Independent transversal domination number of a graph

TL;DR

This paper investigates the independent transversal domination number in graphs, disproves a previous conjecture, confirms a related one, and addresses an open problem in the field.

Contribution

It proves the full validity of Conjecture 2 for connected bipartite graphs, corrects a previous theorem, and solves an open problem on independent transversal total domination.

Findings

Conjecture 1 is false in general.

Conjecture 2 is confirmed for all connected bipartite graphs.

A problem on independent transversal total domination is solved.

Abstract

Let $G=(V, E)$ be a graph. A set $S\subseteq V(G)$ is a {\it dominating set} of $G$ if every vertex in $V\setminus S$ is adjacent to a vertex of $S$. The {\it domination number} of $G$, denoted by $\gamma(G)$, is the cardinality of a minimum dominating set of $G$. Furthermore, a dominating set $S$ is an {\it independent transversal dominating set} of $G$ if it intersects every maximum independent set of $G$. The {\it independent transversal domination number} of $G$, denoted by $\gamma_{it}(G)$, is the cardinality of a minimum independent transversal dominating set of $G$. In 2012, Hamid initiated the study of the independent transversal domination of graphs, and posed the following two conjectures: Conjecture 1. If $G$ is a non-complete connected graph on $n$ vertices, then $\gamma_{it}(G)\leq\lceil\frac{n}{2}\rceil$. Conjecture 2. If G is a connected bipartite graph, then $\gamma_{it}(G)$ is either $\gamma(G)$ or $\gamma(G)+1$. We show that Conjecture 1 is not true in general. Very recently, Conjecture 2 is partially verified to be true by Ahangar, Samodivkin, Yero. Here, we prove the full statement of Conjecture 2. In addition, we give a correct version of a theorem of Hamid. Finally, we answer a problem posed by Mart\'{i}nez, Almira, and Yero on the independent transversal total domination of a graph.

Equations

$$\alpha (G)+\beta (G)=n.$$

$${\alpha }^{\prime }(G)\le \min \{\frac{n}{2},\beta (G)\}.$$

$${\alpha }^{\prime }(G)=\beta (G).$$

$$\gamma (G)\le \alpha (G),$$

$$\gamma (G)\le \beta (G).$$

$$\max \{\gamma (G),{\tau }_i(G)\}\le {\gamma }_{it}(G).$$

$$\alpha (G)\ge \frac{n}{2}\ge {\alpha }^{\prime }(G).$$

$$\alpha (G)={\alpha }^{\prime }(G)=\beta (G)=\frac{n}{2}.$$

$$\alpha (G-x)=|Y|=\frac{n}{2}>\frac{n}{2}-1=|X\setminus \{x\}|\ge {\alpha }^{\prime }(G-x).$$

$$\frac{n}{2}-1\le \gamma (G)\le \frac{n}{2}.$$

$${\gamma }_{tt}(G)\ge {\gamma }_{it}(G)\ge n-9\sqrt{nlogn}.$$

Full text

Independent transversal domination number of a graph

††thanks: Research supported by NSFC (No. 11571294)

Hongting Wang, Baoyindureng Wu 111Corresponding author. Email: baoywu@163.com (B. Wu) , Xinhui An

College of Mathematics and System Sciences, Xinjiang University

Urumqi, Xinjiang 830046, P.R.China

Abstract: Let $G=(V,E)$ be a graph. A set $S\subseteq V(G)$ is a dominating set of $G$ if every vertex in $V\setminus S$ is adjacent to a vertex of $S$. The domination number of $G$, denoted by $\gamma(G)$, is the cardinality of a minimum dominating set of $G$. Furthermore, a dominating set $S$ is an independent transversal dominating set of $G$ if it intersects every maximum independent set of $G$. The independent transversal domination number of $G$, denoted by $\gamma_{it}(G)$, is the cardinality of a minimum independent transversal dominating set of $G$. In 2012, Hamid initiated the study of the independent transversal domination of graphs, and posed the following two conjectures:

Conjecture 1. If $G$ is a non-complete connected graph on $n$ vertices, then $\gamma_{it}(G)\leq\lceil\frac{n}{2}\rceil$.

Conjecture 2. If G is a connected bipartite graph, then $\gamma_{it}(G)$ is either $\gamma(G)$ or $\gamma(G)+1$.

We show that Conjecture 1 is not true in general. Very recently, Conjecture 2 is partially verified to be true by Ahangar, Samodivkin, Yero. Here, we prove the full statement of Conjecture 2. In addition, we give a correct version of a theorem of Hamid. Finally, we answer a problem posed by Martínez, Almira, and Yero on the independent transversal total domination of a graph.

Keywords: Domination number; Independence number; Covering number; Independent transversal domination number

1 Introduction

We consider undirected finite simple graphs only, and follow the notations and terminology in [3]. Let $G=(V(G),E(G))$ be a graph. The order of $G$ is $|V(G)|$. For a vertex $v\in V(G)$, the set of neighbors is denoted by $N(v)$. The degree of $v$, denoted by $d(v)$, is number of edges incident with $v$ in $G$. Since $G$ is simple, $d(v)=|N(v)|$. The minimum degree of $G$, denoted by $\delta(G)$, is $\min\{d(v):\ v\in V(G)\}$. For a set $S\subseteq V(G)$, let $N(S)=\cup_{v\in S}N(v)$. A set $S$ is said to be an independent set of $G$, if no pair of vertices of $S$ are adjacent in $G$. The independence number of $G$, denoted by $\alpha(G)$, is the cardinality of a maximum independent set of $G$. We denote by $\Omega(G)$ the set of all maximum independent sets of $G$.

On the other hand, $S$ is said to be a vertex covering of $G$ if every edge of $G$ is incident with some vertex of $S$ in $G$. It is easy to see that $S$ is an independent set of $G$ if and only if $V(G)\setminus S$ is a vertex covering of $G$. The vertex covering number, denoted by $\beta(G)$, is the cardinality of a minimum vertex covering of $G$. So, for any graph $G$ of order $n$,

[TABLE]

A set $M\subseteq E(G)$ is said to be a matching of $G$ if no pair of edges of $M$ have a common end vertex. The matching number of $G$, denoted by $\alpha^{\prime}(G)$, is the cardinality of a maximum matching of $G$. It is obvious that for a graph $G$ of order $n$,

[TABLE]

The well-known König-Egerváry theorem states that for a bipartite graph $G$,

[TABLE]

A set $S\subseteq V(G)$ is a dominating set of $G$ if every vertex in $V\setminus S$ is adjacent to a vertex of $S$. The domination number of $G$, denoted by $\gamma(G)$, is the minimum cardinality of a dominating set of $G$. A minimum dominating set of a graph $G$ is called a $\gamma(G)$-set of $G$.

It is clear that for a graph $G$,

[TABLE]

and if $G$ has no isolated vertices,

[TABLE]

A set $S\subseteq V(G)$ is said to be an independent transversal of $G$ if $S\cap I\neq\emptyset$ for any $I\in\Omega(G)$. The independent transversal number of $G$, denoted by $\tau_{i}(G)$, is the cardinality of a minimum independent transversal of $G$. A dominating set $S$ of a graph $G$ is said to be an independent transversal dominating set if $S\cap I\neq\emptyset$ for any $I\in\Omega(G)$. By the definitions above, for any graph $G$,

[TABLE]

The notion of independent transversal domination was first introduced by Hamid [8] in 2012. He proved that for a graph $G$ of order $n$, $\gamma_{it}(G)\leq n$, with equality if and only if $G\cong K_{n}$.

Theorem 1.1**.**

(Hamid [8]) For a graph $G$ without isolated vertices, $\gamma_{it}(G)\leq\beta(G)+1$.

Theorem 1.2**.**

(Hamid [8]) If $G$ is a non-complete connected graph of order $n$ with $\alpha(G)\geq\frac{n}{2}$, then $\gamma_{it}(G)\leq\frac{n}{2}$.

In view of the above theorem, Hamid posed the following conjecture in [8].

Conjecture 1.1**.**

(Hamid [8]) If $G$ is a non-complete connected graph on $n$ vertices, then $\gamma_{it}(G)\leq\lceil\frac{n}{2}\rceil$.

We show that the above conjecture is not true in general. Hamid [8] showed that $\gamma(G)\leq\gamma_{it}(G)\leq\gamma(G)+\delta(G)$. In particular, $\gamma_{it}(T)$ is either $\gamma(T)$ or $\gamma(T)+1$ for a tree $T$. So, he proposed the following conjecture.

Conjecture 1.2**.**

(Hamid [8]) If G is a connected bipartite graph, then $\gamma_{it}(G)$ is either $\gamma(G)$ or $\gamma(G)+1$.

Recently, Ahangar, Samodivkin, Yero [1] proved that Conjecture 2 is valid for all unbalanced bipartite graphs.

Theorem 1.3**.**

(Ahangar, Samodivkin, Yero [1]) Let $G$ be a bipartite graph with bipartition $(X,Y)$ such that $|X|\neq|Y|$. Then, $\gamma_{it}(G)\leq\gamma(G)+1$. In particular, this is true when $G$ has odd order.

We show that the full statement of Conjecture 2 in the next section.

Complexity of independent transversal domination problem can be in [1].

2 Disproof of Conjecture 1.1 and Proof of Conjecture 1.2

First we begin with a useful observation.

Lemma 2.1**.**

Let $G$ be non-complete graph of order $n$. If $G$ is the complement of a triangle-free graph, then $\tau_{i}(G)=\beta(\overline{G})$, and thus $\gamma_{it}(G)\geq n-\alpha(\overline{G})$.

Proof.

Since $G$ is the complement of a triangle-free graph, $\alpha(G)=2$. Thus, $\Omega(G)=\{\{u,v\}:\ uv\in E(\overline{G})\}$ and an independent transversal of $G$ is a vertex covering of $\overline{G}$, $\tau_{i}(G)=\beta(\overline{G})$, and thus $\gamma_{it}(G)\geq n-\alpha(\overline{G})$. ∎

To disprove Conjecture 1.1, we recall a celebrated result on the Ramsey theory, due to Kim [10].

Theorem 2.2**.**

(Kim [10]) For sufficiently large $n$, there exists a triangle-free graph $G$ of order $n$ with $\alpha(G)\leq 9\sqrt{nlog\ n}$.

Note that the symbol $``\leq^{\prime\prime}$ in the inequality above was misprinted as $``\geq^{\prime\prime}$ in [10]. The following result is an immediate consequence of Lemma 2.1 and Theorem 2.2.

Corollary 2.3**.**

For sufficiently large $n$, there exists a non-complete graph $G$ (the complement of a triangle-free graph) of order $n$, $\gamma_{it}(G)\geq n-\alpha(\overline{G})\geq n-9\sqrt{nlog\ n}>\lceil\frac{n}{2}\rceil$.

This disproves Conjecture 1. It is straightforward to check that the complement of the Petersen graph $P_{10}$ is also a counterexample to Conjecture 1. Alon [2] gave some explicit construction of triangle-free graphs with relatively small independence numbers contrast to their orders. The complements of these graphs are also counterexamples to Conjecture 1.

Let $core(G)=\cap\{S:S\in\Omega(G)\}$ be the set of vertices belonging to all maximum independent sets, and let $\xi(G)=|core(G)|$.

Theorem 2.4**.**

(Boros, Golumbic, and Levit [4]) If $G$ is a connected graph with $\alpha(G)>\alpha^{\prime}(G)$, then $\xi(G)\geq\alpha(G)-\alpha^{\prime}(G)+1$.

It can be deduced from the theorem above that $\tau_{i}(G)=1$ for any connected graph $G$ with $\alpha(G)>\alpha^{\prime}(G)$. It is an interesting problem for characterizing graphs with $\tau_{i}(G)=1$.

Problem 2.5**.**

What is the best upper bound of $\tau_{i}(G)$ for graphs $G$ in terms of their order $n$ ?

Now, we are ready to show Conjecture 2.

Theorem 2.6**.**

If G is a connected bipartite graph, then $\gamma_{it}(G)$ is either $\gamma(G)$ or $\gamma(G)+1$.

Proof.

Let $S$ be a $\gamma(G)$-set. If $\alpha(G)>\alpha^{\prime}(G)$, then by Theorem 2.4, $\xi(G)\geq\alpha(G)-\alpha^{\prime}(G)+1>0$. So, $S\cup\{v\}$ is an independent transversal dominating set for a vertex $v\in core(G)$, and hence $\gamma_{it}(G)\leq|S\cup\{v\}|=\gamma(G)+1$.

Next, we consider the case $\alpha(G)\leq\alpha^{\prime}(G)$. Since $G$ is bipartite,

[TABLE]

Combining (7) with (1), (2), (3), we have

[TABLE]

If $\gamma(G)=\beta(G)$, then the result follows from Theorem 1.1. So, we assume that $\gamma(G)<\beta(G)$. Let $(X,Y)$ be the bipartition of $G$. By the equation (8), we have $|X|=|Y|=\frac{n}{2}$. Thus, $S\cap X\neq\emptyset$ and $S\cap Y\neq\emptyset$. Take a vertex $x\in S\cap X$, and let $\Omega_{x}$ be the set of all maximum independent set of $G$ containing $x$. In particular, $X\in\Omega_{x}$. We consider $G-x$. It is clear that $\Omega\setminus\Omega_{x}=\Omega(G-x)$ and $Y\in\Omega\setminus\Omega_{x}$. Note that

[TABLE]

By Theorem 2.4, $\xi(G-x)>0$, and let $y\in core(G-x)$. So, $S\cup\{y\}$ is an independent transversal dominating set of $G$. This shows $\gamma_{it}(G)\leq\gamma(G)+1$.

∎

3 Bipartite graphs $G$ with $\gamma_{it}(G)=\frac{n}{2}$

Hamid [8] obtained the following theorem.

Theorem 3.1**.**

(Hamid [8]) For a bipartite graph $G$ with bipartition $(X,Y)$ such that $|X|\leq|Y|$ and $\gamma(G)=|X|$, $\gamma_{it}(G)=\gamma(G)+1$ if and only if every vertex in $X$ is adjacent to at least two pendant vertices.

Actually, the theorem above is not complete, see for the counterexample in Figure 1. It is easy to see that for the graph $G$, $\gamma(G)=2=|X|$ and $\gamma_{it}(G)=3$. However, there is vertex in $X$, which has no two pendent neighbors.

Figure 1. A counterexample $G$

Next we present the correct version of the theorem.

Theorem 3.2**.**

Let $G$ be a bipartite graph with bipartition $(X,Y)$ such that $|X|\leq|Y|$ and $\gamma(G)=|X|$, then $\gamma_{it}(G)=\gamma(G)+1$ if and only if every vertex in $X$ is either a pendent vertex, or is adjacent to at least two pendant vertices.

Proof.

First, we show its necessity. By the assumption that $|X|+1=\gamma(G)+1=\gamma_{it}(G)$ and by Theorem 1.1, we have $|X|\leq\beta(G)$. On the other hand, since $X$ is a vertex covering of $G$, $\beta(G)\leq|X|$. So, $X$ is a minimum covering of $G$, implying that $Y$ is a maximum independent set of $G$.

Next, we show that $\delta(G)=1$. Suppose $\delta(G)\geq 2$. Since $G$ is bipartite and $\gamma(G)=|X|$, $X$ is a $\gamma(G)$-set. Let $u\in X$ and $v\in N(u)$. Since $\delta(G)\geq 2$, it follows that $S=(X\setminus\{u\})\cup\{v\}$ is a dominating set of $G$.

Claim 1. $S$ is an independent transversal dominating set of $G$.

Suppose it is not, and let $I$ be a maximum independent set of $G$ such that $I\cap S=\emptyset$. Since $|I|=\alpha(G)=|Y|$ and $S=(X\setminus\{u\})\cup\{v\}$, we have $I=(Y\setminus\{v\})\cup\{u\}$. However, since $d(u)\geq 2$, $u$ has a neighbor in $I$. It contradicts the fact that $I$ is an independent set of $G$. This proves the claim.

By Claim 1, we have $\gamma_{it}(G)\leq|S|=|X|=\gamma(G)$, contradicting the assumption that $\gamma_{it}(G)=|X|+1$. This proves $\delta(G)=1$.

To complete the proof of the necessity, it suffices to show that if $u\in X$ is not a pendent vertex, then $u$ has at least two pendent neighbors. Suppose, on the contrary, that $u$ has at most one pendent neighbor. Since $d(u)\geq 2$, we choose a neighbor $v$ of $u$ with degree as small as possible. Let $S=(X\setminus\{u\})\cup\{v\}$. We show that $S$ is a dominating set of $G$. If it is not, there exists a vertex $y\in Y\setminus\{v\}$ not dominated by $S$. Then $y$ must be a pendent vertex and be adjacent to $u$. Now, $u$ has two pendent neighbors $v$ and $y$, a contradiction. So, $S$ is a (minimum) dominating set of $G$. Furthermore, by the similar argument as in the proof of Claim 1, we can show that $S$ is an independent transversal dominating set of $G$. Again, we have $\gamma_{it}(G)\leq|S|=|X|=\gamma(G)$, contradicting the assumption that $\gamma_{it}(G)=\gamma(G)+1$. This shows that $u$ has at least two pendent neighbors.

To show its sufficiency, assume that every vertex in $X$ is either a pendent vertex, or is adjacent to at least two pendant vertices. Let $\{x_{1},\ldots,x_{k}\}$ be the set of all pendent vertices in $X$ for integer $k\geq 0$. Let $y_{i}$ be the neighbor of $x_{i}$ in $Y$ for each $i$. Then $\Omega(G)=\{I:I=(Y\setminus Y^{\prime})\cup X^{\prime}\}$, where $X^{\prime}\subseteq\{x_{1},\ldots,x_{k}\}$ and $Y^{\prime}=N(X^{\prime})\cap\{y_{1},\ldots,y_{k}\}$. In particular, if $X^{\prime}=\emptyset$, then $Y^{\prime}=\emptyset$, and $I=Y$.

Note that if $S$ is a minimum dominating set of $G$, then there exists $X^{\prime}\subseteq\{x_{1},\ldots,x_{k}\}$ such that $S=(X\setminus X^{\prime})\cup Y^{\prime}$, where $Y^{\prime}$ is defined as above. However, $I_{S}=(Y\setminus Y^{\prime})\cup X^{\prime}$ is a maximum independent set such that $S\cap I_{S}=\emptyset$. It means that no minimum dominating set of $G$ is an independent transversal dominating set, implying that $\gamma_{it}(G)\geq\gamma(G)+1$. On the other hand, by Theorem 2.1, $\gamma_{it}(G)\leq\beta(G)+1\leq|X|+1=\gamma(G)+1$. We conclude that $\gamma_{it}(G)=\gamma(G)+1$.

The proof is completed.

∎

By Theorem 1.2, $\gamma_{it}(G)\leq\frac{n}{2}$ for any bipartite graph $G$ without a component of order at most 2. Hamid [8] posed the following problem: characterizing all bipartite graphs $G$ for which $\gamma_{it}(G)=\frac{n}{2}$. In what follows, we partially answer the problem. If $G$ is a connected bipartite graph with $\gamma_{it}(G)=\frac{n}{2}$, then by Theorem 2.6,

[TABLE]

The corona of a graph $G$, denoted by $G\circ K_{1}$, is the graph obtained from $G$ by joining a new vertex $v^{\prime}$ to each vertex $v\in V(G)$. In 1982, Payan and Xuong [12] and independently, in 1985, Fink, Jacobson, Kinch and Roberts [7] characterized the graphs $G$ of order $n$ with $\gamma(G)=\frac{n}{2}$.

Theorem 3.3**.**

(Fink et al. [7], Payan et al. [12]) For any graph $G$ with even order $n$ having no isolated vertices $\gamma(G)=\frac{n}{2}$ if and only if the components of $G$ are $C_{4}$ or $H\circ K_{1}$ for any connected graph $H$.

Corollary 3.4**.**

Let $G$ be a connected graph of even order $n\geq 4$. If $\gamma(G)=\frac{n}{2}$, then $\gamma_{it}(G)=\frac{n}{2}$.

Proof.

Since $\gamma(G)=\frac{n}{2}$, by Theorem 3.3, either $G\cong C_{4}$ or $H\circ K_{1}$, where $H$ is a connected graph of order $\frac{n}{2}\geq 2$. It is straightforward to check that $\gamma_{it}(C_{4})=2$. If $G\cong H\circ K_{1}$ for a connected graph of order $\frac{n}{2}$, $\gamma_{it}(G)=\frac{n}{2}$, because the set of pendent vertices is the unique minimum independent transversal dominating set of $G$. So, the result follows. ∎

Theorem 3.5**.**

For a bipartite graph $G$ of even order $n$ with bipartition $(X,Y)$ and without a component of order at most 2, $\gamma_{it}(G)=\frac{n}{2}$, if one of the following cases occurs:

(1) $\gamma(G)=\frac{n}{2}$.

(2) $\gamma(G)=\frac{n}{2}-1$, and $|X|=\frac{n}{2}-1$, $|Y|=\frac{n}{2}+1$, and every vertex in $X$ is either a pendent vertex, or is adjacent to at least two pendant vertices.

Proof.

If $\gamma(G)=\frac{n}{2}$, by Corollary 3.4, $\gamma_{it}(G)=\frac{n}{2}$. If (2) holds, the result follows by Theorem 3.2. ∎

By a long and difficult proof, Hansberg and Vlolkman [9] were able to characterize even order trees $T$ with $\gamma(T)=\frac{n}{2}-1$. By Theorems 3.2, 3.3, and 3.5, the bipartite graphs $G$ with $\gamma_{it}(G)=|X|=|Y|=\frac{n}{2}$ and $\gamma(G)=\frac{n}{2}-1$ remain to be characterized. Some more effort must be used for completing this task.

4 Independent transversal total domination

A new variant of transversal in graphs was introduced very recently by Martínez, Almira, and Yero [11], called independent transversal total domination. A dominating set $S$ of a graph $G$ is called a total dominating set is $G[S]$ has no isolated vertices. The total dominating number of $G$, denoted by $\gamma_{t}(G)$, is the cardinality of a minimum total dominating set of $G$. Further, a total dominating set $S$ is called an independent transversal total dominating set if $S\cap I\neq\emptyset$ for each $I\in\Omega(G)$. The independent transversal total dominating number of $G$, denoted by $\gamma_{tt}(G)$, is the cardinality of a minimum independent transversal total dominating set of $G$. So, for any graph $G$, $\gamma_{tt}(G)\geq\gamma_{it}(G)$. Cockayne, Dawes and Hedetneimi [6] obtained the following upper on the total domination number of a connected graph in terms of the order of the graph.

Theorem 4.1**.**

(Cockayne, Dawes and Hedetneimi [6]) If $G$ is a connected graph of order $n\geq 3$, then $\gamma_{t}(G)\leq\frac{2n}{3}$.

Brigham, Carrington, and Yitary [5] characterized the connected graphs of order at least 3 with total domination number exactly two-thirds their order. Among other things, Martínez, Almira, and Yero showed that $\gamma_{tt}(G)\leq\frac{2n}{3}$ for some classes of graphs of order $n$. Based on these results, they further asked the following open problem:

$\bullet$  Is it true that $\gamma_{tt}(G)\leq\frac{2n}{3}$ for any graph of order $n$ ? If yes, then: Is it true that $\gamma_{tt}(G)=\frac{2n}{3}$ if and only if $\gamma_{t}(G)=\frac{2n}{3}$ ?

The answer for the question above is no. By Corollary 2.3, for sufficiently large $n$, there exists a graph $G$ (the complement of a triangle-free graph) of order $n$ with

[TABLE]

If $\gamma_{tt}(G)\leq\frac{2n}{3}$ holds for any graph $G$ of order $n$, we have $n-9\sqrt{nlog\ n}\leq\frac{2n}{3}$, which is impossible for any sufficiently large $n$.