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/* randist/binomial_tpe.c
*
* Copyright (C) 1996-2003 James Theiler, Brian Gough
*
* This program is free software; you can redistribute it and/or modify
* it under the terms of the GNU General Public License as published by
* the Free Software Foundation; either version 2 of the License, or (at
* your option) any later version.
*
* This program is distributed in the hope that it will be useful, but
* WITHOUT ANY WARRANTY; without even the implied warranty of
* MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
* General Public License for more details.
*
* You should have received a copy of the GNU General Public License
* along with this program; if not, write to the Free Software
* Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301, USA.
*/
#include <config.h>
#include <math.h>
#include <gsl/gsl_rng.h>
#include <gsl/gsl_randist.h>
#include <gsl/gsl_pow_int.h>
#include <gsl/gsl_sf_gamma.h>
/* The binomial distribution has the form,
f(x) = n!/(x!(n-x)!) * p^x (1-p)^(n-x) for integer 0 <= x <= n
= 0 otherwise
This implementation follows the public domain ranlib function
"ignbin", the bulk of which is the BTPE (Binomial Triangle
Parallelogram Exponential) algorithm introduced in
Kachitvichyanukul and Schmeiser[1]. It has been translated to use
modern C coding standards.
If n is small and/or p is near 0 or near 1 (specifically, if
n*min(p,1-p) < SMALL_MEAN), then a different algorithm, called
BINV, is used which has an average runtime that scales linearly
with n*min(p,1-p).
But for larger problems, the BTPE algorithm takes the form of two
functions b(x) and t(x) -- "bottom" and "top" -- for which b(x) <
f(x)/f(M) < t(x), with M = floor(n*p+p). b(x) defines a triangular
region, and t(x) includes a parallelogram and two tails. Details
(including a nice drawing) are in the paper.
[1] Kachitvichyanukul, V. and Schmeiser, B. W. Binomial Random
Variate Generation. Communications of the ACM, 31, 2 (February,
1988) 216.
Note, Bruce Schmeiser (personal communication) points out that if
you want very fast binomial deviates, and you are happy with
approximate results, and/or n and n*p are both large, then you can
just use gaussian estimates: mean=n*p, variance=n*p*(1-p).
This implementation by James Theiler, April 2003, after obtaining
permission -- and some good advice -- from Drs. Kachitvichyanukul
and Schmeiser to use their code as a starting point, and then doing
a little bit of tweaking.
Additional polishing for GSL coding standards by Brian Gough. */
#define SMALL_MEAN 14 /* If n*p < SMALL_MEAN then use BINV
algorithm. The ranlib
implementation used cutoff=30; but
on my computer 14 works better */
#define BINV_CUTOFF 110 /* In BINV, do not permit ix too large */
#define FAR_FROM_MEAN 20 /* If ix-n*p is larger than this, then
use the "squeeze" algorithm.
Ranlib used 20, and this seems to
be the best choice on my machine as
well */
#define LNFACT(x) gsl_sf_lnfact(x)
inline static double
Stirling (double y1)
{
double y2 = y1 * y1;
double s =
(13860.0 -
(462.0 - (132.0 - (99.0 - 140.0 / y2) / y2) / y2) / y2) / y1 / 166320.0;
return s;
}
unsigned int
gsl_ran_binomial_tpe (const gsl_rng * rng, double p, unsigned int n)
{
return gsl_ran_binomial (rng, p, n);
}
unsigned int
gsl_ran_binomial (const gsl_rng * rng, double p, unsigned int n)
{
int ix; /* return value */
int flipped = 0;
double q, s, np;
if (n == 0)
return 0;
if (p > 0.5)
{
p = 1.0 - p; /* work with small p */
flipped = 1;
}
q = 1 - p;
s = p / q;
np = n * p;
/* Inverse cdf logic for small mean (BINV in K+S) */
if (np < SMALL_MEAN)
{
double f0 = gsl_pow_int (q, n); /* f(x), starting with x=0 */
while (1)
{
/* This while(1) loop will almost certainly only loop once; but
* if u=1 to within a few epsilons of machine precision, then it
* is possible for roundoff to prevent the main loop over ix to
* achieve its proper value. following the ranlib implementation,
* we introduce a check for that situation, and when it occurs,
* we just try again.
*/
double f = f0;
double u = gsl_rng_uniform (rng);
for (ix = 0; ix <= BINV_CUTOFF; ++ix)
{
if (u < f)
goto Finish;
u -= f;
/* Use recursion f(x+1) = f(x)*[(n-x)/(x+1)]*[p/(1-p)] */
f *= s * (n - ix) / (ix + 1);
}
/* It should be the case that the 'goto Finish' was encountered
* before this point was ever reached. But if we have reached
* this point, then roundoff has prevented u from decreasing
* all the way to zero. This can happen only if the initial u
* was very nearly equal to 1, which is a rare situation. In
* that rare situation, we just try again.
*
* Note, following the ranlib implementation, we loop ix only to
* a hardcoded value of SMALL_MEAN_LARGE_N=110; we could have
* looped to n, and 99.99...% of the time it won't matter. This
* choice, I think is a little more robust against the rare
* roundoff error. If n>LARGE_N, then it is technically
* possible for ix>LARGE_N, but it is astronomically rare, and
* if ix is that large, it is more likely due to roundoff than
* probability, so better to nip it at LARGE_N than to take a
* chance that roundoff will somehow conspire to produce an even
* larger (and more improbable) ix. If n<LARGE_N, then once
* ix=n, f=0, and the loop will continue until ix=LARGE_N.
*/
}
}
else
{
/* For n >= SMALL_MEAN, we invoke the BTPE algorithm */
int k;
double ffm = np + p; /* ffm = n*p+p */
int m = (int) ffm; /* m = int floor[n*p+p] */
double fm = m; /* fm = double m; */
double xm = fm + 0.5; /* xm = half integer mean (tip of triangle) */
double npq = np * q; /* npq = n*p*q */
/* Compute cumulative area of tri, para, exp tails */
/* p1: radius of triangle region; since height=1, also: area of region */
/* p2: p1 + area of parallelogram region */
/* p3: p2 + area of left tail */
/* p4: p3 + area of right tail */
/* pi/p4: probability of i'th area (i=1,2,3,4) */
/* Note: magic numbers 2.195, 4.6, 0.134, 20.5, 15.3 */
/* These magic numbers are not adjustable...at least not easily! */
double p1 = floor (2.195 * sqrt (npq) - 4.6 * q) + 0.5;
/* xl, xr: left and right edges of triangle */
double xl = xm - p1;
double xr = xm + p1;
/* Parameter of exponential tails */
/* Left tail: t(x) = c*exp(-lambda_l*[xl - (x+0.5)]) */
/* Right tail: t(x) = c*exp(-lambda_r*[(x+0.5) - xr]) */
double c = 0.134 + 20.5 / (15.3 + fm);
double p2 = p1 * (1.0 + c + c);
double al = (ffm - xl) / (ffm - xl * p);
double lambda_l = al * (1.0 + 0.5 * al);
double ar = (xr - ffm) / (xr * q);
double lambda_r = ar * (1.0 + 0.5 * ar);
double p3 = p2 + c / lambda_l;
double p4 = p3 + c / lambda_r;
double var, accept;
double u, v; /* random variates */
TryAgain:
/* generate random variates, u specifies which region: Tri, Par, Tail */
u = gsl_rng_uniform (rng) * p4;
v = gsl_rng_uniform (rng);
if (u <= p1)
{
/* Triangular region */
ix = (int) (xm - p1 * v + u);
goto Finish;
}
else if (u <= p2)
{
/* Parallelogram region */
double x = xl + (u - p1) / c;
v = v * c + 1.0 - fabs (x - xm) / p1;
if (v > 1.0 || v <= 0.0)
goto TryAgain;
ix = (int) x;
}
else if (u <= p3)
{
/* Left tail */
ix = (int) (xl + log (v) / lambda_l);
if (ix < 0)
goto TryAgain;
v *= ((u - p2) * lambda_l);
}
else
{
/* Right tail */
ix = (int) (xr - log (v) / lambda_r);
if (ix > (double) n)
goto TryAgain;
v *= ((u - p3) * lambda_r);
}
/* At this point, the goal is to test whether v <= f(x)/f(m)
*
* v <= f(x)/f(m) = (m!(n-m)! / (x!(n-x)!)) * (p/q)^{x-m}
*
*/
/* Here is a direct test using logarithms. It is a little
* slower than the various "squeezing" computations below, but
* if things are working, it should give exactly the same answer
* (given the same random number seed). */
#ifdef DIRECT
var = log (v);
accept =
LNFACT (m) + LNFACT (n - m) - LNFACT (ix) - LNFACT (n - ix)
+ (ix - m) * log (p / q);
#else /* SQUEEZE METHOD */
/* More efficient determination of whether v < f(x)/f(M) */
k = abs (ix - m);
if (k <= FAR_FROM_MEAN)
{
/*
* If ix near m (ie, |ix-m|<FAR_FROM_MEAN), then do
* explicit evaluation using recursion relation for f(x)
*/
double g = (n + 1) * s;
double f = 1.0;
var = v;
if (m < ix)
{
int i;
for (i = m + 1; i <= ix; i++)
{
f *= (g / i - s);
}
}
else if (m > ix)
{
int i;
for (i = ix + 1; i <= m; i++)
{
f /= (g / i - s);
}
}
accept = f;
}
else
{
/* If ix is far from the mean m: k=ABS(ix-m) large */
var = log (v);
if (k < npq / 2 - 1)
{
/* "Squeeze" using upper and lower bounds on
* log(f(x)) The squeeze condition was derived
* under the condition k < npq/2-1 */
double amaxp =
k / npq * ((k * (k / 3.0 + 0.625) + (1.0 / 6.0)) / npq + 0.5);
double ynorm = -(k * k / (2.0 * npq));
if (var < ynorm - amaxp)
goto Finish;
if (var > ynorm + amaxp)
goto TryAgain;
}
/* Now, again: do the test log(v) vs. log f(x)/f(M) */
#if USE_EXACT
/* This is equivalent to the above, but is a little (~20%) slower */
/* There are five log's vs three above, maybe that's it? */
accept = LNFACT (m) + LNFACT (n - m)
- LNFACT (ix) - LNFACT (n - ix) + (ix - m) * log (p / q);
#else /* USE STIRLING */
/* The "#define Stirling" above corresponds to the first five
* terms in asymptoic formula for
* log Gamma (y) - (y-0.5)log(y) + y - 0.5 log(2*pi);
* See Abramowitz and Stegun, eq 6.1.40
*/
/* Note below: two Stirling's are added, and two are
* subtracted. In both K+S, and in the ranlib
* implementation, all four are added. I (jt) believe that
* is a mistake -- this has been confirmed by personal
* correspondence w/ Dr. Kachitvichyanukul. Note, however,
* the corrections are so small, that I couldn't find an
* example where it made a difference that could be
* observed, let alone tested. In fact, define'ing Stirling
* to be zero gave identical results!! In practice, alv is
* O(1), ranging 0 to -10 or so, while the Stirling
* correction is typically O(10^{-5}) ...setting the
* correction to zero gives about a 2% performance boost;
* might as well keep it just to be pendantic. */
{
double x1 = ix + 1.0;
double w1 = n - ix + 1.0;
double f1 = fm + 1.0;
double z1 = n + 1.0 - fm;
accept = xm * log (f1 / x1) + (n - m + 0.5) * log (z1 / w1)
+ (ix - m) * log (w1 * p / (x1 * q))
+ Stirling (f1) + Stirling (z1) - Stirling (x1) - Stirling (w1);
}
#endif
#endif
}
if (var <= accept)
{
goto Finish;
}
else
{
goto TryAgain;
}
}
Finish:
return (flipped) ? (n - ix) : (unsigned int)ix;
}