in reply to Random Derangement Of An Array

Here's the approach I wanted to code up, but I've been sufficiently distracted.

The number of derangements is d(n) = (n-1)( d(n-1) + d(n-2) ), and there is a combinatorial proof of this at the wikipedia article. That is, there are (n-1) ways to build an n-derangement out of a (n-1)-derangement, and (n-1) ways to build an n-derangement out of a (n-2)-derangment. Furthermore, these correspond uniquely to all the ways to build an n-derangement.

So here is a way to randomly generate an n-derangement:

Unfortunately, I don't have any time to code this up, and the combinatorial proof is not written well. For the case of making an n-derangement out of an (n-1)-derangement, you simply add n to the end, and then swap the last position and a randomly chosen other position. I couldn't quite understand exactly what the wikipedia article was getting at for the other case, though, and it's been too long since I've thought of such things.

Maybe some curious monk can work this into usable code. But this approach is basically recursive: take a random starting derangement and choose a random way to augment it into a larger one. The end result will certainly be randomly distributed, provided you handle the (n-1) and (n-2) cases with the appropriate relative probabilities..

All I managed to get into code was a simple routine to count d(n). It uses a different combinatorial identity that is more amenable to simple computation. It seems like you'd need this to get the relative probabilities for the (n-1) and (n-2) cases to work out:

sub num_d { my ($n) = @_; return 1-$n if $n < 2; my $d = 0; $d += (-1)**$_ + ($_-1)*$d for 2 .. $n; return $d; }