What is the largest prime factor of the number 600851475143(Java)? What could be my mistake?

Question

The code works just fine for the int data type, but 600851475143 seems to be too big of a number. How can I make it work? It just keeps running and never gives me an answer.

public class Main {
    public static void main(String[] args) {
       long a = 600851475143L;
        boolean prime = false;
        long big = 0L;
        for (long i = 1L; i < a; i++){
            if (a % i == 0){
                for (int j = 2; j < i/(float)2; j++){
                    if (i % j == 0){
                        prime = true;
                        break;
                    }
                }
                if(!prime){
                    big = i;
                }
            }
        }
        System.out.println(big);
    }
}

put this `if (i % 1000_000 == 0) System.out.println(i);` at the top of your `for` loop — Scary Wombat, Jan 20 '21 at 06:47
`break;` also only breaks out of the current `for` and as `prime` is not being reset, it seems to be doing nothing. — Scary Wombat, Jan 20 '21 at 06:52
This is Project Euler problem #3. If you search on that you should find some help. — rossum, Jan 20 '21 at 09:04

score 0 · Answer 1 · answered Jan 20 '21 at 06:48

0

You need to use a long also for j.

Your variable naming is also a bit misleading: prime is true when the number is not a prime ...

answered Jan 20 '21 at 06:48

Henry

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This looks like a helpful comment.... but not like an answer. – Yunnosch Jan 20 '21 at 07:04

Juvanis · Answer 2 · 2021-01-20T07:00:57.537

Your code has a lot of problems. My first advice would be to write clean and concise code with well-named variables. Secondly, analyze the runtime complexity of your program even if it works fast for large inputs. The fact that you run an inner loop inside if(a % i == 0) condition, makes your program extremely inefficient.

Here I provide a refactored version of your code with runtime complexity and good variable names in mind:

  public static void main(String[] args) {
    System.out.println(largestPrimeFactorOf(600851475143L));
  }

  public static long largestPrimeFactorOf(long input)
  {
    List<Long> factors = new ArrayList<>();
    // You start from 2, as 1 is not a prime number and also using 1 will cause an infinite loop
    for (long i = 2L; i < input; i++) {
      if (input % i == 0) {
        factors.add(i);
        while (input % i == 0) {
          input /= i;
        }
      }
    }

    // As we always add a bigger number to the factor list, the last element is the greatest factor.
    return factors.get(factors.size() - 1);
  }

Denote that this program will still be slow when the input is a large prime number, e.g. 100000000019. To handle such cases efficiently, it's better to use an efficient primality test algorithm. You can search the web for that. https://en.wikipedia.org/wiki/Primality_test

Helpful. Most of my code is inefficient, I pretty much just started in January, but this website seems like a great resource to help fix that. I just tried solving the problem with the Java I've learned so far, so some of the code you use is new and not yet understandable to me. — Vasila_M, Jan 20 '21 at 07:17
This may be a stupid question, but what does this part of the code do? List factors = new ArrayList<>(); — Vasila_M, Jan 20 '21 at 07:19
The code `List factors = new ArrayList<>();` will initialize `factors` with an empty `List`. There are lots of good tutorials about Java's Generics and Collections for details. — Lars, Jan 22 '21 at 16:06

What is the largest prime factor of the number 600851475143(Java)? What could be my mistake?

2 Answers2