def test_normal():
assert permutations('ABC') == {'ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA'}
assert permutations("PEPE") == {
"EEPP", "EPEP", "EPPE", "PEEP", "PEPE", "PPEE"
}
assert permutations("LMAO") == {
"LMAO", "MLAO", "ALMO", "LAMO", "MALO", "AMLO", "OMLA", "MOLA", "LOMA",
"OLMA", "MLOA", "LMOA", "LAOM", "ALOM", "OLAM", "LOAM", "AOLM", "OALM",
"OAML", "AOML", "MOAL", "OMAL", "AMOL", "MAOL"
}
def bruteForce(string):
permuts = permutations(string)
for p in permuts:
if isPalindrome(p):
return True
return False
def find_mapping(self, digits):
for permutation in permutations(list('abcdefg')):
mapping = dict(zip('abcdefg', permutation))
if all(
self.convert_digit(digit, mapping) in self.digits
for digit in digits):
return mapping
def advantageous_fight_order(breeder1, breeder2):
breeder1_pokemon = breeder1.pokemon_list
breeder2_pokemon = breeder2.pokemon_list
sample_space = [list(zip(x, breeder2_pokemon)) for x in permutations(breeder1_pokemon, len(breeder1_pokemon))]
outcomes = [len(["win" for duel in combination if duel[0].check_odds(duel[1]) == advantage_types[1]])
for combination in sample_space]
wins = [sample_space[index] for index, outcome in enumerate(outcomes) if outcome >= min_wins_to_victory]
return [duel[0] for duel in wins[0]] if len(wins) > 0 else breeder1.pokemon_list
def num_possible_rearrangements(grid):
num = 0
new_list = [x for x in grid]
for perm in permutations([i for i in range(0, len(grid))]):
new_list = apply_permutation(grid, new_list, perm)
if is_magic(new_list):
num += 1
return num
def test_permutations(self):
self.assertEqual(permutations.permutations('HACK', '2'), '''AC
AH
AK
CA
CH
CK
HA
HC
HK
KA
KC
KH''')
def balanced(n):
'''
Given a positive number n, yield all strings of length n, in any order, that only contain balanced brackets. For example:
>>> sorted(list(balanced(6)))
['((()))', '(()())', '(())()', '()(())', '()()()']
'''
if n % 2 != 0 or n < 0:
raise ValueError(
'the input number should be a positive double integer to be balanced'
)
string = ''
for _ in range(n // 2):
string += '('
for _ in range(n // 2):
string += ')'
string = list(permutations(string))
string = list(dict.fromkeys(string)) # remove duplicated string
for bracket in string:
if is_balanced(bracket):
yield bracket
def test_single():
assert list(permutations('a')) == ['a']
def test_single_permutation():
assert sorted(list(permutations('A'))) == ['A']
def test_empty_string():
assert list(permutations('')) == ['']
__licence__ = 'GNU Public Licence (GPL) v3'
import os.path
import sys
# FIXME: the different functions return different sequence structures.
if sys.argv[0] == 'words_iteration.py':
from permutations import permutations_iteration as permutations
elif sys.argv[0] == 'words_comprehension.py':
from permutations import permutations_comprehension as permutations
elif sys.argv[0] == 'words_comprehension_alt.py':
from permutations import permutations_comprehension_alt as permutations
else:
from permutations import permutations_generator as permutations
if __name__ == '__main__':
argCount = len(sys.argv) - 1
if argCount not in (1, 2):
print('Usage: {} <string> [ <dictionary> ]'.format(
os.path.basename(sys.argv[0])))
else:
wordListFileName = sys.argv[
2] if argCount == 2 else '/usr/share/dict/words'
with open(wordListFileName) as file:
wordList = file.read().split()
for trial in permutations(sys.argv[1]):
putative = ''.join(trial)
if putative in wordList:
print(putative)
def test_permutations(self):
self.assertTrue(permutations('act', 'cat'))
self.assertFalse(permutations('', 'foo'))
self.assertFalse(permutations('Nib', 'bin'))
self.assertTrue(permutations('a ct', 'ca t'))
self.assertTrue(permutations('', ''))
def test_repeat():
assert sorted(list(
permutations('ABB'))) == ['ABB', 'ABB', 'BAB', 'BAB', 'BBA', 'BBA']
192 x 1 = 192
192 x 2 = 384
192 x 3 = 576
By concatenating each product we get the 1 to 9 pandigital, 192384576.
We will call 192384576 the concatenated product of 192 and (1,2,3)
The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645,
which is the concatenated product of 9 and (1,2,3,4,5).
What is the largest 1 to 9 pandigital 9-digit number
that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
"""
import permutations
allPermutations = set([permutations.combine_numbers(x) for x in permutations.permutations(range(1, 10))])
highestNumber = 0
for n in xrange(1, 10000): # the was found experimentally
pandigitalMultiply = 0
for i in xrange(1, 10):
pandigitalPart = n * i
pandigitalMultiply *= (10 ** len(str(pandigitalPart)))
pandigitalMultiply += pandigitalPart
if len(str(pandigitalMultiply)) >= 9:
break
if pandigitalMultiply in allPermutations:
if pandigitalMultiply > highestNumber:
def test_permutations():
assert permutations("12") == ["12", "21"]
assert permutations("123") == ['123', '132', '213', '231', '312', '321']
assert permutations("ab ") == ['ab ', 'a b', 'ba ', 'b a', ' ab', ' ba']
def test_case_b(self):
string = 'aabb'
self.assertEqual(permutations(string),
['aabb', 'abab', 'abba', 'baab', 'baba', 'bbaa'])
def test_permutations(self):
self.assertEqual(sorted(["cba", "bca", "bac", "cab", "acb", "abc"]),
sorted(permutations("abc")))
def test_permutations(self):
self.assertEqual(list(permutations([1, 2, 3], 3)),
[(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1),
(3, 1, 2), (3, 2, 1)])
def test_number_permutation():
assert sorted(list(
permutations('123'))) == ['123', '132', '213', '231', '312', '321']
def test_duplicate_permutation():
assert sorted(list(
permutations('ABB'))) == ['ABB', 'ABB', 'BAB', 'BAB', 'BBA', 'BBA']
def test_two_permutation():
assert sorted(list(permutations('AB'))) == ['AB', 'BA']
def test_two():
assert list(permutations('ab')) == ['ab', 'ba']
def test_permutations(self):
self.assertEqual(sorted(permutations('a')), ['a'])
self.assertEqual(sorted(permutations('ab')), ['ab', 'ba'])
self.assertEqual(sorted(permutations('aabb')),
['aabb', 'abab', 'abba', 'baab', 'baba', 'bbaa'])
def test_ABC():
assert list(
permutations('ABC')) == ['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']
import itertools
from permutations import permutations
from primes import primes, is_prime
for p in primes():
if p > 10**4:
break
found = []
for perm in permutations(list(str(p))):
n = int("".join(perm))
if is_prime(n) and len(str(n)) == len(str(p)):
found.append(n)
found = list(set(found))
for f in itertools.combinations(found, 3):
f = list(f)
f.sort()
if f[1] - f[0] == f[2] - f[1]:
print f
def test_all_same():
assert list(
permutations('AAA')) == ['AAA', 'AAA', 'AAA', 'AAA', 'AAA', 'AAA']
from permutations import permutations
from primes import is_prime
digits = list("7654321")
for p in permutations(digits):
if is_prime(int("".join(p))):
print "".join(p)
exit()
""" A permutation is an ordered arrangement of objects. For example, 3124 is one possible permutation of the digits 1, 2, 3 and 4. If all of the permutations are listed numerically or alphabetically, we call it lexicographic order. The lexicographic permutations of 0, 1 and 2 are: 012 021 102 120 201 210 What is the millionth lexicographic permutation of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9? """ import permutations print sorted(permutations.permutations(range(0, 10)))[1000 * 1000 - 1]
__version__ = '1.2'
__copyright__ = 'Copyright © 2007, 2012, 2014 Russel Winder'
__licence__ = 'GNU Public Licence (GPL) v3'
import os.path
import sys
# FIXME: the different functions return different sequence structures.
if sys.argv[0] == 'words_iteration.py':
from permutations import permutations_iteration as permutations
elif sys.argv[0] == 'words_comprehension.py':
from permutations import permutations_comprehension as permutations
elif sys.argv[0] == 'words_comprehension_alt.py':
from permutations import permutations_comprehension_alt as permutations
else:
from permutations import permutations_generator as permutations
if __name__ == '__main__':
argCount = len(sys.argv) - 1
if argCount not in (1, 2):
print('Usage: {} <string> [ <dictionary> ]'.format(os.path.basename(sys.argv[0])))
else:
wordListFileName = sys.argv[2] if argCount == 2 else '/usr/share/dict/words'
with open(wordListFileName) as file:
wordList = file.read().split()
for trial in permutations(sys.argv[1]):
putative = ''.join(trial)
if putative in wordList:
print(putative)
Let d1 be the 1st digit, d2 be the 2nd digit, and so on. In this way, we note the following:
- d2 d3 d4 = 406 is divisible by 2
- d3 d4 d5 = 063 is divisible by 3
- d4 d5 d6 = 635 is divisible by 5
- d5 d6 d7 = 357 is divisible by 7
- d6 d7 d8 = 572 is divisible by 11
- d7 d8 d9 = 728 is divisible by 13
- d8 d9 d10 = 289 is divisible by 17
Find the sum of all 0 to 9 pandigital numbers with this property.
"""
import permutations
allPermutations = permutations.permutations(range(0, 10))
divisibleBy = [2, 3, 5, 7, 11, 13, 17]
sumOfPermutations = 0
def is_special_number(numbers):
i = 1
for d in divisibleBy:
if permutations.combine_numbers(numbers[i : (i + 3)]) % d != 0:
return False
i += 1
return True
for perm in allPermutations:
if is_special_number(perm):
def subpermutations(seq, K):
return sorted(sum([list(permutations.permutations(s)) for s in subsets(seq, K)], []))
def test_case_a(self):
string = 'ab'
self.assertEqual(permutations(string), ['ab', 'ba'])
def test_empty_permutation():
assert sorted(list(permutations(''))) == []
import primes
import permutations
primesToCheck = primes.create_primes(sqrt(987654321))
primeFound = False
def is_prime(number):
limit = sqrt(number)
for p in primesToCheck:
if p > limit:
return True
elif number % p == 0:
return False
return True
for x in xrange(10, 1, -1):
allPermutations = permutations.permutations(range(1, x))
for perm in sorted([permutations.combine_numbers(y) for y in allPermutations], reverse=True):
if is_prime(perm):
print str(perm)
primeFound = True
break
if primeFound:
break
def test_example1():
assert sorted(list(
permutations('ABC'))) == ['ABC', 'ACB', 'BAC', 'BCA', 'CAB', 'CBA']
i = 0
positions = {}
while find_pos(i, map):
positions[i] = (find_pos(i, map))
i += 1
distances = {}
for n, p in positions.items():
for n2, p2 in positions.items():
if n < n2:
distances[(n2, n)] = distances[(n, n2)] = distance(map, p, p2)
for n in positions.keys():
for n2 in positions.keys():
print('{:>3} '.format(distances[(n, n2)] if n != n2 else 0), end='')
print()
to_visit = list(positions.keys())
to_visit.remove(0)
dists = []
for path in permutations.permutations(to_visit):
current = 0
d = 0
for next in path + [0]:
d += distances[(current, next)]
current = next
dists.append(d)
print(min(dists))
def test_example2():
assert sorted(list(
permutations('ABB'))) == ['ABB', 'ABB', 'BAB', 'BAB', 'BBA', 'BBA']
def test_property(n):
prime_factors = list(permutations(n))
for string in prime_factors:
assert len(string) == len(n)
for example, the 5-digit number, 15234, is 1 through 5 pandigital.
The product 7254 is unusual, as the identity, 39 x 186 = 7254,
containing multiplicand, multiplier, and product is 1 through 9 pandigital.
Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.
HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.
"""
import permutations
foundProducts = {}
foundProductsSum = 0
for perm in permutations.permutations(range(1, 10)):
product = permutations.combine_numbers(perm[-4:]) # last four digits
for x in xrange(1, 4):
factor1 = perm[:x] # first x digit
factor2 = perm[x:-4] # starting from x'th digit + 1 (0-based index) all except the last four
factor1 = permutations.combine_numbers(factor1)
factor2 = permutations.combine_numbers(factor2)
if factor1 * factor2 == product:
if product not in foundProducts:
foundProductsSum += product
foundProducts[product] = 1
#print str(factor1) + " x " + str(factor2) + " = " + str(product)