99 questions/Solutions/25
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Remi.berson (Talk  contribs) m (first solution used diff_select instead of rnd_select) 

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Latest revision as of 13:08, 10 May 2014
Generate a random permutation of the elements of a list.
rnd_permu :: [a] > IO [a] rnd_permu xs = rnd_select xs (length xs)
Uses the solution of problem 23 (rnd_select). Choosing N distinct elements from a list of length N will yield a permutation.
Or we can generate the permutation recursively:
import System.Random (randomRIO) rnd_permu :: [a] > IO [a] rnd_permu [] = return [] rnd_permu (x:xs) = do rand < randomRIO (0, (length xs)) rest < rnd_permu xs return $ let (ys,zs) = splitAt rand rest in ys++(x:zs) rnd_permu' [] = return [] rnd_permu' xs = do rand < randomRIO (0, (length xs)1) rest < let (ys,(_:zs)) = splitAt rand xs in rnd_permu' $ ys ++ zs return $ (xs!!rand):rest
permutations
Data.List
import System.Random (getStdGen, randomRIO) import Data.List (permutations) rndElem :: [a] > IO a rndElem xs = do index < randomRIO (0, length xs  1) return $ xs !! index rndPermutation :: [a] > IO [a] rndPermutation xs = rndElem . permutations $ xs
WARNING: this may choke long lists