Perfect Matchings in Total Domination Critical Graphs

Graphs and Combinatorics (2011) 27:685–701DOI 10.1007/s00373-010-1000-3

ORIGINAL PAPER

Perfect Matchings in Total Domination Critical Graphs

Michael A. Henning · Anders Yeo

Received: 5 November 2009 / Revised: 19 November 2010 / Published online: 18 December 2010© Springer 2010

Abstract A graph is total domination edge-critical if the addition of any edgedecreases the total domination number, while a graph with minimum degree at leasttwo is total domination vertex-critical if the removal of any vertex decreases the totaldomination number. A 3t EC graph is a total domination edge-critical graph with totaldomination number 3 and a 3t V C graph is a total domination vertex-critical graphwith total domination number 3. A graph G is factor-critical if G − v has a perfectmatching for every vertex v in G. In this paper, we show that every 3t EC graph of evenorder has a perfect matching, while every 3t EC graph of odd order with no cut-vertexis factor-critical. We also show that every 3t V C graph of even order that is K1,7-freehas a perfect matching, while every 3t V C graph of odd order that is K1,6-free isfactor-critical. We show that these results are tight in the sense that there exist 3t V Cgraphs of even order with no perfect matching that are K1,8-free and 3t V C graphs ofodd order that are K1,7-free but not factor-critical.

Keywords Total domination · Edge-critical · Vertex-critical · Perfect matching ·Factor-critical

Mathematics Subject Classification (2000) 05C69 · 05C70

Research supported in part by the South African National Research Foundation and by a grant from theHarry Oppenheimer Trust.

M. A. HenningDepartment of Mathematics, University of Johannesburg, 2006 Auckland Park, South Africae-mail: [email protected]

A. Yeo (B)Department of Computer Science, Royal Holloway, University of London,Egham, Surrey TW20 OEX, UKe-mail: [email protected]

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1 Introduction

For many graph parameters, criticality is a fundamental question. Much has beenwritten about those graphs where a parameter (such as connectedness or chromaticnumber) goes up or down whenever an edge or vertex is removed or added. In thispaper, we continue the study of total domination edge-critical graphs initiated byvan der Merwe et al. [28] and of total domination vertex-critical graphs introducedby Goddard et al. [5]. Domination in graphs is very well studied in graph theory(see, for example, the recent papers [16,17,24,27]).

A total dominating set, denoted TDS, of a graph G with no isolated vertex is a set Sof vertices of G such that every vertex is adjacent to a vertex in S. The total dominationnumber γt (G) is the minimum cardinality of a TDS. A TDS of G of cardinality γt (G)

is called a γt (G)-set. Total domination in graphs was introduced by Cockayne et al. [2]and is now well studied in graph theory. For more details, the reader is referred to thetwo domination books [7,8] and a recent survey on total domination [13].

For notation and graph theory terminology we in general follow [7]. Specifically,let G = (V, E) be a graph with vertex set V of order n and edge set E . We denotethe degree of v in G by dG(v), or simply by d(v) if the graph G is clear from context.The minimum degree of G is denoted by δ(G). An end-vertex is a vertex of degreeone and a support vertex is one that is adjacent to an end-vertex. If S ⊆ V and v ∈ V ,then we denote dS(v) = |N (v) ∩ S|. In particular, dV (v) = d(v). We say that a graphis F-free if it does not contain F as an induced subgraph. The girth of G is the lengthof a shortest cycle in G.

The open neighborhood of vertex v ∈ V is denoted by N (v) = {u ∈ V | uv ∈ E}while its closed neighborhood is the set N [v] = N (v) ∪ {v}. For a set S ⊆ V , itsopen neighborhood is the set N (S) = ∪v∈S N (v) and its closed neighborhood is theset N [S] = N (S) ∪ S. Further, the subgraph induced by the set S is denoted by G[S].

For sets S, X ⊆ V , if X ⊆ N [S] (X ⊆ N (S), respectively), we say that S dominatesX , written S � X (S totally dominates X , respectively, written S �t X ). If S = {s}or X = {x}, we also write s � X, S �t x , etc. If S � V (S �t V , respectively), wesay that S is a dominating set (total dominating set) of G, and we also write S � G(S �t G, respectively). Thus a set S is a dominating set if N [S] = V , and a totaldominating set if N (S) = V . If u, v, w are vertices in G such that {u, v} �t G − w,then we write uv �→ w.

For two vertices u and v in a connected graph G, the distance dG(u, v) betweenu and v is the length of a shortest u–v path in G. The maximum distance among allpairs of vertices of G is the diameter of G, which is denoted by diam(G). A graphG is diameter 2-critical if its diameter is two, and the deletion of any edge increasesthe diameter. The concept of distance and diameter are fundamental concepts in graphtheory and are well-studied in the literature.

Two edges in a graph G are independent if they are not adjacent in G. A matching ina graph G is a set of independent edges in G, while a matching of maximum cardinalityis a maximum matching. The number of edges in a maximum matching of G is calledthe matching number of G which we denote by α′(G). A perfect matching M in G isa matching in G such that every vertex of G is incident to an edge of M . Thus, G hasa perfect matching if and only if α′(G) = |V (G)|/2. The graph G is factor-critical if

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G − v has a perfect matching for every vertex v in G. An odd component of a graphG is a component of odd order. The number of odd components of G is denoted byoc(G).

A graph G is total domination edge-critical if γt (G + e) < γt (G) for every edgee ∈ E(G) = ∅. Further if γt (G) = k, then we say that G is a kt -edge-critical graph,abbreviated kt EC . Thus if G is kt EC , then its total domination number is k andthe addition of any edge decreases the total domination number. Properties of 3t ECgraphs were the central theme in the Ph.D dissertations of both Van der Merwe [32] andSimmons [22]. Further properties of 3t EC graphs were studied in [6,9–12,15,28–31]and elsewhere.

We say that a vertex v in a graph G is γt -vertex-critical if γt (G −v) < γt (G). Sincetotal domination is undefined for a graph with isolated vertices, we say that a graph Gis total domination vertex-critical, or just γt -vertex-critical, if every vertex of G that isnot adjacent to a vertex of degree one is γt -critical. In particular, if δ(G) ≥ 2, then Gis γt -vertex-critical if every vertex of G is γt -vertex-critical. If G is γt -vertex-critical,and γt (G) = k, then we say that G is kt -vertex-critical, abbreviated kt V C . For exam-ple, the 5-cycle is 3t V C as is the complement of the Petersen graph. Properties of3t V C graphs were studied in [5,14,18,33,34] and elsewhere.

Wang et al. [21,34] studied matching properties in total domination vertex-criticalgraphs. They proved the following result.

Theorem 1 ([34]) Let G be a 3t V C graph that is K1,5-free. If G has even order, thenG has a perfect matching. If G has odd order, then G is factor-critical.

Wang et al. [21,34] pose the following question.

Question 1 Can the hypothesis that G is K1,5-free in Theorem 1 be lowered?

2 Main Results

In this paper, we study matching properties in 3t EC graphs and 3t V C graphs. Ouraim in this paper is twofold: our first aim is to show that every 3t EC graph of evenorder has a perfect matching, while every 3t EC graph of odd order with no cut-vertexis factor-critical. Our second aim is to show that every 3t V C graph of even orderthat is K1,7-free has a perfect matching, while every 3t V C graph of odd order that isK1,6-free is factor-critical.

We note that if G is a 3t EC graph of odd order that contains a cut-vertex, then byTheorem 7, G contains a vertex v adjacent with an end-vertex. Hence, oc(G −v) ≥ 2,and so G − v does not contain a perfect matching. Therefore, G is not factor-critical.However if w is a vertex of G different from v, then we deduce from the charac-terization of 3t EC graphs in Theorem 6 that G − w contains a perfect matching.We henceforth only consider 3t EC graphs with no cut-vertex. We shall prove thefollowing result, a proof of which is presented in Sect. 4.

Theorem 2 Every 3t EC 2-connected graph of even order has a perfect matching,while every 3t EC 2-connected graph of odd order is factor-critical.

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The situation for 3t V C graphs is more complex. As a consequence of Theorem 10,we note that every connected total domination vertex-critical graph with minimumdegree one has a perfect matching. In particular, every 3t V C graph with minimumdegree one has a perfect matching. We henceforth only consider 3t V C graphs withminimum degree at least two. Our second main result is to answer Question 1 in theaffirmative by showing that the K1,5-free condition in Theorem 1 can be relaxed toK1,7-free for even order graphs and relaxed to K1,6-free for odd order graphs, andthese results are best possible. We shall prove the following result, a proof of whichis presented in Sect. 5.

Theorem 3 Let G be a 3t V C graph with minimum degree at least two.

(a) If G has odd order and is K1,6-free, then G is factor-critical.(b) If G has even order and is K1,7-free, then G has a perfect matching.

3 Known Results

3.1 Matching in Graphs

We shall need the following well-known results in matching theory due to Tutte [26]and Lovász and Plummer [19].

Theorem 4 (Tutte’s Matching Theorem [26]) A graph G = (V, E) has a perfectmatching if and only if oc(G − S) ≤ |S| for every S ⊂ V .

Theorem 5 (Factor-Critical Theorem [19]) A graph G = (V, E) is factor-critical ifand only if oc(G − S) ≤ |S| − 1 for every nonempty set S ⊂ V .

3.2 Total Domination Edge-Critical Graphs

Perhaps much of the recent interest in total domination edge-critical graphs arises fromtheir application to diameter 2-critical graphs and in particular with the long-standingdiameter 2-critical graph conjecture due to Murty and Simon which we state below.A graph is diameter 2-critical if its diameter is two, and the deletion of any edgeincreases the diameter. Murty and Simon (see [1]) independently made the followingconjecture:

Conjecture 1 If G is a diameter 2-critical graph of order n and size m, then m ≤ n2/4,with equality if and only if n is even and G is the complete bipartite graph K n

2 , n2.

According to Füredi [4], Erdos said that this conjecture goes back to the work ofOre in the 1960s. Fan [3] proved the first part of the conjecture for n ≤ 24 and forn = 26, while Füredi [4] gave an asymptotic result proving the conjecture is true forn > n0, where n0 is a tower of 2’s of height about 1014. The conjecture remains openfor other values of n. Hanson and Wang [6] were the first to observe a key relationshipbetween diameter 2-critical graphs and total domination edge-critical graphs whichshows that Conjecture 1 is equivalent to the following conjecture.

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Conjecture 2 If G is a 3t EC graph with order n and size m, then m > n(n − 2)/4.

It seems to be a difficult problem to characterize 3t EC graphs. Hence it is importantto study properties of these graphs in order to make further progress on Conjecture 2.Van der Merwe et al. [28] characterized the 3t EC graphs with end-vertices.

Theorem 6 ([28]) Let G be a graph with an end-vertex u that is adjacent to the v.Let A = N (v)\{u} and let B = V \N [v]. Then, G is 3t EC if and only if A is a cliqueand |A| ≥ 2, B is a clique and |B| ≥ 2, and every vertex in A is adjacent to |B| − 1vertices in B and every vertex in B is adjacent to at least one vertex in A.

Theorem 7 ([28]) Every 3t EC graph with no end-vertex is 2-connected.

Let G = (V, E) be a 3t EC graph, and let u and v be non-adjacent vertices inG. We say that uv is a missing edge in G. Further if S ⊆ V and {u, v} ⊆ S, thenwe say that uv is a missing edge in S (rather than “uv is a missing edge in G[S]”).We note that either {u, v} � G or, without loss of generality, vw �→ u for somew ∈ N (v) (recall that, if u, v, w are vertices in G such that {v,w} �t G − u, thenwe write vw �→ u). Further if vw �→ u, then we note that uv /∈ E and uw /∈ E . Asγt (G + uv) = 2, there exists an edge xy ∈ E(G + uv) such that in G + uv, the edgexy totally dominates V , i.e., {x, y} �t V in G + uv. The edge xy may not be unique(possibly, xy = uv). However, we select one such edge xy and call it the quasi-edgefor uv, abbreviated q.e., and denote it by qeG(uv) = xy. Thus for each edge in E(G),we associate a unique quasi-edge. We shall need the following key lemmas, proofs ofwhich are along similar lines to those of Lemmas 2 and 6 in [23] and can be foundin [22]. Since the Ph.D thesis of Simmons [22] may not be readable available to thereader, we present short proofs of these key lemmas for completeness.

Lemma 8 ([22]) Let G = (V, E) be a 3t EC graph and let X be an independent set inG of size k ≥ 3. Then there exists an ordering x1x2 . . . xk of the vertices in X in sucha way that there exists a path v1v2 . . . vk−1 in G − X, where vi xi is the quasi-edge forthe missing edge xi xi+1 for i = 1, 2, . . . , k − 1.

Proof We will construct a tournament, T , with vertex set V (T ) = X as follows. Forevery pair of distinct vertices u and v in X , there exists a quasi-edge, e, incident witheither u or v, but not both, as |X | ≥ 3. If the quasi-edge is incident to v, then orient thearc in T from v to u and otherwise orient it from u to v. Since every tournament containsa hamiltonian (directed) path, there exists an ordering x1x2 . . . xk of the vertices in Xsuch that x1x2 . . . xk is a hamiltonian (directed) path in T . For i = 1, 2, . . . , k − 1,let vi xi be the quasi-edge for xi xi+1. Since |X | ≥ 3, we note that vi ∈ V \X andvi � X\{xi+1}. Hence the vertices v1v2 . . . vk−1 are distinct vertices in V \X . Furtherfor i = 2, 3, . . . , k, we note that vi xi �→ xi+1 and that vi xi+1 /∈ E , implying thatvi−1vi ∈ E . Hence, v1v2 . . . vk−1 is a path in G − X .

��Lemma 9 ([22]) If S is a vertex-cut in a 3t EC graph G, then G−S has at most |S|+1components.

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Proof Assume, to the contrary, that the number of components in G−S exceeds |S|+1.Let |S| = k. If k = 0, then G is disconnected, a contradiction. Hence, k ≥ 1. LetG1, G2, . . . , Gk+2 be k + 2 components of G − S. For i = 1, 2, . . . , k + 2, letxi be a vertex in Gi and let X = {x1x2 . . . xk+2}. Then, X is an independent set ofsize k+2 ≥ 3 in G. By Lemma 8, we may assume, renaming components if necessary,that the vertices of X are ordered in such a way that v1v2 . . . vk+1 is a path in G − X ,where vi xi is the quasi-edge for the missing edge xi xi+1 for i = 1, 2, . . . , k + 1.For i = 1, 2, . . . , k + 1, we note that vi xi ∈ E and that vi � X\{xi+1}, implyingthat vi ∈ S since |X | ≥ 3. Hence, {v1v2 . . . vk+1} ⊆ S, whence |S| ≥ k + 1, acontradiction. ��

3.3 Total Domination Vertex-Critical Graphs

The kt V C graphs with minimum degree one are characterized in [5]. The coronacor(H) of a graph H (denoted H ◦ K1 in [7]) is that graph obtained from H by addinga pendant edge to each vertex of H .

Theorem 10 ([5]) Let G be a connected graph of order at least 3 with δ(G) = 1.Then, G is kt V C if and only if G = cor(H) for some connected graph H of order kwith δ(H) ≥ 2.

Throughout this paper, for a vertex v in a 3t V C graph G we let Dv denote aγt (G − v)-set. The following results are established in [5].

Lemma 11 ([5]) Let G be a k-γt -critical graph with δ(G) ≥ 2, and let u and v bedistinct vertices in G. Then, G has the following properties.

(a) |Dv| = k − 1.(b) N (v) ∩ Dv = ∅.(c) Du = Dv .

4 Proof of Theorem 2

Recall the statement of Theorem 2.Theorem 2 Every 3t EC 2-connected graph of even order has a perfect matching,while every 3t EC 2-connected graph of odd order is factor-critical.

Proof of Theorem 2 Let G = (V, E) be a 3t EC 2-connected graph. Suppose firstthat G has even order. If S is a vertex-cut in G, then by Lemma 9, oc(G − S) ≤ |S|+1.Since G has even order, oc(G − S) ≤ |S|. Hence by Tutte’s Matching Theorem 4, Ghas a perfect matching, as desired. Hence in what follows we may assume that G hasodd order and is 2-connected.

Assume, to the contrary, that G is not factor-critical. By the factor-critical The-orem 5, there exists a nonempty subset S ⊆ V such that oc(G − S) ≥ |S|. SinceG has odd order, oc(G − S) ≥ |S| + 1. By Lemma 9, G − S has at most |S| + 1components. Hence, oc(G − S) = |S| + 1 and G − S has no even components. LetG1, G2, . . . , Gk+1 be the k + 1 odd components of G − S. Since G is 2-connected,

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we note that k ≥ 2. For i = 1, 2, . . . , k + 1, let Vi = V (Gi ) and let xi ∈ Vi . LetX = {x1x2 . . . xk+1}. Then, X is an independent set of size k + 1 ≥ 3 in G. Let Y bethe set of all vertices that belong to an odd component in G − S. Since G − S has noeven component, we note that V = S ∪ Y .

By Lemma 8, we may assume, renaming components if necessary, that the ver-tices of X are ordered in such a way that there exists a path v1v2 . . . vk in G − X ,where vi xi is the quasi-edge for the missing edge xi xi+1 for i = 1, 2, . . . , k. Thusfor i ∈ {1, 2, . . . , k}, we have that vi xi �→ xi+1, and so vi xi ∈ Evi xi+1 /∈ Eand vi � Y\(Vi ∪ {xi+1}). Further since |X | ≥ 3, we note that vi ∈ S. Hence,{v1v2 . . . vk} ⊆ S. Consequently, since |S| = k, we have that S = {v1v2 . . . vk}. Wenote that x1 � S but no vertex in S dominates Y\V1.

Since no vertex in S dominates Y\V1, we note that if V1 is not a clique, then addinga missing edge in V1 cannot decrease the total domination number. Thus, V1 is a clique.As observed earlier, x1 � S. If |V1| > 1, then let x ′

1 ∈ V1\{x1} and consider the miss-ing edge x ′

1x2. We note that {x ′1, x2} � V and that v1x2 /∈ E . Further, {v j , x2} � x j+1

for j = 2, . . . , k. Hence, v1x ′1 must be the quasi-edge for the missing edge x ′

1x2, andso v1x ′

1 �→ x2. In particular, we note that v1x ′1 ∈ E . Since x ′

1 is an arbitrary vertexin V1\{x1}, we deduce that v1 � V1. As observed earlier, vi � V1 for i = 2, . . . , k.Hence, every vertex in S is adjacent to every vertex in V1.

Let i ∈ {1, 2, . . . , k + 1} and suppose that Vi is not a clique. As observed earlier,V1 is a clique, and so i ≥ 2. Let ai bi be a missing edge in Vi , and let ei be the quasi-edge for ai bi . Since {ai , bi } � V , we may assume, renaming the vertices ai and bi ifnecessary, that ei = aiv where v ∈ S. If v = vi−1, then since vbi /∈ E , we note thatv = vi and i ≤ k. But then xi+1 is not dominated by {ai , v} and therefore aiv �→ bi ,a contradiction. Hence, ei = aivi−1 and aivi−1 �→ bi . Since vi−1 � Vi\{xi }, wenote that bi = xi and that aivi−1 ∈ E . We now consider the missing edge ai x1. Wenote that either aiv j �→ x1 or v j x1 �→ ai for some j ∈ {1, 2, . . . , k}. If aiv j �→ x1,then v j x1 /∈ E , and so x1 � S, a contradiction. If v j x1 �→ ai , then aiv j /∈ E , andso j = i − 1. But then v j x j+1 ∈ E , a contradiction. Hence, Vi is a clique for alli ∈ {1, 2, . . . , k + 1}.

We show next that S is a clique. Fix i and j , where 1 ≤ i < j ≤ k and considerthe missing edge xi+1x j+1. We note that either xi+1v �→ x j+1 or x j+1v �→ xi+1 forsome vertex v ∈ S. If xi+1v �→ x j+1, then vx j+1 /∈ E , implying that v = v j . Butthen since vi xi+1 /∈ E , we note that viv j ∈ E . If x j+1v �→ xi+1, then vxi+1 /∈ E ,implying that v = vi . But then since v j x j+1 /∈ E , we note that viv j ∈ E . Hence,viv j ∈ E for all i and j where 1 ≤ i < j ≤ k. Thus, S is a clique.

For i = 1, 2, . . . , k, we show next that vi � V \{xi+1}. Let i ∈ {1, 2, . . . , k}. Fromour earlier observations, we note that vi � V \(Vi ∪ {xi+1}). Suppose that vi � Vi .Since every vertex in S is adjacent to every vertex in V1, we note that v1 � V1, and soi ≥ 2. Since vi xi ∈ E , there exists a vertex yi ∈ Vi\{xi } that is not adjacent to vi . Wenow consider the missing edge xi+1 yi . We note that either vxi+1 �→ yi or vyi �→ xi+1for some vertex v ∈ S. If vxi+1 �→ yi , then vxi+1 ∈ E and vyi /∈ E . However, vi

is the only possible vertex in S that is not adjacent to yi , and so v = vi . But thenvxi+1 /∈ E , a contradiction. If vyi �→ xi+1, then v = vi as vx j+1 ∈ E , but yivi /∈ E , acontradiction. Hence, vi � Vi . Thus, vi � V \{xi+1} for all i = 1, 2, . . . , k. But then{v1, v2} �t V , contradicting the fact that γt (G) = 3. Hence, G is factor-critical. ��

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5 Proof of Theorem 3

Before proceeding with a proof Theorem 3, we present three lemmas about 3t V Cgraph that will prove to be useful. For a graph G and a subset X ⊂ V (G), we letG X = G[X ]. Recall that for a vertex v in a 3t V C graph G, we let Dv denote aγt (G − v)-set.

Lemma 12 Let G = (V, E) be a 3t V C graph with δ(G) ≥ 2 and let X ⊂ V bea nonempty subset of vertices in G. If oc(G − X) > |X |, then the graph G has thefollowing properties.

(a) |Dv ∩ X | ≥ 1 for every vertex v ∈ V .(b) Dv ⊂ X for some vertex v ∈ V \X.(c) 1

2 |X | ≤ �(G X ) ≤ |X | − 2.(d) |X | ≥ 5.

Proof of Lemma 12 Let G = (V, E) be a 3t V C graph of order n with δ(G) ≥ 2.Suppose that oc(G − X) > |X | for some nonempty subset X ⊂ V . We note that Gis connected. If |X | = 1, then G would have a cut-vertex v. Since δ(G) ≥ 2, everycomponent of G −v has order at least 2, and so γt (G −v) ≥ 4, a contradiction. Hence,G is 2-connected and |X | ≥ 2. Let |X | = r and let oc(G − X) = t . We note that r ≥ 2and t ≥ r + 1 ≥ 3.

Let X = {x1x2 . . . xr } and let G X = G[X ]. Let G1, G2, . . . , Gt denote the oddcomponents of G − X . For i = 1, 2, . . . , t , let Vi = V (Gi ) and let V≤i = ∪i

j=1Vj andV≥i = ∪t

j=i V j . Let Y and Z be the set of all vertices that belong to an odd component(respectively, even component) in G − X . Thus, Y = V≥1 and Z = V \(X ∪ Y ). Weare now in a position to present proofs of parts (a)-(d) in the statement of the lemma.

(a) If Dv ∩ X = ∅ for some vertex v ∈ V , then the set Dv , which consists of twoadjacent vertices, does not totally dominate the set Y\{v}, a contradiction. Thisestablishes part (a).

(b) Suppose that |Dv ∩ X | ≤ 1 for every vertex v ∈ Y . Then by part (a), |Dv ∩ X | = 1for every vertex v ∈ Y . Let v1 ∈ V1. Renaming vertices if necessary, we mayassume that Dv1 ∩ X = {x1}. Let Dv1 = {x1, y1}. Renaming components, if nec-essary, we may assume that y1 ∈ V≤2 ∪ Z . If y1 /∈ V2, then x1 � V≥2. If y1 ∈ V2,then x1 y1 ∈ E and x1 � V≥3. In both cases, there exists a vertex v2 ∈ V2 suchthat x1 � V≥3 ∪ {v2}.Since v2x1 ∈ E , Lemma 11(b) implies that x1 /∈ Dv2 . Renaming vertices inX\{x1} if necessary, we may assume that Dv2 ∩ X = {x2}. Let Dv2 = {x2, y2}.Renaming the components G3, . . . , Gt , if necessary, we may assume that y2 ∈V≤3 ∪ Z . If y2 /∈ V3, then x2 � V≥3. If y2 ∈ V3, then x2 y2 ∈ E and x1 � V≥4.In both cases, there exists a vertex v3 ∈ V3 such that x2 � V≥4 ∪ {v3}.Since {x1, x2} ⊆ N (v3), Lemma 11(b) implies that {x1, x2} ∩ Dv3 = ∅. Renam-ing vertices in X\{x1, x2} if necessary, we may assume that Dv3 ∩ X = {x3}. LetDv3 = {x3, y3}. Renaming the components G4, . . . , Gt , if necessary, we mayassume that y3 ∈ V≤4 ∪ Z . If y3 /∈ V4, then x3 � V≥4. If y3 ∈ V4, then x3 y3 ∈ Eand, if t ≥ 5, x3 � V≥5. In both cases, there exists a vertex v4 ∈ V4 such thatx3v4 ∈ E and, if t ≥ 5, x3 � V≥5.

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Continuing in the way, we can rename the vertices in X and Y , if necessary, so thatfor i = 1, 2, . . . , r , we have that xivi+1 ∈ E and, if t ≥ i + 2, xi � V≥i+2. Butthen the vertex w = vr+1 dominates X , and so, by Lemma 11(b), Dw ∩ X = ∅,a contradiction. This establishes part (b).

(c) If |X |/2 > �(G X ), then no two vertices in X totally dominate X . But then|Dv ∩ X | ≤ 1 for every vertex v ∈ V \X , contradicting part (b). Hence, 1

2 |X | ≤�(G X ). If �(G X ) = |X | − 1, then there exists a vertex x ∈ X such thatX\{x} ⊆ N (x). But then by Lemma 11(b), Dx ∩ X = ∅, a contradiction. Hence,�(G X ) ≤ |X | − 2. This establishes part (c).

(d) By part (c), we note that |X | ≥ 4. For the sake of contradiction, suppose that|X | = 4. We note then that r = 4 and t ≥ 5. By part (b), there exists a vertex v ∈ Ysuch that Dv ⊂ X . Renaming vertices if necessary, we may assume that Dv ={x2, x3}. Thus, x2x3 ∈ E and {x2, x3} � V \{v}. By part (c), �(G X ) = 2. Thuswe may assume that either G[X ] is the path x1x2x3x4 or the cycle x1x2x3x4x1.

By Lemma 11(b), we note that Dx3 ∩ X = {x1} and Dx2 ∩ X = {x4}. LetDx3 = {x1, y1} and let Dx2 = {x4, y4}. Renaming components if necessary, wemay assume that y1 ∈ Vt ∪ Z . Thus, x1 � V≤t−1 ∪ {y1}. Renaming the componentsG1, G2, . . . , Gt−1, if necessary, we may assume that y4 ∈ V≥t−1 ∪ Z . Recall thatt ≥ 5. If y4 /∈ V4, then x1 � V≤4 and x4 � V≤4. If y4 ∈ V4, then x1 � V≤4 andx4 � V≤3 ∪ {y4}. In both cases, there exists a vertex v4 ∈ V4 such that x1 � V≤4 andx4 � V≤3 ∪ {v4}.

We now consider the set Dv4 . Since {x1, x4} ⊆ N (v4), we note by Lemma 11(b)that either Dv4 = {x2, x3} or |Dv4 ∩ X | = 1.

Suppose Dv4 = {x2, x3}. Then, {x2, x3} � V≤3. Let v1 ∈ V1. Since {x2, x3} � V1,we may assume that v1x2 ∈ E . Hence, {x1, x2, x4} ⊆ N (v1). Thus, Dv1 ∩ X = {x3}.Renaming the components G2 and G3 if necessary, we may assume there exists a vertexv2 ∈ V2 such that x3 � V3 ∪ {v2} (possibly, v2 ∈ Dv1 ). Then, {x1, x3, x4} ⊆ N (v2).Thus, Dv2 ∩ X = {x2} and there exists a vertex v3 ∈ V3 adjacent to x2 (possibly,v3 ∈ Dv2 ). We note that v3 � X , and so by Lemma 11(b), Dv3 ∩ X = ∅, a contradic-tion. Hence, |Dv4 ∩ X | = 1.

Without loss of generality, we may assume that Dv4 ∩X = {x2}. Thus, x2 dominatesat least two of the three components G1, G2, G3. Renaming components if necessary,we may assume that x2 � V≤2. Let v1 ∈ V1. Since {x1, x2, x4} ⊆ N (v1), we note thatDv1 ∩ X = {x3}, and so at least one vertex, v2 say, in V2 is adjacent to x3 (possibly,v2 ∈ Dv1 ). But then v2 � X , and so, by Lemma 11(b), Dv2 ∩ X = ∅, a contradiction.Hence, |X | ≥ 5. This establishes part (d) and completes the proof of the lemma.

��Lemma 13 Let G = (V, E) be a 3t V C graph that is K1,6-free and let X ⊂ V bea nonempty subset of vertices in G. If oc(G − X) > |X |, then the graph G has thefollowing properties.

(a) �(G X ) ≤ |X | − 3.(b) Dv ⊂ X for every vertex v ∈ V .(c) |X | = 8, oc(G − X) = 9 and |V | is odd.(d) There are no even components in G − X.

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Proof of Lemma 13 Let G = (V, E) be a 3t V C K1,6-free graph of order n withδ(G) ≥ 2. Suppose that oc(G − X) > |X | for some nonempty subset X ⊂ V . Wenote that G is connected. If G has a cut-vertex v, then γt (G − v) ≥ 4, a contradiction.Hence, G is 2-connected and |X | ≥ 2. Let |X | = r and let oc(G − X) = t . ByLemma 12, we note that r ≥ 5, and so t ≥ r + 1 ≥ 6. Let G X = G[X ]. Let Y be theset of all vertices that belong to an odd component in G − X . We are now in a positionto present proofs of parts (a)-(d) in the statement of the lemma.

(a) To prove that �(G X ) ≤ |X | − 3, let x ∈ X and consider the set Dx . ByLemma 12(a), |Dx ∩X | ≥ 1. If |Dx ∩X | = 1, then the vertex in Dx ∩X is the cen-ter of an induced K1,t in G, a contradiction since t ≥ 6 and G is K1,6-free. Hence,|Dx ∩ X | = 2 for every vertex x ∈ X . By Lemma 11(b), N (x)∩ Dx = ∅, and sodX (x) ≤ |X\(Dx ∪{x})| = |X |−3. Thus, �(G X ) = maxx∈X dX (x) ≤ |X |−3.This establishes part (a).

(b) By Lemma 12(c), |X |/2 ≤ �(G X ). By part (a) above, �(G X ) ≤ |X | − 3. Con-sequently, |X | ≥ 6. Thus, t ≥ r +1 ≥ 7. If |Dv ∩ X | = 1 for some vertex v ∈ V ,then the vertex in Dv ∩ X is the center of an induced K1,6, a contradiction. Hence,|Dv ∩ X | = 2 for every vertex v ∈ V . This establishes part (b).

(c) Let v ∈ V and let Dv = {v1, v2}. By part (b), Dv ⊂ X . By Lemma 11(b),we note that the edge v1v2 totally dominates every vertex of G except for thevertex v, that is, {v1, v2} �t V \{v}. By Lemma 11(c), Du = Dv for every vertexu = v. Hence we can uniquely associate the edge v1v2 in G[X ] with the ver-tex v. Thus the number of edges in G X is at least the number of vertices in G.Since n ≥ 2|X | + 1, we note therefore that 2|X | + 1 ≤ |E(G X )|. By part (a),|E(G X )| ≤ 1

2 |X |(|X | − 3). We deduce, therefore, that |X | ≥ 8.To prove that |X | ≤ 8, let x ∈ X and consider the set Dx = {x1, x2}. By part (b),Dx ⊂ X . In particular, we note that Dx �t Y . Since G is K1,6-free, every vertexin X is adjacent to vertices from at most five components of G − X . Hence, Dx

dominates the vertices from at most ten components of G − X , and so t ≤ 10.Suppose t = 10. Then each of x1 and x2 dominate vertices from exactly fiveodd components of G − X . For i = 1, 2, let Yi be the set of all vertices in Ythat are dominated by xi . Since {x1, x2} � Y , we note that Y = Y1 ∪ Y2 andY1 ∩ Y2 = ∅. Let Y ′

1 ⊆ Y1 consist of five vertices of Y1, one from each of thefive odd components dominated by x1. Then, G[Dx ∪ Y ′

1] = K1,6 with x1 thecentral vertex of the resulting induced K1,6, a contradiction. Hence, t ≤ 9. Thussince |X | = r and r + 1 ≤ t , we note that |X | ≤ 8. Consequently, |X | = 8and oc(G − X) = |X | + 1 = 9. If n is even, then oc(G − X) ≥ |X | + 2, acontradiction. Hence, n is odd. This establishes part (c).

(d) Suppose that there is an even component in G − X . By part (c), oc(G − X) = 9.Hence there are at least ten components in G − X . Proceeding now as in theprevious paragraph, we show that there is an induced K1,6 in G, a contradiction.This establishes part (d) and completes the proof of the lemma. ��

Lemma 14 Let G = (V, E) be a 3t V C graph of even order that is K1,7-free and letX ⊂ V be a nonempty subset of vertices in G. If oc(G − X) > |X |, then the graph Ghas the following properties.

(a) �(G X ) ≤ |X | − 3.

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(b) Dv ⊂ X for every vertex v ∈ V .(c) 8 ≤ |X | ≤ 9 and oc(G − X) ≤ 11.

Proof of Lemma 14 Let G = (V, E) be a 3t V C K1,7-free graph of even order nwith δ(G) ≥ 2. We note that G is connected, and so oc(G) = 0. Suppose thatoc(G − X) > |X | for some nonempty subset X ⊂ V . Since n is even, we note thatoc(G − X) ≥ |X | + 2. If G has a cut-vertex v, then γt (G − v) ≥ 4, a contradiction.Hence, G is 2-connected and |X | ≥ 2. Let |X | = r and let oc(G − X) = t . ByLemma 12, we note that r ≥ 5, and so t ≥ r + 2 ≥ 7. Let G X = G[X ]. Let Y be theset of all vertices that belong to an odd component in G − S. We are now in a positionto present proofs of parts (a)-(c) in the statement of the lemma.

(a) Let x ∈ X and consider the set Dx . If |Dx ∩ X | = 1, then the vertex in Dx ∩ Xis the center of an induced K1,t in G, a contradiction since t ≥ 7 and G isK1,7-free. Hence by Lemma 12(a), |Dx ∩ X | = 2 for every vertex x ∈ X .Proceeding now exactly as in the proof of part (a) of Lemma 13, we establishthat �(G X ) ≤ |X | − 3. This establishes part (a).

(b) By Lemma 12(c), |X |/2 ≤ �(G X ). By part (a) above, �(G X ) ≤ |X | − 3. Con-sequently, |X | ≥ 6. Thus, t ≥ r +2 ≥ 8. If |Dv ∩ X | = 1 for some vertex v ∈ V ,then the vertex in Dv ∩ X is the center of an induced K1,7, a contradiction. Hence,|Dv ∩ X | = 2 for every vertex v ∈ V . This establishes part (b).

(c) As in the proof of part (d) of Lemma 13, we note that for every vertex v ∈ V wecan uniquely associate an edge in G[X ]. Since n ≥ 2|X | + 2, we note thereforethat 2|X | + 2 ≤ |E(G X )|. By part (a), |E(G X )| ≤ 1

2 |X |(|X | − 3). We deduce,therefore, that |X | ≥ 8. To prove that |X | ≤ 9, let x ∈ X and consider theset Dx = {x1, x2}. By part (b), Dx ⊂ X . In particular, we note that Dx �t Y .Since G is K1,7-free, every vertex in X is adjacent to vertices from at most sixcomponents of G − X . Hence, Dx dominates the vertices from at most twelvecomponents of G − X , and so t ≤ 12. If t = 12, then each of x1 and x2 is thecentral vertex of an induced K1,7, a contradiction. Hence, t ≤ 11. Thus since|X | = r and r + 2 ≤ t , we note that |X | ≤ 9. This establishes part (c) andcompletes the proof of the lemma. ��

Before proceeding with our proof of Theorem 3, we recall a well-known extremalresult due to Turán [25].

Theorem 15 (Turán’s Theorem [25]) Every graph of order n ≥ 3 and size atleast �n2/4� + 1 contains a triangle. Further, the only triangle-free graph of order nand size �n2/4� is K�n/2�,�n/2�.

Using Lemmas 12, 13 and 14, we are now in a position to prove our second mainresult. Recall the statement of Theorem 3.Theorem 3 Let G be a 3t V C graph with minimum degree at least two.

(a) If G has odd order and is K1,6-free, then G is factor-critical.(b) If G has even order and is K1,7-free, then G has a perfect matching.

Proof of Theorem 3(a) Let G = (V, E) be a 3t V C graph of odd order n withδ(G) ≥ 2 that is K1,6-free. We note that G is connected, and so oc(G) ≤ 1. If G has a

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cut-vertex v, then γt (G − v) ≥ 4, a contradiction. Hence, G is 2-connected. Assume,to the contrary, that G is not factor-critical. By the Factor-Critical Theorem 5, thereexists a nonempty subset X ⊆ V such that oc(G − X) ≥ |X |. Since G has odd order,oc(G − X) ≥ |X | + 1. By Lemma 13, |X | = 8, oc(G − X) = 9 and G − X has noeven component. Let G X = G[X ] and let Y be the set of all vertices that belong to anodd component in G − X . We note that Y = V \X .

We now consider the edges in G X . As observed in the proof of Lemma 13(c), foreach vertex v ∈ V we can uniquely associate an edge ev in G[X ] the ends of whichform the pair Dv which dominates V \{v}. For each vertex v ∈ X , we note that theedge ev dominates Y but not X . We call such an edge in G X a Y -edge. For each vertexv ∈ Y , we note that the edge ev dominates X and dominates Y\{v}. We call such anedge in G X an X -edge. We note that an X -edge dominates every vertex in Y , exceptfor one vertex. Further, we note that the X -edges are distinct from the Y -edges.

We call an edge of G X a good edge if it is a Y -edge or an X -edge. An edge of G X

that is not a good edge we call a bad edge. Every vertex in X is uniquely associatedwith a Y -edge, while every vertex in Y is uniquely associated with an X -edge. Thus,since |X | = 8 and |Y | ≥ 9, the graph G X contains at least eight Y -edges and at leastnine X -edges. Hence, G X contains at least 17 good edges.

We proceed further with the following claim.

Claim A Every triangle in G X contains at least one bad edge.

Proof Let C : xyzx be a 3-cycle in G X . Assume, to the contrary, that C contains nobad edge. Then every edge in C is a good edge. Suppose that all three edges in Care X -edges. Every vertex in X is uniquely associated with a Y -edge. For each vertexv ∈ V (C), let v be uniquely associated with the Y -edge v1v2. We note that the edgev1v2 does not dominate v, and so {v1, v2} ⊂ X\V (C). Let uv be an arbitrary edgeof C . By assumption, the edge uv dominates X . Since v1v2 does not dominate v, thevertex u is adjacent to both v1 and v2. Thus, since u1u2 does not dominate u, we notethat {u1, u2} ∩ {v1, v2} = ∅. This is true for every two vertices u and v in C . Hence,{x1, x2, y1, y2, z1, z2} ⊆ X\V (C), implying that |X | ≥ 9, a contradiction. Hence, Ccontains at least one Y -edge.

Renaming vertices if necessary, we may assume that xy is an Y -edge. Since G isK1,6-free, each of x and y dominate vertices from exactly five components of G − X .Thus one component of G − X is dominated by both x and y, four components ofG − X are dominated by x but no vertex in these components is adjacent to y, andthe remaining four components of G − X are dominated by y but no vertex in thesecomponents is adjacent to x . If yz is a good edge, then z dominates vertices of at leastthree of the four components dominated by x but not by y in G − X . Analogously ifxz is a good edge, then z dominates vertices of at least three of the four componentsdominated by y but not by x in G − X . Therefore, z dominates vertices from at leastsix components in G − X , a contradiction to G being K1,6-free. ��

We now return to our proof of Theorem 3(a). Let H denote the (spanning) subgraphG X after deleting all bad edges. We note that H has order |X | = 8. Since G X containsat least 17 good edges, we note that H contains at least 17 edges which by Turán’sTheorem 15 implies that H contains a triangle. But then there is a triangle in G X

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containing only good edges, contradicting Claim A. We deduce, therefore, that thegraph G is factor-critical. This completes the proof of Theorem 3(a). ��

Proof of Theorem 3(b) Let G = (V, E) be a 3t V C graph of even order n withδ(G) ≥ 2. We note that G is connected, and so oc(G) = 0. If G has a cut-vertexv, then γt (G − v) ≥ 4, a contradiction. Hence, G is 2-connected. Assume, to thecontrary, that G does not have a perfect matching. By Theorem 4, oc(G − X) > |X |for some subset X ⊂ V . Since n is even, we note that oc(G − X) ≥ |X |+2. If X = ∅,then oc(G) ≥ 2, a contradiction. Hence, X = ∅. By Lemma 14(c), 8 ≤ |X | ≤ 9 andoc(G − X) ≤ 11. Let G X = G[X ] and let Y be the set of all vertices that belong to anodd component in G − X . Let the X -edges and Y -edges of G X , and the good and badedges of G X , be defined exactly as in the proof of Theorem 3(a). We proceed furtherwith the following claims.

Claim I G X contains at least 2|X | + 2 good edges.

Proof We note that every vertex in X is uniquely associated with a Y -edge, whileevery vertex in Y is uniquely associated with an X -edge. Thus the graph G X containsat least |X | Y -edges and at least |X |+2 X -edges. Hence, G X contains at least 2|X |+2good edges. ��Claim II If oc(G − X) = 11, then every triangle in G X contains at least one badedge.

Proof Let C : xyzx be a 3-cycle in G X . Assume that C contains no bad edge. Since Gis K1,7-free any vertex in C dominates vertices from at most six components in G − X .As xy is a good edge we note that x dominates vertices from at least four componentsin G − X which are not dominated by y (as at least ten of the eleven components inG − X needs to be dominated by x or y). Therefore there are at most two componentsin G − X which contain vertices dominated by both x and y. As xz is a good edge, thevertex z dominates vertices from at least four of the five components in G − X whichare not dominated by x , which implies that there are at least three components in G−Xadjacent to both y and z. However, analogously to above (considering the good edgeyz in place of the good edge xy), at most two components in G − X contain verticesadjacent to both y and z, a contradiction. This completes the proof of Claim II. ��Claim III |X | = 8.

Proof Suppose that |X | = 9. As observed earlier, |X |+ 2 ≤ oc(G − X) ≤ 11, imply-ing that oc(G − X) = 11. By Claim I, G X contains at least 20 good edges. Let Hdenote the subgraph G X after deleting all bad edges. We note that H has order |X | = 9and size at least 20. Suppose that H contains no triangle. Then, H is a triangle-freegraph of order |X | = 9 and size �|X |2/4� = 20. By Turán’s Theorem 15, H = K4,5.Let X1 and X2 be the two partite sets of H , where |X1| = 4 and |X2| = 5. Wenote that since H is a spanning subgraph of G X , all edges between X1 and X2 arepresent in G X . Let x ∈ X1 be a vertex of maximum degree in G[X1]. By Lemma 14,Dx ⊂ X . By Lemma 11(b), Dx ⊂ X1\{x} and Dx does not dominate x . Hence, x hasdegree at most 1 in G[X1]. However since Dx �t X1\{x}, at least one vertex in Dx

has degree 2 in G[X1], contradicting our choice of the vertex x . Hence, H contains atriangle, contradicting Claim II. ��

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By Claim III, |X | = 8. We note that 10 ≤ oc(G − X) ≤ 11.

Claim IV oc(G − X) = 10.

Proof Suppose that oc(G − X) = 11. By Claim I, G X contains at least 18 good edges.Let H denote the (spanning) subgraph G X after deleting all bad edges. We note thatH has order |X | = 8 and size at least 18. By Turán’s Theorem 15, H contains atriangle. But then there is a triangle in G X containing only good edges, a contradictionto Claim II. ��

Let F denote the complement of G X . That is, V (F) = X and e ∈ E(F) if and onlyif e /∈ E(G X ).

Claim V The graph F has minimum degree 2, |V (F)| = 8 and 8 ≤ |E(F)| ≤ 10.

Proof By Lemma 14(a) we immediately get δ(F) ≥ 2. As observed earlier, |X | = 8,and so |V (F)| = 8. As δ(F) ≥ 2 and |V (F)| = 8, we note that |E(F)| ≥ 8. As thereare at least 18 good edges in G X we note that there can be at most 10 non-edges inG X , which implies that |E(F)| ≤ 10. ��Claim VI F is triangle-free.

Proof Let C : x1x2x3x1 be a 3-cycle in F . For i = 1, 2, 3, let di = dF (xi ). SinceDx1 ⊂ X , we note there is an edge in G X that dominates every vertex in X except forthe vertex x1. If d1 = 2, then every edge in G X dominates x1, a contradiction. Hence,d1 ≥ 3. Analogously, d2 ≥ 3 and d3 ≥ 3. By Claim V, δ(F) ≥ 2. Hence, 2|E(F)| ≥∑

x∈X dF (x) ≥ 19, implying that |E(F)| ≥ 10. Consequently, by Claim V, |E(F)|= 10. For i = 1, 2, 3, let vi ∈ NF (xi )\{x1, x2, x3} and consider the following cases.

Suppose that v1, v2 and v3 are three distinct vertices. If∑3

i=1 di ≥ 11, then2|E(F)| ≥ 21, implying that |E(F)| > 10, a contradiction. Hence,

∑3i=1 di ≤ 10.

Renaming vertices, if necessary, we may assume that d1 = d2 = 3 and 3 ≤ d3 ≤ 4.Since d3 ≤ 4, at most one of the edges v1x3 and v2x3 belong to F . Renaming x1 andx2, if necessary, we may assume that v1x3 is not in F . We note, therefore, that the fiveedges in E∗ = {v1x2, v1x3, v2x1, v3x1, v3x2} belong to G. However if vi x j ∈ E∗,then 1 ≤ i, j ≤ 3, i = j , and the edge vi x j does not dominate the vertex xi in G.Hence if vi x j ∈ E∗ is a good edge, then the vertex xi is uniquely associated withthe good edge vi x j . Thus at most three edges in E∗ are good. So G X contain at leasttwo bad edges and ten non-edges, implying that it has less than 18 good edges, acontradiction.

Suppose |{v1, v2, v3}| = 2. Renaming vertices, if necessary, we may assume thatv1 = v2. If d1 = 3, then an edge of G with both end-points in NF (x1) must dominateG − x1 implying that v1x3 �→ x1 (as {v1x2, x2x3} ⊂ E(F)). However, v1x3 doesnot dominate x2, a contradiction. Therefore, d1 ≥ 4 and analogously d2 ≥ 4. Thus,∑3

i=1 di ≥ 11, and so 2|E(F)| ≥ 21, implying that |E(F)| > 10, a contradiction.If |{v1, v2, v3}| = 1, then analogously to showing d1 ≥ 4 in the previous para-

graph, we can show that di ≥ 4 for i = 1, 2, 3, and so 2|E(F)| ≥ 22, implying that|E(F)| > 10, a contradiction. ��

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Claim VII F has girth at least five.

Proof Assume that Claim VI is false, which by Claim VI implies that there is a 4-cycleC : x1x2x3x4x1 in F . For i = 1, 2, 3, 4, let di = dF (xi ). If d1 = 2, then we note that theonly potential edge in G X that may dominate G −x1 but which does not dominate x1 isx2x4, but x2x4 does not dominate x3, a contradiction. Hence, d1 ≥ 3 and analogouslydi ≥ 3 for i = 2, 3, 4. By Claim V, δ(F) ≥ 2. Hence, 2|E(F)| ≥ ∑

x∈X dF (x) ≥ 20,implying that |E(F)| ≥ 10. Consequently, by Claim V, |E(F)| = 10 and so G X

contains 18 edges. By Claim I, all edges in G X are good. We note that x1x3 ∈ E(G)

by Claim VI. However, x1x3 dominates neither x2 nor x4, implying that it is a badedge, a contradiction to all edges in G X being good. ��Claim VIII F is not 2-regular.

Proof Assume that F is 2-regular. Then by Claim VII, we note that F is a cycleof length 8. Renaming vertices if necessary, we may assume that F is the cyclex1x2 . . . x8x1. Therefore the Y -edges in G X are the eight edges xi xi+2 where i =1, 2, . . . , 8 and where indices are taken modulo 8. Let E ′ = {x1x5, x2x6, x3x7, x4x8}.As there are at least 18 good edges in G X and 8 non-edges in G X , we note that therecan be at most two bad edges in G X . In particular, at most two edges in E ′ are badedges in G X , and so at least two edges in E ′ are X -edges. Without loss of generality,we may assume that x1x5 is an X -edge. As G is K1,7-free, we note that x1 (and x3)are adjacent to the vertices in at most six components in G − X . As x1x3 and x3x5are Y -edges we note that x1 and x5 dominate the vertices in all components in G − Xwhich are not adjacent to x3. As there are at least four such components we note thatthe edge x1x5 dominates at most 6 + 6 − 4 = 8 components in G − X , implying thatx1x5 is a bad edge, a contradiction. ��

We now return to our proof of Proof of Theorem 3(b). Let x ∈ X be arbitrary and letu, v ∈ NF (x) be arbitrary. By Claim VI we note that uv /∈ E(F), and so uv ∈ E(G).Let E(x) = {uv | u, v ∈ NF (x)}, which by the above implies that E(x) ⊆ E(G) forall x ∈ X . Let EX = ∪x∈X E(x). By Claim VII we note that E(x) ∩ E(x ′) = ∅ forall distinct x, x ′ ∈ X . By definition no edge in E(x) is an X -edge for any x ∈ X . Soall X -edges belong to E(G X )\(E(F) ∪ EX ). By Claim VIII we note that either thereare at least two vertices of F with degree at least three or there is a vertex of degreeat least four. In both cases, |EX | ≥ 2 × 3 + 6 = 12. As |E(F)| ≥ 9 by Claim VIII,there are at most 28 − 9 − 12 = 7 X -edges in G X , a contradiction. ��

6 Examples Showing the Tightness of Our Results

In this paper, we have shown that every 3t EC graph of even order has a perfectmatching, while every 3t EC graph of odd order with no cut-vertex is factor-critical.However the situation for 3t V C graphs is more complex. We have shown that theK1,5-free condition in Theorem 1 for odd order graphs can be relaxed to K1,6-free,while the K1,5-free condition in Theorem 1 for even order graphs can be relaxed toK1,7-free. We will now show that our results are best possible.

Proposition 1 There exists a 3t V C graph of odd order with minimum degree at leastfive that is K1,7-free and is not factor-critical.

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Proof Let X = {x1x2 . . . x8} and let Y9 = {y1, y2, . . . , y9}. We now define the graphG17 as follows. Let V (G17) = X ∪ Y9 and let Y9 be an independent set in G17.Let G17[X ] contain all possible edges except x1x2, x2x3, . . . , x7x8, x8x1. The edgesbetween X and Y9 are defined as follows.

Y9\N (x1) = {y4, y5, y6} Y9\N (x5) = {y1, y5, y6}Y9\N (x2) = {y1, y2, y3} Y9\N (x6) = {y1, y3, y4}Y9\N (x3) = {y3, y7, y9} Y9\N (x7) = {y2, y8, y9}Y9\N (x4) = {y6, y7, y8} Y9\N (x8) = {y5, y7, y8}

We note that G17 contains an induced K1,6 (for example x1 is the center of a star withleaves {y1, y2, y3, y7, y8, y9}), but is K1,7-free. We furthermore note that G17 is 3t V Cas the following holds. For all i = 1, 2, . . . , 8, we note that xi−1xi+1 totally dominatesG − xi where indices are taken modulo 8; that is, xi−1xi+1 �→ xi for i = 1, 2, . . . , 8.Furthermore, x2x5 �→ y1, x2x7 �→ y2, x3x6 �→ y3, x1x6 �→ y4, x5x8 �→ y5, x1x4 �→y6, x3x8 �→ y7, x4x7 �→ y8 and x3x7 �→ y9. As γt (G17) = 3, the graph G17 is there-fore 3t V C . Since oc(G17 − X) = |Y | = |X | + 1, the graph G17 is not factor-criticalby Theorem 5. ��Proposition 2 There exists a 3t V C graph of even order with minimum degree at leastfive and with no perfect matching that is K1,8-free.

Proof Let X = {x1x2 . . . x8} and let Y10 = {y1, y2, . . . , y10}. We now define thegraph G18 as follows. Let V (G18) = X ∪ Y10 and let Y10 be an independent set inG18. Let G19[X ] contain all possible edges except x1x2, x2x3, . . . , x7x8, x8x1. Theedges between X and Y10 are defined as follows.

Y10\N (x1) = {y4, y5, y6} Y10\N (x5) = {y1, y5, y6}Y10\N (x2) = {y1, y2, y3} Y10\N (x6) = {y1, y3, y4}Y10\N (x3) = {y3, y7, y9} Y10\N (x7) = {y2, y8, y9}Y10\N (x4) = {y6, y8, y10} Y10\N (x8) = {y5, y7, y10}

We note that G18 contains an induced K1,7, but is K1,8-free. We furthermore notethat G18 is 3t V C as the following holds. For all i = 1, 2, . . . , 8, xi−1xi+1 �→ xi

where indices are taken modulo 8. Furthermore, x2x5 �→ y1, x2x7 �→ y2, x3x6 �→ y3,x1x6 �→ y4, x5x8 �→ y5, x1x4 �→ y6, x3x8 �→ y7, x4x7 �→ y8, x3x7 �→ y9 andx4x8 �→ y10. As γt (G18) = 3, the graph G18 is therefore 3t V C . Since oc(G18 − X) =|X | + 2, the graph G18 contains no perfect matching by Theorem 4. ��

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Perfect Matchings in Total Domination Critical Graphs