Java solution to Project Euler Problem 14

Problem 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.

Running time: 6100ms (6.1 sec)

Assessment: I had to read this problem several times before I really grokked it, and then I thought it was pretty nifty. I’m sure there’s a faster way of working out the answer than my brute force solution.

6 thoughts on “Java solution to Project Euler Problem 14

  1. package CoreJavaAssignments;

    public class Problem14 {

    public static void main(String[] args) {
    long begin = System.currentTimeMillis();
    int maxLength = 1;
    int number = 0;
    for (int n = 1; n maxLength) {
    maxLength = length;
    number = n;
    }

    }
    long end = System.currentTimeMillis();
    System.out.println(“Number:” + number + “Length:” + maxLength);
    System.out.println(end – begin + “ms”);
    }

    private static int findTheChain(int n) {
    long localNumber = n;
    int count = 0;

    while (localNumber != 1) {
    if (localNumber % 2 == 0) {
    localNumber = localNumber / 2;
    count++;

    } else {
    localNumber = (3 * localNumber) + 1;
    count++;
    }

    }

    return count;
    }
    }

  2. public class EulerProject14
    {
    public static void main(String[] args)
    {
    int kingnum = 0;
    int bignum = 0;
    int count =0;
    for( int i=10; i=kingnum)
    {
    kingnum = count;
    bignum = i;
    }

    }
    System.out.println(bignum);
    }
    }

    mine minutes to run how can i make it faster

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