Here's the code I'm trying to convert: the double version of VDT's Pade Exp fast_ex() approx (here's the old repo resource):
inline double fast_exp(double initial_x){
double x = initial_x;
double px=details::fpfloor(details::LOG2E * x +0.5);
const int32_t n = int32_t(px);
x -= px * 6.93145751953125E-1;
x -= px * 1.42860682030941723212E-6;
const double xx = x * x;
// px = x * P(x**2).
px = details::PX1exp;
px *= xx;
px += details::PX2exp;
px *= xx;
px += details::PX3exp;
px *= x;
// Evaluate Q(x**2).
double qx = details::QX1exp;
qx *= xx;
qx += details::QX2exp;
qx *= xx;
qx += details::QX3exp;
qx *= xx;
qx += details::QX4exp;
// e**x = 1 + 2x P(x**2)/( Q(x**2) - P(x**2) )
x = px / (qx - px);
x = 1.0 + 2.0 * x;
// Build 2^n in double.
x *= details::uint642dp(( ((uint64_t)n) +1023)<<52);
if (initial_x > details::EXP_LIMIT)
x = std::numeric_limits<double>::infinity();
if (initial_x < -details::EXP_LIMIT)
x = 0.;
return x;
}
I got this:
__m128d PExpSSE_dbl(__m128d x) {
__m128d initial_x = x;
__m128d half = _mm_set1_pd(0.5);
__m128d one = _mm_set1_pd(1.0);
__m128d log2e = _mm_set1_pd(1.4426950408889634073599);
__m128d p1 = _mm_set1_pd(1.26177193074810590878E-4);
__m128d p2 = _mm_set1_pd(3.02994407707441961300E-2);
__m128d p3 = _mm_set1_pd(9.99999999999999999910E-1);
__m128d q1 = _mm_set1_pd(3.00198505138664455042E-6);
__m128d q2 = _mm_set1_pd(2.52448340349684104192E-3);
__m128d q3 = _mm_set1_pd(2.27265548208155028766E-1);
__m128d q4 = _mm_set1_pd(2.00000000000000000009E0);
__m128d px = _mm_add_pd(_mm_mul_pd(log2e, x), half);
__m128d t = _mm_cvtepi64_pd(_mm_cvttpd_epi64(px));
px = _mm_sub_pd(t, _mm_and_pd(_mm_cmplt_pd(px, t), one));
__m128i n = _mm_cvtpd_epi64(px);
x = _mm_sub_pd(x, _mm_mul_pd(px, _mm_set1_pd(6.93145751953125E-1)));
x = _mm_sub_pd(x, _mm_mul_pd(px, _mm_set1_pd(1.42860682030941723212E-6)));
__m128d xx = _mm_mul_pd(x, x);
px = _mm_mul_pd(xx, p1);
px = _mm_add_pd(px, p2);
px = _mm_mul_pd(px, xx);
px = _mm_add_pd(px, p3);
px = _mm_mul_pd(px, x);
__m128d qx = _mm_mul_pd(xx, q1);
qx = _mm_add_pd(qx, q2);
qx = _mm_mul_pd(xx, qx);
qx = _mm_add_pd(qx, q3);
qx = _mm_mul_pd(xx, qx);
qx = _mm_add_pd(qx, q4);
x = _mm_div_pd(px, _mm_sub_pd(qx, px));
x = _mm_add_pd(one, _mm_mul_pd(_mm_set1_pd(2.0), x));
n = _mm_add_epi64(n, _mm_set1_epi64x(1023));
n = _mm_slli_epi64(n, 52);
// return?
}
But I'm not able to finish the last lines - i.e. this code:
if (initial_x > details::EXP_LIMIT)
x = std::numeric_limits<double>::infinity();
if (initial_x < -details::EXP_LIMIT)
x = 0.;
return x;
How would you convert in SSE2?
Than of course I need to check the whole, since I'm not quite sure I've converted it correctly.
EDIT: I found the SSE conversion of float exp - i.e. from this:
/* multiply by power of 2 */
z *= details::uint322sp((n + 0x7f) << 23);
if (initial_x > details::MAXLOGF) z = std::numeric_limits<float>::infinity();
if (initial_x < details::MINLOGF) z = 0.f;
return z;
to this:
n = _mm_add_epi32(n, _mm_set1_epi32(0x7f));
n = _mm_slli_epi32(n, 23);
return _mm_mul_ps(z, _mm_castsi128_ps(n));
Yup, dividing two polynomials can often give you a better tradeoff between speed and precision than one huge polynomial. As long as there's enough work to hide the divpd throughput. (The latest x86 CPUs have pretty decent FP divide throughput. Still bad vs. multiply, but it's only 1 uop so it doesn't stall the pipeline if you use it rarely enough, i.e. mixed with lots of multiplies. Including in the surrounding code that uses exp)
However, _mm_cvtepi64_pd(_mm_cvttpd_epi64(px)); won't work with SSE2. Packed-conversion intrinsics to/from 64-bit integers requires AVX512DQ.
To do packed rounding to the nearest integer, ideally you'd use SSE4.1 _mm_round_pd(x, _MM_FROUND_TO_NEAREST_INT |_MM_FROUND_NO_EXC), (or truncation towards zero, or floor or ceil towards -+Inf).
But we don't actually need that.
The scalar code ends up with int n and double px both representing the same numeric value. It uses the bad/buggy floor(val+0.5) idiom instead of rint(val) or nearbyint(val) to round to nearest, and then converts that already-integer double to an int (with C++'s truncation semantics, but that doesn't matter because the double value's already an exact integer.)
With SIMD intrinsics, it appears to be easiest to just convert to 32-bit integer and back.
__m128i n = _mm_cvtpd_epi32( _mm_mul_pd(log2e, x) ); // round to nearest
__m128d px = _mm_cvtepi32_pd( n );
Rounding to int with the desired mode, then converting back to double, is equivalent to double->double rounding and then grabbing an int version of that like the scalar version does. (Because you don't care what happens for doubles too large to fit in an int.)
cvtsd2si and si2sd instructions are 2 uops each, and shuffle the 32-bit integers to packed in the low 64 bits of a vector. So to set up for 64-bit integer shifts to stuff the bits into a double again, you'll need to shuffle. The top 64 bits of n will be zeros, so we can use that to create 64-bit integer n lined up with the doubles:
n = _mm_shuffle_epi32(n, _MM_SHUFFLE(3,1,2,0)); // 64-bit integers
But with just SSE2, there are workarounds. Converting to 32-bit integer and back is one option: you don't care about inputs too small or too large. But packed-conversion between double and int costs at least 2 uops on Intel CPUs each way, so a total of 4. But only 2 of those uops need the FMA units, and your code probably doesn't bottleneck on port 5 with all those multiplies and adds.
Or add a very large number and subtract it again: large enough that each double is 1 integer apart, so normal FP rounding does what you want. (This works for inputs that won't fit in 32 bits, but not double > 2^52. So either way that would work.) Also see How to efficiently perform double/int64 conversions with SSE/AVX? which uses that trick. I couldn't find an example on SO, though.
Related:
Fastest Implementation of Exponential Function Using AVX and Fastest Implementation of Exponential Function Using SSE have versions with other speed / precision tradeoffs, for _ps (packed single-precision float).
Fast SSE low precision exponential using double precision operations is at the other end of the spectrum, but still for double.
How many clock cycles does cost AVX/SSE exponentiation on modern x86_64 CPU? discusses some existing libraries like SVML, and Agner Fog's VCL (GPL licensed). And glibc's libmvec.
Then of course I need to check the whole, since I'm not quite sure I've converted it correctly.
iterating over all 2^64 double bit-patterns is impractical, unlike for float where there are only 4 billion, but maybe iterating over all doubles that have the low 32 bits of their mantissa all zero would be a good start. i.e. check in a loop with
bitpatterns = _mm_add_epi64(bitpatterns, _mm_set1_epi64x( 1ULL << 32 ));
doubles = _mm_castsi128_pd(bitpatterns);
https://randomascii.wordpress.com/2014/01/27/theres-only-four-billion-floatsso-test-them-all/
For those last few lines, correcting the input for out-of-range inputs:
The float version you quote just leaves out the range-check entirely. This is obviously the fastest way, if your inputs will always be in range or if you don't care about what happens for out-of-range inputs.
Alternate cheaper range-checking (maybe only for debugging) would be to turn out-of-range values into NaN by ORing the packed-compare result into the result. (An all-ones bit-pattern represents a NaN.)
__m128d out_of_bounds = _mm_cmplt_pd( limit, abs(initial_x) ); // abs = mask off the sign bit
result = _mm_or_pd(result, out_of_bounds);
In general, you can vectorize simple condition setting of a value using branchless compare + blend. Instead of if(x) y=0;, you have the SIMD equivalent of y = (condition) ? 0 : y;, on a per-element basis. SIMD compares produce a mask of all-zero / all-one elements so you can use it to blend.
e.g. in this case cmppd the input and blendvpd the output if you have SSE4.1. Or with just SSE2, and/andnot/or to blend. See SSE intrinsics for comparison (_mm_cmpeq_ps) and assignment operation for a _ps version of both, _pd is identical.
In asm it will look like this:
; result in xmm0 (in need of fixups for out of range inputs)
; initial_x in xmm2
; constants:
; xmm5 = limit
; xmm6 = +Inf
cmpltpd xmm2, xmm5 ; xmm2 = input_x < limit ? 0xffff... : 0
andpd xmm0, xmm2 ; result = result or 0
andnpd xmm2, xmm6 ; xmm2 = 0 or +Inf (In that order because we used ANDN)
orpd xmm0, xmm2 ; result |= 0 or +Inf
; xmm0 = (input < limit) ? result : +Inf
(In an earlier version of the answer, I thought I was maybe saving a movaps to copy a register, but this is just a bog-standard blend. It destroys initial_x, so the compiler needs to copy that register at some point while calculating result, though.)
Optimizations for this special condition
Or in this case, 0.0 is represented by an all-zero bit-pattern, so do a compare that will produce true if in-range, and AND the output with that. (To leave it unchanged or force it to +0.0). This is better than _mm_blendv_pd, which costs 2 uops on most Intel CPUs (and the AVX 128-bit version always costs 2 uops on Intel). And it's not worse on AMD or Skylake.
+-Inf is represented by a bit-pattern of significand=0, exponent=all-ones. (Any other value in the significand represents +-NaN.) Since too-large inputs will presumably still leave non-zero significands, we can't just AND the compare result and OR that into the final result. I think we need to do a regular blend, or something as expensive (3 uops and a vector constant).
It adds 2 cycles of latency to the final result; both the ANDNPD and ORPD are on the critical path. The CMPPD and ANDPD aren't; they can run in parallel with whatever you do to compute the result.
Hopefully your compiler will actually use ANDPS and so on, not PD, for everything except the CMP, because it's 1 byte shorter but identical because they're both just bitwise ops. I wrote ANDPD just so I didn't have to explain this in comments.
You might be able to shorten the critical path latency by combining both fixups before applying to the result, so you only have one blend. But then I think you also need to combine the compare results.
Or since your upper and lower bounds are the same magnitude, maybe you can compare the absolute value? (mask off the sign bit of initial_x and do _mm_cmplt_pd(abs_initial_x, _mm_set1_pd(details::EXP_LIMIT))). But then you have to sort out whether to zero or set to +Inf.
If you had SSE4.1 for _mm_blendv_pd, you could use initial_x itself as the blend control for the fixup that might need applying, because blendv only cares about the sign bit of the blend control (unlike with the AND/ANDN/OR version where all bits need to match.)
__m128d fixup = _mm_blendv_pd( _mm_setzero_pd(), _mm_set1_pd(INFINITY), initial_x ); // fixup = (initial_x signbit) ? 0 : +Inf
// see below for generating fixup with an SSE2 integer arithmetic-shift
const signbit_mask = _mm_castsi128_pd(_mm_set1_epi64x(0x7fffffffffffffff)); // ~ set1(-0.0)
__m128d abs_init_x = _mm_and_pd( initial_x, signbit_mask );
__m128d out_of_range = _mm_cmpgt_pd(abs_init_x, details::EXP_LIMIT);
// Conditionally apply the fixup to result
result = _mm_blendv_pd(result, fixup, out_of_range);
Possibly use cmplt instead of cmpgt and rearrange if you care what happens for initial_x being a NaN. Choosing the compare so false applies the fixup instead of true will mean that an unordered comparison results in either 0 or +Inf for an input of -NaN or +NaN. This still doesn't do NaN propagation. You could _mm_cmpunord_pd(initial_x, initial_x) and OR that into fixup, if you want to make that happen.
Especially on Skylake and AMD Bulldozer/Ryzen where SSE2 blendvpd is only 1 uop, this should be pretty nice. (The VEX encoding, vblendvpd is 2 uops, having 3 inputs and a separate output.)
You might still be able to use some of this idea with only SSE2, maybe creating fixup by doing a compare against zero and then _mm_and_pd or _mm_andnot_pd with the compare result and +Infinity.
Using an integer arithmetic shift to broadcast the sign bit to every position in the double isn't efficient: psraq doesn't exist, only psraw/d. Only logical shifts come in 64-bit element size.
But you could create fixup with just one integer shift and mask, and a bitwise invert
__m128i ix = _mm_castsi128_pd(initial_x);
__m128i ifixup = _mm_srai_epi32(ix, 11); // all 11 bits of exponent field = sign bit
ifixup = _mm_and_si128(ifixup, _mm_set1_epi64x(0x7FF0000000000000ULL) ); // clear other bits
// ix = the bit pattern for 0 (non-negative x) or +Inf (negative x)
__m128d fixup = _mm_xor_si128(ifixup, _mm_set1_epi32(-1)); // bitwise invert
Then blend fixup into result for out-of-range inputs as normal.
Cheaply checking abs(initial_x) > details::EXP_LIMIT
If the exp algorithm was already squaring initial_x, you could compare against EXP_LIMIT squared. But it's not, xx = x*x only happens after some calculation to create x.
If you have AVX512F/VL, VFIXUPIMMPD might be handy here. It's designed for functions where the special case outputs are from "special" inputs like NaN and +-Inf, negative, positive, or zero, saving a compare for those cases. (e.g. for after a Newton-Raphson reciprocal(x) for x=0.)
But both of your special cases need compares. Or do they?
If you square your input and subtract, it only costs one FMA to do initial_x * initial_x - details::EXP_LIMIT * details::EXP_LIMIT to create a result that's negative for abs(initial_x) < details::EXP_LIMIT, and non-negative otherwise.
Agner Fog reports that vfixupimmpd is only 1 uop on Skylake-X.
I'm trying to do Logistic Regression from Coursera in Julia, but it doesn't work.
The Julia code to calculate the Gradient:
sigmoid(z) = 1 / (1 + e ^ -z)
hypotesis(theta, x) = sigmoid(scalar(theta' * x))
function gradient(theta, x, y)
(m, n) = size(x)
h = [hypotesis(theta, x[i,:]') for i in 1:m]
g = Array(Float64, n, 1)
for j in 1:n
g[j] = sum([(h[i] - y[i]) * x[i, j] for i in 1:m])
end
g
end
If this gradient used it produces the wrong results. Can't figure out why, the code seems like the right one.
The full Julia script. In this script the optimal Theta calculated using my Gradient Descent implementation and using the built-in Optim package, and the results are different.
The gradient is correct (up to a scalar multiple, as #roygvib points out). The problem is with the gradient descent.
If you look at the values of the cost function during your gradient descent, you will see a lot of NaN,
which probably come from the exponential:
lowering the step size (e.g., to 1e-5) will avoid the overflow,
but you will have to increase the number of iterations a lot (perhaps to 10_000_000).
A better (faster) solution would be to let the step size vary.
For instance, one could multiply the step size by 1.1
if the cost function improves after a step
(the optimum still looks far away in this direction: we can go faster),
and divide it by 2 if it does not (we went too fast and ended up past the minimum).
One could also do a line search in the direction of the gradient to find the best step size
(but this is time-consuming and can be replaced by approximations, e.g., Armijo's rule).
Rescaling the predictive variables also helps.
I tried comparing gradient() in the OP's code with numerical derivative of cost_j() (which is the objective function of minimization) using the following routine
function grad_num( theta, x, y )
g = zeros( 3 )
eps = 1.0e-6
disp = zeros( 3 )
for k = 1:3
disp[:] = theta[:]
disp[ k ]= theta[ k ] + eps
plus = cost_j( disp, x, y )
disp[ k ]= theta[ k ] - eps
minus = cost_j( disp, x, y )
g[ k ] = ( plus - minus ) / ( 2.0 * eps )
end
return g
end
But the gradient values obtained from the two routines do no seem to agree very well (at least for the initial stage of minimization)... So I manually derived the gradient of cost_j( theta, x, y ), from which it seems that the division by m is missing:
#/ OP's code
# g[j] = sum( [ (h[i] - y[i]) * x[i, j] for i in 1:m ] )
#/ modified code
g[j] = sum( [ (h[i] - y[i]) * x[i, j] for i in 1:m ] ) / m
Because I am not very sure if the above code and expression are really correct, could you check them by yourself...?
But in fact, regardless of whether I use the original or corrected gradients, the program converges to the same minimum value (0.2034977016, almost the same as obtained from Optim), because the two gradients differ only by a multiplicative factor! Because the convergence was very slow, I also modified the stepsize alpha adaptively following the suggestion by Vincent (here I used more moderate values for acceleration/deceleration):
function gradient_descent(x, y, theta, alpha, n_iterations)
...
c = cost_j( theta, x, y )
for i = 1:n_iterations
c_prev = c
c = cost_j( theta, x, y )
if c - c_prev < 0.0
alpha *= 1.01
else
alpha /= 1.05
end
theta[:] = theta - alpha * gradient(theta, x, y)
end
...
end
and called this routine as
optimal_theta = gradient_descent( x, y, [0 0 0]', 1.5e-3, 10^7 )[ 1 ]
The variation of cost_j versus iteration steps is plotted below.
I have hot spots in my code where I'm doing pow() taking up around 10-20% of my execution time.
My input to pow(x,y) is very specific, so I'm wondering if there's a way to roll two pow() approximations (one for each exponent) with higher performance:
I have two constant exponents: 2.4 and 1/2.4.
When the exponent is 2.4, x will be in the range (0.090473935, 1.0].
When the exponent is 1/2.4, x will be in the range (0.0031308, 1.0].
I'm using SSE/AVX float vectors. If platform specifics can be taken advantage of, right on!
A maximum error rate around 0.01% is ideal, though I'm interested in full precision (for float) algorithms as well.
I'm already using a fast pow() approximation, but it doesn't take these constraints into account. Is it possible to do better?
Another answer because this is very different from my previous answer, and this is blazing fast. Relative error is 3e-8. Want more accuracy? Add a couple more Chebychev terms. It's best to keep the order odd as this makes for a small discontinuity between 2^n-epsilon and 2^n+epsilon.
#include <stdlib.h>
#include <math.h>
// Returns x^(5/12) for x in [1,2), to within 3e-8 (relative error).
// Want more precision? Add more Chebychev polynomial coefs.
double pow512norm (
double x)
{
static const int N = 8;
// Chebychev polynomial terms.
// Non-zero terms calculated via
// integrate (2/pi)*ChebyshevT[n,u]/sqrt(1-u^2)*((u+3)/2)^(5/12)
// from -1 to 1
// Zeroth term is similar except it uses 1/pi rather than 2/pi.
static const double Cn[N] = {
1.1758200232996901923,
0.16665763094889061230,
-0.0083154894939042125035,
0.00075187976780420279038,
// Wolfram alpha doesn't want to compute the remaining terms
// to more precision (it times out).
-0.0000832402,
0.0000102292,
-1.3401e-6,
1.83334e-7};
double Tn[N];
double u = 2.0*x - 3.0;
Tn[0] = 1.0;
Tn[1] = u;
for (int ii = 2; ii < N; ++ii) {
Tn[ii] = 2*u*Tn[ii-1] - Tn[ii-2];
}
double y = 0.0;
for (int ii = N-1; ii >= 0; --ii) {
y += Cn[ii]*Tn[ii];
}
return y;
}
// Returns x^(5/12) to within 3e-8 (relative error).
double pow512 (
double x)
{
static const double pow2_512[12] = {
1.0,
pow(2.0, 5.0/12.0),
pow(4.0, 5.0/12.0),
pow(8.0, 5.0/12.0),
pow(16.0, 5.0/12.0),
pow(32.0, 5.0/12.0),
pow(64.0, 5.0/12.0),
pow(128.0, 5.0/12.0),
pow(256.0, 5.0/12.0),
pow(512.0, 5.0/12.0),
pow(1024.0, 5.0/12.0),
pow(2048.0, 5.0/12.0)
};
double s;
int iexp;
s = frexp (x, &iexp);
s *= 2.0;
iexp -= 1;
div_t qr = div (iexp, 12);
if (qr.rem < 0) {
qr.quot -= 1;
qr.rem += 12;
}
return ldexp (pow512norm(s)*pow2_512[qr.rem], 5*qr.quot);
}
Addendum: What's going on here?
Per request, the following explains how the above code works.
Overview
The above code defines two functions, double pow512norm (double x) and double pow512 (double x). The latter is the entry point to the suite; this is the function that user code should call to calculate x^(5/12). The function pow512norm(x) uses Chebyshev polynomials to approximate x^(5/12), but only for x in the range [1,2]. (Use pow512norm(x) for values of x outside that range and the result will be garbage.)
The function pow512(x) splits the incoming x into a pair (double s, int n) such that x = s * 2^n and such that 1≤s<2. A further partitioning of n into (int q, unsigned int r) such that n = 12*q + r and r is less than 12 lets me split the problem of finding x^(5/12) into parts:
x^(5/12)=(s^(5/12))*((2^n)^(5/12)) via (uv)^a=(u^a)(v^a) for positive u,v and real a.
s^(5/12) is calculated via pow512norm(s).
(2^n)^(5/12)=(2^(12*q+r))^(5/12) via substitution.
2^(12*q+r)=(2^(12*q))*(2^r) via u^(a+b)=(u^a)*(u^b) for positive u, real a,b.
(2^(12*q+r))^(5/12)=(2^(5*q))*((2^r)^(5/12)) via some more manipulations.
(2^r)^(5/12) is calculated by the lookup table pow2_512.
Calculate pow512norm(s)*pow2_512[qr.rem] and we're almost there. Here qr.rem is the r value calculated in step 3 above. All that is needed is to multiply this by 2^(5*q) to yield the desired result.
That is exactly what the math library function ldexp does.
Function Approximation
The goal here is to come up with an easily computable approximation of f(x)=x^(5/12) that is 'good enough' for the problem at hand. Our approximation should be close to f(x) in some sense. Rhetorical question: What does 'close to' mean? Two competing interpretations are minimizing the mean square error versus minimizing the maximum absolute error.
I'll use a stock market analogy to describe the difference between these. Suppose you want to save for your eventual retirement. If you are in your twenties, the best thing to do is to invest in stocks or stock market funds. This is because over a long enough span of time, the stock market on average beats any other investment scheme. However, we've all seen times when putting money into stocks is a very bad thing to do. If you are in your fifties or sixties (or forties if you want to retire young) you need to invest a bit more conservatively. Those downswings can wreak have on your retirement portfolio.
Back to function approximation: As the consumer of some approximation, you are typically worried about the worst-case error rather than the performance "on average". Use some approximation constructed to give the best performance "on average" (e.g. least squares) and Murphy's law dictates that your program will spend a whole lot of time using the approximation exactly where the performance is far worse than average. What you want is a minimax approximation, something that minimizes the maximum absolute error over some domain. A good math library will take a minimax approach rather than a least squares approach because this lets the authors of the math library give some guaranteed performance of their library.
Math libraries typically use a polynomial or a rational polynomial to approximate some function f(x) over some domain a≤x≤b. Suppose the function f(x) is analytic over this domain and you want to approximate the function by some polynomial p(x) of degree N. For a given degree N there exists some magical, unique polynomial p(x) such that p(x)-f(x) has N+2 extrema over [a,b] and such that the absolute values of these N+2 extrema are all equal to one another. Finding this magical polynomial p(x) is the holy grail of function approximators.
I did not find that holy grail for you. I instead used a Chebyshev approximation. The Chebyshev polynomials of the first kind are an orthogonal (but not orthonormal) set of polynomials with some very nice features when it comes to function approximation. The Chebyshev approximation oftentimes is very close to that magical polynomial p(x). (In fact, the Remez exchange algorithm that does find that holy grail polynomial typically starts with a Chebyshev approximation.)
pow512norm(x)
This function uses Chebyshev approximation to find some polynomial p*(x) that approximates x^(5/12). Here I'm using p*(x) to distinguish this Chebyshev approximation from the magical polynomial p(x) described above. The Chebyshev approximation p*(x) is easy to find; finding p(x) is a bear. The Chebyshev approximation p*(x) is sum_i Cn[i]*Tn(i,x), where the Cn[i] are the Chebyshev coefficients and Tn(i,x) are the Chebyshev polynomials evaluated at x.
I used Wolfram alpha to find the Chebyshev coefficients Cn for me. For example, this calculates Cn[1]. The first box after the input box has the desired answer, 0.166658 in this case. That's not as many digits as I would like. Click on 'more digits' and voila, you get a whole lot more digits. Wolfram alpha is free; there is a limit on how much computation it will do. It hits that limit on higher order terms. (If you buy or have access to mathematica you will be able to calculate those high-order coefficients to a high degree of precision.)
The Chebyshev polynomials Tn(x) are calculated in the array Tn. Beyond giving something very close to magical polynomial p(x), another reason for using Chebyshev approximation is that the values of those Chebyshev polynomials are easily calculated: Start with Tn[0]=1 and Tn[1]=x, and then iteratively calculate Tn[i]=2*x*Tn[i-1] - Tn[i-2]. (I used 'ii' as the index variable rather than 'i' in my code. I never use 'i' as a variable name. How many words in the English language have an 'i' in the word? How many have two consecutive 'i's?)
pow512(x)
pow512 is the function that user code should be calling. I already described the basics of this function above. A few more details: The math library function frexp(x) returns the significand s and exponent iexp for the input x. (Minor issue: I want s between 1 and 2 for use with pow512norm but frexp returns a value between 0.5 and 1.) The math library function div returns the quotient and remainder for integer division in one swell foop. Finally, I use the math library function ldexp to put the three parts together to form the final answer.
In the IEEE 754 hacking vein, here is another solution which is faster and less "magical." It achieves an error margin of .08% in about a dozen clock cycles (for the case of p=2.4, on an Intel Merom CPU).
Floating point numbers were originally invented as an approximation to logarithms, so you can use the integer value as an approximation of log2. This is somewhat-portably achievable by applying the convert-from-integer instruction to a floating-point value, to obtain another floating-point value.
To complete the pow computation, you can multiply by a constant factor and convert the logarithm back with the convert-to-integer instruction. On SSE, the relevant instructions are cvtdq2ps and cvtps2dq.
It's not quite so simple, though. The exponent field in IEEE 754 is signed, with a bias value of 127 representing an exponent of zero. This bias must be removed before you multiply the logarithm, and re-added before you exponentiate. Furthermore, bias adjustment by subtraction won't work on zero. Fortunately, both adjustments can be achieved by multiplying by a constant factor beforehand.
x^p
= exp2( p * log2( x ) )
= exp2( p * ( log2( x ) + 127 - 127 ) - 127 + 127 )
= cvtps2dq( p * ( log2( x ) + 127 - 127 - 127 / p ) )
= cvtps2dq( p * ( log2( x ) + 127 - log2( exp2( 127 - 127 / p ) ) )
= cvtps2dq( p * ( log2( x * exp2( 127 / p - 127 ) ) + 127 ) )
= cvtps2dq( p * ( cvtdq2ps( x * exp2( 127 / p - 127 ) ) ) )
exp2( 127 / p - 127 ) is the constant factor. This function is rather specialized: it won't work with small fractional exponents, because the constant factor grows exponentially with the inverse of the exponent and will overflow. It won't work with negative exponents. Large exponents lead to high error, because the mantissa bits are mingled with the exponent bits by the multiplication.
But, it's just 4 fast instructions long. Pre-multiply, convert from "integer" (to logarithm), power-multiply, convert to "integer" (from logarithm). Conversions are very fast on this implementation of SSE. We can also squeeze an extra constant coefficient into the first multiplication.
template< unsigned expnum, unsigned expden, unsigned coeffnum, unsigned coeffden >
__m128 fastpow( __m128 arg ) {
__m128 ret = arg;
// std::printf( "arg = %,vg\n", ret );
// Apply a constant pre-correction factor.
ret = _mm_mul_ps( ret, _mm_set1_ps( exp2( 127. * expden / expnum - 127. )
* pow( 1. * coeffnum / coeffden, 1. * expden / expnum ) ) );
// std::printf( "scaled = %,vg\n", ret );
// Reinterpret arg as integer to obtain logarithm.
asm ( "cvtdq2ps %1, %0" : "=x" (ret) : "x" (ret) );
// std::printf( "log = %,vg\n", ret );
// Multiply logarithm by power.
ret = _mm_mul_ps( ret, _mm_set1_ps( 1. * expnum / expden ) );
// std::printf( "powered = %,vg\n", ret );
// Convert back to "integer" to exponentiate.
asm ( "cvtps2dq %1, %0" : "=x" (ret) : "x" (ret) );
// std::printf( "result = %,vg\n", ret );
return ret;
}
A few trials with exponent = 2.4 show this consistently overestimates by about 5%. (The routine is always guaranteed to overestimate.) You could simply multiply by 0.95, but a few more instructions will get us about 4 decimal digits of accuracy, which should be enough for graphics.
The key is to match the overestimate with an underestimate, and take the average.
Compute x^0.8: four instructions, error ~ +3%.
Compute x^-0.4: one rsqrtps. (This is quite accurate enough, but does sacrifice the ability to work with zero.)
Compute x^0.4: one mulps.
Compute x^-0.2: one rsqrtps.
Compute x^2: one mulps.
Compute x^3: one mulps.
x^2.4 = x^2 * x^0.4: one mulps. This is the overestimate.
x^2.4 = x^3 * x^-0.4 * x^-0.2: two mulps. This is the underestimate.
Average the above: one addps, one mulps.
Instruction tally: fourteen, including two conversions with latency = 5 and two reciprocal square root estimates with throughput = 4.
To properly take the average, we want to weight the estimates by their expected errors. The underestimate raises the error to a power of 0.6 vs 0.4, so we expect it to be 1.5x as erroneous. Weighting doesn't add any instructions; it can be done in the pre-factor. Calling the coefficient a: a^0.5 = 1.5 a^-0.75, and a = 1.38316186.
The final error is about .015%, or 2 orders of magnitude better than the initial fastpow result. The runtime is about a dozen cycles for a busy loop with volatile source and destination variables… although it's overlapping the iterations, real-world usage will also see instruction-level parallelism. Considering SIMD, that's a throughput of one scalar result per 3 cycles!
int main() {
__m128 const x0 = _mm_set_ps( 0.01, 1, 5, 1234.567 );
std::printf( "Input: %,vg\n", x0 );
// Approx 5% accuracy from one call. Always an overestimate.
__m128 x1 = fastpow< 24, 10, 1, 1 >( x0 );
std::printf( "Direct x^2.4: %,vg\n", x1 );
// Lower exponents provide lower initial error, but too low causes overflow.
__m128 xf = fastpow< 8, 10, int( 1.38316186 * 1e9 ), int( 1e9 ) >( x0 );
std::printf( "1.38 x^0.8: %,vg\n", xf );
// Imprecise 4-cycle sqrt is still far better than fastpow, good enough.
__m128 xfm4 = _mm_rsqrt_ps( xf );
__m128 xf4 = _mm_mul_ps( xf, xfm4 );
// Precisely calculate x^2 and x^3
__m128 x2 = _mm_mul_ps( x0, x0 );
__m128 x3 = _mm_mul_ps( x2, x0 );
// Overestimate of x^2 * x^0.4
x2 = _mm_mul_ps( x2, xf4 );
// Get x^-0.2 from x^0.4. Combine with x^-0.4 into x^-0.6 and x^2.4.
__m128 xfm2 = _mm_rsqrt_ps( xf4 );
x3 = _mm_mul_ps( x3, xfm4 );
x3 = _mm_mul_ps( x3, xfm2 );
std::printf( "x^2 * x^0.4: %,vg\n", x2 );
std::printf( "x^3 / x^0.6: %,vg\n", x3 );
x2 = _mm_mul_ps( _mm_add_ps( x2, x3 ), _mm_set1_ps( 1/ 1.960131704207789 ) );
// Final accuracy about 0.015%, 200x better than x^0.8 calculation.
std::printf( "average = %,vg\n", x2 );
}
Well… sorry I wasn't able to post this sooner. And extending it to x^1/2.4 is left as an exercise ;v) .
Update with stats
I implemented a little test harness and two x(5⁄12) cases corresponding to the above.
#include <cstdio>
#include <xmmintrin.h>
#include <cmath>
#include <cfloat>
#include <algorithm>
using namespace std;
template< unsigned expnum, unsigned expden, unsigned coeffnum, unsigned coeffden >
__m128 fastpow( __m128 arg ) {
__m128 ret = arg;
// std::printf( "arg = %,vg\n", ret );
// Apply a constant pre-correction factor.
ret = _mm_mul_ps( ret, _mm_set1_ps( exp2( 127. * expden / expnum - 127. )
* pow( 1. * coeffnum / coeffden, 1. * expden / expnum ) ) );
// std::printf( "scaled = %,vg\n", ret );
// Reinterpret arg as integer to obtain logarithm.
asm ( "cvtdq2ps %1, %0" : "=x" (ret) : "x" (ret) );
// std::printf( "log = %,vg\n", ret );
// Multiply logarithm by power.
ret = _mm_mul_ps( ret, _mm_set1_ps( 1. * expnum / expden ) );
// std::printf( "powered = %,vg\n", ret );
// Convert back to "integer" to exponentiate.
asm ( "cvtps2dq %1, %0" : "=x" (ret) : "x" (ret) );
// std::printf( "result = %,vg\n", ret );
return ret;
}
__m128 pow125_4( __m128 arg ) {
// Lower exponents provide lower initial error, but too low causes overflow.
__m128 xf = fastpow< 4, 5, int( 1.38316186 * 1e9 ), int( 1e9 ) >( arg );
// Imprecise 4-cycle sqrt is still far better than fastpow, good enough.
__m128 xfm4 = _mm_rsqrt_ps( xf );
__m128 xf4 = _mm_mul_ps( xf, xfm4 );
// Precisely calculate x^2 and x^3
__m128 x2 = _mm_mul_ps( arg, arg );
__m128 x3 = _mm_mul_ps( x2, arg );
// Overestimate of x^2 * x^0.4
x2 = _mm_mul_ps( x2, xf4 );
// Get x^-0.2 from x^0.4, and square it for x^-0.4. Combine into x^-0.6.
__m128 xfm2 = _mm_rsqrt_ps( xf4 );
x3 = _mm_mul_ps( x3, xfm4 );
x3 = _mm_mul_ps( x3, xfm2 );
return _mm_mul_ps( _mm_add_ps( x2, x3 ), _mm_set1_ps( 1/ 1.960131704207789 * 0.9999 ) );
}
__m128 pow512_2( __m128 arg ) {
// 5/12 is too small, so compute the sqrt of 10/12 instead.
__m128 x = fastpow< 5, 6, int( 0.992245 * 1e9 ), int( 1e9 ) >( arg );
return _mm_mul_ps( _mm_rsqrt_ps( x ), x );
}
__m128 pow512_4( __m128 arg ) {
// 5/12 is too small, so compute the 4th root of 20/12 instead.
// 20/12 = 5/3 = 1 + 2/3 = 2 - 1/3. 2/3 is a suitable argument for fastpow.
// weighting coefficient: a^-1/2 = 2 a; a = 2^-2/3
__m128 xf = fastpow< 2, 3, int( 0.629960524947437 * 1e9 ), int( 1e9 ) >( arg );
__m128 xover = _mm_mul_ps( arg, xf );
__m128 xfm1 = _mm_rsqrt_ps( xf );
__m128 x2 = _mm_mul_ps( arg, arg );
__m128 xunder = _mm_mul_ps( x2, xfm1 );
// sqrt2 * over + 2 * sqrt2 * under
__m128 xavg = _mm_mul_ps( _mm_set1_ps( 1/( 3 * 0.629960524947437 ) * 0.999852 ),
_mm_add_ps( xover, xunder ) );
xavg = _mm_mul_ps( xavg, _mm_rsqrt_ps( xavg ) );
xavg = _mm_mul_ps( xavg, _mm_rsqrt_ps( xavg ) );
return xavg;
}
__m128 mm_succ_ps( __m128 arg ) {
return (__m128) _mm_add_epi32( (__m128i) arg, _mm_set1_epi32( 4 ) );
}
void test_pow( double p, __m128 (*f)( __m128 ) ) {
__m128 arg;
for ( arg = _mm_set1_ps( FLT_MIN / FLT_EPSILON );
! isfinite( _mm_cvtss_f32( f( arg ) ) );
arg = mm_succ_ps( arg ) ) ;
for ( ; _mm_cvtss_f32( f( arg ) ) == 0;
arg = mm_succ_ps( arg ) ) ;
std::printf( "Domain from %g\n", _mm_cvtss_f32( arg ) );
int n;
int const bucket_size = 1 << 25;
do {
float max_error = 0;
double total_error = 0, cum_error = 0;
for ( n = 0; n != bucket_size; ++ n ) {
float result = _mm_cvtss_f32( f( arg ) );
if ( ! isfinite( result ) ) break;
float actual = ::powf( _mm_cvtss_f32( arg ), p );
float error = ( result - actual ) / actual;
cum_error += error;
error = std::abs( error );
max_error = std::max( max_error, error );
total_error += error;
arg = mm_succ_ps( arg );
}
std::printf( "error max = %8g\t" "avg = %8g\t" "|avg| = %8g\t" "to %8g\n",
max_error, cum_error / n, total_error / n, _mm_cvtss_f32( arg ) );
} while ( n == bucket_size );
}
int main() {
std::printf( "4 insn x^12/5:\n" );
test_pow( 12./5, & fastpow< 12, 5, 1059, 1000 > );
std::printf( "14 insn x^12/5:\n" );
test_pow( 12./5, & pow125_4 );
std::printf( "6 insn x^5/12:\n" );
test_pow( 5./12, & pow512_2 );
std::printf( "14 insn x^5/12:\n" );
test_pow( 5./12, & pow512_4 );
}
Output:
4 insn x^12/5:
Domain from 1.36909e-23
error max = inf avg = inf |avg| = inf to 8.97249e-19
error max = 2267.14 avg = 139.175 |avg| = 139.193 to 5.88021e-14
error max = 0.123606 avg = -0.000102963 |avg| = 0.0371122 to 3.85365e-09
error max = 0.123607 avg = -0.000108978 |avg| = 0.0368548 to 0.000252553
error max = 0.12361 avg = 7.28909e-05 |avg| = 0.037507 to 16.5513
error max = 0.123612 avg = -0.000258619 |avg| = 0.0365618 to 1.08471e+06
error max = 0.123611 avg = 8.70966e-05 |avg| = 0.0374369 to 7.10874e+10
error max = 0.12361 avg = -0.000103047 |avg| = 0.0371122 to 4.65878e+15
error max = 0.123609 avg = nan |avg| = nan to 1.16469e+16
14 insn x^12/5:
Domain from 1.42795e-19
error max = inf avg = nan |avg| = nan to 9.35823e-15
error max = 0.000936462 avg = 2.0202e-05 |avg| = 0.000133764 to 6.13301e-10
error max = 0.000792752 avg = 1.45717e-05 |avg| = 0.000129936 to 4.01933e-05
error max = 0.000791785 avg = 7.0132e-06 |avg| = 0.000129923 to 2.63411
error max = 0.000787589 avg = 1.20745e-05 |avg| = 0.000129347 to 172629
error max = 0.000786553 avg = 1.62351e-05 |avg| = 0.000132397 to 1.13134e+10
error max = 0.000785586 avg = 8.25205e-06 |avg| = 0.00013037 to 6.98147e+12
6 insn x^5/12:
Domain from 9.86076e-32
error max = 0.0284339 avg = 0.000441158 |avg| = 0.00967327 to 6.46235e-27
error max = 0.0284342 avg = -5.79938e-06 |avg| = 0.00897913 to 4.23516e-22
error max = 0.0284341 avg = -0.000140706 |avg| = 0.00897084 to 2.77556e-17
error max = 0.028434 avg = 0.000440504 |avg| = 0.00967325 to 1.81899e-12
error max = 0.0284339 avg = -6.11153e-06 |avg| = 0.00897915 to 1.19209e-07
error max = 0.0284298 avg = -0.000140597 |avg| = 0.00897084 to 0.0078125
error max = 0.0284371 avg = 0.000439748 |avg| = 0.00967319 to 512
error max = 0.028437 avg = -7.74294e-06 |avg| = 0.00897924 to 3.35544e+07
error max = 0.0284369 avg = -0.000142036 |avg| = 0.00897089 to 2.19902e+12
error max = 0.0284368 avg = 0.000439183 |avg| = 0.0096732 to 1.44115e+17
error max = 0.0284367 avg = -7.41244e-06 |avg| = 0.00897923 to 9.44473e+21
error max = 0.0284366 avg = -0.000141706 |avg| = 0.00897088 to 6.1897e+26
error max = 0.485129 avg = -0.0401671 |avg| = 0.048422 to 4.05648e+31
error max = 0.994932 avg = -0.891494 |avg| = 0.891494 to 2.65846e+36
error max = 0.999329 avg = nan |avg| = nan to -0
14 insn x^5/12:
Domain from 2.64698e-23
error max = 0.13556 avg = 0.00125936 |avg| = 0.00354677 to 1.73472e-18
error max = 0.000564988 avg = 2.51458e-06 |avg| = 0.000113709 to 1.13687e-13
error max = 0.000565065 avg = -1.49258e-06 |avg| = 0.000112553 to 7.45058e-09
error max = 0.000565143 avg = 1.5293e-06 |avg| = 0.000112864 to 0.000488281
error max = 0.000565298 avg = 2.76457e-06 |avg| = 0.000113713 to 32
error max = 0.000565453 avg = -1.61276e-06 |avg| = 0.000112561 to 2.09715e+06
error max = 0.000565531 avg = 1.42628e-06 |avg| = 0.000112866 to 1.37439e+11
error max = 0.000565686 avg = 2.71505e-06 |avg| = 0.000113715 to 9.0072e+15
error max = 0.000565763 avg = -1.56586e-06 |avg| = 0.000112415 to 1.84467e+19
I suspect accuracy of the more accurate 5/12 is being limited by the rsqrt operation.
Ian Stephenson wrote this code which he claims outperforms pow(). He describes the idea as follows:
Pow is basically implemented using
log's: pow(a,b)=x(logx(a)*b). so we
need a fast log and fast exponent - it
doesn't matter what x is so we use 2.
The trick is that a floating point
number is already in a log style
format:
a=M*2E
Taking the log of both sides gives:
log2(a)=log2(M)+E
or more simply:
log2(a)~=E
In other words if we take the floating
point representation of a number, and
extract the Exponent we've got
something that's a good starting point
as its log. It turns out that when we
do this by massaging the bit patterns,
the Mantissa ends up giving a good
approximation to the error, and it
works pretty well.
This should be good enough for simple
lighting calculations, but if you need
something better, you can then extract
the Mantissa, and use that to
calculate a quadratic correction factor
which is pretty accurate.
First off, using floats isn't going to buy much on most machines nowadays. In fact, doubles can be faster. Your power, 1.0/2.4, is 5/12 or 1/3*(1+1/4). Even though this is calling cbrt (once) and sqrt (twice!) it is still twice as fast as using pow(). (Optimization: -O3, compiler: i686-apple-darwin10-g++-4.2.1).
#include <math.h> // cmath does not provide cbrt; C99 does.
double xpow512 (double x) {
double cbrtx = cbrt(x);
return cbrtx*sqrt(sqrt(cbrtx));
}
This might not answer your question.
The 2.4f and 1/2.4f make me very suspicious, because those are exactly the powers used to convert between sRGB and a linear RGB color space. So you might actually be trying to optimize that, specifically. I don't know, which is why this might not answer your question.
If this is the case, try using a lookup table. Something like:
__attribute__((aligned(64))
static const unsigned short SRGB_TO_LINEAR[256] = { ... };
__attribute__((aligned(64))
static const unsigned short LINEAR_TO_SRGB[256] = { ... };
void apply_lut(const unsigned short lut[256], unsigned char *src, ...
If you are using 16-bit data, change as appropriate. I would make the table 16 bits anyway so you can dither the result if necessary when working with 8-bit data. This obviously won't work very well if your data is floating point to begin with -- but it doesn't really make sense to store sRGB data in floating point, so you might as well convert to 16-bit / 8-bit first and then do the conversion from linear to sRGB.
(The reason sRGB doesn't make sense as floating point is that HDR should be linear, and sRGB is only convenient for storing on disk or displaying on screen, but not convenient for manipulation.)
I shall answer the question you really wanted to ask, which is how to do fast sRGB <-> linear RGB conversion. To do this precisely and efficiently we can use polynomial approximations. The following polynomial approximations have been generated with sollya, and have a worst case relative error of 0.0144%.
inline double poly7(double x, double a, double b, double c, double d,
double e, double f, double g, double h) {
double ab, cd, ef, gh, abcd, efgh, x2, x4;
x2 = x*x; x4 = x2*x2;
ab = a*x + b; cd = c*x + d;
ef = e*x + f; gh = g*x + h;
abcd = ab*x2 + cd; efgh = ef*x2 + gh;
return abcd*x4 + efgh;
}
inline double srgb_to_linear(double x) {
if (x <= 0.04045) return x / 12.92;
// Polynomial approximation of ((x+0.055)/1.055)^2.4.
return poly7(x, 0.15237971711927983387,
-0.57235993072870072762,
0.92097986411523535821,
-0.90208229831912012386,
0.88348956209696805075,
0.48110797889132134175,
0.03563925285274562038,
0.00084585397227064120);
}
inline double linear_to_srgb(double x) {
if (x <= 0.0031308) return x * 12.92;
// Piecewise polynomial approximation (divided by x^3)
// of 1.055 * x^(1/2.4) - 0.055.
if (x <= 0.0523) return poly7(x, -6681.49576364495442248881,
1224.97114922729451791383,
-100.23413743425112443219,
6.60361150127077944916,
0.06114808961060447245,
-0.00022244138470139442,
0.00000041231840827815,
-0.00000000035133685895) / (x*x*x);
return poly7(x, -0.18730034115395793881,
0.64677431008037400417,
-0.99032868647877825286,
1.20939072663263713636,
0.33433459165487383613,
-0.01345095746411287783,
0.00044351684288719036,
-0.00000664263587520855) / (x*x*x);
}
And the sollya input used to generate the polynomials:
suppressmessage(174);
f = ((x+0.055)/1.055)^2.4;
p0 = fpminimax(f, 7, [|D...|], [0.04045;1], relative);
p = fpminimax(f/(p0(1)+1e-18), 7, [|D...|], [0.04045;1], relative);
print("relative:", dirtyinfnorm((f-p)/f, [s;1]));
print("absolute:", dirtyinfnorm((f-p), [s;1]));
print(canonical(p));
s = 0.0523;
z = 3;
f = 1.055 * x^(1/2.4) - 0.055;
p = fpminimax(1.055 * (x^(z+1/2.4) - 0.055*x^z/1.055), 7, [|D...|], [0.0031308;s], relative)/x^z;
print("relative:", dirtyinfnorm((f-p)/f, [0.0031308;s]));
print("absolute:", dirtyinfnorm((f-p), [0.0031308;s]));
print(canonical(p));
p = fpminimax(1.055 * (x^(z+1/2.4) - 0.055*x^z/1.055), 7, [|D...|], [s;1], relative)/x^z;
print("relative:", dirtyinfnorm((f-p)/f, [s;1]));
print("absolute:", dirtyinfnorm((f-p), [s;1]));
print(canonical(p));
Binomial series does account for a constant exponent, but you will be able to use it only if you can normalize all your input to the range [1,2). (Note that it computes (1+x)^a). You'll have to do some analysis to decide how many terms you need for your desired accuracy.
For exponents of 2.4, you could either make a lookup table for all your 2.4 values and lirp or perhaps higher-order function to fill in the in-betweem values if the table wasn't accurate enough (basically a huge log table.)
Or, value squared * value to the 2/5s which could take the initial square value from the first half of the function and then 5th root it. For the 5th root, you could Newton it or do some other fast approximator, though honestly once you get to this point, your probably better off just doing the exp and log functions with the appropriate abbreviated series functions yourself.
The following is an idea you can use with any of the fast calculation methods. Whether it helps things go faster depends on how your data arrives. You can use the fact that if you know x and pow(x, n), you can use the rate of change of the power to compute a reasonable approximation of pow(x + delta, n) for small delta, with a single multiply and add (more or less). If successive values you feed your power functions are close enough together, this would amortize the full cost of the accurate calculation over multiple function calls. Note that you don't need an extra pow calculation to get the derivative. You could extend this to use the second derivative so you can use a quadratic, which would increase the delta you could use and still get the same accuracy.
So traditionally the powf(x, p) = x^p is solved by rewriting x as x=2^(log2(x)) making powf(x,p) = 2^(p*log2(x)), which transforms the problem into two approximations exp2() & log2(). This has the advantage of working with larger powers p, however the downside is that this is not the optimal solution for a constant power p and over a specified input bound 0 ≤ x ≤ 1.
When the power p > 1, the answer is a trivial minimax polynomial over the bound 0 ≤ x ≤ 1, which is the case for p = 12/5 = 2.4 as can be seen below:
float pow12_5(float x){
float mp;
// Minimax horner polynomials for x^(5/12), Note: choose the accurarcy required then implement with fma() [Fused Multiply Accumulates]
// mp = 0x4.a84a38p-12 + x * (-0xd.e5648p-8 + x * (0xa.d82fep-4 + x * 0x6.062668p-4)); // 1.13705697e-3
mp = 0x1.117542p-12 + x * (-0x5.91e6ap-8 + x * (0x8.0f50ep-4 + x * (0xa.aa231p-4 + x * (-0x2.62787p-4)))); // 2.6079002e-4
// mp = 0x5.a522ap-16 + x * (-0x2.d997fcp-8 + x * (0x6.8f6d1p-4 + x * (0xf.21285p-4 + x * (-0x7.b5b248p-4 + x * 0x2.32b668p-4)))); // 8.61377e-5
// mp = 0x2.4f5538p-16 + x * (-0x1.abcdecp-8 + x * (0x5.97464p-4 + x * (0x1.399edap0 + x * (-0x1.0d363ap0 + x * (0xa.a54a3p-4 + x * (-0x2.e8a77cp-4)))))); // 3.524655e-5
return(mp);
}
However when p < 1 the minimax approximation over the bound 0 ≤ x ≤ 1 does not appropriately converge to the desired accuracy. One option [not really] is to rewrite the problem y=x^p=x^(p+m)/x^m where m=1,2,3 is a positive integer, making the new power approximation p > 1 but this introduces division which is inherently slower.
There's however another option which is to decompose the input x as its floating point exponent and mantissa form:
x = mx* 2^(ex) where 1 ≤ mx < 2
y = x^(5/12) = mx^(5/12) * 2^((5/12)*ex), let ey = floor(5*ex/12), k = (5*ex) % 12
= mx^(5/12) * 2^(k/12) * 2^(ey)
The minimax approximation of mx^(5/12) over 1 ≤ mx < 2 now converges much faster than before, without division, but requires 12 point LUT for the 2^(k/12). The code is below:
float powk_12LUT[] = {0x1.0p0, 0x1.0f38fap0, 0x1.1f59acp0, 0x1.306fep0, 0x1.428a3p0, 0x1.55b81p0, 0x1.6a09e6p0, 0x1.7f910ep0, 0x1.965feap0, 0x1.ae89fap0, 0x1.c823ep0, 0x1.e3437ep0};
float pow5_12(float x){
union{float f; uint32_t u;} v, e2;
float poff, m, e, ei;
int xe;
v.f = x;
xe = ((v.u >> 23) - 127);
if(xe < -127) return(0.0f);
// Calculate remainder k in 2^(k/12) to find LUT
e = xe * (5.0f/12.0f);
ei = floorf(e);
poff = powk_12LUT[(int)(12.0f * (e - ei))];
e2.u = ((int)ei + 127) << 23; // Calculate the exponent
v.u = (v.u & ~(0xFFuL << 23)) | (0x7FuL << 23); // Normalize exponent to zero
// Approximate mx^(5/12) on [1,2), with appropriate degree minimax
// m = 0x8.87592p-4 + v.f * (0x8.8f056p-4 + v.f * (-0x1.134044p-4)); // 7.6125e-4
// m = 0x7.582138p-4 + v.f * (0xb.1666bp-4 + v.f * (-0x2.d21954p-4 + v.f * 0x6.3ea0cp-8)); // 8.4522726e-5
m = 0x6.9465cp-4 + v.f * (0xd.43015p-4 + v.f * (-0x5.17b2a8p-4 + v.f * (0x1.6cb1f8p-4 + v.f * (-0x2.c5b76p-8)))); // 1.04091259e-5
// m = 0x6.08242p-4 + v.f * (0xf.352bdp-4 + v.f * (-0x7.d0c1bp-4 + v.f * (0x3.4d153p-4 + v.f * (-0xc.f7a42p-8 + v.f * 0x1.5d840cp-8)))); // 1.367401e-6
return(m * poff * e2.f);
}