Actually, a much better solution is:
variant :: Int -> Gen a -> Gen a
variant v (Gen m) = Gen (\n r -> m n (rands r v))
where
rands r0 0 = r0
rands r0 n = let (r1,r2) = split r0
(n',s) = n `quotRem` 2
in case s of
0 -> rands r1 n'
_ -> rands r2 n'
Patrick
On Aug 14, 2008, at 2:17 AM, Johan Tibell wrote:
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
