[paulgb]

Technically speaking, yes.

But, if you want to uniformly map a random number from set X to set Y where (IIRC) lcm(|X|, |Y|) != |X|, it seems you need an infinite worst-case running time.

Here's an informal proof that you can't have a finite upper bound to the number of iterations. After n iterations, you have |X|^n possible outcomes. But, since lcm(|X|, |Y|) != |X|, |X|^n cannot be divided evenly by |Y| (since its factors are the same). So some outcomes in Y must be more likely than others.

(This is not nearly complete, but hopefully it's enough to show how it might be right.)

Of course, in practice it is highly unlikely that you will get past more than a couple iterations before determining an outcome.

[andreyf]

since lcm(|X|, |Y|) != |X|, |X|^n cannot be divided evenly by |Y| (since its factors are the same)

Can you go into more detail about this part?

[paulgb]

Sure. Sorry, I didn't take the time to work this out on paper before posting or I would have realized that the condition itself is wrong. The condition lcm(|X|,|Y|) != |X| instead should be that |Y| has some prime factor that |X| does not.

Here is an explanation with the new condition:

Let p be any prime factor of |Y| that |X| does not have. It follows from Euclid's First Theorem[1] that p cannot divide |X|^n for any n [2]. Since every integer (> 1) has a unique prime factorization, it follows that |X|^n can't divide |Y|, because the prime p divides |Y| but not |X|^n.

[1] http://mathworld.wolfram.com/EuclidsTheorems.html [2] We are given that p does not divide |X|^1. Suppose that p also does not divide |X|^(n-1) for some n > 1. |X|^n = |X|^(n-1) * |X|^1, so by Euclid's First Theorem, if p divides |X|^n it must divide either |X|^(n-1) or |X|^1. We know it divides neither, so p does not divide |X|^n. By induction, this is true for all n > 0.