Inequalities of Independence Number, Clique Number and Connectivity of Maximal Connected Domination Critical Graphs

TL;DR

This paper investigates inequalities involving independence number, clique number, and connectivity in maximal connected domination critical graphs, establishing bounds and characterizations for these parameters.

Contribution

It proves new bounds on independence and clique numbers in maximal connected domination critical graphs and characterizes graphs achieving these bounds.

Findings

Maximal 3-γc-vertex critical graphs satisfy α ≤ δ, and this bound is optimal.

Such graphs also satisfy α + ω ≤ n - 1, with characterizations of equality cases.

Graphs with connectivity less than minimum degree have Hamiltonian paths between any two vertices.

Abstract

A $k$-$\gamma_{c}$-edge critical graph is a graph $G$ with the connected domination number $\gamma_{c}(G) = k$ and $\gamma_{c}(G + uv) < k$ for every $uv \in E(\overline{G})$. Further, a $2$-connected graph $G$ is said to be $k$-$\gamma_{c}$-vertex critical if $\gamma_{c}(G) = k$ and $\gamma_{c}(G - v) < k$ for all $v \in V(G)$. A maximal $k$-$\gamma_{c}$-vertex critical graph is a graph which are both $k$-$\gamma_{c}$-edge critical and $k$-$\gamma_{c}$-vertex critical. Let $\kappa, \delta, \omega$ and $\alpha$ be respectively connectivity minimum degree, clique number and independence number. In this paper, we prove that every maximal $3$-$\gamma_{c}$-vertex critical graph $G$ satisfies $\alpha \leq \delta$ and this bound is best possible. We prove further that $G$ satisfies $\alpha + \omega \leq n - 1$ and we also characterize all such graphs achieving the upper bounds. We finally show that if $G$ satisfies $\kappa < \delta$, then every two vertices of $G$ are joined by hamiltonian path.

Equations

$$|S-I|+1$$

$$|S-I|={\alpha }_1+{\alpha }_2=p.$$

Full text

**Keywords: connected domination, independence number, connectivity.

AMS subject classification:05C69, 05C40**

1 Introduction

Our basic graph theoretic notation and terminology follows that of Bondy and Murty[3]. Thus $G$ denotes a finite graph with vertex set $V(G)$ and edge set $E(G)$. For $S\subseteq V(G)$, $G[S]$ denotes the subgraph of $G$ induced by $S$. Throughout this paper all graphs are simple and connected. The open neighborhood $N_{G}(v)$ of a vertex $v$ in $G$ is $\{u\in V(G):uv\in E(G)\}$. Further, the closed neighborhood $N_{G}[v]$ of a vertex $v$ in $G$ is $N_{G}(v)\cup\{v\}$. The degree $deg_{G}(v)$ of a vertex $v$ in $G$ is $|N_{G}(v)|$. The minimum degree of a graph $G$ is denoted by $\delta(G)$. $N_{S}(v)$ denotes $N_{G}(v)\cap S$ where $S$ is a vertex subset of $G$. A tree is a connected graph contains no cycle. A star $K_{1,n}$ is a tree containing $n$ vertices of degree $1$ and exactly one vertex of degree $n$. An independent set is a set of pairwise non-adjacent vertices. The independence number $\alpha(G)$ is the maximum cardinality of an independent set. For a connected graph $G$, a cut set is a vertex subset $S\subseteq V(G)$ such that $G-S$ is connected. The minimum cardinality of a vertex cut set of a graph $G$ is called the connectivity and is denoted by $\kappa(G)$. If $G$ has $S=\{a\}$ as a minimum cut set, then $a$ is a cut vertex of $G$ and $\kappa(G)=1$. A graph $G$ is $s$-connected if $\kappa(G)\geq s$. When ambiguity occur, we abbreviate $\delta(G),\alpha(G)$ and $\kappa(G)$ to $\delta,\alpha$ and $\kappa$, respectively. A hamiltonian path is a path containing every vertex of a graph. A graph $G$ is hamiltonian connected if every two vertices of $G$ are joined by a hamiltonian path. Obviously, a graph of connectivity one is not hamiltonian connected. Moreover, a graph of connectivity two does not contain a hamiltonian path between the two vertices in a minimum cut set. Thus we always focus on $3$-connected graphs when we study on the hamiltonian connected property of graphs. For a graph $G$, the Mycielskian $\mu(G)$ of $G$ is the graph with vertex set $V(G)\cup V^{\prime}\cup\{x\}$ where $V^{\prime}=\{u^{\prime}|u\in V(G)\}$ and edge set $E(G)\cup\{uv^{\prime}|uv\in E(G)\}\cup\{v^{\prime}x|v^{\prime}\in V^{\prime}\}$.

For subsets $D,X\subseteq V(G)$, $D$ dominates $X$ if every vertex in $X$ is either in $D$ or adjacent to a vertex in $D$. If $D$ dominates $X$, then we write $D\succ X$, further, we write $a\succ X$ when $D=\{a\}$. If $X=V(G)$, then $D$ is a dominating set of $G$ and we write $D\succ G$ instead of $D\succ V(G)$. A connected dominating set of a graph $G$ is a dominating set $D$ of $G$ such that $G[D]$ is connected. If $D$ is a connected dominating set of $G$, we then write $D\succ_{c}G$. A smallest connected dominating set is call a $\gamma_{c}$-set. The cardinality of a $\gamma_{c}$-set of $G$ is called the connected domination number of $G$ and is denoted by $\gamma_{c}(G)$. A total dominating set of a graph $G$ is a subset $D$ of $V(G)$ such that every vertex in $G$ is adjacent to a vertex in $D$. The minimum cardinality of a total dominating set of $G$ is called the total domination number of $G$ and is denoted by $\gamma_{t}(G)$.

A graph $G$ is $k$-$\gamma_{c}$-edge critical if $\gamma_{c}(G)=k$ and $\gamma_{c}(G+uv)<k$ for each pair of non-adjacent vertices $u$ and $v$ of $G$. A graph $G$ is $k$-$\gamma_{c}$-vertex critical if $\gamma_{c}(G)=k$ and $\gamma_{c}(G-v)<k$ for any vertex $v$ of $G$. A graph $G$ is a maximal $k$-$\gamma_{c}$-vertex critical if $G$ is both $k$-$\gamma_{c}$-edge critical and $k$-$\gamma_{c}$-vertex critical. It is easy to see that a disconnected graph cannot contain a connected dominating set. Thus if a graph $G$ contains a cut vertex $c$, then $G-c$ has no connected dominating set. Hence, in the following, we always assume that $k$-$\gamma_{c}$-vertex critical graphs are $2$-connected. A $k$-$\gamma_{t}$-edge critical graph is similarly defined.

This paper focuses on the relationship of the connectivity and the independence number of $3$-$\gamma_{c}$-edge critical graphs. For related results on $3$-$\gamma$-edge critical graphs, Zhang and Tian[11] showed that the independence number of every $3$-$\gamma$-edge critical graph does not exceed $\kappa+2$ and, moreover if $\alpha=\kappa+2$, then $\kappa=\delta$. They used this result to established, the short proof, that every $3$-$\gamma$-edge critical graph contains a hamiltonian path. Further, Kaemawichanurat and Caccetta [6] showed that every $3$-$\gamma_{c}$-edge critical graph satisfies $\alpha\leq\kappa+2$. Moreover, if $G$ is a $3$-$\gamma_{c}$-edge critical graph such that $\kappa+1\leq\alpha\leq\kappa+2$, then $\kappa=\delta$ with only one exception. They also showed that if $\kappa\geq 3$, then $\alpha=\kappa+p$ if and only if $\alpha=\delta+p$ for all $p\in\{1,2\}$. The condition $\kappa+1\leq\alpha\leq\kappa+2$ is best possible to establish that $\kappa=\delta$.

In this paper, we prove that if a $3$-$\gamma_{c}$-edge critical graph $G$ is $3$-$\gamma_{c}$-vertex critical ($G$ is maximal $3$-$\gamma_{c}$-vertex critical) satisfying $\alpha=\kappa$, then $\kappa=\delta$. We provide a class of maximal $3$-$\gamma_{c}$-vertex critical graphs with $\alpha<\kappa$ and $\kappa<\delta$. Hence, the condition $\alpha=\kappa$ is needed to prove that $\kappa=\delta$ in maximal $3$-$\gamma_{c}$-vertex critical graphs. We finish this work by showing that every $3$-connected maximal $3$-$\gamma_{c}$-vertex critical graph are hamiltonian connected if $\kappa<\delta$.

2 Preliminaries

In this section, we state a number of results that we make use of in establishing our theorems. We begin with a result of Chvátal and Erdös[5] which gives hamiltonian properties of graphs according to the connectivity and the independence number.

Theorem 1

[5]** Let $G$ be an $s$-connected graph containing no independent set of $s$ vertices. Then $G$ is hamiltonian connected.

On $3$-$\gamma_{c}$-edge critical graphs, Ananchuen[1] characterized such graphs with a cut vertex. For positive integers $n_{i}$ and $r\geq 2$, let $H=\cup^{r}_{i=1}K_{1,n_{i}}$. For $1\leq j\leq r$, let $c_{j}$ be the center of $K_{1,n_{j}}$ in $H$ and $w^{j}_{1},w^{j}_{2},...,w^{j}_{n_{j}}$ the end vertices of $K_{1,n_{j}}$ in $H$. The graphs $G_{1}$ and $G_{2}$ are defined as follows. Set $V(G_{1})=V(H)\cup\{x,y\}$ and $E(G_{1})=E(\overline{H})\cup\{xy\}\cup\{xw^{j}_{i}:1\leq i\leq n_{j}$ and $1\leq j\leq r\}$. Set $V(G_{2})=V(H)\cup\{x,y\}\cup U$ where $|U|\geq 1$ and $E(G_{2})=E(\overline{H})\cup\{xy\}\cup\{xw^{j}_{i}:1\leq i\leq n_{j}$ and $1\leq j\leq r\}\cup\{uz:u\in U$ and $z\in V(G_{1})-\{x,y,u\}\}$.

The following two lemmas, observed by Chen et al.[4], give fundamental properties of $3$-$\gamma_{c}$-edge critical graphs.

Lemma 1

[4]** Let $G$ be a $3$-$\gamma_{c}$-edge critical graph and, for a pair of non-adjacent vertices $u$ and $v$ of $G$, let $D_{uv}$ be a $\gamma_{c}$-set of $G+uv$. Then

1. (1)

$|D_{uv}|=2$, 2. (2)

$D_{uv}\cap\{u,v\}\neq\emptyset$* and* 3. (3)

if $u\in D_{uv}$ and $v\notin D_{uv}$, then $N_{G}(v)\cap D_{uv}=\emptyset$.

Lemma 2

[4]** Let $G$ be a $3$-$\gamma_{c}$-edge critical graph and $I$ an independent set with $|I|=p\geq 3$. Then the vertices in $I$ can be ordered as $a_{1},a_{2},...,a_{p}$ and there exists a path $x_{1},x_{2},...,x_{p-1}$ in $G-I$ with $\{a_{i},x_{i}\}\succ_{c}G+a_{i}a_{i+1}$ for $1\leq i\leq p-1$.

On $3$-$\gamma_{c}$-vertex critical graphs, Ananchuen et al.[2] provided fundamental properties of such graphs.

Lemma 3

[2]** Let $G$ be a $3$-$\gamma_{c}$-vertex critical graph and, for a vertex $v$ of $G$, let $D_{v}$ be a $\gamma_{c}$-set of $G-v$. Then

1. (1)

$|D_{v}|=2$* and* 2. (2)

$D_{v}\cap N_{G}[v]=\emptyset$.

In the study of $3$-$\gamma_{t}$-edge critical graphs, Simmons[10] proved that such graphs satisfy $\alpha\leq\delta+2$. It was pointed out in Ananchuen[1] that a $3$-$\gamma_{c}$-edge critical graph is also $3$-$\gamma_{t}$-edge critical and vice versa. Then every $3$-$\gamma_{c}$-edge critical graph satisfies $\alpha\leq\delta+2$. In [7], we established the result of $3$-$\gamma_{c}$-edge critical graphs when $\alpha=\delta+2$.

Theorem 2

([10], [7]) Let $G$ be a $3$-$\gamma_{c}$-edge critical graph with $\delta\geq 2$. Then $\alpha\leq\delta+2$. Moreover if $\alpha=\delta+2$, then $G$ contains exactly one vertex $x$ of degree $\delta$ and $G[N[x]]$ is a clique.

In [8], we established results of maximal $3$-$\gamma_{c}$-vertex critical graphs. Let $G$ be a maximal $3$-$\gamma_{c}$-vertex critical graph with a vertex cut set $S$ and $C_{1},C_{2},...,C_{m}$ be the components of $G-S$.

Lemma 4

[8]** For all $v\in V(G)$, if $m\geq 3$ or $v\in S\cup V(C_{i})$ where $|V(C_{i})|>1$, then $G$ satisfies these following properties.

1. (1)

$D_{v}\cap S\neq\emptyset$. 2. (2)

$v$* does not dominate $S$.*

Lemma 5

[8]** Let $a\in V(C_{i})$ for some $i\in\{1,2,...,m\}$. Then $G$ has these following properties.

1. (1)

Let $b\in V(C_{j})$ for some $j\in\{1,2,...,m\}$ such that $\{a,b\}$ does not dominate $G$, if $m\geq 3$ or $|V(C_{i})|,|V(C_{j})|>1$, then $|D_{ab}\cap\{a,b\}|=1$ and $|D_{ab}\cap S|=1$. 2. (2)

If $c\in D_{a}$ where $c$ is an isolated vertex in $S$, then $m=2$ and $\{w\}=V(C_{j})$ for some $j\in\{1,2\}$ where $\{w\}=D_{a}-\{c\}$.

We characterized all maximal $3$-$\gamma_{c}$-vertex critical graphs of connectivity $s\geq 2$ with a smallest cut set contains no edge

Theorem 3

[8]** Let $G$ be a maximal $3$-$\gamma_{c}$-vertex critical graph of connectivity $s\geq 2$ with a smallest cut set contains no edge. Then $G$ is isomorphic to $G_{3}=\mu(K_{s})$.

$v$Figure 1 : A graph $G_{3}=\mu(K_{s})$

In [9], we established the upper bound of the independence number of maximal $3$-$\gamma_{c}$-vertex critical graphs in term of the minimum degree.

Theorem 4

[9]** Let $G$ be a maximal $3$-$\gamma_{c}$-vertex critical graph. Then $\alpha\leq\delta$.

3 Maximal $3$-$\gamma_{c}$-Vertex Critical Graphs

In this section we show, by using Theorem 4, that every maximal $3$-$\gamma_{c}$-vertex critical graph satisfies $\alpha\leq\kappa$. We also study the such graphs when $\alpha=\kappa$. Since we characterized all maximal $3$-$\gamma_{c}$-vertex critical graphs with $\kappa\leq 3$ in [8], in the following, we focus on $|S|=\kappa\geq 4$. We define $\alpha_{1},\alpha_{2},p,S,H_{1}$ and $H_{2}$ by the same as in the previous section.

Theorem 5

The independence number of any maximal $3$-$\gamma_{c}$-vertex critical graph does not exceed the connectivity.

**Proof. **Suppose to the contrary that $\kappa+1\leq\alpha$. So $|S|+1\leq\alpha_{1}+\alpha_{2}+|S\cap I|$. Then

[TABLE]

Claim 1 : $|V(C_{i})|>1$ for all $i\in\{1,2,...,m\}$, in particular, $|H_{i}|>1$.

Suppose $V(C_{i})=\{c\}$ for some $i\in\{1,2,...,m\}$. Thus $N_{G}(c)\subseteq S$. By Theorem 4,

$\delta\leq deg_{G}(c)\leq|S|<|S|+1=\kappa+1\leq\alpha\leq\delta$,

a contradiction and thus establishing Claim 1.

Let $p=\alpha_{1}+\alpha_{2}$ and $\{a_{1},a_{2},...,a_{p}\}=\cup^{2}_{i=1}I_{i}$. If $p=1$, then, by (1), $|S-I|=0$. It follows that $S=S\cap I$. Thus $S$ is an independent set. By Theorem 3, $G$ is $G_{3}$. Hence, $N_{G_{3}}(x)$ in the graph $G_{3}$ is a minimum cut set such that $G_{3}-N_{G_{3}}(x)$ contains $x$ as a singleton component, contradicting Claim 1. Therefore $p\geq 2$.

Claim 2 : For $a,b\in\cup^{2}_{i=1}I_{i}$, $|D_{ab}\cap\{a,b\}|=1$ and $|D_{ab}\cap(S-I)|=1$.

Since $|S|\geq 4$ and $2\leq p=\alpha_{1}+\alpha_{2}$. If $p\geq 3$, then $\cup^{2}_{i=1}I_{i}-\{a,b\}\neq\emptyset$. If $p=2$, then, by Equation 1, $|S|-|S\cap I|+1\leq 2$. As $|S|\geq 4$, we must have $|S\cap I|\geq 3$, in particular, $S\cap I\neq\emptyset$. Hence $(\cup^{2}_{i=1}I_{i}-\{a,b\})\cup(S\cap I)\neq\emptyset$. So $\{a,b\}$ does not dominate $G$. By Claim 1 and Lemma 5(1), $|D_{ab}\cap\{a,b\}|=1$ and $|D_{ab}\cap S|=1$. Without loss of generality, let $a\in D_{ab}$ and $\{a^{\prime}\}=D_{ab}\cap S$. By the connectedness of $(G+ab)[D_{ab}]$, $a^{\prime}\in S-I$ and thus establishing Claim 2.

Suppose that $p=2$. Consider $G+a_{1}a_{2}$. By Claim 2, $|D_{a_{1}a_{2}}\cap(S-I)|=1$. Since $D_{a_{1}a_{2}}\cap(S-I)\subseteq S-I$, by Equation 1,

$1\leq|S-I|\leq\alpha_{1}+\alpha_{2}-1=p-1=1$.

Therefore, $D_{a_{1}a_{2}}\cap(S-I)=S-I$. If $p\geq 3$, then, by Lemma 2, the vertices $a_{1},a_{2},...,a_{p}$ can be ordered as $x_{1},x_{2},...,x_{p}$ and there exists a path $y_{1},y_{2},...,y_{p-1}$ such that $\{x_{i},y_{i}\}\succ_{c}G+x_{i}x_{i+1}$ for $i\in\{1,2,...,p-1\}$. Since $\{x_{1},x_{2},...,x_{p}\}\subseteq\cup^{2}_{i=1}I_{i}$, it follows by Claim 2 that $\{y_{1},y_{2},...,y_{p-1}\}\subseteq S-I$. Hence, by (1), $p-1\leq|S-I|\leq\alpha_{1}+\alpha_{2}-1=p-1$. In both cases $p=2$ and $p\geq 3$, we have that $\{y_{1},y_{2},...,y_{p-1}\}=S-I$.

If $p=2$, then clearly $G[\{y_{1}\}]$ is a clique. Suppose $p\geq 3$. Consider $G+x_{i}x_{j}$ for $2\leq i\neq j\leq p$. By Claim 2, $|D_{x_{i}x_{j}}\cap\{x_{i},x_{j}\}|=1$ and $|D_{x_{i}x_{j}}\cap(S-I)|=1$. Without loss of generality, let $x_{i}\in D_{x_{i}x_{j}}$. Because $S-I=\{y_{1},y_{2},...,y_{p-1}\}$, by Lemma 1(3), $D_{x_{i}x_{j}}\cap(S-I)=\{y_{j-1}\}$. Since $x_{i}y_{i-1}\notin E(G)$, $y_{i-1}y_{j-1}\in E(G)$. Therefore, $G[\{y_{1},y_{2},...,y_{p-1}\}]$ is a clique. Since $\{x_{1},x_{2},...,x_{p}\}\subseteq I$, $y_{i}\succ(S\cap I)$ for $i\in\{1,2,...,p-1\}$. Therefore $y_{i}\succ S$, contradicting Lemma 4(2). Hence, $\alpha\leq\kappa$. $\Box$

We have by Theorem 3 that the graph $G_{3}=\mu(K_{s})$ has $N_{G_{3}}(x)$ as a minimum cut set and also a maximum independent set. Hence $\alpha(G_{3})=\kappa(G_{3})$. Therefor, the bound in Theorem 5 is best possible. We, further, focus on maximal $3$-$\gamma_{c}$-vertex critical graphs which satisfy $\alpha=\kappa$. We then have $|S-I|+|S\cap I|=|S|=\alpha_{1}+\alpha_{2}+|S\cap I|$. So

[TABLE]

We may assume without loss of generality that $\alpha_{1}\leq\alpha_{2}$. We would like to show that if a maximal $3$-$\gamma_{c}$-vertex critical graph $G$ satisfies $\alpha=\kappa$, then $G-S$ contains at least one singleton component for every minimum cut set $S$. We then suppose to the contrary that there is no singleton component of $G-S$, in particular, $|H_{i}|>1$ for $i=1,2$. We next establish the following lemmas.

Lemma 6

Let $G$ be a maximal $3$-$\gamma_{c}$-vertex critical graph. If $\alpha=\kappa$ and $|V(C_{i})|>1$ for all $i\in\{1,2,...,m\}$, then $p\geq 3$.

**Proof. **By the assumption, $|H_{i}|>1$ for $i=1,2$. Suppose first that $p=0$. Thus $S=S\cap I$. By Theorem 3, $G$ is $G_{3}$. We have $N_{G_{3}}(x)$ is a minimum cut set of $G_{3}$ and $x$ is a singleton component of $G-N_{G_{3}}(x)$, contradicting the assumption. We distinguish $2$ cases.

Case 1 : $p=1$.

By Equation 2, $|S-I|=1$. Let $\{v\}=S-I,\{a_{1}\}=\cup^{2}_{i=1}I_{i}$ and $\{a_{2},a_{3},...,a_{\alpha}\}=S\cap I$. Therefore $\alpha_{1}=0,\alpha_{2}=1$. Therefore, $a_{1}\in H_{2}$. Since $|S|\geq 4$, $|S\cap I|\geq 3$. Lemma 2 yields that the vertices in $\{a_{2},a_{3},...,a_{\alpha}\}$ can be ordered as $x_{1},x_{2},...,x_{\alpha-1}$ and there exists a path $P=y_{1},y_{2},...,y_{\alpha-2}$ with $\{x_{i},y_{i}\}\succ_{c}G+x_{i}x_{i+1}$ for $i\in\{1,..,\alpha-2\}$. Since each $y_{i}$ is adjacent to at least one vertex of $I$ for $i=1,2,...,\alpha-2$, $y_{i}\neq a_{1}$. To dominate $a_{1}$, $y_{i}\in H_{2}\cup\{v\}$.

Subcase 1.1 : $v\notin V(P)$.

So $V(P)\subseteq H_{2}$. Thus $x_{i}\succ H_{1}$ for $i=1,2,...,\alpha-2$. Since $S$ is a minimum cut set, $N_{H_{1}}(v)\neq\emptyset$. Let $u\in N_{H_{1}}(v)$. So $u\succ\{x_{1},x_{2},...,x_{\alpha-2},v\}$. Lemma 4(2) implies that $ux_{\alpha-1}\notin E(G)$. Consider $G+uy_{\alpha-2}$. Since $ux_{\alpha-1},y_{\alpha-2}x_{\alpha-1}\notin E(G)$, by Lemma 5(1), $|D_{uy_{\alpha-2}}\cap\{u,y_{\alpha-2}\}|=1$ and $|D_{uy_{\alpha-2}}\cap S|=1$. So either $y_{\alpha-2}\in D_{uy_{\alpha-2}}$ or $u\in D_{uy_{\alpha-2}}$. In the first case, by Lemma 1(3), $\{x_{1},x_{2},...,x_{\alpha-2},v\}\cap D_{uy_{\alpha-2}}=\emptyset$. Hence $x_{\alpha-1}\in D_{uy_{\alpha-2}}$. But $G[D_{uy_{\alpha-2}}]$ is not connected. Thus $u\in D_{uy_{\alpha-2}}$. By the connectedness of $(G+uy_{\alpha-2})[D_{uy_{\alpha-2}}]$, $x_{\alpha-1}\notin D_{uy_{\alpha-2}}$. If $x_{i}\in D_{uy_{\alpha-2}}$ for $i\in\{1,2,...,\alpha-2\}$, then $D_{uy_{\alpha-2}}$ does not dominate $x_{\alpha-1}$. Therefore $v\in D_{uy_{\alpha-2}}$. Thus $va_{1}\in E(G)$. Consider $G+ua_{1}$. Since $ux_{\alpha-1},a_{1}x_{\alpha-1}\notin E(G)$, by Lemma 5(1), $|D_{ua_{1}}\cap\{u,a_{1}\}|=1$ and $|D_{ua_{1}}\cap S|=1$. Hence either $u\in D_{ua_{1}}$ or $a_{1}\in D_{ua_{1}}$. In the first case, $v\notin D_{ua_{1}}$ by Lemma 1(3). By the connectedness of $(G+ua_{1})[D_{ua_{1}}]$, $x_{\alpha-1}\notin D_{ua_{1}}$. To dominate $x_{\alpha-1}$, $D_{ua_{1}}\cap\{x_{1},x_{2},...,x_{\alpha-2}\}\neq\emptyset$. So $D_{ua_{1}}\cap S=\emptyset$, a contradiction. Hence, $a_{1}\in D_{ua_{1}}$. Lemma 1(3) gives that $v\notin D_{ua_{1}}$. By the connectedness of $(G+ua_{1})[D_{ua_{1}}]$, $\{x_{1},x_{2},...,x_{\alpha-1}\}\cap D_{ua_{1}}=\emptyset$. Clearly $D_{ua_{1}}\cap S=\emptyset$, a contradiction and Case 1.1 cannot occur.

Subcase 1.2 : $v\in V(P)$.

Thus there exists $j\in\{1,2,...,\alpha-2\}$ such that $y_{j}=v$. Thus $x_{i}\succ H_{1}$ for $i\neq j,\alpha-1$ and $va_{1}\in E(G)$. Since $a_{1},x_{\alpha-1}\in I$, $a_{1}x_{\alpha-1}\notin E(G)$. If $x_{\alpha-1}$ is not adjacent to $w\in H_{1}$, then consider $G+wa_{1}$. Lemma 5(1) implies that $|D_{wa_{1}}\cap\{w,a_{1}\}|=1$ and $|D_{wa_{1}}\cap S|=1$. Thus either $w\in D_{wa_{1}}$ or $a_{1}\in D_{wa_{1}}$. In both case, $x_{\alpha-1}\notin D_{wa_{1}}$ because $(G+wa_{1})[D_{wa_{1}}]$ is connected. If $w\in D_{wa_{1}}$, then, by Lemma 1(3), $v\notin D_{wa_{1}}$. To dominate $x_{\alpha-1}$, $\{x_{1},x_{2}...,x_{\alpha-2}\}\cap D_{wa_{1}}=\emptyset$. So $D_{wa_{1}}\cap S=\emptyset$, a contradiction. Hence, $a_{1}\in D_{wa_{1}}$.By the connectedness of $(G+wa_{1})[D_{wa_{1}}]$, $D_{wa_{1}}\cap\{x_{1},x_{2},...,x_{\alpha-1}\}=\emptyset$. To dominate $x_{j+1}$, $v\notin D_{wa_{1}}$. We then have $D_{wa_{1}}\cap S=\emptyset$, a contradiction. Thus $x_{\alpha-1}\succ H_{1}$. Clearly $x_{i}\succ H_{1}$ for $i\neq j$. Since $S$ is a minimum cut set, $N_{H_{1}}(v)\neq\emptyset$. Let $u^{\prime}\in N_{H_{1}}(v)$. Lemma 4(2) yields that $u^{\prime}\succ S-\{x_{j}\}$. Consider $G+u^{\prime}a_{1}$. By the same arguments as considering $G+ua_{1}$, we have a contradiction. Thus Case 1 cannot occur.

Case 2 : $p=2$.

Let $\{a_{1},a_{2}\}=\cup^{2}_{i=1}I_{i}$. By Equation 2, $|S-I|=p=2$. Since $|S|\geq 4$, it follows that $|S\cap I|\geq 2$, in particular, $S\cap I\neq\emptyset$ and $\{a_{1},a_{2}\}$ does not dominate $G$. Consider $G+a_{1}a_{2}$. Lemma 5(1) gives that $|D_{a_{1}a_{2}}\cap\{a_{1},a_{2}\}|=1$ and $|D_{a_{1}a_{2}}\cap S|=1$. Without loss of generality, let $a_{1}\in D_{a_{1}a_{2}}$. By the connectedness of $(G+a_{1}a_{2})[D_{a_{1}a_{2}}]$, $|(S-I)\cap D_{a_{1}a_{2}}|=1$. Let $\{u\}=(S-I)\cap D_{a_{1}a_{2}}$. Thus $ua_{1}\in E(G),ua_{2}\notin E(G)$ and $u\succ S\cap I$. Let $v\in S-(I\cup\{u\})$. Lemma 4(2) implies that $uv\notin E(G)$. Therefore $a_{1}v\in E(G)$

Subcase 2.1 : $\alpha_{1}=1$ and $\alpha_{2}=1$.

Without loss of generality, let $a_{1}\in I_{1}$ and $a_{2}\in I_{2}$. Since $|S\cap I|\geq 2$, there exist $a_{3},a_{4}\in S\cap I$. Consider $G+a_{3}a_{4}$. Lemma 1(2) gives that $D_{a_{3}a_{4}}\cap\{a_{3},a_{4}\}\neq\emptyset$. To dominate $a_{1}$, $D_{a_{3}a_{4}}\neq\{a_{3},a_{4}\}$. Without loss of generality, let $a_{3}\in D_{a_{3}a_{4}}$. Lemma 1(1) implies that $|D_{a_{3}a_{4}}-\{a_{3}\}|=1$. Let $y\in D_{a_{3}a_{4}}-\{a_{3}\}$. To dominate $\{a_{1},a_{2}\}$, $y\notin\cup^{2}_{i=1}H_{i}$. By the connectedness of $(G+a_{3}a_{4})[D_{a_{3}a_{4}}]$, $y\in\{v,u\}$. Since $uv\notin E(G)$, $a_{3}u,a_{3}v\in E(G)$. Consider $G-a_{3}$ . Lemma 3(2) then implies that $D_{a_{3}}\cap\{u,v\}=\emptyset$. Lemma 4(1) implies also that $D_{a_{3}}\cap S\neq\emptyset$. So there exists $z\in D_{a_{3}}\cap(S\cap I)$. Lemma 3(1) yields that $|D_{a_{3}}-\{z\}|=1$. Let $D_{a_{3}}-\{z\}=\{z^{\prime}\}$. Since $z\in S\cap I$, $za_{1}\notin E(G)$. Thus $z^{\prime}\in H_{1}$ to dominate $a_{1}$. So $D_{a_{3}}$ does not dominate $a_{2}$, a contradiction and Subcase 2.1 cannot occur.

Subcase 2.2 : $\alpha_{1}=0$ and $\alpha_{2}=2$.

Thus $u\succ H_{1}$. Let $b_{1}\in H_{1}$. Clearly $\{a_{1},b_{1}\}$ does not dominate $G$. Consider $G+a_{1}b_{1}$. Lemma 5(1) gives that $|D_{a_{1}b_{1}}\cap S|=1$ and either $b_{1}\in D_{a_{1}b_{1}}$ or $a_{1}\in D_{a_{1}b_{1}}$. In the first case, $\{u,v\}\cap D_{a_{1}b_{1}}=\emptyset$ by Lemma 1(3). To dominate $a_{2}$, $D_{a_{1}b_{1}}\cap(S\cap I)=\emptyset$. Hence, $D_{a_{1}b_{1}}\cap S=\emptyset$, a contradiction. Therefore, $a_{1}\in D_{a_{1}b_{1}}$. To dominate $H_{1}-b_{1}$ and by the connectedness of $(G+a_{1}b_{1})[D_{a_{1}b_{1}}]$, $(D_{a_{1}b_{1}}-\{a_{1}\})\subseteq\{u,v\}$. Lemma 1(3) implies that $v\in D_{a_{1}b_{1}}$. Thus $v\succ H_{1}-b_{1}$. Let $b_{2}\in H_{1}-\{b_{1}\}$. Therefore $b_{2}\succ\{u,v\}$. Consider $G+a_{1}b_{2}$. Lemma 5(1) implies that we have $|D_{a_{1}b_{1}}\cap S|=1$ and either $a_{1}\in D_{a_{1}b_{2}}$ or $b_{2}\in D_{a_{1}b_{2}}$. In the first case, $\{u,v\}\cap D_{a_{1}b_{2}}=\emptyset$ by Lemma 1(3). By the connectedness of $(G+a_{1}b_{2})[D_{a_{1}b_{2}}]$, $(S\cap I)\cap D_{a_{1}b_{2}}=\emptyset$. Thus $D_{a_{1}b_{2}}\cap S=\emptyset$, a contradiction. Therefore, $b_{2}\in D_{a_{1}b_{2}}$. To dominate $a_{2}$, $(S\cap I)\cap D_{a_{1}b_{2}}=\emptyset$. Lemma 1(3) yields that $D_{a_{1}b_{2}}\cap\{u,v\}=\emptyset$. Therefore $D_{a_{1}b_{2}}\cap S=\emptyset$, a contradiction and Case 2 cannot occur. So $p\geq 3$ and this completes the proof. $\Box$

In view of Lemma 6, $p\geq 3$. Lemma 2 then implies that the vertices in $\cup^{2}_{i=1}I_{i}$ can be ordered as $x_{1},x_{2},...,x_{p}$ and there exists a path $y_{1},y_{2},...,y_{p-1}$ with $\{x_{i},y_{i}\}\succ_{c}G+x_{i}x_{i+1}$ for $i=1,2,...,p-1$.

Lemma 7

$y_{i}\succ S\cap I$* and $y_{i}\in S-I$ for $i=1,2,...,p-1$.*

**Proof. **Since $\{x_{i},y_{i}\}\succ_{c}G+x_{i}x_{i+1}$ for $i=1,2,...,p-1$ and $x_{i}\in I$, $y_{i}\succ S\cap I$. By the connectedness of $(G+x_{i}x_{i+1})[D_{x_{i}x_{i+1}}]$ and Lemma 5(1), $y_{i}\in S-I$ and this completes the proof. $\Box$

Lemma 7 implies that $\{y_{1},y_{2},...,y_{p-1}\}\subseteq S-I$. By Equation 2, $|(S-I)-\{y_{1},y_{2},$ $...,y_{p-1}\}|=1$. Let $\{y_{p}\}=(S-I)-\{y_{1},y_{2},...,y_{p-1}\}$.

Lemma 8

For $i,j\in\{2,3,...,p\}$, if $y_{p}x_{i},y_{p}x_{j}\in E(G)$, then $y_{i-1}y_{j-1}\in E(G)$.

**Proof. **Consider $G+x_{i}x_{j}$. Lemma 5(1) yields that $|D_{x_{i}x_{j}}\cap\{x_{i},x_{j}\}|=1$ and $|D_{x_{i}x_{j}}\cap S|=1$. Without loss of generality, let $x_{i}\in D_{x_{i}x_{j}}$ and $\{a\}=D_{x_{i}x_{j}}\cap S$. By the connectedness of $(G+x_{i}x_{j})[D_{x_{i}x_{j}}]$, $a\in S-I$. Since $x_{j}\succ(S-I)-\{y_{j-1}\}$, it follows by Lemma 1(3) that $a=y_{j-1}$. Since $y_{i-1}x_{i}\notin E(G)$, $y_{j-1}y_{i-1}\in E(G)$ and this completes the proof. $\Box$

Lemma 9

$\alpha_{1},\alpha_{2}>0$.

**Proof. **By the assumption that $\alpha_{1}\leq\alpha_{2}$, we suppose that $\alpha_{1}=0$. Clearly $\{x_{1},x_{2},...,x_{p}\}\subseteq H_{2}$ and $y_{i}\succ H_{1}$ for $i=1,2,...,p-1$. Since $S$ is a minimum cut set, $N_{H_{1}}(y_{p})\neq\emptyset$. Let $b\in N_{H_{1}}(y_{p})$. Therefore $b\succ S-I$. Consider $G+x_{1}b$. Lemma 5(1) implies that $|D_{x_{1}b}\cap S|=1$ and either $b\in D_{x_{1}b}$ or $x_{1}\in D_{x_{1}b}$. Suppose that $b\in D_{x_{1}b}$. To dominate $x_{2}$, $D_{x_{1}b}\cap(S-I)\neq\emptyset$. Lemmas 2 and 1(3) then imply that $D_{x_{1}b}\cap(S-I)=\{y_{p}\}$. So $y_{p}\succ\{x_{2},x_{3},...,x_{p}\}$. Lemma 8 gives, further, that $G[y_{1},y_{2},...,y_{p-1}]$ is a clique. Lemma 7 then implies that $y_{i}\succ S\cap I$ for $i=1,2,...,p-1$. By Lemma 4(2), $y_{i}y_{p}\notin E(G)$ for $i=1,2,...,p-1$. Therefore $y_{1}y_{p}\notin E(G)$. Because $\{x_{1},y_{1}\}\succ_{c}G+x_{1}x_{2}$, $x_{1}y_{p}\in E(G)$, contradicting Lemma 1(3). Therefore, $x_{1}\in D_{x_{1}b}$. By the connectedness of $(G+x_{1}b)[D_{x_{1}b}]$, $D_{x_{1}b}\cap(S\cap I)=\emptyset$. Lemma 1(3) implies that $D_{x_{1}b}\cap(S-I)=\emptyset$. Thus $D_{x_{1}b}\cap S=\emptyset$ contradicting Lemma 5(1) and this completes the proof. $\Box$

Theorem 6

Let $G$ be a maximal $3$-$\gamma_{c}$-vertex critical graph. If $\alpha=\kappa$, then $G-S$ contains at least one singleton component for all minimum cut set $S$.

**Proof. **Let a graph $G$ be a maximal $3$-$\gamma_{c}$-vertex critical graph with $\alpha=\kappa$. By Equation 2, $|S-I|=\alpha_{1}+\alpha_{2}$. Suppose that there is no singleton component of $G-S$, in particular, $|H_{i}|>1$ for $i=1,2$. Let $\alpha_{1}+\alpha_{2}=p$. Lemma 6 implies that $p\geq 3$. Lemma 9 gives also that $0<\alpha_{1}\leq\alpha_{2}$. We also define $x_{1},x_{2},...,x_{p}$ a path $y_{1},y_{2},...,y_{p-1}$ and a vertex $y_{p}$ by the same as in the previous lemmas.

Suppose there exist $x_{i},x_{j}$ for $i,j\in\{2,3,...,p\}$ such that $y_{p}\in D_{x_{i}x_{j}}$. Lemma 1(1) and (2) then implies that $D_{x_{i}x_{j}}=\{x_{i},y_{p}\}$ or $D_{x_{i}x_{j}}=\{x_{j},y_{p}\}$. Without loss of generality, let $D_{x_{i}x_{j}}=\{x_{j},y_{p}\}$. Thus $y_{p}\succ\{x_{1},x_{2},...,x_{p}\}-\{x_{i}\}$. Since $\{x_{i},y_{i}\}\succ_{c}G+x_{i}x_{i+1}$, $y_{i}y_{p}\in E(G)$. Lemma 8 yields that $G[\{y_{1},y_{2},...,y_{p-1}\}-\{y_{i-1}\}]$ is a clique. Since $y_{i}y_{i-1}\in E(G)$, $y_{i}\succ S-I$. Lemma 7 implies that $y_{i}\succ S\cap I$. Therefore $y_{i}\succ S$, contradicting Lemma 4(2). Hence, $y_{p}\notin D_{x_{i}x_{j}}$ for any $i,j\in\{2,3,...,p\}$. By the same arguments as in the proof of Lemma 8, $G[\{y_{1},y_{2},...,y_{p-1}\}]$ is a clique. As $y_{i}\succ S\cap I$, by Lemma 4(2), we must have $y_{i}y_{p}\notin E(G)$ for $i\in\{1,2,...,p-1\}$. Since $\{x_{i},y_{i}\}\succ_{c}G+x_{i}x_{i+1}$ for $i\in\{1,2,...,p-1\}$, $x_{i}y_{p}\in E(G)$. So $x_{1}\succ S-I$. By Lemma 4(2), $S\cap I\neq\emptyset$ as otherwise $x_{1}\succ S$. Let $x_{1}\in H_{i}$ for some $i\in\{1,2\}$. Consider $G-x_{1}$. Since $|H_{j}|>1$ for $j=1,2$, neither $D_{x_{1}}\subseteq H_{1}$ nor $D_{x_{1}}\subseteq H_{2}$. Lemma 4(1) gives, further, that $D_{x_{1}}\cap S\neq\emptyset$. Lemma 3(2) implies that $D_{x_{1}}\cap(S-I)=\emptyset$. Thus $D_{x_{1}}\cap(S\cap I)\neq\emptyset$. Let $u_{1}\in D_{x_{1}}\cap(S\cap I)$. Lemma 3(1) implies also that $|D_{x_{1}}-\{u_{1}\}|=1$. Let $\{w\}=D_{x_{1}}-\{u_{1}\}$. If $w\in H_{i}$, then $u_{1}\succ H_{3-i}$. Since $u_{1}\in I$, $\alpha_{3-i}=0$ contradicting Lemma 9. So $w\in H_{3-i}$ and $u_{1}\succ H_{i}-x_{1}$. Since $u_{1}\in I$, $I_{i}=\{x_{1}\}$. It follows that $\{x_{2},x_{3},...,x_{p}\}\subseteq H_{3-i}$.

Claim 1 : There is no $u\in S\cap I$ such that $u\succ S-I$.

Suppose $u\succ S-I$. Consider $G-u$. Lemma 4(1) implies that $D_{u}\cap S\neq\emptyset$. Lemma 3(2) implies also that $D_{u}\cap(S-I)=\emptyset$. Thus there exists $u^{\prime}\in D_{u}\cap(S\cap I)$. Lemma 3(1) gives that $|D_{u}-\{u^{\prime}\}|=1$. Let $\{z\}=D_{u}-\{u^{\prime}\}$. To dominate $x_{1}$, $z\in H_{i}$. So $D_{u}$ does not dominate $I_{3-i}$, a contradiction and thus establishing Claim 1.

Claim 1 together with Lemma 7 yield that $y_{p}$ is not adjacent to any vertex in $S\cap I$. We now have $y_{p}$ an isolated vertex in $S$.

Claim 2 : $y_{1}\succ H_{i}$.

Suppose $y_{1}$ is not adjacent to $b_{1}\in H_{i}$. Consider $G+b_{1}x_{2}$. We see that $b_{1}y_{1},x_{2}y_{1}\notin E(G)$. Lemma 5(1) gives that $|D_{b_{1}x_{2}}\cap S|=1$and either $b_{1}\in D_{b_{1}x_{2}}$ or $x_{2}\in D_{b_{1}x_{2}}$. If $b_{1}\in D_{b_{1}x_{2}}$, then $(S-\{y_{1},y_{p}\})\cap D_{b_{1}x_{2}}=\emptyset$ to dominate $I_{3-i}$. Since $y_{p}x_{2}\in E(G)$, by Lemma1(3), $y_{p}\notin D_{b_{1}x_{2}}$. By the connectedness of $(G+b_{1}x_{2})[D_{b_{1}x_{2}}]$, $y_{1}\notin D_{b_{1}x_{2}}$. Therefore $D_{b_{1}x_{2}}\cap S=\emptyset$, a contradiction. Hence, $x_{2}\in D_{b_{1}x_{2}}$. To dominate $I_{3-i}\cup(S\cap I)$, $D_{b_{1}x_{2}}\cap\{y_{2},y_{3},...,y_{p}\}=\emptyset$. By the connectedness of $(G+b_{1}x_{2})[D_{b_{1}x_{2}}]$, $((S\cap I)\cup\{y_{1}\})\cap D_{b_{1}x_{2}}=\emptyset$. Therefore, $D_{b_{1}x_{2}}\cap S=\emptyset$, a contradiction and we settle Claim 2.

Let $b_{1}\in H_{i}-\{x_{1}\}$. Recall that $u_{1}\succ H_{i}-x_{1}$. Clearly $b_{1}u_{1}\in E(G)$. By Claim 2 and Lemma 2, $b_{1}\succ\{y_{1},y_{2},...,y_{p-1}\}\cup\{u_{1}\}$. Consider $G-b_{1}$. Lemma 4(1) implies that $D_{b_{1}}\cap S\neq\emptyset$. Lemma 3(2) gives that $D_{b_{1}}\cap(\{y_{1},y_{2},...,y_{p-1}\}\cup\{u_{1}\})=\emptyset$. If there is $u_{2}\in D_{b_{1}}\cap((S\cap I)-\{u_{1}\})$, then, by Lemma 3(1), let $\{y^{\prime}\}=D_{b_{1}}-\{u_{2}\}$. To dominate $x_{1}$, $y^{\prime}\in H_{i}$. Thus $D_{b_{1}}$ does not dominate $x_{2}$, a contradiction. Therefore, $\{y_{p}\}=D_{b_{1}}\cap S$. Since $y_{p}$ is an isolated vertex in $S$, by Lemma 5(2), at least one of $C_{i}$ is a singleton component contradicting the assumption. We then finish the proof. $\Box$

By Theorem 6, we obtain the following corollary.

Corollary 1

Let $G$ be a maximal $3$-$\gamma_{c}$-vertex critical graph. If $\alpha=\kappa$, then $\kappa=\delta$.

**Proof. **By Theorem 6, we have at least one singleton component $C_{i}$ of $G-S$. Let $\{c\}=V(C_{i})$. Thus $N_{G}(c)\subseteq S$. Hence, $\delta\leq deg_{G}(c)\leq|S|=\kappa\leq\delta$. $\Box$

We next construct the class $\mathcal{G}_{4}(s)$ of maximal $3$-$\gamma_{c}$-vertex critical graphs which $\alpha<\kappa$ and $\kappa<\delta$ in order to show that the condition $\alpha=\kappa$ is necessary to prove Corollary 1. Let $R,T,W$ and $Z$ be non-empty disjoint vertex sets where $R=\{r_{1},r_{2},...,r_{s}\},T=\{t_{1},t_{2},...,t_{s}\},W=\{w_{1},w_{2},...,w_{s}\},Z=\{z_{1},z_{2},...,z_{s}\}$ and $s\geq 3$. A graph $G$ in the class $\mathcal{G}_{4}(s)$ can be constructed from $R,T,W$ and $Z$ by adding edges according the join operations :

• •

for $1\leq i\leq s$, $r_{i}\vee R\cup T\cup W-\{r_{i},t_{i}\}$,

• •

$t_{i}\vee R\cup W\cup Z-\{w_{i},r_{i}\}$,

• •

$w_{i}\vee R\cup T\cup Z-\{t_{i},z_{i}\}$,

• •

$z_{i}\vee Z\cup T\cup W=\{z_{i},w_{i}\}$ and

• •

adding edges so the the vertices in $R$ and $Z$ form cliques.

Observe that, for $1\leq i\leq s$, $N_{G}(r_{i})=R\cup T\cup W-\{r_{i},t_{i}\}$, $N_{G}(t_{i})=R\cup W\cup Z-\{w_{i},r_{i}\}$, $N_{G}(w_{i})=R\cup T\cup Z-\{t_{i},z_{i}\}$ and $N_{G}(z_{i})=Z\cup T\cup W=\{z_{i},w_{i}\}$. Observe also that $T$ and $W$ are independent sets. A graph $G$ is illustrated in Figure 4 where double lines between two sets means joining each vertex in one set to every vertex in the other set.

\begin{picture}(0.0,9.0)\par\put(3.2,6.5){\circle*{0.2}} \put(4.2,6.5){\circle*{0.2}} \put(6.8,6.5){\circle*{0.2}} \par\put(3.2,2.05){\circle*{0.2}} \put(4.2,1.45){\circle*{0.2}} \put(6.8,0.5){\circle*{0.2}} \par\put(10.2,0.5){\circle*{0.2}} \put(12.8,1.45){\circle*{0.2}} \put(13.8,2.05){\circle*{0.2}} \par\put(10.2,6.5){\circle*{0.2}} \put(12.8,6.5){\circle*{0.2}} \put(13.8,6.5){\circle*{0.2}} \par\par\put(3.2,2.05){\line(0,1){0.4}}\put(3.2,2.7){\line(0,1){0.4}}\put(3.2,3.35){\line(0,1){0.4}}\put(3.2,4.0){\line(0,1){0.4}}\put(3.2,4.65){\line(0,1){0.4}}\put(3.2,5.3){\line(0,1){0.4}}\put(3.2,5.95){\line(0,1){0.4}} \put(4.2,1.45){\line(0,1){0.4}}\put(4.2,2.1){\line(0,1){0.4}}\put(4.2,2.75){\line(0,1){0.4}}\put(4.2,3.4){\line(0,1){0.4}}\put(4.2,4.05){\line(0,1){0.4}}\put(4.2,4.7){\line(0,1){0.4}}\put(4.2,5.35){\line(0,1){0.4}}\put(4.2,6.0){\line(0,1){0.4}} \put(6.8,0.5){\line(0,1){0.4}}\put(6.8,1.1){\line(0,1){0.4}}\put(6.8,1.7){\line(0,1){0.4}}\put(6.8,2.3){\line(0,1){0.4}}\put(6.8,2.9){\line(0,1){0.4}}\put(6.8,3.5){\line(0,1){0.4}}\put(6.8,4.1){\line(0,1){0.4}}\put(6.8,4.7){\line(0,1){0.4}}\put(6.8,5.3){\line(0,1){0.4}}\put(6.8,5.9){\line(0,1){0.4}} \par\put(10.2,0.5){\line(0,1){0.4}}\put(10.2,1.1){\line(0,1){0.4}}\put(10.2,1.7){\line(0,1){0.4}}\put(10.2,2.3){\line(0,1){0.4}}\put(10.2,2.9){\line(0,1){0.4}}\put(10.2,3.5){\line(0,1){0.4}}\put(10.2,4.1){\line(0,1){0.4}}\put(10.2,4.7){\line(0,1){0.4}}\put(10.2,5.3){\line(0,1){0.4}}\put(10.2,5.9){\line(0,1){0.4}} \put(12.8,1.45){\line(0,1){0.4}}\put(12.8,2.1){\line(0,1){0.4}}\put(12.8,2.75){\line(0,1){0.4}}\put(12.8,3.4){\line(0,1){0.4}}\put(12.8,4.05){\line(0,1){0.4}}\put(12.8,4.7){\line(0,1){0.4}}\put(12.8,5.35){\line(0,1){0.4}}\put(12.8,6.0){\line(0,1){0.4}} \put(13.8,2.05){\line(0,1){0.4}}\put(13.8,2.7){\line(0,1){0.4}}\put(13.8,3.35){\line(0,1){0.4}}\put(13.8,4.0){\line(0,1){0.4}}\put(13.8,4.65){\line(0,1){0.4}}\put(13.8,5.3){\line(0,1){0.4}}\put(13.8,5.95){\line(0,1){0.4}} \par\put(3.2,2.05){\line(1,0){0.4}}\put(3.8,2.05){\line(1,0){0.4}}\put(4.4,2.05){\line(1,0){0.4}}\put(5.0,2.05){\line(1,0){0.4}}\put(5.6,2.05){\line(1,0){0.4}}\put(6.2,2.05){\line(1,0){0.4}}\put(6.8,2.05){\line(1,0){0.4}}\put(7.4,2.05){\line(1,0){0.4}}\put(8.0,2.05){\line(1,0){0.4}}\put(8.6,2.05){\line(1,0){0.4}}\put(9.2,2.05){\line(1,0){0.4}}\put(9.8,2.05){\line(1,0){0.4}}\put(10.4,2.05){\line(1,0){0.4}}\put(11.0,2.05){\line(1,0){0.4}}\put(11.6,2.05){\line(1,0){0.4}}\put(12.2,2.05){\line(1,0){0.4}}\put(12.8,2.05){\line(1,0){0.4}}\put(13.4,2.05){\line(1,0){0.4}} \put(4.4,1.45){\line(1,0){0.4}}\put(5.0,1.45){\line(1,0){0.4}}\put(5.6,1.45){\line(1,0){0.4}}\put(6.2,1.45){\line(1,0){0.4}}\put(6.8,1.45){\line(1,0){0.4}}\put(7.4,1.45){\line(1,0){0.4}}\put(8.0,1.45){\line(1,0){0.4}}\put(8.6,1.45){\line(1,0){0.4}}\put(9.2,1.45){\line(1,0){0.4}}\put(9.8,1.45){\line(1,0){0.4}}\put(10.4,1.45){\line(1,0){0.4}}\put(11.0,1.45){\line(1,0){0.4}}\put(11.6,1.45){\line(1,0){0.4}}\put(12.2,1.45){\line(1,0){0.4}} \put(6.8,0.5){\line(1,0){0.4}}\put(7.4,0.5){\line(1,0){0.4}}\put(8.0,0.5){\line(1,0){0.4}}\put(8.6,0.5){\line(1,0){0.4}}\put(9.2,0.5){\line(1,0){0.4}}\put(9.8,0.5){\line(1,0){0.4}} \par\put(5.0,6.5){\oval(5.0,1.7)} \put(12.0,6.5){\oval(5.0,1.7)} \par\put(2.5,0.0){\dashbox(0.2)(5.0,2.5)[s]{}} \put(9.5,0.0){\dashbox(0.2)(5.0,2.5)[s]{}} \par\put(5.0,2.5){\line(0,1){3.15}} \put(5.1,2.5){\line(0,1){3.15}} \par\put(12.0,2.5){\line(0,1){3.15}} \put(11.9,2.5){\line(0,1){3.15}} \par\put(7.5,0.8){\line(1,0){2.0}} \put(7.5,0.9){\line(1,0){2.0}} \par\put(7.5,2.5){\line(1,2){2.0}} \put(7.5,2.35){\line(1,2){2.0}} \par\put(9.5,2.5){\line(-1,2){2.0}} \put(9.5,2.35){\line(-1,2){2.0}} \put(11.2,1.0){$...$} \put(11.2,6.5){$...$} \put(5.3,1.0){$...$} \put(5.3,6.5){$...$} \par\put(3.2,6.7){$r_{1}$} \put(4.2,6.7){$r_{2}$} \put(6.8,6.7){$r_{s}$} \par\put(3.0,1.55){$t_{1}$} \put(4.0,0.95){$t_{2}$} \put(6.6,0.0){$t_{s}$} \par\put(14.0,1.55){$w_{1}$} \put(13.0,0.95){$w_{2}$} \put(10.4,0.0){$w_{s}$} \par\put(13.8,6.7){$z_{1}$} \put(12.8,6.7){$z_{2}$} \put(10.2,6.7){$z_{s}$} \par\par\put(5.0,7.5){$R$} \put(5.0,-0.5){$T$} \put(12.0,-0.5){$W$} \put(12.0,7.5){$Z$} \put(5.0,-1.9){\footnotesize{Figure 4} : A graph $G$ in the class $\mathcal{G}_{4}(s)$} \end{picture}

...$$...$$...$$...

r_{1}$$r_{2}$$r_{s}

t_{1}$$t_{2}$$t_{s}

w_{1}$$w_{2}$$w_{s}

z_{1}$$z_{2}$$z_{s}

R$$T$$W$$ZFigure 4 : A graph $G$ in the class $\mathcal{G}_{4}(s)$

Lemma 10

If $G\in\mathcal{G}_{4}(s)$, then $G$ is a maximal $3$-$\gamma_{c}$-vertex critical graph.

**Proof. **We see that $\{r_{1},t_{2},w_{2}\}\succ_{c}G$. Thus $\gamma_{c}(G)\leq 3$. Suppose there exist $u,v\in V(G)$ such that $\{u,v\}\succ_{c}G$. Let $i\in\{1,..,s\}$. Suppose that $u=r_{i}$. To dominate $Z$, $v\notin R$. If $v\in T$, then, by connected, $v\neq t_{i}$. Thus $\{u,v\}$ does not dominate $t_{i}$. To dominate $Z$, $v\notin W$. Clearly $v\in Z$ but $G[\{u,v\}]$ is not connected, a contradiction. Hence, $\{u,v\}\cap R=\emptyset$. By symmetric, $\{u,v\}\cap Z=\emptyset$. Therefore $\{u,v\}\subseteq T\cup W$. Without loss of generality, $u=t_{i}$. By connected, $v\in W-\{w_{i}\}$. Thus $\{u,v\}$ does not dominate $w_{i}$. Therefore $\gamma_{c}(G)=3$.

To establish the criticality, let $u,v$ be a pair of non-adjacent vertices of $G$. For $1\leq i\leq s$, if $\{u,v\}=\{r_{i},t_{i}\}$, then $D_{uv}=\{r_{i},t_{i}\}$. If $\{u,v\}=\{t_{i},w_{i}\}$, then $D_{uv}=\{t_{i},w_{i}\}$. If $\{u,v\}=\{w_{i},z_{i}\}$, then $D_{uv}=\{w_{i},z_{i}\}$. For $1\leq i\neq j\leq s$, if $\{u,v\}=\{t_{i},t_{j}\}$, then $D_{uv}=\{t_{i},r_{j}\}$. If $\{u,v\}=\{w_{i},w_{j}\}$, then $D_{uv}=\{w_{i},z_{j}\}$. If $\{u,v\}=\{r_{i},z_{l}\}$ where $l\in\{1,2,...,s\}$, then $D_{uv}=\{r_{i},z_{l}\}$. Thus $G$ is a $3$-$\gamma_{c}$-edge critical graph. Let $v\in V(G)$. For $1\leq i\neq j\leq s$, if $u=r_{i}$, then $D_{v}=\{t_{i},z_{j}\}$. If $v=t_{i}$, then $D_{v}=\{t_{j},r_{i}\}$. If $v=w_{i}$, then $D_{v}=\{z_{i},w_{j}\}$. Finally, if $v=z_{i}$, then $D_{v}=\{w_{i},r_{j}\}$. Therefore $G$ is a maximal $3$-$\gamma_{c}$-vertex critical graph and this completes the proof. $\Box$

We see that $G$ has $T$ a maximum independent set and $T\cup W$ a mimimum cut set. Thus $\alpha=s<2s=\kappa$. We, further, have that $G$ is a regular graph with $deg_{G}(v)=3s-2$ for all $v\in V(G)$. Since $s\geq 3$, $\delta=3s-2>2s=\kappa$. Hence, the condition $\alpha=\kappa$ is necessary to prove Corollary 1.

We conclude this paper by establishing hamiltonian property of maximal $3$-$\gamma_{c}$-vertex critical graphs. We prove the following result by using Theorem 1.

Corollary 2

For any $3$-connected maximal $3$-$\gamma_{c}$-vertex critical graph $G$. If $\kappa<\delta$, then $G$ is hamiltonian connected.

**Proof. **Suppose that $\kappa<\delta$. Theorem 5 and Corollary 1 then imply that $\alpha<\kappa$. By Theorem 1, $G$ is hamiltonian connected. $\Box$

However, we are sure that every $3$-connected maximal $3$-$\gamma_{c}$-vertex critical graph is hamiltonian connected. To prove this, by Theorem 1 and Corollary 2,we need only prove the following conjecture.

Conjecture 1

Let $G$ be a $3$-connected maximal $3$-$\gamma_{c}$-vertex critical graph with $\alpha=\kappa=\delta$. Then $G$ is hamiltonian connected.