# PythonWise

If it won't be simple, it simply won't be. [Hire me, 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=24Note that Python's 2.6 itertools.permutations return the permutation in order sowe 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, ifilterdef 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, n[1:]))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] + sub    head = next_head(n)    return [head] + sorted([n] + 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)`