[PD] Sieve of erastothen

Hello

I would like to know if there is a way to compute the Sieve of erastothenes with pure data

and spit out numbers which can be used for MIDI notes.

There used to be a simple unix program that i used but it no longer is available.

Has anyone tried this with pd? I found some Java code for it.

pp

/*************************************************************************

• Compilation: javac PrimeSieve.java
• Execution: java -Xmx1100m PrimeSieve N
• Computes the number of primes less than or equal to N using
• the Sieve of Eratosthenes.
• % java PrimeSieve 25
• The number of primes <= 25 is 9
• % java PrimeSieve 100
• The number of primes <= 100 is 25
• % java -Xmx100m PrimeSieve 100000000
• The number of primes <= 100000000 is 5761455
• % java PrimeSieve -Xmx1100m 1000000000
• The number of primes <= 1000000000 is 50847534
• The 110MB and 1100MB is the amount of memory you want to allocate
• to the program. If your computer has less, make this number smaller,
• but it may prevent you from solving the problem for very large
• values of N.

•              N     Primes <= N
  

• ---

•             10               4
  

•            100              25
  

•          1,000             168
  

•         10,000           1,229
  

•        100,000           9,592
  

•      1,000,000          78,498
  

•     10,000,000         664,579
  

•    100,000,000       5,761,455
  

•  1,000,000,000      50,847,534
  

*************************************************************************/

public class PrimeSieve { public static void main(String[] args) { int N = Integer.parseInt(args[0]);

    // initially assume all integers are prime
    boolean[] isPrime = new boolean[N + 1];
    for (int i = 2; i <= N; i++) {
        isPrime[i] = true;
    }

    // mark non-primes <= N using Sieve of Eratosthenes
    for (int i = 2; i*i <= N; i++) {

        // if i is prime, then mark multiples of i as nonprime
        // suffices to consider mutiples i, i+1, ..., N/i
        if (isPrime[i]) {
            for (int j = i; i*j <= N; j++) {
                isPrime[i*j] = false;
            }
        }
    }

    // count primes
    int primes = 0;
    for (int i = 2; i <= N; i++) {
        if (isPrime[i]) primes++;
    }
    System.out.println("The number of primes <= " + N + " is " + primes);
}

}

Patrick Pagano B.S, M.F.A Audio and Projection Design Faculty Digital Worlds Institute University of Florida, USA (352)294-2020