[Haskell-cafe] Re: Church Encoding Function

[Joachim Breitner]

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

-- 
Joachim Breitner
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