On 17 February 2011 19:13, Javier M Mora <jamarier@gmail.com> wrote:

First Step: What I want?
------------------------

In this problem: I think monads as a DSL (Domain Specific Language)

main = do
print $ sumM $ do
makeList 10 -- create candidates list
multiples 3 -- choose multiples of 3
multiples 5 -- choose multiples of 5 (not choosed yet)

Data under de monad is a pair of lists:
(validValues, CandidatesNonValidYet)

Although my suggestion is not to use a monad for this problem, assuming this is a learning exercise, a solution using the state monad is as follows.

I'll keep the main function exactly as you wanted.

sumM x = sum $ fst $ execState ([],[]) x

or, point-free:

sumM = sum . fst . flip execState ([],[])

Here, sumM executes the given state monad, and we end up with the pair of selected and not-selected elements. Then project the fst component, and sum them up.

makeList n = put ([],[1..n])

makeList initialises the state.

multiples n = chooseIf (\ i -> i `mod` n == 0)

multiplies chooses those elements satisfying the given criteria. chooseIf is a helper function I've chosen to define. Obviously, you can do just fine without it.

chooseIf f = do

a <- gets fst

(b,c) <- partition f <$> gets snd

put (a++b,c)

chooseIf partitions the list of candidates into 2, b is the list of elements satisfying the condition, c is the elements not satisfying it. (Remark: ++ is O(n))

And that should be it. If you plug these all together, you'll get 33 as the answer. That is the sum of [3,6,9,5,10]. I don't know why you didn't include 10 in the list of candidates, but if that is very important you can remove it by modifying makeList.

Hope this helps.

Ozgur