Difference between revisions of "99 questions/Solutions/25"
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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