We know free monads are useful, and packages like Operational make it easy to define new monads by only caring about the application-specific effects, not the monadic structure itself.
We can easily define "free arrows" analogous to how free monads are defined:
{-# LANGUAGE GADTs #-}
module FreeA
( FreeA, effect
) where
import Prelude hiding ((.), id)
import Control.Category
import Control.Arrow
import Control.Applicative
import Data.Monoid
data FreeA eff a b where
Pure :: (a -> b) -> FreeA eff a b
Effect :: eff a b -> FreeA eff a b
Seq :: FreeA eff a b -> FreeA eff b c -> FreeA eff a c
Par :: FreeA eff a₁ b₁ -> FreeA eff a₂ b₂ -> FreeA eff (a₁, a₂) (b₁, b₂)
effect :: eff a b -> FreeA eff a b
effect = Effect
instance Category (FreeA eff) where
id = Pure id
(.) = flip Seq
instance Arrow (FreeA eff) where
arr = Pure
first f = Par f id
second f = Par id f
(***) = Par
My question is, what would be the most useful generic operations on free arrows? For my particular application, I needed special cases of these two:
{-# LANGUAGE Rank2Types #-}
{-# LANGUAGE ScopedTypeVariables #-}
analyze :: forall f eff a₀ b₀ r. (Applicative f, Monoid r)
=> (forall a b. eff a b -> f r)
-> FreeA eff a₀ b₀ -> f r
analyze visit = go
where
go :: forall a b. FreeA eff a b -> f r
go arr = case arr of
Pure _ -> pure mempty
Seq f₁ f₂ -> mappend <$> go f₁ <*> go f₂
Par f₁ f₂ -> mappend <$> go f₁ <*> go f₂
Effect eff -> visit eff
evalA :: forall eff arr a₀ b₀. (Arrow arr) => (forall a b. eff a b -> arr a b) -> FreeA eff a₀ b₀ -> arr a₀ b₀
evalA exec = go
where
go :: forall a b. FreeA eff a b -> arr a b
go freeA = case freeA of
Pure f -> arr f
Seq f₁ f₂ -> go f₂ . go f₁
Par f₁ f₂ -> go f₁ *** go f₂
Effect eff -> exec eff
but I don't have any theoretical arguments on why these (and not others) would be the useful ones.
A free functor is left adjoint to a forgetful functor. For the adjunction you need to have the isomorphism (natural in x
and y
):
(Free y :~> x) <-> (y :~> Forget x)
In what category should this be? The forgetful functor forgets the Arrow
instance, so it goes from the category of Arrow
instances to the category of all bifunctors. And the free functor goes the other way, it turns any bifunctor into a free Arrow
instance.
The haskell type of arrows in the category of bifunctors is:
type x :~> y = forall a b. x a b -> y a b
It's the same for arrows in the category of Arrow
instances, but with addition of Arrow
constraints. Since the forgetful functor only forgets the constraint, we don't need to represent it in Haskell. This turns the above isomorphism into two functions:
leftAdjunct :: (FreeA x :~> y) -> x :~> y
rightAdjunct :: Arrow y => (x :~> y) -> FreeA x :~> y
leftAdjunct
should also have an Arrow y
constraint, but it turns out it is never needed in the implementation. There's actually a very simple implementation in terms of the more useful unit
:
unit :: x :~> FreeA x
leftAdjunct f = f . unit
unit
is your effect
and rightAdjunct
is your evalA
. So you have exactly the functions needed for the adjunction! You'd need to show that leftAdjunct
and rightAdjunct
are isomorphic. The easiest way to do that is to prove that rightAdjunct unit = id
, in your case evalA effect = id
, which is straightforward.
What about analyze
? That's evalA
specialized to the constant arrow, with the resulting Monoid
constraint specialized to the applicative monoid. I.e.
analyze visit = getApp . getConstArr . evalA (ConstArr . Ap . visit)
with
newtype ConstArr m a b = ConstArr { getConstArr :: m }
and Ap
from the reducers package.
Edit: I almost forgot, FreeA should be a higher order functor! Edit2: Which, on second thought, can also be implemented with rightAdjunct
and unit
.
hfmap :: (x :~> y) -> FreeA x :~> FreeA y
hfmap f = evalA (effect . f)
By the way: There's another way to define free functors, for which I put a package on Hackage recently. It does not support kind * -> * -> *
(Edit: it does now!), but the code can be adapted to free arrows:
newtype FreeA eff a b = FreeA { runFreeA :: forall arr. Arrow arr => (eff :~> arr) -> arr a b }
evalA f a = runFreeA a f
effect a = FreeA $ \k -> k a
instance Category (FreeA f) where
id = FreeA $ const id
FreeA f . FreeA g = FreeA $ \k -> f k . g k
instance Arrow (FreeA f) where
arr f = FreeA $ const (arr f)
first (FreeA f) = FreeA $ \k -> first (f k)
second (FreeA f) = FreeA $ \k -> second (f k)
FreeA f *** FreeA g = FreeA $ \k -> f k *** g k
FreeA f &&& FreeA g = FreeA $ \k -> f k &&& g k
If you don't need the introspection your FreeA
offers, this FreeA
is probably faster.
来源:https://stackoverflow.com/questions/12001350/useful-operations-on-free-arrows