Question

Here is a way to solve Euler problem 43 (please let me know if this doesn’t give the correct answer). Is there a monad or some other syntatic sugar which could assist with keeping track of the notElem conditions?

toNum xs = foldl (\s d -> s*10+d) 0 xs

numTest xs m = (toNum xs) `mod` m == 0

pandigitals = [ [d0,d1,d2,d3,d4,d5,d6,d7,d8,d9] |
                d7 <- [0..9],
                d8 <- [0..9], d8 `notElem` [d7],
                d9 <- [0..9], d9 `notElem` [d8,d7],
                numTest [d7,d8,d9] 17,
                d5 <- [0,5],  d5 `notElem` [d9,d8,d7],
                d3 <- [0,2,4,6,8], d3 `notElem` [d5,d9,d8,d7],
                d6 <- [0..9], d6 `notElem` [d3,d5,d9,d8,d7],
                numTest [d6,d7,d8] 13,
                numTest [d5,d6,d7] 11,
                d4 <- [0..9], d4 `notElem` [d6,d3,d5,d9,d8,d7],
                numTest [d4,d5,d6] 7,
                d2 <- [0..9], d2 `notElem` [d4,d6,d3,d5,d9,d8,d7],
                numTest [d2,d3,d4] 3,
                d1 <- [0..9], d1 `notElem` [d2,d4,d6,d3,d5,d9,d8,d7],
                d0 <- [1..9], d0 `notElem` [d1,d2,d4,d6,d3,d5,d9,d8,d7]
              ]

main = do
         let nums = map toNum pandigitals
         print $ nums
         putStrLn ""
         print $ sum nums

For instance, in this case the assignment to d3 is not optimal – it really should be moved to just before the numTest [d2,d3,d4] 3 test. Doing that, however, would mean changing some of the notElem tests to remove d3 from the list being checked. Since the successive notElem lists are obtained by just consing the last chosen value to the previous list, it seems like this should be doable – somehow.

UPDATE: Here is the above program re-written with Louis’ UniqueSel monad below:

toNum xs = foldl (\s d -> s*10+d) 0 xs
numTest xs m = (toNum xs) `mod` m == 0

pandigitalUS =
  do d7 <- choose
     d8 <- choose
     d9 <- choose
     guard $ numTest [d7,d8,d9] 17
     d6 <- choose
     guard $ numTest [d6,d7,d8] 13
     d5 <- choose
     guard $ d5 == 0 || d5 == 5
     guard $ numTest [d5,d6,d7] 11
     d4 <- choose
     guard $ numTest [d4,d5,d6] 7
     d3 <- choose
     d2 <- choose
     guard $ numTest [d2,d3,d4] 3
     d1 <- choose
     guard $ numTest [d1,d2,d3] 2
     d0 <- choose
     guard $ d0 /= 0
     return [d0,d1,d2,d3,d4,d5,d6,d7,d8,d9]

pandigitals = map snd $ runUS pandigitalUS [0..9]

main = do print $ pandigitals

Answer 1

Sure.

newtype UniqueSel a = UniqueSel {runUS :: [Int] -> [([Int], a)]}
instance Monad UniqueSel where
  return a = UniqueSel (\ choices -> [(choices, a)])
  m >>= k = UniqueSel (\ choices -> 
    concatMap (\ (choices', a) -> runUS (k a) choices')
      (runUS m choices))

instance MonadPlus UniqueSel where
  mzero = UniqueSel $ \ _ -> []
  UniqueSel m `mplus` UniqueSel k = UniqueSel $ \ choices ->
    m choices ++ k choices

-- choose something that hasn't been chosen before
choose :: UniqueSel Int
choose = UniqueSel $ \ choices ->
  [(pre ++ suc, x) | (pre, x:suc) <- zip (inits choices) (tails choices)]

and then you treat it like the List monad, with guard to enforce choices, except that it won’t choose an item more than once. Once you have a UniqueSel [Int] computation, just do map snd (runUS computation [0..9]) to give it [0..9] as the choices to select from.