Expand a random range from 1–5 to 1–7

Question

Given a function which produces a random integer in the range 1 to 5, write a function which produces a random integer in the range 1 to 7.

1. What is a simple so

Answer 1

Algorithm:

7 can be represented in a sequence of 3 bits

Use rand(5) to randomly fill each bit with 0 or 1.
For e.g: call rand(5) and

if the result is 1 or 2, fill the bit with 0
if the result is 4 or 5, fill the bit with 1
if the result is 3 , then ignore and do it again (rejection)

This way we can fill 3 bits randomly with 0/1 and thus get a number from 1-7.

EDIT: This seems like the simplest and most efficient answer, so here's some code for it:

public static int random_7() {
    int returnValue = 0;
    while (returnValue == 0) {
        for (int i = 1; i <= 3; i++) {
            returnValue = (returnValue << 1) + random_5_output_2();
        }
    }
    return returnValue;
}

private static int random_5_output_2() {
    while (true) {
        int flip = random_5();

        if (flip < 3) {
            return 0;
        }
        else if (flip > 3) {
            return 1;
        }
    }
}