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57 lines
1.5 KiB
Python
57 lines
1.5 KiB
Python
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"""
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Heap's algorithm returns the list of all permutations possible from a list.
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It minimizes movement by generating each permutation from the previous one
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by swapping only two elements.
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More information:
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https://en.wikipedia.org/wiki/Heap%27s_algorithm.
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"""
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def heaps(arr: list) -> list:
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"""
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Pure python implementation of the Heap's algorithm (recursive version),
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returning all permutations of a list.
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>>> heaps([])
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[()]
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>>> heaps([0])
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[(0,)]
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>>> heaps([-1, 1])
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[(-1, 1), (1, -1)]
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>>> heaps([1, 2, 3])
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[(1, 2, 3), (2, 1, 3), (3, 1, 2), (1, 3, 2), (2, 3, 1), (3, 2, 1)]
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>>> from itertools import permutations
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>>> sorted(heaps([1,2,3])) == sorted(permutations([1,2,3]))
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True
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>>> all(sorted(heaps(x)) == sorted(permutations(x))
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... for x in ([], [0], [-1, 1], [1, 2, 3]))
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True
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"""
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if len(arr) <= 1:
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return [tuple(arr)]
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res = []
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def generate(k: int, arr: list):
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if k == 1:
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res.append(tuple(arr[:]))
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return
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generate(k - 1, arr)
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for i in range(k - 1):
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if k % 2 == 0: # k is even
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arr[i], arr[k - 1] = arr[k - 1], arr[i]
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else: # k is odd
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arr[0], arr[k - 1] = arr[k - 1], arr[0]
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generate(k - 1, arr)
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generate(len(arr), arr)
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return res
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if __name__ == "__main__":
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user_input = input("Enter numbers separated by a comma:\n").strip()
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arr = [int(item) for item in user_input.split(",")]
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print(heaps(arr))
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