I have already written about approximations of e^x, log(x) and pow(a, b) in my post Optimized Exponential Functions for Java. Now I have more In particular, the pow() function is now even faster, simpler, and more accurate. Without further ado, I proudly give you the brand new approximation:

Approximation of pow() in Java

public static double pow(final double a, final double b) {
    final int x = (int) (Double.doubleToLongBits(a) >> 32);
    final int y = (int) (b * (x - 1072632447) + 1072632447);
    return Double.longBitsToDouble(((long) y) << 32);
}

This is really very compact. The calculation only requires 2 shifts, 1 mul, 2 add, and 2 register operations. That’s it! In my tests it usually within an error margin of 5% to 12%, in extreme cases sometimes up to 25%. A careful analysis is left as an exercise for the reader. This is very usable for in e.g. metaheuristics or neural nets.

I use Linux, Java 1.6.0-b105 with the server VM, and execute the benchmark with this command:

sudo nice -n -20 java -cp . -server PowTest

The approximation is 27 times faster than Math.pow() on my Pentium-M. On a Pentium 4 it is 41 times faster. Unfortunately, microbenchmarks are difficult to do in Java, so your mileage may vary. You can download the benchmark PowTest.java and have a look, I have tried to prevent overoptimization while still having a low overhead.

Approximation of pow() in C and C++

double pow(double a, double b) {
    int tmp = (*(1 + (int *)&a));
    int tmp2 = (int)(b * (tmp - 1072632447) + 1072632447);
    double p = 0.0;
    *(1 + (int * )&p) = tmp2;
    return p;
}

Compiled on my Pentium-M with gcc 4.1.2:

gcc -O3 -march=pentium-m -fomit-frame-pointer -fno-strict-aliasing

This version is 7.8 times faster than pow() from the standard library.

WARNING! you HAVE to use the -fno-strict-aliasing option, or this does not work!

How the Approximation was Developed

It is quite impossible to understand what is going on in this function, it just magically works. To shine a bit more light on it, here is a detailed description how I have developed this.

Approximation of e^x

As described here, the paper “A Fast, Compact Approximation of the Exponential Function” develops a C macro that does a good job at exploiting the IEEE 754 floating-point representation to calculate e^x. This macro can be transformed into Java code straightforward, which looks like this:

public static double exp(double val) {
    final long tmp = (long) (1512775 * val + (1072693248 - 60801));
    return Double.longBitsToDouble(tmp << 32);
}

Use Exponential Functions for a^b

Thanks to the power of math, we know that a^b can be transformed like this:

1. Take exponential

   a^b = e^(ln(a^b))

2. Extract b

   a^b = e^(ln(a)*b)

Now we have expressed the pow calculation with e^x and ln(x). We already have the e^x approximation, but no good ln(x). The old approximation is very bad, so we need a better one. So what now?

Approximation of ln(x)

Here comes the big trick: Rember that we have the nice e^x approximation? Well, ln(x) is exactly the inverse function! That means we just need to transform the above approximation so that the output of e^x is transformed back into the original input.

That’s not too difficult. Have a look at the above code, we now take the output and move backwards to undo the calculation. First reverse the shift:

final double tmp = (Double.doubleToLongBits(val) >> 32);

Now solve the equation

tmp = (1512775 * val + (1072693248 - 60801))

for val:

1. The original formula

   tmp = (1512775 * val + (1072693248 - 60801))

2. Perform subtraction

   tmp = 1512775 * val + 1072632447

3. Bring value to other side

   tmp - 1072632447 = 1512775 * val

4. Divide by factor

   (tmp - 1072632447) / 1512775 = val

5. Finally, val on the left side

   val = (tmp - 1072632447) / 1512775

Voíla, now we have a nice approximation of ln(x):

public double ln(double val) {
    final double x = (Double.doubleToLongBits(val) >> 32);
    return (x - 1072632447) / 1512775;
}

Combine Both Approximations

Finally we can combine the two approximations into e^(ln(a) * b):

public static double pow1(final double a, final double b) {
    // calculate ln(a)
    final double x = (Double.doubleToLongBits(a) >> 32);
    final double ln_a = (x - 1072632447) / 1512775;

    // ln(a) * b
    final double tmp1 = ln_a * b;

    // e^(ln(a) * b)
    final long tmp2 = (long) (1512775 * tmp1 + (1072693248 - 60801));
    return Double.longBitsToDouble(tmp2 << 32);
}

Between the two shifts, we can simply insert the tmp1 calculation into the tmp2 calculation to get

public static double pow2(final double a, final double b) {
    final double x = (Double.doubleToLongBits(a) >> 32);
    final long tmp2 = (long) (1512775 * (x - 1072632447) / 1512775 * b + (1072693248 - 60801));
    return Double.longBitsToDouble(tmp2 << 32);
}

Now simplify tmp2 calculation:

1. The original formula

   tmp2 = (1512775 * (x - 1072632447) / 1512775 * b + (1072693248 - 60801))

2. We can drop the factor 1512775

   tmp2 = (x - 1072632447) * b + (1072693248 - 60801)

3. And finally, calculate the substraction

   tmp2 = b * (x - 1072632447) + 1072632447

The Result

That’s it! Add some casts, and the complete function is the same as above.

public static double pow(final double a, final double b) {
    final int tmp = (int) (Double.doubleToLongBits(a) >> 32);
    final int tmp2 = (int) (b * (tmp - 1072632447) + 1072632447);
    return Double.longBitsToDouble(((long) tmp2) << 32);
}

This concludes my little tutorial on microoptimization of the pow() function. If you have come this far, I congratulate your presistence

Please post what you think about this!