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