inline double fastPow(double a, double b) {
  union {
    double d;
    int x[2];
  } u = { a };
  u.x[1] = (int)(b * (u.x[1] - 1072632447) + 1072632447);
  u.x[0] = 0;
  return u.d;
}

This new code uses the union trick, instead of the weird casting trick I’ve used before. This means that -fno-strict-aliasing is no more required any more when compiling, and it is also a bit faster because one less temporary variables is needed. When you have a little endian machine, you have to exchange u.x[0] and u.x[1]. On my PC, this version is 4.2 times faster than the much more precise pow().

Besides that, I also have now a slower approximation that has much less error when the exponent is larger than 1. It makes use exponentiation by squaring, which is exact for the integer part of the exponent, and uses only the exponent’s fraction for the approximation:

// should be much more precise with large b
inline double fastPrecisePow(double a, double b) {
  // calculate approximation with fraction of the exponent
  int e = (int) b;
  union {
    double d;
    int x[2];
  } u = { a };
  u.x[1] = (int)((b - e) * (u.x[1] - 1072632447) + 1072632447);
  u.x[0] = 0;

  // exponentiation by squaring with the exponent's integer part
  // double r = u.d makes everything much slower, not sure why
  double r = 1.0;
  while (e) {
    if (e & 1) {
      r *= a;
    }
    a *= a;
    e >>= 1;
  }

  return r * u.d;
}

This code is 3.3 times faster than pow(). Writing a microbenchmark is not easy, so I have posted mine here. Here is also a Java version of the more accurate pow approximation.

Any ideas how this could be improved? Please post them!

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.

UPDATE, December 10, 2011

I just managed to make the above code about 30% faster than the one above on my machine. The error is a tiny fraction different (not better or worse).

public static double pow(final double a, final double b) {
	final long tmp = Double.doubleToLongBits(a);
    final long tmp2 = (long)(b * (tmp - 4606921280493453312L)) + 4606921280493453312L;
    return Double.longBitsToDouble(tmp2);
}

This new approximation is about 23 times as fast as Math.pow() on my machine (Intel Core2 Quad, Q9550, Java 1.7.0_01-b08, 64-Bit Server VM). Unfortunately, microbenchmarks are difficult to do in Java, so your mileage may vary. You can download the benchmark PowBench.java and have a look, I have tried to prevent overoptimization, and substract the overhead introduced due to this preventation.

Approximation of pow() in C and C++

UPDATE, January 25, 2012

The code below is updated with using union, you do not need -fno-strict-aliasing any more for compiling. Also, here is a more precise version of the approximation.

double fastPow(double a, double b) {
  union {
    double d;
    int x[2];
  } u = { a };
  u.x[1] = (int)(b * (u.x[1] - 1072632447) + 1072632447);
  u.x[0] = 0;
  return u.d;
}

Compiled on my Pentium-M with gcc 4.1.2:

gcc -O3 -march=pentium-m -fomit-frame-pointer

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

Approximation of pow() in C#

Jason Jung has posted a port of the this code to C#:

public static double PowerA(double a, double b) {
  int tmp = (int)(BitConverter.DoubleToInt64Bits(a) >> 32);
  int tmp2 = (int)(b * (tmp - 1072632447) + 1072632447);
  return BitConverter.Int64BitsToDouble(((long)tmp2) << 32);
}

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

UPDATE Recently there several other approximative pow calculation methods have been developed, here are some others that I have found through reddit:

• Fast pow() With Adjustable Accuracy — This looks quite a bit more sophisticated and precise than my approximation. Written in C and for float values. A Java port should not be too difficult.
• Fast SSE2 pow: tables or polynomials? — Uses SSE operation and seems to be a bit faster than the table approach from the link above with the potential to scale better when due to less cache usage.

Please post what you think about this!

Should Google Go Nuclear? Clean, cheap, nuclear power (no, really)

Very knowledgeable professor Robert W. Bussard talks about his findings on nuclear fusion which were just recently (November 2006) allowed to be made public. His lab really seems to be on to something big. He also voices big problems in the American research system.

Human Computation

Lots of problems are still very hard to do with the computers, while humans can do them easily. Luis von Ahn gives an entertaining presentation on how to utilize this human power. Best of all, he shows how he can get the people to solve theses problems for free. Or even make them pay for solving your problem. Google is already using this technique, and there are some other games like Peekaboom.

Winning The DARPA Grand Challenge

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Why We Fight

War means big business. This BBC documentation provides an inside look at the anatomy of the American war machine. A must see. It starts with the powerfull farewell speech of Eisenhower.

Alkali Metals

See what happens when you mix alkali metals with water.

The Paradox of Choice – Why More Is Less

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1. The Elegant Universe

The world is very stranger place, so strange that it’s weirdness seems impossible to understand. This video of 3 parts gives an excellent introduction into the String Theory, which currently looks like it can give the deepest understanding of the universe we ever had.

If you like these videos you should have a look at the book The Elegant Universe. I have read half through it already, it is a bit more difficult to understand than the videos, but much more in depth and very well written.

Part 1: Einstein Universe

Part 2: Strings

Part 3: Welcome to the 11th Dimension

Vivek — Homeless Prophet

Vivek is a homeless who really has something to say about life, the universe, and everything. Just ignore the pseudo funny comments of the people who record this.

The Great Dictator — Speech

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