Like I said before (in those posts) when you write an hylomorphism over a particular data type, that means just that the intermediate structure is that data type.
In fact that data will never be stored into that intermediate type or . Because we glue the ana and cata together into a single recursive pattern. and could be some data type your function need. With this post I will try to show you more hylomorphisms over some different data types to show you the power of this field.
The data type that we going to discuss here is the . In Haskell we can represent as:
data LTree a = Leaf a | Fork (LTree a, LTree a)
Is just like a binary tree, but the information is just in the leaf’s. Even more: a leaf tree is a tree that only have leaf’s, no information on the nodes. This is an example of a leaf tree:
To represent all the hylomorphisms over we draw the following diagram:
The example I’m going to give is making the fibonacci function using a hylomorphism over this data type. If you remember the method I used before, I’m going to start by the anamorphism . Before that I’m going to specify the strategy to define factorial. I’m going to use the diagram’s again, remember that type is equivalent to Haskell :
As you can see I’m going to use as my intermediate structure, and I’ve already define the names of my gen functions to the catamorphism and to the anamorphism. The strategy I prefer, is do all the hard work in the anamorphism, so here the gen for the anamorphism is:
fibd n | n < 2 = i1 () | otherwise = i2 (n-1,n-2)
This function combined with the anamorphism, going to generate leaf tree’s with leaf’s, being the result of that fib.
Then we just have to write the gen for the catamorphism. This function (combined with the catamorphism) counts the number of leafs that a leaf tree have.
add = either (const 1) plus where plus = uncurry (+)
The final function, the fibonacci function is the hylomorphism of those two defined before:
fib = hyloLTree add fibd
Here is all the auxiliary functions you need to run this example:
inLTree = either Leaf Fork outLTree :: LTree a -> Either a (LTree a,LTree a) outLTree (Leaf a) = i1 a outLTree (Fork (t1,t2)) = i2 (t1,t2) cataLTree a = a . (recLTree (cataLTree a)) . outLTree anaLTree f = inLTree . (recLTree (anaLTree f) ) . f hyloLTree a c = cataLTree a . anaLTree c baseLTree g f = g -|- (f >< f) recLTree f = baseLTree id f
The lists that I’m going to talk here, are the Haskell lists, wired into the compiler, but is a definition exist, it will be:
data [a] = [ ] | a : [a]
So, our diagram to represent the hylomorphism over this data type is:
The function I’m going to define as a hylomorphism is the factorial function. So, we know that our domain and co-domain is , so now we can make a more specific diagram to represent our solution:
As you can see I’m going to use to represent my intermediate data, and I’ve already define the names of my gen functions to the catamorphism and to the anamorphism. Another time, that I do all the work with the anamorphism, letting the catamorphism with little things to do (just multiply). I’m start to show you the catamorphism first:
mul = either (const 1) mul' where mul' = uncurry (*)
As you can see the only thing it does is multiply all the elements of a list, and multiply by 1 when reach the empty list.
In the other side, the anamorphism is generating a list of all the elements, starting in (the element we want to calculate the factorial) until 1.
nats = (id -|- (split succ id)) . outNat
And finally we combine this together with our hylo, that defines the factorial function:
fac = hylo mul nats
Here is all the code you need to run this example:
inl = either (const ) (uncurry (:)) out  = i1 () out (a:x) = i2(a,x) cata g = g . rec (cata g) . out ana h = inl . (rec (ana h) ) . h hylo g h = cata g . ana h rec f = id -|- id >< f
Here, I’m going to show you the hanoi problem solved with one hylomorphism, first let’s take a look at the structure:
data BTree a = Empty | Node(a, (BTree a, BTree a))
So, our generic diagram representing one hylomorphism over is:
There is a well-known inductive solution to the problem given by the pseudocode below. In this solution we make use of the fact that the given problem is symmetrical with respect to all three poles. Thus it is undesirable to name the individual poles. Instead we visualize the poles as being arranged in a circle; the problem is to move the tower of disks from one pole to the next pole in a speciﬁed direction around the circle. The code deﬁnes to be a sequence of pairs where n is the number of disks, is a disk number and are directions. Disks are numbered from onwards, disk being the smallest. Directions are boolean values, representing a clockwise movement and an anti-clockwise movement. The pair means move the disk numbered from its current position in the direction .
excerpt from R. Backhouse, M. Fokkinga / Information Processing Letters 77 (2001) 71–76
So, here, I will have a diagram like that, type stands for and type for :
I’m going to show all the solution here, because the description of the problem is in this quote, and in the paper:
hanoi = hyloBTree f h f = either (const ) join where join(x,(l,r))=l++[x]++r h(d,0) = Left () h(d,n+1) = Right ((n,d),((not d,n),(not d,n)))
And here it is, all the code you need to run this example:
inBTree :: Either () (b,(BTree b,BTree b)) -> BTree b inBTree = either (const Empty) Node outBTree :: BTree a -> Either () (a,(BTree a,BTree a)) outBTree Empty = Left () outBTree (Node (a,(t1,t2))) = Right(a,(t1,t2)) baseBTree f g = id -|- (f >< g)) cataBTree g = g . (recBTree (cataBTree g)) . outBTree anaBTree g = inBTree . (recBTree (anaBTree g) ) . g hyloBTree h g = cataBTree h . anaBTree g recBTree f = baseBTree id f
Maybe in the future I will talk more about that subject.