Joachim Breitner wrote:
Hi,
Am Samstag, den 10.03.2007, 14:52 -0500 schrieb Stefan Monnier:
I'm pretty sure you can define a catamorphism for any regular algebraic
data type.
Actually, so-called negative occurrences in (regular) data types cause
problems. Try to define the catamorphism of
data Exp = Num Int | Lam (Exp -> Exp) | App Exp Exp
to see the problem,
I guess Robert is still true as this Exp contains a non-algebraic type
((->)), therefore is not a regular algebraic type. (Based on an assumed
meaning of “algebraic data type”). But let’s see...
Am I right to assume that I have found a catamorphism if and only if the
that function, applied to the data type constructors (in the right
order) gives me the identity on this data type?
maybe Nothing Just == id :: Maybe -> Maybe
\b -> if b then True else False == id :: Bool -> Bool
foldr (:) [] == id :: [a] -> [a]
uncurry (,) == id :: (a,b) -> (a,b)
either Left Right == id :: Either a b -> Either a b
Does that also mean that catamorphism for a given type are unique
(besides argument re-ordering)?
Now some brainstorming about the above Exp. I guess we want:
exp Num Lam App e == id e
therefore
exp :: (Int -> Exp) ((Exp -> Exp) -> Exp) (Exp -> Exp -> Exp) Exp
now we want the return type to not matter
exp :: forall b. (Int -> b) ((Exp -> Exp) -> b) (Exp -> Exp -> b) Exp
So my guess would be:
exp f _ _ (Num i) = f i
exp _ f _ (Lam g) = f g
exp _ _ f (App e1 e2) = f e1 e2
Looks a bit stupid, but seems to work, especially as there is not much a
function with type (Exp -> Exp) -> b can do, at least on it’s own. Is
that a catamorphism?
Greetings,
Joachim
Catamorphisms, folds, are the mediating arrows of initial algebras and thus are
unique. I believe those equations combined with the free theorem that such
functions would have, do guarantee that it would be that arrow.
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