I wanted to get proficient in Haskell so I decided to follow An [Essential] Haskell Reading List. 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!

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

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:

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:

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

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


Solution:

 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:

 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:

 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!

5. Is this statement true or false?

The composition of two Functors 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:

 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 Functors:

 main = do
let result :: Maybe (Either String Int) = ffmap (+ 2) (Just . Right $5) print result -- (Just (Right 7))  ## Functor Laws fmap id = id fmap (g . h) = (fmap g) . (fmap h)  ### Exercises 1. Although it is not possible for a Functor instance to satisfy the first Functor law but not the second (excluding undefined), the reverse is possible. Give an example of a (bogus) Functor instance which satisfies the second law but not the first. Solution: This is easy, consider this instance:  instance Functor [] where fmap _ [] = [1] fmap g (x:xs) = g x: fmap g xs  Then, you can test the first and second laws:  λ: fmap id [] -- [1], breaks the first law λ: fmap ((+1) . (+2)) [1,2] -- [4, 5], second law holds λ: fmap (+1) . fmap (+2)$ [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.

 -- 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, 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: $F(f \circ g) = F(f) \circ F(g)$

### Exercises

1. Check the functor laws for the diagram below.

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:

 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

 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:

 pure id <*> v = v

2. Homomorphism:

 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:

 u <*> pure y = pure ($y) <*> u  Intuitively, this says that when evaluating the application of an effectful function to a pure argument, the order in which we evaluate the function and its argument doesn’t matter. 4. Composition:  u <*> (v <*> w) = pure (.) <*> u <*> v <*> w  This one is the trickiest law to gain intuition for. In some sense it is expressing a sort of associativity property of (<*>). The reader may wish to simply convince themselves that this law is type-correct. ### Exercises (Tricky) One might imagine a variant of the interchange law that says something about applying a pure function to an effectful argument. Using the above laws, prove that pure f <*> x = pure (flip ($)) <*> x <*> pure f


Solution:

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 ($)) is something like this (depending on your function f): λ: let f = sqrt λ: :t (flip ($) f) . (flip ($)) (flip ($) f) . (flip ($)) :: Floating c => c -> c  Also see this question on Stack Overflow which includes alternative proofs. ## Instances Applicative instance of lists as a collection of values: newtype ZipList a = ZipList { getZipList :: [a] } instance Applicative ZipList where pure :: a -> ZipList a pure = undefined -- exercise (<*>) :: ZipList (a -> b) -> ZipList a -> ZipList b (ZipList gs) <*> (ZipList xs) = ZipList (zipWith ($) gs xs)


Applicative instance of lists as a non-deterministic computation context:

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:

 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:

 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 ($) gs xs)  You can check the Applicative laws for this implementation. ## Utility functions ### Exercises 1. Implement a function sequenceAL :: Applicative f => [f a] -> f [a] There is a generalized version of this, sequenceA, which works for any Traversable (see the later section on Traversable), but implementing this version specialized to lists is a good exercise. Solution:  createList = replicate 1 sequenceAL :: Applicative f => [f a] -> f [a] sequenceAL = foldr (\x b -> ((++) . createList <$> 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

class Functor f => Monoidal f where
unit :: f ()
(**) :: f a -> f b -> f (a,b)


### Laws:

1. Left identity

 unit ** v ≅ v

2. Right identity

 u ** unit ≅ u

3. Associativity

 u ** (v ** w) ≅ (u ** v) ** w

4. Neutrality

 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:

(x,()) ≅ x ≅ ((),x)
((x,y),z) ≅ (x,(y,z))


### Exercises

1. Implement pure and <*> in terms of unit and **, and vice versa.

 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 type class seems to satisfy these criteria.  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  pure id <*> v = fmap (uncurry ($)) ((unit ** id) ** v)
= fmap (uncurry ($)) (id ** v) = fmap id v = v  2. Homomorphism  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  u <*> pure y = fmap (uncurry ($)) (u ** (unit ** y))
= fmap (uncurry ($)) (u ** y) = fmap (u$) y
= fmap ($y) u = pure ($ y) <*> u

4. Composition

 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 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 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:  instance Monad [] where return a = [a] [] >> _ = [] (x:xs) >>= f = f x : xs >>= f  2. Implement a Monad instance for ((->) e). Solution:  instance Monad ((->) e) where return x = const x g >>= f = f . g  3. Implement Functor and Monad instance for Free f, defined as:  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:  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:  a >>= f = join (fmap f a)  2. Now implement join and fmap (liftM) in terms of (>>=) and return. Solution:  fmap f a = a >>= (return . f) join m = m >>= id  ## Laws Standard: return a >>= k = k a m >>= return = m m >>= (\x -> k x >>= h) = (m >>= k) >>= h  In terms of >=>: 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:  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 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.  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 maybeFix :: (a -> Maybe a) -> Maybe a maybeFix f = ma where ma = f (fromJust ma)  ### Exercises 1. Implement a MonadFix instance for []. Solution:  listFix :: (a -> [a]) -> [a] listFix f = la where la = f (head la)  # Foldable ## Definition 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:  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.  foldMap f = fold . map f  3. Implement foldMap in terms of foldr. Solution:  foldMap f = foldr (\a b -> mappend (f a) b) mempty  4. Implement foldr in terms of foldMap (hint: use the Endo monoid). Solution:  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:  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:  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:  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:  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:  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 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):  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:  (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:  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:  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 Functor’s definition for intuition. 2. Implement Traversable instances for [], Maybe, ((,) a), and Either a. Solution:  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 <$> f a  3. Explain why Set is Foldable but not Traversable. Solution: First, in terms of laws, Set is not a Functor, thus it cannot be made into a Traversable instance, since Traversable instances require Functor superclasses. Second, on an intuitive level: In Foldable, the goal is not to keep the shape/structure of the original container, we are trying to reduce the container into some value, and the shape of the final result doesn’t matter, but in Traversable, we ought to keep the structure of the final result, but we can’t guarantee this while using Sets, because we can define some transformation f :: Set a -> Set a which reduces the length of the Set. See Foldable vs. Traversable and Sets, Functors and Eq confusion. and Foldable and Traversable for more details. 4. Show that Traversable functors compose: that is, implement an instance for Traversable (Compose f g) given Traversable instances for f and g. Solution:  instance (Traversable f, Traversable g) => Traversable (Compose f g) where traverse :: (Applicative f) => (a -> f b) -> Compose g h a -> f (Compose g h b) traverse f (Compose t) = Compose <$> traverse (traverse f) t