100 doors

There are 100 doors in a row that are all initially closed.

You make 100 passes by the doors.

The first time through, visit every door and  ''toggle''  the door  (if the door is closed,  open it;   if it is open,  close it).

The second time, only visit every 2nd door   (door #2, #4, #6, ...),   and toggle it.

The third time, visit every 3rd door   (door #3, #6, #9, ...), etc,   until you only visit the 100th door.

;Task:
Answer the question:   what state are the doors in after the last pass?   Which are open, which are closed?

'''Alternate:'''
As noted in this page's   discussion page,   the only doors that remain open are those whose numbers are perfect squares.

Opening only those doors is an   optimization   that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.

Solution:
class HundredDoors {
    public static void main(String[] args) {
        boolean[] doors = new boolean[101];

        for (int i = 1; i < doors.length; i++) {
            for (int j = i; j < doors.length; j += i) {
                doors[j] = !doors[j];
            }
        }

        for (int i = 1; i < doors.length; i++) {
            if (doors[i]) {
                System.out.printf("Door %d is open.%n", i);
            }
        }
    }
}


100 prisoners

The Problem:
* 100 prisoners are individually numbered 1 to 100
* A room having a cupboard of 100 opaque drawers numbered 1 to 100, that cannot be seen from outside.
* Cards numbered 1 to 100 are placed randomly, one to a drawer, and the drawers all closed; at the start.
* Prisoners start outside the room
:* They can decide some strategy before any enter the room.
:* Prisoners enter the room one by one, can open a drawer, inspect the card number in the drawer, then close the drawer.
:* A prisoner can open no more than 50 drawers.
:* A prisoner tries to find his own number.
:* A prisoner finding his own number is then held apart from the others.
* If '''all''' 100 prisoners find their own numbers then they will all be pardoned. If ''any'' don't then ''all'' sentences stand.

;The task:
# Simulate several thousand instances of the game where the prisoners randomly open drawers
# Simulate several thousand instances of the game where the prisoners use the optimal strategy mentioned in the Wikipedia article, of:
:* First opening the drawer whose outside number is his prisoner number.
:* 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).

Show and compare the computed probabilities of success for the two strategies, here, on this page.

;References:
# The unbelievable solution to the 100 prisoner puzzle standupmaths (Video).
# [[wp:100 prisoners problem]]
# 100 Prisoners Escape Puzzle DataGenetics.
# Random permutation statistics#One hundred prisoners on Wikipedia.

Solution:
import java.util.Collections;
import java.util.List;
import java.util.Objects;
import java.util.function.Function;
import java.util.function.Supplier;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Main {
    private static boolean playOptimal(int n) {
        List<Integer> secretList = IntStream.range(0, n).boxed().collect(Collectors.toList());
        Collections.shuffle(secretList);

        prisoner:
        for (int i = 0; i < secretList.size(); ++i) {
            int prev = i;
            for (int j = 0; j < secretList.size() / 2; ++j) {
                if (secretList.get(prev) == i) {
                    continue prisoner;
                }
                prev = secretList.get(prev);
            }
            return false;
        }
        return true;
    }

    private static boolean playRandom(int n) {
        List<Integer> secretList = IntStream.range(0, n).boxed().collect(Collectors.toList());
        Collections.shuffle(secretList);

        prisoner:
        for (Integer i : secretList) {
            List<Integer> trialList = IntStream.range(0, n).boxed().collect(Collectors.toList());
            Collections.shuffle(trialList);

            for (int j = 0; j < trialList.size() / 2; ++j) {
                if (Objects.equals(trialList.get(j), i)) {
                    continue prisoner;
                }
            }

            return false;
        }
        return true;
    }

    private static double exec(int n, int p, Function<Integer, Boolean> play) {
        int succ = 0;
        for (int i = 0; i < n; ++i) {
            if (play.apply(p)) {
                succ++;
            }
        }
        return (succ * 100.0) / n;
    }

    public static void main(String[] args) {
        final int n = 100_000;
        final int p = 100;
        System.out.printf("# of executions: %d\n", n);
        System.out.printf("Optimal play success rate: %f%%\n", exec(n, p, Main::playOptimal));
        System.out.printf("Random play success rate: %f%%\n", exec(n, p, Main::playRandom));
    }
}


15 puzzle game

Implement the Fifteen Puzzle Game.

The   '''15-puzzle'''   is also known as:
:::*   '''Fifteen Puzzle'''
:::*   '''Gem Puzzle'''
:::*   '''Boss Puzzle'''
:::*   '''Game of Fifteen'''
:::*   '''Mystic Square'''
:::*   '''14-15 Puzzle'''
:::*   and some others.

;Related Tasks:
:*   15 Puzzle Solver
:*   [[16 Puzzle Game]]

Solution:
package fifteenpuzzle;

import java.awt.*;
import java.awt.event.*;
import java.util.Random;
import javax.swing.*;

class FifteenPuzzle extends JPanel {

    private final int side = 4;
    private final int numTiles = side * side - 1;

    private final Random rand = new Random();
    private final int[] tiles = new int[numTiles + 1];
    private final int tileSize;
    private int blankPos;
    private final int margin;
    private final int gridSize;
    private boolean gameOver;

    private FifteenPuzzle() {
        final int dim = 640;

        margin = 80;
        tileSize = (dim - 2 * margin) / side;
        gridSize = tileSize * side;

        setPreferredSize(new Dimension(dim, dim + margin));
        setBackground(Color.WHITE);
        setForeground(new Color(0x6495ED)); // cornflowerblue
        setFont(new Font("SansSerif", Font.BOLD, 60));

        gameOver = true;

        addMouseListener(new MouseAdapter() {
            @Override
            public void mousePressed(MouseEvent e) {
                if (gameOver) {
                    newGame();

                } else {

                    int ex = e.getX() - margin;
                    int ey = e.getY() - margin;

                    if (ex < 0 || ex > gridSize || ey < 0 || ey > gridSize) {
                        return;
                    }

                    int c1 = ex / tileSize;
                    int r1 = ey / tileSize;
                    int c2 = blankPos % side;
                    int r2 = blankPos / side;

                    int clickPos = r1 * side + c1;

                    int dir = 0;
                    if (c1 == c2 && Math.abs(r1 - r2) > 0) {
                        dir = (r1 - r2) > 0 ? 4 : -4;

                    } else if (r1 == r2 && Math.abs(c1 - c2) > 0) {
                        dir = (c1 - c2) > 0 ? 1 : -1;
                    }

                    if (dir != 0) {
                        do {
                            int newBlankPos = blankPos + dir;
                            tiles[blankPos] = tiles[newBlankPos];
                            blankPos = newBlankPos;
                        } while (blankPos != clickPos);
                        tiles[blankPos] = 0;
                    }

                    gameOver = isSolved();
                }
                repaint();
            }
        });

        newGame();
    }

    private void newGame() {
        do {
            reset();
            shuffle();
        } while (!isSolvable());
        gameOver = false;
    }

    private void reset() {
        for (int i = 0; i < tiles.length; i++) {
            tiles[i] = (i + 1) % tiles.length;
        }
        blankPos = tiles.length - 1;
    }

    private void shuffle() {
        // don't include the blank space in the shuffle, leave it
        // in the home position
        int n = numTiles;
        while (n > 1) {
            int r = rand.nextInt(n--);
            int tmp = tiles[r];
            tiles[r] = tiles[n];
            tiles[n] = tmp;
        }
    }

    /*  Only half the permutations of the puzzle are solvable.

        Whenever a tile is preceded by a tile with higher value it counts
        as an inversion. In our case, with the blank space in the home
        position, the number of inversions must be even for the puzzle
        to be solvable.

        See also:
        www.cs.bham.ac.uk/~mdr/teaching/modules04/java2/TilesSolvability.html
     */
    private boolean isSolvable() {
        int countInversions = 0;
        for (int i = 0; i < numTiles; i++) {
            for (int j = 0; j < i; j++) {
                if (tiles[j] > tiles[i]) {
                    countInversions++;
                }
            }
        }
        return countInversions % 2 == 0;
    }

    private boolean isSolved() {
        if (tiles[tiles.length - 1] != 0) {
            return false;
        }
        for (int i = numTiles - 1; i >= 0; i--) {
            if (tiles[i] != i + 1) {
                return false;
            }
        }
        return true;
    }

    private void drawGrid(Graphics2D g) {
        for (int i = 0; i < tiles.length; i++) {
            int r = i / side;
            int c = i % side;
            int x = margin + c * tileSize;
            int y = margin + r * tileSize;

            if (tiles[i] == 0) {
                if (gameOver) {
                    g.setColor(Color.GREEN);
                    drawCenteredString(g, "\u2713", x, y);
                }
                continue;
            }

            g.setColor(getForeground());
            g.fillRoundRect(x, y, tileSize, tileSize, 25, 25);
            g.setColor(Color.blue.darker());
            g.drawRoundRect(x, y, tileSize, tileSize, 25, 25);
            g.setColor(Color.WHITE);

            drawCenteredString(g, String.valueOf(tiles[i]), x, y);
        }
    }

    private void drawStartMessage(Graphics2D g) {
        if (gameOver) {
            g.setFont(getFont().deriveFont(Font.BOLD, 18));
            g.setColor(getForeground());
            String s = "click to start a new game";
            int x = (getWidth() - g.getFontMetrics().stringWidth(s)) / 2;
            int y = getHeight() - margin;
            g.drawString(s, x, y);
        }
    }

    private void drawCenteredString(Graphics2D g, String s, int x, int y) {
        FontMetrics fm = g.getFontMetrics();
        int asc = fm.getAscent();
        int des = fm.getDescent();

        x = x + (tileSize - fm.stringWidth(s)) / 2;
        y = y + (asc + (tileSize - (asc + des)) / 2);

        g.drawString(s, x, y);
    }

    @Override
    public void paintComponent(Graphics gg) {
        super.paintComponent(gg);
        Graphics2D g = (Graphics2D) gg;
        g.setRenderingHint(RenderingHints.KEY_ANTIALIASING,
                RenderingHints.VALUE_ANTIALIAS_ON);

        drawGrid(g);
        drawStartMessage(g);
    }

    public static void main(String[] args) {
        SwingUtilities.invokeLater(() -> {
            JFrame f = new JFrame();
            f.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
            f.setTitle("Fifteen Puzzle");
            f.setResizable(false);
            f.add(new FifteenPuzzle(), BorderLayout.CENTER);
            f.pack();
            f.setLocationRelativeTo(null);
            f.setVisible(true);
        });
    }
}


21 game

'''21''' is a two player game, the game is played by choosing
a number ('''1''', '''2''', or '''3''')  to be added to the ''running total''.

The game is won by the player whose chosen number causes the ''running total''
to reach ''exactly'' '''21'''.

The ''running total'' starts at zero.
One player will be the computer.

Players alternate supplying a number to be added to the ''running total''.

;Task:
Write a computer program that will:
::* do the prompting (or provide a button menu),
::* check for errors and display appropriate error messages,
::* do the additions (add a chosen number to the ''running total''),
::* display the ''running total'',
::* provide a mechanism for the player to quit/exit/halt/stop/close the program,
::* issue a notification when there is a winner, and
::* determine who goes first (maybe a random or user choice, or can be specified when the game begins).

Solution:
import java.util.Random;
import java.util.Scanner;

public class TwentyOneGame {

    public static void main(String[] args) {
        new TwentyOneGame().run(true, 21, new int[] {1, 2, 3});
    }

    public void run(boolean computerPlay, int max, int[] valid) {
        String comma = "";
        for ( int i = 0 ; i < valid.length ; i++ ) {
            comma += valid[i];
            if ( i < valid.length - 2 && valid.length >= 3 ) {
                comma += ", ";
            }
            if ( i == valid.length - 2 ) {
                comma += " or ";
            }
        }
        System.out.printf("The %d game.%nEach player chooses to add %s to a running total.%n" +
                "The player whose turn it is when the total reaches %d will win the game.%n" +
                "Winner of the game starts the next game.  Enter q to quit.%n%n", max, comma, max);
        int cGames = 0;
        int hGames = 0;
        boolean anotherGame = true;
        try (Scanner scanner = new Scanner(System.in);) {
            while ( anotherGame ) {
                Random r = new Random();
                int round = 0;
                int total = 0;
                System.out.printf("Start game %d%n", hGames + cGames + 1);
                DONE:
                    while ( true ) {
                        round++;
                        System.out.printf("ROUND %d:%n%n", round);
                        for ( int play = 0 ; play < 2 ; play++ ) {
                            if ( computerPlay ) {
                                int guess = 0;
                                //  try find one equal
                                for ( int test : valid ) {
                                    if ( total + test == max ) {
                                        guess = test;
                                        break;
                                    }
                                }
                                //  try find one greater than
                                if ( guess == 0 ) {
                                    for ( int test : valid ) {
                                        if ( total + test >= max ) {
                                            guess = test;
                                            break;
                                        }
                                    }
                                }
                                if ( guess == 0 ) {
                                    guess = valid[r.nextInt(valid.length)];
                                }
                                total += guess;
                                System.out.printf("The computer chooses %d%n", guess);
                                System.out.printf("Running total is now %d%n%n", total);
                                if ( total >= max ) {
                                    break DONE;
                                }
                            }
                            else {
                                while ( true ) {
                                    System.out.printf("Your choice among %s: ", comma);
                                    String line = scanner.nextLine();
                                    if ( line.matches("^[qQ].*") ) {
                                        System.out.printf("Computer wins %d game%s, human wins %d game%s.  One game incomplete.%nQuitting.%n", cGames, cGames == 1 ? "" : "s", hGames, hGames == 1 ? "" : "s");
                                        return;
                                    }
                                    try {
                                        int input = Integer.parseInt(line);
                                        boolean inputOk = false;
                                        for ( int test : valid ) {
                                            if ( input == test ) {
                                                inputOk = true;
                                                break;
                                            }
                                        }
                                        if ( inputOk ) {
                                            total += input;
                                            System.out.printf("Running total is now %d%n%n", total);
                                            if ( total >= max ) {
                                                break DONE;
                                            }
                                            break;
                                        }
                                        else {
                                            System.out.printf("Invalid input - must be a number among %s.  Try again.%n", comma);
                                        }
                                    }
                                    catch (NumberFormatException e) {
                                        System.out.printf("Invalid input - must be a number among %s.  Try again.%n", comma);
                                    }
                                }
                            }
                            computerPlay = !computerPlay;
                        }
                    }
                String win;
                if ( computerPlay ) {
                    win = "Computer wins!!";
                    cGames++;
                }
                else {
                    win = "You win and probably had help from another computer!!";
                    hGames++;
                }
                System.out.printf("%s%n", win);
                System.out.printf("Computer wins %d game%s, human wins %d game%s%n%n", cGames, cGames == 1 ? "" : "s", hGames, hGames == 1 ? "" : "s");
                while ( true ) {
                    System.out.printf("Another game (y/n)? ");
                    String line = scanner.nextLine();
                    if ( line.matches("^[yY]$") ) {
                        //  OK
                        System.out.printf("%n");
                        break;
                    }
                    else if ( line.matches("^[nN]$") ) {
                        anotherGame = false;
                        System.out.printf("Quitting.%n");
                        break;
                    }
                    else {
                        System.out.printf("Invalid input - must be a y or n.  Try again.%n");
                    }
                }
            }
        }
    }

}


24 game

The 24 Game tests one's mental arithmetic.

;Task
Write a program that displays four digits, each from 1 --> 9 (inclusive) with repetitions allowed.

The 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.

The goal is for the player to enter an expression that (numerically) evaluates to '''24'''.
*  Only the following operators/functions are allowed: multiplication, division, addition, subtraction
*  Division should use floating point or rational arithmetic, etc, to preserve remainders.
*  Brackets are allowed, if using an infix expression evaluator.
*  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).
*  The order of the digits when given does not have to be preserved.

;Notes
* The type of expression evaluator used is not mandated. An RPN evaluator is equally acceptable for example.
* The task is not for the program to generate the expression, or test whether an expression is even possible.

;Related tasks
* [[24 game/Solve]]

;Reference
* The 24 Game on h2g2.

Solution:
import java.util.*;

public class Game24 {
    static Random r = new Random();

    public static void main(String[] args) {

        int[] digits = randomDigits();
        Scanner in = new Scanner(System.in);

        System.out.print("Make 24 using these digits: ");
        System.out.println(Arrays.toString(digits));
        System.out.print("> ");

        Stack<Float> s = new Stack<>();
        long total = 0;
        for (char c : in.nextLine().toCharArray()) {
            if ('0' <= c && c <= '9') {
                int d = c - '0';
                total += (1 << (d * 5));
                s.push((float) d);
            } else if ("+/-*".indexOf(c) != -1) {
                s.push(applyOperator(s.pop(), s.pop(), c));
            }
        }
        if (tallyDigits(digits) != total)
            System.out.print("Not the same digits. ");
        else if (Math.abs(24 - s.peek()) < 0.001F)
            System.out.println("Correct!");
        else
            System.out.print("Not correct.");
    }

    static float applyOperator(float a, float b, char c) {
        switch (c) {
            case '+':
                return a + b;
            case '-':
                return b - a;
            case '*':
                return a * b;
            case '/':
                return b / a;
            default:
                return Float.NaN;
        }
    }

    static long tallyDigits(int[] a) {
        long total = 0;
        for (int i = 0; i < 4; i++)
            total += (1 << (a[i] * 5));
        return total;
    }

    static int[] randomDigits() {
        int[] result = new int[4];
        for (int i = 0; i < 4; i++)
            result[i] = r.nextInt(9) + 1;
        return result;
    }
}


24 game/Solve

Write a program that takes four digits, either from user input or by random generation, and computes arithmetic expressions following the rules of the [[24 game]].

Show examples of solutions generated by the program.

;Related task:
*   [[Arithmetic Evaluator]]

Solution:
import java.util.*;

public class Game24Player {
    final String[] patterns = {"nnonnoo", "nnonono", "nnnoono", "nnnonoo",
        "nnnnooo"};
    final String ops = "+-*/^";

    String solution;
    List<Integer> digits;

    public static void main(String[] args) {
        new Game24Player().play();
    }

    void play() {
        digits = getSolvableDigits();

        Scanner in = new Scanner(System.in);
        while (true) {
            System.out.print("Make 24 using these digits: ");
            System.out.println(digits);
            System.out.println("(Enter 'q' to quit, 's' for a solution)");
            System.out.print("> ");

            String line = in.nextLine();
            if (line.equalsIgnoreCase("q")) {
                System.out.println("\nThanks for playing");
                return;
            }

            if (line.equalsIgnoreCase("s")) {
                System.out.println(solution);
                digits = getSolvableDigits();
                continue;
            }

            char[] entry = line.replaceAll("[^*+-/)(\\d]", "").toCharArray();

            try {
                validate(entry);

                if (evaluate(infixToPostfix(entry))) {
                    System.out.println("\nCorrect! Want to try another? ");
                    digits = getSolvableDigits();
                } else {
                    System.out.println("\nNot correct.");
                }

            } catch (Exception e) {
                System.out.printf("%n%s Try again.%n", e.getMessage());
            }
        }
    }

    void validate(char[] input) throws Exception {
        int total1 = 0, parens = 0, opsCount = 0;

        for (char c : input) {
            if (Character.isDigit(c))
                total1 += 1 << (c - '0') * 4;
            else if (c == '(')
                parens++;
            else if (c == ')')
                parens--;
            else if (ops.indexOf(c) != -1)
                opsCount++;
            if (parens < 0)
                throw new Exception("Parentheses mismatch.");
        }

        if (parens != 0)
            throw new Exception("Parentheses mismatch.");

        if (opsCount != 3)
            throw new Exception("Wrong number of operators.");

        int total2 = 0;
        for (int d : digits)
            total2 += 1 << d * 4;

        if (total1 != total2)
            throw new Exception("Not the same digits.");
    }

    boolean evaluate(char[] line) throws Exception {
        Stack<Float> s = new Stack<>();
        try {
            for (char c : line) {
                if ('0' <= c && c <= '9')
                    s.push((float) c - '0');
                else
                    s.push(applyOperator(s.pop(), s.pop(), c));
            }
        } catch (EmptyStackException e) {
            throw new Exception("Invalid entry.");
        }
        return (Math.abs(24 - s.peek()) < 0.001F);
    }

    float applyOperator(float a, float b, char c) {
        switch (c) {
            case '+':
                return a + b;
            case '-':
                return b - a;
            case '*':
                return a * b;
            case '/':
                return b / a;
            default:
                return Float.NaN;
        }
    }

    List<Integer> randomDigits() {
        Random r = new Random();
        List<Integer> result = new ArrayList<>(4);
        for (int i = 0; i < 4; i++)
            result.add(r.nextInt(9) + 1);
        return result;
    }

    List<Integer> getSolvableDigits() {
        List<Integer> result;
        do {
            result = randomDigits();
        } while (!isSolvable(result));
        return result;
    }

    boolean isSolvable(List<Integer> digits) {
        Set<List<Integer>> dPerms = new HashSet<>(4 * 3 * 2);
        permute(digits, dPerms, 0);

        int total = 4 * 4 * 4;
        List<List<Integer>> oPerms = new ArrayList<>(total);
        permuteOperators(oPerms, 4, total);

        StringBuilder sb = new StringBuilder(4 + 3);

        for (String pattern : patterns) {
            char[] patternChars = pattern.toCharArray();

            for (List<Integer> dig : dPerms) {
                for (List<Integer> opr : oPerms) {

                    int i = 0, j = 0;
                    for (char c : patternChars) {
                        if (c == 'n')
                            sb.append(dig.get(i++));
                        else
                            sb.append(ops.charAt(opr.get(j++)));
                    }

                    String candidate = sb.toString();
                    try {
                        if (evaluate(candidate.toCharArray())) {
                            solution = postfixToInfix(candidate);
                            return true;
                        }
                    } catch (Exception ignored) {
                    }
                    sb.setLength(0);
                }
            }
        }
        return false;
    }

    String postfixToInfix(String postfix) {
        class Expression {
            String op, ex;
            int prec = 3;

            Expression(String e) {
                ex = e;
            }

            Expression(String e1, String e2, String o) {
                ex = String.format("%s %s %s", e1, o, e2);
                op = o;
                prec = ops.indexOf(o) / 2;
            }
        }

        Stack<Expression> expr = new Stack<>();

        for (char c : postfix.toCharArray()) {
            int idx = ops.indexOf(c);
            if (idx != -1) {

                Expression r = expr.pop();
                Expression l = expr.pop();

                int opPrec = idx / 2;

                if (l.prec < opPrec)
                    l.ex = '(' + l.ex + ')';

                if (r.prec <= opPrec)
                    r.ex = '(' + r.ex + ')';

                expr.push(new Expression(l.ex, r.ex, "" + c));
            } else {
                expr.push(new Expression("" + c));
            }
        }
        return expr.peek().ex;
    }

    char[] infixToPostfix(char[] infix) throws Exception {
        StringBuilder sb = new StringBuilder();
        Stack<Integer> s = new Stack<>();
        try {
            for (char c : infix) {
                int idx = ops.indexOf(c);
                if (idx != -1) {
                    if (s.isEmpty())
                        s.push(idx);
                    else {
                        while (!s.isEmpty()) {
                            int prec2 = s.peek() / 2;
                            int prec1 = idx / 2;
                            if (prec2 >= prec1)
                                sb.append(ops.charAt(s.pop()));
                            else
                                break;
                        }
                        s.push(idx);
                    }
                } else if (c == '(') {
                    s.push(-2);
                } else if (c == ')') {
                    while (s.peek() != -2)
                        sb.append(ops.charAt(s.pop()));
                    s.pop();
                } else {
                    sb.append(c);
                }
            }
            while (!s.isEmpty())
                sb.append(ops.charAt(s.pop()));

        } catch (EmptyStackException e) {
            throw new Exception("Invalid entry.");
        }
        return sb.toString().toCharArray();
    }

    void permute(List<Integer> lst, Set<List<Integer>> res, int k) {
        for (int i = k; i < lst.size(); i++) {
            Collections.swap(lst, i, k);
            permute(lst, res, k + 1);
            Collections.swap(lst, k, i);
        }
        if (k == lst.size())
            res.add(new ArrayList<>(lst));
    }

    void permuteOperators(List<List<Integer>> res, int n, int total) {
        for (int i = 0, npow = n * n; i < total; i++)
            res.add(Arrays.asList((i / npow), (i % npow) / n, i % n));
    }
}


4-rings or 4-squares puzzle

Replace       '''a, b, c, d, e, f,'''   and
  '''g '''       with the decimal
digits   LOW   --->   HIGH
such that the sum of the letters inside of each of the four large squares add up to
the same sum.

            +--------------+      +--------------+
            |              |      |              |
            |      a       |      |      e       |
            |              |      |              |
            |          +---+------+---+      +---+---------+
            |          |   |      |   |      |   |         |
            |          | b |      | d |      | f |         |
            |          |   |      |   |      |   |         |
            |          |   |      |   |      |   |         |
            +----------+---+      +---+------+---+         |
                       |       c      |      |      g      |
                       |              |      |             |
                       |              |      |             |
                       +--------------+      +-------------+

Show all output here.

:*   Show all solutions for each letter being unique with
         LOW=1     HIGH=7
:*   Show all solutions for each letter being unique with
         LOW=3     HIGH=9
:*   Show only the   ''number''   of solutions when each letter can be non-unique
         LOW=0     HIGH=9

;Related task:
* [[Solve the no connection puzzle]]

Solution:
import java.util.Arrays;

public class FourSquares {
    public static void main(String[] args) {
        fourSquare(1, 7, true, true);
        fourSquare(3, 9, true, true);
        fourSquare(0, 9, false, false);
    }

    private static void fourSquare(int low, int high, boolean unique, boolean print) {
        int count = 0;

        if (print) {
            System.out.println("a b c d e f g");
        }
        for (int a = low; a <= high; ++a) {
            for (int b = low; b <= high; ++b) {
                if (notValid(unique, a, b)) continue;

                int fp = a + b;
                for (int c = low; c <= high; ++c) {
                    if (notValid(unique, c, a, b)) continue;
                    for (int d = low; d <= high; ++d) {
                        if (notValid(unique, d, a, b, c)) continue;
                        if (fp != b + c + d) continue;

                        for (int e = low; e <= high; ++e) {
                            if (notValid(unique, e, a, b, c, d)) continue;
                            for (int f = low; f <= high; ++f) {
                                if (notValid(unique, f, a, b, c, d, e)) continue;
                                if (fp != d + e + f) continue;

                                for (int g = low; g <= high; ++g) {
                                    if (notValid(unique, g, a, b, c, d, e, f)) continue;
                                    if (fp != f + g) continue;

                                    ++count;
                                    if (print) {
                                        System.out.printf("%d %d %d %d %d %d %d%n", a, b, c, d, e, f, g);
                                    }
                                }
                            }
                        }
                    }
                }
            }
        }
        if (unique) {
            System.out.printf("There are %d unique solutions in [%d, %d]%n", count, low, high);
        } else {
            System.out.printf("There are %d non-unique solutions in [%d, %d]%n", count, low, high);
        }
    }

    private static boolean notValid(boolean unique, int needle, int... haystack) {
        return unique && Arrays.stream(haystack).anyMatch(p -> p == needle);
    }
}


99 bottles of beer

Display the complete lyrics for the song:     '''99 Bottles of Beer on the Wall'''.

;The beer song:
The lyrics follow this form:

:::              99 bottles of beer on the wall
:::              99 bottles of beer
:::              Take one down, pass it around
:::              98 bottles of beer on the wall

:::              98 bottles of beer on the wall
:::              98 bottles of beer
:::              Take one down, pass it around
:::              97 bottles of beer on the wall

... and so on, until reaching   '''0'''     (zero).

Grammatical support for   ''1 bottle of beer''   is optional.

As with any puzzle, try to do it in as creative/concise/comical a way
as possible (simple, obvious solutions allowed, too).

;See also:
*   http://99-bottles-of-beer.net/
*   [[:Category:99_Bottles_of_Beer]]
*   [[:Category:Programming language families]]
*   Wikipedia 99 bottles of beer

Solution:
public class Beer {
    public static void main(String args[]) {
        song(99);
    }

    public static void song(int bottles) {
        if (bottles >= 0) {
            if (bottles > 1)
                System.out.println(bottles + " bottles of beer on the wall\n" + bottles + " bottles of beer\nTake one down, pass it around\n" + (bottles - 1) + " bottles of beer on the wall.\n");
            else if (bottles == 1)
                System.out.println(bottles + " bottle of beer on the wall\n" + bottles + " bottle of beer\nTake one down, pass it around\n" + (bottles - 1) + " bottles of beer on the wall.\n");
            else
                System.out.println(bottles + " bottles of beer on the wall\n" + bottles + " bottles of beer\nBetter go to the store and buy some more!");
            song(bottles - 1);
        }
    }
}


99 bottles of beer

Solution:
See: [[99 Bottles of Beer/Java/Object Oriented]]

===GUI===


9 billion names of God the integer

This task is a variation of the short story by Arthur C. Clarke.

(Solvers should be aware of the consequences of completing this task.)

In detail, to specify what is meant by a   "name":
:The integer 1 has 1 name       "1".
:The integer 2 has 2 names   "1+1",   and   "2".
:The integer 3 has 3 names   "1+1+1",   "2+1",   and   "3".
:The integer 4 has 5 names   "1+1+1+1",   "2+1+1",   "2+2",   "3+1",   "4".
: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".

;Task
Display the first 25 rows of a number triangle which begins:

                                      1
                                    1   1
                                  1   1   1
                                1   2   1   1
                              1   2   2   1   1
                            1   3   3   2   1   1

Where 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.

A function   G(n)   should return the sum of the   n-th   row.

Demonstrate this function by displaying:   G(23),   G(123),   G(1234),   and   G(12345).

Optionally note that the sum of the   n-th   row   P(n)   is the      integer partition function.

Demonstrate this is equivalent to   G(n)   by displaying:   P(23),   P(123),   P(1234),   and   P(12345).

;Extra credit

If your environment is able, plot   P(n)   against   n   for   n=1\ldots 999.

;Related tasks
* [[Partition function P]]

Solution: Translation of [[9_billion_names_of_God_the_integer#Python|Python]] via [[9_billion_names_of_God_the_integer#D|D]]


9 billion names of God the integer

Solution:
import java.math.BigInteger;
import java.util.*;
import static java.util.Arrays.asList;
import static java.util.stream.Collectors.toList;
import static java.util.stream.IntStream.range;
import static java.lang.Math.min;

public class Test {

    static List<BigInteger> cumu(int n) {
        List<List<BigInteger>> cache = new ArrayList<>();
        cache.add(asList(BigInteger.ONE));

        for (int L = cache.size(); L < n + 1; L++) {
            List<BigInteger> r = new ArrayList<>();
            r.add(BigInteger.ZERO);
            for (int x = 1; x < L + 1; x++)
                r.add(r.get(r.size() - 1).add(cache.get(L - x).get(min(x, L - x))));
            cache.add(r);
        }
        return cache.get(n);
    }

    static List<BigInteger> row(int n) {
        List<BigInteger> r = cumu(n);
        return range(0, n).mapToObj(i -> r.get(i + 1).subtract(r.get(i)))
                .collect(toList());
    }

    public static void main(String[] args) {
        System.out.println("Rows:");
        for (int x = 1; x < 11; x++)
            System.out.printf("%2d: %s%n", x, row(x));

        System.out.println("\nSums:");
        for (int x : new int[]{23, 123, 1234}) {
            List<BigInteger> c = cumu(x);
            System.out.printf("%s %s%n", x, c.get(c.size() - 1));
        }
    }
}


A+B

'''A+B'''   --- a classic problem in programming contests,   it's given so contestants can gain familiarity with the online judging system being used.

;Task:
Given two integers,   '''A''' and '''B'''.

Their sum needs to be calculated.

;Input data:
Two integers are written in the input stream, separated by space(s):
: (-1000 \le A,B \le +1000)

;Output data:
The required output is one integer:   the sum of '''A''' and '''B'''.

;Example:
::{|class="standard"
 ! input
 ! output
 |-
 | 2 2
 | 4
 |-
 | 3 2
 | 5
 |}

Solution:
import java.io.*;
import java.util.*;

public class SumDif {
   StreamTokenizer in;
   PrintWriter out;

   public static void main(String[] args) throws IOException {
      new SumDif().run();
   }

   private int nextInt() throws IOException {
      in.nextToken();
      return (int)in.nval;
   }

   public void run() throws IOException {
      in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in))); // Standard input
      out = new PrintWriter(new OutputStreamWriter(System.out)); // Standard output
      solve();
      out.flush();
   }

   private void solve() throws IOException {
      out.println(nextInt() + nextInt());
   }
}


ABC problem

You are given a collection of ABC blocks   (maybe like the ones you had when you were a kid).

There are twenty blocks with two letters on each block.

A complete alphabet is guaranteed amongst all sides of the blocks.

The sample collection of blocks:
  (B O)
  (X K)
  (D Q)
  (C P)
  (N A)
  (G T)
  (R E)
  (T G)
  (Q D)
  (F S)
  (J W)
  (H U)
  (V I)
  (A N)
  (O B)
  (E R)
  (F S)
  (L Y)
  (P C)
  (Z M)

;Task:
Write a function that takes a string (word) and determines whether the word can be spelled with the given collection of blocks.

The rules are simple:
::#   Once a letter on a block is used that block cannot be used again
::#   The function should be case-insensitive
::#   Show the output on this page for the following 7 words in the following example

;Example:
    >>> can_make_word("A")
    True
    >>> can_make_word("BARK")
    True
    >>> can_make_word("BOOK")
    False
    >>> can_make_word("TREAT")
    True
    >>> can_make_word("COMMON")
    False
    >>> can_make_word("SQUAD")
    True
    >>> can_make_word("CONFUSE")
    True

Solution:
import java.util.Arrays;
import java.util.Collections;
import java.util.List;

public class ABC {

    public static void main(String[] args) {
        List<String> blocks = Arrays.asList(
                "BO", "XK", "DQ", "CP", "NA",
                "GT", "RE", "TG", "QD", "FS",
                "JW", "HU", "VI", "AN", "OB",
                "ER", "FS", "LY", "PC", "ZM");

        for (String word : Arrays.asList("", "A", "BARK", "BOOK", "TREAT", "COMMON", "SQUAD", "CONFUSE")) {
            System.out.printf("%s: %s%n", word.isEmpty() ? "\"\"" : word, canMakeWord(word, blocks));
        }
    }

    public static boolean canMakeWord(String word, List<String> blocks) {
        if (word.isEmpty())
            return true;

        char c = word.charAt(0);
        for (int i = 0; i < blocks.size(); i++) {
            String b = blocks.get(i);
            if (b.charAt(0) != c && b.charAt(1) != c)
                continue;
            Collections.swap(blocks, 0, i);
            if (canMakeWord(word.substring(1), blocks.subList(1, blocks.size())))
                return true;
            Collections.swap(blocks, 0, i);
        }

        return false;
    }
}


ASCII art diagram converter

Given the RFC 1035 message diagram from Section 4.1.1 (Header section format) as a string:
http://www.ietf.org/rfc/rfc1035.txt

 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
 |                      ID                       |
 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
 |QR|   Opcode  |AA|TC|RD|RA|   Z    |   RCODE   |
 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
 |                    QDCOUNT                    |
 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
 |                    ANCOUNT                    |
 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
 |                    NSCOUNT                    |
 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+
 |                    ARCOUNT                    |
 +--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+

Where (every column of the table is 1 bit):

 ID is 16 bits
 QR = Query (0) or Response (1)
 Opcode = Four bits defining kind of query:
   0:    a standard query (QUERY)
   1:    an inverse query (IQUERY)
   2:    a server status request (STATUS)
   3-15: reserved for future use
 AA = Authoritative Answer bit
 TC = Truncation bit
 RD = Recursion Desired bit
 RA = Recursion Available bit
 Z = Reserved
 RCODE = Response code
 QC = Question Count
 ANC = Answer Count
 AUC = Authority Count
 ADC = Additional Count

Write 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.

If 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.

Such "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.

Bonus: perform a thoroughly validation of the input string.

Solution:
import java.math.BigInteger;
import java.util.ArrayList;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;

public class AsciiArtDiagramConverter {

    private static final String TEST = "+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\r\n" +
            "|                      ID                       |\r\n" +
            "+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\r\n" +
            "|QR|   Opcode  |AA|TC|RD|RA|   Z    |   RCODE   |\r\n" +
            "+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\r\n" +
            "|                    QDCOUNT                    |\r\n" +
            "+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\r\n" +
            "|                    ANCOUNT                    |\r\n" +
            "+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\r\n" +
            "|                    NSCOUNT                    |\r\n" +
            "+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+\r\n" +
            "|                    ARCOUNT                    |\r\n" +
            "+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+--+";

    public static void main(String[] args) {
        validate(TEST);
        display(TEST);
        Map<String,List<Integer>> asciiMap = decode(TEST);
        displayMap(asciiMap);
        displayCode(asciiMap, "78477bbf5496e12e1bf169a4");
    }

    private static void displayCode(Map<String,List<Integer>> asciiMap, String hex) {
        System.out.printf("%nTest string in hex:%n%s%n%n", hex);

        String bin = new BigInteger(hex,16).toString(2);

        //  Zero pad in front as needed
        int length = 0;
        for ( String code : asciiMap.keySet() ) {
            List<Integer> pos = asciiMap.get(code);
            length += pos.get(1) - pos.get(0) + 1;
        }
        while ( length > bin.length() ) {
            bin = "0" + bin;
        }
        System.out.printf("Test string in binary:%n%s%n%n", bin);

        System.out.printf("Name      Size  Bit Pattern%n");
        System.out.printf("-------- -----  -----------%n");
        for ( String code : asciiMap.keySet() ) {
            List<Integer> pos = asciiMap.get(code);
            int start = pos.get(0);
            int end   = pos.get(1);
            System.out.printf("%-8s    %2d  %s%n", code, end-start+1, bin.substring(start, end+1));
        }

    }

    private static void display(String ascii) {
        System.out.printf("%nDiagram:%n%n");
        for ( String s : TEST.split("\\r\\n") ) {
            System.out.println(s);
        }
    }

    private static void displayMap(Map<String,List<Integer>> asciiMap) {
        System.out.printf("%nDecode:%n%n");

        System.out.printf("Name      Size  Start    End%n");
        System.out.printf("-------- -----  -----  -----%n");
        for ( String code : asciiMap.keySet() ) {
            List<Integer> pos = asciiMap.get(code);
            System.out.printf("%-8s    %2d     %2d     %2d%n", code, pos.get(1)-pos.get(0)+1, pos.get(0), pos.get(1));
        }

    }

    private static Map<String,List<Integer>> decode(String ascii) {
        Map<String,List<Integer>> map = new LinkedHashMap<>();
        String[] split = TEST.split("\\r\\n");
        int size = split[0].indexOf("+", 1) - split[0].indexOf("+");
        int length = split[0].length() - 1;
        for ( int i = 1 ; i < split.length ; i += 2 ) {
            int barIndex = 1;
            String test = split[i];
            int next;
            while ( barIndex < length && (next = test.indexOf("|", barIndex)) > 0 ) {
                //  List is start and end of code.
                List<Integer> startEnd = new ArrayList<>();
                startEnd.add((barIndex/size) + (i/2)*(length/size));
                startEnd.add(((next-1)/size) + (i/2)*(length/size));
                String code = test.substring(barIndex, next).replace(" ", "");
                map.put(code, startEnd);
                //  Next bar
                barIndex = next + 1;
            }
        }

        return map;
    }

    private static void validate(String ascii) {
        String[] split = TEST.split("\\r\\n");
        if ( split.length % 2 != 1 ) {
            throw new RuntimeException("ERROR 1:  Invalid number of input lines.  Line count = " + split.length);
        }
        int size = 0;
        for ( int i = 0 ; i < split.length ; i++ ) {
            String test = split[i];
            if ( i % 2 == 0 ) {
                //  Start with +, an equal number of -, end with +
                if ( ! test.matches("^\\+([-]+\\+)+$") ) {
                    throw new RuntimeException("ERROR 2:  Improper line format.  Line = " + test);
                }
                if ( size == 0 ) {
                    int firstPlus = test.indexOf("+");
                    int secondPlus = test.indexOf("+", 1);
                    size = secondPlus - firstPlus;
                }
                if ( ((test.length()-1) % size) != 0 ) {
                    throw new RuntimeException("ERROR 3:  Improper line format.  Line = " + test);
                }
                //  Equally spaced splits of +, -
                for ( int j = 0 ; j < test.length()-1 ; j += size ) {
                    if ( test.charAt(j) != '+' ) {
                        throw new RuntimeException("ERROR 4:  Improper line format.  Line = " + test);
                    }
                    for ( int k = j+1 ; k < j + size ; k++ ) {
                        if ( test.charAt(k) != '-' ) {
                            throw new RuntimeException("ERROR 5:  Improper line format.  Line = " + test);
                        }
                    }
                }
            }
            else {
                //  Vertical bar, followed by optional spaces, followed by name, followed by optional spaces, followed by vdrtical bar
                if ( ! test.matches("^\\|(\\s*[A-Za-z]+\\s*\\|)+$") ) {
                    throw new RuntimeException("ERROR 6:  Improper line format.  Line = " + test);
                }
                for ( int j = 0 ; j < test.length()-1 ; j += size ) {
                    for ( int k = j+1 ; k < j + size ; k++ ) {
                        //  Vertical bar only at boundaries
                        if ( test.charAt(k) == '|' ) {
                            throw new RuntimeException("ERROR 7:  Improper line format.  Line = " + test);
                        }
                    }
                }

            }
        }
    }

}


AVL tree

{{wikipedia|AVL tree}}

In 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.

AVL 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.

;Task:
Implement an AVL tree in the language of choice, and provide at least basic operations.

;Related task
[[Red_black_tree_sort]]

Solution:
public class AVLtree {

    private Node root;

    private static class Node {
        private int key;
        private int balance;
        private int height;
        private Node left;
        private Node right;
        private Node parent;

        Node(int key, Node parent) {
            this.key = key;
            this.parent = parent;
        }
    }

    public boolean insert(int key) {
        if (root == null) {
            root = new Node(key, null);
            return true;
        }

        Node n = root;
        while (true) {
            if (n.key == key)
                return false;

            Node parent = n;

            boolean goLeft = n.key > key;
            n = goLeft ? n.left : n.right;

            if (n == null) {
                if (goLeft) {
                    parent.left = new Node(key, parent);
                } else {
                    parent.right = new Node(key, parent);
                }
                rebalance(parent);
                break;
            }
        }
        return true;
    }

    private void delete(Node node) {
        if (node.left == null && node.right == null) {
            if (node.parent == null) {
                root = null;
            } else {
                Node parent = node.parent;
                if (parent.left == node) {
                    parent.left = null;
                } else {
                    parent.right = null;
                }
                rebalance(parent);
            }
            return;
        }

        if (node.left != null) {
            Node child = node.left;
            while (child.right != null) child = child.right;
            node.key = child.key;
            delete(child);
        } else {
            Node child = node.right;
            while (child.left != null) child = child.left;
            node.key = child.key;
            delete(child);
        }
    }

    public void delete(int delKey) {
        if (root == null)
            return;

        Node child = root;
        while (child != null) {
            Node node = child;
            child = delKey >= node.key ? node.right : node.left;
            if (delKey == node.key) {
                delete(node);
                return;
            }
        }
    }

    private void rebalance(Node n) {
        setBalance(n);

        if (n.balance == -2) {
            if (height(n.left.left) >= height(n.left.right))
                n = rotateRight(n);
            else
                n = rotateLeftThenRight(n);

        } else if (n.balance == 2) {
            if (height(n.right.right) >= height(n.right.left))
                n = rotateLeft(n);
            else
                n = rotateRightThenLeft(n);
        }

        if (n.parent != null) {
            rebalance(n.parent);
        } else {
            root = n;
        }
    }

    private Node rotateLeft(Node a) {

        Node b = a.right;
        b.parent = a.parent;

        a.right = b.left;

        if (a.right != null)
            a.right.parent = a;

        b.left = a;
        a.parent = b;

        if (b.parent != null) {
            if (b.parent.right == a) {
                b.parent.right = b;
            } else {
                b.parent.left = b;
            }
        }

        setBalance(a, b);

        return b;
    }

    private Node rotateRight(Node a) {

        Node b = a.left;
        b.parent = a.parent;

        a.left = b.right;

        if (a.left != null)
            a.left.parent = a;

        b.right = a;
        a.parent = b;

        if (b.parent != null) {
            if (b.parent.right == a) {
                b.parent.right = b;
            } else {
                b.parent.left = b;
            }
        }

        setBalance(a, b);

        return b;
    }

    private Node rotateLeftThenRight(Node n) {
        n.left = rotateLeft(n.left);
        return rotateRight(n);
    }

    private Node rotateRightThenLeft(Node n) {
        n.right = rotateRight(n.right);
        return rotateLeft(n);
    }

    private int height(Node n) {
        if (n == null)
            return -1;
        return n.height;
    }

    private void setBalance(Node... nodes) {
        for (Node n : nodes) {
            reheight(n);
            n.balance = height(n.right) - height(n.left);
        }
    }

    public void printBalance() {
        printBalance(root);
    }

    private void printBalance(Node n) {
        if (n != null) {
            printBalance(n.left);
            System.out.printf("%s ", n.balance);
            printBalance(n.right);
        }
    }

    private void reheight(Node node) {
        if (node != null) {
            node.height = 1 + Math.max(height(node.left), height(node.right));
        }
    }

    public static void main(String[] args) {
        AVLtree tree = new AVLtree();

        System.out.println("Inserting values 1 to 10");
        for (int i = 1; i < 10; i++)
            tree.insert(i);

        System.out.print("Printing balance: ");
        tree.printBalance();
    }
}


Abbreviations, automatic

The use of   abbreviations    (also sometimes called synonyms, nicknames, AKAs, or aliases)    can be an
easy way to add flexibility when specifying or using commands, sub-commands, options, etc.

It would make a list of words easier to maintain   (as words are added, changed, and/or deleted)   if
the minimum abbreviation length of that list could be automatically (programmatically) determined.

For 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).

Sunday Monday Tuesday Wednesday Thursday Friday Saturday
Sondag Maandag Dinsdag Woensdag Donderdag Vrydag Saterdag
E_djele E_hene E_marte E_merkure E_enjte E_premte E_shtune
Ehud Segno Maksegno Erob Hamus Arbe Kedame
Al_Ahad Al_Ithinin Al_Tholatha'a Al_Arbia'a Al_Kamis Al_Gomia'a Al_Sabit
Guiragui Yergou_shapti Yerek_shapti Tchorek_shapti Hink_shapti Ourpat Shapat
domingu llunes martes miercoles xueves vienres sabadu
Bazar_gUnU Birinci_gUn Ckinci_gUn UcUncU_gUn DOrdUncU_gUn Bes,inci_gUn Altonco_gUn
Igande Astelehen Astearte Asteazken Ostegun Ostiral Larunbat
Robi_bar Shom_bar Mongal_bar Budhh_bar BRihashpati_bar Shukro_bar Shoni_bar
Nedjelja Ponedeljak Utorak Srijeda Cxetvrtak Petak Subota
Disul Dilun Dimeurzh Dimerc'her Diriaou Digwener Disadorn
nedelia ponedelnik vtornik sriada chetvartak petak sabota
sing_kei_yaht sing_kei_yat sing_kei_yee sing_kei_saam sing_kei_sie sing_kei_ng sing_kei_luk
Diumenge Dilluns Dimarts Dimecres Dijous Divendres Dissabte
Dzeenkk-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
dy_Sul dy_Lun dy_Meurth dy_Mergher dy_You dy_Gwener dy_Sadorn
Dimanch Lendi Madi Mekredi Jedi Vandredi Samdi
nedjelja ponedjeljak utorak srijeda cxetvrtak petak subota
nede^le ponde^li utery str^eda c^tvrtek patek sobota
Sondee Mondee Tiisiday Walansedee TOOsedee Feraadee Satadee
s0ndag mandag tirsdag onsdag torsdag fredag l0rdag
zondag maandag dinsdag woensdag donderdag vrijdag zaterdag
Diman^co Lundo Mardo Merkredo ^Jaudo Vendredo Sabato
pUhapaev esmaspaev teisipaev kolmapaev neljapaev reede laupaev

Diu_prima Diu_sequima Diu_tritima Diu_quartima Diu_quintima Diu_sextima Diu_sabbata
sunnudagur manadagur tysdaguy mikudagur hosdagur friggjadagur leygardagur
Yek_Sham'beh Do_Sham'beh Seh_Sham'beh Cha'har_Sham'beh Panj_Sham'beh Jom'eh Sham'beh
sunnuntai maanantai tiistai keskiviiko torsktai perjantai lauantai
dimanche lundi mardi mercredi jeudi vendredi samedi
Snein Moandei Tiisdei Woansdei Tonersdei Freed Sneon
Domingo Segunda_feira Martes Mercores Joves Venres Sabado
k'vira orshabati samshabati otkhshabati khutshabati p'arask'evi shabati
Sonntag Montag Dienstag Mittwoch Donnerstag Freitag Samstag
Kiriaki' Defte'ra Tri'ti Teta'rti Pe'mpti Paraskebi' Sa'bato
ravivaar somvaar mangalvaar budhvaar guruvaar shukravaar shanivaar
popule po`akahi po`alua po`akolu po`aha po`alima po`aono
Yom_rishon Yom_sheni Yom_shlishi Yom_revi'i Yom_chamishi Yom_shishi Shabat
ravivara somavar mangalavar budhavara brahaspativar shukravara shanivar
vasarnap hetfo kedd szerda csutortok pentek szombat
Sunnudagur Manudagur +ridjudagur Midvikudagar Fimmtudagur FOstudagur Laugardagur
sundio lundio mardio merkurdio jovdio venerdio saturdio
Minggu Senin Selasa Rabu Kamis Jumat Sabtu
Dominica Lunedi Martedi Mercuridi Jovedi Venerdi Sabbato
De_Domhnaigh De_Luain De_Mairt De_Ceadaoin De_ardaoin De_hAoine De_Sathairn
domenica lunedi martedi mercoledi giovedi venerdi sabato
Nichiyou_bi Getzuyou_bi Kayou_bi Suiyou_bi Mokuyou_bi Kin'you_bi Doyou_bi
Il-yo-il Wol-yo-il Hwa-yo-il Su-yo-il Mok-yo-il Kum-yo-il To-yo-il
Dies_Dominica Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Saturni
sve-tdien pirmdien otrdien tresvdien ceturtdien piektdien sestdien
Sekmadienis Pirmadienis Antradienis Trec^iadienis Ketvirtadienis Penktadienis S^es^tadienis
Wangu Kazooba Walumbe Mukasa Kiwanuka Nnagawonye Wamunyi
xing-_qi-_ri xing-_qi-_yi-. xing-_qi-_er xing-_qi-_san-. xing-_qi-_si xing-_qi-_wuv. xing-_qi-_liu
Jedoonee Jelune Jemayrt Jecrean Jardaim Jeheiney Jesam
Jabot Manre Juje Wonje Taije Balaire Jarere
geminrongo minomishi martes mierkoles misheushi bernashi mishabaro
Ahad Isnin Selasa Rabu Khamis Jumaat Sabtu
sphndag mandag tirsdag onsdag torsdag fredag lphrdag
lo_dimenge lo_diluns lo_dimarc lo_dimercres lo_dijous lo_divendres lo_dissabte
djadomingo djaluna djamars djarason djaweps djabierna djasabra
Niedziela Poniedzial/ek Wtorek S,roda Czwartek Pia,tek Sobota
Domingo segunda-feire terca-feire quarta-feire quinta-feire sexta-feira sabado
Domingo Lunes martes Miercoles Jueves Viernes Sabado
Duminica Luni Mart'i Miercuri Joi Vineri Sambata
voskresenie ponedelnik vtornik sreda chetverg pyatnitsa subbota
Sunday Di-luain Di-mairt Di-ciadain Di-ardaoin Di-haoine Di-sathurne
nedjelja ponedjeljak utorak sreda cxetvrtak petak subota
Sontaha Mmantaha Labobedi Laboraro Labone Labohlano Moqebelo
Iridha- Sandhudha- Anga.haruwa-dha- Badha-dha- Brahaspa.thindha- Sikura-dha- Sena.sura-dha-
nedel^a pondelok utorok streda s^tvrtok piatok sobota
Nedelja Ponedeljek Torek Sreda Cxetrtek Petek Sobota
domingo lunes martes miercoles jueves viernes sabado
sonde mundey tude-wroko dride-wroko fode-wroko freyda Saturday
Jumapili Jumatatu Jumanne Jumatano Alhamisi Ijumaa Jumamosi
sondag mandag tisdag onsdag torsdag fredag lordag
Linggo Lunes Martes Miyerkoles Huwebes Biyernes Sabado
Le-pai-jit Pai-it Pai-ji Pai-san Pai-si Pai-gO. Pai-lak
wan-ar-tit wan-tjan wan-ang-kaan wan-phoet wan-pha-ru-hat-sa-boh-die wan-sook wan-sao
Tshipi Mosupologo Labobedi Laboraro Labone Labotlhano Matlhatso
Pazar Pazartesi Sali Car,samba Per,sembe Cuma Cumartesi
nedilya ponedilok vivtorok sereda chetver pyatnytsya subota
Chu?_Nha.t Thu*_Hai Thu*_Ba Thu*_Tu* Thu*_Na'm Thu*_Sau Thu*_Ba?y
dydd_Sul dyds_Llun dydd_Mawrth dyds_Mercher dydd_Iau dydd_Gwener dyds_Sadwrn
Dibeer Altine Talaata Allarba Al_xebes Aljuma Gaaw
iCawa uMvulo uLwesibini uLwesithathu uLuwesine uLwesihlanu uMgqibelo
zuntik montik dinstik mitvokh donershtik fraytik shabes
iSonto uMsombuluko uLwesibili uLwesithathu uLwesine uLwesihlanu uMgqibelo
Dies_Dominica Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Saturni
Bazar_gUnU Bazar_aertaesi Caers,aenbae_axs,amo Caers,aenbae_gUnU CUmae_axs,amo CUmae_gUnU CUmae_Senbae
Sun Moon Mars Mercury Jove Venus Saturn
zondag maandag dinsdag woensdag donderdag vrijdag zaterdag
KoseEraa GyoOraa BenEraa Kuoraa YOwaaraa FeEraa Memenaa
Sonntag Montag Dienstag Mittwoch Donnerstag Freitag Sonnabend
Domingo Luns Terza_feira Corta_feira Xoves Venres Sabado
Dies_Solis Dies_Lunae Dies_Martis Dies_Mercurii Dies_Iovis Dies_Veneris Dies_Sabbatum
xing-_qi-_tian xing-_qi-_yi-. xing-_qi-_er xing-_qi-_san-. xing-_qi-_si xing-_qi-_wuv. xing-_qi-_liu
djadomingu djaluna djamars djarason djaweps djabierne djasabra
Killachau Atichau Quoyllurchau Illapachau Chaskachau Kuychichau Intichau

''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.''

To 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).

Notes concerning the above list of words
::*   each line has a list of days-of-the-week for a language, separated by at least one blank
::*   the words on each line happen to be in order, from Sunday --> Saturday
::*   most lines have words in mixed case and some have all manner of accented words and other characters
::*   some words were translated to the nearest character that was available to ''code page''   '''437'''
::*   the characters in the words are not restricted except that they may not have imbedded blanks
::*   for this example, the use of an underscore ('''_''') was used to indicate a blank in a word

;Task:
::*   The list of words   (days of the week)   needn't be verified/validated.
::*   Write a function to find the (numeric) minimum length abbreviation for each line that would make abbreviations unique.
::*   A blank line   (or a null line)   should return a null string.
::*   Process and show the output for at least the first '''five''' lines of the file.
::*   Show all output here.

Solution:
import java.io.IOException;
import java.nio.file.Files;
import java.nio.file.Path;
import java.nio.file.Paths;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Abbreviations {
    public static void main(String[] args) throws IOException {
        Path path = Paths.get("days_of_week.txt");
        List<String> readAllLines = Files.readAllLines(path);
        for (int i = 0; i < readAllLines.size(); i++) {
            String line = readAllLines.get(i);
            if (line.length() == 0) continue;

            String[] days = line.split(" ");
            if (days.length != 7) throw new RuntimeException("There aren't 7 days on line " + (i + 1));

            Map<String, Integer> temp = new HashMap<>();
            for (String day : days) {
                Integer count = temp.getOrDefault(day, 0);
                temp.put(day, count + 1);
            }
            if (temp.size() < 7) {
                System.out.print(" ∞  ");
                System.out.println(line);
                continue;
            }

            int len = 1;
            while (true) {
                temp.clear();
                for (String day : days) {
                    String sd;
                    if (len >= day.length()) {
                        sd = day;
                    } else {
                        sd = day.substring(0, len);
                    }
                    Integer count = temp.getOrDefault(sd, 0);
                    temp.put(sd, count + 1);
                }
                if (temp.size() == 7) {
                    System.out.printf("%2d  %s\n", len, line);
                    break;
                }
                len++;
            }
        }
    }
}


Abbreviations, easy

This task is an easier (to code) variant of the Rosetta Code task:    [[Abbreviations, simple]].

For this task, the following   ''command table''   will be used:
    Add ALTer  BAckup Bottom  CAppend Change SCHANGE  CInsert CLAst COMPress COpy
    COUnt COVerlay CURsor DELete CDelete Down DUPlicate Xedit EXPand EXTract Find
    NFind NFINDUp NFUp CFind FINdup FUp FOrward GET Help HEXType Input POWerinput
    Join SPlit SPLTJOIN  LOAD  Locate CLocate  LOWercase UPPercase  LPrefix MACRO
    MErge MODify MOve MSG Next Overlay PARSE PREServe PURge PUT PUTD  Query  QUIT
    READ  RECover REFRESH RENum REPeat  Replace CReplace  RESet  RESTore  RGTLEFT
    RIght LEft  SAVE  SET SHift SI  SORT  SOS  STAck STATus  TOP TRAnsfer Type Up

Notes concerning the above   ''command table'':
::*   it can be thought of as one long literal string   (with blanks at end-of-lines)
::*   it may have superfluous blanks
::*   it may be in any case (lower/upper/mixed)
::*   the order of the words in the   ''command table''   must be preserved as shown
::*   the user input(s) may be in any case (upper/lower/mixed)
::*   commands will be restricted to the Latin alphabet   (A --> Z,   a --> z)
::*   A valid abbreviation is a word that has:
:::*   at least the minimum length of the number of capital letters of the word in the ''command table''
:::*   compares equal (regardless of case) to the leading characters of the word in the ''command table''
:::*   a length not longer than the word in the ''command table''
::::*   '''ALT''',   '''aLt''',   '''ALTE''',   and   '''ALTER'''   are all abbreviations of   '''ALTer'''
::::*   '''AL''',   '''ALF''',   '''ALTERS''',   '''TER''',   and   '''A'''    aren't valid abbreviations of   '''ALTer'''
::::*   The number of capital letters in   '''ALTer'''   indicates that any abbreviation for   '''ALTer'''   must be at least three letters
::::*   Any word longer than five characters can't be an abbreviation for   '''ALTer'''
::::*   '''o''',    '''ov''',    '''oVe''',    '''over''',    '''overL''',    '''overla'''   are all acceptable abbreviations for   '''Overlay'''
::*   if there isn't any lowercase letters in the word in the ''command table'',   then there isn't an abbreviation permitted

;Task:
::*   The command table needn't be verified/validated.
::*   Write a function to validate if the user "words"   (given as input)   are valid   (in the ''command table'').
::*   If the word   is   valid,   then return the full uppercase version of that "word".
::*   If the word        isn't     valid,   then return the lowercase string:    '''*error*'''        (7 characters).
::*   A blank input   (or a null input)   should return a null string.
::*   Show all output here.

;An example test case to be used for this task:
For a user string of:
  riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin
the computer program should return the string:
  RIGHT REPEAT *error* PUT MOVE RESTORE *error* *error* *error* POWERINPUT

Solution:
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class AbbreviationsEasy {
    private static final Scanner input = new Scanner(System.in);
    private static final String  COMMAND_TABLE
            =       "  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";

    public static void main(String[] args) {
        String[]             cmdTableArr = COMMAND_TABLE.split("\\s+");
        Map<String, Integer> cmd_table   = new HashMap<String, Integer>();

        for (String word : cmdTableArr) {  //Populate words and number of caps
            cmd_table.put(word, countCaps(word));
        }

        System.out.print("Please enter your command to verify: ");
        String   userInput  = input.nextLine();
        String[] user_input = userInput.split("\\s+");

        for (String s : user_input) {
            boolean match = false; //resets each outer loop
            for (String cmd : cmd_table.keySet()) {
                if (s.length() >= cmd_table.get(cmd) && s.length() <= cmd.length()) {
                    String temp = cmd.toUpperCase();
                    if (temp.startsWith(s.toUpperCase())) {
                        System.out.print(temp + " ");
                        match = true;
                    }
                }
            }
            if (!match) { //no match, print error msg
                System.out.print("*error* ");
            }
        }
    }

    private static int countCaps(String word) {
        int numCaps = 0;
        for (int i = 0; i < word.length(); i++) {
            if (Character.isUpperCase(word.charAt(i))) {
                numCaps++;
            }
        }
        return numCaps;
    }
}


Abbreviations, simple

The use of   abbreviations    (also sometimes called synonyms, nicknames, AKAs, or aliases)    can be an
easy way to add flexibility when specifying or using commands, sub-commands, options, etc.

For this task, the following   ''command table''   will be used:
    add 1  alter 3  backup 2  bottom 1  Cappend 2  change 1  Schange  Cinsert 2  Clast 3
    compress 4 copy 2 count 3 Coverlay 3 cursor 3  delete 3 Cdelete 2  down 1  duplicate
    3 xEdit 1 expand 3 extract 3  find 1 Nfind 2 Nfindup 6 NfUP 3 Cfind 2 findUP 3 fUP 2
    forward 2  get  help 1 hexType 4  input 1 powerInput 3  join 1 split 2 spltJOIN load
    locate 1 Clocate 2 lowerCase 3 upperCase 3 Lprefix 2  macro  merge 2 modify 3 move 2
    msg  next 1 overlay 1 parse preserve 4 purge 3 put putD query 1 quit  read recover 3
    refresh renum 3 repeat 3 replace 1 Creplace 2 reset 3 restore 4 rgtLEFT right 2 left
    2  save  set  shift 2  si  sort  sos  stack 3 status 4 top  transfer 3  type 1  up 1

Notes concerning the above   ''command table'':
::*   it can be thought of as one long literal string   (with blanks at end-of-lines)
::*   it may have superfluous blanks
::*   it may be in any case (lower/upper/mixed)
::*   the order of the words in the   ''command table''   must be preserved as shown
::*   the user input(s) may be in any case (upper/lower/mixed)
::*   commands will be restricted to the Latin alphabet   (A --> Z,   a --> z)
::*   a command is followed by an optional number, which indicates the minimum abbreviation
::*   A valid abbreviation is a word that has:
:::*   at least the minimum length of the word's minimum number in the ''command table''
:::*   compares equal (regardless of case) to the leading characters of the word in the ''command table''
:::*   a length not longer than the word in the ''command table''
::::*   '''ALT''',   '''aLt''',   '''ALTE''',   and   '''ALTER'''   are all abbreviations of   '''ALTER 3'''
::::*   '''AL''',   '''ALF''',   '''ALTERS''',   '''TER''',   and   '''A'''    aren't valid abbreviations of   '''ALTER 3'''
::::*   The   '''3'''   indicates that any abbreviation for   '''ALTER'''   must be at least three characters
::::*   Any word longer than five characters can't be an abbreviation for   '''ALTER'''
::::*   '''o''',    '''ov''',    '''oVe''',    '''over''',    '''overL''',    '''overla'''   are all acceptable abbreviations for   '''overlay 1'''
::*   if there isn't a number after the command,   then there isn't an abbreviation permitted

;Task:
::*   The command table needn't be verified/validated.
::*   Write a function to validate if the user "words"   (given as input)   are valid   (in the ''command table'').
::*   If the word   is   valid,   then return the full uppercase version of that "word".
::*   If the word        isn't     valid,   then return the lowercase string:    '''*error*'''        (7 characters).
::*   A blank input   (or a null input)   should return a null string.
::*   Show all output here.

;An example test case to be used for this task:
For a user string of:
  riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin
the computer program should return the string:
  RIGHT REPEAT *error* PUT MOVE RESTORE *error* *error* *error* POWERINPUT

Solution:
import java.util.*;

public class Abbreviations {
    public static void main(String[] args) {
        CommandList commands = new CommandList(commandTable);
        String input = "riG   rePEAT copies  put mo   rest    types   fup.    6       poweRin";
        System.out.println(" input: " + input);
        System.out.println("output: " + test(commands, input));
    }

    private static String test(CommandList commands, String input) {
        StringBuilder output = new StringBuilder();
        Scanner scanner = new Scanner(input);
        while (scanner.hasNext()) {
            String word = scanner.next();
            if (output.length() > 0)
                output.append(' ');
            Command cmd = commands.findCommand(word);
            if (cmd != null)
                output.append(cmd.cmd);
            else
                output.append("*error*");
        }
        return output.toString();
    }

    private static String commandTable =
        "add 1  alter 3  backup 2  bottom 1  Cappend 2  change 1  Schange  Cinsert 2  Clast 3 " +
        "compress 4 copy 2 count 3 Coverlay 3 cursor 3  delete 3 Cdelete 2  down 1  duplicate " +
        "3 xEdit 1 expand 3 extract 3  find 1 Nfind 2 Nfindup 6 NfUP 3 Cfind 2 findUP 3 fUP 2 " +
        "forward 2  get  help 1 hexType 4  input 1 powerInput 3  join 1 split 2 spltJOIN load " +
        "locate 1 Clocate 2 lowerCase 3 upperCase 3 Lprefix 2  macro  merge 2 modify 3 move 2 " +
        "msg  next 1 overlay 1 parse preserve 4 purge 3 put putD query 1 quit  read recover 3 " +
        "refresh renum 3 repeat 3 replace 1 Creplace 2 reset 3 restore 4 rgtLEFT right 2 left " +
        "2  save  set  shift 2  si  sort  sos  stack 3 status 4 top  transfer 3  type 1  up 1";

    private static class Command {
        private Command(String cmd, int minLength) {
             this.cmd = cmd;
             this.minLength = minLength;
        }
        private boolean match(String str) {
            int olen = str.length();
            return olen >= minLength && olen <= cmd.length()
                && cmd.regionMatches(true, 0, str, 0, olen);
        }
        private String cmd;
        private int minLength;
    }

    private static Integer parseInteger(String word) {
        try {
            return Integer.valueOf(word);
        } catch (NumberFormatException ex) {
            return null;
        }
    }

    private static class CommandList {
        private CommandList(String table) {
            Scanner scanner = new Scanner(table);
            List<String> words = new ArrayList<>();
            while (scanner.hasNext()) {
                String word = scanner.next();
                words.add(word.toUpperCase());
            }
            for (int i = 0, n = words.size(); i < n; ++i) {
                String word = words.get(i);
                // if there's an integer following this word, it specifies the minimum
                // length for the command, otherwise the minimum length is the length
                // of the command string
                int len = word.length();
                if (i + 1 < n) {
                    Integer number = parseInteger(words.get(i + 1));
                    if (number != null) {
                        len = number.intValue();
                        ++i;
                    }
                }
                commands.add(new Command(word, len));
            }
        }
        private Command findCommand(String word) {
            for (Command command : commands) {
                if (command.match(word))
                    return command;
            }
            return null;
        }
        private List<Command> commands = new ArrayList<>();
    }
}


Abelian sandpile model

{{wikipedia|Abelian sandpile model}}
Implement the '''Abelian sandpile model''' also known as '''Bak-Tang-Wiesenfeld model'''. Its history, mathematical definition and properties can be found under its wikipedia article.

The 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.'''

It 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.
Examples up to 2^30, wow!
javascript running on web
'''Examples:'''

0 0 0 0 0    0 0 0 0 0
0 0 0 0 0    0 0 1 0 0
0 0 4 0 0 -> 0 1 0 1 0
0 0 0 0 0    0 0 1 0 0
0 0 0 0 0    0 0 0 0 0

0 0 0 0 0    0 0 0 0 0
0 0 0 0 0    0 0 1 0 0
0 0 6 0 0 -> 0 1 2 1 0
0 0 0 0 0    0 0 1 0 0
0 0 0 0 0    0 0 0 0 0

0  0 0  0  0    0 0 1 0 0
0  0 0  0  0    0 2 1 2 0
0  0 16 0  0 -> 1 1 0 1 1
0  0 0  0  0    0 2 1 2 0
0  0 0  0  0    0 0 1 0 0

Solution:
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;

public class AbelianSandpile {
    public static void main(String[] args) {
        SwingUtilities.invokeLater(new Runnable() {
            public void run() {
                Frame frame = new Frame();
                frame.setVisible(true);
            }
        });
    }

    private static class Frame extends JFrame {
        private Frame() {
            super("Abelian Sandpile Model");
            setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
            Container contentPane = getContentPane();
            JPanel controlPanel = new JPanel(new FlowLayout(FlowLayout.LEFT));
            JButton start = new JButton("Restart Simulation");
            start.addActionListener(e -> restartSimulation());
            JButton stop = new JButton("Stop Simulation");
            stop.addActionListener(e -> stopSimulation());
            controlPanel.add(start);
            controlPanel.add(stop);
            contentPane.add(controlPanel, BorderLayout.NORTH);
            contentPane.add(canvas = new Canvas(), BorderLayout.CENTER);
            timer = new Timer(100, e -> canvas.runAndDraw());
            timer.start();
            pack();
        }

        private void restartSimulation() {
            timer.stop();
            canvas.initGrid();
            timer.start();
        }

        private void stopSimulation() {
            timer.stop();
        }

        private Timer timer;
        private Canvas canvas;
    }

    private static class Canvas extends JComponent {
        private Canvas() {
            setBorder(BorderFactory.createEtchedBorder());
            setPreferredSize(new Dimension(600, 600));
        }

        public void paintComponent(Graphics g) {
            int width = getWidth();
            int height = getHeight();
            g.setColor(Color.WHITE);
            g.fillRect(0, 0, width, height);
            int cellWidth = width/GRID_LENGTH;
            int cellHeight = height/GRID_LENGTH;
            for (int i = 0; i < GRID_LENGTH; ++i) {
                for (int j = 0; j < GRID_LENGTH; ++j) {
                    if (grid[i][j] > 0) {
                        g.setColor(COLORS[grid[i][j]]);
                        g.fillRect(i * cellWidth, j * cellHeight, cellWidth, cellHeight);
                    }
                }
            }
        }

        private void initGrid() {
            for (int i = 0; i < GRID_LENGTH; ++i) {
                for (int j = 0; j < GRID_LENGTH; ++j) {
                    grid[i][j] = 0;
                }
            }
        }

        private void runAndDraw() {
            for (int i = 0; i < 100; ++i)
                addSand(GRID_LENGTH/2, GRID_LENGTH/2);
            repaint();
        }

        private void addSand(int i, int j) {
            int grains = grid[i][j];
            if (grains < 3) {
                grid[i][j]++;
            }
            else {
                grid[i][j] = grains - 3;
                if (i > 0)
                    addSand(i - 1, j);
                if (i < GRID_LENGTH - 1)
                    addSand(i + 1, j);
                if (j > 0)
                    addSand(i, j - 1);
                if (j < GRID_LENGTH - 1)
                    addSand(i, j + 1);
            }
        }

        private int[][] grid = new int[GRID_LENGTH][GRID_LENGTH];
    }

    private static final Color[] COLORS = {
        Color.WHITE,
        new Color(0x00, 0xbf, 0xff),
        new Color(0xff, 0xd7, 0x00),
        new Color(0xb0, 0x30, 0x60)
    };
    private static final int GRID_LENGTH = 300;
}


Abundant odd numbers

An Abundant number is a number '''n''' for which the   ''sum of divisors''   '''s(n) > 2n''',
or,   equivalently,   the   ''sum of proper divisors''   (or aliquot sum)       '''s(n) > n'''.

;E.G.:
'''12'''   is abundant, it has the proper divisors     '''1,2,3,4 & 6'''     which sum to   '''16'''   ( > '''12''' or '''n''');
       or alternately,   has the sigma sum of   '''1,2,3,4,6 & 12'''   which sum to   '''28'''   ( > '''24''' or '''2n''').

Abundant numbers are common, though '''even''' abundant numbers seem to be much more common than '''odd''' abundant numbers.

To make things more interesting, this task is specifically about finding   ''odd abundant numbers''.

;Task
*Find and display here: at least the first 25 abundant odd numbers and either their proper divisor sum or sigma sum.
*Find and display here: the one thousandth abundant odd number and either its proper divisor sum or sigma sum.
*Find and display here: the first abundant odd number greater than one billion (109) and either its proper divisor sum or sigma sum.

;References:
:*   OEIS:A005231: Odd abundant numbers (odd numbers n whose sum of divisors exceeds 2n)
:*   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)

Solution:
import java.util.ArrayList;
import java.util.List;

public class AbundantOddNumbers {
    private static List<Integer> list = new ArrayList<>();
    private static List<Integer> result = new ArrayList<>();

    public static void main(String[] args) {
        System.out.println("First 25: ");
        abundantOdd(1,100000, 25, false);

        System.out.println("\n\nThousandth: ");
        abundantOdd(1,2500000, 1000, true);

        System.out.println("\n\nFirst over 1bn:");
        abundantOdd(1000000001, 2147483647, 1, false);
    }
    private static void abundantOdd(int start, int finish, int listSize, boolean printOne) {
        for (int oddNum = start; oddNum < finish; oddNum += 2) {
            list.clear();
            for (int toDivide = 1; toDivide < oddNum; toDivide+=2) {
                if (oddNum % toDivide == 0)
                    list.add(toDivide);
            }
            if (sumList(list) > oddNum) {
                if(!printOne)
                    System.out.printf("%5d <= %5d \n",oddNum, sumList(list) );
                result.add(oddNum);
            }
            if(printOne && result.size() >= listSize)
                System.out.printf("%5d <= %5d \n",oddNum, sumList(list) );

            if(result.size() >= listSize) break;
        }
    }
    private static int sumList(List list) {
        int sum = 0;
        for (int i = 0; i < list.size(); i++) {
            String temp = list.get(i).toString();
            sum += Integer.parseInt(temp);
        }
        return sum;
    }
}


Accumulator factory

{{requires|Mutable State}}

A 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).

;Rules:
The detailed rules are at http://paulgraham.com/accgensub.html and are reproduced here for simplicity (with additions in ''small italic text'').
:Before you submit an example, make sure the function

:# 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
:# 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)''
:# 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.)''
:# 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.)''
:# 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.)''
: E.g. if after the example, you added the following code (in a made-up language) ''where the factory function is called foo'':
:: x = foo(1);
x(5);
foo(3);
print x(2.3);
: 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.)''

;Task:
Create a function that implements the described rules.

It 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.

Where it is not possible to hold exactly to the constraints above, describe the deviations.

Solution:
Java has no first-class functions, so an accumulator can't use the x(5) syntax.  The standard syntactic workaround is to use a standard method name, like x.call(5) or x.apply(5).  This is a deviation from task.

Our accumulator sums with long integers as far as possible before switching to floats.  This requires the use of the Number class.  The code needs Java 5 to autobox primitive values 1 or 2.3 into instances of Number.  The apply method is ready to implement interface UnaryOperator in Java 8.


Accumulator factory

Solution:
public class Accumulator
    //implements java.util.function.UnaryOperator<Number> // Java 8
{
    private Number sum;

    public Accumulator(Number sum0) {
	sum = sum0;
    }

    public Number apply(Number n) {
	// Acts like sum += n, but chooses long or double.
	// Converts weird types (like BigInteger) to double.
	return (longable(sum) && longable(n)) ?
	    (sum = sum.longValue() + n.longValue()) :
	    (sum = sum.doubleValue() + n.doubleValue());
    }

    private static boolean longable(Number n) {
	return n instanceof Byte || n instanceof Short ||
	    n instanceof Integer || n instanceof Long;
    }

    public static void main(String[] args) {
	Accumulator x = new Accumulator(1);
	x.apply(5);
	new Accumulator(3);
	System.out.println(x.apply(2.3));
    }
}


Accumulator factory

Solution:
import java.util.function.UnaryOperator;

public class AccumulatorFactory {

    public static UnaryOperator<Number> accumulator(Number sum0) {
	// Allows sum[0] = ... inside lambda.
	Number[] sum = { sum0 };

	// Acts like n -> sum[0] += n, but chooses long or double.
	// Converts weird types (like BigInteger) to double.
	return n -> (longable(sum[0]) && longable(n)) ?
	    (sum[0] = sum[0].longValue() + n.longValue()) :
	    (sum[0] = sum[0].doubleValue() + n.doubleValue());
    }

    private static boolean longable(Number n) {
	return n instanceof Byte || n instanceof Short ||
	    n instanceof Integer || n instanceof Long;
    }

    public static void main(String[] args) {
	UnaryOperator<Number> x = accumulator(1);
	x.apply(5);
	accumulator(3);
	System.out.println(x.apply(2.3));
    }
}


Aliquot sequence classifications

An aliquot sequence of a positive integer K is defined recursively as the first member
being K and subsequent members being the sum of the [[Proper divisors]] of the previous term.

:* If the terms eventually reach 0 then the series for K is said to '''terminate'''.

:There are several classifications for non termination:
:* 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'''.
:* If the third term ''would'' be repeating K then the sequence repeats with period 2 and K is called '''amicable'''.
:* 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'''.
:Perfect, amicable and sociable numbers eventually repeat the original number K; there are other repetitions...
:* 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'''.
:* 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'''.

:And finally:
:* 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.

;Task:
# Create routine(s) to generate the aliquot sequence of a positive integer enough to classify it according to the classifications given above.
# Use it to display the classification and sequences of the numbers one to ten inclusive.
# Use it to show the classification and sequences of the following integers, in order:
:: 11, 12, 28, 496, 220, 1184,  12496, 1264460, 790, 909, 562, 1064, 1488, and optionally 15355717786080.

Show all output on this page.

;Related tasks:
*   [[Abundant, deficient and perfect number classifications]]. (Classifications from only the first two members of the whole sequence).
*   [[Proper divisors]]
*   [[Amicable pairs]]

Solution: Translation of [[Aliquot_sequence_classifications#Python|Python]] via [[Aliquot_sequence_classifications#D|D]]


Aliquot sequence classifications

Solution:
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.stream.LongStream;

public class AliquotSequenceClassifications {

    private static Long properDivsSum(long n) {
        return LongStream.rangeClosed(1, (n + 1) / 2).filter(i -> n % i == 0 && n != i).sum();
    }

    static boolean aliquot(long n, int maxLen, long maxTerm) {
        List<Long> s = new ArrayList<>(maxLen);
        s.add(n);
        long newN = n;

        while (s.size() <= maxLen && newN < maxTerm) {

            newN = properDivsSum(s.get(s.size() - 1));

            if (s.contains(newN)) {

                if (s.get(0) == newN) {

                    switch (s.size()) {
                        case 1:
                            return report("Perfect", s);
                        case 2:
                            return report("Amicable", s);
                        default:
                            return report("Sociable of length " + s.size(), s);
                    }

                } else if (s.get(s.size() - 1) == newN) {
                    return report("Aspiring", s);

                } else
                    return report("Cyclic back to " + newN, s);

            } else {
                s.add(newN);
                if (newN == 0)
                    return report("Terminating", s);
            }
        }

        return report("Non-terminating", s);
    }

    static boolean report(String msg, List<Long> result) {
        System.out.println(msg + ": " + result);
        return false;
    }

    public static void main(String[] args) {
        long[] arr = {
                11, 12, 28, 496, 220, 1184, 12496, 1264460,
                790, 909, 562, 1064, 1488};

        LongStream.rangeClosed(1, 10).forEach(n -> aliquot(n, 16, 1L << 47));
        System.out.println();
        Arrays.stream(arr).forEach(n -> aliquot(n, 16, 1L << 47));
    }
}


Anagrams/Deranged anagrams

Two or more words are said to be anagrams if they have the same characters, but in a different order.

By 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.

;Task

Use the word list at unixdict to find and display the longest deranged anagram.

;Related
* [[Permutations/Derangements]]
* Best shuffle

{{Related tasks/Word plays}}

Solution:
import java.io.File;
import java.io.IOException;
import java.nio.file.Files;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class DerangedAnagrams {

    public static void main(String[] args) throws IOException {
        List<String> words = Files.readAllLines(new File("unixdict.txt").toPath());
        printLongestDerangedAnagram(words);
    }

    private static void printLongestDerangedAnagram(List<String> words) {
        words.sort(Comparator.comparingInt(String::length).reversed().thenComparing(String::toString));

        Map<String, ArrayList<String>> map = new HashMap<>();
        for (String word : words) {
            char[] chars = word.toCharArray();
            Arrays.sort(chars);
            String key = String.valueOf(chars);

            List<String> anagrams = map.computeIfAbsent(key, k -> new ArrayList<>());
            for (String anagram : anagrams) {
                if (isDeranged(word, anagram)) {
                    System.out.printf("%s %s%n", anagram, word);
                    return;
                }
            }
            anagrams.add(word);
        }
        System.out.println("no result");
    }

    private static boolean isDeranged(String word1, String word2) {
        for (int i = 0; i < word1.length(); i++) {
            if (word1.charAt(i) == word2.charAt(i)) {
                return false;
            }
        }
        return true;
    }
}