Linear-Time Algorithms for Maximum-Weight Induced Matchings and Minimum Chain Covers in Convex Bipartite Graphs

A bipartite graph $G=(U,V,E)$ is convex if the vertices in $V$ can be linearly ordered such that for each vertex $u\in U$, the neighbors of $u$ are consecutive in the ordering of $V$. An induced matching $H$ of $G$ is a matching such that no edge of $E$ connects endpoints of two different edges of $H$. We show that in a convex bipartite graph with $n$ vertices and $m$ weighted edges, an induced matching of maximum total weight can be computed in $O(n+m)$ time. An unweighted convex bipartite graph has a representation of size $O(n)$ that records for each vertex $u\in U$ the first and last neighbor in the ordering of $V$. Given such a compact representation, we compute an induced matching of maximum cardinality in $O(n)$ time. In convex bipartite graphs, maximum-cardinality induced matchings are dual to minimum chain covers. A chain cover is a covering of the edge set by chain subgraphs, that is, subgraphs that do not contain induced matchings of more than one edge. Given a compact representation, we compute a representation of a minimum chain cover in $O(n)$ time. If no compact representation is given, the cover can be computed in $O(n+m)$ time. All of our algorithms achieve optimal running time for the respective problem and model. Previous algorithms considered only the unweighted case, and the best algorithm for computing a maximum-cardinality induced matching or a minimum chain cover in a convex bipartite graph had a running time of $O(n^2)$

Tight lower bound for the channel assignment problem

We study the complexity of the Channel Assignment problem. A major open problem asks whether Channel Assignment admits an $O(c^n)$-time algorithm, for a constant $c$ independent of the weights on the edges. We answer this question in the negative i.e. we show that there is no $2^{o(n\log n)}$-time algorithm solving Channel Assignment unless the Exponential Time Hypothesis fails. Note that the currently best known algorithm works in time $O^*(n!) = 2^{O(n\log n)}$ so our lower bound is tight