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 <$> x) <*> b`, which results in `f ([a] ++ b)`.
## Alternative formulation
### Definition
```haskell
class Functor f => Monoidal f where
unit :: f ()
(**) :: f a -> f b -> f (a,b)
```
### Laws:
1. Left identity
```haskell
unit ** v ≅ v
```
2. Right identity
```haskell
u ** unit ≅ u
```
3. Associativity
```haskell
u ** (v ** w) ≅ (u ** v) ** w
```
4. Neutrality
```haskell
fmap (g *** h) (u ** v) = fmap g u ** fmap h v
```
### Isomorphism
In the laws above, `≅` refers to isomorphism rather than equality. In particular we consider:
```haskell
(x,()) ≅ x ≅ ((),x)
((x,y),z) ≅ (x,(y,z))
```
### Exercises
instance Applicative [] where
pure :: a -> [a]
pure x = [x]
(<*>) :: [a -> b] -> [a] -> [b]
gs <*> xs = [ g x | g <-gs,x<-xs]
```haskell
pure id <*> v = v
```
```haskell
pure f <*> pure x = pure (f x)
```
```haskell
u <*> pure y = pure ($ y) <*> u
```
```haskell
u <*> (v <*> w) = pure (.) <*> u <*> v <*> w
```
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.
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)))
