## Problem 14: Longest Collatz sequence

*Difficulty: Easy

Problem:

The following iterative sequence is defined for the set of positive integers:

nn/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.

How to solve:

The only way I can think of is doing exactly as above ,that is, I generate and count how long a Collatz sequence of a number is.

Solution took 465ms

Implementation: Java

public class Prob14 {

private static int maxStartingNumber = Integer.MIN_VALUE;

publicstaticvoidmain(String[] args) {

int n = 2;

int max = Integer.MIN_VALUE;

long cur = System.currentTimeMillis();

int countRepeat = 0;

for(int i = 1; i < 1000000; ++i) {

long temp = i;

int chain = 0;

while( temp > 1) {

if (temp % 2 == 0) temp /= 2;

else temp = 3*temp + 1;

++chain;

if(temp <= 1) {

chain++; //add the i value itself

if(chain > max) {

max = chain;

maxStartingNumber= i;

}

break;

}

}

}

System.out.println(max + “, “ + maxStartingNumber);

System.out.println(“Took “ + (System.currentTimeMillis()cur) + ” ms”);

}

}