A Min–Max Property of Chordal Bipartite Graphs with Applications

Graphs and Combinatorics (2010) 26:301–313 DOI 10.1007/s00373-010-0922-0

ORIGINAL PAPER

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 subgraph 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 following 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 comparability 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 bipartite), 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. As the conﬁguration above is symmetric in X and Y , we can thus assume that x3y4 ∈ E. We then obtain the desired contradiction by observing that x1, y2, x2, y3, x3, y4 are the vertices of an induced C6, as indicated below.

Theorem 1 If G is a bipartite graph that does not contain any induced C6, then ch(G) = χ(G∗).

Proof For any bipartite graph G, χ(G∗) ≤ ch(G). We now show that when G does ∗ ∗ not contain any induced C6, ch(G) ≤ χ(G ). Consider any optimal coloring of G , and let I1, I2,...,Ik be the k color classes. Let S j ,1≤ j ≤ k, be a maximal indepen- ∗ dent set of G such that I j ⊆ S j . By Lemma 1, each S j corresponds to the edge-set of a chain subgraph of G. Taken together these chain subgraphs form a chain cover of order k = χ(G∗) and therefore, ch(G) ≤ χ(G∗).

A graph is weakly chordal if neither the graph nor its complement contains any ∗ induced Ck, k ≥ 5. It was shown in [6] that when G is a weakly chordal graph, G is also weakly chordal and that im(G) can be computed in polynomial time. As it can be easily seen that every chordal bipartite graph is weakly chordal, and as weakly chordal graphs are perfect, we have the following corollary.

Corollary 1 For a chordal bipartite graph G, im(G) = ω(G∗) = χ(G∗) = ch(G).

3 Applications

In this section, we consider some algorithmic implications of Theorem 1. 123 Graphs and Combinatorics (2010) 26:301–313 305

3.1 Chordal Bipartite Graphs

As noted in [2], Corollary 1 immediately implies that ch(G) can be computed in polynomial time for chordal bipartite graphs using the same algorithm for im(G) in [6]. We now show that a minimum chain cover can be constructed as well. Proposition 1 When G is chordal bipartite, ch(G) can be computed in O(m3) time. Moreover, a minimum chain subgraph cover can also be found in O(m3) time. Proof For a weakly chordal graph H on n vertices, χ(H), ω(H), an optimal vertex coloring, and a largest clique can all be computed [15]inO(n3) time. As G∗ has m vertices, χ(G∗) can therefore be computed in O(m3) time. By Lemma 1, we can compute a minimum chain subgraph cover for G by simply growing each color class of G∗ into a maximal independent set of G∗. In any graph, using standard techniques, a given independent set can be extended into a maximal independent set in time that is linear in the size of the graph. As the size of G∗ is O(m2) and as G∗ has O(m) color classes, a required minimum chain subgraph cover of G can be constructed in O(m3) time.

3.2 Interval Bigraphs

A bipartite graph B = (X, Y, E) is an interval bigraph if every vertex v ∈ (X ∪ Y ) can be mapped to an interval Iv on the real line such that for x ∈ X and y ∈ Y , xy ∈ E if and only if Ix ∩ Iy =∅. Thus, the intersection, or the lack of intersection, between intervals corresponding to vertices from the same color class is not relevant. This class of graphs was first introduced in [13]. It is well known that the class of interval bigraphs is closely related to the class of interval digraphs [20], and Hell and Huang [16] gave a characterization which relates interval bigraphs to a sub-class of circular-arc graphs. The problem of recognizing interval bigraphs in polynomial time was first posed in [20]. Müller [19] designed an O(n5m6log n) time algorithm to recognize interval bigraphs; his algorithm remains the current best for the problem. Next, we review a characterization of interval bigraphs established in [20]. For a bipartite graph B = (X, Y, E), its bipartite adjacency matrix M is a 0/1 matrix with a row for each vertex of X and a column for each vertex of Y such that xy ∈ E if and only if M[x, y]=1. A 0/1 matrix has a zero-partition if its 0’s admit a partition [20,23] into sets R and C such that every entry of the matrix to the right of an element of R is in R and every entry of the matrix below an element of C is in C. Theorem 2 ([20,23]) G is an interval bigraph if and only if the rows and columns of its bipartite adjacency matrix can be permuted to admit a zero-partition. We note that the proofs in [20,23] translate into O(n2) time algorithms that can construct a zero-partition of the bipartite adjacency matrix of an interval bigraph, given its interval model. It is known [19,20,23] that every convex graph is an interval bigraph and every interval bigraph is a chordal bipartite graph. Therefore, we can conclude from our 123 306 Graphs and Combinatorics (2010) 26:301–313 previous discussions that given an interval bigraph G, ch(G) and a minimum chain subgraph cover for G can be found in O(m3) time. Here, we show that if we are given the zero-partition of the bipartite adjacency matrix of an interval bigraph, these problems can be solved more efficiently. Our basic idea is that an optimal coloring of G∗ can be found without explicitly constructing the graph G∗ itself. Instead, we make use of the properties of a zero-partition of the bipartite adjacency of G and color the edges of G to correspond to an optimal coloring of vertices of G∗. We note that the ideas used here are similar to those used in [2]. = ( , , ) ={ ,..., } = For an interval bigraph G X Y E , where X x1 xn X and Y {y1,...,yn },letM be a zero-partition of the bipartite adjacency matrix of G.As Y a consequence, M cannot have 01 as an induced submatrix, and hence any 2K in 10 2 G must induce the submatrix 10 in M. We refer to the edge x y that corresponds 01 i j to the entry M[i, j]=1aseij.Weuseeij ekl to refer to edges eij and ekl such that i ≤ k and j ≤ l. Proposition 2 If G is an interval bigraph, then G∗ is a comparability graph. Proof Using the structure and notation indicated above, we prove that G∗ is a comparability graph by demonstrating that G∗ can be turned into an ayclic and transitive digraph by appropriately orienting its edges. We orient the edge between eij and ekl ∗ in G to be eij → ekl if and only if eij ekl. It is easily seen that the resulting orientation is acyclic. We now show it is also transitive. Suppose we have directed edges eij → ekl and ekl → epq. Firstly, it is easy to see that i < k < p and j < l < q. Thus, eij epq. Secondly, that M is a zero-partition implies that M[i, l] must be in the set C and M[k, j] is in the set R. Hence, M[p, j]=M[i, q]=0. Therefore, eij ∗ and epq also induce a 2K2 in G and are adjacent in G . Further, as eij epq,wehave the directed edge eij → epq, and hence the orientation is transitive. As a result, we can easily see that a linear extension of a transitive orientation of G∗ can be obtained by simply reading the ones of the matrix row by row. In other words eij precedes epq in the linear order if and only if either (i) i < p or (ii) i = p and j < q. It is clear that G∗ can be constructed in O(m2) time. As comparability graphs can be colored in linear time [3] this already implies that when G is an interval bigraph with a known zero-partition, ch(G) can be computed in O(m2) time, faster than the bound given in Proposition 1 for chordal bipartite graphs. Next, we show that G∗ can be colored (and hence ch(G) can be computed) and a minimum chain subgraph cover for G can be found in O(mn) time. An algorithm [3] to optimally color a comparability graph H is as follows: we start with an arrangement of all the vertices of H that corresponds to a linear extension of a transitive orientation of H, and color the vertices one by one, from left to right. A vertex receives the smallest color not assigned to any of its neighbors that precede it. In view of the proof of Lemma 2, this simply corresponds to scanning the ones in M row-by-row and in each row from left to right, assigning to each edge the smallest color that has not been assigned to any of its previously colored neighbors in G∗.This greedy algorithm, as detailed in [2], is reproduced below. 123 Graphs and Combinatorics (2010) 26:301–313 307

Algorithm Color for k =1to |X| do for l =1to |Y | do if M[k, l]=1 then Assign to ekl the smallest color not assigned to any eij such that (i) eij ekl, and (ii) eij and ekl induce a 2K2 in G endAlgorithm Next, we want to show that this algorithm can be implemented to run in O(mn) time. First, we need to establish some properties of the coloring algorithm. We note that eij ekl implies that eij is colored before ekl.

Proposition 3 (a) Suppose egh and eij are edges of G that induce a 2K2 such that egh eij. Then, any edge ekl with eij ekl also induces a 2K2 with egh. (b) If eij ekl, then the color assigned by algorithm Color to eij is at most the color assigned to ekl. (c) If we scan the edges in a row (resp. column) of M left to right (resp. top to bottom), then the sequence of colors assigned by algorithm Color is nondecreasing.

Proof (a) is easily seen from the fact that M is a zero-partition of the bipartite adjacency matrix of G. (b) follows from (a) and the working of the algorithm Color.(c) is a direct corollary of (b).

As an immediate consequence of the greedy coloring algorithm and property (a), note that an edge with color b is part of an induced matching of size b which includes one edge of each color a ≤ b. With these properties in mind, we are now ready to demonstrate algorithm to ﬁnd im(G) and ch(G) without constructing G∗ when G is an interval bigraph and the zero-partition of M is part of the input.

Proposition 4 Given a zero-partition of the bipartite adjacency matrix of an interval bigraph G, an optimal coloring of G∗ can be found in O(mn) time. Therefore, im(G) and ch(G) can be computed in O(mn) time.

Proof Consider edges eij and ekl that induce a 2K2 in G such that ekl eij. It is clear that k < i and M[k, j]=0. Further, given the structure of M, every entry M[k, p] with p > j is also a 0. Therefore, as far as the edges ekl that impact the color assigned to eij are concerned, we only need to consider those in rows that are before row i and also in each of which the column containing the rightmost one is to the left of column j. Let S be set of all edges ekl such that ekl eij and ekl and eij induce a 2K2. Suppose eij is not the first edge in its row, eir is the edge that precedes eij in row i, and let x be the color assigned to eir.LetS1 be set of all edges ekl such that ekl eir and ekl and eir induce a 2K2. By Proposition 3, S1 ⊆ S.LetS2 = S − S1. For every ekl ∈ S2, M[k, r]=1 and M[k, j]=0. Thus, the rightmost column of row k that contains a one is either column r or it is to the right of column r.Lety be the largest 123 308 Graphs and Combinatorics (2010) 26:301–313 color assigned to a member of S2. Then, it follows that the edge eij should be assigned the color 1 + max(x, y). On the other hand, if eij is the first edge in its row, then it should be assigned the color that is one more than the largest color assigned to a member of S. The method below achieves precisely this. To ensure that all the edges in each row of M can be colored in O(m) time, when we begin processing row i, we build a minimum priority queue Q containing rows 1 through i − 1 where the priority for each row is the column number of the rightmost one in the row. When edge eij is processed, we repeatedly delete a row k from Q whose priority is less than j and record the largest color assigned to an edge on row k that induces a 2K2 with eij. This is done until a row with priority at least j is encountered, and such a row is left behind on Q. We then determine the largest color among those recorded, say, color y.Ifeij corresponds to the first 1 in row i, then it is assigned the color 1 + y. Otherwise, let x be the color assigned to eir, the edge that precedes eij in row i. Edge eij is assigned the color 1 + max(x, y). We then proceed to process the next edge in row i. When we begin processing row i of M, we build a priority queue containing rows 1 through i − 1inO(n) time as a linear list in nondecreasing order using Binsort. Hence, a single deletion of the minimum priority element from the queue can be done in constant time. When a row of M is processed, as each element of the priority queue is deleted at most once, the total cost of the priority queue operations per row of M is O(n). When an edge eij is processed, any row that is scanned is deleted from the priority queue and is never scanned again. Thus, the 1s in the rows preceding row i are scanned just once during the entire processing of row i. Therefore, a row of M is processed in O(m) time and the entire algorithm runs in O(mn) time.

As in the case of convex bipartite graphs [25], it turns out that the set of edges that belong to a color class of G∗ need not form a chain subgraph of G.Next,weshow that the required set of edges can be added to each color class so that we get a chain subgraph of G corresponding to each color class of G∗.

Theorem 3 Given a zero-partition of the bipartite adjacency matrix M of an interval bigraph G, a minimum chain subgraph cover for G can be found in O(mn) time.

Proof We start with an optimal coloring Φ of the 1’s in the matrix M (i.e. an optimal coloring of G∗) as obtained by algorithm Color. We will demonstrate a chain subgraph of G corresponding to each color class in Φ. In particular, if a color class of Φ does not form a chain subgraph of G, we will show that edges can be added to the subgraph to turn it into a chain graph. By Lemma 1, a maximal independent set of G∗ that includes all the edges of a color class will indeed form a chain subgraph of G.But,owingto complexity considerations, we will build chain subgraphs that may not correspond to maximal independent sets of G∗. For a color class b,letCb be the set of edges with color b, and let Mb be the matrix with the same dimensions as the smallest submatrix of M that contains every edge in Cb, where Mb[g, h]=1 if and only if egh ∈ Cb.IfMb contains a 2K2,leti and j be rows with i < j and k and l be columns with k < l that induce the submatrix 10 01 in Mb.Aseik and e jl belong to the same color class, they do not induce a 2K2 in G. 123 Graphs and Combinatorics (2010) 26:301–313 309

Therefore, either M[i, l]=1orM[ j, k]=1. In either case, by Proposition 3,the corresponding edge must also have been colored b, contradicting the existence of such a2K2. Thus, we can assume that any edges eil and e jk which form a 2K2 correspond to a submatrix 01 in M .AsM is a zero-partition, it follows that M[i, k]=1. Our 10 b proposal is to simply add the edge eik to the color class b, for every such rows i, j and columns k, l of Mb. It is clear that such an addition destroys the 2K2 involving rows i, j and columns k, l. However, the set of added edges could potentially introduce a new 2K2 in Mb. We will show later that this is not the case. First, we show that the proposed edges can be added to all the color classes of G∗ foranoverallcostofO(mn). We show that for a given M[i, k]=1, all the color classes to which eik should be added (for the purpose explained above) can be determined in O(n) time, giving us the required overall bound of O(mn). First, let us consider deciding whether eik should be added to a given color class b; recall that we can assume eik has been assigned a color smaller than b.LetR be the bottom most row in column k such that eRk is colored b and let C be the right most column in row i such that eiC is colored b.Ifeik should be added to color class b, there exists a row j > i and a column l > k such that M[ j, k]=M[i, l]=1 and both e jk and eil are colored b,butM[ j, l]=0ore jl is colored with a color b > b.IfM[ j, l]=0, then since M is a zero-partition, M[ j, l] is either in R or in C, and as a result either M[ j, C]=0orM[R, l]=0. On the other hand, if e jl is colored with a color b > b, then either M[ j, C]=0ore jl e jC and by the monotonicity of the colors in row j, e jC received a color greater than b as well. In all cases, either rows i, j and columns k, C or rows i, R and columns k, l induce a2K2 in Mb. Therefore, in order to determine whether or not eik should be added to color class b, it is enough to examine entries M[ j, C], where j > i and e j,k is colored b, and entries M[R, l], where l > k and eil is colored b, and decide whether or not any of these entries equal 0 or are assigned a color larger than b. Thus, whether eik should be added to color class b can be decided in time proportional to rb + cb where rb is the number of 1 entries in column k that are colored b, and cb is the number of 1 entries in row i that are colored b. As colors assigned to 1’s in a row (column) are monotonically increasing, we can therefore determine all the color classes to which a given eik should be added in O(n) time. Let Hb be the graph whose bipartite adjacency matrix is Mb after all the color classes have been augmented as indicated above, and let Eb be the edge-set of Hb. Thus, the set Ab = Eb\Cb is the set of edges that were added to color class b as described, and each such edge is in Ca for some a < b. Next, we will show that Hb is a chain graph. We begin with three claims. The ﬁrst two are immediate consequences of the process described above, and we then show the third claim follows from the other two.

Claim 1 For every edge e ∈ Ab, Hb − e contains an induced 2K2 consisting of two edges with color b.

Claim 2 The graph Hb does not contain any induced 2K2 with both edges from Cb.

Claim 3 The graph Hb does not contain any induced 2K2 which includes any edge of Cb. ∗ ∗ Assume this ﬁnal claim is false, and let ea = xy and eb = x y be edges which induce a 2K2 such that eb has color b. Then by Claim 2, we can assume that ea has 123 310 Graphs and Combinatorics (2010) 26:301–313 color a < b and thus by Claim 1 we can ﬁnd vertices x and y such that xy and xy are both edges of color b and x y is not an edge of Hb. Now, observe that x y ∗ ∗ and x y are both edges colored b, and these edges do not form a 2K2 by Claim 2. ∗ ∗ As x y is not an edge of Hb, we must have the edge x y in Eb, and by an identical ∗ argument, we also conclude that x y is an edge of Hb. We now have the following subgraph in Hb:

b b

Note that opposite edges on this C6 form an induced 2K2 in Hb, and hence by Claim 2, we can assume that all the unlabeled edges have a color less than b.We now show that the coloring algorithm forbids any C6 in G with such a coloring. If we identify the vertex yi whose index is minimal in such a subgraph, and label the neighbors of yi as xs and xt such that xt yi is colored b, then the monotonicity of the colors in column i requires that s < t.Now,lety j be the other neighbor of xt on the cycle and note that by the monotonicity of colors in row t, the edge xt y j has color b ≥ b. But no vertex in the graph shown above is incident with two edges of the cycle whose color is at least b. This contradiction shows that no induced 2K2 in Hb contains any edge with color b, which establishes Claim 3. We now complete the proof of Theorem 3 by showing that Hb is a chain graph. Assume that Hb contains e1 = x1y1 and e2 = x2y2 which form an induced 2K2.By Claim 3,wemusthave{e1, e2}⊆Ab and so by the observation made above, each these edges are the middle edge of an induced P4 in Hb whose pendant edges are colored b.Letyi and xi be the pendant vertices of such a P4 containing ei . = = If x1 x2 x then x y2 and x1y1 form a 2K2 in Hb. Since x y2 has color b,this = contradicts Claim 3. Thus x1 x2, and by an identical argument we conclude that = y1 y2. This gives the following subgraph of Hb (unlabeled edges all have a color less than b and dashed lines are not edges in Hb, but may be present in G):

b b

123 Graphs and Combinatorics (2010) 26:301–313 311

Next, observe that both x1y1 and x2y are edges of Hb and x2y1 is not an edge of 2 Hb. Since x2y2 has color b, we conclude from Claim 3 that, x1y2 is an edge of Hb as shown below.

b b

Finally, we note that x1y2 and x2y2 form a 2K2 in Hb, and x2y2 has color b, contradicting Claim 3. This ﬁnal contradiction establishes that Hb is 2K2 free.

3.3 Partial Order Dimension

A partial order P on the finite set V is an irreflexive, antisymmetric, and transitive binary relation on V . We can think of P as a set of ordered pairs, or equivalently, as a directed, acyclic, and transitively oriented graph on the vertex set V . A partial order P on V is a linear order if, for every pair of distinct elements x, y ∈ V , either (x, y) ∈ P or (y, x) ∈ P.Thedimension of partial order P on V [9], denoted dim(P), is the minimum number of linear orders on V whose intersection is P.Thedimension problem for partial orders is to decide whether a given partial order has dimension at most k. Yannakakis showed [24] that the dimension problem is NP-complete for every fixed k ≥ 3. He also showed that, even for a partial order of height one— namely, one whose underlying comparability graph is bipartite— the dimension problem is NP-complete for every fixed k ≥ 4. Kimble [17] (as credited in [24]) provided a reduction from the dimension problem for partial orders to the chain cover problem for bipartite graphs. We describe a similar reduction, given in [18]. Given a partial order P on V , we can construct a bipartite graph S(P) as follows: for each element vi of V , we add vertex xi to the color-class X of S(P) and vertex yi to the color-class Y of S(P). For each (vi ,vj ) ∈ P,wemakexi and y j adjacent in S(P). Finally, we also make each xi adjacent to yi in S(P). For a bipartite graph G = (X, Y, E), its bipartite complement is the bipartite graph B(G) = (X, Y, E), where for vertices x ∈ X and y ∈ Y , xy ∈ E if and only if xy ∈ E. Theorem 4 ([17,18,24]) If P is a partial order, then dim(P) =ch(B(S(P))). Corollary 2 If P is a partial order such that B(S(P)) is chordal bipartite, then dim(P) can be computed in polynomial time. A partial order P on V is an interval order, if each element of V can be mapped to an interval on the real line such that (x, y) ∈ P if and only if the interval mapped to 123 312 Graphs and Combinatorics (2010) 26:301–313 x is entirely to the left of the interval mapped to y.Theinterval dimension of partial order P on V [21], denoted idim(P), is the minimum number of interval orders on V whose intersection is P. The following lemma of Yannakakis [24] shows that it is NP-complete to determine if the interval dimension of a partial order of height one is at most 3.

Lemma 2 ([24]) If P is a partial order of height one and G is the underlying comparability graph of P, then idim(P) = ch(B(G)).

Thus, Lemma 2 and Proposition 1 combine to give the following corollary.

Corollary 3 If P is a partial order of height one and G is the underlying comparability graph of P such that B(G) is chordal bipartite, then idim(P) can be computed in polynomial time.

4 Conclusion

As far as computing a largest induced matching of a graph G is concerned, the approach of first constructing G∗ and then solving the clique problem in G∗ has been employed in many published papers [5–7,10–12]. Obviously, such an algorithm has the complexity of Ω(m2) owing to the construction of the graph G∗. An approach to improving the complexity of such an algorithm is to simulate what the clique algorithm does on G∗ by working with G itself, i.e., without explicitly constructing G∗. For example, Brandstädt and Hoáng [1] have shown that, when G is a chordal graph, the construction of G∗ can be avoided and the problem can be solved in linear time. In the context of special classes of bipartite graphs, a similar assessment also holds for the minimum chain subgraph cover problem and a coloring algorithm for G∗. For example, the algorithms given in [2] for convex bipartite graphs avoid the computation of G∗ to achieve improvement over the algorithms given in [25]. In this paper, we have further generalized this by showing that a maximum induced matching and a minimum chain subgraph cover can also be computed for an interval bigraph without the explicit construction of G∗. We leave open two related issues. First, whether or not there exist algorithms to find im(G) which avoid explicitly constructing G∗ when G∗ is weakly chordal? A polynomial time algorithm for the problem (via construction of G∗)[6]is already known. We also leave open the issue of whether an improved algorithm to find ch(G) can be constructed for the class of chordal bipartite graphs. Finally, in view of Theorem 1, the complexity of solving the clique and coloring problems on G∗, when G is a bipartite graph containing chordless cycles, but with no induced cycles on six vertices, is also potentially of interest.

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