Can you write this program more efficiently?

Question

Hello :)
I am trying to write a program that does the following:
"The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 317584931803?"
Since we are dealing with such a huge number, I've used the BigInteger class to handle it. But the problem is, the following code is inefficient..... I mean, its been calculating for the past 2.5 hours ;( I its still only on the 8th digit. I was wondering if someone could write this more efficiently?
import java.math.*;
public class euler6 {
     public static void main (String arhs[]){
           BigInteger a[]=new BigInteger[999];
           int j=0;
           BigInteger bi = new BigInteger("317584931803"); //*********
           BigInteger i=new BigInteger("2");
           final BigInteger one=new BigInteger("1");     
           final BigInteger zero=new BigInteger("0");     
           boolean state     ;
     for(;i.compareTo(bi)==-1;i=i.add(one))
          { //System.out.println(i);
            //System.out.println(bi.mod(i)+"*");
            //System.out.println(bi.mod(i).equals(zero)+"**");
               if ((bi.mod(i).equals(zero))&& (findPrime(i)))
                    {a[j]=i;
                     j++;}
     for(int f=0;f<a.length;f++)
          System.out.print(a[f]+" ");
     public static boolean findPrime(BigInteger n){
               final BigInteger one=new BigInteger("1");
               final BigInteger zero=new BigInteger("0");
               BigInteger x=new BigInteger("2");
               for(;n.divide(x).compareTo(one)==1;x=x.add(one))
                    if (n.mod(x).equals(zero))          
                         return false;
                    return true;

Answer 1

Hi,
I agree with prev poster, you don't have to use BigInteger since long is sufficient.
Try this, more optimized than initial version, yet still using brute force ... The drawback is, if the given number 317584931803 is a prime number .. then ... it will also take 24 hours to compute ... can't help that.
My idea is, let's say I have number 30.
So, I will search all possibility between 2 to 15. My reason is the factor will not exceed from 30 div 2 which is 15.
While doing brute, I will decrease the iteration boundary.
Eg:
- 30 is succesfully devided by 2, then iteration will stop bruting force up to iteration #15.
- next iteration, 30 is succesfully devided by 3, then iteration will stop bruting force up to iteration #10.
- next iteration, 30 is succesfully devided by 5, then iteration will stop bruting force up to iteration #6.
So, total there is only 4 iterations to check the factor of 30. And I get these number: 2, 15, 3, 10, 5, 6
The next step would be checking which of the numbers are prime, and we get 2, 3, and 5.
My algorithm is something like this:
import java.util.*;
public class PF // stands from Prime Factors ..
public static void main(String s[]) throws Exception {
        long num = 317584931803l;
        List factors = new ArrayList(255);
        long div2 = (long) (num / 2);
        for (int i=2; i<=div2; i++) {
                if (num % i == 0) {
                        div2 = (int) (num / i);
                        factors.add(new Long(i));
                        factors.add(new Long(div2));
                Thread.yield();
        Iterator iter = factors.iterator();
        while (iter.hasNext()) {
                long factor =  ((Long) iter.next()).longValue();
                if (isPrime(factor)) {
                        System.out.println(factor);
static boolean isPrime(long num) throws Exception {
        long div2 = (long) (num / 2);
        for (int i=2; i<div2; i++)  {
                if (num % i == 0) return false;
                Thread.yield();
        return true;
}The reason I put Thread.yield() is so it won't hang your machine ... this happens if the big number you test is a prime number...
My machine (Intel pentium III 700 MHz, 512 MB SDRAM) takes less than 1 second to compute ... ;-)
Alex