Welcome to this little explanation on how to determine the fold of a Haskell
datatype. First we’ll look at how we define functions over lists, something
everyone starting with Haskell should be sufficiently familiar with, after which
we move on to the datatypes. You’ll see different ways how to calculate the sum
of a list, how to fold over a list, what datatypes are and, how to fold over a
datatype, specifically the
BinTree a datatype. Most importantly, I hope you
will grow to understand what a fold is, and why they are so important and useful
when programming Haskell.
During our functional programming course at Utrecht University I noticed students having anxiety of datatypes and even more for folds. Not because they couldn’t grasp the workings of a single example but more a lack of a view on the complete picture and lack of experience with the Haskell datatype way.
So, what are datatypes? Datatypes are a way of notating the abstract structure of your data. There are several datastructures known to man, such as lists and trees.
ListsLists in Haskell are used in several ways. Today we will look at how to calculate the sum of a list. Intuitively you calculate the sum of a list by adding its elements together. Starting with the first element and then continuing on to the rest. This is how we literally translate that thought into Haskell code: sum :: [Int] -> Int sum  = 0 sum (x:xs) = x + sum xs -- sum [1..4] = 10
Would we be calculating the product of the list, we’d do the same except we multiply instead of adding.product :: [Int] -> Int product  = 1 product (x:xs) = x * product xs -- product [1..4] = 24
Looking at these two examples we can see that we have two similarities. We always have recursion on the tail of the list and we do something with the head of the list and the result of the recursion. In the first example we add them together, and in product we multiply them. Another property of most functions over lists is that there is a base case for the empty list. We call this the identity of our function, sometimes also referred to as unit. The identity of addition is 0 and the identity of multiplication is 1.
Now we look at one of the folds over lists defined in the prelude,
foldr. Make sure you know what each parameter stands for.
Note: There are some downsides to this function, mostly that it
will not work for large lists, you’ll see more on this later.
Now, take a moment to let this function soak in and try to think of how you
product in terms of this
foldr. Crucial at this point is to notice that we do not see any
type hardcoded in the type of
foldr It may be helpful to look at
this fold, which is the identity function for lists:
If we look at our sum function, the operator between each recursive call is
(+) and our base case is 0. So that’s what we are going to use for
our sum in terms of foldr.
Thus far we only have lists of
Int for our examples. However, for
sake of usability we will now move on to lists of numeric elements. Because
(+) is defined for all numbers. (If you wish to read more on this
subject please look up classes and instances.) Notice how the type of sumList
changes while it’s definition remains unaltered.
Now that we have seen how we can determine the function and identity for our
fold from our recursive function to a definition in terms of
And most importantly, the foldr takes care of the recursive nature of the list
for us and the only thing it asks us in return for that is an operator and an
identity for your operation. So the fold can now be used to define several
operations on lists.
I told you about a problem of
foldr. Depending on the size of your
list the above
sumList may not work. Try the following on your
Now try:$ ghci Prelude> let sumList = Data.List.foldl' (+) 0 Prelude> sumList [1..1000000] 500000500000
For a complete overview and analysis on
foldl please read A tutorial on the universality and
expressiveness of fold, by Graham Hutton.
TreesFirst, let’s take a look at how data structures in Haskell can be defined. In short we have the
datakeyword, followed by zero or more
type variables, a
=and then a number of
data constructorsseparated by a
Now that you have familiarized yourself with lists we can proceed to a slightly more complicated datastructure. The Tree. In this example we will use a simple binary tree. A binary tree can be denoted as follows:data BinTree a = Node a (BinTree a) (BinTree a) | Leaf deriving (Show, Eq)
In this case
BinTree is the type constructor and
Leaf the data data constructors.
Remember that the goal of folds is to separate the implementation of the recursion from the actual operation we want to execute on the datatype. So we have one function, the fold, that takes care of the recursion and several other functions that use this fold to specify certain semantics on the datatype. We are going to calculate the sum of all elements in this tree.
The above binary tree has elements in the nodes and nothing in the leaves. You
can notice the recursive occurrence of `
BinTree a’ in the node. We
see two constructors in this datatype,
The Node constructor expects some value of type
a and two subtrees
BinTree a, and the
Leaf constructor has no
parameters. In Haskell:
Notice that our list also has two constructors, namely the
constructor that adds an element and the constructor for the empty list,
. Analogously we have the
constructors. If we were allowed to use the
constructor in our own Haskell code, the datatype for a list would look like
Also, it is customary to keep the arguments for the functions in the same order as the constructors are defined in the datatype, and to name the identifiers containing the functions the same as the constructor function but in lowercase. Applying this we get:foldBinTree :: (a -> BinTree a -> BinTree a -> BinTree a) -> (BinTree a) -> BinTree a -> ?? foldBinTree node leaf = f where f (Node x left right) = node x (f left) (f right) f (Leaf) = leaf
I left out the result type of this fold, try to find it yourself before continuing.
By following our code we can see that every case, the case for
and the one for
Leaf, result in a
Consequently the complete result of the function is a
Now recall that we previously used a fold to calculate the sum of a list. With
that fold we were not restricted to a list. Which is also clearly visible by
looking at the type of
foldr it contains only type variables. As we
want to calculate sum of the elements in this tree, which is something of type
a and not of type
BinTree a we have to revise the type
of our fold. Notice the recurrence of the datatype in it’s declaration,
BinTree a. We will replace all of these occurrences in the types of
our functions with a free type variable, say
Keep in mind that I used ‘eta reduction’ in the above code. This means that the last parameter (in this case the tree) isn’t explicitly specified as it would appear at the very end of the parameter list and ath the very end of the definition. (More on eta reduction.) So, now let’s do something with this fold. Suppose we have the following tree:bintree = Node 1 (Node 2 (Node 3 Leaf Leaf) (Node 3 Leaf Leaf) ) (Node 2 (Node 3 Leaf Leaf) (Node 3 Leaf Leaf) )
A visual representation:
1 /Now we can define several traversals over this tree. Let’s calculate the sum of all values in the tree. sumBinTree :: Num a => BinTree a -> a sumBinTree = foldBinTree (\val res_left res_right -> val + res_left + res_right) 0
2 2 / \ /
3 3 3 3
AlgebrasThe type of our foldBinTree is now more compact. However, there will be datatypes that contain a larger sum of constructors that also may have more or less parameters than in our case. Defining the type of the fold on those datatypes as we have done before will inevitably lead to very clumsy type signatures. The idea is that we split up the section that denotes our functions into a separate type, namely the algebra.
So now we define an algebra for our data structure. To do this we again look at each data constructor and determine it’s type. The fold function takes care of the recursion, so applying this thought consequently gives us the following types and fold.
Keep in mind that we have to change the type of our fold, but not the definition of the fold itself!type BinTreeAlgebra a r = (a -> r -> r -> r, -- Node r -- Leaf ) foldBinTree :: BinTreeAlgebra a r -> BinTree a -> r
Now we can again define a sum on all
Num a trees. Notice that we
went from to seperate parameters for the functions to one tuple with two
That’s it for now. If you did not understand everything by the end of this article, don’t panic. It will sink in eventually. Let this rest a day or two, and then read this article again and you’ll understand folds better. :)