There are many different ways to formalize the intuitive problem of clustering, where the goal is to divide up a collection of objects into groups that are "similar" to one another.
First, a natural way to express the input to a clustering problem is via a set of objectsp1,p2, ...,pn, with a numerical distance d(pt,defined on each pair. (We require only that d(pt, p) = 0; that d(pt, pj)> 0 for distinct pi and pj; and that distances are symmetric: d(pi, pj) = d(pj, pi).)
In Section, earlier in the book, we considered one reasonable formulation of the clustering problem: Divide the objects into k sets so as to maximize the minimum distance between any pair of objects in distinct clusters. This turns out to be solvable by a nice application of the Minimum Spanning Tree Problem.
A different but seemingly related way to formalize the clustering problem would be as follows: Divide the objects into k sets so as to minimize the maximum distance between any pair of objects in the same cluster. Note the change. Where the formulation in the previous paragraph sought clusters so that no two were "close together," this new formulation seeks clusters so that none of them is too "wide"—that is, no cluster contains two points at a large distance from each other.
Given the similarities, it s perhaps surprising that this new formulation is computationally hard to solve optimally. To be able to think about this in terms of NP-completeness, let s write it first as a yes/no decision problem. Given n objects p1, p2,...,pn with distances on them as above, and a bound B, we define the Low-Diameter Clustering Problem as follows: Can the objects be partitioned into k sets, so that no two points in the same set are at a distance greater than B from each other?
Prove that Low-Diameter Clustering is NP-complete.
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