scikit-learn/sklearn/utils/src/gamma.c

156 lines
4.2 KiB
C

/*
* John D. Cook's public domain version of lgamma, from
* https://www.johndcook.com/stand_alone_code.html
*
* Replaces the C99 standard lgamma for stone-age C compilers like the one
* from Redmond.
*
* I removed the test cases and added the cfloat import (Vlad N. <vlad@vene.ro>)
*
* Translated to C by Lars Buitinck. Input validation removed; we handle
* that in the Cython wrapper.
*/
#include <float.h>
#include <math.h>
#include "gamma.h"
/* Euler's gamma constant. */
#define GAMMA 0.577215664901532860606512090
#define HALF_LOG2_PI 0.91893853320467274178032973640562
static double sklearn_gamma(double x)
{
/*
* Split the function domain into three intervals:
* (0, 0.001), [0.001, 12), and (12, infinity).
*/
/*
* First interval: (0, 0.001).
*
* For small x, 1/Gamma(x) has power series x + gamma x^2 - ...
* So in this range, 1/Gamma(x) = x + gamma x^2 with error
* on the order of x^3.
* The relative error over this interval is less than 6e-7.
*/
if (x < 0.001)
return 1.0 / (x * (1.0 + GAMMA * x));
/*
* Second interval: [0.001, 12).
*/
if (x < 12.0) {
/* numerator coefficients for approximation over the interval (1,2) */
static const double p[] = {
-1.71618513886549492533811E+0,
2.47656508055759199108314E+1,
-3.79804256470945635097577E+2,
6.29331155312818442661052E+2,
8.66966202790413211295064E+2,
-3.14512729688483675254357E+4,
-3.61444134186911729807069E+4,
6.64561438202405440627855E+4
};
/* denominator coefficients for approximation over the interval (1,2) */
static const double q[] = {
-3.08402300119738975254353E+1,
3.15350626979604161529144E+2,
-1.01515636749021914166146E+3,
-3.10777167157231109440444E+3,
2.25381184209801510330112E+4,
4.75584627752788110767815E+3,
-1.34659959864969306392456E+5,
-1.15132259675553483497211E+5
};
double den, num, result, z;
/* The algorithm directly approximates gamma over (1,2) and uses
* reduction identities to reduce other arguments to this interval. */
double y = x;
int i, n = 0;
int arg_was_less_than_one = (y < 1.0);
/* Add or subtract integers as necessary to bring y into (1,2)
* Will correct for this below */
if (arg_was_less_than_one)
y += 1.0;
else {
n = (int)floor(y) - 1;
y -= n;
}
num = 0.0;
den = 1.0;
z = y - 1;
for (i = 0; i < 8; i++) {
num = (num + p[i]) * z;
den = den * z + q[i];
}
result = num/den + 1.0;
/* Apply correction if argument was not initially in (1,2) */
if (arg_was_less_than_one)
/* Use identity gamma(z) = gamma(z+1)/z
* The variable "result" now holds gamma of the original y + 1
* Thus we use y-1 to get back the original y. */
result /= (y-1.0);
else
/* Use the identity gamma(z+n) = z*(z+1)* ... *(z+n-1)*gamma(z) */
for (i = 0; i < n; i++, y++)
result *= y;
return result;
}
/*
* Third interval: [12, infinity).
*/
if (x > 171.624)
/* Correct answer too large to display, force +infinity. */
return 2 * DBL_MAX;
return exp(sklearn_lgamma(x));
}
double sklearn_lgamma(double x)
{
/*
* Abramowitz and Stegun 6.1.41
* Asymptotic series should be good to at least 11 or 12 figures
* For error analysis, see Whittiker and Watson
* A Course in Modern Analysis (1927), page 252
*/
static const double c[8] =
{
1.0/12.0,
-1.0/360.0,
1.0/1260.0,
-1.0/1680.0,
1.0/1188.0,
-691.0/360360.0,
1.0/156.0,
-3617.0/122400.0
};
double z, sum;
int i;
if (x < 12.0)
return log(fabs(sklearn_gamma(x)));
z = 1.0 / (x * x);
sum = c[7];
for (i=6; i >= 0; i--) {
sum *= z;
sum += c[i];
}
return (x - 0.5) * log(x) - x + HALF_LOG2_PI + sum / x;
}