Project Euler Problem 014

http://projecteuler.net/index.php?section=problems&id=14

The following iterative sequence is defined for the set of positive integers:
 
    n  -> n/2 (n is even)
    n  -> 3n + 1 (n is odd)

Using the rule above and starting with 13, we generate the following sequence:

    13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1

It can be seen that this sequence (starting at 13 and finishing at 1) contains 10 terms.
Although it has not been proved yet (Collatz Problem), it is thought that all starting numbers finish at 1.
Which starting number, under one million, produces the longest chain?
NOTE: Once the chain starts the terms are allowed to go above one million.

地道に探すしかないのかなぁ。とりあえず。
euler014関数を廃止してトップレベルにコードを書くと、2秒くらい遅くなる。謎。

import datetime


def collatz(n):
    if n & 0x01:
        return 3 * n + 1
    else:
        return n >> 1



def euler014(func, N):
    cache = [ 1 ] * N
    
    maxc = 0
    maxi = 0
    
    for i in xrange(2, N):
        n, c = i, 0
        while n >= i:
            n = func(n)
            c += 1
        else:
            c += cache[n]
            cache[i] = c
            
        if maxc < c:
            maxc, maxi = c, i
            
    return maxi, maxc



begin = datetime.datetime.now()

N = 1000000

print euler014(collatz, 1000000)

end = datetime.datetime.now()
print end - begin

答え: 837799 (長さ=525)
実行時間: 3.085142秒くらい