# Random shuffle

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Revision as of 20:49, 18 October 2007 by ChrisKuklewicz (Talk | contribs)

## 1 The problem

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

## 2 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.