https://www.codewars.com/kata/square-into-squares-protect-trees/

My little sister came back home from school with the following task: given a squared sheet of paper she has to cut it in pieces which, when assembled, give squares the sides of which form an increasing sequence of numbers. At the beginning it was lot of fun but little by little we were tired of seeing the pile of torn paper. So we decided to write a program that could help us and protects trees.

Task

Given a positive integral number n, return a strictly increasing sequence (list/array/string depending on the language) of numbers, so that the sum of the squares is equal to n².

If there are multiple solutions (and there will be), return the result with the largest possible values:

Examples

decompose(11) must return [1,2,4,10]. Note that there are actually two ways to decompose 11², 11² = 121 = 1 + 4 + 16 + 100 = 1² + 2² + 4² + 10² but don't return [2,6,9], since 9 is smaller than 10.

For decompose(50) don't return [1, 1, 4, 9, 49] but [1, 3, 5, 8, 49] since [1, 1, 4, 9, 49] doesn't form a strictly increasing sequence.

Note

Neither [n] nor [1,1,1,…,1] are valid solutions. If no valid solution exists, return nil, null, Nothing, None (depending on the language) or "[]" (C) ,{} (C++), [] (Swift).

The function "decompose" will take a positive integer n and return the decomposition of N = n² as:

[x1 ... xk]

Note for Bash

decompose 50 returns "1,3,5,8,49"
decompose 4  returns "Nothing"

Hint

Very often xk will be n-1.

Best Practices

Py First:

def decompose(n):
    def _recurse(s, i):
        if s < 0:
            return None
        if s == 0:
            return []
        for j in xrange(i-1, 0, -1):
            sub = _recurse(s - j**2, j)
            if sub != None:
                return sub + [j]
    return _recurse(n**2, n)

Py Second:

def decompose(n):
    total = 0
    answer = [n]
    while len(answer):
        temp = answer.pop()
        total += temp ** 2
        for i in range(temp - 1, 0, -1):
            if total - (i ** 2) >= 0:
                total -= i ** 2
                answer.append(i)
                if total == 0:
                    return sorted(answer)
    return None

Py Third:

def dec(r, d):
    for i in range(d-1, 0, -1):
        if r < i * i:
            continue
        if r == i * i:
            return [i]
        a = dec(r - i * i, i)
        if a:
            return a + [i]
    return None
def decompose(n):
    return  dec(n * n, n)

Py Fourth:

from math import sqrt

def decompose(n):
    return decompose_alt(n**2, n-1)

def decompose_alt(n, m):
    if sqrt(n) == int(sqrt(n)) and int(sqrt(n)) <= m : return [ int(sqrt(n)) ]
    for i in range(min(int(sqrt(n)),m),0,-1):
        result = decompose_alt( n-i**2, i-1 )
        if result: return result + [i]
    return None