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!
> ((,) 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,).
> ((->) 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.
Their similarity is in the fact that they both represent types of two values.
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 `*`.
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.
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.
## Introduction to categories
### Category laws:
1. The compositions of morphisms need to be **associative**:
$f \circ (g \circ h) = (f \circ g) \circ h$
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 $h = f \circ g$.
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 $g: A \to B$:
$g \circ id_A = id_B \circ g = g$.
### Exercises
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 $\leq$?
**Solution**:
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 $a \leq c$.
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.
**Solution**:
The first law does not hold:
$f \circ (g \circ h) = (f \circ g) \circ h$
To see that, we can evaluate each side to get an inequality:
$g \circ h = id_B$
$f \circ g = id_A$
$f \circ (g \circ h) = f \circ id_B = f$
$(f \circ g) \circ h = id_A \circ h = h$
$f \neq h$
## Functors
### Functor laws:
1. Given an identity morphism $id_A$ on an object $A$, $F(id_A)$ must be the identity morphism on $F(A)$, so:
$F(id_A) = id_{F(A)}$
2. Functors must distribute over morphism composition:
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.
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 $g$.
(Note: The second law always hold as long as the first one does, as was seen in Typoclassopedia)
2. Check the laws for the Maybe and List functors.
**Solution**:
```haskell
instance Functor [] where
fmap :: (a -> b) -> [a] -> [b]
fmap _ [] = []
fmap g (x:xs) = g x : fmap g xs
-- check the first law for each part:
fmap id [] = []
fmap id (x:xs) = id x : fmap id xs = x : fmap id xs -- the first law holds recursively
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.
3. Interchange:
```haskell
u <*> pure y = pure ($ y) <*> u
```
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.
4. Composition:
```haskell
u <*> (v <*> w) = pure (.) <*> u <*> v <*> w
```
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.
### Exercises
(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
`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`.
`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.
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`.
You can check the type of `(flip ($) f) . (flip ($))` is something like this (depending on your function `f`):
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.
Applicative instance of lists as a non-deterministic computation context:
```haskell
instance Applicative [] where
pure :: a -> [a]
pure x = [x]
(<*>) :: [a -> b] -> [a] -> [b]
gs <*> xs = [ g x | g <-gs,x<-xs]
```
### Exercises
1. Implement an instance of `Applicative` for `Maybe`.
**Solution**:
```haskell
instance Applicative (Maybe a) where
pure :: a -> Maybe a
pure x = Just x
(<*>) :: Maybe (a -> b) -> Maybe a -> Maybe b
_ <*> Nothing = Nothing
Nothing <*> _ = Nothing
(Just f) <*> (Just x) = Just (f x)
```
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 `(<*>)`.
**Solution**:
```haskell
newtype ZipList a = ZipList { getZipList :: [a] }
instance Functor ZipList where
fmap f (ZipList list) = ZipList { getZipList = fmap f list }
You can check the Applicative laws for this implementation.
## Utility functions
### Exercises
1. Implement a function
`sequenceAL :: Applicative f => [f a] -> f [a]`
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.
**Solution**:
```haskell
createList = replicate 1
sequenceAL :: Applicative f => [f a] -> f [a]
sequenceAL = foldr (\x b -> ((++) . createList <$> x) <*> b) (pure [])
```
Explanation:
First, `createList` is a simple function for creating a list of a single element, e.g. `createList 2 == [2]`.
Now let's take `sequenceAL` apart, first, it does a fold over the list `[f a]`, and `b` is initialized to `pure []`, which results in `f [a]` as required by the function's output.
Inside the function, `createList <$> 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 <$> x) <*> b`, which results in `f ([a] ++ b)`.
## Alternative formulation
### Definition
```haskell
class Functor f => Monoidal f where
unit :: f ()
(**) :: f a -> f b -> f (a,b)
```
### Laws:
1. Left identity
```haskell
unit ** v ≅ v
```
2. Right identity
```haskell
u ** unit ≅ u
```
3. Associativity
```haskell
u ** (v ** w) ≅ (u ** v) ** w
```
4. Neutrality
```haskell
fmap (g *** h) (u ** v) = fmap g u ** fmap h v
```
### Isomorphism
In the laws above, `≅` refers to isomorphism rather than equality. In particular we consider:
```haskell
(x,()) ≅ x ≅ ((),x)
((x,y),z) ≅ (x,(y,z))
```
### Exercises
1. Implement `pure` and `<*>` in terms of `unit` and `**`, and vice versa.
```haskell
unit :: f ()
unit = pure ()
(**) :: f a -> f b -> f (a, b)
a ** b = fmap (,) a <*> b
pure :: a -> f a
pure x = unit ** x
(<*>) :: f (a -> b) -> f a -> f b
f <*> a = fmap (uncurry ($)) (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.
2. What would you need in order to implement `foldMap` in terms of `fold`?
**Solution**:
A `map` function should exist for the instance, so we can apply the function `(a -> m)` to the container first.
```haskell
foldMap f = fold . map f
```
3. Implement `foldMap` in terms of `foldr`.
**Solution**:
```haskell
foldMap f = foldr (\a b -> mappend (f a) b) mempty
```
4. Implement `foldr` in terms of `foldMap` (hint: use the `Endo` monoid).
**Solution**:
```haskell
foldr f b c = foldMap (Endo . f) c `appEndo` b
```
5. What is the type of `foldMap . foldMap`? Or `foldMap . foldMap . foldMap`, etc.? What do they do?
**Solution**:
Each composition makes `foldMap` operate on a deeper level, so:
```haskell
foldMap :: Monoid m => (a -> m) -> t a -> m
foldMap . foldMap :: Monoid m => (a -> m) -> t (t a) -> m
foldMap . foldMap . foldMap :: Monoid m => (a -> m) -> t (t (t a)) -> m
foldMap id [1,2,3] :: Sum Int -- 6
(foldMap . foldMap) id [[1,2,3]] :: Sum Int -- 6
(foldMap . foldMap . foldMap) id [[[1,2,3]]] :: Sum Int -- 6
```
## Derived folds
### Exercises
1. Implement `toList :: Foldable f => f a -> [a]` in terms of either `foldr` or `foldMap`.
**Solution**:
```haskell
toList = foldMap (replicate 1)
```
2. Show how one could implement the generic version of `foldr` in terms of `toList`, assuming we had only the list-specific `foldr :: (a -> b -> b) -> b -> [a] -> b`.
**Solution**:
```haskell
foldr f b c = foldr f b (toList c)
```
3. Pick some of the following functions to implement: `concat`, `concatMap`, `and`, `or`, `any`, `all`, `sum`, `product`, `maximum(By)`, `minimum(By)`, `elem`, `notElem`, and `find`. Figure out how they generalize to `Foldable` and come up with elegant implementations using `fold` or `foldMap` along with appropriate `Monoid` instances.
**Solution**:
```haskell
concat :: Foldable t => t [a] -> [a]
concat = foldMap id
concatMap :: Foldable t => (a -> [b]) -> t a -> [b]
concatMap f = foldMap (foldMap (replicate 1 . f))
and :: Foldable t => t Bool -> Bool
and = getAll . foldMap All
or :: Foldable t => t Bool -> Bool
or = getAny . foldMap Any
any :: Foldable t => (a -> Bool) -> t a -> Bool
any f = getAny . foldMap (Any . f)
all :: Foldable t => (a -> Bool) t a -> Bool
all f = getAll . foldMap (All . f)
sum :: (Num a, Foldable t) => t a -> a
sum = getSum . foldMap Sum
product :: (Num a, Foldable t) => t a -> a
product = getProduct . foldMap Product
-- I think there are more elegant implementations for maximumBy, leave a comment
-- if you have a suggestion
maximumBy :: Foldable t => (a -> a -> Ordering) -> t a -> a
maximumBy f c = head $ foldMap (\a -> [a | cmp a]) c
where
cmp a = all (/= LT) (map (f a) lst)
lst = toList c
elem :: (Eq a, Foldable t) => a -> t a -> Bool
elem x c = any (==x) c
find :: Foldable t => (a -> Bool) -> t a -> Maybe a
find f c = listToMaybe $ foldMap (\a -> [a | f a]) c
```
## Utility functions
-`sequenceA_ :: (Applicative f, Foldable t) => t (f a) -> f ()` takes a container full of computations and runs them in sequence, discarding the results (that is, they are used only for their effects). Since the results are discarded, the container only needs to be Foldable. (Compare with `sequenceA :: (Applicative f, Traversable t) => t (f a) -> f (t a)`, which requires a stronger Traversable constraint in order to be able to reconstruct a container of results having the same shape as the original container.)
-`traverse_ :: (Applicative f, Foldable t) => (a -> f b) -> t a -> f ()` applies the given function to each element in a foldable container and sequences the effects (but discards the results).
### Exercises
1. Implement `traverse_` in terms of `sequenceA_` and vice versa. One of these will need an extra constraint. What is it?
**Solution**:
```haskell
sequenceA_ :: (Applicative f, Foldable t) => t (f a) -> f ()
sequenceA_ = traverse_ id
traverse_ :: (Applicative f, Foldable t, Functor t) => (a -> f b) -> t a -> f ()
traverse_ f c = sequenceA_ (fmap f c)
```
The additional constraint for implementing `traverse_` in terms of `sequenceA_` is the requirement of the `Foldable` instance `t` to be a `Functor` as well.
traverse :: Applicative f => (a -> f b) -> t a -> f (t b)
sequenceA :: Applicative f => t (f a) -> f (t a)
```
### Exercises
1. There are at least two natural ways to turn a tree of lists into a list of trees. What are they, and why?
Note: I'm not really sure whether my solution is _natural_, I think the question is rather ambiguous in the sense that it's not clear whether the trees in the final list of trees can have lists as their values, i.e. `Tree [Int] -> [Tree [Int]]` is valid or only `Tree [Int] -> [Tree Int]` is, but let me know if you think otherwise.
**Solution**:
One way is to put each `Node`, `Leaf` or `Empty` in a list in-order, this way the structure of the tree can be recovered from the list, here is a quick sketch (`[]` is an arbitrary list):
```haskell
let tree = Node (Node (Leaf []) Empty) [] (Leaf [])
let list = [Node Empty [] Empty, Node Empty [] Empty, Leaf [], Empty, Leaf []]
```
2. Give a natural way to turn a list of trees into a tree of lists.
**Solution**:
To recover the original tree from the list of trees, whenever we encounter a `Node` in the list, we catch the next three values as left, value, and right nodes of the original node.
3. What is the type of `traverse . traverse`? What does it do?
**Solution**:
```haskell
(traverse . traverse) :: Applicative f => (a -> f b) -> t (t2 a) -> f (t (t2 b))
```
It traverses on a deeper level, retaining the structure of the first level.
4. Implement `traverse` in terms of `sequenceA`, and vice versa.
**Solution**:
```haskell
sequenceA = traverse id
traverseA f c = sequenceA (fmap f c)
```
## Instances and examples
### Exercises
1. Implement `fmap` and `foldMap` using only the `Traversable` methods. (Note that the `Traversable` module provides these implementations as `fmapDefault` and `foldMapDefault`.)
**Solution**:
```haskell
newtype Id a = Id { getId :: a }
instance Functor Id where
fmap f (Id x) = Id (f x)
instance Applicative Id where
pure x = Id x
(Id f) <*> (Id x) = Id (f x)
fmapDefault :: Traversable t => (a -> b) -> t a -> t b
fmapDefault f = getId . traverse (Id . f)
foldMapDefault :: (Monoid m, Traversable t) => (a -> m) -> t a -> m
foldMapDefault f = getConst . traverse (Const . f)
```
See the [Const](https://www.stackage.org/haddock/lts-9.9/base-4.9.1.0/src/Data-Functor-Const.html#Const) Functor's definition for intuition.
2. Implement `Traversable` instances for `[]`, `Maybe`, `((,) a)`, and `Either a`.
**Solution**:
```haskell
instance Traversable [] where
traverse :: Applicative f => (a -> f b) -> [a] -> f [b]
traverse _ [] = pure []
traverse f (x:xs) = (:) <$> f x <*> Main.traverse f xs
instance Traversable Maybe where
traverse :: Applicative f => (a -> f b) -> Maybe a -> f (Maybe b)
traverse _ Nothing = pure Nothing
traverse f (Just x) = Just <$> f x
instance Traversable ((,) c) where
traverse :: Applicative f => (a -> f b) -> (c, a) -> f (c, b)
traverse f (c, a) = (,) c <$> f a
instance Traversable (Either c) where
traverse :: Applicative f => (a -> f b) -> Either c a -> f (Either c b)
traverse _ (Left c) = pure (Left c)
traverse f (Right a) = Right <$> f a
```
3. Explain why `Set` is `Foldable` but not `Traversable`.
**Solution**:
First, in terms of laws, `Set` is not a `Functor`, thus it cannot be made into a `Traversable` instance, since `Traversable` instances require `Functor` superclasses.
Second, on an intuitive level: In `Foldable`, the goal is not to keep the shape/structure of the original container, we are trying to reduce the container into some value, and the shape of the final result doesn't matter, but in `Traversable`, we ought to keep the structure of the final result, but we can't guarantee this while using `Set`s, because we can define some transformation `f :: Set a -> Set a` which reduces the length of the `Set`.
See [Foldable vs. Traversable](https://stackoverflow.com/questions/35857733/foldable-vs-traversable) and [Sets, Functors and Eq confusion](https://stackoverflow.com/questions/19177125/sets-functors-and-eq-confusion). and [Foldable and Traversable](https://wiki.haskell.org/Foldable_and_Traversable) for more details.
4. Show that `Traversable` functors compose: that is, implement an instance for `Traversable (Compose f g)` given `Traversable` instances for `f` and `g`.
**Solution**:
```haskell
instance (Traversable f, Traversable g) => Traversable (Compose f g) where
traverse :: (Applicative f) => (a -> f b) -> Compose g h a -> f (Compose g h b)
traverse f (Compose t) = Compose <$> traverse (traverse f) t
