Question

Question: Suppose you have a random number generator randn() that returns a uniformly distributed random number between 0 and n-1. Given any number m, write a random number generator that returns a uniformly distributed random number between 0 and m-1.

My answer:

-(int)randm() {
    int k=1;
    while (k*n < m) {
        ++k;
    }
    int x = 0;
    for (int i=0; i<k; ++i) {
        x += randn();
    }
    if (x < m) {
        return x;
    } else {
        return randm();
    }
}

Is this correct?

Answer 1

You’re close, but the problem with your answer is that there is more than one way to write a number as a sum of two other numbers.

If m<n, then this works because the numbers 0,1,...,m-1 appear each with equal probability, and the algorithm terminates almost surely.

This answer does not work in general because there is more than one way to write a number as a sum of two other numbers. For instance, there is only one way to get 0 but there are many many ways to get m/2, so the probabilities will not be equal.

Example: n = 2 and m=3

0 = 0+0
1 = 1+0 or 0+1
2 = 1+1

so the probability distribution from your method is

P(0)=1/4
P(1)=1/2
P(2)=1/4

which is not uniform.

---

To fix this, you can use unique factorization. Write m in base n, keeping track of the largest needed exponent, say e. Then, find the biggest multiple of m that is smaller than n^e, call it k. Finally, generate e numbers with randn(), take them as the base n expansion of some number x, if x < k*m, return x, otherwise try again.

Assuming that m < n^2, then

int randm() {

    // find largest power of n needed to write m in base n
    int e=0;
    while (m > n^e) {
        ++e;
    }

    // find largest multiple of m less than n^e
    int k=1;
    while (k*m < n^2) {
        ++k
    }
    --k; // we went one too far

    while (1) {
        // generate a random number in base n
        int x = 0;
        for (int i=0; i<e; ++i) {
            x = x*n + randn(); 
        }
        // if x isn't too large, return it x modulo m
        if (x < m*k) 
            return (x % m);
    }
}