feat(post): typoclassopedia exercise solutions
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_posts/2017-09-27-typoclassopedia-exercise-answers.md
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layout: post
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title: "Typoclassopedia: Exercise solutions"
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date: 2017-09-27 12:12:12
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permalink: typoclassopedia-exercise-solutions/
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categories: programming
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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, 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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{% highlight 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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{% endhighlight %}
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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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{% highlight 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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{% endhighlight %}
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2. Implement `Functor` instances for `((,) e)` and for `Pair`, defined as `data Pair a = Pair a a`. Explain their similarities and differences.
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**Solution**:
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{% highlight 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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{% endhighlight %}
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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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{% highlight haskell %}
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data ITree a = Leaf (Int -> a)
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| Node [ITree a]
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{% endhighlight %}
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**Solution**:
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{% highlight 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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{% endhighlight %}
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To test this instance, I defined a function to apply the tree to an `Int`:
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{% highlight 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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{% endhighlight %}
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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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{% highlight 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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{% endhighlight %}
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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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{% highlight 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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{% endhighlight %}
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You can test this on arbitrary compositions of `Functor`s:
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{% highlight 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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{% endhighlight %}
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### Functor Laws
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{% highlight haskell %}
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fmap id = id
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fmap (g . h) = (fmap g) . (fmap h)
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{% endhighlight %}
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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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{% highlight 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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{% endhighlight %}
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Then, you can test the first and second laws:
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{% highlight 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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{% endhighlight %}
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1. 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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{% highlight 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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{% endhighlight %}
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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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