for loop finding the prime numbers

Question

I am trying to run this code to print the sum of all the prime numbers less than 2 million. This loop is never ending. Can anyone tell me what is wrong with the code? It see

Answer 1

Here is a complete program of Prime using JOptionPane, i.e. Java GUI

import javax.swing.*;

public class ChkPrime {
    public static void main(String[] args) throws NumberFormatException {
        String str = JOptionPane.showInputDialog(null, "Enter any number: ","Input...", 3);

        try {
            int num = Integer.parseInt(str);


            if (num == 1)
                JOptionPane.showMessageDialog(null, "Your inputed no. " + num + " is not prime.","Error!", 0);

            for(int i = 2; i <= Math.sqrt(num); i++) {
                if(num % i == 0) {
                    JOptionPane.showMessageDialog(null, "Your inputed no. " + num + " is not prime.","Error!", 0);
                    System.exit(0);
                }
            }

            JOptionPane.showMessageDialog(null, "Your inputed no. " + num + " is prime.","Yeh! Got it!", 1);
        }

        catch (NumberFormatException e) {
            JOptionPane.showMessageDialog(null, "Please input numbers...","Error!", 0);
            main(null);
        }
    }
}

Answer 2

Tested and bug free Prime check function

static boolean isPrime(int n) {
    if (n == 1) return false;

    for(int i = 2; i <= n/2; i++)
        if(n % i == 0)
            return false;

    return true;
}

Answer 3

You have a bug in isPrime()

The test should be:

if(n%i == 0) { ...

and you need to start counting at 2, not 1, because every number has a remainder of zero when divided by 1!

Also, no need to go past Math.sqrt(n).

You should change it to this:

private static boolean isPrime(long n) {
    long max = (long)Math.sqrt(n);
    for (long i = 2; i < max; i++) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

FYI, with this change, I tested the program on my PC and it completed in under 1 second, giving the result of 143064094810

Answer 4

A naive isPrime function must calculate all the primes up to i (or at least up to sqrt(i)) each time it runs. Make sure your isPrime function caches its results!

Answer 5

In isPrime you are only testing division by 2:

private static boolean isPrime(long n) {
    boolean result = false;

    for(long i=1; i<n/2; i++) {
        if(n%2 == 0) {
            result = false;
            break;
        }
        else result = true;
    }
    return result;

}

it should be division by every i and starting from 2:

for(long i=2; i<n/2; i++) {
    if(n%i == 0) {
      ...

Practically in your current version an odd number n will keep dividing by 2 up to n/2 instead of stopping much sooner. Consider n = 21. You are dividing by 2 from 1 to 10, instead of dividing by 3 at the 3rd step and exiting.

It not only gives incorrect results, but also takes much longer than needed to reach a return statement.

Edit: For faster results check out this sieve of Erathostenes method:

public static long sumOfPrimes(int n) {

    long sum = 0;

    boolean[] sieve = new boolean[n];
    for(int i = 2; i < Math.sqrt(n); i++) {
        if(!sieve[i]) {
            for(int j = i * i; j < n; j += i) {
                sieve[j] = true;
            }
        }
    }

    for(int i = 2; i < n; i++) {
        if(!sieve[i]) {             
            sum += i;
        }
    }

    return sum;
}

Edit #2: Found some bugs with your new version. Here's the corrected one:

private static boolean isPrime(long n) {
    boolean result = false;

    if(n == 2 || n == 3) return true;

    for (long i = 2; i <= (long) Math.sqrt(n); i++) {
        if (n % i == 0) {
            result = false;
            break;
        } else
            result = true;
    }

    System.out.println(n + " " + result);
    return result;
}