# 99 questions/Solutions/25

### From HaskellWiki

< 99 questions | Solutions(Difference between revisions)

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rest <- rnd_permu xs |
rest <- rnd_permu xs |
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return $ let (ys,zs) = splitAt rand rest |
return $ let (ys,zs) = splitAt rand rest |
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− | in ys++(x:zs) |
+ | in ys++(x:zs) |

+ | |||

+ | rnd_permu' [] = return [] |
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+ | rnd_permu' xs = do |
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+ | rand <- randomRIO (0, (length xs)-1) |
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+ | rest <- let (ys,(_:zs)) = splitAt rand xs |
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+ | in rnd_permu' $ ys ++ zs |
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+ | return $ (xs!!rand):rest |
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</haskell> |
</haskell> |

## Revision as of 06:31, 25 November 2010

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