J Comb Optim (2012) 23:507–518DOI 10.1007/s10878-010-9368-9

Perfect matchings in paired domination vertex criticalgraphs

Shenwei Huang · Erfang Shan · Liying Kang

Published online: 10 December 2010© Springer Science+Business Media, LLC 2010

Abstract A vertex subset S of a graph G = (V ,E) is a paired dominating set ifevery vertex of G is adjacent to some vertex in S and the subgraph induced by S con-tains a perfect matching. The paired domination number of G, denoted by γpr(G), isthe minimum cardinality of a paired dominating set of G. A graph with no isolatedvertex is called paired domination vertex critical, or briefly γpr -critical, if for anyvertex v of G that is not adjacent to any vertex of degree one, γpr(G − v) < γpr(G).A γpr -critical graph G is said to be k-γpr -critical if γpr(G) = k. In this paper, wefirstly show that every 4-γpr -critical graph of even order has a perfect matching ifit is K1,5-free and every 4-γpr -critical graph of odd order is factor-critical if it isK1,5-free. Secondly, we show that every 6-γpr -critical graph of even order has a per-fect matching if it is K1,4-free.

Keywords Matching · Perfect matching · Factor-critical · Paired domination vertexcritical

1 Introduction

Domination and its variations in graphs are now well studied. The literature on thissubject has been surveyed and detailed in the two books by Haynes et al. (1998a,1998b). Many authors have studied the concepts of graphs that are critical with re-spect to various operations on graphs. In particular, Sumner and Blitch (1983) in-troduced the concept of a domination critical graph in the sense of edge addition;

This research was partially supported by the National Nature Science Foundation of China(No. 60773078), PuJiang Project of Shanghai (No. 09PJ1405000) and Shanghai Leading AcademicDiscipline Project (No. S30104).

S. Huang · E. Shan (�) · L. KangDepartment of Mathematics, Shanghai University, Shanghai 200444, Chinae-mail: efshan@shu.edu.cn

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In Brigham et al. (1988, 2005), Fulman et al. (1995), the authors studied dominationvertex critical graphs; Goddard et al. (2004) defined and studied total dominationvertex critical graphs; and paired domination critical graphs were first studied by Ed-wards (2006). Recently, further study on paired domination vertex critical has beencontinued in Henning and Mynhardt (2008), Hou and Edward (2008).

In this paper we continue the study of paired domination vertex critical graphs.For notation and graph theory terminology we in general follow (Bondy and Murty1976). Specifically, let G = (V ,E) be a nontrivial simple graph with vertex set V

and the edge set E. The open neighborhood of a vertex v ∈ V is N(v) = {u ∈ V |uv ∈ E} and the closed neighborhood of v is N [v] = N(v) ∪ {v}. For a set S ⊆ V ,N(S) = ⋃

v∈S N(v) and N [S] = N(S) ∪ S. We denote the degree of a vertex v inG by dG(v), or simply d(v) if the graph G is clear from the context. The minimumand maximum degree of the graph G are denoted by δ(G) and �(G), respectively.The subgraph of G induced by S ⊆ V is denoted by G[S]. The complement of G isdenoted by G. A cutset of a connected graph G = (V ,E) is a subset S of V suchthat G − S is disconnected. In particular, if S = {v} is a cutset of G, then v is calleda cutvertex of G. For S ⊆ V , the number of components of G − S and the numberof odd components of G − S are denoted by ω(G − S) and co(G − S), respectively.A graph is said to be K1,k-free if it does not contain the complete bipartite graph K1,k

as an induced graph.A set of vertices or edges is independent if no two of its elements are adjacent.

A matching in a graph G is a set of independent edges in G. A perfect matchingM in a graph G is a matching such that every vertex of G is incident with an edgeof M . A graph G is called factor-critical if G − v has a perfect matching for everyvertex v ∈ V . We denote the maximum cardinality of a matching in G by α′(G), thematching number of G.

A vertex subset D of a graph G is a total dominating set if every vertex of G isadjacent to some vertex in D. The total domination number of G, denoted by γt (G),is the minimum cardinality of a total dominating set of G. A graph with no isolatedvertex is called total domination vertex critical, or briefly γt -critical, if for any vertexv of G that is not adjacent to any vertex of degree one, γt (G − v) < γt (G). If G isγt -critical and γt (G) = k, then we say that G is k-γt -critical.

The concept of paired domination in graphs was first introduced by Haynes andSlater (1998). A vertex subset D of a graph G is a paired dominating set, or abbrevi-ated PDS, if every vertex of G is adjacent to some vertex in D and the subgraph G[D]induced by D contains a perfect matching. Let M be a perfect matching in G[D] fora PDS D of G. Two vertices in D are said to be paired if they are incident to the sameedge in M . The paired domination number of G, denoted by γpr(G), is the minimumcardinality of a paired dominating set of G. For two subsets X and Y of V (G), wesay that X paired-dominates Y if Y ⊆ N [X] and G[X] contains a perfect matching.

An end-vertex of a graph G is a vertex of degree one and a support vertex isone that is adjacent to an end-vertex. Let S(G) be the set of support vertices of G.A graph with no isolated vertex is called paired domination vertex critical, or brieflyγpr -critical, if for every vertex v ∈ V (G)−S(G), γpr(G−v) < γpr(G). If G is γpr -critical and γpr(G) = k, then we say that G is k-γpr -critical. Clearly, the completegraph K2 is trivially 2-γpr -critical. Furthermore, a graph G is 2-γpr -critical if andonly if G is K2.

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2 Preliminary results

For a vertex v ∈ V (G) − S(G), we shall denote by Gv the graph G − v and by Dv

a minimum PDS of G − v. By the definition of k-γpr -critical graph, the followingobservation follows immediately, which is useful in the proof of our main results.

Observation 1 If G is a k-γpr -critical graph, then the following statements are true:

(1) For every vertex v ∈ V (G) − S(G), |Dv| = k − 2 and N(v) ∩ Dv = ∅.(2) For every pair of distinct vertices u,v ∈ V (G) − S(G), Du �= Dv .

The corona cor(H) of a graph H is the graph obtained from H by adding a pen-dent edge to each vertex of H . The next lemma is due to Henning and Mynhardt(2008).

Lemma 1 (Henning and Mynhardt 2008) Let G be a connected graph of order atleast 3 with at least one end-vertex. Then G is γpr -critical if and only if G = cor(H)

for some connected graph H satisfying α′(H) = α′(H − v) for every v ∈ V (H).

Lemma 2 Let G be a 6-γpr -critical graph. If S is a cutset of G such that eitherω(G − S) ≥ 4 or ω(G − S) = 3 and each component has at least 2 vertices, thenS � N(v) for each v ∈ V (G) − S.

Proof Suppose that there is a vertex v ∈ V (G) − S such that S ⊆ N(v). By Obser-vation 1, we have Dv ∩ S = ∅. Note that two paired vertices in Dv are in the samecomponent of G − S. So Dv dose not paired-dominate Gv , a contradiction. �

Next we give the well-known result in matching theory.

Theorem 3 (Berge-Tutte (1958) formula) For every graph G,

α′(G) = 1

2

(|V (G)| − max

X⊆V (G)(co(G − X) − |X|)

).

3 Main results

In this section we start to investigate matching properties in 4-γpr -critical and 6-γpr -critical graphs. Wang et al. (2009) have proved the following two results on matchingproperties in total domination vertex critical graphs.

Theorem 4 (Wang et al. 2009) If G is K1,5-free 3-γt -critical graph of even order,then G has a perfect matching.

Theorem 5 (Wang et al. 2009) If G is K1,5-free 3-γt -critical graph of odd order,then G is factor-critical.

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It is not difficult to see that similar matching properties can be obtained in 4-γpr -critical graphs by the almost same arguments used in the proof of the above twotheorems in Wang et al. (2009). Hence, we directly give the following two matchingproperties in 4-γt -critical graphs and omit their proofs.

Theorem 6 If G is K1,5-free 4-γpr -critical graph of even order, then G has a perfectmatching.

Theorem 7 If G is K1,5-free 4-γpr -critical graph of odd order, then G is factor-critical.

Now we study the matching properties in 6-γpr -critical graphs. For this pur-pose, let us introduce some more terminology and notations. Let S be a cutset ofa graph G and C a component of G − S. For a vertex v ∈ S, we say that v dom-inates C if v is adjacent to each vertex in C, i.e., V (C) ⊆ N(v); v connects C ifN(v)∩V (C) �= ∅; v separates C if N(v)∩V (C) = ∅. For a vertex subset T ⊆ V (G),we say that T paired-dominates the components C1, . . . ,Ck of G − S if T paired-dominates

⋃ki=1 V (Ci). For two disjoint vertex subsets X and Y of V (G), we denote

by E(X,Y ) = {xy ∈ E(G) | x ∈ X,y ∈ Y } the set of edges between X and Y . Nextwe start to give our main results.

Theorem 8 If G is a K1,4-free 6-γpr -critical graph of even order, then G has aperfect matching.

Proof If G is disconnected, then G is either the disjoint union of three copies ofK2 or the disjoint union of a copy of K2 and a connected 4-γpr -critical graph. ByTheorem 6, G has a perfect matching. So we may assume that G is connected. Ifδ(G) = 1, by Lemma 1, it is easy to see that G has a perfect matching. We thereforeassume that δ(G) ≥ 2.

Suppose to the contrary that G has no perfect matching. By Theorem 3, thereexists a cutset S in G such that α′(G) = 1

2 (|V (G)| − (co(G − S) − |S|)). Let S besuch a set with as less vertices as possible. Since G has no perfect matching and|V (G)| is even, co(G − S) − |S| ≥ 2. Clearly, S �= ∅. Let C1,C2, . . . ,Cr be all theodd components of G − S. If |S| ≥ 9, then r ≥ 11. Choose an arbitrary vertex t ∈ S.Clearly, Dt ∩ S �= ∅. Hence, an induced K1,4 centered at one of Dt = Dt ∩ S in G

would arise. Therefore, 1 ≤ |S| ≤ 8.We distinguish the following eight cases depending on the values of |S|.Case 1. |S| = 1. Let S = {v}. Then r ≥ 3. In order to paired-dominate the graph

Gv , obviously |Dv| ≥ 6, which contradicts the fact |Dv| = 4.Case 2. |S| = 2. Let S = {x, y}. Then r ≥ 4. Consider Dy . If Dy ∩ S = ∅,

Dy would not paired-dominate Gy . So Dy ∩ S = {x}. Let x′ be the vertex pairedwith x in Dy . Without loss of generality, assume that x′ ∈ V (C1) and the other twopaired vertices in Dy are in C2. Then x dominates C3 and C4. By Lemma 2, weknow that y separates C3 and C4. Now consider Dx . Then Dx ∩ S = {y}. To paired-dominate V (Ci) in Gx , |Dx ∩ V (Ci)| ≥ 2 for i = 3,4 since y separates C3 and C4.This implies that |Dx | ≥ 5, contradicting Observation 1.

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Case 3. |S| = 3. Let S = {x, y, z}. Then r ≥ 5. Clearly, |Dv ∩ S| = 2 for eachv ∈ S, otherwise an induced K1,4 in G would arise. So S is independent. We considertwo subcases as follows.

Subcase 3.1. For each v ∈ V (G) − S, |Dv ∩ S| = 2. Choose an arbitrary vertexv ∈ V (G)− S. Without loss of generality, assume that Dv ∩ S = {x, y}. Let x′ and y′be the vertices paired with x and y in Dv , respectively. Since z is paired-dominatedby Dv and S is independent, at least one of {x′, y′}, say x′, is adjacent to z. ByObservation 1, |Dx′ ∩ S| ≤ 1, a contradiction.

Subcase 3.2. There exists a vertex v ∈ V (G) − S with |Dv ∩ S| �= 2. Since G

is K1,4-free, |Dv ∩ S| > 0. If |Dv ∩ S| = 3, then since S is independent, we have|Dv| ≥ 6, which contradicts Observation 1. So |Dv ∩S| = 1. Without loss of general-ity, assume that Dv ∩ S = {x}. Then ω(G − S) = co(G − S) = 5 and the componentof G − S containing v is the single vertex {v} otherwise an induced K1,4 centered atx would arise. By δ(G) ≥ 2, vy ∈ E(G) and vz ∈ E(G). Moreover, by δ(G) ≥ 2 andG is K1,4-free, we see that |Du ∩ S| = 1 for a vertex u ∈ V (G) − S if and only if thecomponent of G − S containing u is the single vertex {u}.

Let V (C5) = {v} and Dv = {x, x′,w1,w2}. Without loss of generality, assume thatx′ ∈ V (C1) is paired with x and w1,w2 ∈ V (C2) are paired with each other in Dv .Then x dominates Ci for i = 3,4 and separates C2. Next we show that |V (Ci)| = 1for i = 3,4. If not, assume that |V (C4)| ≥ 2. Then E({y, z},V (C4)) = ∅ since x

dominates C4. Consider Dx . Then Dx ∩S = {y, z}. To paired-dominate V (C4) in Gx ,|Dx ∩ V (C4)| ≥ 2. This implies that |Dx | ≥ 6 since S is independent, contradictingObservation 1. So |V (C3)| = |V (C4)| = 1 and let V (Ci) = {vi} for i = 3,4.

Next we show that y connects C2. If not, we have E({x, y},V (C2)) = ∅. Nowconsider Dz. Then Dz∩S = {x, y}. To paired-dominate V (C2) in Gz, |Dz∩V (C2)| ≥2. This implies that |Dz| ≥ 6, contradicting Observation 1. By similar arguments forz, we have that z connects C2. Let u1 ∈ V (C2) and u2 ∈ V (C2) (possibly u1 = u2)be the vertices adjacent to y and z, respectively.

By δ(G) ≥ 2 and Lemma 2, vi is adjacent to exactly one of {y, z} for i = 3,4. Ifboth v3 and v4 are adjacent to y (resp. z), an induced K1,4 in G centered at y (resp. z)would arise. So we assume that v3y ∈ E(G) and v4z ∈ E(G). Since G is K1,4-free,E({y, z},V (C1)) = ∅. To paired-dominate V (C1) in Gx , |Dx ∩ V (C1)| ≥ 2. Thisimplies that |Dx | ≥ 6, contradicting Observation 1.

Case 4. |S| = 4. Let S = {x, y, z,w}. Then r ≥ 6. Obviously, |Dv ∩ S| ≥ 2 foreach v ∈ V (G) as G is K1,4-free. So by Observation 1, we have |N(v) ∩ S| ≤ 1for each v ∈ S and |N(v) ∩ S| ≤ 2 for each v ∈ V (G) − S. First we claim that S isindependent.

Claim 1 S is independent in G.

Proof Suppose S is not independent and thus we may assume that zw ∈ E(G). Thenby Observation 1, Dw ∩ S = {x, y}. There are two possible cases as follows.

(a) x and y are paired in S. Then the other two paired vertex u1 and u2 in Dw

must lie in the same component of G − S. Without loss of generality, assume thatui ∈ V (C1) for i = 1,2. Then {x, y} paired-dominates

⋃6i=2 V (Ci). Since G is K1,4-

free, we may assume that the following statements are true.

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(1) x dominates C3 and C4 while y separates C3 and C4.(2) y dominates C5 and C6 while x separates C5 and C6.(3) Both x and y dominate C2.

By |N(v) ∩ S| ≤ 2 for each v ∈ V (G) − S, we have E({z,w},V (C2)) = ∅. Nowconsider Dx . Dx ∩ S = {z,w}. To paired-dominate V (C2) in Gx , |Dx ∩ V (C2)| = 2by |Dx | = 4. So {z,w} paired-dominates V (C1) and

⋃6i=3 V (Ci). By |N(v)∩S| ≤ 2

for each v ∈ V (G)− S, the component dominated by both z and w must be C1. Thuswu1 ∈ E(G). But this contradicts Observation 1.

(b) x and y are not paired in S. Let x′ and y′ be the vertices paired with x andy in Dw , respectively. Note that x′ and y′ lie in the different components of G − S

otherwise an induced K1,4 in G centered at x or y would arise. Without loss ofgenerality, assume that x′ ∈ V (C1) and y′ ∈ V (C4).

Since G is K1,4-free, ω(G − S) = co(G − S) = 6 and each of {{x, x′}, {y, y′}}paired-dominates exactly 3 components of G − S. Assume that {x, x′} paired-dominates

⋃3i=1 V (Ci) and {y, y′} paired-dominates

⋃6i=4 V (Ci). Since G is K1,4-

free, xy /∈ E(G) and

E

(

{x},6⋃

i=4

V (Ci)

)

= ∅, E

(

{y},3⋃

i=1

V (Ci)

)

= ∅.

Note also that there is no vertex in V (C2) ∪ V (C3) ∪ V (C5) ∪ V (C6) that is adjacentto both z and w.

Since z is paired-dominated by Dw = {x, x′, y, y′} in Gw , at least one of {x′, y′},say x′, is adjacent to z. Now consider Dx′ . Then Dx′ ∩ S = {y,w}. To paired-dominate V (Ci), w connects Ci for i = 2,3. Since G is K1,4-free, E({w},V (C5) ∪V (C6)) = ∅. If z connects both C5 and C6, an induced K1,4 in G centered at z (x′,w, C5, C6) would arise. So assume that z separates C6. Note that |V (C6)| ≥ 2 byδ(G) ≥ 2 and E({x, z,w},V (C6)) = ∅.

Next we claim that z connects C5. Otherwise, z separates C5 and thus E({x, z,w},V (C5) ∪ V (C6) = ∅. Now C5 and C6 are also odd components of G − S′, whereS′ = {y} with |S′| = 1. Then

co(G − S′) − |S′| ≥ 2 = co(G − S) − |S| = maxX⊆V (G)

(co(G − X) − |X|),

contradicting the choice of S. Hence z connects C5.Choose a vertex v ∈ V (C6) with vy ∈ E(G). Then y /∈ Dv ∩S. So |Dv ∩V (C6)| =

2 and z and w are paired in Dv since Dv is a minimum PDS of cardinality 4. Topaired-dominate y′ in Gv and note that y′w /∈ E(G), y′z ∈ E(G). Then an inducedK1,4 in G centered at z (x′, y′, w, C5) arises. So the claim holds. �

By Claim 1, we immediately see that |Dv ∩ S| = 2 for each v ∈ V (G). Thus wehave ω(G − S) = co(G − S) = 6 otherwise an induced K1,4 in G would arise.

First, we show that there exists a component C of G − S with |V (C)| ≥ 2. Ifnot, then V (Ci) = {vi} and d(vi) = 2 for i = 1,2, . . . ,6. Note that N(vi) �= N(vj )

if i �= j otherwise vj would not be paired-dominated by Dviin Gvi

. So N(vi) (i =

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1,2, . . . ,6) corresponds to all distinct 2-element subsets of S. This results in γpr(G−v) = 6 for each v ∈ S, which contradicts that G is 6-γpr -critical.

Now we assume that |V (C1)| ≥ 2. Let v ∈ V (C1) and consider Dv . Without lossof generality, assume that Dv ∩ S = {x, y}. Let x′ and y′ be the vertices paired withx and y in Dv , respectively. Similar to the proof in Claim 1, assume that {x, x′}paired-dominates

⋃3i=1 V (Ci) and {y, y′} paired-dominates

⋃6i=4 V (Ci).

Since {z,w} is paired-dominated by Dv = {x, x′, y, y′} and |N(v) ∩ S| ≤ 2 foreach v ∈ V (G) − S, each of {x′, y′} is adjacent to at most one of {z,w}. So weassume that x′z ∈ E(G) and y′w ∈ (G).

Next we show that the component of G− S containing x′ is the single vertex {x′}.Consider Dx′ . Then Dx′ ∩ S = {y,w}. By E({y},⋃3

i=1 V (Ci)) = ∅, an induced K1,4centered at w (C1, C2, C3, y′) would arise if the component of G − S containing x′has at least 2 vertices. So assume that V (C2) = {x′}. Similarly, assume that V (C4) ={y′}.

Let w′ and y′′ be the vertices paired with w and y in Dx′ , respectively. Obviously,w′ ∈ V (C1) ∪ V (C3) ∪ {y}. To paired-dominate x in Gx′ , we have w′x ∈ E(G) byClaim 1 and E({y},⋃3

i=1 V (Ci)) = ∅. So w′ �= v and w′ �= y′. Now consider Dy′ .

Dy′ ∩ S = {x, z}. Since E({x},⋃6i=4 V (Ci)) = ∅, z connects C5 and C6. Since G is

K1,4-free, z separates C1 and C3.Finally, we consider Dw′ . Dw′ ∩ S = {y, z}. By E({y, z},V (C1) ∪ V (C3)) = ∅,

|Dw′ ∩ (V (C1)∪V (C3))| ≥ 2. By Claim 1 and Dw′ is a PDS of Gw′ , this implies that|Dw′ | ≥ 6, contradicting Observation 1.

Case 5. |S| = 5. Let S = {x, y, z,w,v}. Then r ≥ 7. Observe that |Du ∩ S| ≥ 3for each u ∈ S otherwise an induced K1,4 would arise by ω(G − S) ≥ r ≥ 7. So�(G[S]) ≤ 1 and thus |E(G[S])| ≤ 2. If |E(G[S])| ≤ 1, there exists a vertex u ∈ S

such that Du ∩ S is independent, which implies that |Du| ≥ 6, contradicting Obser-vation 1. So |E(G[S])| = 2. Without loss of generality, we assume that xy ∈ E(G)

and zw ∈ E(G).Consider Dx . Dx ∩S = {v, z,w} and z and w are paired in Dx . Let vx be the vertex

paired with v in Dx . Since G is K1,4-free, v connects at most 3 components of G−S,say C1, C2 and C3, and thus E({v},⋃7

i=4 V (Ci)) = ∅. Assume that vx ∈ V (C1).So each vertex in

⋃7i=4 V (Ci) is adjacent to at least one of {z,w}. Considering Dz

symmetrically, we have that each vertex in⋃7

i=4 V (Ci) is adjacent to at least oneof {x, y}.

Next we show that |V (Ci)| = 1 for i = 4,5,6,7. If not, assume that |V (C4)| ≥ 2.Choose an arbitrary vertex t ∈ V (C4). Without loss of generality, assume that tx ∈E(G) and tz ∈ E(G). Then Dt ∩ S = {v, y,w} is independent. This implies that|Dt | ≥ 6, contradicting Observation 1.

So let V (Ci) = {vi} for i = 4,5,6,7. We show that vi is adjacent to exactly one of{z,w} for i = 4,5,6,7. If not, we assume that v7 is adjacent to both z and w. If v7x ∈E(G), then Dv7 ∩ S = {v, y}. Since E({v},⋃7

i=4 V (Ci)) = ∅, y is adjacent to v4, v5and v6. Note that yvx ∈ E(G) since y is paired-dominate by Dx = {v, vx, z,w} andy is adjacent to none of {v, z,w}. So an induced K1,4 in G centered at y (v4, v5, v6,vx ) arises. Similarly, an induced K1,4 in G centered at x would arise if v7y ∈ E(G).So vi is adjacent to exactly one of {z,w} for i = 4,5,6,7. Similarly, vi is adjacent toexactly one of {x, y} for i = 4,5,6,7.

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Since G is K1,4-free, two of {v4, v5, v6, v7}, say v4 and v5, are adjacent to x andthe other two vertices, v6 and v7, are adjacent to y. Then an induced K1,4 in G

centered at y (v6, v7, vx , x) arises, a contradiction.Case 6. |S| = 6. Let S = {x, y, z,w,u, v}. Then r ≥ 8. Note that |Dt ∩ S| ≥ 3 for

each t ∈ S by ω(G − S) ≥ r ≥ 8. Next we claim that |Dt ∩ S| = 3 for each t ∈ S.

Claim 2 For each t ∈ S, |Dt ∩ S| = 3.

Proof If not, we assume that |Dv ∩S| = 4 and Dv ∩S = {x, y, z,w}. Without loss ofgenerality, assume that x and z are paired with y and w in Dv , respectively. Since u

is paired-dominated by Dv , u is adjacent to at least one of Dv , say x.Consider Dx . Then Dx ∩ S = {v, z,w} and z and w are paired in Dx . Let vx be

the vertex paired with v in Dx . Since G is K1,4-free, ω(G− S) = co(G− S) = 8 and{v, vx} paired-dominates exactly three components of G − S while {z,w} paired-dominates exactly five components of G − S. Assume that {v, vx} paired-dominates⋃3

i=1 V (Ci) and {z,w} paired-dominates⋃8

i=4 V (Ci). Note that E({v}, {z,w}) = ∅,E({v},⋃8

i=4 V (Ci)) = ∅ and E({z,w},⋃3i=1 V (Ci)) = ∅.

If uz ∈ E(G), then Dz∩S = {v, x, y}. So {x, y} also paired-dominates⋃8

i=4 V (Ci).Since vx is paired-dominated by Dv , an induced K1,4 in G centered at one of Dv

arises. So uz /∈ E(G). Similarly, uw /∈ E(G).Next we show that N(v) ∩ S �= ∅. If not, that is, N(v) ∩ S = ∅. If yz ∈ E(G),

then Dy ∩ S = {v,u,w} is independent. This implies that |Dy | ≥ 6, contradictingObservation 1. So yz /∈ E(G). Similarly, yw /∈ E(G).

Note that E({t},⋃8i=4 V (Ci)) �= ∅ for each t ∈ {x, y,u}. Otherwise, assume that

E({x}, ⋃8i=4 V (Ci)) = ∅. Then C4, . . . ,C8 are also odd components of G − S′,

where S′ = S − {v, x} with |S′| = 4. Thus

co(G − S′) − |S′| ≥ 2 = co(G − S) − |S| = maxX⊆V (G)

(co(G − X) − |X|),

contradicting the choice of S.Next we show that yu ∈ E(G). If yu /∈ E(G), let t ∈ N(x)∩ (

⋃8i=4 V (Ci)). With-

out loss of generality, assume tz ∈ E(G). Then Dt ∩ S ⊆ {v, y,u,w} is independent.This implies that |Dt | ≥ 6, contradicting Observation 1. Hence, yu ∈ E(G). So far,we have shown that G[S] is the disjoint union of the complete graphs K1,K2 andK3, where K1 = G[{v}], K2 = G[{z,w}] and K3 = G[{x, y,u}].

Since {y,u} is paired-dominated by Dx in Gx , vx is adjacent to u and y. Con-sider Dvx . Then Dvx ∩S = {x, z,w}. By E({z,w},⋃3

i=1 V (Ci)) = ∅, we have {x, x′}paired-dominates

⋃3i=1 V (Ci) in Gvx , where x′ is the vertex paired with x in Dvx .

If the component of G − S containing vx has at least 2 vertices, an induced K1,4

in G centered at x (C1, C2, C3, a vertex in N(x) ∩ (⋃8

i=4 V (Ci)) would arise. Soassume that V (C1) = {vx}. By the similar arguments for Dy and Du, we have thatV (C2) = {vy} and V (C3) = {vu}, where vy and vu are the vertices paired with v inDy and Du, respectively. Note that vy is adjacent to each of {x,u} and vu is adjacentto each of {x, y}.

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Now consider Dz. Dz ∩S ⊆ {v,u, x, y}. By E({v},⋃8i=4 V (Ci)) = ∅, there exists

one vertex of {u,x, y}, say x, that connects at least two components of {C4, . . . ,C8}.Thus an induced K1,4 in G centered at x (vy , vu, two of {C4, . . . ,C8}) arises.

Therefore, N(v) ∩ S �= ∅. This implies that vu ∈ E(G). If yz ∈ E(G), thenDz ∩ S = {v,u, x}. Note that u is a vertex of degree 2 in G[Dz ∩ S]. An inducedK1,4 in G centered at one of Dz ∩ S arises. So yz /∈ E(G). Similarly, yw /∈ E(G).Finally, we consider Du. Du ∩ S = {y, z,w} with z and w paired in Du. SinceE({z,w},V (C1) ∪ V (C2) ∪ V (C3)) = ∅, y connects Ci for i = 1,2,3. Thus an in-duced K1,4 in G centered at y (C1, C2, C3, a vertex in N(y) ∩ (

⋃8i=4 V (Ci)) arises.

Consequently, the claim follows. �

By Claim 2, we have ω(G − S) = co(G − S) = 8 and thus co(G − S) − |S| = 2.Note that Dt ∩ S for each t ∈ S corresponds an edge of G[S] and the graph inducedby Dt ∩ S is the disjoint union of a vertex and an edge.

Note that �(G[S]) ≤ 2 and thus |E(G[S])| ≤ 6. If |E(G[S])| = 6, G is either thecycle on 6 vertices or the disjoint union of two copies of the complete graph K3. Ineither case, there exists a vertex t ∈ S such that G[Dt ∩S] contains a vertex of degree2, a contradiction.

Hence, |E(G[S])| ≤ 5. By the pigeonhole principle, there exist two different ver-tices u and v in S such that Du ∩ S = {x, y, z} and Du ∩ S = {x, y, z′} with x andy paired in both Du and Dv . Then {x, y} paired-dominates exactly 5 componentsof {C1, . . . ,C8}, say C1, . . . ,C5. The vertex z (resp., z′) and its paired vertex in Du

(resp., Dv) paired-dominates the other three components C6, C7 and C8.If z �= z′, then E({z, z′},⋃5

i=1 V (Ci)) = ∅. C1, . . . ,C5 are also odd componentsof G − S′, where S′ = S − {z, z′} with |S′| = 4. Then

co(G − S′) − |S′| ≥ 2 = co(G − S) − |S| = maxX⊆V (G)

(co(G − X) − |X|),

contradicting the choice of S. Therefore, z = z′ and Du ∩ S = Dv ∩ S = {x, y, z}.By the minimality of |S|, each of {u,v,w} is adjacent to at least one vertex in⋃5

i=1 V (Ci). Now let t ∈ N(w) ∩ (⋃5

i=1 V (Ci)). Assume that tx ∈ E(G). ThenDt ∩S ⊆ {y, z,u, v}. This implies that uv ∈ E(G). Otherwise |Dt | ≥ 6, contradictingObservation 1.

Now let t ∈ N(u) ∩ (⋃5

i=1 V (Ci)). Without loss of generality, assume that tx ∈E(G). Then Dt ∩ S ⊆ {y, z,w,v}. Since {y, v, z} is independent, w is adjacent to atleast one of {y, v, z}.

If wy ∈ E(G), then Dy ∩ S = {u,v, z} with u and v paired in Dy . SinceE({z},⋃5

i=1 V (Ci)) = ∅, {u,v} paired-dominates {C1, . . . ,C5}. So E({u,v},⋃8

i=6 V (Ci)) = ∅. Then C6, C7 and C8 are also odd components of G − S′, whereS′ = S − {x, y,u, v} with |S′| = 2. Then

co(G − S′) − |S′| ≥ 2 = co(G − S) − |S| = maxX⊆V (G)

(co(G − X) − |X|),

contradicting the choice of S.If wz ∈ E(G), we consider Dw . Note that x and y are not paired in Dw . Without

loss of generality, assume that Dw ∩ S = {u,v, x} with u and v paired in Dw . Since

516 J Comb Optim (2012) 23:507–518

E({z}, {u,v, x}) = ∅, z is adjacent to x′, the vertex paired with x in Dw . Note thatx′ /∈ ⋃8

i=6 V (Ci). Thus an induced K1,4 in G centered at z (C6, C7, C8, x′) arises.Hence, wv ∈ E(G). Furthermore, note that wu ∈ E(G) otherwise |Ds | ≥ 6 for

some vertex s ∈ N(v) ∩ (⋃5

i=1 V (Ci)). By the similar arguments as in the proof ofClaim 2, an induced K1,4 in G centered at one of {u,v,w} arises.

Case 7. |S| = 7. Then r ≥ 9. Clearly, |Dt ∩ S| = 4 for each t ∈ S otherwise aninduced K1,4 in G would arise. Then �(G[S]) ≤ 2. Now we consider two subcasesas follows.

Subcase 7.1. �(G[S]) = 1. Then there are at most three edges in G[S], whichimplies that there is at least one isolated vertex, say v, in G[S]. Choose a vertex t �= v

in S. Then v is not paired-dominated by Dt in Gt , a contradiction.Subcase 7.2. �(G[S]) = 2. Let z be a vertex of degree 2 in G[S] and N(z) ∩ S =

{v3, v4}. Then Dz = Dz ∩ S = S − {z, v3, v4} = {x, v1, y, v2}, where x and y arepaired with v1 and v2 in Dz, respectively. Since {v3, v4} is paired-dominated by Dz,each of {v3, v4} is adjacent to at least one of {x, v1, y, v2}. Without loss of generality,assume that v3 is adjacent to v1.

If v4 is adjacent to x, then E({y, v2}, {v1, v3, v4, x, z}) = ∅. Thus v2 would not bepaired-dominated by Dy in Gy . So assume that v4 is adjacent to v2. If xy /∈ E(G),y would not be paired-dominated by Dv4 in Gv4 . Hence xy ∈ E(G). So far, we haveshown that G[S] is the cycle on 7 vertices. Since G is K1,4-free, we may assume thatthe following statements are true.

(1) v1 dominates C1 and C2 while each of Dz −{v1} separates C1 and C2; v2 dom-inates C8 and C9 while each of Dz − {v2} separates C8 and C9.

(2) x dominates C4 while each of Dz − {x} separates C4; y dominates C6 whileeach of Dz − {y} separates C6.

(3) Each of {v1, x} dominates C3 while each of S − {v1, x} separates C3; each of{v2, y} dominates C7 while each of S −{v2, y} separates C7; each of {x, y} dominatesC5 while each of S − {x, y} separates C5.

Now consider Dv4 . Then Dv4 ∩ S = {v3, v1, x, y}. By E({v1, x, y},V (C8) ∪V (C9)) = ∅, we have that v3 dominates C8 and C9. Since G is K1,4-free,E({v3},⋃7

i=4 V (Ci)) = ∅. Considering Dv3 symmetrically, we have that v4 domi-nates C1 and C2 and E({v4},⋃6

i=3 V (Ci)) = ∅. So E({v3, v4},⋃6i=4 V (Ci)) = ∅.

Finally, we consider Dx . Dx ∩S = {v3, z, v4, v2}. By E({v2, v3, v4},⋃6i=4 V (Ci))

= ∅, z dominates Ci for i = 4,5,6. An induced K1,4 in G centered at z (C4, C5, C6,v3) arises.

Case 8. |S| = 8. Then r ≥ 10. Since G is K1,4-free, ω(G − S) = co(G − S) = 10and thus co(G − S) − |S| = 2. We first give an immediate observation that followsfrom the fact that G is K1,4-free.

Observation 2 (a) |Dt ∩ S| = 4 for each t ∈ S.(b) For some t ∈ S, suppose that Dt ∩ S = {x, y, z,w}, where x and z are paired

with y and w in Dt , respectively. Then each of {{x, y}, {z,w}} paired-dominates ex-actly five components of G − S. Moreover, if {x, y} paired-dominates

⋃5i=1 V (Ci),

then {z,w} paired-dominates⋃10

i=6 V (Ci)) and thus E({x, y}, {z,w}) = ∅, E({x, y},⋃10

i=6 V (Ci)) = ∅ and E({z,w},⋃5i=1 V (Ci)) = ∅.

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By Observation 2(a), �(G[S]) ≤ 3 and thus |E(G[S])| ≤ 12. Note that Dt ∩ S

for each t ∈ S corresponds to two different edges in G[S]. By |S| = 8 and pi-geon hole principle, there exists an edge e = xy in G[S] that belongs to suchtwo sets. So assume that Du ∩ S = {x, y, zu,w1} and Dv ∩ S = {x, y, zv,w2} fortwo different vertices u,v ∈ S. By Observation 2(b), zu (resp. zv) is paired withw1 (resp. w2) in Du (resp. Dv). Furthermore, assume that {x, y} paired-dominates⋃5

i=1 V (Ci). Then both {zu,w1} and {zv,w2} paired-dominate⋃10

i=6 V (Ci). SoE({zu,w1, zv,w2},⋃5

i=1 V (Ci)) = ∅.If {zu,w1} ∩ {zv,w2} = ∅, C1, . . . ,C5 are also odd components of G − S′ where

S′ = S − {zu,w1, zv,w2} with |S′| = 4. Thus

co(G − S′) − |S′| ≥ 2 = co(G − S) − |S| = maxX⊆V (G)

(co(G − X) − |X|),

contradicting the choice of S.Therefore, we may assume that zu = zv = z and w1 �= w2. Then E({z,w1,w2},⋃5

i=1 V (Ci)) = ∅. Note that any two of {x, y, z,w1,w2, u, v} ⊆ S are different ver-tices. Let w be the remaining vertex in S.

Now consider Dz. By |Dz ∩ S| = 4, |Dz ∩ {u,v,w}| ≥ 2. If |Dz ∩ {u,v,w}| = 2,the two vertices in Dz ∩ {u,v,w} must be paired in Dz by Observation 2(b). ThenC1, . . . ,C5 are also odd components of G − S′, where S′ = S − {z,w1,w2} − (Dz ∩{u,v,w}) with |S′| = 3. Thus

co(G − S′) − |S′| ≥ 2 = co(G − S) − |S| = maxX⊆V (G)

(co(G − X) − |X|),

contradicting the choice of S. Consequently, Dz ∩ {u,v,w} = {u,v,w}. Since Dz

is a PDS of Gz with cardinality 4, there is exactly one of {u,v,w} paired with oneof {x, y} in Dz. By E({x, y}, {u,v}) = ∅, u and v are paired in Dz. Without loss ofgenerality, assume that w is paired with x in Dz.

We consider Dx . Clearly Dx ∩ S ⊆ {u,v, z,w1,w2}. By E({z,w1,w2},⋃5i=1 V (Ci)) = ∅, we have that u and v are paired in Dx and {u,v} paired-dominates

⋃5i=1 V (Ci). Thus E({u,v},⋃10

i=6 V (Ci)) = ∅. Then C6, . . . ,C10 are also odd com-ponents of G − S′, where S′ = S − {u,v, x, y} with |S′| = 4. Then

co(G − S′) − |S′| ≥ 2 = co(G − S) − |S| = maxX⊆V (G)

(co(G − X) − |X|),

contradicting the choice of S.In either case, we always arrive at a contradiction. Therefore, we complete the

proof of Theorem 8. �

4 Conclusion

In this paper we have proved that every 6-γpr -critical graph of even order has a perfectmatching if it is K1,4-free. However, we cannot find out an infinite class of graphs toshow that the conditions “K1,4-free” in Theorem 8 is essential. Therefore, a natural“question” is whether the condition “K1,4-free” can be weakened or even dropped.

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