[Haskell-cafe] Existential question
oleg at okmij.org
oleg at okmij.org
Wed Aug 24 09:43:34 CEST 2011
> I had simplified the type to make the plumbing simpler. My intention
> was to include an initial value to use the function as a sequence
> transformer / generator:
>
> data Kl i o = forall s. Kl s (i -> s -> (s, o))
That change makes a world of difference! For example, the above type
(Kl i) is clearly a monad
> instance Monad (Kl i) where
> return x = Kl () (\_ s -> (s,x))
> (Kl s m) >>= f = Kl s (\i s -> case f (snd (m i s)) of
> Kl s' m' -> (s,snd (m' i s')))
It is the Reader monad. Different Kl values have their own, private s.
The quantification ensures that each encapsulated 's' applies only to
its generator. Therefore, we can just as well apply that 's' right at
the very beginning. Therefore,
> data Kl i o = forall s. Kl s (i -> s -> (s, o))
is essentially
> data Kl' i o = Kl' (i -> o)
which is certainly and Arrow and a Monad. We do not need existential
at all. The web page
http://okmij.org/ftp/Computation/Existentials.html
describes other eliminations of Existentials.
> Does the "growing type" s1 -> s2 -> (s1,s2) in the bind-like method
> below support some other abstraction that is monad-like?
>
> data Kl1' s o = Kl1' (s -> (s,o))
>
> bind' :: (Kl1' s1 i) -> (i -> Kl1' s2 o) -> (Kl1' (s1,s2) o)
>
> bind' mi f = Kl1' mo where
> mo (s1,s2) = ((s1',s2'),o) where
> (Kl1' u1) = mi
> (s1', i) = u1 s1
> (Kl1' u2) = f i
> (s2', o) = u2 s2
Not all things are monads; for example, restricted and parameterized
monads are not monads (but do look like them, enough for the
do-notation). Your Kl1' does look like another generalization of
monads. I must say that it seems to me more fitting for a parametrized
applicative:
> class Appish i where
> purish :: a -> i s a
> appish :: i s1 (a ->b) -> i s2 a -> i (s1,s2) b
>
> data KL s a = KL (s -> (s,a))
>
> instance Appish KL where
> purish x = KL (\s -> (s,x))
> appish (KL f1) (KL f2) = KL $ \(s1,s2) ->
> let (s1', ab) = f1 s1
> (s2', a) = f2 s2
> in ((s1',s2'), ab a)
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