Writing zip (longzip) using an anamorphism

Question

Working on a project and trying to write longzip using an anamorphism. I'm having some trouble writing a coalgebra for this use case. I've defined my anamorphism in terms of Fix below:

-- Fixed point of a Functor    
newtype Fix f = In (f (Fix f))     
    
deriving instance (Eq (f (Fix f))) => Eq (Fix f)    
    
out :: Fix f -> f (Fix f)    
out (In f) = f 

-- Anamorphism
type Coalgebra f a = a -> f a

ana :: (Functor f) => Coalgebra f a -> a -> Fix f
ana f = In . fmap (ana f) . f

This is the definition for ana derived from "reversing the arrows" of cata:

-- Catamorphism    
type Algebra f a = f a -> a    
    
cata :: (Functor f) => Algebra f a -> Fix f -> a

cata f = f . fmap (cata f) . out

I've seen zip written using a version of ana that is obviously defined differently (takes a predicate as a parameter):

 zip2 = ana unsp fin
    where
    fin (as,bs) = (as==[]) || (bs ==[]) 
    unsp ((a:as), (b:bs)) = ((a,b),(as,bs))

(Taken from https://en.wikipedia.org/wiki/Anamorphism)

But I'm unsure how to move forward using the version of ana defined above, particularly in regards to writing a Coalgebra of type a -> fa. Like, would it have to take the two list parameters to zip and combine them into a single a?

Wow, the wikipedia page is such a mess. – Abastro Aug 20 '22 at 01:36 — Abastro, Aug 20 '22 at 01:36

Carl · Accepted Answer · 2022-11-11T17:19:06.177

First off, give yourself a starting point. Write down what you're going to do:

-- make this example actually complete
{-# Language StandaloneDeriving, UndecidableInstances, DeriveFunctor #-}
import Prelude hiding (zip)


-- Fixed point of a Functor
newtype Fix f = In (f (Fix f))

deriving instance (Eq (f (Fix f))) => Eq (Fix f)
deriving instance (Show (f (Fix f))) => Show (Fix f)

out :: Fix f -> f (Fix f)
out (In f) = f

-- Anamorphism
type Coalgebra f a = a -> f a

ana :: (Functor f) => Coalgebra f a -> a -> Fix f
ana f = In . fmap (ana f) . f


-- Base functor for a list
data ListF a f = Nil | Cons a f deriving (Eq, Show, Functor)
type List a = Fix (ListF a)

-- write down the type. It helps you think about things
zip :: List a -> List b -> List (a, b)
zip x y = ana undefined undefined

You know zip has to be implemented as a call to ana, so what remains is figuring out the seed and the coalgebra. It should be fairly clear the seed needs to contain x and y. It doesn't seem like any further information should be necessary, so let's just assume the seed will be (x, y) until/unless it poses a problem. That's enough information to pin down the types for this first try:

zip :: List a -> List b -> List (a, b)
zip x y = ana zipCoalgebra (x, y)

zipCoalgebra :: (List a, List b) -> ListF (a, b) (List a, List b)
zipCoalgebra = undefined

I feel like this is the step you've missed: writing down what you're trying to do and following the types to pin down what you need. The rest of this is sort of trivial if you've ever seen any implementation of zip. It's a matter of writing down the most boring thing that type-checks (paying close attention to the difference between List and ListF in the type). I strongly recommend you stop reading here and give it a try yourself. There isn't much else I can say that's actually helpful for learning how to think about this.

If you really have no idea at all:

zipCoalgebra :: (List a, List b) -> ListF (a, b) (List a, List b)
zipCoalgebra (In (Cons a as), In (Cons b bs)) = Cons (a, b) (as, bs)
zipCoalgebra _ = Nil

It really is what you'd expect if you've ever seen an implementation of zip before, with the necessary noise to make it type-check around Fix. Let's give it a spin:

ghci> zip (In (Cons 1 (In Nil))) (In Nil)
In Nil
ghci> zip (In (Cons 1 (In Nil))) (In (Cons 2 (In Nil)))
In (Cons (1,2) (In Nil))

Yep. Behaving as expected. It seems the two lists were sufficient as the seed, and all is well.

Writing zip (longzip) using an anamorphism

1 Answers1