Abstract

The longest path problem asks for a path with the largest number of vertices in a given graph. The first polynomial time algorithm (with running time O(n4)) has been recently developed for interval graphs. Even though interval and circular-arc graphs look superficially similar, they differ substantially, as circular-arc graphs are not perfect. In this paper, we prove that for every path P of a circular-arc graph G, we can appropriately “cut” the circle, such that the obtained (not induced) interval subgraph G′ of G admits a path P′ on the same vertices as P. This non-trivial result is of independent interest, as it suggests a generic reduction of a number of path problems on circular-arc graphs to the case of interval graphs with a multiplicative linear time overhead of O(n). As an application of this reduction, we present the first polynomial algorithm for the longest path problem on circular-arc graphs, which turns out to have the same running time O(n4) with the one on interval graphs, as we manage to get rid of the linear overhead of the reduction. This algorithm computes in the same time an n-approximation of the number of different vertex sets that provide a longest path; in the case where G is an interval graph, we compute the exact number. Moreover, our algorithm can be directly extended with the same running time to the case where every vertex has an arbitrary positive weight.

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