Programming

What does Gödel’s incompletness theorem, Russell’s paradox, Turing’s halting problem, and Cantor’s diagonal argument have to do with the fact that negation has no fixed point? The surprising answer is that they are all related through
Lawvere’s fixed point theorem.

Before we dive deep into category theory, let’s unwrap this statement from the point of view of a (Haskell) programmer. Let’s start with some basics. Negation is a function that flips between True and False:

  not :: Bool -> Bool
not True  = False
not False = True


A fixed point is a value that doesn’t change under the action of a function. Obviously, negation has no fixed point. There are other functions with the same signature that have fixed points. For instance, the constant function:

  true True  = True
true False = True


has True as a fixed point.

All the theorems I listed in the preamble (and a few more) are based on a simple but extremely powerful proof technique invented by Georg Cantor called the diagonal argument. I’ll illustrate this technique first by showing that the set of binary sequences is not countable.

Cantor’s job interview question

A binary sequence is an infinite stream of zeros and ones, which we can represent as Booleans. Here’s the definition a sequence (it’s just like a list, but without the nil constructor), together with two helper functions:

  data Seq a = Cons a (Seq a)
deriving Functor

head :: Seq a -> a
head (Cons a as) = a

tail :: Seq a -> Seq a
tail (Cons a as) = as


And here’s the definition of a binary sequence:

  type BinSeq = Seq Bool


If such sequences were countable, it would mean that you could organize them all into one big (infinite) list. In other words we could implement a sequence generator that generates every possible binary sequence:

  allBinSeq :: Seq BinSeq


Suppose that you gave this problem as a job interview question, and the job candidate came up with an implementation. How would you test it? I’m assuming that you have at your disposal a procedure that can (presumably in infinite time) search and compare infinite sequences. You could throw at it some obvious sequences, like all True, all False, alternating True and False, and a few others that came to your mind.

What Cantor did is to use the candidate’s own contraption to produce a counter-example. First, he extracted the diagonal from the sequence of sequences. This is the code he wrote:

  diag :: Seq (Seq a) -> Seq a

trim :: Seq (Seq a) -> Seq (Seq a)
trim seqs = fmap tail (tail seqs)


You can visualize the sequence of sequences as a two-dimensional table that extends to infinity towards the right and towards the bottom.

  T F F T ...
T F F T ...
F F T T ...
F F F T ...
...


Its diagonal is the sequence that starts with the fist element of the first sequence, followed by the second element of the second sequence, third element of the third sequence, and so on. In our case, it would be a sequence T F T T ....

It’s possible that the sequence, diag allBinSeq has already been listed in allBinSeq. But Cantor came up with a devilish trick: he negated the whole diagonal sequence:

  tricky = fmap not (diag allBinSeq)


and ran his test on the candidate’s solution. In our case, we would get F T F F ... The tricky sequence was obviously not equal to the first sequence because it differed from it in the first position. It was not the second, because it differed (at least) in the second position. Not the third either, because it was different in the third position. You get the gist…

Power sets are big

In reality, Cantor did not screen job applicants and he didn’t even program in Haskell. He used his argument to prove that real numbers cannot be enumerated.

But first let’s see how one could use Cantor’s diagonal argument to show that the set of subsets of natural numbers is not enumerable. Any subset of natural numbers can be represented by a sequence of Booleans, where True means a given number is in the subset, and False that it isn’t. Alternatively, you can think of such a sequence as a function:

  type Chi = Nat -> Bool


called a characteristic function. In fact characteristic functions can be used to define subsets of any set:

  type Chi a = a -> Bool


In particular, we could have defined binary sequences as characteristic functions on naturals:

  type BinSeq = Chi Nat


As programmers we often use this kind of equivalence between functions and data, especially in lazy languages.

The set of all subsets of a given set is called a power set. We have already shown that the power set of natural numbers is not enumerable, that is, there is no function:

  enumerate :: Nat -> Chi Nat


that would cover all characteristic functions. A function that covers its codomain is called surjective. So there is no surjection from natural numbers to all sequences of natural numbers.

In fact Cantor was able to prove a more general theorem: for any set, the power set is always larger than the original set. Let’s reformulate this. There is no surjection from the set $A$ to the set of characteristic functions $A \to 2$ (where $2$ stands for the two-element set of Booleans).

To prove this, let’s be contrarian and assume that there is a surjection:

  enumP :: A -> Chi A


Since we are going to use the diagonal argument, it makes sense to uncurry this function, so it looks more like a table:

  g :: (A, A) -> Bool
g = uncurry enumP


Diagonal entries in the table are indexed using the following function:

  delta :: a -> (a, a)
delta x = (x, x)


We can now define our custom characteristic function by picking diagonal elements and negating them, as we did in the case of natural numbers:

  tricky :: Chi A
tricky = not . g . delta


If enumP is indeed a surjection, then it must produce our function tricky for some value of x :: A. In other words, there exists an x such that tricky is equal to enumP x.
This is an equality of functions, so let’s evaluate both functions at x (which will evaluate g at the diagonal).

  tricky x == (enumP x) x


The left hand side is equal to:

  tricky x = {- definition of tricky -}
not (g (delta x)) = {- definition of g -}
not (uncurry enumP (delta x)) = {- uncurry and delta -}
not ((enumP x) x)


So our assumption that there exists a surjection A -> Chi A led to a contradition!

  not ((enumP x) x) == (enumP x) x


Real numbers are uncountable

You can kind of see how the diagonal argument could be used to show that real numbers are uncountable. Let’s just concentrate on reals that are greater than zero but less than one. Those numbers can be represented as sequences of decimal digits (the digits following the decimal point). If these reals were countable, we could list them one after another, just like we attempted to list all streams of Booleans. We would get some kind of a table infinitely extending to the right and towards the bottom. There is one small complication though. Some numbers have two equivalent decimal representations. For instance $0.48$ is the same as $0.47999...$, with infinitely many nines. So lets remove all rows from our table that end in an infinity of nines. We get something like this:

  3 5 0 5 ...
9 9 0 8 ...
4 0 2 3 ...
0 0 9 9 ...
...


We can now apply our diagonal argument by picking one digit from every row. These digits form our diagonal number. In our example, it would be 3 9 2 9.

In the previous proof, we negated each element of the sequence to get a new sequence. Here we have to come up with some mapping that changes each digit to something else. For instance, we could add one to it, modulo nine. Except that, again, we don’t want to produce nines, because we could end up with a number that ends in an infinity of nines. But something like this will work just fine:

  h n = if n == 8
then 3
else (n + 1) mod 9


The important part is that our function h replaces every digit with a different digit. In other words, h doesn’t have a fixed point.

Lawvere’s fixed point theorem

And this is what Lawvere realized: The diagonal argument establishes the relationship between the existence of a surjection on the one hand, and the existence of a no-fix-point mapping on the other hand. So far it’s been easy to find a no-fix-point functions. But let’s reverse the argument: If there is a surjection $A \to (A \to Y)$ then every function $Y \to Y$ must have a fixed point. That’s because, if we could find a no-fixed-point function, we could use the diagonal argument to show that there is no surjection.

But wait a moment. Except for the trivial case of a function on a one-element set, it’s always possible to find a function that has no fixed point. Just return something else than the argument you’re called with. This is definitely true in $Set$, but when you go to other categories, you can’t just construct morphisms willy-nilly. Even in categories where morphisms are functions, you might have constraints, such as continuity or smoothness. For instance, every continuous function from a real segment to itself has a fixed point (Brouwer’s theorem).

As usual, translating from the language of sets and functions to the language of category theory takes some work. The tricky part is to generalize the definition of a fixed point and surjection.

Points and string diagrams

First, to define a fixed point, we need to be able to define a point. This is normally done by defining a morphism from the terminal object $1$, such a morphism is called a global element. In $Set$, the terminal object is a singleton set, and a morphism from a singleton set just picks an element of a set.

Since things will soon get complicated, I’d like to introduce string diagrams to better visualise things in a cartesian category. In string diagrams lines correspond to objects and dots correspond to morphisms. You read such diagrams bottom up. So a morphism

$\dot a \colon 1 \to A$

can be drawn as:

I will use dotted letters to denote “points” or morphisms originating in the unit. It is also customary to omit the unit from the picture. It turns out that everything works just fine with implicit units.

A fixed point of a morphism $t \colon Y \to Y$ is a global element $\dot y \colon 1 \to Y$ such that $t \circ \dot y = \dot y$. Here’s the traditional diagam:

And here’s the corresponding string diagram that encodes the commuting condition.

In string diagrams, morphisms are composed by stringing them along lines in the bottom up direction.

Surjections can be generalized in many ways. The one that works here is called surjection on points. A morphism $h \colon A \to B$ is surjective on points when for every point $\dot b$ of $B$ (that is a global element $\dot b \colon 1 \to B$) there is a point $\dot a$ of $A$ (the domain of $h$) that is mapped to $\dot b$. In other words $h \circ \dot a = \dot b$

Or string-diagrammatically, for every $\dot b$ there exists an $\dot a$ such that:

Currying and uncurrying

To formulate Lawvere’s theorem, we’ll replace $B$ with the exponential object $Y^A$, that is an object that represents the set of morphisms from $A$ to $Y$. Conceptually, those morphism will correspond to rows in our table (or characteristic functions, when $Y$ is $2$). The mapping:

$\bar g \colon A \to Y^A$

generates these rows. I will use barred symbols, like $\bar g$ for curried morphisms, that is morphisms to exponentials. The object $A$ serves as the source of indexes for enumerating the rows of the table (just like the natural numbers in the previous example). The same object also provides indexing within each row.

This is best seen after uncurrying $\bar g$ (we assume that we are in a cartesian closed category). The uncurried morphism, $g \colon A \times A \to Y$ uses a product $A \times A$ to index simultaneously into rows and columns of the table, just like pairs of natural numbers we used in the previous example.

The currying relationship between these two is given by the universal construction:

with the following commuting condition:

$g = \varepsilon \circ (\bar g \times id_A)$

Here, $\varepsilon$ is the evaluation natural transformation (the couinit of the currying adjunction, or the dollar sign operator in Haskell).

This commuting condition can be visualized as a string diagram. Notice that products of objects correspond to parallel lines going up. Morphisms that operate on products, like $\varepsilon$ or $g$, are drawn as dots that merge such lines.

We are interested in mappings that are point-surjective. In this case, we would like to demand that for every point $\dot f \colon 1 \to Y^A$ there is a point $\dot a \colon 1 \to A$ such that:

$\dot f = \bar g \circ \dot a$

or, diagrammatically, for every $\dot f$ there exists an $\dot a$ such that:

Conceptually, $\dot f$ is a point in $Y^A$, which represents some arbitrary function $A \to Y$. Surjectivity of $\bar g$ means that we can always find this function in our table by indexing into it with some $\dot a$.

This is a very general way of expressing what, in Haskell, would amount to: Every function f :: A -> Y can be obtained by partially applying our g_bar :: X -> A -> Y to some x :: X.

The diagonal

The way to index the diagonal of our table is to use the diagonal morphism $\Delta \colon A \to A \times A$. In a cartesian category, such a morphism always exists. It can be defined using the universal property of the product:

By combining this diagram with the diagram that defines the lifting of a pair of points $\dot a$ we arrive at a useful identity:

$\dot a \times \dot a = \Delta_A \circ \lambda_A \circ (id_A \times \dot a)$

where $\lambda_A$ is the left unitor, which asserts the isomorphism $1 \times A \to A$

Here’s the string diagram representation of this identity:

In string diagrams we ignore unitors (as well as associators). Now you see why I like string diagrams. They make things much simpler.

Lawvere’s fixed point theorem

Theorem. (Lawvere) In a cartesian closed category, if there exists a point-surjective morphism $\bar g \colon A \to Y^A$ then every endomorphism of $Y$ must have a fixed point.

Note: Lawvere actually used a weaker definition of point surjectivity.

The proof is just a generalization of the diagonal argument.

Suppose that there is an endomorphims $t \colon Y \to Y$ that has no fixed point. This generalizes the negation of the original example. We’ll create a special morphism by combining the diagonal entries of our table, and then “negating” them with $t$.

The table is described by (uncurried) $g$; and we access the diagonal through $\Delta_A$. So the tricky morphism $A \to Y$ is just:

$f = t \circ g \circ \Delta_A$

or, in diagramatic form:

Since we were guaranteed that our table $g$ is an exhaustive listing of all morphisms $A \to Y$, our new morphism $f$ must be somewhere there. But in order to search the table, we have to first convert $f$ to a point in the exponential object $Y^A$.

There is a one-to-one correspondence between points $\dot f \colon 1 \to Y^A$ and morphisms $f \colon A \to Y$ given by the universal property of the exponential (noting that $1 \times A$ is isomorphic to $A$ through the left unitor, $\lambda_A \colon 1 \times A \to A$).

In other words, $\dot f$ is the curried form of $f \circ \lambda_A$, and we have the following commuting codition:

$f \circ \lambda_A = \varepsilon \circ (\dot f \times id_A)$

Since $\lambda$ is an isomorphism, we can invert it, and get the following formula for $f$ in terms of $\dot f$:

$f = \varepsilon \circ (\dot f \times id_A) \circ \lambda_A^{-1}$

In the corresponding string diagram we omit the unitor altogether.

Now we can use our assumption that $\bar g$ is point surjective to deduce that there must be a point $\dot x \colon 1 \to A$ that will produce $\dot f$, in other words:

$\dot f = \bar g \circ \dot x$

So $\dot x$ picks the row in which we find our tricky morphism. What we are interested in is “evaluating” this morphism at $\dot x$. That will be our paradoxical diagonal entry. By construction, it should be equal to the corresponding point of $f$, because this row is point-by-point equal to $f$; after all, we have just found it by searching for $f$! On the other hand, it should be different, because we’ve build $f$ by “negating” diagonal entries of our table using $t$. Something has to give and, since we insist on surjectivity, we conclude that $t$ is not doing its job of “negating.” It must have a fixed point at $\dot x$.

Let’s work out the details.

First, let’s apply the function we’ve found in the table at row $\dot x$ to $\dot x$ itself. Except that what we’ve found is a point in $Y^A$. Fortunately we have figured out earlier how to get $f$ from $\dot f$. We apply the result to $\dot x$:

$f \circ \dot x = \varepsilon \circ (\dot f \times id_A) \circ \lambda_A^{-1} \circ \dot x$

Plugging into it the entry $\dot f$ that we have found in the table, we get:

$f \circ \dot x = \varepsilon \circ ((\bar g \circ \dot x) \times id_A) \circ \lambda_A^{-1} \circ \dot x$

Here’s the corresponding string diagram:

We can now uncurry $\bar g$

And replace a pair of $\dot x$ with a $\Delta$:

Compare this with the defining equation for $f$, as applied to $\dot x$:

$f \circ \dot x = t \circ g \circ \Delta_A \circ \dot x$

In other words, the morphism $1 \to Y$:

$g \circ \Delta_A \circ \dot x$

is a fixed point of $t$. This contradicts our assumption that $t$ had no fixed point.

Conclusion

When I started writing this blog post I though it would be a quick and easy job. After all, the proof of Lawvere’s theorem takes just a few lines both in the original paper and in all other sources I looked at. But then the details started piling up. The hardest part was convincing myself that it’s okay to disregard all the unitors. It’s not an easy thing for a programmer, because without unitors the formulas don’t “type check.” The left hand side may be a morphism from $A \times I$ and the right hand side may start at $A$. A compiler would reject such code. I tried to do due diligence as much as I could, but at some point I decided to just go with string diagrams. In the process I got into some interesting discussions and even posted a question on Math Overflow. Hopefully it will be answered by the time you read this post.

One of the things I wasn’t sure about was if it was okay to slide unitors around, as in this diagram:

It turns out this is just naturality condition for $\lambda$, as John Baez was kind to remind me on Twitter (great place to do category theory!).

Acknowledgments

I’m grateful to Derek Elkins for reading the draft of this post.

Literature

1. F. William Lawvere, Diagonal arguments and cartesian closed categories
2. Noson S. Yanofsky, A Universal Approach to Self-Referential Paradoxes, Incompleteness and Fixed Points
3. Qiaochu Yuan, Cartesian closed categories and the Lawvere fixed point theorem

In category theory, as in life, you spend half of your time trying to forget things, and half of the time trying to recover them. A morphism, the basic building block of every category, is like a defective isomorphism. It maps the initial state to the final state, but it provides no guarantees that you can recover the original. But it seems like this lossiness is what makes morphisms useful.

There are people who can memorize mathematical formulas perfectly but have no idea what they mean. And there are those who get just the gist of it, but can derive the rest when needed. Somehow understanding is related to lossy compression.

We can’t recover lost information. Once it’s gone, it’s gone. All we can do is to try to figure out what the original might have been like. In fact, knowing how the information was lost, we might be able to generate all possible inputs that could have led to a given output. It’s the closest we can get to inverting the uninvertible. This is the main idea behind fibrations.

Let me illustrate this concept with an example. Consider the function isEven:

isEven :: Integer -> Bool
isEven n = n mod 2 == 0


This function is definitely not invertible. If I only told you that the output was True, you couldn’t tell me what the input was. You could, however, give me the set of all inputs that could have produced this output: it’s the set of even numbers. We often call this set, which is the inverse image of True, a fiber over True. Similarly, the fiber over False is the set of odd numbers. In this case we only have only two fibers and they happen to be isomorphic.

Here’s a more interesting example. Consider a set of all lists of integers and a function that returns the length of a list: a natural number:

length :: [Integer] -> Nat


This function is not invertible, but it defines fibers over natural numbers. The fiber over zero is a one-element set that contains only the empty list. The fiber over 1 is the set of lists of length one (which is isomorphic to the set of integers). The fiber over 2 is a set of 2-element lists, or pairs of integers, and so on. You may recognize these fibers as length-indexed lists, or vectors. You can find them, for instance, in the Haskell Vec library or as Vect in Idris. These are not your typical data types, though. They are examples of dependent types–types that depend on values (here, natural numbers).

Notice that the name “length-indexed lists” suggests a slightly different interpretation of these types. You may think of them as families of types parameterized by natural numbers. This would suggest a mapping from elements of a set (natural numbers) to types. These two views are equivalent, but in category theory we try to avoid, if possible, talking about sets (and set elements in particular). The interpretation of dependent types as fibrations is more general, so let’s dig into fibrations.

As a generalization of functions like isEven or length, we’ll consider a morphism $\pi \colon e \to b$, and call it a projection, since it projects each fiber down to one element. Our goal, though, is to define a fiber as the pre-image of an element in $b$. But what’s an element? The closest we can get to defining an element in category theory is to consider a morphism $x$ from the terminal object $1$. Such morphism is called a global element and, in $Set$ it really picks a single element from a (non-empty) set. Now we have two morphisms converging on $b$: $\pi$ and $x$. Conceptually, a fiber is a subobject $e_x$ of $e$, which means that there must be a morphism $s$ that embeds $e_x$ in $e$. Moreover, when this embedding is followed by the projection $\pi$, it must produce the same element as $x$. The best such object is given by a universal construction which, in this case, is a pullback.

Fig. 1.

(The exclamation mark stands for the unique morphism to the terminal object.) Incidentally, this is why a pullback is sometimes called a fiber product.

If fibers over all elements $x$ are isomorphic, the pair $(e, \pi)$ is, quite fittingly, called a fiber bundle. The object $b$, from which the fibers sprout, is called the base. (Notice that lenght-indexed lists don’t form a bundle.)

Anything you can do with functions, you can do with functors, only better. So we can have a category $E$, another category $B$, and a functor $p \colon E \to B$. Since a functor acts as a function on objects (modulo size issues), we can define a fiber as a set of objects in $E$ that are mapped to a single object in $B$. The big question is, what do we do with morphisms? We have potentially lots of morphisms in $E$ that go between any two fibers, and which get projected down to a single hom-set in $B$. If we want to invert $p$, we have to design a procedure for lifting morphisms from $B$ to $E$.

Here’s the idea: We would like each fiber to form a subcategory of $E$, and we’d like to pick morphisms between fibers in such a way that $p^{-1}$ becomes a functor from $B$ to $Cat$. In other words, we want $p^{-1}$ to map objects of $B$ to categories (the fibers), and morphisms of $B$ to functors between those categories. If this is too much to ask (which it often is), we’ll settle for $p^{-1}$ to be a pseudo-functor, which is a functor that preserves unit and composition only up to isomorphism. In fact the original construction (attributed to Grothendieck) produces a contravariant pseudo-functor. In this post I’ll describe the covariant version of this construction, which is called opfibration, and which is easier to explain.

The starting point of Grothendieck fibration is the recipe for lifting morphisms from the base category $B$ to the total category $E$. There is a universal construction for doing that. The resulting morphisms are called opcartesian.

Let’s start with a morphism $f \colon a \to b$ in the base category and pick an arbitrary object $s$ (source) in the fiber over $a$ (hence $a = p s$). This will be the source of our opcartesian morphism. We have a lot of choices for the target. Strictly speaking, the target should be one of the objects over $b$, and that’s what we are aiming for. However, a universal construction should look at a much larger pool of candidates, some of them with targets in other fibers. This pool enlargement helps narrow down the final choice with greater accuracy. (Remember, universal constructions are unique only up to isomorphism.)

The opcartesian morphism over $f$, with the specified source $s$, is a morphism $g \colon s \to t$, such that $p g = f$.

Fig. 2.

It must satisfy a universal property that I’m about to describe.

First, we pick an arbitrary object $x$ and a morphism $h \colon s \to x$. This is supposed to be the competition for $g$. When projected down to $B$, it becomes $p h \colon a \to p x$.

Fig. 3.

We are interested in the case when $p h$ factorizes through $f$, that is, there is a morphism $u \colon b \to p x$ such that $p h = u \circ f$. Whenever such factorization is possible in $B$, we demand that there be a unique lifting of it to $E$.

Fig. 4.

In other words, there exists a unique $\nu \colon t \to x$ such that $h = \nu \circ g$ and $p \nu = u$.

If you find this definition a little confusing, you’re in good company. So let’s try a slightly different imagery that has more to do with the original ideas from algebraic topology. Think of objects as shapes. A morphism $f \colon a \to b$ is a proof that $b$ is a proper subset of $a$, or that $a$ contains $b$. A functor between two categories of shapes must map shapes to shapes in a way that preserves inclusion. It may map many shapes to one, so imagine that the shapes in the category $E$ are three-dimensional, and their projections using the functor $p$ are their flat shadows. Functoriality means that, if $s$ contains $t$, then its shadow $a = p s$ contains $b = p t$. Next, we introduce a new object $x$ that is contained inside $s$, and the proof of that is $h \colon s \to x$. It follows from functoriality that $a$ contains the shadow of $x$.

Now suppose that this shadow falls inside the smaller $b$ (with the proof $u \colon b \to p x$).

Fig. 5.

Normally, this would not imply that $x$ is inside $t$. It’s possible that (parts of) $x$ are sticking out below or above $t$. Our universal condition demands that this cannot happen. There can be no room above or below $t$— it’s a cylinder carved into $s$. Universality guarantees that we get the absolutely optimal shape.

Now that we know what an opcartesian morphism is, we might ask the question, does it always exist? Given an arbitrary morphism $f \colon a \to b$ in $B$ and an object $s$ over $a$, can we always find an opcartesian morphism $g \colon s \to t$ such that $t$ is over $b$? If we can, then we call the pair $(E, p \colon E \to B)$ an opfibration.

Here’s an interesting observation. You might wonder whether the definition of an opcartesian morphism isn’t overly complicated. Wouldn’t it be enough to restrict the pool of possible candidates to those morphisms whose target, the $x$ in our picture, was over $b$, the target of $f$? This was, in fact, the original idea in the Grothendieck construction. The problem was that, with such definition, there was no guarantee that a composition of two opcartesian morphisms would be again opcartesian. The current definition makes that automatic.

Given an opfibration, we now face the opposite problem: there may be too many opcartesian morphisms. Remember, we wanted to (a) make fibers into subcategories of $E$ and (b) use opcartesian morphisms to define functors between them. The first part is relatively easy: a fiber $E_a$ has, as objects, those objects of $E$ whose projection is $a$. We select as morphisms in $E_a$ those morphisms that project down to identity, $id_a$ (notice that we ignore other endomorphism $a \to a$). These are called vertical morphisms. But to define functors between fibers we need to map each object of one fiber to exactly one object in the other fiber (and the same for vertical morphisms). Think of this as transporting objects between fibers. In a fibered category, we could use opcartesian morphisms for transport. Any time two objects are connected in the base by a morphism, we have a bunch of opcartesian morphisms over it starting from every single object in the source fiber. We could use them to try to define a functor between fibers.

But in general we have more than one opcartesian morphism between a source object in one fiber and candidate target objects in the other fiber. But we can design a procedure to pick one (if you’re into set theory, you’ll notice that we have to use the Axiom of Choice). Such choice is called an opcleavage, and the resulting construction is called cloven opfibration.

Formally, an opcleavage is described by a function $\kappa (f, s)$

Fig. 6.

It takes a morphism $f \colon a \to b$ and an object $s$ (such that $p s = a$), and produces an object $t$ (such that $p t = b$), which is the target of some opcartesian morphism $s \to t$. This is exactly the morphism selected by opcleavage.

The universal construction of opcartesian morphisms can then be used to define the mapping of vertical morphisms thus completing the definition of a functor between fibers.

A geometric intuition is that an opcleavage provides a way of transporting objects in the horizontal direction. Vertical morphisms transport objects vertically, and the functors defined by the opcleavage transport them horizontally, in such a way that their shadows follow the arrows in the base. The origin of this intuition goes back to differential geometry, where one is able to define continuous paths in the base manifold and use them to transport objects, such as vectors, between fibers. Category theory lets us abstract away continuity (and differentiability) from this picture. You might also see transport used in homotopy type theory, with paths standing for equality proofs.

Now, remember what I said about the composition of opcartesian morphisms resulting in an opcartesian morphism? Unfortunately, once we start picking individual morphisms to construct an opcleavage, this compositionality might be lost. The composition of any two morphisms from the selected pool is still opcartesian, but it’s not necessarily part of the opcleveage. This is why we might have to relax compositionality and embrace pseudofunctors.

But sometimes an opcleavage preserves compositionality. We call this situation split opfibration. It must satisfy these two conditions:

$\kappa (id_a, s) = id_s$

$\kappa (f', s') \circ \kappa(f, s) = \kappa(f' \circ f, s)$

for any $f \colon a \to b$ and $f' \colon b \to c$.

A split opfibration defines a functor $B \to Cat$, which maps objects from the base category to fibers seen as categories; and morphisms from the base category to functors between those fibers. So defined functor may be interpreted as an attempt at inverting the original projection $p \colon E \to B$.

If the splitting conditions are satisfied only up to isomorphism, we get a pseudofunctor $B \to Cat$. This makes things more complicated but also more interesting. It means that horizontal transport depends on the path. In particular, transport along a closed path–a chain of morphisms in the base that compose to identity–may produce an object that’s different from (albeit isomorphic to) the starting object. In differential geometry we would say that the space has non-zero curvature.

Interestingly, this procedure of generating split opfibrations has its inverse. Given a functor $B \to Cat$ it’s possible to reconstruct the total category $E$ and a projection $p \colon E \to B$. This is called the Grothendieck construction.

Since our new slogan is “lenses are everywhere,” it should come as no big surprise that a split opfibration may be seen as a type of a lens. The projection corresponds to view or get. It extracts $a$, the focus of the lens, out of $s$. The opcleavage part of opfibration, $\kappa(f, s)$ corresponds to put or, more precisely to over. It takes a morphism that modifies the focus from $a$ to $b$ and it takes the object $s$, and produces the new object $t$. In programming, get and put are just functions between sets, here they are object mappings of two functors, but the similarity is hard to ignore.

Acknowledment

1. Johnson, Rosebrugh, and Wood, Lenses, fibrations and universal translations
2. Johnson and Rosebrugh, Delta lenses and opfibrations

In my previous blog post, Programming with Universal Constructions, I mentioned in passing that one-to-one mappings between sets of morphisms are often a manifestation of adjunctions between functors. In fact, an adjunction just extends a universal construction over the whole category (or two categories, in general). It combines the mapping-in with the mapping-out conditions. One functor (traditionally called the left adjoint) prepares the input for mapping out, and the other (the right adjoint) prepares the output for mapping in. The trick is to find a pair of functors that complement each other: there are as many mapping-outs from one functor as there are mapping-ins to the other functor.

To gain some insight, let’s dig deeper into the examples from my previous post.

The defining property of a product was the universal mapping-in condition. For every object $c$ equipped with a pair of morphisms going to, respectively, $a$ and $b$, there was a unique morphism $h$ mapping $c$ to the product $a \times b$. The commuting condition ensured that the correspondence went both ways, that is, given an $h$, the two morphisms were uniquely determined.

A pair of morphisms can be seen as a single morphism in a product category $C\times C$. So, really, we have an isomorphism between hom-sets in two categories, one in $C\times C$ and the other in $C$. We can also define two functors going between these categories. An arbitrary object $c$ in $C$ is mapped by the diagonal functor $\Delta$ to a pair $\langle c, c\rangle$ in $C\times C$. That’s our left functor. It prepares the source for mapping out. The right functor maps an arbitrary pair $\langle a, b\rangle$ to the product $a \times b$ in $C$. That’s our target for mapping in.

The adjunction is the (natural) isomorphism of the two hom-sets:

$(C\times C)(\Delta c, \langle a, b\rangle) \cong C(c, a \times b)$

Let’s develop this intuition. As usual in category theory, an object is defined by its morphisms. A product is defined by the mapping-in property, the totality of morphisms incoming from all other objects. Hence we are interested in the hom-set between an arbitrary object $c$ and our product $a \times b$. This is the right hand side of the picture. On the left, we are considering the mapping-out morphism from the object $\langle c, c \rangle$, which is the result of applying the functor $\Delta$ to $c$. Thus an adjunction relates objects that are defined by their mapping-in property and objects defined by their mapping-out property.

Another way of looking at the pair of adjoint functors is to see them as being approximately the inverse of each other. Of course, they can’t, because the two categories in question are not isomorphic. Intuitively, $C\times C$ is “much bigger” than $C$. The functor that assigns the product $a \times b$ to every pair $\langle a, b \rangle$  cannot be injective. It must map many different pairs to the same (up to isomorphism) product. In the process, it “forgets” some of the information, just like the number 12 forgets whether it was obtained by multiplying 2 and 6 or 3 and 4. Common examples of this forgetfulness are isomorphisms such as

$a \times b \cong b \times a$

or

$(a \times b) \times c \cong a \times (b \times c)$

Since the product functor loses some information, its left adjoint must somehow compensate for it, essentially by making stuff up. Because the adjunction is a natural transformation, it must do it uniformly across the whole category. Given a generic object $c$, the only way it can produce a pair of objects is to duplicate $c$. Hence the diagonal functor $\Delta$. You might say that $\Delta$ “freely” generates a pair. In almost every adjunction you can observe this interplay of “freeness” and “forgetfulness.” I’m using these therm loosely, but I can be excused, because there is no formal definition of forgetful (and therefore free or cofree) functors.

Left adjoints often create free stuff. The mnemonic is that “the left” is “liberal.” Right adjoints, on the other hand, are “conservative.” They only use as much data as is strictly necessary and not an iota more (they also preserve limits, which the left adjoints are free to ignore). This is all relative and, as we’ll see later, the same functor may be the left adjoint to one functor and the right adjoint to another.

Because of this lossiness, a round trip using both functors doesn’t produce an identity. It is however “related” to the identity functor. The combination left-after-right produces an object that can be mapped back to the original object. Conversely, right-after-left has a mapping from the identity functor. These two give rise to natural transformations that are called, respectively, the counit $\varepsilon$ and the unit $\eta$.

Here, the combination diagonal functor after the product functor takes a pair $\langle a, b\rangle$ to the pair $\langle a \times b, a \times b\rangle$. The counit $\varepsilon$ then maps it back to $\langle a, b\rangle$ using a pair of projections $\langle \pi_1, \pi_2\rangle$ (which is a single morphism in $C \times C$). It’s easy to see that the family of such morphisms defines a natural transformation.

If we think for a moment in terms of set elements, then for every element of the target object, the counit extracts a pair of elements of the source object (the objects here are pairs of sets). Note that this mapping is not injective and, therefore, not invertible.

The other composition–the product functor after the diagonal functor–maps an object $c$ to $c \times c$. The component of the unit natural transformation, $\eta_c \colon c \to c \times c$, is implemented using the universal property of the product. Indeed, such a morphism is uniquely determined by a pair of identity morphsims $\langle id_c, id_c \rangle$. Again, when we vary $c$, these morphisms combine to form a natural transformation.

Thinking in terms of set elements, the unit inserts an element of the set $c$ in the target set. And again, this is not an injective map, so it cannot be inverted.

Although in an arbitrary category we cannot talk about elements, a lot of intuitions from $Set$ carry over to a more general setting. In a category with a terminal object, for instance, we can talk about global elements as mappings from the terminal object. In the absence of the terminal object, we may use other objects to define generalized elements. This is all in the true spirit of category theory, which defines all properties of objects in terms of morphisms.

Every construction in category theory has its dual, and the product is no exception.

A coproduct is defined by a mapping out property. For every pair of morphisms from, respectively, $a$ and $b$ to the common target $c$ there is a unique mapping out from the coproduct $a + b$ to $c$. In programming, this is called case analysis: a function from a sum type is implemented using two functions corresponding to two cases. Conversely, given a mapping out of a coproduct, the two functions are uniquely determined due to the commuting conditions (this was all discussed in the previous post).

As before, this one-to-one correspondence can be neatly encapsulated as an adjunction. This time, however, the coproduct functor is the left adjoint of the diagonal functor.

The coproduct is still the “forgetful” part of the duo, but now the diagonal functor plays the role of the cofree funtor, relative to the coproduct. Again, I’m using these terms loosely.

The counit now works in the category $C$ and  it “extracts a value” from the symmetric coproduct of  $c$ with $c$. It does it by “pattern matching” and applying the identity morphism.

The unit is more interesting. It’s built from two injections, or two constructors, as we call them in programming.

I find it fascinating that the simple diagonal functor can be used to define both products and coproducts. Moreover, using terse categorical notation, this whole blog post up to this point can be summarized by a single formula.

There is one more very important adjunction that every programmer should know: the exponential, or the currying adjunction. The exponential, a.k.a. the function type, is the right adjoint to the product functor. What’s the product functor? Product is a bifunctor, or a functor from $C \times C$ to $C$. But if you fix one of the arguments, it just becomes a regular functor. We’re interested in the functor $(-) \times b$ or, more explicitly:

$(-) \times b : a \to a \times b$

It’s a functor that multiplies its argument by some fixed object $b$. We are using this functor to define the exponential. The exponential $a^b$ is defined by the mapping-in property.  The mappings out of the product $c \times b$ to $a$ are in one to one correspondence with morphisms from an arbitrary object $c$ to the exponential $a^b$ .

$C(c \times b, a) \cong C(c, a^b)$

The exponential $a^b$ is an object representing the set of morphisms from $b$ to $a$, and the two directions of the isomorphism above are called curry and uncurry.

This is exactly the meaning of the universal property of the exponential I discussed in my previous post.

The counit for this adjunction extracts a value from the product of the function type (the exponential) and its argument. It’s just the evaluation morphism: it applies a function to its argument.

The unit injects a value of type $c$ into a function type $b \to c \times b$. The unit is just the curried version of the product constructor.

I want you to look closely at this formula through your programming glasses. The target of the unit is the type:

b -> (c, b)

You may recognize it as the state monad, with b representing the state. The unit is nothing else but the natural transformation whose component we call return. Coincidence? Not really! Look at the component of the counit:

(b -> a, b) -> a

It’s the extract of the Store comonad.

It seems like, in category theory, if you dig deep enough, everything is related to everything in some meaningful way. And every time you revisit a topic, you discover new insights. That’s what makes category theory so exciting.

Previously we were exploring universal constructions for products, coproducts, and exponentials. In particular, we were able to prove the distributive law:

$(a + b) \times c \cong a \times c + b \times c$

The power of this law is that it relates the mapping-in universal construction (product on the left) with the mapping-out one (coproduct on the right). If you take into account that products and coproducts are just special cases of limits and colimits, you may ask a more general question: under what conditions limits commute with colimits. In a cartesian closed category a product of sums is not equal to the sum of products:

$(a + b) \times (c + d) \ncong a \times c + b \times d$

So, in general, products don’t commute with coproducts. But if you replace coproducts with a special kind of colimits, then it can be shown that:
Theorem.
In $Set$, filtered colimits commute with finite limits.

In this post I’ll try to explain these terms and provide some intuition why it works and how filtered colimits are related to the more traditional notion of limits that we know from calculus.

Limits

Let’s start with limits. They are like products, except that, instead of just two objects at the bottom, you have any number of objects plus a bunch of morphisms between them. That’s called a diagram. Then you have an apex with arrows going down to all the objects in the diagram; and you get what is called a cone. If you have morphisms in your diagram, they form triangles. These triangles must commute. For instance, in Fig 1, we have:

$g \circ \pi_1 = \pi_3$

Fig. 1. A cone

This means that not all projections are independent–that you may obtain one projection from another by post-composing it with a morphism from the diagram. In Fig 1, for instance, you may extract a value of $c$ either directly using $\pi_3$ or by applying $g$ to the result of $\pi_1$.

A limit is defined as the universal cone with the apex $Lim$. It means that, if you have any other cone with some apex $c$, built over the same diagram, there is a unique morphism $h$ from $c$ to $Lim$ that makes all the triangles commute. For instance, in Fig 2, one of the commuting conditions is:

$\pi_1 \circ h = f_1$

and so on. We’ve seen similar commuting conditions in the definition of the product.

Fig. 2. A universal cone

If you think of $Lim$ in this example as a data structure, you would implement it as a product of $a_1$, $a_2$, and $a_3$, together with two functions:

$g_3 : a_1 \to a_3$

$g_2 : a_1 \to a_2$

But because of the commuting conditions, the three values stored in $Lim$ cannot be independent. If you pick a value for $a_1$, then the values for $a_2$ and $a_3$ are uniquely determined.

A limit, just like a product, is defined by a mapping-in property. If you want to define a morphism from some $c$ to $Lim$, you need to provide three morphisms $f_1$, $f_2$, and $f_3$. However, unlike in the case of a product, these morphisms must satisfy some commuting conditions. Here, $f_3$ must be equal to $g_3 \circ f_1$ and $f_2 = g_2 \circ f_1$. So, really, you only need to define $f_1$, and that uniquely determines $h$. This is why the cones in Fig 2 can be simplified, as shown in Fig 3.

Fig. 3. A simplified universal cone

Notice that the diagram essentially forms a subcategory inside the category $C$, even if we don’t explicitly draw all the identity morphisms or all the compositions. This is because triangles built by composing commuting triangles are again commuting. It therefore makes sense to define a diagram as a functor $F$ from an (often much smaller) index category $J$ to $C$. In our case it would be a category with just three objects, $j_1$, $j_2$, $j_3$, and two non-identity morphisms. (The diagram category for the product is even simpler: just two objects, no non-trivial morphisms.)

The properties of the diagram category determine the nature of cones and the nature of the limits. For instance, functors from a finite category will produce finite limits.

Fig. 4. Diagram category $J$

The diagram category $J$ in our example has a very peculiar property: it has a cone for every pair of objects (it’s a cone inside $J$, not to be confused with the cone in $C$). For instance, the pair $j_2$, $j_3$ is part of the cone with the apex $j_1$. This is also the apex for the (somewhat degenerate) cone based on $j_1$ and $j_2$ (with or without the connecting morphism). A category in which there is a cone for every finite subdiagram is called cofiltered. Limits defined by functors from cofiltered categories are called cofiltered limits.

The intuition is that cofiltered categories exhibit some kind of ordering. You may think of $j_1$ as a lower bound of $j_2$ and $j_3$. Following these bounds, you might eventually get to some kind of roots–here it’s the object $j_1$–and these roots will dictate the behavior of cones and the behavior of limits. Things get really interesting when the diagram category is infinite, because then there is no guarantee that you’ll ever reach a root. There is, for instance, no smallest (negative) integer, even though integers are ordered. You can begin to see parallels with traditional limits, like:

$\lim_{j \to -\infty} a_j$

That’s where these ideas originally came from.

Limits in the category of sets have a particularly simple interpretation. In $Set$, we can use functions from the terminal object–the singleton set–to pick individual set elements.

Fig. 5. Elements of the limit

For every selection in Fig 5. of $x_1$, $x_2$, $x_3$ there is a unique $h(x_1, x_2, x_3)$ that picks an element in $Lim$. But a selection of $x_1$, $x_2$, $x_3$ is nothing but a cone with the apex $1$. So there is a one-to-one correspondence between elements of $Lim$ and such cones. In other words, $Lim$ is a set of apex-1 cones.

Colimits

Colimits are dual to limits–you get them by inverting all the arrows. So, instead of projections, you get injections, and the universal condition defines a mapping out of a colimit (see Fig 6).

Fig. 6. A universal cocone

If you look at the colimit as a data structure, it is similar to a coproduct, except that not all the injections are independent. In the example in Fig 6, $i_3$ and $i_2$ are determined by pre-composing $i_1$ with $g_3$ and $g_2$, respectively. It’s not clear how to implement a colimit in Haskell, so here’s a pseudo-Haskell attempt using imaginary dependent-type syntax:

  data Colim a1 a2 a3 (g2 :: a2 -> a1) (g3 :: a3 -> a1) =
= A1 a1 | A2 a2 | A3 a3


To deconstruct this colimit, you only need to provide one function $f_1 : a_1 \to c$.

  h :: (a1 -> c) -> Colim a1 a2 a3 g2 g3 -> c
h f1 (A1 a1)    = f1 a1
h f1 (A2 a2) = f1 (g2 a2)
h f1 (A3 a3) = f1 (g3 a3)


Granted, in a lazy language like Haskell, this would be an overkill way to store essentially just one value.

A colimit in the category of sets simplifies to a disjoint union of sets, in which some elements are identified. Suppose that the colimit $Colim_J F$ is defined by some diagram category $J$ and a functor $F : J \to Set$. Each object $j$ in $J$ produces a set $F j$.

Fig. 7. Colimit in Set. On the left, the diagram category $J$.

The disjoint union of all these sets is a set whose elements are the pairs $(x, j)$ where $x \in F j$. (Notice that the sets may overlap, but each element from the overlap will be counted as many times as the number of sets it belongs to.) Coproduct injections are then functions that take an element $x \in F j$ and map it into an element $(x, j) \in Colim_J F$. But that doesn’t take into account the presence of morphisms in the diagram. These morphisms are mapped to functions between corresponding sets. For instance, in Fig 7, we can take an element $x \in F j_2$. It is injected, using $i_2$, as an element $(x, j_2) \in Colim_J F$. But there is another path from $F j_2$ that uses $F g_2$ followed by $i_1$. That produces $((F g_2) x, j_1)$. If the triangle is to commute, these two must be equal. So in the actual colimit, they must be identified. In general, any two elements of the disjoint union that satisfy this relation:

$(x, j) \rightsquigarrow (x', j')\;\; \text{if}\;\; \exists_{g : j \to j'} (F g) x = x'$

must be identified. This is not an equivalence relation, but it can be extended to one (by first symmetrizing it, and then making it transitive again). A colimit is then a quotient of the disjoint union by this equivalence.

As before, I chose this example to illustrate a special type of a diagram. This is a diagram that can be obtained using a functor from a filtered category. A filtered category has this property that for any finite subdiagram, there is a cocone under it. Here, for instance, the subdiagram formed by $j_2$ and $j_3$ has a cocone with the apex $j_1$. Again, you may think of $j_1$ as a kind of upper bound of $j_2$ and $j_3$. If the filtered category is finite, following upper bounds will eventually lead you to some roots. And in Set, the equivalence relation will allow you shift all the elements down to those roots. But in an infinite case (think natural numbers) there may be no largest element–no root. And that brings filtered colimits closer to the intuition we have for limits in calculus. In fact, all the interesting filtered colimits are based on infinite diagrams.

Commuting Limits and Colimits

What does it mean for a limit to commute with a colimit? A single colimit is generated by a functor from some index category $I \to C$. What we need is a bunch of such colimits so that we can take a limit over those. Therefore we need a bunch of functors $I \to C$. Moreover, those colimits have to form a diagram. So we need another index category $J$ to parameterize those functors. Altogether, we need a functor of two arguments:

$F : I \times J \to C$

It follows that, for any given $j$ in $J$ we have a functor $F(-, j) : I \to C$. We can take a colimit of that. Then we gather those colimits into a diagram whose shape is defined by $J$, and then take its limit. We get:

$Lim_J (Colim_I F)$

Alternatively, when we fix some $i$ in $I$, we get a functor $F(i, -) : J \to C$. We can take a limit of that. Then we can gather all those limits and form a diagram whose shape is defined by $I$. Finally we can take a colimit of that:

$Colim_I (Lim_J F)$

Fig. 8. Commuting limits (red diagram of shape $J$) and colimits (black diagram of shape $I$)

It’s not difficult to construct the mapping:

$Colim_I (Lim_J F) \to Lim_J (Colim_I F)$

using the universal property, since the colimit has the mapping-out property. It’s the other way around that’s tricky. But it always works in the special case when $I$ is filtered, $J$ is finite, and $C$ is $Set$.

Here’s the sketch of this amazing proof, which you can find in Saunders Mac Lane’s Categories for the Working Mathematician.

Since the target of the functor is Set, it might help to visualize its image as a rectangular array of sets. A fixed $j$ picks up a row of such sets, whereas a fixed $i$ picks up a column. Because we are dealing with sets, we can try to define the mapping:

$Lim_J (Colim_I F) \to Colim_I (Lim_J F)$

pointwise. Let’s pick an element of the limit on the left. As we’ve established earlier, a limit in Set is a set of apex-1 cones. So let’s pick one such cone. It’s just a selection of elements from a bunch of colimits.

As we’ve seen before, a colimit in Set is a discriminated union with some identifications. So our apex-1 cone will pick a set of representatives, one per colimit, say $(x_n, i_n)$. Any time there is a morphism $g : i_n \to i'_n$, we can replace one representative with another $(g (x_n), i'_n)$. The intuition is that we can slide the representatives horizontally within each row along morphisms.

If $I$ is a filtered category, then for any finite number of objects $i_n$, we can always find a common root (it will be the apex $i$ of a cocone formed by $i_n$ in $I$). So we can slide all the representatives to a single column. In other words, our cone can be brought to a set of representatives $(y_n, i)$, with a common $i$.

Fig. 9. A single cone after shifting representatives from all colimits to a common column

But that’s just a cone over $J$. It’s an element of $Lim_J F$. And we can inject it into a colimit over $I$ to get an element of $Colim_I (Lim_J F)$. We have thus defined our mapping.

Conclusion

If you didn’t get the proof the first time, don’t get discouraged. Take a break, sleep over it, and then read it slowly again. Make sure you have internalized all the definitions. Draw your own pictures. The two major tricks are: (1) visualizing an element of a limit as a cone originating from the singleton set, and (2) the idea of sliding the elements of multiple colimits to a common column.

The importance of this theorem is that it tells you when and how you can define mappings out of limits. For instance, how to define functions from a product or from an end.

Acknowledgment

I’m grateful to Derek Elkins for correcting mistakes in the original version of this post.

As functional programmers we are interested in functions. Category theorists are similarly interested in morphisms. There is a slight difference in approach, though. A programmer must implement a function, whereas a mathematician is often satisfied with the proof of existence of a morphism (unless said mathematician is a constructivist). Category theory if full of such proofs. It turns out that many of these proofs can be converted to code, often resulting in quite unexpected encodings.

A lot of objects in category theory are defined using universal constructions and universality is used all over the place to show the existence (as a rule: unique, up to unique isomorphism) of morphisms between objects.

There are two major types of universal constructions: the ones asserting the mapping-in property, and the ones asserting the mapping-out property. For instance, the product has the mapping-in property.

Product

Recall that a product of two objects $a$ and $b$ is an object $a \times b$ together with two projections:

$\pi_1 : a \times b \to a$

$\pi_2 : a \times b \to b$

This object must satisfy the universal property: for any other object $c$ with a pair of morphisms:

$f : c \to a$

$g : c \to b$

there exists a unique morphism $h : c \to a \times b$ such that:

$f = \pi_1 \circ h$

$g = \pi_2 \circ h$

In other words, the two triangles in Fig 1 commute.

Fig. 1. Universality of the product

This universal property can be used any time you need to find a morphism that’s mapping into the product, and it can actually produce code.

For instance, let’s say you want to find a morphism from the terminal object $1$ to $a \times b$. All you need is to define two morphisms $x : 1 \to a$ and $y : 1 \to b$. This is not always possible, but if it is, you are guaranteed the existence of a morphism $h : 1 \to a \times b$ (Fig 2).

Fig. 2. Global element of a product

Morphisms from the terminal object are called global elements, so we have just shown that, as long as $a$ and $b$ have global elements, say $x$ and $y$, their product has a global element too. Moreover the projection $\pi_1$ of this global element is the same as $x$, and $\pi_2$ is the same as $y$. In other words, an element of a product is a pair of elements. But you probably knew that.

The universal construction of the product is implemented as an operator in Haskell:

  (&&&) :: (c->a) -> (c->b) -> (c -> (a, b))


We can also go the other way: given a mapping-in $h : c \to a \times b$, we can always extract a pair of morphisms:

$f = \pi_1 \circ h$

$g = \pi_2 \circ h$

This bijection between $h$ and a pair of morphisms $(f, g)$ is in fact an adjunction.

You might think this kind of reasoning is very different from what programmers do, but it’s not. Here’s one possible definition of a product in Haskell (besides the built-in one, (,)):

  data Product a b = MkProduct { fst :: a
, snd :: b }


It is in one-to-one correspondence with what I’ve just explained. The two functions fst and snd are $\pi_1$ and $\pi_2$, and MkProduct corresponds to our $h : 1 \to a \times b$. The categorical definition is just a different, much more general, way of saying the same thing.

Here’s another application of universality: Show that product is functorial. Suppose that you have a pair of morphisms:

$f : a \to a'$

$g : b \to b'$

and you want to lift them to a morphism:

$h : a \times b \to a' \times b'$

Since we are dealing with products, we should use the mapping-in property. So we draw the universality diagram for the target $a' \times b'$, and put the source $a \times b$ at the top. The pair of functions that fits the bill is $(f \circ \pi_1, g \circ \pi_2)$ (Fig 3).

Fig. 3. Functoriality of the product

The universal property gives us, uniquely, the $h$, which is usually written simply as $f \times g$.

Exercise for the reader: Show, using universality, that categorical product is symmetric.

Coproduct

The coproduct, being the dual of the product, is defined by the universal mapping-out property, see Fig 4.

Fig. 4. Universality of the coproduct

So if you need a morphism from a coproduct $a + b$ to some $c$, it’s enough to define two morphisms:

$f : a \to c$

$g : b \to c$

This universal property may also be restated as the isomorphism between pairs of morphisms $(f, g)$ and morphisms of the type $a+b \to c$ (so there is, in fact, a corresponding adjunction).

This is easily illustrated in Haskell:

  h :: Either a b -> c
h (Left a)  = f a
h (Right b) = g b


Here Left and Right correspond to the two injections $i_1$ and $i_2$. There is a convenient function in Haskell that encapsulates this universal construction:

  either :: (a->c) -> (b->c) -> (Either a b -> c)


Exercise for the reader: Show that coproduct is functorial.

So next time you ask yourself, what can I do with a universal construction? the answer is: use it to define a morphism, either mapping in or mapping out of your construct. Why is it useful? Because it decomposes a problem into smaller problems. In the examples above, the problem of constructing one morphism $h$ was nicely decomposed into defining $f$ and $g$ separately.

The flip side of this is that there is no simple way of defining a mapping out of a product or a mapping into a coproduct.

Distributive Law

For instance, you might wonder if the familiar distributive law:

$(a + b) \times c \cong a \times c + b \times c$

holds in an arbitrary category that defines products and coproducts (so called bicartesian category). You can immediately see that defining a morphism from right to left is easy, because it involves the mapping out of a coproduct. All we need is to define a pair of morphisms leading to the common target (Fig 5):

$f : a \times c \to (a + b) \times c$

$g : b \times c \to (a + b) \times c$

Fig. 5. Right to left proof

The trick is to take advantage of the functoriality of the product, which we have already established, and use it to implement $f$ and $g$ as:

$f = i_1 \times id_c$

$g = i_2 \times id_c$

But if you try to construct a proof in the other direction, from left to right, you’re stuck, because it would require the mapping out of a product. So the distributive property does not hold in general.

“Wait a moment!” I hear you say, “I can easily implement it in Haskell.”

  f :: (Either a b, c) -> Either (a, c) (b, c)
f (Left a, c)  = Left  (a, c)
f (Right b, c) = Right (b, c)


Exponential

That’s correct, but Haskell does a little cheating behind the scenes. You can see it clearly when you convert this code to point free notation (I’ll explain later how I figured it out):

  f = uncurry (either (curry Left) (curry Right))


I want to direct your attention to the use of curry and uncurry. Currying is the application of another universal construction, namely that of the exponential object $c^b$, representing the function type b -> c. This is exactly the construction that provides the missing mapping out of a product, (a, b) -> c. Here we go:

  uncurry :: (c -> (a -> b)) -> ((c, a) -> b)


Categorically, we have the bijection between morphisms (again, a sign of an adjunction):

$h : c \to b^a$

$f : c \times a \to b$

Universality tells us that for every $c$ and $f$ there is a unique $h$ in Fig 6 (and vice versa). The arrow $h \times id_a$ is the lifting of the pair $(h, id_a)$ by the product functor (we’ve established the functoriality of the product earlier).

Fig. 6. Universality of the exponential

Not every category has exponentials–the ones that do are called cartesian closed (cartesian, because they must also have products).

So how does the fact that we have exponentials in Haskell help us here? We are trying to define a mapping out of a product:

$f : (a + b) \times c \to a \times c + b \times c$

Here’s where the exponential saves the day. This mapping exists if we can define another mapping:

$h : (a + b) \to (a \times c + b \times c)^c$

see Fig 7.

Fig. 7. Uncurrying

This morphism, in turn, is easy to define, because it involves a mapping out of a sum. We just need a pair of morphisms:

$h_1 : a \to (a \times c + b \times c)^c$

$h_2 : b \to (a \times c + b \times c)^c$

We can define the first morphism using the universal property of the exponential, picking the injection $i_1$:

Fig. 8. Defining $h_1$

This translates to Haskell as h1 = curry Left. Similarly for $h_2$ we get curry Right.

We can now combine all these diagrams into a single point-free definition, and that’s exactly how I came up with the original code:

  f = uncurry (either (curry Left) (curry Right))


Notice that curry is used to get from $f$ to $h$, and uncurry from $h$ to $f$ in the original diagram.

Products and coproducts are examples of more general constructions called limits and colimits. Importantly, the universal property of limits can be used to define the mapping-in morphisms, whereas the universal property of colimits allows us to define the mapping-out morphisms. I’ll talk more about it in the upcoming post.

I’ve been working with profunctors lately. They are interesting beasts, both in category theory and in programming. In Haskell, they form the basis of profunctor optics–in particular the lens library.

Profunctor Recap

The categorical definition of a profunctor doesn’t even begin to describe its richness. You might say that it’s just a functor from a product category $\mathbb{C}^{op}\times \mathbb{D}$ to $Set$ (I’ll stick to $Set$ for simplicity, but there are generalizations to other categories as well).

A profunctor $P$ (a.k.a., a distributor, or bimodule) maps a pair of objects, $c$ from $\mathbb{C}$ and $d$ from $\mathbb{D}$, to a set $P(c, d)$. Being a functor, it also maps any pair of morphisms in $\mathbb{C}^{op}\times \mathbb{D}$:

$f\colon c' \to c$
$g\colon d \to d'$

to a function between those sets:

$P(f, g) \colon P(c, d) \to P(c', d')$

Notice that the first morphism $f$ goes in the opposite direction to what we normally expect for functors. We say that the profunctor is contravariant in its first argument and covariant in the second.

Hom-Profunctor

The key point is to realize that a profunctor generalizes the idea of a hom-functor. Like a profunctor, a hom-functor maps pairs of objects to sets. Indeed, for any two objects in $\mathbb{C}$ we have the set of morphisms between them, $C(a, b)$.

Also, any pair of morphisms in $\mathbb{C}$:

$f\colon a' \to a$
$g\colon b \to b'$

can be lifted to a function, which we will denote by $C(f, g)$, between hom-sets:

$C(f, g) \colon C(a, b) \to C(a', b')$

Indeed, for any $h \in C(a, b)$ we have:

$C(f, g) h = g \circ h \circ f \in C(a', b')$

This (plus functorial laws) completes the definition of a functor from $\mathbb{C}^{op}\times \mathbb{C}$ to $Set$. So a hom-functor is a special case of an endo-profunctor (where $\mathbb{D}$ is the same as $\mathbb{C}$). It’s contravariant in the first argument and covariant in the second.

For Haskell programmers, here’s the definition of a profunctor from Edward Kmett’s Data.Profunctor library:

class Profunctor p where
dimap :: (a' -> a) -> (b -> b') -> p a b -> p a' b'

The function dimap does the lifting of a pair of morphisms.

Here’s the proof that the hom-functor which, in Haskell, is represented by the arrow ->, is a profunctor:

instance Profunctor (->) where
dimap ab cd bc = cd . bc . ab

Not only that: a general profunctor can be considered an extension of a hom-functor that forms a bridge between two categories. Consider a profunctor $P$ spanning two categories $\mathbb{C}$ and $\mathbb{D}$:

$P \colon \mathbb{C}^{op}\times \mathbb{D} \to Set$

For any two objects from one of the categories we have a regular hom-set. But if we take one object $c$ from $\mathbb{C}$ and another object $d$ from $\mathbb{D}$, we can generate a set $P(c, d)$. This set works just like a hom-set. Its elements are called heteromorphisms, because they can be thought of as representing morphism between two different categories. What makes them similar to morphisms is that they can be composed with regular morphisms. Suppose you have a morphism in $\mathbb{C}$:

$f\colon c' \to c$

and a heteromorphism $h \in P(c, d)$. Their composition is another heteromorphism obtained by lifting the pair $(f, id_d)$. Indeed:

$P(f, id_d) \colon P(c, d) \to P(c', d)$

so its action on $h$ produces a heteromorphism from $c'$ to $d$, which we can call the composition $h \circ f$ of a heteromorphism $h$ with a morphism $f$. Similarly, a morphism in $\mathbb{D}$:

$g\colon d \to d'$

can be composed with $h$ by lifting $(id_c, g)$.

In Haskell, this new composition would be implemented by applying dimap f id to precompose p c d with

f :: c' -> c

and dimap id g to postcompose it with

g :: d -> d'

This is how we can use a profunctor to glue together two categories. Two categories connected by a profunctor form a new category known as their collage.

A given profunctor provides unidirectional flow of heteromorphisms from $\mathbb{C}$ to $\mathbb{D}$, so there is no opportunity to compose two heteromorphisms.

Profunctors As Relations

The opportunity to compose heteromorphisms arises when we decide to glue more than two categories. The clue as how to proceed comes from yet another interpretation of profunctors: as proof-relevant relations. In classical logic, a relation between sets assigns a Boolean true or false to each pair of elements. The elements are either related or not, period. In proof-relevant logic, we are not only interested in whether something is true, but also in gathering witnesses to the proofs. So, instead of assigning a single Boolean to each pair of elements, we assign a whole set. If the set is empty, the elements are unrelated. If it’s non-empty, each element is a separate witness to the relation.

This definition of a relation can be generalized to any category. In fact there is already a natural relation between objects in a category–the one defined by hom-sets. Two objects $a$ and $b$ are related this way if the hom-set $C(a, b)$ is non-empty. Each morphism in $C(a, b)$ serves as a witness to this relation.

With profunctors, we can define proof-relevant relations between objects that are taken from different categories. Object $c$ in $\mathbb{C}$ is related to object $d$ in $\mathbb{D}$ if $P(c, d)$ is a non-empty set. Moreover, each element of this set serves as a witness for the relation. Because of functoriality of $P$, this relation is compatible with the categorical structure, that is, it composes nicely with the relation defined by hom-sets.

In general, a composition of two relations $P$ and $Q$, denoted by $P \circ Q$ is defined as a path between objects. Objects $a$ and $c$ are related if there is a go-between object $b$ such that both $P(a, b)$ and $Q(b, c)$ are non-empty. As a witness of this relation we can pick any pair of elements, one from $P(a, b)$ and one from $Q(b, c)$.

By convention, a profunctor $P(a, b)$ is drawn as an arrow (often crossed) from $b$ to $a$, $a \nleftarrow b$.

Composition of profunctors/relations

Profunctor Composition

To create a set of all witnesses of $P \circ Q$ we have to sum over all possible intermediate objects and all pairs of witnesses. Roughly speaking, such a sum (modulo some identifications) is expressed categorically as a coend:

$(P \circ Q)(a, c) = \int^b P(a, b) \times Q(b, c)$

As a refresher, a coend of a profunctor $P$ is a set $\int^a P(a, a)$ equipped with a family of injections

$i_x \colon P(x, x) \to \int^a P(a, a)$

that is universal in the sense that, for any other set $s$ and a family:

$\alpha_x \colon P(x, x) \to s$

there is a unique function $h$ that factorizes them all:

$\alpha_x = h \circ i_x$

Universal property of a coend

Profunctor composition can be translated into pseudo-Haskell as:

type Procompose q p a c = exists b. (p a b, q b c)

where the coend is encoded as an existential data type. The actual implementation (again, see Edward Kmett’s Data.Profunctor.Composition) is:

data Procompose q p a c where
Procompose :: q b c -> p a b -> Procompose q p a c

The existential quantifier is expressed in terms of a GADT (Generalized Algebraic Data Type), with the free occurrence of b inside the data constructor.

Einstein’s Convention

By now you might be getting lost juggling the variances of objects appearing in those formulas. The coend variable, for instance, must appear under the integral sign once in the covariant and once in the contravariant position, and the variances on the right must match the variances on the left. Fortunately, there is a precedent in a different branch of mathematics, tensor calculus in vector spaces, with the kind of notation that takes care of variances. Einstein coopted and expanded this notation in his theory of relativity. Let’s see if we can adapt this technique to the calculus of profunctors.

The trick is to write contravariant indices as superscripts and the covariant ones as subscripts. So, from now on, we’ll write the components of a profunctor $p$ (we’ll switch to lower case to be compatible with Haskell) as $p^c\,_d$. Einstein also came up with a clever convention: implicit summation over a repeated index. In the case of profunctors, the summation corresponds to taking a coend. In this notation, a coend over a profunctor $p$ looks like a trace of a tensor:

$p^a\,_a = \int^a p(a, a)$

The composition of two profunctors becomes:

$(p \circ q)^a\, _c = p^a\,_b \, q^b\,_c = \int^b p(a, b) \times q(b, c)$

The summation convention applies only to adjacent indices. When they are separated by an explicit product sign (or any other operator), the coend is not assumed, as in:

$p^a\,_b \times q^b\,_c$

(no summation).

The hom-functor in a category $\mathbb{C}$ is also a profunctor, so it can be notated appropriately:

$C^a\,_b = C(a, b)$

The co-Yoneda lemma (see Ninja Yoneda) becomes:

$C^c\,_{c'}\,p^{c'}\,_d \cong p^c\,_d \cong p^c\,_{d'}\,D^{d'}\,_d$

suggesting that the hom-functors $C^c\,_{c'}$ and $D^{d'}\,_d$ behave like Kronecker deltas (in tensor-speak) or unit matrices. Here, the profunctor $p$ spans two categories

$p \colon \mathbb{C}^{op}\times \mathbb{D} \to Set$

The lifting of morphisms:

$f\colon c' \to c$
$g\colon d \to d'$

can be written as:

$p^f\,_g \colon p^c\,_d \to p^{c'}\,_{d'}$

There is one more useful identity that deals with mapping out from a coend. It’s the consequence of the fact that the hom-functor is continuous. It means that it maps (co-) limits to limits. More precisely, since the hom-functor is contravariant in the first variable, when we fix the target object, it maps colimits in the first variable to limits. (It also maps limits to limits in the second variable). Since a coend is a colimit, and an end is a limit, continuity leads to the following identity:

$Set(\int^c p(c, c), s) \cong \int_c Set(p(c, c), s)$

for any set $s$. Programmers know this identity as a generalization of case analysis: a function from a sum type is a product of functions (one function per case). If we interpret the coend as an existential quantifier, the end is equivalent to a universal quantifier.

Let’s apply this identity to the mapping out from a composition of two profunctors:

$p^a\,_b \, q^b\,_c \to s = Set\big(\int^b p(a, b) \times q(b, c), s\big)$

This is isomorphic to:

$\int_b Set\Big(p(a,b) \times q(b, c), s\Big)$

or, after currying (using the product/exponential adjunction),

$\int_b Set\Big(p(a, b), q(b, c) \to s\Big)$

This gives us the mapping out formula:

$p^a\,_b \, q^b\,_c \to s \cong p^a\,_b \to q^b\,_c \to s$

with the right hand side natural in $b$. Again, we don’t perform implicit summation on the right, where the repeated indices are separated by an arrow. There, the repeated index $b$ is universally quantified (through the end), giving rise to a natural transformation.

Bicategory Prof

Since profunctors can be composed using the coend formula, it’s natural to ask if there is a category in which they work as morphisms. The only problem is that profunctor composition satisfies the associativity and unit laws (see the co-Yoneda lemma above) only up to isomorphism. Not to worry, there is a name for that: a bicategory. In a bicategory we have objects, which are called 0-cells; morphisms, which are called 1-cells; and morphisms between morphisms, which are called 2-cells. When we say that categorical laws are satisfied up to isomorphism, it means that there is an invertible 2-cell that maps one side of the law to another.

The bicategory $Prof$ has categories as 0-cells, profunctors as 1-cells, and natural transformations as 2-cells. A natural transformation $\alpha$ between profunctors $p$ and $q$

$\alpha \colon p \Rightarrow q$

has components that are functions:

$\alpha^c\,_d \colon p^c\,_d \to q^c\,_d$

satisfying the usual naturality conditions. Natural transformations between profunctors can be composed as functions (this is called vertical composition). In fact 2-cells in any bicategory are composable, and there always is a unit 2-cell. It follows that 1-cells between any two 0-cells form a category called the hom-category.

But there is another way of composing 2-cells that’s called horizontal composition. In $Prof$, this horizontal composition is not the usual horizontal composition of natural transformations, because composition of profunctors is not the usual composition of functors. We have to construct a natural transformation between one composition of profuntors, say $p^a\,_b \, q^b\,_c$ and another, $r^a\,_b \, s^b\,_c$, having at our disposal two natural transformations:

$\alpha \colon p \Rightarrow r$

$\beta \colon q \Rightarrow s$

The construction is a little technical, so I’m moving it to the appendix. We will denote such horizontal composition as:

$(\alpha \circ \beta)^a\,_c \colon p^a\,_b \, q^b\,_c \to r^a\,_b \, s^b\,_c$

If one of the natural transformations is an identity natural transformation, say, from $p^a\,_b$ to $p^a\,_b$, horizontal composition is called whiskering and can be written as:

$(p \circ \beta)^a\,_c \colon p^a\,_b \, q^b\,_c \to p^a\,_b \, s^b\,_c$

The fact that a monad is a monoid in the category of endofunctors is a lucky accident. That’s because, in general, a monad can be defined in any bicategory, and $Cat$ just happens to be a (strict) bicategory. It has (small) categories as 0-cells, functors as 1-cells, and natural transformations as 2-cells. A monad is defined as a combination of a 0-cell (you need a category to define a monad), an endo-1-cell (that would be an endofunctor in that category), and two 2-cells. These 2-cells are variably called multiplication and unit, $\mu$ and $\eta$, or join and return.

Since $Prof$ is a bicategory, we can define a monad in it, and call it a promonad. A promonad consists of a 0-cell $C$, which is a category; an endo-1-cell $p$, which is a profunctor in that category; and two 2-cells, which are natural transformations:

$\mu^a\,_b \colon p^a\,_c \, p^c\,_b \to p^a\,_b$

$\eta^a\,_b \colon C^a\,_b \to p^a\,_b$

Remember that $C^a\,_b$ is the hom-profunctor in the category $C$ which, due to co-Yoneda, happens to be the unit of profunctor composition.

Programmers might recognize elements of the Haskell Arrow in it (see my blog post on monoids).

We can apply the mapping-out identity to the definition of multiplication and get:

$\mu^a\,_b \colon p^a\,_c \to p^c\,_b \to p^a\,_b$

Notice that this looks very much like composition of heteromorphisms. Moreover, the monadic unit $\eta$ maps regular morphisms to heteromorphisms. We can then construct a new category, whose objects are the same as the objects of $\mathbb{C}$, with hom-sets given by the profunctor $p$. That is, a hom set from $a$ to $b$ is the set $p^a\,_b$. We can define an identity-on-object functor $J$ from $\mathbb{C}$ to that category, whose action on hom-sets is given by $\eta$.

Interestingly, this construction also works in the opposite direction (as was brought to my attention by Alex Campbell). Any indentity-on-objects functor defines a promonad. Indeed, given a functor $J$, we can always turn it into a profunctor:

$p(c, d) = D(J\, c, J\, d)$

$p^c\,_d = D^{J\, c}\,_{J\, d}$

Since $J$ is identity on objects, the composition of morphisms in $D$ can be used to define the composition of heteromorphisms. This, in turn, can be used to define $\mu$, thus showing that $p$ is a promonad on $\mathbb{C}$.

Conclusion

I realize that I have touched upon some pretty advanced topics in category theory, like bicategories and promonads, so it’s a little surprising that these concepts can be illustrated in Haskell, some of them being present in popular libraries, like the Arrow library, which has applications in functional reactive programming.

I’ve been experimenting with applying Einstein’s summation convention to profunctors, admittedly with mixed results. This is definitely work in progress and I welcome suggestions to improve it. The main problem is that we sometimes need to apply the sum (coend), and at other times the product (end) to repeated indices. This is in particular awkward in the formulation of the mapping out property. I suggest separating the non-summed indices with product signs or arrows but I’m not sure how well this will work.

Appendix: Horizontal Composition in Prof

We have at our disposal two natural transformations:

$\alpha \colon p \Rightarrow r$

$\beta \colon q \Rightarrow s$

and the following coend, which is the composition of the profunctors $p$ and $q$:

$\int^b p(a, b) \times q(b, c)$

Our goal is to construct an element of the target coend:

$\int^b r(a, b) \times s(b, c)$

Horizontal composition of 2-cells

To construct an element of a coend, we need to provide just one element of $r(a, b') \times s(b', c)$ for some $b'$. We’ll look for a function that would construct such an element in the following hom-set:

$Set\Big(\int^b p(a, b) \times q(b, c), r(a, b') \times s(b', c)\Big)$

Using Einstein notation, we can write it as:

$p^a\,_b \, q^b\,_c \to r^a\,_{b'} \times s^{b'}\,_c$

and then use the mapping out property:

$p^a\,_b \to q^b\,_c \to r^a\,_{b'} \times s^{b'}\,_c$

We can pick $b'$ equal to $b$ and implement the function using the components of the two natural transformations, $\alpha^a\,_{b} \times \beta^{b}\,_c$.

Of course, this is how a programmer might think of it. A mathematician will use the universal property of the coend $(p \circ q)^a\,_c$, as in the diagram below (courtesy Alex Campbell).

Horizontal composition using the universal property of a coend

In Haskell, we can define a natural transformation between two (endo-) profunctors as a polymorphic function:

newtype PNat p q = PNat (forall a b. p a b -> q a b)

Horizontal composition is then given by:

horPNat :: PNat p r -> PNat q s -> Procompose p q a c
-> Procompose r s a c
horPNat (PNat alpha) (PNat beta) (Procompose pbc qdb) =
Procompose (alpha pbc) (beta qdb)


Acknowledgment

I’m grateful to Alex Campbell from Macquarie University in Sydney for extensive help with this blog post.

Yes, it’s this time of the year again! I started a little tradition a year ago with Stalking a Hylomorphism in the Wild. This year I was reminded of the Advent of Code by a tweet with this succint C++ program:

This piece of code is probably unreadable to a regular C++ programmer, but makes perfect sense to a Haskell programmer.

Here’s the description of the problem: You are given a list of equal-length strings. Every string is different, but two of these strings differ only by one character. Find these two strings and return their matching part. For instance, if the two strings were “abcd” and “abxd”, you would return “abd”.

What makes this problem particularly interesting is its potential application to a much more practical task of matching strands of DNA while looking for mutations. I decided to explore the problem a little beyond the brute force approach. And, of course, I had a hunch that I might encounter my favorite wild beast–the hylomorphism.

Brute force approach

First things first. Let’s do the boring stuff: read the file and split it into lines, which are the strings we are supposed to process. So here it is:

main = do
let cs = lines txt
print $findMatch cs The real work is done by the function findMatch, which takes a list of strings and produces the answer, which is a single string. findMatch :: [String] -> String First, let’s define a function that calculates the distance between any two strings. distance :: (String, String) -> Int We’ll define the distance as the count of mismatched characters. Here’s the idea: We have to compare strings (which, let me remind you, are of equal length) character by character. Strings are lists of characters. The first step is to take two strings and zip them together, producing a list of pairs of characters. In fact we can combine the zipping with the next operation–in this case, comparison for inequality, (/=)–using the library function zipWith. However, zipWith is defined to act on two lists, and we will want it to act on a pair of lists–a subtle distinction, which can be easily overcome by applying uncurry: uncurry :: (a -> b -> c) -> ((a, b) -> c) which turns a function of two arguments into a function that takes a pair. Here’s how we use it: uncurry (zipWith (/=)) The comparison operator (/=) produces a Boolean result, True or False. We want to count the number of differences, so we’ll covert True to one, and False to zero: fromBool :: Num a => Bool -> a fromBool False = 0 fromBool True = 1 (Notice that such subtleties as the difference between Bool and Int are blisfully ignored in C++.) Finally, we’ll sum all the ones using sum. Altogether we have: distance = sum . fmap fromBool . uncurry (zipWith (/=))  Now that we know how to find the distance between any two strings, we’ll just apply it to all possible pairs of strings. To generate all pairs, we’ll use list comprehension: let ps = [(s1, s2) | s1 <- ss, s2 <- ss] (In C++ code, this was done by cartesian_product.) Our goal is to find the pair whose distance is exactly one. To this end, we’ll apply the appropriate filter: filter ((== 1) . distance) ps For our purposes, we’ll assume that there is exactly one such pair (if there isn’t one, we are willing to let the program fail with a fatal exception). (s, s') = head$ filter ((== 1) . distance) ps

The final step is to remove the mismatched character:

filter (uncurry (==)) $zip s s' We use our friend uncurry again, because the equality operator (==) expects two arguments, and we are calling it with a pair of arguments. The result of filtering is a list of identical pairs. We’ll fmap fst to pick the first components. findMatch :: [String] -> String findMatch ss = let ps = [(s1, s2) | s1 <- ss, s2 <- ss] (s, s') = head$ filter ((== 1) . distance) ps
in fmap fst $filter (uncurry (==))$ zip s s'

This program produces the correct result and we could stop right here. But that wouldn’t be much fun, would it? Besides, it’s possible that other algorithms could perform better, or be more flexible when applied to a more general problem.

Data-driven approach

The main problem with our brute-force approach is that we are comparing everything with everything. As we increase the number of input strings, the number of comparisons grows like a factorial. There is a standard way of cutting down on the number of comparison: organizing the input into a neat data structure.

We are comparing strings, which are lists of characters, and list comparison is done recursively. Assume that you know that two strings share a prefix. Compare the next character. If it’s equal in both strings, recurse. If it’s not, we have a single character fault. The rest of the two strings must now match perfectly to be considered a solution. So the best data structure for this kind of algorithm should batch together strings with equal prefixes. Such a data structure is called a prefix tree, or a trie (pronounced try).

At every level of our prefix tree we’ll branch based on the current character (so the maximum branching factor is, in our case, 26). We’ll record the character, the count of strings that share the prefix that led us there, and the child trie storing all the suffixes.

data Trie = Trie [(Char, Int, Trie)]
deriving (Show, Eq)

Here’s an example of a trie that stores just two strings, “abcd” and “abxd”. It branches after b.

   a 2
b 2
c 1    x 1
d 1    d 1

When inserting a string into a trie, we recurse both on the characters of the string and the list of branches. When we find a branch with the matching character, we increment its count and insert the rest of the string into its child trie. If we run out of branches, we create a new one based on the current character, give it the count one, and the child trie with the rest of the string:

insertS :: Trie -> String -> Trie
insertS t "" = t
insertS (Trie bs) s = Trie (inS bs s)
where
inS ((x, n, t) : bs) (c : cs) =
if c == x
then (c, n + 1, insertS t cs) : bs
else (x, n, t) : inS bs (c : cs)
inS [] (c : cs) = [(c, 1, insertS (Trie []) cs)]

We convert our input to a trie by inserting all the strings into an (initially empty) trie:

mkTrie :: [String] -> Trie
mkTrie = foldl insertS (Trie [])

Of course, there are many optimizations we could use, if we were to run this algorithm on big data. For instance, we could compress the branches as is done in radix trees, or we could sort the branches alphabetically. I won’t do it here.

I won’t pretend that this implementation is simple and elegant. And it will get even worse before it gets better. The problem is that we are dealing explicitly with recursion in multiple dimensions. We recurse over the input string, the list of branches at each node, as well as the child trie. That’s a lot of recursion to keep track of–all at once.

Now brace yourself: We have to traverse the trie starting from the root. At every branch we check the prefix count: if it’s greater than one, we have more than one string going down, so we recurse into the child trie. But there is also another possibility: we can allow to have a mismatch at the current level. The current characters may be different but, since we allow only one mismatch, the rest of the strings have to match exactly. That’s what the function exact does. Notice that exact t is used inside foldMap, which is a version of fold that works on monoids–here, on strings.

match1 :: Trie -> [String]
match1 (Trie bs) = go bs
where
go :: [(Char, Int, Trie)] -> [String]
go ((x, n, t) : bs) =
let a1s = if n > 1
then fmap (x:) $match1 t else [] a2s = foldMap (exact t) bs a3s = go bs -- recurse over list in a1s ++ a2s ++ a3s go [] = [] exact t (_, _, t') = matchAll t t' Here’s the function that finds all exact matches between two tries. It does it by generating all pairs of branches in which top characters match, and then recursing down. matchAll :: Trie -> Trie -> [String] matchAll (Trie bs) (Trie bs') = mAll bs bs' where mAll :: [(Char, Int, Trie)] -> [(Char, Int, Trie)] -> [String] mAll [] [] = [""] mAll bs bs' = let ps = [ (c, t, t') | (c, _, t) <- bs , (c', _', t') <- bs' , c == c'] in foldMap go ps go (c, t, t') = fmap (c:) (matchAll t t') When mAll reaches the leaves of the trie, it returns a singleton list containing an empty string. Subsequent actions of fmap (c:) will prepend characters to this string. Since we are expecting exactly one solution to the problem, we’ll extract it using head: findMatch1 :: [String] -> String findMatch1 cs = head$ match1 (mkTrie cs)

Recursion schemes

As you hone your functional programming skills, you realize that explicit recursion is to be avoided at all cost. There is a small number of recursive patterns that have been codified, and they can be used to solve the majority of recursion problems (for some categorical background, see F-Algebras). Recursion itself can be expressed in Haskell as a data structure: a fixed point of a functor:

newtype Fix f = In { out :: f (Fix f) }

In particular, our trie can be generated from the following functor:

data TrieF a = TrieF [(Char, a)]
deriving (Show, Functor)

Notice how I have replaced the recursive call to the Trie type constructor with the free type variable a. The functor in question defines the structure of a single node, leaving holes marked by the occurrences of a for the recursion. When these holes are filled with full blown tries, as in the definition of the fixed point, we recover the complete trie.

I have also made one more simplification by getting rid of the Int in every node. This is because, in the recursion scheme I’m going to use, the folding of the trie proceeds bottom-up, rather than top-down, so the multiplicity information can be passed upwards.

The main advantage of recursion schemes is that they let us use simpler, non-recursive building blocks such as algebras and coalgebras. Let’s start with a simple coalgebra that lets us build a trie from a list of strings. A coalgebra is a fancy name for a particular type of function:

type Coalgebra f x = x -> f x

Think of x as a type for a seed from which one can grow a tree. A colagebra tells us how to use this seed to create a single node described by the functor f and populate it with (presumably smaller) seeds. We can then pass this coalgebra to a simple algorithm, which will recursively expand the seeds. This algorithm is called the anamorphism:

ana :: Functor f => Coalgebra f a -> a -> Fix f
ana coa = In . fmap (ana coa) . coa

Let’s see how we can apply it to the task of building a trie. The seed in our case is a list of strings (as per the definition of our problem, we’ll assume they are all equal length). We start by grouping these strings into bunches of strings that start with the same character. There is a library function called groupWith that does exactly that. We have to import the right library:

import GHC.Exts (groupWith)

This is the signature of the function:

groupWith :: Ord b => (a -> b) -> [a] -> [[a]]

It takes a function a -> b that converts each list element to a type that supports comparison (as per the typeclass Ord), and partitions the input into lists that compare equal under this particular ordering. In our case, we are going to extract the first character from a string using head and bunch together all strings that share that first character.

let sss = groupWith head ss

The tails of those strings will serve as seeds for the next tier of the trie. Eventually the strings will be shortened to nothing, triggering the end of recursion.

fromList :: Coalgebra TrieF [String]
fromList ss =
-- are strings empty? (checking one is enough)
then TrieF [] -- leaf
else
let sss = groupWith head ss
in TrieF $fmap mkBranch sss The function mkBranch takes a bunch of strings sharing the same first character and creates a branch seeded with the suffixes of those strings. mkBranch :: [String] -> (Char, [String]) mkBranch sss = let c = head (head sss) -- they're all the same in (c, fmap tail sss) Notice that we have completely avoided explicit recursion. The next step is a little harder. We have to fold the trie. Again, all we have to define is a step that folds a single node whose children have already been folded. This step is defined by an algebra: type Algebra f x = f x -> x Just as the type x described the seed in a coalgebra, here it describes the accumulator–the result of the folding of a recursive data structure. We pass this algebra to a special algorithm called a catamorphism that takes care of the recursion: cata :: Functor f => Algebra f a -> Fix f -> a cata alg = alg . fmap (cata alg) . out Notice that the folding proceeds from the bottom up: the algebra assumes that all the children have already been folded. The hardest part of designing an algebra is figuring out what information needs to be passed up in the accumulator. We obviously need to return the final result which, in our case, is the list of strings with one mismatched character. But when we are in the middle of a trie, we have to keep in mind that the mismatch may still happen above us. So we also need a list of strings that may serve as suffixes when the mismatch occurs. We have to keep them all, because they might be matched later with strings from other branches. In other words, we need to be accumulating two lists of strings. The first list accumulates all suffixes for future matching, the second accumulates the results: strings with one mismatch (after the mismatch has been removed). We therefore should implement the following algebra: Algebra TrieF ([String], [String]) To understand the implementation of this algebra, consider a single node in a trie. It’s a list of branches, or pairs, whose first component is the current character, and the second a pair of lists of strings–the result of folding a child trie. The first list contains all the suffixes gathered from lower levels of the trie. The second list contains partial results: strings that were matched modulo single-character defect. As an example, suppose that you have a node with two branches: [ ('a', (["bcd", "efg"], ["pq"])) , ('x', (["bcd"], []))] First we prepend the current character to strings in both lists using the function prep with the following signature: prep :: (Char, ([String], [String])) -> ([String], [String]) This way we convert each branch to a pair of lists. [ (["abcd", "aefg"], ["apq"]) , (["xbcd"], [])] We then merge all the lists of suffixes and, separately, all the lists of partial results, across all branches. In the example above, we concatenate the lists in the two columns. (["abcd", "aefg", "xbcd"], ["apq"])  Now we have to construct new partial results. To do this, we create another list of accumulated strings from all branches (this time without prefixing them): ss = concat$ fmap (fst . snd) bs

In our case, this would be the list:

["bcd", "efg", "bcd"]

To detect duplicate strings, we’ll insert them into a multiset, which we’ll implement as a map. We need to import the appropriate library:

import qualified Data.Map as M

and define a multiset Counts as:

type Counts a = M.Map a Int

Every time we add a new item, we increment the count:

add :: Ord a => Counts a -> a -> Counts a
add cs c = M.insertWith (+) c 1 cs

To insert all strings from a list, we use a fold:

mset = foldl add M.empty ss

We are only interested in items that have multiplicity greater than one. We can filter them and extract their keys:

dups = M.keys $M.filter (> 1) mset Here’s the complete algebra: accum :: Algebra TrieF ([String], [String]) accum (TrieF []) = ([""], []) accum (TrieF bs) = -- b :: (Char, ([String], [String])) let -- prepend chars to string in both lists pss = unzip$ fmap prep bs
(ss1, ss2) = both concat pss
-- find duplicates
ss = concat $fmap (fst . snd) bs mset = foldl add M.empty ss dups = M.keys$ M.filter (> 1) mset
in (ss1, dups ++ ss2)
where
prep :: (Char, ([String], [String])) -> ([String], [String])
prep (c, pss) = both (fmap (c:)) pss

I used a handy helper function that applies a function to both components of a pair:

both :: (a -> b) -> (a, a) -> (b, b)
both f (x, y) = (f x, f y)

And now for the grand finale: Since we create the trie using an anamorphism only to immediately fold it using a catamorphism, why don’t we cut the middle person? Indeed, there is an algorithm called the hylomorphism that does just that. It takes the algebra, the coalgebra, and the seed, and returns the fully charged accumulator.

hylo :: Functor f => Algebra f a -> Coalgebra f b -> b -> a
hylo alg coa = alg . fmap (hylo alg coa) . coa

And this is how we extract and print the final result:

print $head$ snd \$ hylo accum fromList cs

Conclusion

The advantage of using the hylomorphism is that, because of Haskell’s laziness, the trie is never wholly constructed, and therefore doesn’t require large amounts of memory. At every step enough of the data structure is created as is needed for immediate computation; then it is promptly released. In fact, the definition of the data structure is only there to guide the steps of the algorithm. We use a data structure as a control structure. Since data structures are much easier to visualize and debug than control structures, it’s almost always advantageous to use them to drive computation.

In fact, you may notice that, in the very last step of the computation, our accumulator recreates the original list of strings (actually, because of laziness, they are never fully reconstructed, but that’s not the point). In reality, the characters in the strings are never copied–the whole algorithm is just a choreographed dance of internal pointers, or iterators. But that’s exactly what happens in the original C++ algorithm. We just use a higher level of abstraction to describe this dance.

I haven’t looked at the performance of various implementations. Feel free to test it and report the results. The code is available on github.

Acknowledgments

I’m grateful to the participants of the Seattle Haskell Users’ Group for many helpful comments during my presentation.

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