More Hylomorphisms in Haskell

9 04 2009

If you lost yourself in this post, I advise you to start in catamorphisms, then anamorphisms and then hylomorphisms.

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 C or D. Because we glue the ana and cata together into a single recursive pattern. A and E 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.

Leaf Tree’s

The data type that we going to discuss here is the LTree. In Haskell we can represent LTree 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 Ltree 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 [(h)]. Before that I’m going to specify the strategy to define factorial. I’m going to use the diagram’s again, remember that type 1 is equivalent to Haskell ( ):

As you can see I’m going to use Ltree~1 as my intermediate structure, and I’ve already define the names of my gen functions add to the catamorphism and fibd to the anamorphism. The strategy I prefer, is do all the hard work in the anamorphism, so here the gen fibd 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 n leaf’s, being n the result of that fib.

Then we just have to write the gen add 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 Integers, so now we can make a more specific diagram to represent our solution:

As you can see I’m going to use [Integer] to represent my intermediate data, and I’ve already define the names of my gen functions mul to the catamorphism and nats 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 n (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

Binary Tree’s

Here, I’m going to show you the hanoi problem solved with one hylomorphism, first let’s take a look at the Btree structure:

data BTree a = Empty | Node(a, (BTree a, BTree a))

So, our generic diagram representing one hylomorphism over BTree 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 H_n.d to be a sequence of pairs (k, d) where n is the number of disks, k is a disk number and d are directions. Disks are numbered from 0 onwards, disk 0 being the smallest. Directions are boolean values, true representing a clockwise movement and false an anti-clockwise movement. The pair (k, d) means move the disk numbered k from its current position in the direction d.

excerpt from R. Backhouse, M. Fokkinga / Information Processing Letters 77 (2001) 71–76

So, here, I will have a diagram like that, b type stands for Bool and i type for Integer:

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.


Hylomorphisms in Haskell

9 04 2009

If you miss something in this post, I suggest you to start in Catamorphisms and Anamorphisms.

A Hylomorphism is just the composition of one catamorphism and then one anamorphism.
hylo~f~h~=~cata~f~\circ~ana~h, replacing that by the proper notation we have: [|f,h|]~=~(|f|)~\circ~[(h)]

In this post I will use the structure of a binary tree:

data BTree a = Empty | Node(a, (BTree a, BTree a))

I will use the tuples to don’t have to write uncurry’s. As I will show you, when we say that we are making a hylomorphism on a particular data type T, what we are trying to say is that the intermediate structure of our combination of catamorphism and anamorphism is that data type T. This is the structure throw our morphism will communicate with each other.


So, here I will solve the Quicksort algorithm with a hylomorphism over BTree.

The intermediate structure being a BTree doesn’t mean that my function will receive BTree. My qSort function works over lists. So the first thing to do, is draw the respective anamorphism from [a] to BTree~a:

My strategy here is to do all the work in the anamorphism, so, I need a function h with type:
h : [a] \rightarrow 1 + a \times [a] \times [a], or in Haskell h :: [a] \rightarrow Either () (a, ([a], [a]))

That function is qsep:

qsep :: [a] -> Either () (a, ([a], [a]))
qsep []    = Left ()
qsep (h:t) = Right (h,(s,l))
    where (s,l) = part (<h) t

part:: (a -> Bool) -> [a] -> ([a], [a])
part p []                = ([],[])
part p (h:t) | p h       = let (s,l) = part p t in (h:s,l)
             | otherwise = let (s,l) = part p t in (s,h:l)

This code is very simple, in qsep I chose a pivotal element (first one), and filter the bigger to one side, and the other ones to the other, just like the algorithm. The function that do all that job is part, it process all the list finding the elements that satisfy the condition p, to put them in the left side of the tuple, and the others into the right side.

This function by it self don’t do almost anything, it is only a simple part of the algorithm.


Next step is to see the diagram for catamorphisms from BTree~a to [a]:

As I said before, the heavy duty is on the side of the anamorphism, so here, the catamorphism will be very very simple. In fact it is.

inord :: Either a (a, ([a], [a])) -> [a]
inord = either (const []) join
    where join(x,(l,r))=l++[x]++r

That right! The only thing that the catamorphism do is a inorder passage over the structures a + a \times [a] \times [a], which is very simple, as as shown by the code.


The first thing is to draw the diagram, now for the hylomorphism, the composition of the cata with the ana:

Once having made the two most important parts of the function (the ana and cata), the hylo is very simple to do. You just have to make a function hyloBTree:

hyloBTree h g = cataBTree h . anaBTree g

And our function qSort bacame:

qSort :: Ord a => [a] -> [a]
qSort = hyloBTree inord qsep

And that’s it, now I’m going to show you the all code that you need to put all the things together and working.

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


If you need more explanations feel free to contact me.

Anamorphisms in Haskell

8 04 2009

First I would like to introduce the notation that I use here. The pointfree notation is good to see a program (functions) data flow and as composition of functions, combination of functions, if you prefer. This style is characterized by not using variables in declaration of functions. Haskell allow us to implement that notation natively. The dual of the pointfree notation is the pointwise one.

A simple example of a function in pointwise style:

f n = (n+2)*10 -- pointwise

The dual in pointfree would be:

f = (*10) . (+2) -- pointfree


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

, or .

I will assume that you are familiarized with infix notation, either, and composition (\circ) functions.


For this post I need to explain the data type we will going to use. In Haskell we define it by:

data Tree a = Node a [Tree a]

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

data Tree a = Node (a, [Tree a])

We could see Node as a the following function:

Node :: (a, [Tree a]) -> Tree a

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

Now we can say that (Tree~a) is isomorphic to (a \times~[Tree~a]).
This is something to say that (Tree~a) and (a \times~[Tree~a]) keep the same information without any change. We represent that formally as: (Tree~a) \cong~(a \times~[Tree~a]).


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

ana(h_{Tree}) is the anamorphism of h if the diagram commute.

We use the notation rec_{Tree} to say that function rec in not generic, but only works for data Tree. The same happens with in and ana. We will write ana(h)_{Tree} using the composition of in, ana and rec functions. That way we are breaking our problem in small ones. So, in the end we will have the following definition for ana(h)_{Tree}:

ana(h)_{Tree} = in_{Tree} \circ rec_{Tree} \circ h

The function that we want is ana(h), and that function is over (Tree~a) so we have:

ana :: (A -> B) -> A -> Tree c

Type C is (Tree~c). Maybe this isn’t clear yet, let’s start with function in

function in

The function in_{Tree} is responsible to create the isomorphism between (Tree~a) and (a \times~[Tree~a]), so the code could be something like this:

inTree :: Tree a -> (a, [Tree a])
inTree    = Node

In Haskell we represent the type (\times) as (,). So, type D is (a \times~[Tree~a]). So by now, we already know the following unifications C \sim Tree~c and D \sim c \times~[Tree~c]. So now our graphic is:

function h

The function h is also known as *gen*, here is where we said the step that pattern do. This is the only function we need to take care, if this function is good, our problem is solved. Now image that our problem is:

Suppose that the pair of positive integers (v, p) denotes the number of red balls (v) and black (p) that is inside a bag, the balls which are taking randomly, successively, until the bag is empty.

This is the point-wise version of the function we want to convert to pointfree using anamorphisms. This function represent as a tree, all possible states of the bag over these experiences.

state :: (Int,Int) -> Tree (Int,Int)
state(0,0) = Node (0,0) []
state(v,0) = Node (v,0) [state(v-1,0)]
state(0,p) = Node (0,p) [state(0,p-1)]
state(v,p) = Node (v,p) [state(v-1,p),state(v,p-1)]

If we want that “latex state$ became an anamorphism, we have to say that our type A unify (\sim) with Int \times~Int, and Tree~c became more restrict, and unify with Tree (Int \times~Int). A consequence of changing the co-domain of in_{Tree} is changing the domain of it to (Int \times~Int) \times~[Tree (Int \times~Int)]. We represent ana(h) as [( h )]. Now we can be more specific with our graphic:

function rec

Here we have to get a function rec that co-domain is (Int \times~Int) \times~[Tree~(Int \times~Int)]. Probably the best is to pass the first part of the tuple (part with type (Int \times~Int)) and the rest (part with type [Tree~(Int \times~Int)]) is just a map of the function [(h)]_{Tree}. So, now our graphic is:

As you can see, the second part of the co-domain of h is the type of function map~[(h)]_{Tree}:

map~[(h)]_{Tree}~:~[(Int \times~Int)] \rightarrow~[Tree(Int \times~Int)]

So our final graphic became:

Now, we just have to define the function h and apply them to our anamorphism of Tree.

h :: (Int, Int) -> ( (Int, Int), [ (Int, Int) ] )
h(0,0) = ( (0,0), [] )
h(v,0) = ( (v,0), [ (v-1,0) ] )
h(0,p) = ( (0,p) [ (0,p-1) ] )
h(v,p) = ( (v,p), [ (v-1,p), (v,p-1) ] )

And this is it! Now we can say that:
state \equiv~ana_{Tree} where ana(h)_{Tree} =  in_{Tree} \circ~id~><~map~ana(h)_{Tree} \circ h


Here is all the code you need to run this example in Haskell:

module AnamorphismExample where

infix 5 ><

i1 = Left
i2 = Right
p1 = fst
p2 = snd

data Tree a = Node (a, [Tree a]) deriving Show

split :: (a -> b) -> (a -> c) -> a -> (b,c)
split f g x = (f x, g x)

(><) :: (a -> b) -> (c -> d) -> (a,c) -> (b,d)
f >< g = split (f . p1) (g . p2)

inTree :: (a, [Tree a]) -> Tree a
inTree = Node

anaTree h = inTree . (id >< map (anaTree h)) . h

-- our function
h_gen :: (Int, Int) -> ( (Int, Int), [ (Int, Int) ] )
h_gen(0,0) = ( (0,0), [] )
h_gen(v,0) = ( (v,0), [ (v-1,0) ] )
h_gen(0,p) = ( (0,p) , [ (0,p-1) ] )
h_gen(v,p) = ( (v,p), [ (v-1,p), (v,p-1) ] )

state = anaTree h_gen

Pass a year since I promised this post. The next will be on hylomorphisms I promise not take too that much.

Cryptol the language of cryptography

1 04 2009

Pedro Pereira and I are working on a new project in the Masters. The second half of the Masters is composed of a single project suggested by a company. Some companies are forming partnerships in the Masters formal methods, including: the Critical software, SIG and Galois. We chose the Galois because we also are in the area of cryptography and we already knew some work of some people from this company.

The project suggested by Galois was study the Cryptol as a language of specification of cryptographic algorithms. The cipher we used for this study is the SNOW 3G (The SNOW website), later on I will talk about the specification of this cipher. In this post I am only interested to show the language.

I’m going to show you some details about the language. This post is not intend to be a exhaustive explanation of Cryptol, if you looking for that you can go directly to the manuals. This post only relates my experience, and what I like it most with the language.


Cryptol is a high-level language that is geared to deal with low-level problems. Is a Domain-specific language to design and implement cryptographic algorithms.
This language has a high percentage of correctness of the implementation of a cipher, because it implements type inference, so we can say that a big part of the language implements correctness. This correctness is also achieved thanks to the architecture of the language – functional. We don’t have side effects – a function only return something inside is codomain.
In Cryptol we have this philosophy that says that everything is a sequence. This is very useful because we are working with low level data (array of bits), so we use sequences to represent that arrays. We can have nested sequences to have a more structured representation of data. For example, we can simply transform a 32-bit sequence in a 4 1-byte sequence.
The size of this sequences could be implemented as finite or infinite, as we going to see later in this post. Because Cryptol is a high-level language we can also implement polymorphic functions, most of the primitive functions are implemented in polymorphic mode. The way we have to navigate throw the sequences is using recursion, or sequences comprehension, and with these two techniques we can implement recurrences.

If you are a Haskell programmer you just need the next section to learn Cryptol. This language is so look a like with Haskell that even the philosophy seems to have a lot in commune.

Types in Cryptol

The type [32] means that you have a sequence of 32-bit size. All the types in Cryptol are size oriented. The unit is the Bit, that you can use to represent Bool. To represent a infinite sequence we use the reserved word inf, and we write: [inf] to represent that.

If you want to generate a infinite sequence, we use the syntactic sugar of the sequences like that: [1~..]. Cryptol will infer this sequence as type


That means this sequence have infinite positions of 1-bit words. The type inference mechanism will always optimize the size that he needs, to represent the information.
So, it infer the type of [100~..] as:


Because, it “knows” that needs only 7-bits to represent the decimal 100. But if you need more, you can force the type of your function.
We implement polymorphism in our types, if we have:


This means, that the function f have polymorphism over b, because we say that it domain is one sequence of size a of type b, and it codomain also. Here we could also see: f~:~[a][b]c meaning that f is a constant of sequences of size b of type c, a times.

So, lets talk about some primitive functions in Cryptol, and its types. The tail function have the following type in Cryptol:


As we can see, Cryptol is so size oriented, that we can use arithmetic operators in types. We can probably infer what this function does just from it type: tail works for all a and b such that if we have one sequence os size a+1 of type b it returns one sequence of size a of same type. In fact this function removes the first element of one sequence.

Because of this size oriented philosophy a lot of functions, that change the size of the sequences can be read just from the type.

As you can see in the following list of Cryptol primitive function:

drop~:~\{ a~b~c \}~( fin~a ,~a~\geq~0)~\Rightarrow~(a ,[ a + b ]~c )~\rightarrow~[ b ]~c
take~:~\{ a~b~c \}~( fin~a ,~b~\geq~0)~\Rightarrow~(a ,[ a + b ]~c )~\rightarrow~[ a ]~c
join~:~\{ a~b~c \}~[ a ][ b ] c~\rightarrow~[ a * b ]~c
split~:~\{ a~b~c \}~[ a * b ] c~\rightarrow~[ a ][ b ]~c
tail~:~\{ a~b \}~[ a +1] b~\rightarrow~[ a ]~b

Recursion and Recurrence

Cryptol implements Recursion, just like a lot of functional languages do.

Imagine the fibonacci function definition:

It implementation in Crytol is exactly the same as defined mathematically.

fib : [inf]32 -> [inf]32;
fib n = if n == 0 then 0 else if n == 1 then 1 else fib (n-1) + fib (n-2);

Cryptol uses recursion to permit us to iterate throw sequences.

But, If you prefer you can implement a more functional algorithm of fibonacci function in Cryptol:

fib : [inf]32 -> [inf]32;
fib n = fibs @ n;
   where {
      fibs : [inf]32;
      fibs = [0 1] # [| x + y || x <- drop (1,fibs) || y <- fibs |];

Here, as you can see, we define a infinite list fibs of all the fibonacci numbers, by calling the fibs inside the sequences comprehension fibs, this is called a recurrence, and you can use that too in Cryptol.

Cryptol vs C

I’m going to show you some part of the implementation of SNOW 3G in C. This is a function called MUL_{\alpha}

MULa : [8] -> [32];
MULa(c) = join ( reverse [
   ( MULxPOW(c, 23 :[32], 0xA9) )
   ( MULxPOW(c, 245:[32], 0xA9) )
   ( MULxPOW(c, 48 :[32], 0xA9) )
   ( MULxPOW(c, 239:[32], 0xA9) ) ] );
/* The function MUL alpha.
    Input c: 8-bit input.
    Output : 32-bit output.
    See section 3.4.2 for details.
u32 MULalpha(u8 c) {
  ((((u32)MULxPOW(c,23, 0xa9)) << 24 ) |
  (((u32)MULxPOW(c, 245,0xa9)) << 16 ) |
  (((u32)MULxPOW(c, 48,0xa9)) << 8 ) |
  (((u32)MULxPOW(c, 239,0xa9)))) ;

You can see that in Cryptol we just say that we want to work with a 32-bit word, and we don’t need to do any shift to our parts of the word. We just join them together. We reverse the sequence, because Cryptol stores words in little-endian, and we want to keep the definition like the specification.

This is a very simple function, so the result in C is not so that different. But if we have a more complex function, we were going to start having a nightmare to write that in C.


Well, the conclusion is that Cryptol is a language that really help to write low-level algorithms. With Cryptol the specification is formal and easier to read than other languages. A value of Cryptol is that the code can be converted to other languages, such as VHDL and C.

If you’re interested, take a look at the presentation that we did.