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| 1 | +#!/usr/bin/env python |
| 2 | + |
| 3 | +""" |
| 4 | +This module presents two different solutions for the problem of generating the permutations of a sequence. |
| 5 | +""" |
| 6 | + |
| 7 | +from math import factorial |
| 8 | +import unittest |
| 9 | + |
| 10 | + |
| 11 | +def perm_recursive(S): |
| 12 | + """ |
| 13 | + Return a list with all permutations of the iterable passed as argument. |
| 14 | + Uses the simple recursive solution. This Algorithm does not handle repeated elements well. |
| 15 | + """ |
| 16 | + |
| 17 | + def expand_inserting(c, L): |
| 18 | + return [L[0:i] + [c] + L[i:] for i in range(len(L) + 1)] |
| 19 | + |
| 20 | + if not isinstance(S, list): |
| 21 | + S = list(S) # to handle strings |
| 22 | + if len(S) == 0: |
| 23 | + return [] |
| 24 | + elif len(S) == 1: |
| 25 | + return [S] |
| 26 | + |
| 27 | + c, L = S[0], S[1:] |
| 28 | + res = [] |
| 29 | + for newL in perm_recursive(L): |
| 30 | + res.extend(expand_inserting(c, newL)) |
| 31 | + return res |
| 32 | + |
| 33 | + |
| 34 | +def perm_generator(A): |
| 35 | + """ |
| 36 | + Generates all permutations of elements of the iterable passed as argument. |
| 37 | + Uses Non-recursive lexicographic order (Knuth's L-Algorithm). |
| 38 | + Requires all A elements are comparable to each other (list has to be sortable). |
| 39 | + The algorithm handles repeated elements gracefully. |
| 40 | + Reference: https://en.wikipedia.org/wiki/Permutation#Permutations_in_computing |
| 41 | + """ |
| 42 | + n = len(A) |
| 43 | + a = sorted(list(A)) |
| 44 | + ok = False if n == 0 else True |
| 45 | + |
| 46 | + while ok: |
| 47 | + yield a[:] |
| 48 | + # Find the largest index k such that a[k] < a[k + 1] |
| 49 | + k = next((j for j in range(n-2, -1, -1) if a[j] < a[j+1]), -1) |
| 50 | + if k == -1: # if no such index, is last permutation |
| 51 | + break |
| 52 | + # Find the largest index l greater than k such that a[k] < a[l] |
| 53 | + l = next((j for j in range(n-1, k, -1) if a[k] < a[j]), -1) |
| 54 | + # Swap values |
| 55 | + a[k], a[l] = a[l], a[k] |
| 56 | + # Reverse the sequence from a[k + 1] up to and including the final element a[n] |
| 57 | + a[k+1:] = a[n-1:k:-1] |
| 58 | + |
| 59 | + |
| 60 | +class Test(unittest.TestCase): |
| 61 | + |
| 62 | + def setUp(self): |
| 63 | + self.perm_ABC = (('A', 'B', 'C'), ('B', 'A', 'C'), ('B', 'C', 'A'), ('A', 'C', 'B'), ('C', 'A', 'B'), |
| 64 | + ('C', 'B', 'A')) |
| 65 | + |
| 66 | + def test_recursive1(self): |
| 67 | + """ Permutations of list [A,B,C] correct """ |
| 68 | + A = ['A', 'B', 'C'] |
| 69 | + r = perm_recursive(A) |
| 70 | + self.assertEqual(set(tuple([tuple(l) for l in r])), set(self.perm_ABC)) |
| 71 | + self.assertEqual(len(r), factorial(len(A))) |
| 72 | + |
| 73 | + def test_recursive2(self): |
| 74 | + """ Permutations of str "ABC" correct """ |
| 75 | + A = "ABC" |
| 76 | + r = perm_recursive(A) |
| 77 | + self.assertEqual(set(tuple([tuple(l) for l in r])), set(self.perm_ABC)) |
| 78 | + self.assertEqual(len(r), factorial(len(A))) |
| 79 | + |
| 80 | + def test_recursive3(self): |
| 81 | + """ Empty string produces empty list """ |
| 82 | + A = "" |
| 83 | + r = perm_recursive(A) |
| 84 | + self.assertEqual(r, []) |
| 85 | + |
| 86 | + def test_recursive4(self): |
| 87 | + """ A sigle letter produces only one permutation """ |
| 88 | + A = "A" |
| 89 | + r = perm_recursive(A) |
| 90 | + self.assertEqual(r, [['A']]) |
| 91 | + |
| 92 | + def test_recursive5(self): |
| 93 | + """ Two repeated letter wrongly produce 2 permutations. Algorithm does not handle repeated elements! """ |
| 94 | + A = "AA" |
| 95 | + r = perm_recursive(A) |
| 96 | + self.assertEqual(r, [['A', 'A'], ['A', 'A']]) |
| 97 | + |
| 98 | + def test_generator1(self): |
| 99 | + """ Generated permutations of list [A,B,C] correct """ |
| 100 | + A = ['A', 'B', 'C'] |
| 101 | + r = list(perm_generator(A)) |
| 102 | + self.assertEqual(set(tuple([tuple(l) for l in r])), set(self.perm_ABC)) |
| 103 | + self.assertEqual(len(r), factorial(len(A))) |
| 104 | + |
| 105 | + def test_generator2(self): |
| 106 | + """ Generated Permutations of str "ABC" correct """ |
| 107 | + A = "BAC" |
| 108 | + r = list(perm_generator(A)) |
| 109 | + self.assertEqual(set(tuple([tuple(l) for l in r])), set(self.perm_ABC)) |
| 110 | + self.assertEqual(len(r), factorial(len(A))) |
| 111 | + |
| 112 | + def test_generator3(self): |
| 113 | + """ Two repeated letters generate only one permutation. Algorithm handles repeated elements well. """ |
| 114 | + A = "AA" |
| 115 | + r = list(perm_generator(A)) |
| 116 | + self.assertEqual(r, [['A', 'A']]) |
| 117 | + |
| 118 | + |
| 119 | +if __name__ == '__main__': |
| 120 | + unittest.main() |
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