Re^2: Random Derangement Of An Array

I'm skeptical about this one. It's nice, but it can only give factorial(n-1) possible outputs, since that is the number of choices made in the algorithm. But there are many more derangements than that -- the number is essentially factorial(n)/e, which is larger if n>2. So I don't think it gets the entire range of derangements.

For instance, I think one that it wouldn't be able to get is: (4,3,2,1). In the last step, you must swap the 4 into its final position, so the 4 would have to be swapped with the 1 in the first position. But the 1 would definitely no longer be in the first position at that time.

blokhead

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Re^3: Random Derangement Of An Array by jethro (Monsignor) on Jul 06, 2008 at 12:46 UTC
Excellent example. So that algorithm is sadly lacking, and my proof has a hole as well. That happens when you try to proove something in your head instead of really doing the math.	[reply]