由网友(不问归期)分享简介:给定一个随机源(随机比特流的生成),我怎么产生一个给定的范围内均匀分布的随机浮点值?Given a random source (a generator of random bit stream), how do I generate a uniformly distributed random floating-p...
给定一个随机源(随机比特流的生成),我怎么产生一个给定的范围内均匀分布的随机浮点值?
Given a random source (a generator of random bit stream), how do I generate a uniformly distributed random floating-point value in a given range?
假设我的随机源看起来是这样的:
Assume that my random source looks something like:
unsigned int GetRandomBits(char* pBuf, int nLen);
和我想要实现
double GetRandomVal(double fMin, double fMax);
注:
在我不想要的结果precision被限制(例如,只有5位数字)。 在严格的均匀分布是必须 在我没有要求引用现有的库。我想为知道如何从头开始实现它。 对于伪code / code,C ++是最AP preciated I don't want the result precision to be limited (for example only 5 digits). Strict uniform distribution is a must I'm not asking for a reference to an existing library. I want to know how to implement it from scratch. For pseudo-code / code, C++ would be most appreciated推荐答案
我不认为我会永远确信你确实需要,但它很有趣写。
I don't think I'll ever be convinced that you actually need this, but it was fun to write.
#include <stdint.h>
#include <cmath>
#include <cstdio>
FILE* devurandom;
bool geometric(int x) {
// returns true with probability min(2^-x, 1)
if (x <= 0) return true;
while (1) {
uint8_t r;
fread(&r, sizeof r, 1, devurandom);
if (x < 8) {
return (r & ((1 << x) - 1)) == 0;
} else if (r != 0) {
return false;
}
x -= 8;
}
}
double uniform(double a, double b) {
// requires IEEE doubles and 0.0 < a < b < inf and a normal
// implicitly computes a uniform random real y in [a, b)
// and returns the greatest double x such that x <= y
union {
double f;
uint64_t u;
} convert;
convert.f = a;
uint64_t a_bits = convert.u;
convert.f = b;
uint64_t b_bits = convert.u;
uint64_t mask = b_bits - a_bits;
mask |= mask >> 1;
mask |= mask >> 2;
mask |= mask >> 4;
mask |= mask >> 8;
mask |= mask >> 16;
mask |= mask >> 32;
int b_exp;
frexp(b, &b_exp);
while (1) {
// sample uniform x_bits in [a_bits, b_bits)
uint64_t x_bits;
fread(&x_bits, sizeof x_bits, 1, devurandom);
x_bits &= mask;
x_bits += a_bits;
if (x_bits >= b_bits) continue;
double x;
convert.u = x_bits;
x = convert.f;
// accept x with probability proportional to 2^x_exp
int x_exp;
frexp(x, &x_exp);
if (geometric(b_exp - x_exp)) return x;
}
}
int main() {
devurandom = fopen("/dev/urandom", "r");
for (int i = 0; i < 100000; ++i) {
printf("%.17gn", uniform(1.0 - 1e-15, 1.0 + 1e-15));
}
}
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