Press, Teukolsly, Vetterling, Flannery - Numerical Recipes in C (523184), страница 71
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The factor 0.9 is chosen sothat t never exceeds 1.} while (ran1(idum) > t);}return em;}Binomial DeviatesIf an event occurs with probability q, and we make n trials, then the number oftimes m that it occurs has the binomial distribution,&j+j− n jpn,q (m)dm =q (1 − q)n−jj(7.3.7)The binomial distribution is integer valued, with m taking on possible valuesfrom 0 to n. It depends on two parameters, n and q, so is correspondingly abit harder to implement than our previous examples.
Nevertheless, the techniquesalready illustrated are sufficiently powerful to do the job:#include <math.h>#define PI 3.141592654float bnldev(float pp, int n, long *idum)Returns as a floating-point number an integer value that is a random deviate drawn froma binomial distribution of n trials each of probability pp, using ran1(idum) as a source ofuniform random deviates.{float gammln(float xx);float ran1(long *idum);int j;static int nold=(-1);float am,em,g,angle,p,bnl,sq,t,y;static float pold=(-1.0),pc,plog,pclog,en,oldg;p=(pp <= 0.5 ? pp : 1.0-pp);The binomial distribution is invariant under changing pp to 1-pp, if we also change theanswer to n minus itself; we’ll remember to do this below.am=n*p;This is the mean of the deviate to be produced.if (n < 25) {Use the direct method while n is not too large.bnl=0.0;This can require up to 25 calls to ran1.for (j=1;j<=n;j++)if (ran1(idum) < p) ++bnl;} else if (am < 1.0) {If fewer than one event is expected out of 25g=exp(-am);or more trials, then the distribution is quitet=1.0;accurately Poisson.
Use direct Poisson method.for (j=0;j<=n;j++) {t *= ran1(idum);if (t < g) break;}bnl=(j <= n ? j : n);} else {Use the rejection method.296Chapter 7.Random Numbersif (n != nold) {If n has changed, then compute useful quantien=n;ties.oldg=gammln(en+1.0);nold=n;} if (p != pold) {If p has changed, then compute useful quantipc=1.0-p;ties.plog=log(p);pclog=log(pc);pold=p;}sq=sqrt(2.0*am*pc);The following code should by now seem familiar:do {rejection method with a Lorentzian compardo {ison function.angle=PI*ran1(idum);y=tan(angle);em=sq*y+am;} while (em < 0.0 || em >= (en+1.0));Reject.em=floor(em);Trick for integer-valued distribution.t=1.2*sq*(1.0+y*y)*exp(oldg-gammln(em+1.0)-gammln(en-em+1.0)+em*plog+(en-em)*pclog);} while (ran1(idum) > t);Reject. This happens about 1.5 times per devibnl=em;ate, on average.}if (p != pp) bnl=n-bnl;return bnl;Remember to undo the symmetry transformation.}See Devroye [2] and Bratley [3] for many additional algorithms.CITED REFERENCES AND FURTHER READING:Knuth, D.E.
1981, Seminumerical Algorithms, 2nd ed., vol. 2 of The Art of Computer Programming(Reading, MA: Addison-Wesley), pp. 120ff. [1]Devroye, L. 1986, Non-Uniform Random Variate Generation (New York: Springer-Verlag), §X.4.[2]Bratley, P., Fox, B.L., and Schrage, E.L. 1983, A Guide to Simulation (New York: SpringerVerlag). [3].7.4 Generation of Random BitsThe C language gives you useful access to some machine-level bitwise operationssuch as << (left shift). This section will show you how to put such abilities to good use.The problem is how to generate single random bits, with 0 and 1 equallyprobable.
Of course you can just generate uniform random deviates between zeroand one and use their high-order bit (i.e., test if they are greater than or less than0.5). However this takes a lot of arithmetic; there are special-purpose applications,such as real-time signal processing, where you want to generate bits very muchfaster than that.One method for generating random bits, with two variant implementations, isbased on “primitive polynomials modulo 2.” The theory of these polynomials isbeyond our scope (although §7.7 and §20.3 will give you small tastes of it). Here,7.4 Generation of Random Bits297suffice it to say that there are special polynomials among those whose coefficientsare zero or one. An example isx18 + x5 + x2 + x1 + x0(7.4.1)which we can abbreviate by just writing the nonzero powers of x, e.g.,(18, 5, 2, 1, 0)Every primitive polynomial modulo 2 of order n (=18 above) defines arecurrence relation for obtaining a new random bit from the n preceding ones.
Therecurrence relation is guaranteed to produce a sequence of maximal length, i.e.,cycle through all possible sequences of n bits (except all zeros) before it repeats.Therefore one can seed the sequence with any initial bit pattern (except all zeros),and get 2n − 1 random bits before the sequence repeats.Let the bits be numbered from 1 (most recently generated) through n (generatedn steps ago), and denoted a1 , a2 , .
. . , an . We want to give a formula for a new bita0 . After generating a0 we will shift all the bits by one, so that the old an is finallylost, and the new a0 becomes a1 . We then apply the formula again, and so on.“Method I” is the easiest to implement in hardware, requiring only a singleshift register n bits long and a few XOR (“exclusive or” or bit addition mod 2)gates, the operation denoted in C by “∧”. For the primitive polynomial given above,the recurrence formula isa0 = a18 ∧ a5 ∧ a2 ∧ a1(7.4.2)The terms that are ∧’d together can be thought of as “taps” on the shift register,∧’d into the register’s input.
More generally, there is precisely one term for eachnonzero coefficient in the primitive polynomial except the constant (zero bit) term.So the first term will always be an for a primitive polynomial of degree n, while thelast term might or might not be a1 , depending on whether the primitive polynomialhas a term in x1 .While it is simple in hardware, Method I is somewhat cumbersome in C, becausethe individual bits must be collected by a sequence of full-word masks:int irbit1(unsigned long *iseed)Returns as an integer a random bit, based on the 18 low-significance bits in iseed (which ismodified for the next call).{unsigned long newbit;The accumulated XOR’s.newbit = (*iseed >> 17) & 1^ (*iseed >> 4) & 1^ (*iseed >> 1) & 1^ (*iseed & 1);*iseed=(*iseed << 1) | newbit;return (int) newbit;}Get bit 18.XOR with bit 5.XOR with bit 2.XOR with bit 1.Leftshift the seed and put the result of theXOR’s in its bit 1.298Chapter 7.181754Random Numbers3210shift left(a)1817543210shift left(b)Figure 7.4.1.Two related methods for obtaining random bits from a shift register and a primitivepolynomial modulo 2.
(a) The contents of selected taps are combined by exclusive-or (addition modulo2), and the result is shifted in from the right. This method is easiest to implement in hardware. (b)Selected bits are modified by exclusive-or with the leftmost bit, which is then shifted in from the right.This method is easiest to implement in software.“Method II” is less suited to direct hardware implementation (though stillpossible), but is beautifully suited to C. It modifies more than one bit among thesaved n bits as each new bit is generated (Figure 7.4.1). It generates the maximallength sequence, but not in the same order as Method I. The prescription for theprimitive polynomial (7.4.1) is:a0 = a18a5 = a5 ∧ a 0a2 = a2 ∧ a 0(7.4.3)a1 = a1 ∧ a 0In general there will be an exclusive-or for each nonzero term in the primitivepolynomial except 0 and n.
The nice feature about Method II is that all theexclusive-or’s can usually be done as a single full-word exclusive-or operation:#define#define#define#define#defineIB1 1IB2 2IB5 16IB18 131072MASK (IB1+IB2+IB5)Powers of 2.int irbit2(unsigned long *iseed)Returns as an integer a random bit, based on the 18 low-significance bits in iseed (which ismodified for the next call).{if (*iseed & IB18) {Change all masked bits, shift, and put 1 into bit 1.*iseed=((*iseed ^ MASK) << 1) | IB1;return 1;} else {Shift and put 0 into bit 1.2997.4 Generation of Random Bits*iseed <<= 1;return 0;}}Some Primitive Polynomials Modulo 2 (after Watson)(1,(2,(3,(4,(5,(6,(7,(8,(9,(10,(11,(12,(13,(14,(15,(16,(17,(18,(19,(20,(21,(22,(23,(24,(25,(26,(27,(28,(29,(30,(31,(32,(33,(34,(35,(36,(37,(38,(39,(40,(41,(42,(43,(44,(45,(46,(47,(48,(49,(50,0)1,1,1,2,1,1,4,4,3,2,6,4,5,1,5,3,5,5,3,2,1,5,4,3,6,5,3,2,6,3,7,6,7,2,6,5,6,4,5,3,5,6,6,4,8,5,7,6,4,0)0)0)0)0)0)3,0)0)0)4,3,3,0)3,0)2,2,0)0)0)0)3,0)2,2,0)0)4,0)5,4,6,0)5,4,5,0)40)4,4,5,3,5,0)5,5,3,2, 0)1, 0)1, 0)1, 0)2, 0)1, 0)1, 0)1, 0)1, 0)1, 0)1, 0)3, 2, 1, 0)1, 0)5, 2, 1, 0)4, 2, 1, 0)3, 2, 1, 0)1, 0)3, 0)3,3,2,1,3,2, 1, 0)0)0)0)2, 1, 0)4, 2, 1, 0)4, 0)2, 0)(51,(52,(53,(54,(55,(56,(57,(58,(59,(60,(61,(62,(63,(64,(65,(66,(67,(68,(69,(70,(71,(72,(73,(74,(75,(76,(77,(78,(79,(80,(81,(82,(83,(84,(85,(86,(87,(88,(89,(90,(91,(92,(93,(94,(95,(96,(97,(98,(99,(100,6, 3,3, 0)6, 2,6, 5,6, 2,7, 4,5, 3,6, 5,6, 5,1, 0)5, 2,6, 5,1, 0)4, 3,4, 3,8, 6,5, 2,7, 5,6, 5,5, 3,5, 3,6, 4,4, 3,7, 4,6, 3,5, 4,6, 5,7, 2,4, 3,7, 5,4 0)8, 7,7, 4,8, 7,8, 2,6, 5,7, 5,8, 5,6, 5,5, 3,7, 6,6, 5,2, 0)6, 5,6, 5,7, 6,6, 0)7, 4,7, 5,8, 7,1, 0)1,4,1,2,2,1,4,0)3, 2, 0)0)0)0)0)3, 1, 0)1, 0)3, 0)1,1,5,1,1,2,1,1,3,2,3,1,2,2,1,2,3,0)0)3, 2, 0)0)0)0)0)0)2, 1, 0)0)0)0)0)0)0)0)2, 1, 0)6,2,5,1,2,1,4,3,2,5,2,4,0)3,0)0)0)3,0)0)3,0)1, 0)1, 0)1, 0)2, 0)1, 0)4, 2, 1, 0)4, 3, 2, 0)3, 2, 1, 0)4, 0)2, 0)A word of caution is: Don’t use sequential bits from these routines as the bitsof a large, supposedly random, integer, or as the bits in the mantissa of a supposedly300Chapter 7.Random Numbersrandom floating-point number.
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