Graphs and Combinatorics (2010) 26:301–313 DOI 10.1007/s00373-010-0922-0 ORIGINAL PAPER A Min–Max Property of Chordal Bipartite Graphs with Applications Atif Abueida · Arthur H. Busch · R. Sritharan Received: 19 February 2008 / Revised: 2 February 2009 / Published online: 3 April 2010 © Springer 2010 Abstract We show that if G is a bipartite graph with no induced cycles on exactly 6 vertices, then the minimum number of chain subgraphs of G needed to cover E(G) equals the chromatic number of the complement of the square of line graph of G. Using this, we establish that for a chordal bipartite graph G, the minimum number of chain subgraphs of G needed to cover E(G) equals the size of a largest induced matching in G, and also that a minimum chain subgraph cover can be computed in polynomial time. The problems of computing a minimum chain cover and a largest induced matching are NP-hard for general bipartite graphs. Finally, we show that our results can be used to efficiently compute a minimum chain subgraph cover when the input is an interval bigraph. Keywords Chain cover · Induced matching · Bipartite graphs 1 Introduction and Motivation In this paper, all graphs G = (V, E) are simple and finite and consist of a set of vertices V and a set of edges E, which are unordered pairs of vertices. We follow the notation and terminology used in [22], and refer readers there for any unfamiliar terms. We use Ck to refer to a cycle on k vertices, n to refer to the number of vertices, and m to refer to the number of edges in a graph. An induced matching inagraphis R. Sritharan is supported by a grant from the NSA. A. Abueida · A. H. Busch (B) Department of Mathematics, The University of Dayton, Dayton, OH 45469-2316, USA e-mail: [email protected] R. Sritharan Computer Science Department, The University of Dayton, Dayton, OH 45469, USA 123 302 Graphs and Combinatorics (2010) 26:301–313 a matching that is also an induced subgraph, i.e., no two edges of the matching are joined by an edge in the graph. We use im(G) to denote the size of a largest induced matching in G.GivenG and integer k, the problem of deciding whether im(G) ≥ k is NP-complete [4] even when G is bipartite. A bipartite graph G = (X, Y, E) is a chain graph if it does not have a 2K2 as an induced subgraph, i.e., if im(G) = 1. Bipartite graph G = (X , Y , E) is a chain sub- graph of bipartite graph G = (X, Y, E),ifG is a subgraph of G and im(G) = 1. We refer to a set of chain subgraphs of a bipartite graph G = (X, Y, E) as a chain subgraph cover or simply chain cover if the union of the edge sets of the subgraphs is E, and we use ch(G) to denote the minimum cardinality of a chain subgraph cover. Yannakakis showed [24] that when k ≥ 3, deciding whether ch(G) ≤ k for a given bipartite graph G is NP-complete. An efficient algorithm to determine whether ch(G) ≤ 2 for a given bipartite graph G is known [18]. For several restricted families of graphs, the problem of computing a largest induced matching has been shown [5–7,10–12] to be solvable in polynomial time using the fol- lowing transformation. Given graph G = (V, E), consider the graph G∗ constructed as follows: V (G∗) = E(G), and edges wx and yz of G are adjacent in G∗ if and only ∗ if {w, x, y, z} induces a 2K2 in G. Note that G is the complement of the square of the line graph of G. It is clear that any induced matching in G corresponds to a clique in G∗ and vice versa. Therefore, a largest induced matching of G can be computed by finding a largest clique in G∗, and thus im(G) = ω(G∗). Similarly, the edge set of any chain subgraph H of G corresponds to an independent set of vertices in G∗. Therefore, the edge-sets of any chain cover of G can be viewed as a proper multi-coloring of the vertices of G∗ and hence χ(G∗) ≤ ch(G). Combining these results, im(G) = ω(G∗) ≤ χ(G∗) ≤ ch(G) for any bipartite graph G. A natural question is whether or not the three parameters can differ. It is shown in [8] that there exist bipartite graphs G with ch(G) arbitrarily larger than χ(G∗). The discussion in [8] is about threshold dimension of split graphs.Asevery threshold graph can be derived from a chain graph by turning one side of the bipartition into a clique [3] and as a split graph is a bipartite graph with one side of the bipartition turned into a clique, the same construction applies in the context of the chain cover number of bipartite graphs. A bipartite graph G = (X, Y, E) is convex if the vertices in Y can be linearly ordered so that for each vertex x ∈ X the neighbors of x occur consecutively. The graph G is a comparability graph if each edge of G can be oriented so that the resulting directed graph is acyclic and transitive. It was shown in [25] that when G is a convex bipartite graph, G∗ is a comparability graph and also ch(G) = χ(G∗). As compa- rability graphs are perfect, we have [25] that when G is a convex bipartite graph, im(G) = ω(G∗) = χ(G∗) = ch(G). This relationship was used in [25] to design an O(m2) time algorithm to compute ch(G) as well as a minimum chain subgraph cover for G, when G is a convex bipartite graph; it was subsequently shown in [2] that the algorithm from [25] can be implemented to run in O(n2) time. A bipartite graph is chordal bipartite if it does not contain any induced Ck, k ≥ 6. The class of chordal bipartite graphs is a well known [3] generalization of the class of convex bipartite graphs. The problem of determining the complexity of computing ch(G), when G is chordal bipartite, was posed in [25]. More specifically, noting that 123 Graphs and Combinatorics (2010) 26:301–313 303 the graphs constructed in [8] contain induced C6s (and hence are not chordal bipar- tite), the issue of whether ch(G) equals χ(G∗), when G is chordal bipartite graph, was raised in [2]. Further, as observed in [2], if ch(G) = χ(G∗) when G is chordal bipartite, then it would imply that im(G) and ch(G) can be computed in polynomial time for chordal bipartite graphs. We confirm this equality by proving that when G is a ∗ bipartite graph that does not contain any induced C6, ch(G) = χ(G ). We also address the efficient computation of ch(G) and im(G) when G is an interval bigraph.The class of convex bipartite graphs is a proper subclass of the class of interval bigraphs, and the class of interval bigraphs, in turn, is a proper subclass of the class of chordal bipartite graphs. 2 A Min–Max Theorem Lemma 1 Suppose G is a bipartite graph that does not contain any induced C6. Then every maximal independent set of G∗ is the edge-set of a chain subgraph of G. Proof Let G = (X, Y, E) be a bipartite graph that contains no induced C6. Assume the result is false, and consider a maximal independent set S of G∗ which is not the edge-set of a chain subgraph of G. Then there must be two edges x1y1 ∈ S and x2y2 ∈ S that form an induced 2K2 in the graph G = (X, Y, S) (hence x1 = x2 and ∗ y1 = y2). Since these two edges are non-adjacent in G , we immediately conclude that either x1y2 ∈ E\S or x2y1 ∈ E\S, or both. Without loss of generality, assume ∗ x1y2 ∈ E\S.AsS is maximal, S∪{x1y2} is not an independent set in G , so there is an edge x3y3 ∈ S such that x1y2 and x3y3 form an induced 2K2 in G. Thus, x3 = x1, x2, y3 = y1, y2, and furthermore, x3y2 ∈/ E and x1y3 ∈/ E as indicated below (dashed lines indicate non-edges of G). Next, note that as {x1y1, x2y2, x3y3}⊆S, no two of these edges induce a 2K2 in G,sowemustalsohave{x2y3, x3y1}⊂E. Note the edges x1y2, x2y2, x2y3, x3y3, x3y1 and x1y1 are the edge-set of a C6 in G. Since G contains no induced C6,atleast one chord of this cycle must be present, and as shown above, the only possible chord is x2y1 and so x2y1 ∈ E. Now, note that since x1y1 and x2y2 induce a 2K2 in G = (X, Y, S), x2y1 ∈/ S. Repeating the above argument, we note that S ∪{x2y1} is not an independent set of ∗ G , and so there is some edge x4y4 ∈ S such that x2y1 and x4y4 form an induced 2K2 in G, meaning that x4 = x2 and y4 = y1 as well as x4y1 ∈/ E and x2y4 ∈/ E. From these last two observations, we conclude that x4 = x1, x3 and that y4 = y2, y3. 123 304 Graphs and Combinatorics (2010) 26:301–313 Additionally, as x4y4 ∈ S, neither x1y1 and x4y4, nor x2y2 and x4y4 form an induced 2K2 in G. Thus, {x1y4, x4y2}⊂E which gives the following configuration in G. Finally, we note that both x3y3 ∈ S and x4y4 ∈ S, and so these two edges do not induce a 2K2 in G.