Conal Elliott wrote:
Thanks for the reply. Here's the decomposition I had in mind. Start with
type List a = Maybe (a, List a)
Rewrite a bit
type List a = Maybe (Id a, List a)
Then make the type *constructor* pairing explicit
type List a = Maybe ((Id :*: List) a)
where
newtype (f :*: g) a = Prod { unProd :: (f a, g a) }
Then make the type-constructor composition explicit
type List = Maybe :. (Id :*: List)
(which isn't legal Haskell, due to the type synonym cycle). From there use the Functor and Applicative instances for composition and pairing of type constructors and for Id. I think the result is equivalent to ZipList.
Ah, I didn't think of feeding a to both f and g in the product f :* g . Your argument cheats a bit because of its circularity: assuming List is an applicative functor, you deduce that List is an applicative functor. But in this case, the recursion is (co-)inductive, so things work out. Here's the formalization: -- higher-order functors g :: (* -> *) -> (* -> *) -- (not sure how to do these classes directly in Haskell, -- but you know what I want to do here) class Functor2 g where forall f . Functor f => Functor (g f) class Applicative2 g where forall f . Applicative f => Applicative (g f) -- higher-order composition type (f :.. g) h = f :. (g :. h) -- fixed points for higher-order functors newtype Mu g a = In { out :: g (Mu g) a } type List a = Mu ((Maybe :.) :.. (Id :*)) a instance Applicative2 g => Applicative (Mu g) where pure x = In (pure x) (In f) <*> (In x) = In (f <*> g) This last class instance looks ridiculous of course, but does nothing more than use the assertion Applicative (Mu g) in its own definition. But fortunately, this definition terminates.
Is there some construction simpler than lists (non-recursive) that introduces cross products?
To clarify my "cross products" question, I mean fs <*> xs = [f x | f <- fs, x <- xs], as with lists.
I'm not sure how to decouple the notion of cross products from lists. Maybe the other characterization of applicative functors sheds some light on it: applicative functors f can also be defined with the following two primitive operations pure :: a -> f a cross :: (f a, f b) -> f (a,b) f <*> x = fmap eval (cross (f,x)) where eval (f,x) = f x Then, the choice pure x = repeat x [1,2] `cross` [3,4] = [(1,3), (2,4)] yields zip lists whereas the choice pure x = [x] [1,2] `cross` [3,4] = [(1,3), (1,4), (2,3), (2,4)] yields backtracking lists. I'm not sure whether other choices are possible too, they probably violate the laws mentioned in chapter 7 of the applicative functor paper. Regards, apfelmus