## Problem:

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.

837799

## Code:The solution may include methods that will be found here: Library.java .

public interface EulerSolution{

public String run();

}
/*
* Solution to Project Euler problem 14
* By Nayuki Minase
*
* http://nayuki.eigenstate.org/page/project-euler-solutions
* https://github.com/nayuki/Project-Euler-solutions
*/

import java.math.BigInteger;

public final class p014 implements EulerSolution {

public static void main(String[] args) {
System.out.println(new p014().run());
}

private static final int LIMIT = Library.pow(10, 6);
private static final BigInteger CACHE_SIZE = BigInteger.valueOf(LIMIT);  // Any non-negative number, though there are diminishing returns

public String run() {
int maxArg = -1;
int maxChain = 0;
for (int i = 1; i < LIMIT; i++) {
int chainLen = collatzChainLength(BigInteger.valueOf(i));
if (chainLen > maxChain) {
maxArg = i;
maxChain = chainLen;
}
}
return Integer.toString(maxArg);
}

// Memoization
private int[] collatzChainLength = new int[CACHE_SIZE.intValue()];

private int collatzChainLength(BigInteger n) {
if (n.signum() < 0)
throw new IllegalArgumentException();

if (n.compareTo(CACHE_SIZE) >= 0)  // Caching not available
return collatzChainLengthDirect(n);

int index = n.intValue();  // Index in the cache
if (collatzChainLength[index] == 0)
collatzChainLength[index] = collatzChainLengthDirect(n);
return collatzChainLength[index];
}

private int collatzChainLengthDirect(BigInteger n) {
if (n.equals(BigInteger.ONE))  // Base case
return 1;
else if (!n.testBit(0))  // If n is even
return collatzChainLength(n.shiftRight(1)) + 1;
else  // Else n is odd
}

}