Heaps algorithm (#2475)

* heaps_algorithm: new algo

* typo

* correction doctest

* doctests: compare with itertools.permutations

* doctest: sorted instead of set

* doctest

* doctest

* rebuild
This commit is contained in:
Guillaume Rochedix 2020-09-29 12:39:07 +02:00 committed by GitHub
parent 04322e67e5
commit d95d643351
No known key found for this signature in database
GPG Key ID: 4AEE18F83AFDEB23

View File

@ -0,0 +1,56 @@
"""
Heap's algorithm returns the list of all permutations possible from a list.
It minimizes movement by generating each permutation from the previous one
by swapping only two elements.
More information:
https://en.wikipedia.org/wiki/Heap%27s_algorithm.
"""
def heaps(arr: list) -> list:
"""
Pure python implementation of the Heap's algorithm (recursive version),
returning all permutations of a list.
>>> heaps([])
[()]
>>> heaps([0])
[(0,)]
>>> heaps([-1, 1])
[(-1, 1), (1, -1)]
>>> heaps([1, 2, 3])
[(1, 2, 3), (2, 1, 3), (3, 1, 2), (1, 3, 2), (2, 3, 1), (3, 2, 1)]
>>> from itertools import permutations
>>> sorted(heaps([1,2,3])) == sorted(permutations([1,2,3]))
True
>>> all(sorted(heaps(x)) == sorted(permutations(x))
... for x in ([], [0], [-1, 1], [1, 2, 3]))
True
"""
if len(arr) <= 1:
return [tuple(arr)]
res = []
def generate(k: int, arr: list):
if k == 1:
res.append(tuple(arr[:]))
return
generate(k - 1, arr)
for i in range(k - 1):
if k % 2 == 0: # k is even
arr[i], arr[k - 1] = arr[k - 1], arr[i]
else: # k is odd
arr[0], arr[k - 1] = arr[k - 1], arr[0]
generate(k - 1, arr)
generate(len(arr), arr)
return res
if __name__ == "__main__":
user_input = input("Enter numbers separated by a comma:\n").strip()
arr = [int(item) for item in user_input.split(",")]
print(heaps(arr))