# Typoclassopedia: Exercise solutions

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!

In each section below, I left some reference material for the exercises and then
the solutions.

Note: The post will be updated as I progress in Typoclassopedia myself

Functor
==========
## Instances

```haskell
instance Functor [] where
  fmap :: (a -> b) -> [a] -> [b]
  fmap _ []     = []
  fmap g (x:xs) = g x : fmap g xs
  -- or we could just say fmap = map
 
instance Functor Maybe where
  fmap :: (a -> b) -> Maybe a -> Maybe b
  fmap _ Nothing  = Nothing
  fmap g (Just a) = Just (g a)
```

> ((,) 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. 

### Exercises

 1. Implement `Functor` instances for `Either e` and `((->) e)`.

    **Solution**:
    ```haskell
    instance Functor (Either e) where
      fmap _ (Left e) = Left e
      fmap g (Right a) = Right (g a)
      
    instance Functor ((->) e) where
      fmap g f = g . f
    ```

 2. Implement `Functor` instances for `((,) e)` and for `Pair`, defined as below. Explain their similarities and differences.

    **Solution**:
    ```haskell
    instance Functor ((,) e) where
      fmap g (a, b) = (a, g b) 
      
      
    data Pair a = Pair a a
    instance Functor Pair where
      fmap g (Pair a b) = Pair (g a) (g b)
    ```

    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 `*`.

3. Implement a `Functor` instance for the type `ITree`, defined as

    ```haskell
    data ITree a = Leaf (Int -> a)
                | Node [ITree a]
    ```

    **Solution**:
    ```haskell
    instance Functor ITree where
      fmap g (Leaf f) = Leaf (g . f)
      fmap g (Node xs) = Node (fmap (fmap g) xs)
    ```

    To test this instance, I defined a function to apply the tree to an `Int`:

    ```haskell
    applyTree :: ITree a -> Int -> [a]
    applyTree (Leaf g) i = [g i]
    applyTree (Node []) _ = []
    applyTree (Node (x:xs)) i = applyTree x i ++ applyTree (Node xs) i
    ```

    This is not a standard tree traversing algorithm, I just wanted it to be simple for testing.

    Now test the instance:

    ```haskell
    λ: let t = Node [Node [Leaf (+5), Leaf (+1)], Leaf (*2)]
    λ: applyTree t 1
    [6,2,2]
    λ: applyTree (fmap id t) 1
    [6,2,2]
    λ: applyTree (fmap (+10) t) 1
    [16, 12, 12]
    ```

4. Give an example of a type of kind `* -> *` which cannot be made an instance
   of `Functor` (without using `undefined`).

    I don't know the answer to this one yet!

6. Is this statement true or false?
 
    > The composition of two `Functor`s is also a `Functor`.

    If false, give a counterexample; if true, prove it by exhibiting some
    appropriate Haskell code.

    **Solution**:

    It's true, and can be proved by the following function:

    ```haskell
    ffmap :: (Functor f, Functor j) => (a -> b) -> f (j a) -> f (j b)
    ffmap g = fmap (fmap g)
    ```

    You can test this on arbitrary compositions of `Functor`s:

    ```haskell
    main = do
      let result :: Maybe (Either String Int) = ffmap (+ 2) (Just . Right $ 5)
      print result -- (Just (Right 7))
    ```

## Functor Laws

```haskell
fmap id = id
fmap (g . h) = (fmap g) . (fmap h)
```

### Exercises

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. 

    **Solution**:

    This is easy, consider this instance:

    ```haskell
    instance Functor [] where
      fmap _ [] = [1]
      fmap g (x:xs) = g x: fmap g xs
    ```

    Then, you can test the first and second laws:

    ```haskell
    λ: fmap id [] -- [1], breaks the first law
    λ: fmap ((+1) . (+2)) [1,2] -- [4, 5], second law holds
    λ: fmap (+1) . fmap (+2) $ [1,2] -- [4, 5]
    ```

2. Which laws are violated by the evil Functor instance for list shown above:
   both laws, or the first law alone? Give specific counterexamples. 

    ```haskell
    -- Evil Functor instance
    instance Functor [] where
      fmap :: (a -> b) -> [a] -> [b]
      fmap _ [] = []
      fmap g (x:xs) = g x : g x : fmap g xs
    ```

    **Solution**:

    The instance defined breaks the first law (`fmap id [1] -- [1,1]`), but holds for the second law.
    
Category Theory
===============

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.
   
   ![not a category, an additional h: B -> A](/img/typoclassopedia/not-a-cat.png)

    **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:
    $F(f \circ g) = F(f) \circ F(g)$
    
### Exercises

1. Check the functor laws for the diagram below.

    ![functor diagram](/img/typoclassopedia/functor-diagram.png)

    **Solution**:

    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
    
    -- check the second law for each part:
    fmap (f . g) [] = []
    fmap (f . g) (x:xs) = (f . g) x : fmap (f . g) xs = f (g x) : fmap (f . g) xs
    fmap f (fmap g (x:xs)) = fmap f (g x : fmap g xs) = f (g x) : fmap (f . g) xs
    ```

    ```haskell
    instance Functor Maybe where
      fmap :: (a -> b) -> Maybe a -> Maybe b
      fmap _ Nothing  = Nothing
      fmap g (Just a) = Just (g a)
    
    -- check the first law for each part:
    fmap id Nothing = Nothing
    fmap id (Just a) = Just (id a) = Just a
    
    -- check the second law for each part:
    fmap (f . g) Nothing = Nothing
    fmap (f . g) (Just x) = Just ((f . g) x) = Just (f (g x))
    fmap f (fmap g (Just x)) = Just (f (g x)) = Just ((f . g) x)
    ```
    
Applicative
===========

## Laws

1. The identity law:
   
    ```haskell
    pure id <*> v = v
    ```

2. Homomorphism:

    ```haskell
    pure f <*> pure x = pure (f x)
    ```

    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

```haskell
pure f <*> x = pure (flip ($)) <*> x <*> pure f
```

**Solution**:

```haskell
pure (flip ($)) <*> x <*> pure f
  = (pure (flip ($)) <*> x) <*> pure f -- <*> is left-associative
  = pure ($ f) <*> (pure (flip ($)) <*> x) -- interchange
  = pure (.) <*> pure ($ f) <*> pure (flip ($)) <*> x -- composition
  = pure (($ f) . (flip ($))) <*> x -- homomorphism
  = pure ((flip ($) f) . (flip ($))) <*> x -- identical
  = pure f <*> x
```
Explanation of the last transformation:

`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`):

```haskell
λ: let f = sqrt
λ: :t (flip ($) f) . (flip ($))
(flip ($) f) . (flip ($)) :: Floating c => c -> c
```

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.

## Instances

Applicative instance of lists as a collection of values:

```haskell
newtype ZipList a = ZipList { getZipList :: [a] }
 
instance Applicative ZipList where
  pure :: a -> ZipList a
  pure = undefined   -- exercise
 
  (<*>) :: ZipList (a -> b) -> ZipList a -> ZipList b
  (ZipList gs) <*> (ZipList xs) = ZipList (zipWith ($) gs xs)
```

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 }
    
    instance Applicative ZipList where
      pure = ZipList . pure
    
      (ZipList gs) <*> (ZipList xs) = ZipList (zipWith ($) gs xs)
    ```

    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. 

    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 =
        ```

Monad
=====

## Definition

```haskell
class Applicative m => Monad m where
  return :: a -> m a
  (>>=)  :: m a -> (a -> m b) -> m b
  (>>)   :: m a -> m b -> m b
  m >> n = m >>= \_ -> n
 
  fail   :: String -> m a
```

## Instances

```haskell
instance Monad Maybe where
  return :: a -> Maybe a
  return = Just
 
  (>>=) :: Maybe a -> (a -> Maybe b) -> Maybe b
  (Just x) >>= g = g x
  Nothing  >>= _ = Nothing
```

### Exercises

1. Implement a `Monad` instance for the list constructor, `[]`. Follow the types!
   
   **Solution**:
   
    ```haskell
    instance Monad [] where
      return a = [a]
      
      [] >> _ = []
      (x:xs) >>= f = f x : xs >>= f
    ```
    
2. Implement a `Monad` instance for `((->) e)`.
   
   **Solution**:
   
    ```haskell
    instance Monad ((->) e) where
      return x = const x
      
      g >>= f = f . g
    ```

3. Implement `Functor` and `Monad` instance for `Free f`, defined as:

    ```haskell
    data Free f a = Var a
                  | Node (f (Free f a))
    ```
    
    You may assume that `f` has a `Functor` instance. This is known as the _free
    monad_ built from the functor f. 
    
    **Solution**:
    
    ```haskell
    instance Functor (Free f) where
      fmap f (Var a) = Var (f a)
      fmap f (Node x) = Node (f x)
      
    instance Monad (Free f) where
      return x = Var x
      
      (Var x) >>= f = Var (f x)
      (Node x) >>= f = Node (fmap f x)
    ```

## Intuition

### Exercises

1. Implement `(>>=)` in terms of `fmap` (or `liftM`) and `join`.
  
    **Solution**:
    
    ```haskell
    a >>= f = join (fmap f a)
    ```
    
2. Now implement `join` and `fmap` (`liftM`) in terms of `(>>=)` and `return`.

    **Solution**:
    
    ```haskell
    fmap f a = a >>= (return . f)
    join m = m >>= id
    ```
    
## Laws

Standard:

```haskell
return a >>= k  =  k a
m >>= return    =  m
m >>= (\x -> k x >>= h)  =  (m >>= k) >>= h
```

In terms of `>=>`:

```haskell
return >=> g  =  g
g >=> return  =  g
(g >=> h) >=> k  =  g >=> (h >=> k)
```

### Exercises

1. Given the definition `g >=> h = \x -> g x >>= h`, prove the equivalence of
   the above laws and the standard monad laws.

    **Solution**:
    
    ```haskell
    return >=> g
      = \x -> return x >>= g
      = \x -> g x
      = g
      
    g >=> return
      = \x -> g x >>= return
      = \x -> g x
      = g
    
    g >=> (h >=> k)
      = \y -> g y >>= (\x -> h x >>= k)
      = \y -> (g y >>= h) >>= k
      = \y -> (\x -> g x >>= h) y >>= k
      = (\x -> g x >>= h) >=> k
      = (g >=> h) >=> k
    ```

Monad Transformers
==================

## Definition and laws

```haskell
class MonadTrans t where
  lift :: Monad m => m a -> t m a
```

### Exercises

1. What is the kind of `t` in the declaration of `MonadTrans`?

    **Solution**: 
    
    `t` is of the kind `(* -> *) -> * -> *`, as we see in `(t m) a`, `t` accepts
    a `Monad` first, which is of type `* -> *`, and then another argument of
    kind `*`.
    
## Composing Monads

### Exercises

1. Implement `join :: M (N (M (N a))) -> M (N a)` given `distrib :: N (M a) -> M
   (N a)` and assuming `M` and `N` are instances of `Monad`.

    ```haskell
    join :: M (N (M (N a))) -> M (N a)
    join m = distrib ((distrib m) >>= join) >>= join
    
    -- one by one
    let m :: M (N (M (N a)))
        a = distrib m  :: N (M (M (N a)))
        b = a >>= join :: N (M (N a))
        c = distrib b  :: M (N (N a))
    in c >>= join      :: M (N a)
    ```
    
MonadFix
========

## Examples and intuition

```haskell
maybeFix :: (a -> Maybe a) -> Maybe a
maybeFix f = ma
  where ma = f (fromJust ma)
```

### Exercises

1. Implement a MonadFix instance for []. 
    
    **Solution**:
    
    ```haskell
    listFix :: (a -> [a]) -> [a]
    listFix f = la
      where la = f (head la)
    ```
    
Foldable
========

## Definition

```haskell
class Foldable t where
  fold    :: Monoid m => t m -> m
  foldMap :: Monoid m => (a -> m) -> t a -> m
  foldr   :: (a -> b -> b) -> b -> t a -> b
  foldr'  :: (a -> b -> b) -> b -> t a -> b
  foldl   :: (b -> a -> b) -> b -> t a -> b
  foldl'  :: (b -> a -> b) -> b -> t a -> b
  foldr1  :: (a -> a -> a) -> t a -> a
  foldl1  :: (a -> a -> a) -> t a -> a
  toList  :: t a -> [a]
  null    :: t a -> Bool
  length  :: t a -> Int
  elem    :: Eq a => a -> t a -> Bool
  maximum :: Ord a => t a -> a
  minimum :: Ord a => t a -> a
  sum     :: Num a => t a -> a
  product :: Num a => t a -> a
```

## Instances and examples

### Exercises

1. Implement `fold` in terms of `foldMap`.

    **Solution**:
    
    ```haskell
    fold = foldMap id
    ```
    
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.
    
Traversable
===========

## Intuition

```haskell
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):
    
    ![tree to list](/img/typoclassopedia/tree.jpg)
    
    ```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
    ```