Sunday, June 28, 2009

Clustering: The "I don't like you" view

(this is part of an occasional series of essays on clustering: for all posts in this topic, click here)

Any clustering problem starts with a data set. And you can basically tell what algorithm you're going to need by asking, 'what structure do I have on the data ?'. The bare minimum you need is some way of taking two items and returning a number that tells you how similar they are, or more usefully, how far apart they are.

Of course, if I know that A and B are a certain distance apart, and B and C are a certain distance apart, I might like to infer something about the distance between A and C. This is what the triangle inequality gives you, and so the most elementary form of clustering takes items and a metric defined on these items (humans actually DON'T think about distances in this manner, as many psychological studies have shown: more on that later on).

I find it convenient to view clustering algorithms from a pop-psychology self-help perspective, and so I'll deem algorithms that operate with a distance metric the 'I don't like you' methods. The general goal with such methods is to group the items into "clusters" where there isn't too much hatred :).

There are many ways to quantify this "lack of hatred". The one most natural to theoreticians is the 'minimize the worst-case angst' viewpoint: in other words, find k clusters such that over all clusters, the maximum pairwise distance between points in a cluster is minimized. This is the well-known k-center problem.

If you don't like the lack of robustness (one point can mess up the cost), you could instead sum up all the pairwise distances to assign the cost for a cluster. This gives a variant of what's normally considered the k-median problem.

A third way of measuring cluster cost is to sum the squares of distances, rather than the distances themselves. This gives you a cost function that looks a lot like what k-means attempts to solve.

It's often useful to define a representative of a cluster: in the k-center view of the world, the center of a cluster is a point whose maximum distance to all points in the cluster is minimum. Note that by triangle inequality, this is within a factor of two of the k-center cluster cost, so it's often convenient to use this as the definition of the cluster cost (the cluster radius, instead of diameter, if you will). In the sum-of-all-costs view, the center correspondingly can be defined as as the point whose sum of all distances to all other points is minimized. Similarly for the case of sum-of-square-costs.

Remarks on complexity:
There is an elegant 2-approximation for the k-center problem by Gonzalez. The algorithm itself is the essence of simplicity: pick any point in the space, take its furthest neighbor, take the point furthest from these two, and so on. The first k points picked in this manner yield the desired cluster centers, and a simple triangle inequality argument yields the 2-approximation.

The k-median problem is a little harder to solve. The best known bound in a general metric space is a 3+eps approximation, and it's known that you can't get a PTAS. The approximation algorithm itself is not too hard: it's a local search heuristic that gets closer and closer to 3 as you permit larger and larger sets of centers to be swapped out.

Going to Euclidean space, or what's in a name ?
General metrics come either from some kind of similarity function between pairs of elements or from the induced shortest path metric on a graph (there are other metrics one can induce from a graph, and we'll talk about them later). But what if you have more information ? There's a general principle of clustering that's worth enunciating here: more structure on the data can lead you to more targeted algorithms that will probably work better, so resist the urge to be too general.

The most natural space to consider is a Euclidean space, and it is this space that explains the names 'median' and 'mean'. It's well known that on the line, the point minimizing the sum of distances to a set of points (the 1-median) is given by the actual median of the numbers. This explains why the problem of minimizing sum of distances is called the k-median problem. Similarly, the point that minimizes the sum of squared Euclidean distance is the centroid or the 'mean', giving rise to the k-means problem.

There's a long line of results on clustering under various measures in Euclidean space. In brief, for most of the above formulations, you can get a PTAS in Euclidean space, but you end up playing whack-a-mole with the parameters n, d, epsilon, and k (that is to say, something ends up exhibiting exponential dependence, and which parameter it is depends on what you care most about).

The Voronoi trick
But there's a general principle that governs much of the algorithm design for these problems. It's the so-called Voronoi trick, and can be summarized by pointing out that you can always improve the clustering cost by making sure that each point is assigned to its nearest neighbor. Effectively this means that the clusters define a Voronoi partition, with the center as a site, and all points within a Voronoi cell being assigned to the corresponding site. First of all, this means that the number of possible outcomes reduces from k^n (the number of ways of partitioning n things into k groups) to n^{kd} (since the number of Voronoi partitions is much smaller than the number of partitions). Secondly, it suggests a simple heuristic for improvement: Take a given clustering: reassign points so that each point is assigned to its nearest cluster center, and then recompute the center for the cluster associated with all points in the (erstwhile) Voronoi cell. Rinse and repeat....

This heuristic is the basis of the k-means algorithm, and is also the basis for a heuristic for the k-median problem clumsily named the k-medoids algorithm.

Notes:
  • all the above algorithms have the parameter k for the number of clusters. It's obvious why: all the above cost measures can be minimized by placing each item in a cluster by itself, but the resulting answer is meaningless, and "no man is an island"
  • In a general metric space defined over a finite set of points, the cluster centers will by default by elements of the input. This is often a desirable property, if you want representatives to be from the input. But in a smooth space like a Euclidean space, there's no reason to limit yourselves to data points, and this creates a dichotomy between the so-called discrete and continuous variants of a clustering problem. Often, you can use triangle inequality to argue that the answers don't differ significantly in cost, but it's important to ask yourself which variant you care about. The discrete problem can often be "easier" to solve, because you can enumerate over the choices, but this "easier" can be expensive: for example, minimizing the sum of squares is easy in a (continuous) Euclidean space, but is more expensive in a (discrete) metric space.
  • For some unknown reason, it's common to describe clustering problems by the algorithm used to "solve" them, which causes no end of confusion. I lost count of the number of times I had to point out that k-means is an algorithm that attempts to optimize a specific cost function. This is more than just pedantry, because unless you realize the difference between a problem and a specific solution, you'll never be able to conceive of an alternate approach, and you'll never even try to abstract out what's special about your problem. To theoreticians, this might be obvious, but it isn't really obvious IRL.
Next up: everything you ever wanted to know about k-means.

p.s this is written in a much more breezy style than I'd use for any formal document, but the organization and thinking reflects how I'd like to structure the material. Comments, as always, are welcome.
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