为什么快速反平方根在Java上如此奇怪和缓慢？

时间：2013-05-14 19:19:46

标签： java math sqrt math.sqrt

我正在尝试在java上实现Fast Inverse Square Root以加快向量规范化。但是，当我在Java中实现单精度版本时，我首先获得与1F / (float)Math.sqrt()大致相同的速度，然后迅速降低到速度的一半。这很有意思，因为虽然Math.sqrt使用（我推测）一个本机方法，但这涉及浮点除法，我听说它实在很慢。我计算数字的代码如下：

public static float fastInverseSquareRoot(float x){
    float xHalf = 0.5F * x;
    int temp = Float.floatToRawIntBits(x);
    temp = 0x5F3759DF - (temp >> 1);
    float newX = Float.intBitsToFloat(temp);
    newX = newX * (1.5F - xHalf * newX * newX);
    return newX;
}

使用我编写的短程序，每次迭代1600万次，然后汇总结果，然后重复，我得到这样的结果：

1F / Math.sqrt() took 65209490 nanoseconds.
Fast Inverse Square Root took 65456128 nanoseconds.
Fast Inverse Square Root was 0.378224 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 64131293 nanoseconds.
Fast Inverse Square Root took 26214534 nanoseconds.
Fast Inverse Square Root was 59.123647 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 27312205 nanoseconds.
Fast Inverse Square Root took 56234714 nanoseconds.
Fast Inverse Square Root was 105.895914 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 26493281 nanoseconds.
Fast Inverse Square Root took 56004783 nanoseconds.
Fast Inverse Square Root was 111.392402 percent slower than 1F / Math.sqrt()

我始终获得两者速度大致相同的数字，然后进行迭代，快速反向平方根节省1F / Math.sqrt()所需时间的约60％，然后进行几次迭代，大约需要两倍的时间用于快速反向平方根作为控件运行。我很困惑为什么FISR会来自Same - ＆gt;快60％ - ＆gt;速度慢100％，每次运行程序时都会发生这种情况。

编辑：以上数据是我在eclipse中运行它的时候。当我使用javac/java运行程序时，我得到完全不同的数据：

1F / Math.sqrt() took 57870498 nanoseconds.
Fast Inverse Square Root took 88206794 nanoseconds.
Fast Inverse Square Root was 52.421004 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 54982400 nanoseconds.
Fast Inverse Square Root took 83777562 nanoseconds.
Fast Inverse Square Root was 52.371599 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 21115822 nanoseconds.
Fast Inverse Square Root took 76705152 nanoseconds.
Fast Inverse Square Root was 263.259133 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 20159210 nanoseconds.
Fast Inverse Square Root took 80745616 nanoseconds.
Fast Inverse Square Root was 300.539585 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 21814675 nanoseconds.
Fast Inverse Square Root took 85261648 nanoseconds.
Fast Inverse Square Root was 290.845374 percent slower than 1F / Math.sqrt()

EDIT2：经过几次回复后，似乎速度在几次迭代后趋于稳定，但它稳定的数字是高度不稳定的。任何人都知道为什么？

这是我的代码（不完全简洁，但这是完整的事情）：

public class FastInverseSquareRootTest {

    public static FastInverseSquareRootTest conductTest() {
        float result = 0F;
        long startTime, endTime, midTime;
        startTime = System.nanoTime();
        for (float x = 1F; x < 4_000_000F; x += 0.25F) {
            result = 1F / (float) Math.sqrt(x);
        }
        midTime = System.nanoTime();
        for (float x = 1F; x < 4_000_000F; x += 0.25F) {
            result = fastInverseSquareRoot(x);
        }
        endTime = System.nanoTime();
        return new FastInverseSquareRootTest(midTime - startTime, endTime
                - midTime);
    }

    public static float fastInverseSquareRoot(float x) {
        float xHalf = 0.5F * x;
        int temp = Float.floatToRawIntBits(x);
        temp = 0x5F3759DF - (temp >> 1);
        float newX = Float.intBitsToFloat(temp);
        newX = newX * (1.5F - xHalf * newX * newX);
        return newX;
    }

    public static void main(String[] args) throws Exception {
        for (int i = 0; i < 7; i++) {
            System.out.println(conductTest().toString());
        }
    }

    private long controlDiff;

    private long experimentalDiff;

    private double percentError;

    public FastInverseSquareRootTest(long controlDiff, long experimentalDiff) {
        this.experimentalDiff = experimentalDiff;
        this.controlDiff = controlDiff;
        this.percentError = 100D * (experimentalDiff - controlDiff)
                / controlDiff;
    }

    @Override
    public String toString() {
        StringBuilder sb = new StringBuilder();
        sb.append(String.format("1F / Math.sqrt() took %d nanoseconds.%n",
                controlDiff));
        sb.append(String.format(
                "Fast Inverse Square Root took %d nanoseconds.%n",
                experimentalDiff));
        sb.append(String
                .format("Fast Inverse Square Root was %f percent %s than 1F / Math.sqrt()%n",
                        Math.abs(percentError), percentError > 0D ? "slower"
                                : "faster"));
        return sb.toString();
    }
}

3 个答案:

答案 0 :(得分：11)

JIT优化器似乎已经将调用发送到Math.sqrt。

使用未经修改的代码，我得到了

1F / Math.sqrt() took 65358495 nanoseconds.
Fast Inverse Square Root took 77152791 nanoseconds.
Fast Inverse Square Root was 18,045544 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 52872498 nanoseconds.
Fast Inverse Square Root took 75242075 nanoseconds.
Fast Inverse Square Root was 42,308531 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 23386359 nanoseconds.
Fast Inverse Square Root took 73532080 nanoseconds.
Fast Inverse Square Root was 214,422951 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 23790209 nanoseconds.
Fast Inverse Square Root took 76254902 nanoseconds.
Fast Inverse Square Root was 220,530610 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 23885467 nanoseconds.
Fast Inverse Square Root took 74869636 nanoseconds.
Fast Inverse Square Root was 213,452678 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 23473514 nanoseconds.
Fast Inverse Square Root took 73063699 nanoseconds.
Fast Inverse Square Root was 211,260168 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 23738564 nanoseconds.
Fast Inverse Square Root took 71917013 nanoseconds.
Fast Inverse Square Root was 202,954353 percent slower than 1F / Math.sqrt()

fastInverseSquareRoot的时间一直较慢，而且时间都在同一个球场，而Math.sqrt则会大幅加快。

更改代码，以便无法避免对Math.sqrt的调用，

    for (float x = 1F; x < 4_000_000F; x += 0.25F) {
        result += 1F / (float) Math.sqrt(x);
    }
    midTime = System.nanoTime();
    for (float x = 1F; x < 4_000_000F; x += 0.25F) {
        result -= fastInverseSquareRoot(x);
    }
    endTime = System.nanoTime();
    if (result == 0) System.out.println("Wow!");

我得到了

1F / Math.sqrt() took 184884684 nanoseconds.
Fast Inverse Square Root took 85298761 nanoseconds.
Fast Inverse Square Root was 53,863804 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 182183542 nanoseconds.
Fast Inverse Square Root took 83040574 nanoseconds.
Fast Inverse Square Root was 54,419278 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 165269658 nanoseconds.
Fast Inverse Square Root took 81922280 nanoseconds.
Fast Inverse Square Root was 50,431143 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 163272877 nanoseconds.
Fast Inverse Square Root took 81906141 nanoseconds.
Fast Inverse Square Root was 49,834815 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 165314846 nanoseconds.
Fast Inverse Square Root took 81124465 nanoseconds.
Fast Inverse Square Root was 50,927296 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 164079534 nanoseconds.
Fast Inverse Square Root took 80453629 nanoseconds.
Fast Inverse Square Root was 50,966689 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 162350821 nanoseconds.
Fast Inverse Square Root took 79854355 nanoseconds.
Fast Inverse Square Root was 50,813704 percent faster than 1F / Math.sqrt()
Math.sqrt的

多次慢，而fastInverseSqrt的时间只有中等速度（现在每次迭代都需要减法）。

答案 1 :(得分：0)

我发布的代码输出是：

1F / Math.sqrt() took 165769968 nanoseconds.
Fast Inverse Square Root took 251809517 nanoseconds.
Fast Inverse Square Root was 51.902977 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 162953919 nanoseconds.
Fast Inverse Square Root took 251212721 nanoseconds.
Fast Inverse Square Root was 54.161816 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 161524902 nanoseconds.
Fast Inverse Square Root took 36242909 nanoseconds.
Fast Inverse Square Root was 77.562030 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 162289014 nanoseconds.
Fast Inverse Square Root took 36552036 nanoseconds.
Fast Inverse Square Root was 77.477196 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 163157620 nanoseconds.
Fast Inverse Square Root took 36152720 nanoseconds.
Fast Inverse Square Root was 77.841844 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 162511997 nanoseconds.
Fast Inverse Square Root took 36426705 nanoseconds.
Fast Inverse Square Root was 77.585221 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 162302698 nanoseconds.
Fast Inverse Square Root took 36797410 nanoseconds.
Fast Inverse Square Root was 77.327912 percent faster than 1F / Math.sqrt()
似乎JIT被踢了，表演提升了近十倍。希望有更好地掌握JIT的人会来解释这一点。我的环境：Java 6，Eclipse。

答案 2 :(得分：0)

我的jit有两个加快步骤的步骤：第一步可能是算法优化，第二步可能是装配优化。

1F / Math.sqrt() took 78202645 nanoseconds.
Fast Inverse Square Root took 79248400 nanoseconds.
Fast Inverse Square Root was 1,337237 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 76856008 nanoseconds.
Fast Inverse Square Root took 24788247 nanoseconds.
Fast Inverse Square Root was 67,747158 percent faster than 1F / Math.sqrt()

1F / Math.sqrt() took 24162119 nanoseconds.
Fast Inverse Square Root took 70651968 nanoseconds.
Fast Inverse Square Root was 192,407996 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 24163301 nanoseconds.
Fast Inverse Square Root took 70598983 nanoseconds.
Fast Inverse Square Root was 192,174414 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 24201621 nanoseconds.
Fast Inverse Square Root took 70667344 nanoseconds.
Fast Inverse Square Root was 191,994259 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 24219835 nanoseconds.
Fast Inverse Square Root took 70698568 nanoseconds.
Fast Inverse Square Root was 191,903591 percent slower than 1F / Math.sqrt()

1F / Math.sqrt() took 24231663 nanoseconds.
Fast Inverse Square Root took 70633991 nanoseconds.
Fast Inverse Square Root was 191,494608 percent slower than 1F / Math.sqrt()
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