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FunctionalWorks

More on Anamorphism: Unfolding More Than List in Haskell

Marty Stumpf

7 Apr 2022

7 April 2022

•

3 min read

More on Anamorphism: Unfolding More Than List in Haskell
  • Functional Programming
  • Haskell
  • Typed Functional Programming

This is the seventh of a series of articles that illustrates functional programming (FP) concepts. As you go through these articles with code examples in Haskell (a very popular FP language), you gain the grounding for picking up any FP language quickly. Why should you care about FP? See this post.

In the last post, I mentioned that Haskell's unfold only works for lists. But why can't it be like fold which can work on a whole type class? Recall that fold works for any Foldable type.

In this post, I will show you an unfold function that can work on more than list. I first explain the Monoid type class, then I show a more general unfold than Haskell's default unfold. In the next post, I will show examples of a few things you can do with this unfold function that you can't do with Haskell's default unfold.

A Type Class for Unfold

Like the Foldable type class for fold, the type class for unfold needs to fulfill some requirements such that the structure can be unfolded. What requirements do we need? Let's revisit the formal definition of unfold for list:

Given a predicate and a function f which returns a pair (a, b’). I.e., f b = (a, b’). A list anamorphism unfold is: When the predicate is true, unfold b = [], otherwise, unfold b = a : unfold b’.

Notice that we need to be able to perform two things depending on whether the predicate is true or not:

  1. Return an empty list. This implies that we need a structure for which empty is defined.
  2. Return a result which cons (:) or appends one value to another.

That's it! We can unfold any structure that fulfills these two requirements! It so happens that Monoid describes exactly such a structure!

The Monoid Type Class

We talked about typeclasses in an earlier post - readers who are not familiar with them may want to review it before proceeding.

Monoid is not to be confused with Monad. However, they are related, as depicted from the below graph from Typeclassopedia:

  • Solid arrows point from the general to the specific; that is, if there is an arrow from Foo to Bar it means that every Bar is (or should be, or can be made into) a Foo.
  • Dotted lines indicate some other sort of relationship.

We can see that Monoid and Foldable are related too! We'll cover more of these type classes in later posts. For now, we'll focus on Semigroup and Monoid.

Semigroup

Semigroup is a superclass of Monoid, i.e., anything that belongs to the Monoid type class also belongs to the Semigroup type class. This is because the requirement for a type to be a Semigroup is also a requirement for the Monoid type class!

Looking at the [reference](https://hackage.haskell.org/package/base-4.15.0.0/docs/Prelude.html# g:9) for semigroup, you will see that there's only one method:

(<>) :: a -> a -> a

The function (<>) has to satisfy associativity. I.e., the following has to hold for <>:

 x <> (y <> z) = (x <> y) <> z

And that's it! For example, list is an instance of semigroup because the append function ((++)) is associative:

(++) :: [a] -> [a] -> [a]

Because we know that:

[1,2] ++ ([3,4] ++ [5,6]) 
= [1,2,3,4,5,6]
= ([1,2] ++ [3,4]) ++ [5,6]

Because list's (++) satisfies the requirement for semigroup, it is an instance of semigroup.

If a type is an instance of the semigroup we know that we can **append a value to another**, which is what we need for unfolds!

Monoid

We still need the requirement that **empty is defined**. Monoid has such requirement on top of the semigroup requirement. From the [reference](https://hackage.haskell.org/package/base-4.15.0.0/docs/Data-Monoid.html# g:1), Monoid has the following methods:

mempty :: a

mappend :: a -> a -> a

The requirement for mappend is exactly as the (<>) method of Semigroup. That's why the reference mentions that this method is redundant and has the default implementation mappend = (<>).

These are the requirements for mempty:

Right identity
    x <> mempty = x
Left identity
    mempty <> x = x

That is, it has to be the case that when you append anything to mempty from the left or right, you get the same thing. This is exactly what we need for unfolds.

For list, the empty list fullfills the requirements for mempty because we know that [] <> [a] = [a] and [a] <> [] = [a]. List is therefore an instance of the Monoid type class.

A More General Unfold

Now we are ready to write our unfold by requiring the Monoid type class:

unfoldm :: Monoid m => (t -> Maybe (m, t)) -> t -> m 
unfoldm f a = 
  case f a of 
    Just (first, second) ->
      first `mappend` (unfoldm f second)
    Nothing -> mempty

We implement the same function as the list version but replace the empty list with mempty and the (:) with mappend. Doing so lets us generalize the function to work for all types that belong to the Monoid type class.

In the next post, I will go into detail about how we use this, along with some examples to illustrate how powerful this is. Stay tuned!

Did you like this article?

Marty Stumpf

Software engineer. Loves FP Haskell Coq Agda PLT. Always learning. Prior: Economist. Vegan, WOC in solidarity with POC.

See other articles by Marty

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