寻找600851475143中最大的素数？

Question

I'm trying to solve problem 3 from http://projecteuler.net . However, when I run thing program nothing prints out. What am I doing wrong? Problem: What is the largest prime factor of the number 600851475143 ?

public class project_3 
{
    public boolean prime(long x)   // if x is prime return true
    {
        boolean bool = false;

        for(long count=1L; count<x; count++)
        {
            if( x%count==0 )
            {
                bool = false;
                break;
            }
            else { bool = true; }
        }
        return bool;
    }

    public static void main(String[] args)
    {
        long ultprime = 0L;  // largest prime value
        project_3 object = new project_3();

        for(long x=1L; x <= 600851475143L; x++)
        {
            if( object.prime(x)==true )
            {
                ultprime = ((x>ultprime) ? x : ultprime);
            }
        }
        System.out.println(ultprime);
    }
}

Answer 1

Not only does your prime checking function always return false ; even if it were functioning properly, your main loop does not seek the input number's factors at all, but rather just the largest prime smaller or equal to it. In pseudocode, your code is equivalent to:

foo(n):
    x := 0 ;
    foreach d from 1 to n step 1:
        if is_prime(d):          // always false
            x := d
    return x                     // always 0

is_prime(d):
    not( d % 1 == 0 )            // always false

But you don't need the prime checking function here at all. The following finds all factors of a number, by trial division :

factors(n):
    fs := []
    d  := 2
    while ( d <= n/d ):
        if ( n % d == 0 ): { n := n/d ; fs := append(fs,d) }
        else:              { d := d+1 }
    if ( n > 1 ): { fs := append(fs, n) }
    return fs

The testing for divisibility is done only up to the square root of the number. Each factor, as it is found, is divided out of the number being factorized, thus further reducing the run time. Factorization of the number in question runs instantly, taking just 1473 iterations.

By construction all the factors thus found are guaranteed to be prime (that's why no prime checking is needed). It is crucial to enumerate the possible divisors in ascending order for this to happen ¹ . Ascending order is also the most efficient , because any given number is more likely to have smaller prime factor than larger one. Enumerating the primes instead of odds, though not necessary, will be more efficient if you have an efficient way of getting those primes, to test divide by.

It is trivial to augment the above to find the largest factor: just implement append as

append(fs,d):
    return d

---

¹ because then for any composite divisor d of the original number being factorized, when we'll reach d , we will have already divided its prime factors out of the original number, and so the reduced number will have no common prime factors with it, ie d won't divide the reduced number even though it divides the original.

Answer 2

Two things:

1) You are starting count at 1 instead of 2. All integers are divisible by 1.

2) You are running an O(n^2) algorithm against a rather large N (or at least you will be once you fix point #1). The runtime will be quite long.

Answer 3

The whole point of Project Euler is that the most obvious approaches to finding the answer will take so long to compute that they aren't worth running. That way you learn to look for the less obvious, more efficient approaches.

Your approach is technically correct in terms of whether or not it is capable of computing the largest prime of some number. The reason you aren't seeing anything print out is that your algorithm is not capable of solving the problem quickly.

The way you've designed this, it'll take somewhere around 4,000,000 years to finish.

If you replaced the 600851475143 number with say 20 it would be able to finish fairly quickly. But you have the 600 billion number, so it's not that simple.