Question

Originally, I was having some issues getting this code to function, but after a little tweaking I got it debugged and ready to go.

I have gone through several revisions of this program. I started with integer values only to find that the number was too large to fit into an int. I then changed to BigIntegers, which proved to be a hassle, but workable. From there, I switched to longs (as should have done from the beginning) and cut the runtime of my code 8-fold (or more).

Here’s the code as it is now:

long qNum = 600851475143L;

for (long i = qNum - 1L; i * i >= qNum; i -= 2L)
    if (qNum % i == 0 && isPrime(i)) {
        System.out.println("Solution:" + i); // for debugging
        return i;
    }
    else
        System.out.println(i);// for debugging

return 0L;

And

public static boolean isPrime(long num) {
    // unnecessary if statement for this problem (b/c of for loop), but useful for others 
    if (num % 2 == 0)
        return false;

    for (long i = 3; i * i <= num; i += 2)
        if (num % i == 0)
            return false;

    return true;
}

It’s been running for multiple hours and it still hasn’t found anything. I saw online that solving this puzzle the typical way is like parsing 560GB of data =/.

Any tips for speeding this up?

Many thanks,

Justian

EDIT:

Optimized code:

public static long greatestPrimeFactor(ArrayList<Long> factors, long num) {
    for (long i = 2; i <= Math.sqrt(num); i++) {
        if (num % i == 0) {
            factors.add(i);
            return greatestPrimeFactor(factors, num / i);
        }
    }

    for (int i = factors.size()-1; i > 0; i--)
        if (isPrime(factors.get(i)))
            return num;

    return 0;
}

AND

public static boolean isPrime(long num) {
if (num % 2 == 0)
    return false;

for (long i = 3; i * i <= num; i += 2)
    if (num % i == 0)
        return false;

    return true;
}

RUN WITH

greatestPrimeFactor(new ArrayList<Long>(), 600851475143L);

Answer 1

My solution hits in less than a hundredth of a second. Each time you find a divisor of the number, divide the number by that divisor and start again. The highest number you divide by is your target.