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

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< 99 questions | Solutions

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Uses the solution for the previous problem. Choosing N distinct elements from a list of length N will yield a permutation. |
Uses the solution for the previous problem. Choosing N distinct elements from a list of length N will yield a permutation. |
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+ | |||

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

+ | <haskell> |
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+ | import System.Random (randomRIO) |
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+ | |||

+ | rnd_permu :: [a] -> IO [a] |
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+ | rnd_permu [] = return [] |
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+ | rnd_permu (x:xs) = do |
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+ | rand <- randomRIO (0, (length xs)) |
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+ | rest <- rnd_permu xs |
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+ | return $ let (ys,zs) = splitAt rand rest |
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+ | in ys++(x:zs) |
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+ | </haskell> |

## Revision as of 06:24, 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)
```