Question

There is known Random(0,1) function, it is a uniformed random function, which means, it will give 0 or 1, with probability 50%. Implement Random(a, b) that only makes calls to Random(0,1)

What I though so far is, put the range a-b in a 0 based array, then I have index 0, 1, 2…b-a.

then call the RANDOM(0,1) b-a times, sum the results as generated idx. and return the element.

However since there is no answer in the book, I don’t know if this way is correct or the best. How to prove that the probability of returning each element is exactly same and is 1/(b-a+1) ?

And what is the right/better way to do this?

Answer 1

If your RANDOM(0, 1) returns either 0 or 1, each with probability 0.5 then you can generate bits until you have enough to represent the number (b-a+1) in binary. This gives you a random number in a slightly too large range: you can test and repeat if it fails. Something like this (in Python).

def rand_pow2(bit_count):
    """Return a random number with the given number of bits."""
    result = 0
    for i in xrange(bit_count):
        result = 2 * result + RANDOM(0, 1)
    return result

def random_range(a, b):
    """Return a random integer in the closed interval [a, b]."""
    bit_count = math.ceil(math.log2(b - a + 1))
    while True:
        r = rand_pow2(bit_count)
        if a + r <= b:
            return a + r