Catamorphisms in Haskell

19 12 2007

Lately I have had many works to do at university, therefore I’m sorry for the non regularity of my post’s.
By the way, I also liked to announce that this is my first conscientious post at Planet Haskell.

Why Catamorphisms and Point-Free

Catamorphisms is the way that we can explain in one function how recursive patterns works in some data type.
Here I will use Point-Free notation because what matters here is to show the data flow and the composition of functions.
Point-Free style is used to think in data flow terms, and very useful to program verification, applying formalism to our code.

In point-free style of programming programs are expressed as combinations of simpler functions. This notation is known as write functions without their arguments. Pointwise is the normal form how we write a function.

Couple of examples of point-free notation:

sum = foldr (+) 0 -- point-free
sum l = foldr (+) 0 l -- pointwise


f = (*10).(+2) -- point-free
f n = (n+2)*10 -- pointwise


First of all to define a function, for example f, I say:


I will assume that you are familiarized with infix notation, const, either, uncurry and composition \circ function.


In Haskell we have this definition for lists:

data [a] = [] | a : [a]

Let’s create the same, but more convenient. Consider the following isomorphic type for lists:

data List a = Empty | Node(a,List a) deriving Show

To represent [1,2,3] we wrote Node(1,Node(2,Node(3,Empty))).

As you can see, to construct a (List a) we have two options, Empty or Node. Formally we represent the constructor Empty as 1. And we use (+) to say that our two possibilities are 1 or Node. We could see Node as a the following function:

Node :: (a,List a) -> List a

So typologically we have 1 + (a,List~a). We use (\times) to define that two things occurs in parallel, like tuples do, so we can redefine it: 1 + (a \times~List~a)

Now we can say that (List~a) is isomorphic to (1 + a \times~List~a).
This is something to say that (List~a) and (1 + a \times~List~a) keep the same information without any change or any loss.

Catamorphisms as composition of functions

Let A, B, C, D be Inductive data types (sets) and out, cata, rec functions.

We will write cata(g)_{List} using the composition of out, cata, rec functions. That way we are breaking our problem in small ones. So, in the end we will have the following definition for cata(g)_{List}:

cata(g)_{List} = g \circ rec_{List} \circ out_{List}

The function that we want is cata(g), and that function is over (List~a) so we have:

cata :: (D -> C) -> List a -> C

Type A is (List~a). Maybe this isn’t clear yet, let’s start with function out


The function outList is responsible to create the isomorphism between (1 + a \times~List~a) and (List~a), so the code could be something like this:

outList :: List a -> Either () (a,List a)
outList Empty    = Left ()
outList (Node t) = Right t

In Haskell we represent the type 1 as (), (+) as Either and (\times) as (,).

So, type B is (1 + a \times~List~a).

function g

The function g is also known as *gen*, here is where we said the step that pattern do. Imagine that we want to insert all the values of (List~a) into [a]:

-- pointwise
g :: Either () (a,[a]) -> [a]
g (Left()) = []
g (Right(a,h)) = a:h

-- pointfree
g = either (const []) (uncurry (:))

We represent cata(g) as (| g |).
Now we can be more specific with our graphic:


Here we have to get a function rec that transform 1 + (a \times~List~a) into 1 + (a \times~[a]). That function, general rec, will be:

recg f g h = f -|- (g ><  g) x = ((f . fst) x , (g . snd) x)

With that function we can say exactly what to do with type 1, a, and List~a in domain of rec.
So we want something like this:

rec g = recG id id g

like that we said that (1 + (a \times~\_)) will be the same in the counter domain (1 + (a \times~\_)) of rec. Now we need a function that receive a List~a and give us a [a]
Yes, that function is (| g |)! So, the final graphic became:


Finally we gonna to define the function cata(g):

cata g = outList . rec (cata g) . g

where g is our *gen* function,

g = either (const []) (uncurry (:))

And our objective:

List2HaskellList = cata $ either (const []) (uncurry (:))

More catamorphisms

Imagine we have the following data type:

data NList a where
    Leaf  :: a -> NList a
    NNode :: a -> [NList a] -> NList a

out, rec, and cata became:

out (Leaf a) = Left a
out (NNode a l) = Right (a,l)

Using the previous definitions of (-|-) and (><)

rec f = id -|- (id >< map f)
cata g = g . rec (cata g) . out

Imagging that g has type:

g :: Either a (a,[[a]]) -> [a]

And the graphic for this cata became:


I’ve talked about cata’s without any formalism, the idea was to explain to someone who didn’t know.

I will talk more about catamorphisms and how to calculate programs with them.
In the future I will like to talk about anamorphisms too. And Later on I will talk about point-free over non recursive functions.

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9 responses

19 12 2007

Cool post ! I like these kind of tutorials even if I already know the topic. That kind of tutorial was very useful to me when I started learning Haskell one year ago.

I think that without blogs I would not have been able to understand Haskell.

20 12 2007

The definition of “recg f g h” shadows the f and g in both the (-|-) and (><) definitions. This makes for a much more difficult to read tutorial.

also, spelling: “madders” should be “matters”

20 12 2007

thank you


f a b = a + b * (g b)
    where g a = a + 1

this is the same as:

f a b = a + b * (g b)
g a = a + 1
5 01 2008


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