterable.\n    '''\n    return ''.join(xs)\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been\n       mapped.\n       The list monad can be derived by using a function f\n       which wraps its output in a list, (using an empty\n       list to represent computational failure).\n    '''\n    def go(xs):\n        return chain.from_iterable(map(f, xs))\n    return go\n\n\n# inits :: [a] -> [[a]]\ndef inits(xs):\n    '''all initial segments of xs, shortest first.'''\n    return accumulate(chain([[]], xs), lambda a, x: a + [x])\n\n\n# intersect :: [a] -> [a] -> [a]\ndef intersect(xs, ys):\n    '''The ordered intersection of xs and ys.\n       intersect([1,2,2,3,4])([6,4,4,2]) -> [2,2,4]\n    '''\n    s = set(ys)\n    return (x for x in xs if x in s)\n\n\n# scanl :: (b -> a -> b) -> b -> [a] -> [b]\ndef scanl(f):\n    '''scanl is like reduce, but defines a succession of\n       intermediate values, building from the left.\n    '''\n    def go(a):\n        def g(xs):\n            return accumulate(chain([a], xs), f)\n        return g\n    return go\n\n\n# tail :: [a] -> [a]\n# tail :: Gen [a] -> [a]\ndef tail(xs):\n    '''The elements following the head of a\n       (non-empty) list.\n    '''\n    return xs[1:]\n\n\n# tails :: [a] -> [[a]]\ndef tails(xs):\n    '''All final segments of xs,\n       longest first.\n    '''\n    return [\n        xs[i:] for i in\n        range(0, 1 + len(xs))\n    ]\n\n\n# MAIN ---\nmain()\n"}
{"title": "Longest increasing subsequence", "language": "Python", "task": "Calculate and show here a longest increasing subsequence of the list:\n:\\{3, 2, 6, 4, 5, 1\\}\nAnd of the list:\n:\\{0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15\\}\n\nNote that a list may have more than one subsequence that is of the maximum length.\n\n\n\n;Ref:\n# Dynamic Programming #1: Longest Increasing Subsequence on YouTube\n# An efficient solution can be based on Patience sorting.\n\n", "solution": "def longest_increasing_subsequence(X):\n    \"\"\"Returns the Longest Increasing Subsequence in the Given List/Array\"\"\"\n    N = len(X)\n    P = [0] * N\n    M = [0] * (N+1)\n    L = 0\n    for i in range(N):\n       lo = 1\n       hi = L\n       while lo <= hi:\n           mid = (lo+hi)//2\n           if (X[M[mid]] < X[i]):\n               lo = mid+1\n           else:\n               hi = mid-1\n    \n       newL = lo\n       P[i] = M[newL-1]\n       M[newL] = i\n    \n       if (newL > L):\n           L = newL\n    \n    S = []\n    k = M[L]\n    for i in range(L-1, -1, -1):\n        S.append(X[k])\n        k = P[k]\n    return S[::-1]\n\nif __name__ == '__main__':\n    for d in [[3,2,6,4,5,1], [0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13, 3, 11, 7, 15]]:\n        print('a L.I.S. of %s is %s' % (d, longest_increasing_subsequence(d)))"}
{"title": "Lucky and even lucky numbers", "language": "Python", "task": "Note that in the following explanation list indices are assumed to start at ''one''.\n\n;Definition of lucky numbers\n''Lucky numbers'' are positive integers that are formed by:\n\n# Form a list of all the positive odd integers > 01, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39 ...\n# Return the first number from the list (which is '''1''').\n# (Loop begins here)\n#* Note then return the second number from the list (which is '''3''').\n#* Discard every third, (as noted), number from the list to form the new list1, 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, 45, 49, 51, 55, 57 ...\n# (Expanding the loop a few more times...)\n#* Note then return the third number from the list (which is '''7''').\n#* Discard every 7th, (as noted), number from the list to form the new list1, 3, 7, 9, 13, 15, 21, 25, 27, 31, 33, 37, 43, 45, 49, 51, 55, 57, 63, 67 ...\n#* Note then return the 4th number from the list (which is '''9''').\n#* Discard every 9th, (as noted), number from the list to form the new list1, 3, 7, 9, 13, 15, 21, 25, 31, 33, 37, 43, 45, 49, 51, 55, 63, 67, 69, 73 ...\n#* Take the 5th, i.e. '''13'''.  Remove every 13th.\n#* Take the 6th, i.e. '''15'''.  Remove every 15th.\n#* Take the 7th, i.e. '''21'''.  Remove every 21th.\n#* Take the 8th, i.e. '''25'''.  Remove every 25th.\n# (Rule for the loop)\n#* Note the nth, which is m.\n#* Remove every mth.\n#* Increment n.\n\n;Definition of even lucky numbers\nThis follows the same rules as the definition of lucky numbers above ''except for the very first step'':\n# Form a list of all the positive '''even''' integers > 02, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40 ...\n# Return the first number from the list (which is '''2''').\n# (Loop begins here)\n#* Note then return the second number from the list (which is '''4''').\n#* Discard every 4th, (as noted), number from the list to form the new list2, 4, 6, 10, 12, 14, 18, 20, 22, 26, 28, 30, 34, 36, 38, 42, 44, 46, 50, 52 ...\n# (Expanding the loop a few more times...)\n#* Note then return the third number from the list (which is '''6''').\n#* Discard every 6th, (as noted), number from the list to form the new list2, 4, 6, 10, 12, 18, 20, 22, 26, 28, 34, 36, 38, 42, 44, 50, 52, 54, 58, 60 ...\n#* Take the 4th, i.e. '''10'''.  Remove every 10th.\n#* Take the 5th, i.e. '''12'''.  Remove every 12th.\n# (Rule for the loop)\n#* Note the nth, which is m.\n#* Remove every mth.\n#* Increment n.\n\n;Task requirements\n* Write one or two subroutines (functions) to generate ''lucky numbers'' and ''even lucky numbers'' \n* Write a command-line interface to allow selection of which kind of numbers and which number(s). Since input is from the command line, tests should be made for the common errors:\n** missing arguments\n** too many arguments\n** number (or numbers) aren't legal\n** misspelled argument ('''lucky''' or '''evenLucky''')\n* The command line handling should:\n** support mixed case handling of the (non-numeric) arguments\n** support printing a particular number\n** support printing a range of numbers by their index\n** support printing a range of numbers by their values\n* The resulting list of numbers should be printed on a single line.\n\nThe program should support the arguments:\n\n                             what is displayed  (on a single line)\n       argument(s)              (optional verbiage is encouraged)\n  +-------------------+----------------------------------------------------+\n  |  j                |  Jth       lucky number                            |\n  |  j  ,      lucky  |  Jth       lucky number                            |\n  |  j  ,  evenLucky  |  Jth  even lucky number                            |\n  |                   |                                                    |\n  |  j  k             |  Jth  through  Kth (inclusive)       lucky numbers |\n  |  j  k      lucky  |  Jth  through  Kth (inclusive)       lucky numbers |\n  |  j  k  evenLucky  |  Jth  through  Kth (inclusive)  even lucky numbers |\n  |                   |                                                    |\n  |  j -k             |  all       lucky numbers in the range  j --> |k|   |\n  |  j -k      lucky  |  all       lucky numbers in the range  j --> |k|   |\n  |  j -k  evenLucky  |  all  even lucky numbers in the range  j --> |k|   |\n  +-------------------+----------------------------------------------------+\n                           where    |k|    is the absolute value of   k\n\nDemonstrate the program by:\n* showing the first twenty ''lucky'' numbers\n* showing the first twenty ''even lucky'' numbers\n* showing all ''lucky'' numbers between 6,000 and 6,100 (inclusive)\n* showing all ''even lucky'' numbers in the same range as above\n* showing the 10,000th ''lucky'' number (extra credit)\n* showing the 10,000th ''even lucky'' number (extra credit)\n\n;See also:\n* This task is related to the [[Sieve of Eratosthenes]] task.\n* OEIS Wiki Lucky numbers.\n* Sequence A000959 lucky numbers on The On-Line Encyclopedia of Integer Sequences.\n* Sequence A045954 even lucky numbers or ELN on The On-Line Encyclopedia of Integer Sequences.\n* Entry lucky numbers on The Eric Weisstein's World of Mathematics.\n", "solution": "from itertools import islice\nimport sys, re\n\nclass ArgumentError(Exception):\n    pass\ndef arghandler(argstring):\n    match_obj = re.match( r\"\"\"(?mx)\n    (?:\n      (?P<SINGLE>\n         (?: ^ (?P<SINGLEL> \\d+ ) (?:  | \\s , \\s lucky ) \\s* $ )\n        |(?: ^ (?P<SINGLEE> \\d+ ) (?:  | \\s , \\s evenLucky ) \\s* $ )\n      )\n     |(?P<KTH>\n         (?: ^ (?P<KTHL> \\d+ \\s \\d+ ) (?:  | \\s lucky ) \\s* $ )\n        |(?: ^ (?P<KTHE> \\d+ \\s \\d+ ) (?:  | \\s evenLucky ) \\s* $ )\n      )\n     |(?P<RANGE>\n         (?: ^ (?P<RANGEL> \\d+ \\s -\\d+ ) (?:  | \\s lucky ) \\s* $ )\n        |(?: ^ (?P<RANGEE> \\d+ \\s -\\d+ ) (?:  | \\s evenLucky ) \\s* $ )\n      )\n    )\"\"\", argstring)\n    \n    if match_obj:\n        # Retrieve group(s) by name\n        SINGLEL = match_obj.group('SINGLEL')\n        SINGLEE = match_obj.group('SINGLEE')\n        KTHL = match_obj.group('KTHL')\n        KTHE = match_obj.group('KTHE')\n        RANGEL = match_obj.group('RANGEL')\n        RANGEE = match_obj.group('RANGEE')\n        if SINGLEL: \n            j = int(SINGLEL)\n            assert 0 < j < 10001, \"Argument out of range\"\n            print(\"Single %i'th lucky number:\" % j, end=' ')\n            print( list(islice(lgen(), j-1, j))[0] )\n        elif SINGLEE: \n            j = int(SINGLEE)\n            assert 0 < j < 10001, \"Argument out of range\"\n            print(\"Single %i'th even lucky number:\" % j, end=' ')\n            print( list(islice(lgen(even=True), j-1, j))[0] )\n        elif KTHL: \n            j, k = [int(num) for num in KTHL.split()]\n            assert 0 < j < 10001, \"first argument out of range\"\n            assert 0 < k < 10001 and k > j, \"second argument out of range\"\n            print(\"List of %i ... %i lucky numbers:\" % (j, k), end=' ')\n            for n, luck in enumerate(lgen(), 1):\n                if n > k: break\n                if n >=j: print(luck, end = ', ')\n            print('')\n        elif KTHE: \n            j, k = [int(num) for num in KTHE.split()]\n            assert 0 < j < 10001, \"first argument out of range\"\n            assert 0 < k < 10001 and k > j, \"second argument out of range\"\n            print(\"List of %i ... %i even lucky numbers:\" % (j, k), end=' ')\n            for n, luck in enumerate(lgen(even=True), 1):\n                if n > k: break\n                if n >=j: print(luck, end = ', ')\n            print('')\n        elif RANGEL: \n            j, k = [int(num) for num in RANGEL.split()]\n            assert 0 < j < 10001, \"first argument out of range\"\n            assert 0 < -k < 10001 and -k > j, \"second argument out of range\"\n            k = -k\n            print(\"List of lucky numbers in the range %i ... %i :\" % (j, k), end=' ')\n            for n in lgen():\n                if n > k: break\n                if n >=j: print(n, end = ', ')\n            print('')\n        elif RANGEE: \n            j, k = [int(num) for num in RANGEE.split()]\n            assert 0 < j < 10001, \"first argument out of range\"\n            assert 0 < -k < 10001 and -k > j, \"second argument out of range\"\n            k = -k\n            print(\"List of even lucky numbers in the range %i ... %i :\" % (j, k), end=' ')\n            for n in lgen(even=True):\n                if n > k: break\n                if n >=j: print(n, end = ', ')\n            print('')\n    else:\n        raise ArgumentError('''\n        \n  Error Parsing Arguments!\n  \n  Expected Arguments of the form (where j and k are integers):\n      \n      j                #  Jth       lucky number\n      j  ,      lucky  #  Jth       lucky number\n      j  ,  evenLucky  #  Jth  even lucky number\n                       #\n      j  k             #  Jth  through  Kth (inclusive)       lucky numbers\n      j  k      lucky  #  Jth  through  Kth (inclusive)       lucky numbers\n      j  k  evenLucky  #  Jth  through  Kth (inclusive)  even lucky numbers\n                       #\n      j -k             #  all       lucky numbers in the range  j --? |k|\n      j -k      lucky  #  all       lucky numbers in the range  j --? |k|\n      j -k  evenLucky  #  all  even lucky numbers in the range  j --? |k|\n        ''')\n\nif __name__ == '__main__':\n    arghandler(' '.join(sys.argv[1:]))"}
{"title": "Lychrel numbers", "language": "Python", "task": "::#   Take an integer n, greater than zero.\n::#   Form the next n of its series by reversing the digits of the current n and adding the result to the current n.\n::#   Stop when n becomes palindromic - i.e. the digits of n in reverse order == n.\n\n\nThe above recurrence relation when applied to most starting numbers n = 1, 2, ... terminates in a palindrome quite quickly.\n\n\n;Example:\nIf n0 = 12 we get\n\n       12\n       12 +  21 =  33,   a palindrome!\n\n\nAnd if n0 = 55 we get\n\n       55\n       55 +  55 = 110\n      110 + 011 = 121,   a palindrome!\n\n\nNotice that the check for a palindrome happens   ''after''   an addition.\n\n\nSome starting numbers seem to go on forever; the recurrence relation for 196 has been calculated for millions of repetitions forming numbers with millions of digits, without forming a palindrome. \n\nThese numbers that do not end in a palindrome are called '''Lychrel numbers'''.\n\nFor the purposes of this task a Lychrel number is any starting number that does not form a palindrome within 500 (or more) iterations.\n\n\n;Seed and related Lychrel numbers:\nAny integer produced in the sequence of a Lychrel number is also a Lychrel number.\n\nIn general, any sequence from one Lychrel number ''might'' converge to join the sequence from a prior Lychrel number candidate; for example the sequences for the numbers 196 and then 689 begin:\n\n        196\n        196 +    691 =   887\n        887 +    788 =  1675\n       1675 +   5761 =  7436\n       7436 +   6347 = 13783\n      13783 +  38731 = 52514\n      52514 +  41525 = 94039\n       ...\n\n\n        689\n        689 +    986 =  1675\n       1675 +   5761 =  7436\n       ...\n\nSo we see that the sequence starting with 689 converges to, and continues with the same numbers as that for 196. \n\nBecause of this we can further split the Lychrel numbers into true '''Seed''' Lychrel number candidates, and '''Related''' numbers that produce no palindromes but have integers in their sequence seen as part of the sequence generated from a lower Lychrel number.\n\n\n;Task:\n*   Find the number of seed Lychrel number candidates and related numbers for n in the range 1..10000 inclusive. (With that iteration limit of 500).\n*    Print the number of seed Lychrels found; the actual seed Lychrels; and just the ''number'' of relateds found.\n*    Print any seed Lychrel or related number that is itself a palindrome.\n\n\nShow all output here.\n\n\n;References:\n*   What's special about 196? Numberphile video.\n*   A023108 Positive integers which apparently never result in a palindrome under repeated applications of the function f(x) = x + (x with digits reversed).\n*   Status of the 196 conjecture? Mathoverflow.\n\n", "solution": "from __future__ import print_function\n\ndef add_reverse(num, max_iter=1000):\n    i, nums = 0, {num}\n    while True:\n        i, num = i+1, num + reverse_int(num)\n        nums.add(num)\n        if reverse_int(num) == num or i >= max_iter:\n            break\n    return nums\n    \n#@functools.lru_cache(maxsize=2**20)\ndef reverse_int(num):\n    return int(str(num)[::-1])\n\ndef split_roots_from_relateds(roots_and_relateds):\n    roots = roots_and_relateds[::]\n    i = 1\n    while i < len(roots):\n        this = roots[i]\n        if any(this.intersection(prev) for prev in roots[:i]):\n            del roots[i]\n        else:\n            i += 1\n    root = [min(each_set) for each_set in roots]\n    related = [min(each_set) for each_set in roots_and_relateds]\n    related = [n for n in related if n not in root]\n    return root, related\n\ndef find_lychrel(maxn, max_reversions):\n    'Lychrel number generator'\n    series = [add_reverse(n, max_reversions*2) for n in range(1, maxn + 1)]\n    roots_and_relateds = [s for s in series if len(s) > max_reversions]\n    return split_roots_from_relateds(roots_and_relateds)\n\n\nif __name__ == '__main__':\n    maxn, reversion_limit = 10000, 500\n    print(\"Calculations using n = 1..%i and limiting each search to 2*%i reverse-digits-and-adds\"\n          % (maxn, reversion_limit))\n    lychrel, l_related = find_lychrel(maxn, reversion_limit)\n    print('  Number of Lychrel numbers:', len(lychrel))\n    print('    Lychrel numbers:', ', '.join(str(n) for n in lychrel))\n    print('  Number of Lychrel related:', len(l_related))\n    #print('    Lychrel related:', ', '.join(str(n) for n in l_related))\n    pals = [x for x in lychrel + l_related  if x == reverse_int(x)]\n    print('  Number of Lychrel palindromes:', len(pals))\n    print('    Lychrel palindromes:', ', '.join(str(n) for n in pals))"}
{"title": "Mad Libs", "language": "Python", "task": "{{wikipedia}}\n\n\nMad Libs is a phrasal template word game where one player prompts another for a list of words to substitute for blanks in a story, usually with funny results. \n\n\n;Task;\nWrite a program to create a Mad Libs like story. \n\nThe program should read an arbitrary multiline story from input.\n\nThe story will be terminated with a blank line. \n\nThen, find each replacement to be made within the story, ask the user for a word to replace it with, and make all the replacements.\n\nStop when there are none left and print the final story.\n\n\nThe input should be an arbitrary story in the form:\n\n went for a walk in the park. \nfound a .  decided to take it home.\n\nGiven this example, it should then ask for a name, a he or she and a noun ( gets replaced both times with the same value).\n\n\n\n", "solution": "import re\n\n# Optional Python 2.x compatibility\n#try: input = raw_input\n#except: pass\n\ntemplate = '''<name> went for a walk in the park. <he or she>\nfound a <noun>. <name> decided to take it home.'''\n\ndef madlibs(template):\n    print('The story template is:\\n' + template)\n    fields = sorted(set( re.findall('<[^>]+>', template) ))\n    values = input('\\nInput a comma-separated list of words to replace the following items'\n                   '\\n  %s: ' % ','.join(fields)).split(',')\n    story = template\n    for f,v in zip(fields, values):\n        story = story.replace(f, v)\n    print('\\nThe story becomes:\\n\\n' + story)\n\nmadlibs(template)"}
{"title": "Magic 8-ball", "language": "Python", "task": "Create Magic 8-Ball. \n\n\nSee details at:   Magic 8-Ball.\n\n\n\n", "solution": "import random\n\ns = ('It is certain', 'It is decidedly so', 'Without a doubt', 'Yes, definitely',\n 'You may rely on it', 'As I see it, yes', 'Most likely', 'Outlook good',\n 'Signs point to yes', 'Yes', 'Reply hazy, try again', 'Ask again later',\n 'Better not tell you now', 'Cannot predict now', 'Concentrate and ask again',\n \"Don't bet on it\", 'My reply is no', 'My sources say no', 'Outlook not so good',\n 'Very doubtful')\n\nq_and_a = {}\n\nwhile True:\n    question = input('Ask your question:')\n    if len(question) == 0: break\n        \n    if question in q_and_a:\n        print('Your question has already been answered')\n    else:\n        answer = random.choice(s)\n        q_and_a[question] = answer\n        print(answer)"}
{"title": "Magic constant", "language": "Python", "task": "A magic square is a square grid containing consecutive integers from 1 to N2, arranged so that every row, column and diagonal adds up to the same number. That number is a constant. There is no way to create a valid N x N magic square that does not sum to the associated constant. \n\n;EG\n\nA 3 x 3 magic square ''always'' sums to 15. \n\n    +---+---+---+\n    | 2 | 7 | 6 |\n    +---+---+---+\n    | 9 | 5 | 1 |\n    +---+---+---+\n    | 4 | 3 | 8 |\n    +---+---+---+\n\nA 4 x 4 magic square ''always'' sums to 34.\n\nTraditionally, the sequence leaves off terms for n = 0 and n = 1 as the magic squares of order 0 and 1 are trivial; and a term for n = 2 because it is impossible to form a magic square of order 2.\n\n\n;Task\n\n* Starting at order 3, show the first 20 magic constants.\n* Show the 1000th magic constant. (Order 1003)\n* Find and show the order of the smallest N x N magic square whose constant is greater than 101 through 1010.\n\n\n;Stretch\n\n* Find and show the order of the smallest N x N magic square whose constant is greater than 1011 through 1020.\n\n\n;See also\n\n* Wikipedia: Magic constant\n* OEIS: A006003 (Similar sequence, though it includes terms for 0, 1 & 2.)\n* [[Magic squares of odd order]]\n* [[Magic squares of singly even order]]\n* [[Magic squares of doubly even order]]\n\n\n\n", "solution": "#!/usr/bin/python\n\ndef a(n):\n    n += 2\n    return n*(n**2 + 1)/2\n \ndef inv_a(x):\n    k = 0\n    while k*(k**2+1)/2+2 < x:\n        k+=1\n    return k\n\n \nif __name__ == '__main__':\n    print(\"The first 20 magic constants are:\");\n    for n in range(1, 20):\n        print(int(a(n)), end = \" \");\n    print(\"\\nThe 1,000th magic constant is:\",int(a(1000)));\n     \n    for e in range(1, 20):\n        print(f'10^{e}: {inv_a(10**e)}');"}
{"title": "Magic squares of doubly even order", "language": "Python", "task": "A magic square is an   '''NxN'''  square matrix whose numbers consist of consecutive numbers arranged so that the sum of each row and column,   ''and''   both diagonals are equal to the same sum   (which is called the ''magic number'' or ''magic constant'').  \n\nA magic square of doubly even order has a size that is a multiple of four   (e.g.     '''4''', '''8''', '''12'''). \n\nThis means that the subsquares also have an even size, which plays a role in the construction.\n\n \n{| class=\"wikitable\" style=\"float:right;border: 2px solid black; background:lightblue; color:black; margin-left:auto;margin-right:auto;text-align:center;width:22em;height:15em;table-layout:fixed;font-size:100%\"\n|-\n|'''1'''||'''2'''||'''62'''||'''61'''||'''60'''||'''59'''||'''7'''||'''8'''\n|-\n|'''9'''||'''10'''||'''54'''||'''53'''||'''52'''||'''51'''||'''15'''||'''16'''\n|-\n|'''48'''||'''47'''||'''19'''||'''20'''||'''21'''||'''22'''||'''42'''||'''41'''\n|-\n|'''40'''||'''39'''||'''27'''||'''28'''||'''29'''||'''30'''||'''34'''||'''33'''\n|-\n|'''32'''||'''31'''||'''35'''||'''36'''||'''37'''||'''38'''||'''26'''||'''25'''\n|-\n|'''24'''||'''23'''||'''43'''||'''44'''||'''45'''||'''46'''||'''18'''||'''17'''\n|-\n|'''49'''||'''50'''||'''14'''||'''13'''||'''12'''||'''11'''||'''55'''||'''56'''\n|-\n|'''57'''||'''58'''||'''6'''||'''5'''||'''4'''||'''3'''||'''63'''||'''64'''\n|}\n\n;Task\nCreate a magic square of    '''8 x 8'''.\n\n\n;Related tasks\n* [[Magic squares of odd order]]\n* [[Magic squares of singly even order]]\n\n\n;See also:\n* Doubly Even Magic Squares (1728.org)\n\n", "solution": "def MagicSquareDoublyEven(order):\n    sq = [range(1+n*order,order + (n*order)+1) for n in range(order) ]\n    n1 = order/4\n    for r in range(n1):\n        r1 = sq[r][n1:-n1]\n        r2 = sq[order -r - 1][n1:-n1]\n        r1.reverse()\n        r2.reverse()\n        sq[r][n1:-n1] = r2\n        sq[order -r - 1][n1:-n1] = r1\n    for r in range(n1, order-n1):\n        r1 = sq[r][:n1]\n        r2 = sq[order -r - 1][order-n1:]\n        r1.reverse()\n        r2.reverse()\n        sq[r][:n1] = r2\n        sq[order -r - 1][order-n1:] = r1\n    return sq\n\ndef printsq(s):\n    n = len(s)\n    bl = len(str(n**2))+1\n    for i in range(n):\n        print ''.join( [ (\"%\"+str(bl)+\"s\")%(str(x)) for x in s[i]] )\n    print \"\\nMagic constant = %d\"%sum(s[0])\n\nprintsq(MagicSquareDoublyEven(8))\n"}
{"title": "Magic squares of doubly even order", "language": "Python 3.7", "task": "A magic square is an   '''NxN'''  square matrix whose numbers consist of consecutive numbers arranged so that the sum of each row and column,   ''and''   both diagonals are equal to the same sum   (which is called the ''magic number'' or ''magic constant'').  \n\nA magic square of doubly even order has a size that is a multiple of four   (e.g.     '''4''', '''8''', '''12'''). \n\nThis means that the subsquares also have an even size, which plays a role in the construction.\n\n \n{| class=\"wikitable\" style=\"float:right;border: 2px solid black; background:lightblue; color:black; margin-left:auto;margin-right:auto;text-align:center;width:22em;height:15em;table-layout:fixed;font-size:100%\"\n|-\n|'''1'''||'''2'''||'''62'''||'''61'''||'''60'''||'''59'''||'''7'''||'''8'''\n|-\n|'''9'''||'''10'''||'''54'''||'''53'''||'''52'''||'''51'''||'''15'''||'''16'''\n|-\n|'''48'''||'''47'''||'''19'''||'''20'''||'''21'''||'''22'''||'''42'''||'''41'''\n|-\n|'''40'''||'''39'''||'''27'''||'''28'''||'''29'''||'''30'''||'''34'''||'''33'''\n|-\n|'''32'''||'''31'''||'''35'''||'''36'''||'''37'''||'''38'''||'''26'''||'''25'''\n|-\n|'''24'''||'''23'''||'''43'''||'''44'''||'''45'''||'''46'''||'''18'''||'''17'''\n|-\n|'''49'''||'''50'''||'''14'''||'''13'''||'''12'''||'''11'''||'''55'''||'''56'''\n|-\n|'''57'''||'''58'''||'''6'''||'''5'''||'''4'''||'''3'''||'''63'''||'''64'''\n|}\n\n;Task\nCreate a magic square of    '''8 x 8'''.\n\n\n;Related tasks\n* [[Magic squares of odd order]]\n* [[Magic squares of singly even order]]\n\n\n;See also:\n* Doubly Even Magic Squares (1728.org)\n\n", "solution": "'''Magic squares of doubly even order'''\n\nfrom itertools import chain, repeat\nfrom functools import reduce\nfrom math import log\n\n\n# doublyEvenMagicSquare :: Int -> [[Int]]\ndef doublyEvenMagicSquare(n):\n    '''Magic square of order n'''\n\n    # magic :: Int -> [Bool]\n    def magic(n):\n        '''Truth-table series'''\n        if 0 < n:\n            xs = magic(n - 1)\n            return xs + [not x for x in xs]\n        else:\n            return [True]\n\n    sqr = n * n\n    power = log(sqr, 2)\n    scale = replicate(n / 4)\n    return chunksOf(n)([\n        succ(i) if bln else sqr - i for i, bln in\n        enumerate(magic(power) if isInteger(power) else (\n            flatten(scale(\n                map(scale, chunksOf(4)(magic(4)))\n            ))\n        ))\n    ])\n\n\n# TEST ----------------------------------------------------\n# main :: IO()\ndef main():\n    '''Tests'''\n\n    order = 8\n    magicSquare = doublyEvenMagicSquare(order)\n\n    print(\n        'Row sums: ',\n        [sum(xs) for xs in magicSquare],\n        '\\nCol sums:',\n        [sum(xs) for xs in transpose(magicSquare)],\n        '\\n1st diagonal sum:',\n        sum(magicSquare[i][i] for i in range(0, order)),\n        '\\n2nd diagonal sum:',\n        sum(magicSquare[i][(order - 1) - i] for i in range(0, order)),\n        '\\n'\n    )\n    print(wikiTable({\n        'class': 'wikitable',\n        'style': cssFromDict({\n            'text-align': 'center',\n            'color': '#605B4B',\n            'border': '2px solid silver'\n        }),\n        'colwidth': '3em'\n    })(magicSquare))\n\n\n# GENERIC -------------------------------------------------\n\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n,\n       subdividing the contents of xs.\n       Where the length of xs is not evenly divible\n       the final list will be shorter than n.'''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''Concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output a in list\n       (using an empty list to represent computational failure).'''\n    return lambda xs: list(\n        chain.from_iterable(\n            map(f, xs)\n        )\n    )\n\n\n# cssFromDict :: Dict -> String\ndef cssFromDict(dct):\n    '''CSS string serialized from key values in a Dictionary.'''\n    return reduce(\n        lambda a, k: a + k + ':' + dct[k] + '; ', dct.keys(), ''\n    )\n\n\n# flatten :: NestedList a -> [a]\ndef flatten(x):\n    '''A list of atoms resulting from fully flattening\n       an arbitrarily nested list.'''\n    return concatMap(flatten)(x) if isinstance(x, list) else [x]\n\n\n# isInteger :: Num -> Bool\ndef isInteger(n):\n    '''Divisible by one without remainder ?'''\n    return 0 == (n - int(n))\n\n\n# replicate :: Int -> a -> [a]\ndef replicate(n):\n    '''A list of length n in which every element\n       has the value x.'''\n    return lambda x: list(repeat(x, n))\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value. For numeric types, (1 +).'''\n    return 1 + x if isinstance(x, int) else (\n        chr(1 + ord(x))\n    )\n\n\n# transpose :: Matrix a -> Matrix a\ndef transpose(m):\n    '''The rows and columns of the argument transposed.\n       (The matrix containers and rows can be lists or tuples).'''\n    if m:\n        inner = type(m[0])\n        z = zip(*m)\n        return (type(m))(\n            map(inner, z) if tuple != inner else z\n        )\n    else:\n        return m\n\n\n# wikiTable :: Dict -> [[a]] -> String\ndef wikiTable(opts):\n    '''List of lists rendered as a wiki table string.'''\n    def colWidth():\n        return 'width:' + opts['colwidth'] + '; ' if (\n            'colwidth' in opts\n        ) else ''\n\n    def cellStyle():\n        return opts['cell'] if 'cell' in opts else ''\n\n    return lambda rows: '{| ' + reduce(\n        lambda a, k: (\n            a + k + '=\"' + opts[k] + '\" ' if k in opts else a\n        ),\n        ['class', 'style'],\n        ''\n    ) + '\\n' + '\\n|-\\n'.join(\n        '\\n'.join(\n            ('|' if (0 != i and ('cell' not in opts)) else (\n                '|style=\"' + colWidth() + cellStyle() + '\"|'\n            )) + (\n                str(x) or ' '\n            ) for x in row\n        ) for i, row in enumerate(rows)\n    ) + '\\n|}\\n\\n'\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Magic squares of odd order", "language": "Python", "task": "A magic square is an   '''NxN'''   square matrix whose numbers (usually integers) consist of consecutive numbers arranged so that the sum of each row and column,   ''and''   both long (main) diagonals are equal to the same sum (which is called the   ''magic number''   or   ''magic constant''). \n\nThe numbers are usually (but not always) the first   '''N'''2   positive integers.\n\nA magic square whose rows and columns add up to a magic number but whose main diagonals do not, is known as a ''semimagic square''.\n\n{| class=\"wikitable\" style=\"float:right;border: 4px solid blue; background:lightgreen; color:black; margin-left:auto;margin-right:auto;text-align:center;width:15em;height:15em;table-layout:fixed;font-size:150%\"\n|-\n| '''8''' || '''1''' || '''6'''\n|-\n| '''3''' || '''5''' || '''7'''\n|-\n| '''4''' || '''9''' || '''2'''\n|}\n\n\n;Task\nFor any odd   '''N''',   generate a magic square with the integers   ''' 1''' --> '''N''',   and show the results here. \n\nOptionally, show the ''magic number''.  \n\nYou should demonstrate the generator by showing at least a magic square for   '''N''' = '''5'''.\n\n\n; Related tasks\n* [[Magic squares of singly even order]]\n* [[Magic squares of doubly even order]]\n\n\n; See also:\n* MathWorld(tm) entry: Magic_square \n* Odd Magic Squares (1728.org)\n\n", "solution": ">>> def magic(n):\n    for row in range(1, n + 1):\n        print(' '.join('%*i' % (len(str(n**2)), cell) for cell in\n                       (n * ((row + col - 1 + n // 2) % n) +\n                       ((row + 2 * col - 2) % n) + 1\n                       for col in range(1, n + 1))))\n    print('\\nAll sum to magic number %i' % ((n * n + 1) * n // 2))\n\n    \n>>> for n in (5, 3, 7):\n\tprint('\\nOrder %i\\n=======' % n)\n\tmagic(n)\n\n\t\n\nOrder 5\n=======\n17 24  1  8 15\n23  5  7 14 16\n 4  6 13 20 22\n10 12 19 21  3\n11 18 25  2  9\n\nAll sum to magic number 65\n\nOrder 3\n=======\n8 1 6\n3 5 7\n4 9 2\n\nAll sum to magic number 15\n\nOrder 7\n=======\n30 39 48  1 10 19 28\n38 47  7  9 18 27 29\n46  6  8 17 26 35 37\n 5 14 16 25 34 36 45\n13 15 24 33 42 44  4\n21 23 32 41 43  3 12\n22 31 40 49  2 11 20\n\nAll sum to magic number 175\n>>> "}
{"title": "Magic squares of odd order", "language": "Python 3.7", "task": "A magic square is an   '''NxN'''   square matrix whose numbers (usually integers) consist of consecutive numbers arranged so that the sum of each row and column,   ''and''   both long (main) diagonals are equal to the same sum (which is called the   ''magic number''   or   ''magic constant''). \n\nThe numbers are usually (but not always) the first   '''N'''2   positive integers.\n\nA magic square whose rows and columns add up to a magic number but whose main diagonals do not, is known as a ''semimagic square''.\n\n{| class=\"wikitable\" style=\"float:right;border: 4px solid blue; background:lightgreen; color:black; margin-left:auto;margin-right:auto;text-align:center;width:15em;height:15em;table-layout:fixed;font-size:150%\"\n|-\n| '''8''' || '''1''' || '''6'''\n|-\n| '''3''' || '''5''' || '''7'''\n|-\n| '''4''' || '''9''' || '''2'''\n|}\n\n\n;Task\nFor any odd   '''N''',   generate a magic square with the integers   ''' 1''' --> '''N''',   and show the results here. \n\nOptionally, show the ''magic number''.  \n\nYou should demonstrate the generator by showing at least a magic square for   '''N''' = '''5'''.\n\n\n; Related tasks\n* [[Magic squares of singly even order]]\n* [[Magic squares of doubly even order]]\n\n\n; See also:\n* MathWorld(tm) entry: Magic_square \n* Odd Magic Squares (1728.org)\n\n", "solution": "'''Magic squares of odd order N'''\n\nfrom itertools import cycle, islice, repeat\nfrom functools import reduce\n\n\n# magicSquare :: Int -> [[Int]]\ndef magicSquare(n):\n    '''Magic square of odd order n.'''\n    return applyN(2)(\n        compose(transposed)(cycled)\n    )(plainSquare(n)) if 1 == n % 2 else []\n\n\n# plainSquare :: Int -> [[Int]]\ndef plainSquare(n):\n    '''The sequence of integers from 1 to N^2,\n       subdivided into N sub-lists of equal length,\n       forming N rows, each of N integers.\n    '''\n    return chunksOf(n)(\n        enumFromTo(1)(n ** 2)\n    )\n\n\n# cycled :: [[Int]] -> [[Int]]\ndef cycled(rows):\n    '''A table in which the rows are\n       rotated by descending deltas.\n    '''\n    n = len(rows)\n    d = n // 2\n    return list(map(\n        lambda d, xs: take(n)(\n            drop(n - d)(cycle(xs))\n        ),\n        enumFromThenTo(d)(d - 1)(-d),\n        rows\n    ))\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Magic squares of order 3, 5, 7'''\n    print(\n        fTable(__doc__ + ':')(lambda x: '\\n' + repr(x))(\n            showSquare\n        )(magicSquare)([3, 5, 7])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# applyN :: Int -> (a -> a) -> a -> a\ndef applyN(n):\n    '''n applications of f.\n       (Church numeral n).\n    '''\n    def go(f):\n        return lambda x: reduce(\n            lambda a, g: g(a), repeat(f, n), x\n        )\n    return lambda f: go(f)\n\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n,\n       subdividing the contents of xs.\n       Where the length of xs is not evenly divible,\n       the final list will be shorter than n.'''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The sublist of xs beginning at\n       (zero-based) index n.'''\n    def go(xs):\n        if isinstance(xs, (list, tuple, str)):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return lambda xs: go(xs)\n\n\n# enumFromThenTo :: Int -> Int -> Int -> [Int]\ndef enumFromThenTo(m):\n    '''Integer values enumerated from m to n\n       with a step defined by nxt-m.\n    '''\n    def go(nxt, n):\n        d = nxt - m\n        return range(m, n - 1 if d < 0 else 1 + n, d)\n    return lambda nxt: lambda n: list(go(nxt, n))\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, (list, tuple))\n        else list(islice(xs, n))\n    )\n\n\n# transposed :: Matrix a -> Matrix a\ndef transposed(m):\n    '''The rows and columns of the argument transposed.\n       (The matrix containers and rows can be lists or tuples).\n    '''\n    if m:\n        inner = type(m[0])\n        z = zip(*m)\n        return (type(m))(\n            map(inner, z) if tuple != inner else z\n        )\n    else:\n        return m\n\n\n# DISPLAY -------------------------------------------------\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# indented :: Int -> String -> String\ndef indented(n):\n    '''String indented by n multiples\n       of four spaces\n    '''\n    return lambda s: (n * 4 * ' ') + s\n\n\n# showSquare :: [[Int]] -> String\ndef showSquare(rows):\n    '''Lines representing rows of lists.'''\n    w = 1 + len(str(reduce(max, map(max, rows), 0)))\n    return '\\n' + '\\n'.join(\n        map(\n            lambda row: indented(1)(''.join(\n                map(lambda x: str(x).rjust(w, ' '), row)\n            )),\n            rows\n        )\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Magic squares of singly even order", "language": "Python from Lua", "task": "A magic square is an '''NxN''' square matrix whose numbers consist of consecutive numbers arranged so that the sum of each row and column, ''and'' both diagonals are equal to the same sum (which is called the ''magic number'' or ''magic constant''). \n\nA  magic square of singly even order has a size that is a multiple of 4, plus 2 (e.g. 6, 10, 14). This means that the subsquares have an odd size, which plays a role in the construction.\n\n;Task\nCreate a magic square of 6 x 6.\n\n\n; Related tasks\n* [[Magic squares of odd order]]\n* [[Magic squares of doubly even order]]\n\n; See also\n* Singly Even Magic Squares (1728.org)\n\n", "solution": "import math\nfrom sys import stdout\n\nLOG_10 = 2.302585092994\n\n\n# build odd magic square\ndef build_oms(s):\n    if s % 2 == 0:\n        s += 1\n    q = [[0 for j in range(s)] for i in range(s)]\n    p = 1\n    i = s // 2\n    j = 0\n    while p <= (s * s):\n        q[i][j] = p\n        ti = i + 1\n        if ti >= s: ti = 0\n        tj = j - 1\n        if tj < 0: tj = s - 1\n        if q[ti][tj] != 0:\n            ti = i\n            tj = j + 1\n        i = ti\n        j = tj\n        p = p + 1\n\n    return q, s\n\n\n# build singly even magic square\ndef build_sems(s):\n    if s % 2 == 1:\n        s += 1\n    while s % 4 == 0:\n        s += 2\n\n    q = [[0 for j in range(s)] for i in range(s)]\n    z = s // 2\n    b = z * z\n    c = 2 * b\n    d = 3 * b\n    o = build_oms(z)\n\n    for j in range(0, z):\n        for i in range(0, z):\n            a = o[0][i][j]\n            q[i][j] = a\n            q[i + z][j + z] = a + b\n            q[i + z][j] = a + c\n            q[i][j + z] = a + d\n\n    lc = z // 2\n    rc = lc\n    for j in range(0, z):\n        for i in range(0, s):\n            if i < lc or i > s - rc or (i == lc and j == lc):\n                if not (i == 0 and j == lc):\n                    t = q[i][j]\n                    q[i][j] = q[i][j + z]\n                    q[i][j + z] = t\n\n    return q, s\n\n\ndef format_sqr(s, l):\n    for i in range(0, l - len(s)):\n        s = \"0\" + s\n    return s + \" \"\n\n\ndef display(q):\n    s = q[1]\n    print(\" - {0} x {1}\\n\".format(s, s))\n    k = 1 + math.floor(math.log(s * s) / LOG_10)\n    for j in range(0, s):\n        for i in range(0, s):\n            stdout.write(format_sqr(\"{0}\".format(q[0][i][j]), k))\n        print()\n    print(\"Magic sum: {0}\\n\".format(s * ((s * s) + 1) // 2))\n\n\nstdout.write(\"Singly Even Magic Square\")\ndisplay(build_sems(6))\n"}
{"title": "Map range", "language": "Python", "task": "Given two ranges: \n:::*   [a_1,a_2]   and \n:::*   [b_1,b_2];  \n:::*   then a value   s   in range   [a_1,a_2]\n:::*   is linearly mapped to a value   t   in range   [b_1,b_2]\n   where: \n\n\n:::*   t = b_1 + {(s - a_1)(b_2 - b_1) \\over (a_2 - a_1)}\n\n\n;Task:\nWrite a function/subroutine/... that takes two ranges and a real number, and returns the mapping of the real number from the first to the second range. \n\nUse this function to map values from the range    [0, 10]    to the range    [-1, 0]. \n\n\n;Extra credit:\nShow additional idiomatic ways of performing the mapping, using tools available to the language.\n\n", "solution": ">>> def maprange( a, b, s):\n\t(a1, a2), (b1, b2) = a, b\n\treturn  b1 + ((s - a1) * (b2 - b1) / (a2 - a1))\n\n>>> for s in range(11):\n\tprint(\"%2g maps to %g\" % (s, maprange( (0, 10), (-1, 0), s)))\n\n\t\n 0 maps to -1\n 1 maps to -0.9\n 2 maps to -0.8\n 3 maps to -0.7\n 4 maps to -0.6\n 5 maps to -0.5\n 6 maps to -0.4\n 7 maps to -0.3\n 8 maps to -0.2\n 9 maps to -0.1\n10 maps to 0"}
{"title": "Matrix digital rain", "language": "Python", "task": "Implement the Matrix Digital Rain visual effect from the movie \"The Matrix\" as described in Wikipedia.\n\nProvided is a reference implementation in Common Lisp to be run in a terminal.\n\n", "solution": "import curses\nimport random\nimport time\n\n\"\"\"\n\nBased on C ncurses version\n\nhttp://rosettacode.org/wiki/Matrix_Digital_Rain#NCURSES_version\n\n\"\"\"\n\n\"\"\"\nTime between row updates in seconds\nControls the speed of the digital rain effect.\n\"\"\"\n\nROW_DELAY=.0001\n\ndef get_rand_in_range(min, max):\n    return random.randrange(min,max+1)\n\ntry:\n    # Characters to randomly appear in the rain sequence.\n    chars = ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']\n    \n    total_chars = len(chars)\n        \n    stdscr = curses.initscr()\n    curses.noecho()\n    curses.curs_set(False)\n        \n    curses.start_color()\n    curses.init_pair(1, curses.COLOR_GREEN, curses.COLOR_BLACK)\n    stdscr.attron(curses.color_pair(1))\n    \n    max_x = curses.COLS - 1\n    max_y = curses.LINES - 1\n    \n         \n    # Create arrays of columns based on screen width.\n    \n    # Array containing the current row of each column.\n    \n    columns_row = []\n    \n    # Array containing the active status of each column.\n    # A column draws characters on a row when active.\n    \n    columns_active = []\n    \n    for i in range(max_x+1):\n        columns_row.append(-1)\n        columns_active.append(0)\n        \n    while(True):\n        for i in range(max_x):\n            if columns_row[i] == -1:\n                # If a column is at the top row, pick a\n                # random starting row and active status.\n                columns_row[i] = get_rand_in_range(0, max_y)\n                columns_active[i] = get_rand_in_range(0, 1)\n     \n        # Loop through columns and draw characters on rows\n        \n        for i in range(max_x):\n            if columns_active[i] == 1:\n                # Draw a random character at this column's current row.\n                char_index = get_rand_in_range(0, total_chars-1)\n                #mvprintw(columns_row[i], i, \"%c\", chars[char_index])                \n                stdscr.addstr(columns_row[i], i, chars[char_index])\n            else:\n                # Draw an empty character if the column is inactive.\n                #mvprintw(columns_row[i], i, \" \");\n                stdscr.addstr(columns_row[i], i, \" \");\n                \n     \n            columns_row[i]+=1\n     \n            # When a column reaches the bottom row, reset to top.\n            if columns_row[i] >= max_y:\n                columns_row[i] = -1\n    \n            # Randomly alternate the column's active status.\n            if get_rand_in_range(0, 1000) == 0:\n                if columns_active[i] == 0:      \n                    columns_active[i] = 1\n                else:\n                    columns_active[i] = 0\n     \n            time.sleep(ROW_DELAY)\n            stdscr.refresh()\n    \nexcept KeyboardInterrupt as err:\n    curses.endwin()    \n"}
{"title": "Maximum triangle path sum", "language": "Python", "task": "Starting from the top of a pyramid of numbers like this, you can walk down going one step on the right or on the left, until you reach the bottom row:\n\n                          55\n                        94 48\n                       95 30 96\n                     77 71 26 67\n\n\nOne of such walks is 55 - 94 - 30 - 26. \nYou can compute the total of the numbers you have seen in such walk, \nin this case it's 205.\n\nYour problem is to find the maximum total among all possible paths from the top to the bottom row of the triangle. In the little example above it's 321.\n\n\n;Task:\nFind the maximum total in the triangle below:\n\n                          55\n                        94 48\n                       95 30 96\n                     77 71 26 67\n                    97 13 76 38 45\n                  07 36 79 16 37 68\n                 48 07 09 18 70 26 06\n               18 72 79 46 59 79 29 90\n              20 76 87 11 32 07 07 49 18\n            27 83 58 35 71 11 25 57 29 85\n           14 64 36 96 27 11 58 56 92 18 55\n         02 90 03 60 48 49 41 46 33 36 47 23\n        92 50 48 02 36 59 42 79 72 20 82 77 42\n      56 78 38 80 39 75 02 71 66 66 01 03 55 72\n     44 25 67 84 71 67 11 61 40 57 58 89 40 56 36\n   85 32 25 85 57 48 84 35 47 62 17 01 01 99 89 52\n  06 71 28 75 94 48 37 10 23 51 06 48 53 18 74 98 15\n27 02 92 23 08 71 76 84 15 52 92 63 81 10 44 10 69 93\n\n\nSuch numbers can be included in the solution code, or read from a \"triangle.txt\" file.\n\nThis task is derived from the Euler Problem #18.\n\n", "solution": "'''Maximum triangle path sum'''\n\nfrom functools import (reduce)\n\n\n# maxPathSum :: [[Int]] -> Int\ndef maxPathSum(rows):\n    '''The maximum total among all possible\n       paths from the top to the bottom row.\n    '''\n    return reduce(\n        lambda xs, ys: [\n            a + max(b, c) for (a, b, c)\n            in zip(ys, xs, xs[1:])\n        ],\n        reversed(rows[:-1]), rows[-1]\n    )[0]\n\n\n# ------------------------- TEST -------------------------\nprint(\n    maxPathSum([\n        [55],\n        [94, 48],\n        [95, 30, 96],\n        [77, 71, 26, 67],\n        [97, 13, 76, 38, 45],\n        [7, 36, 79, 16, 37, 68],\n        [48, 7, 9, 18, 70, 26, 6],\n        [18, 72, 79, 46, 59, 79, 29, 90],\n        [20, 76, 87, 11, 32, 7, 7, 49, 18],\n        [27, 83, 58, 35, 71, 11, 25, 57, 29, 85],\n        [14, 64, 36, 96, 27, 11, 58, 56, 92, 18, 55],\n        [2, 90, 3, 60, 48, 49, 41, 46, 33, 36, 47, 23],\n        [92, 50, 48, 2, 36, 59, 42, 79, 72, 20, 82, 77, 42],\n        [56, 78, 38, 80, 39, 75, 2, 71, 66, 66, 1, 3, 55, 72],\n        [44, 25, 67, 84, 71, 67, 11, 61, 40, 57, 58, 89, 40, 56, 36],\n        [85, 32, 25, 85, 57, 48, 84, 35, 47, 62, 17, 1, 1, 99, 89, 52],\n        [6, 71, 28, 75, 94, 48, 37, 10, 23, 51, 6, 48, 53, 18, 74, 98, 15],\n        [27, 2, 92, 23, 8, 71, 76, 84, 15, 52, 92, 63, 81, 10, 44, 10, 69, 93]\n    ])\n)"}
{"title": "Mayan calendar", "language": "Python 3.6", "task": "The ancient Maya people had two somewhat distinct calendar systems.\n\nIn somewhat simplified terms, one is a cyclical calendar known as '''The Calendar Round''',\nthat meshes several sacred and civil cycles; the other is an offset calendar\nknown as '''The Long Count''', similar in many ways to the Gregorian calendar.\n\n'''The Calendar Round'''\n\n'''The Calendar Round''' has several intermeshing sacred and civil cycles that uniquely identify a specific date in an approximately 52 year cycle.\n\n'''The Tzolk'in'''\n\nThe sacred cycle in the Mayan calendar round was called the '''Tzolk'in'''. The '''Tzolk'in''' has a cycle of 20 day names:\n\n    Imix'\n    Ik'\n    Ak'bal\n    K'an\n    Chikchan\n    Kimi\n    Manik'\n    Lamat\n    Muluk\n    Ok\n    Chuwen\n    Eb\n    Ben\n    Hix\n    Men\n    K'ib'\n    Kaban\n    Etz'nab'\n    Kawak\n    Ajaw\n\nIntermeshed with the named days, the '''Tzolk'in''' has a cycle of 13 numbered days; 1\nthrough 13. Every day has both a number and a name that repeat in a 260 day cycle.\n\nFor example:\n\n    1 Imix'\n    2 Ik'\n    3 Ak'bal\n    ...\n    11 Chuwen\n    12 Eb\n    13 Ben\n    1 Hix\n    2 Men\n    3 K'ib'\n    ... and so on.\n\n'''The Haab''''\n\nThe Mayan civil calendar is called the '''Haab''''. This calendar has 365 days per\nyear, and is sometimes called the 'vague year.' It is substantially the same as\nour year, but does not make leap year adjustments, so over long periods of time,\ngets out of synchronization with the seasons. It consists of 18 months with 20 days each, \nand the end of the year, a special month of only 5 days, giving a total of 365. \nThe 5 days of the month of '''Wayeb'''' (the last month), are usually considered to be \na time of bad luck.\n\nEach month in the '''Haab'''' has a name. The Mayan names for the civil months are:\n\n    Pop\n    Wo'\n    Sip\n    Sotz'\n    Sek\n    Xul\n    Yaxk'in\n    Mol\n    Ch'en\n    Yax\n    Sak'\n    Keh\n    Mak\n    K'ank'in\n    Muwan\n    Pax\n    K'ayab\n    Kumk'u\n    Wayeb' (Short, \"unlucky\" month)\n\nThe months function very much like ours do. That is, for any given month we\ncount through all the days of that month, and then move on to the next month.\n\nNormally, the day '''1 Pop''' is considered the first day of the civil year, just as '''1 January''' \nis the first day of our year. In 2019, '''1 Pop''' falls on '''April 2nd'''. But,\nbecause of the leap year in 2020, '''1 Pop''' falls on '''April 1st''' in the years 2020-2023.\n\nThe only really unusual aspect of the '''Haab'''' calendar is that, although there are\n20 (or 5) days in each month, the last day of the month is not called\nthe 20th (5th). Instead, the last day of the month is referred to as the\n'seating,' or 'putting in place,' of the next month. (Much like how in our\nculture, December 24th is Christmas Eve and December 31st is 'New-Years Eve'.) In\nthe language of the ancient Maya, the word for seating  was '''chum''', So you might\nhave:\n\n    ...\n    18 Pop (18th day of the first month)\n    19 Pop (19th day of the first month)\n    Chum Wo' (20th day of the first month)\n    1 Wo' (1st day of the second month)\n    ... and so on.\n\nDates for any particular day are a combination of the '''Tzolk'in''' sacred date,\nand '''Haab'''' civil date. When put together we get the '''\"Calendar Round.\"'''\n\n'''Calendar Round''' dates always have two numbers and two names, and they are\nalways written in the same order:\n\n    (1) the day number in the '''Tzolk'in'''\n    (2) the day name in the  '''Tzolk'in'''\n    (3) the day of the month in the '''Haab''''\n    (4) the month name in the '''Haab''''\n\nA calendar round is a repeating cycle with a period of just short of 52\nGregorian calendar years. To be precise: it is 52 x 365 days. (No leap days)\n\n'''Lords of the Night'''\n\nA third cycle of nine days honored the nine '''Lords of the Night'''; nine deities\nthat were associated with each day in turn. The names of the nine deities are\nlost; they are now commonly referred to as '''G1''' through '''G9'''. The Lord of the Night\nmay or may not be included in a Mayan date, if it is, it is typically\njust the appropriate '''G(''x'')''' at the end.\n\n'''The Long Count'''\n\nMayans had a separate independent way of measuring time that did not run in\ncycles. (At least, not on normal human scales.) Instead, much like our yearly\ncalendar, each day just gets a little further from the starting point. For the\nancient Maya, the starting point was the 'creation date' of the current world.\nThis date corresponds to our date of August 11, 3114 B.C. Dates are calculated\nby measuring how many days have transpired since this starting date; This is\ncalled '''\"The Long Count.\"''' Rather than just an absolute count of days, the long\ncount is broken up into unit blocks, much like our calendar has months, years,\ndecades and centuries.\n\n    The basic unit is a '''k'in''' - one day.\n    A 20 day month is a '''winal'''.\n    18 '''winal''' (360 days) is a '''tun''' - sometimes referred to as a short year.\n    20 short years ('''tun''') is a '''k'atun'''\n    20 '''k'atun''' is a '''bak'tun'''\n\nThere are longer units too:\n\n    '''Piktun''' == 20 '''Bak'tun''' (8,000 short years)\n    '''Kalabtun'''  == 20 '''Piktun''' (160,000 short years)\n    '''Kinchiltun'''  == 20 '''Kalabtun''' (3,200,000 short years)\n\nFor the most part, the Maya only used the blocks of time up to the '''bak'tun'''. One\n'''bak'tun''' is around 394 years, much more than a human life span, so that was all\nthey usually needed to describe dates in this era, or this world. It is worth\nnoting, the two calendars working together allow much more accurate and reliable\nnotation for dates than is available in many other calendar systems; mostly due\nto the pragmatic choice to make the calendar simply track days, rather than\ntrying to make it align with seasons and/or try to conflate it with the notion\nof time.\n\n'''Mayan Date correlations'''\n\nThere is some controversy over finding a correlation point between the Gregorian\nand Mayan calendars. The Gregorian calendar is full of jumps and skips to keep\nthe calendar aligned with the seasons so is much more difficult to work with.\nThe most commonly used correlation\nfactor is The '''GMT: 584283'''. Julian 584283 is a day count corresponding '''Mon, Aug 11, 3114 BCE''' \nin the Gregorian calendar, and the final day in the last Mayan long count\ncycle: 13.0.0.0.0 which is referred to as \"the day of creation\" in the Mayan\ncalendar. There is nothing in known Mayan writing or history that suggests that\na long count \"cycle\" resets ''every'' 13 '''bak'tun'''. Judging by their other practices,\nit would make much more sense for it to reset at 20, if at all.\n\nThe reason there was much interest at all, outside historical scholars, in\nthe Mayan calendar is that the long count recently (relatively speaking) rolled over to 13.0.0.0.0 (the same as the historic day of creation Long Count date) on Fri,\nDec 21, 2012 (using the most common GMT correlation factor), prompting conspiracy\ntheorists to predict a cataclysmic \"end-of-the-world\" scenario.\n\n'''Excerpts taken from, and recommended reading:'''\n\n*''From the website of the Foundation for the Advancement of Meso-America Studies, Inc.'' Pitts, Mark. The complete Writing in Maya Glyphs Book 2 - Maya Numbers and the Maya Calendar. 2009. Accessed 2019-01-19. http://www.famsi.org/research/pitts/MayaGlyphsBook2.pdf\n\n*wikipedia: Maya calendar\n\n*wikipedia: Mesoamerican Long Count calendar\n\n\n'''The Task:'''\n\nWrite a reusable routine that takes a Gregorian date and returns the equivalent date in Mayan in the Calendar Round and the Long Count. At a minimum, use the GMT correlation. ''If desired, support other correlations.''\n\nUsing the GMT correlation, the following Gregorian and Mayan dates are equivalent:  \n\n   '''Dec 21, 2012''' (Gregorian)\n   '''4 Ajaw 3 K'ank'in G9''' (Calendar round)\n   '''13.0.0.0.0''' (Long count)\n\nSupport looking up dates for at least 50 years before and after the Mayan Long Count '''13 bak'tun''' rollover: ''Dec 21, 2012''. ''(Will require taking into account Gregorian leap days.)''\n\nShow the output here, on this page, for ''at least'' the following dates:\n\n''(Note that these are in ISO-8601 format: YYYY-MM-DD. There is no requirement to use ISO-8601 format in your program, but if you differ, make a note of the expected format.)''\n\n    2004-06-19\n    2012-12-18\n    2012-12-21\n    2019-01-19\n    2019-03-27\n    2020-02-29\n    2020-03-01\n", "solution": "import datetime\n\n\ndef g2m(date, gtm_correlation=True):\n    \"\"\"\n    Translates Gregorian date into Mayan date, see\n    https://rosettacode.org/wiki/Mayan_calendar\n\n    Input arguments:\n\n     date: string date in ISO-8601 format: YYYY-MM-DD\n     gtm_correlation: GTM correlation to apply if True, Astronomical correlation otherwise (optional, True by default)\n\n    Output arguments:\n\n     long_date: Mayan date in Long Count system as string\n     round_date: Mayan date in Calendar Round system as string\n    \"\"\"\n\n    # define some parameters and names\n\n    correlation = 584283 if gtm_correlation else 584285\n\n    long_count_days = [144000, 7200, 360, 20, 1]\n\n    tzolkin_months = ['Imix\u2019', 'Ik\u2019', 'Ak\u2019bal', 'K\u2019an', 'Chikchan', 'Kimi', 'Manik\u2019', 'Lamat', 'Muluk', 'Ok', 'Chuwen',\n                      'Eb', 'Ben', 'Hix', 'Men', 'K\u2019ib\u2019', 'Kaban', 'Etz\u2019nab\u2019', 'Kawak', 'Ajaw']  # tzolk'in\n\n    haad_months = ['Pop', 'Wo\u2019', 'Sip', 'Sotz\u2019', 'Sek', 'Xul', 'Yaxk\u2019in', 'Mol', 'Ch\u2019en', 'Yax', 'Sak\u2019', 'Keh', 'Mak',\n                   'K\u2019ank\u2019in', 'Muwan', 'Pax', 'K\u2019ayab', 'Kumk\u2019u', 'Wayeb\u2019']  # haab'\n\n    gregorian_days = datetime.datetime.strptime(date, '%Y-%m-%d').toordinal()\n    julian_days = gregorian_days + 1721425\n\n    # 1. calculate long count date\n\n    long_date = list()\n    remainder = julian_days - correlation\n\n    for days in long_count_days:\n\n        result, remainder = divmod(remainder, days)\n        long_date.append(int(result))\n\n    long_date = '.'.join(['{:02d}'.format(d) for d in long_date])\n\n    # 2. calculate round calendar date\n\n    tzolkin_month = (julian_days + 16) % 20\n    tzolkin_day = ((julian_days + 5) % 13) + 1\n\n    haab_month = int(((julian_days + 65) % 365) / 20)\n    haab_day = ((julian_days + 65) % 365) % 20\n    haab_day = haab_day if haab_day else 'Chum'\n\n    lord_number = (julian_days - correlation) % 9\n    lord_number = lord_number if lord_number else 9\n\n    round_date = f'{tzolkin_day} {tzolkin_months[tzolkin_month]} {haab_day} {haad_months[haab_month]} G{lord_number}'\n\n    return long_date, round_date\n\nif __name__ == '__main__':\n\n    dates = ['2004-06-19', '2012-12-18', '2012-12-21', '2019-01-19', '2019-03-27', '2020-02-29', '2020-03-01']\n\n    for date in dates:\n\n        long, round = g2m(date)\n        print(date, long, round)\n"}
{"title": "McNuggets problem", "language": "Python", "task": "From Wikipedia:\n The McNuggets version of the coin problem was introduced by Henri Picciotto,\n who included it in his algebra textbook co-authored with Anita Wah. Picciotto\n thought of the application in the 1980s while dining with his son at\n McDonald's, working the problem out on a napkin. A McNugget number is\n the total number of McDonald's Chicken McNuggets in any number of boxes.\n In the United Kingdom, the original boxes (prior to the introduction of\n the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.\n\n;Task:\nCalculate (from 0 up to a limit of 100) the largest non-McNuggets\nnumber (a number ''n'' which cannot be expressed with ''6x + 9y + 20z = n''\nwhere ''x'', ''y'' and ''z'' are natural numbers).\n\n", "solution": ">>> from itertools import product\n>>> nuggets = set(range(101))\n>>> for s, n, t in product(range(100//6+1), range(100//9+1), range(100//20+1)):\n\tnuggets.discard(6*s + 9*n + 20*t)\n\n\t\n>>> max(nuggets)\n43\n>>> "}
{"title": "McNuggets problem", "language": "Python from FSharp", "task": "From Wikipedia:\n The McNuggets version of the coin problem was introduced by Henri Picciotto,\n who included it in his algebra textbook co-authored with Anita Wah. Picciotto\n thought of the application in the 1980s while dining with his son at\n McDonald's, working the problem out on a napkin. A McNugget number is\n the total number of McDonald's Chicken McNuggets in any number of boxes.\n In the United Kingdom, the original boxes (prior to the introduction of\n the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.\n\n;Task:\nCalculate (from 0 up to a limit of 100) the largest non-McNuggets\nnumber (a number ''n'' which cannot be expressed with ''6x + 9y + 20z = n''\nwhere ''x'', ''y'' and ''z'' are natural numbers).\n\n", "solution": "#Wherein I observe that Set Comprehension is not intrinsically dysfunctional. Nigel Galloway: October 28th., 2018\nn = {n for x in range(0,101,20) for y in range(x,101,9) for n in range(y,101,6)}\ng = {n for n in range(101)}\nprint(max(g.difference(n)))\n"}
{"title": "McNuggets problem", "language": "Python 3.7", "task": "From Wikipedia:\n The McNuggets version of the coin problem was introduced by Henri Picciotto,\n who included it in his algebra textbook co-authored with Anita Wah. Picciotto\n thought of the application in the 1980s while dining with his son at\n McDonald's, working the problem out on a napkin. A McNugget number is\n the total number of McDonald's Chicken McNuggets in any number of boxes.\n In the United Kingdom, the original boxes (prior to the introduction of\n the Happy Meal-sized nugget boxes) were of 6, 9, and 20 nuggets.\n\n;Task:\nCalculate (from 0 up to a limit of 100) the largest non-McNuggets\nnumber (a number ''n'' which cannot be expressed with ''6x + 9y + 20z = n''\nwhere ''x'', ''y'' and ''z'' are natural numbers).\n\n", "solution": "'''mcNuggets list monad'''\n\nfrom itertools import (chain, dropwhile)\n\n\n# mcNuggetsByListMonad :: Int -> Set Int\ndef mcNuggetsByListMonad(limit):\n    '''McNugget numbers up to limit.'''\n\n    box = size(limit)\n    return set(\n        bind(\n            box(6)\n        )(lambda x: bind(\n            box(9)\n        )(lambda y: bind(\n            box(20)\n        )(lambda z: (\n            lambda v=sum([x, y, z]): (\n                [] if v > limit else [v]\n            )\n        )())))\n    )\n\n\n# Which, for comparison, is equivalent to:\n\n# mcNuggetsByComprehension :: Int -> Set Int\ndef mcNuggetsByComprehension(limit):\n    '''McNuggets numbers up to limit'''\n    box = size(limit)\n    return {\n        v for v in (\n            sum([x, y, z])\n            for x in box(6)\n            for y in box(9)\n            for z in box(20)\n        ) if v <= limit\n    }\n\n\n# size :: Int -> Int -> [Int]\ndef size(limit):\n    '''Multiples of n up to limit.'''\n    return lambda n: enumFromThenTo(0)(n)(limit)\n\n\n# -------------------------- TEST --------------------------\ndef main():\n    '''List monad and set comprehension - parallel routes'''\n\n    def test(limit):\n        def go(nuggets):\n            ys = list(dropwhile(\n                lambda x: x in nuggets,\n                enumFromThenTo(limit)(limit - 1)(1)\n            ))\n            return str(ys[0]) if ys else (\n                'No unreachable targets in this range.'\n            )\n        return lambda nuggets: go(nuggets)\n\n    def fName(f):\n        return f.__name__\n\n    limit = 100\n    print(\n        fTable(main.__doc__ + ':\\n')(fName)(test(limit))(\n            lambda f: f(limit)\n        )([mcNuggetsByListMonad, mcNuggetsByComprehension])\n    )\n\n\n# ------------------------ GENERIC -------------------------\n\n# bind (>>=) :: [a] -> (a -> [b]) -> [b]\ndef bind(xs):\n    '''List monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.\n    '''\n    return lambda f: chain.from_iterable(\n        map(f, xs)\n    )\n\n\n# enumFromThenTo :: Int -> Int -> Int -> [Int]\ndef enumFromThenTo(m):\n    '''Integer values enumerated from m to n\n       with a step defined by nxt-m.\n    '''\n    def go(nxt, n):\n        d = nxt - m\n        return range(m, n - 1 if d < 0 else 1 + n, d)\n    return lambda nxt: lambda n: go(nxt, n)\n\n\n# ------------------------ DISPLAY -------------------------\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def gox(xShow):\n        def gofx(fxShow):\n            def gof(f):\n                def goxs(xs):\n                    ys = [xShow(x) for x in xs]\n                    w = max(map(len, ys))\n\n                    def arrowed(x, y):\n                        return y.rjust(w, ' ') + ' -> ' + fxShow(f(x))\n                    return s + '\\n' + '\\n'.join(\n                        map(arrowed, xs, ys)\n                    )\n                return goxs\n            return gof\n        return gofx\n    return gox\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Meissel\u2013Mertens constant", "language": "Python from FreeBASIC", "task": "Calculate Meissel-Mertens constant up to a precision your language can handle.\n\n\n;Motivation:\nAnalogous to Euler's constant, which is important in determining the sum of reciprocal natural numbers, Meissel-Mertens' constant is important in calculating the sum of reciprocal primes.\n\n\n;Example:\nWe consider the finite sum of reciprocal natural numbers:\n\n''1 + 1/2 + 1/3 + 1/4 + 1/5 ... 1/n''\n\nthis sum can be well approximated with:\n\n''log(n) + E''  \n\nwhere ''E'' denotes Euler's constant: 0.57721...\n \n''log(n)'' denotes the natural logarithm of ''n''.\n\n\nNow consider the finite sum of reciprocal primes:\n\n''1/2 + 1/3 + 1/5 + 1/7 + 1/11 ... 1/p''\n\nthis sum can be well approximated with:\n\n''log( log(p) ) + M''\n\nwhere ''M'' denotes Meissel-Mertens constant: 0.26149...\n\n\n;See:\n:*   Details in the Wikipedia article:  Meissel-Mertens constant\n\n", "solution": "#!/usr/bin/python\n\nfrom math import log\n\ndef isPrime(n):\n    for i in range(2, int(n**0.5) + 1):\n        if n % i == 0:\n            return False        \n    return True\n\n\nif __name__ == '__main__':\n    Euler = 0.57721566490153286\n    m = 0\n    for x in range(2, 10_000_000):\n        if isPrime(x):\n            m += log(1-(1/x)) + (1/x)\n\n    print(\"MM =\", Euler + m)"}
{"title": "Memory layout of a data structure", "language": "Python", "task": "It is often useful to control the memory layout of fields in a data structure to match an interface control definition, or to interface with hardware. Define a data structure matching the RS-232 Plug Definition. Use the 9-pin definition for brevity.\n Pin Settings for Plug\n (Reverse order for socket.)\n __________________________________________\n 1  2  3  4  5  6  7  8  9  10 11 12 13\n  14 15 16 17 18 19 20 21 22 23 24 25\n _________________\n 1  2  3  4  5\n 6  7  8  9\n 25 pin                        9 pin\n 1 - PG   Protective ground\n 2 - TD   Transmitted data     3\n 3 - RD   Received data        2\n 4 - RTS  Request to send      7\n 5 - CTS  Clear to send        8\n 6 - DSR  Data set ready       6\n 7 - SG   Signal ground        5\n 8 - CD   Carrier detect       1\n 9 - + voltage (testing)\n 10 - - voltage (testing)\n 11 -\n 12 - SCD  Secondary CD\n 13 - SCS  Secondary CTS\n 14 - STD  Secondary TD\n 15 - TC   Transmit clock\n 16 - SRD  Secondary RD\n 17 - RC   Receiver clock\n 18 -\n 19 - SRS  Secondary RTS            \n 20 - DTR  Data terminal ready      4\n 21 - SQD  Signal quality detector\n 22 - RI   Ring indicator           9\n 23 - DRS  Data rate select\n 24 - XTC  External clock\n 25 -\n", "solution": "from ctypes import Structure, c_int\n\nrs232_9pin  = \"_0 CD RD TD DTR SG DSR RTS CTS RI\".split()\nrs232_25pin = ( \"_0  PG  TD  RD  RTS CTS DSR SG  CD  pos neg\"\n                \"_11 SCD SCS STD TC  SRD RC\"\n                \"_18 SRS DTR SQD RI DRS XTC\" ).split()\n\nclass RS232_9pin(Structure):\n    _fields_ = [(__, c_int, 1) for __ in rs232_9pin]\n\n\t\nclass RS232_25pin(Structure):\n    _fields_ = [(__, c_int, 1) for __ in rs232_25pin]"}
{"title": "Metallic ratios", "language": "Python", "task": "Many people have heard of the '''Golden ratio''', phi ('''ph'''). Phi is just one of a series\nof related ratios that are referred to as the \"'''Metallic ratios'''\".\n\nThe '''Golden ratio''' was discovered and named by ancient civilizations as it was\nthought to be the most pure and beautiful (like Gold). The '''Silver ratio''' was was\nalso known to the early Greeks, though was not named so until later as a nod to\nthe '''Golden ratio''' to which it is closely related. The series has been extended to\nencompass all of the related ratios and was given the general name '''Metallic ratios''' (or ''Metallic means'').\n''Somewhat incongruously as the original Golden ratio referred to the adjective \"golden\" rather than the metal \"gold\".''\n\n'''Metallic ratios''' are the real roots of the general form equation:\n\n          x2 - bx - 1 = 0 \n\nwhere the integer '''b''' determines which specific one it is.\n\nUsing the quadratic equation:\n\n          ( -b +- (b2 - 4ac) ) / 2a = x \n\nSubstitute in (from the top equation) '''1''' for '''a''', '''-1''' for '''c''', and recognising that -b is negated we get:\n\n          ( b +- (b2 + 4) ) ) / 2 = x \n\nWe only want the real root:\n\n          ( b + (b2 + 4) ) ) / 2 = x \n\nWhen we set '''b''' to '''1''', we get an irrational number: the '''Golden ratio'''.\n\n          ( 1 + (12 + 4) ) / 2  =  (1 + 5) / 2 = ~1.618033989... \n\nWith '''b''' set to '''2''', we get a different irrational number: the '''Silver ratio'''.\n\n          ( 2 + (22 + 4) ) / 2  =  (2 + 8) / 2 = ~2.414213562... \n\nWhen the ratio '''b''' is '''3''', it is commonly referred to as the '''Bronze''' ratio, '''4''' and '''5'''\nare sometimes called the '''Copper''' and '''Nickel''' ratios, though they aren't as\nstandard. After that there isn't really any attempt at standardized names. They\nare given names here on this page, but consider the names fanciful rather than\ncanonical.\n\nNote that technically, '''b''' can be '''0''' for a \"smaller\" ratio than the '''Golden ratio'''.\nWe will refer to it here as the '''Platinum ratio''', though it is kind-of a\ndegenerate case.\n\n'''Metallic ratios''' where '''b''' > '''0''' are also defined by the irrational continued fractions:\n\n          [b;b,b,b,b,b,b,b,b,b,b,b,b,b,b,b,b...] \n\n\nSo, The first ten '''Metallic ratios''' are:\n\n:::::: {| class=\"wikitable\" style=\"text-align: center;\"\n|+ Metallic ratios\n!Name!!'''b'''!!Equation!!Value!!Continued fraction!!OEIS link\n|-\n|Platinum||0||(0 + 4) / 2|| 1||-||-\n|-\n|Golden||1||(1 + 5) / 2|| 1.618033988749895...||[1;1,1,1,1,1,1,1,1,1,1...]||[[OEIS:A001622]]\n|-\n|Silver||2||(2 + 8) / 2|| 2.414213562373095...||[2;2,2,2,2,2,2,2,2,2,2...]||[[OEIS:A014176]]\n|-\n|Bronze||3||(3 + 13) / 2|| 3.302775637731995...||[3;3,3,3,3,3,3,3,3,3,3...]||[[OEIS:A098316]]\n|-\n|Copper||4||(4 + 20) / 2|| 4.23606797749979...||[4;4,4,4,4,4,4,4,4,4,4...]||[[OEIS:A098317]]\n|-\n|Nickel||5||(5 + 29) / 2|| 5.192582403567252...||[5;5,5,5,5,5,5,5,5,5,5...]||[[OEIS:A098318]]\n|-\n|Aluminum||6||(6 + 40) / 2|| 6.16227766016838...||[6;6,6,6,6,6,6,6,6,6,6...]||[[OEIS:A176398]]\n|-\n|Iron||7||(7 + 53) / 2|| 7.140054944640259...||[7;7,7,7,7,7,7,7,7,7,7...]||[[OEIS:A176439]]\n|-\n|Tin||8||(8 + 68) / 2|| 8.123105625617661...||[8;8,8,8,8,8,8,8,8,8,8...]||[[OEIS:A176458]]\n|-\n|Lead||9||(9 + 85) / 2|| 9.109772228646444...||[9;9,9,9,9,9,9,9,9,9,9...]||[[OEIS:A176522]]\n|}\n\n\n\n\nThere are other ways to find the '''Metallic ratios'''; one, (the focus of this task)\nis through '''successive approximations of Lucas sequences'''.\n\nA traditional '''Lucas sequence''' is of the form:\n\n    x''n'' = P * x''n-1'' - Q * x''n-2''\n\nand starts with the first 2 values '''0, 1'''. \n\nFor our purposes in this task, to find the metallic ratios we'll use the form:\n\n    x''n'' = b * x''n-1'' + x''n-2''\n\n( '''P''' is set to '''b''' and '''Q''' is set to '''-1'''. ) To avoid \"divide by zero\" issues we'll start the sequence with the first two terms '''1, 1'''. The initial starting value has very little effect on the final ratio or convergence rate. ''Perhaps it would be more accurate to call it a Lucas-like sequence.'' \n\nAt any rate, when '''b = 1''' we get:\n\n    x''n'' = x''n-1'' + x''n-2''\n\n    1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144...\n\nmore commonly known as the Fibonacci sequence.\n\nWhen '''b = 2''':\n\n    x''n'' = 2 * x''n-1'' + x''n-2''\n\n    1, 1, 3, 7, 17, 41, 99, 239, 577, 1393...\n\n\nAnd so on.\n\n\nTo find the ratio by successive approximations, divide the ('''n+1''')th term by the\n'''n'''th. As '''n''' grows larger, the ratio will approach the '''b''' metallic ratio.\n\nFor '''b = 1''' (Fibonacci sequence):\n\n    1/1   = 1\n    2/1   = 2\n    3/2   = 1.5\n    5/3   = 1.666667\n    8/5   = 1.6\n    13/8  = 1.625\n    21/13 = 1.615385\n    34/21 = 1.619048\n    55/34 = 1.617647\n    89/55 = 1.618182\n    etc.\n\nIt converges, but pretty slowly. In fact, the '''Golden ratio''' has the slowest\npossible convergence for any irrational number.\n\n\n;Task\n\nFor each of the first '''10 Metallic ratios'''; '''b''' = '''0''' through '''9''':\n\n* Generate the corresponding \"Lucas\" sequence.\n* Show here, on this page, at least the first '''15''' elements of the \"Lucas\" sequence.\n* Using successive approximations, calculate the value of the ratio accurate to '''32''' decimal places.\n* Show the '''value''' of the '''approximation''' at the required accuracy.\n* Show the '''value''' of '''n''' when the approximation reaches the required accuracy (How many iterations did it take?).\n\nOptional, stretch goal - Show the '''value''' and number of iterations '''n''', to approximate the '''Golden ratio''' to '''256''' decimal places.\n\nYou may assume that the approximation has been reached when the next iteration does not cause the value (to the desired places) to change.\n\n;See also\n* Wikipedia: Metallic mean\n* Wikipedia: Lucas sequence\n\n", "solution": "from itertools import count, islice\nfrom _pydecimal import getcontext, Decimal\n\ndef metallic_ratio(b):\n    m, n = 1, 1\n    while True:\n        yield m, n\n        m, n = m*b + n, m\n\ndef stable(b, prec):\n    def to_decimal(b):\n        for m,n in metallic_ratio(b):\n            yield Decimal(m)/Decimal(n)\n\n    getcontext().prec = prec\n    last = 0\n    for i,x in zip(count(), to_decimal(b)):\n        if x == last:\n            print(f'after {i} iterations:\\n\\t{x}')\n            break\n        last = x\n\nfor b in range(4):\n    coefs = [n for _,n in islice(metallic_ratio(b), 15)]\n    print(f'\\nb = {b}: {coefs}')\n    stable(b, 32)\n\nprint(f'\\nb = 1 with 256 digits:')\nstable(1, 256)"}
{"title": "Metaprogramming", "language": "Python", "task": "{{omit from|BBC BASIC}}\nName and briefly demonstrate any support your language has for metaprogramming. Your demonstration may take the form of cross-references to other tasks on Rosetta Code. When possible, provide links to relevant documentation. \n\nFor the purposes of this task, \"support for metaprogramming\" means any way the user can effectively modify the language's syntax that's built into the language (like Lisp macros) or that's conventionally used with the language (like the C preprocessor). Such facilities need not be very powerful: even user-defined infix operators count. On the other hand, in general, neither operator overloading nor eval count. The task author acknowledges that what qualifies as metaprogramming is largely a judgment call.\n\n", "solution": "Metaprogramming is frowned on in Python and considered un-pythonic. The only widely known example of metaprogramming in Python was an implementation of a goto (and comefrom) keyword done as an [http://entrian.com/goto/ April-fools joke].\n\nAnother more recent library that shows it can be done in Python is [https://github.com/lihaoyi/macropy MacroPy].\n\n"}
{"title": "Metronome", "language": "Python", "task": "The task is to implement a   metronome. \n\nThe metronome should be capable of producing high and low audio beats, accompanied by a visual beat indicator, and the beat pattern and tempo should be configurable.\n\nFor the purpose of this task, it is acceptable to play sound files for production of the beat notes, and an external player may be used. \n\nHowever, the playing of the sounds should not interfere with the timing of the metronome.\n\nThe visual indicator can simply be a blinking red or green area of the screen (depending on whether a high or low beat is being produced), and the metronome can be implemented using a terminal display, or optionally, a graphical display, depending on the language capabilities. \n\nIf the language has no facility to output sound, then it is permissible for this to implemented using just the visual indicator.\n\n", "solution": "#lang Python\nimport time\n\ndef main(bpm = 72, bpb = 4):\n    sleep = 60.0 / bpm\n    counter = 0\n    while True:\n        counter += 1\n        if counter % bpb:\n            print 'tick'\n        else:\n            print 'TICK'\n        time.sleep(sleep)\n        \n\n\nmain()\n\n"}
{"title": "Mian-Chowla sequence", "language": "Python", "task": "The Mian-Chowla sequence is an integer sequence defined recursively.\n\n\nMian-Chowla is an infinite instance of a Sidon sequence, and belongs to the class known as B2 sequences.\n\n\nThe sequence starts with:\n\n::a1 = 1\n\nthen for n > 1, an is the smallest positive integer such that every pairwise sum\n\n::ai + aj \n\nis distinct, for all i and j less than or equal to n.\n\n;The Task:\n\n:* Find and display, here, on this page the first 30 terms of the Mian-Chowla sequence.\n:* Find and display, here, on this page the 91st through 100th terms of the Mian-Chowla sequence.\n\n\nDemonstrating working through the first few terms longhand:\n\n::a1 = 1\n\n::1 + 1 = 2\n\nSpeculatively try a2 = 2\n\n::1 + 1 = 2\n::1 + 2 = 3\n::2 + 2 = 4\n\nThere are no repeated sums so '''2''' is the next number in the sequence.\n\nSpeculatively try a3 = 3\n\n::1 + 1 = 2\n::1 + 2 = 3  \n::1 + 3 = 4\n::2 + 2 = 4\n::2 + 3 = 5\n::3 + 3 = 6\n\nSum of '''4''' is repeated so '''3''' is rejected.\n\nSpeculatively try a3 = 4\n\n::1 + 1 = 2\n::1 + 2 = 3\n::1 + 4 = 5\n::2 + 2 = 4\n::2 + 4 = 6\n::4 + 4 = 8\n\nThere are no repeated sums so '''4''' is the next number in the sequence.\n\nAnd so on...\n\n;See also:\n\n:* OEIS:A005282 Mian-Chowla sequence\n\n", "solution": "from itertools import count, islice, chain\nimport time\n\ndef mian_chowla():\n    mc = [1]\n    yield mc[-1]\n    psums = set([2])\n    newsums = set([])\n    for trial in count(2):\n        for n in chain(mc, [trial]):\n            sum = n + trial\n            if sum in psums:\n                newsums.clear()\n                break\n            newsums.add(sum)\n        else:\n            psums |= newsums\n            newsums.clear()\n            mc.append(trial)\n            yield trial\n\ndef pretty(p, t, s, f):\n    print(p, t, \" \".join(str(n) for n in (islice(mian_chowla(), s, f))))\n\nif __name__ == '__main__':\n    st = time.time()\n    ts = \"of the Mian-Chowla sequence are:\\n\"\n    pretty(\"The first 30 terms\", ts, 0, 30)\n    pretty(\"\\nTerms 91 to 100\", ts, 90, 100)\n    print(\"\\nComputation time was\", (time.time()-st) * 1000, \"ms\")"}
{"title": "Mian-Chowla sequence", "language": "Python 3.7", "task": "The Mian-Chowla sequence is an integer sequence defined recursively.\n\n\nMian-Chowla is an infinite instance of a Sidon sequence, and belongs to the class known as B2 sequences.\n\n\nThe sequence starts with:\n\n::a1 = 1\n\nthen for n > 1, an is the smallest positive integer such that every pairwise sum\n\n::ai + aj \n\nis distinct, for all i and j less than or equal to n.\n\n;The Task:\n\n:* Find and display, here, on this page the first 30 terms of the Mian-Chowla sequence.\n:* Find and display, here, on this page the 91st through 100th terms of the Mian-Chowla sequence.\n\n\nDemonstrating working through the first few terms longhand:\n\n::a1 = 1\n\n::1 + 1 = 2\n\nSpeculatively try a2 = 2\n\n::1 + 1 = 2\n::1 + 2 = 3\n::2 + 2 = 4\n\nThere are no repeated sums so '''2''' is the next number in the sequence.\n\nSpeculatively try a3 = 3\n\n::1 + 1 = 2\n::1 + 2 = 3  \n::1 + 3 = 4\n::2 + 2 = 4\n::2 + 3 = 5\n::3 + 3 = 6\n\nSum of '''4''' is repeated so '''3''' is rejected.\n\nSpeculatively try a3 = 4\n\n::1 + 1 = 2\n::1 + 2 = 3\n::1 + 4 = 5\n::2 + 2 = 4\n::2 + 4 = 6\n::4 + 4 = 8\n\nThere are no repeated sums so '''4''' is the next number in the sequence.\n\nAnd so on...\n\n;See also:\n\n:* OEIS:A005282 Mian-Chowla sequence\n\n", "solution": "'''Mian-Chowla series'''\n\nfrom itertools import (islice)\nfrom time import time\n\n\n# mianChowlas :: Gen [Int]\ndef mianChowlas():\n    '''Mian-Chowla series - Generator constructor\n    '''\n    mcs = [1]\n    sumSet = set([2])\n    x = 1\n    while True:\n        yield x\n        (sumSet, mcs, x) = nextMC(sumSet, mcs, x)\n\n\n# nextMC :: (Set Int, [Int], Int) -> (Set Int, [Int], Int)\ndef nextMC(setSums, mcs, n):\n    '''(Set of sums, series so far, current term) ->\n        (updated sum set, updated series, next term)\n    '''\n    def valid(x):\n        for m in mcs:\n            if x + m in setSums:\n                return False\n        return True\n\n    x = until(valid)(succ)(n)\n    setSums.update(\n        [x + y for y in mcs] + [2 * x]\n    )\n    return (setSums, mcs + [x], x)\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Tests'''\n\n    start = time()\n    genMianChowlas = mianChowlas()\n    print(\n        'First 30 terms of the Mian-Chowla series:\\n',\n        take(30)(genMianChowlas)\n    )\n    drop(60)(genMianChowlas)\n    print(\n        '\\n\\nTerms 91 to 100 of the Mian-Chowla series:\\n',\n        take(10)(genMianChowlas),\n        '\\n'\n    )\n    print(\n        '(Computation time c. ' + str(round(\n            1000 * (time() - start)\n        )) + ' ms)'\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# drop :: Int -> [a] -> [a]\n# drop :: Int -> String -> String\ndef drop(n):\n    '''The suffix of xs after the\n       first n elements, or [] if n > length xs'''\n    def go(xs):\n        if isinstance(xs, list):\n            return xs[n:]\n        else:\n            take(n)(xs)\n            return xs\n    return lambda xs: go(xs)\n\n\n# succ :: Int -> Int\ndef succ(x):\n    '''The successor of a numeric value (1 +)'''\n    return 1 + x\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of applying f until p holds.\n       The initial seed value is x.'''\n    def go(f, x):\n        v = x\n        while not p(v):\n            v = f(v)\n        return v\n    return lambda f: lambda x: go(f, x)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Middle three digits", "language": "Python", "task": "Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.\n\nNote: The order of the middle digits should be preserved.\n\nYour function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:\n\n123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345\n1, 2, -1, -10, 2002, -2002, 0\n\nShow your output on this page.\n\n", "solution": ">>> def middle_three_digits(i):\n\ts = str(abs(i))\n\tlength = len(s)\n\tassert length >= 3 and length % 2 == 1, \"Need odd and >= 3 digits\"\n\tmid = length // 2\n\treturn s[mid-1:mid+2]\n\n>>> passing = [123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345]\n>>> failing = [1, 2, -1, -10, 2002, -2002, 0]\n>>> for x in passing + failing:\n\ttry:\n\t\tanswer = middle_three_digits(x)\n\texcept AssertionError as error:\n\t\tanswer = error\n\tprint(\"middle_three_digits(%s) returned: %r\" % (x, answer))\n\n\t\nmiddle_three_digits(123) returned: '123'\nmiddle_three_digits(12345) returned: '234'\nmiddle_three_digits(1234567) returned: '345'\nmiddle_three_digits(987654321) returned: '654'\nmiddle_three_digits(10001) returned: '000'\nmiddle_three_digits(-10001) returned: '000'\nmiddle_three_digits(-123) returned: '123'\nmiddle_three_digits(-100) returned: '100'\nmiddle_three_digits(100) returned: '100'\nmiddle_three_digits(-12345) returned: '234'\nmiddle_three_digits(1) returned: AssertionError('Need odd and >= 3 digits',)\nmiddle_three_digits(2) returned: AssertionError('Need odd and >= 3 digits',)\nmiddle_three_digits(-1) returned: AssertionError('Need odd and >= 3 digits',)\nmiddle_three_digits(-10) returned: AssertionError('Need odd and >= 3 digits',)\nmiddle_three_digits(2002) returned: AssertionError('Need odd and >= 3 digits',)\nmiddle_three_digits(-2002) returned: AssertionError('Need odd and >= 3 digits',)\nmiddle_three_digits(0) returned: AssertionError('Need odd and >= 3 digits',)\n>>> "}
{"title": "Middle three digits", "language": "Python from Haskell", "task": "Write a function/procedure/subroutine that is called with an integer value and returns the middle three digits of the integer if possible or a clear indication of an error if this is not possible.\n\nNote: The order of the middle digits should be preserved.\n\nYour function should be tested with the following values; the first line should return valid answers, those of the second line should return clear indications of an error:\n\n123, 12345, 1234567, 987654321, 10001, -10001, -123, -100, 100, -12345\n1, 2, -1, -10, 2002, -2002, 0\n\nShow your output on this page.\n\n", "solution": "'''Middle 3 digits'''\n\n\n# mid3digits :: Int -> Either String String\ndef mid3digits(n):\n    '''Either the middle three digits,\n       or an explanatory string.'''\n    m = abs(n)\n    s = str(m)\n    return Left('Less than 3 digits') if (100 > m) else (\n        Left('Even digit count') if even(len(s)) else Right(\n            s[(len(s) - 3) // 2:][0:3]\n        )\n    )\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''Test'''\n\n    def bracketed(x):\n        return '(' + str(x) + ')'\n\n    print(\n        tabulated('Middle three digits, where defined:\\n')(str)(\n            either(bracketed)(str)\n        )(mid3digits)([\n            123, 12345, 1234567, 987654321, 10001, -10001, -123,\n            -100, 100, -12345, 1, 2, -1, -10, 2002, -2002, 0\n        ])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# Left :: a -> Either a b\ndef Left(x):\n    '''Constructor for an empty Either (option type) value\n       with an associated string.'''\n    return {'type': 'Either', 'Right': None, 'Left': x}\n\n\n# Right :: b -> Either a b\ndef Right(x):\n    '''Constructor for a populated Either (option type) value'''\n    return {'type': 'Either', 'Left': None, 'Right': x}\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# either :: (a -> c) -> (b -> c) -> Either a b -> c\ndef either(fl):\n    '''The application of fl to e if e is a Left value,\n       or the application of fr to e if e is a Right value.'''\n    return lambda fr: lambda e: fl(e['Left']) if (\n        None is e['Right']\n    ) else fr(e['Right'])\n\n\n# even :: Int -> Bool\ndef even(x):\n    '''Is x even ?'''\n    return 0 == x % 2\n\n\n# tabulated :: String -> (b -> String) -> (a -> b) -> [a] -> String\ndef tabulated(s):\n    '''Heading -> x display function -> fx display function ->\n                f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(str), xs))\n        return s + '\\n' + '\\n'.join(\n            xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs\n        )\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Mind boggling card trick", "language": "Python", "task": "Matt Parker of the \"Stand Up Maths channel\" has a   YouTube video   of a card trick that creates a semblance of order from chaos.\n\nThe task is to simulate the trick in a way that mimics the steps shown in the video.\n\n; 1. Cards.\n# Create a common deck of cards of 52 cards    (which are half red, half black).\n# Give the pack a good shuffle.\n; 2. Deal from the shuffled deck, you'll be creating three piles.\n#   Assemble the cards face down.\n##   Turn up the   ''top card''   and hold it in your hand. \n###    if the card is         black,        then add the   ''next''   card (unseen) to the       \"black\"      pile. \n###    If the card is     red,    then add the   ''next''   card (unseen) to the   \"red\"  pile.\n##   Add the   ''top card''   that you're holding to the '''discard''' pile.   (You might optionally show these discarded cards to get an idea of the randomness).\n# Repeat the above for the rest of the shuffled deck.\n; 3. Choose a random number   (call it '''X''')   that will be used to swap cards from the \"red\" and \"black\" piles.\n# Randomly choose   '''X'''   cards from the   \"red\"  pile (unseen), let's call this the   \"red\"  bunch. \n# Randomly choose   '''X'''   cards from the       \"black\"      pile (unseen), let's call this the      \"black\"       bunch.\n# Put the     \"red\"    bunch into the         \"black\"      pile.\n# Put the        \"black\"        bunch into the      \"red\"  pile.\n# (The above two steps complete the swap of   '''X'''   cards of the \"red\" and \"black\" piles.  (Without knowing what those cards are --- they could be red or black, nobody knows).\n; 4. Order from randomness?\n# Verify (or not) the mathematician's assertion that: \n      '''The number of black cards in the \"black\" pile equals the number of red cards in the \"red\" pile.''' \n\n\n(Optionally, run this simulation a number of times, gathering more evidence of the truthfulness of the assertion.)\n\nShow output on this page.\n\n", "solution": "import random\n\n## 1. Cards\nn = 52\nBlack, Red = 'Black', 'Red'\nblacks = [Black] * (n // 2) \nreds = [Red] * (n // 2)\npack = blacks + reds\n# Give the pack a good shuffle.\nrandom.shuffle(pack)\n\n## 2. Deal from the randomised pack into three stacks\nblack_stack, red_stack, discard = [], [], []\nwhile pack:\n    top = pack.pop()\n    if top == Black:\n        black_stack.append(pack.pop())\n    else:\n        red_stack.append(pack.pop())\n    discard.append(top)\nprint('(Discards:', ' '.join(d[0] for d in discard), ')\\n')\n\n## 3. Swap the same, random, number of cards between the two stacks.\n# We can't swap more than the number of cards in a stack.\nmax_swaps = min(len(black_stack), len(red_stack))\n# Randomly choose the number of cards to swap.\nswap_count = random.randint(0, max_swaps)\nprint('Swapping', swap_count)\n# Randomly choose that number of cards out of each stack to swap.\ndef random_partition(stack, count):\n    \"Partition the stack into 'count' randomly selected members and the rest\"\n    sample = random.sample(stack, count)\n    rest = stack[::]\n    for card in sample:\n        rest.remove(card)\n    return rest, sample\n\nblack_stack, black_swap = random_partition(black_stack, swap_count)\nred_stack, red_swap = random_partition(red_stack, swap_count)\n\n# Perform the swap.\nblack_stack += red_swap\nred_stack += black_swap\n\n## 4. Order from randomness?\nif black_stack.count(Black) == red_stack.count(Red):\n    print('Yeha! The mathematicians assertion is correct.')\nelse:\n    print('Whoops - The mathematicians (or my card manipulations) are flakey')"}
{"title": "Minimal steps down to 1", "language": "Python", "task": "Given:\n* A starting, positive integer (greater than one), N.\n* A selection of possible integer perfect divisors, D.\n* And a selection of possible subtractors, S.\nThe goal is find the minimum number of steps  necessary to reduce N down to one.\n\nAt any step, the number may be:\n* Divided by any member of D if it is perfectly divided by D, (remainder zero).\n* OR have one of S subtracted from it, if N is greater than the member of S.\n\n\nThere may be many ways to reduce the initial N down to 1. Your program needs to:\n* Find the ''minimum'' number of ''steps'' to reach 1.\n* Show '''one''' way of getting fron N to 1 in those minimum steps.\n\n;Examples:\nNo divisors, D. a single subtractor of 1.\n:Obviousely N will take N-1 subtractions of 1 to reach 1\n\nSingle divisor of 2; single subtractor of 1:\n:N = 7 Takes 4 steps N -1=> 6, /2=> 3, -1=> 2, /2=> 1\n:N = 23 Takes 7 steps N -1=>22, /2=>11, -1=>10, /2=> 5, -1=> 4, /2=> 2, /2=> 1\n\nDivisors 2 and 3; subtractor 1:\n:N = 11 Takes 4 steps N -1=>10, -1=> 9, /3=> 3, /3=> 1\n\n;Task:\nUsing the possible divisors D, of 2 and 3; together with a possible subtractor S, of 1:\n\n:1. Show the number of steps and possible way of diminishing the numbers 1 to 10 down to 1.\n:2. Show a count of, and the numbers that: have the maximum minimal_steps_to_1, in the range 1 to 2,000.\n\nUsing the possible divisors D, of 2 and 3; together with a possible subtractor S, of 2:\n\n:3. Show the number of steps and possible way of diminishing the numbers 1 to 10 down to 1.\n:4. Show a count of, and the numbers that: have the maximum minimal_steps_to_1, in the range 1 to 2,000.\n\n\n;Optional stretch goal:\n:2a, and 4a: As in 2 and 4 above, but for N in the range 1 to 20_000\n\n\n;Reference:\n* Learn Dynamic Programming (Memoization & Tabulation) Video of similar task.\n\n", "solution": "from functools import lru_cache\n\n\n#%%\n\nDIVS = {2, 3}\nSUBS = {1}\n\nclass Minrec():\n    \"Recursive, memoised minimised steps to 1\"\n\n    def __init__(self, divs=DIVS, subs=SUBS):\n        self.divs, self.subs = divs, subs\n\n    @lru_cache(maxsize=None)\n    def _minrec(self, n):\n        \"Recursive, memoised\"\n        if n == 1:\n            return 0, ['=1']\n        possibles = {}\n        for d in self.divs:\n            if n % d == 0:\n                possibles[f'/{d}=>{n // d:2}'] = self._minrec(n // d)\n        for s in self.subs:\n            if n > s:\n                possibles[f'-{s}=>{n - s:2}'] = self._minrec(n - s)\n        thiskind, (count, otherkinds) = min(possibles.items(), key=lambda x: x[1])\n        ret = 1 + count, [thiskind] + otherkinds\n        return ret\n\n    def __call__(self, n):\n        \"Recursive, memoised\"\n        ans = self._minrec(n)[1][:-1]\n        return len(ans), ans\n\n\nif __name__ == '__main__':\n    for DIVS, SUBS in [({2, 3}, {1}), ({2, 3}, {2})]:\n        minrec = Minrec(DIVS, SUBS)\n        print('\\nMINIMUM STEPS TO 1: Recursive algorithm')\n        print('  Possible divisors:  ', DIVS)\n        print('  Possible decrements:', SUBS)\n        for n in range(1, 11):\n            steps, how = minrec(n)\n            print(f'    minrec({n:2}) in {steps:2} by: ', ', '.join(how))\n\n        upto = 2000\n        print(f'\\n    Those numbers up to {upto} that take the maximum, \"minimal steps down to 1\":')\n        stepn = sorted((minrec(n)[0], n) for n in range(upto, 0, -1))\n        mx = stepn[-1][0]\n        ans = [x[1] for x in stepn if x[0] == mx]\n        print('      Taking', mx, f'steps is/are the {len(ans)} numbers:',\n              ', '.join(str(n) for n in sorted(ans)))\n        #print(minrec._minrec.cache_info())\n        print()"}
{"title": "Minimum multiple of m where digital sum equals m", "language": "Python", "task": "Generate the sequence '''a(n)''' when each element is the minimum integer multiple '''m''' such that the digit sum of '''n''' times '''m''' is equal to '''n'''.\n\n\n;Task\n\n* Find the first 40 elements of the sequence.\n\n\n\n;Stretch\n\n* Find the next 30 elements of the sequence.\n\n\n;See also\n\n;* OEIS:A131382 - Minimal number m such that Sum_digits(n*m)=n\n\n\n\n\n", "solution": "'''A131382'''\n\nfrom itertools import count, islice\n\n\n# a131382 :: [Int]\ndef a131382():\n    '''An infinite series of the terms of A131382'''\n    return (\n        elemIndex(x)(\n            productDigitSums(x)\n        ) for x in count(1)\n    )\n\n\n# productDigitSums :: Int -> [Int]\ndef productDigitSums(n):\n    '''The sum of the decimal digits of n'''\n    return (digitSum(n * x) for x in count(0))\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''First 40 terms of A131382'''\n\n    print(\n        table(10)([\n            str(x) for x in islice(\n                a131382(),\n                40\n            )\n        ])\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n, subdividing the\n       contents of xs. Where the length of xs is not evenly\n       divisible, the final list will be shorter than n.\n    '''\n    def go(xs):\n        return (\n            xs[i:n + i] for i in range(0, len(xs), n)\n        ) if 0 < n else None\n    return go\n\n\n# digitSum :: Int -> Int\ndef digitSum(n):\n    '''The sum of the digital digits of n.\n    '''\n    return sum(int(x) for x in list(str(n)))\n\n\n# elemIndex :: a -> [a] -> (None | Int)\ndef elemIndex(x):\n    '''Just the first index of x in xs,\n       or None if no elements match.\n    '''\n    def go(xs):\n        try:\n            return next(\n                i for i, v in enumerate(xs) if x == v\n            )\n        except StopIteration:\n            return None\n    return go\n\n\n# table :: Int -> [String] -> String\ndef table(n):\n    '''A list of strings formatted as\n       right-justified rows of n columns.\n    '''\n    def go(xs):\n        w = len(xs[-1])\n        return '\\n'.join(\n            ' '.join(row) for row in chunksOf(n)([\n                s.rjust(w, ' ') for s in xs\n            ])\n        )\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Minimum positive multiple in base 10 using only 0 and 1", "language": "Python from Kotlin", "task": "Every positive integer has infinitely many base-10 multiples that only use the digits '''0''' and '''1'''. The goal of this task is to find and display the '''minimum''' multiple that has this property.\n\nThis is simple to do, but can be challenging to do efficiently.\n\nTo avoid repeating long, unwieldy phrases, the operation \"minimum positive multiple of a positive integer n in base 10 that only uses the digits 0 and 1\" will hereafter be referred to as \"'''B10'''\".\n\n;Task:\n\nWrite a routine to find the B10 of a given integer.  \n\nE.G. \n                   \n       '''n'''                  '''B10'''      '''n'''  x '''multiplier'''\n       1                    1    ( 1  x 1         )\n       2                   10    ( 2  x 5         )\n       7                 1001    ( 7  x 143       )\n       9            111111111    ( 9  x 12345679  )\n      10                   10    ( 10 x 1         )\n\nand so on.\n\nUse the routine to find and display here, on this page, the '''B10''' value for: \n\n    1 through 10, 95 through 105, 297, 576, 594, 891, 909, 999\n\nOptionally find '''B10''' for:\n\n    1998, 2079, 2251, 2277\n\nStretch goal; find '''B10''' for:\n\n    2439, 2997, 4878\n\nThere are many opportunities for optimizations, but avoid using magic numbers as much as possible. If you ''do'' use magic numbers, explain briefly why and what they do for your implementation.\n\n\n;See also:\n\n:* OEIS:A004290 Least positive multiple of n that when written in base 10 uses only 0's and 1's.\n:* How to find Minimum Positive Multiple in base 10 using only 0 and 1\n\n", "solution": "def getA004290(n):\n    if n < 2:\n        return 1\n    arr = [[0 for _ in range(n)] for _ in range(n)]\n    arr[0][0] = 1\n    arr[0][1] = 1\n    m = 0\n    while True:\n        m += 1\n        if arr[m - 1][-10 ** m % n] == 1:\n            break\n        arr[m][0] = 1\n        for k in range(1, n):\n            arr[m][k] = max([arr[m - 1][k], arr[m - 1][k - 10 ** m % n]])\n    r = 10 ** m\n    k = -r % n\n    for j in range((m - 1), 0, -1):\n        if arr[j - 1][k] == 0:\n            r = r + 10 ** j\n            k = (k - 10 ** j) % n   \n    if k == 1:\n        r += 1\n    return r\n \nfor n in [i for i in range(1, 11)] + \\\n          [i for i in range(95, 106)] + \\\n          [297, 576, 594, 891, 909, 999, 1998, 2079, 2251, 2277, 2439, 2997, 4878]:\n    result = getA004290(n)\n    print(f\"A004290({n}) = {result} = {n} * {result // n})\")\n"}
{"title": "Minkowski question-mark function", "language": "Python from Go", "task": "The '''Minkowski question-mark function''' converts the continued fraction representation {{math|[a0; a1, a2, a3, ...]}} of a number into a binary decimal representation in which the integer part {{math|a0}} is unchanged and the {{math|a1, a2, ...}} become alternating runs of binary zeroes and ones of those lengths. The decimal point takes the place of the first zero. \n\nThus, {{math|?}}(31/7) = 71/16 because 31/7 has the continued fraction representation {{math|[4;2,3]}} giving the binary expansion {{math|4 + 0.01112}}.\n\nAmong its interesting properties is that it maps roots of quadratic equations, which have repeating continued fractions, to rational numbers, which have repeating binary digits.\n\nThe question-mark function is continuous and monotonically increasing, so it has an inverse.\n\n* Produce a function for {{math|?(x)}}.   Be careful: rational numbers have two possible continued fraction representations:\n:::*   {{math|[a0;a1,... an-1,an]}}      and \n:::*   {{math|[a0;a1,... an-1,an-1,1]}} \n* Choose one of the above that will give a binary expansion ending with a   '''1'''.\n* Produce the inverse function {{math|?-1(x)}}\n* Verify that {{math|?(ph)}} = 5/3, where {{math|ph}} is the Greek golden ratio.\n* Verify that {{math|?-1(-5/9)}} = (13 - 7)/6\n* Verify that the two functions are inverses of each other by showing that {{math|?-1(?(x))}}={{math|x}} and {{math|?(?-1(y))}}={{math|y}} for {{math|x, y}} of your choice \n\n\nDon't worry about precision error in the last few digits. \n\n\n;See also:\n* Wikipedia entry: Minkowski's question-mark function\n\n\n", "solution": "import math\n\nMAXITER = 151\n\n\ndef minkowski(x):\n    if x > 1 or x < 0:\n        return math.floor(x) + minkowski(x - math.floor(x))\n\n    p = int(x)\n    q = 1\n    r = p + 1\n    s = 1\n    d = 1.0\n    y = float(p)\n\n    while True:\n        d /= 2\n        if y + d == y:\n            break\n\n        m = p + r\n        if m < 0 or p < 0:\n            break\n\n        n = q + s\n        if n < 0:\n            break\n\n        if x < m / n:\n            r = m\n            s = n\n        else:\n            y += d\n            p = m\n            q = n\n\n    return y + d\n\n\ndef minkowski_inv(x):\n    if x > 1 or x < 0:\n        return math.floor(x) + minkowski_inv(x - math.floor(x))\n\n    if x == 1 or x == 0:\n        return x\n\n    cont_frac = [0]\n    current = 0\n    count = 1\n    i = 0\n\n    while True:\n        x *= 2\n\n        if current == 0:\n            if x < 1:\n                count += 1\n            else:\n                cont_frac.append(0)\n                cont_frac[i] = count\n\n                i += 1\n                count = 1\n                current = 1\n                x -= 1\n        else:\n            if x > 1:\n                count += 1\n                x -= 1\n            else:\n                cont_frac.append(0)\n                cont_frac[i] = count\n\n                i += 1\n                count = 1\n                current = 0\n\n        if x == math.floor(x):\n            cont_frac[i] = count\n            break\n\n        if i == MAXITER:\n            break\n\n    ret = 1.0 / cont_frac[i]\n    for j in range(i - 1, -1, -1):\n        ret = cont_frac[j] + 1.0 / ret\n\n    return 1.0 / ret\n\n\nif __name__ == \"__main__\":\n    print(\n        \"{:19.16f} {:19.16f}\".format(\n            minkowski(0.5 * (1 + math.sqrt(5))),\n            5.0 / 3.0,\n        )\n    )\n\n    print(\n        \"{:19.16f} {:19.16f}\".format(\n            minkowski_inv(-5.0 / 9.0),\n            (math.sqrt(13) - 7) / 6,\n        )\n    )\n\n    print(\n        \"{:19.16f} {:19.16f}\".format(\n            minkowski(minkowski_inv(0.718281828)),\n            minkowski_inv(minkowski(0.1213141516171819)),\n        )\n    )\n"}
{"title": "Modified random distribution", "language": "Python", "task": "Given a random number generator, (rng), generating numbers in the range 0.0 .. 1.0 called rgen, for example; and a function modifier(x)\ntaking an number in the same range and generating the probability that the input should be generated, in the same range 0..1; then implement the following algorithm for generating random numbers to the probability given by  function modifier:\nwhile True:\n    random1 = rgen()\n    random2 = rgen()\n    if random2 < modifier(random1):\n        answer = random1\n        break\n    endif\nendwhile\n\n;Task:\n* Create a modifier function that generates a 'V' shaped probability of number generation using something like, for example:\n                       modifier(x)  =  2*(0.5 - x)  if x < 0.5  else 2*(x - 0.5) \n* Create a generator of random numbers with probabilities modified by the above function.\n* Generate >= 10,000 random numbers subject to the probability modification.\n* Output a textual histogram with from 11 to 21 bins showing the distribution of the random numbers generated.\n\n\nShow your output here, on this page.\n\n", "solution": "import random\nfrom typing import List, Callable, Optional\n\n\ndef modifier(x: float) -> float:\n    \"\"\"\n    V-shaped, modifier(x) goes from 1 at 0 to 0 at 0.5 then back to 1 at 1.0 .\n\n    Parameters\n    ----------\n    x : float\n        Number, 0.0 .. 1.0 .\n\n    Returns\n    -------\n    float\n        Target probability for generating x; between 0 and 1.\n\n    \"\"\"\n    return 2*(.5 - x) if x < 0.5 else 2*(x - .5)\n\n\ndef modified_random_distribution(modifier: Callable[[float], float],\n                                 n: int) -> List[float]:\n    \"\"\"\n    Generate n random numbers between 0 and 1 subject to modifier.\n\n    Parameters\n    ----------\n    modifier : Callable[[float], float]\n        Target random number gen. 0 <= modifier(x) < 1.0 for 0 <= x < 1.0 .\n    n : int\n        number of random numbers generated.\n\n    Returns\n    -------\n    List[float]\n        n random numbers generated with given probability.\n\n    \"\"\"\n    d: List[float] = []\n    while len(d) < n:\n        r1 = prob = random.random()\n        if random.random() < modifier(prob):\n            d.append(r1)\n    return d\n\n\nif __name__ == '__main__':\n    from collections import Counter\n\n    data = modified_random_distribution(modifier, 50_000)\n    bins = 15\n    counts = Counter(d // (1 / bins) for d in data)\n    #\n    mx = max(counts.values())\n    print(\"   BIN, COUNTS, DELTA: HISTOGRAM\\n\")\n    last: Optional[float] = None\n    for b, count in sorted(counts.items()):\n        delta = 'N/A' if last is None else str(count - last)\n        print(f\"  {b / bins:5.2f},  {count:4},  {delta:>4}: \"\n              f\"{'#' * int(40 * count / mx)}\")\n        last = count"}
{"title": "Modular arithmetic", "language": "Python 3.x", "task": "equivalence relation called ''congruence''.  \n\nFor any positive integer p called the ''congruence modulus'', \ntwo numbers a and b are said to be ''congruent modulo p'' whenever there exists an integer k such that:\n:a = b + k\\,p\n\nThe corresponding set of multiplicative inverse for this task.\n\nAddition and multiplication on this ring have the same algebraic structure as in usual arithmetic, so that a function such as a polynomial expression could receive a ring element as argument and give a consistent result.\n\nThe purpose of this task is to show, if your programming language allows it, \nhow to redefine operators so that they can be used transparently on modular integers.  \nYou can do it either by using a dedicated library, or by implementing your own class.\n\nYou will use the following function for demonstration:\n:f(x) = x^{100} + x + 1\nYou will use 13 as the congruence modulus and you will compute f(10).\n\nIt is important that the function f is agnostic about whether or not its argument is modular; it should behave the same way with normal and modular integers.  \nIn other words, the function is an algebraic expression that could be used with any ring, not just integers.\n\n", "solution": "import operator\nimport functools\n\n@functools.total_ordering\nclass Mod:\n    __slots__ = ['val','mod']\n\n    def __init__(self, val, mod):\n        if not isinstance(val, int):\n            raise ValueError('Value must be integer')\n        if not isinstance(mod, int) or mod<=0:\n            raise ValueError('Modulo must be positive integer')\n        self.val = val % mod\n        self.mod = mod\n\n    def __repr__(self):\n        return 'Mod({}, {})'.format(self.val, self.mod)\n\n    def __int__(self):\n        return self.val\n\n    def __eq__(self, other):\n        if isinstance(other, Mod):\n            if self.mod == other.mod:\n                return self.val==other.val\n            else:\n                return NotImplemented\n        elif isinstance(other, int):\n            return self.val == other\n        else:\n            return NotImplemented\n\n    def __lt__(self, other):\n        if isinstance(other, Mod):\n            if self.mod == other.mod:\n                return self.val<other.val\n            else:\n                return NotImplemented\n        elif isinstance(other, int):\n            return self.val < other\n        else:\n            return NotImplemented\n\n    def _check_operand(self, other):\n        if not isinstance(other, (int, Mod)):\n            raise TypeError('Only integer and Mod operands are supported')\n        if isinstance(other, Mod) and self.mod != other.mod:\n            raise ValueError('Inconsistent modulus: {} vs. {}'.format(self.mod, other.mod))\n\n    def __pow__(self, other):\n        self._check_operand(other)\n        # We use the built-in modular exponentiation function, this way we can avoid working with huge numbers.\n        return Mod(pow(self.val, int(other), self.mod), self.mod)\n\n    def __neg__(self):\n        return Mod(self.mod - self.val, self.mod)\n\n    def __pos__(self):\n        return self # The unary plus operator does nothing.\n\n    def __abs__(self):\n        return self # The value is always kept non-negative, so the abs function should do nothing.\n\n# Helper functions to build common operands based on a template.\n# They need to be implemented as functions for the closures to work properly.\ndef _make_op(opname):\n    op_fun = getattr(operator, opname)  # Fetch the operator by name from the operator module\n    def op(self, other):\n        self._check_operand(other)\n        return Mod(op_fun(self.val, int(other)) % self.mod, self.mod)\n    return op\n\ndef _make_reflected_op(opname):\n    op_fun = getattr(operator, opname)\n    def op(self, other):\n        self._check_operand(other)\n        return Mod(op_fun(int(other), self.val) % self.mod, self.mod)\n    return op\n\n# Build the actual operator overload methods based on the template.\nfor opname, reflected_opname in [('__add__', '__radd__'), ('__sub__', '__rsub__'), ('__mul__', '__rmul__')]:\n    setattr(Mod, opname, _make_op(opname))\n    setattr(Mod, reflected_opname, _make_reflected_op(opname))\n\ndef f(x):\n    return x**100+x+1\n\nprint(f(Mod(10,13)))\n# Output: Mod(1, 13)"}
{"title": "Modular exponentiation", "language": "Python", "task": "Find the last   '''40'''   decimal digits of   a^b,   where\n\n::*   a = 2988348162058574136915891421498819466320163312926952423791023078876139\n::*   b = 2351399303373464486466122544523690094744975233415544072992656881240319\n\n\nA computer is too slow to find the entire value of   a^b. \n\nInstead, the program must use a fast algorithm for modular exponentiation:   a^b \\mod m.\n\nThe algorithm must work for any integers   a, b, m,     where   b \\ge 0   and   m > 0.\n\n", "solution": "a = 2988348162058574136915891421498819466320163312926952423791023078876139\nb = 2351399303373464486466122544523690094744975233415544072992656881240319\nm = 10 ** 40\nprint(pow(a, b, m))"}
{"title": "Modular inverse", "language": "Python", "task": "From Wikipedia:\n\nIn modulo    ''m''    is an integer    ''x''    such that\n\n::a\\,x \\equiv 1 \\pmod{m}.\n\nOr in other words, such that:\n\n::\\exists k \\in\\Z,\\qquad a\\, x = 1 + k\\,m\n\nIt can be shown that such an inverse exists   if and only if    ''a''    and    ''m''    are coprime,   but we will ignore this for this task.\n\n\n;Task:\nEither by implementing the algorithm, by using a dedicated library or by using a built-in function in \nyour language,   compute the modular inverse of   42 modulo 2017.\n\n", "solution": "from functools import (reduce)\nfrom itertools import (chain)\n\n\n# modInv :: Int -> Int -> Maybe Int\ndef modInv(a):\n    return lambda m: (\n        lambda ig=gcdExt(a)(m): (\n            lambda i=ig[0]: (\n                Just(i + m if 0 > i else i) if 1 == ig[2] else (\n                    Nothing()\n                )\n            )\n        )()\n    )()\n\n\n# gcdExt :: Int -> Int -> (Int, Int, Int)\ndef gcdExt(x):\n    def go(a, b):\n        if 0 == b:\n            return (1, 0, a)\n        else:\n            (q, r) = divmod(a, b)\n            (s, t, g) = go(b, r)\n        return (t, s - q * t, g)\n    return lambda y: go(x, y)\n\n\n#  TEST ---------------------------------------------------\n\n# Numbers between 2010 and 2015 which do yield modular inverses for 42:\n\n# main :: IO ()\ndef main():\n    print (\n        mapMaybe(\n            lambda y: bindMay(modInv(42)(y))(\n                lambda mInv: Just((y, mInv))\n            )\n        )(\n            enumFromTo(2010)(2025)\n        )\n    )\n\n# -> [(2011, 814), (2015, 48), (2017, 1969), (2021, 1203)]\n\n\n# GENERIC ABSTRACTIONS ------------------------------------\n\n\n# enumFromTo :: Int -> Int -> [Int]\ndef enumFromTo(m):\n    return lambda n: list(range(m, 1 + n))\n\n\n# bindMay (>>=) :: Maybe  a -> (a -> Maybe b) -> Maybe b\ndef bindMay(m):\n    return lambda mf: (\n        m if m.get('Nothing') else mf(m.get('Just'))\n    )\n\n\n# Just :: a -> Maybe a\ndef Just(x):\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# mapMaybe :: (a -> Maybe b) -> [a] -> [b]\ndef mapMaybe(mf):\n    return lambda xs: reduce(\n        lambda a, x: maybe(a)(lambda j: a + [j])(mf(x)),\n        xs,\n        []\n    )\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    return lambda f: lambda m: v if m.get('Nothing') else (\n        f(m.get('Just'))\n    )\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# MAIN ---\nmain()"}
{"title": "Monads/List monad", "language": "Python", "task": "A Monad is a combination of a data-type with two helper functions written for that type. \n\nThe data-type can be of any kind which can contain values of some other type - common examples are lists, records, sum-types, even functions or IO streams. The two special functions, mathematically known as '''eta''' and '''mu''', but usually given more expressive names like 'pure', 'return', or 'yield' and 'bind', abstract away some boilerplate needed for pipe-lining or enchaining sequences of computations on values held in the containing data-type.\n\nThe bind operator in the List monad enchains computations which return their values wrapped in lists. One application of this is the representation of indeterminacy, with returned lists representing a set of possible values. An empty list can be returned to express incomputability, or computational failure.\n\nA sequence of two list monad computations (enchained with the use of bind) can be understood as the computation of a cartesian product. \n\nThe natural implementation of bind for the List monad is a composition of '''concat''' and '''map''', which, used with a function which returns its value as a (possibly empty) list, provides for filtering in addition to transformation or mapping.\n\n\nDemonstrate in your programming language the following:\n\n#Construct a List Monad by writing the 'bind' function and the 'pure' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)\n#Make two functions, each which take a number and return a monadic number, e.g. Int -> List Int and Int -> List String\n#Compose the two functions with bind\n\n", "solution": "\"\"\"A List Monad. Requires Python >= 3.7 for type hints.\"\"\"\nfrom __future__ import annotations\nfrom itertools import chain\n\nfrom typing import Any\nfrom typing import Callable\nfrom typing import Iterable\nfrom typing import List\nfrom typing import TypeVar\n\n\nT = TypeVar(\"T\")\n\n\nclass MList(List[T]):\n    @classmethod\n    def unit(cls, value: Iterable[T]) -> MList[T]:\n        return cls(value)\n\n    def bind(self, func: Callable[[T], MList[Any]]) -> MList[Any]:\n        return MList(chain.from_iterable(map(func, self)))\n\n    def __rshift__(self, func: Callable[[T], MList[Any]]) -> MList[Any]:\n        return self.bind(func)\n\n\nif __name__ == \"__main__\":\n    # Chained int and string functions\n    print(\n        MList([1, 99, 4])\n        .bind(lambda val: MList([val + 1]))\n        .bind(lambda val: MList([f\"${val}.00\"]))\n    )\n\n    # Same, but using `>>` as the bind operator.\n    print(\n        MList([1, 99, 4])\n        >> (lambda val: MList([val + 1]))\n        >> (lambda val: MList([f\"${val}.00\"]))\n    )\n\n    # Cartesian product of [1..5] and [6..10]\n    print(\n        MList(range(1, 6)).bind(\n            lambda x: MList(range(6, 11)).bind(lambda y: MList([(x, y)]))\n        )\n    )\n\n    # Pythagorean triples with elements between 1 and 25\n    print(\n        MList(range(1, 26)).bind(\n            lambda x: MList(range(x + 1, 26)).bind(\n                lambda y: MList(range(y + 1, 26)).bind(\n                    lambda z: MList([(x, y, z)])\n                    if x * x + y * y == z * z\n                    else MList([])\n                )\n            )\n        )\n    )\n"}
{"title": "Monads/Maybe monad", "language": "Python", "task": "Demonstrate in your programming language the following:\n\n#Construct a Maybe Monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that Monad (or just use what the language already has implemented)\n#Make two functions, each which take a number and return a monadic number, e.g. Int -> Maybe Int and Int -> Maybe String\n#Compose the two functions with bind\n\n\nA Monad is a single type which encapsulates several other types, eliminating boilerplate code. In practice it acts like a dynamically typed computational sequence, though in many cases the type issues can be resolved at compile time.\n\nA Maybe Monad is a monad which specifically encapsulates the type of an undefined value.\n\n", "solution": "\"\"\"A Maybe Monad. Requires Python >= 3.7 for type hints.\"\"\"\nfrom __future__ import annotations\n\nfrom typing import Any\nfrom typing import Callable\nfrom typing import Generic\nfrom typing import Optional\nfrom typing import TypeVar\nfrom typing import Union\n\n\nT = TypeVar(\"T\")\n\n\nclass Maybe(Generic[T]):\n    def __init__(self, value: Union[Optional[T], Maybe[T]] = None):\n        if isinstance(value, Maybe):\n            self.value: Optional[T] = value.value\n        else:\n            self.value = value\n\n    def __rshift__(self, func: Callable[[Optional[T]], Maybe[Any]]):\n        return self.bind(func)\n\n    def bind(self, func: Callable[[Optional[T]], Maybe[Any]]) -> Maybe[Any]:\n        return func(self.value)\n\n    def __str__(self):\n        return f\"{self.__class__.__name__}({self.value!r})\"\n\n\ndef plus_one(value: Optional[int]) -> Maybe[int]:\n    if value is not None:\n        return Maybe[int](value + 1)\n    return Maybe[int](None)\n\n\ndef currency(value: Optional[int]) -> Maybe[str]:\n    if value is not None:\n        return Maybe[str](f\"${value}.00\")\n    return Maybe[str](None)\n\n\nif __name__ == \"__main__\":\n    test_cases = [1, 99, None, 4]\n\n    for case in test_cases:\n        m_int = Maybe[int](case)\n        result = m_int >> plus_one >> currency\n        # or..\n        # result = m_int.bind(plus_one).bind(currency)\n        print(f\"{str(case):<4} -> {result}\")\n"}
{"title": "Monads/Writer monad", "language": "Python", "task": "The Writer monad is a programming design pattern which makes it possible to compose functions which return their result values paired with a log string. The final result of a composed function yields both a value, and a concatenation of the logs from each component function application.\n\nDemonstrate in your programming language the following:\n\n# Construct a Writer monad by writing the 'bind' function and the 'unit' (sometimes known as 'return') function for that monad (or just use what the language already provides)\n# Write three simple functions: root, addOne, and half\n# Derive Writer monad versions of each of these functions\n# Apply a composition of the Writer versions of root, addOne, and half to the integer 5, deriving both a value for the Golden Ratio ph, and a concatenated log of the function applications (starting with the initial value, and followed by the application of root, etc.)\n\n", "solution": "\"\"\"A Writer Monad. Requires Python >= 3.7 for type hints.\"\"\"\nfrom __future__ import annotations\n\nimport functools\nimport math\nimport os\n\nfrom typing import Any\nfrom typing import Callable\nfrom typing import Generic\nfrom typing import List\nfrom typing import TypeVar\nfrom typing import Union\n\n\nT = TypeVar(\"T\")\n\n\nclass Writer(Generic[T]):\n    def __init__(self, value: Union[T, Writer[T]], *msgs: str):\n        if isinstance(value, Writer):\n            self.value: T = value.value\n            self.msgs: List[str] = value.msgs + list(msgs)\n        else:\n            self.value = value\n            self.msgs = list(f\"{msg}: {self.value}\" for msg in msgs)\n\n    def bind(self, func: Callable[[T], Writer[Any]]) -> Writer[Any]:\n        writer = func(self.value)\n        return Writer(writer, *self.msgs)\n\n    def __rshift__(self, func: Callable[[T], Writer[Any]]) -> Writer[Any]:\n        return self.bind(func)\n\n    def __str__(self):\n        return f\"{self.value}\\n{os.linesep.join(reversed(self.msgs))}\"\n\n    def __repr__(self):\n        return f\"Writer({self.value}, \\\"{', '.join(reversed(self.msgs))}\\\")\"\n\n\ndef lift(func: Callable, msg: str) -> Callable[[Any], Writer[Any]]:\n    \"\"\"Return a writer monad version of the simple function `func`.\"\"\"\n\n    @functools.wraps(func)\n    def wrapped(value):\n        return Writer(func(value), msg)\n\n    return wrapped\n\n\nif __name__ == \"__main__\":\n    square_root = lift(math.sqrt, \"square root\")\n    add_one = lift(lambda x: x + 1, \"add one\")\n    half = lift(lambda x: x / 2, \"div two\")\n\n    print(Writer(5, \"initial\") >> square_root >> add_one >> half)\n"}
{"title": "Move-to-front algorithm", "language": "Python", "task": "Given a symbol table of a ''zero-indexed'' array of all possible input symbols\nthis algorithm reversibly transforms a sequence\nof input symbols into an array of output numbers (indices).\n\nThe transform in many cases acts to give frequently repeated input symbols\nlower indices which is  useful in some compression algorithms.\n\n\n;Encoding algorithm:\n\n    for each symbol of the input sequence:\n        output the index of the symbol in the symbol table\n        move that symbol to the front of the symbol table\n\n\n\n;Decoding algorithm:\n\n    # Using the same starting symbol table\n    for each index of the input sequence:\n        output the symbol at that index of the symbol table\n        move that symbol to the front of the symbol table\n\n\n\n;Example:\nEncoding the string of character symbols 'broood' using a symbol table of the lowercase characters   '''a'''-to-'''z'''\n\n:::::{| class=\"wikitable\" border=\"1\"\n|-\n! Input\n! Output\n! SymbolTable\n|-\n| '''b'''roood\n| 1\n| 'abcdefghijklmnopqrstuvwxyz'\n|-\n| b'''r'''oood\n| 1 17\n| 'bacdefghijklmnopqrstuvwxyz'\n|-\n| br'''o'''ood\n| 1 17 15\n| 'rbacdefghijklmnopqstuvwxyz'\n|-\n| bro'''o'''od\n| 1 17 15 0\n| 'orbacdefghijklmnpqstuvwxyz'\n|-\n| broo'''o'''d\n| 1 17 15 0 0\n| 'orbacdefghijklmnpqstuvwxyz'\n|-\n| brooo'''d'''\n| 1 17 15 0 0 5\n| 'orbacdefghijklmnpqstuvwxyz'\n|}\n\n\nDecoding the indices back to the original symbol order:\n:::::{| class=\"wikitable\" border=\"1\"\n|-\n! Input\n! Output\n! SymbolTable\n|-\n|  '''1''' 17 15 0 0 5\n| b\n| 'abcdefghijklmnopqrstuvwxyz'\n|-\n| 1 '''17''' 15 0 0 5\n| br\n| 'bacdefghijklmnopqrstuvwxyz'\n|-\n| 1 17 '''15''' 0 0 5\n| bro\n| 'rbacdefghijklmnopqstuvwxyz'\n|-\n| 1 17 15 '''0''' 0 5\n| broo\n| 'orbacdefghijklmnpqstuvwxyz'\n|-\n| 1 17 15 0 '''0''' 5\n| brooo\n| 'orbacdefghijklmnpqstuvwxyz'\n|-\n| 1 17 15 0 0 '''5''' \n| broood\n| 'orbacdefghijklmnpqstuvwxyz'\n|}\n\n\n;Task:\n:*   Encode and decode the following three strings of characters using the symbol table of the lowercase characters   '''a'''-to-'''z'''   as above. \n:*   Show the strings and their encoding here.\n:*   Add a check to ensure that the decoded string is the same as the original.\n\n\nThe strings are: \n\n    broood       \n bananaaa     \n hiphophiphop \n(Note the misspellings in the above strings.)\n\n", "solution": "from __future__ import print_function\nfrom string import ascii_lowercase\n\nSYMBOLTABLE = list(ascii_lowercase)\n\ndef move2front_encode(strng, symboltable):\n    sequence, pad = [], symboltable[::]\n    for char in strng:\n        indx = pad.index(char)\n        sequence.append(indx)\n        pad = [pad.pop(indx)] + pad\n    return sequence\n\ndef move2front_decode(sequence, symboltable):\n    chars, pad = [], symboltable[::]\n    for indx in sequence:\n        char = pad[indx]\n        chars.append(char)\n        pad = [pad.pop(indx)] + pad\n    return ''.join(chars)\n\nif __name__ == '__main__':\n    for s in ['broood', 'bananaaa', 'hiphophiphop']:\n        encode = move2front_encode(s, SYMBOLTABLE)\n        print('%14r encodes to %r' % (s, encode), end=', ')\n        decode = move2front_decode(encode, SYMBOLTABLE)\n        print('which decodes back to %r' % decode)\n        assert s == decode, 'Whoops!'"}
{"title": "Multi-dimensional array", "language": "Python", "task": "For the purposes of this task, the actual memory layout or access method of this data structure is not mandated. \n\nIt is enough to:\n# State the number and extent of each index to the array.\n# Provide specific, ordered, integer indices for all dimensions of the array together with a new value to update the indexed value.\n# Provide specific, ordered, numeric indices for all dimensions of the array to obtain the arrays value at that indexed position.\n\n\n;Task:\n* State if the language supports multi-dimensional arrays in its syntax and usual implementation.\n* State whether the language uses row-major or column major order for multi-dimensional array storage, or any other relevant kind of storage.\n* Show how to create a four dimensional array in your language and set, access, set to another value; and access the new value of an integer-indexed item of the array. The idiomatic method for the language is preferred.\n:* The array should allow a range of five, four, three and two (or two three four five if convenient), in each of the indices, in order. (For example, ''if'' indexing starts at zero for the first index then a range of 0..4 inclusive would suffice).\n* State if memory allocation is optimised for the array - especially if contiguous memory is likely to be allocated.\n* If the language has exceptional native multi-dimensional array support such as optional bounds checking, reshaping, or being able to state both the lower and upper bounds of index ranges, then this is the task to mention them.\n\n\nShow all output here, (but you may judiciously use ellipses to shorten repetitive output text).\n\n", "solution": ">>> from numpy import *\n>>> \n>>> mdarray = zeros((2, 3, 4, 5), dtype=int8, order='F')\n\n>>> mdarray\narray([[[[0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0]],\n\n        [[0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0]],\n\n        [[0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0]]],\n\n\n       [[[0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0]],\n\n        [[0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0]],\n\n        [[0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0],\n         [0, 0, 0, 0, 0]]]], dtype=int8)\n>>> mdarray[0, 1, 2, 3]\n0\n>>> mdarray[0, 1, 2, 3] = 123\n>>> mdarray[0, 1, 2, 3]\n123\n>>> mdarray[0, 1, 2, 3] = 666\n>>> mdarray[0, 1, 2, 3]\n-102\n>>> mdarray[0, 1, 2, 3] = 255\n>>> mdarray[0, 1, 2, 3]\n-1\n>>> mdarray[0, 1, 2, 3] = -128\n>>> mdarray[0, 1, 2, 3]\n-128\n>>> mdarray\narray([[[[   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0]],\n\n        [[   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0, -128,    0],\n         [   0,    0,    0,    0,    0]],\n\n        [[   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0]]],\n\n\n       [[[   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0]],\n\n        [[   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0]],\n\n        [[   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0],\n         [   0,    0,    0,    0,    0]]]], dtype=int8)\n>>> "}
{"title": "Multifactorial", "language": "Python", "task": "The factorial of a number, written as n!, is defined as n! = n(n-1)(n-2)...(2)(1).\n\nMultifactorials generalize factorials as follows:\n: n! = n(n-1)(n-2)...(2)(1)\n: n!! = n(n-2)(n-4)...\n: n!! ! = n(n-3)(n-6)...\n: n!! !! = n(n-4)(n-8)...\n: n!! !! ! = n(n-5)(n-10)...\n\nIn all cases, the terms in the products are positive integers.\n\nIf we define the degree of the multifactorial as the difference in successive terms that are multiplied together for a multifactorial (the number of exclamation marks), then the task is twofold: \n# Write a function that given n and the degree, calculates the multifactorial.\n# Use the function to generate and display here a table of the first ten members (1 to 10) of the first five degrees of multifactorial.\n\n\n'''Note:''' The wikipedia entry on multifactorials gives a different formula. This task uses the Wolfram mathworld definition.\n\n", "solution": ">>> from functools import reduce\n>>> from operator import mul\n>>> def mfac(n, m): return reduce(mul, range(n, 0, -m))\n\n>>> for m in range(1, 11): print(\"%2i: %r\" % (m, [mfac(n, m) for n in range(1, 11)]))\n\n 1: [1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800]\n 2: [1, 2, 3, 8, 15, 48, 105, 384, 945, 3840]\n 3: [1, 2, 3, 4, 10, 18, 28, 80, 162, 280]\n 4: [1, 2, 3, 4, 5, 12, 21, 32, 45, 120]\n 5: [1, 2, 3, 4, 5, 6, 14, 24, 36, 50]\n 6: [1, 2, 3, 4, 5, 6, 7, 16, 27, 40]\n 7: [1, 2, 3, 4, 5, 6, 7, 8, 18, 30]\n 8: [1, 2, 3, 4, 5, 6, 7, 8, 9, 20]\n 9: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n10: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n>>> "}
{"title": "Multiple distinct objects", "language": "Python", "task": "Create a [[sequence]] (array, list, whatever) consisting of n distinct, initialized items of the same type. n should be determined at runtime.\n\nBy ''distinct'' we mean that if they are mutable, changes to one do not affect all others; if there is an appropriate equality operator they are considered unequal; etc. The code need not specify a particular kind of distinction, but do not use e.g. a numeric-range generator which does not generalize.\n\nBy ''initialized'' we mean that each item must be in a well-defined state appropriate for its type, rather than e.g. arbitrary previous memory contents in an array allocation. Do not show only an initialization technique which initializes only to \"zero\" values (e.g. calloc() or int a[n] = {}; in C), unless user-defined types can provide definitions of \"zero\" for that type.\n\nThis task was inspired by the common error of intending to do this, but instead creating a sequence of n references to the ''same'' mutable object; it might be informative to show the way to do that as well, both as a negative example and as how to do it when that's all that's actually necessary.\n\nThis task is most relevant to languages operating in the pass-references-by-value style (most object-oriented, garbage-collected, and/or 'dynamic' languages).\n\nSee also: [[Closures/Value capture]]\n\n", "solution": "[Foo()] * n # here Foo() can be any expression that returns a new object"}
{"title": "Multisplit", "language": "Python", "task": "It is often necessary to split a string into pieces \nbased on several different (potentially multi-character) separator strings, \nwhile still retaining the information about which separators were present in the input. \n\nThis is particularly useful when doing small parsing tasks. \nThe task is to write code to demonstrate this.\n\nThe function (or procedure or method, as appropriate) should \ntake an input string and an ordered collection of separators. \n\nThe order of the separators is significant: \nThe delimiter order represents priority in matching, with the first defined delimiter having the highest priority. \nIn cases where there would be an ambiguity as to \nwhich separator to use at a particular point \n(e.g., because one separator is a prefix of another) \nthe separator with the highest priority should be used. \nDelimiters can be reused and the output from the function should be an ordered sequence of substrings.\n\nTest your code using the input string \"a!===b=!=c\" and the separators \"==\", \"!=\" and \"=\".\n\nFor these inputs the string should be parsed as \"a\" (!=) \"\" (==) \"b\" (=) \"\" (!=) \"c\", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is \"a\", empty string, \"b\", empty string, \"c\". \nNote that the quotation marks are shown for clarity and do not form part of the output.\n\n'''Extra Credit:''' provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.\n\n", "solution": "def min_pos(List):\n\treturn List.index(min(List))\n\ndef find_all(S, Sub, Start = 0, End = -1, IsOverlapped = 0):\n\tRes = []\n\tif End == -1:\n\t\tEnd = len(S)\n\tif IsOverlapped:\n\t\tDeltaPos = 1\n\telse:\n\t\tDeltaPos = len(Sub)\n\tPos = Start\n\twhile True:\n\t\tPos = S.find(Sub, Pos, End)\n\t\tif Pos == -1:\n\t\t\tbreak\n\t\tRes.append(Pos)\n\t\tPos += DeltaPos\n\treturn Res\n\ndef multisplit(S, SepList):\n\tSepPosListList = []\n\tSLen = len(S)\n\tSepNumList = []\n\tListCount = 0\n\tfor i, Sep in enumerate(SepList):\n\t\tSepPosList = find_all(S, Sep, 0, SLen, IsOverlapped = 1)\n\t\tif SepPosList != []:\n\t\t\tSepNumList.append(i)\n\t\t\tSepPosListList.append(SepPosList)\n\t\t\tListCount += 1\n\tif ListCount == 0:\n\t\treturn [S]\n\tMinPosList = []\n\tfor i in range(ListCount):\n\t\tMinPosList.append(SepPosListList[i][0])\n\tSepEnd = 0\n\tMinPosPos = min_pos(MinPosList)\n\tRes = []\n\twhile True:\n\t\tRes.append( S[SepEnd : MinPosList[MinPosPos]] )\n\t\tRes.append([SepNumList[MinPosPos], MinPosList[MinPosPos]])\n\t\tSepEnd = MinPosList[MinPosPos] + len(SepList[SepNumList[MinPosPos]])\n\t\twhile True:\n\t\t\tMinPosPos = min_pos(MinPosList)\n\t\t\tif MinPosList[MinPosPos] < SepEnd:\n\t\t\t\tdel SepPosListList[MinPosPos][0]\n\t\t\t\tif len(SepPosListList[MinPosPos]) == 0:\n\t\t\t\t\tdel SepPosListList[MinPosPos]\n\t\t\t\t\tdel MinPosList[MinPosPos]\n\t\t\t\t\tdel SepNumList[MinPosPos]\n\t\t\t\t\tListCount -= 1\n\t\t\t\t\tif ListCount == 0:\n\t\t\t\t\t\tbreak\n\t\t\t\telse:\n\t\t\t\t\tMinPosList[MinPosPos] = SepPosListList[MinPosPos][0]\n\t\t\telse:\n\t\t\t\tbreak\n\t\tif ListCount == 0:\n\t\t\tbreak\n\tRes.append(S[SepEnd:])\n\treturn Res\n\n\nS = \"a!===b=!=c\"\nmultisplit(S, [\"==\", \"!=\", \"=\"]) # output: ['a', [1, 1], '', [0, 3], 'b', [2, 6], '', [1, 7], 'c']\nmultisplit(S, [\"=\", \"!=\", \"==\"]) # output: ['a', [1, 1], '', [0, 3], '', [0, 4], 'b', [0, 6], '', [1, 7], 'c']"}
{"title": "Multisplit", "language": "Python 3.7", "task": "It is often necessary to split a string into pieces \nbased on several different (potentially multi-character) separator strings, \nwhile still retaining the information about which separators were present in the input. \n\nThis is particularly useful when doing small parsing tasks. \nThe task is to write code to demonstrate this.\n\nThe function (or procedure or method, as appropriate) should \ntake an input string and an ordered collection of separators. \n\nThe order of the separators is significant: \nThe delimiter order represents priority in matching, with the first defined delimiter having the highest priority. \nIn cases where there would be an ambiguity as to \nwhich separator to use at a particular point \n(e.g., because one separator is a prefix of another) \nthe separator with the highest priority should be used. \nDelimiters can be reused and the output from the function should be an ordered sequence of substrings.\n\nTest your code using the input string \"a!===b=!=c\" and the separators \"==\", \"!=\" and \"=\".\n\nFor these inputs the string should be parsed as \"a\" (!=) \"\" (==) \"b\" (=) \"\" (!=) \"c\", where matched delimiters are shown in parentheses, and separated strings are quoted, so our resulting output is \"a\", empty string, \"b\", empty string, \"c\". \nNote that the quotation marks are shown for clarity and do not form part of the output.\n\n'''Extra Credit:''' provide information that indicates which separator was matched at each separation point and where in the input string that separator was matched.\n\n", "solution": "'''Multisplit'''\n\n\nfrom functools import reduce\n\n\n# multiSplit :: [String] -> String -> [(String, String, Int)]\ndef multiSplit(separators):\n    '''List of triples:\n       [(token, separator, start index of separator].\n    '''\n    def go(s):\n        def f(tokensPartsOffset, ic):\n            tokens, parts, offset = tokensPartsOffset\n            i, c = ic\n            inDelim = offset > i\n            return maybe(\n                (\n                    tokens if inDelim\n                    else c + tokens, parts, offset\n                )\n            )(\n                lambda x: (\n                    '',\n                    [(tokens, x, i)] + parts,\n                    i + len(x)\n                )\n            )(\n                None if inDelim else find(\n                    s[i:].startswith\n                )(separators)\n            )\n        ts, ps, _ = reduce(f, enumerate(s), ('', [], 0))\n        return list(reversed(ps)) + [(ts, '', len(s))]\n    return go\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''String split on three successive separators.'''\n    print(\n        multiSplit(['==', '!=', '='])(\n            'a!===b=!=c'\n        )\n    )\n\n\n# ------------------ GENERIC FUNCTIONS -------------------\n\n# find :: (a -> Bool) -> [a] -> (a | None)\ndef find(p):\n    '''Just the first element in the list that matches p,\n       or None if no elements match.\n    '''\n    def go(xs):\n        try:\n            return next(x for x in xs if p(x))\n        except StopIteration:\n            return None\n    return go\n\n\n# maybe :: b -> (a -> b) -> (a | None) -> b\ndef maybe(v):\n    '''Either the default value v, if m is None,\n       or the application of f to x.\n    '''\n    return lambda f: lambda m: v if (\n        None is m\n    ) else f(m)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Munchausen numbers", "language": "Python", "task": "A Munchausen number is a natural number ''n'' the sum of whose digits (in base 10), each raised to the power of itself, equals ''n''.\n\n('''Munchausen''' is also spelled: '''Munchhausen'''.) \n\nFor instance:    3435 = 33 + 44 + 33 + 55 \n\n\n;Task\nFind all Munchausen numbers between   '''1'''   and   '''5000'''.\n\n\n;Also see:\n:* The OEIS entry:  A046253\n:* The Wikipedia entry:  Perfect digit-to-digit invariant, redirected from ''Munchausen Number''\n\n", "solution": "for i in range(5000):\n    if i == sum(int(x) ** int(x) for x in str(i)):\n        print(i)"}
{"title": "Munchausen numbers", "language": "Python 3", "task": "A Munchausen number is a natural number ''n'' the sum of whose digits (in base 10), each raised to the power of itself, equals ''n''.\n\n('''Munchausen''' is also spelled: '''Munchhausen'''.) \n\nFor instance:    3435 = 33 + 44 + 33 + 55 \n\n\n;Task\nFind all Munchausen numbers between   '''1'''   and   '''5000'''.\n\n\n;Also see:\n:* The OEIS entry:  A046253\n:* The Wikipedia entry:  Perfect digit-to-digit invariant, redirected from ''Munchausen Number''\n\n", "solution": "'''Munchausen numbers'''\n\nfrom functools import (reduce)\n\n\n# isMunchausen :: Int -> Bool\ndef isMunchausen(n):\n    '''True if n equals the sum of\n       each of its digits raised to\n       the power of itself.'''\n    def powerOfSelf(d):\n        i = digitToInt(d)\n        return i**i\n    return n == reduce(\n        lambda n, c: n + powerOfSelf(c),\n        str(n), 0\n    )\n\n\n# main :: IO ()\ndef main():\n    '''Test'''\n    print(list(filter(\n        isMunchausen,\n        enumFromTo(1)(5000)\n    )))\n\n\n# GENERIC -------------------------------------------------\n\n# digitToInt :: Char -> Int\ndef digitToInt(c):\n    '''The integer value of any digit character\n       drawn from the 0-9, A-F or a-f ranges.'''\n    oc = ord(c)\n    if 48 > oc or 102 < oc:\n        return None\n    else:\n        dec = oc - 48   # ord('0')\n        hexu = oc - 65  # ord('A')\n        hexl = oc - 97  # ord('a')\n    return dec if 9 >= dec else (\n        10 + hexu if 0 <= hexu <= 5 else (\n            10 + hexl if 0 <= hexl <= 5 else None\n        )\n    )\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Musical scale", "language": "Python", "task": "Output the 8 notes of the C major diatonic scale to the default musical sound device on the system. Specifically, pitch must be tuned to 12-tone equal temperament (12TET) with the modern standard A=440Hz.\n\nThese are the notes \"C, D, E, F, G, A, B, C(1 octave higher)\", or \"Do, Re, Mi, Fa, Sol, La, Si/Ti, Do(1 octave higher)\" on Fixed do Solfege.\n\nFor the purpose of this task, Middle C (in the case of the above tuning, around 261.63 Hz) should be used as the starting note, and any note duration is allowed.\n\nFor languages that cannot utilize a sound device, it is permissible to output to a musical score sheet (or midi file),  or the task can be omitted.\n\n", "solution": ">>> import winsound\n>>> for note in [261.63, 293.66, 329.63, 349.23, 392.00, 440.00, 493.88, 523.25]:\n\twinsound.Beep(int(note+.5), 500)\t\n>>> "}
{"title": "Mutex", "language": "Python", "task": "{{requires|Concurrency}}\n\n\nA '''mutex''' (''abbreviated'' '''Mut'''ually '''Ex'''clusive access) is a synchronization object, a variant of [[semaphore]] with ''k''=1. \nA mutex is said to be seized by a [[task]] decreasing ''k''. \nIt is released when the task restores ''k''. Mutexes are typically used to protect a shared resource from concurrent access. \nA [[task]] seizes (or acquires) the mutex, then accesses the resource, and after that releases the mutex.\n\nA mutex is a low-level synchronization primitive exposed to deadlocking. A deadlock can occur with just two tasks and two mutexes (if each task attempts to acquire both mutexes, but in the opposite order). \nEntering the deadlock is usually aggravated by a [[race condition]] state, which leads to sporadic hangups, which are very difficult to track down. \n\n=Variants of mutexes=\n\n==Global and local mutexes==\nUsually the [[OS]] provides various implementations of mutexes corresponding to the variants of [[task]]s available in the OS. For example, system-wide mutexes can be used by [[process]]es. Local mutexes can be used only by [[threads]] etc. This distinction is maintained because, depending on the hardware, seizing a global mutex might be a thousand times slower than seizing a local one.\n\n==Reentrant mutex==\nA reentrant mutex can be seized by the same [[task]] multiple times. Each seizing of the mutex is matched by releasing it, in order to allow another task to seize it.\n\n==Read write mutex==\nA read write mutex can be seized at two levels for ''read'' and for ''write''. The mutex can be seized for ''read'' by any number of tasks. Only one task may seize it for '''write''. Read write mutexes are usually used to protect resources which can be accessed in mutable and immutable ways. Immutable (read) access is granted concurrently for many tasks because they do not change the resource state. Read write mutexes can be reentrant, global or local. Further, promotion operations may be provided. That's when a [[task]] that has seized the mutex for ''write'' releases it while keeping seized for ''read''. Note that the reverse operation is potentially deadlocking and requires some additional access policy control.\n\n=Deadlock prevention=\nThere exists a simple technique of deadlock prevention when mutexes are seized in some fixed order. This is discussed in depth in the [[Dining philosophers]] problem.\n\n=Sample implementations / APIs=\n", "solution": "import threading\nfrom time import sleep\n\n# res: max number of resources. If changed to 1, it functions\n# identically to a mutex/lock object\nres = 2\nsema = threading.Semaphore(res)\n\nclass res_thread(threading.Thread):\n    def run(self):\n        global res\n        n = self.getName()\n        for i in range(1, 4):\n            # acquire a resource if available and work hard\n            # for 2 seconds.  if all res are occupied, block\n            # and wait\n            sema.acquire()\n            res = res - 1\n            print n, \"+  res count\", res\n            sleep(2)\n\n                        # after done with resource, return it to pool and flag so\n            res = res + 1\n            print n, \"-  res count\", res\n            sema.release()\n\n# create 4 threads, each acquire resorce and work\nfor i in range(1, 5):\n    t = res_thread()\n    t.start()"}
{"title": "N-queens problem", "language": "Python", "task": "right\n\nSolve the eight queens puzzle. \n\n\nYou can extend the problem to solve the puzzle with a board of size   '''N'''x'''N'''.\n \nFor the number of solutions for small values of   '''N''',   see   OEIS: A000170.\n\n\n;Related tasks:\n* [[A* search algorithm]]\n* [[Solve a Hidato puzzle]]\n* [[Solve a Holy Knight's tour]]\n* [[Knight's tour]]\n* [[Peaceful chess queen armies]]\n* [[Solve a Hopido puzzle]]\n* [[Solve a Numbrix puzzle]]\n* [[Solve the no connection puzzle]]\n\n", "solution": "from itertools import permutations\n\nn = 8\ncols = range(n)\nfor vec in permutations(cols):\n    if n == len(set(vec[i]+i for i in cols)) \\\n         == len(set(vec[i]-i for i in cols)):\n        print ( vec )"}
{"title": "N-queens problem", "language": "Python 2.6, 3.x", "task": "right\n\nSolve the eight queens puzzle. \n\n\nYou can extend the problem to solve the puzzle with a board of size   '''N'''x'''N'''.\n \nFor the number of solutions for small values of   '''N''',   see   OEIS: A000170.\n\n\n;Related tasks:\n* [[A* search algorithm]]\n* [[Solve a Hidato puzzle]]\n* [[Solve a Holy Knight's tour]]\n* [[Knight's tour]]\n* [[Peaceful chess queen armies]]\n* [[Solve a Hopido puzzle]]\n* [[Solve a Numbrix puzzle]]\n* [[Solve the no connection puzzle]]\n\n", "solution": "def queens_lex(n):\n    a = list(range(n))\n    up = [True]*(2*n - 1)\n    down = [True]*(2*n - 1)\n    def sub(i):\n        if i == n:\n            yield tuple(a)\n        else:\n            for k in range(i, n):\n                a[i], a[k] = a[k], a[i]\n                j = a[i]\n                p = i + j\n                q = i - j + n - 1\n                if up[p] and down[q]:\n                    up[p] = down[q] = False\n                    yield from sub(i + 1)\n                    up[p] = down[q] = True\n            x = a[i]\n            for k in range(i + 1, n):\n                a[k - 1] = a[k]\n            a[n - 1] = x\n    yield from sub(0)\n\nnext(queens(31))\n(0, 2, 4, 1, 3, 8, 10, 12, 14, 6, 17, 21, 26, 28, 25, 27, 24, 30, 7, 5, 29, 15, 13, 11, 9, 18, 22, 19, 23, 16, 20)\n\nnext(queens_lex(31))\n(0, 2, 4, 1, 3, 8, 10, 12, 14, 5, 17, 22, 25, 27, 30, 24, 26, 29, 6, 16, 28, 13, 9, 7, 19, 11, 15, 18, 21, 23, 20)\n\n#Compare to A065188\n#1, 3, 5, 2, 4, 9, 11, 13, 15, 6, 8, 19, 7, 22, 10, 25, 27, 29, 31, 12, 14, 35, 37, ..."}
{"title": "N-queens problem", "language": "Python 3.7", "task": "right\n\nSolve the eight queens puzzle. \n\n\nYou can extend the problem to solve the puzzle with a board of size   '''N'''x'''N'''.\n \nFor the number of solutions for small values of   '''N''',   see   OEIS: A000170.\n\n\n;Related tasks:\n* [[A* search algorithm]]\n* [[Solve a Hidato puzzle]]\n* [[Solve a Holy Knight's tour]]\n* [[Knight's tour]]\n* [[Peaceful chess queen armies]]\n* [[Solve a Hopido puzzle]]\n* [[Solve a Numbrix puzzle]]\n* [[Solve the no connection puzzle]]\n\n", "solution": "'''N Queens problem'''\n\nfrom functools import reduce\nfrom itertools import chain\n\n\n# queenPuzzle :: Int -> Int -> [[Int]]\ndef queenPuzzle(nCols):\n    '''All board patterns of this dimension\n       in which no two Queens share a row,\n       column, or diagonal.\n    '''\n    def go(nRows):\n        lessRows = nRows - 1\n        return reduce(\n            lambda a, xys: a + reduce(\n                lambda b, iCol: b + [xys + [iCol]] if (\n                    safe(lessRows, iCol, xys)\n                ) else b,\n                enumFromTo(1)(nCols),\n                []\n            ),\n            go(lessRows),\n            []\n        ) if 0 < nRows else [[]]\n    return go\n\n\n# safe :: Int -> Int -> [Int] -> Bool\ndef safe(iRow, iCol, pattern):\n    '''True if no two queens in the pattern\n       share a row, column or diagonal.\n    '''\n    def p(sc, sr):\n        return (iCol == sc) or (\n            sc + sr == (iCol + iRow)\n        ) or (sc - sr == (iCol - iRow))\n    return not any(map(p, pattern, range(0, iRow)))\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Number of solutions for boards of various sizes'''\n\n    n = 5\n    xs = queenPuzzle(n)(n)\n\n    print(\n        str(len(xs)) + ' solutions for a {n} * {n} board:\\n'.format(n=n)\n    )\n    print(showBoards(10)(xs))\n\n    print(\n        fTable(\n            '\\n\\n' + main.__doc__ + ':\\n'\n        )(str)(lambda n: str(n).rjust(3, ' '))(\n            lambda n: len(queenPuzzle(n)(n))\n        )(enumFromTo(1)(10))\n    )\n\n\n# ---------------------- FORMATTING ----------------------\n\n# showBoards :: Int -> [[Int]] -> String\ndef showBoards(nCols):\n    '''String representation, with N columns\n       of a set of board patterns.\n    '''\n    def showBlock(b):\n        return '\\n'.join(map(intercalate('  '), zip(*b)))\n\n    def go(bs):\n        return '\\n\\n'.join(map(\n            showBlock,\n            chunksOf(nCols)([\n                showBoard(b) for b in bs\n            ])\n        ))\n    return go\n\n\n# showBoard :: [Int] -> String\ndef showBoard(xs):\n    '''String representation of a Queens board.'''\n    lng = len(xs)\n\n    def showLine(n):\n        return ('.' * (n - 1)) + '\u265b' + ('.' * (lng - n))\n    return map(showLine, xs)\n\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: range(m, 1 + n)\n\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n, subdividing the\n       contents of xs. Where the length of xs is not evenly\n       divible, the final list will be shorter than n.\n    '''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# intercalate :: [a] -> [[a]] -> [a]\n# intercalate :: String -> [String] -> String\ndef intercalate(x):\n    '''The concatenation of xs\n       interspersed with copies of x.\n    '''\n    return lambda xs: x.join(xs) if isinstance(x, str) else list(\n        chain.from_iterable(\n            reduce(lambda a, v: a + [x, v], xs[1:], [xs[0]])\n        )\n    ) if xs else []\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Narcissist", "language": "Python 2", "task": "Quoting from the Esolangs wiki page:\n\n\nA '''narcissist''' (or '''Narcissus program''') is the decision-problem version of a [[quine]].\n\nA quine, when run, takes no input, but produces a copy of its own source code at its output. In contrast, a narcissist reads a string of symbols from its input, and produces no output except a \"1\" or \"accept\" if that string matches its own source code, or a \"0\" or \"reject\" if it does not.\n\n\nFor concreteness, in this task we shall assume that symbol = character. \n\nThe narcissist should be able to cope with any finite input, whatever its length. \n\nAny form of output is allowed, as long as the program always halts, and \"accept\", \"reject\" and \"not yet finished\" are distinguishable.\n\n", "solution": "import sys\nwith open(sys.argv[0]) as quine:\n    code = raw_input(\"Enter source code: \")\n    if code == quine.read():\n        print(\"Accept\")\n    else:\n        print(\"Reject\")\n"}
{"title": "Narcissistic decimal number", "language": "Python", "task": "A   Narcissistic decimal number   is a non-negative integer,   n,   that is equal to the sum of the   m-th   powers of each of the digits in the decimal representation of   n,   where   m   is the number of digits in the decimal representation of   n.\n\n\nNarcissistic (decimal) numbers are sometimes called   '''Armstrong'''   numbers, named after Michael F. Armstrong. \nThey are also known as   '''Plus Perfect'''   numbers.\n\n\n;An example:\n::::*   if   n   is   '''153''' \n::::*   then   m,   (the number of decimal digits)   is   '''3''' \n::::*   we have    13 + 53 + 33   =   1 + 125 + 27   =   '''153'''  \n::::*   and so   '''153'''   is a narcissistic decimal number\n\n\n;Task:\nGenerate and show here the first   '''25'''   narcissistic decimal numbers.\n\n\n\nNote:   0^1 = 0,   the first in the series. \n\n\n;See also:\n*   the  OEIS entry:     Armstrong (or Plus Perfect, or narcissistic) numbers.\n*   MathWorld entry:   Narcissistic Number.\n*   Wikipedia entry:     Narcissistic number.\n\n", "solution": "from __future__ import print_function\nfrom itertools import count, islice\n\ndef narcissists():\n    for digits in count(0):\n        digitpowers = [i**digits for i in range(10)]\n        for n in range(int(10**(digits-1)), 10**digits):\n            div, digitpsum = n, 0\n            while div:\n                div, mod = divmod(div, 10)\n                digitpsum += digitpowers[mod]\n            if n == digitpsum:\n                yield n\n\nfor i, n in enumerate(islice(narcissists(), 25), 1):\n    print(n, end=' ')\n    if i % 5 == 0: print() \nprint()"}
{"title": "Narcissistic decimal number", "language": "Python from D", "task": "A   Narcissistic decimal number   is a non-negative integer,   n,   that is equal to the sum of the   m-th   powers of each of the digits in the decimal representation of   n,   where   m   is the number of digits in the decimal representation of   n.\n\n\nNarcissistic (decimal) numbers are sometimes called   '''Armstrong'''   numbers, named after Michael F. Armstrong. \nThey are also known as   '''Plus Perfect'''   numbers.\n\n\n;An example:\n::::*   if   n   is   '''153''' \n::::*   then   m,   (the number of decimal digits)   is   '''3''' \n::::*   we have    13 + 53 + 33   =   1 + 125 + 27   =   '''153'''  \n::::*   and so   '''153'''   is a narcissistic decimal number\n\n\n;Task:\nGenerate and show here the first   '''25'''   narcissistic decimal numbers.\n\n\n\nNote:   0^1 = 0,   the first in the series. \n\n\n;See also:\n*   the  OEIS entry:     Armstrong (or Plus Perfect, or narcissistic) numbers.\n*   MathWorld entry:   Narcissistic Number.\n*   Wikipedia entry:     Narcissistic number.\n\n", "solution": "try:\n    import psyco\n    psyco.full()\nexcept:\n    pass\n\nclass Narcissistics:\n    def __init__(self, max_len):\n        self.max_len = max_len\n        self.power = [0] * 10\n        self.dsum = [0] * (max_len + 1)\n        self.count = [0] * 10\n        self.len = 0\n        self.ord0 = ord('0')\n\n    def check_perm(self, out = [0] * 10):\n        for i in xrange(10):\n            out[i] = 0\n\n        s = str(self.dsum[0])\n        for d in s:\n            c = ord(d) - self.ord0\n            out[c] += 1\n            if out[c] > self.count[c]:\n                return\n\n        if len(s) == self.len:\n            print self.dsum[0],\n\n    def narc2(self, pos, d):\n        if not pos:\n            self.check_perm()\n            return\n\n        while True:\n            self.dsum[pos - 1] = self.dsum[pos] + self.power[d]\n            self.count[d] += 1\n            self.narc2(pos - 1, d)\n            self.count[d] -= 1\n            if d == 0:\n                break\n            d -= 1\n\n    def show(self, n):\n        self.len = n\n        for i in xrange(len(self.power)):\n            self.power[i] = i ** n\n        self.dsum[n] = 0\n        print \"length %d:\" % n,\n        self.narc2(n, 9)\n        print\n\ndef main():\n    narc = Narcissistics(14)\n    for i in xrange(1, narc.max_len + 1):\n        narc.show(i)\n\nmain()"}
{"title": "Narcissistic decimal number", "language": "Python 3.7", "task": "A   Narcissistic decimal number   is a non-negative integer,   n,   that is equal to the sum of the   m-th   powers of each of the digits in the decimal representation of   n,   where   m   is the number of digits in the decimal representation of   n.\n\n\nNarcissistic (decimal) numbers are sometimes called   '''Armstrong'''   numbers, named after Michael F. Armstrong. \nThey are also known as   '''Plus Perfect'''   numbers.\n\n\n;An example:\n::::*   if   n   is   '''153''' \n::::*   then   m,   (the number of decimal digits)   is   '''3''' \n::::*   we have    13 + 53 + 33   =   1 + 125 + 27   =   '''153'''  \n::::*   and so   '''153'''   is a narcissistic decimal number\n\n\n;Task:\nGenerate and show here the first   '''25'''   narcissistic decimal numbers.\n\n\n\nNote:   0^1 = 0,   the first in the series. \n\n\n;See also:\n*   the  OEIS entry:     Armstrong (or Plus Perfect, or narcissistic) numbers.\n*   MathWorld entry:   Narcissistic Number.\n*   Wikipedia entry:     Narcissistic number.\n\n", "solution": "'''Narcissistic decimal numbers'''\n\nfrom itertools import chain\nfrom functools import reduce\n\n\n# main :: IO ()\ndef main():\n    '''Narcissistic numbers of digit lengths 1 to 7'''\n    print(\n        fTable(main.__doc__ + ':\\n')(str)(str)(\n            narcissiOfLength\n        )(enumFromTo(1)(7))\n    )\n\n\n# narcissiOfLength :: Int -> [Int]\ndef narcissiOfLength(n):\n    '''List of Narcissistic numbers of\n       (base 10) digit length n.\n    '''\n    return [\n        x for x in digitPowerSums(n)\n        if isDaffodil(n)(x)\n    ]\n\n\n# digitPowerSums :: Int -> [Int]\ndef digitPowerSums(e):\n    '''The subset of integers of e digits that are potential narcissi.\n       (Flattened leaves of a tree of unique digit combinations, in which\n       order is not significant. The sum is independent of the sequence.)\n    '''\n    powers = [(x, x ** e) for x in enumFromTo(0)(9)]\n\n    def go(n, parents):\n        return go(\n            n - 1,\n            chain.from_iterable(map(\n                lambda pDigitSum: (\n                    map(\n                        lambda lDigitSum: (\n                            lDigitSum[0],\n                            lDigitSum[1] + pDigitSum[1]\n                        ),\n                        powers[0: 1 + pDigitSum[0]]\n                    )\n                ),\n                parents\n            )) if parents else powers\n        ) if 0 < n else parents\n\n    return [xs for (_, xs) in go(e, [])]\n\n\n# isDaffodil :: Int -> Int -> Bool\ndef isDaffodil(e):\n    '''True if n is a narcissistic number\n       of decimal digit length e.\n    '''\n    def go(n):\n        ds = digitList(n)\n        return e == len(ds) and n == powerSum(e)(ds)\n    return lambda n: go(n)\n\n\n# powerSum :: Int -> [Int] -> Int\ndef powerSum(e):\n    '''The sum of a list obtained by raising\n       each element of xs to the power of e.\n    '''\n    return lambda xs: reduce(\n        lambda a, x: a + x ** e,\n        xs, 0\n    )\n\n\n# -----------------------FORMATTING------------------------\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n       f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# digitList :: Int -> [Int]\ndef digitList(n):\n    '''A decomposition of n into a\n       list of single-digit integers.\n    '''\n    def go(x):\n        return go(x // 10) + [x % 10] if x else []\n    return go(n) if n else [0]\n\n\n# enumFromTo :: Int -> Int -> [Int]\ndef enumFromTo(m):\n    '''Enumeration of integer values [m..n]'''\n    def go(n):\n        return list(range(m, 1 + n))\n    return lambda n: go(n)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Nautical bell", "language": "Python", "task": "Task\n\nWrite a small program that emulates a nautical bell producing a ringing bell pattern at certain times throughout the day.\n\nThe bell timing should be in accordance with Greenwich Mean Time, unless locale dictates otherwise.\n\nIt is permissible for the program to daemonize, or to slave off a scheduler, and it is permissible to use alternative notification methods (such as producing a written notice \"Two Bells Gone\"), if these are more usual for the system type. \n\n\n;Related task:\n* [[Sleep]]\n\n", "solution": "import time, calendar, sched, winsound\n\nduration = 750      # Bell duration in ms\nfreq = 1280         # Bell frequency in hertz\nbellchar = \"\\u2407\"\nwatches = 'Middle,Morning,Forenoon,Afternoon,First/Last dog,First'.split(',')\n\ndef gap(n=1):\n    time.sleep(n * duration / 1000)\noff = gap\n \ndef on(n=1):\n    winsound.Beep(freq, n * duration)\n \ndef bong():\n    on(); off(0.5)\n\ndef bongs(m):\n    for i in range(m):\n        print(bellchar, end=' ')\n        bong()\n        if i % 2:\n            print('  ', end='')\n            off(0.5)\n    print('')\n        \nscheds =  sched.scheduler(time.time, time.sleep)\n\ndef ships_bell(now=None):\n    def adjust_to_half_hour(atime):\n        atime[4] = (atime[4] // 30) * 30\n        atime[5] = 0\n        return atime\n\n    debug = now is not None\n    rightnow = time.gmtime()\n    if not debug:\n        now = adjust_to_half_hour( list(rightnow) )\n    then = now[::]\n    then[4] += 30\n    hr, mn = now[3:5]\n    watch, b = divmod(int(2 * hr + mn // 30 - 1), 8)\n    b += 1\n    bells = '%i bell%s' % (b, 's' if b > 1 else ' ')\n    if debug:\n        print(\"%02i:%02i, %-20s %s\" % (now[3], now[4], watches[watch] + ' watch', bells), end=' ')\n    else:\n        print(\"%02i:%02i, %-20s %s\" % (rightnow[3], rightnow[4], watches[watch] + ' watch', bells), end=' ')\n    bongs(b)\n    if not debug:\n        scheds.enterabs(calendar.timegm(then), 0, ships_bell)\n        #print(time.struct_time(then))\n        scheds.run()\n\ndef dbg_tester():\n    for h in range(24):\n        for m in (0, 30):\n            if (h,m) == (24,30): break\n            ships_bell( [2013, 3, 2, h, m, 15, 5, 61, 0] )\n        \n    \nif __name__ == '__main__':\n    ships_bell()"}
{"title": "Negative base numbers", "language": "Python", "task": "Negative base numbers are an alternate way to encode numbers without the need for a minus sign. Various negative bases may be used including negadecimal (base -10), negabinary (-2) and negaternary (-3).Negabinary on Wolfram MathworldNegative base on Wikipedia\n\n\n;Task:\n\n*Encode the decimal number 10 as negabinary (expect 11110)\n*Encode the decimal number 146 as negaternary (expect 21102)\n*Encode the decimal number 15 as negadecimal (expect 195)\n*In each of the above cases, convert the encoded number back to decimal.\n\n\n;extra credit:\n\n* supply an integer, that when encoded to base   -62   (or something \"higher\"),   expresses the name of the language being used   (with correct capitalization).   If the computer language has non-alphanumeric characters,   try to encode them into the negatory numerals,   or use other characters instead.\n\n", "solution": "#!/bin/python\nfrom __future__ import print_function\n\ndef EncodeNegBase(n, b): #Converts from decimal\n\tif n == 0:\n\t\treturn \"0\"\n\tout = []\n\twhile n != 0:\n\t\tn, rem = divmod(n, b)\n\t\tif rem < 0:\n\t\t\tn += 1\n\t\t\trem -= b\n\t\tout.append(rem)\n\treturn \"\".join(map(str, out[::-1]))\n\ndef DecodeNegBase(nstr, b): #Converts to decimal\n\tif nstr == \"0\":\n\t\treturn 0\n\t\n\ttotal = 0\n\tfor i, ch in enumerate(nstr[::-1]):\n\t\ttotal += int(ch) * b**i\n\treturn total\n\nif __name__==\"__main__\":\n\t\n\tprint (\"Encode 10 as negabinary (expect 11110)\")\n\tresult = EncodeNegBase(10, -2)\n\tprint (result)\n\tif DecodeNegBase(result, -2) == 10: print (\"Converted back to decimal\")\n\telse: print (\"Error converting back to decimal\")\n\n\tprint (\"Encode 146 as negaternary (expect 21102)\")\n\tresult = EncodeNegBase(146, -3)\n\tprint (result)\n\tif DecodeNegBase(result, -3) == 146: print (\"Converted back to decimal\")\n\telse: print (\"Error converting back to decimal\")\n\n\tprint (\"Encode 15 as negadecimal (expect 195)\")\n\tresult = EncodeNegBase(15, -10)\n\tprint (result)\n\tif DecodeNegBase(result, -10) == 15: print (\"Converted back to decimal\")\n\telse: print (\"Error converting back to decimal\")"}
{"title": "Nested function", "language": "Python 3+", "task": "In many languages, functions can be nested, resulting in outer functions and inner functions. The inner function can access variables from the outer function. In most languages, the inner function can also modify variables in the outer function.\n\n\n;Task: \nWrite a program consisting of two nested functions that prints the following text.\n\n 1. first\n 2. second\n 3. third\n\nThe outer function (called MakeList or equivalent) is responsible for creating the list as a whole and is given the separator \". \" as argument. It also defines a counter variable to keep track of the item number. This demonstrates how the inner function can influence the variables in the outer function.\n\nThe inner function (called MakeItem or equivalent) is responsible for creating a list item. It accesses the separator from the outer function and modifies the counter.\n\n\n;References:\n:* Nested function\n\n", "solution": "def makeList(separator):\n    counter = 1\n\n    def makeItem(item):\n        nonlocal counter\n        result = str(counter) + separator + item + \"\\n\"\n        counter += 1\n        return result\n\n    return makeItem(\"first\") + makeItem(\"second\") + makeItem(\"third\")\n\nprint(makeList(\". \"))"}
{"title": "Nested templated data", "language": "Python", "task": "A template for data is an arbitrarily nested tree of integer indices.\n    \nData payloads are given as a separate mapping, array or other simpler, flat,\nassociation of indices to individual items of data, and are strings.\nThe idea is to create a data structure with the templates' nesting, and the \npayload corresponding to each index appearing at the position of each index.\n\nAnswers using simple string replacement or regexp are to be avoided. The idea is \nto use the native, or usual implementation of lists/tuples etc of the language\nand to hierarchically traverse the template to generate the output.\n\n;Task Detail:\nGiven the following input template t  and list of payloads p:\n# Square brackets are used here to denote nesting but may be changed for other,\n# clear, visual representations of nested data appropriate to ones programming \n# language.\nt = [\n    [[1, 2],\n     [3, 4, 1], \n     5]]\n\np = 'Payload#0' ... 'Payload#6'\n\nThe correct output should have the following structure, (although spacing and \nlinefeeds may differ, the nesting and order should follow):\n[[['Payload#1', 'Payload#2'],\n  ['Payload#3', 'Payload#4', 'Payload#1'],\n  'Payload#5']]\n\n1. Generate the output for the above template, t.\n\n;Optional Extended tasks:\n2. Show which payloads remain unused.\n3. Give some indication/handling of indices without a payload.\n\n''Show output on this page.''\n\n", "solution": "from pprint import pprint as pp\n\nclass Template():\n    def __init__(self, structure):\n        self.structure = structure\n        self.used_payloads, self.missed_payloads = [], []\n    \n    def inject_payload(self, id2data):\n        \n        def _inject_payload(substruct, i2d, used, missed):\n            used.extend(i2d[x] for x in substruct if type(x) is not tuple and x in i2d)\n            missed.extend(f'??#{x}' \n                          for x in substruct if type(x) is not tuple and x not in i2d)\n            return tuple(_inject_payload(x, i2d, used, missed) \n                           if type(x) is tuple \n                           else i2d.get(x, f'??#{x}') \n                         for x in substruct)\n                           \n        ans = _inject_payload(self.structure, id2data, \n                              self.used_payloads, self.missed_payloads)\n        self.unused_payloads = sorted(set(id2data.values()) \n                                      - set(self.used_payloads))\n        self.missed_payloads = sorted(set(self.missed_payloads))\n        return ans\n\nif __name__ == '__main__':\n    index2data = {p: f'Payload#{p}' for p in range(7)}\n    print(\"##PAYLOADS:\\n  \", end='')\n    print('\\n  '.join(list(index2data.values())))\n    for structure in [\n     (((1, 2),\n       (3, 4, 1),\n       5),),\n    \n     (((1, 2),\n       (10, 4, 1),\n       5),)]:\n        print(\"\\n\\n# TEMPLATE:\")\n        pp(structure, width=13)\n        print(\"\\n TEMPLATE WITH PAYLOADS:\")\n        t = Template(structure)\n        out = t.inject_payload(index2data)\n        pp(out)\n        print(\"\\n UNUSED PAYLOADS:\\n  \", end='')\n        unused = t.unused_payloads\n        print('\\n  '.join(unused) if unused else '-')\n        print(\" MISSING PAYLOADS:\\n  \", end='')\n        missed = t.missed_payloads\n        print('\\n  '.join(missed) if missed else '-')"}
{"title": "Next highest int from digits", "language": "Python", "task": "Given a zero or positive integer, the task is to generate the next largest\ninteger using only the given digits*1.\n\n*   Numbers will not be padded to the left with zeroes.\n*   Use all given digits, with their given multiplicity. (If a digit appears twice in the input number, it should appear twice in the result).\n*   If there is no next highest integer return zero.\n\n\n:*1   Alternatively phrased as:   \"Find the smallest integer larger than the (positive or zero) integer   '''N''' \n:: which can be obtained by reordering the (base ten) digits of   '''N'''\".\n\n\n;Algorithm 1:\n#   Generate all the permutations of the digits and sort into numeric order.\n#   Find the number in the list.\n#   Return the next highest number from the list.\n\n\nThe above could prove slow and memory hungry for numbers with large numbers of\ndigits, but should be easy to reason about its correctness.\n\n\n;Algorithm 2:\n#   Scan right-to-left through the digits of the number until you find a digit with a larger digit somewhere to the right of it.\n#   Exchange that digit with the digit on the right that is ''both'' more than it, and closest to it.\n#   Order the digits to the right of this position, after the swap; lowest-to-highest, left-to-right. (I.e. so they form the lowest numerical representation)\n\n\n'''E.g.:'''\n\n    n = 12453\n\n    12_4_53\n\n    12_5_43\n\n    12_5_34\n\n    return: 12534\n\n\nThis second algorithm is faster and more memory efficient, but implementations\nmay be harder to test. \n\nOne method of testing, (as used in developing the task),   is to compare results from both\nalgorithms for random numbers generated from a range that the first algorithm can handle.\n\n\n;Task requirements:\nCalculate the next highest int from the digits of the following numbers:\n:*   0\n:*   9\n:*   12\n:*   21\n:*   12453\n:*   738440\n:*   45072010\n:*   95322020\n\n\n;Optional stretch goal:\n:*   9589776899767587796600\n\n", "solution": "def closest_more_than(n, lst):\n    \"(index of) closest int from lst, to n that is also > n\"\n    large = max(lst) + 1\n    return lst.index(min(lst, key=lambda x: (large if x <= n else x)))\n\ndef nexthigh(n):\n    \"Return nxt highest number from n's digits using scan & re-order\"\n    assert n == int(abs(n)), \"n >= 0\"\n    this = list(int(digit) for digit in str(int(n)))[::-1]\n    mx = this[0]\n    for i, digit in enumerate(this[1:], 1):\n        if digit < mx:\n            mx_index = closest_more_than(digit, this[:i + 1])\n            this[mx_index], this[i] = this[i], this[mx_index]\n            this[:i] = sorted(this[:i], reverse=True)\n            return int(''.join(str(d) for d in this[::-1]))\n        elif digit > mx:\n            mx, mx_index = digit, i\n    return 0\n\n\nif __name__ == '__main__':\n    for x in [0, 9, 12, 21, 12453, 738440, 45072010, 95322020,\n              9589776899767587796600]:\n        print(f\"{x:>12_d} -> {nexthigh(x):>12_d}\")"}
{"title": "Nim game", "language": "Python", "task": "Nim is a simple game where the second player-if they know the trick-will always win.\n\n\nThe game has only 3 rules:\n::*   start with   '''12'''   tokens\n::*   each player takes    '''1,  2,  or  3'''   tokens in turn\n::*  the player who takes the last token wins.\n\n\nTo win every time,   the second player simply takes 4 minus the number the first player took.   So if the first player takes 1,   the second takes 3; if the first player takes 2,   the second should take 2; and if the first player takes 3,   the second player will take 1. \n\n;Task:\nDesign a simple Nim game where the human player goes first, and the computer always wins. The game should enforce the rules.\n\n", "solution": "print(\"Py Nim\\n\")\n\ndef getTokens(curTokens):\n\tglobal tokens\n\t\n\tprint(\"How many tokens would you like to take? \", end='')\n\ttake = int(input())\n\t\n\tif (take < 1 or take > 3):\n\t\tprint(\"Number must be between 1 and 3.\\n\")\n\t\tgetTokens(curTokens)\n\t\treturn\n\t\n\ttokens = curTokens - take\n\tprint(f'You take {take} tokens.')\n\tprint(f'{tokens} tokens remaining.\\n')\n\ndef compTurn(curTokens):\n\tglobal tokens\n\t\n\ttake = curTokens % 4\n\ttokens = curTokens - take\n\tprint (f'Computer takes {take} tokens.')\n\tprint (f'{tokens} tokens remaining.\\n')\n\t\n\ntokens = 12\nwhile (tokens > 0):\n\tgetTokens(tokens)\n\tcompTurn(tokens)\n\nprint(\"Computer wins!\")\n"}
{"title": "Non-transitive dice", "language": "Python", "task": "Let our dice select numbers on their faces with equal probability, i.e. fair dice.\nDice may have more or less than six faces. (The possibility of there being a \n3D physical shape that has that many \"faces\" that allow them to be fair dice,\nis ignored for this task - a die with 3 or 33 defined sides is defined by the \nnumber of faces and the numbers on each face).\n\nThrowing dice will randomly select a face on each die with equal probability.\nTo show which die of dice thrown multiple times is more likely to win over the \nothers:\n# calculate all possible combinations of different faces from each die\n# Count how many times each die wins a combination\n# Each ''combination'' is equally likely so the die with more winning face combinations is statistically more likely to win against the other dice.\n\n\n'''If two dice X and Y are thrown against each other then X likely to: win, lose, or break-even against Y can be shown as: X > Y, X < Y, or X = Y respectively.\n'''\n;Example 1:\nIf X is the three sided die with 1, 3, 6 on its faces and Y has 2, 3, 4 on its\nfaces then the equal possibility outcomes from throwing both, and the winners \nis:\n    \n    X   Y   Winner\n    =   =   ======\n    1   2   Y\n    1   3   Y\n    1   4   Y\n    3   2   X\n    3   3   -\n    3   4   Y\n    6   2   X\n    6   3   X\n    6   4   X\n    \n    TOTAL WINS: X=4, Y=4\n\nBoth die will have the same statistical probability of winning, i.e.their comparison can be written as X = Y\n\n;Transitivity:\nIn mathematics transitivity are rules like: \n    if a op b and b op c then a op c\nIf, for example, the op, (for operator), is the familiar less than, <, and it's applied to integers\nwe get the familiar if a < b and b < c then a < c\n\n;Non-transitive dice\nThese are an ordered list of dice where the '>' operation between successive \ndice pairs applies but a comparison between the first and last of the list\nyields the opposite result, '<'.\n''(Similarly '<' successive list comparisons with a final '>' between first and last is also non-transitive).''\n\n\nThree dice S, T, U with appropriate face values could satisfy\n\n    S < T, T < U and yet S > U\n\nTo be non-transitive.\n\n;Notes:\n\n* The order of numbers on the faces of a die is not relevant. For example, three faced die described with face numbers of 1, 2, 3 or 2, 1, 3 or any other permutation are equivalent. For the purposes of the task '''show only the permutation in lowest-first sorted order i.e. 1, 2, 3''' (and remove any of its perms).\n* A die can have more than one instance of the same number on its faces, e.g. 2, 3, 3, 4\n* '''Rotations''': Any rotation of non-transitive dice from an answer is also an answer. You may ''optionally'' compute and show only one of each such rotation sets, ideally the first when sorted in a natural way. If this option is used then prominently state in the output that rotations of results are also solutions.\n\n\n;Task:\n;====\nFind all the ordered lists of ''three'' non-transitive dice S, T, U of the form \nS < T, T < U and yet S > U; where the dice are selected from all ''four-faced die''\n, (unique w.r.t the notes), possible by having selections from the integers \n''one to four'' on any dies face.\n\nSolution can be found by generating all possble individual die then testing all\npossible permutations, (permutations are ordered), of three dice for \nnon-transitivity.\n\n;Optional stretch goal:\nFind lists of '''four''' non-transitive dice selected from the same possible dice from the non-stretch goal.\n\n\nShow the results here, on this page.\n\n;References:\n* The Most Powerful Dice - Numberphile Video.\n* Nontransitive dice - Wikipedia.\n\n", "solution": "from itertools import combinations_with_replacement as cmbr\nfrom time import time\n \ndef dice_gen(n, faces, m):\n    dice = list(cmbr(faces, n))\n \n    succ = [set(j for j, b in enumerate(dice)\n                    if sum((x>y) - (x<y) for x in a for y in b) > 0)\n                for a in dice]\n \n    def loops(seq):\n        s = succ[seq[-1]]\n\n        if len(seq) == m:\n            if seq[0] in s: yield seq\n            return\n\n        for d in (x for x in s if x > seq[0] and not x in seq):\n            yield from loops(seq + (d,))\n \n    yield from (tuple(''.join(dice[s]) for s in x)\n                    for i, v in enumerate(succ)\n                    for x in loops((i,)))\n \nt = time()\nfor n, faces, loop_len in [(4, '1234', 3), (4, '1234', 4), (6, '123456', 3), (6, '1234567', 3)]:\n    for i, x in enumerate(dice_gen(n, faces, loop_len)): pass\n \n    print(f'{n}-sided, markings {faces}, loop length {loop_len}:')\n    print(f'\\t{i + 1}*{loop_len} solutions, e.g. {\" > \".join(x)} > [loop]')\n    t, t0 = time(), t\n    print(f'\\ttime: {t - t0:.4f} seconds\\n')"}
{"title": "Nonoblock", "language": "Python", "task": "Nonogram puzzle.\n\n\n;Given:\n* The number of cells in a row.\n* The size of each, (space separated), connected block of cells to fit in the row, in left-to right order.\n\n\n;Task: \n* show all possible positions. \n* show the number of positions of the blocks for the following cases within the row. \n* show all output on this page. \n* use a \"neat\" diagram of the block positions.\n\n\n;Enumerate the following configurations:\n#   '''5'''   cells   and   '''[2, 1]'''   blocks\n#   '''5'''   cells   and   '''[]'''   blocks   (no blocks)\n#   '''10'''   cells   and   '''[8]'''   blocks\n#   '''15'''   cells   and   '''[2, 3, 2, 3]'''   blocks\n#   '''5'''   cells   and   '''[2, 3]'''   blocks   (should give some indication of this not being possible)\n\n\n;Example:\nGiven a row of five cells and a block of two cells followed by a block of one cell - in that order, the example could be shown as:\n\n    |_|_|_|_|_| # 5 cells and [2, 1] blocks\n\nAnd would expand to the following 3 possible rows of block positions:\n\n    |A|A|_|B|_|\n    |A|A|_|_|B|\n    |_|A|A|_|B|\n\n\nNote how the sets of blocks are always separated by a space.\n\nNote also that it is not necessary for each block to have a separate letter. \nOutput approximating\n\nThis:\n                        |#|#|_|#|_|\n                        |#|#|_|_|#|\n                        |_|#|#|_|#|\n\nThis would also work:\n                        ##.#.\n                        ##..#\n                       .##.#\n\n\n;An algorithm:\n* Find the minimum space to the right that is needed to legally hold all but the leftmost block of cells (with a space between blocks remember).\n* The leftmost cell can legitimately be placed in all positions from the LHS up to a RH position that allows enough room for the rest of the blocks.\n* for each position of the LH block recursively compute the position of the rest of the blocks in the ''remaining'' space to the right of the current placement of the LH block.\n\n(This is the algorithm used in the [[Nonoblock#Python]] solution).    \n\n\n;Reference:\n* The blog post Nonogram puzzle solver (part 1) Inspired this task and donated its [[Nonoblock#Python]] solution.\n\n", "solution": "def nonoblocks(blocks, cells):\n    if not blocks or blocks[0] == 0:\n        yield [(0, 0)]\n    else:\n        assert sum(blocks) + len(blocks)-1 <= cells, \\\n            'Those blocks will not fit in those cells'\n        blength, brest = blocks[0], blocks[1:]      # Deal with the first block of length\n        minspace4rest = sum(1+b for b in brest)     # The other blocks need space\n        # Slide the start position from left to max RH index allowing for other blocks.\n        for bpos in range(0, cells - minspace4rest - blength + 1):\n            if not brest:\n                # No other blocks to the right so just yield this one.\n                yield [(bpos, blength)]\n            else:\n                # More blocks to the right so create a *sub-problem* of placing\n                # the brest blocks in the cells one space to the right of the RHS of \n                # this block.\n                offset = bpos + blength +1\n                nonoargs = (brest, cells - offset)  # Pre-compute arguments to nonoargs\n                # recursive call to nonoblocks yields multiple sub-positions\n                for subpos in nonoblocks(*nonoargs):\n                    # Remove the offset from sub block positions\n                    rest = [(offset + bp, bl) for bp, bl in subpos]\n                    # Yield this block plus sub blocks positions\n                    vec = [(bpos, blength)] + rest\n                    yield vec\n\ndef pblock(vec, cells):\n    'Prettyprints each run of blocks with a different letter A.. for each block of filled cells'\n    vector = ['_'] * cells\n    for ch, (bp, bl) in enumerate(vec, ord('A')):\n        for i in range(bp, bp + bl):\n            vector[i] = chr(ch) if vector[i] == '_' else'?'\n    return '|' + '|'.join(vector) + '|'\n\n\nif __name__ == '__main__':\n    for blocks, cells in (\n            ([2, 1], 5),\n            ([], 5),\n            ([8], 10),\n            ([2, 3, 2, 3], 15),\n           # ([4, 3], 10),\n           # ([2, 1], 5),\n           # ([3, 1], 10),\n            ([2, 3], 5),\n            ):\n        print('\\nConfiguration:\\n    %s # %i cells and %r blocks' % (pblock([], cells), cells, blocks))        \n        print('  Possibilities:')\n        for i, vector in enumerate(nonoblocks(blocks, cells)):\n            print('   ', pblock(vector, cells))\n        print('  A total of %i Possible configurations.' % (i+1))"}
{"title": "Nonogram solver", "language": "Python", "task": "nonogram is a puzzle that provides \nnumeric clues used to fill in a grid of cells, \nestablishing for each cell whether it is filled or not. \nThe puzzle solution is typically a picture of some kind.\n\nEach row and column of a rectangular grid is annotated with the lengths \nof its distinct runs of occupied cells. \nUsing only these lengths you should find one valid configuration \nof empty and occupied cells, or show a failure message.\n\n;Example\nProblem:                 Solution:\n\n. . . . . . . .  3       . # # # . . . .  3\n. . . . . . . .  2 1     # # . # . . . .  2 1\n. . . . . . . .  3 2     . # # # . . # #  3 2\n. . . . . . . .  2 2     . . # # . . # #  2 2\n. . . . . . . .  6       . . # # # # # #  6\n. . . . . . . .  1 5     # . # # # # # .  1 5\n. . . . . . . .  6       # # # # # # . .  6\n. . . . . . . .  1       . . . . # . . .  1\n. . . . . . . .  2       . . . # # . . .  2\n1 3 1 7 5 3 4 3          1 3 1 7 5 3 4 3\n2 1 5 1                  2 1 5 1   \nThe problem above could be represented by two lists of lists:\nx = [[3], [2,1], [3,2], [2,2], [6], [1,5], [6], [1], [2]]\ny = [[1,2], [3,1], [1,5], [7,1], [5], [3], [4], [3]]\nA more compact representation of the same problem uses strings, \nwhere the letters represent the numbers, A=1, B=2, etc:\nx = \"C BA CB BB F AE F A B\"\ny = \"AB CA AE GA E C D C\"\n\n;Task\nFor this task, try to solve the 4 problems below, read from a \"nonogram_problems.txt\" file that has this content \n(the blank lines are separators):\nC BA CB BB F AE F A B\nAB CA AE GA E C D C\n\nF CAC ACAC CN AAA AABB EBB EAA ECCC HCCC\nD D AE CD AE A DA BBB CC AAB BAA AAB DA AAB AAA BAB AAA CD BBA DA\n\nCA BDA ACC BD CCAC CBBAC BBBBB BAABAA ABAD AABB BBH BBBD ABBAAA CCEA AACAAB BCACC ACBH DCH ADBE ADBB DBE ECE DAA DB CC\nBC CAC CBAB BDD CDBDE BEBDF ADCDFA DCCFB DBCFC ABDBA BBF AAF BADB DBF AAAAD BDG CEF CBDB BBB FC\n\nE BCB BEA BH BEK AABAF ABAC BAA BFB OD JH BADCF Q Q R AN AAN EI H G\nE CB BAB AAA AAA AC BB ACC ACCA AGB AIA AJ AJ ACE AH BAF CAG DAG FAH FJ GJ ADK ABK BL CM\n\n'''Extra credit''': generate nonograms with unique solutions, of desired height and width.\n\nThis task is the problem n.98 of the \"99 Prolog Problems\" by Werner Hett (also thanks to Paul Singleton for the idea and the examples).\n\n; Related tasks\n* [[Nonoblock]].\n\n;See also\n* Arc Consistency Algorithm\n* http://www.haskell.org/haskellwiki/99_questions/Solutions/98 (Haskell)\n* http://twanvl.nl/blog/haskell/Nonograms (Haskell)\n* http://picolisp.com/5000/!wiki?99p98 (PicoLisp)\n\n", "solution": "from itertools import izip\n\ndef gen_row(w, s):\n    \"\"\"Create all patterns of a row or col that match given runs.\"\"\"\n    def gen_seg(o, sp):\n        if not o:\n            return [[2] * sp]\n        return [[2] * x + o[0] + tail\n                for x in xrange(1, sp - len(o) + 2)\n                for tail in gen_seg(o[1:], sp - x)]\n\n    return [x[1:] for x in gen_seg([[1] * i for i in s], w + 1 - sum(s))]\n\n\ndef deduce(hr, vr):\n    \"\"\"Fix inevitable value of cells, and propagate.\"\"\"\n    def allowable(row):\n        return reduce(lambda a, b: [x | y for x, y in izip(a, b)], row)\n\n    def fits(a, b):\n        return all(x & y for x, y in izip(a, b))\n\n    def fix_col(n):\n        \"\"\"See if any value in a given column is fixed;\n        if so, mark its corresponding row for future fixup.\"\"\"\n        c = [x[n] for x in can_do]\n        cols[n] = [x for x in cols[n] if fits(x, c)]\n        for i, x in enumerate(allowable(cols[n])):\n            if x != can_do[i][n]:\n                mod_rows.add(i)\n                can_do[i][n] &= x\n\n    def fix_row(n):\n        \"\"\"Ditto, for rows.\"\"\"\n        c = can_do[n]\n        rows[n] = [x for x in rows[n] if fits(x, c)]\n        for i, x in enumerate(allowable(rows[n])):\n            if x != can_do[n][i]:\n                mod_cols.add(i)\n                can_do[n][i] &= x\n\n    def show_gram(m):\n        # If there's 'x', something is wrong.\n        # If there's '?', needs more work.\n        for x in m:\n            print \" \".join(\"x#.?\"[i] for i in x)\n        print\n\n    w, h = len(vr), len(hr)\n    rows = [gen_row(w, x) for x in hr]\n    cols = [gen_row(h, x) for x in vr]\n    can_do = map(allowable, rows)\n\n    # Initially mark all columns for update.\n    mod_rows, mod_cols = set(), set(xrange(w))\n\n    while mod_cols:\n        for i in mod_cols:\n            fix_col(i)\n        mod_cols = set()\n        for i in mod_rows:\n            fix_row(i)\n        mod_rows = set()\n\n    if all(can_do[i][j] in (1, 2) for j in xrange(w) for i in xrange(h)):\n        print \"Solution would be unique\" # but could be incorrect!\n    else:\n        print \"Solution may not be unique, doing exhaustive search:\"\n\n    # We actually do exhaustive search anyway. Unique solution takes\n    # no time in this phase anyway, but just in case there's no\n    # solution (could happen?).\n    out = [0] * h\n\n    def try_all(n = 0):\n        if n >= h:\n            for j in xrange(w):\n                if [x[j] for x in out] not in cols[j]:\n                    return 0\n            show_gram(out)\n            return 1\n        sol = 0\n        for x in rows[n]:\n            out[n] = x\n            sol += try_all(n + 1)\n        return sol\n\n    n = try_all()\n    if not n:\n        print \"No solution.\"\n    elif n == 1:\n        print \"Unique solution.\"\n    else:\n        print n, \"solutions.\"\n    print\n\n\ndef solve(p, show_runs=True):\n    s = [[[ord(c) - ord('A') + 1 for c in w] for w in l.split()]\n         for l in p.splitlines()]\n    if show_runs:\n        print \"Horizontal runs:\", s[0]\n        print \"Vertical runs:\", s[1]\n    deduce(s[0], s[1])\n\n\ndef main():\n    # Read problems from file.\n    fn = \"nonogram_problems.txt\"\n    for p in (x for x in open(fn).read().split(\"\\n\\n\") if x):\n        solve(p)\n\n    print \"Extra example not solvable by deduction alone:\"\n    solve(\"B B A A\\nB B A A\")\n\n    print \"Extra example where there is no solution:\"\n    solve(\"B A A\\nA A A\")\n\nmain()"}
{"title": "Numbers which are the cube roots of the product of their proper divisors", "language": "Python", "task": "Example\nConsider the number 24. Its proper divisors are: 1, 2, 3, 4, 6, 8 and 12. Their product is 13,824 and the cube root of this is 24. So 24 satisfies the definition in the task title.\n\n;Task\nCompute and show here the first '''50''' positive integers which are the cube roots of the product of their proper divisors.\n\nAlso show the '''500th''' and '''5,000th''' such numbers.\n\n;Stretch\nCompute and show the '''50,000th''' such number.\n\n;Reference\n* OEIS:A111398 - Numbers which are the cube roots of the product of their proper divisors.\n\n;Note\nOEIS considers 1 to be the first number in this sequence even though, strictly speaking, it has no proper divisors. Please therefore do likewise.\n\n", "solution": "''' Rosetta code rosettacode.org/wiki/Numbers_which_are_the_cube_roots_of_the_product_of_their_proper_divisors '''\n\nfrom functools import reduce\nfrom sympy import divisors\n\n\nFOUND = 0\nfor num in range(1, 1_000_000):\n    divprod = reduce(lambda x, y: x * y, divisors(num)[:-1])if num > 1 else 1\n    if num * num * num == divprod:\n        FOUND += 1\n        if FOUND <= 50:\n            print(f'{num:5}', end='\\n' if FOUND % 10 == 0 else '')\n        if FOUND == 500:\n            print(f'\\nFive hundreth: {num:,}')\n        if FOUND == 5000:\n            print(f'\\nFive thousandth: {num:,}')\n        if FOUND == 50000:\n            print(f'\\nFifty thousandth: {num:,}')\n            break\n"}
{"title": "Numbers with equal rises and falls", "language": "Python", "task": "When a number is written in base 10,   adjacent digits may \"rise\" or \"fall\" as the number is read   (usually from left to right).\n\n\n;Definition:\nGiven the decimal digits of the number are written as a series   d:\n:*   A   ''rise''         is an index   i   such that    d(i)  <  d(i+1)\n:*   A   ''fall''    is an index   i   such that    d(i)  >  d(i+1)\n\n\n;Examples:\n:*   The number          '''726,169'''   has   '''3'''   rises and   '''2'''   falls,   so it isn't in the sequence.\n:*   The number     '''83,548'''    has   '''2'''   rises and   '''2'''   falls,   so it   is   in the sequence.\n\n\n;Task:\n:*   Print the first   '''200'''    numbers in the sequence \n:*   Show that the   '''10 millionth'''   (10,000,000th)   number in the sequence is   '''41,909,002'''\n\n\n;See also:\n*   OEIS Sequence  A296712   describes numbers whose digit sequence in base 10 have equal \"rises\" and \"falls\".\n\n\n;Related tasks:\n*   Esthetic numbers\n\n", "solution": "import itertools\n\ndef riseEqFall(num):\n    \"\"\"Check whether a number belongs to sequence A296712.\"\"\"\n    height = 0\n    d1 = num % 10\n    num //= 10\n    while num:\n        d2 = num % 10\n        height += (d1<d2) - (d1>d2)\n        d1 = d2\n        num //= 10\n    return height == 0\n    \ndef sequence(start, fn):\n    \"\"\"Generate a sequence defined by a function\"\"\"\n    num=start-1\n    while True:\n        num += 1\n        while not fn(num): num += 1\n        yield num\n\na296712 = sequence(1, riseEqFall)\n\n# Generate the first 200 numbers\nprint(\"The first 200 numbers are:\")\nprint(*itertools.islice(a296712, 200))\n\n# Generate the 10,000,000th number\nprint(\"The 10,000,000th number is:\")\nprint(*itertools.islice(a296712, 10000000-200-1, 10000000-200))\n# It is necessary to subtract 200 from the index, because 200 numbers\n# have already been consumed.\n"}
{"title": "Numeric error propagation", "language": "Python", "task": "If   '''f''',   '''a''',   and   '''b'''   are values with uncertainties   sf,   sa,   and   sb,   and   '''c'''   is a constant; \nthen if   '''f'''   is derived from   '''a''',   '''b''',   and   '''c'''   in the following ways, \nthen   sf   can be calculated as follows:\n\n:;Addition/Subtraction\n:* If   f = a +- c,   or   f = c +- a   then   '''sf = sa'''\n:* If   f = a +- b   then   '''sf2 = sa2 + sb2'''\n\n:;Multiplication/Division\n:* If   f = ca   or   f = ac       then   '''sf = |csa|'''\n:* If   f = ab   or   f = a / b              then   '''sf2 = f2( (sa / a)2 + (sb / b)2)'''\n\n:;Exponentiation\n:* If   f = ac   then   '''sf = |fc(sa / a)|'''\n\n\nCaution:\n::This implementation of error propagation does not address issues of dependent and independent values.   It is assumed that   '''a'''   and   '''b'''   are independent and so the formula for multiplication should not be applied to   '''a*a'''   for example.   See   the talk page   for some of the implications of this issue.\n\n\n;Task details:\n# Add an uncertain number type to your language that can support addition, subtraction, multiplication, division, and exponentiation between numbers with an associated error term together with 'normal' floating point numbers without an associated error term. Implement enough functionality to perform the following calculations.\n# Given coordinates and their errors:x1 = 100 +- 1.1y1 =  50 +- 1.2x2 = 200 +- 2.2y2 = 100 +- 2.3 if point p1 is located at (x1, y1) and p2 is at (x2, y2); calculate the distance between the two points using the classic Pythagorean formula:   d =    (x1 - x2)2   +   (y1 - y2)2    \n# Print and display both    '''d'''    and its error.\n\n  \n\n;References:\n* A Guide to Error Propagation B. Keeney, 2005.\n* Propagation of uncertainty Wikipedia.\n\n\n;Related task:\n*   [[Quaternion type]]\n\n", "solution": "from collections import namedtuple\nimport math\n \nclass I(namedtuple('Imprecise', 'value, delta')):\n    'Imprecise type: I(value=0.0, delta=0.0)' \n \n    __slots__ = () \n \n    def __new__(_cls, value=0.0, delta=0.0):\n        'Defaults to 0.0 \u00b1 delta'\n        return super().__new__(_cls, float(value), abs(float(delta)))\n \n    def reciprocal(self):\n        return I(1. / self.value, self.delta / (self.value**2)) \n \n    def __str__(self):\n        'Shorter form of Imprecise as string'\n        return 'I(%g, %g)' % self\n \n    def __neg__(self):\n        return I(-self.value, self.delta)\n \n    def __add__(self, other):\n        if type(other) == I:\n            return I( self.value + other.value, (self.delta**2 + other.delta**2)**0.5 )\n        try:\n            c = float(other)\n        except:\n            return NotImplemented\n        return I(self.value + c, self.delta)\n\n    def __sub__(self, other):\n        return self + (-other)\n \n    def __radd__(self, other):\n        return I.__add__(self, other)\n \n    def __mul__(self, other):\n        if type(other) == I:\n            #if id(self) == id(other):    \n            #    return self ** 2\n            a1,b1 = self\n            a2,b2 = other\n            f = a1 * a2\n            return I( f, f * ( (b1 / a1)**2 + (b2 / a2)**2 )**0.5 )\n        try:\n            c = float(other)\n        except:\n            return NotImplemented\n        return I(self.value * c, self.delta * c)\n \n    def __pow__(self, other):\n        if type(other) == I:\n            return NotImplemented\n        try:\n            c = float(other)\n        except:\n            return NotImplemented\n        f = self.value ** c\n        return I(f, f * c * (self.delta / self.value))\n \n    def __rmul__(self, other):\n        return I.__mul__(self, other)\n \n    def __truediv__(self, other):\n        if type(other) == I:\n            return self.__mul__(other.reciprocal())\n        try:\n            c = float(other)\n        except:\n            return NotImplemented\n        return I(self.value / c, self.delta / c)\n \n    def __rtruediv__(self, other):\n        return other * self.reciprocal()\n \n    __div__, __rdiv__ = __truediv__, __rtruediv__\n \nImprecise = I\n\ndef distance(p1, p2):\n    x1, y1 = p1\n    x2, y2 = p2\n    return ((x1 - x2)**2 + (y1 - y2)**2)**0.5\n \nx1 = I(100, 1.1)\nx2 = I(200, 2.2)\ny1 = I( 50, 1.2)\ny2 = I(100, 2.3)\n\np1, p2 = (x1, y1), (x2, y2)\nprint(\"Distance between points\\n  p1: %s\\n  and p2: %s\\n  = %r\" % (\n      p1, p2, distance(p1, p2)))"}
{"title": "Odd word problem", "language": "Python", "task": "Write a program that solves the odd word problem with the restrictions given below.\n\n\n;Description:\nYou are promised an input stream consisting of English letters and punctuations.  \n\nIt is guaranteed that:\n* the words (sequence of consecutive letters) are delimited by one and only one punctuation,\n* the stream will begin with a word,\n* the words will be at least one letter long,   and \n* a full stop (a period, [.]) appears after, and only after, the last word.\n\n\n;Example:\nA stream with six words: \n:: what,is,the;meaning,of:life.   \n\n\nThe task is to reverse the letters in every other word while leaving punctuations intact, producing:\n:: what,si,the;gninaem,of:efil. \nwhile observing the following restrictions:\n# Only I/O allowed is reading or writing one character at a time, which means: no reading in a string, no peeking ahead, no pushing characters back into the stream, and no storing characters in a global variable for later use;\n# You '''are not''' to explicitly save characters in a collection data structure, such as arrays, strings, hash tables, etc, for later reversal;\n# You '''are''' allowed to use recursions, closures, continuations, threads, co-routines, etc., even if their use implies the storage of multiple characters.\n\n\n;Test cases:\nWork on both the   \"life\"   example given above, and also the text:\n:: we,are;not,in,kansas;any,more.\n\n", "solution": "from sys import stdin, stdout\n\ndef char_in(): return stdin.read(1)\ndef char_out(c): stdout.write(c)\n\ndef odd(prev = lambda: None):\n\ta = char_in()\n\tif not a.isalpha():\n\t\tprev()\n\t\tchar_out(a)\n\t\treturn a != '.'\n\n\t# delay action until later, in the shape of a closure\n\tdef clos():\n\t\tchar_out(a)\n\t\tprev()\n\n\treturn odd(clos)\n\ndef even():\n\twhile True:\n\t\tc = char_in()\n\t\tchar_out(c)\n\t\tif not c.isalpha(): return c != '.'\n\ne = False\nwhile odd() if e else even():\n\te = not e"}
{"title": "Odd word problem", "language": "Python from Scheme", "task": "Write a program that solves the odd word problem with the restrictions given below.\n\n\n;Description:\nYou are promised an input stream consisting of English letters and punctuations.  \n\nIt is guaranteed that:\n* the words (sequence of consecutive letters) are delimited by one and only one punctuation,\n* the stream will begin with a word,\n* the words will be at least one letter long,   and \n* a full stop (a period, [.]) appears after, and only after, the last word.\n\n\n;Example:\nA stream with six words: \n:: what,is,the;meaning,of:life.   \n\n\nThe task is to reverse the letters in every other word while leaving punctuations intact, producing:\n:: what,si,the;gninaem,of:efil. \nwhile observing the following restrictions:\n# Only I/O allowed is reading or writing one character at a time, which means: no reading in a string, no peeking ahead, no pushing characters back into the stream, and no storing characters in a global variable for later use;\n# You '''are not''' to explicitly save characters in a collection data structure, such as arrays, strings, hash tables, etc, for later reversal;\n# You '''are''' allowed to use recursions, closures, continuations, threads, co-routines, etc., even if their use implies the storage of multiple characters.\n\n\n;Test cases:\nWork on both the   \"life\"   example given above, and also the text:\n:: we,are;not,in,kansas;any,more.\n\n", "solution": "from sys import stdin, stdout\n\ndef char_in(): return stdin.read(1)\ndef char_out(c): stdout.write(c)\n\ndef odd():\n\ta = char_in()\n\tif a.isalpha():\n\t\tr = odd()\n\t\tchar_out(a)\n\t\treturn r\n\n\t# delay printing terminator until later, in the shape of a closure\n\tdef clos():\n\t\tchar_out(a)\n\t\treturn a != '.'\n\n\treturn clos\n\ndef even():\n\twhile True:\n\t\tc = char_in()\n\t\tchar_out(c)\n\t\tif not c.isalpha(): return c != '.'\n\ne = False\nwhile odd()() if e else even():\n\te = not e"}
{"title": "Odd word problem", "language": "Python 3.3+", "task": "Write a program that solves the odd word problem with the restrictions given below.\n\n\n;Description:\nYou are promised an input stream consisting of English letters and punctuations.  \n\nIt is guaranteed that:\n* the words (sequence of consecutive letters) are delimited by one and only one punctuation,\n* the stream will begin with a word,\n* the words will be at least one letter long,   and \n* a full stop (a period, [.]) appears after, and only after, the last word.\n\n\n;Example:\nA stream with six words: \n:: what,is,the;meaning,of:life.   \n\n\nThe task is to reverse the letters in every other word while leaving punctuations intact, producing:\n:: what,si,the;gninaem,of:efil. \nwhile observing the following restrictions:\n# Only I/O allowed is reading or writing one character at a time, which means: no reading in a string, no peeking ahead, no pushing characters back into the stream, and no storing characters in a global variable for later use;\n# You '''are not''' to explicitly save characters in a collection data structure, such as arrays, strings, hash tables, etc, for later reversal;\n# You '''are''' allowed to use recursions, closures, continuations, threads, co-routines, etc., even if their use implies the storage of multiple characters.\n\n\n;Test cases:\nWork on both the   \"life\"   example given above, and also the text:\n:: we,are;not,in,kansas;any,more.\n\n", "solution": "from sys import stdin, stdout\n\ndef fwd(c):\n    if c.isalpha():\n        return [stdout.write(c), (yield from fwd((yield f)))][1]\n    else:\n        return c\n\ndef rev(c):\n    if c.isalpha():\n        return [(yield from rev((yield r))), stdout.write(c)][0]\n    else:\n        return c\n\ndef fw():\n    while True:\n        stdout.write((yield from fwd((yield r))))\n\ndef re():\n    while True:\n        stdout.write((yield from rev((yield f))))\n\nf = fw()\nr = re()\nnext(f)\nnext(r)\n\ncoro = f\nwhile True:\n    c = stdin.read(1)\n    if not c:\n        break\n    coro = coro.send(c)"}
{"title": "Old Russian measure of length", "language": "Python", "task": "Write a program to perform a conversion of the old Russian measures of length to the metric system   (and vice versa).\n\n\nIt is an example of a linear transformation of several variables. \n\n\nThe program should accept a single value in a selected unit of measurement, and convert and return it to the other units: \n''vershoks'', ''arshins'', ''sazhens'', ''versts'', ''meters'', ''centimeters'' and ''kilometers''.\n\n\n;Also see:\n:*   Old Russian measure of length\n\n", "solution": "from sys import argv\n \nunit2mult = {\"arshin\": 0.7112, \"centimeter\": 0.01,     \"diuym\":   0.0254,\n             \"fut\":    0.3048, \"kilometer\":  1000.0,   \"liniya\":  0.00254,\n             \"meter\":  1.0,    \"milia\":      7467.6,   \"piad\":    0.1778,\n             \"sazhen\": 2.1336, \"tochka\":     0.000254, \"vershok\": 0.04445,\n             \"versta\": 1066.8}\n \nif __name__ == '__main__':\n    assert len(argv) == 3, 'ERROR. Need two arguments - number then units'\n    try:\n        value = float(argv[1])\n    except:\n        print('ERROR. First argument must be a (float) number')\n        raise\n    unit = argv[2]\n    assert unit in unit2mult, ( 'ERROR. Only know the following units: ' \n                                + ' '.join(unit2mult.keys()) )\n\n    print(\"%g %s to:\" % (value, unit))\n    for unt, mlt in sorted(unit2mult.items()):\n        print('  %10s: %g' % (unt, value * unit2mult[unit] / mlt))"}
{"title": "Old lady swallowed a fly", "language": "Python", "task": "Present a program which emits the lyrics to the song   ''I Knew an Old Lady Who Swallowed a Fly'',   taking advantage of the repetitive structure of the song's lyrics. \n\nThis song has multiple versions with slightly different lyrics, so all these programs might not emit identical output.\n\n\n\n", "solution": "animals = [\n        (\"fly\", \"I don't know why she swallowed a fly, perhaps she'll die.\"),\n        (\"spider\", \"It wiggled and jiggled and tickled inside her.\"),\n        (\"bird\", \"How absurd, to swallow a bird.\"),\n        (\"cat\", \"Imagine that, she swallowed a cat.\"),\n        (\"dog\", \"What a hog, to swallow a dog.\"),\n        (\"goat\", \"She just opened her throat and swallowed a goat.\"),\n        (\"cow\", \"I don't know how she swallowed a cow.\"),\n        (\"horse\", \"She's dead, of course.\")]\n\nfor i, (animal, lyric) in enumerate(animals):\n    print(\"There was an old lady who swallowed a {}.\\n{}\".format(animal, lyric))\n\n    if animal == \"horse\": break\n\n    for (predator, _), (prey, _) in zip(animals[i:0:-1], animals[i-1::-1]):\n        print(\"\\tShe swallowed the {} to catch the {}\".format(predator, prey))\n\n    if animal != \"fly\": print(animals[0][1])  # fly lyric\n    print()  # new line"}
{"title": "One-time pad", "language": "Python", "task": "One-time pad, for encrypting and decrypting messages.\nTo keep it simple, we will be using letters only.\n\n;Sub-Tasks:\n* '''Generate''' the data for a One-time pad (user needs to specify a filename and length)\n: The important part is to get \"true random\" numbers, e.g. from /dev/random\n* '''encryption / decryption''' ( basically the same operation, much like [[Rot-13]] )\n: For this step, much of [[Vigenere cipher]] could be reused,with the key to be read from the file containing the One-time pad.\n* optional: '''management''' of One-time pads: list, mark as used, delete, etc.\n: Somehow, the users needs to keep track which pad to use for which partner.\n\nTo support the management of pad-files:\n* Such files have a file-extension \".1tp\"\n* Lines starting with \"#\" may contain arbitary meta-data (i.e. comments)\n* Lines starting with \"-\" count as \"used\"\n* Whitespace within the otp-data is ignored\n\n\n\nFor example, here is the data from Wikipedia:\n\n# Example data - Wikipedia - 2014-11-13\n-ZDXWWW EJKAWO FECIFE WSNZIP PXPKIY URMZHI JZTLBC YLGDYJ \n-HTSVTV RRYYEG EXNCGA GGQVRF FHZCIB EWLGGR BZXQDQ DGGIAK \n YHJYEQ TDLCQT HZBSIZ IRZDYS RBYJFZ AIRCWI UCVXTW YKPQMK \n CKHVEX VXYVCS WOGAAZ OUVVON GCNEVR LMBLYB SBDCDC PCGVJX \n QXAUIP PXZQIJ JIUWYH COVWMJ UZOJHL DWHPER UBSRUJ HGAAPR \n CRWVHI FRNTQW AJVWRT ACAKRD OZKIIB VIQGBK IJCWHF GTTSSE \n EXFIPJ KICASQ IOUQTP ZSGXGH YTYCTI BAZSTN JKMFXI RERYWE \n\n\n;See also\n* one time pad encryption in Python\n* snapfractalpop -  One-Time-Pad Command-Line-Utility (C).\n* Crypt-OTP-2.00 on CPAN (Perl)\n\n", "solution": "\"\"\"One-time pad using an XOR cipher. Requires Python >=3.6.\"\"\"\n\nimport argparse\nimport itertools\nimport pathlib\nimport re\nimport secrets\nimport sys\n\n\n# One-time pad file signature.\nMAGIC = \"#one-time pad\"\n\n\ndef make_keys(n, size):\n    \"\"\"Generate ``n`` secure, random keys of ``size`` bytes.\"\"\"\n    # We're generating and storing keys in their hexadecimal form to make\n    # one-time pad files a little more human readable and to ensure a key\n    # can not start with a hyphen.\n    return (secrets.token_hex(size) for _ in range(n))\n\n\ndef make_pad(name, pad_size, key_size):\n    \"\"\"Create a new one-time pad identified by the given name.\n\n    Args:\n        name (str): Unique one-time pad identifier.\n        pad_size (int): The number of keys (or pages) in the pad.\n        key_size (int): The number of bytes per key.\n    Returns:\n        The new one-time pad as a string.\n    \"\"\"\n    pad = [\n        MAGIC,\n        f\"#name={name}\",\n        f\"#size={pad_size}\",\n        *make_keys(pad_size, key_size),\n    ]\n\n    return \"\\n\".join(pad)\n\n\ndef xor(message, key):\n    \"\"\"Return ``message`` XOR-ed with ``key``.\n\n    Args:\n        message (bytes): Plaintext or cyphertext to be encrypted or decrypted.\n        key (bytes): Encryption and decryption key.\n    Returns:\n        Plaintext or cyphertext as a byte string.\n    \"\"\"\n    return bytes(mc ^ kc for mc, kc in zip(message, itertools.cycle(key)))\n\n\ndef use_key(pad):\n    \"\"\"Use the next available key from the given one-time pad.\n\n    Args:\n        pad (str): A one-time pad.\n    Returns:\n        (str, str) A two-tuple of updated pad and key.\n    \"\"\"\n    match = re.search(r\"^[a-f0-9]+$\", pad, re.MULTILINE)\n    if not match:\n        error(\"pad is all used up\")\n\n    key = match.group()\n    pos = match.start()\n\n    return (f\"{pad[:pos]}-{pad[pos:]}\", key)\n\n\ndef log(msg):\n    \"\"\"Log a message.\"\"\"\n    sys.stderr.write(msg)\n    sys.stderr.write(\"\\n\")\n\n\ndef error(msg):\n    \"\"\"Exit with an error message.\"\"\"\n    sys.stderr.write(msg)\n    sys.stderr.write(\"\\n\")\n    sys.exit(1)\n\n\ndef write_pad(path, pad_size, key_size):\n    \"\"\"Write a new one-time pad to the given path.\n\n    Args:\n        path (pathlib.Path): Path to write one-time pad to.\n        length (int): Number of keys in the pad.\n    \"\"\"\n    if path.exists():\n        error(f\"pad '{path}' already exists\")\n\n    with path.open(\"w\") as fd:\n        fd.write(make_pad(path.name, pad_size, key_size))\n\n    log(f\"New one-time pad written to {path}\")\n\n\ndef main(pad, message, outfile):\n    \"\"\"Encrypt or decrypt ``message`` using the given pad.\n\n    Args:\n        pad (pathlib.Path): Path to one-time pad.\n        message (bytes): Plaintext or ciphertext message to encrypt or decrypt.\n        outfile: File-like object to write to.\n    \"\"\"\n    if not pad.exists():\n        error(f\"no such pad '{pad}'\")\n\n    with pad.open(\"r\") as fd:\n        if fd.readline().strip() != MAGIC:\n            error(f\"file '{pad}' does not look like a one-time pad\")\n\n    # Rewrites the entire one-time pad every time\n    with pad.open(\"r+\") as fd:\n        updated, key = use_key(fd.read())\n\n        fd.seek(0)\n        fd.write(updated)\n\n    outfile.write(xor(message, bytes.fromhex(key)))\n\n\nif __name__ == \"__main__\":\n    # Command line interface\n    parser = argparse.ArgumentParser(description=\"One-time pad.\")\n\n    parser.add_argument(\n        \"pad\",\n        help=(\n            \"Path to one-time pad. If neither --encrypt or --decrypt \"\n            \"are given, will create a new pad.\"\n        ),\n    )\n\n    parser.add_argument(\n        \"--length\",\n        type=int,\n        default=10,\n        help=\"Pad size. Ignored if --encrypt or --decrypt are given. Defaults to 10.\",\n    )\n\n    parser.add_argument(\n        \"--key-size\",\n        type=int,\n        default=64,\n        help=\"Key size in bytes. Ignored if --encrypt or --decrypt are given. Defaults to 64.\",\n    )\n\n    parser.add_argument(\n        \"-o\",\n        \"--outfile\",\n        type=argparse.FileType(\"wb\"),\n        default=sys.stdout.buffer,\n        help=(\n            \"Write encoded/decoded message to a file. Ignored if --encrypt or \"\n            \"--decrypt is not given. Defaults to stdout.\"\n        ),\n    )\n\n    group = parser.add_mutually_exclusive_group()\n\n    group.add_argument(\n        \"--encrypt\",\n        metavar=\"FILE\",\n        type=argparse.FileType(\"rb\"),\n        help=\"Encrypt FILE using the next available key from pad.\",\n    )\n    group.add_argument(\n        \"--decrypt\",\n        metavar=\"FILE\",\n        type=argparse.FileType(\"rb\"),\n        help=\"Decrypt FILE using the next available key from pad.\",\n    )\n\n    args = parser.parse_args()\n\n    if args.encrypt:\n        message = args.encrypt.read()\n    elif args.decrypt:\n        message = args.decrypt.read()\n    else:\n        message = None\n\n    # Sometimes necessary if message came from stdin\n    if isinstance(message, str):\n        message = message.encode()\n\n    pad = pathlib.Path(args.pad).with_suffix(\".1tp\")\n\n    if message:\n        main(pad, message, args.outfile)\n    else:\n        write_pad(pad, args.length, args.key_size)\n"}
{"title": "One of n lines in a file", "language": "Python", "task": "A method of choosing a line randomly from a file:\n::* Without reading the file more than once\n::* When substantial parts of the file cannot be held in memory\n::* Without knowing how many lines are in the file\nIs to:\n::* keep the first line of the file as a possible choice, then\n::* Read the second line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/2.\n::* Read the third line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/3.\n::* ...\n::* Read the Nth line of the file if possible and make it the possible choice if a uniform random value between zero and one is less than 1/N\n\n::* Return the computed possible choice when no further lines exist in the file.\n\n\n;Task:\n# Create a function/method/routine called one_of_n that given n, the number of actual lines in a file, follows the algorithm above to return an integer - the line number of the line chosen from the file. The number returned can vary, randomly, in each run.\n# Use one_of_n in a ''simulation'' to find what would be the chosen line of a 10-line file simulated 1,000,000 times.\n# Print and show how many times each of the 10 lines is chosen as a rough measure of how well the algorithm works.\n\n\nNote: You may choose a smaller number of repetitions if necessary, but mention this up-front.\n\nNote: This is a specific version of a Reservoir Sampling algorithm: https://en.wikipedia.org/wiki/Reservoir_sampling\n\n", "solution": "from random import randrange\ntry:\n    range = xrange\nexcept: pass\n\ndef one_of_n(lines): # lines is any iterable\n    choice = None\n    for i, line in enumerate(lines):\n        if randrange(i+1) == 0:\n            choice = line\n    return choice\n            \ndef one_of_n_test(n=10, trials=1000000):\n    bins = [0] * n\n    if n:\n        for i in range(trials):\n            bins[one_of_n(range(n))] += 1\n    return bins\n\nprint(one_of_n_test())"}
{"title": "OpenWebNet password", "language": "Python", "task": "Calculate the password requested by ethernet gateways from the Legrand / Bticino MyHome OpenWebNet home automation system when the user's ip address is not in the gateway's whitelist\n\n'''Note:''' Factory default password is '12345'. Changing it is highly recommended !\n\nconversation goes as follows\n\n- *#*1##\n- *99*0##\n- *#603356072##\n\nat which point a password should be sent back, calculated from the \"password open\" that is set in the gateway, and the nonce that was just sent\n\n- *#25280520##\n- *#*1##\n", "solution": "def ownCalcPass (password, nonce, test=False) :\n    start = True    \n    num1 = 0\n    num2 = 0\n    password = int(password)\n    if test:\n        print(\"password: %08x\" % (password))\n    for c in nonce :\n        if c != \"0\":\n            if start:\n                num2 = password\n            start = False\n        if test:\n            print(\"c: %s num1: %08x num2: %08x\" % (c, num1, num2))\n        if c == '1':\n            num1 = (num2 & 0xFFFFFF80) >> 7\n            num2 = num2 << 25\n        elif c == '2':\n            num1 = (num2 & 0xFFFFFFF0) >> 4\n            num2 = num2 << 28\n        elif c == '3':\n            num1 = (num2 & 0xFFFFFFF8) >> 3\n            num2 = num2 << 29\n        elif c == '4':\n            num1 = num2 << 1\n            num2 = num2 >> 31\n        elif c == '5':\n            num1 = num2 << 5\n            num2 = num2 >> 27\n        elif c == '6':\n            num1 = num2 << 12\n            num2 = num2 >> 20\n        elif c == '7':\n            num1 = num2 & 0x0000FF00 | (( num2 & 0x000000FF ) << 24 ) | (( num2 & 0x00FF0000 ) >> 16 )\n            num2 = ( num2 & 0xFF000000 ) >> 8\n        elif c == '8':\n            num1 = (num2 & 0x0000FFFF) << 16 | ( num2 >> 24 )\n            num2 = (num2 & 0x00FF0000) >> 8\n        elif c == '9':\n            num1 = ~num2\n        else :\n            num1 = num2\n\n        num1 &= 0xFFFFFFFF\n        num2 &= 0xFFFFFFFF\n        if (c not in \"09\"):\n            num1 |= num2\n        if test:\n            print(\"     num1: %08x num2: %08x\" % (num1, num2))\n        num2 = num1\n    return num1\n\ndef test_passwd_calc(passwd, nonce, expected):\n    res = ownCalcPass(passwd, nonce, False)\n    m = passwd+' '+nonce+' '+str(res)+' '+str(expected)\n    if res == int(expected) :\n        print('PASS '+m)\n    else :\n        print('FAIL '+m)\n\nif __name__ == '__main__':\n    test_passwd_calc('12345','603356072','25280520')\n    test_passwd_calc('12345','410501656','119537670')\n    test_passwd_calc('12345','630292165','4269684735')\n\n"}
{"title": "Operator precedence", "language": "Python", "task": "Operators in C and C++}}\n\n\n;Task:\nProvide a list of   [[wp:order of operations|precedence   and   associativity   of all the operators and constructs that the language utilizes in descending order of precedence such that an operator which is listed on some row will be evaluated prior to any operator that is listed on a row further below it. \n\nOperators that are in the same cell (there may be several rows of operators listed in a cell) are evaluated with the same level of precedence, in the given direction. \n\nState whether arguments are passed by value or by reference.\n\n", "solution": "See [http://docs.python.org/py3k/reference/expressions.html?highlight=precedence#summary this table] and the whole page for details on Python version 3.x\nAn excerpt of which is this table:\n\n{| class=\"wikitable\"\n|-\n! Precedence\n! Operator\n! Description\n|-\n! lowest\n| '''lambda'''\n| Lambda expression\n|-\n!\n| '''if \u2013 else'''\n| Conditional expression\n|-\n!\n| '''or'''\n| <nowiki>Boolean OR</nowiki>\n|-\n!\n| '''<nowiki>and</nowiki>'''\n| Boolean AND\n|-\n!\n| '''not x'''\n| Boolean NOT\n|-\n!\n| '''in, not in, is, is not, <, <=, >, >=, !=, =='''\n| Comparisons, including membership tests and identity tests,\n|-\n!\n| '''<nowiki>|</nowiki>'''\n| Bitwise OR\n|-\n!\n| '''^'''\n| Bitwise XOR\n|-\n!\n| '''<nowiki>&</nowiki>'''\n| <nowiki>Bitwise AND</nowiki>\n|-\n!\n| '''<nowiki><<, >></nowiki>'''\n| <nowiki>Shifts</nowiki>\n|-\n!\n| '''<nowiki>+, -</nowiki>'''\n| Addition and subtraction\n|-\n!\n| '''<nowiki>*, /, //, %</nowiki>'''\n| <nowiki>Multiplication, division, remainder [1]</nowiki>\n|-\n!\n| '''+x, -x, ~x'''\n| Positive, negative, bitwise NOT\n|-\n!\n| '''**'''\n| Exponentiation [2]\n|-\n!\n| '''x[index], x[index:index], x(arguments...), x.attribute'''\n| <nowiki>Subscription, slicing, call, attribute reference</nowiki>\n|-\n! highest\n| '''<nowiki>(expressions...), [expressions...], {key:datum...}, {expressions...}</nowiki>'''\n| Binding or tuple display, list display, dictionary display, set display\n|}\n\n;Footnotes:\n# The <code>%</code> operator is also used for string formatting; the same precedence applies.\n# The power operator <code>**</code> binds less tightly than an arithmetic or bitwise unary operator on its right, that is, <code>2**-1</code> is <code>0.5</code>.\n\n"}
{"title": "Palindrome dates", "language": "Python 3.7", "task": "Today   ('''2020-02-02''',   at the time of this writing)   happens to be a palindrome,   without the hyphens,   not only for those countries which express their dates in the   '''yyyy-mm-dd'''   format but,   unusually,   also for countries which use the   '''dd-mm-yyyy'''   format.\n\n\n;Task\nWrite a program which calculates and shows the next 15 palindromic dates for those countries which express their dates in the   '''yyyy-mm-dd'''   format.\n\n", "solution": "'''Palindrome dates'''\n\nfrom functools import reduce\nfrom itertools import chain\nfrom datetime import date\n\n\n# palinDay :: Integer -> [ISO Date]\ndef palinDay(y):\n    '''A possibly empty list containing the palindromic\n       date for the given year, if such a date exists.\n    '''\n    [m, d] = [undigits(pair) for pair in chunksOf(2)(\n        reversedDecimalDigits(y)\n    )]\n    return [] if (\n        1 > m or m > 12 or 31 < d\n    ) else validISODate((y, m, d))\n\n\n# --------------------------TEST---------------------------\n# main :: IO ()\ndef main():\n    '''Count and samples of palindromic dates [2021..9999]\n    '''\n    palinDates = list(chain.from_iterable(\n        map(palinDay, range(2021, 10000))\n    ))\n    for x in [\n            'Count of palindromic dates [2021..9999]:',\n            len(palinDates),\n            '\\nFirst 15:',\n            '\\n'.join(palinDates[0:15]),\n            '\\nLast 15:',\n            '\\n'.join(palinDates[-15:])\n    ]:\n        print(x)\n\n\n# -------------------------GENERIC-------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n, subdividing the\n       contents of xs. Where the length of xs is not evenly\n       divible, the final list will be shorter than n.\n    '''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# reversedDecimalDigits :: Int -> [Int]\ndef reversedDecimalDigits(n):\n    '''A list of the decimal digits of n,\n       in reversed sequence.\n    '''\n    return unfoldr(\n        lambda x: Nothing() if (\n            0 == x\n        ) else Just(divmod(x, 10))\n    )(n)\n\n\n# unDigits :: [Int] -> Int\ndef undigits(xs):\n    '''An integer derived from a list of decimal digits\n    '''\n    return reduce(lambda a, x: a * 10 + x, xs, 0)\n\n\n# unfoldr(lambda x: Just((x, x - 1)) if 0 != x else Nothing())(10)\n# -> [10, 9, 8, 7, 6, 5, 4, 3, 2, 1]\n# unfoldr :: (b -> Maybe (a, b)) -> b -> [a]\ndef unfoldr(f):\n    '''Dual to reduce or foldr.\n       Where catamorphism reduces a list to a summary value,\n       the anamorphic unfoldr builds a list from a seed value.\n       As long as f returns Just(a, b), a is prepended to the list,\n       and the residual b is used as the argument for the next\n       application of f.\n       When f returns Nothing, the completed list is returned.\n    '''\n    def go(v):\n        xr = v, v\n        xs = []\n        while True:\n            mb = f(xr[0])\n            if mb.get('Nothing'):\n                return xs\n            else:\n                xr = mb.get('Just')\n                xs.append(xr[1])\n        return xs\n    return go\n\n\n# validISODate :: (Int, Int, Int) -> [Date]\ndef validISODate(ymd):\n    '''A possibly empty list containing the\n       ISO8601 string for a date, if that date exists.\n    '''\n    try:\n        return [date(*ymd).isoformat()]\n    except ValueError:\n        return []\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Palindromic gapful numbers", "language": "Python", "task": "Numbers   (positive integers expressed in base ten)   that are (evenly) divisible by the number formed by the\nfirst and last digit are known as   '''gapful numbers'''.\n\n\n''Evenly divisible''   means divisible with   no   remainder.\n\n\nAll   one-   and two-digit   numbers have this property and are trivially excluded.   Only\nnumbers    >=  '''100'''   will be considered for this Rosetta Code task.\n\n\n;Example:\n'''1037'''   is a   '''gapful'''   number because it is evenly divisible by the\nnumber   '''17'''   which is formed by the first and last decimal digits\nof    '''1037'''. \n\n\nA palindromic number is   (for this task, a positive integer expressed in base ten),   when the number is \nreversed,   is the same as the original number.\n\n\n;Task:\n:*   Show   (nine sets)   the first   '''20'''   palindromic gapful numbers that   ''end''   with:\n:::*   the digit   '''1'''\n:::*   the digit   '''2'''\n:::*   the digit   '''3'''\n:::*   the digit   '''4'''\n:::*   the digit   '''5'''\n:::*   the digit   '''6'''\n:::*   the digit   '''7'''\n:::*   the digit   '''8'''\n:::*   the digit   '''9'''\n:*   Show   (nine sets, like above)   of palindromic gapful numbers:\n:::*   the last   '''15'''   palindromic gapful numbers   (out of      '''100''')\n:::*   the last   '''10'''   palindromic gapful numbers   (out of                '''1,000''')       {optional}\n\n\nFor other ways of expressing the (above) requirements, see the   ''discussion''   page.\n\n\n;Note:\nAll palindromic gapful numbers are divisible by eleven. \n\n\n;Related tasks:\n:*   palindrome detection.\n:*   gapful numbers.\n\n\n;Also see:\n:*   The OEIS entry:   A108343 gapful numbers.\n\n", "solution": "from itertools import count\nfrom pprint import pformat\nimport re\nimport heapq\n\n\ndef pal_part_gen(odd=True):\n    for i in count(1):\n        fwd = str(i)\n        rev = fwd[::-1][1:] if odd else fwd[::-1]\n        yield int(fwd + rev)\n\ndef pal_ordered_gen():\n    yield from heapq.merge(pal_part_gen(odd=True), pal_part_gen(odd=False))\n\ndef is_gapful(x):\n    return (x % (int(str(x)[0]) * 10 + (x % 10)) == 0)\n\nif __name__ == '__main__':\n    start = 100\n    for mx, last in [(20, 20), (100, 15), (1_000, 10)]:\n        print(f\"\\nLast {last} of the first {mx} binned-by-last digit \" \n              f\"gapful numbers >= {start}\")\n        bin = {i: [] for i in range(1, 10)}\n        gen = (i for i in pal_ordered_gen() if i >= start and is_gapful(i))\n        while any(len(val) < mx for val in bin.values()):\n            g = next(gen)\n            val = bin[g % 10]\n            if len(val) < mx:\n                val.append(g)\n        b = {k:v[-last:] for k, v in bin.items()}\n        txt = pformat(b, width=220)\n        print('', re.sub(r\"[{},\\[\\]]\", '', txt))"}
{"title": "Pancake numbers", "language": "Python 3.7", "task": "Adrian Monk has problems and an assistant,  Sharona Fleming. Sharona can deal with most of Adrian's problems except his lack of punctuality paying her remuneration. 2 pay checks down and she prepares him pancakes for breakfast. Knowing that he will be unable to eat them unless they are stacked in ascending order of size she leaves him only a skillet which he can insert at any point in the pile and flip all the above pancakes, repeating until the pile is sorted. Sharona has left the pile of n pancakes such that the maximum number of flips is required. Adrian is determined to do this in as few flips as possible. This sequence n->p(n) is known as the Pancake numbers.\n\nThe task is to determine p(n) for n = 1 to 9, and for each show an example requiring p(n) flips.\n\n[[Sorting_algorithms/Pancake_sort]] actually performs the sort some giving the number of flips used. How do these compare with p(n)?\n\nFew people know p(20), generously I shall award an extra credit for anyone doing more than p(16).\n\n\n;References\n# Bill Gates and the pancake problem\n# A058986\n\n", "solution": "\"\"\"Pancake numbers. Requires Python>=3.7.\"\"\"\nimport time\n\nfrom collections import deque\nfrom operator import itemgetter\nfrom typing import Tuple\n\nPancakes = Tuple[int, ...]\n\n\ndef flip(pancakes: Pancakes, position: int) -> Pancakes:\n    \"\"\"Flip the stack of pancakes at the given position.\"\"\"\n    return tuple([*reversed(pancakes[:position]), *pancakes[position:]])\n\n\ndef pancake(n: int) -> Tuple[Pancakes, int]:\n    \"\"\"Return the nth pancake number.\"\"\"\n    init_stack = tuple(range(1, n + 1))\n    stack_flips = {init_stack: 0}\n    queue = deque([init_stack])\n\n    while queue:\n        stack = queue.popleft()\n        flips = stack_flips[stack] + 1\n\n        for i in range(2, n + 1):\n            flipped = flip(stack, i)\n            if flipped not in stack_flips:\n                stack_flips[flipped] = flips\n                queue.append(flipped)\n\n    return max(stack_flips.items(), key=itemgetter(1))\n\n\nif __name__ == \"__main__\":\n    start = time.time()\n\n    for n in range(1, 10):\n        pancakes, p = pancake(n)\n        print(f\"pancake({n}) = {p:>2}. Example: {list(pancakes)}\")\n\n    print(f\"\\nTook {time.time() - start:.3} seconds.\")\n"}
{"title": "Pangram checker", "language": "Python", "task": "A pangram is a sentence that contains all the letters of the English alphabet at least once.\n\nFor example:   ''The quick brown fox jumps over the lazy dog''.\n\n\n;Task:\nWrite a function or method to check a sentence to see if it is a   pangram   (or not)   and show its use.\n\n\n;Related tasks:\n:*   determine if a string has all the same characters\n:*   determine if a string has all unique characters\n\n", "solution": "import string, sys\nif sys.version_info[0] < 3:\n    input = raw_input\n\ndef ispangram(sentence, alphabet=string.ascii_lowercase):\n    alphaset = set(alphabet)\n    return alphaset <= set(sentence.lower())\n\nprint ( ispangram(input('Sentence: ')) )"}
{"title": "Paraffins", "language": "Python", "task": "This organic chemistry task is essentially to implement a tree enumeration algorithm.\n\n\n;Task:\nEnumerate, without repetitions and in order of increasing size, all possible paraffin molecules (also known as alkanes).  \n\n\nParaffins are built up using only carbon atoms, which has four bonds, and hydrogen, which has one bond.   All bonds for each atom must be used, so it is easiest to think of an alkane as linked carbon atoms forming the \"backbone\" structure, with adding hydrogen atoms linking the remaining unused bonds.\n\nIn a paraffin, one is allowed neither double bonds (two bonds between the same pair of atoms), nor cycles of linked carbons.   So all paraffins with   '''n'''   carbon atoms share the empirical formula     CnH2n+2\n\nBut for all   '''n''' >= 4   there are several distinct molecules (\"isomers\") with the same formula but different structures. \n\nThe number of isomers rises rather rapidly when   '''n'''   increases. \n\nIn counting isomers it should be borne in mind that the four bond positions on a given carbon atom can be freely interchanged and bonds rotated (including 3-D \"out of the paper\" rotations when it's being observed on a flat diagram),   so rotations or re-orientations of parts of the molecule (without breaking bonds) do not give different isomers.   So what seem at first to be different molecules may in fact turn out to be different orientations of the same molecule.\n\n\n;Example:\nWith   '''n''' = 3   there is only one way of linking the carbons despite the different orientations the molecule can be drawn;   and with   '''n''' = 4   there are two configurations: \n:::*   a   straight   chain:     (CH3)(CH2)(CH2)(CH3)   \n:::*   a        branched chain:       (CH3)(CH(CH3))(CH3)\n\nDue to bond rotations, it doesn't matter which direction the branch points in. \n\nThe phenomenon of \"stereo-isomerism\" (a molecule being different from its mirror image due to the actual 3-D arrangement of bonds) is ignored for the purpose of this task.\n\nThe input is the number   '''n'''   of carbon atoms of a molecule (for instance '''17'''). \n\nThe output is how many different different paraffins there are with   '''n'''   carbon atoms  (for instance   24,894   if   '''n''' = 17).\n\nThe sequence of those results is visible in the OEIS entry:   \n:::   A00602: number of n-node unrooted quartic trees; number of n-carbon alkanes C(n)H(2n+2) ignoring stereoisomers. \n\nThe sequence is (the index starts from zero, and represents the number of carbon atoms):\n\n 1, 1, 1, 1, 2, 3, 5, 9, 18, 35, 75, 159, 355, 802, 1858, 4347, 10359,\n 24894, 60523, 148284, 366319, 910726, 2278658, 5731580, 14490245,\n 36797588, 93839412, 240215803, 617105614, 1590507121, 4111846763,\n 10660307791, 27711253769, ...\n\n\n;Extra credit:\nShow the paraffins in some way. \n\nA flat 1D representation, with arrays or lists is enough, for instance:\n\n*Main> all_paraffins 1\n                        [CCP H H H H]\n*Main> all_paraffins 2\n                        [BCP (C H H H) (C H H H)]\n*Main> all_paraffins 3\n                        [CCP H H (C H H H) (C H H H)]\n*Main> all_paraffins 4\n                        [BCP (C H H (C H H H)) (C H H (C H H H)),\n                         CCP H (C H H H) (C H H H) (C H H H)]\n*Main> all_paraffins 5\n                        [CCP H H (C H H (C H H H)) (C H H (C H H H)),\n                         CCP H (C H H H) (C H H H) (C H H (C H H H)),\n                         CCP (C H H H) (C H H H) (C H H H) (C H H H)]\n*Main> all_paraffins 6\n                        [BCP (C H H (C H H (C H H H))) (C H H (C H H (C H H H))),\n                         BCP (C H H (C H H (C H H H))) (C H (C H H H) (C H H H)),\n                         BCP (C H (C H H H) (C H H H)) (C H (C H H H) (C H H H)),\n                         CCP H (C H H H) (C H H (C H H H)) (C H H (C H H H)),\n                         CCP (C H H H) (C H H H) (C H H H) (C H H (C H H H))]\nShowing a basic 2D ASCII-art representation of the paraffins is better; for instance (molecule names aren't necessary):\n       methane          ethane               propane               isobutane\n                     \n          H              H   H              H   H   H              H   H   H\n          |              |   |              |   |   |              |   |   |\n      H - C - H      H - C - C - H      H - C - C - C - H      H - C - C - C - H\n          |              |   |              |   |   |              |   |   |\n          H              H   H              H   H   H              H   |   H\n                                                                       |\n                                                                   H - C - H\n                                                                       |\n                                                                       H  \n\n;Links:\n*   A paper that explains the problem and its solution in a functional language:\nhttp://www.cs.wright.edu/~tkprasad/courses/cs776/paraffins-turner.pdf\n\n*   A Haskell implementation:\nhttps://github.com/ghc/nofib/blob/master/imaginary/paraffins/Main.hs\n\n*   A Scheme implementation:\nhttp://www.ccs.neu.edu/home/will/Twobit/src/paraffins.scm\n\n*   A Fortress implementation:         (this site has been closed)\nhttp://java.net/projects/projectfortress/sources/sources/content/ProjectFortress/demos/turnersParaffins0.fss?rev=3005\n\n", "solution": "This version only counts different paraffins. The multi-precision integers of Python avoid overflows.\n"}
{"title": "Paraffins", "language": "Python from C", "task": "This organic chemistry task is essentially to implement a tree enumeration algorithm.\n\n\n;Task:\nEnumerate, without repetitions and in order of increasing size, all possible paraffin molecules (also known as alkanes).  \n\n\nParaffins are built up using only carbon atoms, which has four bonds, and hydrogen, which has one bond.   All bonds for each atom must be used, so it is easiest to think of an alkane as linked carbon atoms forming the \"backbone\" structure, with adding hydrogen atoms linking the remaining unused bonds.\n\nIn a paraffin, one is allowed neither double bonds (two bonds between the same pair of atoms), nor cycles of linked carbons.   So all paraffins with   '''n'''   carbon atoms share the empirical formula     CnH2n+2\n\nBut for all   '''n''' >= 4   there are several distinct molecules (\"isomers\") with the same formula but different structures. \n\nThe number of isomers rises rather rapidly when   '''n'''   increases. \n\nIn counting isomers it should be borne in mind that the four bond positions on a given carbon atom can be freely interchanged and bonds rotated (including 3-D \"out of the paper\" rotations when it's being observed on a flat diagram),   so rotations or re-orientations of parts of the molecule (without breaking bonds) do not give different isomers.   So what seem at first to be different molecules may in fact turn out to be different orientations of the same molecule.\n\n\n;Example:\nWith   '''n''' = 3   there is only one way of linking the carbons despite the different orientations the molecule can be drawn;   and with   '''n''' = 4   there are two configurations: \n:::*   a   straight   chain:     (CH3)(CH2)(CH2)(CH3)   \n:::*   a        branched chain:       (CH3)(CH(CH3))(CH3)\n\nDue to bond rotations, it doesn't matter which direction the branch points in. \n\nThe phenomenon of \"stereo-isomerism\" (a molecule being different from its mirror image due to the actual 3-D arrangement of bonds) is ignored for the purpose of this task.\n\nThe input is the number   '''n'''   of carbon atoms of a molecule (for instance '''17'''). \n\nThe output is how many different different paraffins there are with   '''n'''   carbon atoms  (for instance   24,894   if   '''n''' = 17).\n\nThe sequence of those results is visible in the OEIS entry:   \n:::   A00602: number of n-node unrooted quartic trees; number of n-carbon alkanes C(n)H(2n+2) ignoring stereoisomers. \n\nThe sequence is (the index starts from zero, and represents the number of carbon atoms):\n\n 1, 1, 1, 1, 2, 3, 5, 9, 18, 35, 75, 159, 355, 802, 1858, 4347, 10359,\n 24894, 60523, 148284, 366319, 910726, 2278658, 5731580, 14490245,\n 36797588, 93839412, 240215803, 617105614, 1590507121, 4111846763,\n 10660307791, 27711253769, ...\n\n\n;Extra credit:\nShow the paraffins in some way. \n\nA flat 1D representation, with arrays or lists is enough, for instance:\n\n*Main> all_paraffins 1\n                        [CCP H H H H]\n*Main> all_paraffins 2\n                        [BCP (C H H H) (C H H H)]\n*Main> all_paraffins 3\n                        [CCP H H (C H H H) (C H H H)]\n*Main> all_paraffins 4\n                        [BCP (C H H (C H H H)) (C H H (C H H H)),\n                         CCP H (C H H H) (C H H H) (C H H H)]\n*Main> all_paraffins 5\n                        [CCP H H (C H H (C H H H)) (C H H (C H H H)),\n                         CCP H (C H H H) (C H H H) (C H H (C H H H)),\n                         CCP (C H H H) (C H H H) (C H H H) (C H H H)]\n*Main> all_paraffins 6\n                        [BCP (C H H (C H H (C H H H))) (C H H (C H H (C H H H))),\n                         BCP (C H H (C H H (C H H H))) (C H (C H H H) (C H H H)),\n                         BCP (C H (C H H H) (C H H H)) (C H (C H H H) (C H H H)),\n                         CCP H (C H H H) (C H H (C H H H)) (C H H (C H H H)),\n                         CCP (C H H H) (C H H H) (C H H H) (C H H (C H H H))]\nShowing a basic 2D ASCII-art representation of the paraffins is better; for instance (molecule names aren't necessary):\n       methane          ethane               propane               isobutane\n                     \n          H              H   H              H   H   H              H   H   H\n          |              |   |              |   |   |              |   |   |\n      H - C - H      H - C - C - H      H - C - C - C - H      H - C - C - C - H\n          |              |   |              |   |   |              |   |   |\n          H              H   H              H   H   H              H   |   H\n                                                                       |\n                                                                   H - C - H\n                                                                       |\n                                                                       H  \n\n;Links:\n*   A paper that explains the problem and its solution in a functional language:\nhttp://www.cs.wright.edu/~tkprasad/courses/cs776/paraffins-turner.pdf\n\n*   A Haskell implementation:\nhttps://github.com/ghc/nofib/blob/master/imaginary/paraffins/Main.hs\n\n*   A Scheme implementation:\nhttp://www.ccs.neu.edu/home/will/Twobit/src/paraffins.scm\n\n*   A Fortress implementation:         (this site has been closed)\nhttp://java.net/projects/projectfortress/sources/sources/content/ProjectFortress/demos/turnersParaffins0.fss?rev=3005\n\n", "solution": "from itertools import count, chain, tee, islice, cycle\nfrom fractions import Fraction\nfrom sys import setrecursionlimit\nsetrecursionlimit(5000)\n\ndef frac(a,b): return a//b if a%b == 0 else Fraction(a,b)\n\n# infinite polynomial class\nclass Poly:\n    def __init__(self, gen = None):\n        self.gen, self.source = (None, gen) if type(gen) is Poly \\\n            else (gen, None)\n\n    def __iter__(self):\n        # We're essentially tee'ing it everytime the iterator\n        # is, well, iterated.  This may be excessive.\n        return Poly(self)\n\n    def getsource(self):\n        if self.gen == None:\n            s = self.source\n            s.getsource()\n            s.gen, self.gen = tee(s.gen, 2)\n\n    def next(self):\n        self.getsource()\n        return next(self.gen)\n\n    __next__ = next\n\n    # Overload \"<<\" as stream input operator. Hey, C++ does it.\n    def __lshift__(self, a): self.gen = a\n\n    # The other operators are pretty much what one would expect\n    def __neg__(self): return Poly(-x for x in self)\n\n    def __sub__(a, b): return a + (-b)\n\n    def __rsub__(a, n):\n        a = Poly(a)\n        def gen():\n            yield(n - next(a))\n            for x in a: yield(-x)\n        return Poly(gen())\n\n    def __add__(a, b):\n        if type(b) is Poly:\n            return Poly(x + y for (x,y) in zip(a,b))\n\n        a = Poly(a)\n        def gen():\n            yield(next(a) + b)\n            for x in a: yield(x)\n\n        return Poly(gen())\n\n    def __radd__(a,b):\n        return a + b\n\n    def __mul__(a,b):\n        if not type(b) is Poly:\n            return Poly(x*b for x in a)\n\n        def gen():\n            s = Poly(cycle([0]))\n            for y in b:\n                s += y*a\n                yield(next(s))\n\n        return Poly(gen())\n\n    def __rmul__(a,b): return a*b\n\n    def __truediv__(a,b):\n        if not type(b) is Poly:\n            return Poly(frac(x, b) for x in a)\n\n        a, b = Poly(a), Poly(b)\n        def gen():\n            r, bb = a,next(b)\n            while True:\n                aa = next(r)\n                q = frac(aa, bb)\n                yield(q)\n                r -= q*b\n\n        return Poly(gen())\n\n    def repl(self, n):\n        def gen():\n            for x in self:\n                yield(x)\n                for i in range(n-1): yield(0)\n        return Poly(gen())\n\n    def __pow__(self, n):\n        return Poly(self) if n == 1 else self * self**(n-1)\n\ndef S2(a,b): return (a*a + b)/2\ndef S4(a,b,c,d): return a**4/24 + a**2*b/4 + a*c/3 + b**2/8 + d/4\n\nx1 = Poly()\nx2 = x1.repl(2)\nx3 = x1.repl(3)\nx4 = x1.repl(4)\nx1 << chain([1], (x1**3 + 3*x1*x2 + 2*x3)/6)\n\na598 = x1\na678 = Poly(chain([0], S4(x1, x2, x3, x4)))\na599 = S2(x1 - 1, x2 - 1)\na602 = a678 - a599 + x2\n\nfor n,x in zip(count(0), islice(a602, 500)): print(n,x)"}
{"title": "Parse an IP Address", "language": "Python 3.5", "task": "The purpose of this task is to demonstrate parsing of text-format IP addresses, using IPv4 and IPv6.\n \n\nTaking the following as inputs:\n::: {| border=\"5\" cellspacing=\"0\" cellpadding=2\n|-\n|127.0.0.1\n|The \"localhost\" IPv4 address\n|-\n|127.0.0.1:80\n|The \"localhost\" IPv4 address, with a specified port (80)\n|-\n|::1\n|The \"localhost\" IPv6 address\n|-\n|[::1]:80\n|The \"localhost\" IPv6 address, with a specified port (80)\n|-\n|2605:2700:0:3::4713:93e3\n|Rosetta Code's primary server's public IPv6 address\n|-\n|[2605:2700:0:3::4713:93e3]:80\n|Rosetta Code's primary server's public IPv6 address, with a specified port (80)\n|}\n\n\n;Task:\nEmit each described IP address as a hexadecimal integer representing the address, the address space, and the port number specified, if any.  \n\nIn languages where variant result types are clumsy, the result should be ipv4 or ipv6 address number, something which says which address space was represented, port number and something that says if the port was specified.\n\n\n;Example:\n'''127.0.0.1'''   has the address number   '''7F000001'''   (2130706433 decimal) \nin the ipv4 address space.  \n\n'''::ffff:127.0.0.1'''   represents the same address in the ipv6 address space where it has the \naddress number    '''FFFF7F000001'''   (281472812449793 decimal).  \n\n'''::1'''   has address number   '''1'''   and serves the same purpose in the ipv6 address \nspace that   '''127.0.0.1'''   serves in the ipv4 address space.\n\n", "solution": "from ipaddress import ip_address\nfrom urllib.parse import urlparse\n\ntests = [\n    \"127.0.0.1\",\n    \"127.0.0.1:80\",\n    \"::1\",\n    \"[::1]:80\",\n    \"::192.168.0.1\",\n    \"2605:2700:0:3::4713:93e3\",\n    \"[2605:2700:0:3::4713:93e3]:80\" ]\n\ndef parse_ip_port(netloc):\n    try:\n        ip = ip_address(netloc)\n        port = None\n    except ValueError:\n        parsed = urlparse('//{}'.format(netloc))\n        ip = ip_address(parsed.hostname)\n        port = parsed.port\n    return ip, port\n\nfor address in tests:\n    ip, port = parse_ip_port(address)\n    hex_ip = {4:'{:08X}', 6:'{:032X}'}[ip.version].format(int(ip))\n    print(\"{:39s}  {:>32s}  IPv{}  port={}\".format(\n        str(ip), hex_ip, ip.version, port ))"}
{"title": "Parsing/RPN calculator algorithm", "language": "Python", "task": "Create a stack-based evaluator for an expression in   reverse Polish notation (RPN)   that also shows the changes in the stack as each individual token is processed ''as a table''.\n\n\n* Assume an input of a correct, space separated, string of tokens of an RPN expression\n* Test with the RPN expression generated from the   [[Parsing/Shunting-yard algorithm]]   task: \n         3 4 2 * 1 5 - 2 3 ^ ^ / +  \n* Print or display the output here\n\n\n;Notes:\n*    '''^'''    means exponentiation in the expression above.\n*    '''/'''    means division.\n\n\n;See also:\n*   [[Parsing/Shunting-yard algorithm]] for a method of generating an RPN from an infix expression.\n*   Several solutions to [[24 game/Solve]] make use of RPN evaluators (although tracing how they work is not a part of that task).\n*   [[Parsing/RPN to infix conversion]].\n*   [[Arithmetic evaluation]].\n\n", "solution": "def op_pow(stack):\n    b = stack.pop(); a = stack.pop()\n    stack.append( a ** b )\ndef op_mul(stack):\n    b = stack.pop(); a = stack.pop()\n    stack.append( a * b )\ndef op_div(stack):\n    b = stack.pop(); a = stack.pop()\n    stack.append( a / b )\ndef op_add(stack):\n    b = stack.pop(); a = stack.pop()\n    stack.append( a + b )\ndef op_sub(stack):\n    b = stack.pop(); a = stack.pop()\n    stack.append( a - b )\ndef op_num(stack, num):\n    stack.append( num )\n    \nops = {\n '^': op_pow,\n '*': op_mul,\n '/': op_div,\n '+': op_add,\n '-': op_sub,\n }\n\ndef get_input(inp = None):\n    'Inputs an expression and returns list of tokens'\n    \n    if inp is None:\n        inp = input('expression: ')\n    tokens = inp.strip().split()\n    return tokens\n\ndef rpn_calc(tokens):\n    stack = []\n    table = ['TOKEN,ACTION,STACK'.split(',')]\n    for token in tokens:\n        if token in ops:\n            action = 'Apply op to top of stack'\n            ops[token](stack)\n            table.append( (token, action, ' '.join(str(s) for s in stack)) )\n        else:\n            action = 'Push num onto top of stack'\n            op_num(stack, eval(token))\n            table.append( (token, action, ' '.join(str(s) for s in stack)) )\n    return table\n\nif __name__ == '__main__':\n    rpn = '3 4 2 * 1 5 - 2 3 ^ ^ / +'\n    print( 'For RPN expression: %r\\n' % rpn )\n    rp = rpn_calc(get_input(rpn))\n    maxcolwidths = [max(len(y) for y in x) for x in zip(*rp)]\n    row = rp[0]\n    print( ' '.join('{cell:^{width}}'.format(width=width, cell=cell) for (width, cell) in zip(maxcolwidths, row)))\n    for row in rp[1:]:\n        print( ' '.join('{cell:<{width}}'.format(width=width, cell=cell) for (width, cell) in zip(maxcolwidths, row)))\n\n    print('\\n The final output value is: %r' % rp[-1][2])"}
{"title": "Parsing/RPN to infix conversion", "language": "Python", "task": "Create a program that takes an infix notation.\n\n* Assume an input of a correct, space separated, string of tokens\n* Generate a space separated output string representing the same expression in infix notation\n* Show how the major datastructure of your algorithm changes with each new token parsed.\n* Test with the following input RPN strings then print and display the output here.\n:::::{| class=\"wikitable\"\n! RPN input !! sample output\n|- || align=\"center\"\n| 3 4 2 * 1 5 - 2 3 ^ ^ / +|| 3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3\n|- || align=\"center\"\n| 1 2 + 3 4 + ^ 5 6 + ^|| ( ( 1 + 2 ) ^ ( 3 + 4 ) ) ^ ( 5 + 6 )\n|}\n\n* Operator precedence and operator associativity is given in this table:\n::::::::{| class=\"wikitable\"\n\n! operator !! associativity !! operation\n|- || align=\"center\" \n|                    ^   ||  4  ||  right  ||  exponentiation\n|- || align=\"center\" \n|                    *   ||  3  ||   left  ||  multiplication\n|- || align=\"center\" \n|                    /   ||  3  ||   left  ||     division\n|- || align=\"center\" \n|                    +   ||  2  ||   left  ||     addition\n|- || align=\"center\" \n|                    -   ||  2  ||   left  ||   subtraction\n|}\n\n\n;See also:\n*   [[Parsing/Shunting-yard algorithm]]   for a method of generating an RPN from an infix expression.\n*   [[Parsing/RPN calculator algorithm]]   for a method of calculating a final value from this output RPN expression.\n*   Postfix to infix   from the RubyQuiz site.\n\n", "solution": "\"\"\"\n>>> # EXAMPLE USAGE\n>>> result = rpn_to_infix('3 4 2 * 1 5 - 2 3 ^ ^ / +', VERBOSE=True)\nTOKEN  STACK\n3      ['3']\n4      ['3', '4']\n2      ['3', '4', '2']\n*      ['3', Node('2','*','4')]\n1      ['3', Node('2','*','4'), '1']\n5      ['3', Node('2','*','4'), '1', '5']\n-      ['3', Node('2','*','4'), Node('5','-','1')]\n2      ['3', Node('2','*','4'), Node('5','-','1'), '2']\n3      ['3', Node('2','*','4'), Node('5','-','1'), '2', '3']\n^      ['3', Node('2','*','4'), Node('5','-','1'), Node('3','^','2')]\n^      ['3', Node('2','*','4'), Node(Node('3','^','2'),'^',Node('5','-','1'))]\n/      ['3', Node(Node(Node('3','^','2'),'^',Node('5','-','1')),'/',Node('2','*','4'))]\n+      [Node(Node(Node(Node('3','^','2'),'^',Node('5','-','1')),'/',Node('2','*','4')),'+','3')]\n\"\"\"\n\nprec_dict =  {'^':4, '*':3, '/':3, '+':2, '-':2}\nassoc_dict = {'^':1, '*':0, '/':0, '+':0, '-':0}\n \nclass Node:\n    def __init__(self,x,op,y=None):\n        self.precedence = prec_dict[op]\n        self.assocright = assoc_dict[op]\n        self.op = op\n        self.x,self.y = x,y\n \n    def __str__(self):\n        \"\"\"\n        Building an infix string that evaluates correctly is easy.\n        Building an infix string that looks pretty and evaluates\n        correctly requires more effort.\n        \"\"\"\n        # easy case, Node is unary\n        if self.y == None:\n            return '%s(%s)'%(self.op,str(self.x))\n \n        # determine left side string\n        str_y = str(self.y)\n        if  self.y < self or \\\n            (self.y == self and self.assocright) or \\\n            (str_y[0] is '-' and self.assocright):\n \n            str_y = '(%s)'%str_y\n        # determine right side string and operator\n        str_x = str(self.x)\n        str_op = self.op\n        if self.op is '+' and not isinstance(self.x, Node) and str_x[0] is '-':\n            str_x = str_x[1:]\n            str_op = '-'\n        elif self.op is '-' and not isinstance(self.x, Node) and str_x[0] is '-':\n            str_x = str_x[1:]\n            str_op = '+'\n        elif self.x < self or \\\n             (self.x == self and not self.assocright and \\\n              getattr(self.x, 'op', 1) != getattr(self, 'op', 2)):\n \n            str_x = '(%s)'%str_x\n        return ' '.join([str_y, str_op, str_x])\n \n    def __repr__(self):\n        \"\"\"\n        >>> repr(Node('3','+','4')) == repr(eval(repr(Node('3','+','4'))))\n        True\n        \"\"\"\n        return 'Node(%s,%s,%s)'%(repr(self.x), repr(self.op), repr(self.y))\n \n    def __lt__(self, other):\n        if isinstance(other, Node):\n            return self.precedence < other.precedence\n        return self.precedence < prec_dict.get(other,9)\n \n    def __gt__(self, other):\n        if isinstance(other, Node):\n            return self.precedence > other.precedence\n        return self.precedence > prec_dict.get(other,9)\n\n    def __eq__(self, other):\n        if isinstance(other, Node):\n            return self.precedence == other.precedence\n        return self.precedence > prec_dict.get(other,9)\n \n \n \ndef rpn_to_infix(s, VERBOSE=False):\n    \"\"\"\n    \n    converts rpn notation to infix notation for string s\n \n    \"\"\"\n    if VERBOSE : print('TOKEN  STACK')\n \n    stack=[]\n    for token in s.replace('^','^').split():\n        if token in prec_dict:\n            stack.append(Node(stack.pop(),token,stack.pop()))\n        else:\n            stack.append(token)\n \n        # can't use \\t in order to make global docstring pass doctest\n        if VERBOSE : print(token+' '*(7-len(token))+repr(stack)) \n \n    return str(stack[0])\n  \nstrTest = \"3 4 2 * 1 5 - 2 3 ^ ^ / +\"\nstrResult = rpn_to_infix(strTest, VERBOSE=False)\nprint (\"Input: \",strTest)\nprint (\"Output:\",strResult)\n\nprint()\n\nstrTest = \"1 2 + 3 4 + ^ 5 6 + ^\"\nstrResult = rpn_to_infix(strTest, VERBOSE=False)\nprint (\"Input: \",strTest)\nprint (\"Output:\",strResult)\n"}
{"title": "Parsing/Shunting-yard algorithm", "language": "Python", "task": "Given the operator characteristics and input from the Shunting-yard algorithm page and tables, use the algorithm to show the changes in the operator stack and RPN output\nas each individual token is processed.\n\n* Assume an input of a correct, space separated, string of tokens representing an infix expression\n* Generate a space separated output string representing the RPN\n* Test with the input string:\n::::  3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3  \n* print and display the output here.\n* Operator precedence is given in this table:\n:{| class=\"wikitable\"\n\n! operator !! associativity !! operation\n|- || align=\"center\"\n|                     ^   ||  4  ||  right  ||  exponentiation \n|- || align=\"center\" \n|                     *   ||  3  ||  left   ||  multiplication\n|- || align=\"center\" \n|                     /   ||  3  ||  left   ||  division\n|- || align=\"center\" \n|                     +   ||  2  ||  left   ||  addition\n|- || align=\"center\"\n|                     -   ||  2  ||  left   ||  subtraction\n|}\n\n\n;Extra credit\nAdd extra text explaining the actions and an optional comment for the action on receipt of each token.\n\n\n;Note\nThe handling of functions and arguments is not required.\n\n\n;See also:\n* [[Parsing/RPN calculator algorithm]] for a method of calculating a final value from this output RPN expression.\n* [[Parsing/RPN to infix conversion]].\n\n", "solution": "from collections import namedtuple\nfrom pprint import pprint as pp\n\nOpInfo = namedtuple('OpInfo', 'prec assoc')\nL, R = 'Left Right'.split()\n\nops = {\n '^': OpInfo(prec=4, assoc=R),\n '*': OpInfo(prec=3, assoc=L),\n '/': OpInfo(prec=3, assoc=L),\n '+': OpInfo(prec=2, assoc=L),\n '-': OpInfo(prec=2, assoc=L),\n '(': OpInfo(prec=9, assoc=L),\n ')': OpInfo(prec=0, assoc=L),\n }\n\nNUM, LPAREN, RPAREN = 'NUMBER ( )'.split()\n\n\ndef get_input(inp = None):\n    'Inputs an expression and returns list of (TOKENTYPE, tokenvalue)'\n    \n    if inp is None:\n        inp = input('expression: ')\n    tokens = inp.strip().split()\n    tokenvals = []\n    for token in tokens:\n        if token in ops:\n            tokenvals.append((token, ops[token]))\n        #elif token in (LPAREN, RPAREN):\n        #    tokenvals.append((token, token))\n        else:    \n            tokenvals.append((NUM, token))\n    return tokenvals\n\ndef shunting(tokenvals):\n    outq, stack = [], []\n    table = ['TOKEN,ACTION,RPN OUTPUT,OP STACK,NOTES'.split(',')]\n    for token, val in tokenvals:\n        note = action = ''\n        if token is NUM:\n            action = 'Add number to output'\n            outq.append(val)\n            table.append( (val, action, ' '.join(outq), ' '.join(s[0] for s in stack), note) )\n        elif token in ops:\n            t1, (p1, a1) = token, val\n            v = t1\n            note = 'Pop ops from stack to output' \n            while stack:\n                t2, (p2, a2) = stack[-1]\n                if (a1 == L and p1 <= p2) or (a1 == R and p1 < p2):\n                    if t1 != RPAREN:\n                        if t2 != LPAREN:\n                            stack.pop()\n                            action = '(Pop op)'\n                            outq.append(t2)\n                        else:    \n                            break\n                    else:        \n                        if t2 != LPAREN:\n                            stack.pop()\n                            action = '(Pop op)'\n                            outq.append(t2)\n                        else:    \n                            stack.pop()\n                            action = '(Pop & discard \"(\")'\n                            table.append( (v, action, ' '.join(outq), ' '.join(s[0] for s in stack), note) )\n                            break\n                    table.append( (v, action, ' '.join(outq), ' '.join(s[0] for s in stack), note) )\n                    v = note = ''\n                else:\n                    note = ''\n                    break\n                note = '' \n            note = '' \n            if t1 != RPAREN:\n                stack.append((token, val))\n                action = 'Push op token to stack'\n            else:\n                action = 'Discard \")\"'\n            table.append( (v, action, ' '.join(outq), ' '.join(s[0] for s in stack), note) )\n    note = 'Drain stack to output'\n    while stack:\n        v = ''\n        t2, (p2, a2) = stack[-1]\n        action = '(Pop op)'\n        stack.pop()\n        outq.append(t2)\n        table.append( (v, action, ' '.join(outq), ' '.join(s[0] for s in stack), note) )\n        v = note = ''\n    return table\n\nif __name__ == '__main__':\n    infix = '3 + 4 * 2 / ( 1 - 5 ) ^ 2 ^ 3'\n    print( 'For infix expression: %r\\n' % infix )\n    rp = shunting(get_input(infix))\n    maxcolwidths = [len(max(x, key=len)) for x in zip(*rp)]\n    row = rp[0]\n    print( ' '.join('{cell:^{width}}'.format(width=width, cell=cell) for (width, cell) in zip(maxcolwidths, row)))\n    for row in rp[1:]:\n        print( ' '.join('{cell:<{width}}'.format(width=width, cell=cell) for (width, cell) in zip(maxcolwidths, row)))\n\n    print('\\n The final output RPN is: %r' % rp[-1][2])"}
{"title": "Pascal's triangle/Puzzle", "language": "Python 2.4+", "task": "This puzzle involves a Pascals Triangle, also known as a Pyramid of Numbers.\n\n           [ 151]\n          [  ][  ]\n        [40][  ][  ]\n      [  ][  ][  ][  ]\n    [ X][11][ Y][ 4][ Z]\n\nEach brick of the pyramid is the sum of the two bricks situated below it.\nOf the three missing numbers at the base of the pyramid, \nthe middle one is the sum of the other two (that is, Y = X + Z).\n\n\n;Task:\nWrite a program to find a solution to this puzzle.\n\n", "solution": "# Pyramid solver\n#            [151]\n#         [   ] [   ]\n#      [ 40] [   ] [   ]\n#   [   ] [   ] [   ] [   ]\n#[ X ] [ 11] [ Y ] [ 4 ] [ Z ]\n#  X -Y + Z = 0\n\ndef combine( snl, snr ):\n\n\tcl = {}\n\tif isinstance(snl, int):\n\t\tcl['1'] = snl\n\telif isinstance(snl, string):\n\t\tcl[snl] = 1\n\telse:\n\t\tcl.update( snl)\n\n\tif isinstance(snr, int):\n\t\tn = cl.get('1', 0)\n\t\tcl['1'] = n + snr\n\telif isinstance(snr, string):\n\t\tn = cl.get(snr, 0)\n\t\tcl[snr] = n + 1\n\telse:\n\t\tfor k,v in snr.items():\n\t\t\tn = cl.get(k, 0)\n\t\t\tcl[k] = n+v\n\treturn cl\n\n\ndef constrain(nsum, vn ):\n\tnn = {}\n\tnn.update(vn)\n\tn = nn.get('1', 0)\n\tnn['1'] = n - nsum\n\treturn nn\n\ndef makeMatrix( constraints ):\n\tvmap = set()\n\tfor c in constraints:\n\t\tvmap.update( c.keys())\n\tvmap.remove('1')\n\tnvars = len(vmap)\n\tvmap = sorted(vmap)\t\t# sort here so output is in sorted order\n\tmtx = []\n\tfor c in constraints:\n\t\trow = []\n\t\tfor vv in vmap:\n\t\t\trow.append(float(c.get(vv, 0)))\n\t\trow.append(-float(c.get('1',0)))\n\t\tmtx.append(row)\n\t\n\tif len(constraints) == nvars:\n\t\tprint 'System appears solvable'\n\telif len(constraints) < nvars:\n\t\tprint 'System is not solvable - needs more constraints.'\n\treturn mtx, vmap\n\n\ndef SolvePyramid( vl, cnstr ):\n\n\tvl.reverse()\n\tconstraints = [cnstr]\n\tlvls = len(vl)\n\tfor lvln in range(1,lvls):\n\t\tlvd = vl[lvln]\n\t\tfor k in range(lvls - lvln):\n\t\t\tsn = lvd[k]\n\t\t\tll = vl[lvln-1]\n\t\t\tvn = combine(ll[k], ll[k+1])\n\t\t\tif sn is None:\n\t\t\t\tlvd[k] = vn\n\t\t\telse:\n\t\t\t\tconstraints.append(constrain( sn, vn ))\n\n\tprint 'Constraint Equations:'\n\tfor cstr in constraints:\n\t\tfset = ('%d*%s'%(v,k) for k,v in cstr.items() )\n\t\tprint ' + '.join(fset), ' = 0'\n\n\tmtx,vmap = makeMatrix(constraints)\n\n\tMtxSolve(mtx)\n\n\td = len(vmap)\n\tfor j in range(d):\n\t\tprint vmap[j],'=', mtx[j][d]\n\n\ndef MtxSolve(mtx):\n\t# Simple Matrix solver...\n\n\tmDim = len(mtx)\t\t\t# dimension---\n\tfor j in range(mDim):\n\t\trw0= mtx[j]\n\t\tf = 1.0/rw0[j]\n\t\tfor k in range(j, mDim+1):\n\t\t\trw0[k] *= f\n\t\t\n\t\tfor l in range(1+j,mDim):\n\t\t\trwl = mtx[l]\n\t\t\tf = -rwl[j]\n\t\t\tfor k in range(j, mDim+1):\n\t\t\t\trwl[k] += f * rw0[k]\n\n\t# backsolve part ---\n\tfor j1 in range(1,mDim):\n\t\tj = mDim - j1\n\t\trw0= mtx[j]\n\t\tfor l in range(0, j):\n\t\t\trwl = mtx[l]\n\t\t\tf = -rwl[j]\n\t\t\trwl[j]    += f * rw0[j]\n\t\t\trwl[mDim] += f * rw0[mDim]\n\n\treturn mtx\n\n\np = [ [151], [None,None], [40,None,None], [None,None,None,None], ['X', 11, 'Y', 4, 'Z'] ]\naddlConstraint = { 'X':1, 'Y':-1, 'Z':1, '1':0 }\nSolvePyramid( p, addlConstraint)"}
{"title": "Pascal matrix generation", "language": "Python", "task": "A pascal matrix is a two-dimensional square matrix holding numbers from   binomial coefficients   and which can be shown as    nCr.\n\nShown below are truncated   5-by-5   matrices   M[i, j]   for   i,j   in range   0..4. \n\nA Pascal upper-triangular matrix that is populated with   jCi:\n\n[[1, 1, 1, 1, 1],\n [0, 1, 2, 3, 4],\n [0, 0, 1, 3, 6],\n [0, 0, 0, 1, 4],\n [0, 0, 0, 0, 1]]\n\n\nA Pascal lower-triangular matrix that is populated with   iCj   (the transpose of the upper-triangular matrix):\n\n[[1, 0, 0, 0, 0],\n [1, 1, 0, 0, 0],\n [1, 2, 1, 0, 0],\n [1, 3, 3, 1, 0],\n [1, 4, 6, 4, 1]]\n\n\nA Pascal symmetric matrix that is populated with   i+jCi:\n\n[[1, 1, 1, 1, 1],\n [1, 2, 3, 4, 5],\n [1, 3, 6, 10, 15],\n [1, 4, 10, 20, 35],\n [1, 5, 15, 35, 70]]\n\n\n\n;Task:\nWrite functions capable of generating each of the three forms of   n-by-n   matrices.\n\nUse those functions to display upper, lower, and symmetric Pascal   5-by-5   matrices on this page. \n\nThe output should distinguish between different matrices and the rows of each matrix   (no showing a list of 25 numbers assuming the reader should split it into rows).\n\n\n;Note:  \nThe   [[Cholesky decomposition]]   of a Pascal symmetric matrix is the Pascal lower-triangle matrix of the same size. \n\n", "solution": "from pprint import pprint as pp\n\ndef pascal_upp(n):\n    s = [[0] * n for _ in range(n)]\n    s[0] = [1] * n\n    for i in range(1, n):\n        for j in range(i, n):\n            s[i][j] = s[i-1][j-1] + s[i][j-1]\n    return s\n\ndef pascal_low(n):\n    # transpose of pascal_upp(n)\n    return [list(x) for x in zip(*pascal_upp(n))]\n\ndef pascal_sym(n):\n    s = [[1] * n for _ in range(n)]\n    for i in range(1, n):\n        for j in range(1, n):\n            s[i][j] = s[i-1][j] + s[i][j-1]\n    return s\n    \n\nif __name__ == \"__main__\":\n    n = 5\n    print(\"\\nUpper:\")\n    pp(pascal_upp(n))\n    print(\"\\nLower:\")\n    pp(pascal_low(n))\n    print(\"\\nSymmetric:\")\n    pp(pascal_sym(n))"}
{"title": "Password generator", "language": "Python", "task": "Create a password generation program which will generate passwords containing random ASCII characters from the following groups:\n          lower-case letters:  a --> z\n          upper-case letters:  A --> Z\n                      digits:  0 --> 9\n  other printable characters:  !\"#$%&'()*+,-./:;<=>?@[]^_{|}~ \n  (the above character list excludes white-space, backslash and grave) \n\n\nThe generated password(s) must include   ''at least one''   (of each of the four groups):\n    lower-case letter, \n    upper-case letter,\n    digit  (numeral),   and \n    one  \"other\"  character. \n\n\nThe user must be able to specify the password length and the number of passwords to generate. \n\nThe passwords should be displayed or written to a file, one per line.\n\nThe randomness should be from a system source or library. \n\nThe program should implement a help option or button which should describe the program and options when invoked. \n\nYou may also allow the user to specify a seed value, and give the option of excluding visually similar characters.\n\nFor example:            Il1     O0     5S     2Z            where the characters are: \n::::*   capital eye, lowercase ell, the digit one\n::::*   capital oh, the digit zero \n::::*   the digit five, capital ess\n::::*   the digit two, capital zee\n\n\n", "solution": "import random\n\nlowercase = 'abcdefghijklmnopqrstuvwxyz' # same as string.ascii_lowercase\nuppercase = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' # same as string.ascii_uppercase\ndigits = '0123456789'                    # same as string.digits\npunctuation = '!\"#$%&\\'()*+,-./:;<=>?@[]^_{|}~' # like string.punctuation but without backslash \\ nor grave `\n\nallowed = lowercase + uppercase + digits + punctuation\n\nvisually_similar = 'Il1O05S2Z'\n\n\ndef new_password(length:int, readable=True) -> str:\n    if length < 4:\n        print(\"password length={} is too short,\".format(length),\n            \"minimum length=4\")\n        return ''\n    choice = random.SystemRandom().choice\n    while True:\n        password_chars = [\n            choice(lowercase),\n            choice(uppercase),\n            choice(digits),\n            choice(punctuation)\n            ] + random.sample(allowed, length-4)\n        if (not readable or \n                all(c not in visually_similar for c in password_chars)):\n            random.SystemRandom().shuffle(password_chars)\n            return ''.join(password_chars)\n\n\ndef password_generator(length, qty=1, readable=True):\n    for i in range(qty):\n        print(new_password(length, readable))\n"}
{"title": "Pathological floating point problems", "language": "Python", "task": "Most programmers are familiar with the inexactness of floating point calculations in a binary processor. \n\nThe classic example being:\n\n0.1 + 0.2 =  0.30000000000000004\n\n\nIn many situations the amount of error in such calculations is very small and can be overlooked or eliminated with rounding.\n\nThere are pathological problems however, where seemingly simple, straight-forward calculations are extremely sensitive to even tiny amounts of imprecision.\n\nThis task's purpose is to show how your language deals with such classes of problems.\n\n\n'''A sequence that seems to converge to a wrong limit.''' \n\nConsider the sequence:\n::::::   v1 =  2                                                                     \n::::::   v2 = -4                                                                     \n::::::   vn = 111   -   1130   /   vn-1   +   3000  /   (vn-1 * vn-2)    \n\n\nAs   '''n'''   grows larger, the series should converge to   '''6'''   but small amounts of error will cause it to approach   '''100'''.\n\n\n;Task 1:\nDisplay the values of the sequence where   n =   3, 4, 5, 6, 7, 8, 20, 30, 50 & 100   to at least '''16''' decimal places.\n\n    n = 3     18.5\n    n = 4      9.378378\n    n = 5      7.801153\n    n = 6      7.154414\n    n = 7      6.806785\n    n = 8      6.5926328\n    n = 20     6.0435521101892689\n    n = 30     6.006786093031205758530554\n    n = 50     6.0001758466271871889456140207471954695237\n    n = 100    6.000000019319477929104086803403585715024350675436952458072592750856521767230266\n\n\n\n;Task 2:\n'''The Chaotic Bank Society'''   is offering a new investment account to their customers. \n\nYou first deposit   $e - 1   where   e   is   2.7182818...   the base of natural logarithms.\n\nAfter each year, your account balance will be multiplied by the number of years that have passed, and $1 in service charges will be removed. \n\nSo ...\n::* after 1 year, your balance will be multiplied by 1 and $1 will be removed for service charges.\n::* after 2 years your balance will be doubled and $1 removed.\n::* after 3 years your balance will be tripled and $1 removed.\n::*  ... \n::* after 10 years, multiplied by 10 and $1 removed,  and so on. \n\n\nWhat will your balance be after   25   years?\n    Starting balance: $e-1\n    Balance = (Balance * year) - 1 for 25 years\n    Balance after 25 years: $0.0399387296732302\n\n\n;Task 3, extra credit:\n'''Siegfried Rump's example.'''   Consider the following function, designed by Siegfried Rump in 1988.\n::::::   f(a,b) = 333.75b6 + a2( 11a2b2 - b6 - 121b4 - 2 ) + 5.5b8 + a/(2b)          \n::::::  compute    f(a,b)    where    a=77617.0    and    b=33096.0  \n::::::   f(77617.0, 33096.0)   =   -0.827396059946821          \n\n\nDemonstrate how to solve at least one of the first two problems, or both, and the third if you're feeling particularly jaunty.\n\n\n;See also;\n*   Floating-Point Arithmetic   Section 1.3.2 Difficult problems.\n\n", "solution": "from fractions import Fraction\n\ndef muller_seq(n:int) -> float:\n    seq = [Fraction(0), Fraction(2), Fraction(-4)]\n    for i in range(3, n+1):\n        next_value = (111 - 1130/seq[i-1]\n            + 3000/(seq[i-1]*seq[i-2]))\n        seq.append(next_value)\n    return float(seq[n])\n\nfor n in [3, 4, 5, 6, 7, 8, 20, 30, 50, 100]:\n    print(\"{:4d} -> {}\".format(n, muller_seq(n)))"}
{"title": "Peaceful chess queen armies", "language": "Python", "task": "In chess, a queen attacks positions from where it is, in straight lines up-down and left-right as well as on both its diagonals. It attacks only pieces ''not'' of its own colour.\n\n\n\n\n\\\n|\n/\n\n\n\n=\n=\n\n=\n=\n\n\n\n/\n|\n\\\n\n\n\n/\n\n|\n\n\\\n\n\n\n\n|\n\n\n\n\n\n\nThe goal of Peaceful chess queen armies is to arrange m black queens and m white queens on an n-by-n square grid, (the board), so that ''no queen attacks another of a different colour''.\n\n\n;Task:\n# Create a routine to represent two-colour queens on a 2-D board. (Alternating black/white background colours, Unicode chess pieces and other embellishments are not necessary, but may be used at your discretion).\n# Create a routine to generate at least one solution to placing m equal numbers of black and white queens on an n square board.\n# Display here results for the m=4, n=5 case.\n\n\n;References:\n* Peaceably Coexisting Armies of Queens (Pdf) by Robert A. Bosch. Optima, the Mathematical Programming Socity newsletter, issue 62.\n* A250000 OEIS\n\n", "solution": "from itertools import combinations, product, count\nfrom functools import lru_cache, reduce\n\n\n_bbullet, _wbullet = '\\u2022\\u25E6'\n_or = set.__or__\n\ndef place(m, n):\n    \"Place m black and white queens, peacefully, on an n-by-n board\"\n    board = set(product(range(n), repeat=2))  # (x, y) tuples\n    placements = {frozenset(c) for c in combinations(board, m)}\n    for blacks in placements:\n        black_attacks = reduce(_or, \n                               (queen_attacks_from(pos, n) for pos in blacks), \n                               set())\n        for whites in {frozenset(c)     # Never on blsck attacking squares\n                       for c in combinations(board - black_attacks, m)}:\n            if not black_attacks & whites:\n                return blacks, whites\n    return set(), set()\n\n@lru_cache(maxsize=None)\ndef queen_attacks_from(pos, n):\n    x0, y0 = pos\n    a = set([pos])    # Its position\n    a.update((x, y0) for x in range(n))    # Its row\n    a.update((x0, y) for y in range(n))    # Its column\n    # Diagonals\n    for x1 in range(n):\n        # l-to-r diag\n        y1 = y0 -x0 +x1\n        if 0 <= y1 < n: \n            a.add((x1, y1))\n        # r-to-l diag\n        y1 = y0 +x0 -x1\n        if 0 <= y1 < n: \n            a.add((x1, y1))\n    return a\n\ndef pboard(black_white, n):\n    \"Print board\"\n    if black_white is None: \n        blk, wht = set(), set()\n    else:\n        blk, wht = black_white\n    print(f\"## {len(blk)} black and {len(wht)} white queens \"\n          f\"on a {n}-by-{n} board:\", end='')\n    for x, y in product(range(n), repeat=2):\n        if y == 0:\n            print()\n        xy = (x, y)\n        ch = ('?' if xy in blk and xy in wht \n              else 'B' if xy in blk\n              else 'W' if xy in wht\n              else _bbullet if (x + y)%2 else _wbullet)\n        print('%s' % ch, end='')\n    print()\n\nif __name__ == '__main__':\n    n=2\n    for n in range(2, 7):\n        print()\n        for m in count(1):\n            ans = place(m, n)\n            if ans[0]:\n                pboard(ans, n)\n            else:\n                print (f\"# Can't place {m} queens on a {n}-by-{n} board\")\n                break\n    #\n    print('\\n')\n    m, n = 5, 7\n    ans = place(m, n)\n    pboard(ans, n)"}
{"title": "Pentagram", "language": "Python 3.4.1", "task": "A pentagram is a star polygon, consisting of a central pentagon of which each side forms the base of an isosceles triangle. The vertex of each triangle, a point of the star, is 36 degrees.\n\n\n;Task:\nDraw (or print) a regular pentagram, in any orientation. Use a different color (or token) for stroke and fill, and background. For the fill it should be assumed that all points inside the triangles and the pentagon are inside the pentagram.\n\n\n;See also\n* Angle sum of a pentagram\n\n", "solution": "import turtle\n\nturtle.bgcolor(\"green\")\nt = turtle.Turtle()\nt.color(\"red\", \"blue\")\nt.begin_fill()\nfor i in range(0, 5):\n    t.forward(200)\n    t.right(144)\nt.end_fill()"}
{"title": "Pentomino tiling", "language": "Python", "task": "A pentomino is a polyomino that consists of 5 squares. There are 12 pentomino shapes, \nif you don't count rotations and reflections. Most pentominoes can form their own mirror image through \nrotation, but some of them have to be flipped over.\n\n        I                                                                        \n        I     L       N                                                 Y        \n FF     I     L      NN     PP     TTT              V       W     X    YY      ZZ\nFF      I     L      N      PP      T     U U       V      WW    XXX    Y      Z \n F      I     LL     N      P       T     UUU     VVV     WW      X     Y     ZZ \n\n\nA Pentomino tiling is an example of an exact cover problem and can take on many forms. \nA traditional tiling presents an 8 by 8 grid, where 4 cells are left uncovered. The other cells are covered \nby the 12 pentomino shapes, without overlaps, with every shape only used once.\n\nThe 4 uncovered cells should be chosen at random. Note that not all configurations are solvable.\n\n\n;Task\nCreate an 8 by 8 tiling and print the result.\n\n\n;Example\n\nF I I I I I L N\nF F F L L L L N\nW F - X Z Z N N\nW W X X X Z N V\nT W W X - Z Z V\nT T T P P V V V\nT Y - P P U U U\nY Y Y Y P U - U\n\n\n;Related tasks\n* Free polyominoes enumeration\n\n", "solution": "from itertools import product\n\nminos = (((197123, 7, 6), (1797, 6, 7), (1287, 6, 7), (196867, 7, 6)),\n        ((263937, 6, 6), (197126, 6, 6), (393731, 6, 6), (67332, 6, 6)),\n        ((16843011, 7, 5), (2063, 5, 7), (3841, 5, 7), (271, 5, 7), (3848, 5, 7), (50463234, 7, 5), (50397441, 7, 5), (33686019, 7, 5)),\n        ((131843, 7, 6), (1798, 6, 7), (775, 6, 7), (1795, 6, 7), (1543, 6, 7), (197377, 7, 6), (197378, 7, 6), (66307, 7, 6)),\n        ((132865, 6, 6), (131846, 6, 6), (198146, 6, 6), (132611, 6, 6), (393986, 6, 6), (263938, 6, 6), (67330, 6, 6), (132868, 6, 6)),\n        ((1039, 5, 7), (33751554, 7, 5), (16843521, 7, 5), (16974081, 7, 5), (33686274, 7, 5), (3842, 5, 7), (3844, 5, 7), (527, 5, 7)),\n        ((1804, 5, 7), (33751297, 7, 5), (33686273, 7, 5), (16974338, 7, 5), (16843522, 7, 5), (782, 5, 7), (3079, 5, 7), (3587, 5, 7)),\n        ((263683, 6, 6), (198148, 6, 6), (66310, 6, 6), (393985, 6, 6)),\n        ((67329, 6, 6), (131591, 6, 6), (459266, 6, 6), (263940, 6, 6)),\n        ((459780, 6, 6), (459009, 6, 6), (263175, 6, 6), (65799, 6, 6)),\n        ((4311810305, 8, 4), (31, 4, 8)),\n        ((132866, 6, 6),))\n\nboxchar_double_width = ' \u2576\u257a\u2575\u2514\u2515\u2579\u2516\u2517\u2574\u2500\u257c\u2518\u2534\u2536\u251a\u2538\u253a\u2578\u257e\u2501\u2519\u2535\u2537\u251b\u2539\u253b\u2577\u250c\u250d\u2502\u251c\u251d\u257f\u251e\u2521\u2510\u252c\u252e\u2524\u253c\u253e\u2526\u2540\u2544\u2511\u252d\u252f\u2525\u253d\u253f\u2529\u2543\u2547\u257b\u250e\u250f\u257d\u251f\u2522\u2503\u2520\u2523\u2512\u2530\u2532\u2527\u2541\u2546\u2528\u2542\u254a\u2513\u2531\u2533\u252a\u2545\u2548\u252b\u2549\u254b'\nboxchar_single_width = [c + ' \u2500\u2501'[i%3] for i, c in enumerate(boxchar_double_width)]\n\n# choose drawing alphabet based on terminal font\npatterns = boxchar_single_width\n\ntiles = []\nfor row in reversed(minos):\n    tiles.append([])\n    for n, x, y in row:\n        for shift in (b*8 + a for a, b in product(range(x), range(y))):\n            tiles[-1].append(n << shift)\n\ndef img(seq):\n    b = [[0]*10 for _ in range(10)]\n\n    for i, s in enumerate(seq):\n        for j, k in product(range(8), range(8)):\n            if s & (1<<(j*8 + k)):\n                b[j + 1][k + 1] = i + 1\n\n    idices = [[0]*9 for _ in range(9)]\n    for i, j in product(range(9), range(9)):\n        n = (b[i+1][j+1], b[i][j+1], b[i][j], b[i+1][j], b[i+1][j+1])\n        idices[i][j] = sum((a != b)*(1 + (not a or not b))*3**i for i, (a,b) in enumerate(zip(n, n[1:])))\n\n    return '\\n'.join(''.join(patterns[i] for i in row) for row in idices)\n\ndef tile(board=0, seq=tuple(), tiles=tiles):\n    if not tiles:\n        yield img(seq)\n        return\n\n    for c in tiles[0]:\n        b = board | c\n\n        tnext = [] # not using list comprehension ...\n        for t in tiles[1:]:\n            tnext.append(tuple(n for n in t if not n&b))\n            if not tnext[-1]: break # because early break is faster\n        else:\n            yield from tile(b, seq + (c,), tnext)\n\nfor x in tile():\n    print(x)"}
{"title": "Perfect shuffle", "language": "Python", "task": "A perfect shuffle (or faro/weave shuffle) means splitting a deck of cards into equal halves, and perfectly interleaving them - so that you end up with the first card from the left half, followed by the first card from the right half, and so on:\n\n\n\n::: 7 8 9 J Q K-7  8  9  J  Q  K-7 J 8 Q 9 K\n\n\n\nWhen you repeatedly perform perfect shuffles on an even-sized deck of unique cards, it will at some point arrive back at its original order. How many shuffles this takes, depends solely on the number of cards in the deck - for example for a deck of eight cards it takes three shuffles:\n\n\n\n::::: {| style=\"border-spacing:0.5em 0;border-collapse:separate;margin:0 1em;text-align:right\"\n|-\n| ''original:'' ||\n1\n2\n3\n4\n5\n6\n7\n8\n|-\n| ''after 1st shuffle:'' ||\n1\n5\n2\n6\n3\n7\n4\n8\n|-\n| ''after 2nd shuffle:'' ||\n1\n3\n5\n7\n2\n4\n6\n8\n|-\n| ''after 3rd shuffle:'' ||\n1\n2\n3\n4\n5\n6\n7\n8\n|}\n\n\n'''''The Task'''''\n\n# Write a function that can perform a perfect shuffle on an even-sized list of values.\n# Call this function repeatedly to count how many shuffles are needed to get a deck back to its original order, for each of the deck sizes listed under \"Test Cases\" below.\n#* You can use a list of numbers (or anything else that's convenient) to represent a deck; just make sure that all \"cards\" are unique within each deck.\n#* Print out the resulting shuffle counts, to demonstrate that your program passes the test-cases.\n'''''Test Cases'''''\n\n::::: {| class=\"wikitable\"\n|-\n! input ''(deck size)'' !! output ''(number of shuffles required)''\n|-\n| 8 || 3\n|-\n| 24 || 11\n|-\n| 52 || 8\n|-\n| 100 || 30\n|-\n| 1020 || 1018\n|-\n| 1024 || 10\n|-\n| 10000 || 300\n|}\n\n", "solution": "\"\"\"\nBrute force solution for the Perfect Shuffle problem.\nSee http://oeis.org/A002326 for possible improvements\n\"\"\"\nfrom functools import partial\nfrom itertools import chain\nfrom operator import eq\nfrom typing import (Callable,\n                    Iterable,\n                    Iterator,\n                    List,\n                    TypeVar)\n\nT = TypeVar('T')\n\n\ndef main():\n    print(\"Deck length | Shuffles \")\n    for length in (8, 24, 52, 100, 1020, 1024, 10000):\n        deck = list(range(length))\n        shuffles_needed = spin_number(deck, shuffle)\n        print(f\"{length:<11} | {shuffles_needed}\")\n\n\ndef shuffle(deck: List[T]) -> List[T]:\n    \"\"\"[1, 2, 3, 4] -> [1, 3, 2, 4]\"\"\"\n    half = len(deck) // 2\n    return list(chain.from_iterable(zip(deck[:half], deck[half:])))\n\n\ndef spin_number(source: T,\n                function: Callable[[T], T]) -> int:\n    \"\"\"\n    Applies given function to the source\n    until the result becomes equal to it,\n    returns the number of calls \n    \"\"\"\n    is_equal_source = partial(eq, source)\n    spins = repeat_call(function, source)\n    return next_index(is_equal_source,\n                      spins,\n                      start=1)\n\n\ndef repeat_call(function: Callable[[T], T],\n                value: T) -> Iterator[T]:\n    \"\"\"(f, x) -> f(x), f(f(x)), f(f(f(x))), ...\"\"\"\n    while True:\n        value = function(value)\n        yield value\n\n\ndef next_index(predicate: Callable[[T], bool],\n               iterable: Iterable[T],\n               start: int = 0) -> int:\n    \"\"\"\n    Returns index of the first element of the iterable\n    satisfying given condition\n    \"\"\"\n    for index, item in enumerate(iterable, start=start):\n        if predicate(item):\n            return index\n\n\nif __name__ == \"__main__\":\n    main()\n"}
{"title": "Perfect totient numbers", "language": "Python", "task": "Generate and show here, the first twenty Perfect totient numbers.\n\n\n;Related task:\n::*   [[Totient function]]\n\n\n;Also see:\n::*   the OEIS      entry for   perfect totient numbers.\n::*   mrob            list of the first 54\n\n", "solution": "from math import gcd\nfrom functools import lru_cache\nfrom itertools import islice, count\n\n@lru_cache(maxsize=None)\ndef  \u03c6(n):\n    return sum(1 for k in range(1, n + 1) if gcd(n, k) == 1)\n\ndef perfect_totient():\n    for n0 in count(1):\n        parts, n = 0, n0\n        while n != 1:\n            n = \u03c6(n)\n            parts += n\n        if parts == n0:\n            yield n0\n        \n\nif __name__ == '__main__':\n    print(list(islice(perfect_totient(), 20)))"}
{"title": "Periodic table", "language": "Python", "task": "Display the row and column in the periodic table of the given atomic number.\n\n;The periodic table:\nLet us consider the following periodic table representation.\n\n\n     __________________________________________________________________________ \n    |   1   2   3   4   5   6   7   8   9   10  11  12  13  14  15  16  17  18 |\n    |                                                                          |\n    |1  H                                                                   He |\n    |                                                                          |\n    |2  Li  Be                                          B   C   N   O   F   Ne |\n    |                                                                          |\n    |3  Na  Mg                                          Al  Si  P   S   Cl  Ar |\n    |                                                                          |\n    |4  K   Ca  Sc  Ti  V   Cr  Mn  Fe  Co  Ni  Cu  Zn  Ga  Ge  As  Se  Br  Kr |\n    |                                                                          |\n    |5  Rb  Sr  Y   Zr  Nb  Mo  Tc  Ru  Rh  Pd  Ag  Cd  In  Sn  Sb  Te  I   Xe |\n    |                                                                          |\n    |6  Cs  Ba  *   Hf  Ta  W   Re  Os  Ir  Pt  Au  Hg  Tl  Pb  Bi  Po  At  Rn |\n    |                                                                          |\n    |7  Fr  Ra  deg   Rf  Db  Sg  Bh  Hs  Mt  Ds  Rg  Cn  Nh  Fl  Mc  Lv  Ts  Og |\n    |__________________________________________________________________________|\n    |                                                                          |\n    |                                                                          |\n    |8  Lantanoidi* La  Ce  Pr  Nd  Pm  Sm  Eu  Gd  Tb  Dy  Ho  Er  Tm  Yb  Lu |\n    |                                                                          |\n    |9   Aktinoidideg Ak  Th  Pa  U   Np  Pu  Am  Cm  Bk  Cf  Es  Fm  Md  No  Lr |\n    |__________________________________________________________________________|\n\n\n;Example test cases;\n\n*   1 -> 1 1\n*   2 -> 1 18\n*   29 -> 4 11\n*   42 -> 5 6\n*   57 -> 8 4\n*   58 -> 8 5\n*   72 -> 6 4\n*   89 -> 9 4\n\n;Details;\n\nThe representation of the periodic table may be represented in various way. The one presented in this challenge does have the following property : Lantanides and Aktinoides are all in a dedicated row, hence there is no element that is placed at 6, 3 nor 7, 3.\n\nYou may take a look at the atomic number repartitions here.\n\nThe atomic number is at least 1, at most 118.\n\n\n;See also: \n*   the periodic table\n*   This task was an idea from CompSciFact\n*   The periodic table in ascii that was used as template\n\n", "solution": "def perta(atomic) -> (int, int):\n\n    NOBLES = 2, 10, 18, 36, 54, 86, 118\n    INTERTWINED = 0, 0, 0, 0, 0, 57, 89\n    INTERTWINING_SIZE = 14\n    LINE_WIDTH = 18\n\n    prev_noble = 0\n    for row, noble in enumerate(NOBLES):\n        if atomic <= noble:  # we are at the good row. We now need to determine the column\n            nb_elem = noble - prev_noble  # number of elements on that row\n            rank =  atomic - prev_noble  # rank of the input element among elements\n            if INTERTWINED[row] and INTERTWINED[row] <= atomic <= INTERTWINED[row] + INTERTWINING_SIZE:  # lantanides or actinides\n                row += 2\n                col = rank + 1\n            else:  # not a lantanide nor actinide\n                # handle empty spaces between 1-2, 4-5 and 12-13.\n                nb_empty = LINE_WIDTH - nb_elem  # spaces count as columns\n                inside_left_element_rank = 2 if noble > 2 else 1\n                col = rank + (nb_empty if rank > inside_left_element_rank else 0)\n            break\n        prev_noble = noble\n    return row+1, col\n\n\n\n# small test suite\n\nTESTS = {\n    1: (1, 1),\n    2: (1, 18),\n    29: (4,11),\n    42: (5, 6),\n    58: (8, 5),\n    59: (8, 6),\n    57: (8, 4),\n    71: (8, 18),\n    72: (6, 4),\n    89: (9, 4),\n    90: (9, 5),\n    103: (9, 18),\n}\n\nfor input, out in TESTS.items():\n    found = perta(input)\n    print('TEST:{:3d} -> '.format(input) + str(found) + (f' ; ERROR: expected {out}' if found != out else ''))\n\n"}
{"title": "Perlin noise", "language": "Python from Java", "task": "The   '''computer graphics,   most notably to procedurally generate textures or heightmaps. \n\nThe Perlin noise is basically a   pseudo-random   mapping of   \\R^d   into   \\R   with an integer   d   which can be arbitrarily large but which is usually   2,   3,   or   4.\n\nEither by using a dedicated library or by implementing the algorithm, show that the Perlin noise   (as defined in 2002 in the Java implementation below)   of the point in 3D-space with coordinates     3.14,   42,   7     is     0.13691995878400012.\n\n\n''Note: this result assumes 64 bit IEEE-754 floating point calculations. If your language uses a different floating point representation, make a note of it and calculate the value accurate to 15 decimal places, or your languages accuracy threshold if it is less. Trailing zeros need not be displayed.''\n\n", "solution": "import math\n\ndef perlin_noise(x, y, z):\n    X = math.floor(x) & 255                  # FIND UNIT CUBE THAT\n    Y = math.floor(y) & 255                  # CONTAINS POINT.\n    Z = math.floor(z) & 255\n    x -= math.floor(x)                                # FIND RELATIVE X,Y,Z\n    y -= math.floor(y)                                # OF POINT IN CUBE.\n    z -= math.floor(z)\n    u = fade(x)                                # COMPUTE FADE CURVES\n    v = fade(y)                                # FOR EACH OF X,Y,Z.\n    w = fade(z)\n    A = p[X  ]+Y; AA = p[A]+Z; AB = p[A+1]+Z      # HASH COORDINATES OF\n    B = p[X+1]+Y; BA = p[B]+Z; BB = p[B+1]+Z      # THE 8 CUBE CORNERS,\n \n    return lerp(w, lerp(v, lerp(u, grad(p[AA  ], x  , y  , z   ),  # AND ADD\n                                   grad(p[BA  ], x-1, y  , z   )), # BLENDED\n                           lerp(u, grad(p[AB  ], x  , y-1, z   ),  # RESULTS\n                                   grad(p[BB  ], x-1, y-1, z   ))),# FROM  8\n                   lerp(v, lerp(u, grad(p[AA+1], x  , y  , z-1 ),  # CORNERS\n                                   grad(p[BA+1], x-1, y  , z-1 )), # OF CUBE\n                           lerp(u, grad(p[AB+1], x  , y-1, z-1 ),\n                                   grad(p[BB+1], x-1, y-1, z-1 ))))\n                                   \ndef fade(t): \n    return t ** 3 * (t * (t * 6 - 15) + 10)\n    \ndef lerp(t, a, b):\n    return a + t * (b - a)\n    \ndef grad(hash, x, y, z):\n    h = hash & 15                      # CONVERT LO 4 BITS OF HASH CODE\n    u = x if h<8 else y                # INTO 12 GRADIENT DIRECTIONS.\n    v = y if h<4 else (x if h in (12, 14) else z)\n    return (u if (h&1) == 0 else -u) + (v if (h&2) == 0 else -v)\n\np = [None] * 512\npermutation = [151,160,137,91,90,15,\n   131,13,201,95,96,53,194,233,7,225,140,36,103,30,69,142,8,99,37,240,21,10,23,\n   190, 6,148,247,120,234,75,0,26,197,62,94,252,219,203,117,35,11,32,57,177,33,\n   88,237,149,56,87,174,20,125,136,171,168, 68,175,74,165,71,134,139,48,27,166,\n   77,146,158,231,83,111,229,122,60,211,133,230,220,105,92,41,55,46,245,40,244,\n   102,143,54, 65,25,63,161, 1,216,80,73,209,76,132,187,208, 89,18,169,200,196,\n   135,130,116,188,159,86,164,100,109,198,173,186, 3,64,52,217,226,250,124,123,\n   5,202,38,147,118,126,255,82,85,212,207,206,59,227,47,16,58,17,182,189,28,42,\n   223,183,170,213,119,248,152, 2,44,154,163, 70,221,153,101,155,167, 43,172,9,\n   129,22,39,253, 19,98,108,110,79,113,224,232,178,185, 112,104,218,246,97,228,\n   251,34,242,193,238,210,144,12,191,179,162,241, 81,51,145,235,249,14,239,107,\n   49,192,214, 31,181,199,106,157,184, 84,204,176,115,121,50,45,127, 4,150,254,\n   138,236,205,93,222,114,67,29,24,72,243,141,128,195,78,66,215,61,156,180]\nfor i in range(256):\n    p[256+i] = p[i] = permutation[i]\n\nif __name__ == '__main__':\n    print(\"%1.17f\" % perlin_noise(3.14, 42, 7))"}
{"title": "Permutations/Derangements", "language": "Python", "task": "A derangement is a permutation of the order of distinct items in which ''no item appears in its original place''.\n\nFor example, the only two derangements of the three items (0, 1, 2) are (1, 2, 0), and (2, 0, 1).\n\nThe number of derangements of ''n'' distinct items is known as the subfactorial of ''n'', sometimes written as !''n''. \nThere are various ways to calculate !''n''.\n\n\n;Task:\n# Create a named function/method/subroutine/... to generate derangements of the integers ''0..n-1'', (or ''1..n'' if you prefer). \n# Generate ''and show'' all the derangements of 4 integers using the above routine.\n# Create a function that calculates the subfactorial of ''n'', !''n''.\n# Print and show a table of the ''counted'' number of derangements of ''n'' vs. the calculated !''n'' for n from 0..9 inclusive.\n\n\n;Optional stretch goal:\n*   Calculate    !''20'' \n\n\n;Related tasks:\n*   [[Anagrams/Deranged anagrams]]\n*   [[Best shuffle]]\n*   [[Left_factorials]]\n\n\n\n", "solution": "from itertools import permutations\nimport math\n\n\ndef derangements(n):\n    'All deranged permutations of the integers 0..n-1 inclusive'\n    return ( perm for perm in permutations(range(n))\n             if all(indx != p for indx, p in enumerate(perm)) )\n\ndef subfact(n):\n    if n == 2 or n == 0:\n        return 1\n    elif n == 1:\n        return 0\n    elif  1 <= n <=18:\n        return round(math.factorial(n) / math.e)\n    elif n.imag == 0 and n.real == int(n.real) and n > 0:\n        return (n-1) * ( subfact(n - 1) + subfact(n - 2) )\n    else:\n        raise ValueError()\n\ndef _iterlen(iter):\n    'length of an iterator without taking much memory'\n    l = 0\n    for x in iter:\n        l += 1\n    return l\n\nif __name__ == '__main__':\n    n = 4\n    print(\"Derangements of %s\" % (tuple(range(n)),))\n    for d in derangements(n):\n        print(\"  %s\" % (d,))\n\n    print(\"\\nTable of n vs counted vs calculated derangements\")\n    for n in range(10):\n        print(\"%2i %-5i %-5i\" %\n              (n, _iterlen(derangements(n)), subfact(n)))\n\n    n = 20\n    print(\"\\n!%i = %i\" % (n, subfact(n)))"}
{"title": "Permutations/Rank of a permutation", "language": "Python", "task": "A particular ranking of a permutation associates an integer with a particular ordering of all the permutations of a set of distinct items. \nFor our purposes the ranking will assign integers 0 .. (n! - 1) to an ordering of all the permutations of the integers 0 .. (n - 1).\n\nFor example, the permutations of the digits zero to 3 arranged lexicographically have the following rank:\n\n  PERMUTATION      RANK\n  (0, 1, 2, 3) ->  0\n  (0, 1, 3, 2) ->  1\n  (0, 2, 1, 3) ->  2\n  (0, 2, 3, 1) ->  3\n  (0, 3, 1, 2) ->  4\n  (0, 3, 2, 1) ->  5\n  (1, 0, 2, 3) ->  6\n  (1, 0, 3, 2) ->  7\n  (1, 2, 0, 3) ->  8\n  (1, 2, 3, 0) ->  9\n  (1, 3, 0, 2) -> 10\n  (1, 3, 2, 0) -> 11\n  (2, 0, 1, 3) -> 12\n  (2, 0, 3, 1) -> 13\n  (2, 1, 0, 3) -> 14\n  (2, 1, 3, 0) -> 15\n  (2, 3, 0, 1) -> 16\n  (2, 3, 1, 0) -> 17\n  (3, 0, 1, 2) -> 18\n  (3, 0, 2, 1) -> 19\n  (3, 1, 0, 2) -> 20\n  (3, 1, 2, 0) -> 21\n  (3, 2, 0, 1) -> 22\n  (3, 2, 1, 0) -> 23\n\nAlgorithms exist that can generate a rank from a permutation for some particular ordering of permutations, and that can generate the same rank from the given individual permutation (i.e. given a rank of 17 produce (2, 3, 1, 0) in the example above).\n\nOne use of such algorithms could be in generating a small, random, sample of permutations of n items without duplicates when the total number of permutations is large. Remember that the total number of permutations of n items is given by n! which grows large very quickly: A 32 bit integer can only hold 12!, a 64 bit integer only 20!. It becomes difficult to take the straight-forward approach of generating all permutations then taking a random sample of them.\n\nA question on the Stack Overflow site asked how to generate one million random and indivudual permutations of 144 items.\n\n\n;Task:\n# Create a function to generate a permutation from a rank.\n# Create the inverse function that given the permutation generates its rank.\n# Show that for n=3 the two functions are indeed inverses of each other.\n# Compute and show here 4 random, individual, samples of permutations of 12 objects.\n\n\n;Stretch goal:\n* State how reasonable it would be to use your program to address the limits of the Stack Overflow question.\n\n\n;References:\n# Ranking and Unranking Permutations in Linear Time by Myrvold & Ruskey. (Also available via Google here).\n# Ranks on the DevData site.\n# Another answer on Stack Overflow to a different question that explains its algorithm in detail.\n\n;Related tasks:\n#[[Factorial_base_numbers_indexing_permutations_of_a_collection]]\n\n", "solution": "from math import factorial as fact\nfrom random import randrange\nfrom textwrap import wrap\n\ndef identity_perm(n): \n    return list(range(n))\n\ndef unranker1(n, r, pi):\n    while n > 0:\n        n1, (rdivn, rmodn) = n-1, divmod(r, n)\n        pi[n1], pi[rmodn] = pi[rmodn], pi[n1]\n        n = n1\n        r = rdivn\n    return pi\n\ndef init_pi1(n, pi): \n    pi1 = [-1] * n\n    for i in range(n): \n        pi1[pi[i]] = i\n    return pi1\n\ndef ranker1(n, pi, pi1):\n    if n == 1: \n        return 0\n    n1 = n-1\n    s = pi[n1]\n    pi[n1], pi[pi1[n1]] = pi[pi1[n1]], pi[n1]\n    pi1[s], pi1[n1] = pi1[n1], pi1[s]\n    return s + n * ranker1(n1, pi, pi1)\n\ndef unranker2(n, r, pi):\n    while n > 0:\n        n1 = n-1\n        s, rmodf = divmod(r, fact(n1))\n        pi[n1], pi[s] = pi[s], pi[n1]\n        n = n1\n        r = rmodf\n    return pi\n\ndef ranker2(n, pi, pi1):\n    if n == 1: \n        return 0\n    n1 = n-1\n    s = pi[n1]\n    pi[n1], pi[pi1[n1]] = pi[pi1[n1]], pi[n1]\n    pi1[s], pi1[n1] = pi1[n1], pi1[s]\n    return s * fact(n1) + ranker2(n1, pi, pi1)\n\ndef get_random_ranks(permsize, samplesize):    \n    perms = fact(permsize)\n    ranks = set()\n    while len(ranks) < samplesize:\n        ranks |= set( randrange(perms) \n                      for r in range(samplesize - len(ranks)) )\n    return ranks    \n\ndef test1(comment, unranker, ranker):    \n    n, samplesize, n2 = 3, 4, 12\n    print(comment)\n    perms = []\n    for r in range(fact(n)):\n        pi = identity_perm(n)\n        perm = unranker(n, r, pi)\n        perms.append((r, perm))\n    for r, pi in perms:\n        pi1 = init_pi1(n, pi)\n        print('  From rank %2i to %r back to %2i' % (r, pi, ranker(n, pi[:], pi1)))\n    print('\\n  %i random individual samples of %i items:' % (samplesize, n2))\n    for r in get_random_ranks(n2, samplesize):\n        pi = identity_perm(n2)\n        print('    ' + ' '.join('%2i' % i for i in unranker(n2, r, pi)))\n    print('')\n\ndef test2(comment, unranker):    \n    samplesize, n2 = 4, 144\n    print(comment)\n    print('  %i random individual samples of %i items:' % (samplesize, n2))\n    for r in get_random_ranks(n2, samplesize):\n        pi = identity_perm(n2)\n        print('    ' + '\\n      '.join(wrap(repr(unranker(n2, r, pi)))))\n    print('')\n\nif __name__ == '__main__':\n    test1('First ordering:', unranker1, ranker1)\n    test1('Second ordering:', unranker2, ranker2)\n    test2('First ordering, large number of perms:', unranker1)"}
{"title": "Permutations by swapping", "language": "Python", "task": "Generate permutations of n items in which successive permutations differ from each other by the swapping of any two items. \n\nAlso generate the sign of the permutation which is +1 when the permutation is generated from an even number of swaps from the initial state, and -1 for odd. \n\nShow the permutations and signs of three items, in order of generation ''here''.\n\nSuch data are of use in generating the determinant of a square matrix and any functions created should bear this in mind.\n\nNote: The Steinhaus-Johnson-Trotter algorithm generates successive permutations where ''adjacent'' items are swapped, but from this discussion adjacency is not a requirement.\n\n\n;References:\n* Steinhaus-Johnson-Trotter algorithm\n* Johnson-Trotter Algorithm Listing All Permutations\n* Heap's algorithm\n* Tintinnalogia\n\n\n;Related tasks:\n*   [[Matrix arithmetic]\n*   [[Gray code]]\n\n", "solution": "from operator import itemgetter\n \nDEBUG = False # like the built-in __debug__\n \ndef spermutations(n):\n    \"\"\"permutations by swapping. Yields: perm, sign\"\"\"\n    sign = 1\n    p = [[i, 0 if i == 0 else -1] # [num, direction]\n         for i in range(n)]\n \n    if DEBUG: print ' #', p\n    yield tuple(pp[0] for pp in p), sign\n \n    while any(pp[1] for pp in p): # moving\n        i1, (n1, d1) = max(((i, pp) for i, pp in enumerate(p) if pp[1]),\n                           key=itemgetter(1))\n        sign *= -1\n        if d1 == -1:\n            # Swap down\n            i2 = i1 - 1\n            p[i1], p[i2] = p[i2], p[i1]\n            # If this causes the chosen element to reach the First or last\n            # position within the permutation, or if the next element in the\n            # same direction is larger than the chosen element:\n            if i2 == 0 or p[i2 - 1][0] > n1:\n                # The direction of the chosen element is set to zero\n                p[i2][1] = 0\n        elif d1 == 1:\n            # Swap up\n            i2 = i1 + 1\n            p[i1], p[i2] = p[i2], p[i1]\n            # If this causes the chosen element to reach the first or Last\n            # position within the permutation, or if the next element in the\n            # same direction is larger than the chosen element:\n            if i2 == n - 1 or p[i2 + 1][0] > n1:\n                # The direction of the chosen element is set to zero\n                p[i2][1] = 0\n        if DEBUG: print ' #', p\n        yield tuple(pp[0] for pp in p), sign\n \n        for i3, pp in enumerate(p):\n            n3, d3 = pp\n            if n3 > n1:\n                pp[1] = 1 if i3 < i2 else -1\n                if DEBUG: print ' # Set Moving'\n \n \nif __name__ == '__main__':\n    from itertools import permutations\n \n    for n in (3, 4):\n        print '\\nPermutations and sign of %i items' % n\n        sp = set()\n        for i in spermutations(n):\n            sp.add(i[0])\n            print('Perm: %r Sign: %2i' % i)\n            #if DEBUG: raw_input('?')\n        # Test\n        p = set(permutations(range(n)))\n        assert sp == p, 'Two methods of generating permutations do not agree'"}
{"title": "Phrase reversals", "language": "Python", "task": "Given a string of space separated words containing the following phrase:\n  rosetta code phrase reversal\n\n:# Reverse the characters of the string.\n:# Reverse the characters of each individual word in the string, maintaining original word order within the string.\n:# Reverse the order of each word of the string, maintaining the order of characters in each word.\n\nShow your output here.\n \n\n\n", "solution": "'''String reversals at different levels.'''\n\n\n# reversedCharacters :: String -> String\ndef reversedCharacters(s):\n    '''All characters in reversed sequence.'''\n    return reverse(s)\n\n\n# wordsWithReversedCharacters :: String -> String\ndef wordsWithReversedCharacters(s):\n    '''Characters within each word in reversed sequence.'''\n    return unwords(map(reverse, words(s)))\n\n\n# reversedWordOrder :: String -> String\ndef reversedWordOrder(s):\n    '''Sequence of words reversed.'''\n    return unwords(reverse(words(s)))\n\n\n# TESTS -------------------------------------------------\n# main :: IO()\ndef main():\n    '''Tests'''\n\n    s = 'rosetta code phrase reversal'\n    print(\n        tabulated(s + ':\\n')(\n            lambda f: f.__name__\n        )(lambda s: \"'\" + s + \"'\")(\n            lambda f: f(s)\n        )([\n            reversedCharacters,\n            wordsWithReversedCharacters,\n            reversedWordOrder\n        ])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# reverse :: [a] -> [a]\n# reverse :: String -> String\ndef reverse(xs):\n    '''The elements of xs in reverse order.'''\n    return xs[::-1] if isinstance(xs, str) else (\n        list(reversed(xs))\n    )\n\n\n# tabulated :: String -> (a -> String) ->\n#                        (b -> String) ->\n#                        (a -> b) -> [a] -> String\ndef tabulated(s):\n    '''Heading -> x display function -> fx display function ->\n                f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + '\\n'.join(\n            xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs\n        )\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# unwords :: [String] -> String\ndef unwords(xs):\n    '''A space-separated string derived from a list of words.'''\n    return ' '.join(xs)\n\n\n# words :: String -> [String]\ndef words(s):\n    '''A list of words delimited by characters\n       representing white space.'''\n    return s.split()\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Pig the dice game", "language": "Python", "task": "The   game of Pig   is a multiplayer game played with a single six-sided die.   The\nobject of the game is to reach   '''100'''   points or more.   \nPlay is taken in turns.   On each person's turn that person has the option of either:\n\n:# '''Rolling the dice''':   where a roll of two to six is added to their score for that turn and the player's turn continues as the player is given the same choice again;   or a roll of   '''1'''   loses the player's total points   ''for that turn''   and their turn finishes with play passing to the next player.\n:# '''Holding''':   the player's score for that round is added to their total and becomes safe from the effects of throwing a   '''1'''   (one).   The player's turn finishes with play passing to the next player.\n\n\n;Task:\nCreate a program to score for, and simulate dice throws for, a two-person game. \n\n\n;Related task:\n*   [[Pig the dice game/Player]]\n\n", "solution": "#!/usr/bin/python3\n\n'''\nSee: http://en.wikipedia.org/wiki/Pig_(dice)\n\nThis program scores and throws the dice for a two player game of Pig\n\n'''\n\nfrom random import randint\n\nplayercount = 2\nmaxscore = 100\nsafescore = [0] * playercount\nplayer = 0\nscore=0\n\nwhile max(safescore) < maxscore:\n    rolling = input(\"Player %i: (%i, %i) Rolling? (Y) \"\n                    % (player, safescore[player], score)).strip().lower() in {'yes', 'y', ''}\n    if rolling:\n        rolled = randint(1, 6)\n        print('  Rolled %i' % rolled)\n        if rolled == 1:\n            print('  Bust! you lose %i but still keep your previous %i'\n                  % (score, safescore[player]))\n            score, player = 0, (player + 1) % playercount\n        else:\n            score += rolled\n    else:\n        safescore[player] += score\n        if safescore[player] >= maxscore:\n            break\n        print('  Sticking with %i' % safescore[player])\n        score, player = 0, (player + 1) % playercount\n        \nprint('\\nPlayer %i wins with a score of %i' %(player, safescore[player]))"}
{"title": "Pig the dice game/Player", "language": "Python", "task": "Create a dice simulator and scorer of [[Pig the dice game]] and add to it the ability to play the game to at least one strategy.\n* State here the play strategies involved.\n* Show play during a game here.\n\n\nAs a stretch goal:\n* Simulate playing the game a number of times with two players of given strategies and report here summary statistics such as, but not restricted to, the influence of going first or which strategy seems stronger.\n\n\n;Game Rules:\nThe game of Pig is a multiplayer game played with a single six-sided die. The\nobject of the game is to reach 100 points or more.\nPlay is taken in turns. On each person's turn that person has the option of either\n\n# '''Rolling the dice''': where a roll of two to six is added to their score for that turn and the player's turn continues as the player is given the same choice again; or a roll of 1 loses the player's total points ''for that turn'' and their turn finishes with play passing to the next player.\n# '''Holding''': The player's score for that round is added to their total and becomes safe from the effects of throwing a one. The player's turn finishes with play passing to the next player.\n\n\n;References\n* Pig (dice)\n* The Math of Being a Pig and Pigs (extra) - Numberphile videos featuring Ben Sparks.\n\n", "solution": "#!/usr/bin/python3\n\n'''\nSee: http://en.wikipedia.org/wiki/Pig_(dice)\n\nThis program scores, throws the dice, and plays for an N player game of Pig.\n\n'''\n\nfrom random import randint\nfrom collections import namedtuple\nimport random\nfrom pprint import pprint as pp\nfrom collections import Counter\n\n\nplayercount = 2\nmaxscore = 100\nmaxgames = 100000\n\n\nGame = namedtuple('Game', 'players, maxscore, rounds')\nRound = namedtuple('Round', 'who, start, scores, safe')\n\n\nclass Player():\n    def __init__(self, player_index):\n        self.player_index = player_index\n\n    def __repr__(self):\n        return '%s(%i)' % (self.__class__.__name__, self.player_index)\n\n    def __call__(self, safescore, scores, game):\n        'Returns boolean True to roll again'\n        pass\n\nclass RandPlay(Player):\n    def __call__(self, safe, scores, game):\n        'Returns random boolean choice of whether to roll again'\n        return bool(random.randint(0, 1))\n\nclass RollTo20(Player):\n    def __call__(self, safe, scores, game):\n        'Roll again if this rounds score < 20'\n        return (((sum(scores) + safe[self.player_index]) < maxscore)    # Haven't won yet\n                and(sum(scores) < 20))                                  # Not at 20 this round\n\nclass Desparat(Player):\n    def __call__(self, safe, scores, game):\n        'Roll again if this rounds score < 20 or someone is within 20 of winning'\n        return (((sum(scores) + safe[self.player_index]) < maxscore)    # Haven't won yet\n                and( (sum(scores) < 20)                                 # Not at 20 this round\n                     or max(safe) >= (maxscore - 20)))                  # Someone's close\n\n\ndef game__str__(self):\n    'Pretty printer for Game class'\n    return (\"Game(players=%r, maxscore=%i,\\n  rounds=[\\n    %s\\n  ])\"\n            % (self.players, self.maxscore,\n               ',\\n    '.join(repr(round) for round in self.rounds)))\nGame.__str__ = game__str__\n\n\ndef winningorder(players, safescores):\n    'Return (players in winning order, their scores)'\n    return tuple(zip(*sorted(zip(players, safescores),\n                            key=lambda x: x[1], reverse=True)))\n\ndef playpig(game):\n    '''\n    Plays the game of pig returning the players in winning order\n    and their scores whilst updating argument game with the details of play.\n    '''\n    players, maxscore, rounds = game\n    playercount = len(players)\n    safescore = [0] * playercount   # Safe scores for each player\n    player = 0                      # Who plays this round\n    scores=[]                       # Individual scores this round\n\n    while max(safescore) < maxscore:\n        startscore = safescore[player]\n        rolling = players[player](safescore, scores, game)\n        if rolling:\n            rolled = randint(1, 6)\n            scores.append(rolled)\n            if rolled == 1:\n                # Bust! \n                round = Round(who=players[player],\n                              start=startscore,\n                              scores=scores,\n                              safe=safescore[player])\n                rounds.append(round)\n                scores, player = [], (player + 1) % playercount\n        else:\n            # Stick\n            safescore[player] += sum(scores)\n            round = Round(who=players[player],\n                          start=startscore,\n                          scores=scores,\n                          safe=safescore[player])\n            rounds.append(round)\n            if safescore[player] >= maxscore:\n                break\n            scores, player = [], (player + 1) % playercount\n\n    # return players in winning order and all scores\n    return winningorder(players, safescore)\n\nif __name__ == '__main__':\n    game = Game(players=tuple(RandPlay(i) for i in range(playercount)),\n                maxscore=20,\n                rounds=[])\n    print('ONE GAME')\n    print('Winning order: %r; Respective scores: %r\\n' % playpig(game))\n    print(game)\n    game = Game(players=tuple(RandPlay(i) for i in range(playercount)),\n                maxscore=maxscore,\n                rounds=[])\n    algos = (RollTo20, RandPlay, Desparat)\n    print('\\n\\nMULTIPLE STATISTICS using %r\\n  for %i GAMES'\n          % (', '.join(p.__name__ for p in algos), maxgames,))\n    winners = Counter(repr(playpig(game._replace(players=tuple(random.choice(algos)(i)\n                                                               for i in range(playercount)),\n                                                 rounds=[]))[0])\n                      for i in range(maxgames))\n    print('  Players(position) winning on left; occurrences on right:\\n    %s'\n          % ',\\n    '.join(str(w) for w in winners.most_common()))"}
{"title": "Plasma effect", "language": "Python from Raku", "task": "The plasma effect is a visual effect created by applying various functions, notably sine and cosine, to the color values of screen pixels. When animated (not a task requirement) the effect may give the impression of a colorful flowing liquid.\n\n\n;Task\nCreate a plasma effect.\n\n\n;See also\n* Computer Graphics Tutorial (lodev.org)\n* Plasma (bidouille.org)\n\n", "solution": "import math\nimport colorsys\nfrom PIL import Image\n\ndef plasma (w, h):\n\tout = Image.new(\"RGB\", (w, h))\n\tpix = out.load()\n\tfor x in range (w):\n\t\tfor y in range(h):\n\t\t\thue = 4.0 + math.sin(x / 19.0) + math.sin(y / 9.0) \\\n\t\t\t\t+ math.sin((x + y) / 25.0) + math.sin(math.sqrt(x**2.0 + y**2.0) / 8.0)\n\t\t\thsv = colorsys.hsv_to_rgb(hue/8.0, 1, 1)\n\t\t\tpix[x, y] = tuple([int(round(c * 255.0)) for c in hsv])\n\treturn out\n\nif __name__==\"__main__\":\n\tim = plasma(400, 400)\n\tim.show()"}
{"title": "Playfair cipher", "language": "Python from Raku", "task": "Implement a  Playfair cipher for encryption and decryption.\n\n\nThe user must be able to choose   '''J'''  =  '''I'''      or   no   '''Q'''   in the alphabet.\n\nThe output of the encrypted and decrypted message must be in capitalized digraphs, separated by spaces.\n\n\n;Output example:\n                 HI DE TH EG OL DI NT HE TR EX ES TU MP\n\n", "solution": "from string import ascii_uppercase\nfrom itertools import product\nfrom re import findall\n\ndef uniq(seq):\n    seen = {}\n    return [seen.setdefault(x, x) for x in seq if x not in seen]\n\ndef partition(seq, n):\n    return [seq[i : i + n] for i in xrange(0, len(seq), n)]\n\n\n\"\"\"Instantiate a specific encoder/decoder.\"\"\"\ndef playfair(key, from_ = 'J', to = None):\n    if to is None:\n        to = 'I' if from_ == 'J' else ''\n\n    def canonicalize(s):\n        return filter(str.isupper, s.upper()).replace(from_, to)\n\n    # Build 5x5 matrix.\n    m = partition(uniq(canonicalize(key + ascii_uppercase)), 5)\n\n    # Pregenerate all forward translations.\n    enc = {}\n\n    # Map pairs in same row.\n    for row in m:\n        for i, j in product(xrange(5), repeat=2):\n            if i != j:\n                enc[row[i] + row[j]] = row[(i + 1) % 5] + row[(j + 1) % 5]\n\n    # Map pairs in same column.\n    for c in zip(*m):\n        for i, j in product(xrange(5), repeat=2):\n            if i != j:\n                enc[c[i] + c[j]] = c[(i + 1) % 5] + c[(j + 1) % 5]\n\n    # Map pairs with cross-connections.\n    for i1, j1, i2, j2 in product(xrange(5), repeat=4):\n        if i1 != i2 and j1 != j2:\n            enc[m[i1][j1] + m[i2][j2]] = m[i1][j2] + m[i2][j1]\n\n    # Generate reverse translations.\n    dec = dict((v, k) for k, v in enc.iteritems())\n\n    def sub_enc(txt):\n        lst = findall(r\"(.)(?:(?!\\1)(.))?\", canonicalize(txt))\n        return \" \".join(enc[a + (b if b else 'X')] for a, b in lst)\n\n    def sub_dec(encoded):\n        return \" \".join(dec[p] for p in partition(canonicalize(encoded), 2))\n\n    return sub_enc, sub_dec\n\n\n(encode, decode) = playfair(\"Playfair example\")\norig = \"Hide the gold in...the TREESTUMP!!!\"\nprint \"Original:\", orig\nenc = encode(orig)\nprint \"Encoded:\", enc\nprint \"Decoded:\", decode(enc)"}
{"title": "Plot coordinate pairs", "language": "Python", "task": "Plot a function represented as     x,  y     numerical arrays.\n\nPost the resulting image for the following input arrays (taken from Python's Example section on ''Time a function''):\n       x = {0,   1,    2,    3,    4,    5,     6,     7,     8,     9};\n       y = {2.7, 2.8, 31.4, 38.1, 58.0, 76.2, 100.5, 130.0, 149.3, 180.0};\n\nThis task is intended as a subtask for [[Measure relative performance of sorting algorithms implementations]].\n\n", "solution": "from visual import *\nfrom visual.graph import *\n\nplot1 = gdisplay( title='VPython Plot-Demo', \n                  xtitle='x',\n                  ytitle='y    (click and drag mouse to see coordinates)',\n                  foreground=color.black,\n                  background=color.white, \n                  x=0, y=0,\n                  width=400, height=400,\n                  xmin=0, xmax=10, \n                  ymin=0, ymax=200 )\n\nf1 = gdots(color=color.red)                 # create plot-object\n\nf1.plot(pos= (0,   2.7), color=color.blue ) # add a single point\nf1.plot(pos=[(1,   2.8),                    # add a list of points\n             (2,  31.4),\n             (3,  38.1), \n             (4,  58.0),\n             (5,  76.2),\n             (6, 100.5),\n             (7, 130.0),\n             (8, 149.3),\n             (9, 180.0) ]\n        )\nlabel(display=plot1.display, text=\"Look here\",\n      pos=(6,100.5), xoffset=30,yoffset=-20 )\n"}
{"title": "Poker hand analyser", "language": "Python", "task": "Create a program to parse a single five card poker hand and rank it according to this list of poker hands.\n\n\nA poker hand is specified as a space separated list of five playing cards. \n\nEach input card has two characters indicating face and suit. \n\n\n;Example:\n::::'''2d'''       (two of diamonds).\n\n\n\nFaces are:    '''a''', '''2''', '''3''', '''4''', '''5''', '''6''', '''7''', '''8''', '''9''', '''10''', '''j''', '''q''', '''k'''\n\nSuits are:    '''h''' (hearts),   '''d''' (diamonds),   '''c''' (clubs),   and   '''s''' (spades),   or \nalternatively,   the unicode card-suit characters:         \n\n\nDuplicate cards are illegal.\n\nThe program should analyze a single hand and produce one of the following outputs:\n  straight-flush\n  four-of-a-kind\n  full-house\n  flush\n  straight\n  three-of-a-kind\n  two-pair\n  one-pair\n  high-card\n  invalid\n\n\n;Examples:\n    2 2 2 k q:   three-of-a-kind\n    2 5 7 8 9:   high-card\n    a 2 3 4 5:   straight\n    2 3 2 3 3:   full-house\n    2 7 2 3 3:   two-pair\n    2 7 7 7 7:   four-of-a-kind \n    10 j q k a:  straight-flush\n    4 4 k 5 10:  one-pair\n    q 10 7 6 q:  invalid\n\nThe programs output for the above examples should be displayed here on this page.\n\n\n;Extra credit:\n# use the playing card characters introduced with Unicode 6.0 (U+1F0A1 - U+1F0DE).\n# allow two jokers\n::* use the symbol   '''joker'''\n::* duplicates would be allowed (for jokers only)\n::* five-of-a-kind would then be the highest hand\n\n\n;More extra credit examples:\n    joker  2  2  k  q:     three-of-a-kind\n    joker  5  7  8  9:     straight\n    joker  2  3  4  5:     straight\n    joker  3  2  3  3:     four-of-a-kind\n    joker  7  2  3  3:     three-of-a-kind\n    joker  7  7  7  7:     five-of-a-kind\n    joker  j  q  k  A:     straight-flush\n    joker  4  k  5 10:     one-pair\n    joker  k  7  6  4:     flush\n    joker  2  joker  4  5:  straight\n    joker  Q  joker  A 10:  straight\n    joker  Q  joker  A 10:  straight-flush\n    joker  2  2  joker  q:  four-of-a-kind\n\n\n;Related tasks:\n* [[Playing cards]]\n* [[Card shuffles]]\n* [[Deal cards_for_FreeCell]]\n* [[War Card_Game]]\n* [[Go Fish]]\n\n", "solution": "from collections import namedtuple\n\nclass Card(namedtuple('Card', 'face, suit')):\n    def __repr__(self):\n        return ''.join(self)\n\n\nsuit = '\u2665 \u2666 \u2663 \u2660'.split()\n# ordered strings of faces\nfaces   = '2 3 4 5 6 7 8 9 10 j q k a'\nlowaces = 'a 2 3 4 5 6 7 8 9 10 j q k'\n# faces as lists\nface   = faces.split()\nlowace = lowaces.split()\n\n\ndef straightflush(hand):\n    f,fs = ( (lowace, lowaces) if any(card.face == '2' for card in hand)\n             else (face, faces) )\n    ordered = sorted(hand, key=lambda card: (f.index(card.face), card.suit))\n    first, rest = ordered[0], ordered[1:]\n    if ( all(card.suit == first.suit for card in rest) and\n         ' '.join(card.face for card in ordered) in fs ):\n        return 'straight-flush', ordered[-1].face\n    return False\n\ndef fourofakind(hand):\n    allfaces = [f for f,s in hand]\n    allftypes = set(allfaces)\n    if len(allftypes) != 2:\n        return False\n    for f in allftypes:\n        if allfaces.count(f) == 4:\n            allftypes.remove(f)\n            return 'four-of-a-kind', [f, allftypes.pop()]\n    else:\n        return False\n\ndef fullhouse(hand):\n    allfaces = [f for f,s in hand]\n    allftypes = set(allfaces)\n    if len(allftypes) != 2:\n        return False\n    for f in allftypes:\n        if allfaces.count(f) == 3:\n            allftypes.remove(f)\n            return 'full-house', [f, allftypes.pop()]\n    else:\n        return False\n\ndef flush(hand):\n    allstypes = {s for f, s in hand}\n    if len(allstypes) == 1:\n        allfaces = [f for f,s in hand]\n        return 'flush', sorted(allfaces,\n                               key=lambda f: face.index(f),\n                               reverse=True)\n    return False\n\ndef straight(hand):\n    f,fs = ( (lowace, lowaces) if any(card.face == '2' for card in hand)\n             else (face, faces) )\n    ordered = sorted(hand, key=lambda card: (f.index(card.face), card.suit))\n    first, rest = ordered[0], ordered[1:]\n    if ' '.join(card.face for card in ordered) in fs:\n        return 'straight', ordered[-1].face\n    return False\n\ndef threeofakind(hand):\n    allfaces = [f for f,s in hand]\n    allftypes = set(allfaces)\n    if len(allftypes) <= 2:\n        return False\n    for f in allftypes:\n        if allfaces.count(f) == 3:\n            allftypes.remove(f)\n            return ('three-of-a-kind', [f] +\n                     sorted(allftypes,\n                            key=lambda f: face.index(f),\n                            reverse=True))\n    else:\n        return False\n\ndef twopair(hand):\n    allfaces = [f for f,s in hand]\n    allftypes = set(allfaces)\n    pairs = [f for f in allftypes if allfaces.count(f) == 2]\n    if len(pairs) != 2:\n        return False\n    p0, p1 = pairs\n    other = [(allftypes - set(pairs)).pop()]\n    return 'two-pair', pairs + other if face.index(p0) > face.index(p1) else pairs[::-1] + other\n\ndef onepair(hand):\n    allfaces = [f for f,s in hand]\n    allftypes = set(allfaces)\n    pairs = [f for f in allftypes if allfaces.count(f) == 2]\n    if len(pairs) != 1:\n        return False\n    allftypes.remove(pairs[0])\n    return 'one-pair', pairs + sorted(allftypes,\n                                      key=lambda f: face.index(f),\n                                      reverse=True)\n\ndef highcard(hand):\n    allfaces = [f for f,s in hand]\n    return 'high-card', sorted(allfaces,\n                               key=lambda f: face.index(f),\n                               reverse=True)\n\nhandrankorder =  (straightflush, fourofakind, fullhouse,\n                  flush, straight, threeofakind,\n                  twopair, onepair, highcard)\n              \ndef rank(cards):\n    hand = handy(cards)\n    for ranker in handrankorder:\n        rank = ranker(hand)\n        if rank:\n            break\n    assert rank, \"Invalid: Failed to rank cards: %r\" % cards\n    return rank\n\ndef handy(cards='2\u2665 2\u2666 2\u2663 k\u2663 q\u2666'):\n    hand = []\n    for card in cards.split():\n        f, s = card[:-1], card[-1]\n        assert f in face, \"Invalid: Don't understand card face %r\" % f\n        assert s in suit, \"Invalid: Don't understand card suit %r\" % s\n        hand.append(Card(f, s))\n    assert len(hand) == 5, \"Invalid: Must be 5 cards in a hand, not %i\" % len(hand)\n    assert len(set(hand)) == 5, \"Invalid: All cards in the hand must be unique %r\" % cards\n    return hand\n\n\nif __name__ == '__main__':\n    hands = [\"2\u2665 2\u2666 2\u2663 k\u2663 q\u2666\",\n     \"2\u2665 5\u2665 7\u2666 8\u2663 9\u2660\",\n     \"a\u2665 2\u2666 3\u2663 4\u2663 5\u2666\",\n     \"2\u2665 3\u2665 2\u2666 3\u2663 3\u2666\",\n     \"2\u2665 7\u2665 2\u2666 3\u2663 3\u2666\",\n     \"2\u2665 7\u2665 7\u2666 7\u2663 7\u2660\",\n     \"10\u2665 j\u2665 q\u2665 k\u2665 a\u2665\"] + [\n     \"4\u2665 4\u2660 k\u2660 5\u2666 10\u2660\",\n     \"q\u2663 10\u2663 7\u2663 6\u2663 4\u2663\",\n     ]\n    print(\"%-18s %-15s %s\" % (\"HAND\", \"CATEGORY\", \"TIE-BREAKER\"))\n    for cards in hands:\n        r = rank(cards)\n        print(\"%-18r %-15s %r\" % (cards, r[0], r[1]))"}
{"title": "Population count", "language": "Python", "task": "The   ''population count''   is the number of   '''1'''s   (ones)   in the binary representation of a non-negative integer.\n\n''Population count''   is also known as:\n::::*   ''pop count''\n::::*   ''popcount'' \n::::*   ''sideways sum''\n::::*   ''bit summation'' \n::::*   ''Hamming weight'' \n\n\nFor example,   '''5'''   (which is   '''101'''   in binary)   has a population count of   '''2'''.\n\n\n''Evil numbers''   are non-negative integers  that have an   ''even''   population count.\n\n''Odious numbers''      are  positive integers that have an    ''odd''   population count.\n\n\n;Task:\n* write a function (or routine) to return the population count of a non-negative integer.\n* all computation of the lists below should start with   '''0'''   (zero indexed).\n:* display the   ''pop count''   of the   1st   thirty powers of   '''3'''       ('''30''',   '''31''',   '''32''',   '''33''',   '''34''',      '''329''').\n:* display the   1st   thirty     ''evil''        numbers.\n:* display the   1st   thirty          ''odious''           numbers.\n* display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified.\n\n\n;See also\n* The On-Line Encyclopedia of Integer Sequences:   A000120 population count.\n* The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers.\n* The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.\n\n", "solution": ">>> def popcount(n): return bin(n).count(\"1\")\n... \n>>> [popcount(3**i) for i in range(30)]\n[1, 2, 2, 4, 3, 6, 6, 5, 6, 8, 9, 13, 10, 11, 14, 15, 11, 14, 14, 17, 17, 20, 19, 22, 16, 18, 24, 30, 25, 25]\n>>> evil, odious, i = [], [], 0\n>>> while len(evil) < 30 or len(odious) < 30:\n...     p = popcount(i)\n...     if p % 2: odious.append(i)\n...     else: evil.append(i)\n...     i += 1\n... \n>>> evil[:30]\n[0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20, 23, 24, 27, 29, 30, 33, 34, 36, 39, 40, 43, 45, 46, 48, 51, 53, 54, 57, 58]\n>>> odious[:30]\n[1, 2, 4, 7, 8, 11, 13, 14, 16, 19, 21, 22, 25, 26, 28, 31, 32, 35, 37, 38, 41, 42, 44, 47, 49, 50, 52, 55, 56, 59]\n>>> "}
{"title": "Population count", "language": "Python 3", "task": "The   ''population count''   is the number of   '''1'''s   (ones)   in the binary representation of a non-negative integer.\n\n''Population count''   is also known as:\n::::*   ''pop count''\n::::*   ''popcount'' \n::::*   ''sideways sum''\n::::*   ''bit summation'' \n::::*   ''Hamming weight'' \n\n\nFor example,   '''5'''   (which is   '''101'''   in binary)   has a population count of   '''2'''.\n\n\n''Evil numbers''   are non-negative integers  that have an   ''even''   population count.\n\n''Odious numbers''      are  positive integers that have an    ''odd''   population count.\n\n\n;Task:\n* write a function (or routine) to return the population count of a non-negative integer.\n* all computation of the lists below should start with   '''0'''   (zero indexed).\n:* display the   ''pop count''   of the   1st   thirty powers of   '''3'''       ('''30''',   '''31''',   '''32''',   '''33''',   '''34''',      '''329''').\n:* display the   1st   thirty     ''evil''        numbers.\n:* display the   1st   thirty          ''odious''           numbers.\n* display each list of integers on one line   (which may or may not include a title),   each set of integers being shown should be properly identified.\n\n\n;See also\n* The On-Line Encyclopedia of Integer Sequences:   A000120 population count.\n* The On-Line Encyclopedia of Integer Sequences:   A000069 odious numbers.\n* The On-Line Encyclopedia of Integer Sequences:   A001969 evil numbers.\n\n", "solution": "'''Population count'''\n\nfrom functools import reduce\n\n\n# popCount :: Int -> Int\ndef popCount(n):\n    '''The count of non-zero digits in the binary\n       representation of the positive integer n.'''\n    def go(x):\n        return Just(divmod(x, 2)) if 0 < x else Nothing()\n    return sum(unfoldl(go)(n))\n\n\n# -------------------------- TEST --------------------------\ndef main():\n    '''Tests'''\n\n    print('Population count of first 30 powers of 3:')\n    print('    ' + showList(\n        [popCount(pow(3, x)) for x in enumFromTo(0)(29)]\n    ))\n\n    evilNums, odiousNums = partition(\n        compose(even, popCount)\n    )(enumFromTo(0)(59))\n\n    print(\"\\nFirst thirty 'evil' numbers:\")\n    print('    ' + showList(evilNums))\n\n    print(\"\\nFirst thirty 'odious' numbers:\")\n    print('    ' + showList(odiousNums))\n\n\n# ------------------------ GENERIC -------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# compose :: ((a -> a), ...) -> (a -> a)\ndef compose(*fs):\n    '''Composition, from right to left,\n       of a series of functions.\n    '''\n    def go(f, g):\n        def fg(x):\n            return f(g(x))\n        return fg\n    return reduce(go, fs, lambda x: x)\n\n\n# enumFromTo :: Int -> Int -> [Int]\ndef enumFromTo(m):\n    '''Enumeration of integer values [m..n]'''\n    return lambda n: range(m, 1 + n)\n\n\n# even :: Int -> Bool\ndef even(x):\n    '''True if x is an integer\n       multiple of two.\n    '''\n    return 0 == x % 2\n\n\n# partition :: (a -> Bool) -> [a] -> ([a], [a])\ndef partition(p):\n    '''The pair of lists of those elements in xs\n       which respectively do, and don't\n       satisfy the predicate p.\n    '''\n\n    def go(a, x):\n        ts, fs = a\n        return (ts + [x], fs) if p(x) else (ts, fs + [x])\n    return lambda xs: reduce(go, xs, ([], []))\n\n\n# showList :: [a] -> String\ndef showList(xs):\n    '''Stringification of a list.'''\n    return '[' + ','.join(repr(x) for x in xs) + ']'\n\n\n# unfoldl(lambda x: Just(((x - 1), x)) if 0 != x else Nothing())(10)\n# -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n# unfoldl :: (b -> Maybe (b, a)) -> b -> [a]\ndef unfoldl(f):\n    '''Dual to reduce or foldl.\n       Where these reduce a list to a summary value, unfoldl\n       builds a list from a seed value.\n       Where f returns Just(a, b), a is appended to the list,\n       and the residual b is used as the argument for the next\n       application of f.\n       When f returns Nothing, the completed list is returned.\n    '''\n    def go(v):\n        x, r = v, v\n        xs = []\n        while True:\n            mb = f(x)\n            if mb.get('Nothing'):\n                return xs\n            else:\n                x, r = mb.get('Just')\n                xs.insert(0, r)\n        return xs\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Pragmatic directives", "language": "Python", "task": "Pragmatic directives cause the language to operate in a specific manner,   allowing support for operational variances within the program code   (possibly by the loading of specific or alternative modules).\n\n\n;Task:\nList any pragmatic directives supported by the language,   and demonstrate how to activate and deactivate the pragmatic directives and to describe or demonstrate the scope of effect that the pragmatic directives have within a program.\n\n", "solution": "Python 3.2 (r32:88445, Feb 20 2011, 21:30:00) [MSC v.1500 64 bit (AMD64)] on win32\nType \"copyright\", \"credits\" or \"license()\" for more information.\n>>> import __future__\n>>> __future__.all_feature_names\n['nested_scopes', 'generators', 'division', 'absolute_import', 'with_statement', 'print_function', 'unicode_literals', 'barry_as_FLUFL']\n>>> "}
{"title": "Prime triangle", "language": "Python", "task": "You will require a function f which when given an integer S will return a list of the arrangements of the integers 1 to S such that g1=1 gS=S and generally for n=1 to n=S-1 gn+gn+1 is prime. S=1 is undefined. For S=2 to S=20 print f(S) to form a triangle. Then again for S=2 to S=20 print the number of possible arrangements of 1 to S meeting these requirements.\n\n\n;See also\n:* OEIS:A036440\n\n\n", "solution": "from numpy import array\n# for Rosetta Code by MG - 20230312\ndef is_prime(n: int) -> bool:\n    assert n < 64\n    return ((1 << n) & 0x28208a20a08a28ac) != 0\n\ndef prime_triangle_row(a: array, start: int, length: int) -> bool:\n    if length == 2:\n        return is_prime(a[0] + a[1])\n    for i in range(1, length - 1, 1):\n        if is_prime(a[start] + a[start + i]):\n            a[start + i], a[start + 1] = a[start + 1], a[start + i]\n            if prime_triangle_row(a, start + 1, length - 1):\n                return True\n            a[start + i], a[start + 1] = a[start + 1], a[start + i]\n    return False\n\ndef prime_triangle_count(a: array, start: int, length: int) -> int:\n    count: int = 0\n    if length == 2:\n        if is_prime(a[start] + a[start + 1]):\n            count += 1\n    else:\n        for i in range(1, length - 1, 1):\n            if is_prime(a[start] + a[start + i]):\n                a[start + i], a[start + 1] = a[start + 1], a[start + i]\n                count += prime_triangle_count(a, start + 1, length - 1)\n                a[start + i], a[start + 1] = a[start + 1], a[start + i]\n    return count\n\ndef print_row(a: array):\n    if a == []:\n        return\n    print(\"%2d\"% a[0], end=\" \")\n    for x in a[1:]:\n        print(\"%2d\"% x, end=\" \")\n    print()\n\nfor n in range(2, 21):\n    tr: array = [_ for _ in range(1, n + 1)]\n    if prime_triangle_row(tr, 0, n):\n        print_row(tr)\nprint()\nfor n in range(2, 21):\n    tr: array = [_ for _ in range(1, n + 1)]\n    if n > 2:\n        print(\" \", end=\"\")\n    print(prime_triangle_count(tr, 0, n), end=\"\")\nprint()\n"}
{"title": "Priority queue", "language": "Python", "task": "A queue, with an important distinction: each item is added to a priority queue with a priority level, and will be later removed from the queue with the highest priority element first. That is, the items are (conceptually) stored in the queue in priority order instead of in insertion order.\n\n\n;Task:\nCreate a priority queue.   The queue must support at least two operations:\n:#   Insertion.   An element is added to the queue with a priority (a numeric value).\n:#   Top item removal.   Deletes the element or one of the elements with the current top priority and return it.\n\n\nOptionally, other operations may be defined, such as peeking (find what current top priority/top element is), merging (combining two priority queues into one), etc.\n\n\nTo test your implementation, insert a number of elements into the queue, each with some random priority. \n\nThen dequeue them sequentially; now the elements should be sorted by priority. \n\nYou can use the following task/priority items as input data:\n         '''Priority'''         '''Task'''\n        ----------   ----------------\n            3         Clear drains\n            4         Feed cat\n            5         Make tea\n            1         Solve RC tasks\n            2         Tax return\n\n\nThe implementation should try to be efficient.   A typical implementation has   '''O(log n)'''   insertion and extraction time,   where   '''n'''   is the number of items in the queue.  \n\nYou may choose to impose certain limits such as small range of allowed priority levels, limited capacity, etc.   If so, discuss the reasons behind it.\n\n", "solution": ">>> import queue\n>>> help(queue.PriorityQueue)\nHelp on class PriorityQueue in module queue:\n\nclass PriorityQueue(Queue)\n |  Variant of Queue that retrieves open entries in priority order (lowest first).\n |  \n |  Entries are typically tuples of the form:  (priority number, data).\n |  \n |  Method resolution order:\n |      PriorityQueue\n |      Queue\n |      builtins.object\n |  \n |  Methods inherited from Queue:\n |  \n |  __init__(self, maxsize=0)\n |  \n |  empty(self)\n |      Return True if the queue is empty, False otherwise (not reliable!).\n |      \n |      This method is likely to be removed at some point.  Use qsize() == 0\n |      as a direct substitute, but be aware that either approach risks a race\n |      condition where a queue can grow before the result of empty() or\n |      qsize() can be used.\n |      \n |      To create code that needs to wait for all queued tasks to be\n |      completed, the preferred technique is to use the join() method.\n |  \n |  full(self)\n |      Return True if the queue is full, False otherwise (not reliable!).\n |      \n |      This method is likely to be removed at some point.  Use qsize() >= n\n |      as a direct substitute, but be aware that either approach risks a race\n |      condition where a queue can shrink before the result of full() or\n |      qsize() can be used.\n |  \n |  get(self, block=True, timeout=None)\n |      Remove and return an item from the queue.\n |      \n |      If optional args 'block' is true and 'timeout' is None (the default),\n |      block if necessary until an item is available. If 'timeout' is\n |      a positive number, it blocks at most 'timeout' seconds and raises\n |      the Empty exception if no item was available within that time.\n |      Otherwise ('block' is false), return an item if one is immediately\n |      available, else raise the Empty exception ('timeout' is ignored\n |      in that case).\n |  \n |  get_nowait(self)\n |      Remove and return an item from the queue without blocking.\n |      \n |      Only get an item if one is immediately available. Otherwise\n |      raise the Empty exception.\n |  \n |  join(self)\n |      Blocks until all items in the Queue have been gotten and processed.\n |      \n |      The count of unfinished tasks goes up whenever an item is added to the\n |      queue. The count goes down whenever a consumer thread calls task_done()\n |      to indicate the item was retrieved and all work on it is complete.\n |      \n |      When the count of unfinished tasks drops to zero, join() unblocks.\n |  \n |  put(self, item, block=True, timeout=None)\n |      Put an item into the queue.\n |      \n |      If optional args 'block' is true and 'timeout' is None (the default),\n |      block if necessary until a free slot is available. If 'timeout' is\n |      a positive number, it blocks at most 'timeout' seconds and raises\n |      the Full exception if no free slot was available within that time.\n |      Otherwise ('block' is false), put an item on the queue if a free slot\n |      is immediately available, else raise the Full exception ('timeout'\n |      is ignored in that case).\n |  \n |  put_nowait(self, item)\n |      Put an item into the queue without blocking.\n |      \n |      Only enqueue the item if a free slot is immediately available.\n |      Otherwise raise the Full exception.\n |  \n |  qsize(self)\n |      Return the approximate size of the queue (not reliable!).\n |  \n |  task_done(self)\n |      Indicate that a formerly enqueued task is complete.\n |      \n |      Used by Queue consumer threads.  For each get() used to fetch a task,\n |      a subsequent call to task_done() tells the queue that the processing\n |      on the task is complete.\n |      \n |      If a join() is currently blocking, it will resume when all items\n |      have been processed (meaning that a task_done() call was received\n |      for every item that had been put() into the queue).\n |      \n |      Raises a ValueError if called more times than there were items\n |      placed in the queue.\n |  \n |  ----------------------------------------------------------------------\n |  Data descriptors inherited from Queue:\n |  \n |  __dict__\n |      dictionary for instance variables (if defined)\n |  \n |  __weakref__\n |      list of weak references to the object (if defined)\n\n>>> "}
{"title": "Pseudo-random numbers/Combined recursive generator MRG32k3a", "language": "Python", "task": "MRG32k3a Combined recursive generator (pseudo-code):\n\n    /* Constants */\n    /* First generator */\n    a1 = [0, 1403580, -810728]\n    m1 = 2**32 - 209\n    /* Second Generator */\n    a2 = [527612, 0, -1370589]\n    m2 = 2**32 - 22853\n     \n    d = m1 + 1\n    \n    class MRG32k3a\n        x1 = [0, 0, 0]  /* list of three last values of gen #1 */\n        x2 = [0, 0, 0]  /* list of three last values of gen #2 */\n        \n        method seed(u64 seed_state)\n            assert seed_state in range >0 and < d \n            x1 = [seed_state, 0, 0]\n            x2 = [seed_state, 0, 0]\n        end method\n            \n        method next_int()\n            x1i = (a1[0]*x1[0] + a1[1]*x1[1] + a1[2]*x1[2]) mod m1\n            x2i = (a2[0]*x2[0] + a2[1]*x2[1] + a2[2]*x2[2]) mod m2\n            x1 = [x1i, x1[0], x1[1]]    /* Keep last three */\n            x2 = [x2i, x2[0], x2[1]]    /* Keep last three */\n            z = (x1i - x2i) % m1\n            answer = (z + 1)\n            \n            return answer\n        end method\n        \n        method next_float():\n            return float next_int() / d\n        end method\n        \n    end class\n \n:MRG32k3a Use:\n    random_gen = instance MRG32k3a\n    random_gen.seed(1234567)\n    print(random_gen.next_int())   /* 1459213977 */\n    print(random_gen.next_int())   /* 2827710106 */\n    print(random_gen.next_int())   /* 4245671317 */\n    print(random_gen.next_int())   /* 3877608661 */\n    print(random_gen.next_int())   /* 2595287583 */\n    \n        \n;Task\n\n* Generate a class/set of functions that generates pseudo-random\nnumbers as shown above.\n\n* Show that the first five integers generated with the seed `1234567`\nare as shown above       \n\n* Show that for an initial seed of '987654321' the counts of 100_000\nrepetitions of\n\n    floor(random_gen.next_float() * 5)\n\nIs as follows:\n    \n    0: 20002, 1: 20060, 2: 19948, 3: 20059, 4: 19931\n\n* Show your output here, on this page.\n\n\n", "solution": "# Constants\na1 = [0, 1403580, -810728]\nm1 = 2**32 - 209\n#\na2 = [527612, 0, -1370589]\nm2 = 2**32 - 22853\n#\nd = m1 + 1\n\nclass MRG32k3a():\n    \n    def __init__(self, seed_state=123):\n        self.seed(seed_state)\n    \n    def seed(self, seed_state):\n        assert 0 <seed_state < d, f\"Out of Range 0 x < {d}\"\n        self.x1 = [seed_state, 0, 0]\n        self.x2 = [seed_state, 0, 0]\n        \n    def next_int(self):\n        \"return random int in range 0..d\"\n        x1i = sum(aa * xx  for aa, xx in zip(a1, self.x1)) % m1\n        x2i = sum(aa * xx  for aa, xx in zip(a2, self.x2)) % m2\n        self.x1 = [x1i] + self.x1[:2]\n        self.x2 = [x2i] + self.x2[:2]\n\n        z = (x1i - x2i) % m1\n        answer = (z + 1)\n        \n        return answer\n    \n    def  next_float(self):\n        \"return random float between 0 and 1\"\n        return self.next_int() / d\n    \n\nif __name__ == '__main__':\n    random_gen = MRG32k3a()\n    random_gen.seed(1234567)\n    for i in range(5):\n        print(random_gen.next_int())\n        \n    random_gen.seed(987654321)\n    hist = {i:0 for i in range(5)}\n    for i in range(100_000):\n        hist[int(random_gen.next_float() *5)] += 1\n    print(hist)"}
{"title": "Pseudo-random numbers/Middle-square method", "language": "Python", "task": "{{Wikipedia|Middle-square method|en}}\n; The Method:\nTo generate a sequence of n-digit pseudorandom numbers, an n-digit starting value is created and squared, producing a 2n-digit number. If the result has fewer than 2n digits, leading zeroes are added to compensate. The middle n digits of the result would be the next number in the sequence and returned as the result. This process is then repeated to generate more numbers. \n; Pseudo code:\n\nvar seed = 675248\nfunction random()\n    var s = str(seed * seed) 'str: turn a number into string\n    do while not len(s) = 12\n        s = \"0\" + s          'add zeroes before the string\n    end do\n    seed = val(mid(s, 4, 6)) 'mid: string variable, start, length\n                             'val: turn a string into number\n    return seed\nend function\n\n; Middle-square method use:\n\nfor i = 1 to 5\n    print random()\nend for\n\n;Task:\n\n* Generate a class/set of functions that generates pseudo-random\nnumbers (6 digits) as shown above.\n\n* Show the first five integers generated with the seed 675248 as shown above.\n\n* Show your output here, on this page.\n\n", "solution": "seed = 675248\ndef random():\n    global seed\n    seed = int(str(seed ** 2).zfill(12)[3:9])\n    return seed\nfor _ in range(5):\n    print(random())\n"}
{"title": "Pseudo-random numbers/PCG32", "language": "Python", "task": "Some definitions to help in the explanation:\n\n:Floor operation\n::https://en.wikipedia.org/wiki/Floor_and_ceiling_functions\n::Greatest integer less than or equal to a real number.\n\n:Bitwise Logical shift operators (c-inspired)\n::https://en.wikipedia.org/wiki/Bitwise_operation#Bit_shifts\n::Binary bits of value shifted left or right, with zero bits shifted in where appropriate. \n::Examples are shown for 8 bit binary numbers; most significant bit to the left.\n    \n:: '''<<''' Logical shift left by given number of bits.\n:::E.g Binary 00110101 '''<<''' 2 == Binary 11010100\n    \n:: '''>>''' Logical shift right by given number of bits.\n:::E.g Binary 00110101 '''>>''' 2 == Binary 00001101\n    \n:'''^''' Bitwise exclusive-or operator\n::https://en.wikipedia.org/wiki/Exclusive_or\n::Bitwise comparison for if bits differ\n:::E.g Binary 00110101 '''^''' Binary 00110011 == Binary 00000110\n    \n:'''|''' Bitwise or operator\n::https://en.wikipedia.org/wiki/Bitwise_operation#OR\n::Bitwise comparison gives 1 if any of corresponding bits are 1\n:::E.g Binary 00110101 '''|''' Binary 00110011 == Binary 00110111\n\n\n;PCG32 Generator (pseudo-code):\n\nPCG32 has two unsigned 64-bit integers of internal state:\n# '''state''': All 2**64 values may be attained.\n# '''sequence''': Determines which of 2**63 sequences that state iterates through. (Once set together with state at time of seeding will stay constant for this generators lifetime).\n\nValues of sequence allow 2**63 ''different'' sequences of random numbers from the same state.\n\nThe algorithm is given 2 U64 inputs called seed_state, and seed_sequence. The algorithm proceeds in accordance with the following pseudocode:-\n\nconst N<-U64 6364136223846793005\nconst inc<-U64 (seed_sequence << 1) | 1\nstate<-U64 ((inc+seed_state)*N+inc\ndo forever\n  xs<-U32 (((state>>18)^state)>>27)\n  rot<-INT (state>>59)\n  OUTPUT U32 (xs>>rot)|(xs<<((-rot)&31))\n  state<-state*N+inc\nend do\n\nNote that this an anamorphism - dual to catamorphism, and encoded in some languages as a general higher-order `unfold` function, dual to `fold` or `reduce`.\n\n;Task:\n\n* Generate a class/set of functions that generates pseudo-random\nnumbers using the above.\n\n* Show that the first five integers generated with the seed 42, 54\nare: 2707161783 2068313097 3122475824 2211639955 3215226955\n\n   \n\n* Show that for an initial seed of 987654321, 1 the counts of 100_000 repetitions of\n\n    floor(random_gen.next_float() * 5)\n\n:Is as follows:\n    \n    0: 20049, 1: 20022, 2: 20115, 3: 19809, 4: 20005\n\n* Show your output here, on this page.\n\n\n", "solution": "mask64 = (1 << 64) - 1\nmask32 = (1 << 32) - 1\nCONST = 6364136223846793005\n\n\nclass PCG32():\n    \n    def __init__(self, seed_state=None, seed_sequence=None):\n        if all(type(x) == int for x in (seed_state, seed_sequence)):\n            self.seed(seed_state, seed_sequence)\n        else:\n            self.state = self.inc = 0\n    \n    def seed(self, seed_state, seed_sequence):\n        self.state = 0\n        self.inc = ((seed_sequence << 1) | 1) & mask64\n        self.next_int()\n        self.state = (self.state + seed_state)\n        self.next_int()\n        \n    def next_int(self):\n        \"return random 32 bit unsigned int\"\n        old = self.state\n        self.state = ((old * CONST) + self.inc) & mask64\n        xorshifted = (((old >> 18) ^ old) >> 27) & mask32\n        rot = (old >> 59) & mask32\n        answer = (xorshifted >> rot) | (xorshifted << ((-rot) & 31))\n        answer = answer &mask32\n        \n        return answer\n    \n    def  next_float(self):\n        \"return random float between 0 and 1\"\n        return self.next_int() / (1 << 32)\n    \n\nif __name__ == '__main__':\n    random_gen = PCG32()\n    random_gen.seed(42, 54)\n    for i in range(5):\n        print(random_gen.next_int())\n        \n    random_gen.seed(987654321, 1)\n    hist = {i:0 for i in range(5)}\n    for i in range(100_000):\n        hist[int(random_gen.next_float() *5)] += 1\n    print(hist)"}
{"title": "Pseudo-random numbers/Xorshift star", "language": "Python", "task": "Some definitions to help in the explanation:\n\n:Floor operation\n::https://en.wikipedia.org/wiki/Floor_and_ceiling_functions\n::Greatest integer less than or equal to a real number.\n\n:Bitwise Logical shift operators (c-inspired)\n::https://en.wikipedia.org/wiki/Bitwise_operation#Bit_shifts\n::Binary bits of value shifted left or right, with zero bits shifted in where appropriate. \n::Examples are shown for 8 bit binary numbers; most significant bit to the left.\n    \n:: '''<<''' Logical shift left by given number of bits.\n:::E.g Binary 00110101 '''<<''' 2 == Binary 11010100\n    \n:: '''>>''' Logical shift right by given number of bits.\n:::E.g Binary 00110101 '''>>''' 2 == Binary 00001101\n    \n:'''^''' Bitwise exclusive-or operator\n::https://en.wikipedia.org/wiki/Exclusive_or\n::Bitwise comparison for if bits differ\n:::E.g Binary 00110101 '''^''' Binary 00110011 == Binary 00000110\n    \n;Xorshift_star Generator (pseudo-code):\n\n    /* Let u64 denote an unsigned 64 bit integer type. */\n    /* Let u32 denote an unsigned 32 bit integer type. */\n    \n    class Xorshift_star\n        u64 state       /* Must be seeded to non-zero initial value */\n        u64 const = HEX '2545F4914F6CDD1D'\n    \n        method seed(u64 num):\n            state =  num\n        end method\n        \n        method next_int():\n            u64 x = state\n            x = x ^ (x >> 12)\n            x = x ^ (x << 25)\n            x = x ^ (x >> 27)\n            state = x\n            u32 answer = ((x * const) >> 32)\n            \n            return answer\n        end method\n        \n        method next_float():\n            return float next_int() / (1 << 32)\n        end method\n        \n    end class\n        \n:;Xorshift use:\n\n    random_gen = instance Xorshift_star\n    random_gen.seed(1234567)\n    print(random_gen.next_int())   /* 3540625527 */\n    print(random_gen.next_int())   /* 2750739987 */\n    print(random_gen.next_int())   /* 4037983143 */\n    print(random_gen.next_int())   /* 1993361440 */\n    print(random_gen.next_int())   /* 3809424708 */\n\n;Task:\n\n* Generate a class/set of functions that generates pseudo-random\nnumbers as shown above.\n\n* Show that the first five integers genrated with the seed 1234567\nare as shown above       \n\n* Show that for an initial seed of 987654321, the counts of 100_000 repetitions of\n\n    floor(random_gen.next_float() * 5)\n\n:Is as follows:\n    \n    0: 20103, 1: 19922, 2: 19937, 3: 20031, 4: 20007\n\n* Show your output here, on this page.\n\n\n\n", "solution": "mask64 = (1 << 64) - 1\nmask32 = (1 << 32) - 1\nconst = 0x2545F4914F6CDD1D\n\n\n\nclass Xorshift_star():\n    \n    def __init__(self, seed=0):\n        self.state = seed & mask64\n\n    def seed(self, num):\n        self.state =  num & mask64\n    \n    def next_int(self):\n        \"return random int between 0 and 2**32\"\n        x = self.state\n        x = (x ^ (x >> 12)) & mask64\n        x = (x ^ (x << 25)) & mask64\n        x = (x ^ (x >> 27)) & mask64\n        self.state = x\n        answer = (((x * const) & mask64) >> 32) & mask32 \n        return answer\n    \n    def  next_float(self):\n        \"return random float between 0 and 1\"\n        return self.next_int() / (1 << 32)\n    \n\nif __name__ == '__main__':\n    random_gen = Xorshift_star()\n    random_gen.seed(1234567)\n    for i in range(5):\n        print(random_gen.next_int())\n        \n    random_gen.seed(987654321)\n    hist = {i:0 for i in range(5)}\n    for i in range(100_000):\n        hist[int(random_gen.next_float() *5)] += 1\n    print(hist)"}
{"title": "Pythagoras tree", "language": "Python", "task": "The Pythagoras tree is a fractal tree constructed from squares.  It is named after Pythagoras because each triple of touching squares encloses a right triangle, in a configuration traditionally used to represent the Pythagorean theorem. \n\n;Task\nConstruct a Pythagoras tree of order 7 using only vectors (no rotation or trigonometric functions).\n\n;Related tasks\n* Fractal tree\n\n", "solution": "from turtle import goto, pu, pd, color, done\n\ndef level(ax, ay, bx, by, depth=0):\n    if depth > 0:\n        dx,dy = bx-ax, ay-by\n        x3,y3 = bx-dy, by-dx\n        x4,y4 = ax-dy, ay-dx\n        x5,y5 = x4 + (dx - dy)/2, y4 - (dx + dy)/2\n        goto(ax, ay), pd()\n        for x, y in ((bx, by), (x3, y3), (x4, y4), (ax, ay)):\n            goto(x, y)\n        pu()\n        level(x4,y4, x5,y5, depth - 1)\n        level(x5,y5, x3,y3, depth - 1)\n\nif __name__ == '__main__':\n    color('red', 'yellow')\n    pu()\n    level(-100, 500, 100, 500, depth=8)\n    done()"}
{"title": "Pythagorean quadruples", "language": "Python from Julia", "task": "One form of   '''Pythagorean quadruples'''   is   (for positive integers   '''a''',   '''b''',   '''c''',   and   '''d'''): \n\n\n::::::::    a2   +   b2   +   c2     =     d2 \n\n\nAn example:\n\n::::::::    22   +   32   +   62     =     72 \n\n::::: which is:\n\n::::::::    4    +   9    +   36     =     49 \n\n\n;Task:\n\nFor positive integers up   '''2,200'''   (inclusive),   for all values of   '''a''',  \n'''b''',   '''c''',   and   '''d''', \nfind   (and show here)   those values of   '''d'''   that   ''can't''   be represented.\n\nShow the values of   '''d'''   on one line of output   (optionally with a title).\n\n\n;Related tasks:\n*   [[Euler's sum of powers conjecture]]. \n*   [[Pythagorean triples]].\n\n\n;Reference:\n:*   the Wikipedia article:   Pythagorean quadruple.\n\n", "solution": "def quad(top=2200):\n    r = [False] * top\n    ab = [False] * (top * 2)**2\n    for a in range(1, top):\n        for b in range(a, top):\n            ab[a * a + b * b] = True\n    s = 3\n    for c in range(1, top):\n        s1, s, s2 = s, s + 2, s + 2\n        for d in range(c + 1, top):\n            if ab[s1]:\n                r[d] = True\n            s1 += s2\n            s2 += 2\n    return [i for i, val in enumerate(r) if not val and i]\n    \nif __name__ == '__main__':\n    n = 2200\n    print(f\"Those values of d in 1..{n} that can't be represented: {quad(n)}\")"}
{"title": "Pythagorean quadruples", "language": "Python from AppleScript", "task": "One form of   '''Pythagorean quadruples'''   is   (for positive integers   '''a''',   '''b''',   '''c''',   and   '''d'''): \n\n\n::::::::    a2   +   b2   +   c2     =     d2 \n\n\nAn example:\n\n::::::::    22   +   32   +   62     =     72 \n\n::::: which is:\n\n::::::::    4    +   9    +   36     =     49 \n\n\n;Task:\n\nFor positive integers up   '''2,200'''   (inclusive),   for all values of   '''a''',  \n'''b''',   '''c''',   and   '''d''', \nfind   (and show here)   those values of   '''d'''   that   ''can't''   be represented.\n\nShow the values of   '''d'''   on one line of output   (optionally with a title).\n\n\n;Related tasks:\n*   [[Euler's sum of powers conjecture]]. \n*   [[Pythagorean triples]].\n\n\n;Reference:\n:*   the Wikipedia article:   Pythagorean quadruple.\n\n", "solution": "'''Pythagorean Quadruples'''\n\nfrom itertools import islice, takewhile\n\n\n# unrepresentables :: () -> [Int]\ndef unrepresentables():\n    '''A non-finite stream of powers of two which can\n       not be represented as a Pythagorean quadruple.\n    '''\n    return merge(\n        powersOfTwo()\n    )(\n        5 * x for x in powersOfTwo()\n    )\n\n\n# powersOfTwo :: Gen [Int]\ndef powersOfTwo():\n    '''A non-finite stream of successive powers of two.\n    '''\n    def double(x):\n        return 2 * x\n\n    return iterate(double)(1)\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''For positive integers up to 2,200 (inclusive)\n    '''\n    def p(x):\n        return 2200 >= x\n\n    print(\n        list(\n            takewhile(p, unrepresentables())\n        )\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated\n       applications of f to x.\n    '''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return go\n\n\n# merge :: Gen [Int] -> Gen [Int] -> Gen [Int]\ndef merge(ga):\n    '''An ordered stream of values drawn from two\n       other ordered streams.\n    '''\n    def go(gb):\n        def f(ma, mb):\n            a, b = ma, mb\n            while a and b:\n                ta, tb = a, b\n                if ta[0] < tb[0]:\n                    yield ta[0]\n                    a = uncons(ta[1])\n                else:\n                    yield tb[0]\n                    b = uncons(tb[1])\n        return f(uncons(ga), uncons(gb))\n    return go\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    def go(xs):\n        return (\n            xs[0:n]\n            if isinstance(xs, (list, tuple))\n            else list(islice(xs, n))\n        )\n    return go\n\n\n# uncons :: [a] -> Maybe (a, [a])\ndef uncons(xs):\n    '''The deconstruction of a non-empty list\n       (or generator stream) into two parts:\n       a head value, and the remaining values.\n    '''\n    if isinstance(xs, list):\n        return (xs[0], xs[1:]) if xs else None\n    else:\n        nxt = take(1)(xs)\n        return (nxt[0], xs) if nxt else None\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Pythagorean triples", "language": "Python", "task": "A Pythagorean triple is defined as three positive integers (a, b, c) where a < b < c, and a^2+b^2=c^2.\n\nThey are called primitive triples if a, b, c are co-prime, that is, if their pairwise greatest common divisors {\\rm gcd}(a, b) = {\\rm gcd}(a, c) = {\\rm gcd}(b, c) = 1. \n\nBecause of their relationship through the Pythagorean theorem, a, b, and c are co-prime if a and b are co-prime ({\\rm gcd}(a, b) = 1).   \n\nEach triple forms the length of the sides of a right triangle, whose perimeter is P=a+b+c.\n\n\n;Task:\nThe task is to determine how many Pythagorean triples there are with a perimeter no larger than 100 and the number of these that are primitive.\n\n\n;Extra credit: \nDeal with large values.   Can your program handle a maximum perimeter of 1,000,000?   What about 10,000,000?   100,000,000?\n\nNote: the extra credit is not for you to demonstrate how fast your language is compared to others;   you need a proper algorithm to solve them in a timely manner.\n\n\n;Related tasks:\n*   [[Euler's sum of powers conjecture]]  \n*   [[List comprehensions]]\n*   [[Pythagorean quadruples]] \n\n", "solution": "from fractions import gcd\n\n\ndef pt1(maxperimeter=100):\n    '''\n# Naive method\n    '''\n    trips = []\n    for a in range(1, maxperimeter):\n        aa = a*a\n        for b in range(a, maxperimeter-a+1):\n            bb = b*b\n            for c in range(b, maxperimeter-b-a+1):\n                cc = c*c\n                if a+b+c > maxperimeter or cc > aa + bb: break\n                if aa + bb == cc:\n                    trips.append((a,b,c, gcd(a, b) == 1))\n    return trips\n\ndef pytrip(trip=(3,4,5),perim=100, prim=1):\n    a0, b0, c0 = a, b, c = sorted(trip)\n    t, firstprim = set(), prim>0\n    while a + b + c <= perim:\n        t.add((a, b, c, firstprim>0))\n        a, b, c, firstprim = a+a0, b+b0, c+c0, False\n    #\n    t2 = set()\n    for a, b, c, firstprim in t:\n        a2, a5, b2, b5, c2, c3, c7 = a*2, a*5, b*2, b*5, c*2, c*3, c*7\n        if  a5 - b5 + c7 <= perim:\n            t2 |= pytrip(( a - b2 + c2,  a2 - b + c2,  a2 - b2 + c3), perim, firstprim)\n        if  a5 + b5 + c7 <= perim:\n            t2 |= pytrip(( a + b2 + c2,  a2 + b + c2,  a2 + b2 + c3), perim, firstprim)\n        if -a5 + b5 + c7 <= perim:\n            t2 |= pytrip((-a + b2 + c2, -a2 + b + c2, -a2 + b2 + c3), perim, firstprim)\n    return t | t2\n\ndef pt2(maxperimeter=100):\n    '''\n# Parent/child relationship method:\n# http://en.wikipedia.org/wiki/Formulas_for_generating_Pythagorean_triples#XI.\n    '''\n    trips = pytrip((3,4,5), maxperimeter, 1)\n    return trips\n\ndef printit(maxperimeter=100, pt=pt1):\n    trips = pt(maxperimeter)\n    print(\"  Up to a perimeter of %i there are %i triples, of which %i are primitive\"\n          % (maxperimeter,\n             len(trips),\n             len([prim for a,b,c,prim in trips if prim])))\n  \nfor algo, mn, mx in ((pt1, 250, 2500), (pt2, 500, 20000)):\n    print(algo.__doc__)\n    for maxperimeter in range(mn, mx+1, mn):\n        printit(maxperimeter, algo)\n"}
{"title": "Quaternion type", "language": "Python", "task": "complex numbers.\n\nA complex number has a real and complex part,   sometimes written as     a + bi,  \nwhere     a     and     b     stand for real numbers, and     i     stands for the square root of minus 1.\n\nAn example of a complex number might be     -3 + 2i,     \nwhere the real part,     a     is     '''-3.0'''     and the complex part,     b     is     '''+2.0'''.   \n\nA quaternion has one real part and ''three'' imaginary parts,     i,       j,     and     k.   \n\nA quaternion might be written as     a + bi + cj + dk.  \n\nIn the quaternion numbering system:\n:::*     ii = jj = kk = ijk = -1,         or more simply,\n:::*     ii  = jj  = kk  = ijk   = -1.  \n\nThe order of multiplication is important, as, in general, for two quaternions:\n::::     q1     and     q2:         q1q2  q2q1.  \n\nAn example of a quaternion might be     1 +2i +3j +4k  \n\nThere is a list form of notation where just the numbers are shown and the imaginary multipliers    i,       j,     and     k     are assumed by position. \n\nSo the example above would be written as     (1, 2, 3, 4)   \n\n\n;Task:\nGiven the three quaternions and their components:  \n    q  = (1, 2, 3, 4) = (a,  b,  c,  d)\n    q1 = (2, 3, 4, 5) = (a1, b1, c1, d1)\n    q2 = (3, 4, 5, 6) = (a2, b2, c2, d2) \nAnd a wholly real number     r = 7.  \n\n\nCreate functions   (or classes)   to perform simple maths with quaternions including computing:\n# The norm of a quaternion: = \\sqrt{a^2 + b^2 + c^2 + d^2}  \n# The negative of a quaternion:    = (-a, -b, -c, -d)  \n# The conjugate of a quaternion:    = ( a, -b, -c, -d)  \n# Addition of a real number     r     and a quaternion     q:      r + q = q + r = (a+r, b, c, d)  \n# Addition of two quaternions:    q1 + q2 = (a1+a2, b1+b2, c1+c2, d1+d2)  \n# Multiplication of a real number and a quaternion:    qr = rq = (ar, br, cr, dr)  \n# Multiplication of two quaternions     q1     and   q2     is given by:    ( a1a2 - b1b2 - c1c2 - d1d2,      a1b2 + b1a2 + c1d2 - d1c2,      a1c2 - b1d2 + c1a2 + d1b2,      a1d2 + b1c2 - c1b2 + d1a2 )  \n# Show that, for the two quaternions     q1     and     q2:  q1q2  q2q1  \n\nIf a language has built-in support for quaternions, then use it.\n\n\n;C.f.:\n*   [[Vector products]]\n*   On Quaternions;   or on a new System of Imaginaries in Algebra.   By Sir William Rowan Hamilton LL.D, P.R.I.A., F.R.A.S., Hon. M. R. Soc. Ed. and Dub., Hon. or Corr. M. of the Royal or Imperial Academies of St. Petersburgh, Berlin, Turin and Paris, Member of the American Academy of Arts and Sciences, and of other Scientific Societies at Home and Abroad, Andrews' Prof. of Astronomy in the University of Dublin, and Royal Astronomer of Ireland.\n\n", "solution": "from collections import namedtuple\nimport math\n\nclass Q(namedtuple('Quaternion', 'real, i, j, k')):\n    'Quaternion type: Q(real=0.0, i=0.0, j=0.0, k=0.0)' \n\n    __slots__ = () \n\n    def __new__(_cls, real=0.0, i=0.0, j=0.0, k=0.0):\n        'Defaults all parts of quaternion to zero'\n        return super().__new__(_cls, float(real), float(i), float(j), float(k))\n\n    def conjugate(self):\n        return Q(self.real, -self.i, -self.j, -self.k)\n\n    def _norm2(self):\n        return sum( x*x for x in self)\n\n    def norm(self):\n        return math.sqrt(self._norm2())\n\n    def reciprocal(self):\n        n2 = self._norm2()\n        return Q(*(x / n2 for x in self.conjugate())) \n\n    def __str__(self):\n        'Shorter form of Quaternion as string'\n        return 'Q(%g, %g, %g, %g)' % self\n\n    def __neg__(self):\n        return Q(-self.real, -self.i, -self.j, -self.k)\n\n    def __add__(self, other):\n        if type(other) == Q:\n            return Q( *(s+o for s,o in zip(self, other)) )\n        try:\n            f = float(other)\n        except:\n            return NotImplemented\n        return Q(self.real + f, self.i, self.j, self.k)\n\n    def __radd__(self, other):\n        return Q.__add__(self, other)\n\n    def __mul__(self, other):\n        if type(other) == Q:\n            a1,b1,c1,d1 = self\n            a2,b2,c2,d2 = other\n            return Q(\n                 a1*a2 - b1*b2 - c1*c2 - d1*d2,\n                 a1*b2 + b1*a2 + c1*d2 - d1*c2,\n                 a1*c2 - b1*d2 + c1*a2 + d1*b2,\n                 a1*d2 + b1*c2 - c1*b2 + d1*a2 )\n        try:\n            f = float(other)\n        except:\n            return NotImplemented\n        return Q(self.real * f, self.i * f, self.j * f, self.k * f)\n\n    def __rmul__(self, other):\n        return Q.__mul__(self, other)\n\n    def __truediv__(self, other):\n        if type(other) == Q:\n            return self.__mul__(other.reciprocal())\n        try:\n            f = float(other)\n        except:\n            return NotImplemented\n        return Q(self.real / f, self.i / f, self.j / f, self.k / f)\n\n    def __rtruediv__(self, other):\n        return other * self.reciprocal()\n\n    __div__, __rdiv__ = __truediv__, __rtruediv__\n\nQuaternion = Q       \n\nq  = Q(1, 2, 3, 4)\nq1 = Q(2, 3, 4, 5)\nq2 = Q(3, 4, 5, 6)\nr  = 7"}
{"title": "Quine", "language": "Python 2.x and 3.x", "task": "A quine is a self-referential program that can, \nwithout any external access, output its own source. \n\n\nA   '''quine'''   (named after Willard Van Orman Quine)   is also known as:\n::*   ''self-reproducing automata''        (1972)\n::*   ''self-replicating program''                             or   ''self-replicating computer program''\n::*   ''self-reproducing program''                                  or   ''self-reproducing computer program''\n::*   ''self-copying program''                       or   ''self-copying computer program''\n\n\n\nIt is named after the philosopher and logician \nwho studied self-reference and quoting in natural language, \nas for example in the paradox \"'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation.\"\n\n\"Source\" has one of two meanings. It can refer to the text-based program source.  \nFor languages in which program source is represented as a data structure, \"source\" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.\n\nThe usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once ''quoted'' in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.\n\n\n;Task:\nWrite a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out.  Empty programs producing no output are not allowed.\n\nThere are several difficulties that one runs into when writing a quine, mostly dealing with quoting:\n* Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.\n** Some languages have a function for getting the \"source code representation\" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.\n** Another solution is to construct the quote character from its [[character code]], without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.\n* Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. \"\\n\"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.\n** If the language has a way of getting the \"source code representation\", it usually handles the escaping of characters, so this is not a problem.\n** Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.\n** Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.\n\n\n'''Next to the Quines presented here, many other versions can be found on the Quine page.'''\n\n\n;Related task:\n:*   print itself.\n\n", "solution": "import sys,inspect;sys.stdout.write(inspect.getsource(inspect.currentframe()))"}
{"title": "Quine", "language": "Python 3.8+", "task": "A quine is a self-referential program that can, \nwithout any external access, output its own source. \n\n\nA   '''quine'''   (named after Willard Van Orman Quine)   is also known as:\n::*   ''self-reproducing automata''        (1972)\n::*   ''self-replicating program''                             or   ''self-replicating computer program''\n::*   ''self-reproducing program''                                  or   ''self-reproducing computer program''\n::*   ''self-copying program''                       or   ''self-copying computer program''\n\n\n\nIt is named after the philosopher and logician \nwho studied self-reference and quoting in natural language, \nas for example in the paradox \"'Yields falsehood when preceded by its quotation' yields falsehood when preceded by its quotation.\"\n\n\"Source\" has one of two meanings. It can refer to the text-based program source.  \nFor languages in which program source is represented as a data structure, \"source\" may refer to the data structure: quines in these languages fall into two categories: programs which print a textual representation of themselves, or expressions which evaluate to a data structure which is equivalent to that expression.\n\nThe usual way to code a quine works similarly to this paradox: The program consists of two identical parts, once as plain code and once ''quoted'' in some way (for example, as a character string, or a literal data structure). The plain code then accesses the quoted code and prints it out twice, once unquoted and once with the proper quotation marks added. Often, the plain code and the quoted code have to be nested.\n\n\n;Task:\nWrite a program that outputs its own source code in this way. If the language allows it, you may add a variant that accesses the code directly. You are not allowed to read any external files with the source code. The program should also contain some sort of self-reference, so constant expressions which return their own value which some top-level interpreter will print out.  Empty programs producing no output are not allowed.\n\nThere are several difficulties that one runs into when writing a quine, mostly dealing with quoting:\n* Part of the code usually needs to be stored as a string or structural literal in the language, which needs to be quoted somehow. However, including quotation marks in the string literal itself would be troublesome because it requires them to be escaped, which then necessitates the escaping character (e.g. a backslash) in the string, which itself usually needs to be escaped, and so on.\n** Some languages have a function for getting the \"source code representation\" of a string (i.e. adds quotation marks, etc.); in these languages, this can be used to circumvent the quoting problem.\n** Another solution is to construct the quote character from its [[character code]], without having to write the quote character itself. Then the character is inserted into the string at the appropriate places. The ASCII code for double-quote is 34, and for single-quote is 39.\n* Newlines in the program may have to be reproduced as newlines in the string, which usually requires some kind of escape sequence (e.g. \"\\n\"). This causes the same problem as above, where the escaping character needs to itself be escaped, etc.\n** If the language has a way of getting the \"source code representation\", it usually handles the escaping of characters, so this is not a problem.\n** Some languages allow you to have a string literal that spans multiple lines, which embeds the newlines into the string without escaping.\n** Write the entire program on one line, for free-form languages (as you can see for some of the solutions here, they run off the edge of the screen), thus removing the need for newlines. However, this may be unacceptable as some languages require a newline at the end of the file; and otherwise it is still generally good style to have a newline at the end of a file. (The task is not clear on whether a newline is required at the end of the file.) Some languages have a print statement that appends a newline; which solves the newline-at-the-end issue; but others do not.\n\n\n'''Next to the Quines presented here, many other versions can be found on the Quine page.'''\n\n\n;Related task:\n:*   print itself.\n\n", "solution": "data = (\n\t'ZGF0YSA9ICgKCSc=',\n\t'JywKCSc=',\n\t'JwopCnByZWZpeCwgc2VwYXJhdG9yLCBzdWZmaXggPSAoZC5kZWNvZGUoJ2Jhc2U2NCcpIGZvciBkIGluIGRhdGEpCnByaW50IHByZWZpeCArIGRhdGFbMF0gKyBzZXBhcmF0b3IgKyBkYXRhWzFdICsgc2VwYXJhdG9yICsgZGF0YVsyXSArIHN1ZmZpeA=='\n)\nprefix, separator, suffix = (d.decode('base64') for d in data)\nprint prefix + data[0] + separator + data[1] + separator + data[2] + suffix"}
{"title": "RIPEMD-160", "language": "Python", "task": "'''RIPEMD-160''' is another hash function; it computes a 160-bit message digest.\n\nThere is a RIPEMD-160 home page, with test vectors and pseudocode for RIPEMD-160. \nFor padding the message, RIPEMD-160 acts like [[MD4]] (RFC 1320).\n\nFind the RIPEMD-160 message digest of a string of [[octet]]s. \nUse the ASCII encoded string \"Rosetta Code\". \nYou may either call an RIPEMD-160 library, or implement RIPEMD-160 in your language.\n\n", "solution": "Python 3.3.0 (v3.3.0:bd8afb90ebf2, Sep 29 2012, 10:57:17) [MSC v.1600 64 bit (AMD64)] on win32\nType \"copyright\", \"credits\" or \"license()\" for more information.\n>>> import hashlib\n>>> h = hashlib.new('ripemd160')\n>>> h.update(b\"Rosetta Code\")\n>>> h.hexdigest()\n'b3be159860842cebaa7174c8fff0aa9e50a5199f'\n>>> "}
{"title": "RPG attributes generator", "language": "Python", "task": "'''RPG'''   =   Role Playing Game.\n\n\n\nYou're running a tabletop RPG, and your players are creating characters.\n\nEach character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.\n\nOne way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.\n\nSome players like to assign values to their attributes in the order they're rolled.\n\nTo ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:\n\n* The total of all character attributes must be at least 75.\n* At least two of the attributes must be at least 15.\n\nHowever, this can require a lot of manual dice rolling. A programatic solution would be much faster.\n\n\n;Task:\nWrite a program that:\n# Generates 4 random, whole values between 1 and 6.\n# Saves the sum of the 3 largest values.\n# Generates a total of 6 values this way.\n# Displays the total, and all 6 values once finished.\n\n* The order in which each value was generated must be preserved.\n* The total of all 6 values must be at least 75.\n* At least 2 of the values must be 15 or more.\n\n", "solution": "import random\nrandom.seed()\nattributes_total = 0\ncount = 0\n\nwhile attributes_total < 75 or count < 2:\n    attributes = []\n\n    for attribute in range(0, 6):\n        rolls = []\n        \n        for roll in range(0, 4):\n            result = random.randint(1, 6)\n            rolls.append(result)\n        \n        sorted_rolls = sorted(rolls)\n        largest_3 = sorted_rolls[1:]\n        rolls_total = sum(largest_3)\n        \n        if rolls_total >= 15:\n            count += 1\n        \n        attributes.append(rolls_total)\n\n    attributes_total = sum(attributes)\n    \nprint(attributes_total, attributes)"}
{"title": "RPG attributes generator", "language": "Python 3.7", "task": "'''RPG'''   =   Role Playing Game.\n\n\n\nYou're running a tabletop RPG, and your players are creating characters.\n\nEach character has six core attributes: strength, dexterity, constitution, intelligence, wisdom, and charisma.\n\nOne way of generating values for these attributes is to roll four, 6-sided dice (d6) and sum the three highest rolls, discarding the lowest roll.\n\nSome players like to assign values to their attributes in the order they're rolled.\n\nTo ensure generated characters don't put players at a disadvantage, the following requirements must be satisfied:\n\n* The total of all character attributes must be at least 75.\n* At least two of the attributes must be at least 15.\n\nHowever, this can require a lot of manual dice rolling. A programatic solution would be much faster.\n\n\n;Task:\nWrite a program that:\n# Generates 4 random, whole values between 1 and 6.\n# Saves the sum of the 3 largest values.\n# Generates a total of 6 values this way.\n# Displays the total, and all 6 values once finished.\n\n* The order in which each value was generated must be preserved.\n* The total of all 6 values must be at least 75.\n* At least 2 of the values must be 15 or more.\n\n", "solution": "'''RPG Attributes Generator'''\n\nfrom itertools import islice\nfrom random import randint\nfrom operator import eq\n\n\n# heroes :: Gen IO [(Int, Int, Int, Int, Int, Int)]\ndef heroes(p):\n    '''Non-finite list of heroes matching\n       the requirements of predicate p.\n    '''\n    while True:\n        yield tuple(\n            until(p)(character)([])\n        )\n\n\n# character :: () -> IO [Int]\ndef character(_):\n    '''A random character with six\n       integral attributes.\n    '''\n    return [\n        sum(sorted(map(\n            randomRInt(1)(6),\n            enumFromTo(1)(4)\n        ))[1:])\n        for _ in enumFromTo(1)(6)\n    ]\n\n\n# ------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Test :: Sample of 10'''\n\n    # seventyFivePlusWithTwo15s :: [Int] -> Bool\n    def seventyFivePlusIncTwo15s(xs):\n        '''Sums to 75 or more,\n           and includes at least two 15s.\n        '''\n        return 75 <= sum(xs) and (\n            1 < len(list(filter(curry(eq)(15), xs)))\n        )\n\n    print('A sample of 10:\\n')\n    print(unlines(\n        str(sum(x)) + ' -> ' + str(x) for x\n        in take(10)(heroes(\n            seventyFivePlusIncTwo15s\n        ))\n    ))\n\n\n# ------------------------- GENERIC -------------------------\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    '''A curried function derived\n       from an uncurried function.\n    '''\n    return lambda x: lambda y: f(x, y)\n\n\n# enumFromTo :: Int -> Int -> [Int]\ndef enumFromTo(m):\n    '''Enumeration of integer values [m..n]'''\n    return lambda n: range(m, 1 + n)\n\n\n# randomRInt :: Int -> Int -> IO () -> Int\ndef randomRInt(m):\n    '''The return value of randomRInt is itself\n       a function. The returned function, whenever\n       called, yields a a new pseudo-random integer\n       in the range [m..n].\n    '''\n    return lambda n: lambda _: randint(m, n)\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, (list, tuple))\n        else list(islice(xs, n))\n    )\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string formed by the intercalation\n       of a list of strings with the newline character.\n    '''\n    return '\\n'.join(xs)\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f, x):\n        v = x\n        while not p(v):\n            v = f(v)\n        return v\n    return lambda f: lambda x: go(f, x)\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Random Latin squares", "language": "Python", "task": "A Latin square of size n is an arrangement of n symbols  in an n-by-n square in such a way that each row and column has each symbol appearing exactly once.\nFor the purposes of this task, a random Latin square of size n is a Latin square constructed or generated by a probabilistic procedure such that the probability of any particular Latin square of size n being produced is non-zero.\n\n;Example n=4 randomised Latin square:\n0 2 3 1\n2 1 0 3\n3 0 1 2\n1 3 2 0\n\n;Task:\n# Create a function/routine/procedure/method/... that given n generates a randomised Latin square of size n.\n# Use the function to generate ''and show here'', two randomly generated squares of size 5.\n\n;Note: \nStrict ''uniformity'' in the random generation is a hard problem and '''not''' a requirement of the task.\n\n;Related tasks:\n* [[Latin Squares in reduced form/Randomizing using Jacobson and Matthews' Technique]]\n* [[Latin Squares in reduced form]]\n\n;Reference:\n* Wikipedia: Latin square\n* OEIS: A002860\n\n", "solution": "from random import choice, shuffle\nfrom copy import deepcopy\n\ndef rls(n):\n    if n <= 0:\n        return []\n    else:\n        symbols = list(range(n))\n        square = _rls(symbols)\n        return _shuffle_transpose_shuffle(square)\n\n\ndef _shuffle_transpose_shuffle(matrix):\n    square = deepcopy(matrix)\n    shuffle(square)\n    trans = list(zip(*square))\n    shuffle(trans)\n    return trans\n\n\ndef _rls(symbols):\n    n = len(symbols)\n    if n == 1:\n        return [symbols]\n    else:\n        sym = choice(symbols)\n        symbols.remove(sym)\n        square = _rls(symbols)\n        square.append(square[0].copy())\n        for i in range(n):\n            square[i].insert(i, sym)\n        return square\n\ndef _to_text(square):\n    if square:\n        width = max(len(str(sym)) for row in square for sym in row)\n        txt = '\\n'.join(' '.join(f\"{sym:>{width}}\" for sym in row)\n                        for row in square)\n    else:\n        txt = ''\n    return txt\n\ndef _check(square):\n    transpose = list(zip(*square))\n    assert _check_rows(square) and _check_rows(transpose), \\\n        \"Not a Latin square\"\n\ndef _check_rows(square):\n    if not square:\n        return True\n    set_row0 = set(square[0])\n    return all(len(row) == len(set(row)) and set(row) == set_row0\n               for row in square)\n\n\nif __name__ == '__main__':\n    for i in [3, 3,  5, 5, 12]:\n        square = rls(i)\n        print(_to_text(square))\n        _check(square)\n        print()"}
{"title": "Random number generator (device)", "language": "Python", "task": "If your system has a means to generate random numbers involving not only a software algorithm   (like the /dev/urandom devices in Unix),   then:\n\nshow how to obtain a random 32-bit number from that mechanism.\n\n;Related task\n* [[Random_number_generator_(included)]]\n\n", "solution": "import random\nrand = random.SystemRandom()\nrand.randint(1,10)"}
{"title": "Random number generator (included)", "language": "Python", "task": "The task is to:\n: State the type of random number generator algorithm used in a language's built-in random number generator. If the language or its immediate libraries don't provide a random number generator, skip this task.\n: If possible, give a link to a wider explanation of the algorithm used.\n\nNote: the task is ''not'' to create an RNG, but to report on the languages in-built RNG that would be the most likely RNG used.\n\nThe main types of pseudo-random number generator (Mersenne twister generator is a subclass). The last main type is where the output of one of the previous ones (typically a Mersenne twister) is fed through a [[cryptographic hash function]] to maximize unpredictability of individual bits. \n\nNote that neither LCGs nor GFSRs should be used for the most demanding applications (cryptography) without additional steps.\n\n", "solution": "Python uses the [[wp:Mersenne twister|Mersenne twister]] algorithm accessed via the built-in [http://docs.python.org/library/random.html random module].\n\n"}
{"title": "Random sentence from book", "language": "Python", "task": "* Read in the book \"The War of the Worlds\", by H. G. Wells.\n* Skip to the start of the book, proper.\n* Remove extraneous punctuation, but keep at least sentence-ending punctuation characters . ! and ?\n* Keep account of what words follow words and how many times it is seen, (treat sentence terminators as words too).\n* Keep account of what words follow ''two'' words and how many times it is seen, (again treating sentence terminators as words too).\n* Assume that a sentence starts with a not to be shown full-stop character then ''use a weighted random choice'' of the possible words that may follow a full-stop to add to the sentence.\n* Then repeatedly add words to the sentence based on weighted random choices of what words my follow the last two words to extend the sentence.\n* Stop after adding a sentence ending punctuation character.\n* Tidy and then print the sentence.\n\n\nShow examples of random sentences generated.\n\n;Related task\n* [[Markov_chain_text_generator]]\n\n\n", "solution": "from urllib.request import urlopen\nimport re\nfrom string import punctuation\nfrom collections import Counter, defaultdict\nimport random\n\n\n# The War of the Worlds, by H. G. Wells\ntext_url = 'http://www.gutenberg.org/files/36/36-0.txt'\ntext_start = 'No one would have believed'\n\nsentence_ending = '.!?'\nsentence_pausing = ',;:'\n\ndef read_book(text_url, text_start) -> str:\n    with urlopen(text_url) as book:\n        text = book.read().decode('utf-8')\n    return text[text.index(text_start):]\n\ndef remove_punctuation(text: str, keep=sentence_ending+sentence_pausing)-> str:\n    \"Remove punctuation, keeping some\"\n    to_remove = ''.join(set(punctuation) - set(keep))\n    text = text.translate(str.maketrans(to_remove, ' ' * len(to_remove))).strip()\n    text = re.sub(fr\"[^a-zA-Z0-9{keep}\\n ]+\", ' ', text)\n    # Remove duplicates and put space around remaining punctuation\n    if keep:\n        text = re.sub(f\"([{keep}])+\", r\" \\1 \", text).strip()\n    if text[-1] not in sentence_ending:\n        text += ' .'\n    return text.lower()\n\ndef word_follows_words(txt_with_pauses_and_endings):\n    \"return dict of freq of words following one/two words\"\n    words = ['.'] + txt_with_pauses_and_endings.strip().split()\n\n    # count of what word follows this\n    word2next = defaultdict(lambda :defaultdict(int))\n    word2next2 = defaultdict(lambda :defaultdict(int))\n    for lh, rh in zip(words, words[1:]):\n        word2next[lh][rh] += 1\n    for lh, mid, rh in zip(words, words[1:], words[2:]):\n        word2next2[(lh, mid)][rh] += 1\n\n    return dict(word2next), dict(word2next2)\n\ndef gen_sentence(word2next, word2next2) -> str:\n\n    s = ['.']\n    s += random.choices(*zip(*word2next[s[-1]].items()))\n    while True:\n        s += random.choices(*zip(*word2next2[(s[-2], s[-1])].items()))\n        if s[-1] in sentence_ending:\n            break\n\n    s  = ' '.join(s[1:]).capitalize()\n    s = re.sub(fr\" ([{sentence_ending+sentence_pausing}])\", r'\\1', s)\n    s = re.sub(r\" re\\b\", \"'re\", s)\n    s = re.sub(r\" s\\b\", \"'s\", s)\n    s = re.sub(r\"\\bi\\b\", \"I\", s)\n\n    return s\n\nif __name__ == \"__main__\":\n    txt_with_pauses_and_endings = remove_punctuation(read_book(text_url, text_start))\n    word2next, word2next2 = word_follows_words(txt_with_pauses_and_endings)\n    #%%\n    sentence = gen_sentence(word2next, word2next2)\n    print(sentence)"}
{"title": "Range consolidation", "language": "Python", "task": "Define a range of numbers   '''R''',   with bounds   '''b0'''   and   '''b1'''   covering all numbers ''between and including both bounds''. \n\n\nThat range can be shown as:\n::::::::: '''[b0, b1]'''\n::::::::    or equally as:\n::::::::: '''[b1, b0]'''\n\n\nGiven two ranges, the act of consolidation between them compares the two ranges:\n*   If one range covers all of the other then the result is that encompassing range.\n*   If the ranges touch or intersect then the result is   ''one''   new single range covering the overlapping ranges.\n*   Otherwise the act of consolidation is to return the two non-touching ranges.\n\n\nGiven   '''N'''   ranges where   '''N > 2'''   then the result is the same as repeatedly replacing all combinations of two ranges by their consolidation until no further consolidation between range pairs is possible. \n\nIf   '''N < 2'''   then range consolidation has no strict meaning and the input can be returned. \n\n\n;Example 1:\n:   Given the two ranges   '''[1, 2.5]'''   and   '''[3, 4.2]'''   then \n:   there is no common region between the ranges and the result is the same as the input.\n\n\n;Example 2:\n:   Given the two ranges   '''[1, 2.5]'''   and   '''[1.8, 4.7]'''   then \n:   there is :   an overlap   '''[2.5, 1.8]'''   between the ranges and \n:   the result is the single range   '''[1, 4.7]'''.  \n:   Note that order of bounds in a range is not (yet) stated.\n\n\n;Example 3:\n:   Given the two ranges   '''[6.1, 7.2]'''   and   '''[7.2, 8.3]'''   then \n:   they touch at   '''7.2'''   and \n:   the result is the single range   '''[6.1, 8.3]'''. \n\n\n;Example 4:\n:   Given the three ranges   '''[1, 2]'''   and   '''[4, 8]'''   and   '''[2, 5]''' \n:   then there is no intersection of the ranges   '''[1, 2]'''   and   '''[4, 8]''' \n:   but the ranges   '''[1, 2]'''   and   '''[2, 5]'''   overlap and \n:   consolidate to produce the range   '''[1, 5]'''. \n:   This range, in turn, overlaps the other range   '''[4, 8]''',   and \n:   so consolidates to the final output of the single range   '''[1, 8]'''.\n\n\n;Task:\nLet a normalized range display show the smaller bound to the left;   and show the\nrange with the smaller lower bound to the left of other ranges when showing multiple ranges.\n\nOutput the ''normalized'' result of applying consolidation to these five sets of ranges: \n            [1.1, 2.2]\n            [6.1, 7.2], [7.2, 8.3]\n            [4, 3], [2, 1]\n            [4, 3], [2, 1], [-1, -2], [3.9, 10]\n            [1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]  \nShow all output here.\n\n\n;See also:\n* [[Set consolidation]]\n* [[Set of real numbers]]\n\n", "solution": "def normalize(s):\n    return sorted(sorted(bounds) for bounds in s if bounds)\n\ndef consolidate(ranges):\n    norm = normalize(ranges)\n    for i, r1 in enumerate(norm):\n        if r1:\n            for r2 in norm[i+1:]:\n                if r2 and r1[-1] >= r2[0]:     # intersect?\n                    r1[:] = [r1[0], max(r1[-1], r2[-1])]\n                    r2.clear()\n    return [rnge for rnge in norm if rnge]\n\nif __name__ == '__main__':\n    for s in [\n            [[1.1, 2.2]],\n            [[6.1, 7.2], [7.2, 8.3]],\n            [[4, 3], [2, 1]],\n            [[4, 3], [2, 1], [-1, -2], [3.9, 10]],\n            [[1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]],\n            ]:\n        print(f\"{str(s)[1:-1]} => {str(consolidate(s))[1:-1]}\")\n"}
{"title": "Range consolidation", "language": "Python 3.7", "task": "Define a range of numbers   '''R''',   with bounds   '''b0'''   and   '''b1'''   covering all numbers ''between and including both bounds''. \n\n\nThat range can be shown as:\n::::::::: '''[b0, b1]'''\n::::::::    or equally as:\n::::::::: '''[b1, b0]'''\n\n\nGiven two ranges, the act of consolidation between them compares the two ranges:\n*   If one range covers all of the other then the result is that encompassing range.\n*   If the ranges touch or intersect then the result is   ''one''   new single range covering the overlapping ranges.\n*   Otherwise the act of consolidation is to return the two non-touching ranges.\n\n\nGiven   '''N'''   ranges where   '''N > 2'''   then the result is the same as repeatedly replacing all combinations of two ranges by their consolidation until no further consolidation between range pairs is possible. \n\nIf   '''N < 2'''   then range consolidation has no strict meaning and the input can be returned. \n\n\n;Example 1:\n:   Given the two ranges   '''[1, 2.5]'''   and   '''[3, 4.2]'''   then \n:   there is no common region between the ranges and the result is the same as the input.\n\n\n;Example 2:\n:   Given the two ranges   '''[1, 2.5]'''   and   '''[1.8, 4.7]'''   then \n:   there is :   an overlap   '''[2.5, 1.8]'''   between the ranges and \n:   the result is the single range   '''[1, 4.7]'''.  \n:   Note that order of bounds in a range is not (yet) stated.\n\n\n;Example 3:\n:   Given the two ranges   '''[6.1, 7.2]'''   and   '''[7.2, 8.3]'''   then \n:   they touch at   '''7.2'''   and \n:   the result is the single range   '''[6.1, 8.3]'''. \n\n\n;Example 4:\n:   Given the three ranges   '''[1, 2]'''   and   '''[4, 8]'''   and   '''[2, 5]''' \n:   then there is no intersection of the ranges   '''[1, 2]'''   and   '''[4, 8]''' \n:   but the ranges   '''[1, 2]'''   and   '''[2, 5]'''   overlap and \n:   consolidate to produce the range   '''[1, 5]'''. \n:   This range, in turn, overlaps the other range   '''[4, 8]''',   and \n:   so consolidates to the final output of the single range   '''[1, 8]'''.\n\n\n;Task:\nLet a normalized range display show the smaller bound to the left;   and show the\nrange with the smaller lower bound to the left of other ranges when showing multiple ranges.\n\nOutput the ''normalized'' result of applying consolidation to these five sets of ranges: \n            [1.1, 2.2]\n            [6.1, 7.2], [7.2, 8.3]\n            [4, 3], [2, 1]\n            [4, 3], [2, 1], [-1, -2], [3.9, 10]\n            [1, 3], [-6, -1], [-4, -5], [8, 2], [-6, -6]  \nShow all output here.\n\n\n;See also:\n* [[Set consolidation]]\n* [[Set of real numbers]]\n\n", "solution": "'''Range consolidation'''\n\nfrom functools import reduce\n\n\n# consolidated :: [(Float, Float)] -> [(Float, Float)]\ndef consolidated(xs):\n    '''A consolidated list of\n       [(Float, Float)] ranges.'''\n\n    def go(abetc, xy):\n        '''A copy of the accumulator abetc,\n           with its head range ab either:\n           1. replaced by or\n           2. merged with\n           the next range xy, or\n           with xy simply prepended.'''\n        if abetc:\n            a, b = abetc[0]\n            etc = abetc[1:]\n            x, y = xy\n            return [xy] + etc if y >= b else (   # ab replaced.\n                [(x, b)] + etc if y >= a else (  # xy + ab merged.\n                    [xy] + abetc                 # xy simply prepended.\n                )\n            )\n        else:\n            return [xy]\n\n    def tupleSort(ab):\n        a, b = ab\n        return ab if a <= b else (b, a)\n\n    return reduce(\n        go,\n        sorted(map(tupleSort, xs), reverse=True),\n        []\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Tests'''\n\n    print(\n        tabulated('Consolidation of numeric ranges:')(str)(str)(\n            consolidated\n        )([\n            [(1.1, 2.2)],\n            [(6.1, 7.2), (7.2, 8.3)],\n            [(4, 3), (2, 1)],\n            [(4, 3), (2, 1), (-1, -2), (3.9, 10)],\n            [(1, 3), (-6, -1), (-4, -5), (8, 2), (-6, -6)]\n        ])\n    )\n\n\n# GENERIC FUNCTIONS FOR DISPLAY ---------------------------\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# tabulated :: String -> (a -> String) ->\n#                        (b -> String) ->\n#                        (a -> b) -> [a] -> String\ndef tabulated(s):\n    '''Heading -> x display function -> fx display function ->\n          f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + '\\n'.join([\n            xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs\n        ])\n    return lambda xShow: lambda fxShow: (\n        lambda f: lambda xs: go(\n            xShow, fxShow, f, xs\n        )\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Range expansion", "language": "Python", "task": "{{:Range extraction/Format}}\n\n\n;Task:\nExpand the range description: \n   -6,-3--1,3-5,7-11,14,15,17-20 \nNote that the second element above, \nis the '''range from minus 3 to ''minus'' 1'''. \n\n\n;Related task:\n*   [[Range extraction]]\n\n", "solution": "def rangeexpand(txt):\n    lst = []\n    for r in txt.split(','):\n        if '-' in r[1:]:\n            r0, r1 = r[1:].split('-', 1)\n            lst += range(int(r[0] + r0), int(r1) + 1)\n        else:\n            lst.append(int(r))\n    return lst\n\nprint(rangeexpand('-6,-3--1,3-5,7-11,14,15,17-20'))"}
{"title": "Range expansion", "language": "Python 3.7", "task": "{{:Range extraction/Format}}\n\n\n;Task:\nExpand the range description: \n   -6,-3--1,3-5,7-11,14,15,17-20 \nNote that the second element above, \nis the '''range from minus 3 to ''minus'' 1'''. \n\n\n;Related task:\n*   [[Range extraction]]\n\n", "solution": "'''Range expansion'''\n\nfrom functools import (reduce)\n\n\n# ------------------- EXPANSION FUNCTION -------------------\n\n# rangeExpansion :: String -> [Int]\ndef rangeExpansion(s):\n    '''List of integers expanded from a\n       comma-delimited string of individual\n       numbers and hyphenated ranges.\n    '''\n    def go(a, x):\n        tpl = breakOn('-')(x[1:])\n        r = tpl[1]\n        return a + (\n            [int(x)] if not r\n            else enumFromTo(int(x[0] + tpl[0]))(\n                int(r[1:])\n            )\n        )\n    return reduce(go, s.split(','), [])\n\n\n# -------------------------- TEST --------------------------\ndef main():\n    '''Expansion test'''\n\n    print(\n        fTable(__doc__ + ':')(\n            lambda x: \"\\n'\" + str(x) + \"'\"\n        )(lambda x: '\\n\\n\\t' + showList(x))(\n            rangeExpansion\n        )([\n            '-6,-3--1,3-5,7-11,14,15,17-20'\n        ])\n    )\n\n\n# ------------------- GENERIC FUNCTIONS --------------------\n\n# breakOn :: String -> String -> (String, String)\ndef breakOn(needle):\n    '''A tuple of:\n       1. the prefix of haystack before needle,\n       2. the remainder of haystack, starting\n          with needle.\n    '''\n    def go(haystack):\n        xs = haystack.split(needle)\n        return (xs[0], haystack[len(xs[0]):]) if (\n            1 < len(xs)\n        ) else (haystack, '')\n    return lambda haystack: go(haystack) if (\n        needle\n    ) else None\n\n\n# enumFromTo :: Int -> Int -> [Int]\ndef enumFromTo(m):\n    '''Enumeration of integer values [m..n]\n    '''\n    return lambda n: list(range(m, 1 + n))\n\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> \n       fx display function -> f -> xs -> tabular string.\n    '''\n    def gox(xShow):\n        def gofx(fxShow):\n            def gof(f):\n                def goxs(xs):\n                    ys = [xShow(x) for x in xs]\n                    w = max(map(len, ys))\n\n                    def arrowed(x, y):\n                        return y.rjust(w, ' ') + ' -> ' + (\n                            fxShow(f(x))\n                        )\n                    return s + '\\n' + '\\n'.join(\n                        map(arrowed, xs, ys)\n                    )\n                return goxs\n            return gof\n        return gofx\n    return gox\n\n\n# showList :: [a] -> String\ndef showList(xs):\n    '''Stringification of a list.\n    '''\n    return '[' + ','.join(str(x) for x in xs) + ']'\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Range extraction", "language": "Python", "task": "{{:Range extraction/Format}}\n\n;Task:\n* Create a function that takes a list of integers in increasing order and returns a correctly formatted string in the range format. \n* Use the function to compute and print the range formatted version of the following ordered list of integers. (The correct answer is: 0-2,4,6-8,11,12,14-25,27-33,35-39).\n\n     0,  1,  2,  4,  6,  7,  8, 11, 12, 14,\n    15, 16, 17, 18, 19, 20, 21, 22, 23, 24,\n    25, 27, 28, 29, 30, 31, 32, 33, 35, 36,\n    37, 38, 39\n* Show the output of your program.\n\n\n;Related task:\n*   [[Range expansion]]\n\n", "solution": "def range_extract(lst):\n    'Yield 2-tuple ranges or 1-tuple single elements from list of increasing ints'\n    lenlst = len(lst)\n    i = 0\n    while i< lenlst:\n        low = lst[i]\n        while i <lenlst-1 and lst[i]+1 == lst[i+1]: i +=1\n        hi = lst[i]\n        if   hi - low >= 2:\n            yield (low, hi)\n        elif hi - low == 1:\n            yield (low,)\n            yield (hi,)\n        else:\n            yield (low,)\n        i += 1\n\ndef printr(ranges):\n    print( ','.join( (('%i-%i' % r) if len(r) == 2 else '%i' % r)\n                     for r in ranges ) )\n\nif __name__ == '__main__':\n    for lst in [[-8, -7, -6, -3, -2, -1, 0, 1, 3, 4, 5, 7,\n                 8, 9, 10, 11, 14, 15, 17, 18, 19, 20],\n                [0, 1, 2, 4, 6, 7, 8, 11, 12, 14, 15, 16, 17, 18, 19, 20, 21, 22,\n                 23, 24, 25, 27, 28, 29, 30, 31, 32, 33, 35, 36, 37, 38, 39]]:\n        #print(list(range_extract(lst)))\n        printr(range_extract(lst))"}
{"title": "Range extraction", "language": "Python 3.7", "task": "{{:Range extraction/Format}}\n\n;Task:\n* Create a function that takes a list of integers in increasing order and returns a correctly formatted string in the range format. \n* Use the function to compute and print the range formatted version of the following ordered list of integers. (The correct answer is: 0-2,4,6-8,11,12,14-25,27-33,35-39).\n\n     0,  1,  2,  4,  6,  7,  8, 11, 12, 14,\n    15, 16, 17, 18, 19, 20, 21, 22, 23, 24,\n    25, 27, 28, 29, 30, 31, 32, 33, 35, 36,\n    37, 38, 39\n* Show the output of your program.\n\n\n;Related task:\n*   [[Range expansion]]\n\n", "solution": "'''Range extraction'''\n\nfrom functools import reduce\n\n\n# rangeFormat :: [Int] -> String\ndef rangeFormat(xs):\n    '''Range-formatted display string for\n       a list of integers.\n    '''\n    return ','.join([\n        rangeString(x) for x\n        in splitBy(lambda a, b: 1 < b - a)(xs)\n    ])\n\n\n# rangeString :: [Int] -> String\ndef rangeString(xs):\n    '''Start and end of xs delimited by hyphens\n       if there are more than two integers.\n       Otherwise, comma-delimited xs.\n    '''\n    ys = [str(x) for x in xs]\n    return '-'.join([ys[0], ys[-1]]) if 2 < len(ys) else (\n        ','.join(ys)\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test'''\n\n    xs = [\n        0, 1, 2, 4, 6, 7, 8, 11, 12, 14,\n        15, 16, 17, 18, 19, 20, 21, 22, 23, 24,\n        25, 27, 28, 29, 30, 31, 32, 33, 35, 36,\n        37, 38, 39\n    ]\n    print(\n        __doc__ + ':\\n[' + '\\n'.join(map(\n            lambda x: ' ' + repr(x)[1:-1],\n            chunksOf(11)(xs)\n        )) + \" ]\\n\\n        -> '\" + rangeFormat(xs) + \"'\\n\"\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n,\n       subdividing the contents of xs.\n       Where the length of xs is not evenly divible,\n       the final list will be shorter than n.'''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# splitBy :: (a -> a -> Bool) -> [a] -> [[a]]\ndef splitBy(p):\n    '''A list split wherever two consecutive\n       items match the binary predicate p.\n    '''\n    # step :: ([[a]], [a], a) -> a -> ([[a]], [a], a)\n    def step(acp, x):\n        acc, active, prev = acp\n        return (acc + [active], [x], x) if p(prev, x) else (\n            (acc, active + [x], x)\n        )\n\n    # go :: [a] -> [[a]]\n    def go(xs):\n        if 2 > len(xs):\n            return xs\n        else:\n            h = xs[0]\n            ys = reduce(step, xs[1:], ([], [h], h))\n            # The accumulated sublists, and the current group.\n            return ys[0] + [ys[1]]\n\n    return lambda xs: go(xs)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Rate counter", "language": "Python", "task": "Counting the frequency at which something occurs is a common activity in measuring performance and managing resources. In this task, we assume that there is some job which we want to perform repeatedly, and we want to know how quickly these jobs are being performed.\nOf interest is the code that performs the actual measurements. Any other code (such as job implementation or dispatching) that is required to demonstrate the rate tracking is helpful, but not the focus.\n\nMultiple approaches are allowed (even preferable), so long as they can accomplish these goals:\n\n* Run N seconds worth of jobs and/or Y jobs.\n* Report at least three distinct times.\n\nBe aware of the precision and accuracy limitations of your timing mechanisms, and document them if you can.\n\n'''See also:''' [[System time]], [[Time a function]]\n\n", "solution": "import subprocess\nimport time\n\nclass Tlogger(object):\n    def __init__(self):\n        self.counts = 0\n        self.tottime = 0.0\n        self.laststart = 0.0\n        self.lastreport = time.time()\n\n    def logstart(self):\n        self.laststart = time.time()\n\n    def logend(self):\n        self.counts +=1\n        self.tottime += (time.time()-self.laststart)\n        if (time.time()-self.lastreport)>5.0:   # report once every 5 seconds\n           self.report()\n\n    def report(self):\n        if ( self.counts > 4*self.tottime):\n            print \"Subtask execution rate: %f times/second\"% (self.counts/self.tottime);\n        else:\n            print \"Average execution time: %f seconds\"%(self.tottime/self.counts);\n        self.lastreport = time.time()\n\n\ndef taskTimer( n, subproc_args ):\n    logger = Tlogger()\n\n    for x in range(n):\n        logger.logstart()\n        p = subprocess.Popen(subproc_args)\n        p.wait()\n        logger.logend()\n    logger.report()\n\n\nimport timeit\nimport sys\n\ndef main( ):\n\n    # for accurate timing of code segments \n    s = \"\"\"j = [4*n for n in range(50)]\"\"\"\n    timer = timeit.Timer(s)\n    rzlts = timer.repeat(5, 5000)\n    for t in rzlts:\n        print \"Time for 5000 executions of statement = \",t\n    \n    # subprocess execution timing\n    print \"#times:\",sys.argv[1]\n    print \"Command:\",sys.argv[2:]\n    print \"\"\n    for k in range(3):\n       taskTimer( int(sys.argv[1]), sys.argv[2:])\n\nmain()"}
{"title": "Read a specific line from a file", "language": "Python", "task": "Some languages have special semantics for obtaining a known line number from a file.\n\n\n;Task:\nDemonstrate how to obtain the contents of a specific line within a file. \n\nFor the purpose of this task demonstrate how the contents of the seventh line of a file can be obtained,   and store it in a variable or in memory   (for potential future use within the program if the code were to become embedded). \n\nIf the file does not contain seven lines,   or the seventh line is empty,   or too big to be retrieved,   output an appropriate message. \n\nIf no special semantics are available for obtaining the required line,   it is permissible to read line by line. \n\nNote that empty lines are considered and should still be counted.\n\nAlso note that for functional languages or languages without variables or storage,   it is permissible to output the extracted data to standard output.\n\n", "solution": "from itertools import islice\n\nwith open('xxx.txt') as f:\n    try:\n        line = next(islice(f, 6, 7))\n    except StopIteration:\n        print('Not 7 lines in file')"}
{"title": "Recaman's sequence", "language": "Python", "task": "The '''Recaman's sequence''' generates Natural numbers.\n\nStarting from a(0)=0, the n'th term a(n), where n>0, is the previous term minus n i.e a(n) = a(n-1) - n but only if this is '''both''' positive ''and'' has not been previousely generated.\n\nIf the conditions ''don't'' hold then a(n) = a(n-1) + n.\n\n\n;Task:\n# Generate and show here the first 15 members of the sequence.\n# Find and show here, the first duplicated number in the sequence.\n# '''Optionally''': Find and show here, how many terms of the sequence are needed until all the integers 0..1000, inclusive, are generated.\n\n\n;References:\n* A005132,  The On-Line Encyclopedia of Integer Sequences.\n* The Slightly Spooky Recaman Sequence, Numberphile video.\n* Recaman's sequence, on Wikipedia.\n\n", "solution": "from itertools import islice\n\nclass Recamans():\n    \"Recam\u00e1n's sequence generator callable class\"\n    def __init__(self):\n        self.a = None   # Set of results so far\n        self.n = None   # n'th term (counting from zero)\n    \n    def __call__(self):\n        \"Recam\u00e1n's sequence  generator\"\n        nxt = 0\n        a, n = {nxt}, 0\n        self.a = a\n        self.n = n\n        yield nxt\n        while True:\n            an1, n = nxt, n + 1\n            nxt = an1 - n\n            if nxt < 0 or nxt in a:\n                nxt = an1 + n\n            a.add(nxt)\n            self.n = n\n            yield nxt\n\nif __name__ == '__main__':\n    recamans = Recamans()\n    print(\"First fifteen members of Recamans sequence:\", \n          list(islice(recamans(), 15)))\n\n    so_far = set()\n    for term in recamans():\n        if term in so_far:\n            print(f\"First duplicate number in series is: a({recamans.n}) = {term}\")\n            break\n        so_far.add(term)\n    \n    n = 1_000\n    setn = set(range(n + 1))    # The target set of numbers to be covered\n    for _ in recamans():\n        if setn.issubset(recamans.a):\n            print(f\"Range 0 ..{n} is covered by terms up to a({recamans.n})\")\n            break"}
{"title": "Remove lines from a file", "language": "Python", "task": "Remove a specific line or a number of lines from a file. \n\nThis should be implemented as a routine that takes three parameters (filename, starting line, and the number of lines to be removed). \n\nFor the purpose of this task, line numbers and the number of lines start at one, so to remove the first two lines from the file foobar.txt, the parameters should be: foobar.txt, 1, 2\n\nEmpty lines are considered and should still be counted, and if the specified line is empty, it should still be removed. \n\nAn appropriate message should appear if an attempt is made to remove lines beyond the end of the file.\n\n", "solution": "#!/usr/bin/env python\n\nimport fileinput, sys\n\nfname, start, count = sys.argv[1:4]\nstart, count = int(start), int(count)\n\nfor line in fileinput.input(fname, inplace=1, backup='.orig'):\n    if start <= fileinput.lineno() < start + count:\n        pass\n    else:\n        print line.rstrip(\"\\n\")\nfileinput.close()"}
{"title": "Rep-string", "language": "Python", "task": "Given a series of ones and zeroes in a string, define a repeated string or ''rep-string'' as a string which is created by repeating a substring of the ''first'' N characters of the string ''truncated on the right to the length of the input string, and in which the substring appears repeated at least twice in the original''.\n\nFor example, the string '''10011001100''' is a rep-string as the leftmost four characters of '''1001''' are repeated three times and truncated on the right to give the original string.\n\nNote that the requirement for having the repeat occur two or more times means that the repeating unit is ''never'' longer than half the length of the input string.\n\n\n;Task:\n* Write a function/subroutine/method/... that takes a string and returns an indication of if it is a rep-string and the repeated string.   (Either the string that is repeated, or the number of repeated characters would suffice).  \n* There may be multiple sub-strings that make a string a rep-string - in that case an indication of all, or the longest, or the shortest would suffice.\n* Use the function to indicate the repeating substring if any, in the following:\n\n\n1001110011\n1110111011\n0010010010\n1010101010\n1111111111\n0100101101\n0100100\n101\n11\n00\n1\n\n\n* Show your output on this page.\n\n\n\n", "solution": "def is_repeated(text):\n    'check if the first part of the string is repeated throughout the string'\n    for x in range(len(text)//2, 0, -1):\n        if text.startswith(text[x:]): return x\n    return 0\n\nmatchstr = \"\"\"\\\n1001110011\n1110111011\n0010010010\n1010101010\n1111111111\n0100101101\n0100100\n101\n11\n00\n1\n\"\"\"\nfor line in matchstr.split():\n    ln = is_repeated(line)\n    print('%r has a repetition length of %i i.e. %s' \n           % (line, ln, repr(line[:ln]) if ln else '*not* a rep-string'))"}
{"title": "Rep-string", "language": "Python 3.7", "task": "Given a series of ones and zeroes in a string, define a repeated string or ''rep-string'' as a string which is created by repeating a substring of the ''first'' N characters of the string ''truncated on the right to the length of the input string, and in which the substring appears repeated at least twice in the original''.\n\nFor example, the string '''10011001100''' is a rep-string as the leftmost four characters of '''1001''' are repeated three times and truncated on the right to give the original string.\n\nNote that the requirement for having the repeat occur two or more times means that the repeating unit is ''never'' longer than half the length of the input string.\n\n\n;Task:\n* Write a function/subroutine/method/... that takes a string and returns an indication of if it is a rep-string and the repeated string.   (Either the string that is repeated, or the number of repeated characters would suffice).  \n* There may be multiple sub-strings that make a string a rep-string - in that case an indication of all, or the longest, or the shortest would suffice.\n* Use the function to indicate the repeating substring if any, in the following:\n\n\n1001110011\n1110111011\n0010010010\n1010101010\n1111111111\n0100101101\n0100100\n101\n11\n00\n1\n\n\n* Show your output on this page.\n\n\n\n", "solution": "'''Rep-strings'''\n\nfrom itertools import (accumulate, chain, cycle, islice)\n\n\n# repCycles :: String -> [String]\ndef repCycles(s):\n    '''Repeated sequences of characters in s.'''\n    n = len(s)\n    cs = list(s)\n\n    return [\n        x for x in\n        tail(inits(take(n // 2)(s)))\n        if cs == take(n)(cycle(x))\n    ]\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Tests - longest cycle (if any) in each string.'''\n    print(\n        fTable('Longest cycles:\\n')(repr)(\n            lambda xs: ''.join(xs[-1]) if xs else '(none)'\n        )(repCycles)([\n            '1001110011',\n            '1110111011',\n            '0010010010',\n            '1010101010',\n            '1111111111',\n            '0100101101',\n            '0100100',\n            '101',\n            '11',\n            '00',\n            '1',\n        ])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# inits :: [a] -> [[a]]\ndef inits(xs):\n    '''all initial segments of xs, shortest first.'''\n    return accumulate(chain([[]], xs), lambda a, x: a + [x])\n\n\n# tail :: [a] -> [a]\n# tail :: Gen [a] -> [a]\ndef tail(xs):\n    '''The elements following the head of a\n       (non-empty) list or generator stream.'''\n    if isinstance(xs, list):\n        return xs[1:]\n    else:\n        list(islice(xs, 1))  # First item dropped.\n        return xs\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, (list, tuple))\n        else list(islice(xs, n))\n    )\n\n\n# OUTPUT FORMATTING ---------------------------------------\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function ->\n                 fx display function ->\n          f -> value list -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Repeat", "language": "Python", "task": "Write a procedure which accepts as arguments another procedure and a positive integer. \n\nThe latter procedure is executed a number of times equal to the accepted integer.\n\n", "solution": "#!/usr/bin/python\ndef repeat(f,n):\n  for i in range(n):\n    f();\n\ndef procedure():\n  print(\"Example\");\n\nrepeat(procedure,3); #prints \"Example\" (without quotes) three times, separated by newlines."}
{"title": "Repeat", "language": "Python 3.7", "task": "Write a procedure which accepts as arguments another procedure and a positive integer. \n\nThe latter procedure is executed a number of times equal to the accepted integer.\n\n", "solution": "'''Application of a given function, repeated N times'''\n\nfrom itertools import repeat\nfrom functools import reduce\nfrom inspect import getsource\n\n\n# applyN :: Int -> (a -> a) -> a -> a\ndef applyN(n):\n    '''n compounding applications of the supplied\n       function f. Equivalent to Church numeral n.\n    '''\n    def go(f):\n        return lambda x: reduce(\n            lambda a, g: g(a), repeat(f, n), x\n        )\n    return lambda f: go(f)\n\n\n# MAIN ----------------------------------------------------\ndef main():\n    '''Tests - compounding repetition\n       of function application.\n    '''\n    def f(x):\n        return x + 'Example\\n'\n\n    def g(x):\n        return 2 * x\n\n    def h(x):\n        return 1.05 * x\n\n    print(\n        fTable(__doc__ + ':')(\n            lambda fx: '\\nRepeated * 3:\\n (' + (\n                getsource(fst(fx)).strip() + ')(' +\n                repr(snd(fx)) + ')'\n            )\n        )(str)(\n            liftA2(applyN(3))(fst)(snd)\n        )([(f, '\\n'), (g, 1), (h, 100)])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# liftA2 :: (a0 -> b -> c) -> (a -> a0) -> (a -> b) -> a -> c\ndef liftA2(op):\n    '''Lift a binary function to a composition\n       over two other functions.\n       liftA2 (*) (+ 2) (+ 3) 7 == 90\n    '''\n    def go(f, g):\n        return lambda x: op(\n            f(x)\n        )(g(x))\n    return lambda f: lambda g: go(f, g)\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second member of a pair.'''\n    return tpl[1]\n\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Repunit primes", "language": "Python", "task": "portmanteau of the words \"repetition\" and \"unit\", with unit being \"unit value\"... or in laymans terms, '''1'''.  So 1, 11, 111, 1111 & 11111 are all repunits.\n\nEvery standard integer base has repunits since every base has the digit 1. This task involves finding the repunits in different bases that are prime.\n\nIn base two, the repunits 11, 111, 11111, 1111111, etc. are prime. (These correspond to the Mersenne primes.) \n\nIn base three: 111, 1111111, 1111111111111, etc.\n\n''Repunit primes, by definition, are also [[circular primes]].''\n\nAny repunit in any base having a composite number of digits is necessarily composite. Only repunits (in any base) having a prime number of digits ''might'' be prime.\n\n\nRather than expanding the repunit out as a giant list of '''1'''s or converting to base 10, it is common to just list the ''number'' of '''1'''s in the repunit; effectively the digit count. The base two repunit primes listed above would be represented as: 2, 3, 5, 7, etc.\n\nMany of these sequences exist on OEIS, though they aren't specifically listed as \"repunit prime digits\" sequences. \n\nSome bases have very few repunit primes. Bases 4, 8, and likely 16 have only one. Base 9 has none at all. Bases above 16 may have repunit primes as well... but this task is getting large enough already.\n\n\n;Task\n\n* For bases 2 through 16, Find and show, here on this page, the repunit primes as digit counts, up to a limit of 1000.\n\n\n;Stretch\n\n* Increase the limit to 2700 (or as high as you have patience for.)\n\n\n;See also\n\n;* Wikipedia: Repunit primes\n;* OEIS:A000043 - Mersenne exponents: primes p such that 2^p - 1 is prime. Then 2^p - 1 is called a Mersenne prime (base 2)\n;* OEIS:A028491 - Numbers k such that (3^k - 1)/2 is prime (base 3)\n;* OEIS:A004061 - Numbers n such that (5^n - 1)/4 is prime (base 5)\n;* OEIS:A004062 - Numbers n such that (6^n - 1)/5 is prime (base 6)\n;* OEIS:A004063 - Numbers k such that (7^k - 1)/6 is prime (base 7)\n;* OEIS:A004023 - Indices of prime repunits: numbers n such that 11...111 (with n 1's) = (10^n - 1)/9 is prime (base 10)\n;* OEIS:A005808 - Numbers k such that (11^k - 1)/10 is prime (base 11)\n;* OEIS:A004064 - Numbers n such that (12^n - 1)/11 is prime (base 12)\n;* OEIS:A016054 - Numbers n such that (13^n - 1)/12 is prime (base 13)\n;* OEIS:A006032 - Numbers k such that (14^k - 1)/13 is prime (base 14)\n;* OEIS:A006033 - Numbers n such that (15^n - 1)/14 is prime (base 15)\n;* Related task: Circular primes\n\n\n\n", "solution": "from sympy import isprime\nfor b in range(2, 17):\n    print(b, [n for n in range(2, 1001) if isprime(n) and isprime(int('1'*n, base=b))])"}
{"title": "Resistor mesh", "language": "Python from D", "task": "Given    10x10    grid nodes   (as shown in the image)   interconnected by    1O    resistors as shown,\nfind the resistance between points   '''A'''   and   '''B'''.\n\n\n;See also:\n*   (humor, nerd sniping)   xkcd.com cartoon (you can solve that for extra credits)\n*   An article on how to calculate this and an implementation in Mathematica\n\n", "solution": "DIFF_THRESHOLD = 1e-40\n\nclass Fixed:\n    FREE = 0\n    A = 1\n    B = 2\n\nclass Node:\n    __slots__ = [\"voltage\", \"fixed\"]\n    def __init__(self, v=0.0, f=Fixed.FREE):\n        self.voltage = v\n        self.fixed = f\n\ndef set_boundary(m):\n    m[1][1] = Node( 1.0, Fixed.A)\n    m[6][7] = Node(-1.0, Fixed.B)\n\ndef calc_difference(m, d):\n    h = len(m)\n    w = len(m[0])\n    total = 0.0\n\n    for i in xrange(h):\n        for j in xrange(w):\n            v = 0.0\n            n = 0\n            if i != 0:  v += m[i-1][j].voltage; n += 1\n            if j != 0:  v += m[i][j-1].voltage; n += 1\n            if i < h-1: v += m[i+1][j].voltage; n += 1\n            if j < w-1: v += m[i][j+1].voltage; n += 1\n            v = m[i][j].voltage - v / n\n\n            d[i][j].voltage = v\n            if m[i][j].fixed == Fixed.FREE:\n                total += v ** 2\n    return total\n\ndef iter(m):\n    h = len(m)\n    w = len(m[0])\n    difference = [[Node() for j in xrange(w)] for i in xrange(h)]\n\n    while True:\n        set_boundary(m) # Enforce boundary conditions.\n        if calc_difference(m, difference) < DIFF_THRESHOLD:\n            break\n        for i, di in enumerate(difference):\n            for j, dij in enumerate(di):\n                m[i][j].voltage -= dij.voltage\n\n    cur = [0.0] * 3\n    for i, di in enumerate(difference):\n        for j, dij in enumerate(di):\n            cur[m[i][j].fixed] += (dij.voltage *\n                (bool(i) + bool(j) + (i < h-1) + (j < w-1)))\n\n    return (cur[Fixed.A] - cur[Fixed.B]) / 2.0\n\ndef main():\n    w = h = 10\n    mesh = [[Node() for j in xrange(w)] for i in xrange(h)]\n    print \"R = %.16f\" % (2 / iter(mesh))\n\nmain()"}
{"title": "Resistor mesh", "language": "Python from Maxima", "task": "Given    10x10    grid nodes   (as shown in the image)   interconnected by    1O    resistors as shown,\nfind the resistance between points   '''A'''   and   '''B'''.\n\n\n;See also:\n*   (humor, nerd sniping)   xkcd.com cartoon (you can solve that for extra credits)\n*   An article on how to calculate this and an implementation in Mathematica\n\n", "solution": "import sys, copy\nfrom fractions import Fraction\n\ndef gauss(a, b):\n    n, p = len(a), len(a[0])\n    for i in range(n):\n        t = abs(a[i][i])\n        k = i\n        for j in range(i + 1, n):\n            if abs(a[j][i]) > t:\n                t = abs(a[j][i])\n                k = j\n        if k != i:\n            for j in range(i, n):\n                a[i][j], a[k][j] = a[k][j], a[i][j]\n            b[i], b[k] = b[k], b[i]\n        t = 1/a[i][i]\n        for j in range(i + 1, n):\n            a[i][j] *= t\n        b[i] *= t\n        for j in range(i + 1, n):\n            t = a[j][i]\n            for k in range(i + 1, n):\n                a[j][k] -= t*a[i][k]\n            b[j] -= t * b[i]\n    for i in range(n - 1, -1, -1):\n        for j in range(i):\n            b[j] -= a[j][i]*b[i]\n    return b\n\ndef resistor_grid(p, q, ai, aj, bi, bj):\n    n = p*q\n    I = Fraction(1, 1)\n    v = [0*I]*n\n    a = [copy.copy(v) for i in range(n)]\n    for i in range(p):\n        for j in range(q):\n            k = i*q + j\n            if i == ai and j == aj:\n                a[k][k] = I\n            else:\n                c = 0\n                if i + 1 < p:\n                    c += 1\n                    a[k][k + q] = -1\n                if i >= 1:\n                    c += 1\n                    a[k][k - q] = -1\n                if j + 1 < q:\n                    c += 1\n                    a[k][k + 1] = -1\n                if j >= 1:\n                    c += 1\n                    a[k][k - 1] = -1\n                a[k][k] = c*I\n    b = [0*I]*n\n    k = bi*q + bj\n    b[k] = 1\n    return gauss(a, b)[k]\n\ndef main(arg):\n    r = resistor_grid(int(arg[0]), int(arg[1]), int(arg[2]), int(arg[3]), int(arg[4]), int(arg[5]))\n    print(r)\n    print(float(r))\n\nmain(sys.argv[1:])\n\n# Output:\n# python grid.py 10 10 1 1 7 6\n# 455859137025721/283319837425200\n# 1.6089912417307297"}
{"title": "Reverse words in a string", "language": "Python", "task": "Reverse the order of all tokens in each of a number of strings and display the result;   the order of characters within a token should not be modified.\n\n\n;Example:\nHey you, Bub!    would be shown reversed as:    Bub! you, Hey \n\n\nTokens are any non-space characters separated by spaces (formally, white-space);   the visible punctuation form part of the word within which it is located and should not be modified. \n\nYou may assume that there are no significant non-visible characters in the input.   Multiple or superfluous spaces may be compressed into a single space.\n\nSome strings have no tokens, so an empty string   (or one just containing spaces)   would be the result.\n\n'''Display''' the strings in order   (1st, 2nd, 3rd, ***),   and one string per line. \n\n(You can consider the ten strings as ten lines, and the tokens as words.)\n\n\n;Input data\n\n             (ten lines within the box)\n line\n     +----------------------------------------+\n   1 |  ---------- Ice and Fire ------------  |\n   2 |                                        |  <--- a blank line here.\n   3 |  fire, in end will world the say Some  |\n   4 |  ice. in say Some                      |\n   5 |  desire of tasted I've what From       |\n   6 |  fire. favor who those with hold I     |\n   7 |                                        |  <--- a blank line here.\n   8 |  ... elided paragraph last ...         |\n   9 |                                        |  <--- a blank line here.\n  10 |  Frost Robert -----------------------  |\n     +----------------------------------------+\n\n\n;Cf.\n* [[Phrase reversals]]\n\n", "solution": "text = '''\\\n---------- Ice and Fire ------------\n\nfire, in end will world the say Some\nice. in say Some\ndesire of tasted I've what From\nfire. favor who those with hold I\n\n... elided paragraph last ...\n\nFrost Robert -----------------------'''\n\nfor line in text.split('\\n'): print(' '.join(line.split()[::-1]))"}
{"title": "Roman numerals/Decode", "language": "Python", "task": "Create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer. \n\nYou don't need to validate the form of the Roman numeral.\n\nModern Roman numerals are written by expressing each decimal digit of the number to be encoded separately, \nstarting with the leftmost decimal digit and skipping any '''0'''s   (zeroes). \n\n'''1990''' is rendered as   '''MCMXC'''     (1000 = M,   900 = CM,   90 = XC)     and \n'''2008''' is rendered as   '''MMVIII'''       (2000 = MM,   8 = VIII).\n \nThe Roman numeral for '''1666''',   '''MDCLXVI''',   uses each letter in descending order.\n\n", "solution": "_rdecode = dict(zip('MDCLXVI', (1000, 500, 100, 50, 10, 5, 1)))\n\ndef decode( roman ):\n    result = 0\n    for r, r1 in zip(roman, roman[1:]):\n        rd, rd1 = _rdecode[r], _rdecode[r1]\n        result += -rd if rd < rd1 else rd\n    return result + _rdecode[roman[-1]]\n\nif __name__ == '__main__':\n    for r in 'MCMXC MMVIII MDCLXVI'.split():\n        print( r, decode(r) )"}
{"title": "Roman numerals/Decode", "language": "Python 3", "task": "Create a function that takes a Roman numeral as its argument and returns its value as a numeric decimal integer. \n\nYou don't need to validate the form of the Roman numeral.\n\nModern Roman numerals are written by expressing each decimal digit of the number to be encoded separately, \nstarting with the leftmost decimal digit and skipping any '''0'''s   (zeroes). \n\n'''1990''' is rendered as   '''MCMXC'''     (1000 = M,   900 = CM,   90 = XC)     and \n'''2008''' is rendered as   '''MMVIII'''       (2000 = MM,   8 = VIII).\n \nThe Roman numeral for '''1666''',   '''MDCLXVI''',   uses each letter in descending order.\n\n", "solution": "'''Roman numerals decoded'''\n\nfrom operator import mul\nfrom functools import reduce\nfrom collections import defaultdict\nfrom itertools import accumulate, chain, cycle\n\n\n# intFromRoman :: String -> Maybe Int\ndef intFromRoman(s):\n    '''Just the integer represented by a Roman\n       numeral string, or Nothing if any\n       characters are unrecognised.\n    '''\n    dct = defaultdict(\n        lambda: None,\n        zip(\n            'IVXLCDM',\n            accumulate(chain([1], cycle([5, 2])), mul)\n        )\n    )\n\n    def go(mb, x):\n        '''Just a letter value added to or\n           subtracted from a total, or Nothing\n           if no letter value is defined.\n        '''\n        if None in (mb, x):\n            return None\n        else:\n            r, total = mb\n            return x, total + (-x if x < r else x)\n\n    return bindMay(reduce(\n        go,\n        [dct[k.upper()] for k in reversed(list(s))],\n        (0, 0)\n    ))(snd)\n\n\n# ------------------------- TEST -------------------------\ndef main():\n    '''Testing a sample of dates.'''\n\n    print(\n        fTable(__doc__ + ':\\n')(str)(\n            maybe('(Contains unknown character)')(str)\n        )(\n            intFromRoman\n        )([\n            \"MDCLXVI\", \"MCMXC\", \"MMVIII\",\n            \"MMXVI\", \"MMXVIII\", \"MMZZIII\"\n        ])\n    )\n\n\n# ----------------------- GENERIC ------------------------\n\n# bindMay (>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b\ndef bindMay(m):\n    '''Injection operator for the Maybe monad.\n       If m is Nothing, it is passed straight through.\n       If m is Just(x), the result is an application\n       of the (a -> Maybe b) function (mf) to x.'''\n    return lambda mf: (\n        m if None is m else mf(m)\n    )\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if m is Nothing,\n       or the application of f to x,\n       where m is Just(x).\n    '''\n    return lambda f: lambda m: v if None is m else (\n        f(m)\n    )\n\n\n# snd :: (a, b) -> b\ndef snd(ab):\n    '''Second member of a pair.'''\n    return ab[1]\n\n\n# ---------------------- FORMATTING ----------------------\n\n# fTable :: String -> (a -> String) ->\n# (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function ->\n       fx display function -> f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: (\n                f'{y.rjust(w, \" \")} -> {fxShow(f(x))}'\n            ),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: (\n        lambda xs: go(xShow, fxShow, f, xs)\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Roman numerals/Encode", "language": "Python", "task": "Create a function taking a positive integer as its parameter and returning a string containing the Roman numeral representation of that integer. Modern Roman numerals are written by expressing each digit separately, starting with the left most digit and skipping any digit with a value of zero. \n\n\nIn Roman numerals:\n* 1990  is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC\n* 2008  is written as 2000=MM, 8=VIII; or MMVIII\n* 1666  uses each Roman symbol in descending order: MDCLXVI\n\n", "solution": "def toRoman(n):\n    res=''\t\t#converts int to str(Roman numeral)\n    reg=n\t\t#using the numerals (M,D,C,L,X,V,I)\n    if reg<4000:#no more than three repetitions\n        while reg>=1000:\t#thousands up to MMM\n            res+='M'\t\t#MAX is MMMCMXCIX\n            reg-=1000\t\t\n        if reg>=900:\t\t#nine hundreds in 900-999\n            res+='CM'\n            reg-=900\n        if reg>=500:\t\t#five hudreds in 500-899\n            res+='D'\n            reg-=500\n        if reg>=400:\t\t#four hundreds in 400-499\n            res+='CD'\n            reg-=400\n        while reg>=100:\t\t#hundreds in 100-399\n            res+='C'\n            reg-=100\n        if reg>=90:\t\t\t#nine tens in 90-99\n            res+='XC'\n            reg-=90\n        if reg>=50:\t\t\t#five Tens in 50-89\n            res+='L'\n            reg-=50\n        if reg>=40:\n            res+='XL'\t\t#four Tens\n            reg-=40\n        while reg>=10:\n            res+=\"X\"\t\t#tens\n            reg-=10\n        if reg>=9:\n            res+='IX'\t\t#nine Units\n            reg-=9\n        if reg>=5:\n            res+='V'\t\t#five Units\n            reg-=5\n        if reg>=4:\n            res+='IV'\t\t#four Units\n            reg-=4\n        while reg>0:\t\t#three or less Units\n            res+='I'\n            reg-=1\n    return res"}
{"title": "Roman numerals/Encode", "language": "Python from Haskell", "task": "Create a function taking a positive integer as its parameter and returning a string containing the Roman numeral representation of that integer. Modern Roman numerals are written by expressing each digit separately, starting with the left most digit and skipping any digit with a value of zero. \n\n\nIn Roman numerals:\n* 1990  is rendered: 1000=M, 900=CM, 90=XC; resulting in MCMXC\n* 2008  is written as 2000=MM, 8=VIII; or MMVIII\n* 1666  uses each Roman symbol in descending order: MDCLXVI\n\n", "solution": "'''Encoding Roman Numerals'''\n\nfrom functools import reduce\nfrom itertools import chain\n\n\n# romanFromInt ::  Int -> String\ndef romanFromInt(n):\n    '''A string of Roman numerals encoding an integer.'''\n    def go(a, ms):\n        m, s = ms\n        q, r = divmod(a, m)\n        return (r, s * q)\n\n    return concat(snd(mapAccumL(go)(n)(\n        zip([\n            1000, 900, 500, 400, 100, 90, 50,\n            40, 10, 9, 5, 4, 1\n        ], [\n            'M', 'CM', 'D', 'CD', 'C', 'XC', 'L',\n            'XL', 'X', 'IX', 'V', 'IV', 'I'\n        ])\n    )))\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Sample of years'''\n    for s in [\n            romanFromInt(x) for x in [\n                1666, 1990, 2008, 2016, 2018, 2020\n            ]\n    ]:\n        print(s)\n\n\n# ------------------ GENERIC FUNCTIONS -------------------\n\n# concat :: [[a]] -> [a]\n# concat :: [String] -> String\ndef concat(xxs):\n    '''The concatenation of all the elements in a list.'''\n    xs = list(chain.from_iterable(xxs))\n    unit = '' if isinstance(xs, str) else []\n    return unit if not xs else (\n        ''.join(xs) if isinstance(xs[0], str) else xs\n    )\n\n\n# mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])\ndef mapAccumL(f):\n    '''A tuple of an accumulation and a list derived by a\n       combined map and fold,\n       with accumulation from left to right.'''\n    def go(a, x):\n        tpl = f(a[0], x)\n        return (tpl[0], a[1] + [tpl[1]])\n    return lambda acc: lambda xs: (\n        reduce(go, xs, (acc, []))\n    )\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second component of a tuple.'''\n    return tpl[1]\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Runge-Kutta method", "language": "Python", "task": "Given the example Differential equation:\n:y'(t) = t \\times \\sqrt {y(t)}\nWith initial condition:\n:t_0 = 0 and y_0 = y(t_0) = y(0) = 1\nThis equation has an exact solution:\n:y(t) = \\tfrac{1}{16}(t^2 +4)^2\n\n\n;Task\nDemonstrate the commonly used explicit   fourth-order Runge-Kutta method   to solve the above differential equation.\n* Solve the given differential equation over the range t = 0 \\ldots 10 with a step value of \\delta t=0.1 (101 total points, the first being given)\n* Print the calculated values of y at whole numbered t's (0.0, 1.0, \\ldots 10.0) along with error as compared to the exact solution.\n\n\n;Method summary\nStarting with a given y_n and t_n calculate:\n:\\delta y_1 = \\delta t\\times y'(t_n, y_n)\\quad\n:\\delta y_2 = \\delta t\\times y'(t_n + \\tfrac{1}{2}\\delta t , y_n + \\tfrac{1}{2}\\delta y_1)\n:\\delta y_3 = \\delta t\\times y'(t_n + \\tfrac{1}{2}\\delta t , y_n + \\tfrac{1}{2}\\delta y_2)\n:\\delta y_4 = \\delta t\\times y'(t_n + \\delta t , y_n + \\delta y_3)\\quad\nthen:\n:y_{n+1} = y_n + \\tfrac{1}{6} (\\delta y_1 + 2\\delta y_2 + 2\\delta y_3 + \\delta y_4)\n:t_{n+1} = t_n + \\delta t\\quad\n\n", "solution": "from math import sqrt\n \ndef rk4(f, x0, y0, x1, n):\n    vx = [0] * (n + 1)\n    vy = [0] * (n + 1)\n    h = (x1 - x0) / float(n)\n    vx[0] = x = x0\n    vy[0] = y = y0\n    for i in range(1, n + 1):\n        k1 = h * f(x, y)\n        k2 = h * f(x + 0.5 * h, y + 0.5 * k1)\n        k3 = h * f(x + 0.5 * h, y + 0.5 * k2)\n        k4 = h * f(x + h, y + k3)\n        vx[i] = x = x0 + i * h\n        vy[i] = y = y + (k1 + k2 + k2 + k3 + k3 + k4) / 6\n    return vx, vy\n \ndef f(x, y):\n    return x * sqrt(y)\n \nvx, vy = rk4(f, 0, 1, 10, 100)\nfor x, y in list(zip(vx, vy))[::10]:\n    print(\"%4.1f %10.5f %+12.4e\" % (x, y, y - (4 + x * x)**2 / 16))\n\n 0.0    1.00000  +0.0000e+00\n 1.0    1.56250  -1.4572e-07\n 2.0    4.00000  -9.1948e-07\n 3.0   10.56250  -2.9096e-06\n 4.0   24.99999  -6.2349e-06\n 5.0   52.56249  -1.0820e-05\n 6.0   99.99998  -1.6595e-05\n 7.0  175.56248  -2.3518e-05\n 8.0  288.99997  -3.1565e-05\n 9.0  451.56246  -4.0723e-05\n10.0  675.99995  -5.0983e-05"}
{"title": "Runtime evaluation/In an environment", "language": "Python", "task": "Given a program in the language (as a string or AST) with a free variable named x (or another name if that is not valid syntax), evaluate it with x bound to a provided value, then evaluate it again with x bound to another provided value, then subtract the result of the first from the second and return or print it.\n\nDo so in a way which:\n* does not involve string manipulation of the input source code\n* is plausibly extensible to a runtime-chosen set of bindings rather than just x\n* does not make x a ''global'' variable\n\n\nor note that these are impossible.\n\n\n;See also:\n* For more general examples and language-specific details, see [[Eval]].\n* [[Dynamic variable names]] is a similar task.\n\n", "solution": ">>> def eval_with_args(code, **kwordargs):\n\treturn eval(code, kwordargs)\n\n>>> code = '2 ** x'\n>>> eval_with_args(code, x=5) - eval_with_args(code, x=3)\n24\n>>> code = '3 * x + y'\n>>> eval_with_args(code, x=5, y=2) - eval_with_args(code, x=3, y=1)\n7"}
{"title": "SHA-1", "language": "Python", "task": "'''SHA-1''' or '''SHA1''' is a one-way hash function; \nit computes a 160-bit message digest. \nSHA-1 often appears in security protocols; for example, \nmany HTTPS websites use RSA with SHA-1 to secure their connections. \nBitTorrent uses SHA-1 to verify downloads. \nGit and Mercurial use SHA-1 digests to identify commits.\n\nA US government standard, FIPS 180-1, defines SHA-1.\n\nFind the SHA-1 message digest for a string of [[octet]]s. You may either call a SHA-1 library, or implement SHA-1 in your language. Both approaches interest Rosetta Code.\n\n{{alertbox|lightgray|'''Warning:''' SHA-1 has known weaknesses. Theoretical attacks may find a collision after 252 operations, or perhaps fewer. \nThis is much faster than a brute force attack of 280 operations. USgovernment deprecated SHA-1. \nFor production-grade cryptography, users may consider a stronger alternative, such as SHA-256 (from the SHA-2 family) or the upcoming SHA-3.}}\n", "solution": "import hashlib\nh = hashlib.sha1()\nh.update(bytes(\"Ars longa, vita brevis\", encoding=\"ASCII\"))\nh.hexdigest()\n# \"e640d285242886eb96ab80cbf858389b3df52f43\""}
{"title": "Sailors, coconuts and a monkey problem", "language": "Python", "task": "Five sailors are shipwrecked on an island and collect a large pile of coconuts during the day. \n\nThat night the first sailor wakes up and decides to take his first share early  so tries to divide the pile of coconuts equally into five piles but finds that there is one coconut left over, so he tosses it to a monkey and then hides \"his\" one of the five equally sized piles of coconuts and pushes the other four piles together to form a single visible pile of coconuts again and goes to bed.\n\nTo cut a long story short, each of the sailors in turn gets up once during the night and performs the same actions of dividing the coconut pile into five, finding that one coconut is left over and giving that single remainder coconut to the monkey.\n\nIn the morning (after the surreptitious and separate action of each of the five sailors during the night), the remaining coconuts are divided into five equal piles for each of the sailors, whereupon it is found that the pile of coconuts divides equally amongst the sailors with no remainder. (Nothing for the monkey in the morning.)\n\n\n;The task:\n# Calculate the minimum possible size of the initial pile of coconuts collected during the first day.\n# Use a method that assumes an answer is possible, and then applies the constraints of the tale to see if it is correct. (I.e. no applying some formula that generates the correct answer without integer divisions and remainders and tests on remainders; but constraint solvers ''are'' allowed.)\n# Calculate the size of the initial pile of coconuts if six sailors were marooned and went through a similar process (but split into six piles instead of five of course).\n# Show your answers here.\n\n\n;Extra credit (optional):\n* Give some indication of the number of coconuts each sailor hides during the night.\n\n\n;Note:\n* Of course the tale is told in a world where the collection of any amount of coconuts in a day and multiple divisions of the pile, etc can occur in time fitting the story line, so as not to affect the mathematics.\n* The tale is also told in a version where the monkey also gets a coconut in the morning. This is ''not'' that tale!\n\n\n;C.f:\n* Monkeys and Coconuts - Numberphile (Video) Analytical solution.\n* A002021: Pile of coconuts problem The On-Line Encyclopedia of Integer Sequences. (Although some of its references may use the alternate form of the tale).\n\n", "solution": "def monkey_coconuts(sailors=5):\n    \"Parameterised the number of sailors using an inner loop including the last mornings case\"    \n    nuts = sailors\n    while True:\n        n0, wakes = nuts, []\n        for sailor in range(sailors + 1):\n            portion, remainder = divmod(n0, sailors)\n            wakes.append((n0, portion, remainder))\n            if portion <= 0 or remainder != (1 if sailor != sailors else 0):\n                nuts += 1\n                break\n            n0 = n0 - portion - remainder\n        else:\n            break\n    return nuts, wakes\n\nif __name__ == \"__main__\":\n    for sailors in [5, 6]:\n        nuts, wake_stats = monkey_coconuts(sailors)\n        print(\"\\nFor %i sailors the initial nut count is %i\" % (sailors, nuts))\n        print(\"On each waking, the nut count, portion taken, and monkeys share are:\\n \", \n              ',\\n  '.join(repr(ws) for ws in wake_stats))"}
{"title": "Same fringe", "language": "Python", "task": "Write a routine that will compare the leaves (\"fringe\") of two binary trees to determine whether they are the same list of leaves when visited left-to-right.  The structure or balance of the trees does not matter; only the number, order, and value of the leaves is important.\n\nAny solution is allowed here, but many computer scientists will consider it inelegant to collect either fringe in its entirety before starting to collect the other one.  In fact, this problem is usually proposed in various forums as a way to show off various forms of concurrency (tree-rotation algorithms have also been used to get around the need to collect one tree first).  Thinking of it a slightly different way, an elegant solution is one that can perform the minimum amount of work to falsify the equivalence of the fringes when they differ somewhere in the middle, short-circuiting the unnecessary additional traversals and comparisons.\n\nAny representation of a binary tree is allowed, as long as the nodes are orderable, and only downward links are used (for example, you may not use parent or sibling pointers to avoid recursion).\n\n", "solution": "try:\n    from itertools import zip_longest as izip_longest # Python 3.x\nexcept:\n    from itertools import izip_longest                # Python 2.6+\n\ndef fringe(tree):\n    \"\"\"Yield tree members L-to-R depth first,\n    as if stored in a binary tree\"\"\"\n    for node1 in tree:\n        if isinstance(node1, tuple):\n            for node2 in fringe(node1):\n                yield node2\n        else:\n            yield node1\n\ndef same_fringe(tree1, tree2):\n    return all(node1 == node2 for node1, node2 in\n               izip_longest(fringe(tree1), fringe(tree2)))\n\nif __name__ == '__main__':\n    a = 1, 2, 3, 4, 5, 6, 7, 8\n    b = 1, (( 2, 3 ), (4, (5, ((6, 7), 8))))\n    c = (((1, 2), 3), 4), 5, 6, 7, 8\n\n    x = 1, 2, 3, 4, 5, 6, 7, 8, 9\n    y = 0, 2, 3, 4, 5, 6, 7, 8\n    z = 1, 2, (4, 3), 5, 6, 7, 8\n\n    assert same_fringe(a, a)\n    assert same_fringe(a, b)\n    assert same_fringe(a, c)\n\n    assert not same_fringe(a, x)\n    assert not same_fringe(a, y)\n    assert not same_fringe(a, z)"}
{"title": "Search in paragraph's text", "language": "Python", "task": "The goal is to verify the presence of a word or regular expression within several paragraphs of text (structured or not) and to format the output of the relevant paragraphs before putting them on the standard output.\n\nSo here, let's imagine that we are trying to verify the presence of a keyword \"SystemError\" within what I want to call \"the paragraphs\" \"Traceback (most recent call last):\" in the file Traceback.txt\n\n\ncat Traceback.txt :\n\n2018-06-19 23:19:34,877 ERROR Processes plugin raised an exception.\nTraceback (most recent call last):\n  File \"/usr/lib/python3/dist-packages/landscape/sysinfo/sysinfo.py\", line 99, in run\n    result = plugin.run()\n  File \"/usr/lib/python3/dist-packages/landscape/sysinfo/processes.py\", line 18, in run\n    for process_info in info.get_all_process_info():\n  File \"/usr/lib/python3/dist-packages/landscape/lib/process.py\", line 39, in get_all_process_info\n    process_info = self.get_process_info(process_id)\n  File \"/usr/lib/python3/dist-packages/landscape/lib/process.py\", line 61, in get_process_info\n    cmd_line = file.readline()\n  File \"/usr/lib/python3.6/encodings/ascii.py\", line 26, in decode\n    return codecs.ascii_decode(input, self.errors)[0]\nUnicodeDecodeError: 'ascii' codec can't decode byte 0xc5 in position 152: ordinal not in range(128)\n\n2018-06-19 23:19:34,877 ERROR Processes plugin raised an exception.\nTraceback (most recent call last):\n    vmodl.fault.SystemError: (vmodl.fault.SystemError) {\n    dynamicType = ,\n    dynamicProperty = (vmodl.DynamicProperty) [],\n    msg = 'A general system error occurred: Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.',\n    faultCause = ,\n    faultMessage = (vmodl.LocalizableMessage) [],\n    reason = 'Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.'\n    }\n\n[Tue Jan 21 16:16:19.250245 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] Traceback (most recent call last):\n[Tue Jan 21 16:16:19.252221 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] SystemError: unable to access /home/dir\n[Tue Jan 21 16:16:19.249067 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] mod_wsgi (pid=6515): Failed to exec Python script file '/home/pi/RaspBerryPiAdhan/www/sysinfo.wsgi'.\n[Tue Jan 21 16:16:19.249609 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] mod_wsgi (pid=6515): Exception occurred processing WSGI script '/home/pi/RaspBerryPiAdhan/www/sysinfo.wsgi'.\n\n[Tue Jan 21 17:16:19.250245 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] Traceback (most recent call last):\n[Tue Jan 21 17:16:19.250679 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757]   File \"/home/pi/RaspBerryPiAdhan/www/sysinfo.wsgi\", line 5, in \n[Tue Jan 21 17:16:19.251735 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757]     from test import app as application\n[Tue Jan 21 17:16:19.252221 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] ImportError: cannot import name app\n\n2021-04-23 17:13:14,425 ERROR    Network plugin raised an exception.\nTraceback (most recent call last):\n  File \"/usr/lib/python3/dist-packages/landscape/sysinfo/sysinfo.py\", line 99, \nin run\n    result = plugin.run()\n  File \"/usr/lib/python3/dist-packages/landscape/sysinfo/network.py\", line 36, \nin run\n    device_info = self._get_device_info()\n  File \"/usr/lib/python3/dist-packages/landscape/lib/network.py\", line 163, in \nget_active_device_info\n    speed, duplex = get_network_interface_speed(\n  File \"/usr/lib/python3/dist-packages/landscape/lib/network.py\", line 249, in \nget_network_interface_speed\n    res = status_cmd.tostring()\nAttributeError: 'array.array' object has no attribute 'tostring'\n\n11/01 18:24:57.726 ERROR| log:0072| post-test sysinfo error: 11/01 18:24:57.727 ERROR| traceback:0013| Traceback (most recent call last): 11/01 18:24:57.728 ERROR| traceback:0013| File \"/tmp/sysinfo/autoserv-0tMj3m/common_lib/log.py\", line 70, in decorated_func 11/01 18:24:57.729 ERROR| traceback:0013| fn(*args, **dargs) 11/01 18:24:57.730 ERROR| traceback:0013| File \"/tmp/sysinfo/autoserv-0tMj3m/bin/base_sysinfo.py\", line 286, in log_after_each_test 11/01 18:24:57.731 ERROR| traceback:0013| old_packages = set(self._installed_packages) 11/01 18:24:57.731 ERROR| traceback:0013| TypeError: 'NoneType' object is not iterable\n\n12/01 19:24:57.726 ERROR| log:0072| post-test sysinfo error: 11/01 18:24:57.727 ERROR| traceback:0013| Traceback (most recent call last): 11/01 18:24:57.728 ERROR| traceback:0013| File \"/tmp/sysinfo/autoserv-0tMj3m/common_lib/log.py\", line 70, in decorated_func 11/01 18:24:57.729 ERROR| traceback:0013| fn(*args, **dargs) 11/01 18:24:57.730 ERROR| traceback:0013| File \"/tmp/sysinfo/autoserv-0tMj3m/bin/base_sysinfo.py\", line 286, in log_after_each_test 11/01 18:24:57.731 ERROR| traceback:0013| old_packages = set(self._installed_packages) 11/01 18:24:57.731 ERROR| traceback:0013| SystemError: no such file or directory\n\nTraceback (most recent call last):\n    File \"/usr/lib/vmware-vpx/vsan-health/pyMoVsan/VsanClusterPrototypeImpl.py\", line 1492, in WaitForUpdateTask\n    WaitForTask(task)\n    File \"/usr/lib/vmware-vpx/pyJack/pyVim/task.py\", line 123, in WaitForTask\n    raise task.info.error\n    vmodl.fault.SystemError: (vmodl.fault.SystemError) {\n    dynamicType = ,\n    dynamicProperty = (vmodl.DynamicProperty) [],\n    msg = 'A general system error occurred: Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.',\n    faultCause = ,\n    faultMessage = (vmodl.LocalizableMessage) [],\n    reason = 'Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.'\n    }\n\n\nThe expected result must be formated with ---------------- for paragraph's separator AND \"Traceback (most recent call last):\" as the beginning of each relevant's paragraph :\n\n\nTraceback (most recent call last):\n    vmodl.fault.SystemError: (vmodl.fault.SystemError) {\n    dynamicType = ,\n    dynamicProperty = (vmodl.DynamicProperty) [],\n    msg = 'A general system error occurred: Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.',\n    faultCause = ,\n    faultMessage = (vmodl.LocalizableMessage) [],\n    reason = 'Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.'\n    }\n----------------\nTraceback (most recent call last):\n[Tue Jan 21 16:16:19.252221 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] SystemError: unable to access /home/dir\n[Tue Jan 21 16:16:19.249067 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] mod_wsgi (pid=6515): Failed to exec Python script file '/home/pi/RaspBerryPiAdhan/www/sysinfo.wsgi'.\n[Tue Jan 21 16:16:19.249609 2020] [wsgi:error] [pid 6515:tid 3041002528] [remote 10.0.0.12:50757] mod_wsgi (pid=6515): Exception occurred processing WSGI script '/home/pi/RaspBerryPiAdhan/www/sysinfo.wsgi'.\n----------------\nTraceback (most recent call last): 11/01 18:24:57.728 ERROR| traceback:0013| File \"/tmp/sysinfo/autoserv-0tMj3m/common_lib/log.py\", line 70, in decorated_func 11/01 18:24:57.729 ERROR| traceback:0013| fn(*args, **dargs) 11/01 18:24:57.730 ERROR| traceback:0013| File \"/tmp/sysinfo/autoserv-0tMj3m/bin/base_sysinfo.py\", line 286, in log_after_each_test 11/01 18:24:57.731 ERROR| traceback:0013| old_packages = set(self._installed_packages) 11/01 18:24:57.731 ERROR| traceback:0013| SystemError: no such file or directory\n----------------\nTraceback (most recent call last):\n    File \"/usr/lib/vmware-vpx/vsan-health/pyMoVsan/VsanClusterPrototypeImpl.py\", line 1492, in WaitForUpdateTask\n    WaitForTask(task)\n    File \"/usr/lib/vmware-vpx/pyJack/pyVim/task.py\", line 123, in WaitForTask\n    raise task.info.error\n    vmodl.fault.SystemError: (vmodl.fault.SystemError) {\n    dynamicType = ,\n    dynamicProperty = (vmodl.DynamicProperty) [],\n    msg = 'A general system error occurred: Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.',\n    faultCause = ,\n    faultMessage = (vmodl.LocalizableMessage) [],\n    reason = 'Unable to complete Sysinfo operation. Please see the VMkernel log file for more details.: Sysinfo error: Bad parameterSee VMkernel log for details.'\n    }\n----------------\n\n", "solution": "with open('Traceback.txt', 'r' ) as f:\n    rawText = f.read()\n\nparagraphs = rawText.split( \"\\n\\n\" )\n\nfor p in paragraphs:\n    if \"SystemError\" in p:\n\n        index = p.find( \"Traceback (most recent call last):\" )\n\n        if -1 != index:\n            print( p[index:] )\n            print( \"----------------\" )\n"}
{"title": "Selectively replace multiple instances of a character within a string", "language": "Python from Julia", "task": "Task\nThis is admittedly a trivial task but I thought it would be interesting to see how succinctly (or otherwise) different languages can handle it.\n\nGiven the string: \"abracadabra\", replace programatically:\n\n* the first 'a' with 'A'\n* the second 'a' with 'B'\n* the fourth 'a' with 'C'\n* the fifth 'a' with 'D'\n* the first 'b' with 'E'\n* the second 'r' with 'F'\n\nNote that there is no replacement for the third 'a', second 'b' or first 'r'.\n\nThe answer should, of course, be : \"AErBcadCbFD\".\n\n\n", "solution": "import functools\n\nfrom typing import Iterable\nfrom typing import Tuple\n\n\n@functools.cache\ndef find_nth(s: str, sub: str, n: int) -> int:\n    assert n >= 1\n    if n == 1:\n        return s.find(sub)\n    return s.find(sub, find_nth(s, sub, n - 1) + 1)\n\n\ndef selective_replace(s: str, ops: Iterable[Tuple[int, str, str]]) -> str:\n    chars = list(s)\n    for n, old, new in ops:\n        chars[find_nth(s, old, n)] = new\n    return \"\".join(chars)\n\n\nprint(\n    selective_replace(\n        \"abracadabra\",\n        [\n            (1, \"a\", \"A\"),  # the first 'a' with 'A'\n            (2, \"a\", \"B\"),  # the second 'a' with 'B'\n            (4, \"a\", \"C\"),  # the fourth 'a' with 'C'\n            (5, \"a\", \"D\"),  # the fifth 'a' with 'D'\n            (1, \"b\", \"E\"),  # the first 'b' with 'E'\n            (2, \"r\", \"F\"),  # the second 'r' with 'F'\n        ],\n    )\n)"}
{"title": "Self-describing numbers", "language": "Python", "task": "{{task}}There are several so-called \"self-describing\" or \"self-descriptive\" integers.\n\nAn integer is said to be \"self-describing\" if it has the property that, when digit positions are labeled 0 to N-1, the digit in each position is equal to the number of times that that digit appears in the number.\n\nFor example,   '''2020'''   is a four-digit self describing number:\n\n*   position   0   has value   2   and there are two 0s in the number;\n*   position   1   has value   0   and there are no 1s in the number;\n*   position   2   has value   2   and there are two 2s;\n*   position   3   has value   0   and there are zero 3s.\n\n\nSelf-describing numbers < 100.000.000  are:     1210,   2020,   21200,   3211000,   42101000.\n\n\n;Task Description\n# Write a function/routine/method/... that will check whether a given positive integer is self-describing.\n# As an optional stretch goal - generate and display the set of self-describing numbers.\n\n\n;Related tasks:\n*   [[Fours is the number of letters in the ...]]\n*   [[Look-and-say sequence]]\n*   [[Number names]]\n*   [[Self-referential sequence]]\n*   [[Spelling of ordinal numbers]]\n\n", "solution": ">>> def isSelfDescribing(n):\n\ts = str(n)\n\treturn all(s.count(str(i)) == int(ch) for i, ch in enumerate(s))\n\n>>> [x for x in range(4000000) if isSelfDescribing(x)]\n[1210, 2020, 21200, 3211000]\n>>> [(x, isSelfDescribing(x)) for x in (1210, 2020, 21200, 3211000, 42101000, 521001000, 6210001000)]\n[(1210, True), (2020, True), (21200, True), (3211000, True), (42101000, True), (521001000, True), (6210001000, True)]"}
{"title": "Self numbers", "language": "Python 2.7", "task": "A number n is a self number if there is no number g such that g + the sum of g's digits = n. So 18 is not a self number because 9+9=18, 43 is not a self number because 35+5+3=43.\n\nThe task is:\n  Display the first 50 self numbers;\n  I believe that the 100000000th self number is 1022727208. You should either confirm or dispute my conjecture.\n\n224036583-1 is a Mersenne prime, claimed to also be a self number. Extra credit to anyone proving it.\n\n;See also: \n\n;*OEIS: A003052 - Self numbers or Colombian numbers\n;*Wikipedia: Self numbers\n\n", "solution": "class DigitSumer :\n    def __init__(self): \n        sumdigit = lambda n : sum( map( int,str( n )))\n        self.t = [sumdigit( i ) for i in xrange( 10000 )]\n    def __call__ ( self,n ):\n        r = 0\n        while n >= 10000 :\n            n,q = divmod( n,10000 )\n            r += self.t[q]\n        return r + self.t[n] \n\n\ndef self_numbers ():\n    d = DigitSumer()\n    s = set([])\n    i = 1\n    while 1 :\n        n = i + d( i )\n        if i in s :\n            s.discard( i )\n        else:\n            yield i\n        s.add( n )\n        i += 1\n\nimport time\np = 100\nt = time.time()\nfor i,s in enumerate( self_numbers(),1 ):\n    if i <= 50 : \n        print s,\n        if i == 50 : print\n    if i == p :\n        print '%7.1f sec  %9dth = %d'%( time.time()-t,i,s )\n        p *= 10"}
{"title": "Semordnilap", "language": "Python", "task": "A semordnilap is a word (or phrase) that spells a different word (or phrase) backward. \"Semordnilap\" is a word that itself is a semordnilap.\n\nExample: ''lager'' and ''regal''\n \n;Task\nThis task does not consider semordnilap phrases, only single words.\nUsing only words from this list, report the total number of unique semordnilap pairs, and print 5 examples.\nTwo matching semordnilaps, such as ''lager'' and ''regal'', should be counted as one unique pair.\n(Note that the word \"semordnilap\" is not in the above dictionary.)\n\n\n\n", "solution": ">>> with open('unixdict.txt') as f:\n\twordset = set(f.read().strip().split())\n\n>>> revlist = (''.join(word[::-1]) for word in wordset)\n>>> pairs   = set((word, rev) for word, rev in zip(wordset, revlist) \n                  if word < rev and rev in wordset)\n>>> len(pairs)\n158\n>>> sorted(pairs, key=lambda p: (len(p[0]), p))[-5:]\n[('damon', 'nomad'), ('lager', 'regal'), ('leper', 'repel'), ('lever', 'revel'), ('kramer', 'remark')]\n>>> "}
{"title": "Semordnilap", "language": "Python 3.7", "task": "A semordnilap is a word (or phrase) that spells a different word (or phrase) backward. \"Semordnilap\" is a word that itself is a semordnilap.\n\nExample: ''lager'' and ''regal''\n \n;Task\nThis task does not consider semordnilap phrases, only single words.\nUsing only words from this list, report the total number of unique semordnilap pairs, and print 5 examples.\nTwo matching semordnilaps, such as ''lager'' and ''regal'', should be counted as one unique pair.\n(Note that the word \"semordnilap\" is not in the above dictionary.)\n\n\n\n", "solution": "'''Dictionary words paired by equivalence under reversal'''\n\nfrom functools import (reduce)\nfrom itertools import (chain)\nimport urllib.request\n\n\n# semordnilaps :: [String] -> [String]\ndef semordnilaps(xs):\n    '''The subset of words in a list which\n       are paired (by equivalence under reversal)\n       with other words in that list.\n    '''\n    def go(tpl, w):\n        (s, ws) = tpl\n        if w[::-1] in s:\n            return (s, ws + [w])\n        else:\n            s.add(w)\n            return (s, ws)\n    return reduce(go, xs, (set(), []))[1]\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''Test'''\n\n    url = 'http://wiki.puzzlers.org/pub/wordlists/unixdict.txt'\n    ws = semordnilaps(\n        urllib.request.urlopen(\n            url\n        ).read().splitlines()\n    )\n    print(\n        fTable(\n            __doc__ + ':\\n\\n(longest of ' +\n            str(len(ws)) + ' in ' + url + ')\\n'\n        )(snd)(fst)(identity)(\n            sorted(\n                concatMap(\n                    lambda x: (\n                        lambda s=x.decode('utf8'): [\n                            (s, s[::-1])\n                        ] if 4 < len(x) else []\n                    )()\n                )(ws),\n                key=compose(len)(fst),\n                reverse=True\n            )\n        )\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).'''\n    return lambda xs: list(\n        chain.from_iterable(map(f, xs))\n    )\n\n\n# FORMATTING ----------------------------------------------\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) -> (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n                     f -> xs -> tabular string.\n    '''\n    def go(xShow, fxShow, f, xs):\n        ys = [xShow(x) for x in xs]\n        w = max(map(len, ys))\n        return s + '\\n' + '\\n'.join(map(\n            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),\n            xs, ys\n        ))\n    return lambda xShow: lambda fxShow: lambda f: lambda xs: go(\n        xShow, fxShow, f, xs\n    )\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# identity :: a -> a\ndef identity(x):\n    '''The identity function.'''\n    return x\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second member of a pair.'''\n    return tpl[1]\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Sequence: nth number with exactly n divisors", "language": "Python", "task": "Calculate the sequence where each term an is the nth that has '''n''' divisors.\n\n;Task\n\nShow here, on this page, at least the first '''15''' terms of the sequence.\n\n;See also\n\n:*OEIS:A073916\n\n;Related tasks\n\n:*[[Sequence: smallest number greater than previous term with exactly n divisors]]\n:*[[Sequence: smallest number with exactly n divisors]]\n\n", "solution": "def divisors(n):\n    divs = [1]\n    for ii in range(2, int(n ** 0.5) + 3):\n        if n % ii == 0:\n            divs.append(ii)\n            divs.append(int(n / ii))\n    divs.append(n)\n    return list(set(divs))\n\n\ndef is_prime(n):\n    return len(divisors(n)) == 2\n\n\ndef primes():\n    ii = 1\n    while True:\n        ii += 1\n        if is_prime(ii):\n            yield ii\n\n\ndef prime(n):\n    generator = primes()\n    for ii in range(n - 1):\n        generator.__next__()\n    return generator.__next__()\n\n\ndef n_divisors(n):\n    ii = 0\n    while True:\n        ii += 1\n        if len(divisors(ii)) == n:\n            yield ii\n\n\ndef sequence(max_n=None):\n    if max_n is not None:\n        for ii in range(1, max_n + 1):\n            if is_prime(ii):\n                yield prime(ii) ** (ii - 1)\n            else:\n                generator = n_divisors(ii)\n                for jj, out in zip(range(ii - 1), generator):\n                    pass\n                yield generator.__next__()\n    else:\n        ii = 1\n        while True:\n            ii += 1\n            if is_prime(ii):\n                yield prime(ii) ** (ii - 1)\n            else:\n                generator = n_divisors(ii)\n                for jj, out in zip(range(ii - 1), generator):\n                    pass\n                yield generator.__next__()\n\n\nif __name__ == '__main__':\n    for item in sequence(15):\n        print(item)\n"}
{"title": "Sequence: smallest number greater than previous term with exactly n divisors", "language": "Python", "task": "Calculate the sequence where each term an is the '''smallest natural number''' greater than the previous term, that has exactly '''n''' divisors.\n\n\n;Task\nShow here, on this page, at least the first '''15''' terms of the sequence.\n\n\n;See also\n:* OEIS:A069654\n\n\n;Related tasks\n:* [[Sequence: smallest number with exactly n divisors]]\n:* [[Sequence: nth number with exactly n divisors]]\n\n", "solution": "def divisors(n):\n    divs = [1]\n    for ii in range(2, int(n ** 0.5) + 3):\n        if n % ii == 0:\n            divs.append(ii)\n            divs.append(int(n / ii))\n    divs.append(n)\n    return list(set(divs))\n\n\ndef sequence(max_n=None):\n    previous = 0\n    n = 0\n    while True:\n        n += 1\n        ii = previous\n        if max_n is not None:\n            if n > max_n:\n                break\n        while True:\n            ii += 1\n            if len(divisors(ii)) == n:\n                yield ii\n                previous = ii\n                break\n\n\nif __name__ == '__main__':\n    for item in sequence(15):\n        print(item)\n"}
{"title": "Sequence: smallest number with exactly n divisors", "language": "Python", "task": "Calculate the sequence where each term   an   is the '''smallest natural number''' that has exactly   '''n'''   divisors.\n\n\n;Task\nShow here, on this page, at least the first  '''15'''  terms of the sequence.\n\n\n;Related tasks:\n:* [[Sequence: smallest number greater than previous term with exactly n divisors]]\n:* [[Sequence: nth number with exactly n divisors]]\n\n\n;See also:\n:* OEIS:A005179\n\n", "solution": "def divisors(n):\n    divs = [1]\n    for ii in range(2, int(n ** 0.5) + 3):\n        if n % ii == 0:\n            divs.append(ii)\n            divs.append(int(n / ii))\n    divs.append(n)\n    return list(set(divs))\n\n\ndef sequence(max_n=None):\n    n = 0\n    while True:\n        n += 1\n        ii = 0\n        if max_n is not None:\n            if n > max_n:\n                break\n        while True:\n            ii += 1\n            if len(divisors(ii)) == n:\n                yield ii\n                break\n\n\nif __name__ == '__main__':\n    for item in sequence(15):\n        print(item)"}
{"title": "Set consolidation", "language": "Python", "task": "Given two sets of items then if any item is common to any set then the result of applying ''consolidation'' to those sets is a set of sets whose contents is:\n*  The two input sets if no common item exists between the two input sets of items.\n*  The single set that is the union of the two input sets if they share a common item.\n\nGiven N sets of items where N>2 then the result is the same as repeatedly replacing all combinations of two sets by their consolidation until no further consolidation between set pairs is possible.\nIf N<2 then consolidation has no strict meaning and the input can be returned.\n\n;'''Example 1:'''\n:Given the two sets {A,B} and {C,D} then there is no common element between the sets and the result is the same as the input.\n;'''Example 2:'''\n:Given the two sets {A,B} and {B,D} then there is a common element B between the sets and the result is the single set {B,D,A}.  (Note that order of items in a set is immaterial: {A,B,D} is the same as {B,D,A} and {D,A,B}, etc).\n;'''Example 3:'''\n:Given the three sets {A,B} and {C,D} and {D,B} then there is no common element between the sets {A,B} and {C,D} but the sets {A,B} and {D,B} do share a common element that consolidates to produce the result {B,D,A}. On examining this result with the remaining set, {C,D}, they share a common element and so consolidate to the final output of the single set {A,B,C,D}\n;'''Example 4:'''\n:The consolidation of the five sets:\n::{H,I,K}, {A,B}, {C,D}, {D,B}, and {F,G,H}\n:Is the two sets:\n::{A, C, B, D}, and {G, F, I, H, K}\n\n'''See also'''\n* Connected component (graph theory)\n* [[Range consolidation]]\n\n", "solution": "def _test(consolidate=consolidate):\n    \n    def freze(list_of_sets):\n        'return a set of frozensets from the list of sets to allow comparison'\n        return set(frozenset(s) for s in list_of_sets)\n        \n    # Define some variables\n    A,B,C,D,E,F,G,H,I,J,K = 'A,B,C,D,E,F,G,H,I,J,K'.split(',')\n    # Consolidate some lists of sets\n    assert (freze(consolidate([{A,B}, {C,D}])) == freze([{'A', 'B'}, {'C', 'D'}]))\n    assert (freze(consolidate([{A,B}, {B,D}])) == freze([{'A', 'B', 'D'}]))\n    assert (freze(consolidate([{A,B}, {C,D}, {D,B}])) == freze([{'A', 'C', 'B', 'D'}]))\n    assert (freze(consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}])) ==\n             freze([{'A', 'C', 'B', 'D'}, {'G', 'F', 'I', 'H', 'K'}]))\n    assert (freze(consolidate([{A,H}, {H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}])) ==\n             freze([{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}]))\n    assert (freze(consolidate([{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}])) ==\n             freze([{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}]))\n    # Confirm order-independence\n    from copy import deepcopy\n    import itertools\n    sets = [{H,I,K}, {A,B}, {C,D}, {D,B}, {F,G,H}, {A,H}]\n    answer = consolidate(deepcopy(sets))\n    for perm in itertools.permutations(sets):\n            assert consolidate(deepcopy(perm)) == answer\n \n    assert (answer == [{'A', 'C', 'B', 'D', 'G', 'F', 'I', 'H', 'K'}])\n    assert (len(list(itertools.permutations(sets))) == 720)\n    \n    print('_test(%s) complete' % consolidate.__name__)\n\nif __name__ == '__main__':\n    _test(consolidate)\n    _test(conso)"}
{"title": "Set consolidation", "language": "Python 3.7", "task": "Given two sets of items then if any item is common to any set then the result of applying ''consolidation'' to those sets is a set of sets whose contents is:\n*  The two input sets if no common item exists between the two input sets of items.\n*  The single set that is the union of the two input sets if they share a common item.\n\nGiven N sets of items where N>2 then the result is the same as repeatedly replacing all combinations of two sets by their consolidation until no further consolidation between set pairs is possible.\nIf N<2 then consolidation has no strict meaning and the input can be returned.\n\n;'''Example 1:'''\n:Given the two sets {A,B} and {C,D} then there is no common element between the sets and the result is the same as the input.\n;'''Example 2:'''\n:Given the two sets {A,B} and {B,D} then there is a common element B between the sets and the result is the single set {B,D,A}.  (Note that order of items in a set is immaterial: {A,B,D} is the same as {B,D,A} and {D,A,B}, etc).\n;'''Example 3:'''\n:Given the three sets {A,B} and {C,D} and {D,B} then there is no common element between the sets {A,B} and {C,D} but the sets {A,B} and {D,B} do share a common element that consolidates to produce the result {B,D,A}. On examining this result with the remaining set, {C,D}, they share a common element and so consolidate to the final output of the single set {A,B,C,D}\n;'''Example 4:'''\n:The consolidation of the five sets:\n::{H,I,K}, {A,B}, {C,D}, {D,B}, and {F,G,H}\n:Is the two sets:\n::{A, C, B, D}, and {G, F, I, H, K}\n\n'''See also'''\n* Connected component (graph theory)\n* [[Range consolidation]]\n\n", "solution": "'''Set consolidation'''\n\nfrom functools import (reduce)\n\n\n# consolidated :: Ord a => [Set a] -> [Set a]\ndef consolidated(sets):\n    '''A consolidated list of sets.'''\n    def go(xs, s):\n        if xs:\n            h = xs[0]\n            return go(xs[1:], h.union(s)) if (\n                h.intersection(s)\n            ) else [h] + go(xs[1:], s)\n        else:\n            return [s]\n    return reduce(go, sets, [])\n\n\n# TESTS ---------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Illustrative consolidations.'''\n\n    print(\n        tabulated('Consolidation of sets of characters:')(\n            lambda x: str(list(map(compose(concat)(list), x)))\n        )(str)(\n            consolidated\n        )(list(map(lambda xs: list(map(set, xs)), [\n            ['ab', 'cd'],\n            ['ab', 'bd'],\n            ['ab', 'cd', 'db'],\n            ['hik', 'ab', 'cd', 'db', 'fgh']\n        ])))\n    )\n\n\n# DISPLAY OF RESULTS --------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# concat :: [String] -> String\ndef concat(xs):\n    '''Concatenation of strings in xs.'''\n    return ''.join(xs)\n\n\n# tabulated :: String -> (a -> String) ->\n#                        (b -> String) ->\n#                        (a -> b) -> [a] -> String\ndef tabulated(s):\n    '''Heading -> x display function -> fx display function ->\n          f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + '\\n'.join([\n            xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs\n        ])\n    return lambda xShow: lambda fxShow: (\n        lambda f: lambda xs: go(\n            xShow, fxShow, f, xs\n        )\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Set of real numbers", "language": "Python", "task": "All real numbers form the uncountable set R.  Among its subsets, relatively simple are the convex sets, each expressed as a range between two real numbers ''a'' and ''b'' where ''a'' <= ''b''.  There are actually four cases for the meaning of \"between\", depending on open or closed boundary:\n* [''a'', ''b'']: {''x'' | ''a'' <= ''x'' and ''x'' <= ''b'' }\n* (''a'', ''b''): {''x'' | ''a'' < ''x'' and ''x'' < ''b'' }\n* [''a'', ''b''): {''x'' | ''a'' <= ''x'' and ''x'' < ''b'' }\n* (''a'', ''b'']: {''x'' | ''a'' < ''x'' and ''x'' <= ''b'' }\nNote that if ''a'' = ''b'', of the four only [''a'', ''a''] would be non-empty.\n\n'''Task'''\n* Devise a way to represent any set of real numbers, for the definition of 'any' in the implementation notes below.\n* Provide methods for these common set operations (''x'' is a real number; ''A'' and ''B'' are sets):\n:* ''x''  ''A'': determine if ''x'' is an element of ''A''\n:: example: 1 is in [1, 2), while 2, 3, ... are not.\n:* ''A''  ''B'': union of ''A'' and ''B'', i.e. {''x'' | ''x''  ''A'' or ''x''  ''B''}\n:: example: [0, 2)  (1, 3) = [0, 3); [0, 1)  (2, 3] = well, [0, 1)  (2, 3]\n:* ''A''  ''B'': intersection of ''A'' and ''B'', i.e. {''x'' | ''x''  ''A'' and ''x''  ''B''}\n:: example: [0, 2)  (1, 3) = (1, 2); [0, 1)  (2, 3] = empty set\n:* ''A'' - ''B'': difference between ''A'' and ''B'', also written as ''A'' \\ ''B'', i.e. {''x'' | ''x''  ''A'' and ''x''  ''B''}\n:: example: [0, 2) - (1, 3) = [0, 1]\n* Test your implementation by checking if numbers 0, 1, and 2 are in any of the following sets:\n:* (0, 1]  [0, 2)\n:* [0, 2)  (1, 2]\n:* [0, 3) - (0, 1)\n:* [0, 3) - [0, 1]\n\n'''Implementation notes'''\n* 'Any' real set means 'sets that can be expressed as the union of a finite number of convex real sets'.  Cantor's set needs not apply.\n* Infinities should be handled gracefully; indeterminate numbers (NaN) can be ignored.\n* You can use your machine's native real number representation, which is probably IEEE floating point, and assume it's good enough (it usually is).\n\n'''Optional work'''\n* Create a function to determine if a given set is empty (contains no element).\n* Define ''A'' = {''x'' | 0 < ''x'' < 10 and |sin(p ''x''2)| > 1/2 }, ''B'' = {''x'' | 0 < ''x'' < 10 and |sin(p ''x'')| > 1/2}, calculate the length of the real axis covered by the set ''A'' - ''B''.  Note that \n|sin(p ''x'')| > 1/2 is the same as ''n'' + 1/6 < ''x'' < ''n'' + 5/6 for all integers ''n''; your program does not need to derive this by itself.\n\n", "solution": "class Setr():\n    def __init__(self, lo, hi, includelo=True, includehi=False):\n        self.eqn = \"(%i<%sX<%s%i)\" % (lo,\n                                      '=' if includelo else '',\n                                      '=' if includehi else '',\n                                      hi)\n\n    def __contains__(self, X):\n        return eval(self.eqn, locals())\n\n    # union\n    def __or__(self, b):\n        ans = Setr(0,0)\n        ans.eqn = \"(%sor%s)\" % (self.eqn, b.eqn)\n        return ans\n\n    # intersection\n    def __and__(self, b):\n        ans = Setr(0,0)\n        ans.eqn = \"(%sand%s)\" % (self.eqn, b.eqn)\n        return ans\n\n    # difference\n    def __sub__(self, b):\n        ans = Setr(0,0)\n        ans.eqn = \"(%sand not%s)\" % (self.eqn, b.eqn)\n        return ans\n\n    def __repr__(self):\n        return \"Setr%s\" % self.eqn\n\n\nsets = [\n    Setr(0,1, 0,1) | Setr(0,2, 1,0),\n    Setr(0,2, 1,0) & Setr(1,2, 0,1),\n    Setr(0,3, 1,0) - Setr(0,1, 0,0),\n    Setr(0,3, 1,0) - Setr(0,1, 1,1),\n]\nsettexts = '(0, 1] \u222a [0, 2);[0, 2) \u2229 (1, 2];[0, 3) \u2212 (0, 1);[0, 3) \u2212 [0, 1]'.split(';')\n\nfor s,t in zip(sets, settexts):\n    print(\"Set %s %s. %s\" % (t,\n                             ', '.join(\"%scludes %i\"\n                                     % ('in' if v in s else 'ex', v)\n                                     for v in range(3)),\n                             s.eqn))"}
{"title": "Set right-adjacent bits", "language": "Python", "task": "Given a left-to-right ordered collection of e bits, b, where 1 <= e <= 10000, \nand a zero or more integer n :\n* Output the result of setting the n bits to the right of any set bit in b \n(if those bits are present in b and therefore also preserving the width, e).\n    \n'''Some examples:''' \n\n    Set of examples showing how one bit in a nibble gets changed:\n        \n        n = 2; Width e = 4:\n        \n             Input b: 1000\n              Result: 1110\n        \n             Input b: 0100\n              Result: 0111\n        \n             Input b: 0010\n              Result: 0011\n        \n             Input b: 0000\n              Result: 0000\n    \n    Set of examples with the same input with set bits of varying distance apart:\n    \n        n = 0; Width e = 66:\n        \n             Input b: 010000000000100000000010000000010000000100000010000010000100010010\n              Result: 010000000000100000000010000000010000000100000010000010000100010010\n        \n        n = 1; Width e = 66:\n        \n             Input b: 010000000000100000000010000000010000000100000010000010000100010010\n              Result: 011000000000110000000011000000011000000110000011000011000110011011\n        \n        n = 2; Width e = 66:\n        \n             Input b: 010000000000100000000010000000010000000100000010000010000100010010\n              Result: 011100000000111000000011100000011100000111000011100011100111011111\n        \n        n = 3; Width e = 66:\n        \n             Input b: 010000000000100000000010000000010000000100000010000010000100010010\n              Result: 011110000000111100000011110000011110000111100011110011110111111111\n\n'''Task:'''\n\n* Implement a routine to perform the setting of right-adjacent bits on representations of bits that will scale over the given range of input width e.\n* Use it to show, here, the results for the input examples above.\n* Print the output aligned in a way that allows easy checking by eye of the binary input vs output.\n\n", "solution": "from operator import or_\nfrom functools import reduce\n\ndef set_right_adjacent_bits(n: int, b: int) -> int:\n    return reduce(or_, (b >> x for x in range(n+1)), 0)\n\n\nif __name__ == \"__main__\":\n    print(\"SAME n & Width.\\n\")\n    n = 2  # bits to the right of set bits, to also set\n    bits = \"1000 0100 0010 0000\"\n    first = True\n    for b_str in bits.split():\n        b = int(b_str, 2)\n        e = len(b_str)\n        if first:\n            first = False\n            print(f\"n = {n}; Width e = {e}:\\n\")\n        result = set_right_adjacent_bits(n, b)\n        print(f\"     Input b: {b:0{e}b}\")\n        print(f\"      Result: {result:0{e}b}\\n\")\n        \n    print(\"SAME Input & Width.\\n\")\n    #bits = \"01000010001001010110\"\n    bits = '01' + '1'.join('0'*x for x in range(10, 0, -1))\n    for n in range(4):\n        first = True\n        for b_str in bits.split():\n            b = int(b_str, 2)\n            e = len(b_str)\n            if first:\n                first = False\n                print(f\"n = {n}; Width e = {e}:\\n\")\n            result = set_right_adjacent_bits(n, b)\n            print(f\"     Input b: {b:0{e}b}\")\n            print(f\"      Result: {result:0{e}b}\\n\")"}
{"title": "Shoelace formula for polygonal area", "language": "Python", "task": "Given the n + 1 vertices x[0], y[0] .. x[N], y[N] of a simple polygon described in a clockwise direction, then the polygon's area can be calculated by:\n\nabs( (sum(x[0]*y[1] + ... x[n-1]*y[n]) + x[N]*y[0]) -\n     (sum(x[1]*y[0] + ... x[n]*y[n-1]) + x[0]*y[N])\n   ) / 2\n(Where abs returns the absolute value)\n\n;Task:\nWrite a function/method/routine to use the the Shoelace formula to calculate the area of the polygon described by the ordered points:\n      (3,4), (5,11), (12,8), (9,5), and (5,6) \n\n\nShow the answer here, on this page.\n\n", "solution": "# Even simpler:\n# In python we can take an advantage of that x[-1] refers to the last element in an array, same as x[N-1].\n# Introducing the index i=[0,1,2,...,N-1]; i-1=[-1,0,...,N-2]; N is the number of vertices of a polygon.\n# Thus x[i] is a sequence of the x-coordinate of the polygon vertices, x[i-1] is the sequence shifted by 1 index.\n# Note that the shift must be negative. The positive shift x[i+1] results in an error: x[N] index out of bound.\n\nimport numpy as np\n# x,y are arrays containing coordinates of the polygon vertices\nx=np.array([3,5,12,9,5]) \ny=np.array([4,11,8,5,6]) \ni=np.arange(len(x))\n#Area=np.sum(x[i-1]*y[i]-x[i]*y[i-1])*0.5 # signed area, positive if the vertex sequence is counterclockwise\nArea=np.abs(np.sum(x[i-1]*y[i]-x[i]*y[i-1])*0.5) # one line of code for the shoelace formula\n\n# Remember that applying the Shoelace formula\n# will result in a loss of precision if x,y have big offsets.\n# Remove the offsets first, e.g. \n# x=x-np.mean(x);y=y-np.mean(y)\n# or\n# x=x-x[0];y=y-y[0]\n# before applying the Shoelace formula.   \n\n"}
{"title": "Shoelace formula for polygonal area", "language": "Python 3.7", "task": "Given the n + 1 vertices x[0], y[0] .. x[N], y[N] of a simple polygon described in a clockwise direction, then the polygon's area can be calculated by:\n\nabs( (sum(x[0]*y[1] + ... x[n-1]*y[n]) + x[N]*y[0]) -\n     (sum(x[1]*y[0] + ... x[n]*y[n-1]) + x[0]*y[N])\n   ) / 2\n(Where abs returns the absolute value)\n\n;Task:\nWrite a function/method/routine to use the the Shoelace formula to calculate the area of the polygon described by the ordered points:\n      (3,4), (5,11), (12,8), (9,5), and (5,6) \n\n\nShow the answer here, on this page.\n\n", "solution": "'''Polygonal area by shoelace formula'''\n\nfrom itertools import cycle, islice\nfrom functools import reduce\nfrom operator import sub\n\n# --------- SHOELACE FORMULA FOR POLYGONAL AREA ----------\n\n# shoelaceArea :: [(Float, Float)] -> Float\ndef shoelaceArea(xys):\n    '''Area of polygon with vertices\n       at (x, y) points in xys.\n    '''\n    def go(a, tpl):\n        l, r = a\n        (x, y), (dx, dy) = tpl\n        return l + x * dy, r + y * dx\n\n    return abs(sub(*reduce(\n        go,\n        zip(\n            xys,\n            islice(cycle(xys), 1, None)\n        ),\n        (0, 0)\n    ))) / 2\n\n\n# ------------------------- TEST -------------------------\n# main :: IO()\ndef main():\n    '''Sample calculation'''\n\n    ps = [(3, 4), (5, 11), (12, 8), (9, 5), (5, 6)]\n    print(__doc__ + ':')\n    print(repr(ps) + '  ->  ' + str(shoelaceArea(ps)))\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Shortest common supersequence", "language": "Python", "task": "The   '''shortest common supersequence'''   is a problem closely related to the   [[longest common subsequence]],   which you can use as an external function for this task.\n\n\n;;Task:\nGiven two strings u and v, find the shortest possible sequence s, which is the shortest common super-sequence of u and v where both u and v are a subsequence of s.  Defined as such, s is not necessarily unique.\n\nDemonstrate this by printing s where u = \"abcbdab\" and v = \"bdcaba\".\n\n\n\n;Also see:\n* Wikipedia: shortest common supersequence \n\n", "solution": "# Use the Longest Common Subsequence algorithm\n\ndef shortest_common_supersequence(a, b):\n    lcs = longest_common_subsequence(a, b)\n    scs = \"\"\n    # Consume lcs\n    while len(lcs) > 0:\n        if a[0]==lcs[0] and b[0]==lcs[0]:\n        # Part of the LCS, so consume from all strings\n            scs += lcs[0]\n            lcs = lcs[1:]\n            a = a[1:]\n            b = b[1:]\n        elif a[0]==lcs[0]:\n            scs += b[0]\n            b = b[1:]\n        else:\n            scs += a[0]\n            a = a[1:]\n    # append remaining characters\n    return scs + a + b\n"}
{"title": "Show ASCII table", "language": "Python from Go", "task": "Show  the ASCII character set  from values   '''32'''   to   '''127'''   (decimal)   in a table format.\n\n\n\n", "solution": "for i in range(16):\n    for j in range(32+i, 127+1, 16):\n        if j == 32:\n            k = 'Spc'\n        elif j == 127:\n            k = 'Del'\n        else:\n            k = chr(j)\n        print(\"%3d : %-3s\" % (j,k), end=\"\")\n    print()\n"}
{"title": "Show the epoch", "language": "Python", "task": "Choose popular date libraries used by your language and show the   epoch   those libraries use. \n\nA demonstration is preferable   (e.g. setting the internal representation of the date to 0 ms/ns/etc.,   or another way that will still show the epoch even if it is changed behind the scenes by the implementers),   but text from (with links to) documentation is also acceptable where a demonstration is impossible/impractical. \n\nFor consistency's sake, show the date in UTC time where possible.\n\n\n;Related task:\n*   [[Date format]]\n\n", "solution": ">>> import time\n>>> time.asctime(time.gmtime(0))\n'Thu Jan  1 00:00:00 1970'\n>>>"}
{"title": "Sierpinski pentagon", "language": "Python", "task": "Produce a graphical or ASCII-art representation of a Sierpinski pentagon (aka a Pentaflake) of order 5. Your code should also be able to correctly generate representations of lower orders: 1 to 4.\n\n\n;See also\n* Sierpinski pentagon\n\n", "solution": "from turtle import *\nimport math\nspeed(0)      # 0 is the fastest speed. Otherwise, 1 (slow) to 10 (fast)\nhideturtle()  # hide the default turtle\n\npart_ratio = 2 * math.cos(math.radians(72))\nside_ratio = 1 / (part_ratio + 2)\n\nhide_turtles = True   # show/hide turtles as they draw\npath_color = \"black\"  # path color\nfill_color = \"black\"  # fill color\n\n# turtle, size\ndef pentagon(t, s):\n  t.color(path_color, fill_color)\n  t.pendown()\n  t.right(36)\n  t.begin_fill()\n  for i in range(5):\n    t.forward(s)\n    t.right(72)\n  t.end_fill()\n\n# iteration, turtle, size\ndef sierpinski(i, t, s):\n  t.setheading(0)\n  new_size = s * side_ratio\n  \n  if i > 1:\n    i -= 1\n    \n    # create four more turtles\n    for j in range(4):\n      t.right(36)\n      short = s * side_ratio / part_ratio\n      dist = [short, s, s, short][j]\n      \n      # spawn a turtle\n      spawn = Turtle()\n      if hide_turtles:spawn.hideturtle()\n      spawn.penup()\n      spawn.setposition(t.position())\n      spawn.setheading(t.heading())\n      spawn.forward(dist)\n      \n      # recurse for spawned turtles\n      sierpinski(i, spawn, new_size)\n    \n    # recurse for parent turtle\n    sierpinski(i, t, new_size)\n    \n  else:\n    # draw a pentagon\n    pentagon(t, s)\n    # delete turtle\n    del t\n\ndef main():\n  t = Turtle()\n  t.hideturtle()\n  t.penup()\n  screen = t.getscreen()\n  y = screen.window_height()\n  t.goto(0, y/2-20)\n  \n  i = 5       # depth. i >= 1\n  size = 300  # side length\n  \n  # so the spawned turtles move only the distance to an inner pentagon\n  size *= part_ratio\n  \n  # begin recursion\n  sierpinski(i, t, size)\n\nmain()"}
{"title": "Simple turtle graphics", "language": "Python from Quackery", "task": "The first turtle graphic discussed in Mindstorms: Children, Computers, and Powerful Ideas by Seymour Papert is a simple drawing of a house. It is a square with a triangle on top for the roof. \n\nFor a slightly more advanced audience, a more practical introduction to turtle graphics might be to draw a bar chart. \n\nthumb\n\n;Task:\n::* Create a function (or subroutine) that uses turtle graphics to draw a house of a specified size as depicted. Optionally make it lovely by adding details such as, for example, doors and windows.\n\n::* Create a function (or subroutine) that takes a list (array, vector) of non-negative numbers and draws a bar chart from them, scaled to fit exactly in a square of a specified size. The enclosing square need not be drawn. \n\n::* Both functions should return the turtle to the location it was at and facing in the same direction  as it was immediately before the function was executed. \n\n", "solution": "from turtle import *\n\ndef rectangle(width, height):\n    for _ in range(2):\n        forward(height)\n        left(90)\n        forward(width)\n        left(90)\n \ndef square(size):\n    rectangle(size, size)\n\ndef triangle(size):\n    for _ in range(3):\n        forward(size)\n        right(120)\n    \ndef house(size):\n    right(180)\n    square(size)\n    triangle(size)\n    right(180)\n    \ndef barchart(lst, size):\n    scale = size/max(lst)\n    width = size/len(lst)\n    for i in lst:\n        rectangle(i*scale, width)\n        penup()\n        forward(width)\n        pendown()\n    penup()\n    back(size)\n    pendown()\n    \nclearscreen()\nhideturtle()\nhouse(150)\npenup()\nforward(10)\npendown()\nbarchart([0.5, (1/3), 2, 1.3, 0.5], 200)\npenup()\nback(10)\npendown()"}
{"title": "Smith numbers", "language": "Python", "task": "sum of the decimal digits of the integers that make up that number is the same as the sum of the decimal digits of its prime factors excluding 1. \n\nBy definition, all primes are ''excluded'' as they (naturally) satisfy this condition!\n\nSmith numbers are also known as   ''joke''   numbers.\n\n\n;Example\nUsing the number '''166'''\nFind the prime factors of '''166''' which are: '''2''' x '''83'''\nThen, take those two prime factors and sum all their decimal digits: '''2 + 8 + 3''' which is '''13'''\nThen, take the decimal digits of '''166''' and add their decimal digits: '''1 + 6 + 6''' which is '''13'''\nTherefore, the number '''166''' is a Smith number.\n\n\n;Task\nWrite a program to find all Smith numbers ''below'' 10000.\n\n\n;See also\n* from Wikipedia:   [Smith number].\n* from MathWorld:   [Smith number]. \n* from OEIS A6753:   [OEIS sequence A6753].\n* from OEIS A104170:   [Number of Smith numbers below 10^n]. \n* from The Prime pages:   [Smith numbers].\n\n", "solution": "from sys import stdout\n\n\ndef factors(n):\n    rt = []\n    f = 2\n    if n == 1:\n        rt.append(1);\n    else:\n        while 1:\n            if 0 == ( n % f ):\n                rt.append(f);\n                n //= f\n                if n == 1:\n                    return rt\n            else:\n                f += 1\n    return rt\n\n\ndef sum_digits(n):\n    sum = 0\n    while n > 0:\n        m = n % 10\n        sum += m\n        n -= m\n        n //= 10\n\n    return sum\n\n\ndef add_all_digits(lst):\n    sum = 0\n    for i in range (len(lst)):\n        sum += sum_digits(lst[i])\n\n    return sum\n\n\ndef list_smith_numbers(cnt):\n    for i in range(4, cnt):\n        fac = factors(i)\n        if len(fac) > 1:\n            if sum_digits(i) == add_all_digits(fac):\n                stdout.write(\"{0} \".format(i) )\n\n# entry point\nlist_smith_numbers(10_000)"}
{"title": "Smith numbers", "language": "Python 3.7", "task": "sum of the decimal digits of the integers that make up that number is the same as the sum of the decimal digits of its prime factors excluding 1. \n\nBy definition, all primes are ''excluded'' as they (naturally) satisfy this condition!\n\nSmith numbers are also known as   ''joke''   numbers.\n\n\n;Example\nUsing the number '''166'''\nFind the prime factors of '''166''' which are: '''2''' x '''83'''\nThen, take those two prime factors and sum all their decimal digits: '''2 + 8 + 3''' which is '''13'''\nThen, take the decimal digits of '''166''' and add their decimal digits: '''1 + 6 + 6''' which is '''13'''\nTherefore, the number '''166''' is a Smith number.\n\n\n;Task\nWrite a program to find all Smith numbers ''below'' 10000.\n\n\n;See also\n* from Wikipedia:   [Smith number].\n* from MathWorld:   [Smith number]. \n* from OEIS A6753:   [OEIS sequence A6753].\n* from OEIS A104170:   [Number of Smith numbers below 10^n]. \n* from The Prime pages:   [Smith numbers].\n\n", "solution": "'''Smith numbers'''\n\nfrom itertools import dropwhile\nfrom functools import reduce\nfrom math import floor, sqrt\n\n\n# isSmith :: Int -> Bool\ndef isSmith(n):\n    '''True if n is a Smith number.'''\n    pfs = primeFactors(n)\n    return (1 < len(pfs) or n != pfs[0]) and (\n        sumDigits(n) == reduce(\n            lambda a, x: a + sumDigits(x),\n            pfs, 0\n        )\n    )\n\n\n# primeFactors :: Int -> [Int]\ndef primeFactors(x):\n    '''List of prime factors of x'''\n    def go(n):\n        fs = list(dropwhile(\n            mod(n),\n            range(2, 1 + floor(sqrt(n)))\n        ))[0:1]\n\n        return fs + go(floor(n / fs[0])) if fs else [n]\n    return go(x)\n\n\n# sumDigits :: Int -> Int\ndef sumDigits(n):\n    '''The sum of the decimal digits of n'''\n    def f(x):\n        return Just(divmod(x, 10)) if x else Nothing()\n    return sum(unfoldl(f)(n))\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Count and samples of Smith numbers below 10k'''\n\n    lowSmiths = [x for x in range(2, 10000) if isSmith(x)]\n    lowSmithCount = len(lowSmiths)\n\n    print('\\n'.join([\n        'Count of Smith Numbers below 10k:',\n        str(lowSmithCount),\n        '\\nFirst 15 Smith Numbers:',\n        ' '.join(str(x) for x in lowSmiths[0:15]),\n        '\\nLast 12 Smith Numbers below 10000:',\n        ' '.join(str(x) for x in lowSmiths[lowSmithCount - 12:])\n    ]))\n\n\n# GENERIC -------------------------------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# mod :: Int -> Int -> Int\ndef mod(n):\n    '''n modulo d'''\n    return lambda d: n % d\n\n\n# unfoldl(lambda x: Just(((x - 1), x)) if 0 != x else Nothing())(10)\n# -> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]\n# unfoldl :: (b -> Maybe (b, a)) -> b -> [a]\ndef unfoldl(f):\n    '''Dual to reduce or foldl.\n       Where these reduce a list to a summary value, unfoldl\n       builds a list from a seed value.\n       Where f returns Just(a, b), a is appended to the list,\n       and the residual b is used as the argument for the next\n       application of f.\n       When f returns Nothing, the completed list is returned.\n    '''\n    def go(v):\n        x, r = v, v\n        xs = []\n        while True:\n            mb = f(x)\n            if mb.get('Nothing'):\n                return xs\n            else:\n                x, r = mb.get('Just')\n                xs.insert(0, r)\n        return xs\n    return lambda x: go(x)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Soloway's recurring rainfall", "language": "Python", "task": "Soloway's Recurring Rainfall is commonly used to assess general programming knowledge by requiring basic program structure, input/output, and program exit procedure.\n\n'''The problem:'''\n\nWrite a program that will read in integers and output their average. Stop reading when the value 99999 is input.\n\nFor languages that aren't traditionally interactive, the program can read in values as makes sense and stopping once 99999 is encountered.  The classic rainfall problem comes from identifying success of Computer Science programs with their students, so the original problem statement is written above -- though it may not strictly apply to a given language in the modern era.\n\n'''Implementation Details:'''\n* Only Integers are to be accepted as input\n* Output should be floating point\n* Rainfall can be negative (https://www.geographyrealm.com/what-is-negative-rainfall/)\n* For languages where the user is inputting data, the number of data inputs can be \"infinite\"\n* A complete implementation should handle error cases reasonably (asking the user for more input, skipping the bad value when encountered, etc)\n\nThe purpose of this problem, as originally proposed in the 1980's through its continued use today, is to just show fundamentals of CS: iteration, branching, program structure, termination, management of data types, input/output (where applicable), etc with things like input validation or management of numerical limits being more \"advanced\".  It isn't meant to literally be a rainfall calculator so implementations should strive to implement the solution clearly and simply.\n\n'''References:'''\n* http://cs.brown.edu/~kfisler/Pubs/icer14-rainfall/icer14.pdf\n* https://www.curriculumonline.ie/getmedia/8bb01bff-509e-48ed-991e-3ab5dad74a78/Seppletal-2015-DoweknowhowdifficulttheRainfallProblemis.pdf\n* https://en.wikipedia.org/wiki/Moving_average#Cumulative_average\n\n\n", "solution": "import sys\n\ndef get_next_input():\n    try:\n        num = int(input(\"Enter rainfall int, 99999 to quit: \"))\n    except:\n        print(\"Invalid input\")\n        return get_next_input()\n    return num\n\ncurrent_average = 0.0\ncurrent_count = 0\n\nwhile True:\n    next = get_next_input()\n\n    if next == 99999:\n        sys.exit()\n    else:\n        current_count += 1\n        current_average = current_average + (1.0/current_count)*next - (1.0/current_count)*current_average\n        \n        print(\"New average: \", current_average)\n\n"}
{"title": "Solve a Hidato puzzle", "language": "Python", "task": "The task is to write a program which solves Hidato (aka Hidoku) puzzles.\n\nThe rules are:\n* You are given a grid with some numbers placed in it. The other squares in the grid will be blank.\n** The grid is not necessarily rectangular.\n** The grid may have holes in it.\n** The grid is always connected.\n** The number \"1\" is always present, as is another number that is equal to the number of squares in the grid. Other numbers are present so as to force the solution to be unique.\n** It may be assumed that the difference between numbers present on the grid is not greater than lucky 13.\n* The aim is to place a natural number in each blank square so that in the sequence of numbered squares from \"1\" upwards, each square is in the [[wp:Moore neighborhood]] of the squares immediately before and after it in the sequence (except for the first and last squares, of course, which only have one-sided constraints).\n** Thus, if the grid was overlaid on a chessboard, a king would be able to make legal moves along the path from first to last square in numerical order.\n** A square may only contain one number.\n* In a proper Hidato puzzle, the solution is unique.\n\nFor example the following problem\nSample Hidato problem, from Wikipedia\n\nhas the following solution, with path marked on it:\n\nSolution to sample Hidato problem\n\n\n;Related tasks:\n* [[A* search algorithm]]\n* [[N-queens problem]]\n* [[Solve a Holy Knight's tour]]\n* [[Solve a Knight's tour]]\n* [[Solve a Hopido puzzle]]\n* [[Solve a Numbrix puzzle]]\n* [[Solve the no connection puzzle]];\n\n", "solution": "board = []\ngiven = []\nstart = None\n\ndef setup(s):\n    global board, given, start\n    lines = s.splitlines()\n    ncols = len(lines[0].split())\n    nrows = len(lines)\n    board = [[-1] * (ncols + 2) for _ in xrange(nrows + 2)]\n\n    for r, row in enumerate(lines):\n        for c, cell in enumerate(row.split()):\n            if cell == \"__\" :\n                board[r + 1][c + 1] = 0\n                continue\n            elif cell == \".\":\n                continue # -1\n            else:\n                val = int(cell)\n                board[r + 1][c + 1] = val\n                given.append(val)\n                if val == 1:\n                    start = (r + 1, c + 1)\n    given.sort()\n\ndef solve(r, c, n, next=0):\n    if n > given[-1]:\n        return True\n    if board[r][c] and board[r][c] != n:\n        return False\n    if board[r][c] == 0 and given[next] == n:\n        return False\n\n    back = 0\n    if board[r][c] == n:\n        next += 1\n        back = n\n\n    board[r][c] = n\n    for i in xrange(-1, 2):\n        for j in xrange(-1, 2):\n            if solve(r + i, c + j, n + 1, next):\n                return True\n    board[r][c] = back\n    return False\n\ndef print_board():\n    d = {-1: \"  \", 0: \"__\"}\n    bmax = max(max(r) for r in board)\n    form = \"%\" + str(len(str(bmax)) + 1) + \"s\"\n    for r in board[1:-1]:\n        print \"\".join(form % d.get(c, str(c)) for c in r[1:-1])\n\nhi = \"\"\"\\\n__ 33 35 __ __  .  .  .\n__ __ 24 22 __  .  .  .\n__ __ __ 21 __ __  .  .\n__ 26 __ 13 40 11  .  .\n27 __ __ __  9 __  1  .\n .  . __ __ 18 __ __  .\n .  .  .  . __  7 __ __\n .  .  .  .  .  .  5 __\"\"\"\n\nsetup(hi)\nprint_board()\nsolve(start[0], start[1], 1)\nprint\nprint_board()"}
{"title": "Solve a Holy Knight's tour", "language": "Python", "task": "Chess coaches have been known to inflict a kind of torture on beginners by taking a chess board, placing pennies on some squares and requiring that a Knight's tour be constructed that avoids the squares with pennies. \n\nThis kind of knight's tour puzzle is similar to   Hidato.\n\nThe present task is to produce a solution to such problems. At least demonstrate your program by solving the following:\n\n\n;Example:\n\n  0 0 0 \n  0   0 0 \n  0 0 0 0 0 0 0\n0 0 0     0   0\n0   0     0 0 0\n1 0 0 0 0 0 0\n    0 0   0\n      0 0 0\n\n\nNote that the zeros represent the available squares, not the pennies.\n\nExtra credit is available for other interesting examples.\n\n\n;Related tasks:\n* [[A* search algorithm]]\n* [[Knight's tour]]\n* [[N-queens problem]]\n* [[Solve a Hidato puzzle]]\n* [[Solve a Hopido puzzle]]\n* [[Solve a Numbrix puzzle]]\n* [[Solve the no connection puzzle]]\n\n", "solution": "from sys import stdout\nmoves = [\n    [-1, -2], [1, -2], [-1, 2], [1, 2],\n    [-2, -1], [-2, 1], [2, -1], [2, 1]\n]\n\n\ndef solve(pz, sz, sx, sy, idx, cnt):\n    if idx > cnt:\n        return 1\n\n    for i in range(len(moves)):\n        x = sx + moves[i][0]\n        y = sy + moves[i][1]\n        if sz > x > -1 and sz > y > -1 and pz[x][y] == 0:\n            pz[x][y] = idx\n            if 1 == solve(pz, sz, x, y, idx + 1, cnt):\n                return 1\n            pz[x][y] = 0\n\n    return 0\n\n\ndef find_solution(pz, sz):\n    p = [[-1 for j in range(sz)] for i in range(sz)]\n    idx = x = y = cnt = 0\n    for j in range(sz):\n        for i in range(sz):\n            if pz[idx] == \"x\":\n                p[i][j] = 0\n                cnt += 1\n            elif pz[idx] == \"s\":\n                p[i][j] = 1\n                cnt += 1\n                x = i\n                y = j\n            idx += 1\n\n    if 1 == solve(p, sz, x, y, 2, cnt):\n        for j in range(sz):\n            for i in range(sz):\n                if p[i][j] != -1:\n                    stdout.write(\" {:0{}d}\".format(p[i][j], 2))\n                else:\n                    stdout.write(\"   \")\n            print()\n    else:\n        print(\"Cannot solve this puzzle!\")\n\n\n# entry point\nfind_solution(\".xxx.....x.xx....xxxxxxxxxx..x.xx.x..xxxsxxxxxx...xx.x.....xxx..\", 8)\nprint()\nfind_solution(\".....s.x..........x.x.........xxxxx.........xxx.......x..x.x..x..xxxxx...xxxxx..xx.....xx..xxxxx...xxxxx..x..x.x..x.......xxx.........xxxxx.........x.x..........x.x.....\", 13)\n"}
{"title": "Solve a Hopido puzzle", "language": "Python", "task": "Hopido puzzles are similar to  Hidato. The most important difference is that the only moves allowed are:  hop over one tile diagonally; and over two tiles horizontally and vertically. It should be possible to start anywhere in the path, the end point isn't indicated and there are no intermediate clues. Hopido Design Post Mortem contains the following:\n\n\"Big puzzles represented another problem. Up until quite late in the project our puzzle solver was painfully slow with most puzzles above 7x7 tiles. Testing the solution from each starting point could take hours. If the tile layout was changed even a little, the whole puzzle had to be tested again. We were just about to give up the biggest puzzles entirely when our programmer suddenly came up with a magical algorithm that cut the testing process down to only minutes. Hooray!\"\n\nKnowing the kindness in the heart of every contributor to Rosetta Code, I know that we shall feel that as an act of humanity we must solve these puzzles for them in let's say milliseconds.\n\nExample:\n\n . 0 0 . 0 0 .\n 0 0 0 0 0 0 0\n 0 0 0 0 0 0 0\n . 0 0 0 0 0 .\n . . 0 0 0 . .\n . . . 0 . . .\n\nExtra credits are available for other interesting designs.\n\n\n;Related tasks:\n* [[A* search algorithm]]\n* [[Solve a Holy Knight's tour]]\n* [[Knight's tour]]\n* [[N-queens problem]]\n* [[Solve a Hidato puzzle]]\n* [[Solve a Holy Knight's tour]]\n* [[Solve a Numbrix puzzle]]\n* [[Solve the no connection puzzle]]\n\n", "solution": "from sys import stdout\n\nneighbours = [[2, 2], [-2, 2], [2, -2], [-2, -2], [3, 0], [0, 3], [-3, 0], [0, -3]]\ncnt = 0\npWid = 0\npHei = 0\n\n\ndef is_valid(a, b):\n    return -1 < a < pWid and -1 < b < pHei\n\n\ndef iterate(pa, x, y, v):\n    if v > cnt:\n        return 1\n\n    for i in range(len(neighbours)):\n        a = x + neighbours[i][0]\n        b = y + neighbours[i][1]\n        if is_valid(a, b) and pa[a][b] == 0:\n            pa[a][b] = v\n            r = iterate(pa, a, b, v + 1)\n            if r == 1:\n                return r\n            pa[a][b] = 0\n    return 0\n\n\ndef solve(pz, w, h):\n    global cnt, pWid, pHei\n\n    pa = [[-1 for j in range(h)] for i in range(w)]\n    f = 0\n    pWid = w\n    pHei = h\n    for j in range(h):\n        for i in range(w):\n            if pz[f] == \"1\":\n                pa[i][j] = 0\n                cnt += 1\n            f += 1\n\n    for y in range(h):\n        for x in range(w):\n            if pa[x][y] == 0:\n                pa[x][y] = 1\n                if 1 == iterate(pa, x, y, 2):\n                    return 1, pa\n                pa[x][y] = 0\n\n    return 0, pa\n\nr = solve(\"011011011111111111111011111000111000001000\", 7, 6)\nif r[0] == 1:\n    for j in range(6):\n        for i in range(7):\n            if r[1][i][j] == -1:\n                stdout.write(\"   \")\n            else:\n                stdout.write(\" {:0{}d}\".format(r[1][i][j], 2))\n        print()\nelse:\n    stdout.write(\"No solution!\")\n"}
{"title": "Solve a Numbrix puzzle", "language": "Python", "task": "Numbrix puzzles are similar to Hidato. \nThe most important difference is that it is only possible to move 1 node left, right, up, or down (sometimes referred to as the Von Neumann neighborhood). \nPublished puzzles also tend not to have holes in the grid and may not always indicate the end node. \nTwo examples follow:\n\n;Example 1\nProblem.\n\n 0  0  0  0  0  0  0  0  0\n 0  0 46 45  0 55 74  0  0\n 0 38  0  0 43  0  0 78  0\n 0 35  0  0  0  0  0 71  0\n 0  0 33  0  0  0 59  0  0\n 0 17  0  0  0  0  0 67  0\n 0 18  0  0 11  0  0 64  0\n 0  0 24 21  0  1  2  0  0\n 0  0  0  0  0  0  0  0  0\n\nSolution.\n\n 49 50 51 52 53 54 75 76 81\n 48 47 46 45 44 55 74 77 80\n 37 38 39 40 43 56 73 78 79\n 36 35 34 41 42 57 72 71 70\n 31 32 33 14 13 58 59 68 69\n 30 17 16 15 12 61 60 67 66\n 29 18 19 20 11 62 63 64 65\n 28 25 24 21 10  1  2  3  4\n 27 26 23 22  9  8  7  6  5\n\n;Example 2\nProblem.\n\n 0  0  0  0  0  0  0  0  0\n 0 11 12 15 18 21 62 61  0\n 0  6  0  0  0  0  0 60  0\n 0 33  0  0  0  0  0 57  0\n 0 32  0  0  0  0  0 56  0\n 0 37  0  1  0  0  0 73  0\n 0 38  0  0  0  0  0 72  0\n 0 43 44 47 48 51 76 77  0\n 0  0  0  0  0  0  0  0  0\n\nSolution.\n\n  9 10 13 14 19 20 63 64 65\n  8 11 12 15 18 21 62 61 66\n  7  6  5 16 17 22 59 60 67\n 34 33  4  3 24 23 58 57 68\n 35 32 31  2 25 54 55 56 69\n 36 37 30  1 26 53 74 73 70\n 39 38 29 28 27 52 75 72 71\n 40 43 44 47 48 51 76 77 78\n 41 42 45 46 49 50 81 80 79\n\n;Task\nWrite a program to solve puzzles of this ilk, \ndemonstrating your program by solving the above examples. \nExtra credit for other interesting examples.\n\n\n;Related tasks:\n* [[A* search algorithm]]\n* [[Solve a Holy Knight's tour]]\n* [[Knight's tour]]\n* [[N-queens problem]]\n* [[Solve a Hidato puzzle]]\n* [[Solve a Holy Knight's tour]]\n* [[Solve a Hopido puzzle]]\n* [[Solve the no connection puzzle]]\n\n", "solution": "from sys import stdout\nneighbours = [[-1, 0], [0, -1], [1, 0], [0, 1]]\nexists = []\nlastNumber = 0\nwid = 0\nhei = 0\n\n\ndef find_next(pa, x, y, z):\n    for i in range(4):\n        a = x + neighbours[i][0]\n        b = y + neighbours[i][1]\n        if wid > a > -1 and hei > b > -1:\n            if pa[a][b] == z:\n                return a, b\n\n    return -1, -1\n\n\ndef find_solution(pa, x, y, z):\n    if z > lastNumber:\n        return 1\n    if exists[z] == 1:\n        s = find_next(pa, x, y, z)\n        if s[0] < 0:\n            return 0\n        return find_solution(pa, s[0], s[1], z + 1)\n\n    for i in range(4):\n        a = x + neighbours[i][0]\n        b = y + neighbours[i][1]\n        if wid > a > -1 and hei > b > -1:\n            if pa[a][b] == 0:\n                pa[a][b] = z\n                r = find_solution(pa, a, b, z + 1)\n                if r == 1:\n                    return 1\n                pa[a][b] = 0\n\n    return 0\n\n\ndef solve(pz, w, h):\n    global lastNumber, wid, hei, exists\n\n    lastNumber = w * h\n    wid = w\n    hei = h\n    exists = [0 for j in range(lastNumber + 1)]\n\n    pa = [[0 for j in range(h)] for i in range(w)]\n    st = pz.split()\n    idx = 0\n    for j in range(h):\n        for i in range(w):\n            if st[idx] == \".\":\n                idx += 1\n            else:\n                pa[i][j] = int(st[idx])\n                exists[pa[i][j]] = 1\n                idx += 1\n\n    x = 0\n    y = 0\n    t = w * h + 1\n    for j in range(h):\n        for i in range(w):\n            if pa[i][j] != 0 and pa[i][j] < t:\n                t = pa[i][j]\n                x = i\n                y = j\n\n    return find_solution(pa, x, y, t + 1), pa\n\n\ndef show_result(r):\n    if r[0] == 1:\n        for j in range(hei):\n            for i in range(wid):\n                stdout.write(\" {:0{}d}\".format(r[1][i][j], 2))\n            print()\n    else:\n        stdout.write(\"No Solution!\\n\")\n\n    print()\n\n\nr = solve(\". . . . . . . . . . . 46 45 . 55 74 . . . 38 . . 43 . . 78 . . 35 . . . . . 71 . . . 33 . . . 59 . . . 17\"\n          \" . . . . . 67 . . 18 . . 11 . . 64 . . . 24 21 . 1  2 . . . . . . . . . . .\", 9, 9)\nshow_result(r)\n\nr = solve(\". . . . . . . . . . 11 12 15 18 21 62 61 . .  6 . . . . . 60 . . 33 . . . . . 57 . . 32 . . . . . 56 . . 37\"\n          \" .  1 . . . 73 . . 38 . . . . . 72 . . 43 44 47 48 51 76 77 . . . . . . . . . .\", 9, 9)\nshow_result(r)\n\nr = solve(\"17 . . . 11 . . . 59 . 15 . . 6 . . 61 . . . 3 . . .  63 . . . . . . 66 . . . . 23 24 . 68 67 78 . 54 55\"\n          \" . . . . 72 . . . . . . 35 . . . 49 . . . 29 . . 40 . . 47 . 31 . . . 39 . . . 45\", 9, 9)\nshow_result(r)\n"}
{"title": "Sort an outline at every level", "language": "Python", "task": "Write and test a function over an indented plain text outline which either:\n# Returns a copy of the outline in which the sub-lists at every level of indentation are sorted, or\n# reports that the indentation characters or widths are not consistent enough to make the outline structure clear.\n\n\n\nYour code should detect and warn of at least two types of inconsistent indentation:\n* inconsistent use of whitespace characters (e.g. mixed use of tabs and spaces)\n* inconsistent indent widths. For example, an indentation with an odd number of spaces in an outline in which the unit indent appears to be 2 spaces, or 4 spaces.\n\n\nYour code should be able to detect and handle both tab-indented, and space-indented (e.g. 4 space, 2 space etc) outlines, without being given any advance warning of the indent characters used, or the size of the indent units.\n\nYou should also be able to specify different types of sort, for example, as a minimum, both ascending and descending lexical sorts.\n\nYour sort should not alter the type or size of the indentation units used in the input outline.\n\n\n(For an application of Indent Respectful Sort, see the Sublime Text package of that name. The Python source text is available for inspection on Github).\n\n\n'''Tests'''\n\n* Sort every level of the (4 space indented) outline below lexically, once ascending and once descending.\nzeta\n    beta\n    gamma\n        lambda\n        kappa\n        mu\n    delta\nalpha\n    theta\n    iota\n    epsilon\n\n* Do the same with a tab-indented equivalent of the same outline.\n\nzeta\n\tgamma\n\t\tmu\n\t\tlambda\n\t\tkappa\n\tdelta\n\tbeta\nalpha\n\ttheta\n\tiota\n\tepsilon\n\n\nThe output sequence of an ascending lexical sort of each level should be:\n\nalpha\n    epsilon\n    iota\n    theta\nzeta\n    beta\n    delta\n    gamma\n        kappa\n        lambda\n        mu\n\nThe output sequence of a descending lexical sort of each level should be:\n\nzeta\n    gamma\n        mu\n        lambda\n        kappa\n    delta\n    beta\nalpha\n    theta\n    iota\n    epsilon\n\n* Attempt to separately sort each of the following two outlines, reporting any inconsistencies detected in their indentations by your validation code.\n\nalpha\n    epsilon\n\tiota\n    theta\nzeta\n    beta\n    delta\n    gamma\n    \tkappa\n        lambda\n        mu\nzeta\n    beta\n   gamma\n        lambda\n         kappa\n        mu\n    delta\nalpha\n    theta\n    iota\n    epsilon\n\n\n\n;Related tasks:\n\n:*   [[Functional_coverage_tree]\n:*   [[Display_an_outline_as_a_nested_table]]\n\n\n", "solution": "'''Sort an outline at every level'''\n\n\nfrom itertools import chain, product, takewhile, tee\nfrom functools import cmp_to_key, reduce\n\n\n# ------------- OUTLINE SORTED AT EVERY LEVEL --------------\n\n# sortedOutline :: (Tree String -> Tree String -> Ordering)\n#                     -> String\n#                     -> Either String String\ndef sortedOutline(cmp):\n    '''Either a message reporting inconsistent\n       indentation, or an outline sorted at every\n       level by the supplied comparator function.\n    '''\n    def go(outlineText):\n        indentTuples = indentTextPairs(\n            outlineText.splitlines()\n        )\n        return bindLR(\n            minimumIndent(enumerate(indentTuples))\n        )(lambda unitIndent: Right(\n            outlineFromForest(\n                unitIndent,\n                nest(foldTree(\n                    lambda x: lambda xs: Node(x)(\n                        sorted(xs, key=cmp_to_key(cmp))\n                    )\n                )(Node('')(\n                    forestFromIndentLevels(\n                        indentLevelsFromLines(\n                            unitIndent\n                        )(indentTuples)\n                    )\n                )))\n            )\n        ))\n    return go\n\n\n# -------------------------- TEST --------------------------\n# main :: IO ()\ndef main():\n    '''Ascending and descending sorts attempted on\n       space-indented and tab-indented outlines, both\n       well-formed and ill-formed.\n    '''\n\n    ascending = comparing(root)\n    descending = flip(ascending)\n\n    spacedOutline = '''\nzeta\n    beta\n    gamma\n        lambda\n        kappa\n        mu\n    delta\nalpha\n    theta\n    iota\n    epsilon'''\n\n    tabbedOutline = '''\nzeta\n\tbeta\n\tgamma\n\t\tlambda\n\t\tkappa\n\t\tmu\n\tdelta\nalpha\n\ttheta\n\tiota\n\tepsilon'''\n\n    confusedOutline = '''\nalpha\n    epsilon\n\tiota\n    theta\nzeta\n    beta\n    delta\n    gamma\n    \tkappa\n        lambda\n        mu'''\n\n    raggedOutline = '''\nzeta\n    beta\n   gamma\n        lambda\n         kappa\n        mu\n    delta\nalpha\n    theta\n    iota\n    epsilon'''\n\n    def displaySort(kcmp):\n        '''Sort function output with labelled comparator\n           for a set of four labelled outlines.\n        '''\n        k, cmp = kcmp\n        return [\n            tested(cmp, k, label)(\n                outline\n            ) for (label, outline) in [\n                ('4-space indented', spacedOutline),\n                ('tab indented', tabbedOutline),\n                ('Unknown 1', confusedOutline),\n                ('Unknown 2', raggedOutline)\n            ]\n        ]\n\n    def tested(cmp, cmpName, outlineName):\n        '''Print either message or result.\n        '''\n        def go(outline):\n            print('\\n' + outlineName, cmpName + ':')\n            either(print)(print)(\n                sortedOutline(cmp)(outline)\n            )\n        return go\n\n    # Tests applied to two comparators:\n    ap([\n        displaySort\n    ])([\n        (\"(A -> Z)\", ascending),\n        (\"(Z -> A)\", descending)\n    ])\n\n\n# ------------- OUTLINE PARSING AND RENDERING --------------\n\n# forestFromIndentLevels :: [(Int, a)] -> [Tree a]\ndef forestFromIndentLevels(tuples):\n    '''A list of trees derived from a list of values paired\n       with integers giving their levels of indentation.\n    '''\n    def go(xs):\n        if xs:\n            intIndent, v = xs[0]\n            firstTreeLines, rest = span(\n                lambda x: intIndent < x[0]\n            )(xs[1:])\n            return [Node(v)(go(firstTreeLines))] + go(rest)\n        else:\n            return []\n    return go(tuples)\n\n\n# indentLevelsFromLines :: String -> [(String, String)]\n# -> [(Int, String)]\ndef indentLevelsFromLines(indentUnit):\n    '''Each input line stripped of leading\n       white space, and tupled with a preceding integer\n       giving its level of indentation from 0 upwards.\n    '''\n    def go(xs):\n        w = len(indentUnit)\n        return [\n            (len(x[0]) // w, x[1])\n            for x in xs\n        ]\n    return go\n\n\n# indentTextPairs :: [String] -> (String, String)\ndef indentTextPairs(xs):\n    '''A list of (indent, bodyText) pairs.'''\n    def indentAndText(s):\n        pfx = list(takewhile(lambda c: c.isspace(), s))\n        return (pfx, s[len(pfx):])\n    return [indentAndText(x) for x in xs]\n\n\n# outlineFromForest :: String -> [Tree String] -> String\ndef outlineFromForest(tabString, forest):\n    '''An indented outline serialisation of forest,\n       using tabString as the unit of indentation.\n    '''\n    def go(indent):\n        def serial(node):\n            return [indent + root(node)] + list(\n                concatMap(\n                    go(tabString + indent)\n                )(nest(node))\n            )\n        return serial\n    return '\\n'.join(\n        concatMap(go(''))(forest)\n    )\n\n\n# --------------- MINIMUM INDENT, OR ANOMALY ---------------\n\n# minimumIndent :: [(Int, [Char])]\n#       -> Either String String\ndef minimumIndent(indexedPrefixes):\n    '''Either a message, if indentation characters are\n       mixed, or indentation widths are inconsistent,\n       or the smallest consistent non-empty indentation.\n    '''\n    (xs, ts) = tee(indexedPrefixes)\n    (ys, zs) = tee(ts)\n\n    def mindentLR(charSet):\n        if list(charSet):\n            def w(x):\n                return len(x[1][0])\n\n            unit = min(filter(w, ys), key=w)[1][0]\n            unitWidth = len(unit)\n\n            def widthCheck(a, ix):\n                '''Is there a line number at which\n                   an anomalous indent width is seen?\n                '''\n                wx = len(ix[1][0])\n                return a if (a or 0 == wx) else (\n                    ix[0] if 0 != wx % unitWidth else a\n                )\n            oddLine = reduce(widthCheck, zs, None)\n            return Left(\n                'Inconsistent indentation width at line ' + (\n                    str(1 + oddLine)\n                )\n            ) if oddLine else Right(''.join(unit))\n        else:\n            return Right('')\n\n    def tabSpaceCheck(a, ics):\n        '''Is there a line number at which a\n           variant indent character is used?\n        '''\n        charSet = a[0].union(set(ics[1][0]))\n        return a if a[1] else (\n            charSet, ics[0] if 1 < len(charSet) else None\n        )\n\n    indentCharSet, mbAnomalyLine = reduce(\n        tabSpaceCheck, xs, (set([]), None)\n    )\n    return bindLR(\n        Left(\n            'Mixed indent characters found in line ' + str(\n                1 + mbAnomalyLine\n            )\n        ) if mbAnomalyLine else Right(list(indentCharSet))\n    )(mindentLR)\n\n\n# ------------------------ GENERIC -------------------------\n\n# Left :: a -> Either a b\ndef Left(x):\n    '''Constructor for an empty Either (option type) value\n       with an associated string.\n    '''\n    return {'type': 'Either', 'Right': None, 'Left': x}\n\n\n# Right :: b -> Either a b\ndef Right(x):\n    '''Constructor for a populated Either (option type) value'''\n    return {'type': 'Either', 'Left': None, 'Right': x}\n\n\n# Node :: a -> [Tree a] -> Tree a\ndef Node(v):\n    '''Constructor for a Tree node which connects a\n       value of some kind to a list of zero or\n       more child trees.\n    '''\n    return lambda xs: {'type': 'Tree', 'root': v, 'nest': xs}\n\n\n# ap (<*>) :: [(a -> b)] -> [a] -> [b]\ndef ap(fs):\n    '''The application of each of a list of functions,\n       to each of a list of values.\n    '''\n    def go(xs):\n        return [\n            f(x) for (f, x)\n            in product(fs, xs)\n        ]\n    return go\n\n\n# bindLR (>>=) :: Either a -> (a -> Either b) -> Either b\ndef bindLR(m):\n    '''Either monad injection operator.\n       Two computations sequentially composed,\n       with any value produced by the first\n       passed as an argument to the second.\n    '''\n    def go(mf):\n        return (\n            mf(m.get('Right')) if None is m.get('Left') else m\n        )\n    return go\n\n\n# comparing :: (a -> b) -> (a -> a -> Ordering)\ndef comparing(f):\n    '''An ordering function based on\n       a property accessor f.\n    '''\n    def go(x, y):\n        fx = f(x)\n        fy = f(y)\n        return -1 if fx < fy else (1 if fx > fy else 0)\n    return go\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).\n    '''\n    def go(xs):\n        return chain.from_iterable(map(f, xs))\n    return go\n\n\n# either :: (a -> c) -> (b -> c) -> Either a b -> c\ndef either(fl):\n    '''The application of fl to e if e is a Left value,\n       or the application of fr to e if e is a Right value.\n    '''\n    return lambda fr: lambda e: fl(e['Left']) if (\n        None is e['Right']\n    ) else fr(e['Right'])\n\n\n# flip :: (a -> b -> c) -> b -> a -> c\ndef flip(f):\n    '''The binary function f with its\n       arguments reversed.\n    '''\n    return lambda a, b: f(b, a)\n\n\n# foldTree :: (a -> [b] -> b) -> Tree a -> b\ndef foldTree(f):\n    '''The catamorphism on trees. A summary\n       value defined by a depth-first fold.\n    '''\n    def go(node):\n        return f(root(node))([\n            go(x) for x in nest(node)\n        ])\n    return go\n\n\n# nest :: Tree a -> [Tree a]\ndef nest(t):\n    '''Accessor function for children of tree node.'''\n    return t.get('nest')\n\n\n# root :: Tree a -> a\ndef root(t):\n    '''Accessor function for data of tree node.'''\n    return t.get('root')\n\n\n# span :: (a -> Bool) -> [a] -> ([a], [a])\ndef span(p):\n    '''The longest (possibly empty) prefix of xs\n       that contains only elements satisfying p,\n       tupled with the remainder of xs.\n       span p xs is equivalent to\n       (takeWhile p xs, dropWhile p xs).\n    '''\n    def match(ab):\n        b = ab[1]\n        return not b or not p(b[0])\n\n    def f(ab):\n        a, b = ab\n        return a + [b[0]], b[1:]\n\n    def go(xs):\n        return until(match)(f)(([], xs))\n    return go\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f):\n        def g(x):\n            v = x\n            while not p(v):\n                v = f(v)\n            return v\n        return g\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Sparkline in unicode", "language": "Python", "task": "A sparkline is a graph of successive values laid out horizontally \nwhere the height of the line is proportional to the values in succession.\n\n\n;Task:\nUse the following series of Unicode characters to create a program \nthat takes a series of numbers separated by one or more whitespace or comma characters \nand generates a sparkline-type bar graph of the values on a single line of output.\n\nThe eight characters: '########'\n(Unicode values U+2581 through U+2588).\n\nUse your program to show sparklines for the following input, \nhere on this page:\n# 1 2 3 4 5 6 7 8 7 6 5 4 3 2 1\n# 1.5, 0.5 3.5, 2.5 5.5, 4.5 7.5, 6.5 \n:(note the mix of separators in this second case)!\n\n;Notes: \n* A space is not part of the generated sparkline.\n* The sparkline may be accompanied by simple statistics of the data such as its range.\n* A suggestion emerging in later discussion (see Discussion page) is that the bounds between bins should ideally be set to yield the following results for two particular edge cases:\n\n:: \"0, 1, 19, 20\" -> ####\n:: (Aiming to use just two spark levels)\n\n:: \"0, 999, 4000, 4999, 7000, 7999\" -> ######\n:: (Aiming to use just three spark levels)\n\n:: It may be helpful to include these cases in output tests.\n* You may find that the unicode sparklines on this page are rendered less noisily by Google Chrome than by Firefox or Safari.\n\n", "solution": "# -*- coding: utf-8 -*-\n\n# Unicode: 9601, 9602, 9603, 9604, 9605, 9606, 9607, 9608\nbar = '\u2581\u2582\u2583\u2584\u2585\u2586\u2587\u2588'\nbarcount = len(bar)\n\ndef sparkline(numbers):\n    mn, mx = min(numbers), max(numbers)\n    extent = mx - mn\n    sparkline = ''.join(bar[min([barcount - 1,\n                                 int((n - mn) / extent * barcount)])]\n                        for n in numbers)\n    return mn, mx, sparkline\n\nif __name__ == '__main__':\n    import re\n    \n    for line in (\"0 0 1 1; 0 1 19 20; 0 999 4000 4999 7000 7999;\"\n                 \"1 2 3 4 5 6 7 8 7 6 5 4 3 2 1;\"\n                 \"1.5, 0.5 3.5, 2.5 5.5, 4.5 7.5, 6.5 \").split(';'):\n        print(\"\\nNumbers:\", line)\n        numbers = [float(n) for n in re.split(r'[\\s,]+', line.strip())]\n        mn, mx, sp = sparkline(numbers)\n        print('  min: %5f; max: %5f' % (mn, mx))\n        print(\"  \" + sp)"}
{"title": "Spelling of ordinal numbers", "language": "Python", "task": "'''Ordinal numbers'''   (as used in this Rosetta Code task),   are numbers that describe the   ''position''   of something in a list.\n\nIt is this context that ordinal numbers will be used, using an English-spelled name of an ordinal number.\n\n\nThe ordinal numbers are   (at least, one form of them):\n   1st  2nd  3rd  4th  5th  6th  7th  ***  99th  100th  ***  1000000000th  ***  etc\n\nsometimes expressed as:\n   1st  2nd  3rd  4th  5th  6th  7th  ***  99th  100th  ***  1000000000th  ***\n\n\nFor this task, the following (English-spelled form) will be used:\n   first second third fourth fifth sixth seventh ninety-nineth one hundredth one billionth\n\n\nFurthermore, the American version of numbers will be used here   (as opposed to the British).\n\n'''2,000,000,000'''   is two billion,   ''not''   two milliard.\n\n\n;Task:\nWrite a driver and a function (subroutine/routine ***) that returns the English-spelled ordinal version of a specified number   (a positive integer).\n\nOptionally, try to support as many forms of an integer that can be expressed:   '''123'''   '''00123.0'''   '''1.23e2'''   all are forms of the same integer.\n\nShow all output here.\n\n\n;Test cases:\nUse (at least) the test cases of:\n   1  2  3  4  5  11  65  100  101  272  23456  8007006005004003\n\n\n;Related tasks:\n*   [[Number names]]\n*   [[N'th]]\n\n", "solution": "irregularOrdinals = {\n\t\"one\":    \"first\",\n\t\"two\":    \"second\",\n\t\"three\":  \"third\",\n\t\"five\":   \"fifth\",\n\t\"eight\":  \"eighth\",\n\t\"nine\":   \"ninth\",\n\t\"twelve\": \"twelfth\",\n}\n\ndef num2ordinal(n):\n    conversion = int(float(n))\n    num = spell_integer(conversion)\n    hyphen = num.rsplit(\"-\", 1)\n    num = num.rsplit(\" \", 1)\n    delim = \" \"\n    if len(num[-1]) > len(hyphen[-1]):\n        num = hyphen\n        delim = \"-\"\n    \n    if num[-1] in irregularOrdinals:\n        num[-1] = delim + irregularOrdinals[num[-1]]\n    elif num[-1].endswith(\"y\"):\n        num[-1] = delim + num[-1][:-1] + \"ieth\"\n    else:\n        num[-1] = delim + num[-1] + \"th\"\n    return \"\".join(num)\n    \nif __name__ == \"__main__\":\n    tests = \"1  2  3  4  5  11  65  100  101  272  23456  8007006005004003 123   00123.0   1.23e2\".split()\n    for num in tests:\n        print(\"{} => {}\".format(num, num2ordinal(num)))\n\n\n#This is a copy of the code from https://rosettacode.org/wiki/Number_names#Python\n\nTENS = [None, None, \"twenty\", \"thirty\", \"forty\",\n        \"fifty\", \"sixty\", \"seventy\", \"eighty\", \"ninety\"]\nSMALL = [\"zero\", \"one\", \"two\", \"three\", \"four\", \"five\",\n         \"six\", \"seven\", \"eight\", \"nine\", \"ten\", \"eleven\",\n         \"twelve\", \"thirteen\", \"fourteen\", \"fifteen\",\n         \"sixteen\", \"seventeen\", \"eighteen\", \"nineteen\"]\nHUGE = [None, None] + [h + \"illion\" \n                       for h in (\"m\", \"b\", \"tr\", \"quadr\", \"quint\", \"sext\", \n                                  \"sept\", \"oct\", \"non\", \"dec\")]\n \ndef nonzero(c, n, connect=''):\n    return \"\" if n == 0 else connect + c + spell_integer(n)\n \ndef last_and(num):\n    if ',' in num:\n        pre, last = num.rsplit(',', 1)\n        if ' and ' not in last:\n            last = ' and' + last\n        num = ''.join([pre, ',', last])\n    return num\n \ndef big(e, n):\n    if e == 0:\n        return spell_integer(n)\n    elif e == 1:\n        return spell_integer(n) + \" thousand\"\n    else:\n        return spell_integer(n) + \" \" + HUGE[e]\n \ndef base1000_rev(n):\n    # generates the value of the digits of n in base 1000\n    # (i.e. 3-digit chunks), in reverse.\n    while n != 0:\n        n, r = divmod(n, 1000)\n        yield r\n \ndef spell_integer(n):\n    if n < 0:\n        return \"minus \" + spell_integer(-n)\n    elif n < 20:\n        return SMALL[n]\n    elif n < 100:\n        a, b = divmod(n, 10)\n        return TENS[a] + nonzero(\"-\", b)\n    elif n < 1000:\n        a, b = divmod(n, 100)\n        return SMALL[a] + \" hundred\" + nonzero(\" \", b, ' and')\n    else:\n        num = \", \".join([big(e, x) for e, x in\n                         enumerate(base1000_rev(n)) if x][::-1])\n        return last_and(num)"}
{"title": "Sphenic numbers", "language": "Python", "task": "Definitions\nA '''sphenic number''' is a positive integer that is the product of three distinct prime numbers. More technically it's a square-free 3-almost prime (see Related tasks below).\n\nFor the purposes of this task, a '''sphenic triplet''' is a group of three sphenic numbers which are consecutive. \n\nNote that sphenic quadruplets are not possible because every fourth consecutive positive integer is divisible by 4 (= 2 x 2) and its prime factors would not therefore be distinct.\n\n;Examples\n30 (= 2 x 3 x 5) is a sphenic number and is also clearly the first one.\n\n[1309, 1310, 1311] is a sphenic triplet because 1309 (= 7 x 11 x 17), 1310 (= 2 x 5 x 31) and 1311 (= 3 x 19 x 23) are 3 consecutive sphenic numbers.\n\n;Task\nCalculate and show here:\n\n1. All sphenic numbers less than 1,000.\n\n2. All sphenic triplets less than 10,000.\n\n;Stretch\n\n3. How many sphenic numbers are there less than 1 million?\n\n4. How many sphenic triplets are there less than 1 million?\n\n5. What is the 200,000th sphenic number and its 3 prime factors?\n\n6. What is the 5,000th sphenic triplet?\n\nHint: you only need to consider sphenic numbers less than 1 million to answer 5. and 6.\n\n;References\n\n* Wikipedia: Sphenic number\n* OEIS:A007304 - Sphenic numbers\n* OEIS:A165936 - Sphenic triplets (in effect)\n\n;Related tasks\n* [[Almost prime]]\n* [[Square-free integers]]\n\n", "solution": "\"\"\" rosettacode.org task Sphenic_numbers \"\"\"\n\n\nfrom sympy import factorint\n\nsphenics1m, sphenic_triplets1m = [], []\n\nfor i in range(3, 1_000_000):\n    d = factorint(i)\n    if len(d) == 3 and sum(d.values()) == 3:\n        sphenics1m.append(i)\n        if len(sphenics1m) > 2 and i - sphenics1m[-3] == 2 and i - sphenics1m[-2] == 1:\n            sphenic_triplets1m.append(i)\n\nprint('Sphenic numbers less than 1000:')\nfor i, n in enumerate(sphenics1m):\n    if n < 1000:\n        print(f'{n : 5}', end='\\n' if (i + 1) % 15 == 0 else '')\n    else:\n        break\n\nprint('\\n\\nSphenic triplets less than 10_000:')\nfor i, n in enumerate(sphenic_triplets1m):\n    if n < 10_000:\n        print(f'({n - 2} {n - 1} {n})', end='\\n' if (i + 1) % 3 == 0 else '  ')\n    else:\n        break\n\nprint('\\nThere are', len(sphenics1m), 'sphenic numbers and', len(sphenic_triplets1m),\n      'sphenic triplets less than 1 million.')\n\nS2HK = sphenics1m[200_000 - 1]\nT5K = sphenic_triplets1m[5000 - 1]\nprint(f'The 200_000th sphenic number is {S2HK}, with prime factors {list(factorint(S2HK).keys())}.')\nprint(f'The 5000th sphenic triplet is ({T5K - 2} {T5K - 1} {T5K}).')\n"}
{"title": "Split a character string based on change of character", "language": "Python", "task": "Split a (character) string into comma (plus a blank) delimited\nstrings based on a change of character   (left to right).\n\nShow the output here   (use the 1st example below).\n\n\nBlanks should be treated as any other character   (except\nthey are problematic to display clearly).   The same applies\nto commas.\n\n\nFor instance, the string: \n  gHHH5YY++///\\ \nshould be split and show:  \n  g, HHH, 5, YY, ++, ///, \\ \n\n", "solution": "from itertools import groupby\n\ndef splitter(text):\n    return ', '.join(''.join(group) for key, group in groupby(text))\n\nif __name__ == '__main__':\n    txt = 'gHHH5YY++///\\\\'      # Note backslash is the Python escape char.\n    print(f'Input: {txt}\\nSplit: {splitter(txt)}')"}
{"title": "Square-free integers", "language": "Python", "task": "Write a function to test if a number is   ''square-free''.\n\n\nA   ''square-free''   is an integer which is divisible by no perfect square other\nthan   '''1'''   (unity).\n\nFor this task, only positive square-free numbers will be used.\n\n\n\nShow here (on this page) all square-free integers (in a horizontal format) that are between:\n::*   '''1'''            --->   '''145'''                        (inclusive)\n::*   '''1''' trillion   --->   '''1''' trillion + '''145'''     (inclusive)\n\n\n(One trillion = 1,000,000,000,000)\n\n\nShow here (on this page) the count of square-free integers from:\n::*   '''1'''            --->   one hundred                      (inclusive)\n::*   '''1'''            --->   one thousand                     (inclusive)\n::*   '''1'''            --->   ten thousand                     (inclusive)\n::*   '''1'''            --->   one hundred thousand             (inclusive)\n::*   '''1'''            --->   one million                      (inclusive)\n\n\n;See also:\n:*   the Wikipedia entry:   square-free integer\n\n", "solution": "import math\n\ndef SquareFree ( _number ) :\n\tmax = (int) (math.sqrt ( _number ))\n\n\tfor root in range ( 2, max+1 ):\t\t\t\t\t# Create a custom prime sieve\n\t\tif 0 == _number % ( root * root ):\n\t\t\treturn False\n\n\treturn True\n\ndef ListSquareFrees( _start, _end ):\n\tcount = 0\n\tfor i in range ( _start, _end+1 ):\n\t\tif True == SquareFree( i ):\n\t\t\tprint ( \"{}\\t\".format(i), end=\"\" )\n\t\t\tcount += 1\n\n\tprint ( \"\\n\\nTotal count of square-free numbers between {} and {}: {}\".format(_start, _end, count))\n\nListSquareFrees( 1, 100 )\nListSquareFrees( 1000000000000, 1000000000145 )\n"}
{"title": "Square but not cube", "language": "Python", "task": "Show the first   '''30'''   positive integers which are squares but not cubes of such integers.\n\nOptionally, show also the first   '''3'''   positive integers which are both squares and cubes,   and mark them as such.\n\n", "solution": "# nonCubeSquares :: Int -> [(Int, Bool)]\ndef nonCubeSquares(n):\n    upto = enumFromTo(1)\n    ns = upto(n)\n    setCubes = set(x ** 3 for x in ns)\n    ms = upto(n + len(set(x * x for x in ns).intersection(\n        setCubes\n    )))\n    return list(tuple([x * x, x in setCubes]) for x in ms)\n\n\n# squareListing :: [(Int, Bool)] -> [String]\ndef squareListing(xs):\n    justifyIdx = justifyRight(len(str(1 + len(xs))))(' ')\n    justifySqr = justifyRight(1 + len(str(xs[-1][0])))(' ')\n    return list(\n        '(' + str(1 + idx) + '^2 = ' + str(n) +\n        ' = ' + str(round(n ** (1 / 3))) + '^3)' if bln else (\n            justifyIdx(1 + idx) + ' ->' +\n            justifySqr(n)\n        )\n        for idx, (n, bln) in enumerate(xs)\n    )\n\n\ndef main():\n    print(\n        unlines(\n            squareListing(\n                nonCubeSquares(30)\n            )\n        )\n    )\n\n\n# GENERIC ------------------------------------------------------------------\n\n# enumFromTo :: Int -> Int -> [Int]\ndef enumFromTo(m):\n    return lambda n: list(range(m, 1 + n))\n\n\n# justifyRight :: Int -> Char -> String -> String\ndef justifyRight(n):\n    return lambda cFiller: lambda a: (\n        ((n * cFiller) + str(a))[-n:]\n    )\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    return '\\n'.join(xs)\n\n\nmain()"}
{"title": "Stair-climbing puzzle", "language": "Python", "task": "From Chung-Chieh Shan (LtU):\n\nYour stair-climbing robot has a very simple low-level API: the \"step\" function takes no argument and attempts to climb one step as a side effect. Unfortunately, sometimes the attempt fails and the robot clumsily falls one step instead. The \"step\" function detects what happens and returns a boolean flag: true on success, false on failure. \n\nWrite a function \"step_up\" that climbs one step up [from the initial position] (by repeating \"step\" attempts if necessary). Assume that the robot is not already at the top of the stairs, and neither does it ever reach the bottom of the stairs. How small can you make \"step_up\"? Can you avoid using variables (even immutable ones) and numbers?\n\nHere's a pseudo-code of a simple recursive solution without using variables:\n\nfunc step_up()\n{\n    if not step() {\n        step_up();\n        step_up();\n    }\n}\n\nInductive proof that step_up() steps up one step, if it terminates:\n* Base case (if the step() call returns true): it stepped up one step. QED\n* Inductive case (if the step() call returns false): Assume that recursive calls to step_up() step up one step. It stepped down one step (because step() returned false), but now we step up two steps using two step_up() calls. QED\n\n\nThe second (tail) recursion above can be turned into an iteration, as follows:\n\nfunc step_up()\n{\n    while not step() {\n        step_up();\n    }\n}\n\n", "solution": "def step_up1():\n  \"\"\"Straightforward implementation: keep track of how many level we\n     need to ascend, and stop when this count is zero.\"\"\"\n  deficit = 1\n  while deficit > 0:\n    if step():\n      deficit -= 1\n    else:\n      deficit += 1"}
{"title": "Statistics/Normal distribution", "language": "Python", "task": "The derive normally distributed random numbers from a uniform generator.\n\n\n;The task:\n# Take a uniform random number generator and create a large (you decide how large) set of numbers that follow a normal (Gaussian) distribution.  Calculate the dataset's mean and standard deviation, and show a histogram of the data.\n# Mention any native language support for the generation of normally distributed random numbers.\n\n\n;Reference:\n* You may refer to code in [[Statistics/Basic]] if available.\n\n", "solution": "from __future__ import division\nimport matplotlib.pyplot as plt \nimport random\n\nmean, stddev, size = 50, 4, 100000\ndata = [random.gauss(mean, stddev) for c in range(size)]\n\nmn = sum(data) / size\nsd = (sum(x*x for x in data) / size \n      - (sum(data) / size) ** 2) ** 0.5\n\nprint(\"Sample mean = %g; Stddev = %g; max = %g; min = %g for %i values\" \n      % (mn, sd, max(data), min(data), size))\n\nplt.hist(data,bins=50)"}
{"title": "Stern-Brocot sequence", "language": "Python", "task": "For this task, the Stern-Brocot sequence is to be generated by an algorithm similar to that employed in generating the [[Fibonacci sequence]].\n\n# The first and second members of the sequence are both 1:\n#*     1, 1\n# Start by considering the second member of the sequence\n# Sum the considered member of the sequence and its precedent, (1 + 1) = 2, and append it to the end of the sequence:\n#*     1, 1, 2\n# Append the considered member of the sequence to the end of the sequence:\n#*     1, 1, 2, 1\n# Consider the next member of the series, (the third member i.e. 2)\n# GOTO 3\n#*\n#*         --- Expanding another loop we get: ---\n#*\n# Sum the considered member of the sequence and its precedent, (2 + 1) = 3, and append it to the end of the sequence:\n#*     1, 1, 2, 1, 3\n# Append the considered member of the sequence to the end of the sequence:\n#*     1, 1, 2, 1, 3, 2\n# Consider the next member of the series, (the fourth member i.e. 1)\n\n\n;The task is to:\n* Create a function/method/subroutine/procedure/... to generate the Stern-Brocot sequence of integers using the method outlined above.\n* Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4)\n* Show the (1-based) index of where the numbers 1-to-10 first appear in the sequence.\n* Show the (1-based) index of where the number 100 first appears in the sequence.\n* Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.\n\nShow your output on this page.\n\n\n;Related tasks:\n:*   [[Fusc sequence]].\n:*   [[Continued fraction/Arithmetic]]\n\n\n;Ref:\n* Infinite Fractions - Numberphile (Video).\n* Trees, Teeth, and Time: The mathematics of clock making.\n* A002487 The On-Line Encyclopedia of Integer Sequences.\n\n", "solution": "def stern_brocot(predicate=lambda series: len(series) < 20):\n    \"\"\"\\\n    Generates members of the stern-brocot series, in order, returning them when the predicate becomes false\n\n    >>> print('The first 10 values:',\n              stern_brocot(lambda series: len(series) < 10)[:10])\n    The first 10 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3]\n    >>>\n    \"\"\"\n\n    sb, i = [1, 1], 0\n    while predicate(sb):\n        sb += [sum(sb[i:i + 2]), sb[i + 1]]\n        i += 1\n    return sb\n\n\nif __name__ == '__main__':\n    from fractions import gcd\n\n    n_first = 15\n    print('The first %i values:\\n  ' % n_first,\n          stern_brocot(lambda series: len(series) < n_first)[:n_first])\n    print()\n    n_max = 10\n    for n_occur in list(range(1, n_max + 1)) + [100]:\n        print('1-based index of the first occurrence of %3i in the series:' % n_occur,\n              stern_brocot(lambda series: n_occur not in series).index(n_occur) + 1)\n              # The following would be much faster. Note that new values always occur at odd indices\n              # len(stern_brocot(lambda series: n_occur != series[-2])) - 1)\n\n    print()\n    n_gcd = 1000\n    s = stern_brocot(lambda series: len(series) < n_gcd)[:n_gcd]\n    assert all(gcd(prev, this) == 1\n               for prev, this in zip(s, s[1:])), 'A fraction from adjacent terms is reducible'"}
{"title": "Stern-Brocot sequence", "language": "Python 3.7", "task": "For this task, the Stern-Brocot sequence is to be generated by an algorithm similar to that employed in generating the [[Fibonacci sequence]].\n\n# The first and second members of the sequence are both 1:\n#*     1, 1\n# Start by considering the second member of the sequence\n# Sum the considered member of the sequence and its precedent, (1 + 1) = 2, and append it to the end of the sequence:\n#*     1, 1, 2\n# Append the considered member of the sequence to the end of the sequence:\n#*     1, 1, 2, 1\n# Consider the next member of the series, (the third member i.e. 2)\n# GOTO 3\n#*\n#*         --- Expanding another loop we get: ---\n#*\n# Sum the considered member of the sequence and its precedent, (2 + 1) = 3, and append it to the end of the sequence:\n#*     1, 1, 2, 1, 3\n# Append the considered member of the sequence to the end of the sequence:\n#*     1, 1, 2, 1, 3, 2\n# Consider the next member of the series, (the fourth member i.e. 1)\n\n\n;The task is to:\n* Create a function/method/subroutine/procedure/... to generate the Stern-Brocot sequence of integers using the method outlined above.\n* Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4)\n* Show the (1-based) index of where the numbers 1-to-10 first appear in the sequence.\n* Show the (1-based) index of where the number 100 first appears in the sequence.\n* Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.\n\nShow your output on this page.\n\n\n;Related tasks:\n:*   [[Fusc sequence]].\n:*   [[Continued fraction/Arithmetic]]\n\n\n;Ref:\n* Infinite Fractions - Numberphile (Video).\n* Trees, Teeth, and Time: The mathematics of clock making.\n* A002487 The On-Line Encyclopedia of Integer Sequences.\n\n", "solution": "'''Stern-Brocot sequence'''\n\nimport math\nimport operator\nfrom itertools import count, dropwhile, islice, takewhile\n\n\n# sternBrocot :: Generator [Int]\ndef sternBrocot():\n    '''Non-finite list of the Stern-Brocot\n       sequence of integers.\n    '''\n    def go(xs):\n        [a, b] = xs[:2]\n        return (a, xs[1:] + [a + b, b])\n    return unfoldr(go)([1, 1])\n\n\n# ------------------------ TESTS -------------------------\n# main :: IO ()\ndef main():\n    '''Various tests'''\n\n    [eq, ne, gcd] = map(\n        curry,\n        [operator.eq, operator.ne, math.gcd]\n    )\n\n    sbs = take(1200)(sternBrocot())\n    ixSB = zip(sbs, enumFrom(1))\n\n    print(unlines(map(str, [\n\n        # First 15 members of the sequence.\n        take(15)(sbs),\n\n        # Indices of where the numbers [1..10] first appear.\n        take(10)(\n            nubBy(on(eq)(fst))(\n                sorted(\n                    takewhile(\n                        compose(ne(12))(fst),\n                        ixSB\n                    ),\n                    key=fst\n                )\n            )\n        ),\n\n        #  Index of where the number 100 first appears.\n        take(1)(dropwhile(compose(ne(100))(fst), ixSB)),\n\n        # Is the gcd of any two consecutive members,\n        # up to the 1000th member, always one ?\n        every(compose(eq(1)(gcd)))(\n            take(1000)(zip(sbs, tail(sbs)))\n        )\n    ])))\n\n\n# ----------------- GENERIC ABSTRACTIONS -----------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# curry :: ((a, b) -> c) -> a -> b -> c\ndef curry(f):\n    '''A curried function derived\n       from an uncurried function.'''\n    return lambda a: lambda b: f(a, b)\n\n\n# enumFrom :: Enum a => a -> [a]\ndef enumFrom(x):\n    '''A non-finite stream of enumerable values,\n       starting from the given value.'''\n    return count(x) if isinstance(x, int) else (\n        map(chr, count(ord(x)))\n    )\n\n\n# every :: (a -> Bool) -> [a] -> Bool\ndef every(p):\n    '''True if p(x) holds for every x in xs'''\n    return lambda xs: all(map(p, xs))\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# head :: [a] -> a\ndef head(xs):\n    '''The first element of a non-empty list.'''\n    return xs[0]\n\n\n# nubBy :: (a -> a -> Bool) -> [a] -> [a]\ndef nubBy(p):\n    '''A sublist of xs from which all duplicates,\n       (as defined by the equality predicate p)\n       are excluded.\n    '''\n    def go(xs):\n        if not xs:\n            return []\n        x = xs[0]\n        return [x] + go(\n            list(filter(\n                lambda y: not p(x)(y),\n                xs[1:]\n            ))\n        )\n    return go\n\n\n# on :: (b -> b -> c) -> (a -> b) -> a -> a -> c\ndef on(f):\n    '''A function returning the value of applying\n      the binary f to g(a) g(b)'''\n    return lambda g: lambda a: lambda b: f(g(a))(g(b))\n\n\n# tail :: [a] -> [a]\n# tail :: Gen [a] -> [a]\ndef tail(xs):\n    '''The elements following the head of a\n       (non-empty) list or generator stream.'''\n    if isinstance(xs, list):\n        return xs[1:]\n    else:\n        islice(xs, 1)  # First item dropped.\n        return xs\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# unfoldr :: (b -> Maybe (a, b)) -> b -> Gen [a]\ndef unfoldr(f):\n    '''A lazy (generator) list unfolded from a seed value\n       by repeated application of f until no residue remains.\n       Dual to fold/reduce.\n       f returns either None or just (value, residue).\n       For a strict output list, wrap the result with list()\n    '''\n    def go(x):\n        valueResidue = f(x)\n        while valueResidue:\n            yield valueResidue[0]\n            valueResidue = f(valueResidue[1])\n    return go\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.'''\n    return '\\n'.join(xs)\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Stirling numbers of the first kind", "language": "Python from Java", "task": "Stirling numbers of the first kind, or Stirling cycle numbers, count permutations according to their number\nof cycles (counting fixed points as cycles of length one).\n\nThey may be defined directly to be the number of permutations of '''n'''\nelements with '''k''' disjoint cycles.\n\nStirling numbers of the first kind express coefficients of polynomial expansions of falling or rising factorials.\n\nDepending on the application, Stirling numbers of the first kind may be \"signed\"\nor \"unsigned\". Signed Stirling numbers of the first kind arise when the\npolynomial expansion is expressed in terms of falling factorials; unsigned when\nexpressed in terms of rising factorials. The only substantial difference is that,\nfor signed Stirling numbers of the first kind, values of S1(n, k) are negative\nwhen n + k is odd.\n\nStirling numbers of the first kind follow the simple identities:\n\n    S1(0, 0) = 1\n    S1(n, 0) = 0 if n > 0\n    S1(n, k) = 0 if k > n\n    S1(n, k) = S1(n - 1, k - 1) + (n - 1) * S1(n - 1, k) # For unsigned\n      ''or''\n    S1(n, k) = S1(n - 1, k - 1) - (n - 1) * S1(n - 1, k) # For signed\n\n\n;Task:\n\n:* Write a routine (function, procedure, whatever) to find '''Stirling numbers of the first kind'''. There are several methods to generate Stirling numbers of the first kind. You are free to choose the most appropriate for your language. If your language has a built-in, or easily, publicly available library implementation, it is acceptable to use that.\n\n:* Using the routine, generate and show here, on this page, a table (or triangle) showing the Stirling numbers of the first kind, '''S1(n, k)''', up to '''S1(12, 12)'''. it is optional to show the row / column for n == 0 and k == 0. It is optional to show places where S1(n, k) == 0 (when k > n). You may choose to show signed or unsigned Stirling numbers of the first kind, just make a note of which was chosen.\n\n:* If your language supports large integers, find and show here, on this page, the maximum value of '''S1(n, k)''' where '''n == 100'''.\n\n\n;See also:\n\n:* '''Wikipedia - Stirling numbers of the first kind'''\n:* '''OEIS:A008275 - Signed Stirling numbers of the first kind'''\n:* '''OEIS:A130534 - Unsigned Stirling numbers of the first kind'''\n\n\n;Related Tasks:\n\n:* '''Stirling numbers of the second kind'''\n:* '''Lah numbers'''\n\n\n", "solution": "computed = {}\n\ndef sterling1(n, k):\n\tkey = str(n) + \",\" + str(k)\n\n\tif key in computed.keys():\n\t\treturn computed[key]\n\tif n == k == 0:\n\t\treturn 1\n\tif n > 0 and k == 0:\n\t\treturn 0\n\tif k > n:\n\t\treturn 0\n\tresult = sterling1(n - 1, k - 1) + (n - 1) * sterling1(n - 1, k)\n\tcomputed[key] = result\n\treturn result\n\nprint(\"Unsigned Stirling numbers of the first kind:\")\nMAX = 12\nprint(\"n/k\".ljust(10), end=\"\")\nfor n in range(MAX + 1):\n\tprint(str(n).rjust(10), end=\"\")\nprint()\nfor n in range(MAX + 1):\n\tprint(str(n).ljust(10), end=\"\")\n\tfor k in range(n + 1):\n\t\tprint(str(sterling1(n, k)).rjust(10), end=\"\")\n\tprint()\nprint(\"The maximum value of S1(100, k) = \")\nprevious = 0\nfor k in range(1, 100 + 1):\n\tcurrent = sterling1(100, k)\n\tif current > previous:\n\t\tprevious = current\n\telse:\n\t\tprint(\"{0}\\n({1} digits, k = {2})\\n\".format(previous, len(str(previous)), k - 1))\n\t\tbreak\n"}
{"title": "Stirling numbers of the second kind", "language": "Python from Java", "task": "Stirling numbers of the second kind, or Stirling partition numbers, are the\nnumber of ways to partition a set of n objects into k non-empty subsets. They are\nclosely related to [[Bell numbers]], and may be derived from them.\n\n\nStirling numbers of the second kind obey the recurrence relation:\n\n    S2(n, 0) and S2(0, k) = 0 # for n, k > 0\n    S2(n, n) = 1\n    S2(n + 1, k) = k * S2(n, k) + S2(n, k - 1)\n\n\n\n;Task:\n\n:* Write a routine (function, procedure, whatever) to find '''Stirling numbers of the second kind'''. There are several methods to generate Stirling numbers of the second kind. You are free to choose the most appropriate for your language. If your language has a built-in, or easily, publicly available library implementation, it is acceptable to use that.\n\n:* Using the routine, generate and show here, on this page, a table (or triangle) showing the Stirling numbers of the second kind, '''S2(n, k)''', up to '''S2(12, 12)'''. it is optional to show the row / column for n == 0 and k == 0. It is optional to show places where S2(n, k) == 0 (when k > n).\n\n:* If your language supports large integers, find and show here, on this page, the maximum value of '''S2(n, k)''' where '''n == 100'''.\n\n\n;See also:\n\n:* '''Wikipedia - Stirling numbers of the second kind'''\n:* '''OEIS:A008277 - Stirling numbers of the second kind'''\n\n\n;Related Tasks:\n\n:* '''Stirling numbers of the first kind'''\n:* '''Bell numbers'''\n:* '''Lah numbers'''\n\n\n", "solution": "computed = {}\n\ndef sterling2(n, k):\n\tkey = str(n) + \",\" + str(k)\n\n\tif key in computed.keys():\n\t\treturn computed[key]\n\tif n == k == 0:\n\t\treturn 1\n\tif (n > 0 and k == 0) or (n == 0 and k > 0):\n\t\treturn 0\n\tif n == k:\n\t\treturn 1\n\tif k > n:\n\t\treturn 0\n\tresult = k * sterling2(n - 1, k) + sterling2(n - 1, k - 1)\n\tcomputed[key] = result\n\treturn result\n\nprint(\"Stirling numbers of the second kind:\")\nMAX = 12\nprint(\"n/k\".ljust(10), end=\"\")\nfor n in range(MAX + 1):\n\tprint(str(n).rjust(10), end=\"\")\nprint()\nfor n in range(MAX + 1):\n\tprint(str(n).ljust(10), end=\"\")\n\tfor k in range(n + 1):\n\t\tprint(str(sterling2(n, k)).rjust(10), end=\"\")\n\tprint()\nprint(\"The maximum value of S2(100, k) = \")\nprevious = 0\nfor k in range(1, 100 + 1):\n\tcurrent = sterling2(100, k)\n\tif current > previous:\n\t\tprevious = current\n\telse:\n\t\tprint(\"{0}\\n({1} digits, k = {2})\\n\".format(previous, len(str(previous)), k - 1))\n\t\tbreak\n"}
{"title": "Strassen's algorithm", "language": "Python", "task": "Description\nIn linear algebra, the Strassen algorithm   (named after Volker Strassen),   is an algorithm for matrix multiplication. \n\nIt is faster than the standard matrix multiplication algorithm and is useful in practice for large matrices,   but would be slower than the fastest known algorithms for extremely large matrices.\n\n\n;Task\nWrite a routine, function, procedure etc. in your language to implement the Strassen algorithm for matrix multiplication.\n\nWhile practical implementations of Strassen's algorithm usually switch to standard methods of matrix multiplication for small enough sub-matrices (currently anything less than   512x512   according to Wikipedia),   for the purposes of this task you should not switch until reaching a size of 1 or 2.\n\n\n;Related task\n:* [[Matrix multiplication]]\n\n\n;See also\n:* Wikipedia article\n\n", "solution": "\"\"\"Matrix multiplication using Strassen's algorithm. Requires Python >= 3.7.\"\"\"\n\nfrom __future__ import annotations\nfrom itertools import chain\nfrom typing import List\nfrom typing import NamedTuple\nfrom typing import Optional\n\n\nclass Shape(NamedTuple):\n    rows: int\n    cols: int\n\n\nclass Matrix(List):\n    \"\"\"A matrix implemented as a two-dimensional list.\"\"\"\n\n    @classmethod\n    def block(cls, blocks) -> Matrix:\n        \"\"\"Return a new Matrix assembled from nested blocks.\"\"\"\n        m = Matrix()\n        for hblock in blocks:\n            for row in zip(*hblock):\n                m.append(list(chain.from_iterable(row)))\n\n        return m\n\n    def dot(self, b: Matrix) -> Matrix:\n        \"\"\"Return a new Matrix that is the product of this matrix and matrix `b`.\n        Uses 'simple' or 'naive' matrix multiplication.\"\"\"\n        assert self.shape.cols == b.shape.rows\n        m = Matrix()\n        for row in self:\n            new_row = []\n            for c in range(len(b[0])):\n                col = [b[r][c] for r in range(len(b))]\n                new_row.append(sum(x * y for x, y in zip(row, col)))\n            m.append(new_row)\n        return m\n\n    def __matmul__(self, b: Matrix) -> Matrix:\n        return self.dot(b)\n\n    def __add__(self, b: Matrix) -> Matrix:\n        \"\"\"Matrix addition.\"\"\"\n        assert self.shape == b.shape\n        rows, cols = self.shape\n        return Matrix(\n            [[self[i][j] + b[i][j] for j in range(cols)] for i in range(rows)]\n        )\n\n    def __sub__(self, b: Matrix) -> Matrix:\n        \"\"\"Matrix subtraction.\"\"\"\n        assert self.shape == b.shape\n        rows, cols = self.shape\n        return Matrix(\n            [[self[i][j] - b[i][j] for j in range(cols)] for i in range(rows)]\n        )\n\n    def strassen(self, b: Matrix) -> Matrix:\n        \"\"\"Return a new Matrix that is the product of this matrix and matrix `b`.\n        Uses strassen algorithm.\"\"\"\n        rows, cols = self.shape\n\n        assert rows == cols, \"matrices must be square\"\n        assert self.shape == b.shape, \"matrices must be the same shape\"\n        assert rows and (rows & rows - 1) == 0, \"shape must be a power of 2\"\n\n        if rows == 1:\n            return self.dot(b)\n\n        p = rows // 2  # partition\n\n        a11 = Matrix([n[:p] for n in self[:p]])\n        a12 = Matrix([n[p:] for n in self[:p]])\n        a21 = Matrix([n[:p] for n in self[p:]])\n        a22 = Matrix([n[p:] for n in self[p:]])\n\n        b11 = Matrix([n[:p] for n in b[:p]])\n        b12 = Matrix([n[p:] for n in b[:p]])\n        b21 = Matrix([n[:p] for n in b[p:]])\n        b22 = Matrix([n[p:] for n in b[p:]])\n\n        m1 = (a11 + a22).strassen(b11 + b22)\n        m2 = (a21 + a22).strassen(b11)\n        m3 = a11.strassen(b12 - b22)\n        m4 = a22.strassen(b21 - b11)\n        m5 = (a11 + a12).strassen(b22)\n        m6 = (a21 - a11).strassen(b11 + b12)\n        m7 = (a12 - a22).strassen(b21 + b22)\n\n        c11 = m1 + m4 - m5 + m7\n        c12 = m3 + m5\n        c21 = m2 + m4\n        c22 = m1 - m2 + m3 + m6\n\n        return Matrix.block([[c11, c12], [c21, c22]])\n\n    def round(self, ndigits: Optional[int] = None) -> Matrix:\n        return Matrix([[round(i, ndigits) for i in row] for row in self])\n\n    @property\n    def shape(self) -> Shape:\n        cols = len(self[0]) if self else 0\n        return Shape(len(self), cols)\n\n\ndef examples():\n    a = Matrix(\n        [\n            [1, 2],\n            [3, 4],\n        ]\n    )\n    b = Matrix(\n        [\n            [5, 6],\n            [7, 8],\n        ]\n    )\n    c = Matrix(\n        [\n            [1, 1, 1, 1],\n            [2, 4, 8, 16],\n            [3, 9, 27, 81],\n            [4, 16, 64, 256],\n        ]\n    )\n    d = Matrix(\n        [\n            [4, -3, 4 / 3, -1 / 4],\n            [-13 / 3, 19 / 4, -7 / 3, 11 / 24],\n            [3 / 2, -2, 7 / 6, -1 / 4],\n            [-1 / 6, 1 / 4, -1 / 6, 1 / 24],\n        ]\n    )\n    e = Matrix(\n        [\n            [1, 2, 3, 4],\n            [5, 6, 7, 8],\n            [9, 10, 11, 12],\n            [13, 14, 15, 16],\n        ]\n    )\n    f = Matrix(\n        [\n            [1, 0, 0, 0],\n            [0, 1, 0, 0],\n            [0, 0, 1, 0],\n            [0, 0, 0, 1],\n        ]\n    )\n\n    print(\"Naive matrix multiplication:\")\n    print(f\"  a * b = {a @ b}\")\n    print(f\"  c * d = {(c @ d).round()}\")\n    print(f\"  e * f = {e @ f}\")\n\n    print(\"Strassen's matrix multiplication:\")\n    print(f\"  a * b = {a.strassen(b)}\")\n    print(f\"  c * d = {c.strassen(d).round()}\")\n    print(f\"  e * f = {e.strassen(f)}\")\n\n\nif __name__ == \"__main__\":\n    examples()\n"}
{"title": "Stream merge", "language": "Python", "task": " 2-stream merge\n: Read two sorted streams of items from external source (e.g. disk, or network), and write one stream of sorted items to external sink.\n: Common algorithm: keep 1 buffered item from each source, select minimal of them, write it, fetch another item from that stream from which the written item was.\n\n; ''N''-stream merge\n: The same as above, but reading from    ''N''    sources.\n: Common algorithm: same as above, but keep buffered items and their source descriptors in a [[heap]].\n\n\nAssume streams are very big. You must not suck them whole in the memory, but read them as streams.\n\n", "solution": "import heapq\nimport sys\n\nsources = sys.argv[1:]\nfor item in heapq.merge(open(source) for source in sources):\n    print(item)"}
{"title": "Strip control codes and extended characters from a string", "language": "Python", "task": "Strip control codes and extended characters from a string. \n\n\nThe solution should demonstrate how to achieve each of the following results:\n:*   a string with control codes stripped (but extended characters not stripped)\n:*   a string with control codes and extended characters stripped\n\n\nIn ASCII, the control codes have decimal codes 0 through to 31 and 127. \n\nOn an ASCII based system, if the control codes are stripped, the resultant string would have all of its characters within the range of 32 to 126 decimal on the ASCII table.\n\nOn a non-ASCII based system, we consider characters that do not have a corresponding glyph on the ASCII table (within the ASCII range of 32 to 126 decimal) to be an extended character for the purpose of this task.\n\n\n", "solution": "stripped = lambda s: \"\".join(i for i in s if 31 < ord(i) < 127)\n\nprint(stripped(\"\\ba\\x00b\\n\\rc\\fd\\xc3\"))"}
{"title": "Subleq", "language": "Python", "task": "One-Instruction Set Computer (OISC). \n\nIt is named after its only instruction, which is '''SU'''btract and '''B'''ranch if '''L'''ess than or '''EQ'''ual to zero.  \n\n;Task\nYour task is to create an interpreter which emulates a SUBLEQ machine.\n\nThe machine's memory consists of an array of signed integers.   These integers may be interpreted in three ways:\n::::*   simple numeric values \n::::*   memory addresses \n::::*   characters for input or output\n\nAny reasonable word size that accommodates all three of the above uses is fine. \n\nThe program should load the initial contents of the emulated machine's memory, set the instruction pointer to the first address (which is defined to be address 0), and begin emulating the machine, which works as follows:\n:#   Let '''A''' be the value in the memory location identified by the instruction pointer;   let '''B''' and '''C''' be the values stored in the next two consecutive addresses in memory.\n:#   Advance the instruction pointer three words, to point at the address ''after'' the address containing '''C'''.\n:#   If '''A''' is   '''-1'''   (negative unity),   then a character is read from the machine's input and its numeric value stored in the address given by '''B'''.   '''C''' is unused.\n:#   If '''B''' is   '''-1'''   (negative unity),   then the number contained in the address given by '''A''' is interpreted as a character and written to the machine's output.   '''C''' is unused.\n:#   Otherwise, both '''A''' and '''B''' are treated as addresses.   The number contained in address '''A''' is subtracted from the number in address '''B''' (and the difference left in address '''B''').   If the result is positive, execution continues uninterrupted; if the result is zero or negative, the number in '''C''' becomes the new instruction pointer.\n:#   If the instruction pointer becomes negative, execution halts.\n\nYour solution may initialize the emulated machine's memory in any convenient manner, but if you accept it as input, it should be a separate input stream from the one fed to the emulated machine once it is running. And if fed as text input, it should be in the form of raw subleq \"machine code\" - whitespace-separated decimal numbers, with no symbolic names or other assembly-level extensions, to be loaded into memory starting at address   '''0'''   (zero).\n\nFor purposes of this task, show the output of your solution when fed the below   \"Hello, world!\"   program. \n\nAs written, this example assumes ASCII or a superset of it, such as any of the Latin-N character sets or Unicode;   you may translate the numbers representing characters (starting with 72=ASCII 'H') into another character set if your implementation runs in a non-ASCII-compatible environment. If 0 is not an appropriate terminator in your character set, the program logic will need some adjustment as well.\n\n15 17 -1 17 -1 -1 16 1 -1 16 3 -1 15 15 0 0 -1 72 101 108 108 111 44 32 119 111 114 108 100 33 10 0\n\nThe above \"machine code\" corresponds to something like this in a hypothetical assembler language for a signed 8-bit version of the machine:\n\nstart:\n    0f 11 ff subleq (zero), (message), -1\n    11 ff ff subleq (message), -1, -1    ; output character at message\n    10 01 ff subleq (neg1), (start+1), -1\n    10 03 ff subleq (neg1), (start+3), -1\n    0f 0f 00 subleq (zero), (zero), start\n; useful constants\nzero: \n    00      .data 0  \nneg1: \n    ff      .data -1\n; the message to print\nmessage: .data \"Hello, world!\\n\\0\"\n    48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 21 0a 00\n\n", "solution": "import sys\n\ndef subleq(a):\n    i = 0\n    try:\n        while i >= 0:\n            if a[i] == -1:\n                a[a[i + 1]] = ord(sys.stdin.read(1))\n            elif a[i + 1] == -1:\n                print(chr(a[a[i]]), end=\"\")\n            else:\n                a[a[i + 1]] -= a[a[i]]\n                if a[a[i + 1]] <= 0:\n                    i = a[i + 2]\n                    continue\n            i += 3\n    except (ValueError, IndexError, KeyboardInterrupt):\n        print(\"abort\")\n        print(a)\n\nsubleq([15, 17, -1, 17, -1, -1, 16, 1, -1, 16, 3, -1, 15, 15,\n        0, 0, -1, 72, 101, 108, 108, 111, 44, 32, 119, 111,\n        114, 108, 100, 33, 10, 0])"}
{"title": "Substring/Top and tail", "language": "Python", "task": "The task is to demonstrate how to remove the first and last characters from a string. \n\nThe solution should demonstrate how to obtain the following results:\n\n* String with first character removed\n* String with last character removed\n* String with both the first and last characters removed\n\n\nIf the program uses UTF-8 or UTF-16, it must work on any valid Unicode code point, whether in the Basic Multilingual Plane or above it. \n\nThe program must reference logical characters (code points), not 8-bit code units for UTF-8 or 16-bit code units for UTF-16. \n\nPrograms for other encodings (such as 8-bit ASCII, or EUC-JP) are not required to handle all Unicode characters.\n\n\n\n", "solution": "from functools import (reduce)\n\n\ndef main():\n    for xs in transpose(\n        (chunksOf(3)(\n            ap([tail, init, compose(init)(tail)])(\n                ['knights', 'socks', 'brooms']\n            )\n        ))\n    ):\n        print(xs)\n\n\n# GENERIC -------------------------------------------------\n\n# tail :: [a] -> [a]\ndef tail(xs):\n    return xs[1:]\n\n\n# init::[a] - > [a]\ndef init(xs):\n    return xs[:-1]\n\n\n# ap (<*>) :: [(a -> b)] -> [a] -> [b]\ndef ap(fs):\n    return lambda xs: reduce(\n        lambda a, f: a + reduce(\n            lambda a, x: a + [f(x)], xs, []\n        ), fs, []\n    )\n\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    return lambda f: lambda x: g(f(x))\n\n\n# transpose :: [[a]] -> [[a]]\ndef transpose(xs):\n    return list(map(list, zip(*xs)))\n\n\nif __name__ == '__main__':\n    main()"}
{"title": "Sum and product puzzle", "language": "Python", "task": "*   Wikipedia:   Sum and Product Puzzle\n\n", "solution": "#!/usr/bin/env python\n\nfrom collections import Counter\n\ndef decompose_sum(s):\n    return [(a,s-a) for a in range(2,int(s/2+1))]\n\n# Generate all possible pairs\nall_pairs = set((a,b) for a in range(2,100) for b in range(a+1,100) if a+b<100)\n\n# Fact 1 --> Select pairs for which all sum decompositions have non-unique product\nproduct_counts = Counter(c*d for c,d in all_pairs)\nunique_products = set((a,b) for a,b in all_pairs if product_counts[a*b]==1)\ns_pairs = [(a,b) for a,b in all_pairs if\n    all((x,y) not in unique_products for (x,y) in decompose_sum(a+b))]\n\n# Fact 2 --> Select pairs for which the product is unique\nproduct_counts = Counter(c*d for c,d in s_pairs)\np_pairs = [(a,b) for a,b in s_pairs if product_counts[a*b]==1]\n\n# Fact 3 --> Select pairs for which the sum is unique\nsum_counts = Counter(c+d for c,d in p_pairs)\nfinal_pairs = [(a,b) for a,b in p_pairs if sum_counts[a+b]==1]\n\nprint(final_pairs)"}
{"title": "Sum digits of an integer", "language": "Python", "task": "Take a   Natural Number   in a given base and return the sum of its digits:\n:*   '''1'''10                  sums to   '''1'''\n:*   '''1234'''10       sums to   '''10'''\n:*   '''fe'''16              sums to   '''29'''\n:*   '''f0e'''16           sums to   '''29'''\n\n", "solution": "def sumDigits(num, base=10):\n    if base < 2:\n        print(\"Error: base must be at least 2\")\n        return\n    num, sum = abs(num), 0\n    while num >= base:\n        num, rem = divmod(num, base)\n        sum += rem\n    return sum + num\n\nprint(sumDigits(1))\nprint(sumDigits(12345))\nprint(sumDigits(-123045))\nprint(sumDigits(0xfe, 16))\nprint(sumDigits(0xf0e, 16))"}
{"title": "Sum multiples of 3 and 5", "language": "Python", "task": "The objective is to write a function that finds the sum of all positive multiples of 3 or 5 below ''n''. \n\nShow output for ''n'' = 1000.\n\nThis is is the same as Project Euler problem 1.\n\n'''Extra credit:''' do this efficiently for ''n'' = 1e20 or higher.\n\n", "solution": "def sum35a(n):\n    'Direct count'\n    # note: ranges go to n-1\n    return sum(x for x in range(n) if x%3==0 or x%5==0)\n\ndef sum35b(n): \n    \"Count all the 3's; all the 5's; minus double-counted 3*5's\"\n    # note: ranges go to n-1\n    return sum(range(3, n, 3)) + sum(range(5, n, 5)) - sum(range(15, n, 15))\n    \ndef sum35c(n):\n    'Sum the arithmetic progressions: sum3 + sum5 - sum15'\n    consts = (3, 5, 15)\n    # Note: stop at n-1\n    divs = [(n-1) // c for c in consts]\n    sums = [d*c*(1+d)/2 for d,c in zip(divs, consts)]\n    return sums[0] + sums[1] - sums[2]\n\n#test\nfor n in range(1001):\n    sa, sb, sc = sum35a(n), sum35b(n), sum35c(n)\n    assert sa == sb == sc  # python tests aren't like those of c.\n\nprint('For n = %7i -> %i\\n' % (n, sc))\n\n# Pretty patterns\nfor p in range(7):\n    print('For n = %7i -> %i' % (10**p, sum35c(10**p)))\n\n# Scalability \np = 20\nprint('\\nFor n = %20i -> %i' % (10**p, sum35c(10**p)))"}
{"title": "Sum multiples of 3 and 5", "language": "Python 3.7", "task": "The objective is to write a function that finds the sum of all positive multiples of 3 or 5 below ''n''. \n\nShow output for ''n'' = 1000.\n\nThis is is the same as Project Euler problem 1.\n\n'''Extra credit:''' do this efficiently for ''n'' = 1e20 or higher.\n\n", "solution": "'''Summed multiples of 3 and 5 up to n'''\n\n\n# sum35 :: Int -> Int\ndef sum35(n):\n    '''Sum of all positive multiples\n       of 3 or 5 below n.\n    '''\n    f = sumMults(n)\n    return f(3) + f(5) - f(15)\n\n\n# sumMults :: Int -> Int -> Int\ndef sumMults(n):\n    '''Area under a straight line between\n       the first multiple and the last.\n    '''\n    def go(n, m):\n        n1 = (n - 1) // m\n        return (m * n1 * (n1 + 1)) // 2\n    return lambda x: go(n, x)\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''Tests for [10^1 .. 10^5], and [10^8 .. 10^25]\n    '''\n    print(\n        fTable(__doc__ + ':\\n')(lambda x: '10E' + str(x))(\n            str\n        )(compose(sum35)(lambda x: 10**x))(\n            enumFromTo(1)(5) + enumFromTo(18)(25)\n        )\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# enumFromTo :: (Int, Int) -> [Int]\ndef enumFromTo(m):\n    '''Integer enumeration from m to n.'''\n    return lambda n: list(range(m, 1 + n))\n\n\n# fTable :: String -> (a -> String) ->\n#                     (b -> String) ->\n#        (a -> b) -> [a] -> String\ndef fTable(s):\n    '''Heading -> x display function -> fx display function ->\n          f -> value list -> tabular string.'''\n    def go(xShow, fxShow, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + '\\n'.join([\n            xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs\n        ])\n    return lambda xShow: lambda fxShow: (\n        lambda f: lambda xs: go(\n            xShow, fxShow, f, xs\n        )\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Sum of elements below main diagonal of matrix", "language": "Python", "task": "Find and display the sum of elements that are below the main diagonal of a matrix.\n\nThe matrix should be a square matrix.\n\n\n::: ---   Matrix to be used:   ---\n\n      [[1,3,7,8,10],\n       [2,4,16,14,4],\n       [3,1,9,18,11],\n       [12,14,17,18,20],\n       [7,1,3,9,5]] \n\n", "solution": "'''Lower triangle of a matrix'''\n\nfrom itertools import chain, islice\nfrom functools import reduce\n\n\n# lowerTriangle :: [[a]] -> None | [[a]]\ndef lowerTriangle(matrix):\n    '''Either None, if the matrix is not square, or\n       the rows of the matrix, each containing only\n       those values that form part of the lower triangle.\n    '''\n    def go(n_rows, xs):\n        n, rows = n_rows\n        return 1 + n, rows + [list(islice(xs, n))]\n\n    return reduce(\n        go,\n        matrix,\n        (0, [])\n    )[1] if isSquare(matrix) else None\n\n\n# isSquare :: [[a]] -> Bool\ndef isSquare(matrix):\n    '''True if all rows of the matrix share\n       the length of the matrix itself.\n    '''\n    n = len(matrix)\n    return all([n == len(x) for x in matrix])\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Sum of integers in the lower triangle of a matrix.\n    '''\n    rows = lowerTriangle([\n        [1, 3, 7, 8, 10],\n        [2, 4, 16, 14, 4],\n        [3, 1, 9, 18, 11],\n        [12, 14, 17, 18, 20],\n        [7, 1, 3, 9, 5]\n    ])\n\n    print(\n        \"Not a square matrix.\" if None is rows else (\n            sum(chain(*rows))\n        )\n    )\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Sum to 100", "language": "Python", "task": "Find solutions to the    ''sum to one hundred''    puzzle.\n\n\nAdd (insert) the mathematical\noperators      '''+'''    or    '''-'''      (plus\nor minus)   before any of the digits in the\ndecimal numeric string    '''123456789'''    such that the\nresulting mathematical expression adds up to a\nparticular sum    (in this iconic case,   '''100''').\n\n\nExample:   \n              123 + 4 - 5 + 67 - 89   =   100     \n\nShow all output here.\n\n\n:*   Show all solutions that sum to    '''100''' \n:*   Show the sum that has the maximum   ''number''   of solutions   (from zero to infinity++)\n:*   Show the lowest positive sum that   ''can't''   be expressed   (has no solutions),   using the rules for this task\n:*   Show the ten highest numbers that can be expressed using the rules for this task   (extra credit)\n\n++   (where   ''infinity''   would be a relatively small   123,456,789)\n\n\nAn example of a sum that can't be expressed   (within the rules of this task)   is:   '''5074'''\n(which,   of course,   isn't the lowest positive sum that can't be expressed).\n\n", "solution": "from itertools import product, islice\n\n\ndef expr(p):\n    return \"{}1{}2{}3{}4{}5{}6{}7{}8{}9\".format(*p)\n\n\ndef gen_expr():\n    op = ['+', '-', '']\n    return [expr(p) for p in product(op, repeat=9) if p[0] != '+']\n\n\ndef all_exprs():\n    values = {}\n    for expr in gen_expr():\n        val = eval(expr)\n        if val not in values:\n            values[val] = 1\n        else:\n            values[val] += 1\n    return values\n\n\ndef sum_to(val):\n    for s in filter(lambda x: x[0] == val, map(lambda x: (eval(x), x), gen_expr())):\n        print(s)\n\n\ndef max_solve():\n    print(\"Sum {} has the maximum number of solutions: {}\".\n          format(*max(all_exprs().items(), key=lambda x: x[1])))\n\n\ndef min_solve():\n    values = all_exprs()\n    for i in range(123456789):\n        if i not in values:\n            print(\"Lowest positive sum that can't be expressed: {}\".format(i))\n            return\n\n\ndef highest_sums(n=10):\n    sums = map(lambda x: x[0],\n               islice(sorted(all_exprs().items(), key=lambda x: x[0], reverse=True), n))\n    print(\"Highest Sums: {}\".format(list(sums)))\n\n\nsum_to(100)\nmax_solve()\nmin_solve()\nhighest_sums()"}
{"title": "Summarize and say sequence", "language": "Python", "task": "There are several ways to generate a self-referential sequence. One very common one (the [[Look-and-say sequence]]) is to start with a positive integer, then generate the next term by concatenating enumerated groups of adjacent alike digits:\n        0, 10, 1110, 3110, 132110, 1113122110, 311311222110 ...\nThe terms generated grow in length geometrically and never converge.\n\nAnother way to generate a self-referential sequence is to summarize the previous term.\n\nCount how many of each alike digit there is, then concatenate the sum and digit for each of the sorted enumerated digits. Note that the first five terms are the same as for the previous sequence.\n        0, 10, 1110, 3110, 132110, 13123110, 23124110 ... \nSort the digits largest to smallest. Do not include counts of digits that do not appear in the previous term.\n\nDepending on the seed value, series generated this way always either converge to a stable value or to a short cyclical pattern. (For our purposes, I'll use converge to mean an element matches a previously seen element.) The sequence shown, with a seed value of 0, converges to a stable value of 1433223110 after 11 iterations. The seed value that converges most quickly is 22. It goes stable after the first element. (The next element is 22, which has been seen before.)\n\n\n;Task:\nFind all the positive integer seed values under 1000000, for the above convergent self-referential sequence, that takes the largest number of iterations before converging. Then print out the number of iterations and the sequence they return. Note that different permutations of the digits of the seed will yield the same sequence. For this task, assume leading zeros are not permitted.  \n\nSeed Value(s): 9009 9090 9900\n\nIterations: 21 \n\nSequence: (same for all three seeds except for first element)\n9009\n2920\n192210\n19222110\n19323110\n1923123110\n1923224110\n191413323110\n191433125110\n19151423125110\n19251413226110\n1916151413325110\n1916251423127110\n191716151413326110\n191726151423128110\n19181716151413327110\n19182716151423129110\n29181716151413328110\n19281716151423228110\n19281716151413427110\n19182716152413228110\n\n\n;Related tasks:\n*   [[Fours is the number of letters in the ...]]\n*   [[Look-and-say sequence]]\n*   [[Number names]]\n*   [[Self-describing numbers]]\n*   [[Spelling of ordinal numbers]]\n\n\n\n\n\n;Also see:\n*   The On-Line Encyclopedia of Integer Sequences.\n\n", "solution": "from itertools import groupby, permutations\n\ndef A036058(number):\n    return ''.join( str(len(list(g))) + k\n                    for k,g in groupby(sorted(str(number), reverse=True)) )\n\ndef A036058_length(numberstring='0', printit=False):\n    iterations, last_three, queue_index = 1, ([None] * 3), 0\n\n    def A036058(number):\n        # rely on external reverse-sort of digits of number\n        return ''.join( str(len(list(g))) + k\n                        for k,g in groupby(number) )\n\n    while True:\n        if printit:\n            print(\"  %2i %s\" % (iterations, numberstring))\n        numberstring = ''.join(sorted(numberstring, reverse=True))\n        if numberstring in last_three:\n            break\n        assert iterations < 1000000\n        last_three[queue_index], numberstring = numberstring, A036058(numberstring)\n        iterations += 1\n        queue_index +=1\n        queue_index %=3\n    return iterations\n    \ndef max_A036058_length( start_range=range(11) ):\n    already_done = set()\n    max_len = (-1, [])\n    for n in start_range:\n        sn = str(n)\n        sns = tuple(sorted(sn, reverse=True))\n        if sns not in already_done:\n            already_done.add(sns)\n            size = A036058_length(sns)\n            if size > max_len[0]:\n                max_len = (size, [n])\n            elif size == max_len[0]:\n                max_len[1].append(n)\n    return max_len\n\nlenmax, starts = max_A036058_length( range(1000000) )\n\n# Expand\nallstarts = []\nfor n in starts:\n    allstarts += [int(''.join(x))\n                  for x in set(k\n                               for k in permutations(str(n), 4)\n                               if k[0] != '0')]\nallstarts = [x for x in sorted(allstarts) if x < 1000000]\n\nprint ( '''\\\nThe longest length, followed by the number(s) with the longest sequence length\nfor starting sequence numbers below 1000000 are:\n  Iterations = %i and sequence-starts = %s.''' % (lenmax, allstarts)   )\n\nprint ( '''\nNote that only the first of any sequences with the same digits is printed below.\n(The others will differ only in their first term)''' )\n\nfor n in starts:\n    print()\n    A036058_length(str(n), printit=True)"}
{"title": "Summarize primes", "language": "Python", "task": "Considering in order of length, n, all sequences of consecutive\nprimes, p, from 2 onwards, where p < 1000 and n>0, select those\nsequences whose sum is prime, and for these display the length of the\nsequence, the last item in the sequence, and the sum.\n\n", "solution": "'''Prime sums of primes up to 1000'''\n\n\nfrom itertools import accumulate, chain, takewhile\n\n\n# primeSums :: [(Int, (Int, Int))]\ndef primeSums():\n    '''Non finite stream of enumerated tuples,\n       in which the first value is a prime,\n       and the second the sum of that prime and all\n       preceding primes.\n    '''\n    return (\n        x for x in enumerate(\n            accumulate(\n                chain([(0, 0)], primes()),\n                lambda a, p: (p, p + a[1])\n            )\n        ) if isPrime(x[1][1])\n    )\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''Prime sums of primes below 1000'''\n    for x in takewhile(\n            lambda t: 1000 > t[1][0],\n            primeSums()\n    ):\n        print(f'{x[0]} -> {x[1][1]}')\n\n\n# ----------------------- GENERIC ------------------------\n\n# isPrime :: Int -> Bool\ndef isPrime(n):\n    '''True if n is prime.'''\n    if n in (2, 3):\n        return True\n    if 2 > n or 0 == n % 2:\n        return False\n    if 9 > n:\n        return True\n    if 0 == n % 3:\n        return False\n\n    def p(x):\n        return 0 == n % x or 0 == n % (2 + x)\n\n    return not any(map(p, range(5, 1 + int(n ** 0.5), 6)))\n\n\n# primes :: [Int]\ndef primes():\n    ''' Non finite sequence of prime numbers.\n    '''\n    n = 2\n    dct = {}\n    while True:\n        if n in dct:\n            for p in dct[n]:\n                dct.setdefault(n + p, []).append(p)\n            del dct[n]\n        else:\n            yield n\n            dct[n * n] = [n]\n        n = 1 + n\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()\n"}
{"title": "Super-d numbers", "language": "Python", "task": "A super-d number is a positive, decimal (base ten) integer   '''n'''   such that   '''d x nd'''   has at least   '''d'''   consecutive digits   '''d'''   where\n    2 <= d <= 9\nFor instance, 753 is a super-3 number because 3 x 7533 = 1280873331.\n\n\n'''Super-d'''   numbers are also shown on   '''MathWorld'''(tm)   as   '''super-''d'' '''   or   '''super-''d'''''.\n\n\n;Task:\n:* Write a function/procedure/routine to find super-d numbers.\n:* For   '''d=2'''   through   '''d=6''',   use the routine to show the first   '''10'''   super-d numbers.\n\n\n;Extra credit:\n:* Show the first   '''10'''   super-7, super-8, and/or super-9 numbers   (optional).\n\n\n;See also:\n:*   Wolfram MathWorld - Super-d Number.\n:*   OEIS: A014569 - Super-3 Numbers.\n\n", "solution": "from itertools import islice, count\n\ndef superd(d):\n    if d != int(d) or not 2 <= d <= 9:\n        raise ValueError(\"argument must be integer from 2 to 9 inclusive\")\n    tofind = str(d) * d\n    for n in count(2):\n        if tofind in str(d * n ** d):\n            yield n\n\nif __name__ == '__main__':\n    for d in range(2, 9):\n        print(f\"{d}:\", ', '.join(str(n) for n in islice(superd(d), 10)))"}
{"title": "Superellipse", "language": "Python", "task": "A superellipse is a geometric figure defined as the set of all points (x, y) with \n\n\n::: \\left|\\frac{x}{a}\\right|^n\\! + \\left|\\frac{y}{b}\\right|^n\\! = 1,\n\nwhere ''n'', ''a'', and ''b'' are positive numbers.\n\n\n;Task\nDraw a superellipse with n = 2.5, and a = b = 200\n\n", "solution": "# Superellipse drawing in Python 2.7.9\n# pic can see at http://www.imgup.cz/image/712\n\nimport matplotlib.pyplot as plt\nfrom math import sin, cos, pi\n\ndef sgn(x):\n\treturn ((x>0)-(x<0))*1\n\na,b,n=200,200,2.5 # param n making shape\nna=2/n\nstep=100 # accuracy\npiece=(pi*2)/step\nxp=[];yp=[]\n\nt=0\nfor t1 in range(step+1):\n\t# because sin^n(x) is mathematically the same as (sin(x))^n...\n\tx=(abs((cos(t)))**na)*a*sgn(cos(t))\n\ty=(abs((sin(t)))**na)*b*sgn(sin(t))\n\txp.append(x);yp.append(y)\n\tt+=piece\n\nplt.plot(xp,yp) # plotting all point from array xp, yp\nplt.title(\"Superellipse with parameter \"+str(n))\nplt.show()\n"}
{"title": "Superpermutation minimisation", "language": "Python", "task": "A superpermutation of N different characters is a string consisting of an arrangement of multiple copies of those N different characters in which every permutation of those characters can be found as a substring.\n\nFor example, representing the characters as A..Z, using N=2 we choose to use the first two characters 'AB'. \nThe permutations of 'AB' are the two, (i.e. two-factorial), strings: 'AB' and 'BA'.\n\nA too obvious method of generating a superpermutation is to just join all the permutations together forming 'ABBA'.\n\nA little thought will produce the shorter (in fact the shortest) superpermutation of 'ABA' - it contains 'AB' at the beginning and contains 'BA' from the middle to the end.\n\nThe \"too obvious\" method of creation generates a string of length N!*N. Using this as a yardstick, the task is to investigate other methods of generating superpermutations of N from 1-to-7 characters, that never generate larger superpermutations.\n\nShow descriptions and comparisons of algorithms used here, and select the \"Best\" algorithm as being the one generating shorter superpermutations.\n\nThe problem of generating the shortest superpermutation for each N ''might'' be NP complete, although the minimal strings for small values of N have been found by brute -force searches.\n\n\n\n\n;Reference:\n* The Minimal Superpermutation Problem. by Nathaniel Johnston.\n* oeis A180632 gives 0-5 as 0, 1, 3, 9, 33, 153. 6 is thought to be 872.\n* Superpermutations - Numberphile. A video\n* Superpermutations: the maths problem solved by 4chan - Standupmaths. A video of recent (2018) mathematical progress.\n* New Superpermutations Discovered! Standupmaths & Numberphile.\n\n", "solution": "\"Generate a short Superpermutation of n characters A... as a string using various algorithms.\"\n\n\nfrom __future__ import print_function, division\n\nfrom itertools import permutations\nfrom math import factorial\nimport string\nimport datetime\nimport gc\n\n\n\nMAXN = 7\n\n\ndef s_perm0(n):\n    \"\"\"\n    Uses greedy algorithm of adding another char (or two, or three, ...)\n    until an unseen perm is formed in the last n chars\n    \"\"\"\n    allchars = string.ascii_uppercase[:n]\n    allperms = [''.join(p) for p in permutations(allchars)]\n    sp, tofind = allperms[0], set(allperms[1:])\n    while tofind:\n        for skip in range(1, n):\n            for trial_add in (''.join(p) for p in permutations(sp[-n:][:skip])):\n                #print(sp, skip, trial_add)\n                trial_perm = (sp + trial_add)[-n:]\n                if trial_perm in tofind:\n                    #print(sp, skip, trial_add)\n                    sp += trial_add\n                    tofind.discard(trial_perm)\n                    trial_add = None    # Sentinel\n                    break\n            if trial_add is None:\n                break\n    assert all(perm in sp for perm in allperms) # Check it is a superpermutation\n    return sp\n\ndef s_perm1(n):\n    \"\"\"\n    Uses algorithm of concatenating all perms in order if not already part\n    of concatenation.\n    \"\"\"\n    allchars = string.ascii_uppercase[:n]\n    allperms = [''.join(p) for p in sorted(permutations(allchars))]\n    perms, sp = allperms[::], ''\n    while perms:\n        nxt = perms.pop()\n        if nxt not in sp:\n            sp += nxt\n    assert all(perm in sp for perm in allperms)\n    return sp\n\ndef s_perm2(n):\n    \"\"\"\n    Uses algorithm of concatenating all perms in order first-last-nextfirst-\n    nextlast... if not already part of concatenation.\n    \"\"\"\n    allchars = string.ascii_uppercase[:n]\n    allperms = [''.join(p) for p in sorted(permutations(allchars))]\n    perms, sp = allperms[::], ''\n    while perms:\n        nxt = perms.pop(0)\n        if nxt not in sp:\n            sp += nxt\n        if perms:\n            nxt = perms.pop(-1)\n            if nxt not in sp:\n                sp += nxt\n    assert all(perm in sp for perm in allperms)\n    return sp\n\ndef _s_perm3(n, cmp):\n    \"\"\"\n    Uses algorithm of concatenating all perms in order first,\n    next_with_LEASTorMOST_chars_in_same_position_as_last_n_chars, ...\n    \"\"\"\n    allchars = string.ascii_uppercase[:n]\n    allperms = [''.join(p) for p in sorted(permutations(allchars))]\n    perms, sp = allperms[::], ''\n    while perms:\n        lastn = sp[-n:]\n        nxt = cmp(perms,\n                  key=lambda pm:\n                    sum((ch1 == ch2) for ch1, ch2 in zip(pm, lastn)))\n        perms.remove(nxt)\n        if nxt not in sp:\n            sp += nxt\n    assert all(perm in sp for perm in allperms)\n    return sp\n\ndef s_perm3_max(n):\n    \"\"\"\n    Uses algorithm of concatenating all perms in order first,\n    next_with_MOST_chars_in_same_position_as_last_n_chars, ...\n    \"\"\"\n    return _s_perm3(n, max)\n\ndef s_perm3_min(n):\n    \"\"\"\n    Uses algorithm of concatenating all perms in order first,\n    next_with_LEAST_chars_in_same_position_as_last_n_chars, ...\n    \"\"\"\n    return _s_perm3(n, min)\n\n\nlongest = [factorial(n) * n for n in range(MAXN + 1)]\nweight, runtime = {}, {}\nprint(__doc__)\nfor algo in [s_perm0, s_perm1, s_perm2, s_perm3_max, s_perm3_min]:\n    print('\\n###\\n### %s\\n###' % algo.__name__)\n    print(algo.__doc__)\n    weight[algo.__name__], runtime[algo.__name__] = 1, datetime.timedelta(0)\n    for n in range(1, MAXN + 1):\n        gc.collect()\n        gc.disable()\n        t = datetime.datetime.now()\n        sp = algo(n)\n        t = datetime.datetime.now() - t\n        gc.enable()\n        runtime[algo.__name__] += t\n        lensp = len(sp)\n        wt = (lensp / longest[n]) ** 2\n        print('  For N=%i: SP length %5i Max: %5i Weight: %5.2f'\n              % (n, lensp, longest[n], wt))\n        weight[algo.__name__] *= wt\n    weight[algo.__name__] **= 1 / n  # Geometric mean\n    weight[algo.__name__] = 1 / weight[algo.__name__]\n    print('%*s Overall Weight: %5.2f in %.1f seconds.'\n          % (29, '', weight[algo.__name__], runtime[algo.__name__].total_seconds()))\n\nprint('\\n###\\n### Algorithms ordered by shortest superpermutations first\\n###')\nprint('\\n'.join('%12s (%.3f)' % kv for kv in\n                sorted(weight.items(), key=lambda keyvalue: -keyvalue[1])))\n      \nprint('\\n###\\n### Algorithms ordered by shortest runtime first\\n###')\nprint('\\n'.join('%12s (%.3f)' % (k, v.total_seconds()) for k, v in\n                sorted(runtime.items(), key=lambda keyvalue: keyvalue[1])))\n"}
{"title": "Superpermutation minimisation", "language": "Python from D", "task": "A superpermutation of N different characters is a string consisting of an arrangement of multiple copies of those N different characters in which every permutation of those characters can be found as a substring.\n\nFor example, representing the characters as A..Z, using N=2 we choose to use the first two characters 'AB'. \nThe permutations of 'AB' are the two, (i.e. two-factorial), strings: 'AB' and 'BA'.\n\nA too obvious method of generating a superpermutation is to just join all the permutations together forming 'ABBA'.\n\nA little thought will produce the shorter (in fact the shortest) superpermutation of 'ABA' - it contains 'AB' at the beginning and contains 'BA' from the middle to the end.\n\nThe \"too obvious\" method of creation generates a string of length N!*N. Using this as a yardstick, the task is to investigate other methods of generating superpermutations of N from 1-to-7 characters, that never generate larger superpermutations.\n\nShow descriptions and comparisons of algorithms used here, and select the \"Best\" algorithm as being the one generating shorter superpermutations.\n\nThe problem of generating the shortest superpermutation for each N ''might'' be NP complete, although the minimal strings for small values of N have been found by brute -force searches.\n\n\n\n\n;Reference:\n* The Minimal Superpermutation Problem. by Nathaniel Johnston.\n* oeis A180632 gives 0-5 as 0, 1, 3, 9, 33, 153. 6 is thought to be 872.\n* Superpermutations - Numberphile. A video\n* Superpermutations: the maths problem solved by 4chan - Standupmaths. A video of recent (2018) mathematical progress.\n* New Superpermutations Discovered! Standupmaths & Numberphile.\n\n", "solution": "from array import array\nfrom string import ascii_uppercase, digits\nfrom operator import mul\n\ntry:\n    import psyco\n    psyco.full()\nexcept:\n    pass\n\nN_MAX = 12\n\n# fact_sum(n) = 1! + 2! + ... + n!\ndef fact_sum(n):\n    return sum(reduce(mul, xrange(1, m + 1), 1) for m in xrange(1, n + 1))\n\n\ndef r(n, superperm, pos, count):\n    if not n:\n        return False\n\n    c = superperm[pos - n]\n    count[n] -= 1\n    if not count[n]:\n        count[n] = n\n        if not r(n - 1, superperm, pos, count):\n            return False\n\n    superperm[pos] = c\n    pos += 1\n    return True\n\n\ndef super_perm(n, superperm, pos, count, chars = digits + ascii_uppercase):\n    assert len(chars) >= N_MAX\n    pos = n\n    superperm += array(\"c\", \" \") * (fact_sum(n) - len(superperm))\n\n    for i in xrange(n + 1):\n        count[i] = i\n    for i in xrange(1, n + 1):\n        superperm[i - 1] = chars[i]\n\n    while r(n, superperm, pos, count):\n        pass\n\n\ndef main():\n    superperm = array(\"c\", \"\")\n    pos = 0\n    count = array(\"l\", [0]) * N_MAX\n\n    for n in xrange(N_MAX):\n        super_perm(n, superperm, pos, count)\n        print \"Super perm(%2d) len = %d\" % (n, len(superperm)),\n        #print superperm.tostring(),\n        print\n\nmain()"}
{"title": "Sylvester's sequence", "language": "Python", "task": "{{Wikipedia|Sylvester's sequence}}\n\n\nIn number theory, '''Sylvester's sequence''' is an integer sequence in which each term of the sequence is the product of the previous terms,  plus one.\n\nIts values grow doubly exponentially, and the sum of its reciprocals forms a series of unit fractions that converges to '''1''' more rapidly than any other series of unit fractions with the same number of terms. \n\nFurther, the sum of the first '''k''' terms of the infinite series of reciprocals provides the closest possible underestimate of  '''1''' by any k-term Egyptian fraction.\n\n\n;Task:\n* Write a routine (function, procedure, generator, whatever) to calculate '''Sylvester's sequence'''.\n* Use that routine to show the values of the first '''10''' elements in the sequence.\n* Show the sum of the reciprocals of the first '''10''' elements on the sequence, ideally as an exact fraction.\n\n\n;Related tasks:\n* [[Egyptian fractions]]\n* [[Harmonic series]]\n\n\n;See also: \n* OEIS A000058 - Sylvester's sequence\n\n", "solution": "'''Sylvester's sequence'''\n\nfrom functools import reduce\nfrom itertools import count, islice\n\n\n# sylvester :: [Int]\ndef sylvester():\n    '''Non-finite stream of the terms\n       of Sylvester's sequence.\n       (OEIS A000058)\n    '''\n    def go(n):\n        return 1 + reduce(\n            lambda a, x: a * go(x),\n            range(0, n),\n            1\n        ) if 0 != n else 2\n\n    return map(go, count(0))\n\n\n# ------------------------- TEST -------------------------\n# main :: IO ()\ndef main():\n    '''First terms, and sum of reciprocals.'''\n\n    print(\"First 10 terms of OEIS A000058:\")\n    xs = list(islice(sylvester(), 10))\n    print('\\n'.join([\n        str(x) for x in xs\n    ]))\n\n    print(\"\\nSum of the reciprocals of the first 10 terms:\")\n    print(\n        reduce(lambda a, x: a + 1 / x, xs, 0)\n    )\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Tarjan", "language": "Python", "task": "{{wikipedia|Graph}}\n\n\n\nTarjan's algorithm is an algorithm in graph theory for finding the strongly connected components of a graph. \n\nIt runs in linear time, matching the time bound for alternative methods including Kosaraju's algorithm and the path-based strong component algorithm. \n\nTarjan's Algorithm is named for its discoverer, Robert Tarjan.\n\n\n;References:\n* The article on Wikipedia.\n\nSee also: [[Kosaraju]]\n\n", "solution": "from collections import defaultdict\n\ndef from_edges(edges):\n    '''translate list of edges to list of nodes'''\n\n    class Node:\n        def __init__(self):\n            # root is one of:\n            #   None: not yet visited\n            #   -1: already processed\n            #   non-negative integer: what Wikipedia pseudo code calls 'lowlink'\n            self.root = None\n            self.succ = []\n\n    nodes = defaultdict(Node)\n    for v,w in edges:\n        nodes[v].succ.append(nodes[w])\n\n    for i,v in nodes.items(): # name the nodes for final output\n        v.id = i\n\n    return nodes.values()\n\ndef trajan(V):\n    def strongconnect(v, S):\n        v.root = pos = len(S)\n        S.append(v)\n\n        for w in v.succ:\n            if w.root is None:  # not yet visited\n                yield from strongconnect(w, S)\n\n            if w.root >= 0:  # still on stack\n                v.root = min(v.root, w.root)\n\n        if v.root == pos:  # v is the root, return everything above\n            res, S[pos:] = S[pos:], []\n            for w in res:\n                w.root = -1\n            yield [r.id for r in res]\n\n    for v in V:\n        if v.root is None:\n            yield from strongconnect(v, [])\n\n\ntables = [  # table 1\n            [(1,2), (3,1), (3,6), (6,7), (7,6), (2,3), (4,2),\n             (4,3), (4,5), (5,6), (5,4), (8,5), (8,7), (8,6)],\n\n            # table 2\n            [('A', 'B'), ('B', 'C'), ('C', 'A'), ('A', 'Other')]]\n\nfor table in (tables):\n    for g in trajan(from_edges(table)):\n        print(g)\n    print()"}
{"title": "Tau function", "language": "Python", "task": "Given a positive integer, count the number of its positive divisors.\n\n\n;Task\nShow the result for the first   '''100'''   positive integers.\n\n\n;Related task\n*  [[Tau number]]\n\n", "solution": "def factorize(n):\n    assert(isinstance(n, int))\n    if n < 0: \n        n = -n \n    if n < 2: \n        return \n    k = 0 \n    while 0 == n%2: \n        k += 1 \n        n //= 2 \n    if 0 < k: \n        yield (2,k) \n    p = 3 \n    while p*p <= n: \n        k = 0 \n        while 0 == n%p: \n            k += 1 \n            n //= p \n        if 0 < k: \n            yield (p,k)\n        p += 2 \n    if 1 < n: \n        yield (n,1) \n\ndef tau(n): \n    assert(n != 0) \n    ans = 1 \n    for (p,k) in factorize(n): \n        ans *= 1 + k\n    return ans\n\nif __name__ == \"__main__\":\n    print(*map(tau, range(1, 101)))"}
{"title": "Teacup rim text", "language": "Python", "task": "On a set of coasters we have, there's a picture of a teacup.   On the rim of the teacup the word   '''TEA'''   appears a number of times separated by bullet characters   (*). \n\nIt occurred to me that if the bullet were removed and the words run together,   you could start at any letter and still end up with a meaningful three-letter word. \n\nSo start at the   '''T'''   and read   '''TEA'''.   Start at the   '''E'''   and read   '''EAT''',   or start at the   '''A'''   and read   '''ATE'''.\n\nThat got me thinking that maybe there are other words that could be used rather that   '''TEA'''.   And that's just English.   What about Italian or Greek or ... um ... Telugu. \n\nFor English, we will use the unixdict (now) located at:   unixdict.txt. \n\n(This will maintain continuity with other Rosetta Code tasks that also use it.)\n\n\n;Task:\nSearch for a set of words that could be printed around the edge of a teacup.   The words in each set are to be of the same length, that length being greater than two (thus precluding   '''AH'''   and   '''HA''',   for example.) \n\nHaving listed a set, for example   ['''ate tea eat'''],   refrain from displaying permutations of that set, e.g.:   ['''eat tea ate''']   etc. \n\nThe words should also be made of more than one letter   (thus precluding   '''III'''   and   '''OOO'''   etc.) \n\nThe relationship between these words is (using ATE as an example) that the first letter of the first becomes the last letter of the second.   The first letter of the second becomes the last letter of the third.   So   '''ATE'''   becomes   '''TEA'''   and   '''TEA'''   becomes   '''EAT'''. \n\nAll of the possible permutations, using this particular permutation technique, must be words in the list. \n\nThe set you generate for   '''ATE'''   will never included the word   '''ETA'''   as that cannot be reached via the first-to-last movement method. \n\nDisplay one line for each set of teacup rim words.\n\n\n\n", "solution": "'''Teacup rim text'''\n\nfrom itertools import chain, groupby\nfrom os.path import expanduser\nfrom functools import reduce\n\n\n# main :: IO ()\ndef main():\n    '''Circular anagram groups, of more than one word,\n       and containing words of length > 2, found in:\n       https://www.mit.edu/~ecprice/wordlist.10000\n    '''\n    print('\\n'.join(\n        concatMap(circularGroup)(\n            anagrams(3)(\n                # Reading from a local copy.\n                lines(readFile('~/mitWords.txt'))\n            )\n        )\n    ))\n\n\n# anagrams :: Int -> [String] -> [[String]]\ndef anagrams(n):\n    '''Groups of anagrams, of minimum group size n,\n       found in the given word list.\n    '''\n    def go(ws):\n        def f(xs):\n            return [\n                [snd(x) for x in xs]\n            ] if n <= len(xs) >= len(xs[0][0]) else []\n        return concatMap(f)(groupBy(fst)(sorted(\n            [(''.join(sorted(w)), w) for w in ws],\n            key=fst\n        )))\n    return go\n\n\n# circularGroup :: [String] -> [String]\ndef circularGroup(ws):\n    '''Either an empty list, or a list containing\n       a string showing any circular subset found in ws.\n    '''\n    lex = set(ws)\n    iLast = len(ws) - 1\n    # If the set contains one word that is circular,\n    # then it must contain all of them.\n    (i, blnCircular) = until(\n        lambda tpl: tpl[1] or (tpl[0] > iLast)\n    )(\n        lambda tpl: (1 + tpl[0], isCircular(lex)(ws[tpl[0]]))\n    )(\n        (0, False)\n    )\n    return [' -> '.join(allRotations(ws[i]))] if blnCircular else []\n\n\n# isCircular :: Set String -> String -> Bool\ndef isCircular(lexicon):\n    '''True if all of a word's rotations\n       are found in the given lexicon.\n    '''\n    def go(w):\n        def f(tpl):\n            (i, _, x) = tpl\n            return (1 + i, x in lexicon, rotated(x))\n\n        iLast = len(w) - 1\n        return until(\n            lambda tpl: iLast < tpl[0] or (not tpl[1])\n        )(f)(\n            (0, True, rotated(w))\n        )[1]\n    return go\n\n\n# allRotations :: String -> [String]\ndef allRotations(w):\n    '''All rotations of the string w.'''\n    return takeIterate(len(w) - 1)(\n        rotated\n    )(w)\n\n\n# GENERIC -------------------------------------------------\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list over which a function has been mapped.\n       The list monad can be derived by using a function f which\n       wraps its output in a list,\n       (using an empty list to represent computational failure).\n    '''\n    def go(xs):\n        return chain.from_iterable(map(f, xs))\n    return go\n\n\n# fst :: (a, b) -> a\ndef fst(tpl):\n    '''First member of a pair.'''\n    return tpl[0]\n\n\n# groupBy :: (a -> b) -> [a] -> [[a]]\ndef groupBy(f):\n    '''The elements of xs grouped,\n       preserving order, by equality\n       in terms of the key function f.\n    '''\n    def go(xs):\n        return [\n            list(x[1]) for x in groupby(xs, key=f)\n        ]\n    return go\n\n\n# lines :: String -> [String]\ndef lines(s):\n    '''A list of strings,\n       (containing no newline characters)\n       derived from a single new-line delimited string.\n    '''\n    return s.splitlines()\n\n\n# mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])\ndef mapAccumL(f):\n    '''A tuple of an accumulation and a list derived by a\n       combined map and fold,\n       with accumulation from left to right.\n    '''\n    def go(a, x):\n        tpl = f(a[0], x)\n        return (tpl[0], a[1] + [tpl[1]])\n    return lambda acc: lambda xs: (\n        reduce(go, xs, (acc, []))\n    )\n\n\n# readFile :: FilePath -> IO String\ndef readFile(fp):\n    '''The contents of any file at the path\n       derived by expanding any ~ in fp.\n    '''\n    with open(expanduser(fp), 'r', encoding='utf-8') as f:\n        return f.read()\n\n\n# rotated :: String -> String\ndef rotated(s):\n    '''A string rotated 1 character to the right.'''\n    return s[1:] + s[0]\n\n\n# snd :: (a, b) -> b\ndef snd(tpl):\n    '''Second member of a pair.'''\n    return tpl[1]\n\n\n# takeIterate :: Int -> (a -> a) -> a -> [a]\ndef takeIterate(n):\n    '''Each value of n iterations of f\n       over a start value of x.\n    '''\n    def go(f):\n        def g(x):\n            def h(a, i):\n                v = f(a) if i else x\n                return (v, v)\n            return mapAccumL(h)(x)(\n                range(0, 1 + n)\n            )[1]\n        return g\n    return go\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.\n    '''\n    def go(f):\n        def g(x):\n            v = x\n            while not p(v):\n                v = f(v)\n            return v\n        return g\n    return go\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Temperature conversion", "language": "Python", "task": "There are quite a number of temperature scales. For this task we will concentrate on four of the perhaps best-known ones: \nRankine.\n\nThe Celsius and Kelvin scales have the same magnitude, but different null points.\n \n: 0 degrees Celsius corresponds to 273.15 kelvin.\n: 0 kelvin is absolute zero.\n\nThe Fahrenheit and Rankine scales also have the same magnitude, but different null points.\n\n: 0 degrees Fahrenheit corresponds to 459.67 degrees Rankine.\n: 0 degrees Rankine is absolute zero.\n\nThe Celsius/Kelvin and Fahrenheit/Rankine scales have a ratio of 5 : 9.\n\n\n;Task\nWrite code that accepts a value of kelvin, converts it to values of the three other scales, and prints the result. \n\n\n;Example:\n\nK  21.00\n\nC  -252.15\n\nF  -421.87\n\nR  37.80\n\n\n", "solution": ">>> toK = {'C': (lambda c: c + 273.15),\n           'F': (lambda f: (f + 459.67) / 1.8),\n           'R': (lambda r: r / 1.8),\n           'K': (lambda k: k) }\n>>> while True:\n    magnitude, unit = input('<value> <K/R/F/C> ? ').split()\n    k = toK[unit](float(magnitude))\n    print(\"%g Kelvin = %g Celsius = %g Fahrenheit = %g Rankine degrees.\"\n          % (k, k - 273.15, k * 1.8 - 459.67, k * 1.8))\n\n    \n<value> <K/R/F/C> ? 222.2 K\n222.2 Kelvin = -50.95 Celsius = -59.71 Fahrenheit = 399.96 Rankine degrees.\n<value> <K/R/F/C> ? -50.95 C\n222.2 Kelvin = -50.95 Celsius = -59.71 Fahrenheit = 399.96 Rankine degrees.\n<value> <K/R/F/C> ? -59.71 F\n222.2 Kelvin = -50.95 Celsius = -59.71 Fahrenheit = 399.96 Rankine degrees.\n<value> <K/R/F/C> ? 399.96 R\n222.2 Kelvin = -50.95 Celsius = -59.71 Fahrenheit = 399.96 Rankine degrees.\n<value> <K/R/F/C> ? "}
{"title": "Terminal control/Unicode output", "language": "Python", "task": "The task is to check that the terminal supports Unicode output, before outputting a Unicode character. If the terminal supports Unicode, then the terminal should output a Unicode delta (U+25b3). If the terminal does not support Unicode, then an appropriate error should be raised.\n\nNote that it is permissible to use system configuration data to determine terminal capabilities if the system provides such a facility.\n\n", "solution": "import sys\n\nif \"UTF-8\" in sys.stdout.encoding:\n    print(\"\u25b3\")\nelse:\n    raise Exception(\"Terminal can't handle UTF-8\")"}
{"title": "Test integerness", "language": "Python", "task": "{| class=\"wikitable\"\n|-\n! colspan=2 | Input\n! colspan=2 | Output\n! rowspan=2 | Comment\n|-\n! Type\n! Value\n! exact\n! tolerance = 0.00001\n|-\n| rowspan=3 | decimal\n| 25.000000\n| colspan=2 | true\n| \n|-\n| 24.999999\n| false\n| true\n| \n|-\n| 25.000100\n| colspan=2 | false\n| \n|-\n| rowspan=4 | floating-point\n| -2.1e120\n| colspan=2 | true\n| This one is tricky, because in most languages it is too large to fit into a native integer type.It is, nonetheless, mathematically an integer, and your code should identify it as such.\n|-\n| -5e-2\n| colspan=2 | false\n| \n|-\n| NaN\n| colspan=2 | false\n| \n|-\n| Inf\n| colspan=2 | false\n| This one is debatable. If your code considers it an integer, that's okay too.\n|-\n| rowspan=2 | complex\n| 5.0+0.0i\n| colspan=2 | true\n| \n|-\n| 5-5i\n| colspan=2 | false\n| \n|}\n\n(The types and notations shown in these tables are merely examples - you should use the native data types and number literals of your programming language and standard library. Use a different set of test-cases, if this one doesn't demonstrate all relevant behavior.)\n\n\n", "solution": ">>> def isint(f): \n    return complex(f).imag == 0 and complex(f).real.is_integer()\n\n>>> [isint(f) for f in (1.0, 2, (3.0+0.0j), 4.1, (3+4j), (5.6+0j))]\n[True, True, True, False, False, False]\n\n>>> # Test cases\n...\n>>> isint(25.000000)\nTrue\n>>> isint(24.999999)\nFalse\n>>> isint(25.000100)\nFalse\n>>> isint(-2.1e120)\nTrue\n>>> isint(-5e-2)\nFalse\n>>> isint(float('nan'))\nFalse\n>>> isint(float('inf'))\nFalse\n>>> isint(5.0+0.0j)\nTrue\n>>> isint(5-5j)\nFalse\n"}
{"title": "Textonyms", "language": "Python", "task": "When entering text on a phone's digital pad it is possible that a particular combination of digits corresponds to more than one word. Such are called textonyms.\n\nAssuming the digit keys are mapped to letters as follows:\n     2 -> ABC\n     3 -> DEF\n     4 -> GHI\n     5 -> JKL\n     6 -> MNO\n     7 -> PQRS\n     8 -> TUV\n     9 -> WXYZ  \n\n\n;Task:\nWrite a program that finds textonyms in a list of words such as   \n[[Textonyms/wordlist]]   or   \nunixdict.txt.\n\nThe task should produce a report:\n\n There are #{0} words in #{1} which can be represented by the digit key mapping.\n They require #{2} digit combinations to represent them.\n #{3} digit combinations represent Textonyms.\n\nWhere:\n #{0} is the number of words in the list which can be represented by the digit key mapping.\n #{1} is the URL of the wordlist being used.\n #{2} is the number of digit combinations required to represent the words in #{0}.\n #{3} is the number of #{2} which represent more than one word.\n\nAt your discretion show a couple of examples of your solution displaying Textonyms. \n\nE.G.:\n\n  2748424767 -> \"Briticisms\", \"criticisms\"\n\n\n;Extra credit:\nUse a word list and keypad mapping other than English.\n\n\n\n", "solution": "from collections import defaultdict\nimport urllib.request\n\nCH2NUM = {ch: str(num) for num, chars in enumerate('abc def ghi jkl mno pqrs tuv wxyz'.split(), 2) for ch in chars}\nURL = 'http://www.puzzlers.org/pub/wordlists/unixdict.txt'\n\n\ndef getwords(url):\n return urllib.request.urlopen(url).read().decode(\"utf-8\").lower().split()\n\ndef mapnum2words(words):\n    number2words = defaultdict(list)\n    reject = 0\n    for word in words:\n        try:\n            number2words[''.join(CH2NUM[ch] for ch in word)].append(word)\n        except KeyError:\n            # Reject words with non a-z e.g. '10th'\n            reject += 1\n    return dict(number2words), reject\n\ndef interactiveconversions():\n    global inp, ch, num\n    while True:\n        inp = input(\"\\nType a number or a word to get the translation and textonyms: \").strip().lower()\n        if inp:\n            if all(ch in '23456789' for ch in inp):\n                if inp in num2words:\n                    print(\"  Number {0} has the following textonyms in the dictionary: {1}\".format(inp, ', '.join(\n                        num2words[inp])))\n                else:\n                    print(\"  Number {0} has no textonyms in the dictionary.\".format(inp))\n            elif all(ch in CH2NUM for ch in inp):\n                num = ''.join(CH2NUM[ch] for ch in inp)\n                print(\"  Word {0} is{1} in the dictionary and is number {2} with textonyms: {3}\".format(\n                    inp, ('' if inp in wordset else \"n't\"), num, ', '.join(num2words[num])))\n            else:\n                print(\"  I don't understand %r\" % inp)\n        else:\n            print(\"Thank you\")\n            break\n\n\nif __name__ == '__main__':\n    words = getwords(URL)\n    print(\"Read %i words from %r\" % (len(words), URL))\n    wordset = set(words)\n    num2words, reject = mapnum2words(words)\n    morethan1word = sum(1 for w in num2words if len(num2words[w]) > 1)\n    maxwordpernum = max(len(values) for values in num2words.values())\n    print(\"\"\"\nThere are {0} words in {1} which can be represented by the Textonyms mapping.\nThey require {2} digit combinations to represent them.\n{3} digit combinations represent Textonyms.\\\n\"\"\".format(len(words) - reject, URL, len(num2words), morethan1word))\n\n    print(\"\\nThe numbers mapping to the most words map to %i words each:\" % maxwordpernum)\n    maxwpn = sorted((key, val) for key, val in num2words.items() if len(val) == maxwordpernum)\n    for num, wrds in maxwpn:\n        print(\"  %s maps to: %s\" % (num, ', '.join(wrds)))\n\n    interactiveconversions()"}
{"title": "The Name Game", "language": "Python", "task": "Write a program that accepts a name as input and outputs the lyrics to the Shirley Ellis song \"The Name Game\".\n\n\nThe regular verse\n\nUnless your name begins with a vowel (A, E, I, O, U), 'B', 'F' or 'M' you don't have to care about special rules.\nThe verse for the name 'Gary' would be like this:\n\n    Gary, Gary, bo-bary\n    Banana-fana fo-fary\n    Fee-fi-mo-mary\n    Gary! \n\nAt the end of every line, the name gets repeated without the first letter: Gary becomes ary\nIf we take (X) as the full name (Gary) and (Y) as the name without the first letter (ary) the verse would look like this:\n\n    (X), (X), bo-b(Y)\n    Banana-fana fo-f(Y)\n    Fee-fi-mo-m(Y)\n    (X)! \n\nVowel as first letter of the name\n\nIf you have a vowel as the first letter of your name (e.g. Earl) you do not truncate the name.\nThe verse looks like this:\n\n    Earl, Earl, bo-bearl\n    Banana-fana fo-fearl\n    Fee-fi-mo-mearl\n    Earl! \n\n'B', 'F' or 'M' as first letter of the name\n\nIn case of a 'B', an 'F' or an 'M' (e.g. Billy, Felix, Mary) there is a special rule.\nThe line which would 'rebuild' the name (e.g. bo-billy) is sung without the first letter of the name.\nThe verse for the name Billy looks like this:\n\n    Billy, Billy, bo-illy\n    Banana-fana fo-filly\n    Fee-fi-mo-milly\n    Billy! \n\nFor the name 'Felix', this would be right:\n\n    Felix, Felix, bo-belix\n    Banana-fana fo-elix\n    Fee-fi-mo-melix\n    Felix!\n\n\n", "solution": "def print_verse(n):\n    l = ['b', 'f', 'm']\n    s = n[1:]\n    if str.lower(n[0]) in l:\n        l[l.index(str.lower(n[0]))] = ''\n    elif n[0] in ['A', 'E', 'I', 'O', 'U']:\n        s = str.lower(n)\n    print('{0}, {0}, bo-{2}{1}\\nBanana-fana fo-{3}{1}\\nFee-fi-mo-{4}{1}\\n{0}!\\n'.format(n, s, *l))\n\n# Assume that the names are in title-case and they're more than one character long\nfor n in ['Gary', 'Earl', 'Billy', 'Felix', 'Mary']:\n    print_verse(n)"}
{"title": "The Twelve Days of Christmas", "language": "Python", "task": "Write a program that outputs the lyrics of the Christmas carol ''The Twelve Days of Christmas''. \nThe lyrics can be found here.  \n\n(You must reproduce the words in the correct order, but case, format, and punctuation are left to your discretion.)\n\n\n\n", "solution": "gifts = '''\\\nA partridge in a pear tree.\nTwo turtle doves\nThree french hens\nFour calling birds\nFive golden rings\nSix geese a-laying\nSeven swans a-swimming\nEight maids a-milking\nNine ladies dancing\nTen lords a-leaping\nEleven pipers piping\nTwelve drummers drumming'''.split('\\n')\n\ndays = '''first second third fourth fifth\n          sixth seventh eighth ninth tenth\n          eleventh twelfth'''.split()\n\nfor n, day in enumerate(days, 1):\n    g = gifts[:n][::-1]\n    print(('\\nOn the %s day of Christmas\\nMy true love gave to me:\\n' % day) +\n          '\\n'.join(g[:-1]) +\n          (' and\\n' + g[-1] if n > 1 else g[-1].capitalize()))"}
{"title": "Thue-Morse", "language": "Python", "task": "Create a Thue-Morse sequence.\n\n\n;See also\n*   YouTube entry: The Fairest Sharing Sequence Ever\n*   YouTube entry: Math and OCD - My story with the Thue-Morse sequence\n*   Task: [[Fairshare between two and more]]\n\n", "solution": "m='0'\nprint(m)\nfor i in range(0,6):\n     m0=m\n     m=m.replace('0','a')\n     m=m.replace('1','0')\n     m=m.replace('a','1')\n     m=m0+m\n     print(m)\n"}
{"title": "Tonelli-Shanks algorithm", "language": "Python 3", "task": "{{wikipedia}}\n\n\nIn computational number theory, the Tonelli-Shanks algorithm is a technique for solving for '''x''' in a congruence of the form:\n\n:: x2  n (mod p)\n\nwhere '''n''' is an integer which is a quadratic residue (mod p), '''p''' is an odd prime, and '''x,n  Fp''' where Fp = {0, 1, ..., p - 1}.\n\nIt is used in cryptography techniques.\n\n\nTo apply the algorithm, we need the Legendre symbol:\n\nThe Legendre symbol '''(a | p)''' denotes the value of a(p-1)/2 (mod p).\n* '''(a | p)  1'''    if '''a''' is a square (mod p)\n* '''(a | p)  -1'''    if '''a''' is not a square (mod p)\n* '''(a | p)  0'''    if '''a'''  0 (mod p) \n\n\n;Algorithm pseudo-code:\n\nAll  are taken to mean (mod p) unless stated otherwise.\n\n* Input: '''p''' an odd prime, and an integer '''n''' .\n* Step 0: Check that '''n''' is indeed a square: (n | p) must be  1 .\n* Step 1: By factoring out powers of 2 from p - 1, find '''q''' and '''s''' such that p - 1 = q2s with '''q''' odd .\n** If p  3 (mod 4) (i.e. s = 1), output the two solutions r  +- n(p+1)/4 .\n* Step 2: Select a non-square '''z''' such that (z | p)  -1 and set c   zq .\n* Step 3: Set r  n(q+1)/2, t  nq, m = s .\n* Step 4: Loop the following:\n** If t  1, output '''r''' and '''p - r''' .\n** Otherwise find, by repeated squaring, the lowest '''i''', 0 < i < m , such that t2i  1 .\n** Let b  c2(m - i - 1), and set r  rb, t  tb2, c  b2 and m = i .\n\n\n\n;Task:\nImplement the above algorithm. \n\nFind solutions (if any) for \n* n = 10 p = 13\n* n = 56 p = 101\n* n = 1030 p = 10009\n* n = 1032 p = 10009\n* n = 44402 p = 100049  \n\n;Extra credit:\n* n  =    665820697     p  =   1000000009      \n* n  =   881398088036     p   =  1000000000039    \n* n  =  41660815127637347468140745042827704103445750172002   p  = 10^50 + 577   \t\n\n\n;See also:\n* [[Modular exponentiation]]\n* [[Cipolla's algorithm]]\n\n", "solution": "def legendre(a, p):\n    return pow(a, (p - 1) // 2, p)\n\ndef tonelli(n, p):\n    assert legendre(n, p) == 1, \"not a square (mod p)\"\n    q = p - 1\n    s = 0\n    while q % 2 == 0:\n        q //= 2\n        s += 1\n    if s == 1:\n        return pow(n, (p + 1) // 4, p)\n    for z in range(2, p):\n        if p - 1 == legendre(z, p):\n            break\n    c = pow(z, q, p)\n    r = pow(n, (q + 1) // 2, p)\n    t = pow(n, q, p)\n    m = s\n    t2 = 0\n    while (t - 1) % p != 0:\n        t2 = (t * t) % p\n        for i in range(1, m):\n            if (t2 - 1) % p == 0:\n                break\n            t2 = (t2 * t2) % p\n        b = pow(c, 1 << (m - i - 1), p)\n        r = (r * b) % p\n        c = (b * b) % p\n        t = (t * c) % p\n        m = i\n    return r\n\nif __name__ == '__main__':\n    ttest = [(10, 13), (56, 101), (1030, 10009), (44402, 100049),\n\t     (665820697, 1000000009), (881398088036, 1000000000039),\n             (41660815127637347468140745042827704103445750172002, 10**50 + 577)]\n    for n, p in ttest:\n        r = tonelli(n, p)\n        assert (r * r - n) % p == 0\n        print(\"n = %d p = %d\" % (n, p))\n        print(\"\\t  roots : %d %d\" % (r, p - r))"}
{"title": "Top rank per group", "language": "Python", "task": "Find the top   ''N''   salaries in each department,   where   ''N''   is provided as a parameter.\n\nUse this data as a formatted internal data structure (adapt it to your language-native idioms, rather than parse at runtime), or identify your external data source:\n\nEmployee Name,Employee ID,Salary,Department\nTyler Bennett,E10297,32000,D101\nJohn Rappl,E21437,47000,D050\nGeorge Woltman,E00127,53500,D101\nAdam Smith,E63535,18000,D202\nClaire Buckman,E39876,27800,D202\nDavid McClellan,E04242,41500,D101\nRich Holcomb,E01234,49500,D202\nNathan Adams,E41298,21900,D050\nRichard Potter,E43128,15900,D101\nDavid Motsinger,E27002,19250,D202\nTim Sampair,E03033,27000,D101\nKim Arlich,E10001,57000,D190\nTimothy Grove,E16398,29900,D190\n\n\n", "solution": "from collections import defaultdict\nfrom heapq import nlargest\n \ndata = [('Employee Name', 'Employee ID', 'Salary', 'Department'),\n        ('Tyler Bennett', 'E10297', 32000, 'D101'),\n        ('John Rappl', 'E21437', 47000, 'D050'),\n        ('George Woltman', 'E00127', 53500, 'D101'),\n        ('Adam Smith', 'E63535', 18000, 'D202'),\n        ('Claire Buckman', 'E39876', 27800, 'D202'),\n        ('David McClellan', 'E04242', 41500, 'D101'),\n        ('Rich Holcomb', 'E01234', 49500, 'D202'),\n        ('Nathan Adams', 'E41298', 21900, 'D050'),\n        ('Richard Potter', 'E43128', 15900, 'D101'),\n        ('David Motsinger', 'E27002', 19250, 'D202'),\n        ('Tim Sampair', 'E03033', 27000, 'D101'),\n        ('Kim Arlich', 'E10001', 57000, 'D190'),\n        ('Timothy Grove', 'E16398', 29900, 'D190')]\n \ndepartments = defaultdict(list)\nfor rec in data[1:]:\n    departments[rec[-1]].append(rec)\n \nN = 3\nformat = \" %-15s \" * len(data[0])\nfor department, recs in sorted(departments.items()):\n    print (\"Department %s\" % department)\n    print (format % data[0])\n    for rec in nlargest(N, recs, key=lambda rec: rec[-2]):\n        print (format % rec)\n    print('')"}
{"title": "Top rank per group", "language": "Python 3.7", "task": "Find the top   ''N''   salaries in each department,   where   ''N''   is provided as a parameter.\n\nUse this data as a formatted internal data structure (adapt it to your language-native idioms, rather than parse at runtime), or identify your external data source:\n\nEmployee Name,Employee ID,Salary,Department\nTyler Bennett,E10297,32000,D101\nJohn Rappl,E21437,47000,D050\nGeorge Woltman,E00127,53500,D101\nAdam Smith,E63535,18000,D202\nClaire Buckman,E39876,27800,D202\nDavid McClellan,E04242,41500,D101\nRich Holcomb,E01234,49500,D202\nNathan Adams,E41298,21900,D050\nRichard Potter,E43128,15900,D101\nDavid Motsinger,E27002,19250,D202\nTim Sampair,E03033,27000,D101\nKim Arlich,E10001,57000,D190\nTimothy Grove,E16398,29900,D190\n\n\n", "solution": "from collections import namedtuple\nfrom itertools import groupby\n\nN = 2\n\ndb = '''Employee Name,Employee ID,Salary,Department\nTyler Bennett,E10297,32000,D101\nJohn Rappl,E21437,47000,D050\nGeorge Woltman,E00127,53500,D101\nAdam Smith,E63535,18000,D202\nClaire Buckman,E39876,27800,D202\nDavid McClellan,E04242,41500,D101\nRich Holcomb,E01234,49500,D202\nNathan Adams,E41298,21900,D050\nRichard Potter,E43128,15900,D101\nDavid Motsinger,E27002,19250,D202\nTim Sampair,E03033,27000,D101\nKim Arlich,E10001,57000,D190\nTimothy Grove,E16398,29900,D190'''\n\nrows = db.split('\\n')\nDBRecord = namedtuple('DBRecord', rows[0].replace(' ', '_'))\n\nrecords = [DBRecord(*row.split(',')) for row in rows[1:]]\n\nrecords.sort(key=lambda record: (record.Department, -float(record.Salary)))\n\nprint('\\n\\n'.join(\n    '\\n  '.join([dpt] + [str(g) for g in grp][:N])\n    for dpt, grp in groupby(\n        records,\n        lambda record: record.Department\n    )\n))"}
{"title": "Topswops", "language": "Python", "task": "Topswops is a card game created by John Conway in the 1970's.\n\n\nAssume you have a particular permutation of a set of   n   cards numbered   1..n   on both of their faces, for example the arrangement of four cards given by   [2, 4, 1, 3]   where the leftmost card is on top. \n\nA round is composed of reversing the first   m   cards where   m   is the value of the topmost card. \n\nRounds are repeated until the topmost card is the number   1   and the number of swaps is recorded. \n\n\nFor our example the swaps produce:\n    \n    [2, 4, 1, 3]    # Initial shuffle\n    [4, 2, 1, 3]\n    [3, 1, 2, 4]\n    [2, 1, 3, 4]\n    [1, 2, 3, 4]\n\n\nFor a total of four swaps from the initial ordering to produce the terminating case where   1   is on top.\n\n\nFor a particular number    n    of cards,    topswops(n)    is the maximum swaps needed for any starting permutation of the   n   cards.\n\n\n;Task:\nThe task is to generate and show here a table of    n    vs    topswops(n)    for    n    in the range   1..10   inclusive.\n\n\n;Note:\nTopswops   is also known as   Fannkuch   from the German word   ''Pfannkuchen''    meaning   pancake.\n\n\n;Related tasks:\n*   [[Number reversal game]]\n*   [[Sorting algorithms/Pancake sort]]\n\n", "solution": ">>> from itertools import permutations\n>>> def f1(p):\n\ti = 0\n\twhile True:\n\t\tp0  = p[0]\n\t\tif p0 == 1: break\n\t\tp[:p0] = p[:p0][::-1]\n\t\ti  += 1\n\treturn i\n\n>>> def fannkuch(n):\n\treturn max(f1(list(p)) for p in permutations(range(1, n+1)))\n\n>>> for n in range(1, 11): print(n,fannkuch(n))\n\n1 0\n2 1\n3 2\n4 4\n5 7\n6 10\n7 16\n8 22\n9 30\n10 38\n>>> "}
{"title": "Topswops", "language": "Python from C", "task": "Topswops is a card game created by John Conway in the 1970's.\n\n\nAssume you have a particular permutation of a set of   n   cards numbered   1..n   on both of their faces, for example the arrangement of four cards given by   [2, 4, 1, 3]   where the leftmost card is on top. \n\nA round is composed of reversing the first   m   cards where   m   is the value of the topmost card. \n\nRounds are repeated until the topmost card is the number   1   and the number of swaps is recorded. \n\n\nFor our example the swaps produce:\n    \n    [2, 4, 1, 3]    # Initial shuffle\n    [4, 2, 1, 3]\n    [3, 1, 2, 4]\n    [2, 1, 3, 4]\n    [1, 2, 3, 4]\n\n\nFor a total of four swaps from the initial ordering to produce the terminating case where   1   is on top.\n\n\nFor a particular number    n    of cards,    topswops(n)    is the maximum swaps needed for any starting permutation of the   n   cards.\n\n\n;Task:\nThe task is to generate and show here a table of    n    vs    topswops(n)    for    n    in the range   1..10   inclusive.\n\n\n;Note:\nTopswops   is also known as   Fannkuch   from the German word   ''Pfannkuchen''    meaning   pancake.\n\n\n;Related tasks:\n*   [[Number reversal game]]\n*   [[Sorting algorithms/Pancake sort]]\n\n", "solution": "try:\n    import psyco\n    psyco.full()\nexcept ImportError:\n    pass\n\nbest = [0] * 16\n\ndef try_swaps(deck, f, s, d, n):\n    if d > best[n]:\n        best[n] = d\n\n    i = 0\n    k = 1 << s\n    while s:\n        k >>= 1\n        s -= 1\n        if deck[s] == -1 or deck[s] == s:\n            break\n        i |= k\n        if (i & f) == i and d + best[s] <= best[n]:\n            return d\n    s += 1\n\n    deck2 = list(deck)\n    k = 1\n    for i2 in xrange(1, s):\n        k <<= 1\n        if deck2[i2] == -1:\n            if f & k: continue\n        elif deck2[i2] != i2:\n            continue\n\n        deck[i2] = i2\n        deck2[:i2 + 1] = reversed(deck[:i2 + 1])\n        try_swaps(deck2, f | k, s, 1 + d, n)\n\ndef topswops(n):\n    best[n] = 0\n    deck0 = [-1] * 16\n    deck0[0] = 0\n    try_swaps(deck0, 1, n, 0, n)\n    return best[n]\n\nfor i in xrange(1, 13):\n    print \"%2d: %d\" % (i, topswops(i))"}
{"title": "Total circles area", "language": "Python", "task": "Example circles\nExample circles filtered\n\nGiven some partially overlapping circles on the plane, compute and show the total area covered by them, with four or six (or a little more) decimal digits of precision. The area covered by two or more disks needs to be counted only once. \n\nOne point of this Task is also to compare and discuss the relative merits of various solution strategies, their performance, precision and simplicity. This means keeping both slower and faster solutions for a language (like C) is welcome.\n\nTo allow a better comparison of the different implementations, solve the problem with this standard dataset, each line contains the '''x''' and '''y''' coordinates of the centers of the disks and their radii   (11 disks are fully contained inside other disks):\n\n         xc              yc          radius\n     1.6417233788   1.6121789534   0.0848270516\n    -1.4944608174   1.2077959613   1.1039549836\n     0.6110294452  -0.6907087527   0.9089162485\n     0.3844862411   0.2923344616   0.2375743054\n    -0.2495892950  -0.3832854473   1.0845181219\n     1.7813504266   1.6178237031   0.8162655711\n    -0.1985249206  -0.8343333301   0.0538864941\n    -1.7011985145  -0.1263820964   0.4776976918\n    -0.4319462812   1.4104420482   0.7886291537\n     0.2178372997  -0.9499557344   0.0357871187\n    -0.6294854565  -1.3078893852   0.7653357688\n     1.7952608455   0.6281269104   0.2727652452\n     1.4168575317   1.0683357171   1.1016025378\n     1.4637371396   0.9463877418   1.1846214562\n    -0.5263668798   1.7315156631   1.4428514068\n    -1.2197352481   0.9144146579   1.0727263474\n    -0.1389358881   0.1092805780   0.7350208828\n     1.5293954595   0.0030278255   1.2472867347\n    -0.5258728625   1.3782633069   1.3495508831\n    -0.1403562064   0.2437382535   1.3804956588\n     0.8055826339  -0.0482092025   0.3327165165\n    -0.6311979224   0.7184578971   0.2491045282\n     1.4685857879  -0.8347049536   1.3670667538\n    -0.6855727502   1.6465021616   1.0593087096\n     0.0152957411   0.0638919221   0.9771215985\n\n\nThe result is   21.56503660... .\n\n\n;Related task:\n*   [[Circles of given radius through two points]].\n\n\n;See also:\n* http://www.reddit.com/r/dailyprogrammer/comments/zff9o/9062012_challenge_96_difficult_water_droplets/\n* http://stackoverflow.com/a/1667789/10562\n\n", "solution": "from collections import namedtuple\n\nCircle = namedtuple(\"Circle\", \"x y r\")\n\ncircles = [\n    Circle( 1.6417233788,  1.6121789534, 0.0848270516),\n    Circle(-1.4944608174,  1.2077959613, 1.1039549836),\n    Circle( 0.6110294452, -0.6907087527, 0.9089162485),\n    Circle( 0.3844862411,  0.2923344616, 0.2375743054),\n    Circle(-0.2495892950, -0.3832854473, 1.0845181219),\n    Circle( 1.7813504266,  1.6178237031, 0.8162655711),\n    Circle(-0.1985249206, -0.8343333301, 0.0538864941),\n    Circle(-1.7011985145, -0.1263820964, 0.4776976918),\n    Circle(-0.4319462812,  1.4104420482, 0.7886291537),\n    Circle( 0.2178372997, -0.9499557344, 0.0357871187),\n    Circle(-0.6294854565, -1.3078893852, 0.7653357688),\n    Circle( 1.7952608455,  0.6281269104, 0.2727652452),\n    Circle( 1.4168575317,  1.0683357171, 1.1016025378),\n    Circle( 1.4637371396,  0.9463877418, 1.1846214562),\n    Circle(-0.5263668798,  1.7315156631, 1.4428514068),\n    Circle(-1.2197352481,  0.9144146579, 1.0727263474),\n    Circle(-0.1389358881,  0.1092805780, 0.7350208828),\n    Circle( 1.5293954595,  0.0030278255, 1.2472867347),\n    Circle(-0.5258728625,  1.3782633069, 1.3495508831),\n    Circle(-0.1403562064,  0.2437382535, 1.3804956588),\n    Circle( 0.8055826339, -0.0482092025, 0.3327165165),\n    Circle(-0.6311979224,  0.7184578971, 0.2491045282),\n    Circle( 1.4685857879, -0.8347049536, 1.3670667538),\n    Circle(-0.6855727502,  1.6465021616, 1.0593087096),\n    Circle( 0.0152957411,  0.0638919221, 0.9771215985)]\n\ndef main():\n    # compute the bounding box of the circles\n    x_min = min(c.x - c.r for c in circles)\n    x_max = max(c.x + c.r for c in circles)\n    y_min = min(c.y - c.r for c in circles)\n    y_max = max(c.y + c.r for c in circles)\n\n    box_side = 500\n\n    dx = (x_max - x_min) / box_side\n    dy = (y_max - y_min) / box_side\n\n    count = 0\n\n    for r in xrange(box_side):\n        y = y_min + r * dy\n        for c in xrange(box_side):\n            x = x_min + c * dx\n            if any((x-circle.x)**2 + (y-circle.y)**2 <= (circle.r ** 2)\n                   for circle in circles):\n                count += 1\n\n    print \"Approximated area:\", count * dx * dy\n\nmain()"}
{"title": "Total circles area", "language": "Python from Haskell", "task": "Example circles\nExample circles filtered\n\nGiven some partially overlapping circles on the plane, compute and show the total area covered by them, with four or six (or a little more) decimal digits of precision. The area covered by two or more disks needs to be counted only once. \n\nOne point of this Task is also to compare and discuss the relative merits of various solution strategies, their performance, precision and simplicity. This means keeping both slower and faster solutions for a language (like C) is welcome.\n\nTo allow a better comparison of the different implementations, solve the problem with this standard dataset, each line contains the '''x''' and '''y''' coordinates of the centers of the disks and their radii   (11 disks are fully contained inside other disks):\n\n         xc              yc          radius\n     1.6417233788   1.6121789534   0.0848270516\n    -1.4944608174   1.2077959613   1.1039549836\n     0.6110294452  -0.6907087527   0.9089162485\n     0.3844862411   0.2923344616   0.2375743054\n    -0.2495892950  -0.3832854473   1.0845181219\n     1.7813504266   1.6178237031   0.8162655711\n    -0.1985249206  -0.8343333301   0.0538864941\n    -1.7011985145  -0.1263820964   0.4776976918\n    -0.4319462812   1.4104420482   0.7886291537\n     0.2178372997  -0.9499557344   0.0357871187\n    -0.6294854565  -1.3078893852   0.7653357688\n     1.7952608455   0.6281269104   0.2727652452\n     1.4168575317   1.0683357171   1.1016025378\n     1.4637371396   0.9463877418   1.1846214562\n    -0.5263668798   1.7315156631   1.4428514068\n    -1.2197352481   0.9144146579   1.0727263474\n    -0.1389358881   0.1092805780   0.7350208828\n     1.5293954595   0.0030278255   1.2472867347\n    -0.5258728625   1.3782633069   1.3495508831\n    -0.1403562064   0.2437382535   1.3804956588\n     0.8055826339  -0.0482092025   0.3327165165\n    -0.6311979224   0.7184578971   0.2491045282\n     1.4685857879  -0.8347049536   1.3670667538\n    -0.6855727502   1.6465021616   1.0593087096\n     0.0152957411   0.0638919221   0.9771215985\n\n\nThe result is   21.56503660... .\n\n\n;Related task:\n*   [[Circles of given radius through two points]].\n\n\n;See also:\n* http://www.reddit.com/r/dailyprogrammer/comments/zff9o/9062012_challenge_96_difficult_water_droplets/\n* http://stackoverflow.com/a/1667789/10562\n\n", "solution": "from collections import namedtuple\nfrom functools import partial\nfrom itertools import repeat, imap, izip\nfrom decimal import Decimal, getcontext\n\n# Requires the egg: https://pypi.python.org/pypi/dmath/\nfrom dmath import atan2, asin, sin, cos, pi as piCompute\n\ngetcontext().prec = 40 # Set FP precision.\nsqrt = Decimal.sqrt\npi = piCompute()\nD2 = Decimal(2)\n\nVec = namedtuple(\"Vec\", \"x y\")\nvcross = lambda (a, b), (c, d): a*d - b*c\nvdot   = lambda (a, b), (c, d): a*c + b*d\nvadd   = lambda (a, b), (c, d): Vec(a + c, b + d)\nvsub   = lambda (a, b), (c, d): Vec(a - c, b - d)\nvlen   = lambda x: sqrt(vdot(x, x))\nvdist  = lambda a, b: vlen(vsub(a, b))\nvscale = lambda s, (x, y): Vec(x * s, y * s)\n\ndef vnorm(v):\n    l = vlen(v)\n    return Vec(v.x / l, v.y / l)\n\nvangle = lambda (x, y): atan2(y, x)\n\ndef anorm(a):\n    if a > pi:  return a - pi * D2\n    if a < -pi: return a + pi * D2\n    return             a\n\nCircle = namedtuple(\"Circle\", \"x y r\")\n\ndef circle_cross((x0, y0, r0), (x1, y1, r1)):\n    d = vdist(Vec(x0, y0), Vec(x1, y1))\n    if d >= r0 + r1 or d <= abs(r0 - r1):\n        return []\n\n    s = (r0 + r1 + d) / D2\n    a = sqrt(s * (s - d) * (s - r0) * (s - r1))\n    h = D2 * a / d\n    dr = Vec(x1 - x0, y1 - y0)\n    dx = vscale(sqrt(r0 ** 2 - h ** 2), vnorm(dr))\n    ang = vangle(dr) if \\\n          r0 ** 2 + d ** 2 > r1 ** 2 \\\n          else pi + vangle(dr)\n    da = asin(h / r0)\n    return map(anorm, [ang - da, ang + da])\n\n# Angles of the start and end points of the circle arc.\nAngle2 = namedtuple(\"Angle2\", \"a1 a2\")\n\nArc = namedtuple(\"Arc\", \"c aa\")\n\narcPoint = lambda (x, y, r), a: \\\n    vadd(Vec(x, y), Vec(r * cos(a), r * sin(a)))\n\narc_start  = lambda (c, (a0, a1)):  arcPoint(c, a0)\narc_mid    = lambda (c, (a0, a1)):  arcPoint(c, (a0 + a1) / D2)\narc_end    = lambda (c, (a0, a1)):  arcPoint(c, a1)\narc_center = lambda ((x, y, r), _): Vec(x, y)\n\narc_area = lambda ((_0, _1, r), (a0, a1)):  r ** 2 * (a1 - a0) / D2\n\ndef split_circles(cs):\n    cSplit = lambda (c, angs): \\\n        imap(Arc, repeat(c), imap(Angle2, angs, angs[1:]))\n\n    # If an arc that was part of one circle is inside *another* circle,\n    # it will not be part of the zero-winding path, so reject it.\n    in_circle = lambda ((x0, y0), c), (x, y, r): \\\n        c != Circle(x, y, r) and vdist(Vec(x0, y0), Vec(x, y)) < r\n\n    def in_any_circle(arc):\n        return any(in_circle((arc_mid(arc), arc.c), c) for c in cs)\n\n    concat_map = lambda f, xs: [y for x in xs for y in f(x)]\n\n    f = lambda c: \\\n        (c, sorted([-pi, pi] +\n                   concat_map(partial(circle_cross, c), cs)))\n    cAngs = map(f, cs)\n    arcs = concat_map(cSplit, cAngs)\n    return filter(lambda ar: not in_any_circle(ar), arcs)\n\n# Given a list of arcs, build sets of closed paths from them.\n# If one arc's end point is no more than 1e-4 from another's\n# start point, they are considered connected.  Since these\n# start/end points resulted from intersecting circles earlier,\n# they *should* be exactly the same, but floating point\n# precision may cause small differences, hence the 1e-4 error\n# margin.  When there are genuinely different intersections\n# closer than this margin, the method will backfire, badly.\ndef make_paths(arcs):\n    eps = Decimal(\"0.0001\")\n    def join_arcs(a, xxs):\n        if not xxs:\n            return [a]\n        x, xs = xxs[0], xxs[1:]\n        if not a:\n            return join_arcs([x], xs)\n        if vdist(arc_start(a[0]), arc_end(a[-1])) < eps:\n            return [a] + join_arcs([], xxs)\n        if vdist(arc_end(a[-1]), arc_start(x)) < eps:\n            return join_arcs(a + [x], xs)\n        return join_arcs(a, xs + [x])\n    return join_arcs([], arcs)\n\n# Slice N-polygon into N-2 triangles.\ndef polyline_area(vvs):\n    tri_area = lambda a, b, c: vcross(vsub(b, a), vsub(c, b)) / D2\n    v, vs = vvs[0], vvs[1:]\n    return sum(tri_area(v, v1, v2) for v1, v2 in izip(vs, vs[1:]))\n\ndef path_area(arcs):\n    f = lambda (a, e), arc: \\\n        (a + arc_area(arc), e + [arc_center(arc), arc_end(arc)])\n    (a, e) = reduce(f, arcs, (0, []))\n    return a + polyline_area(e)\n\ncircles_area = lambda cs: \\\n    sum(imap(path_area, make_paths(split_circles(cs))))\n\ndef main():\n    raw_circles = \"\"\"\\\n         1.6417233788  1.6121789534 0.0848270516\n        -1.4944608174  1.2077959613 1.1039549836\n         0.6110294452 -0.6907087527 0.9089162485\n         0.3844862411  0.2923344616 0.2375743054\n        -0.2495892950 -0.3832854473 1.0845181219\n         1.7813504266  1.6178237031 0.8162655711\n        -0.1985249206 -0.8343333301 0.0538864941\n        -1.7011985145 -0.1263820964 0.4776976918\n        -0.4319462812  1.4104420482 0.7886291537\n         0.2178372997 -0.9499557344 0.0357871187\n        -0.6294854565 -1.3078893852 0.7653357688\n         1.7952608455  0.6281269104 0.2727652452\n         1.4168575317  1.0683357171 1.1016025378\n         1.4637371396  0.9463877418 1.1846214562\n        -0.5263668798  1.7315156631 1.4428514068\n        -1.2197352481  0.9144146579 1.0727263474\n        -0.1389358881  0.1092805780 0.7350208828\n         1.5293954595  0.0030278255 1.2472867347\n        -0.5258728625  1.3782633069 1.3495508831\n        -0.1403562064  0.2437382535 1.3804956588\n         0.8055826339 -0.0482092025 0.3327165165\n        -0.6311979224  0.7184578971 0.2491045282\n         1.4685857879 -0.8347049536 1.3670667538\n        -0.6855727502  1.6465021616 1.0593087096\n         0.0152957411  0.0638919221 0.9771215985\"\"\".splitlines()\n\n    circles = [Circle(*imap(Decimal, row.split()))\n               for row in raw_circles]\n    print \"Total Area:\", circles_area(circles)\n\nmain()"}
{"title": "Trabb Pardo\u2013Knuth algorithm", "language": "Python", "task": "The TPK algorithm is an early example of a programming chrestomathy. \nIt was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages. \nThe report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.\n\nFrom the wikipedia entry:\n\n '''ask''' for 11 numbers to be read into a sequence ''S''\n '''reverse''' sequence ''S''\n '''for each''' ''item'' '''in''' sequence ''S''\n     ''result'' ''':=''' '''call''' a function to do an ''operation''\n     '''if''' ''result'' overflows\n         '''alert''' user\n     '''else'''\n         '''print''' ''result''\n\nThe task is to implement the algorithm:\n# Use the function:     f(x) = |x|^{0.5} + 5x^3\n# The overflow condition is an answer of greater than 400.\n# The 'user alert' should not stop processing of other items of the sequence.\n# Print a prompt before accepting '''eleven''', textual, numeric inputs.\n# You may optionally print the item as well as its associated result, but the results must be in reverse order of input.\n# The sequence   ''' ''S'' '''   may be 'implied' and so not shown explicitly.\n# ''Print and show the program in action from a typical run here''. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).\n\n", "solution": "Python 3.2.2 (default, Sep  4 2011, 09:51:08) [MSC v.1500 32 bit (Intel)] on win32\nType \"copyright\", \"credits\" or \"license()\" for more information.\n>>> def f(x): return abs(x) ** 0.5 + 5 * x**3\n\n>>> print(', '.join('%s:%s' % (x, v if v<=400 else \"TOO LARGE!\")\n\t           for x,v in ((y, f(float(y))) for y in input('\\nnumbers: ').strip().split()[:11][::-1])))\n\n11 numbers: 1 2 3 4 5 6 7 8 9 10 11\n11:TOO LARGE!, 10:TOO LARGE!, 9:TOO LARGE!, 8:TOO LARGE!, 7:TOO LARGE!, 6:TOO LARGE!, 5:TOO LARGE!, 4:322.0, 3:136.73205080756887, 2:41.41421356237309, 1:6.0\n>>> "}
{"title": "Trabb Pardo\u2013Knuth algorithm", "language": "Python 3.10", "task": "The TPK algorithm is an early example of a programming chrestomathy. \nIt was used in Donald Knuth and Luis Trabb Pardo's Stanford tech report The Early Development of Programming Languages. \nThe report traces the early history of work in developing computer languages in the 1940s and 1950s, giving several translations of the algorithm.\n\nFrom the wikipedia entry:\n\n '''ask''' for 11 numbers to be read into a sequence ''S''\n '''reverse''' sequence ''S''\n '''for each''' ''item'' '''in''' sequence ''S''\n     ''result'' ''':=''' '''call''' a function to do an ''operation''\n     '''if''' ''result'' overflows\n         '''alert''' user\n     '''else'''\n         '''print''' ''result''\n\nThe task is to implement the algorithm:\n# Use the function:     f(x) = |x|^{0.5} + 5x^3\n# The overflow condition is an answer of greater than 400.\n# The 'user alert' should not stop processing of other items of the sequence.\n# Print a prompt before accepting '''eleven''', textual, numeric inputs.\n# You may optionally print the item as well as its associated result, but the results must be in reverse order of input.\n# The sequence   ''' ''S'' '''   may be 'implied' and so not shown explicitly.\n# ''Print and show the program in action from a typical run here''. (If the output is graphical rather than text then either add a screendump or describe textually what is displayed).\n\n", "solution": "import math\n\ndef f(x):\n    return math.sqrt(abs(x)) + 5 * x**3\n\ndef ask_numbers(n=11):\n    print(f'Enter {n} numbers:')\n    return (float(input('>')) for _ in range(n))\n\nif __name__ == '__main__':\n    for x in ask_numbers().reverse():\n        if (result := f(x)) > 400:\n            print(f'f({x}): overflow')\n        else:\n            print(f'f({x}) = {result}')"}
{"title": "Tree datastructures", "language": "Python", "task": "The following shows a tree of data with nesting denoted by visual levels of indentation:\nRosettaCode\n    rocks\n        code\n        comparison\n        wiki\n    mocks\n        trolling\n\nA common datastructure for trees is to define node structures having a name and a, (possibly empty), list of child nodes. The nesting of nodes captures the indentation of the tree. Lets call this '''the nest form'''.\n# E.g. if child nodes are surrounded by brackets\n#      and separated by commas then:\nRosettaCode(rocks(code, ...), ...)\n# But only an _example_\n\nAnother datastructure for trees is to construct from the root an ordered list of the nodes level of indentation and the name of that node. The indentation for the root node is zero; node 'rocks is indented by one level from the left, and so on. Lets call this '''the indent form'''.\n0 RosettaCode\n1 rocks\n2 code\n...\n\n;Task:\n# Create/use a nest datastructure format and textual representation for arbitrary trees.\n# Create/use an indent datastructure format and textual representation for arbitrary trees.\n# Create methods/classes/proceedures/routines etc to:\n## Change from a nest tree datastructure to an indent one.\n## Change from an indent tree datastructure to a nest one\n# Use the above to encode the example at the start into the nest format, and show it.\n# transform the initial nest format to indent format and show it.\n# transform the indent format to final nest format and show it.\n# Compare initial and final nest formats which should be the same.\n\n;Note:\n* It's all about showing aspects of the contrasting datastructures as they hold the tree.\n* Comparing nested datastructures is secondary - saving formatted output as a string then a string compare would suffice for this task, if its easier.\n\n\nShow all output on this page.\n\n", "solution": "from pprint import pprint as pp\n\ndef to_indent(node, depth=0, flat=None):\n    if flat is None:\n        flat = []\n    if node:\n        flat.append((depth, node[0]))\n    for child in node[1]:\n        to_indent(child, depth + 1, flat)\n    return flat\n\ndef to_nest(lst, depth=0, level=None):\n    if level is None:\n        level = []\n    while lst:\n        d, name = lst[0]\n        if d == depth:\n            children = []\n            level.append((name, children))\n            lst.pop(0)\n        elif d > depth:  # down\n            to_nest(lst, d, children)\n        elif d < depth:  # up\n            return\n    return level[0] if level else None\n                    \nif __name__ == '__main__':\n    print('Start Nest format:')\n    nest = ('RosettaCode', [('rocks', [('code', []), ('comparison', []), ('wiki', [])]), \n                            ('mocks', [('trolling', [])])])\n    pp(nest, width=25)\n\n    print('\\n... To Indent format:')\n    as_ind = to_indent(nest)\n    pp(as_ind, width=25)\n\n    print('\\n... To Nest format:')\n    as_nest = to_nest(as_ind)\n    pp(as_nest, width=25)\n\n    if nest != as_nest:\n        print(\"Whoops round-trip issues\")"}
{"title": "Tree from nesting levels", "language": "Python", "task": "Given a flat list of integers greater than zero, representing object nesting\nlevels, e.g. [1, 2, 4],\ngenerate a tree formed from nested lists of those nesting level integers where:\n* Every int appears, in order, at its depth of nesting.\n* If the next level int is greater than the previous then it appears in a sub-list of the list containing the previous item\n\n\nThe generated tree data structure should ideally be in a languages nested list format that can\nbe used for further calculations rather than something just calculated for printing.\n\n\nAn input of [1, 2, 4] should produce the equivalent of: [1, [2, [[4]]]]\nwhere 1 is at depth1, 2 is two deep and 4 is nested 4 deep.\n\n[1, 2, 4, 2, 2, 1] should produce [1, [2, [[4]], 2, 2], 1]. \nAll the nesting integers are in the same order but at the correct nesting\nlevels.\n\nSimilarly [3, 1, 3, 1] should generate [[[3]], 1, [[3]], 1]\n\n;Task:\nGenerate and show here the results for the following inputs:\n:* []\n:* [1, 2, 4]\n:* [3, 1, 3, 1]\n:* [1, 2, 3, 1]\n:* [3, 2, 1, 3]\n:* [3, 3, 3, 1, 1, 3, 3, 3]\n", "solution": "def to_tree(x, index=0, depth=1):\n   so_far = []\n   while index < len(x):\n       this = x[index]\n       if this == depth:\n           so_far.append(this)\n       elif this > depth:\n           index, deeper = to_tree(x, index, depth + 1)\n           so_far.append(deeper)\n       else: # this < depth:\n           index -=1\n           break\n       index += 1\n   return (index, so_far) if depth > 1 else so_far\n\n\nif __name__ ==  \"__main__\":\n    from pprint import pformat\n\n    def pnest(nest:list, width: int=9) -> str:\n        text = pformat(nest, width=width).replace('\\n', '\\n    ')\n        print(f\" OR {text}\\n\")\n\n    exercises = [\n        [],\n        [1, 2, 4],\n        [3, 1, 3, 1],\n        [1, 2, 3, 1],\n        [3, 2, 1, 3],\n        [3, 3, 3, 1, 1, 3, 3, 3],\n        ]\n    for flat in exercises:\n        nest = to_tree(flat)\n        print(f\"{flat} NESTS TO: {nest}\")\n        pnest(nest)"}
{"title": "Tropical algebra overloading", "language": "Python", "task": "In algebra, a max tropical semiring (also called a max-plus algebra) is the semiring\n(R  -Inf, , ) containing the ring of real numbers R augmented by negative infinity,\nthe max function (returns the greater of two real numbers), and addition.\n\nIn max tropical algebra, x  y = max(x, y) and x  y = x + y. The identity for \nis -Inf (the max of any number with -infinity is that number), and the identity for  is 0.\n\n;Task:\n\n* Define functions or, if the language supports the symbols as operators, operators for  and  that fit the above description. If the language does not support  and  as operators but allows overloading operators for a new object type, you may instead overload + and * for a new min tropical albrbraic type. If you cannot overload operators in the language used, define ordinary functions for the purpose. \n\nShow that 2  -2 is 0, -0.001  -Inf is -0.001, 0  -Inf is -Inf, 1.5  -1 is 1.5, and -0.5  0 is -0.5.\n\n* Define exponentiation as serial , and in general that a to the power of b is a * b, where a is a real number and b must be a positive integer. Use either | or similar up arrow or the carat ^, as an exponentiation operator if this can be used to overload such \"exponentiation\" in the language being used. Calculate 5 | 7 using this definition.\n\n* Max tropical algebra is distributive, so that\n\n   a  (b  c) equals a  b  b  c, \n\nwhere  has precedence over . Demonstrate that 5  (8  7) equals  5  8  5  7.\n\n* If the language used does not support operator overloading, you may use ordinary function names such as tropicalAdd(x, y) and tropicalMul(x, y).\n\n\n;See also\n:;*[Tropical algebra]\n:;*[Tropical geometry review article]\n:;*[Operator overloading]\n\n\n", "solution": "from numpy import Inf\n\nclass MaxTropical:\n    \"\"\"\n    Class for max tropical algebra.\n    x + y is max(x, y) and X * y is x + y\n    \"\"\"\n    def __init__(self, x=0):\n        self.x = x\n\n    def __str__(self):\n        return str(self.x)\n\n    def __add__(self, other):\n        return MaxTropical(max(self.x, other.x))\n\n    def __mul__(self, other):\n        return MaxTropical(self.x + other.x)\n\n    def __pow__(self, other):\n        assert other.x // 1 == other.x and other.x > 0, \"Invalid Operation\" \n        return MaxTropical(self.x * other.x)\n\n    def __eq__(self, other):\n        return self.x == other.x\n\n\nif __name__ == \"__main__\":\n    a = MaxTropical(-2)\n    b = MaxTropical(-1)\n    c = MaxTropical(-0.5)\n    d = MaxTropical(-0.001)\n    e = MaxTropical(0)\n    f = MaxTropical(0.5)\n    g = MaxTropical(1)\n    h = MaxTropical(1.5)\n    i = MaxTropical(2)\n    j = MaxTropical(5)\n    k = MaxTropical(7)\n    l = MaxTropical(8)\n    m = MaxTropical(-Inf)\n\n    print(\"2 * -2 == \", i * a)\n    print(\"-0.001 + -Inf == \", d + m)\n    print(\"0 * -Inf == \", e * m)\n    print(\"1.5 + -1 == \", h + b)\n    print(\"-0.5 * 0 == \", c * e)\n    print(\"5**7 == \", j**k)\n    print(\"5 * (8 + 7)) == \", j * (l + k))\n    print(\"5 * 8 + 5 * 7 == \", j * l + j * k)\n    print(\"5 * (8 + 7) == 5 * 8 + 5 * 7\", j * (l + k) == j * l + j * k)\n\n"}
{"title": "Truncate a file", "language": "Python", "task": "Truncate a file to a specific length.   This should be implemented as a routine that takes two parameters: the filename and the required file length (in bytes).\n\n\nTruncation can be achieved using system or library calls intended for such a task, if such methods exist, or by creating a temporary file of a reduced size and renaming it, after first deleting the original file, if no other method is available.   The file may contain non human readable binary data in an unspecified format, so the routine should be \"binary safe\", leaving the contents of the untruncated part of the file unchanged.\n\nIf the specified filename does not exist, or the provided length is not less than the current file length, then the routine should raise an appropriate error condition. \n\nOn some systems, the provided file truncation facilities might not change the file or may extend the file, if the specified length is greater than the current length of the file. \n\nThis task permits the use of such facilities.   However, such behaviour should be noted, or optionally a warning message relating to an non change or increase in file size may be implemented.\n\n", "solution": "def truncate_file(name, length):\n    if not os.path.isfile(name):\n        return False\n    if length >= os.path.getsize(name):\n        return False\n    with open(name, 'ab') as f:\n        f.truncate(length)\n    return True\n"}
{"title": "Truth table", "language": "Python", "task": "A truth table is a display of the inputs to, and the output of a Boolean function organized as a table where each row gives one combination of input values and the corresponding value of the function.\n\n\n;Task:\n# Input a Boolean function from the user as a string then calculate and print a formatted truth table for the given function. (One can assume that the user input is correct).\n# Print and show output for Boolean functions of two and three input variables, but any program should not be limited to that many variables in the function. \n# Either reverse-polish or infix notation expressions are allowed.\n\n\n;Related tasks:\n*   [[Boolean values]]\n*   [[Ternary logic]]\n\n\n;See also:\n*   Wolfram MathWorld entry on truth tables.\n*   some \"truth table\" examples from Google.\n\n", "solution": "from itertools import product\n\nwhile True:\n    bexp = input('\\nBoolean expression: ')\n    bexp = bexp.strip()\n    if not bexp:\n        print(\"\\nThank you\")\n        break\n    code = compile(bexp, '<string>', 'eval')\n    names = code.co_names\n    print('\\n' + ' '.join(names), ':', bexp)\n    for values in product(range(2), repeat=len(names)):\n        env = dict(zip(names, values))\n        print(' '.join(str(v) for v in values), ':', eval(code, env))\n"}
{"title": "Two bullet roulette", "language": "Python", "task": "The following is supposedly a question given to mathematics graduates seeking jobs on Wall Street:\n\n  A revolver handgun has a revolving cylinder with six chambers for bullets.\n    \n  It is loaded with the following procedure:\n\n      1. Check the first chamber to the right of the trigger for a bullet.  If a bullet\n         is seen, the cylinder is rotated one chamber clockwise and the next chamber\n         checked until an empty chamber is found.\n\n      2. A cartridge containing a bullet is placed in the empty chamber.\n\n      3. The cylinder is then rotated one chamber clockwise.\n    \n  To randomize the cylinder's position, the cylinder is spun, which causes the cylinder to take\n  a random position from 1 to 6 chamber rotations clockwise from its starting position.\n    \n  When the trigger is pulled the gun will fire if there is a bullet in position 0, which is just\n  counterclockwise from the loading position.\n    \n  The gun is unloaded by removing all cartridges from the cylinder.\n    \n  According to the legend, a suicidal Russian imperial military officer plays a game of Russian\n  roulette by putting two bullets in a six-chamber cylinder and pulls the trigger twice.\n  If the gun fires with a trigger pull, this is considered a successful suicide.\n    \n  The cylinder is always spun before the first shot, but it may or may not be spun after putting\n  in the first bullet and may or may not be spun after taking the first shot.\n    \n  Which of the following situations produces the highest probability of suicide?\n    \n    A. Spinning the cylinder after loading the first bullet, and spinning again after the first shot.\n\n    B. Spinning the cylinder after loading the first bullet only.\n\n    C. Spinning the cylinder after firing the first shot only.\n\n    D. Not spinning the cylinder either after loading the first bullet or after the first shot.\n\n    E. The probability is the same for all cases.\n\n\n;Task:\n# Run a repeated simulation of each of the above scenario, calculating the percentage of suicide with a randomization of the four spinning, loading and firing order scenarios.\n# Show the results as a percentage of deaths for each type of scenario.\n# The hand calculated probabilities are 5/9, 7/12, 5/9, and 1/2. A correct program should produce results close enough to those to allow a correct response to the interview question.  \n\n\n;Reference:\nYoutube video on the Russian 1895 Nagant revolver [[https://www.youtube.com/watch?v=Dh1mojMaEtM]]\n\n", "solution": "\"\"\" Russian roulette problem \"\"\"\nimport numpy as np\n\nclass Revolver:\n    \"\"\" simulates 6-shot revolving cylinger pistol \"\"\"\n\n    def __init__(self):\n        \"\"\" start unloaded \"\"\"\n        self.cylinder = np.array([False] * 6)\n\n    def unload(self):\n        \"\"\" empty all chambers of cylinder \"\"\"\n        self.cylinder[:] = False\n\n    def load(self):\n        \"\"\" load a chamber (advance til empty if full already), then advance once \"\"\"\n        while self.cylinder[1]:\n            self.cylinder[:] = np.roll(self.cylinder, 1)\n        self.cylinder[1] = True\n\n    def spin(self):\n        \"\"\" spin cylinder, randomizing position of chamber to be fired \"\"\"\n        self.cylinder[:] = np.roll(self.cylinder, np.random.randint(1, high=7))\n\n    def fire(self):\n        \"\"\" pull trigger of revolver, return True if fired, False if did not fire \"\"\"\n        shot = self.cylinder[0]\n        self.cylinder[:] = np.roll(self.cylinder, 1)\n        return shot\n\n    def LSLSFSF(self):\n        \"\"\" load, spin, load, spin, fire, spin, fire \"\"\"\n        self.unload()\n        self.load()\n        self.spin()\n        self.load()\n        self.spin()\n        if self.fire():\n            return True\n        self.spin()\n        if self.fire():\n            return True\n        return False\n\n    def LSLSFF(self):\n        \"\"\" load, spin, load, spin, fire, fire \"\"\"\n        self.unload()\n        self.load()\n        self.spin()\n        self.load()\n        self.spin()\n        if self.fire():\n            return True\n        if self.fire():\n            return True\n        return False\n\n    def LLSFSF(self):\n        \"\"\" load, load, spin, fire, spin, fire \"\"\"\n        self.unload()\n        self.load()\n        self.load()\n        self.spin()\n        if self.fire():\n            return True\n        self.spin()\n        if self.fire():\n            return True\n        return False\n\n    def LLSFF(self):\n        \"\"\" load, load, spin, fire, fire \"\"\"\n        self.unload()\n        self.load()\n        self.load()\n        self.spin()\n        if self.fire():\n            return True\n        if self.fire():\n            return True\n        return False\n\n\nif __name__ == '__main__':\n\n    REV = Revolver()\n    TESTCOUNT = 100000\n    for (name, method) in [['load, spin, load, spin, fire, spin, fire', REV.LSLSFSF],\n                           ['load, spin, load, spin, fire, fire', REV.LSLSFF],\n                           ['load, load, spin, fire, spin, fire', REV.LLSFSF],\n                           ['load, load, spin, fire, fire', REV.LLSFF]]:\n\n        percentage = 100 * sum([method() for _ in range(TESTCOUNT)]) / TESTCOUNT\n        print(\"Method\", name, \"produces\", percentage, \"per cent deaths.\")\n"}
{"title": "UPC", "language": "Python", "task": "Goal: \nConvert UPC bar codes to decimal.\n\n\nSpecifically:\n\nThe UPC standard is actually a collection of standards -- physical standards, data format standards, product reference standards... \n\nHere,   in this task,   we will focus on some of the data format standards,   with an imaginary physical+electrical implementation which converts physical UPC bar codes to ASCII   (with spaces and   '''#'''   characters representing the presence or absence of ink).\n\n\n;Sample input:\nBelow, we have a representation of ten different UPC-A bar codes read by our imaginary bar code reader:\n\n         # #   # ##  #  ## #   ## ### ## ### ## #### # # # ## ##  #   #  ##  ## ###  # ##  ## ### #  # #       \n        # # #   ##   ## # #### #   # ## #   ## #   ## # # # ###  # ###  ##  ## ###  # #  ### ###  # # #         \n         # #    # # #  ###  #   #    # #  #   #    # # # # ## #   ## #   ## #   ##   # # #### ### ## # #         \n       # # ##  ## ##  ##   #  #   #  # ###  # ##  ## # # #   ## ##  #  ### ## ## #   # #### ## #   # #        \n         # # ### ## #   ## ## ###  ##  # ##   #   # ## # # ### #  ## ##  #    # ### #  ## ##  #      # #          \n          # #  #   # ##  ##  #   #   #  # ##  ##  #   # # # # #### #  ##  # #### #### # #  ##  # #### # #         \n         # #  #  ##  ##  # #   ## ##   # ### ## ##   # # # #  #   #   #  #  ### # #    ###  # #  #   # #        \n        # # #    # ##  ##   #  # ##  ##  ### #   #  # # # ### ## ## ### ## ### ### ## #  ##  ### ## # #         \n         # # ### ##   ## # # #### #   ## # #### # #### # # #   #  # ###  #    # ###  # #    # ###  # # #       \n        # # # #### ##   # #### # #   ## ## ### #### # # # #  ### # ###  ###  # # ###  #    # #  ### # #         \n\nSome of these were entered upside down,   and one entry has a timing error.\n\n\n;Task:\nImplement code to find the corresponding decimal representation of each, rejecting the error. \n\nExtra credit for handling the rows entered upside down   (the other option is to reject them).\n\n\n;Notes:\nEach digit is represented by 7 bits:\n\n    0:  0 0 0 1 1 0 1\n    1:  0 0 1 1 0 0 1\n    2:  0 0 1 0 0 1 1\n    3:  0 1 1 1 1 0 1\n    4:  0 1 0 0 0 1 1\n    5:  0 1 1 0 0 0 1\n    6:  0 1 0 1 1 1 1\n    7:  0 1 1 1 0 1 1\n    8:  0 1 1 0 1 1 1\n    9:  0 0 0 1 0 1 1\n\nOn the left hand side of the bar code a space represents a '''0''' and a '''#''' represents a '''1'''. \nOn the right hand side of the bar code, a '''#''' represents a '''0''' and a space represents a '''1''' \nAlternatively (for the above):   spaces always represent zeros and '''#''' characters always represent ones, but the representation is logically negated -- '''1'''s and '''0'''s are flipped -- on the right hand side of the bar code.\n\n\n\n;The UPC-A bar code structure:\n::*   It begins with at least 9 spaces   (which our imaginary bar code reader unfortunately doesn't always reproduce properly), \n::*   then has a     ''' # # '''     sequence marking the start of the sequence, \n::*   then has the six \"left hand\" digits, \n::*   then has a   ''' # # '''   sequence in the middle, \n::*   then has the six \"right hand digits\",  \n::*   then has another   ''' # # '''   (end sequence),   and finally, \n::*   then ends with nine trailing spaces   (which might be eaten by wiki edits, and in any event, were not quite captured correctly by our imaginary bar code reader).\n\n\nFinally, the last digit is a checksum digit which may be used to help detect errors. \n\n\n;Verification:\nMultiply each digit in the represented 12 digit sequence by the corresponding number in   (3,1,3,1,3,1,3,1,3,1,3,1)   and add the products.\n\nThe sum (mod 10) must be '''0'''   (must have a zero as its last digit)   if the UPC number has been read correctly.\n\n", "solution": "\"\"\"UPC-A barcode reader. Requires Python =>3.6\"\"\"\nimport itertools\nimport re\n\nRE_BARCODE = re.compile(\n    r\"^(?P<s_quiet> +)\"  # quiet zone\n    r\"(?P<s_guard># #)\"  # start guard\n    r\"(?P<left>[ #]{42})\"  # left digits\n    r\"(?P<m_guard> # # )\"  # middle guard\n    r\"(?P<right>[ #]{42})\"  # right digits\n    r\"(?P<e_guard># #)\"  # end guard\n    r\"(?P<e_quiet> +)$\"  # quiet zone\n)\n\nLEFT_DIGITS = {\n    (0, 0, 0, 1, 1, 0, 1): 0,\n    (0, 0, 1, 1, 0, 0, 1): 1,\n    (0, 0, 1, 0, 0, 1, 1): 2,\n    (0, 1, 1, 1, 1, 0, 1): 3,\n    (0, 1, 0, 0, 0, 1, 1): 4,\n    (0, 1, 1, 0, 0, 0, 1): 5,\n    (0, 1, 0, 1, 1, 1, 1): 6,\n    (0, 1, 1, 1, 0, 1, 1): 7,\n    (0, 1, 1, 0, 1, 1, 1): 8,\n    (0, 0, 0, 1, 0, 1, 1): 9,\n}\n\nRIGHT_DIGITS = {\n    (1, 1, 1, 0, 0, 1, 0): 0,\n    (1, 1, 0, 0, 1, 1, 0): 1,\n    (1, 1, 0, 1, 1, 0, 0): 2,\n    (1, 0, 0, 0, 0, 1, 0): 3,\n    (1, 0, 1, 1, 1, 0, 0): 4,\n    (1, 0, 0, 1, 1, 1, 0): 5,\n    (1, 0, 1, 0, 0, 0, 0): 6,\n    (1, 0, 0, 0, 1, 0, 0): 7,\n    (1, 0, 0, 1, 0, 0, 0): 8,\n    (1, 1, 1, 0, 1, 0, 0): 9,\n}\n\n\nMODULES = {\n    \" \": 0,\n    \"#\": 1,\n}\n\nDIGITS_PER_SIDE = 6\nMODULES_PER_DIGIT = 7\n\n\nclass ParityError(Exception):\n    \"\"\"Exception raised when a parity error is found.\"\"\"\n\n\nclass ChecksumError(Exception):\n    \"\"\"Exception raised when check digit does not match.\"\"\"\n\n\ndef group(iterable, n):\n    \"\"\"Chunk the iterable into groups of size ``n``.\"\"\"\n    args = [iter(iterable)] * n\n    return tuple(itertools.zip_longest(*args))\n\n\ndef parse(barcode):\n    \"\"\"Return the 12 digits represented by the given barcode. Raises a\n    ParityError if any digit fails the parity check.\"\"\"\n    match = RE_BARCODE.match(barcode)\n\n    # Translate bars and spaces to 1s and 0s so we can do arithmetic\n    # with them. Group \"modules\" into chunks of 7 as we go.\n    left = group((MODULES[c] for c in match.group(\"left\")), MODULES_PER_DIGIT)\n    right = group((MODULES[c] for c in match.group(\"right\")), MODULES_PER_DIGIT)\n\n    # Parity check\n    left, right = check_parity(left, right)\n\n    # Lookup digits\n    return tuple(\n        itertools.chain(\n            (LEFT_DIGITS[d] for d in left),\n            (RIGHT_DIGITS[d] for d in right),\n        )\n    )\n\n\ndef check_parity(left, right):\n    \"\"\"Check left and right parity. Flip left and right if the barcode\n    was scanned upside down.\"\"\"\n    # When reading from left to right, each digit on the left should\n    # have odd parity, and each digit on the right should have even\n    # parity.\n    left_parity = sum(sum(d) % 2 for d in left)\n    right_parity = sum(sum(d) % 2 for d in right)\n\n    # Use left and right parity to check if the barcode was scanned\n    # upside down. Flip it if it was.\n    if left_parity == 0 and right_parity == DIGITS_PER_SIDE:\n        _left = tuple(tuple(reversed(d)) for d in reversed(right))\n        right = tuple(tuple(reversed(d)) for d in reversed(left))\n        left = _left\n    elif left_parity != DIGITS_PER_SIDE or right_parity != 0:\n        # Error condition. Mixed parity.\n        error = tuple(\n            itertools.chain(\n                (LEFT_DIGITS.get(d, \"_\") for d in left),\n                (RIGHT_DIGITS.get(d, \"_\") for d in right),\n            )\n        )\n        raise ParityError(\" \".join(str(d) for d in error))\n\n    return left, right\n\n\ndef checksum(digits):\n    \"\"\"Return the check digit for the given digits. Raises a\n    ChecksumError if the check digit does not match.\"\"\"\n    odds = (digits[i] for i in range(0, 11, 2))\n    evens = (digits[i] for i in range(1, 10, 2))\n\n    check_digit = (sum(odds) * 3 + sum(evens)) % 10\n\n    if check_digit != 0:\n        check_digit = 10 - check_digit\n\n    if digits[-1] != check_digit:\n        raise ChecksumError(str(check_digit))\n\n    return check_digit\n\n\ndef main():\n    barcodes = [\n        \"         # #   # ##  #  ## #   ## ### ## ### ## #### # # # ## ##  #   #  ##  ## ###  # ##  ## ### #  # #       \",\n        \"        # # #   ##   ## # #### #   # ## #   ## #   ## # # # ###  # ###  ##  ## ###  # #  ### ###  # # #         \",\n        \"         # #    # # #  ###  #   #    # #  #   #    # # # # ## #   ## #   ## #   ##   # # #### ### ## # #         \",\n        \"       # # ##  ## ##  ##   #  #   #  # ###  # ##  ## # # #   ## ##  #  ### ## ## #   # #### ## #   # #        \",\n        \"         # # ### ## #   ## ## ###  ##  # ##   #   # ## # # ### #  ## ##  #    # ### #  ## ##  #      # #          \",\n        \"          # #  #   # ##  ##  #   #   #  # ##  ##  #   # # # # #### #  ##  # #### #### # #  ##  # #### # #         \",\n        \"         # #  #  ##  ##  # #   ## ##   # ### ## ##   # # # #  #   #   #  #  ### # #    ###  # #  #   # #        \",\n        \"        # # #    # ##  ##   #  # ##  ##  ### #   #  # # # ### ## ## ### ## ### ### ## #  ##  ### ## # #         \",\n        \"         # # ### ##   ## # # #### #   ## # #### # #### # # #   #  # ###  #    # ###  # #    # ###  # # #       \",\n        \"        # # # #### ##   # #### # #   ## ## ### #### # # # #  ### # ###  ###  # # ###  #    # #  ### # #         \",\n        \"        # # # #### ##   # #### # #   ## ## ### #### # # # #  ### # ###  ###  # # ###  ## ##  #  ### # #         \",\n    ]\n\n    for barcode in barcodes:\n        try:\n            digits = parse(barcode)\n        except ParityError as err:\n            print(f\"{err} parity error!\")\n            continue\n\n        try:\n            check_digit = checksum(digits)\n        except ChecksumError as err:\n            print(f\"{' '.join(str(d) for d in digits)} checksum error! ({err})\")\n            continue\n\n        print(f\"{' '.join(str(d) for d in digits)}\")\n\n\nif __name__ == \"__main__\":\n    main()\n"}
{"title": "URL decoding", "language": "Python", "task": "This task   (the reverse of   [[URL encoding]]   and distinct from   [[URL parser]])   is to provide a function \nor mechanism to convert an URL-encoded string into its original unencoded form.\n\n\n;Test cases:\n*   The encoded string   \"http%3A%2F%2Ffoo%20bar%2F\"   should be reverted to the unencoded form   \"http://foo bar/\".\n\n*   The encoded string   \"google.com/search?q=%60Abdu%27l-Bah%C3%A1\"   should revert to the unencoded form   \"google.com/search?q=`Abdu'l-Baha\".\n\n*   The encoded string   \"%25%32%35\"   should revert to the unencoded form   \"%25\" and '''not''' \"%\".\n\n", "solution": "#Python 2.X\nimport urllib\nprint urllib.unquote(\"http%3A%2F%2Ffoo%20bar%2F\")\n#Python 3.5+\nfrom urllib.parse import unquote\nprint(unquote('http%3A%2F%2Ffoo%20bar%2F'))\n"}
{"title": "URL parser", "language": "Python", "task": "URLs are strings with a simple syntax:\n   scheme://[username:password@]domain[:port]/path?query_string#fragment_id\n\n\n;Task:\nParse a well-formed URL to retrieve the relevant information:   '''scheme''', '''domain''', '''path''', ...\n\n\nNote:   this task has nothing to do with [[URL encoding]] or [[URL decoding]].\n\n\nAccording to the standards, the characters:\n::::       ! * ' ( ) ; : @ & = + $ , / ? % # [ ]    \nonly need to be percent-encoded   ('''%''')   in case of possible confusion. \n\nAlso note that the '''path''', '''query''' and '''fragment''' are case sensitive, even if the '''scheme''' and '''domain''' are not.\n\nThe way the returned information is provided (set of variables, array, structured, record, object,...) \nis language-dependent and left to the programmer, but the code should be clear enough to reuse.\n\nExtra credit is given for clear error diagnostics.\n\n*   Here is the official standard:     https://tools.ietf.org/html/rfc3986, \n*   and here is a simpler   BNF:     http://www.w3.org/Addressing/URL/5_URI_BNF.html.\n\n\n;Test cases:\nAccording to T. Berners-Lee\n \n'''foo://example.com:8042/over/there?name=ferret#nose'''     should parse into:\n::*   scheme = foo\n::*   domain = example.com\n::*   port = :8042\n::*   path = over/there\n::*   query = name=ferret\n::*   fragment = nose\n\n\n'''urn:example:animal:ferret:nose'''     should parse into:\n::*   scheme = urn\n::*   path = example:animal:ferret:nose\n\n\n'''other URLs that must be parsed include:'''\n:*    jdbc:mysql://test_user:ouupppssss@localhost:3306/sakila?profileSQL=true  \n:*    ftp://ftp.is.co.za/rfc/rfc1808.txt                                      \n:*    http://www.ietf.org/rfc/rfc2396.txt#header1                             \n:*    ldap://[2001:db8::7]/c=GB?objectClass=one&objectClass=two               \n:*    mailto:John.Doe@example.com                                             \n:*    news:comp.infosystems.www.servers.unix                                  \n:*    tel:+1-816-555-1212                                                     \n:*    telnet://192.0.2.16:80/                                                 \n:*    urn:oasis:names:specification:docbook:dtd:xml:4.1.2                     \n\n", "solution": "import urllib.parse as up # urlparse for Python v2\n\nurl = up.urlparse('http://user:pass@example.com:8081/path/file.html;params?query1=1#fragment')\n\nprint('url.scheme = ', url.scheme)\nprint('url.netloc = ', url.netloc)\nprint('url.hostname = ', url.hostname)\nprint('url.port = ', url.port)\nprint('url.path = ', url.path)\nprint('url.params = ', url.params)\nprint('url.query = ', url.query)\nprint('url.fragment = ', url.fragment)\nprint('url.username = ', url.username)\nprint('url.password = ', url.password)\n"}
{"title": "UTF-8 encode and decode", "language": "Python", "task": "As described in Wikipedia, UTF-8 is a popular encoding of (multi-byte) [[Unicode]] code-points into eight-bit octets.\n\nThe goal of this task is to write a encoder that takes a unicode code-point (an integer representing a unicode character) and returns a sequence of 1-4 bytes representing that character in the UTF-8 encoding. \n\nThen you have to write the corresponding decoder that takes a sequence of 1-4 UTF-8 encoded bytes and return the corresponding unicode character.\n\nDemonstrate the functionality of your encoder and decoder on the following five characters:\n\n\nCharacter   Name                                  Unicode    UTF-8 encoding (hex)\n---------------------------------------------------------------------------------\nA           LATIN CAPITAL LETTER A                U+0041     41\no           LATIN SMALL LETTER O WITH DIAERESIS   U+00F6     C3 B6\nZh           CYRILLIC CAPITAL LETTER ZHE           U+0416     D0 96\nEUR           EURO SIGN                             U+20AC     E2 82 AC\n           MUSICAL SYMBOL G CLEF                 U+1D11E    F0 9D 84 9E\n\n\nProvided below is a reference implementation in Common Lisp.\n\n", "solution": "#!/usr/bin/env python3\nfrom unicodedata import name\n\n\ndef unicode_code(ch):\n    return 'U+{:04x}'.format(ord(ch))\n\n\ndef utf8hex(ch):\n    return \" \".join([hex(c)[2:] for c in ch.encode('utf8')]).upper()\n\n\nif __name__ == \"__main__\":\n    print('{:<11} {:<36} {:<15} {:<15}'.format('Character', 'Name', 'Unicode', 'UTF-8 encoding (hex)'))\n    chars = ['A', '\u00f6', '\u0416', '\u20ac', '\ud834\udd1e']\n    for char in chars:\n        print('{:<11} {:<36} {:<15} {:<15}'.format(char, name(char), unicode_code(char), utf8hex(char)))"}
{"title": "Unbias a random generator", "language": "Python", "task": "Given a weighted one-bit generator of random numbers where the probability of a one occurring, P_1, is not the same as P_0, the probability of a zero occurring, the probability of the occurrence of a one followed by a zero is P_1 x P_0, assuming independence. This is the same as the probability of a zero followed by a one: P_0 x P_1.\n\n;Task details:\n* Use your language's random number generator to create a function/method/subroutine/... '''randN''' that returns a one or a zero, but with one occurring, on average, 1 out of N times, where N is an integer from the range 3 to 6 inclusive.\n* Create a function '''unbiased''' that uses only randN as its source of randomness to become an unbiased generator of random ones and zeroes.\n* For N over its range, generate and show counts of the outputs of randN and unbiased(randN).\n\n\nThe actual unbiasing should be done by generating two numbers at a time from randN and only returning a 1 or 0 if they are different. As long as you always return the first number or always return the second number, the probabilities discussed above should take over the biased probability of randN.\n\nThis task is an implementation of Von Neumann debiasing, first described in a 1951 paper.\n\n", "solution": "from __future__ import print_function\nimport random\n\ndef randN(N):\n    \" 1,0 random generator factory with 1 appearing 1/N'th of the time\"\n    return lambda: random.randrange(N) == 0\n\ndef unbiased(biased):\n    'uses a biased() generator of 1 or 0, to create an unbiased one'\n    this, that = biased(), biased()\n    while this == that: # Loop until 10 or 01\n        this, that = biased(), biased()\n    return this         # return the first\n\nif __name__ == '__main__':\n    from collections import namedtuple\n\n    Stats = namedtuple('Stats', 'count1 count0 percent')\n\n    for N in range(3, 7):\n        biased = randN(N)\n        v = [biased() for x in range(1000000)]\n        v1, v0 = v.count(1), v.count(0)\n        print ( \"Biased(%i)  = %r\" % (N, Stats(v1, v0, 100. * v1/(v1 + v0))) )\n\n        v = [unbiased(biased) for x in range(1000000)]\n        v1, v0 = v.count(1), v.count(0)\n        print ( \"  Unbiased = %r\" % (Stats(v1, v0, 100. * v1/(v1 + v0)), ) )"}
{"title": "Unicode strings", "language": "Python", "task": "As the world gets smaller each day, internationalization becomes more and more important.   For handling multiple languages, [[Unicode]] is your best friend. \n\nIt is a very capable tool, but also quite complex compared to older single- and double-byte character encodings. \n\nHow well prepared is your programming language for Unicode? \n\n\n;Task:\nDiscuss and demonstrate its unicode awareness and capabilities. \n\n\nSome suggested topics:\n:*   How easy is it to present Unicode strings in source code? \n:*   Can Unicode literals be written directly, or be part of identifiers/keywords/etc?\n:*   How well can the language communicate with the rest of the world? \n:*   Is it good at input/output with Unicode?\n:*   Is it convenient to manipulate Unicode strings in the language?\n:*   How broad/deep does the language support Unicode? \n:*   What encodings (e.g. UTF-8, UTF-16, etc) can be used? \n:*   Does it support normalization?\n\n\n;Note:\nThis task is a bit unusual in that it encourages general discussion rather than clever coding.\n\n\n;See also:\n*   [[Unicode variable names]]\n*   [[Terminal control/Display an extended character]]\n\n", "solution": "#!/usr/bin/env python\n# -*- coding: latin-1 -*-\n\nu = 'abcd\u00e9'\nprint(ord(u[-1]))"}
{"title": "Universal Turing machine", "language": "Python from Perl", "task": "One of the foundational mathematical constructs behind computer science \nis the universal Turing Machine.  \n\n\n(Alan Turing introduced the idea of such a machine in 1936-1937.)\n\nIndeed one way to definitively prove that a language \nis turing-complete \nis to implement a universal Turing machine in it.\n\n\n;Task:\nSimulate such a machine capable \nof taking the definition of any other Turing machine and executing it. \n \nOf course, you will not have an infinite tape, \nbut you should emulate this as much as is possible.  \n\nThe three permissible actions on the tape are \"left\", \"right\" and \"stay\".\n\nTo test your universal Turing machine (and prove your programming language \nis Turing complete!), you should execute the following two Turing machines \nbased on the following definitions.\n\n\n'''Simple incrementer'''\n* '''States:''' q0, qf\n* '''Initial state:''' q0\n* '''Terminating states:''' qf\n* '''Permissible symbols:''' B, 1\n* '''Blank symbol:''' B\n* '''Rules:'''\n** (q0, 1, 1, right, q0)\n** (q0, B, 1, stay,  qf)\n\n\nThe input for this machine should be a tape of 1 1 1\n\n\n'''Three-state busy beaver'''\n* '''States:''' a, b, c, halt\n* '''Initial state:''' a\n* '''Terminating states:''' halt\n* '''Permissible symbols:''' 0, 1\n* '''Blank symbol:''' 0\n* '''Rules:'''\n** (a, 0, 1, right, b)\n** (a, 1, 1, left,  c)\n** (b, 0, 1, left,  a)\n** (b, 1, 1, right, b)\n** (c, 0, 1, left,  b)\n** (c, 1, 1, stay,  halt)\n\n\nThe input for this machine should be an empty tape.\n\n\n'''Bonus:'''\n\n'''5-state, 2-symbol probable Busy Beaver machine from Wikipedia'''\n* '''States:''' A, B, C, D, E, H\n* '''Initial state:''' A\n* '''Terminating states:''' H\n* '''Permissible symbols:''' 0, 1\n* '''Blank symbol:''' 0\n* '''Rules:'''\n** (A, 0, 1, right, B)\n** (A, 1, 1, left,  C)\n** (B, 0, 1, right, C)\n** (B, 1, 1, right, B)\n** (C, 0, 1, right, D)\n** (C, 1, 0, left,  E)\n** (D, 0, 1, left,  A)\n** (D, 1, 1, left,  D)\n** (E, 0, 1, stay,  H)\n** (E, 1, 0, left,  A)\n\n\nThe input for this machine should be an empty tape.\n\nThis machine runs for more than 47 millions steps.\n\n", "solution": "from __future__ import print_function\n\ndef run_utm(\n        state = None,\n        blank = None,\n        rules = [],\n        tape = [],\n        halt = None,\n        pos = 0):\n    st = state\n    if not tape: tape = [blank]\n    if pos < 0: pos += len(tape)\n    if pos >= len(tape) or pos < 0: raise Error( \"bad init position\")\n    rules = dict(((s0, v0), (v1, dr, s1)) for (s0, v0, v1, dr, s1) in rules)\n\n    while True:\n        print(st, '\\t', end=\" \")\n        for i, v in enumerate(tape):\n            if i == pos: print(\"[%s]\" % (v,), end=\" \")\n            else: print(v, end=\" \")\n        print()\n\n        if st == halt: break\n        if (st, tape[pos]) not in rules: break\n\n        (v1, dr, s1) = rules[(st, tape[pos])]\n        tape[pos] = v1\n        if dr == 'left':\n            if pos > 0: pos -= 1\n            else: tape.insert(0, blank)\n        if dr == 'right':\n            pos += 1\n            if pos >= len(tape): tape.append(blank) \n        st = s1\n    \n\n# EXAMPLES\n            \nprint(\"incr machine\\n\")\nrun_utm(\n    halt = 'qf',\n\tstate = 'q0',\n\ttape = list(\"111\"),\n\tblank = 'B',\n\trules = map(tuple, \n               [\"q0 1 1 right q0\".split(),\n\t\t        \"q0 B 1 stay  qf\".split()]\n        )\n    )\n\nprint(\"\\nbusy beaver\\n\")\nrun_utm(\n    halt = 'halt',\n\tstate = 'a',\n\tblank = '0',\n\trules = map(tuple,\n        [\"a 0 1 right b\".split(),\n         \"a 1 1 left  c\".split(),\n         \"b 0 1 left  a\".split(),\n         \"b 1 1 right b\".split(),\n         \"c 0 1 left  b\".split(),\n         \"c 1 1 stay  halt\".split()]\n        )\n    )\n\nprint(\"\\nsorting test\\n\")\nrun_utm(halt = 'STOP',\n\tstate = 'A',\n\tblank = '0',\n\ttape = \"2 2 2 1 2 2 1 2 1 2 1 2 1 2\".split(),\n\trules = map(tuple,\n       [\"A 1 1 right A\".split(),\n\t\t\"A 2 3 right B\".split(),\n\t\t\"A 0 0 left  E\".split(),\n\t\t\"B 1 1 right B\".split(),\n\t\t\"B 2 2 right B\".split(),\n\t\t\"B 0 0 left  C\".split(),\n\t\t\"C 1 2 left  D\".split(),\n\t\t\"C 2 2 left  C\".split(),\n\t\t\"C 3 2 left  E\".split(),\n\t\t\"D 1 1 left  D\".split(),\n\t\t\"D 2 2 left  D\".split(),\n\t\t\"D 3 1 right A\".split(),\n\t\t\"E 1 1 left  E\".split(),\n\t\t\"E 0 0 right STOP\".split()]\n        )\n    )\n"}
{"title": "Unix/ls", "language": "Python", "task": "Write a program that will list everything in the current folder,   similar to:\n:::*   the Unix utility   \"ls\"   [http://man7.org/linux/man-pages/man1/ls.1.html]       or \n:::*   the Windows terminal command   \"DIR\" \n\n\nThe output must be sorted, but printing extended details and producing multi-column output is not required.\n\n\n;Example output\nFor the list of paths:\n\n/foo/bar\n/foo/bar/1\n/foo/bar/2\n/foo/bar/a\n/foo/bar/b\n\n\n\nWhen the program is executed in   `/foo`,   it should print:\n\nbar\n\nand when the program is executed in   `/foo/bar`,   it should print:\n\n1\n2\na\nb\n\n\n", "solution": ">>> import os\n>>> print('\\n'.join(sorted(os.listdir('.'))))\nDLLs\nDoc\nLICENSE.txt\nLib\nNEWS.txt\nREADME.txt\nScripts\nTools\ninclude\nlibs\npython.exe\npythonw.exe\ntcl\n>>> "}
{"title": "Validate International Securities Identification Number", "language": "Python", "task": "An International Securities Identification Number (ISIN) is a unique international identifier for a financial security such as a stock or bond. \n\n\n;Task:\nWrite a function or program that takes a string as input, and checks whether it is a valid ISIN.\n\nIt is only valid if it has the correct format,   ''and''   the embedded checksum is correct.\n\nDemonstrate that your code passes the test-cases listed below.\n\n\n;Details:\nThe format of an ISIN is as follows:\n\n\n\n+------------- a 2-character ISO country code (A-Z)\n| +----------- a 9-character security code (A-Z, 0-9)\n| |        +-- a checksum digit (0-9)\nAU0000XVGZA3\n\n\n\n\nFor this task, you may assume that any 2-character alphabetic sequence is a valid country code.\n\nThe checksum can be validated as follows:\n# '''Replace letters with digits''', by converting each character from base 36 to base 10, e.g. AU0000XVGZA3 -1030000033311635103.\n# '''Perform the Luhn test on this base-10 number.'''There is a separate task for this test: ''[[Luhn test of credit card numbers]]''.You don't have to replicate the implementation of this test here   ---   you can just call the existing function from that task.   (Add a comment stating if you did this.)\n\n\n;Test cases:\n:::: {| class=\"wikitable\"\n! ISIN\n! Validity\n! Comment\n|-\n| US0378331005  || valid || \n|-\n| US0373831005  || not valid || The transposition typo is caught by the checksum constraint.\n|-\n| U50378331005  || not valid || The substitution typo is caught by the format constraint.\n|-\n| US03378331005 || not valid || The duplication typo is caught by the format constraint.\n|-\n| AU0000XVGZA3  || valid || \n|-\n| AU0000VXGZA3  || valid || Unfortunately, not ''all'' transposition typos are caught by the checksum constraint.\n|-\n| FR0000988040  || valid || \n|}\n\n(The comments are just informational.   Your function should simply return a Boolean result.   See [[#Raku]] for a reference solution.)\n\n\nRelated task:\n* [[Luhn test of credit card numbers]]\n\n\n;Also see:\n* Interactive online ISIN validator\n* Wikipedia article: International Securities Identification Number\n\n", "solution": "def check_isin(a):\n    if len(a) != 12 or not all(c.isalpha() for c in a[:2]) or not all(c.isalnum() for c in a[2:]):\n        return False\n    s = \"\".join(str(int(c, 36)) for c in a)\n    return 0 == (sum(sum(divmod(2 * (ord(c) - 48), 10)) for c in s[-2::-2]) +\n                 sum(ord(c) - 48 for c in s[::-2])) % 10\n\n# A more readable version \ndef check_isin_alt(a):\n    if len(a) != 12:\n        return False\n    s = []\n    for i, c in enumerate(a):\n        if c.isdigit():\n            if i < 2:\n                return False\n            s.append(ord(c) - 48)\n        elif c.isupper():\n            if i == 11:\n                return False\n            s += divmod(ord(c) - 55, 10)\n        else:\n            return False\n    v = sum(s[::-2])\n    for k in s[-2::-2]:\n        k = 2 * k\n        v += k - 9 if k > 9 else k\n    return v % 10 == 0\n\n[check_isin(s) for s in [\"US0378331005\", \"US0373831005\", \"U50378331005\", \"US03378331005\",\n                         \"AU0000XVGZA3\", \"AU0000VXGZA3\", \"FR0000988040\"]]\n\n# [True, False, False, False, True, True, True]"}
{"title": "Van Eck sequence", "language": "Python", "task": "The sequence is generated by following this pseudo-code:\n\nA:  The first term is zero.\n    Repeatedly apply:\n        If the last term is *new* to the sequence so far then:\nB:          The next term is zero.\n        Otherwise:\nC:          The next term is how far back this last term occured previously.\n\n\n\n;Example:\nUsing A:\n:0\nUsing B:\n:0 0\nUsing C:\n:0 0 1\nUsing B:\n:0 0 1 0\nUsing C:  (zero last occurred two steps back - before the one)\n:0 0 1 0 2\nUsing B:\n:0 0 1 0 2 0\nUsing C:  (two last occurred two steps back - before the zero)\n:0 0 1 0 2 0 2 2\nUsing C:  (two last occurred one step back)\n:0 0 1 0 2 0 2 2 1\nUsing C:  (one last appeared six steps back)\n:0 0 1 0 2 0 2 2 1 6\n...\n\n\n;Task:\n# Create a function/procedure/method/subroutine/... to generate the Van Eck sequence of numbers.\n# Use it to display here, on this page:\n:# The first ten terms of the sequence.\n:# Terms 991 - to - 1000 of the sequence.\n\n\n;References:\n* Don't Know (the Van Eck Sequence) - Numberphile video.\n* Wikipedia Article: Van Eck's Sequence.\n*  OEIS sequence: A181391.\n\n", "solution": "def van_eck():\n    n, seen, val = 0, {}, 0\n    while True:\n        yield val\n        last = {val: n}\n        val = n - seen.get(val, n)\n        seen.update(last)\n        n += 1\n#%%\nif __name__ == '__main__':\n    print(\"Van Eck: first 10 terms:  \", list(islice(van_eck(), 10)))\n    print(\"Van Eck: terms 991 - 1000:\", list(islice(van_eck(), 1000))[-10:])"}
{"title": "Van Eck sequence", "language": "Python 3.7", "task": "The sequence is generated by following this pseudo-code:\n\nA:  The first term is zero.\n    Repeatedly apply:\n        If the last term is *new* to the sequence so far then:\nB:          The next term is zero.\n        Otherwise:\nC:          The next term is how far back this last term occured previously.\n\n\n\n;Example:\nUsing A:\n:0\nUsing B:\n:0 0\nUsing C:\n:0 0 1\nUsing B:\n:0 0 1 0\nUsing C:  (zero last occurred two steps back - before the one)\n:0 0 1 0 2\nUsing B:\n:0 0 1 0 2 0\nUsing C:  (two last occurred two steps back - before the zero)\n:0 0 1 0 2 0 2 2\nUsing C:  (two last occurred one step back)\n:0 0 1 0 2 0 2 2 1\nUsing C:  (one last appeared six steps back)\n:0 0 1 0 2 0 2 2 1 6\n...\n\n\n;Task:\n# Create a function/procedure/method/subroutine/... to generate the Van Eck sequence of numbers.\n# Use it to display here, on this page:\n:# The first ten terms of the sequence.\n:# Terms 991 - to - 1000 of the sequence.\n\n\n;References:\n* Don't Know (the Van Eck Sequence) - Numberphile video.\n* Wikipedia Article: Van Eck's Sequence.\n*  OEIS sequence: A181391.\n\n", "solution": "'''Van Eck sequence'''\n\nfrom functools import reduce\nfrom itertools import repeat\n\n\n# vanEck :: Int -> [Int]\ndef vanEck(n):\n    '''First n terms of the van Eck sequence.'''\n\n    return churchNumeral(n)(\n        lambda xs: cons(\n            maybe(0)(succ)(\n                elemIndex(xs[0])(xs[1:])\n            )\n        )(xs) if xs else [0]\n    )([])[::-1]\n\n\n# TEST ----------------------------------------------------\ndef main():\n    '''Terms of the Van Eck sequence'''\n    print(\n        main.__doc__ + ':\\n\\n' +\n        'First 10: '.rjust(18, ' ') + repr(vanEck(10)) + '\\n' +\n        '991 - 1000: '.rjust(18, ' ') + repr(vanEck(1000)[990:])\n    )\n\n\n# GENERIC -------------------------------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# churchNumeral :: Int -> (a -> a) -> a -> a\ndef churchNumeral(n):\n    '''n applications of a function\n    '''\n    return lambda f: lambda x: reduce(\n        lambda a, g: g(a), repeat(f, n), x\n    )\n\n\n# cons :: a -> [a] -> [a]\ndef cons(x):\n    '''Construction of a list from a head and a tail.\n    '''\n    return lambda xs: [x] + xs\n\n\n# elemIndex :: Eq a => a -> [a] -> Maybe Int\ndef elemIndex(x):\n    '''Just the index of the first element in xs\n       which is equal to x,\n       or Nothing if there is no such element.\n    '''\n    def go(xs):\n        try:\n            return Just(xs.index(x))\n        except ValueError:\n            return Nothing()\n    return go\n\n\n# maybe :: b -> (a -> b) -> Maybe a -> b\ndef maybe(v):\n    '''Either the default value v, if m is Nothing,\n       or the application of f to x,\n       where m is Just(x).\n    '''\n    return lambda f: lambda m: v if None is m or m.get('Nothing') else (\n        f(m.get('Just'))\n    )\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value.\n       For numeric types, (1 +).\n    '''\n    return 1 + x\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Van der Corput sequence", "language": "Python", "task": "When counting integers in binary, if you put a (binary) point to the right of the count then the column immediately to the left denotes a digit with a multiplier of 2^0; the digit in the next column to the left has a multiplier of 2^1; and so on.\n\nSo in the following table:\n  0.\n  1.\n 10.\n 11.\n ...\nthe binary number \"10\" is 1 \\times 2^1 + 0 \\times 2^0.\n\nYou can also have binary digits to the right of the \"point\", just as in the decimal number system. In that case, the digit in the place immediately to the right of the point has a weight of 2^{-1}, or 1/2. \nThe weight for the second column to the right of the point is 2^{-2} or 1/4. And so on.\n\nIf you take the integer binary count of the first table, and ''reflect'' the digits about the binary point, you end up with '''the van der Corput sequence of numbers in base 2'''.\n\n  .0\n  .1\n  .01\n  .11\n  ...\n\nThe third member of the sequence, binary 0.01, is therefore 0 \\times 2^{-1} + 1 \\times 2^{-2} or 1/4.\n\n Monte Carlo simulations. \nThis sequence is also a superset of the numbers representable by the \"fraction\" field of an old IEEE floating point standard. In that standard, the \"fraction\" field represented the fractional part of a binary number beginning with \"1.\" e.g. 1.101001101.\n\n'''Hint'''\n\nA ''hint'' at a way to generate members of the sequence is to modify a routine used to change the base of an integer:\n>>> def base10change(n, base):\n\tdigits = []\n\twhile n:\n\t\tn,remainder = divmod(n, base)\n\t\tdigits.insert(0, remainder)\n\treturn digits\n\n>>> base10change(11, 2)\n[1, 0, 1, 1]\nthe above showing that 11 in decimal is 1\\times 2^3 + 0\\times 2^2 + 1\\times 2^1 + 1\\times 2^0.\nReflected this would become .1101 or 1\\times 2^{-1} + 1\\times 2^{-2} + 0\\times 2^{-3} + 1\\times 2^{-4}\n\n\n;Task description:\n* Create a function/method/routine that given ''n'', generates the ''n'''th term of the van der Corput sequence in base 2.\n* Use the function to compute ''and display'' the first ten members of the sequence. (The first member of the sequence is for ''n''=0).\n\n* As a stretch goal/extra credit, compute and show members of the sequence for bases other than 2.\n\n\n\n;See also:\n* The Basic Low Discrepancy Sequences\n* [[Non-decimal radices/Convert]]\n* Van der Corput sequence\n\n", "solution": ">>> [vdc(i) for i in range(10)]\n[0, 0.5, 0.25, 0.75, 0.125, 0.625, 0.375, 0.875, 0.0625, 0.5625]\n>>> [vdc(i, 3) for i in range(10)]\n[0, 0.3333333333333333, 0.6666666666666666, 0.1111111111111111, 0.4444444444444444, 0.7777777777777777, 0.2222222222222222, 0.5555555555555556, 0.8888888888888888, 0.037037037037037035]\n>>> "}
{"title": "Variable declaration reset", "language": "Python", "task": "A decidely non-challenging task to highlight a potential difference between programming languages.\n\nUsing a straightforward longhand loop as in the JavaScript and Phix examples below, show the locations of elements which are identical to the immediately preceding element in {1,2,2,3,4,4,5}. The (non-blank) results may be 2,5 for zero-based or 3,6 if one-based. \nThe purpose is to determine whether variable declaration (in block scope) resets the contents on every iteration.\nThere is no particular judgement of right or wrong here, just a plain-speaking statement of subtle differences.\nShould your first attempt bomb with \"unassigned variable\" exceptions, feel free to code it as (say)\n  // int prev // crashes with unassigned variable\n     int prev = -1 // predictably no output\nIf your programming language does not support block scope (eg assembly) it should be omitted from this task.\n\n", "solution": "s = [1, 2, 2, 3, 4, 4, 5]\n \nfor i in range(len(s)):\n    curr = s[i]\n    if i > 0 and curr == prev:\n        print(i)\n    prev = curr\n"}
{"title": "Vector products", "language": "Python", "task": "A vector is defined as having three dimensions as being represented by an ordered collection of three numbers:   (X, Y, Z). \n\nIf you imagine a graph with the   '''x'''   and   '''y'''   axis being at right angles to each other and having a third,   '''z'''   axis coming out of the page, then a triplet of numbers,   (X, Y, Z)   would represent a point in the region,   and a vector from the origin to the point.\n\nGiven the vectors:\n         A = (a1,  a2,  a3) \n B = (b1,  b2,  b3) \n C = (c1,  c2,  c3) \nthen the following common vector products are defined:\n* '''The dot product'''       (a scalar quantity)\n::::  A * B = a1b1   +   a2b2   +    a3b3  \n* '''The cross product'''       (a vector quantity)\n::::  A x B = (a2b3   -   a3b2,     a3b1   -   a1b3,     a1b2   -   a2b1)  \n* '''The scalar triple product'''       (a scalar quantity)\n::::  A * (B x C) \n* '''The vector triple product'''       (a vector quantity)\n::::  A x (B x C) \n\n\n;Task:\nGiven the three vectors: \n         a = ( 3,    4,    5)\n         b = ( 4,    3,    5)\n         c = (-5,  -12,  -13)\n# Create a named function/subroutine/method to compute the dot product of two vectors.\n# Create a function to compute the cross product of two vectors.\n# Optionally create a function to compute the scalar triple product of three vectors.\n# Optionally create a function to compute the vector triple product of three vectors.\n# Compute and display: a * b\n# Compute and display: a x b\n# Compute and display: a * (b x c), the scalar triple product.\n# Compute and display: a x (b x c), the vector triple product.\n\n\n;References:\n*   A starting page on Wolfram MathWorld is   {{Wolfram|Vector|Multiplication}}.\n*   Wikipedia   dot product. \n*   Wikipedia   cross product. \n*   Wikipedia   triple product.\n\n\n;Related tasks:\n*   [[Dot product]]\n*   [[Quaternion type]]\n\n", "solution": "def crossp(a, b):\n    '''Cross product of two 3D vectors'''\n    assert len(a) == len(b) == 3, 'For 3D vectors only'\n    a1, a2, a3 = a\n    b1, b2, b3 = b\n    return (a2*b3 - a3*b2, a3*b1 - a1*b3, a1*b2 - a2*b1)\n \ndef dotp(a,b):\n    '''Dot product of two eqi-dimensioned vectors'''\n    assert len(a) == len(b), 'Vector sizes must match'\n    return sum(aterm * bterm for aterm,bterm in zip(a, b))\n \ndef scalartriplep(a, b, c):\n    '''Scalar triple product of three vectors: \"a . (b x c)\"'''\n    return dotp(a, crossp(b, c))\n \ndef vectortriplep(a, b, c):\n    '''Vector triple product of three vectors: \"a x (b x c)\"'''\n    return crossp(a, crossp(b, c))\n \nif __name__ == '__main__':\n    a, b, c = (3, 4, 5), (4, 3, 5), (-5, -12, -13)\n    print(\"a = %r;  b = %r;  c = %r\" % (a, b, c))\n    print(\"a . b = %r\" % dotp(a,b))\n    print(\"a x b = %r\"  % (crossp(a,b),))\n    print(\"a . (b x c) = %r\" % scalartriplep(a, b, c))\n    print(\"a x (b x c) = %r\" % (vectortriplep(a, b, c),))"}
{"title": "Verhoeff algorithm", "language": "Python", "task": "Description\nThe Verhoeff algorithm is a checksum formula for error detection developed by the Dutch mathematician Jacobus Verhoeff and first published in 1969. It was the first decimal check digit algorithm which detects all single-digit errors, and all transposition errors involving two adjacent digits, which was at the time thought impossible with such a code. \n\nAs the workings of the algorithm are clearly described in the linked Wikipedia article they will not be repeated here. \n\n;Task:\nWrite routines, methods, procedures etc. in your language to generate a Verhoeff checksum digit for non-negative integers of any length and to validate the result. A combined routine is also acceptable.\n\nThe more mathematically minded may prefer to generate the 3 tables required from the description provided rather than to hard-code them.\n\nWrite your routines in such a way that they can optionally display digit by digit calculations as in the Wikipedia example.\n\nUse your routines to calculate check digits for the integers: '''236''', '''12345''' and '''123456789012''' and then validate them. Also attempt to validate the same integers if the check digits in all cases were '''9''' rather than what they actually are.\n\nDisplay digit by digit calculations for the first two integers but not for the third. \n\n;Related task:\n*   [[Damm algorithm]]\n\n", "solution": "MULTIPLICATION_TABLE = [\n    (0, 1, 2, 3, 4, 5, 6, 7, 8, 9),\n    (1, 2, 3, 4, 0, 6, 7, 8, 9, 5),\n    (2, 3, 4, 0, 1, 7, 8, 9, 5, 6),\n    (3, 4, 0, 1, 2, 8, 9, 5, 6, 7),\n    (4, 0, 1, 2, 3, 9, 5, 6, 7, 8),\n    (5, 9, 8, 7, 6, 0, 4, 3, 2, 1),\n    (6, 5, 9, 8, 7, 1, 0, 4, 3, 2),\n    (7, 6, 5, 9, 8, 2, 1, 0, 4, 3),\n    (8, 7, 6, 5, 9, 3, 2, 1, 0, 4),\n    (9, 8, 7, 6, 5, 4, 3, 2, 1, 0),\n]\n\nINV = (0, 4, 3, 2, 1, 5, 6, 7, 8, 9)\n\nPERMUTATION_TABLE = [\n    (0, 1, 2, 3, 4, 5, 6, 7, 8, 9),\n    (1, 5, 7, 6, 2, 8, 3, 0, 9, 4),\n    (5, 8, 0, 3, 7, 9, 6, 1, 4, 2),\n    (8, 9, 1, 6, 0, 4, 3, 5, 2, 7),\n    (9, 4, 5, 3, 1, 2, 6, 8, 7, 0),\n    (4, 2, 8, 6, 5, 7, 3, 9, 0, 1),\n    (2, 7, 9, 3, 8, 0, 6, 4, 1, 5),\n    (7, 0, 4, 6, 9, 1, 3, 2, 5, 8),\n]\n\ndef verhoeffchecksum(n, validate=True, terse=True, verbose=False):\n    \"\"\"\n    Calculate the Verhoeff checksum over `n`.\n    Terse mode or with single argument: return True if valid (last digit is a correct check digit).\n    If checksum mode, return the expected correct checksum digit.\n    If validation mode, return True if last digit checks correctly.\n    \"\"\"\n    if verbose:\n        print(f\"\\n{'Validation' if validate else 'Check digit'}\",\\\n            f\"calculations for {n}:\\n\\n i  n\u1d62  p[i,n\u1d62]   c\\n------------------\")\n    # transform number list\n    c, dig = 0, list(str(n if validate else 10 * n))\n    for i, ni in enumerate(dig[::-1]):\n        p = PERMUTATION_TABLE[i % 8][int(ni)]\n        c = MULTIPLICATION_TABLE[c][p]\n        if verbose:\n            print(f\"{i:2}  {ni}      {p}    {c}\")\n\n    if verbose and not validate:\n        print(f\"\\ninv({c}) = {INV[c]}\")\n    if not terse:\n        print(f\"\\nThe validation for '{n}' is {'correct' if c == 0 else 'incorrect'}.\"\\\n              if validate else f\"\\nThe check digit for '{n}' is {INV[c]}.\")\n    return c == 0 if validate else INV[c]\n\nif __name__ == '__main__':\n\n    for n, va, t, ve in [\n        (236, False, False, True), (2363, True, False, True), (2369, True, False, True),\n        (12345, False, False, True), (123451, True, False, True), (123459, True, False, True),\n        (123456789012, False, False, False), (1234567890120, True, False, False),\n        (1234567890129, True, False, False)]:\n        verhoeffchecksum(n, va, t, ve)\n"}
{"title": "Visualize a tree", "language": "Python", "task": "A tree structure   (i.e. a rooted, connected acyclic graph)   is often used in programming.  \n\nIt's often helpful to visually examine such a structure. \n\nThere are many ways to represent trees to a reader, such as:\n:::*   indented text   (a la unix  tree  command)\n:::*   nested HTML tables\n:::*   hierarchical GUI widgets\n:::*   2D   or   3D   images\n:::*   etc.\n\n;Task:\nWrite a program to produce a visual representation of some tree.  \n\nThe content of the tree doesn't matter, nor does the output format, the only requirement being that the output is human friendly. \n\nMake do with the vague term \"friendly\" the best you can.\n\n", "solution": "Python 3.2.3 (default, May  3 2012, 15:54:42) \n[GCC 4.6.3] on linux2\nType \"copyright\", \"credits\" or \"license()\" for more information.\n>>> help('pprint.pprint')\nHelp on function pprint in pprint:\n\npprint.pprint = pprint(object, stream=None, indent=1, width=80, depth=None)\n    Pretty-print a Python object to a stream [default is sys.stdout].\n\n>>> from pprint import pprint\n>>> for tree in [ (1, 2, 3, 4, 5, 6, 7, 8),\n\t          (1, (( 2, 3 ), (4, (5, ((6, 7), 8))))),\n\t          ((((1, 2), 3), 4), 5, 6, 7, 8) ]:\n\tprint(\"\\nTree %r can be pprint'd as:\" % (tree, ))\n\tpprint(tree, indent=1, width=1)\n\n\t\n\nTree (1, 2, 3, 4, 5, 6, 7, 8) can be pprint'd as:\n(1,\n 2,\n 3,\n 4,\n 5,\n 6,\n 7,\n 8)\n\nTree (1, ((2, 3), (4, (5, ((6, 7), 8))))) can be pprint'd as:\n(1,\n ((2,\n   3),\n  (4,\n   (5,\n    ((6,\n      7),\n     8)))))\n\nTree ((((1, 2), 3), 4), 5, 6, 7, 8) can be pprint'd as:\n((((1,\n    2),\n   3),\n  4),\n 5,\n 6,\n 7,\n 8)\n>>> "}
{"title": "Visualize a tree", "language": "Python 3", "task": "A tree structure   (i.e. a rooted, connected acyclic graph)   is often used in programming.  \n\nIt's often helpful to visually examine such a structure. \n\nThere are many ways to represent trees to a reader, such as:\n:::*   indented text   (a la unix  tree  command)\n:::*   nested HTML tables\n:::*   hierarchical GUI widgets\n:::*   2D   or   3D   images\n:::*   etc.\n\n;Task:\nWrite a program to produce a visual representation of some tree.  \n\nThe content of the tree doesn't matter, nor does the output format, the only requirement being that the output is human friendly. \n\nMake do with the vague term \"friendly\" the best you can.\n\n", "solution": "'''Visualize a tree'''\n\nfrom itertools import (chain, repeat, starmap)\nfrom operator import (add)\n\n\n# drawTree :: Tree a -> String\ndef drawTree(tree):\n    '''ASCII diagram of a tree.'''\n    return '\\n'.join(draw(tree))\n\n\n# draw :: Tree a -> [String]\ndef draw(node):\n    '''List of the lines of an ASCII\n       diagram of a tree.'''\n    def shift(first, other, xs):\n        return list(starmap(\n            add,\n            zip(\n                chain([first], repeat(other, len(xs) - 1)),\n                xs\n            )\n        ))\n\n    def drawSubTrees(xs):\n        return (\n            (\n                ['\u2502'] + shift(\n                    '\u251c\u2500 ', '\u2502  ', draw(xs[0])\n                ) + drawSubTrees(xs[1:])\n            ) if 1 < len(xs) else ['\u2502'] + shift(\n                '\u2514\u2500 ', '   ', draw(xs[0])\n            )\n        ) if xs else []\n\n    return (str(root(node))).splitlines() + (\n        drawSubTrees(nest(node))\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Test'''\n\n    # tree :: Tree Int\n    tree = Node(1)([\n        Node(2)([\n            Node(4)([\n                Node(7)([])\n            ]),\n            Node(5)([])\n        ]),\n        Node(3)([\n            Node(6)([\n                Node(8)([]),\n                Node(9)([])\n            ])\n        ])\n    ])\n\n    print(drawTree(tree))\n\n\n# GENERIC -------------------------------------------------\n\n\n# Node :: a -> [Tree a] -> Tree a\ndef Node(v):\n    '''Contructor for a Tree node which connects a\n       value of some kind to a list of zero or\n       more child trees.'''\n    return lambda xs: {'root': v, 'nest': xs}\n\n\n# nest :: Tree a -> [Tree a]\ndef nest(tree):\n    '''Accessor function for children of tree node.'''\n    return tree['nest'] if 'nest' in tree else None\n\n\n# root :: Dict -> a\ndef root(dct):\n    '''Accessor function for data of tree node.'''\n    return dct['root'] if 'root' in dct else None\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Vogel's approximation method", "language": "Python from Ruby", "task": "Vogel's Approximation Method (VAM) is a technique for finding a good initial feasible solution to an allocation problem.\n\nThe powers that be have identified 5 tasks that need to be solved urgently. Being imaginative chaps, they have called them \"A\", \"B\", \"C\", \"D\", and \"E\". They estimate that:\n* A will require 30 hours of work,\n* B will require 20 hours of work,\n* C will require 70 hours of work,\n* D will require 30 hours of work, and\n* E will require 60 hours of work.\n\nThey have identified 4 contractors willing to do the work, called \"W\", \"X\", \"Y\", and \"Z\".\n* W has 50 hours available to commit to working,\n* X has 60 hours available,\n* Y has 50 hours available, and\n* Z has 50 hours available.\nThe cost per hour for each contractor for each task is summarized by the following table:\n\n\n   A  B  C  D  E\nW 16 16 13 22 17\nX 14 14 13 19 15\nY 19 19 20 23 50\nZ 50 12 50 15 11\n\n\nThe task is to use VAM to allocate contractors to tasks. It scales to large problems, so ideally keep sorts out of the iterative cycle. It works as follows:\n\n:Step 1: Balance the given transportation problem if either (total supply>total demand) or (total supply\n        A       B       C       D       E       W       X       Y       Z\n1       2       2       0       4       4       3       1       0       1   E-Z(50)\n\n\nDetermine the largest difference (D or E above). In the case of ties I shall choose the one with the lowest price (in this case E because the lowest price for D is Z=15, whereas for E it is Z=11). For your choice determine the minimum cost (chosen E above so Z=11 is chosen now). Allocate as much as possible from Z to E (50 in this case limited by Z's supply).\nAdjust the supply and demand accordingly. If demand or supply becomes 0 for a given task or contractor it plays no further part. In this case Z is out of it. If you choose arbitrarily, and chose D see here for the working.\n\nRepeat until all supply and demand is met:\n\n2       2       2       0       3       2       3       1       0       -   C-W(50)\n3       5       5       7       4      35       -       1       0       -   E-X(10)\n4       5       5       7       4       -       -       1       0       -   C-X(20)\n5       5       5       -       4       -       -       0       0       -   A-X(30)\n6       -      19       -      23       -       -       -       4       -   D-Y(30)\n        -       -       -       -       -       -       -       -       -   B-Y(20)\n\nFinally calculate the cost of your solution. In the example given it is PS3100:\n\n   A  B  C  D  E\nW       50\nX 30    20    10\nY    20    30\nZ             50\n\n\nThe optimal solution determined by GLPK is PS3100:\n\n   A  B  C  D  E\nW       50\nX 10 20 20    10\nY 20       30\nZ             50\n\n\n;Cf.\n* Transportation problem\n\n", "solution": "from collections import defaultdict\n\ncosts  = {'W': {'A': 16, 'B': 16, 'C': 13, 'D': 22, 'E': 17},\n          'X': {'A': 14, 'B': 14, 'C': 13, 'D': 19, 'E': 15},\n          'Y': {'A': 19, 'B': 19, 'C': 20, 'D': 23, 'E': 50},\n          'Z': {'A': 50, 'B': 12, 'C': 50, 'D': 15, 'E': 11}}\ndemand = {'A': 30, 'B': 20, 'C': 70, 'D': 30, 'E': 60}\ncols = sorted(demand.iterkeys())\nsupply = {'W': 50, 'X': 60, 'Y': 50, 'Z': 50}\nres = dict((k, defaultdict(int)) for k in costs)\ng = {}\nfor x in supply:\n    g[x] = sorted(costs[x].iterkeys(), key=lambda g: costs[x][g])\nfor x in demand:\n    g[x] = sorted(costs.iterkeys(), key=lambda g: costs[g][x])\n\nwhile g:\n    d = {}\n    for x in demand:\n        d[x] = (costs[g[x][1]][x] - costs[g[x][0]][x]) if len(g[x]) > 1 else costs[g[x][0]][x]\n    s = {}\n    for x in supply:\n        s[x] = (costs[x][g[x][1]] - costs[x][g[x][0]]) if len(g[x]) > 1 else costs[x][g[x][0]]\n    f = max(d, key=lambda n: d[n])\n    t = max(s, key=lambda n: s[n])\n    t, f = (f, g[f][0]) if d[f] > s[t] else (g[t][0], t)\n    v = min(supply[f], demand[t])\n    res[f][t] += v\n    demand[t] -= v\n    if demand[t] == 0:\n        for k, n in supply.iteritems():\n            if n != 0:\n                g[k].remove(t)\n        del g[t]\n        del demand[t]\n    supply[f] -= v\n    if supply[f] == 0:\n        for k, n in demand.iteritems():\n            if n != 0:\n                g[k].remove(f)\n        del g[f]\n        del supply[f]\n\nfor n in cols:\n    print \"\\t\", n,\nprint\ncost = 0\nfor g in sorted(costs):\n    print g, \"\\t\",\n    for n in cols:\n        y = res[g][n]\n        if y != 0:\n            print y,\n        cost += y * costs[g][n]\n        print \"\\t\",\n    print\nprint \"\\n\\nTotal Cost = \", cost"}
{"title": "Voronoi diagram", "language": "Python", "task": "A Voronoi diagram is a diagram consisting of a number of sites. \n\nEach Voronoi site ''s''  also has a Voronoi cell consisting of all points closest to ''s''.\n\n\n;Task:\nDemonstrate how to generate and display a Voroni diagram.  \n\n\nSee algo [[K-means++ clustering]].\n\n", "solution": "from PIL import Image\nimport random\nimport math\n\ndef generate_voronoi_diagram(width, height, num_cells):\n\timage = Image.new(\"RGB\", (width, height))\n\tputpixel = image.putpixel\n\timgx, imgy = image.size\n\tnx = []\n\tny = []\n\tnr = []\n\tng = []\n\tnb = []\n\tfor i in range(num_cells):\n\t\tnx.append(random.randrange(imgx))\n\t\tny.append(random.randrange(imgy))\n\t\tnr.append(random.randrange(256))\n\t\tng.append(random.randrange(256))\n\t\tnb.append(random.randrange(256))\n\tfor y in range(imgy):\n\t\tfor x in range(imgx):\n\t\t\tdmin = math.hypot(imgx-1, imgy-1)\n\t\t\tj = -1\n\t\t\tfor i in range(num_cells):\n\t\t\t\td = math.hypot(nx[i]-x, ny[i]-y)\n\t\t\t\tif d < dmin:\n\t\t\t\t\tdmin = d\n\t\t\t\t\tj = i\n\t\t\tputpixel((x, y), (nr[j], ng[j], nb[j]))\n\timage.save(\"VoronoiDiagram.png\", \"PNG\")\n        image.show()\n\t\ngenerate_voronoi_diagram(500, 500, 25)"}
{"title": "Wagstaff primes", "language": "Python", "task": "Definition\nA ''Wagstaff prime'' is a prime number of the form ''(2^p + 1)/3'' where the exponent ''p'' is an odd prime.\n\n;Example \n(2^5 + 1)/3 = 11 is a Wagstaff prime because both 5 and 11 are primes.\n\n;Task\nFind and show here the first ''10'' Wagstaff primes and their corresponding exponents ''p''.\n\n;Stretch (requires arbitrary precision integers)\nFind and show here the exponents ''p'' corresponding to the next ''14'' Wagstaff primes (not the primes themselves) and any more that you have the patience for. \n\nWhen testing for primality, you may use a method which determines that a large number is probably prime with reasonable certainty.\n\n;Note\nIt can be shown (see talk page) that ''(2^p + 1)/3'' is always integral if ''p'' is odd. So there's no need to check for that prior to checking for primality.\n\n;References\n\n* Wikipedia - Wagstaff prime\n* OEIS:A000979 - Wagstaff primes\n\n", "solution": "\"\"\" Rosetta code Wagstaff_primes task \"\"\"\n\nfrom sympy import isprime\n\ndef wagstaff(N):\n    \"\"\" find first N Wagstaff primes \"\"\"\n    pri, wcount = 1, 0\n    while wcount < N:\n        pri += 2\n        if isprime(pri):\n            wag = (2**pri + 1) // 3\n            if isprime(wag):\n                wcount += 1\n                print(f'{wcount: 3}: {pri: 5} => ', \n                      f'{wag:,}' if wcount < 11 else f'[{len(str(wag))} digit number]')\n\n\nwagstaff(24)\n"}
{"title": "War card game", "language": "Python from Julia", "task": "War Card Game\n\nSimulate the card game War.  Use the Bicycle playing card manufacturer's rules. Show a game as played.  User input is optional.\n\nReferences:\n\n:* Bicycle card company: War game site\n\n:* Wikipedia: War (card game)\n\nRelated tasks:\n* [[Playing cards]]\n* [[Card shuffles]]\n* [[Deal cards for FreeCell]]\n* [[Poker hand_analyser]]\n* [[Go Fish]]\n\n\n", "solution": "\"\"\" https://bicyclecards.com/how-to-play/war/ \"\"\"\n\nfrom numpy.random import shuffle\n\nSUITS = ['\u2663', '\u2666', '\u2665', '\u2660']\nFACES = ['2', '3', '4', '5', '6', '7', '8', '9', '10', 'J', 'Q', 'K', 'A']\nDECK = [f + s for f in FACES for s in SUITS]\nCARD_TO_RANK = dict((DECK[i], (i + 3) // 4) for i in range(len(DECK)))\n\n\nclass WarCardGame:\n    \"\"\" card game War \"\"\"\n    def __init__(self):\n        deck = DECK.copy()\n        shuffle(deck)\n        self.deck1, self.deck2 = deck[:26], deck[26:]\n        self.pending = []\n\n    def turn(self):\n        \"\"\" one turn, may recurse on tie \"\"\"\n        if len(self.deck1) == 0 or len(self.deck2) == 0:\n            return self.gameover()\n\n        card1, card2 = self.deck1.pop(0), self.deck2.pop(0)\n        rank1, rank2 = CARD_TO_RANK[card1], CARD_TO_RANK[card2]\n        print(\"{:10}{:10}\".format(card1, card2), end='')\n        if rank1 > rank2:\n            print('Player 1 takes the cards.')\n            self.deck1.extend([card1, card2])\n            self.deck1.extend(self.pending)\n            self.pending = []\n        elif rank1 < rank2:\n            print('Player 2 takes the cards.')\n            self.deck2.extend([card2, card1])\n            self.deck2.extend(self.pending)\n            self.pending = []\n        else:  #  rank1 == rank2\n            print('Tie!')\n            if len(self.deck1) == 0 or len(self.deck2) == 0:\n                return self.gameover()\n\n            card3, card4 = self.deck1.pop(0), self.deck2.pop(0)\n            self.pending.extend([card1, card2, card3, card4])\n            print(\"{:10}{:10}\".format(\"?\", \"?\"), 'Cards are face down.', sep='')\n            return self.turn()\n\n        return True\n\n    def gameover(self):\n        \"\"\" game over who won message \"\"\"\n        if len(self.deck2) == 0:\n            if len(self.deck1) == 0:\n                print('\\nGame ends as a tie.')\n            else:\n                print('\\nPlayer 1 wins the game.')\n        else:\n            print('\\nPlayer 2 wins the game.')\n\n        return False\n\n\nif __name__ == '__main__':\n    WG = WarCardGame()\n    while WG.turn():\n        continue\n"}
{"title": "Water collected between towers", "language": "Python", "task": "In a two-dimensional world, we begin with any bar-chart (or row of close-packed 'towers', each of unit width), and then it rains, \ncompletely filling all convex enclosures in the chart with water. \n\n                                                        \n9               ##           9               ##    \n8               ##           8               ##    \n7     ##        ##           7     ####    \n6     ##  ##    ##           6     ######    \n5 ##  ##  ##  ####           5 ##########    \n4 ##  ##  ########           4 ############    \n3 ######  ########           3 ##############    \n2 ################  ##       2 ##################\n1 ####################       1 ####################\n                                                        \n\nIn the example above, a bar chart representing the values [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] has filled, collecting 14 units of water.\n\nWrite a function, in your language, from a given array of heights, to the number of water units that can be held in this way, by a corresponding bar chart.\n\nCalculate the number of water units that could be collected by bar charts representing each of the following seven series:\n\n   [[1, 5, 3, 7, 2],\n    [5, 3, 7, 2, 6, 4, 5, 9, 1, 2],\n    [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],\n    [5, 5, 5, 5],\n    [5, 6, 7, 8],\n    [8, 7, 7, 6],\n    [6, 7, 10, 7, 6]]\n\n\nSee, also:\n\n* Four Solutions to a Trivial Problem - a Google Tech Talk by Guy Steele\n* Water collected between towers on Stack Overflow, from which the example above is taken)\n* An interesting Haskell solution, using the Tardis monad, by Phil Freeman in a Github gist.\n\n\n\n", "solution": "def water_collected(tower):\n    N = len(tower)\n    highest_left = [0] + [max(tower[:n]) for n in range(1,N)]\n    highest_right = [max(tower[n:N]) for n in range(1,N)] + [0]\n    water_level = [max(min(highest_left[n], highest_right[n]) - tower[n], 0)\n        for n in range(N)]\n    print(\"highest_left:  \", highest_left)\n    print(\"highest_right: \", highest_right)\n    print(\"water_level:   \", water_level)\n    print(\"tower_level:   \", tower)\n    print(\"total_water:   \", sum(water_level))\n    print(\"\")\n    return sum(water_level)\n\ntowers = [[1, 5, 3, 7, 2],\n    [5, 3, 7, 2, 6, 4, 5, 9, 1, 2],\n    [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],\n    [5, 5, 5, 5],\n    [5, 6, 7, 8],\n    [8, 7, 7, 6],\n    [6, 7, 10, 7, 6]]\n\n[water_collected(tower) for tower in towers]"}
{"title": "Weird numbers", "language": "Python 3", "task": "In number theory, a perfect either).\n\nIn other words, the sum of the proper divisors of the number (divisors including 1 but not itself) is greater than the number itself (the number is ''abundant''), but no subset of those divisors sums to the number itself (the number is not ''semiperfect'').\n\nFor example:\n\n* '''12''' is ''not'' a weird number.\n** It is abundant; its proper divisors '''1, 2, 3, 4, 6''' sum to '''16''' (which ''is'' > 12),\n** but it ''is'' semiperfect, e.g.:     '''6 + 4 + 2 == 12'''.\n* '''70''' ''is'' a weird number.\n** It is abundant; its proper divisors '''1, 2, 5, 7, 10, 14, 35''' sum to '''74''' (which ''is'' > 70),\n** and there is no subset of proper divisors that sum to '''70'''.\n\n\n;Task:\nFind and display, here on this page, the first '''25''' weird numbers.\n\n\n;Related tasks:\n:* Abundant, deficient and perfect number classifications\n:* Proper divisors\n\n\n;See also:\n:* OEIS: A006037 weird numbers\n:* Wikipedia: weird number\n:* MathWorld: weird number\n\n", "solution": "'''Weird numbers'''\n\nfrom itertools import chain, count, islice, repeat\nfrom functools import reduce\nfrom math import sqrt\nfrom time import time\n\n\n# weirds :: Gen [Int]\ndef weirds():\n    '''Non-finite stream of weird numbers.\n       (Abundant, but not semi-perfect)\n       OEIS: A006037\n    '''\n    def go(n):\n        ds = descPropDivs(n)\n        d = sum(ds) - n\n        return [n] if 0 < d and not hasSum(d, ds) else []\n    return concatMap(go)(count(1))\n\n\n# hasSum :: Int -> [Int] -> Bool\ndef hasSum(n, xs):\n    '''Does any subset of xs sum to n ?\n       (Assuming xs to be sorted in descending\n       order of magnitude)'''\n    def go(n, xs):\n        if xs:\n            h, t = xs[0], xs[1:]\n            if n < h:  # Head too big. Forget it. Tail ?\n                return go(n, t)\n            else:\n                # The head IS the target ?\n                # Or the tail contains a sum for the\n                # DIFFERENCE between the head and the target ?\n                # Or the tail contains some OTHER sum for the target ?\n                return n == h or go(n - h, t) or go(n, t)\n        else:\n            return False\n    return go(n, xs)\n\n\n# descPropDivs :: Int -> [Int]\ndef descPropDivs(n):\n    '''Descending positive divisors of n,\n       excluding n itself.'''\n    root = sqrt(n)\n    intRoot = int(root)\n    blnSqr = root == intRoot\n    lows = [x for x in range(1, 1 + intRoot) if 0 == n % x]\n    return [\n        n // x for x in (\n            lows[1:-1] if blnSqr else lows[1:]\n        )\n    ] + list(reversed(lows))\n\n\n# --------------------------TEST---------------------------\n\n# main :: IO ()\ndef main():\n    '''Test'''\n\n    start = time()\n    n = 50\n    xs = take(n)(weirds())\n\n    print(\n        (tabulated('First ' + str(n) + ' weird numbers:\\n')(\n            lambda i: str(1 + i)\n        )(str)(5)(\n            index(xs)\n        )(range(0, n)))\n    )\n    print(\n        '\\nApprox computation time: ' +\n        str(int(1000 * (time() - start))) + ' ms'\n    )\n\n\n# -------------------------GENERIC-------------------------\n\n# chunksOf :: Int -> [a] -> [[a]]\ndef chunksOf(n):\n    '''A series of lists of length n,\n       subdividing the contents of xs.\n       Where the length of xs is not evenly divible,\n       the final list will be shorter than n.'''\n    return lambda xs: reduce(\n        lambda a, i: a + [xs[i:n + i]],\n        range(0, len(xs), n), []\n    ) if 0 < n else []\n\n\n# compose (<<<) :: (b -> c) -> (a -> b) -> a -> c\ndef compose(g):\n    '''Right to left function composition.'''\n    return lambda f: lambda x: g(f(x))\n\n\n# concatMap :: (a -> [b]) -> [a] -> [b]\ndef concatMap(f):\n    '''A concatenated list or string over which a function f\n       has been mapped.\n       The list monad can be derived by using an (a -> [b])\n       function which wraps its output in a list (using an\n       empty list to represent computational failure).\n    '''\n    return lambda xs: chain.from_iterable(map(f, xs))\n\n\n# index (!!) :: [a] -> Int -> a\ndef index(xs):\n    '''Item at given (zero-based) index.'''\n    return lambda n: None if 0 > n else (\n        xs[n] if (\n            hasattr(xs, \"__getitem__\")\n        ) else next(islice(xs, n, None))\n    )\n\n\n# paddedMatrix :: a -> [[a]] -> [[a]]\ndef paddedMatrix(v):\n    ''''A list of rows padded to equal length\n        (where needed) with instances of the value v.'''\n    def go(rows):\n        return paddedRows(\n            len(max(rows, key=len))\n        )(v)(rows)\n    return lambda rows: go(rows) if rows else []\n\n\n# paddedRows :: Int -> a -> [[a]] -[[a]]\ndef paddedRows(n):\n    '''A list of rows padded (but never truncated)\n       to length n with copies of value v.'''\n    def go(v, xs):\n        def pad(x):\n            d = n - len(x)\n            return (x + list(repeat(v, d))) if 0 < d else x\n        return list(map(pad, xs))\n    return lambda v: lambda xs: go(v, xs) if xs else []\n\n\n# showColumns :: Int -> [String] -> String\ndef showColumns(n):\n    '''A column-wrapped string\n       derived from a list of rows.'''\n    def go(xs):\n        def fit(col):\n            w = len(max(col, key=len))\n\n            def pad(x):\n                return x.ljust(4 + w, ' ')\n            return ''.join(map(pad, col))\n\n        q, r = divmod(len(xs), n)\n        return unlines(map(\n            fit,\n            transpose(paddedMatrix('')(\n                chunksOf(q + int(bool(r)))(\n                    xs\n                )\n            ))\n        ))\n    return lambda xs: go(xs)\n\n\n# succ :: Enum a => a -> a\ndef succ(x):\n    '''The successor of a value. For numeric types, (1 +).'''\n    return 1 + x if isinstance(x, int) else (\n        chr(1 + ord(x))\n    )\n\n\n# tabulated :: String -> (a -> String) ->\n#                        (b -> String) ->\n#                        Int ->\n#                        (a -> b) -> [a] -> String\ndef tabulated(s):\n    '''Heading -> x display function -> fx display function ->\n          number of columns -> f -> value list -> tabular string.'''\n    def go(xShow, fxShow, intCols, f, xs):\n        w = max(map(compose(len)(xShow), xs))\n        return s + '\\n' + showColumns(intCols)([\n            xShow(x).rjust(w, ' ') + ' -> ' + fxShow(f(x)) for x in xs\n        ])\n    return lambda xShow: lambda fxShow: lambda nCols: (\n        lambda f: lambda xs: go(\n            xShow, fxShow, nCols, f, xs\n        )\n    )\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.'''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, list)\n        else list(islice(xs, n))\n    )\n\n\n# transpose :: Matrix a -> Matrix a\ndef transpose(m):\n    '''The rows and columns of the argument transposed.\n       (The matrix containers and rows can be lists or tuples).'''\n    if m:\n        inner = type(m[0])\n        z = zip(*m)\n        return (type(m))(\n            map(inner, z) if tuple != inner else z\n        )\n    else:\n        return m\n\n\n# unlines :: [String] -> String\ndef unlines(xs):\n    '''A single string derived by the intercalation\n       of a list of strings with the newline character.'''\n    return '\\n'.join(xs)\n\n\n# until :: (a -> Bool) -> (a -> a) -> a -> a\ndef until(p):\n    '''The result of repeatedly applying f until p holds.\n       The initial seed value is x.'''\n    def go(f, x):\n        v = x\n        while not p(v):\n            v = f(v)\n        return v\n    return lambda f: lambda x: go(f, x)\n\n\n# MAIN ----------------------------------------------------\nif __name__ == '__main__':\n    main()"}
{"title": "Word frequency", "language": "Python", "task": "Given a text file and an integer   '''n''',   print/display the   '''n'''   most\ncommon words in the file   (and the number of their occurrences)   in decreasing frequency.\n\n\nFor the purposes of this task:\n*   A word is a sequence of one or more contiguous letters.\n*   You are free to define what a   ''letter''   is. \n*   Underscores, accented letters, apostrophes, hyphens, and other special characters can be handled at your discretion.\n*   You may treat a compound word like   '''well-dressed'''   as either one word or two. \n*   The word   '''it's'''   could also be one or two words as you see fit. \n*   You may also choose not to support non US-ASCII characters. \n*   Assume words will not span multiple lines.\n*   Don't worry about normalization of word spelling differences. \n*   Treat   '''color'''   and   '''colour'''   as two distinct words.\n*   Uppercase letters are considered equivalent to their lowercase counterparts.\n*   Words of equal frequency can be listed in any order.\n*   Feel free to explicitly state the thoughts behind the program decisions.\n\n\nShow example output using Les Miserables from Project Gutenberg as the text file input and display the top   '''10'''   most used words.\n\n\n;History:\nThis task was originally taken from programming pearls from Communications of the ACM June 1986 Volume 29 Number 6\nwhere this problem is solved by Donald Knuth using literate programming and then critiqued by Doug McIlroy,\ndemonstrating solving the problem in a 6 line Unix shell script (provided as an example below).\n\n\n;References:\n*McIlroy's program\n\n\n\n\n", "solution": "import collections\nimport re\nimport string\nimport sys\n\ndef main():\n  counter = collections.Counter(re.findall(r\"\\w+\",open(sys.argv[1]).read().lower()))\n  print counter.most_common(int(sys.argv[2]))\n\nif __name__ == \"__main__\":\n  main()"}
{"title": "Word frequency", "language": "Python 3.7", "task": "Given a text file and an integer   '''n''',   print/display the   '''n'''   most\ncommon words in the file   (and the number of their occurrences)   in decreasing frequency.\n\n\nFor the purposes of this task:\n*   A word is a sequence of one or more contiguous letters.\n*   You are free to define what a   ''letter''   is. \n*   Underscores, accented letters, apostrophes, hyphens, and other special characters can be handled at your discretion.\n*   You may treat a compound word like   '''well-dressed'''   as either one word or two. \n*   The word   '''it's'''   could also be one or two words as you see fit. \n*   You may also choose not to support non US-ASCII characters. \n*   Assume words will not span multiple lines.\n*   Don't worry about normalization of word spelling differences. \n*   Treat   '''color'''   and   '''colour'''   as two distinct words.\n*   Uppercase letters are considered equivalent to their lowercase counterparts.\n*   Words of equal frequency can be listed in any order.\n*   Feel free to explicitly state the thoughts behind the program decisions.\n\n\nShow example output using Les Miserables from Project Gutenberg as the text file input and display the top   '''10'''   most used words.\n\n\n;History:\nThis task was originally taken from programming pearls from Communications of the ACM June 1986 Volume 29 Number 6\nwhere this problem is solved by Donald Knuth using literate programming and then critiqued by Doug McIlroy,\ndemonstrating solving the problem in a 6 line Unix shell script (provided as an example below).\n\n\n;References:\n*McIlroy's program\n\n\n\n\n", "solution": "\"\"\"\nWord count task from Rosetta Code\nhttp://www.rosettacode.org/wiki/Word_count#Python\n\"\"\"\nfrom itertools import (groupby,\n                       starmap)\nfrom operator import itemgetter\nfrom pathlib import Path\nfrom typing import (Iterable,\n                    List,\n                    Tuple)\n\n\nFILEPATH = Path('lesMiserables.txt')\nCOUNT = 10\n\n\ndef main():\n    words_and_counts = most_frequent_words(FILEPATH)\n    print(*words_and_counts[:COUNT], sep='\\n')\n\n\ndef most_frequent_words(filepath: Path,\n                        *,\n                        encoding: str = 'utf-8') -> List[Tuple[str, int]]:\n    \"\"\"\n    A list of word-frequency pairs sorted by their occurrences.\n    The words are read from the given file.\n    \"\"\"\n    def word_and_frequency(word: str,\n                           words_group: Iterable[str]) -> Tuple[str, int]:\n        return word, capacity(words_group)\n\n    file_contents = filepath.read_text(encoding=encoding)\n    words = file_contents.lower().split()\n    grouped_words = groupby(sorted(words))\n    words_and_frequencies = starmap(word_and_frequency, grouped_words)\n    return sorted(words_and_frequencies, key=itemgetter(1), reverse=True)\n\n\ndef capacity(iterable: Iterable) -> int:\n    \"\"\"Returns a number of elements in an iterable\"\"\"\n    return sum(1 for _ in iterable)\n\n\nif __name__ == '__main__':\n    main()\n"}
{"title": "Word ladder", "language": "Python", "task": "Yet another shortest path problem. Given two words of equal length the task is to transpose the first into the second.\n\nOnly one letter may be changed at a time and the change must result in a word in unixdict, the minimum number of intermediate words should be used.\n\nDemonstrate the following:\n\nA boy can be made into a man: boy -> bay -> ban -> man\n\nWith a little more difficulty a girl can be made into a lady: girl -> gill -> gall -> gale -> gaze -> laze -> lazy -> lady\n\nA john can be made into a jane: john -> cohn -> conn -> cone -> cane -> jane \n\nA child can not be turned into an adult.\n\nOptional transpositions of your choice.\n\n\n\n", "solution": "import os,sys,zlib,urllib.request\n\ndef h ( str,x=9 ):\n    for c in str :\n        x = ( x*33 + ord( c )) & 0xffffffffff\n    return x  \n\ndef cache ( func,*param ):\n    n = 'cache_%x.bin'%abs( h( repr( param )))\n    try    : return eval( zlib.decompress( open( n,'rb' ).read()))\n    except : pass\n    s = func( *param )\n    open( n,'wb' ).write( zlib.compress( bytes( repr( s ),'ascii' )))\n    return s\n\ndico_url  = 'https://raw.githubusercontent.com/quinnj/Rosetta-Julia/master/unixdict.txt'\nread_url  = lambda url   : urllib.request.urlopen( url ).read()\nload_dico = lambda url   : tuple( cache( read_url,url ).split( b'\\n'))\nisnext    = lambda w1,w2 : len( w1 ) == len( w2 ) and len( list( filter( lambda l : l[0]!=l[1] , zip( w1,w2 )))) == 1\n\ndef build_map ( words ):\n    map = [(w.decode('ascii'),[]) for w in words]\n    for i1,(w1,n1) in enumerate( map ):\n        for i2,(w2,n2) in enumerate( map[i1+1:],i1+1 ):\n            if isnext( w1,w2 ):\n                n1.append( i2 )\n                n2.append( i1 )\n    return map\n\ndef find_path ( words,w1,w2 ):\n    i = [w[0] for w in words].index( w1 )\n    front,done,res  = [i],{i:-1},[]\n    while front :\n        i = front.pop(0)\n        word,next = words[i]\n        for n in next :\n            if n in done : continue\n            done[n] = i\n            if words[n][0] == w2 :\n                while n >= 0 :\n                    res = [words[n][0]] + res\n                    n = done[n]\n                return ' '.join( res )\n            front.append( n )\n    return '%s can not be turned into %s'%( w1,w2 )\n\nfor w in ('boy man','girl lady','john jane','alien drool','child adult'):\n    print( find_path( cache( build_map,load_dico( dico_url )),*w.split()))"}
{"title": "Word search", "language": "Python 3.x", "task": "A word search puzzle typically consists of a grid of letters in which words are hidden.\n\nThere are many varieties of word search puzzles. For the task at hand we will use a rectangular grid in which the words may be placed horizontally, vertically, or diagonally. The words may also be spelled backwards.\n\nThe words may overlap but are not allowed to zigzag, or wrap around.\n\n;Task \nCreate a 10 by 10 word search and fill it using words from the unixdict. Use only words that are longer than 2, and contain no non-alphabetic characters.\n\nThe cells not used by the hidden words should contain the message: ''Rosetta Code'', read from left to right, top to bottom. These letters should be somewhat evenly distributed over the grid, not clumped together. The message should be in upper case, the hidden words in lower case. All cells should either contain letters from the hidden words or from the message. \n\nPack a minimum of 25 words into the grid.\n\nPrint the resulting grid and the solutions.\n\n;Example\n\n\n     0  1  2  3  4  5  6  7  8  9\n\n0    n  a  y  r  y  R  e  l  m  f \n1    y  O  r  e  t  s  g  n  a  g \n2    t  n  e  d  i  S  k  y  h  E \n3    n  o  t  n  c  p  c  w  t  T \n4    a  l  s  u  u  n  T  m  a  x \n5    r  o  k  p  a  r  i  s  h  h \n6    a  A  c  f  p  a  e  a  c  C \n7    u  b  u  t  t  t  O  l  u  n \n8    g  y  h  w  a  D  h  p  m  u \n9    m  i  r  p  E  h  o  g  a  n \n\nparish     (3,5)(8,5)   gangster   (9,1)(2,1)\npaucity    (4,6)(4,0)   guaranty   (0,8)(0,1)\nprim       (3,9)(0,9)   huckster   (2,8)(2,1)\nplasm      (7,8)(7,4)   fancy      (3,6)(7,2)\nhogan      (5,9)(9,9)   nolo       (1,2)(1,5)\nunder      (3,4)(3,0)   chatham    (8,6)(8,0)\nate        (4,8)(6,6)   nun        (9,7)(9,9)\nbutt       (1,7)(4,7)   hawk       (9,5)(6,2)\nwhy        (3,8)(1,8)   ryan       (3,0)(0,0)\nfay        (9,0)(7,2)   much       (8,8)(8,5)\ntar        (5,7)(5,5)   elm        (6,0)(8,0)\nmax        (7,4)(9,4)   pup        (5,3)(3,5)\nmph        (8,8)(6,8)\n\n\n", "solution": "import re\nfrom random import shuffle, randint\n\ndirs = [[1, 0], [0, 1], [1, 1], [1, -1], [-1, 0], [0, -1], [-1, -1], [-1, 1]]\nn_rows = 10\nn_cols = 10\ngrid_size = n_rows * n_cols\nmin_words = 25\n\n\nclass Grid:\n    def __init__(self):\n        self.num_attempts = 0\n        self.cells = [['' for _ in range(n_cols)] for _ in range(n_rows)]\n        self.solutions = []\n\n\ndef read_words(filename):\n    max_len = max(n_rows, n_cols)\n\n    words = []\n    with open(filename, \"r\") as file:\n        for line in file:\n            s = line.strip().lower()\n            if re.match(r'^[a-z]{3,' + re.escape(str(max_len)) + r'}$', s) is not None:\n                words.append(s)\n\n    return words\n\n\ndef place_message(grid, msg):\n    msg = re.sub(r'[^A-Z]', \"\", msg.upper())\n\n    message_len = len(msg)\n    if 0 < message_len < grid_size:\n        gap_size = grid_size // message_len\n\n        for i in range(0, message_len):\n            pos = i * gap_size + randint(0, gap_size)\n            grid.cells[pos // n_cols][pos % n_cols] = msg[i]\n\n        return message_len\n\n    return 0\n\n\ndef try_location(grid, word, direction, pos):\n    r = pos // n_cols\n    c = pos % n_cols\n    length = len(word)\n\n    # check bounds\n    if (dirs[direction][0] == 1 and (length + c) > n_cols) or \\\n       (dirs[direction][0] == -1 and (length - 1) > c) or \\\n       (dirs[direction][1] == 1 and (length + r) > n_rows) or \\\n       (dirs[direction][1] == -1 and (length - 1) > r):\n        return 0\n\n    rr = r\n    cc = c\n    i = 0\n    overlaps = 0\n\n    # check cells\n    while i < length:\n        if grid.cells[rr][cc] != '' and grid.cells[rr][cc] != word[i]:\n            return 0\n        cc += dirs[direction][0]\n        rr += dirs[direction][1]\n        i += 1\n\n    rr = r\n    cc = c\n    i = 0\n    # place\n    while i < length:\n        if grid.cells[rr][cc] == word[i]:\n            overlaps += 1\n        else:\n            grid.cells[rr][cc] = word[i]\n\n        if i < length - 1:\n            cc += dirs[direction][0]\n            rr += dirs[direction][1]\n\n        i += 1\n\n    letters_placed = length - overlaps\n    if letters_placed > 0:\n        grid.solutions.append(\"{0:<10} ({1},{2})({3},{4})\".format(word, c, r, cc, rr))\n\n    return letters_placed\n\n\ndef try_place_word(grid, word):\n    rand_dir = randint(0, len(dirs))\n    rand_pos = randint(0, grid_size)\n\n    for direction in range(0, len(dirs)):\n        direction = (direction + rand_dir) % len(dirs)\n\n        for pos in range(0, grid_size):\n            pos = (pos + rand_pos) % grid_size\n\n            letters_placed = try_location(grid, word, direction, pos)\n            if letters_placed > 0:\n                return letters_placed\n\n    return 0\n\n\ndef create_word_search(words):\n    grid = None\n    num_attempts = 0\n\n    while num_attempts < 100:\n        num_attempts += 1\n        shuffle(words)\n\n        grid = Grid()\n        message_len = place_message(grid, \"Rosetta Code\")\n        target = grid_size - message_len\n\n        cells_filled = 0\n        for word in words:\n            cells_filled += try_place_word(grid, word)\n            if cells_filled == target:\n                if len(grid.solutions) >= min_words:\n                    grid.num_attempts = num_attempts\n                    return grid\n                else:\n                    break # grid is full but we didn't pack enough words, start over\n\n    return grid\n\n\ndef print_result(grid):\n    if grid is None or grid.num_attempts == 0:\n        print(\"No grid to display\")\n        return\n\n    size = len(grid.solutions)\n\n    print(\"Attempts: {0}\".format(grid.num_attempts))\n    print(\"Number of words: {0}\".format(size))\n\n    print(\"\\n     0  1  2  3  4  5  6  7  8  9\\n\")\n    for r in range(0, n_rows):\n        print(\"{0}   \".format(r), end='')\n        for c in range(0, n_cols):\n            print(\" %c \" % grid.cells[r][c], end='')\n        print()\n    print()\n\n    for i in range(0, size - 1, 2):\n        print(\"{0}   {1}\".format(grid.solutions[i], grid.solutions[i+1]))\n\n    if size % 2 == 1:\n        print(grid.solutions[size - 1])\n\n\nif __name__ == \"__main__\":\n    print_result(create_word_search(read_words(\"unixdict.txt\")))\n"}
{"title": "Word wrap", "language": "Python", "task": "Even today, with proportional fonts and complex layouts, there are still cases where you need to wrap text at a specified column.  \n\n\n;Basic task:\nThe basic task is to wrap a paragraph of text in a simple way in your language.  \n\nIf there is a way to do this that is built-in, trivial, or provided in a standard library, show that.  Otherwise implement the minimum length greedy algorithm from Wikipedia.\n\nShow your routine working on a sample of text at two different wrap columns.\n\n\n;Extra credit:\nWrap text using a more sophisticated algorithm such as the Knuth and Plass TeX algorithm.  \nIf your language provides this, you get easy extra credit, \nbut you ''must reference documentation'' indicating that the algorithm \nis something better than a simple minimum length algorithm.\n\nIf you have both basic and extra credit solutions, show an example where \nthe two algorithms give different results.\n\n\n\n", "solution": ">>> import textwrap\n>>> help(textwrap.fill)\nHelp on function fill in module textwrap:\n\nfill(text, width=70, **kwargs)\n    Fill a single paragraph of text, returning a new string.\n    \n    Reformat the single paragraph in 'text' to fit in lines of no more\n    than 'width' columns, and return a new string containing the entire\n    wrapped paragraph.  As with wrap(), tabs are expanded and other\n    whitespace characters converted to space.  See TextWrapper class for\n    available keyword args to customize wrapping behaviour.\n\n>>> txt = '''\\\nReformat the single paragraph in 'text' to fit in lines of no more\nthan 'width' columns, and return a new string containing the entire\nwrapped paragraph.  As with wrap(), tabs are expanded and other\nwhitespace characters converted to space.  See TextWrapper class for\navailable keyword args to customize wrapping behaviour.'''\n>>> print(textwrap.fill(txt, width=75))\nReformat the single paragraph in 'text' to fit in lines of no more than\n'width' columns, and return a new string containing the entire wrapped\nparagraph.  As with wrap(), tabs are expanded and other whitespace\ncharacters converted to space.  See TextWrapper class for available keyword\nargs to customize wrapping behaviour.\n>>> print(textwrap.fill(txt, width=45))\nReformat the single paragraph in 'text' to\nfit in lines of no more than 'width' columns,\nand return a new string containing the entire\nwrapped paragraph.  As with wrap(), tabs are\nexpanded and other whitespace characters\nconverted to space.  See TextWrapper class\nfor available keyword args to customize\nwrapping behaviour.\n>>> print(textwrap.fill(txt, width=85))\nReformat the single paragraph in 'text' to fit in lines of no more than 'width'\ncolumns, and return a new string containing the entire wrapped paragraph.  As with\nwrap(), tabs are expanded and other whitespace characters converted to space.  See\nTextWrapper class for available keyword args to customize wrapping behaviour.\n>>> "}
{"title": "World Cup group stage", "language": "Python", "task": "It's World Cup season (or at least it was when this page was created)! \n\nThe World Cup is an international football/soccer tournament that happens every 4 years.   Countries put their international teams together in the years between tournaments and qualify for the tournament based on their performance in other international games.   Once a team has qualified they are put into a group with 3 other teams. \n\nFor the first part of the World Cup tournament the teams play in \"group stage\" games where each of the four teams in a group plays all three other teams once.   The results of these games determine which teams will move on to the \"knockout stage\" which is a standard single-elimination tournament.   The two teams from each group with the most standings points move on to the knockout stage. \n\nEach game can result in a win for one team and a loss for the other team or it can result in a draw/tie for each team. \n:::*   A win is worth three points.\n:::*   A draw/tie is worth one point. \n:::*   A loss is worth zero points.\n\n\n;Task:\n:*   Generate all possible outcome combinations for the six group stage games.   With three possible outcomes for each game there should be 36 = 729 of them. \n:*   Calculate the standings points for each team with each combination of outcomes. \n:*   Show a histogram (graphical,   ASCII art, or straight counts--whichever is easiest/most fun) of the standings points for all four teams over all possible outcomes.\n\n\nDon't worry about tiebreakers as they can get complicated.   We are basically looking to answer the question \"if a team gets x standings points, where can they expect to end up in the group standings?\".\n\n''Hint: there should be no possible way to end up in second place with less than two points as well as no way to end up in first with less than three.   Oddly enough, there is no way to get 8 points at all.''\n\n", "solution": "from itertools import product, combinations, izip\n\nscoring = [0, 1, 3]\nhisto = [[0] * 10 for _ in xrange(4)]\n\nfor results in product(range(3), repeat=6):\n    s = [0] * 4\n    for r, g in izip(results, combinations(range(4), 2)):\n        s[g[0]] += scoring[r]\n        s[g[1]] += scoring[2 - r]\n\n    for h, v in izip(histo, sorted(s)):\n        h[v] += 1\n\nfor x in reversed(histo):\n    print x"}
{"title": "Write float arrays to a text file", "language": "Python 2.6", "task": "Write two equal-sized numerical arrays 'x' and 'y' to \na two-column text file named 'filename'.\n\nThe first column of the file contains values from an 'x'-array with a \ngiven 'xprecision', the second -- values from 'y'-array with 'yprecision'.\n\nFor example, considering:\n    x = {1, 2, 3, 1e11};\n    y = {1, 1.4142135623730951, 1.7320508075688772, 316227.76601683791}; \n                                                           /* sqrt(x) */\n    xprecision = 3;\n    yprecision = 5;\n\nThe file should look like:\n    1    1\n    2    1.4142\n    3    1.7321\n    1e+011   3.1623e+005\n\nThis task is intended as a subtask for [[Measure relative performance of sorting algorithms implementations]].\n\n", "solution": ">>> import math\n>>> x = [1, 2, 3, 1e11]\n>>> y = map(math.sqrt, x)\n>>> y\n[1.0, 1.4142135623730951, 1.7320508075688772, 316227.76601683791]\n>>> writedat(\"sqrt.dat\", x, y)\n>>> # check \n...\n>>> for line in open('sqrt.dat'):\n...   print line,\n...\n1       1\n2       1.4142\n3       1.7321\n1e+011  3.1623e+005"}
{"title": "Write float arrays to a text file", "language": "Python 3", "task": "Write two equal-sized numerical arrays 'x' and 'y' to \na two-column text file named 'filename'.\n\nThe first column of the file contains values from an 'x'-array with a \ngiven 'xprecision', the second -- values from 'y'-array with 'yprecision'.\n\nFor example, considering:\n    x = {1, 2, 3, 1e11};\n    y = {1, 1.4142135623730951, 1.7320508075688772, 316227.76601683791}; \n                                                           /* sqrt(x) */\n    xprecision = 3;\n    yprecision = 5;\n\nThe file should look like:\n    1    1\n    2    1.4142\n    3    1.7321\n    1e+011   3.1623e+005\n\nThis task is intended as a subtask for [[Measure relative performance of sorting algorithms implementations]].\n\n", "solution": "def writedat(filename, x, y, xprecision=3, yprecision=5):\n    with open(filename,'w') as f:\n        for a, b in zip(x, y):\n            print(\"%.*g\\t%.*g\" % (xprecision, a, yprecision, b), file=f)\n            #or, using the new-style formatting:\n            #print(\"{1:.{0}g}\\t{3:.{2}g}\".format(xprecision, a, yprecision, b), file=f)"}
{"title": "Write language name in 3D ASCII", "language": "Python", "task": "Write/display a language's name in '''3D''' ASCII. \n\n\n(We can leave the definition of \"3D ASCII\" fuzzy, \nso long as the result is interesting or amusing, \nnot a cheap hack to satisfy the task.)\n\n\n;Related tasks:\n* draw a sphere\n* draw a cuboid\n* draw a rotating cube\n* draw a Deathstar\n\n", "solution": "py = '''\\\n ####   #   #  #####  #   #   ###   #   #\n #   #   # #     #    #   #  #   #  ##  #\n ####     #      #    #####  #   #  # # #\n #        #      #    #   #  #   #  #  ##\n #        #      #    #   #   ###   #   #'''\n\nlines = py.replace('#', '<<<').replace(' ','X') \\\n        .replace('X', '   ').replace('\\n', ' Y') \\\n        .replace('< ', '<>').split('Y')\n\nfor i, l in enumerate(lines): \n    print('  ' * (len(lines) - i) + l)"}
{"title": "Yellowstone sequence", "language": "Python 3.7", "task": "The '''Yellowstone sequence''', also called the '''Yellowstone permutation''', is defined as:\n\nFor n <= 3,\n\n    a(n) = n\n\nFor n >= 4,\n\n    a(n) = the smallest number not already in sequence such that a(n) is relatively prime to a(n-1) and \n           is not relatively prime to a(n-2).\n\n\nThe sequence is a permutation of the natural numbers, and gets its name from what its authors felt was a spiking, geyser like appearance of a plot of the sequence.\n\n\n;Example:\na(4) is 4 because 4 is the smallest number following 1, 2, 3 in the sequence that is relatively prime to the entry before it (3), and is not relatively prime to the number two entries before it (2).\n\n\n;Task\n: Find and show as output the first  '''30'''  Yellowstone numbers.\n\n\n;Extra\n:  Demonstrate how to plot, with x = n and y coordinate a(n), the first 100 Yellowstone numbers.\n\n\n;Related tasks:\n:*   Greatest common divisor.\n:*   Plot coordinate pairs.\n\n\n;See also:\n:*   The OEIS entry:   A098550 The Yellowstone permutation.\n:*   Applegate et al, 2015: The Yellowstone Permutation [https://arxiv.org/abs/1501.01669].\n\n", "solution": "'''Yellowstone permutation OEIS A098550'''\n\nfrom itertools import chain, count, islice\nfrom operator import itemgetter\nfrom math import gcd\n\nfrom matplotlib import pyplot\n\n\n# yellowstone :: [Int]\ndef yellowstone():\n    '''A non-finite stream of terms from\n       the Yellowstone permutation.\n       OEIS A098550.\n    '''\n    # relativelyPrime :: Int -> Int -> Bool\n    def relativelyPrime(a):\n        return lambda b: 1 == gcd(a, b)\n\n    # nextWindow :: (Int, Int, [Int]) -> (Int, Int, [Int])\n    def nextWindow(triple):\n        p2, p1, rest = triple\n        [rp2, rp1] = map(relativelyPrime, [p2, p1])\n\n        # match :: [Int] -> (Int, [Int])\n        def match(xxs):\n            x, xs = uncons(xxs)['Just']\n            return (x, xs) if rp1(x) and not rp2(x) else (\n                second(cons(x))(\n                    match(xs)\n                )\n            )\n        n, residue = match(rest)\n        return (p1, n, residue)\n\n    return chain(\n        range(1, 3),\n        map(\n            itemgetter(1),\n            iterate(nextWindow)(\n                (2, 3, count(4))\n            )\n        )\n    )\n\n\n# TEST ----------------------------------------------------\n# main :: IO ()\ndef main():\n    '''Terms of the Yellowstone permutation.'''\n\n    print(showList(\n        take(30)(yellowstone())\n    ))\n    pyplot.plot(\n        take(100)(yellowstone())\n    )\n    pyplot.xlabel(main.__doc__)\n    pyplot.show()\n\n\n# GENERIC -------------------------------------------------\n\n# Just :: a -> Maybe a\ndef Just(x):\n    '''Constructor for an inhabited Maybe (option type) value.\n       Wrapper containing the result of a computation.\n    '''\n    return {'type': 'Maybe', 'Nothing': False, 'Just': x}\n\n\n# Nothing :: Maybe a\ndef Nothing():\n    '''Constructor for an empty Maybe (option type) value.\n       Empty wrapper returned where a computation is not possible.\n    '''\n    return {'type': 'Maybe', 'Nothing': True}\n\n\n# cons :: a -> [a] -> [a]\ndef cons(x):\n    '''Construction of a list from x as head,\n       and xs as tail.\n    '''\n    return lambda xs: [x] + xs if (\n        isinstance(xs, list)\n    ) else x + xs if (\n        isinstance(xs, str)\n    ) else chain([x], xs)\n\n\n# iterate :: (a -> a) -> a -> Gen [a]\ndef iterate(f):\n    '''An infinite list of repeated\n       applications of f to x.\n    '''\n    def go(x):\n        v = x\n        while True:\n            yield v\n            v = f(v)\n    return go\n\n\n# second :: (a -> b) -> ((c, a) -> (c, b))\ndef second(f):\n    '''A simple function lifted to a function over a tuple,\n       with f applied only to the second of two values.\n    '''\n    return lambda xy: (xy[0], f(xy[1]))\n\n\n# showList :: [a] -> String\ndef showList(xs):\n    '''Stringification of a list.'''\n    return '[' + ','.join(repr(x) for x in xs) + ']'\n\n\n# take :: Int -> [a] -> [a]\n# take :: Int -> String -> String\ndef take(n):\n    '''The prefix of xs of length n,\n       or xs itself if n > length xs.\n    '''\n    return lambda xs: (\n        xs[0:n]\n        if isinstance(xs, (list, tuple))\n        else list(islice(xs, n))\n    )\n\n\n# uncons :: [a] -> Maybe (a, [a])\ndef uncons(xs):\n    '''The deconstruction of a non-empty list\n       (or generator stream) into two parts:\n       a head value, and the remaining values.\n    '''\n    if isinstance(xs, list):\n        return Just((xs[0], xs[1:])) if xs else Nothing()\n    else:\n        nxt = take(1)(xs)\n        return Just((nxt[0], xs)) if nxt else Nothing()\n\n\n# MAIN ---\nif __name__ == '__main__':\n    main()"}
{"title": "Zeckendorf number representation", "language": "Python", "task": "Just as numbers can be represented in a positional notation as sums of multiples of the powers of ten (decimal) or two (binary); all the positive integers can be represented as the sum of one or zero times the distinct members of the Fibonacci series.\n\nRecall that the first six distinct Fibonacci numbers are:  1, 2, 3, 5, 8, 13. \n\nThe decimal number eleven can be written as 0*13 + 1*8 + 0*5 + 1*3 + 0*2 + 0*1 or 010100 in positional notation where the columns represent multiplication by a particular member of the sequence. Leading zeroes are dropped so that 11 decimal becomes 10100.\n\n10100 is not the only way to make 11 from the Fibonacci numbers however;  0*13 + 1*8 + 0*5 + 0*3 + 1*2 + 1*1 or 010011 would also represent decimal 11. For a true Zeckendorf number there is the added restriction that ''no two consecutive Fibonacci numbers can be used'' which leads to the former unique solution.\n\n\n;Task:\nGenerate and show here a table of the Zeckendorf number representations of the decimal numbers zero to twenty, in order.  \n\nThe intention in this task to find the Zeckendorf form of an arbitrary integer.  The Zeckendorf form can be iterated by some bit twiddling rather than calculating each value separately but leave that to another separate task. \n\n\n;Also see:\n*   OEIS A014417   for the the sequence of required results.\n*   Brown's Criterion - Numberphile\n\n\n;Related task:\n*   [[Fibonacci sequence]]\n\n", "solution": "def fib():\n    memo = [1, 2]\n    while True:\n        memo.append(sum(memo))\n        yield memo.pop(0)\n\ndef sequence_down_from_n(n, seq_generator):\n    seq = []\n    for s in seq_generator():\n        seq.append(s)\n        if s >= n: break\n    return seq[::-1]\n\ndef zeckendorf(n):\n    if n == 0: return [0]\n    seq = sequence_down_from_n(n, fib)\n    digits, nleft = [], n\n    for s in seq:\n        if s <= nleft:\n            digits.append(1)\n            nleft -= s\n        else:\n            digits.append(0)\n    assert nleft == 0, 'Check all of n is accounted for'\n    assert sum(x*y for x,y in zip(digits, seq)) == n, 'Assert digits are correct'\n    while digits[0] == 0:\n        # Remove any zeroes padding L.H.S.\n        digits.pop(0)\n    return digits\n\nn = 20\nprint('Fibonacci digit multipliers: %r' % sequence_down_from_n(n, fib))\nfor i in range(n + 1):\n    print('%3i: %8s' % (i, ''.join(str(d) for d in zeckendorf(i))))"}
{"title": "Zero to the zero power", "language": "Python", "task": "Some computer programming languages are not exactly consistent   (with other computer programming languages)   \nwhen   ''raising zero to the zeroth power'':     00\n\n\n;Task:\nShow the results of raising   zero   to the   zeroth   power.\n\n\nIf your computer language objects to      '''0**0'''      or      '''0^0'''      at compile time,   you may also try something like:\n            x = 0\n            y = 0\n            z = x**y\n            say  'z='  z\n\n\n'''Show the result here.'''\nAnd of course use any symbols or notation that is supported in your computer programming language for exponentiation. \n\n\n;See also:\n* The Wiki entry: Zero to the power of zero. \n* The Wiki entry: Zero to the power of zero: History.\n* The MathWorld(tm) entry: exponent laws.\n** Also, in the above MathWorld(tm) entry, see formula ('''9'''): x^0=1.\n* The OEIS entry: The special case of zero to the zeroth power\n\n", "solution": "from decimal import Decimal\nfrom fractions import Fraction\nfrom itertools import product\n\nzeroes = [0, 0.0, 0j, Decimal(0), Fraction(0, 1), -0.0, -0.0j, Decimal(-0.0)]\nfor i, j in product(zeroes, repeat=2):\n    try:\n        ans = i**j\n    except:\n        ans = '<Exception raised>'\n    print(f'{i!r:>15} ** {j!r:<15} = {ans!r}')"}
{"title": "Zhang-Suen thinning algorithm", "language": "Python", "task": "This is an algorithm used to thin a black and white i.e. one bit per pixel images. \n\nFor example, with an input image of:\n                                                           \n #################                   #############         \n ##################               ################         \n ###################            ##################         \n ########     #######          ###################         \n   ######     #######         #######       ######         \n   ######     #######        #######                       \n   #################         #######                       \n   ################          #######                       \n   #################         #######                       \n   ######     #######        #######                       \n   ######     #######        #######                       \n   ######     #######         #######       ######         \n ########     #######          ###################         \n ########     ####### ######    ################## ######  \n ########     ####### ######      ################ ######  \n ########     ####### ######         ############# ######  \n                                                           \nIt produces the thinned output:\n                                                           \n                                                           \n    # ##########                       #######             \n     ##        #                   ####       #            \n     #          #                 ##                       \n     #          #                #                         \n     #          #                #                         \n     #          #                #                         \n     ############               #                          \n     #          #               #                          \n     #          #                #                         \n     #          #                #                         \n     #          #                #                         \n     #                            ##                       \n     #                             ############            \n                       ###                          ###    \n                                                           \n                                                           \n\n;Algorithm:\nAssume black pixels are one and white pixels zero, and that the input image is a rectangular N by M array of ones and zeroes.\n\nThe algorithm operates on all black pixels P1 that can have eight neighbours. \n\nThe neighbours are, in order, arranged as:\n\n\n   P9          P2           P3   \n   P8       P1       P4   \n   P7          P6           P5   \n\n\n\nObviously the boundary pixels of the image cannot have the full eight neighbours.\n\n* Define A(P1) = the number of transitions from white to black, (0 -> 1) in the sequence P2,P3,P4,P5,P6,P7,P8,P9,P2. (Note the extra P2 at the end - it is circular).\n* Define B(P1) = The number of black pixel neighbours of P1. ( = sum(P2 .. P9) )\n\n\n;Step 1:\nAll pixels are tested and pixels satisfying all the following conditions (simultaneously) are just noted at this stage.\n* (0) The pixel is black and has eight neighbours\n* (1) 2 <= B(P1) <= 6\n* (2) A(P1) = 1\n* (3) At least one of P2 and P4 and P6 is white\n* (4) At least one of P4 and P6 and P8 is white\n\nAfter iterating over the image and collecting all the pixels satisfying all step 1 conditions, all these condition satisfying pixels are set to white.\n\n\n;Step 2:\nAll pixels are again tested and pixels satisfying all the following conditions are just noted at this stage.\n* (0) The pixel is black and has eight neighbours\n* (1) 2 <= B(P1) <= 6\n* (2) A(P1) = 1\n* (3) At least one of P2 and P4 and '''P8''' is white\n* (4) At least one of '''P2''' and P6 and P8 is white\n\nAfter iterating over the image and collecting all the pixels satisfying all step 2 conditions, all these condition satisfying pixels are again set to white.\n\n\n;Iteration:\nIf any pixels were set in this round of either step 1 or step 2 then all steps are repeated until no image pixels are so changed.\n\n\n;Task:\n# Write a routine to perform Zhang-Suen thinning on an image matrix of ones and zeroes.\n# Use the routine to thin the following image and show the output here on this page as either a matrix of ones and zeroes, an image, or an ASCII-art image of space/non-space characters. \n\n           00000000000000000000000000000000\n           01111111110000000111111110000000\n           01110001111000001111001111000000\n           01110000111000001110000111000000\n           01110001111000001110000000000000\n           01111111110000001110000000000000\n           01110111100000001110000111000000\n           01110011110011101111001111011100\n           01110001111011100111111110011100\n           00000000000000000000000000000000\n\n\n;Reference:\n* Zhang-Suen Thinning Algorithm, Java Implementation by Nayef Reza.\n* \"Character Recognition Systems: A Guide for Students and Practitioners\" By Mohamed Cheriet, Nawwaf Kharma, Cheng-Lin Liu, Ching Suen\n\n", "solution": "# -*- coding: utf-8 -*-\n\n# Example from [http://nayefreza.wordpress.com/2013/05/11/zhang-suen-thinning-algorithm-java-implementation/ this blog post].\nbeforeTxt = '''\\\n1100111\n1100111\n1100111\n1100111\n1100110\n1100110\n1100110\n1100110\n1100110\n1100110\n1100110\n1100110\n1111110\n0000000\\\n'''\n\n# Thanks to [http://www.network-science.de/ascii/ this site] and vim for these next two examples\nsmallrc01 = '''\\\n00000000000000000000000000000000\n01111111110000000111111110000000\n01110001111000001111001111000000\n01110000111000001110000111000000\n01110001111000001110000000000000\n01111111110000001110000000000000\n01110111100000001110000111000000\n01110011110011101111001111011100\n01110001111011100111111110011100\n00000000000000000000000000000000\\\n'''\n\nrc01 = '''\\\n00000000000000000000000000000000000000000000000000000000000\n01111111111111111100000000000000000001111111111111000000000\n01111111111111111110000000000000001111111111111111000000000\n01111111111111111111000000000000111111111111111111000000000\n01111111100000111111100000000001111111111111111111000000000\n00011111100000111111100000000011111110000000111111000000000\n00011111100000111111100000000111111100000000000000000000000\n00011111111111111111000000000111111100000000000000000000000\n00011111111111111110000000000111111100000000000000000000000\n00011111111111111111000000000111111100000000000000000000000\n00011111100000111111100000000111111100000000000000000000000\n00011111100000111111100000000111111100000000000000000000000\n00011111100000111111100000000011111110000000111111000000000\n01111111100000111111100000000001111111111111111111000000000\n01111111100000111111101111110000111111111111111111011111100\n01111111100000111111101111110000001111111111111111011111100\n01111111100000111111101111110000000001111111111111011111100\n00000000000000000000000000000000000000000000000000000000000\\\n'''\n\ndef intarray(binstring):\n    '''Change a 2D matrix of 01 chars into a list of lists of ints'''\n    return [[1 if ch == '1' else 0 for ch in line] \n            for line in binstring.strip().split()]\n\ndef chararray(intmatrix):\n    '''Change a 2d list of lists of 1/0 ints into lines of 1/0 chars'''\n    return '\\n'.join(''.join(str(p) for p in row) for row in intmatrix)\n\ndef toTxt(intmatrix):\n    '''Change a 2d list of lists of 1/0 ints into lines of '#' and '.' chars'''\n    return '\\n'.join(''.join(('#' if p else '.') for p in row) for row in intmatrix)\n\ndef neighbours(x, y, image):\n    '''Return 8-neighbours of point p1 of picture, in order'''\n    i = image\n    x1, y1, x_1, y_1 = x+1, y-1, x-1, y+1\n    #print ((x,y))\n    return [i[y1][x],  i[y1][x1],   i[y][x1],  i[y_1][x1],  # P2,P3,P4,P5\n            i[y_1][x], i[y_1][x_1], i[y][x_1], i[y1][x_1]]  # P6,P7,P8,P9\n\ndef transitions(neighbours):\n    n = neighbours + neighbours[0:1]    # P2, ... P9, P2\n    return sum((n1, n2) == (0, 1) for n1, n2 in zip(n, n[1:]))\n\ndef zhangSuen(image):\n    changing1 = changing2 = [(-1, -1)]\n    while changing1 or changing2:\n        # Step 1\n        changing1 = []\n        for y in range(1, len(image) - 1):\n            for x in range(1, len(image[0]) - 1):\n                P2,P3,P4,P5,P6,P7,P8,P9 = n = neighbours(x, y, image)\n                if (image[y][x] == 1 and    # (Condition 0)\n                    P4 * P6 * P8 == 0 and   # Condition 4\n                    P2 * P4 * P6 == 0 and   # Condition 3\n                    transitions(n) == 1 and # Condition 2\n                    2 <= sum(n) <= 6):      # Condition 1\n                    changing1.append((x,y))\n        for x, y in changing1: image[y][x] = 0\n        # Step 2\n        changing2 = []\n        for y in range(1, len(image) - 1):\n            for x in range(1, len(image[0]) - 1):\n                P2,P3,P4,P5,P6,P7,P8,P9 = n = neighbours(x, y, image)\n                if (image[y][x] == 1 and    # (Condition 0)\n                    P2 * P6 * P8 == 0 and   # Condition 4\n                    P2 * P4 * P8 == 0 and   # Condition 3\n                    transitions(n) == 1 and # Condition 2\n                    2 <= sum(n) <= 6):      # Condition 1\n                    changing2.append((x,y))\n        for x, y in changing2: image[y][x] = 0\n        #print changing1\n        #print changing2\n    return image\n            \n\nif __name__ == '__main__':\n    for picture in (beforeTxt, smallrc01, rc01):\n        image = intarray(picture)\n        print('\\nFrom:\\n%s' % toTxt(image))\n        after = zhangSuen(image)\n        print('\\nTo thinned:\\n%s' % toTxt(after))"}