### Benjamin Kovach

We’ve seen in the previous post what monads are and how they work. We saw a few monad instances of some common Haskell structures and toyed around with them a little bit. Towards the end of the post, I asked about making a Monad instance for a generalized tree data type. My guess is that it was relatively difficult. But why? Well, one thing that I didn’t touch on in my last post was that the monadic >>= operation can also be represented as (join . fmap f), that is, a >>= f = join $fmap f a, as long as a has a Functor instance. join is a generalized function of the following type: join :: (Monad m) => m (m a) -> m a Basically, this function takes a nested monad and “joins” it with the top level in order to get a regular m a out of it. Since you bind to functions that take as and produce m as, fmap f actually makes m (m a)s and join is just the tool to fix up the >>= function to produce what we want. Keep in mind that join is not a part of the Monad typeclass, and therefore the above definition for >>= will not work. However, if we are able to make a specific join function (named something else, since join is taken!) for whatever type we are making a Monad instance for, we can certainly use the above definition. I don’t want to spend too much time on this, but I would like to direct the reader to the Monad instance for [] that I mentioned in the last post – can you see any similarities between this and the way I structured >>= above? Now back to the Tree type. Can you devise some way to join Trees? That is, can you think of a way to flatten a Tree of Tree as into a Tree a? This actually turns out to be quite difficult, and up to interpretation as to how you want to do it. It’s not as straightforward as moving Maybe (Maybe a)s into Maybe as or [[a]]s into [a]s. Maybe if we put a Monoid restriction on our a type, as such… data Tree a = Node a [Tree a] | Leaf a deriving (Show, Eq) instance (Monoid a) => Monad (Tree a) where ... …we could use mappend in some way in order to concatenate all of the nested elements using some joining function. While this is a valid way to define a Tree Monad, it seems a little bit “unnatural.” I won’t delve too deeply into that, though, because that’s not the point of this post. Let’s instead take a look at a structure related to monads that may make more sense for our generalized Tree type. #### What is a comonad? Recall the type signature of >>= for Monads: (>>=) :: (Monad m) => (m a) -> (a -> m b) -> m b That is, we’re taking an m a and converting it to an m b by means of some function that operates on the contained type. In other words, we’re producing a new value using elements contained inside the Monad. Comonads have a similar function: (=>>) :: (Comonad w) => w a -> (w a -> b) -> w b The difference here is that the function that we use to produce a new value operates on the whole – we’re not operating on the elements contained inside the Comonad, but the Comonad itself, to produce a new value. We also have a function similar to the Monadic return: coreturn :: (Comonad w) => w a -> a Whereas return puts a value into a Monadic context, coreturn extracts a value from a Comonadic context. I mentioned the Monadic join function above because I would also like to mention that there is a similar operation for Comonads: cojoin :: (Comonad w) => w a -> w (w a) Instead of “removing a layer,” we’re “adding” a layer. And, as it turns out, just as a >>= f can be represented as join$ fmap f a, =>> can be representaed as : a =>> f = fmap f $cojoin a. The full Comonad typeclass (as I like to define it) is as follows: class (Functor w) => Comonad w where coreturn :: (Comonad w) => w a -> a cojoin :: (Comonad w) => w a -> w (w a) a =>> f = fmap f$ cojoin a 

By now, it should at least be clear that Monads and Comonads are related – it shouldn’t be hard to see why they are so similar in name! Note: There is a package called Control.Comonad on Hackage. It uses different names for the Comonad operations, but they do the same things. It’s a good package, but I wanted to show how Comonads are built and use the operation names I used to make things clearer. **

#### What can I do with Comonads?

As it turns out, the Tree a structure that I’ve been mentioning fits into the Comonadic context quite well, and provides a simple example as to how Comonads work.

class (Functor w) => Comonad w where
coreturn :: w a -> a
cojoin   :: w a -> w (w a)
(=>>)    ::  w a -> (w a -> b) -> w b

Then we’ll go ahead and make a Functor instance of our Tree a data type:

data Tree a = Node a [Tree a] | Leaf a deriving (Show, Eq)

instance Functor Tree where
fmap f (Leaf a) = Leaf $f a fmap f (Node a b) = Node (f a) (map (fmap f) b) From here, we’re able to make a Comonadic Tree like so: instance Comonad Tree where coreturn (Leaf a) = a coreturn (Node a _) = a cojoin l@(Leaf a) = Leaf l cojoin n@(Node a b) = Node n (map cojoin b) x =>> f = fmap f$ cojoin x

The only real point of confusion here is in the cojoin function for Nodes. But, all we are doing is wrapping the entire node in a new node, and then mapping cojoin over every child node of the current node to produce its children. In other words, we’re turning a Node a into a Node (Node a), which is exactly what we want to do. So what can we do with a comonadic tree? Let’s tackle a simple problem. Say we’re hanging out in California, and we want to get to the East coast as quickly as possible. We don’t really care where on the East coast we end up – we just want to get there. We can map out the different paths that we can take, along with the time it takes to get to each one. In other words, we’re going to model spots on the map as Nodes on a Tree and give them a weight corresponding to the time it takes to get there. We can model this situation with a Tree as follows:

trip =
Node 0
[Node 6
[Node 2
[Node 6
[Leaf 1],
Leaf 14],
Node 1
[Leaf 19],
Node 8
[Leaf 21,
Node 4
[Leaf 2, Leaf 6]
]
],
Node 3
[Node 9
[Leaf 12],
Node 14
[Leaf 6,
Node 3
[Leaf 1]
]
]
]

The number of each Node marks its distance from the previous Node. The root Node of the Tree is the starting point, so it is 0 distance away from itself. What we want to do to find the shortest path through the country is essentially, as follows. First, we’re going to need to check the deepest nodes, and find their minimum distance children. We will add the distance of the Node closest to the one we’re examining to its own distance, to find the shortest way to get from the node we’re examining to the destination. Once all of that has been done, we’ll need to traverse up the tree, repeating this as we go. By the end of the algorithm, the root Node will be marked with the shortest distance to the destination. Now, that may sound somewhat iterative in nature, but we’re going to morph this into a comonadic operation. First, let’s take a look at a function we can use to find the minimum distance to the next level of our tree:

shortest :: (Num a, Ord a) => Tree a -> a
shortest (Leaf x) = x
shortest (Node x xs) = x + (minimum \$ map shortest xs)

This is relatively simple, and does precisely what was mentioned above: adds the minimum value contained in the connected Nodes to the parent Node. Next, take a look at the type signature. We can see that this function produces a new number from a tree full of numbers. This coincides precisely with the function type we need to use with =>> so we’ll be able to use it to get exactly what we want. The rest is very simple:

minimumDist = trip =>> shortest

This produces a tree full of the minimum distances from each node to the East coast. Pulling the actual value out is as easy as calling coreturn on the resultant tree.