4 ms·
I have used an alternate shuffling approach in the past of my own naive creation, which is to go from 1-N items, each pass selecting a random number to swap the
by Aron 17y ago
I have used an alternate shuffling approach in the past of my own naive creation, which is to go from 1-N items, each pass selecting a random number to swap the current position with. This would not require counting over missing items (those copied to result) on each loop. What is the flaw in this approach I wonder?
- paulgb 17y agoI've used that method too. The flaw is that it is biased; some permutations will be more likely than others. Each shuffle will consume N random numbers between 1 and N (inclusive). This gives a total of N^N equally probable N-tuples of random numbers, all corresponding to exactly one output permutation. But there are only N! possible permutations, so some of the N^N N-tuples must share the same output permutation. Since N! does not usually divide N^N evenly, some permutations must be more probable than others.
- Aron 17y agoAh....clever. thx.