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Line 18: Line 18:
 
rest <- rnd_permu xs
 
rest <- rnd_permu xs
 
return $ let (ys,zs) = splitAt rand rest
 
return $ let (ys,zs) = splitAt rand rest
in ys++(x:zs)
+
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
 
</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