Ozgur Akgun ozgurakgun at gmail.com
Thu Feb 17 20:54:53 CET 2011

```On 17 February 2011 19:13, Javier M Mora <jamarier at 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
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