Random N out of M
Someone asked about randomly picking N elements out of an array of M, and I remembered reading about one way of doing that, so here it is: private static IEnumerable<int> Generate(Random rnd, int n, int m) { var i = 0; while (n > 0) { var r = rnd.NextDouble(); if (r < (double) n / m) { yield return i; n--; } m--; i++; } } The above C# method assumes 0 < N <= M and will return an IEnumerable<int> containing exactly N numbers in the 0 .. M - 1 interval. This method can even be used to pick items out of a stream without waiting for the whole stream to be produced first. A simpler method would be to just do this: for (var i = 0; i < n; i++) { do { var next = rnd.Next(m); while (!alreadyGenerated(next)); yield return next; } but this one requires 1) saving all numbers (to check...