# 99 questions/Solutions/25

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

(Added an alternative solution) |
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in rnd_permu' $ ys ++ zs |
in rnd_permu' $ ys ++ zs |
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return $ (xs!!rand):rest |
return $ (xs!!rand):rest |
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+ | </haskell> |
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+ | |||

+ | Or we can use the <hask>permutations</hask> function from <hask>Data.List</hask>: |
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+ | <haskell> |
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+ | import System.Random (getStdGen, randomRIO) |
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+ | import Data.List (permutations) |
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+ | |||

+ | rndElem :: [a] -> IO a |
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+ | rndElem xs = do |
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+ | gen <- getStdGen |
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+ | index <- randomRIO (0, length xs - 1) |
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+ | return $ xs !! index |
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+ | |||

+ | rndPermutation :: [a] -> IO [a] |
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+ | rndPermutation xs = rndElem . permutations $ xs |
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</haskell> |
</haskell> |

## Revision as of 06:18, 15 August 2011

Generate a random permutation of the elements of a list.

rnd_permu xs = diff_select' (length xs) xs

Uses the solution for the previous problem. 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 gen <- getStdGen index <- randomRIO (0, length xs - 1) return $ xs !! index rndPermutation :: [a] -> IO [a] rndPermutation xs = rndElem . permutations $ xs