# Random shuffle

From HaskellWiki

Revision as of 20:49, 18 October 2007 by ChrisKuklewicz (talk | contribs) (Add pure version using ST)

## The problem

Shuffling a list, i.e. creating a random permutation, is not easy to do correctly. Each permutation should have the same probability.

## Imperative algorithm

The standard imperative algorithm can be implemented as follows:

```
{-# LANGUAGE ScopedTypeVariables #-}
import System.Random
import Data.Array.IO
import Control.Monad
-- | Randomly shuffle a list
-- /O(N)/
shuffle :: forall a. [a] -> IO [a]
shuffle xs = do
let n = length xs
ar <- newListArray (1,n) xs :: IO (IOArray Int a)
forM [1..n] $ \i -> do
j <- randomRIO (i,n)
vi <- readArray ar i
vj <- readArray ar j
writeArray ar j vi
return vj
```

Or one can use ST to avoid needing IO:

```
-- | Randomly shuffle a list without the IO Monad
-- /O(N)/
shuffle' :: [a] -> StdGen -> ([a],StdGen)
shuffle' xs gen = runST (do
g <- newSTRef gen
let randomRST lohi = do
(a,s') <- liftM (randomR lohi) (readSTRef g)
writeSTRef g s'
return a
ar <- newArray n xs
xs' <- forM [1..n] $ \i -> do
j <- randomRST (i,n)
vi <- readArray ar i
vj <- readArray ar j
writeArray ar j vi
return vj
gen' <- readSTRef g
return (xs',gen'))
where
n = length xs
newArray :: Int -> [a] -> ST s (STArray s Int a)
newArray n xs = newListArray (1,n) xs
```

And if you are using IO's hidden StdGen you can wrap this as usual:

```
shuffleIO :: [a] -> IO [a]
shuffleIO xs = getStdRandom (shuffle' xs)
```

This is a lot simpler than the purely functional algorithm linked below.