On Mon, Jul 30, 2012 at 10:38 PM, Ryan Ingram <ryani.spam@gmail.com> wrote:

To take this a step further, if what you really want is the syntax sugar for do-notation (and I understand that, I love sweet, sweet syntactical sugar), you are probably implementing a Writer monad over some monoid.

Here's two data structures that can encode this type;

data Replacer1 k a = Replacer1 (k -> Maybe k) a
data Replacer2 k a = Replacer2 [(k,k)] a

instance Monad Replacer1 where
    return x = Replacer1 (\_ -> Nothing) x
    Replacer1 ka a >>= f = result where
        Replacer1 kb b = f a
      result = Replacer1 (\x -> ka x `mplus` kb x) b

(!>) :: Eq k => k -> k -> Replacer1 k ()
x !> y = Replacer1 (\k -> if k == x then Just y else Nothing) ()

replace1 :: Replacer1 k () -> [k] -> [k]    -- look ma, no Eq requirement!
replace1 (Replacer k ()) = map (\x -> fromMaybe x $ k x) -- from Data.Maybe

table1 :: Replacer1 Char ()
table1 = do
    'a' !> 'b'
    'A' !> 'B'

test = replace1 table1 "All I want"

-- Exercise: what changes if we switch ka and kb in the result of (>>=)? When does it matter?

-- Exercises for you to implement:
instance Monad Replacer2 k where
replacer :: Eq k => Replacer2 k -> [k] -> [k]
($>) :: k -> k -> Replacer2 k

-- Exercise: Lets make use of the fact that we're a monad!
--
-- What if the operator !> had a different type?
-- (!>) :: Eq k => k -> k -> Replacer k Integer
-- which returns the count of replacements done.
--
-- table3 = do
--     count <- 'a' !> 'b'
--     when (count > 3) ('A' !> 'B')
--     return ()
--
-- Do any of the data structures I've given work? Why or why not?
-- Can you come up with a way to implement this?

-- ryan

On Sat, Jul 28, 2012 at 10:07 AM, Steffen Schuldenzucker <sschuldenzucker@uni-bonn.de> wrote:

On 07/28/2012 03:35 PM, Thiago Negri wrote:
> [...]

As Monads are used for sequencing, first thing I did was to define the
following data type:

data TableDefinition a = Match a a (TableDefinition a) | Restart

So TableDefinition a is like [(a, a)].

[...]

>

So, to create a replacement table:

table' :: TableDefinition Char
table' =
Match 'a' 'b'
(Match 'A' 'B'
Restart)

It look like a Monad (for me), as I can sequence any number of
replacement values:

table'' :: TableDefinition Char
table'' = Match 'a' 'c'
(Match 'c' 'a'
(Match 'b' 'e'
(Match 'e' 'b'
Restart)))

Yes, but monads aren't just about sequencing. I like to see a monad as a generalized computation (e.g. nondeterministic, involving IO, involving state etc). Therefore, you should ask yourself if TableDefinition can be seen as some kind of abstract "computation". In particular, can you "execute" a computation and "extract" its result? as in

do
r <- Match 'a' 'c' Restart
if r == 'y' then Restart else Match 2 3 (Match 3 4 Restart)

Doesn't immediately make sense to me. In particular think about the different possible result types of a TableDefinition computation.

If all you want is sequencing, you might be looking for a Monoid instance instead, corresponding to the Monoid instance of [b], where b=(a,a) here.

[...]

>

I'd like to define the same data structure as:

newTable :: TableDefinition Char
newTable = do
'a' :> 'b'
'A' :> 'B'

But I can't figure a way to define a Monad instance for that. :(

The desugaring of the example looks like this:

('a' :> 'b') >> ('A' :> 'B')

Only (>>) is used, but not (>>=) (i.e. results are always discarded). If this is the only case that makes sense, you're probably looking for a Monoid instead (see above)

-- Steffen

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