meetforever/GRPO_CL huggingface dataset:
user: You have a collection of lovely poems. Unfortuantely they aren't formatted very well. They're all on one line, like this:

```
Beautiful is better than ugly. Explicit is better than implicit. Simple is better than complex. Complex is better than complicated.
```
What you want is to present each sentence on a new line, so that it looks like this:
```
Beautiful is better than ugly.
Explicit is better than implicit.
Simple is better than complex.
Complex is better than complicated.
```
Write a function, `format_poem()` that takes a string like the one line example as an argument and returns a new string that is formatted across multiple lines with each new sentence starting on a new line when you print it out.

Try to solve this challenge with the [str.split()](https://docs.python.org/3/library/stdtypes.html#str.split) and the [str.join()](https://docs.python.org/3/library/stdtypes.html#str.join) string methods.

Every sentence will end with a period, and every new sentence will have one space before the previous period. Be careful about trailing whitespace in your solution.

Answer: {"fn_name": "format_poem", "inputs": [["Beautiful is better than ugly. Explicit is better than implicit. Simple is better than complex. Complex is better than complicated."], ["Flat is better than nested. Sparse is better than dense. Readability counts. Special cases aren't special enough to break the rules."], ["Although practicality beats purity. Errors should never pass silently. Unless explicitly silenced. In the face of ambiguity, refuse the temptation to guess."], ["There should be one-- and preferably only one --obvious way to do it. Although that way may not be obvious at first unless you're Dutch. Now is better than never. Although never is often better than *right* now."], ["If the implementation is hard to explain, it's a bad idea. If the implementation is easy to explain, it may be a good idea. Namespaces are one honking great idea -- let's do more of those!"]], "outputs": [["Beautiful is better than ugly.\nExplicit is better than implicit.\nSimple is better than complex.\nComplex is better than complicated."], ["Flat is better than nested.\nSparse is better than dense.\nReadability counts.\nSpecial cases aren't special enough to break the rules."], ["Although practicality beats purity.\nErrors should never pass silently.\nUnless explicitly silenced.\nIn the face of ambiguity, refuse the temptation to guess."], ["There should be one-- and preferably only one --obvious way to do it.\nAlthough that way may not be obvious at first unless you're Dutch.\nNow is better than never.\nAlthough never is often better than *right* now."], ["If the implementation is hard to explain, it's a bad idea.\nIf the implementation is easy to explain, it may be a good idea.\nNamespaces are one honking great idea -- let's do more of those!"]]}
user: Lexicographic permutations are ordered combinations of a set of items ordered in a specific way.

For instance, the first 8 permutations of the digits 0123, in lexicographic order, are:

```
1st 0123
2nd 0132
3rd 0213
4th 0231
5th 0312
6th 0321
7th 1023
8th 1032
```

Your task is to write a function ```L( n, d )``` that will return a `string` representing the `nth` permutation of the `d` digit, starting with 0 (i.e. d=10 means all digits 0123456789).

So for `d = 4`, `L(7,4)` should return `'1023'`, and `L(4,4)` should return `'0231'`
.


Some things to bear in mind:

• The function should return a `string`, otherwise permutations beginning with a 0 will have it removed. 

• Test cases will not exceed the highest possible valid values for `n`

• The function should work for any `d` between `1` and `10`.

• A value of 1 for `n` means the 1st permutation, so `n = 0` is not a valid input.

• Oh, and no itertools ;)

Answer: {"fn_name": "nth_perm", "inputs": [[1, 1], [12, 5], [1000, 7], [1000, 8], [1000000, 10], [874, 7], [100, 5], [400, 6], [900, 8], [3000000, 10], [1234567, 10], [654321, 10], [100000, 9], [33333, 8]], "outputs": [["0"], ["02431"], ["1325460"], ["02436571"], ["2783915460"], ["1234560"], ["40231"], ["314250"], ["02354761"], ["8241697530"], ["3469702158"], ["1827653409"], ["247815360"], ["64157203"]]}
user: Given a side length `n`, traveling only right and down how many ways are there to get from the top left corner to the bottom right corner of an `n by n` grid?

Your mission is to write a program to do just that!

Add code to `route(n)` that returns the number of routes for a grid `n by n` (if n is less than 1 return 0).

Examples:
    
    -100 -> 0
    
    1 -> 2
     
    2 -> 6
     
    20 -> 137846528820
    
Note: 
you're traveling on the edges of the squares in the grid not the squares themselves.

PS.If anyone has any suggestions of how to improve this kata please let me know.

Answer: {"fn_name": "routes", "inputs": [[156], [106], [108], [126], [165], [125], [103]], "outputs": [[376594020312425061595746241557406711201605603899894320577733841460621358204120168599683135056], [360262512886894560870925723424985925545493469115736766525399280], [5710703280600670271749409312650477833283445281465962347881848400], [363385715856739898597174879118845755433251497432024467186246395557048813504], [95995892383488599901870820732368671856306475542669347911637204757217297906428740013991568571288240], [91208366928185711600087718663295946582847985411225264672245111235434562752], [5710294458198606715524045745816008575257432967999860738082400]]}
user: _Based on [Project Euler problem 35](https://projecteuler.net/problem=35)_

A circular prime is a prime in which every circular permutation of that number is also prime. Circular permutations are created by rotating the digits of the number, for example: `197, 971, 719`. One-digit primes are circular primes by definition.

Complete the function that dertermines if a number is a circular prime.

There are 100 random tests for numbers up to 10000.

Answer: {"fn_name": "circular_prime", "inputs": [[197], [179], [971], [222], [9377], [7], [213], [35], [1]], "outputs": [[true], [false], [true], [false], [true], [true], [false], [false], [false]]}
user: ## Terminal game move function

In this game, the hero moves from left to right. The player rolls the dice and moves the number of spaces indicated by the dice **two times**.

~~~if-not:sql
Create a function for the terminal game that takes the current position of the hero and the roll (1-6) and return the new position.
~~~
~~~if:sql
In SQL, you will be given a table `moves` with columns `position` and `roll`. Return a table which uses the current position of the hero and the roll (1-6) and returns the new position in a column `res`.
~~~

### Example:
```python
move(3, 6) should equal 15
```

```if:bf
### BF:

Since this is an 8kyu kata, you are provided a modified runBF function, numericRunBF, that automatically parses input and output for your ease.

See the sample test cases to see what I mean: You can simply input two numbers and get a number as output (unless you're doing something wrong), so it should be convenient for you to modify the tests as you wish.

Oh, and you won't have to worry about overflow, the correct answer will never be higher than 255. :)
```

Answer: {"fn_name": "move", "inputs": [[0, 4], [3, 6], [2, 5]], "outputs": [[8], [15], [12]]}
user: In this kata the function returns an array/list of numbers without its last element. The function is already written for you and the basic tests pass, but random tests fail. Your task is to figure out why and fix it.

Good luck!

Hint: watch out for side effects.

~~~if:javascript
Some good reading: [MDN Docs about arrays](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array)
~~~
~~~if:python
Some good reading: [Python Lists](http://www.compciv.org/guides/python/fundamentals/lists-mutability/)
~~~

Answer: {"fn_name": "without_last", "inputs": [[[1, 2, 3, 4, 5]], [[9, 7, 6, 9]]], "outputs": [[[1, 2, 3, 4]], [[9, 7, 6]]]}
user: Implement `String.eight_bit_signed_number?` (Ruby), `String.eightBitSignedNumber()` (Python), `eight_bit_signed_number()` (JS) or `StringUtils.isSignedEightBitNumber(String)` (Java) which should return `true/True` if given object is a number representable by 8 bit signed integer (-128 to -1 or 0 to 127), `false/False` otherwise.

It should only accept numbers in canonical representation, so no leading `+`, extra `0`s, spaces etc.

Answer: {"fn_name": "signed_eight_bit_number", "inputs": [[""], ["0"], ["00"], ["-0"], ["55"], ["-23"], ["042"], ["127"], ["128"], ["999"], ["-128"], ["-129"], ["-999"], ["1\n"], ["1 "], [" 1"], ["1\n2"], ["+1"], ["--1"], ["01"], ["-01"]], "outputs": [[false], [true], [false], [false], [true], [true], [false], [true], [false], [false], [true], [false], [false], [false], [false], [false], [false], [false], [false], [false], [false]]}
user: Given two words and a letter, return a single word that's a combination of both words, merged at the point where the given letter first appears in each word. The returned word should have the beginning of the first word and the ending of the second, with the dividing letter in the middle. You can assume both words will contain the dividing letter.

## Examples

```python
string_merge("hello", "world", "l")      ==>  "held"
string_merge("coding", "anywhere", "n")  ==>  "codinywhere"
string_merge("jason", "samson", "s")     ==>  "jasamson"
string_merge("wonderful", "people", "e") ==>  "wondeople"
```

Answer: {"fn_name": "string_merge", "inputs": [["hello", "world", "l"], ["coding", "anywhere", "n"], ["jason", "samson", "s"], ["wonderful", "people", "e"], ["person", "here", "e"], ["apowiejfoiajsf", "iwahfeijouh", "j"], ["abcdefxxxyzz", "abcxxxyyyxyzz", "x"], ["12345654321", "123456789", "6"], ["JiOdIdA4", "oopopopoodddasdfdfsd", "d"], ["incredible", "people", "e"]], "outputs": [["held"], ["codinywhere"], ["jasamson"], ["wondeople"], ["pere"], ["apowiejouh"], ["abcdefxxxyyyxyzz"], ["123456789"], ["JiOdddasdfdfsd"], ["increople"]]}
user: Write a function that accepts two square (`NxN`) matrices (two dimensional arrays), and returns the product of the two. Only square matrices will be given.

How to multiply two square matrices: 

We are given two matrices, A and B, of size 2x2 (note: tests are not limited to 2x2). Matrix C, the solution, will be equal to the product of A and B. To fill in cell `[0][0]` of matrix C, you need to compute: `A[0][0] * B[0][0] + A[0][1] * B[1][0]`.

More general: To fill in cell `[n][m]` of matrix C, you need to first multiply the elements in the nth row of matrix A by the elements in the mth column of matrix B, then take the sum of all those products. This will give you the value for cell `[m][n]` in matrix C. 

## Example
```
  A         B          C
|1 2|  x  |3 2|  =  | 5 4|
|3 2|     |1 1|     |11 8|
```

Detailed calculation:
```
C[0][0] = A[0][0] * B[0][0] + A[0][1] * B[1][0] = 1*3 + 2*1 =  5
C[0][1] = A[0][0] * B[0][1] + A[0][1] * B[1][1] = 1*2 + 2*1 =  4
C[1][0] = A[1][0] * B[0][0] + A[1][1] * B[1][0] = 3*3 + 2*1 = 11
C[1][1] = A[1][0] * B[0][1] + A[1][1] * B[1][1] = 3*2 + 2*1 =  8
```
Link to Wikipedia explaining matrix multiplication (look at the square matrix example): 
http://en.wikipedia.org/wiki/Matrix_multiplication

A more visual explanation of matrix multiplication: http://matrixmultiplication.xyz

~~~if:c
**Note:** In **C**, the dimensions of both square matrices `n` will be passed into your function.  However, since the dimensions of your returned "matrix" is expected to be the same as that of the inputs, you will not need to keep track of the dimensions of your matrix in another variable.
~~~

Answer: {"fn_name": "matrix_mult", "inputs": [[[[1, 2], [3, 2]], [[3, 2], [1, 1]]], [[[9, 7], [0, 1]], [[1, 1], [4, 12]]], [[[1, 2, 3], [3, 2, 1], [2, 1, 3]], [[4, 5, 6], [6, 5, 4], [4, 6, 5]]]], "outputs": [[[[5, 4], [11, 8]]], [[[37, 93], [4, 12]]], [[[28, 33, 29], [28, 31, 31], [26, 33, 31]]]]}
user: You certainly can tell which is the larger number between 2^(10) and 2^(15).

But what about, say, 2^(10) and 3^(10)? You know this one too.

Things tend to get a bit more complicated with **both** different bases and exponents: which is larger between 3^(9) and 5^(6)?

Well, by now you have surely guessed that you have to build a function to compare powers, returning -1 if the first member is larger, 0 if they are equal, 1 otherwise; powers to compare will be provided in the `[base, exponent]` format:
```python
compare_powers([2,10],[2,15])==1
compare_powers([2,10],[3,10])==1
compare_powers([2,10],[2,10])==0
compare_powers([3,9],[5,6])==-1
compare_powers([7,7],[5,8])==-1
```
```if:nasm
int compare_powers(const int n1[2], const int n2[2])
```
Only positive integers will be tested, incluing bigger numbers - you are warned now, so be diligent try to implement an efficient solution not to drain too much on CW resources ;)!

Answer: {"fn_name": "compare_powers", "inputs": [[[2, 10], [2, 15]], [[1, 10], [3, 10]], [[2, 10], [2, 10]], [[3, 9], [1, 6]], [[1, 7], [1, 8]], [[2, 100], [2, 150]], [[2, 100], [3, 100]], [[2, 100], [2, 100]], [[34, 98], [51, 67]], [[1024, 97], [1024, 81]]], "outputs": [[1], [1], [0], [-1], [0], [1], [1], [0], [-1], [-1]]}
user: Consider the following numbers (where `n!` is `factorial(n)`):
```
u1 = (1 / 1!) * (1!)
u2 = (1 / 2!) * (1! + 2!)
u3 = (1 / 3!) * (1! + 2! + 3!)
...
un = (1 / n!) * (1! + 2! + 3! + ... + n!)
```

Which will win: `1 / n!` or `(1! + 2! + 3! + ... + n!)`?

Are these numbers going to `0` because of `1/n!` or to infinity due
to the sum of factorials or to another number?

## Task
Calculate `(1 / n!) * (1! + 2! + 3! + ... + n!)` 
for a given `n`, where `n` is an integer greater or equal to `1`.

To avoid discussions about rounding, return the result **truncated** to 6 decimal places, for example:
```
1.0000989217538616 will be truncated to 1.000098
1.2125000000000001 will be truncated to 1.2125
```

## Remark
Keep in mind that factorials grow rather rapidly, and you need to handle large inputs.

## Hint
You could try to simplify the expression.

Answer: {"fn_name": "going", "inputs": [[5], [6], [7], [8], [20], [30], [50], [113], [200], [523], [1011], [10110]], "outputs": [[1.275], [1.2125], [1.173214], [1.146651], [1.052786], [1.034525], [1.020416], [1.008929], [1.005025], [1.001915], [1.00099], [1.000098]]}
user: The squarefree part of a positive integer is the largest divisor of that integer which itself has no square factors (other than 1). For example, the squareefree part of 12 is 6, since the only larger divisor is 12, and 12 has a square factor (namely, 4).

Your challenge, should you choose to accept it, is to implement a `squareFreePart` function which accepts a number `n` and returns the squarefree part of `n`. 

In the case that `n = 1`, your function should return 1. Also, if the input doesn't make sense (e.g. if it is not a positive integer), the function should return `null/None`. 

Here are some examples:

```python
square_free_part(2) # returns 2
square_free_part(4) # returns 2
square_free_part(24) # returns 6, since any larger divisor is divisible by 4, which is a square
square_free_part("hi") # returns None
square_free_part(0) # returns None
```

Good luck!

Answer: {"fn_name": "square_free_part", "inputs": [[0], [-100], [2.5]], "outputs": [[null], [null], [null]]}
user: Given time in 24-hour format, convert it to words. 

```
For example:
13:00 = one o'clock 
13:09 = nine minutes past one 
13:15 = quarter past one 
13:29 = twenty nine minutes past one
13:30 = half past one 
13:31 = twenty nine minutes to two
13:45 = quarter to two 
00:48 = twelve minutes to one
00:08 = eight minutes past midnight
12:00 = twelve o'clock
00:00 = midnight

Note: If minutes == 0, use 'o'clock'. If minutes <= 30, use 'past', and for minutes > 30, use 'to'. 

```

More examples in test cases. Good luck!

Answer: {"fn_name": "solve", "inputs": [["13:00"], ["13:09"], ["13:15"], ["13:29"], ["13:30"], ["13:31"], ["13:45"], ["13:59"], ["00:48"], ["00:08"], ["12:00"], ["00:00"], ["19:01"], ["07:01"], ["01:59"], ["12:01"], ["00:01"], ["11:31"], ["23:31"], ["11:45"], ["11:59"], ["23:45"], ["23:59"], ["01:45"]], "outputs": [["one o'clock"], ["nine minutes past one"], ["quarter past one"], ["twenty nine minutes past one"], ["half past one"], ["twenty nine minutes to two"], ["quarter to two"], ["one minute to two"], ["twelve minutes to one"], ["eight minutes past midnight"], ["twelve o'clock"], ["midnight"], ["one minute past seven"], ["one minute past seven"], ["one minute to two"], ["one minute past twelve"], ["one minute past midnight"], ["twenty nine minutes to twelve"], ["twenty nine minutes to midnight"], ["quarter to twelve"], ["one minute to twelve"], ["quarter to midnight"], ["one minute to midnight"], ["quarter to two"]]}
user: For building the encrypted string:Take every 2nd char from the string, then the other chars, that are not every 2nd char, and concat them as new String.
Do this n times!

Examples:
```
"This is a test!", 1 -> "hsi  etTi sats!"
"This is a test!", 2 -> "hsi  etTi sats!" -> "s eT ashi tist!"
```

Write two methods:
```python
def encrypt(text, n)
def decrypt(encrypted_text, n)
```

```Fsharp
let encrypt (str:string) (n:int) -> string
let decrypt (str:string) (n:int) -> string
```

For both methods:
If the input-string is null or empty return exactly this value!
If n is <= 0 then return the input text.

This kata is part of the Simple Encryption Series:
Simple Encryption #1 - Alternating Split
Simple Encryption #2 - Index-Difference
Simple Encryption #3 - Turn The Bits Around
Simple Encryption #4 - Qwerty

Have fun coding it and please don't forget to vote and rank this kata! :-)

Answer: {"fn_name": "decrypt", "inputs": [["This is a test!", 0], ["hsi  etTi sats!", 1], ["s eT ashi tist!", 2], [" Tah itse sits!", 3], ["This is a test!", 4], ["This is a test!", -1], ["hskt svr neetn!Ti aai eyitrsig", 1], ["", 0], [null, 0]], "outputs": [["This is a test!"], ["This is a test!"], ["This is a test!"], ["This is a test!"], ["This is a test!"], ["This is a test!"], ["This kata is very interesting!"], [""], [null]]}
user: You have been recruited by an unknown organization for your cipher encrypting/decrypting skills.  
Being new to the organization they decide to test your skills.  
Your first test is to write an algorithm that encrypts the given string in the following steps.

1. The first step of the encryption is a standard ROT13 cipher.
This is a special case of the caesar cipher where the letter is encrypted with its key that is thirteen letters down the alphabet,  
i.e. `A => N, B => O, C => P, etc..`

1. Part two of the encryption is to take the ROT13 output and replace each letter with its exact opposite. `A => Z, B => Y, C => X`.  
The return value of this should be the encrypted message.


Do not worry about capitalization or punctuation. All encrypted messages should be lower case and punctuation free.  
As an example, the string `"welcome to our organization"` should return `"qibkyai ty ysv yvgmzenmteyz"`.

Good luck, and congratulations on the new position.

Answer: {"fn_name": "encrypter", "inputs": [["amz"], ["welcome to the organization"], ["hello"], ["my name is"], ["goodbye"]], "outputs": [["man"], ["qibkyai ty tfi yvgmzenmteyz"], ["fibby"], ["ao zmai eu"], ["gyyjloi"]]}
user: The look and say sequence is a sequence in which each number is the result of a "look and say" operation on the previous element.

Considering for example the classical version startin with `"1"`: `["1", "11", "21, "1211", "111221", ...]`. You can see that the second element describes the first as `"1(times number)1"`, the third is `"2(times number)1"` describing the second, the fourth is `"1(times number)2(and)1(times number)1"` and so on.

Your goal is to create a function which takes a starting string (not necessarily the classical `"1"`, much less a single character start) and return the nth element of the series.

## Examples

```python
look_and_say_sequence("1", 1)   == "1"
look_and_say_sequence("1", 3)   == "21"
look_and_say_sequence("1", 5)   == "111221"
look_and_say_sequence("22", 10) == "22"
look_and_say_sequence("14", 2)  == "1114"
```
Trivia: `"22"` is the only element that can keep the series constant.

Answer: {"fn_name": "look_and_say_sequence", "inputs": [["1", 1], ["1", 3], ["1", 5], ["22", 10], ["14", 2]], "outputs": [["1"], ["21"], ["111221"], ["22"], ["1114"]]}
user: Americans are odd people: in their buildings, the first floor is actually the ground floor and there is no 13th floor (due to superstition).

Write a function that given a floor in the american system returns the floor in the european system.

With the 1st floor being replaced by the ground floor and the 13th floor being removed, the numbers move down to take their place. In case of above 13, they move down by two because there are two omitted numbers below them.

Basements (negatives) stay the same as the universal level.

[More information here](https://en.wikipedia.org/wiki/Storey#European_scheme)

## Examples

```
1  =>  0 
0  =>  0
5  =>  4
15  =>  13
-3  =>  -3
```

Answer: {"fn_name": "get_real_floor", "inputs": [[1], [0], [5], [10], [12], [14], [15], [37], [200], [-2], [-5]], "outputs": [[0], [0], [4], [9], [11], [12], [13], [35], [198], [-2], [-5]]}
user: Your task is simply to count the total number of lowercase letters in a string.

## Examples

Answer: {"fn_name": "lowercase_count", "inputs": [["abc"], ["abcABC123"], ["abcABC123!@#$%^&*()_-+=}{[]|':;?/>.<,~"], [""], ["ABC123!@#$%^&*()_-+=}{[]|':;?/>.<,~"], ["abcdefghijklmnopqrstuvwxyz"]], "outputs": [[3], [3], [3], [0], [0], [26]]}
user: Is the number even?

If the numbers is even return `true`. If it's odd, return `false`. 


Oh yeah... the following symbols/commands have been disabled!

 use of ```%```
 use of ```.even?``` in Ruby
 use of ```mod``` in Python

Answer: {"fn_name": "is_even", "inputs": [[2], [3], [14], [15], [26], [27]], "outputs": [[true], [false], [true], [false], [true], [false]]}
user: General primality test are often computationally expensive, so in the biggest prime number race the idea is to study special sub-families of prime number and develop more effective tests for them. 

[Mersenne Primes](https://en.wikipedia.org/wiki/Mersenne_prime) are prime numbers of the form: Mn = 2^(n) - 1. So far, 49 of them was found between M2 (3) and M74,207,281 which contains 22,338,618 digits and is (by September 2015) the biggest known prime. It has been found by GIMPS (Great Internet Mersenne Prime Search), wich use the test srudied here. (plus heuristics mentionned below)

Lucas and Lehmer have developed a [simple test](https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test) for Mersenne primes: 

> for `p` (an odd prime), the Mersenne number Mp is prime if and only if `S(p − 1) = 0 mod Mp`,

>where `S(n + 1) = S(n) * S(n) − 2` and `S(1) = 4`

The goal of this kata is to implement this test. Given an integer `>=2`, the program should calculate the sequence `S(n)` up to `p-1` and calculate the remainder modulo `Mp`, then return `True` or `False` as a result of the test. The case `n = 2` should return `True`.

You should take advantage of the fact that:

```k = (k mod 2^n + floor(k/2^n)) mod Mn```

Or in other words, if we take the least significant `n` bits of `k`, and add the remaining bits of `k`, and then do this repeatedly until at most `n` bits remain, we can compute the remainder after dividing `k` by the Mersenne number `2^n−1` without using division.

This test can be improved using the fact that if `n` is not prime, `2^n-1` will not be prime. So in theory you can test for the primality of `n` beforehand. But in practice the test for the primality of `n` will rapidly outgrow in difficulty the Lucas-Lehmer test. So we can only use heuristics to rule out `n` with small factors but not complete factorization algorithm. You don't need to implement such heuristics here. 

The rapid growth of `s^2` makes javascript reach its integer limit quickly (coherent result for `n = 13`, but not `17`). Whereas python seems to be only limited by the execution time limit and let us see that M11,213 is prime (contains 3376 digits).

Answer: {"fn_name": "lucas_lehmer", "inputs": [[2], [3], [4], [5], [7], [10], [33], [101], [11213], [11215]], "outputs": [[true], [true], [false], [true], [true], [false], [false], [false], [true], [false]]}