Fast sqrt in Java at the expense of accuracy

只愿长相守 提交于 2019-11-28 23:50:50
Martin Thurau

I don't believe (without a benchmark to prove this wrong) that a pure Java implementation could me much faster than Math.sqrt(). Both the Oracle JRE implementation and the OpenJDK implementation are native implementations.

Once you give the code time to warm up. Math.sqrt() can be pretty fast

static double[] values = new double[500 * 1000];

public static void main(String... args) {
    for (int i = 0; i < values.length; i++) values[i] = i;

    for (int j = 0; j < 5; j++) {
        long start = System.nanoTime();

        for (int i = 1; i < values.length; i++) {
            values[i] = Math.sqrt(values[i]);
        }
        long time = System.nanoTime() - start;

        System.out.printf("Took %d ns to Math.sqrt on average%n", time / values.length);
    }
}

prints

Took 20 ns to Math.sqrt on average
Took 22 ns to Math.sqrt on average
Took 9 ns to Math.sqrt on average
Took 9 ns to Math.sqrt on average
Took 9 ns to Math.sqrt on average

Try this

double d = 289358932.0;
double sqrt = Double.longBitsToDouble( ( ( Double.doubleToLongBits( d )-(1l<<52) )>>1 ) + ( 1l<<61 ) );

I haven't benchmarked it, but I'd expect it to be faster. The accuracy isn't extremely good, but try it out and see if it meets your needs. I think you can add an additional bias term a to the end of the expression to make it more accurate.

EDIT: You can drastically improve the accuracy by passing it through a round or two of Newton's method

double better = (sqrt + d/sqrt)/2.0;
double evenbetter = (better + d/better)/2.0;

The second pass gives you almost the exact value of the square root.

sqrt            17022.533813476562
better          17010.557763511835
evenbetter      17010.553547724947
Math.sqrt()     17010.553547724423
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!