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Thread: Need a more efficient algorithm for prime numbers.

  1. #1
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    Default Need a more efficient algorithm for prime numbers.

    Here is my current algorithm, the TimeExec is just a class to print out how long it takes to run the code:

    public class Primes {
     
    	/**
    	 * Get a set of prime numbers.
    	 * @param no the number of primes to create
    	 * @return an array containing the requested number
    	 * of primes
    	 */
    	public static int[] getPrimes(int no) {
    		int[] primes = new int[no];
    		int primeInx = 0;
    		int i = 2;
     
    		if (primeInx < no) {
    			primes[primeInx++] = 1;
    		}
     
    		while (primeInx < no) {
    			boolean prime = true;
    			for (int j = 2; j < i; j++) {
     
    				if (i == i / j * j) {
    					prime = false;
    				}
    			}
    			if (prime) {
    				primes[primeInx++] = i;
    			}
    			i++;
    		}
     
    		return primes;
    	}
     
    	public static void main(String[] args) {
    		new TimeExec(new Runnable() {
    			public void run() {
    				int[] primes = getPrimes(1000);
    			}
    		}, "Get 1,000 primes", System.out).start();
     
    		//the 10,000 primes took the library's lab comp 19.125s
    		new TimeExec(new Runnable() {
    			public void run() {
    				int[] primes = getPrimes(10000);
    			}
    		}, "Get 10,000 primes", System.out).start();
     
    		new TimeExec(new Runnable() {
    			public void run() {
    				int[] primes = getPrimes(100000);
    			}
    		}, "Get 100,000 primes", System.out).start();
     
    //		new TimeExec(new Runnable() {
    //			public void run() {
    //				int[] primes = getPrimes(1000000);
    //			}
    //		}, "Get 1,000,000 primes", System.out).start();
    	}
    }

    I need to get it to be able to print out the 1,000,000 primes fairly quickly. Any help would be awesome, thank you!
    Last edited by helloworld922; September 29th, 2012 at 11:39 AM. Reason: please use code tags


  2. #2
    Super Moderator Norm's Avatar
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    Default Re: Need a more efficient algorithm for prime numbers.

    There must be algorithms that google could find. Try that.
    If you don't understand my answer, don't ignore it, ask a question.

  3. #3
    Super Moderator helloworld922's Avatar
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    Default Re: Need a more efficient algorithm for prime numbers.

    Try using a sieve. These are more memory intensive but they're very fast at finding prime numbers less than x.

    For something fairly simple see Wikipedia: Sieve of Eratosthenese.

    A little bit of self-promotion, but for moderately heavy prime generation (basically an optimized Sieve of Eratosthenes with a small hand-coded wheel factorization), see Optimizing the Sieve of Eratosthenes.

    And for your balls-out bonkers version, see Prime Sieve, which I think is the current record holder for consecutive prime number generation. From what I can tell it's a multi-threaded Sieve of Eratosthenes with advanced wheel factorization. Unfortunately it's implemented in C++, not Java.
    Last edited by helloworld922; September 29th, 2012 at 12:26 PM.

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    Default Re: Need a more efficient algorithm for prime numbers.

    Sieves work great for finding lists of primes up to n, much faster than the method you are currently using (for a list of primes to 1000000, sieving returns the list in 64ms on my machine, while brute force returns it in 643 ms). However, I can give you some pointers if you want with your currents algorithm:

    for (int j = 2; j < i; j++) {
     
    				if (i == i / j * j) {
    					prime = false;
    				}
    			}

    First, you upper bound is way to high. j need only go to the square root of i to check all possible factors. Also

    if (i == i / j * j) {
    					prime = false;
    				}

    can be changed to

    if (i % j == 0) {
    					prime = false;
    				}

    That percent sign thingy is called a modulo. What it does is it, in this case, divides i by j and returns the remainder. So, for example, 7 % 2 = 1 because when you long divide, you get 3 remainder 1 or simply 3 and 1/2. If i evenly divides into j, then it will always be zero, if not, then it will be some integer > 0.

    One other thing with building the list your way, using the ArrayList class will be much, much easier than using that array.

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