Euler 43 - is there a monad to help write this list comprehension?
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
Solution 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.
Solution 2:
Before jumping to monads, let's consider guided unique selection from finite domains first:
-- all possibilities:
pick_any [] = []
pick_any (x:xs) = (xs,x) : [ (x:dom,y) | (dom,y) <- pick_any xs ]
-- guided selection (assume there's no repetitions in the domain):
one_of ns xs = [ (dom,y) | let choices = pick_any xs, n <- ns,
(dom,y) <- take 1 $ filter ((==n).snd) choices ]
With this a list comprehension can be written without the use of elem
calls:
p43 = sum [ fromDigits [d0,d1,d2,d3,d4,d5,d6,d7,d8,d9]
| (dom5,d5) <- one_of [0,5] [0..9]
, (dom6,d6) <- pick_any dom5
, (dom7,d7) <- pick_any dom6
, rem (100*d5+10*d6+d7) 11 == 0
....
fromDigits :: (Integral a) => [a] -> Integer
fromDigits ds = foldl' (\s d-> s*10 + fromIntegral d) 0 ds
The monad from Louis Wasserman's answer can be further augmented with additional operations based on the functions above:
import Control.Monad
newtype UniqueSel a = UniqueSel { runUS :: [Int] -> [([Int], a)] }
instance Monad UniqueSel where
-- as in Louis's answer
instance MonadPlus UniqueSel where
-- as in Louis's answer
choose = UniqueSel pick_any
choose_one_of xs = UniqueSel $ one_of xs
choose_n n = replicateM n choose
set_choices cs = UniqueSel (\ _ -> [(cs, ())])
get_choices = UniqueSel (\cs -> [(cs, cs)])
So that we can write
numTest xs m = fromDigits xs `rem` m == 0
pandigitalUS :: UniqueSel [Int]
pandigitalUS = do
set_choices [0..9]
[d7,d8,d9] <- choose_n 3
guard $ numTest [d7,d8,d9] 17
d6 <- choose
guard $ numTest [d6,d7,d8] 13
d5 <- choose_one_of [0,5]
guard $ numTest [d5,d6,d7] 11
d4 <- choose
guard $ numTest [d4,d5,d6] 7
d3 <- choose_one_of [0,2..8]
d2 <- choose
guard $ rem (d2+d3+d4) 3 == 0
[d1,d0] <- choose_n 2
guard $ d0 /= 0
return [d0,d1,d2,d3,d4,d5,d6,d7,d8,d9]
pandigitals = map (fromDigits.snd) $ runUS pandigitalUS []
main = do print $ sum pandigitals
Solution 3:
The UniqueSel
monad suggested by Louis Wasserman is exactly StateT [Integer] []
(I'm using Integer
everywhere for simplicity).
The state keeps the available digits and every computation is nondeterministic - from a given state we can select different digits to continue with. Now the choose
function can be implemented as
import Control.Monad
import Control.Monad.State
import Control.Monad.Trans
import Data.List
choose :: PanM Integer
choose = do
xs <- get
x <- lift xs -- pick one of `xs`
let xs' = x `delete` xs
put xs'
return x
And then the monad is run by evalStateT
as
main = do
let nums = evalStateT pandigitals [0..9]
-- ...