如何在C中以概率随机生成0和1的mxn矩阵

Question

I wrote C program that defined a 2D matrix with m rows and n columns with random numbers (Either 0 or 1). The code is as following:

int i,j;
    int original_matrix[m][n];
    for (i=0; i<=m-1; i++){
        for (j=0; j<=n-1; j++){
            original_matrix[i][j] = rand() % 2;
        }
    }

It worked. For the next step, I want to create the matrix with a probability. For example, 1 is written into a cell with probability p and 0 is written with probability 1-p. Could you please share any ideas on this if you have?

Answer 1

Since rand() gives you a value between 0 and RAND_MAX , you can get a value at particular perentage simply by choosing an appropriate threshold. For example, if RAND_MAX was 999 , 42% of all values would be expected to be less than 420 .

So you can use code like in the following complete program, to set up an appropriate threshold and test the distribution of your values:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <time.h>

int main(int argc, char *argv[]) {
    // Get threshold (defaults to ~PI%), seed random numbers.

    double percent = (argc > 1) ? atof(argv[1]) : .0314159;
    int threshold = round(RAND_MAX * percent);
    srand(time(0));

    // Work out distribution (millions of samples).

    int below = 0, total = 0;
    for (int i = 0 ; i < 1000000; ++i) {
        ++total;
        if (rand() < threshold) ++below;
    }

    // Output stats.

    printf("Using probability of %f, below was %d / %d, %f%%\n",
        percent, below, total, below * 100.0 / total);
}

Some sample runs, with varying desired probabilities:

Using probability of 0.031416, below was 31276 / 1000000, 3.127600%
Using probability of 0.031416, below was 31521 / 1000000, 3.152100%

Using probability of 0.421230, below was 420936 / 1000000, 42.093600%
Using probability of 0.421230, below was 421634 / 1000000, 42.163400%

Using probability of 0.175550, below was 175441 / 1000000, 17.544100%
Using probability of 0.175550, below was 176031 / 1000000, 17.603100%

Using probability of 0.980000, below was 979851 / 1000000, 97.985100%
Using probability of 0.980000, below was 980032 / 1000000, 98.003200%

Using probability of 0.000000, below was 0 / 1000000, 0.000000%
Using probability of 1.000000, below was 1000000 / 1000000, 100.000000%

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So, the bottom line is: to acheive your desire of one having a probabilty p (a double value) and zero having the probability 1 - p , you need the following:

srand(time(0));                               // done once, seed generator.
int threshold = round(RAND_MAX * p);          // done once.
int oneOrZero = (rand() < threshold) ? 1 : 0; // done for each cell.

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Just keep in mind the limits of rand() , the difference between (for example) probabilities 0.0000000000 and 0.0000000001 will most likely be non-existent, unless RAND_MAX is large enough to make a difference. I doubt you'll be using probabilities that fine but I thought I'd better mention it just in case.

Answer 2

rand() % 2 gives you a probability of 0.5.

p is a float, so you'll look at How to generate random float number in C to generate a random value in a real range. The top answer gives us: float x = (float)rand()/(float)(RAND_MAX/a);

We want a equal to 1 for probabilities. So, to get 0 with a probability of p , the formula is:

int zeroWithAProbabilityOfP = (float)rand()/(float)RAND_MAX <= p;

Which can be also be written:

int zeroWithAProbabilityOfP = rand() <= p * RAND_MAX;

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ps: if available, for precision reasons, you should favor arc4random() or arc4random_buf() instead of rand() :

• rand() precision is 1 / 0x7FFFFFFF (on macOS)
• arc4random() precision is 1 / 0xFFFFFFFF (so twice better)

In that case, formula would be:

int zeroWithAProbabilityOfP = arc4random() <= p * UINT32_MAX;