/* 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; }

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