Sieve Of Atkin, Improved

February 19, 2010

We examined the Sieve of Atkin for enumerating prime numbers in a recent exercise. Though the Sieve of Atkin has the potential to be faster than the Sieve of Eratosthenes, our implementation was not; the Sieve of Eratosthenes takes 1.219 seconds to enumerate the first million primes, our implementation of the Sieve of Atkins takes 22.954 seconds, nineteen times as long, to perform the same task.

The problem is that the naive implementation we used performs useless work. For instance, consider that you are sieving the numbers to a million, with the x and y variables of the naive implementation ranging from one to a thousand. In the first case, with 4x² + y² = k, 4·1000² + 1000² = 5000000, which is far beyond the end of the range being sieved. Reducing the limits means a lot less work and a much faster program.

Your task is to rewrite the naive Sieve of Atkins to eliminate useless work. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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4 Responses to “Sieve Of Atkin, Improved”

  1. Mike said

    Here is my python version of an improved Sieve of Atkins. On my laptop it calculates the primes up to 1 million in .16 seconds, and primes to 10 million in about 1.5 seconds. That is 3.5 times faster than a basic Sieve of Eratosthenes and about 5.6 times faster than a naive Sieve of Atkins.

    To imptove the naive sieve, I first calculated appropriate loop limits. Second I replaced squaring of x and y with additions using the well known formula:(x+1)^2 = x^2 + (2x + 1), and changed the loop counters to return the difference between the squares (i.e., the 2x+1). Obviously, the factor on the x^2 term could also be included in the loop counter. For example, for the 4x^2 term goes 4 16 36 … , so the loop was set up to return the differences 4, 12, 20, ….

    Because of the way the loops counters incremented, I suspected there would be a pattern to the results of the n%12 calculations in the y^2 loops. After confirming the pattern, I changed the loop variables and split some loops in two, so that the loops only iterate over values of x and y for which the n%12 test would pass. This greatly reduced the number of loop iterations and eliminated the n%12 test. The start, stop, and step values for the loops were tricky to figure out.

    Although I plan further optimization of the 3x^2 – y^2 section when I have time, I thought I’d post what I have now. Also, I didn’t optimize memory usage at all, so I’m sure that can be reduced.

    def atkins13(limit=1000000):
        '''use sieve of atkins to find primes <= limit.'''
        primes = [0] * limit
    
        # n = 3x^2 + y^2 section
        xx3 = 3
        for dxx in xrange( 0, 12*int( sqrt( ( limit - 1 ) / 3 ) ), 24 ):
            xx3 += dxx
            y_limit = int(12*sqrt(limit - xx3) - 36)
            n = xx3 + 16
            for dn in xrange( -12, y_limit + 1, 72 ):
                n += dn
                primes[n] = not primes[n]
    
            n = xx3 + 4
            for dn in xrange( 12, y_limit + 1, 72 ):
                n += dn
                primes[n] = not primes[n]
    
        # n = 4x^2 + y^2 section
        xx4 = 0
        for dxx4 in xrange( 4, 8*int(sqrt((limit - 1 )/4)) + 4, 8 ):
            xx4 += dxx4
            n = xx4+1
    
            if xx4%3:
                for dn in xrange( 0, 4*int( sqrt( limit - xx4 ) ) - 3, 8 ):
                    n += dn
                    primes[n] = not primes[n]
    
            else:
                y_limit = 12 * int( sqrt( limit - xx4 ) ) - 36
    
                n = xx4 + 25
                for dn in xrange( -24, y_limit + 1, 72 ):
                    n += dn
                    primes[n] = not primes[n]
    
                n = xx4+1
                for dn in xrange( 24, y_limit + 1, 72 ):
                    n += dn
                    primes[n] = not primes[n]
    
        # n = 3x^2 - y^2 section
        xx = 1
        for x in xrange( 3, int( sqrt( limit / 2 ) ) + 1, 2 ):
            xx += 4*x - 4
            n = 3*xx
    
            if n > limit:
                min_y = ((int(sqrt(n - limit))>>2)<<2)
                yy = min_y*min_y
                n -= yy
                s = 4*min_y + 4
    
            else:
                s = 4
    
            for dn in xrange( s, 4*x, 8 ):
                n -= dn
                if n <= limit and n%12 == 11:
                        primes[n] = not primes[n]
    
        xx = 0
        for x in xrange( 2, int( sqrt( limit / 2 ) ) + 1, 2 ):
            xx += 4*x - 4
            n = 3*xx
    
            if n > limit:
                min_y = ((int(sqrt(n - limit))>>2)<<2)-1
                yy = min_y*min_y
                n -= yy
                s = 4*min_y + 4
    
            else:
                n -= 1
                s = 0
    
            for dn in xrange( s, 4*x, 8 ):
                n -= dn
                if n <= limit and n%12 == 11:
                        primes[n] = not primes[n]
    
        # eliminate squares        
        for n in xrange(5,int(sqrt(limit))+1,2):
            if primes[n]:
                for k in range(n*n, limit, n*n):
                    primes[k] = False
    
        return [2,3] + filter(primes.__getitem__, xrange(5,limit,2))
    
  2. Garrett said

    I made the above code MUCH faster by simply replacing the line:

    primes = [0] * limit

    with:

    primes = numpy.zeros(limit, dtype = ‘bool’)

    Obviously this requires the module numpy. This also helps considerably with the “memory management” he was talking about I would think!

    I’m still trying to wrap my head around the rest of of the code (really trying to wrap my head around the sieve in genera). Will post when I find out more!

  3. geeksumm said

    It’s simple. We can cut the loop by using these condition:
    if (xx + yy >= limit) {
    break;
    }

    Using JavaScript running on Chrome’s V8, it took about 2.8 seconds to generate primes with limit of 10 million. Here’s the code:

    function sieveOfAtkin(limit){
    limitSqrt = Math.sqrt(limit);
    sieve = [];

    //prime start from 2, and 3
    sieve[2] = true;
    sieve[3] = true;

    for (x = 1; x <= limitSqrt; x++) {
    xx = x*x;
    for (y = 1; y = limit) {
    break;
    }
    // first quadratic using m = 12 and r in R1 = {r : 1, 5}
    n = (4 * xx) + (yy);
    if (n <= limit && (n % 12 == 1 || n % 12 == 5)) {
    sieve[n] = !sieve[n];
    }
    // second quadratic using m = 12 and r in R2 = {r : 7}
    n = (3 * xx) + (yy);
    if (n y && n <= limit && (n % 12 == 11)) {
    sieve[n] = !sieve[n];
    }
    }
    }

    for (n = 5; n <= limitSqrt; n++) {
    if (sieve[n]) {
    x = n * n;
    for (i = x; i <= limit; i += x) {
    sieve[i] = false;
    }
    }
    }

    // primes are the one which sieve[n] returns true
    return sieve;
    }

  4. Harsh said

    Could you please explain how the you got the limits?

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