Heinrich Apfelmus wrote:
Jon Fairbairn wrote:
Heinrich Apfelmus writes:
The answer is a resounding "yes" and the main idea is that shuffling a list is *essentially the same* as sorting a list; the minor difference being that the former chooses a permutation at random while the latter chooses a very particular permutation, namely the one that sorts the input.
For the full exposition, see
http://apfelmus.nfshost.com/random-permutations.html I haven't been following the thread, but my initial reaction would have been something like use System.Random.randoms to get a list rs and then do (roughly)
randomPerm = map snd . sortBy (compare `on` fst) . zip rs
How bad is that? I mean, how unfair does it get?
It's fair, but may duplicate elements, i.e. it doesn't necessarily create a permutation. For example, rs could be something like
rs = [5,3,3,3,2,4]
How about using random doubles? randomPerm xs = fmap (map snd . sort . flip zip xs) rs where rs = fmap (randoms . mkStdGen) randomIO :: IO [Double]