Problem 14 - Find the longest sequence using a starting number under one million.

링크

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.

그 유명한 3n+1 문제구나.

다 계산해보는 수 밖에 별다른 수가 보이지 않는다.

python

d = {1:1} def func(n): if d.has_key(n): return d[n] if n & 1 == 0: d[n] = func(n / 2) + 1 else: d[n] = func(3 * n + 1) + 1 return d[n] longest = 0 startnum = 0 for i in range(1, 1000000): length = func(i) if length > longest: longest = length startnum = i print startnum

파이썬으로 memoization도 해보는구나