On Thu, Aug 14, 2008 at 1:58 AM, Patrick Perry <[EMAIL PROTECTED]> wrote: > variant :: Int -> Gen a -> Gen a > variant v (Gen m) = Gen (\n r -> m n (rands r !! v')) > where > v' = abs (v+1) `mod` 10000 > rands r0 = r1 : rands r2 where (r1, r2) = split r0
Note that if you have a uniformly distributed random value in [0, n) and take its value mod k you don't end up with another random uniformly distributed value unless n `mod` k == 0. Consider n = 3 and k = 2. What you can do is to throw away all random numbers larger than n - (n `mod` k) and just generate a new number. This will terminate with a high probability if n >> k. Cheers, Johan _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
