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

From HaskellWiki

< 99 questions | Solutions

(delete unneeded codes) |
|||

Line 43: | Line 43: | ||

WARNING: this may choke long lists |
WARNING: this may choke long lists |
||

+ | |||

+ | |||

+ | [[Category:Programming exercise spoilers]] |

## Revision as of 19:39, 18 January 2014

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
index <- randomRIO (0, length xs - 1)
return $ xs !! index
rndPermutation :: [a] -> IO [a]
rndPermutation xs = rndElem . permutations $ xs
```

WARNING: this may choke long lists