577 lines
18 KiB
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577 lines
18 KiB
Markdown
---
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layout: post
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title: "Typoclassopedia: Exercise solutions"
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date: 2017-09-27
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permalink: typoclassopedia-exercise-solutions/
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categories: programming
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math: true
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toc: true
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---
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I wanted to get proficient in Haskell so I decided to follow [An [Essential] Haskell Reading List](http://www.stephendiehl.com/posts/essential_haskell.html). There I stumbled upon [Typoclassopedia](https://wiki.haskell.org/Typeclassopedia), while the material is great, I couldn't find solutions for the exercises to check against, so I decided I would write my own and hopefully the solutions would get fixed in case I have gone wrong by others. So if you think a solution is wrong, let me know in the comments!
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In each section below, I left some reference material for the exercises and then the solutions.
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Functor
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==========
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## Instances
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```haskell
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instance Functor [] where
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fmap :: (a -> b) -> [a] -> [b]
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fmap _ [] = []
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fmap g (x:xs) = g x : fmap g xs
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-- or we could just say fmap = map
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instance Functor Maybe where
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fmap :: (a -> b) -> Maybe a -> Maybe b
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fmap _ Nothing = Nothing
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fmap g (Just a) = Just (g a)
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```
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> ((,) e) represents a container which holds an “annotation” of type e along with the actual value it holds. It might be clearer to write it as (e,), by analogy with an operator section like (1+), but that syntax is not allowed in types (although it is allowed in expressions with the TupleSections extension enabled). However, you can certainly think of it as (e,).
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> ((->) e) (which can be thought of as (e ->); see above), the type of functions which take a value of type e as a parameter, is a Functor. As a container, (e -> a) represents a (possibly infinite) set of values of a, indexed by values of e. Alternatively, and more usefully, ((->) e) can be thought of as a context in which a value of type e is available to be consulted in a read-only fashion. This is also why ((->) e) is sometimes referred to as the reader monad; more on this later.
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### Exercises
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1. Implement `Functor` instances for `Either e` and `((->) e)`.
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**Solution**:
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```haskell
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instance Functor (Either e) where
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fmap _ (Left e) = Left e
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fmap g (Right a) = Right (g a)
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instance Functor ((->) e) where
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fmap g f = g . f
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```
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2. Implement `Functor` instances for `((,) e)` and for `Pair`, defined as below. Explain their similarities and differences.
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**Solution**:
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```haskell
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instance Functor ((,) e) where
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fmap g (a, b) = (a, g b)
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data Pair a = Pair a a
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instance Functor Pair where
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fmap g (Pair a b) = Pair (g a) (g b)
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```
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Their similarity is in the fact that they both represent types of two values.
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Their difference is that `((,) e)` (tuples of two) can have values of different types (kind of `(,)` is `* -> *`) while both values of `Pair` have the same type `a`, so `Pair` has kind `*`.
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3. Implement a `Functor` instance for the type `ITree`, defined as
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```haskell
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data ITree a = Leaf (Int -> a)
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| Node [ITree a]
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```
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**Solution**:
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```haskell
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instance Functor ITree where
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fmap g (Leaf f) = Leaf (g . f)
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fmap g (Node xs) = Node (fmap (fmap g) xs)
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```
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To test this instance, I defined a function to apply the tree to an `Int`:
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```haskell
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applyTree :: ITree a -> Int -> [a]
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applyTree (Leaf g) i = [g i]
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applyTree (Node []) _ = []
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applyTree (Node (x:xs)) i = applyTree x i ++ applyTree (Node xs) i
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```
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This is not a standard tree traversing algorithm, I just wanted it to be simple for testing.
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Now test the instance:
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```haskell
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λ: let t = Node [Node [Leaf (+5), Leaf (+1)], Leaf (*2)]
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λ: applyTree t 1
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[6,2,2]
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λ: applyTree (fmap id t) 1
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[6,2,2]
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λ: applyTree (fmap (+10) t) 1
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[16, 12, 12]
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```
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4. Give an example of a type of kind `* -> *` which cannot be made an instance of `Functor` (without using `undefined`).
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I don't know the answer to this one yet!
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6. Is this statement true or false?
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> The composition of two `Functor`s is also a `Functor`.
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If false, give a counterexample; if true, prove it by exhibiting some appropriate Haskell code.
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**Solution**:
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It's true, and can be proved by the following function:
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```haskell
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ffmap :: (Functor f, Functor j) => (a -> b) -> f (j a) -> f (j b)
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ffmap g = fmap (fmap g)
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```
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You can test this on arbitrary compositions of `Functor`s:
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```haskell
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main = do
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let result :: Maybe (Either String Int) = ffmap (+ 2) (Just . Right $ 5)
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print result -- (Just (Right 7))
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```
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## Functor Laws
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```haskell
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fmap id = id
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fmap (g . h) = (fmap g) . (fmap h)
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```
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### Exercises
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1. Although it is not possible for a Functor instance to satisfy the first Functor law but not the second (excluding undefined), the reverse is possible. Give an example of a (bogus) Functor instance which satisfies the second law but not the first.
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**Solution**:
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This is easy, consider this instance:
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```haskell
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instance Functor [] where
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fmap _ [] = [1]
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fmap g (x:xs) = g x: fmap g xs
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```
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Then, you can test the first and second laws:
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```haskell
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λ: fmap id [] -- [1], breaks the first law
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λ: fmap ((+1) . (+2)) [1,2] -- [4, 5], second law holds
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λ: fmap (+1) . fmap (+2) $ [1,2] -- [4, 5]
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```
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2. Which laws are violated by the evil Functor instance for list shown above: both laws, or the first law alone? Give specific counterexamples.
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```haskell
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-- Evil Functor instance
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instance Functor [] where
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fmap :: (a -> b) -> [a] -> [b]
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fmap _ [] = []
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fmap g (x:xs) = g x : g x : fmap g xs
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```
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**Solution**:
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The instance defined breaks the first law (`fmap id [1] -- [1,1]`), but holds for the second law.
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Category Theory
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===============
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The Functor section links to [Category Theory](https://en.wikibooks.org/wiki/Haskell/Category_theory), so here I'm going to cover the exercises of that page, too.
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## Introduction to categories
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### Category laws:
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1. The compositions of morphisms need to be **associative**:
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$f \circ (g \circ h) = (f \circ g) \circ h$
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2. The category needs to be **closed** under the composition operator. So if $f : B \to C$ and $g: A \to B$, then there must be some $h: A \to C$ in the category such that $h = f \circ g$.
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3. Every object $A$ in a category must have an identity morphism, $id_A : A \to A$ that is an identity of composition with other morphisms. So for every morphism $g: A \to B$:
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$g \circ id_A = id_B \circ g = g$.
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### Exercises
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1. As was mentioned, any partial order $(P, \leq)$ is a category with objects as the elements of P and a morphism between elements a and b iff $a \leq b$. Which of the above laws guarantees the transitivity of $\leq$?
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**Solution**:
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The second law, which states that the category needs to be closed under the composition operator guarantess that because we have a morphism $a \leq b$, and another morphism $b \leq c$, there must also be some other morphism such that $a \leq c$.
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2. If we add another morphism to the above example, as illustrated below, it fails to be a category. Why? Hint: think about associativity of the composition operation.
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**Solution**:
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The first law does not hold:
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$f \circ (g \circ h) = (f \circ g) \circ h$
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To see that, we can evaluate each side to get an inequality:
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$g \circ h = id_B$
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$f \circ g = id_A$
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$f \circ (g \circ h) = f \circ id_B = f$
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$(f \circ g) \circ h = id_A \circ h = h$
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$f \neq h$
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## Functors
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### Functor laws:
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1. Given an identity morphism $id_A$ on an object $A$, $F(id_A)$ must be the identity morphism on $F(A)$, so:
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$F(id_A) = id_{F(A)}$
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2. Functors must distribute over morphism composition:
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$F(f \circ g) = F(f) \circ F(g)$
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### Exercises
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1. Check the functor laws for the diagram below.
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**Solution**:
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The first law is obvious as it's directly written, the pale blue dotted arrows from $id_C$ to $F(id_C) = id_{F(C)}$ and $id_A$ and $id_B$ to $F(id_A) = F(id_B) = id_{F(A)} = id_{F(B)}$ show this.
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The second law also holds, the only compositions in category $C$ are between $f$ and identities, and $g$ and identities, there is no composition between $f$ and $g$.
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(Note: The second law always hold as long as the first one does, as was seen in Typoclassopedia)
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2. Check the laws for the Maybe and List functors.
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**Solution**:
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```haskell
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instance Functor [] where
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fmap :: (a -> b) -> [a] -> [b]
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fmap _ [] = []
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fmap g (x:xs) = g x : fmap g xs
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-- check the first law for each part:
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fmap id [] = []
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fmap id (x:xs) = id x : fmap id xs = x : fmap id xs -- the first law holds recursively
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-- check the second law for each part:
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fmap (f . g) [] = []
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fmap (f . g) (x:xs) = (f . g) x : fmap (f . g) xs = f (g x) : fmap (f . g) xs
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fmap f (fmap g (x:xs)) = fmap f (g x : fmap g xs) = f (g x) : fmap (f . g) xs
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```
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```haskell
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instance Functor Maybe where
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fmap :: (a -> b) -> Maybe a -> Maybe b
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fmap _ Nothing = Nothing
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fmap g (Just a) = Just (g a)
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-- check the first law for each part:
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fmap id Nothing = Nothing
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fmap id (Just a) = Just (id a) = Just a
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-- check the second law for each part:
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fmap (f . g) Nothing = Nothing
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fmap (f . g) (Just x) = Just ((f . g) x) = Just (f (g x))
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fmap f (fmap g (Just x)) = Just (f (g x)) = Just ((f . g) x)
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```
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Applicative
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===========
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## Laws
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1. The identity law:
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```haskell
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pure id <*> v = v
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```
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2. Homomorphism:
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```haskell
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pure f <*> pure x = pure (f x)
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```
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Intuitively, applying a non-effectful function to a non-effectful argument in an effectful context is the same as just applying the function to the argument and then injecting the result into the context with pure.
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3. Interchange:
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```haskell
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u <*> pure y = pure ($ y) <*> u
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```
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Intuitively, this says that when evaluating the application of an effectful function to a pure argument, the order in which we evaluate the function and its argument doesn't matter.
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4. Composition:
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```haskell
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u <*> (v <*> w) = pure (.) <*> u <*> v <*> w
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```
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This one is the trickiest law to gain intuition for. In some sense it is expressing a sort of associativity property of (`<*>`). The reader may wish to simply convince themselves that this law is type-correct.
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### Exercises
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(Tricky) One might imagine a variant of the interchange law that says something about applying a pure function to an effectful argument. Using the above laws, prove that
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```haskell
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pure f <*> x = pure (flip ($)) <*> x <*> pure f
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```
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**Solution**:
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```haskell
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pure (flip ($)) <*> x <*> pure f
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= (pure (flip ($)) <*> x) <*> pure f -- <*> is left-associative
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= pure ($ f) <*> (pure (flip ($)) <*> x) -- interchange
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= pure (.) <*> pure ($ f) <*> pure (flip ($)) <*> x -- composition
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= pure (($ f) . (flip ($))) <*> x -- homomorphism
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= pure ((flip ($) f) . (flip ($))) <*> x -- identical
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= pure f <*> x
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```
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Explanation of the last transformation:
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`flip ($)` has type `a -> (a -> c) -> c`, intuitively, it first takes an argument of type `a`, then a function that accepts that argument, and in the end it calls the function with the first argument. So `(flip ($) 5)` takes as argument a function which gets called with `5` as it's argument. If we pass `(+ 2)` to `(flip ($) 5)`, we get `(flip ($) 5) (+2)` which is equivalent to the expression `(+2) $ 5`, evaluating to `7`.
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`flip ($) f` is equivalent to `\x -> x $ f`, that means, it takes as input a function and calls it with the function `f` as argument.
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The composition of these functions works like this: First `flip ($)` takes `x` as it's first argument, and returns a function `(flip ($) x)`, this function is awaiting a function as it's last argument, which will be called with `x` as it's argument. Now this function `(flip ($) x)` is passed to `flip ($) f`, or to write it's equivalent `(\x -> x $ f) (flip ($) x)`, this results in the expression `(flip ($) x) f`, which is equivalent to `f $ x`.
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You can check the type of `(flip ($) f) . (flip ($))` is something like this (depending on your function `f`):
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```haskell
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λ: let f = sqrt
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λ: :t (flip ($) f) . (flip ($))
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(flip ($) f) . (flip ($)) :: Floating c => c -> c
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```
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Also see [this question on Stack Overflow](https://stackoverflow.com/questions/46503793/applicative-prove-pure-f-x-pure-flip-x-pure-f/46505868#46505868) which includes alternative proofs.
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## Instances
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Applicative instance of lists as a collection of values:
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```haskell
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newtype ZipList a = ZipList { getZipList :: [a] }
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instance Applicative ZipList where
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pure :: a -> ZipList a
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pure = undefined -- exercise
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(<*>) :: ZipList (a -> b) -> ZipList a -> ZipList b
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(ZipList gs) <*> (ZipList xs) = ZipList (zipWith ($) gs xs)
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```
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Applicative instance of lists as a non-deterministic computation context:
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```haskell
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instance Applicative [] where
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pure :: a -> [a]
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pure x = [x]
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(<*>) :: [a -> b] -> [a] -> [b]
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gs <*> xs = [ g x | g <- gs, x <- xs ]
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```
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### Exercises
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1. Implement an instance of `Applicative` for `Maybe`.
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**Solution**:
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```haskell
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instance Applicative (Maybe a) where
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pure :: a -> Maybe a
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pure x = Just x
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(<*>) :: Maybe (a -> b) -> Maybe a -> Maybe b
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_ <*> Nothing = Nothing
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Nothing <*> _ = Nothing
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(Just f) <*> (Just x) = Just (f x)
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```
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2. Determine the correct definition of `pure` for the `ZipList` instance of `Applicative`—there is only one implementation that satisfies the law relating `pure` and `(<*>)`.
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**Solution**:
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```haskell
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newtype ZipList a = ZipList { getZipList :: [a] }
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instance Functor ZipList where
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fmap f (ZipList list) = ZipList { getZipList = fmap f list }
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instance Applicative ZipList where
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pure = ZipList . pure
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(ZipList gs) <*> (ZipList xs) = ZipList (zipWith ($) gs xs)
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```
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You can check the Applicative laws for this implementation.
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## Utility functions
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### Exercises
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1. Implement a function
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`sequenceAL :: Applicative f => [f a] -> f [a]`
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There is a generalized version of this, `sequenceA`, which works for any `Traversable` (see the later section on `Traversable`), but implementing this version specialized to lists is a good exercise.
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**Solution**:
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```haskell
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createList = replicate 1
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sequenceAL :: Applicative f => [f a] -> f [a]
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sequenceAL = foldr (\x b -> ((++) . createList <$> x) <*> b) (pure [])
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```
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Explanation:
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First, `createList` is a simple function for creating a list of a single element, e.g. `createList 2 == [2]`.
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Now let's take `sequenceAL` apart, first, it does a fold over the list `[f a]`, and `b` is initialized to `pure []`, which results in `f [a]` as required by the function's output.
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Inside the function, `createList <$> x` applies `createList` to the value inside `f a`, resulting in `f [a]`, and then `(++)` is applied to the value again, so it becomes `f ((++) [a])`, now we can apply the function `(++) [a]` to `b` by `((++) . createList <$> x) <*> b`, which results in `f ([a] ++ b)`.
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## Alternative formulation
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### Definition
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```haskell
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class Functor f => Monoidal f where
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unit :: f ()
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(**) :: f a -> f b -> f (a,b)
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```
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### Laws:
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1. Left identity
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```haskell
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unit ** v ≅ v
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```
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2. Right identity
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```haskell
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u ** unit ≅ u
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```
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3. Associativity
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```haskell
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u ** (v ** w) ≅ (u ** v) ** w
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```
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4. Neutrality
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```haskell
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fmap (g *** h) (u ** v) = fmap g u ** fmap h v
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```
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### Isomorphism
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In the laws above, `≅` refers to isomorphism rather than equality. In particular we consider:
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```haskell
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(x,()) ≅ x ≅ ((),x)
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((x,y),z) ≅ (x,(y,z))
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```
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### Exercises
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instance Applicative [] where
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pure :: a -> [a]
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pure x = [x]
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(<*>) :: [a -> b] -> [a] -> [b]
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gs <*> xs = [ g x | g <- gs, x <- xs ]
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```haskell
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pure id <*> v = v
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```
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```haskell
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pure f <*> pure x = pure (f x)
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```
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```haskell
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u <*> pure y = pure ($ y) <*> u
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|
```
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|
|
|
```haskell
|
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u <*> (v <*> w) = pure (.) <*> u <*> v <*> w
|
|
```
|
|
|
|
1. Implement `pure` and `<*>` in terms of `unit` and `**`, and vice versa.
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|
|
|
```haskell
|
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unit :: f ()
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unit = pure ()
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|
|
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(**) :: f a -> f b -> f (a, b)
|
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a ** b = fmap (,) a <*> b
|
|
|
|
pure :: a -> f a
|
|
pure x = unit ** x
|
|
|
|
(<*>) :: f (a -> b) -> f a -> f b
|
|
f <*> a = fmap (uncurry ($)) (f ** a) = fmap (\(f, a) -> f a) (f ** a)
|
|
```
|
|
|
|
2. Are there any `Applicative` instances for which there are also functions `f () -> ()` and `f (a,b) -> (f a, f b)`, satisfying some "reasonable" laws?
|
|
|
|
The [`Arrow`](https://wiki.haskell.org/Typeclassopedia#Arrow) type class seems to satisfy these criteria.
|
|
|
|
```haskell
|
|
first unit = ()
|
|
|
|
(id *** f) (a, b) = (f a, f b)
|
|
```
|
|
|
|
3. (Tricky) Prove that given your implementations from the first exercise, the usual Applicative laws and the Monoidal laws stated above are equivalent.
|
|
|
|
1. Identity Law
|
|
|
|
```haskell
|
|
pure id <*> v
|
|
= fmap (uncurry ($)) ((unit ** id) ** v)
|
|
= fmap (uncurry ($)) (id ** v)
|
|
= fmap id v
|
|
= v
|
|
```
|
|
|
|
2. Homomorphism
|
|
|
|
```haskell
|
|
pure f <*> pure x
|
|
= (unit ** f) <*> (unit ** x)
|
|
= fmap (\(f, a) -> f a) (unit ** f) (unit ** x)
|
|
= fmap (\(f, a) -> f a) (f ** x)
|
|
= fmap f x
|
|
= pure (f x)
|
|
```
|
|
|
|
3. Interchange
|
|
|
|
```haskell
|
|
u <*> pure y
|
|
= fmap (uncurry ($)) (u ** (unit ** y))
|
|
= fmap (uncurry ($)) (u ** y)
|
|
= fmap (u $) y
|
|
= fmap ($ y) u
|
|
= pure ($ y) <*> u
|
|
|
|
4. Composition
|
|
|
|
```haskell
|
|
u <*> (v <*> w)
|
|
= fmap (uncurry ($)) (u ** (fmap (uncurry ($)) (v ** w)))
|
|
= fmap (uncurry ($)) (u ** (fmap v w))
|
|
= fmap u (fmap v w)
|
|
= fmap (u . v) w
|
|
= pure (.) <*> u <*> v <*> w =
|
|
```
|