Division By Zero Error For Project Euler?

Question

I am trying to solve some project Euler problems and for whatever reason the following code gives me a division by zero error whenever I try to run it with large numbers. Can anyone tell me why?

import java.util.Scanner;

    public class Problem3LargestPrimeFactor {
        public static void main(String[] args) {
            Scanner input = new Scanner(System.in);

            System.out.println("Please enter the number to find the largest prime factor for.");
            long num = input.nextLong();
            int largest = 1;
            boolean isPrime = true;

            for(int i = 2; i < num; i++) {
                if(num % i == 0) {
                    for(int u = 2; u < i; u++){
                if(i % u == 0)
                    isPrime = false;
            }
            if(isPrime)
                largest = i;
            }
        }
    }
}

Now I'm aware that this is not the most efficient way to design the algorithm but can anyone tell me what is going on here?

"...with large numbers". Sounds like you're overflowing the capacity of `int`. — woz, Dec 17 '13 at 21:05
`num` is a `long`, but `i` is only an `int`. So if `num` is bigger than the maximum integer, `i++` will eventually overflow. I guess it eventually reaches 0, whereupon `num % i` will raise a division by zero error. — Gareth Rees, Dec 17 '13 at 21:06
thank you! I can't believe something so simple slipped my mind. — , Dec 17 '13 at 21:08
Have you seen this algorithm [sieve of eratosthenes](http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes) since you already know your limit may be useful — Felix Castor, Dec 17 '13 at 21:09

woz · Answer 1 · 2013-12-17T21:25:14.620

0

You're overflowing int. In the loop for(int i = 2; i < num; i++), num is a long and i is only an int. When i reaches it's capacity, it wraps around to -2,147,483,648 and keeps being incremented until it gets to 0, at which point you get your error.

edited Dec 17 '13 at 21:25

answered Dec 17 '13 at 21:08

woz

10,888
3
34
64

woz, I am using the code with numbers in the range of long and it is still giving me a division by zero error. – Dec 17 '13 at 21:11
Even if you are in the range of a long, when you do `for(int i = 2; i < num; i++)`, `i` can get as big as `num`, and `i` is only an `int`. – woz Dec 17 '13 at 21:18
"any numbers larger than that could give you an error" -> Not necessarily, this would overflow back around to -2,147,483,648 and continue along all the way up until 0, at which it will error. – Sinkingpoint Dec 17 '13 at 21:19

score 0 · Answer 2 · answered Dec 18 '13 at 03:15

Other responses have already pointed out your error. Let me give you a better algorithm for factoring a composite integer using trial division. The basic idea is to simply iterate through the possible factors, reducing n each time you find one. Here's pseudocode:

function factors(n)
    f, fs := 2, {}
    while f * f <= n
        while n % f == 0
            append f to fs
            n := n / f
        f := f + 1
    if n > 1
        append n to fs
    return fs

If you're interested in programming with prime numbers, I modestly recommend this essay at my blog.

Division By Zero Error For Project Euler?

2 Answers2