The question of Set monad comes up quite regularly, most recently at
        http://www.ittc.ku.edu/csdlblog/?p=134

Indeed, we cannot make Data.Set.Set to be the instance of Monad type
class -- not immediately, that it. That does not mean that there is no
Set Monad, a non-determinism monad that returns the set of answers,
rather than a list. I mean genuine *monad*, rather than a restricted,
generalized, etc. monad.

It is surprising that the question of the Set monad still comes up
given how simple it is to implement it. Footnote 4 in the ICFP
2009 paper on ``Purely Functional Lazy Non-deterministic Programming''
described the idea, which is probably folklore. Just in case, here is
the elaboration, a Set monad transformer.

{-# LANGUAGE TypeSynonymInstances, FlexibleInstances #-}

module SetMonad where

import qualified Data.Set as S
import Control.Monad.Cont

-- Since ContT is a bona fide monad transformer, so is SetMonadT r.
type SetMonadT r = ContT (S.Set r)

-- These are the only two places the Ord constraint shows up

instance (Ord r, Monad m) => MonadPlus (SetMonadT r m) where
    mzero = ContT $ \k -> return S.empty
    mplus m1 m2 = ContT $ \k -> liftM2 S.union (runContT m1 k) (runContT m2 k)

runSet :: (Monad m, Ord r) => SetMonadT r m r -> m (S.Set r)
runSet m = runContT m (return . S.singleton)

choose :: MonadPlus m => [a] -> m a
choose = msum . map return

test1 = print =<< runSet (do
  n1 <- choose [1..5]
  n2 <- choose [1..5]
  let n = n1 + n2
  guard $ n < 7
  return n)
-- fromList [2,3,4,5,6]

-- Values to choose from might be higher-order or actions
test1' = print =<< runSet (do
  n1 <- choose . map return $ [1..5]
  n2 <- choose . map return $ [1..5]
  n  <- liftM2 (+) n1 n2
  guard $ n < 7
  return n)
-- fromList [2,3,4,5,6]

test2 = print =<< runSet (do
  i <- choose [1..10]
  j <- choose [1..10]
  k <- choose [1..10]
  guard $ i*i + j*j == k * k
  return (i,j,k))
-- fromList [(3,4,5),(4,3,5),(6,8,10),(8,6,10)]

test3 = print =<< runSet (do
  i <- choose [1..10]
  j <- choose [1..10]
  k <- choose [1..10]
  guard $ i*i + j*j == k * k
  return k)
-- fromList [5,10]



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