By: Jeff

Jeff — Mon, 08 Dec 2008 21:51:00 +0000

I agree with Shawn. This looks like, here is something wrong with the position variable. But items[available-1] does not fit too, because then you would pick the same element twice if your random condition is true twice in a row:
if( random() < needed / available )

So you would need something like this:
items[available-k+needed]

But way of selection is not uniform distributed: the probability would increase to the end of the while-loop, so the elements at the end have allways better chance to get into the array than the first element.

Also this approach has a very simple problem: the items are only picked randomly, but the order stays the same. I mean, if i use this to take 3 elements from:
1 4 5 8 9
i could get:
5 4 1
or
9 5 1
but never: 1 9 5

I would suggest to generate a random number for every element from you list, and the sort the array by this number. After that you can just take the first k elements from your list. If you use round(256*rand()) as your random number (or any other constant number * rand()) and radixsort as your sortalgorithm, you can arrange this in linear time, and it stays deterministic.

Of course, it is not that easy to implement as your snippet, so i would use my suggestion only if you really need an uniform distributed random numbers array. (=you work for a lottery) ;)

By: Shawn Poulson

Shawn Poulson — Sat, 28 Jun 2008 16:15:30 +0000

Hmm, it looks like you’re not incrementing position. This code will retrieve the first item in each iteration.

I’d recommend replacing items[position] with items[available-1] and ditch position altogether.

Very good code snippet. I think I’ll keep it.

Comments on: A random selection algorithm

By: Jeff

By: Shawn Poulson