-- ah, randomization. Always fun in Haskell.
-- this selects only one letter. Not what we want...
rnd_select_one :: [a] -> IO [a]
rnd_select_one xs = do
    let len = (length xs) - 1
    n <- randomRIO (0, len)
    return [(xs!!n)]

-- this version allows resampling. 
rnd_select :: [a] -> Int -> IO [a]
rnd_select xs n = do
    g <- sequence (replicate n (rnd_select_one xs))
    return (concat g)

-- Here's how to prevent re-sampling, using partition from Data.List
-- like so: 
-- partition (`elem` "g") "abcdefghijklmn"
--  --> ("g", "abcdefhijklmn")
--
-- It would also be fairly trivial to use removeAt from Problem 20


rnd_select_one' :: Eq a => ([a], [a]) -> Int -> IO ([a], [a])
rnd_select_one' (acc, xs) 0 = return (acc, xs)
rnd_select_one' (acc, []) _ = return (acc, [])
rnd_select_one' (acc, xs) n = do
    let len = (length xs) - 1
    d <- randomRIO (0, len)
    let el = (xs!!d)
    let (g, ys) = partition (`elem` [el]) xs
    let ret = (acc ++ g, ys)
    rnd_select_one' ret (n-1) 

rnd_select' :: Eq a => [a] -> Int -> IO [a]
rnd_select' xs n = do
    j <- rnd_select_one' ([], xs) n
    return $ fst j

-- ah, randomization. Always fun in Haskell.
rnd_select_one :: [a] -> IO [a]
rnd_select_one xs = do
    let len = (length xs) - 1
    n' <- randomRIO (0, len)
    return [(xs!!n')]

-- this version allows resampling. 
rnd_select n xs = sequence (replicate n (rnd_select_one xs))

-- Ideally, to prevent re-sampling, we would use partition from Data.List
-- like so: 
-- partition (`elem` "g") "abcdefghijklmn"
--  --> ("g", "abcdefhijklmn")
-- I'll have to revisit this one when I have a chance to 
-- play around with partitioning.

http://haskell.org/haskellwiki/99_questions/21_to_28

problem 23, Extract a given number of randomly selected elements from a list. 
    Example:

    Prelude System.Random>rnd_select "abcdefgh" 3 >>= putStrLn

        "eda"
        
    Two problems: 1) How to return a list, and 2) how to sample without duplication

Haskell 99 Problems - Problem 23

Haskell