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High-performance coding by Eddie Edwards

Finding the Nth Root

28 Apr 2016, 16:06 UTC

The Nth-root Algorithm

The Nth-root Algorithm is described on Wikipedia. It requires an initial guess, and then Newton-Raphson iterations are taken to improve that guess.

For instance, to refine a cube root, we take an initial guess and then apply this refinement procedure to it: 

float refine3(float x, float est)
{
    return (1.0f/3.0f)*(2*est + x/(est*est)); 
}

Where x is the value we're finding the cube-root of, and est is the current estimate. The refined estimate is returned.

Newton-Raphson generally doubles the number of valid bits in each iteration. So if we have only 3 valid bits (relative error < 12.5%) then we only need to apply NR three times to refine it from 3 -> 6 -> 12 -> 24 bits, which is the limit of precision for a 32-bit float.

But how do we obtain a good initial estimate that's no more than 12.5% out?

That is the main subject of this article, and my solution requires floating-point bit-twiddling hacks.

The integer representation of floats

It's easy to grab the integer representation of a float in C++, and to convert back from an integer representation to a float (in assembly, it's even easier!): 

union IntFloat {
  int i;
  float f;
};

int getIntFromFloat(float f) {
  IntFloat u;
  u.f = f;
  return u.i;
}

float getFloatFromInt(int i) {
  IntFloat u;
  u.i = i;
  return u.f;
}

(Note that because of C++ strict-aliasing rules, this is the only correct way to perform this conversion.)

What does the format look like? Well, the bit layout of a 32-bit float is as follows: 

SEEEEEEE EMMMMMMM MMMMMMMM MMMMMMMM

S = sign (0 = +ve, 1 = -ve)
E = exponent, biased by 127
M = 23-bit mantissa

Represents 2^(E-127)*1.MMMMMMMMMMMMMMMMMMMMMMM

When E=127, this simply represents the binary fraction 1.MMMMMMMMMMMMMMMMMMMMMMM, which is a number between 1 (M=0) and just slightly under 2 (M=11111111111111111111111).

The representation for the "slightly under 2" number is 00111111 11111111 11111111 11111111. If we add 1 to this we get the representation for the next number: 01000000 00000000 00000000 00000000. This has E=128, M=0, and represents the number 2: 2^1*1.00000000000000000000000 = 10.0000000000000000000000. In general, float with E=128 represents numbers between 2 (M=0) and just slightly under 4 (M=11111111111111111111111).

E=129 then represents numbers between 4 and slightly under 8, and so on.

Over the whole range, this is a continuous mapping, with the represented number going up in steps each time M increments, until M wraps, E increments, and then the step size doubles (so the range for that exponent is also doubled, when covering all possible values of M).

Not only is it a continuous mapping, but every time M goes to 0, the value of the integer representation of x is *exactly* (log2(x) + 127) << 23. Which means that (modulo the sign bit, and the exponent bias of 127) the integer representation of a float is *approximately* equal to a s9:23 fixed-point representation of log2(x). The approximation is exact whenever M=0. Effectively we're taking the curve of log2(x), taking points on that curve at integer locations (in y), and drawing straight-line segments between those points. The approximation is globally very good (viewed from a distance, the two curves blur together) but locally there are small errors as the straight line doesn't represent the actual log2 curve when viewed close-up.

The following code masks off the sign bit and fixes the exponent bias, giving an s9:23 fixed-point estimate for log2(|f|) (log2(f) isn't defined for f <= 0): 

int log2_approx_9_23 = (getIntFromFloat(f) & 0x7FFFFFFF) - 0x3F800000;

We can also go the other way. Given an approximate s9.23 log like this, we can go back to a float. This is the exact inverse of the above operation; we are approximating the function pow(2.0f, i / float(1 << 23)): 

float pow2_approx_from_9_23 = getFloatFromInt(i + 0x3F800000);

It's now very easy to approximate the Nth root of a float: 

int log2 = (getIntFromFloat(f) & 0x7FFFFFFF) - 0x3F800000;
int log2_nthroot = log2 / N;
return getFloatFromInt(log2_nthroot + 0x3F800000);

And that's all there is to it! There are no range issues - denormals might not be very accurate, including zero, and infinities and NaNs give finite Nth roots here. But for normalized floats the overall float range is compressed, so we don't have to worry about integer overflow on any of these operations.

We can refine this a little with sign bits (if N is even, we should take the Nth root of |x| anyway; if N is odd we should save the sign of x, and then restore it to the result, so e.g. cube root of -8 gives -2.)

The astounding thing is how well this works. Over the range of all normalized floats (E=0 and E=255 are special cases, which we ignore) the error in the approximation is only about 6%, regardless of the value of N. This means we have 4 good bits in the result, so applying Newton-Raphson refinement three times gives us full float precision (modulo rounding errors in the NR algorithm itself).

In fact, we can be even more approximate. For cube-root, instead of dividing by 3, we can approximate a divide by 3 as follows: 

log2_nthroot = (log2 >> 20) * 0x55555;

Amazingly this hardly affects the accuracy at all! Similar approximations can be made for other Nth roots than 3.

Finally, instead of doing (int - 0x3F800000) / 3 + 0x3F800000 we can notice that this is equal to int/3 + 2*0x3F800000/3 = int/3 + 0x2A555555.

Our reduced algorithm (for cube-root) is then: 

float cuberootapprox(float x)
{
    int i = getIntFromFloat(x);
    int sign = i & 0x80000000;
    i &= 0x7FFFFFFF;
    i = (i >> 20) * 0x55555;
    i += 0x2A555555;
    return floatFromInt(i | sign);
}

Denormals

A denormal is a number not in the above format. For a denormal, E = 0, and the number has no "hidden 1 bit": 

S0000000 0MMMMMMM MMMMMMMM MMMMMMMM

S = sign (0 = +ve, 1 = -ve)
M = 23-bit mantissa

Represents 2^-126*0.MMMMMMMMMMMMMMMMMMMMMMM

This is essentially the segment E = 1, continued backwards without changing the step size, until it hits zero. The smallest possible non-zero value has M=1 and this has value 2^-149.

We probably don't care about denormals very much, except for zero. We would like the Nth root of zero to be zero. The following simple fix flushes all denormals to zero: 

float cuberootapprox(float x)
{
    int i = getIntFromFloat(x);
    int sign = i & 0x80000000;
    i &= 0x7FFFFFFF;
    if (i <= 0x007FFFFF) {
        return 0;
    } else {
        i = (i >> 20) * 0x55555;
        i += 0x2A555555;
        return floatFromInt(i | sign);
    }
}

Alternatively we can handle denormals properly. To do this we use the same estimate, but on the number M instead of the number x. x is M * 2^-149, so we then multiply the estimate from M by the Nth root of 2^-149. For a cuberoot this is equal to cbrt(2)*cbrt(2^-150) = cbrt(2)*2^-50: 

float cuberootapprox(float x)
{
    int i = getIntFromFloat(x);
    int sign = i & 0x80000000;
    i &= 0x7FFFFFFF;
    if (!i) {
        return 0;
    } else if (i <= 0x007FFFFF) {
        i = getIntFromFloat(float(i));
        i = (i >> 20) * 0x55555;
        i += 0x2A555555;
        i -= 50 << 23;
        return floatFromInt(i | sign) * 1.2599210498948731647672106072782f;
    } else {
        i = (i >> 20) * 0x55555;
        i += 0x2A555555;
        return floatFromInt(i | sign);
    }
}

For SIMDification, all these branches can be combined into one basic sequence (note that 0x11555555 = 0x2A555555 - (50 << 23)): 

float cuberootapprox(float x)
{
    int i = getIntFromFloat(x);
    int sign = i & 0x80000000;
    i &= 0x7FFFFFFF;
    bool denormal = (i <= 0x007FFFFF);
    bool zero = (i == 0);
    unsigned int adder = denormal ? 0x11555555 : 0x2A555555;
    float multiplier = denormal ? 1.2599210498948731647672106072782f : 1;
    multiplier = zero ? 0 : multiplier;
    i = denormal ? getIntFromFloat(i) : i;
    i = (i >> 20) * 0x55555;
    i += adder;
    return floatFromInt(i | sign) * multiplier;
}

And, since the error is always less than 6%, we can now get an accurate result just by blindly applying Newton-Raphson three times: 

float cuberoot(float x)
{
    float est = cuberootapprox(x);
    est = refine3(x, est);
    est = refine3(x, est);
    est = refine3(x, est);
    return est;
}

This gives "OK" results for very small denormals (worst-case relative error 0.000115%) but for all other floating-point numbers it gives a worst-case error of around 0.000045%. This is not as good as the C library sequence pow(x, 1.0f/3.0f), but it is only a few LSBs different, which is good enough for most purposes, and it is fast - especially if implemented using SIMD to act on 4 or 8 floats at a time.

Similar reduced algorithms can be determined for any Nth-root. The only differences are:

  1. Nth root of 2^-149 must be split into a 2^-P part and a part between 1 and 2. P is rolled into the constant 0x11555555 = 0x2A555555 - (P << 23). The fractional part is rolled into the multiplier.
  2. (i >> 20) * 0x55555 must be changed to approximate 1/N instead of 1/3. Or just use i/N.

And there you have it. A similar approximation can also be made for reciprocal roots (pow(x, -1/N)) - now the format would be i = adder - i/N. If you look at the famous Quake inverse square root approximation, this is of the same form, except adder is hand-tweaked to absolutely minimize the estimation error - this is required since Quake only did one iteration of NR, instead of the required three. With three iterations, we don't need to worry about this refinement, but the basic algorithm is fast enough that you could do a brute-force search for an optimal value here in a few hours on a desktop PC, if that's the kind of thing that interests you!

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