Shuffle a deck without local variables

Haskell

Here is a point-free implementation. No variables, formal parameters, or explicit recursion. I used lambdabot's @pl ("pointless") refactoring feature quite a bit.

import Data.List
import Control.Applicative
import Control.Monad
import System.Random

shuffle :: [a] -> IO [a]
shuffle = liftM2 (<$>) ((fst .) . foldl' (uncurry ((. flip splitAt) . (.) .
          (`ap` snd) . (. fst) . flip flip tail . (ap .) . flip flip head .
          ((.) .) . (. (++)) . flip . (((.) . (,)) .) . flip (:))) . (,) [])
          (sequence . map (randomRIO . (,) 0 . subtract 1) . reverse .
          enumFromTo 1 . length)

main = print =<< shuffle [1..52]

Here's my test procedure to make sure the numbers were uniformly distributed:

main = print . foldl' (zipWith (+)) (replicate 52 0)
       =<< replicateM 1000 (shuffle [1..52])

Here is the original algorithm:

shuffle :: [a] -> IO [a]
shuffle xs = shuffleWith xs <$>
             sequence [randomRIO (0, i - 1) | i <- reverse [1..length xs]]

shuffleWith :: [a] -> [Int] -> [a]
shuffleWith xs ns = fst $ foldl' f ([], xs) ns where
    f (a,b) n = (x:a, xs++ys) where
        (xs, x:ys) = splitAt n b

JavaScript

I believe this is the intended form of solution, I use the card in position 0 to keep track of progress, only shuffling the cards that have already been used as counter, this achieves the standard 52! permutations with a perfect equal distribution. The procedure is complicated by XOR swap not allowing that an element is swapped by itself.

Edit: I built in a sorting that sorts each element into place just before it is used, thus allowing this to work with an unsorted array. I also dropped recursive calling in favour of a while loop.

deck=[]
for(a=0;a<52;a++){
    deck[a]=a
}
function swap(a,b){
    deck[a]=deck[b]^deck[a]
    deck[b]=deck[b]^deck[a]
    deck[a]=deck[b]^deck[a]
}
function rand(a,b){
    return Math.floor(Math.random()*(1+b-a))+a
}
function shuffle(){
    while(deck[0]!=0){ //Sort 0 into element 0
        swap(0,deck[0])
    }
    while(deck[0]<51){ //Run 51 times
        while(deck[deck[0]+1]!=deck[0]+1){ //Sort element deck[0]+1 into position deck[0]+1
            swap(deck[deck[0]+1],deck[0]+1)
        }
        swap(0,deck[0]+1) //Swap element deck[0]+1 into position 0, thus increasing the value of deck[0] by 1
        if(rand(0,deck[0]-1)){ //Swap the element at position deck[0] to a random position in the range 1 to deck[0]
            swap(deck[0],rand(1,deck[0]-1))
        }
    }
    if(rand(0,51)){ //Swap the element at position 0 to a random position
        swap(0,rand(1,51))
    }
}
for(c=0;c<100;c++){
    shuffle()
    document.write(deck+"<br>")
}

J

Ignoring that deck is a variable, there's the obvious...

52 ? 52

Of course, if you really want a function, there's this, which will work even if you forget to remove the jokers (or try to shuffle something other than cards).

{~ (# ? #)

So that...

shuffle =: {~ (# ? #)
deck =: i. 52
shuffle deck

This is probably outside the intent of the question, which would be to implement the shuffle yourself from rand (?). I might do that later when I'm not supposed to be working.

Explanation

Explanation of 52 ? 52:

• x ? y is x random unique items from y.

Explanation of {~ (# ? #) is harder because of forks and hooks. Basically, it is the same as shuffle =: 3 : '((# y) ? (# y)) { y', which has one implicit argument (y).

• # y gives the length of y
• This gives 52 ? 52 like before, which is a random permutation of 0..51
• x { y is the item of y in index x, or (in this case) items in the indexes in x.
• This lets you shuffle whatever is passed in, not just integers.

See J Vocabulary for details of operators, though the syntax and semantics are quite a bit tricky because of rank and tacit programming.