Category Theory for Programmers Challenges

Part One

1 Category: The Essence of Composition

  1. Implement, as best as you can, the identity function in your favorite language (or the second favorite, if your favorite language happens to be Haskell).

    See here.

  2. Implement the composition function in your favorite language. It takes two functions as arguments and returns a function that is their composition.

    See here.

  3. Write a program that tries to test that your composition function respects identity.

    See here.

  4. Is the world-wide web a category in any sense? Are links morphisms?

    Not sure.

  5. Is Facebook a category, with people as objects and friendships as morphisms?

    Not sure.

  6. When is a directed graph a category?

    When for every node, there is a edge from it to itself, and for every pair of edges a -> b and b -> c, there is an edge a -> c.

2 Types and Functions

  1. Define a higher-order function (or a function object) memoize in your favorite language. This function takes a pure function f as an argument and returns a function that behaves almost exactly like f, except that it only calls the original function once for every argument, stores the result internally, and subsequently returns this stored result every time it’s called with the same argument. You can tell the memoized function from the original by watching its performance. For instance, try to memoize a function that takes a long time to evaluate. You’ll have to wait for the result the first time you call it, but on subsequent calls, with the same argument, you should get the result immediately.

    See here.

  2. Try to memoize a function from your standard library that you normally use to produce random numbers. Does it work?

    No, it does not work.

    See here.

  3. Most random number generators can be initialized with a seed. Implement a function that takes a seed, calls the random number generator with that seed, and returns the result. Memoize that function. Does it work?

    Yes it works.

    See here.

  4. Which of these C++ functions are pure? Try to memoize them and observe what happens when you call them multiple times: memoized and not.

    1. The factorial function from the example in the text.
    2. std::getchar()
    3. bool f() {
          std::cout << "Hello!" << std::endl;
          return true;
    4. int f(int x) {
          static int y = 0;
          y += x;
          return y;
    1. Pure.
    2. Not pure.
    3. Not pure.
    4. Not pure.
  5. How many different functions are there from Bool to Bool? Can you implement them all?

    There are 4 pure functions from Bool to Bool.

    See here.

  6. Draw a picture of a category whose only objects are the types Void, () (unit), and Bool; with arrows corresponding to all possible functions between these types. Label the arrows with the names of the functions.

    Not sure.

3 Categories Great and Small

  1. Generate a free category from:

    1. A graph with one node and no edges
    2. A graph with one node and one (directed) edge (hint: this edge can be composed with itself)
    3. A graph with two nodes and a single arrow between them
    4. A graph with a single node and 26 arrows marked with the letters of the alphabet: a, b, c … z.
    1. Add an edge from the node to itself.
    2. Do nothing.
    3. Add an edge to each node from that node to itself.
    4. Not sure.
  2. What kind of order is this?

    1. A set of sets with the inclusion relation: A is included in B if every element of A is also an element of B.
    2. C++ types with the following subtyping relation: T1 is a subtype of T2 if a pointer to T1 can be passed to a function that expects a pointer to T2 without triggering a compilation error.
    1. Partial order.
    2. Partial order.
  3. Considering that Bool is a set of two values True and False, show that it forms two (set-theoretical) monoids with respect to, respectively, operator && (AND) and || (OR).

    1. For &&:
      • Neutral value is True.
      • a && True == True && a.
      • (a && b) && c == a && (b && c).
    2. For ||:
      • Neutral value is False.
      • a || False == False || a.
      • (a || b) || c == a || (b || c).
  4. Represent the Bool monoid with the AND operator as a category: List the morphisms and their rules of composition.


    • \x -> x && False
    • \x -> x && True

    Rules of composition:

    • (\x -> x && False) . (\x -> x && False) == \x -> x && False
    • (\x -> x && False) . (\x -> x && True) == \x -> x && False
    • (\x -> x && True) . (\x -> x && False) == \x -> x && False
    • (\x -> x && True) . (\x -> x && True) == \x -> x && True
  5. Represent addition modulo 3 as a monoid category.


    • \x -> mod (x + 0) 3
    • \x -> mod (x + 1) 3
    • \x -> mod (x + 2) 3

    Rules of composition:

    • (\x -> mod (x + m) 3) . (\x -> mod (x + n) 3) == \x -> mod (x + (m + n)) 3

4 Kleisli Categories

A function that is not defined for all possible values of its argument is called a partial function. It’s not really a function in the mathematical sense, so it doesn’t fit the standard categorical mold. It can, however, be represented by a function that returns an embellished type optional:

template<class A> class optional {
    bool _isValid;
    A _value;
    optional() : _isValid(false) {}
    optional(A v) : _isValid(true), _value(v) {}
    bool isValid() const { return _isValid; }
    A value() const { return _value; }

As an example, here’s the implementation of the embellished function safe_root:

optional<double> safe_root(double x) {
    if (x >= 0) return optional<double>{sqrt(x)};
    else return optional<double>{};

Here’s the challenge:

  1. Construct the Kleisli category for partial functions (define composition and identity).
  2. Implement the embellished function safe_reciprocal that returns a valid reciprocal of its argument, if it’s different from zero.
  3. Compose safe_root and safe_reciprocal to implement safe_root_reciprocal that calculates sqrt(1/x) whenever possible.
  1. Composition:

    • (A -> optional<B>) . (B -> optional<C>)A -> optional<C>.


    • The optional constructor, which has the type A -> optional<A>.
  2. See here.

  3. See here.

5 Products and Coproducts

  1. Show that the terminal object is unique up to unique isomorphism.

    Suppose that we have two terminal objects t1 and t2. Since t1 is terminal, there is a unique morphism f from t2 to t1. By the same token, since t2 is terminal, there is a unique morphism g from t1 to t2. The composition gf must be a morphism from t2 to t2. But t2 is terminal so there can only be one morphism going from t2 to t2. Since we are in a category, we know that there is an identity morphism from t2 to t2, and since there is room from only one, that must be it. Therefore gf is equal to identity. Similarly, fg must be equal to identity, because there can be only one morphism from t1 back to t1. This proves that f and g must be the inverse of each other. Therefore any two terminal objects are isomorphic.

  2. What is a product of two objects in a poset? Hint: Use the universal construction.

    A product of two objects a and b in a poset is the object c that ca and cb such that for any other object c′ that c′ ≤ a and c′ ≤ b, c′ ≤ c.

  3. What is a coproduct of two objects in a poset?

    A coproduct of two objects a and b in a poset is the object c that ac and bc such that for any other object c′ that ac′ and bc′, cc′.

  4. Implement the equivalent of Haskell Either as a generic type in your favorite language (other than Haskell).

    See here.

  5. Show that Either is a “better” coproduct than int equipped with two injections:

    int i(int n) { return n; }
    int j(bool b) { return b ? 0: 1; }

    Hint: Define a function

    int m(Either const & e);

    that factorizes i and j.

    See here.

  6. Continuing the previous problem: How would you argue that int with the two injections i and j cannot be “better” than Either?

    If int with the two injections i and j is better than Either, there should be a function f that maps int to Either where fi == Left and fj == Right. But after applying i or j, We can not know the origin object is int or bool, so we can not determine whether to use Left or Right to construct Either.

  7. Still continuing: What about these injections?

    int i(int n) {
        if (n < 0) return n;
        return n + 2;
    int j(bool b) { return b ? 0: 1; }

    No, because the codomain of i is smaller than the domain. For each result from i, there may be more than one input corresponds to it. So information is lost. We cannot construct an Either object without losing information.

  8. Come up with an inferior candidate for a coproduct of int and bool that cannot be better than Either because it allows multiple acceptable morphisms from it to Either.

    See here.

6 Simple Algebraic Data Types

  1. Show the isomorphism between Maybe a and Either () a.

    See here.

  2. Hereʼs a sum type defined in Haskell:

    data Shape = Circle Float
               | Rect Float Float

    When we want to define a function like area that acts on a Shape, we do it by pattern matching on the two constructors:

    area :: Shape -> Float
    area (Circle r) = pi * r * r
    area (Rect d h) = d * h

    Implement Shape in C++ or Java as an interface and create two classes: Circle and Rect. Implement area as a virtual function.

    See here.

  3. Continuing with the previous example: We can easily add a new function circ that calculates the circumference of a Shape. We can do it without touching the definition of Shape:

    circ :: Shape -> Float
    circ (Circle r) = 2.0 * pi * r
    circ (Rect d h) = 2.0 * (d + h)

    Add circ to your C++ or Java implementation. What parts of the original code did you have to touch?

    See here.

  4. Continuing further: Add a new shape, Square, to Shape and make all the necessary updates. What code did you have to touch in Haskell vs. C++ or Java? (Even if youʼre not a Haskell programmer, the modifications should be pretty obvious.)

    See here.

  5. Show that a + a = 2 × a holds for types (up to isomorphism). Remember that 2 corresponds to Bool, according to our translation table.

    • a + aEither a a
    • 2 × a(Bool, a)

    See here for definitions for morphisms.

7 Functors

  1. Can we turn the Maybe type constructor into a functor by defining:

    fmap _ _ = Nothing

    which ignores both of its arguments? (Hint: Check the functor laws.)

    No. Because fmap id (Just 4) = Nothing, while id (Just 4) = Just 4, you can see that this version of fmap does not preserve identity.

  2. Prove functor laws for the reader functor. Hint: itʼs really simple.

    • fmap id x = id . x = x, so identity is preserved.
    • fmap (f . g) x = (f . g) . x, (fmap f . fmap g) x = fmap f (fmap g x) = fmap f (g . x) = f . (g . x) = (f . g) . x, so fmap (f . g) x = (fmap f . fmap g) x, so composition is preserved.
  3. Implement the reader functor in your second favorite language (the first being Haskell, of course).

    See here.

  4. Prove the functor laws for the list functor. Assume that the laws are true for the tail part of the list youʼre applying it to (in other words, use induction).

    • Nil case:
      • Preservation of identity:

        fmap id Nil = Nil.

      • Preservation of composition:

        fmap (f . g) Nil = Nil, and (fmap f . fmap g) Nil = fmap f (fmap g Nil) = fmap f Nil = Nil, so fmap (f . g) Nil = (fmap f . fmap g) Nil

    • Cons x t case:
      • Preservation of identity:

        fmap id (Cons x t) = Cons (id x) (fmap id t). Since id x = x, and by induction, fmap id t = t, so Cons (id x) (fmap id t) = Cons x t, so fmap id (Cons x t) = Cons x t

      • Preservation of composition:

        fmap (f . g) (Cons x t) = Cons ((f . g) x) (fmap (f . g) t), and (fmap f . fmap g) (Cons x t) = fmap f (fmap g (Cons x t)) = fmap f (Cons (g x) (fmap g t)) = Cons (f (g x)) (fmap f (fmap g t)) = Cons ((f . g) x) ((fmap f . fmap g) t). By induction, fmap (f . g) t = (fmap f . fmap g) t, so Cons ((f . g) x) (fmap (f . g) t) = Cons ((f . g) x) ((fmap f . fmap g) t), so fmap (f . g) (Cons x t) = (fmap f . fmap g) (Cons x t).

8 Functoriality

  1. Show that the data type:

    data Pair a b = Pair a b

    is a bifunctor. For additional credit implement all three methods of Bifunctor and use equational reasoning to show that these definitions are compatible with the default implementations whenever they can be applied.

    Haskell implementations:

    instance Bifunctor Pair where
        bimap f g (a, b) = (f a, g b)
        first f (a, b) = (f a, b)
        second g (a, b) = (a, g b)

    Proofs that my implementations are compatible with the default implementations:

    • Proof that bimap f g (a, b) = (first f . second g) (a, b):

      • bimap f g (a, b) = (f a, g b)
      • (first f . second g) (a, b) = first f (second g (a, b)) = first f (a, g b) = (f a, g b)
    • Proof that first f (a, b) = bimap f id (a, b):

      • first f (a, b) = (f a, b)
      • bimap f id (a, b) = (f a, id b) = (f a, b)
    • Proof that second g (a, b) = bimap id g (a, b):

      • second g (a, b) = (a, g b)
      • bimap id g (a, b) = (id a, g b) = (a, g b)
  2. Show the isomorphism between the standard definition of Maybe and this desugaring:

    type Maybe' a = Either (Const () a) (Identity a)

    Hint: Define two mappings between the two implementations. For additional credit, show that they are the inverse of each other using equational reasoning.

    standardToDesugaring Nothing = Left (Const ())
    standardToDesugaring (Just a) = Right a
    desugaringToStandard Left (Const ()) = Nothing
    desugaringToStandard Right a = Just a

    Proofs that standardToDesugaring and desugaringToStandard are the inverse of each other.

    • Proof that standardToDesugaring (desugaringToStandard a) = a:

      • standardToDesugaring (desugaringToStandard (Left (Const ()))) = standardToDesugaring Nothing = Left (Const ())
      • standardToDesugaring (desugaringToStandard (Right a)) = standardToDesugaring (Just a)) = Right a
    • Proof that desugaringToStandard (standardToDesugaring a) = a:

      • desugaringToStandard (standardToDesugaring Nothing) = desugaringToStandard (Left (Const ())) = Nothing
      • desugaringToStandard (standardToDesugaring (Just a)) = desugaringToStandard (Right a)) = Just a
  3. Letʼs try another data structure. I call it a PreList because itʼs a precursor to a List. It replaces recursion with a type parameter b.

    data PreList a b = Nil | Cons a b

    You could recover our earlier definition of a List by recursively applying PreList to itself (weʼll see how itʼs done when we talk about fixed points).

    Show that PreList is an instance of Bifunctor.

    First, I will implement bimap:

    bimap f g Nil = Nil
    bimap f g (Cons a b) = Cons (f a) (g b)

    Then, I will show that (bimap f g . bimap m n) a = bimap (f . m) (g . n) a:

    • Case Nil:

      • (bimap f g . bimap m n) Nil = bimap f g (bimap m n Nil) = bimap f g Nil = Nil
      • bimap (f . m) (g . n) Nil = Nil
    • Case (Cons a b):

      • (bimap f g . bimap m n) (Cons a b) = bimap f g (bimap m n (Cons a b)) = bimap f g (Cons (m a) (n b)) = Cons (f (m a)) (g (n b))
      • bimap (f . m) (g . n) (Cons a b) = Cons ((f . m) a) ((g . n) b) = Cons (f (m a)) (g (n b))

    So in both cases, composition is preserved.

    Finally, I will show that bimap id id a = a:

    • Case Nil:

      • bimap id id Nil = Nil
    • Case (Cons a b):

      • bimap id id (Cons a b) = Cons (id a) (id b) = (Cons a b)

    So in both cases, identity is preserved.

    So PreList is an instance of Bifunctor.

  4. Show that the following data types define bifunctors in a and b:

    data K2 c a b = K2 c
    data Fst a b = Fst a
    data Snd a b = Snd b

    For additional credit, check your solutions against Conor McBrideʼs paper Clowns to the Left of me, Jokers to the Right.

    For K2, define bimap as follows:

    bimap f g = id

    So bimap f g . bimap m n = id . id = id, and bimap (f . m) (g . n) = id, so bimap f g . bimap m n = bimap (f . m) (g . n). Also, bimap id id = id, so K2 c is a bifunctor.

    For Fst, define bimap as follows:

    bimap f g (Fst a) = Fst (f a)

    So (bimap f g . bimap m n) (Fst a) = bimap f g (bimap m n (Fst a)) = bimap f g (Fst (m a)) = Fst (f (m a)), and bimap (f . m) (g . n) (Fst a) = Fst ((f . m) a) = Fst (f (m a)). So (bimap f g . bimap m n) (Fst a) = bimap (f . m) (g . n) (Fst a). Also, bimap id id (Fst a) = Fst (id a) = Fst a, so Fst is a bifunctor. The same for Snd.

  5. Define a bifunctor in a language other than Haskell. Implement bimap for a generic pair in that language.

    See here.

  6. Should std::map be considered a bifunctor or a profunctor in the two template arguments Key and T? How would you redesign this data type to make it so?

    Not sure.

9 Function Types

10 Natural Transformations

  1. Define a natural transformation from the Maybe functor to the list functor. Prove the naturality condition for it.

    alpha Nothing = Nil
    alpha (Just a) = Cons a Nil

    We need to prove (fmap f . alpha) x = (alpha . fmap f) x:

    • Nothing case:
      • (fmap f . alpha) Nothing
        = fmap f (alpha Nothing)
        = fmap f Nil
        = Nil
      • (alpha . fmap f) Nothing
        = alpha (fmap f Nothing)
        = alpha Nothing
        = Nil
    • Just a case:
      • (fmap f . alpha) (Just a)
        = fmap f (alpha (Just a))
        = fmap f (Cons a Nil)
        = Cons (f a) Nil
      • (alpha . fmap f) (Just a)
        = alpha (fmap f (Just a))
        = alpha (Just (f a))
        = Cons (f a) Nil
  2. Define at least two different natural transformations between Reader () and the list functor. How many different lists of () are there?

    alpha1 f = Cons (f ()) Nil
    alpha2 f = Cons (f ()) (Cons (f ()) Nil)

    There are infinite countable number of lists of ().

  3. Continue the previous exercise with Reader Bool and Maybe.

    alpha1 f = Nothing
    alpha2 f = Just (f False)

    I can also define alpha3 f = Just (f True).

  4. Show that horizontal composition of natural transformation satisfies the naturality condition (hint: use components). Itʼs a good exercise in diagram chasing.

    We need to prove \(\left(G' \circ F'\right) f \circ \left(β \circ α\right)_a = \left(β \circ α\right)_b \circ \left(G \circ F\right) f\).

    Let \(f : a → b\), then
    \(\left(\left(G' \circ F'\right) f \circ \left(β \circ α\right)_a\right) \left(\left(G \circ F\right) a\right) = \left(\left(G' \circ F'\right) f\right) \left(\left(G' \circ F'\right) a\right) = \left(G' \circ F'\right) b\), Also,
    \(\left(\left(β \circ α\right)_b \circ \left(G \circ F\right) f\right) \left(\left(G \circ F\right) a\right) = \left(β \circ α\right)_b \left(\left(G \circ F\right) b\right) = \left(G' \circ F'\right) b\).

  5. Write a short essay about how you may enjoy writing down the evident diagrams needed to prove the interchange law.


  6. Create a few test cases for the opposite naturality condition of transformations between different Op functors. Hereʼs one choice:

    op :: Op Bool Int
    op = Op (\x -> x > 0)


    f :: String -> Int
    f x = read x
    op2 :: Op () Int
    op2 = Op (\x -> ())
    g :: String -> Int
    g x = read x