Information Processing Letters 81 (2002) 7–11

On maximum induced matchingsin bipartite graphs

V.V. LozinRUTCOR, Rutgers University, 640 Bartholomew Road, Piscataway, NJ 08854-8003, USA

Received 17 October 2000; received in revised form 7 March 2001Communicated by K. Iwama

Abstract

The problem of finding a maximum induced matching is known to be NP-hard in general bipartite graphs. We strengthen thisresult by reducing the problem to some special classes of bipartite graphs such as bipartite graphs with maximum degree 3 orC4-free bipartite graphs. On the other hand, we describe a new polynomially solvable case for the problem in bipartite graphswhich deals with a generalization of bi-complement reducible graphs. 2002 Elsevier Science B.V. All rights reserved.

Keywords:Graph algorithms; Computational complexity; Maximum induced matching; Bipartite graphs

1. Introduction

A matching in a graphG = (V ,E) is a subsetof edgesM ⊆ E no two of which have a vertex incommon. A matching is called induced if the subgraphof G induced by the endpoints of edges inM is 1-regular. We study the problem of finding inG aninduced matching of maximum cardinality, denotediµ(G). This problem has been introduced in [1],where the author has proved its NP-hardness in theclass of bipartite graphs. In the present paper westrengthen this result in the following way.

Let C be a class of graphs defined by a finite setof forbidden induced subgraphsF . In Section 2, wecharacterize a condition for the setF under whichthe maximum induced matching problem is NP-hardfor bipartite graphs in classC. In particular, due tothis condition the problem is NP-hard in classes ofK1,4-free orC4-free bipartite graphs. Note thatK1,4-

E-mail address:lozin@rutcor.rutgers.edu (V.V. Lozin).

free bipartite graphs are exactly bipartite graphs withmaximum degree 3. A similar result has been obtainedin [6], where the authors have proved the NP-hardnessfor the problem under consideration in the class ofplanar graphs with maximum degree 3.

In spite of general NP-hardness, there are knownseveral classes of graphs where the problem canbe solved in polynomial time: chordal and intervalgraphs [1], trees [2], circular-arc graphs [4], trapezoidgraphs,k-interval-dimension graphs and cocompara-bility graphs [5]. In the present paper we show thepolynomial solvability of the problem in a subclassof bipartite graphs which generalizes bi-complementreducible graphs (bi-cographsfor short) introducedin [3] as the bipartite analog of cographs.

For a graphG, we denote byVG and EG thevertex set and the edge set ofG, respectively. Theneighbourhood of a vertexx ∈ VG is denotedN(x).The degree ofx ∈ VG is the number of vertices inN(x). Given a subset of verticesU ⊆ VG, we denote

0020-0190/02/$ – see front matter 2002 Elsevier Science B.V. All rights reserved.PII: S0020-0190(01)00185-5

8 V.V. Lozin / Information Processing Letters 81 (2002) 7–11

Fig. 1. GraphHn.

byG[U ] the subgraph ofG induced byU , andG−U

= G[VG− U ].As usual,Pn andCn denote the chordless path and

the chordless cycle withn vertices, respectively, andKn,m is the complete bipartite graph with parts ofcardinalityn andm. By Hn we denote the graph whichcan be obtained from two copies ofP3 by joining theircentral vertices by a chordless path of lengthn (Fig. 1).

Given two graphsG andH , we say thatG is H -free if G does not containH as an induced subgraph.We use special notations for some particular classes ofgraphs:

Xk , the class of(C3,C4, . . . ,Ck)-free graphs,

Yl , the class of(H1,H2, . . . ,Hl)-free graphs,

Z3, the class of graphs with maximum degree 3,

B, the class of bipartite graphs.

2. NP-hardness

Let G be a graph andx be a vertex inG. We definea graph transformation illustrated in Fig. 2 as follows:(1) partition the neighbourhoodN(x) of vertexx into

two subsetsY andZ in arbitrary way;(2) delete vertexx from the graph together with

incident edges;(3) add aP4 = (y, a, b, z) to the rest of the graph;(4) connect vertexy of theP4 to each vertex inY , and

connectz to each vertex inZ.We denote the transformed graph byG′ and say that

G′ is obtained from graphG by vertex stretching withrespect tox.

Lemma 1. iµ(G′) = iµ(G)+ 1.

Proof. At first, let us consider an induced matchingM in G and show that graphG′ contains an inducedmatchingM ′ of cardinality|M|+1. If x is not coveredby an edge inM, then M ′ = M ∪ {(a, b)} is the

Fig. 2. Stretching operation.

required matching. IfM contains an edge(x, c), wemay assume, without loss of generality, thatc ∈ Y .Then, clearly, no vertex inZ is covered by an edgeof M, and henceM ′ = (M −{(x, c)})∪{(y, c), (b, z)}is the matching we are looking for. Thus,iµ(G′) �iµ(G)+ 1.

Conversely, letM ′ be an induced matching inG′with at least one edge. Our purpose now is to find inG an induced matchingM of cardinality|M ′| − 1. IfM ′ contains edge(a, b), then obviouslyM = M ′ −{(a, b)} meets the purpose. Assume next(a, y) ∈ M ′.If M ′ does not cover vertexz, thenM = M ′ − {(a, y)}is what we need. If(z, c) ∈ M ′ with somec ∈ Z, thenthe desired matchingM can be obtained as follows:

M = (M ′ − {(a, y), (z, c)})∪ {

(x, c)}.

Similarly if (b, z) ∈ M ′.Now let M ′ do not contain edges of theP4 =

(y, a, b, z). If M ′ covers neithery nor z, then forany e ∈ M ′, M = M ′ − {e} is the sought for match-ing. Assumey is covered by an edge(c, y). If M ′does not cover vertexz, then M = M ′ − {(c, y)}satisfies our purpose. If(z, d) ∈ M ′ with somed ∈Z, then we may be complied with matchingM =(M ′ − {(c, y), (z, d)})∪ {(x, d)}. Therefore,iµ(G)�iµ(G′)− 1. ✷Lemma 2. Any graphG can be transformed by asequence of stretching operations into a graph in classXk ∩ Yl ∩Z3 ∩B with any integerk � 3 andl � 1.

Proof. Assume first that graphG has a vertexx ofdegree at least four. For vertex stretching with respectto x, let us choose asY any two vertices adjacent tox, and letZ contain all the remaining vertices in theneighbourhood ofx. Under the vertex stretching weobtain a graph with four new verticesy, a, b, z, wherey is of degree three,a andb are of degree two, andz is

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of degree exactly one less than that ofx. If the degreeof z is still greater than three, we can decrease it ina similar way by application of vertex stretching withrespect toz. Thus, repeatedly applying the stretchingoperation we can obtain a graph in which every vertexhas degree at most three, i.e., a graph inZ3.

Now let us consider another variant of vertexstretching for which setY consists of a single vertex.In this case, the stretching operation is equivalent toa triple subdivision of an edge in the graph. In otherwords, the edge is replaced by a chordless path oflength four. Application of this operation to each edgeof the graph is called the total stretching ofG. Underthe total stretching the length of every induced cycleincreases four times, and therefore, the resulting graphcontains no cycles of odd length, i.e., it is bipartite.Moreover, applying the total stretching a necessarynumber of times, we get rid of induced cyclesCi withi � k and induced graphs of formHi with i � l. ✷

It is not hard to see that the transformation describedin Lemma 2 can be carried out in time boundedby a polynomial in the size of the input graph. Inconjunction with Lemma 1 and the result in [1] thisimplies

Corollary 1. For any integersk � 3 and l � 1, themaximum induced matching problem isNP-hard in theclass of graphsXk ∩ Yl ∩ Z3 ∩ B.

Proof. To prove the corollary we only have to showthat the maximum induced matching problem is poly-nomially equivalent to the problem of determiningiµ(G). This is a consequence of the following sim-ple observation. Let(x, y) be an edge in a graphG.If iµ(G − x) = iµ(G) or iµ(G − y) = iµ(G), thenclearly there is a maximum induced matching inGwhich is contained inG − x or G − y, respectively.And if iµ(G − x) = iµ(G − y) = iµ(G) − 1, thenevery maximum induced matchingM in G is of formM = {(x, y)} ∪ M ′, whereM ′ is a maximum inducedmatching in graphG− ({x, y} ∪N(x)∪N(y)). Thus,one can determine a maximum induced matching inG

by computingiµ(H) for induced subgraphsH of Gat most|VG| times. ✷

Denote byT (l1, l2, l3) the graph in Fig. 3, wherel1, l2, and l3 are nonnegative integers. In particular,

Fig. 3.T (l1, l2, l3).

T (1,1,1) = K1,3 andT (l1, l2,0) = Pl1+l2+1. And letT be the class of graphs whose every connected com-ponent is isomorphic to a graph of formT (l1, l2, l3).

Theorem 1. Let C be a class of graphs defined by afinite setF of forbidden induced subgraphs. IfF ∩T = ∅, then the maximum induced matching problemis NP-hard in the classC ∩ B.

Proof. Let k be an integer greater than the numberof vertices in a largest graph inF . And suppose thata graphG ∈ Xk ∩ Yk ∩ Z3 ∩ B does not belong toC ∩ B. ThenG must contain a graphA ∈ F as aninduced subgraph. SinceG ∈ Xk, graphA containsno induced cyclesCj of length j � k. Moreover,A can not contain a cycleCj with j > k, because|VA| < k due to the choice ofk. Therefore,A containsno cycles, i.e.,A is a forest. Analogously, sinceG ∈Yk and |VA| < k, A contains no induced subgraphsof form Hi , i.e., every connected component ofA

has at most one vertex of degree at least three. Andfurthermore, sinceG ∈ Z3, the maximum degree ofAis at most three. But thenA is in T that contradicts thehypothesis of the theorem. We thus have proved thatXk ∩Yk ∩Z3 ∩B ⊆ C ∩B. Now the conclusion of thetheorem follows from Corollary 1. ✷

As an immediate consequence from the theorem,we obtain the NP-hardness of the maximum inducedmatching problem in a large family of subclasses ofbipartite graphs such asK1,4-free bipartite graphsor C4-free bipartite graphs. On the other hand, thistheorem characterizes graph classes which have thepotential for accepting polynomial algorithms. In thenext section we consider one of such classes and provethat the maximum induced matching problem can besolved in polynomial time in this class.

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3. A polynomially solvable case

All graphs in this section will be bipartite. A bi-partite graphG = (V1,V2,E) consists of set of ver-tices V1 ∪ V2 and set of edgesE ⊆ V1 × V2. For abipartite graphG = (V1,V2,E), we denote byG thebipartite complement toG, i.e.,

G = (V1,V2, (V1 × V2)− E

).

By mK2 we denote the regular graph of degree 1with 2m vertices. Clearly,mK2 is a bipartite graph.Furthermore, form = 2 we have2K2 = 2K2.

Let us call two vertices of a bipartite graphsimilarif they have the same neighbourhood. Clearly thesimilarity is an equivalence relation. A bipartite graphwhose every class of similarity is of size 1 will becalled prime. It is not hard to see that any bipartitegraphG has a unique (up to isomorphism) maximalprime induced subgraph that can be obtained bychoosing exactly one vertex in each similarity classof G.

Throughout the section we denote byC the classof bipartite graphs containing no induced subgraphsdepicted in Fig. 4.

Let us note that classC is an extension of bi-complement reducible graphs (bi-cographs) introducedrecently in [3]. Bi-cographs are exactly those graphsin C that areP7-free. By definition, a bipartite graphis a bi-cograph if any its induced subgraph with atleast two vertices is either disconnected or the bipartitecomplement to a disconnected graph.

The structure of graphs in classC has been charac-terized in [7] as follows.

Theorem 2. LetG = (V1,V2,E) be a prime bipartite(Star1,2,3, Sun4)-free graph. IfG is connected andGis connected, then eitherG or G is K1,3-free.

Fig. 4. Forbidden graphs for classC.

Now we use this characterization in order to provethe polynomial solvability of the maximum inducedmatching problem in classC. To this end, the follow-ing simple lemmas will be useful.

Lemma 3. If G1, . . . ,Gk are connected componentsof a graphG, theniµ(G)= ∑k

i=1 iµ(Gi).

Lemma 4. If H is a maximal prime induced subgraphof a graphG, theniµ(H)= iµ(G).

Proof. Lemma 3 is obvious. To prove Lemma 4, let usnote that any induced matchingM in G covers at mostone vertex in each similarity class ofG, otherwise thematching is not induced. Without loss of generalitywe may assume that vertices covered byM belong toH . Hence,iµ(H) � iµ(G). The converse inequalityis trivial. ✷

Due to Lemmas 3 and 4, we assume from nowon that a bipartite graphG is connected and prime.Moreover, we assume thatG contains an induced 2K2,otherwiseiµ(G) � 1 and the problem is trivial. Letus partition the vertex set ofG into subsetsU1, . . . ,

Uk in such a way that eachUi induces a connectedcomponent in the bipartite complement toG. WedenoteHi = G[Ui] and callHi co-component ofG.

Lemma 5. If H1, . . . ,Hk are co-components of agraphG = (V1,V2,E) that contains an induced2K2,theniµ(G)= max{2, iµ(H1), . . . , iµ(Hk)}.Proof. Let M be a maximum induced matching inG,and (a, b) be an edge inM with a ∈ Ui ∩ V1 andb ∈ Uj ∩ V2, whereUi = VHi (i = 1, . . . , k).

Assume first thati �= j . Since G contains aninduced 2K2, there must be another edge inM, say(c, d) with c ∈ V1 andd ∈ V2. Thenc ∈ Uj , otherwisec is adjacent tob, and similarlyd ∈ Ui . But now anyvertexf in V1 is adjacent either tob (if f /∈ Uj ) orto d (if f /∈ Ui ). Hence(a, b) and(c, d) are the onlyedges inM, i.e.,iµ(G) = 2.

Now let i = j . In that case, every vertex outside ofUi is adjacent either toa or to b. Therefore, all theremaining edges ofM belong toG[Ui], i.e.,iµ(G) =iµ(Hi). ✷

Lemmas 3, 4, 5 together with Theorem 2 allow usto reduce the problem under consideration from class

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C to K1,3-free bipartite graphs or their complements.Clearly any connectedK1,3-free bipartite graph iseither an even cycle or a path. Denoting by[p] theintegral part of a real numberp, we get the followingimmediate conclusion for these graphs.

Lemma 6. iµ(Ck) = [k/3], iµ(Pk) = [(k + 1)/3].

If a graphG ∈ C is the bipartite complement to acycleCk or to a pathPk , then we may be restricted tocasek � 7, otherwiseG is P7-free (a bi-cograph), andtherefore is disconnected due to results in [3].

Lemma 7. Let k � 7, theniµ(Ck) = iµ(Pk) = 2.

Proof. To prove the lemma, let us note that thebipartite complement to a 3K2 is aC6. ClearlyCk andPk with k � 7 areC6-free. HenceCk andPk with k �7 are 3K2-free, i.e.,iµ(Ck) � 2 andiµ(Pk) � 2. Onthe other hand, bothCk andPk with k � 7 contain aninduced 2K2, and moreover,2K2 = 2K2. Therefore,iµ(Ck) = iµ(Pk) = 2. ✷

Summarizing the above arguments, we conclude thefollowing. The maximum induced matching problemis trivial for a graphG ∈ C if(1) G is 2K2-free (iµ(G)� 1) or(2) bothG andG are connected and prime (Lemmas 6

and 7).The problem for a general graphG in C can be re-duced to the trivial cases as follows. We first find amaximal prime induced subgraphH of G (Lemma 4),

and then partitionH into disjoint induced subgraphsby recursively applying decomposition into connectedcomponents (Lemma 3) or co-components (Lemma 5).The first stage can be trivially implemented in timeO(|VG|2). In the second stage, we decomposeH

as long as we get subgraphs satisfying (1) or (2).In the worst case, we have to apply verification of2K2-freeness or decomposition operations|VG| times.Each of the operations can be carried out in O(|VG|2)steps. Hence we have proved

Theorem 3. Given a graphG ∈ C with n vertices,one can find a maximum induced matching inG inO(n3) time.

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