If it won't be simple, it simply won't be. [source code] by Miki Tebeka, CEO, 353Solutions

Wednesday, April 22, 2009

Solving Euler Question 24

I'm learning Clojure by solving project Euler. Here is a python version of one of the solutions.

#!/usr/bin/env python
''' Solving http://projecteuler.net/index.php?section=problems&id=24

Note that Python's 2.6 itertools.permutations return the permutation in order so
we can just write:
from itertools import islice, permutations
print islice(permutations(range(10)), 999999, None).next()

And it'll work much faster :)
'''

from itertools import islice, ifilter

def is_last_permutation(n):
return n == sorted(n, reverse=1)

def next_head(n):
'''Find next number to be 'head'.

It is smallest number if the tail that is bigger than the head.
In the case of (2 4 3 1) it will pick 3 to get the next permutation
of (3 1 2 4)
'''
return sorted(filter(lambda i: i > n[0], n[1:]))[0]

def remove(element, items):
return filter(lambda i: i != element, items)

def next_permutation(n):
if is_last_permutation(n):
return None

sub = next_permutation(n[1:])
if sub:
return [n[0]] + sub

head = next_head(n)
return [head] + sorted([n[0]] + remove(head, n[1:]))

def nth(it, n):
'''Return the n'th element of an iterator'''
return islice(it, n, None).next()

def iterate(func, n):
'''iterate(func, n) -> n, func(n), func(func(n)) ...'''
while 1:
yield n
n = func(n)

def permutations(n):
return ifilter(None, iterate(next_permutation, n))

if __name__ == "__main__":
n = range(10)
m = 1000000
print "Calculateing the %d permutation of %s" % (m, n)
print nth(permutations(n), m-1)
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