Random rational numbers generation

Question

Rationals are enumerable. For example this code finds k-th rational in open interval 0..1, with ordering that {n1, d1} is before {n2, d2} if (d1&

Answer 1

Here are some random thoughts on the problem you raise. I haven't carefully checked the math so I could be off by 1 here or there. But it represents the sort of reasoning I would follow.

Let's only consider fractions in the interval (0,1). It's much easier that way. We can deal later with 1/1 and improper fractions.

The Stern - Brocot Tree uniquely lists each reduced positive common fraction (and hence each positive rational number less than or equal to one) once, in order, and in reduced form, as a node in the tree. In this binary tree, any node and thus any fraction can be reached by a finite sequence of left-right turns starting from the uppermost level (for convenience let's call it level -1), containing 0/1 and 1/0. [Yes, 1/0. That's not a misprint!]

Given a denominator, k, you would need to take at most k turns to reach any reduced fraction j/k, where j is less than k. For example, if the denominator were 101, all possible fractions with a denominator of 101 or less will be in the tree somewhere between Level 1 (containing 1/1) and Level 101 (containing 1/101 in the leftmost position).

Let's assume we have a number generator that generate 0's and 1's. (Please don't ask me how to do that; I have no idea.) Lef's arbitrarily decide that Left=0 and Right=1.

Assume that we have another number generator that can randomly generate integers between 1 and n. Assume further that the first number generated is 0, ie. turn left: this guarantees that the fraction will fall in the interval (0,1).

Select the maximum denominator, k. Randomly generate a number, m, between 1 and k. Then generate a random list of R's and L's. Traverse (i.e. descend) the Stern-Brocot tree, following the list of turns. Stop when you reach the destination fraction.

If that fraction has a denominator equal to or less than k, stop, that's your number.

If the denominator is greater than k, ascend the tree (along the same path you descended) until you reach a fraction with a denominator no greater than k.

I don't know that the number generation is truly random. I wouldn't even know how to tell. But for what it's worthe, I don't detect any obvious source of bias.