Project Euler Problem 14

Longest Collatz sequence

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. Link to original description
Source code examples on Github

Erlang solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

#!/usr/bin/env escript
%% -*- erlang -*-
%%! -smp enable -sname p14
% vim:syn=erlang

-mode(compile).

main(_) ->
    io:format("Answer: ~p ~n", [test14()]).

test14()                          -> test14i(2, 1, 1).
test14i(N, C, L) when N < 1000000 ->
    case lseq(N, 0) of
        Length when Length > L -> test14i(N+1, N, Length);
        _                      -> test14i(N+1, C, L)
    end;
test14i(_, C, _)                  -> C.

lseq(1, N)                    -> N;
lseq(P, N) when P rem 2 =:= 0 -> lseq(P div 2, N+1);
lseq(P, N)                    -> lseq(P*3 + 1, N+1).

Python solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

#!/usr/bin/python

cache = { 1: 1 }

def chain(cache, n):
    if not cache.get(n,0):
        if n % 2: cache[n] = 1 + chain(cache, 3*n + 1)
        else: cache[n] = 1 + chain(cache, n/2)
    return cache[n]

m,n = 0,0
for i in xrange(1, 1000000):
    c = chain(cache, i)
    if c > m: m,n = c,i

print n

Performance

1
2
3
4
5
6
7
8
9
10
11
12

mkh@mkh-xps:~/work/mblog/pr_euler/p14$ time ./p14.erl
Answer: 837799

real    0m5.928s
user    0m5.931s
sys     0m0.035s
mkh@mkh-xps:~/work/mblog/pr_euler/p14$ time ./p14.py
837799

real    0m1.600s
user    0m1.543s
sys     0m0.056s