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 solution?
2. What is an effective solution to reduce memory usage or run on a slower CPU?

Answer 1

Are homework problems allowed here?

This function does crude "base 5" math to generate a number between 0 and 6.

function rnd7() {
    do {
        r1 = rnd5() - 1;
        do {
            r2=rnd5() - 1;
        } while (r2 > 1);
        result = r2 * 5 + r1;
    } while (result > 6);
    return result + 1;
}

Answer 2

just scale your output from your first function

0) you have a number in range 1-5
1) subtract 1 to make it in range 0-4
2) multiply by (7-1)/(5-1) to make it in range 0-6
3) add 1 to increment the range: Now your result is in between 1-7

Answer 3

int rand7() {
    int value = rand5()
              + rand5() * 2
              + rand5() * 3
              + rand5() * 4
              + rand5() * 5
              + rand5() * 6;
    return value%7;
}

Unlike the chosen solution, the algorithm will run in constant time. It does however make 2 more calls to rand5 than the average run time of the chosen solution.

Note that this generator is not perfect (the number 0 has 0.0064% more chance than any other number), but for most practical purposes the guarantee of constant time probably outweighs this inaccuracy.

Explanation

This solution is derived from the fact that the number 15,624 is divisible by 7 and thus if we can randomly and uniformly generate numbers from 0 to 15,624 and then take mod 7 we can get a near-uniform rand7 generator. Numbers from 0 to 15,624 can be uniformly generated by rolling rand5 6 times and using them to form the digits of a base 5 number as follows:

rand5 * 5^5 + rand5 * 5^4 + rand5 * 5^3 + rand5 * 5^2 + rand5 * 5 + rand5

Properties of mod 7 however allow us to simplify the equation a bit:

5^5 = 3 mod 7
5^4 = 2 mod 7
5^3 = 6 mod 7
5^2 = 4 mod 7
5^1 = 5 mod 7

So

rand5 * 5^5 + rand5 * 5^4 + rand5 * 5^3 + rand5 * 5^2 + rand5 * 5 + rand5

becomes

rand5 * 3 + rand5 * 2 + rand5 * 6 + rand5 * 4 + rand5 * 5 + rand5

Theory

The number 15,624 was not chosen randomly, but can be discovered using fermat's little theorem, which states that if p is a prime number then

a^(p-1) = 1 mod p

So this gives us,

(5^6)-1 = 0 mod 7

(5^6)-1 is equal to

4 * 5^5 + 4 * 5^4 + 4 * 5^3 + 4 * 5^2 + 4 * 5 + 4

This is a number in base 5 form and thus we can see that this method can be used to go from any random number generator to any other random number generator. Though a small bias towards 0 is always introduced when using the exponent p-1.

Answer 4

in php

function rand1to7() {
    do {
        $output_value = 0;
        for ($i = 0; $i < 28; $i++) {
            $output_value += rand1to5();
        }
    while ($output_value != 140);
    $output_value -= 12;
    return floor($output_value / 16);
}

loops to produce a random number between 16 and 127, divides by sixteen to create a float between 1 and 7.9375, then rounds down to get an int between 1 and 7. if I am not mistaken, there is a 16/112 chance of getting any one of the 7 outcomes.

Answer 5

This is equivalent to Adam Rosenfield's solution, but may be a bit more clear for some readers. It assumes rand5() is a function that returns a statistically random integer in the range 1 through 5 inclusive.

int rand7()
{
    int vals[5][5] = {
        { 1, 2, 3, 4, 5 },
        { 6, 7, 1, 2, 3 },
        { 4, 5, 6, 7, 1 },
        { 2, 3, 4, 5, 6 },
        { 7, 0, 0, 0, 0 }
    };

    int result = 0;
    while (result == 0)
    {
        int i = rand5();
        int j = rand5();
        result = vals[i-1][j-1];
    }
    return result;
}

How does it work? Think of it like this: imagine printing out this double-dimension array on paper, tacking it up to a dart board and randomly throwing darts at it. If you hit a non-zero value, it's a statistically random value between 1 and 7, since there are an equal number of non-zero values to choose from. If you hit a zero, just keep throwing the dart until you hit a non-zero. That's what this code is doing: the i and j indexes randomly select a location on the dart board, and if we don't get a good result, we keep throwing darts.

Like Adam said, this can run forever in the worst case, but statistically the worst case never happens. :)

Answer 6

There is no (exactly correct) solution which will run in a constant amount of time, since 1/7 is an infinite decimal in base 5. One simple solution would be to use rejection sampling, e.g.:

int i;
do
{
  i = 5 * (rand5() - 1) + rand5();  // i is now uniformly random between 1 and 25
} while(i > 21);
// i is now uniformly random between 1 and 21
return i % 7 + 1;  // result is now uniformly random between 1 and 7

This has an expected runtime of 25/21 = 1.19 iterations of the loop, but there is an infinitesimally small probability of looping forever.