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(Added an alternative solution)
Line 26: Line 26:
 
in rnd_permu' $ ys ++ zs
 
in rnd_permu' $ ys ++ zs
 
return $ (xs!!rand):rest
 
return $ (xs!!rand):rest
  +
</haskell>
  +
  +
Or we can use the <hask>permutations</hask> function from <hask>Data.List</hask>:
  +
<haskell>
  +
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
 
</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
Or we can use the
permutations
function from
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