## Thursday, January 24, 2013

### The Handshake Problem

The Handshake Problem is something of a classic in mathematics. I first heard about it in an algebra course I took in high school, and it's stuck with me through the years. The question is this:

In a room containing $n$ people, how many handshakes must take place for everyone to have shaken hands with everyone else?
The goal of this short blog post will be to present a solution to this problem using the concept of graphs.

### The Complete Graph

 fig. I: A complete graph with 7 vertices
First, we must understand the idea of a complete graph. A complete graph is a graph such that each node is connected to every other node. A graphical example can be seen in fig. I.

We can use the model of a complete graph to reason about the problem at hand. The nodes in the graph may represent the people in the room, while the connecting edges represent the handshakes that must take place. As such, the key to solving the Handshake Problem is to count the edges in a complete graph.

But it would be silly to draw out a graph and count the edges one-by-one in order to solve this problem (and would take a very long time for large values of $n$), so we'll use math!

### The Solution

To find the solution, we have to make a few key observations. The easiest way to start out in this case is to map out the first few complete graphs and try to find a pattern:

Let $n$ = the number of nodes, $e$ = the number of edges.
• $n = 1 \Rightarrow e = 0$
• $n = 2 \Rightarrow e = 1$
• $n = 3 \Rightarrow e = 3$
• $n = 4 \Rightarrow e = 6$
At this point, you may be noticing a pattern. As $n$ increases, we're adding $n -1$ edges. This makes sense -- each time a new node is introduced, it must connect to each node other than itself. In other words, $n-1$ connections must be made.

It follows that the number of the edges ($e$) in a complete graph with $n$ vertices can be represented by the sum $(n-1) + (n-2) + \dots + 1 + 0$, or more compactly as:
$$\sum_{i=1}^n i-1$$
You may already know from summation laws that this evaluates to $\frac{n(n-1)}{2}$. Let's prove it.

### The Proof

Theorem: $\forall n \in \mathbb{N} : \sum_{i=1}^{n} i-1 = \frac{n(n-1)}{2}$

Proof:  Base case: Let $n = 1$. Then, $\frac{1(1-1)}{2} = \sum_{i=1}^1 i-1 = 0$.
Inductive case: Let $n \in \mathbb{N}$. We need to prove that $\frac{n(n-1)}{2} + n = \frac{n(n+1)}{2}$.
$$\begin{array} {lcl} \frac{n(n-1)}{2} + n & = & \frac{n(n-1)+2n}{2} \\ & = & \frac{n^2 - n + 2n}{2} \\ & = & \frac{n^2 + n}{2} \\ & = & \frac{n(n+1)}{2}■\end{array}$$
We can now use the theorem knowing it's correct, and, in fact, provide a solution to the original problem. The answer to the Handshake Problem involving $n$ people in a room is simply $\frac{n(n-1)}{2}$. So, the next time someone asks you how many handshakes would have to take place in order for everyone in a room with 1029 people to shake each other's hands, you can proudly answer "528906 handshakes!"

Until next time,

Ben

## Wednesday, January 16, 2013

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: $$join :: (Monad ~m) \Rightarrow m ~(m ~a) \rightarrow m ~a$$ 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... $$instance ~(Monoid ~a) \Rightarrow Monad ~(Tree ~a) ~where ~\dots$$ ...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: $$(>>=) :: (Monad ~m) \Rightarrow m ~a \rightarrow ~(a \rightarrow m ~b) \rightarrow m ~b$$ 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: $$(=>>) :: (Comonad ~w) \Rightarrow w ~a \rightarrow (w ~a \rightarrow ~b) \rightarrow w ~b$$ 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: $$coreturn :: (Comonad ~w) \Rightarrow w ~a \rightarrow a$$ 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: $$cojoin :: (Comonad ~w) \Rightarrow w ~a \rightarrow w ~(w ~a)$$ 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 represented as:
$$a ~=>> ~f = fmap ~f ~\ ~cojoin ~a$$

The full Comonad typeclass (as I like to define it) is as follows:

$$class ~(Functor ~w) \Rightarrow Comonad ~w ~a ~where \\ \hspace{22pt}coreturn :: (Comonad ~w) \Rightarrow w ~a \rightarrow a \\ \hspace{38pt}cojoin :: (Comonad ~w) \Rightarrow w ~a \rightarrow w ~(w ~a) \\ a ~=>> ~f = fmap ~f ~\ ~cojoin ~a$$

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.

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

From here, we're able to make a Comonadic Tree like so:

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:

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:

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:

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.

Ben

## Wednesday, January 2, 2013

If you've used Haskell before, chances are that you've heard the term "monad" a good bit. But, you might not know what they are, or what they are really used for. It is my goal in this blog post to shed some light on what monads are, introduce and define few simple ones, and touch on their usefulness.

I will assume basic familiarity with Haskell syntax in this blog post. A working knowledge of monoids and functors will also be beneficial, although not strictly necessary.

We can see a Monad has two operations, return and >>= (this is commonly pronounced "bind"). You may suspect that you know what return does from other programming languages, but be warned: Haskell's return is very different! Haskell's return only operates on Monads, and essentially acts as a "wrapping" function. That is, if you call return on a value, it will turn it into a monadic value of that type. We will look at an example of this shortly. The second operation, >>=, takes two arguments: an a wrapped in a Monad m (I will refer to this as m a from now on), and a function that converts an a to b wrapped in the same type of Monadm. It produces a value of type b, wrapped in a Monad m (I will call this m b from now on). This may sound complicated, but I'll do my best to explain it after showing the most basic Monad type, namely, the Identity Monad:

The Identity data declaration defines only one type: Identity a. Basically, this is just a wrapper for a value, and nothing more. return is simply defined as Identity. We can, for example, call return 3 and get an Identity 3. Turning values into Identitys is as simple as that. The bind function may look a little bit more obscure, though. Let's look closely: We first use pattern matching to be able to operate on the type that the Identity Monad wraps (Identity a), bind it (>>=) to a function (f). Since f converts normal values into monadic ones (look at the type declaration), we can simply apply f to a, and we've produced a new Identity. I've provided a couple of examples of how this works. addThree adds 3 to an Identity Int (m1) and getLength turns an Identity [a]  (m2) into an Identity Int. Both of the bind examples produce the value Identity 6. Can you see why?

I want to take a little sidebar here and try to explain how I understand Monads before moving on. Basically the real tough part of understanding Monads is just understanding >>=. A little bit of dissection can explain a lot, but it took me months to really grasp the concept. On the surface, you're morphing an m a into an m b. At first, I thought that the m's in the type signature were allowed to different types of Monads. For instance, I thought that Monads made it possible to bind Identity Ints to functions that produced [String]s (we'll look at the list Monad in a minute). This is wrong, and for good reason! It was incredibly confusing to think about a function that could generalize to this degree and it is, to my knowledge, not possible to do so. The Monad (wrapper) type is retained, but the wrapped type can be changed by means of the function you bind to.

The second thing that I want to stress is what >>= really does. What we're doing with the >>= function is turning an m a into an m b, but it's not so direct as you might want to think. The function argument in >>=  ( (a -> m b) ) was really confusing to me to begin with, so I'm going to try to explain it. The function you're binding your m a to must take an a and return an m b. This means that you're acting on the wrapped part of the Monad in order to produce an entirely new Monad. It took me a while to realize what exactly was happening, that I wasn't supposed to be directly converting m a to m b by means of my argument function (that's actually what >>= is for). Just remember that the function you bind to should operate on an a to produce an m b.

With that knowledge, let's take a look at some more examples of Monads:

If you're a Haskell user, you're likely already familiar with the Maybe data type:

Defined in the Prelude

The Maybe Monad is hardly an extension of the aforementioned Identity Monad. All we are really adding is the functionality to fail -- that is, to produce a Nothing value if necessary. In fact, the Just a type is exactly the same as Identity a, except that a Nothing value anywhere in the computation will produce a Nothing as a result. Next, we'll look at the Monad instance for lists:

Defined in the Prelude

The list Monad is a little more complicated, but it's really nothing too new. For return, we just have to wrap our value inside a list. Now, let's look at >>=concatMap actually is the >>= instance for lists. Why? Well, concatMap is, simply, (concat . map). We know that map takes a list of as and maps a function over them, and concat is a function that flattens lists, that is, turns [[a]]s into [a]s. When we compose these two functions (or use concatMap), we're making a function that must produce [[a]]s out of [a]s, and then flatten them back to [a]s. That is exactly how >>=  works for lists! Now, let's take a look at a slightly more complicated Monad:

This one requires some explanation, because it's a bit of a leap from the last three, since we're defining a Monad instance for functions. I have added a couple of comments in the above code in order to further explain how everything is working in the (->) r Monad.

Well, first things's first. We define return as const, which takes some arbitrary argument and produces a function that produces a constant value (what we pass in to return). This is exactly what we want; a minimal context for functions.

(>>=) is a bit more complex. First, let's take a look at the type signature, specific to this Monad. We're binding a function (r -> a) to a function (a -> (r -> b)) to produce a function of type (r -> b). So in the declaration of the function, f is of type (r -> a) and g is of type (a -> (r -> b)). I've conveniently named the argument our function is going to take r, for clarity. This is what the user will pass into the function that (>>=) produces.

f takes an r, so we apply the value from the lambda into it to produce an a. We then use that a to produce an (r -> b), using g. This produces our function, but it needs to be fed some value, so we finally apply it to r in order to complete everything. If any of this sounds confusing, try to work through it on paper and read the comments in my code -- after a bit of staring, it should start to make sense.

I've also implemented the mean function here, using the (->) r Monad with and without do-notation. I think it's a pretty elegant way to express it.

We've only seen the tip of the iceberg when it comes to Monads, but this should serve as a good basic introduction. Once you get used to seeing Monads, they become a bit less scary. The more time you spend with them, the more comfortable you'll feel with them.

Before I end this post, I want to leave a question for the reader. Given the general Tree data type represented below, can you intuitively make a Monad instance?

I'll explain why this is difficult when we talk about a similar mathematical type in the next post.