A 3/2-approximation algorithm for some minimum-cost graph problems

Abstract

We consider a class of graph problems introduced in a paper of Goemans and Williamson that involve finding forests of minimum edge cost. This class includes a number of location/routing problems; it also includes a problem in which we are given as input a parameter $k$, and want to find a forest such that each component has at least $k$ vertices. Goemans and Williamson gave a 2-approximation algorithm for this class of problems. We give an improved 3/2-approximation algorithm.

Appendix

We illustrate a run of Algorithm 1 on the problem in which we want to find a minimum-cost set of edges such that each component has at least four vertices. We give an instance of the problem in which the graph contains edges of cost 0, 1, 2, or $2+\varepsilon$ to keep things simple.

The instance is described in the Fig. 3; we give an optimal solution and a solution produced by the algorithm. We draw good edges with a double line.

Fig. 3

The graph on the left gives an optimal solution of cost $6+\varepsilon$ and the graph on the right gives a solution produced by the algorithm of cost 8

In Fig. 4, in the first four iterations, the algorithm picks all four edges of cost 0. In the next iteration, the cheapest good edge ($e_{5}$) has cost 2, and the remaining bad edges all have cost 1, so the algorithm chooses the good edge $e_5$.

Fig. 4

The first four iterations of the algorithm pick the four edges of cost 0 (shown on the left). On the right, in the next iteration the algorithm picks $e_5$ as a good edge

In Fig. 5, the algorithm now picks a bad edge of cost 1, $e_6$, since this is less than half the cost of the cheapest good edge ($e_{12}$, of cost $2+\varepsilon$). Adding this edge to the solution makes the edge $e_7$ a good edge. Since $e_7$ is of cost 1, the algorithm adds it as a good edge. After this, $e_{12}$ is no longer a good edge.

Fig. 5

The algorithm adds $e_6$ and $e_7$ to the partial solution

In Fig. 6, there are only bad edges; the algorithm picks $e_8$, which is of cost 1. The addition of $e_8$ makes $e_{13}$ a good edge. Then the algorithm picks $e_9$, after which $e_{13}$ is no longer a good edge.

Fig. 6

The algorithm adds $e_8$ and $e_9$ to the partial solution

In Fig. 7, edge $e_{10}$ is the last bad edge of cost 1 and, when it is examined, its addition would link two big components, therefore it isn’t added to the partial solution. Edge $e_{11}$ is one of the bad edges of cost $2$ remaining; the algorithm adds it to the solution. Now edge $e_{12}$ would form a cycle with the partial solution and edge $e_{13}$ would link two big components, so neither of them are added to the solution. The algorithm returns the indicated set of edges, and each component has at least 4 vertices in it.

Fig. 7

On the left, $e_{10}$ and $e_{11}$ has been examined. On the right the two last edges $e_{12}$ and $e_{13}$ has been examined and removed