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

From HaskellWiki

< 99 questions | Solutions

(Added an alternative solution) |
Remi.berson (talk | contribs) m (first solution used diff_select instead of rnd_select) |
||

(3 intermediate revisions by 2 users not shown) | |||

Line 2: | Line 2: | ||

<haskell> |
<haskell> |
||

− | rnd_permu |
+ | rnd_permu :: [a] -> IO [a] |

+ | rnd_permu xs = rnd_select xs (length xs) |
||

</haskell> |
</haskell> |
||

− | Uses the solution |
+ | Uses the solution of problem 23 (rnd_select). Choosing N distinct elements from a list of length N will yield a permutation. |

Or we can generate the permutation recursively: |
Or we can generate the permutation recursively: |
||

Line 35: | Line 35: | ||

rndElem :: [a] -> IO a |
rndElem :: [a] -> IO a |
||

rndElem xs = do |
rndElem xs = do |
||

− | gen <- getStdGen |
||

index <- randomRIO (0, length xs - 1) |
index <- randomRIO (0, length xs - 1) |
||

return $ xs !! index |
return $ xs !! index |
||

Line 42: | Line 41: | ||

rndPermutation xs = rndElem . permutations $ xs |
rndPermutation xs = rndElem . permutations $ xs |
||

</haskell> |
</haskell> |
||

+ | |||

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

+ | |||

+ | |||

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

## Latest revision as of 13:08, 10 May 2014

Generate a random permutation of the elements of a list.

```
rnd_permu :: [a] -> IO [a]
rnd_permu xs = rnd_select xs (length xs)
```

Uses the solution of problem 23 (rnd_select). 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