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22:54, 18 February 2012 MN (talk | contribs) New thread created  
10:12, 19 February 2012 Skilgannon (talk | contribs) New reply created (Reply to Fast gaussian kernel function)
18:21, 19 February 2012 MN (talk | contribs) New reply created (Reply to Fast gaussian kernel function)

Fast gaussian kernel function

Patched a gaussian kernel to work together with that fast exp.

private static final double SQRT_2_PI = Math.sqrt(2 * Math.PI);
private static final double EXP_LIMIT = 700;
private static final double GAUSSIAN_LIMIT = Math.sqrt(EXP_LIMIT * 2);

public static double gaussian(final double u) {
    if (u > GAUSSIAN_LIMIT || u < -GAUSSIAN_LIMIT) {
        return 0;
    }
    return exp(-(u * u) / 2) / SQRT_2_PI;
}

public static double exp(double val) {
    final long tmp = (long) (1512775 * val + 1072632447);
    return Double.longBitsToDouble(tmp << 32);
}
MN22:54, 18 February 2012

A suggestion: get rid of divisions, they are almost as slow as sqrt and slow the code down a lot. Also, sometimes it is important that the gaussian doesn't return 0, so change that to something a tiny bit bigger =)


private static final double SQRT_2_PI = Math.sqrt(2 * Math.PI);
private static final double EXP_LIMIT = 700;
private static final double GAUSSIAN_LIMIT = Math.sqrt(EXP_LIMIT * 2);

public static double gaussian(final double u) {
    if (u > GAUSSIAN_LIMIT || u < -GAUSSIAN_LIMIT) {
        return 2e-200;
    }
    return exp((u * u) * -0.5) * (1 / SQRT_2_PI);
}

public static double exp(double val) {
    final long tmp = (long) (1512775 * val + 1072632447);
    return Double.longBitsToDouble(tmp << 32);
}
Skilgannon10:12, 19 February 2012

That "get rid of divisions" tip doubled the speed of my bot. O.o'

And for a mathematical purism, returning the value at the boundaries as minimum to keep the function smooth:

private static final double SQRT_2_PI_INVERSE = 1 / Math.sqrt(2 * Math.PI);
private static final double EXP_LIMIT = 700;
private static final double GAUSSIAN_LIMIT = Math.sqrt(EXP_LIMIT * 2);
private static final double MIN_GAUSSIAN_VALUE = gaussian(GAUSSIAN_LIMIT);

public static double gaussian(final double u) {
    if (u > GAUSSIAN_LIMIT || u < -GAUSSIAN_LIMIT) {
        return MIN_GAUSSIAN_VALUE;
    }
    return exp(u * u * -0.5) * SQRT_2_PI_INVERSE;
}
 
public static double exp(final double val) {
    final long tmp = (long) (1512775 * val + 1072632447);
    return Double.longBitsToDouble(tmp << 32);
}
MN18:21, 19 February 2012