47 lines
1.4 KiB
Python
47 lines
1.4 KiB
Python
from lib_misc import permutations
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def euler_024_library():
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from itertools import permutations
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return int("".join(list(permutations("0123456789"))[1000000 - 1]))
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def nth(generator, n):
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for i in range(n):
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next(generator)
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return next(generator)
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def nth_permutation(iterable, n):
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""" Returns the nth permutation of the iterable. """
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def iterate_permutations(accumulator, iterable, count=[0]):
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if not iterable:
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if count[0] == n:
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return accumulator
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count[0] += 1
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return False
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for i in range(len(iterable)):
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elem = iterable[i:i + 1]
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rest = iterable[:i] + iterable[i + 1:]
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p = iterate_permutations(accumulator + elem, rest)
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if p:
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return p
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return False
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return iterate_permutations([], iterable)
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def euler_024():
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""" My first solution used a generator to get all permutations till
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the desired one. My improved solution uses a function that returns only
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the nth iteration which is the same algorithm, but does not required to
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return all the permutations that are discarded anyway. """
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return int("".join(nth_permutation(list("0123456789"), 1000000 - 1)))
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# return int(nth(permutations("0123456789"), 1000000 - 1))
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if __name__ == "__main__":
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print("e024.py: {}".format(euler_024()))
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assert(euler_024() == 2783915460)
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