Is using Random and OrderBy a good shuffle algorithm?

Question

I have read an article about various shuffle algorithms over at Coding Horror. I have seen that somewhere people have done this to shuffle a list:

var r = ne         


        

Answer 1

This is based on Jon Skeet's answer.

In that answer, the array is shuffled, then returned using yield. The net result is that the array is kept in memory for the duration of foreach, as well as objects necessary for iteration, and yet the cost is all at the beginning - the yield is basically an empty loop.

This algorithm is used a lot in games, where the first three items are picked, and the others will only be needed later if at all. My suggestion is to yield the numbers as soon as they are swapped. This will reduce the start-up cost, while keeping the iteration cost at O(1) (basically 5 operations per iteration). The total cost would remain the same, but the shuffling itself would be quicker. In cases where this is called as collection.Shuffle().ToArray() it will theoretically make no difference, but in the aforementioned use cases it will speed start-up. Also, this would make the algorithm useful for cases where you only need a few unique items. For example, if you need to pull out three cards from a deck of 52, you can call deck.Shuffle().Take(3) and only three swaps will take place (although the entire array would have to be copied first).

public static IEnumerable Shuffle(this IEnumerable source, Random rng)
{
    T[] elements = source.ToArray();
    // Note i > 0 to avoid final pointless iteration
    for (int i = elements.Length - 1; i > 0; i--)
    {
        // Swap element "i" with a random earlier element it (or itself)
        int swapIndex = rng.Next(i + 1);
        yield return elements[swapIndex];
        elements[swapIndex] = elements[i];
        // we don't actually perform the swap, we can forget about the
        // swapped element because we already returned it.
    }

    // there is one item remaining that was not returned - we return it now
    yield return elements[0]; 
}