Bounds on the Domination Numbers of δ-Complement Graphs

Abstract

This study examines the $\delta$-complements of graphs—a specific type of graph complement whose adjacency depends on the adjacency of the vertices with identical degrees in the original graph. More specifically, we study this type of complement regarding the domination number. We provide sharp Nordhaus–Gaddum-type bounds on the domination number of a graph and its $\delta$-complement. We also provide sharp bounds on the domination numbers of the $\delta$-complements of joined graphs and Cartesian product graphs.

1. Introduction

A dominating set D of a simple undirected graph $G=(V,E)$ is a subset of V such that each vertex not in D is adjacent to a vertex in D. The smallest cardinality among the dominating sets of G is called the domination number of G, denoted by $\gamma \left(G\right)$. A dominating set with the smallest cardinality is called a minimum dominating set. Applications of dominating sets and domination numbers exist in various fields. For instance, in ad hoc wireless networks, devices form a dominating set that ensures message delivery [1,2,3]; in social networks, we identify influential users based on their connections [4,5]; and in facility location, we can select optimal locations to cover a region [6,7]. Moreover, domination-related structural properties are also used to improve network reliability and fault diagnosis in multiprocessor systems [8].

Many properties of the domination number have been investigated; for example, Nordhaus–Gaddum-type bounds on the domination number in terms of the number of vertices were given individually by Borowiecki [9] and Jaeger [10]. Desormeaux et al. [11] provided a Nordhaus–Gaddum-type relation of the domination number in terms of minimum degrees of graphs. They also proved several results on the domination number in various parameters.

In 2022, Pai et al. [12] introduced a variant of graph complements in which the complement takes place only among the vertices of the same degree. They called it the $\delta$-complement graph $G_{\delta }$ of G. The $\delta$-complement of a graph extends the classical complement by incorporating the degree information rather than depending solely on adjacency. In 2023, Vichitkunakorn et al. [13] gave a Nordhaus–Gaddum-type relation on the chromatic number of a graph and its $\delta$-complement. The Nordhaus–Gaddum-type bounds are given in various parameters, including the number of vertices, degrees, and clique number. The chromatic numbers on the $\delta$-complement of the Cartesian product graphs were further discussed in [14].

In this work, we further study the $\delta$-complement by investigating the domination numbers of the $\delta$-complements of graphs. The domination properties in $\delta$-complements can exhibit structural behaviors that are not apparent in the original graph or its standard complement, providing further insight into the role of degree conditions in graph structure. In particular, we provide Nordhaus–Gaddum-type bounds, bounds on the join of graphs, and bounds on Cartesian products of graphs. Examples showing that most of our bounds are sharp are also given.

The followings provide an outline of the paper. Section 2 contains necessary notations, definitions, and some theorems about the domination number of a graph and its complement. Section 3, Section 4 and Section 5 contain our results. Specifically, a Nordhaus–Gaddum-type bound on the domination number is provided in Section 3, some upper bounds and exact values on the domination number of the $\delta$-complement of the join of two graphs are provided in Section 4, and some upper bounds and exact values of the $\delta$-complement of the Cartesian product of two graphs are provided in Section 5. We also provide examples relevant to each case.

2. Background

In this section, we provide necessary notations and definitions of a graph, its complement and $\delta$-complement, a dominating set, and the domination number. We also present some results on bounds and Nordhaus-Gaddum-type bounds on the domination number of the complement of a graph.

Let $G=\left(V\right(G),E(G\left)\right)$ (or $(V,E)$) be a simple undirected graph where V is the vertex set and E is the edge set. The subgraph of G induced by a subset S of V is denoted by $G\left[S\right]$. For each vertex $u\in V$, the neighborhood of u in G is denoted by $N_G\left(u\right)$, and the degree of u in G is denoted by ${deg}_G\left(u\right)$. Given two graphs G and H, the join of G and H, denoted by $G\vee H$, is a graph with $V(G\vee H)=V\left(G\right)\cup V\left(H\right)$ and $E(G\vee H)=E\left(G\right)\cup E\left(H\right)\cup \left\{\right\{u,v\}:u\in V(G)\mathrm{and}v\in V(H\left)\right\}$. The disjoint union of G and H, denoted by $G+H$, is a graph with $V(G+H)=V\left(G\right)\cup V\left(H\right)$ and $E(G+H)=E\left(G\right)\cup E\left(H\right)$. The Cartesian product of G and H, denoted by $G\square H$, is a graph with $V(G\square H)=V\left(G\right)\times V\left(H\right)$ and $E(G\square H)=\{\{w_1,w_2\}:w_1=(u_1,v_1)\in V(G\square H),w_2=(u_2,v_2)\in V(G\square H)$ where $u_1=u_2$ and $\{v_1,v_2\}\in E\left(H\right)$, or $v_1=v_2$ and $\{u_1,u_2\}\in E\left(G\right)\}$. In this paper, we denote a path and a cycle of order n by $P_n$ and $C_n$, respectively. The complement of G is denoted by $\overline{G}$.

Many results on the bounds of the domination numbers of a graph and its complement have been discovered. The findings listed below serve as inspiration for our work.

Theorem 1

([11]). If G is a graph with $\gamma \left(G\right)\ge 2$, then $\gamma \left(\overline{G}\right)\le ⌈{\displaystyle \frac{\delta \left(G\right)}{\gamma \left(G\right)-1}}⌉+1$ where $\delta \left(G\right)$ is the minimum degree of G.

Theorem 2

([11]). If G is a graph with $\gamma \left(G\right)=\gamma \left(\overline{G}\right)$, then $\gamma \left(G\right)+\gamma \left(\overline{G}\right)<4+\sqrt{\delta \left(G\right)}+\sqrt{\delta \left(\overline{G}\right)}$ where $\delta \left(G\right)$ is the minimum degree of G.

Theorem 3

([11]). Let G be a graph. If $\gamma \left(G\right)\le \gamma \left(\overline{G}\right)-2$, then G contains $K_{\gamma \left(G\right),\gamma \left(G\right)}$.

Theorem 4

([10]). For a graph G with $n\ge 2$ vertices, we have

$$3\le \gamma \left(G\right)+\gamma \left(\overline{G}\right)\le n+1$$

and

$$2\le \gamma \left(G\right)·\gamma \left(\overline{G}\right)\le n.$$

Theorem 5

([9]). For a graph G with $n\ge 1$ vertices, we have

$$2\le \gamma \left(G\right)+\gamma \left(\overline{G}\right)\le n+1$$

and

$$1\le \gamma \left(G\right)·\gamma \left(\overline{G}\right)\le n.$$

With a new variant of graph complements introduced by Pai et al. [12], we explore bounds on the domination numbers of this particular complement. A formal definition of the $\delta$-complement of a graph is given below.

Definition 1

([12]). The δ-complement of a graph G, denoted by $G_{\delta }$, is a graph obtained from G by using the same vertex set and the following edge conditions: $\{u,v\}\in E\left(G_{\delta }\right)$ if

• $deg\left(u\right)=deg\left(v\right)$ in G and $\{u,v\}\notin E\left(G\right)$;
• $deg\left(u\right)\ne deg\left(v\right)$ in G and $\{u,v\}\in E\left(G\right)$.

Some properties of the $\delta$-complement of a graph were studied in [12,13,14]. For instance, the following theorem will be used to prove one of our results in Section 5.

Theorem 6

([14]). For graphs G and H, we have ${(G\square H)}_{\delta }=(V,E)$ where $V=V(G\square H)$ and $E=E(G_{\delta }\square H_{\delta })\cup S$ where $S=\{\{w_1,w_2\}:w_1=(u_1,v_1)\in V(G\square H)$ and $w_2=(u_2,v_2)\in V(G\square H)$ where $u_1\ne u_2,v_1\ne v_2$ and ${deg}_{G\square H}\left(w_1\right)={deg}_{G\square H}\left(w_2\right)\}$.

3. Results on a Nordhaus–Gaddum-Type Bound on the Domination Number

This section contains our results on a Nordhaus–Gaddum-type bound on the domination number. Throughout the rest of the paper, we denote $n=\left|V\right(G\left)\right|$ the number of vertices in G and denote $k=k\left(G\right)$ the number of distinct degrees of vertices in G. We also denote $V_d\left(G\right)$ the set of vertices of degree d in G and $\mathcal{P}\left(G\right)$ the partition of the vertex set of G by vertex degree, that is,

$$\mathcal{P}\left(G\right)=\{V_d\left(G\right):d\mathrm{is}\mathrm{the}\mathrm{degree}\mathrm{of}\mathrm{a}\mathrm{vertex}\mathrm{in}G\}.$$

We observe that $\gamma \left(G_{\delta }\right)=1$ if and only if there exists a vertex v such that $N_G\left(v\right)=\{u\in V\left(G\right):deg\left(u\right)\ne deg\left(v\right)\}$.

Lemma 1.

There are infinitely many graphs G that $\gamma \left(G\right)=\gamma \left(G_{\delta }\right)=1$.

Proof. 

For $n\ge 5$, let $G=u\vee C_{n-1}$. Since u is adjacent to all other vertices in G, we have $\gamma \left(G\right)=1$. In addition, u is the only vertex of degree $n-1$ in G, so u is also adjacent to all other vertices in $G_{\delta }$. Hence, $\gamma \left(G_{\delta }\right)=1$. See Figure 1 for an example. □

Lemma 2.

There are infinitely many graphs G that $\gamma \left(G\right)=\gamma \left(G_{\delta }\right)={\displaystyle \frac{n+k}{2}}$ where n is an order of G and k is the number of distinct degrees of G.

Proof. 

For an even number $n\ge 4$, let $G=K_r+\overline{K_r}$ where $r={\displaystyle \frac{n}{2}}$. We have $G_{\delta }=\overline{K_r}+K_r\cong G$. Notice that G has r vertices of degree 0 and r vertices of degree $r-1>0$. So, $k=2$. It is easy to see that $\gamma \left(G\right)=\gamma \left(G_{\delta }\right)=r+1={\displaystyle \frac{n+k}{2}}$. □

Theorem 7.

Let G be a graph of order n having k distinct degrees. We have $2\le \gamma \left(G\right)+\gamma \left(G_{\delta }\right)\le n+k$ and $1\le \gamma \left(G\right)·\gamma \left(G_{\delta }\right)\le {\left({\displaystyle \frac{n+k}{2}}\right)}^2$. Furthermore, the bounds are sharp for infinitely many n.

Proof. 

We write $\mathcal{P}\left(G\right)=\{V_{d_1}\left(G\right),V_{d_2}\left(G\right),V_{d_3}\left(G\right),\dots ,V_{d_k}\left(G\right)\}$. Let ${\gamma }_i=\gamma \left(G\left[V_{d_i}\left(G\right)\right]\right)$, ${\overline{\gamma }}_i=\gamma \left(\overline{G\left[V_{d_i}\left(G\right)\right]}\right)$, and $n_i=\left|V_{d_i}\left(G\right)\right|$ for $i=1,\dots ,k$. Since $G_{\delta }\left[V_{d_i}\left(G\right)\right]=\overline{G\left[V_{d_i}\left(G\right)\right]}$, we have $2\le {\gamma }_i+\gamma \left(G_{\delta }\left[V_i\left(G\right)\right]\right)={\gamma }_i+{\overline{\gamma }}_i\le n_i+1$ for $i=1,\dots ,k$, by Theorem 5. The union of dominating sets of $G\left[V_{d_i}\left(G\right)\right]$ gives a dominating set of G. Similarly, the union of dominating sets of $G_{\delta }\left[V_{d_i}\left(G\right)\right]$ gives a dominating set of $G_{\delta }$. Thus, $\gamma \left(G\right)+\gamma \left(G_{\delta }\right)\le {\sum }_{i=1}^k({\gamma }_i+{\overline{\gamma }}_i)\le n+k$. Since $0\le {(\gamma \left(G\right)-\gamma \left(G_{\delta }\right))}^2$, it follows that $4\gamma \left(G\right)·\gamma \left(G_{\delta }\right)\le {(\gamma \left(G\right)+\gamma \left(G_{\delta }\right))}^2$. Hence, we have $\gamma \left(G\right)·\gamma \left(G_{\delta }\right)\le {\left({\displaystyle \frac{n+k}{2}}\right)}^2$. By Lemmas 1 and 2, we complete the proof. □

Theorem 8.

If G is not a complete graph and $\gamma \left(G\right)=1$, then $\gamma \left(G_{\delta }\right)\le 2$.

Proof. 

Consider $n\ge 2$. Since $\gamma \left(G\right)=1$, there is a vertex v which is adjacent to all other vertices in G.

Case 1

v is the only vertex of degree $n-1$ in G. Then, v also has degree $n-1$ in $G_{\delta }$. Hence, $\gamma \left(G_{\delta }\right)=1$.

Case 2

v is not the only vertex of degree $n-1$ in G. Pick $u\in V\left(G\right)\setminus V_{n-1}\left(G\right)$. In $G_{\delta }$, we have v dominates $V\left(G\right)\setminus V_{n-1}\left(G\right)$ and u dominates $V_{n-1}\left(G\right)$. So, the vertex subset $\{u,v\}$ dominates $V\left(G\right)$ in $G_{\delta }$. Hence, $\gamma \left(G_{\delta }\right)\le 2$.

From these two cases, we have $\gamma \left(G_{\delta }\right)\le 2$. □

The converse of Theorem 8 is not necessarily true, as in the following example.

Example 1.

If $G=K_3+K_3$, then $G_{\delta }=K_{3,3}$. Hence, $\gamma \left(G\right)=2$ and $\gamma \left(G_{\delta }\right)=2$.

4. Join of Graphs

This section contains our results on upper bounds and exact values on the domination number of the $\delta$-complement of the join of two graphs. Here is an overview of this section. Given two graphs G and H, if they satisfy Condition (1) in Theorem 9, then $\gamma \left({(G\vee H)}_{\delta }\right)\le 2$. If not, Theorem 10 provides the exact domination number of ${(G\vee H)}_{\delta }$ when both G and H are regular graphs. If one of the graphs is nonregular, a sharp upper bound is given in Theorem 11. Finally, Corollary 1 provides an upper bound when the two are nonregular. We summarize these results in

Table 1. The summary of bounds on the join of two graphs. To use the table, first check whether G and H satisfy Condition (1) and then proceed with the types of the two graphs.

Condition (1) in Theorem 9	G	H	Upper Bound of $\mathit{\gamma }\left({(\mathit{G}\vee \mathit{H})}_{\mathit{\delta }}\right)$	Reference
true	any graph	any graph	2	Theorem 9
false	regular	regular	$\gamma \left(G_{\delta }\right)+\gamma \left(H_{\delta }\right)$	Theorem 10
false	regular	nonregular	$\gamma \left(H_{\delta }\right)+1$	Theorem 11
false	nonregular	regular	$\gamma \left(G_{\delta }\right)+1$	Theorem 11
false	nonregular	nonregular	$min\{\gamma \left(G_{\delta }\right),\gamma \left(H_{\delta }\right)\}+1$	Corollary 1

at the end of this section.

Theorem 9.

Let G and H be graphs. If

$${deg}_G\left(u\right)+\left|V\left(H\right)\right|\ne {deg}_H\left(v\right)+\left|V\left(G\right)\right|forallu\in V\left(G\right)andv\in V\left(H\right),$$

(1)

then $\gamma \left({(G\vee H)}_{\delta }\right)\le 2$.

Proof. 

Since, in ${(G\vee H)}_{\delta }$, each vertex in $V\left(G\right)$ covers all vertices in $V\left(H\right)$, and vice versa, it follows that $\gamma \left({(G\vee H)}_{\delta }\right)\le 2$. □

Example 2.

Let $G=P_3$ be a path graph with $V\left(G\right)=\{u_1,u_2,u_3\}$ and $E\left(G\right)=\{\{u_1,u_2\},\{u_2,u_3\}\}$ and $H=P_5$ be a path graph with $V\left(H\right)=\{v_1,v_2,v_3,v_4,v_5\}$ and $E\left(H\right)=\{\{v_1,v_2\},\{v_2,v_3\},\{v_3,v_4\},\{v_4,v_5\}\}$. We see that

$${deg}_G\left(u_i\right)+|V\left(H\right)|=\left\{\begin{array}{cc}6\hfill & ifi=1,3,\hfill \\ 7\hfill & ifi=2,\hfill \end{array}\right.and{deg}_H\left(v_i\right)+\left|V\left(G\right)\right|=\left\{\begin{array}{cc}4\hfill & ifi=1,5,\hfill \\ 5\hfill & ifi=2,3,4.\hfill \end{array}\right.$$

Graphs G and H satisfy the condition in Theorem 9. Hence, $\gamma \left({(P_3\vee P_5)}_{\delta }\right)\le 2$. In fact, $\gamma \left({(P_3\vee P_5)}_{\delta }\right)=1$ as vertex $u_2$ dominates ${(P_3\vee P_5)}_{\delta }$. See Figure 2 for an illustration.

When G and H are regular, $G\vee H$ is also regular, which implies that ${(G\vee H)}_{\delta }=\overline{G\vee H}$. Thus, $\gamma \left({(G\vee H)}_{\delta }\right)=\gamma \left(\overline{G\vee H}\right)$. This yields the result in Theorem 10. Despite the simplicity of its proof, we have not found this result in any of the literature, to our knowledge. Hence, we present it here. (This is the same for the case of the Cartesian product where ${(G\square H)}_{\delta }=\overline{G\square H}$ when G and H are regular and the results are present in Lemmas 5–7 and Corollary 2.)

Theorem 10.

Let G be an a-regular and H be a b-regular. If $a+\left|V\right(H\left)\right|=b+\left|V\right(G\left)\right|$, then $\gamma \left({(G\vee H)}_{\delta }\right)=\gamma \left(G_{\delta }\right)+\gamma \left(H_{\delta }\right)$.

Proof. 

Since all vertices in $G\vee H$ have the same degree, it follows that ${(G\vee H)}_{\delta }\cong G_{\delta }+H_{\delta }$. This completes the proof. □

In the following example, we construct two graphs G and H that exhibit the sharp upper bound given in Theorem 11.

Example 3.

For $n\ge 2$, let G be a graph such that $V\left(G\right)=U\cup W$ where $U=\{u_1,\dots ,u_n\}$ and $W=\{w_1,\dots ,w_{2n}\}$ in which U and W induce complete graphs $K_n$ and $K_{2n}$, respectively, and $u_i$ is adjacent to $w_{2i-1}$ and $w_{2i}$ for each $i=1,\dots ,n$. See Figure 3 for an example when $n=2$. With this construction, we have $G_{\delta }=n{P}_3$, a disjoint union of n paths of order 3. Also, $\gamma \left(G_{\delta }\right)=n$. Let H be an $(n+1)$-regular graph with $3n$ vertices. We have $H_{\delta }=\overline{H}$.

In $G\vee H$, the degree of $u_i$ is $4n+1$, the degree of $w_i$ is $5n$, and the degree of $v\in V\left(H\right)$ is $4n+1$. Then,

$$\begin{array}{c}E\left({(G\vee H)}_{\delta }\right)\hfill \\ =E\left(G_{\delta }\right)\cup E\left(H_{\delta }\right)\cup \{\{w,v\}:w\in Wandv\in V\left(H\right)\}\hfill \\ =\{\{u_i,w_{2i-1}\},\{u_i,w_{2i}\}:i=1,\dots ,n\}\cup E\left(\overline{H}\right)\cup \{\{w,v\}:w\in Wandv\in V\left(H\right)\}.\hfill \end{array}$$

We see that $U\cup \left\{w_1\right\}$ is a dominating set of ${(G\vee H)}_{\delta }$ of size $n+1$.

Let S be a dominating set of ${(G\vee H)}_{\delta }$. We will show that $\left|S\right|\ge n+1$. Suppose that $\left|S\right|=n$. To cover $u_i$, at least one of $u_i,w_{2i-1},w_{2i}$ must be in S. Since $\left|S\right|=n$, exactly one of $u_i,w_{2i-1},w_{2i}$ is in S. If $S=U$, then S does not cover $V\left(H\right)$. If $S\ne U$, we assume that $u_1\notin S$ and $w_1\in S$. Then, $w_2$ is not covered by S. Hence, S cannot be a dominating set, which is a contradiction.

Thus, $\gamma \left({(G\vee H)}_{\delta }\right)=n+1=\gamma \left(G_{\delta }\right)+1$.

Figure 3. An example where $\gamma \left({(G\vee H)}_{\delta }\right)=\gamma \left(G_{\delta }\right)+1$ as discussed in Example 3.

Figure 3. An example where $\gamma \left({(G\vee H)}_{\delta }\right)=\gamma \left(G_{\delta }\right)+1$ as discussed in Example 3.

Theorem 11.

For any graph G and H, if G is nonregular, then $\gamma \left({(G\vee H)}_{\delta }\right)\le \gamma \left(G_{\delta }\right)+1$. Furthermore, the bound is sharp for infinitely many graphs.

Proof. 

Let $k=k\left(G\right)$ and $\ell =k\left(H\right)$, and let $\mathcal{P}\left(G\right)=\{V_{d_1}\left(G\right),\dots ,V_{d_k}\left(G\right)\}$ and $\mathcal{P}\left(H\right)=\{V_{d_1^{\prime }}\left(H\right),\dots ,V_{d_{\ell }^{\prime }}\left(H\right)\}$ be the partitions of the vertex sets of G and H, respectively. A vertex in $V_{d_i}\left(G\right)$ and a vertex in $V_{d_j^{\prime }}\left(H\right)$ have the same degree in $G\vee H$ if and only if $d_i+\left|V\left(H\right)\right|=d_j^{\prime }+\left|V\left(G\right)\right|$. Without loss of generality, we order $d_i$ so that $d_i+\left|V\left(H\right)\right|=d_i^{\prime }+\left|V\left(G\right)\right|$ for all $i=1,\dots ,p$ for some $p\le min\{k,\ell \}$.

We note that ${(G\vee H)}_{\delta }\left[V\left(G\right)\right]=G_{\delta }$ and ${(G\vee H)}_{\delta }\left[V\left(H\right)\right]=H_{\delta }$. In addition, in graph ${(G\vee H)}_{\delta }$, the vertices in $V_{d_i}\left(G\right)$ are adjacent to the vertices in $V_{d_j^{\prime }}\left(H\right)$ for any $j\ne i$.

Let D be a dominating set of $G_{\delta }$. Since G is nonregular, $V\left(G\right)\setminus V_{d_i}\left(G\right)$ is nonempty for all $i=1,\dots ,p$. If $D\subseteq V_{d_i}\left(G\right)$ for some $i=1,\dots ,p$, then $D\cup \left\{v\right\}$ is a dominating set of ${(G\vee H)}_{\delta }$ where $v\in V\left(G\right)\setminus V_{d_i}\left(G\right)$. This is because D dominates $V(G\vee H)\setminus V_{d_i^{\prime }}\left(H\right)$ and v dominates $V_{d_i^{\prime }}\left(H\right)$. Otherwise, D is a dominating set of ${(G\vee H)}_{\delta }$.

The bound is sharp by Example 3. □

Corollary 1.

Let G and H be nonregular graphs. Then,

$$\gamma \left({(G\vee H)}_{\delta }\right)\le min\{\gamma \left(G_{\delta }\right),\gamma \left(H_{\delta }\right)\}+1.$$

5. Cartesian Product of Graphs

This section contains our results on upper bounds on the $\delta$-complement of the Cartesian product of two graphs, in particular, when one of them is a regular graph. Sharp bounds are obtained when both of the graphs are regular or one of them is a cycle. Recall that, for a graph G, the vertex partition of $V\left(G\right)$ by the degrees is $\mathcal{P}\left(G\right)$.

There are a number of results of bounds on the domination number of the Cartesian product of graphs [15,16,17]. We include one of the bounds below as Lemma 3, which, combining with Theorem 6, produces a rough bound on the $\delta$-complement of the Cartesian product of any two graphs shown in Lemma 4.

Lemma 3

([17]). Let G and H be graphs. Then,

$$\gamma (G\square H)\le min\left\{\gamma \right(G)·|V\left(H\right)|,\gamma (H)·|V\left(G\right)\left|\right\}.$$

Using Theorem 6, we get the following bound for ${(G\square H)}_{\delta }$.

Lemma 4.

Let G and H be graphs. Then,

$$\gamma \left({(G\square H)}_{\delta }\right)\le min\{\gamma \left(G_{\delta }\right)·|V\left(H\right)|,\gamma \left(H_{\delta }\right)·|V\left(G\right)\left|\right\}.$$

Proof. 

From Theorem 6, we have $G_{\delta }\square H_{\delta }\subset {(G\square H)}_{\delta }$ and $V(G_{\delta }\square H_{\delta })=V\left({(G\square H)}_{\delta }\right)$. We have $\gamma \left({(G\square H)}_{\delta }\right)\le \gamma (G_{\delta }\square H_{\delta })\le min\{\gamma \left(G_{\delta }\right)·|V\left(H\right)|,\gamma \left(H_{\delta }\right)·|V\left(G\right)\left|\right\}$. The last inequality follows from Lemma 3. □

We note that the bound in Lemma 4 is sharp. Some examples include $G=P_3$ and $H=P_2$ or when one of the graphs is $K_1$.

When one of the graphs in the product is regular, the bounds can be improved as shown in Theorem 12. We first give the following lemmas which will be used in Theorem 12.

Lemma 5.

For $r\ge 3$ and $n\ge 3$, then $\gamma \left({(K_r\square \overline{K_n})}_{\delta }\right)=2$.

Proof. 

Two vertices $(u,v)$ and $(u^{\prime },v^{\prime })$ are adjacent in ${(K_r\square \overline{K_n})}_{\delta }$ if and only if $v\ne v^{\prime }$. Let $u\in V\left(K_r\right)$ and $v_1,v_2\in V\left(K_n\right)$ where $v_1\ne v_2$. Then, $(u,v_i)$ dominates $V\left(K_r\right)\times (V\left(K_n\right)\setminus \left\{v_i\right\})$. Hence, the vertex subset $\{(u,v_1),(u,v_2)\}$ is a dominating set. Since every vertex of ${(K_r\square \overline{K_n})}_{\delta }$ is of degree $r(n-1)<rn-1$, there is no dominating set of size 1. Hence, $\gamma \left({(K_r\square \overline{K_n})}_{\delta }\right)=2$. □

Lemma 6.

For $r\ge 1$ and $n\ge 3$, we have

$$\gamma \left({(K_r\square C_n)}_{\delta }\right)=\left\{\begin{array}{cc}3\hfill & ifr\ne 2andn=3,\hfill \\ 2\hfill & otherwise.\hfill \end{array}\right.$$

Proof. 

Consider $r=1$. Then, $K_r\square C_n\cong C_n$. Hence, $\gamma \left({(K_r\square C_3)}_{\delta }\right)=\gamma \left(\overline{C_3}\right)=3$ and $\gamma \left({(K_r\square C_n)}_{\delta }\right)=\gamma \left(\overline{C_n}\right)=2$ when $n\ge 4$.

Consider $r=2$. Let $u_1,u_2\in V\left(K_2\right)$ and $v\in V\left(C_n\right)$. Then, the vertex subset $\{(u_1,v),(u_2,v)\}$ is a minimum dominating set of ${(K_r\square C_n)}_{\delta }$, see Figure 4 for an example.

Consider $r\ge 3$. Let $u\in V\left(K_r\right)$. For $n=3$, we let $v_1,v_2,v_3\in V\left(C_3\right)$. We have that the vertex subset $\{(u,v_1),(u,v_2),(u,v_3)\}$ is a minimum dominating set of ${(K_r\square C_3)}_{\delta }$, see Figure 4. For $n\ge 4$, we let $C_n$ be a cycle with the vertex set $\{w_1,w_2\dots ,w_n\}$ and the edge set $\{\{w_1,w_2\},\{w_2,w_3\},\dots ,\{w_{n-1},w_n\},\{w_n,w_1\}\}$. Then, the vertex subset $\{(u,w_1),(u,w_2)\}$ is a minimum dominating set of ${(K_r\square C_n)}_{\delta }$. □

Figure 4. (Top figures) $K_2\square C_4$ and $K_4\square C_3$. (Bottom figures) The $\delta$-complement of $K_2\square C_4$ and $K_4\square C_3$ with vertices of their minimum dominating sets indicated in red color.

Figure 4. (Top figures) $K_2\square C_4$ and $K_4\square C_3$. (Bottom figures) The $\delta$-complement of $K_2\square C_4$ and $K_4\square C_3$ with vertices of their minimum dominating sets indicated in red color.

Lemma 7.

For $r\ge 3$ and $n\ge 3$, we have $\gamma \left({(K_r\square K_n)}_{\delta }\right)=3$.

Proof. 

Two vertices $(u,v)$ and $(u^{\prime },v^{\prime })$ are adjacent in ${(K_r\square K_n)}_{\delta }$ if and only if $u\ne u^{\prime }$ and $v\ne v^{\prime }$. Let $u_1,u_2\in V\left(K_r\right)$ and $v_1,v_2\in V\left(K_n\right)$ be such that $u_1\ne u_2$ and $v_1\ne v_2$. Suppose $D=\{(u_1,v_1),(u_1,v_2),(u_2,v_1)\}$. Consider $(u,v)\notin D$. If $u=u_1$, then $v\ne v_1$. So, $(u,v)$ is adjacent to $(u_2,v_1)$. If $u\ne u_1$, then $(u,v)$ is adjacent to $(u_1,v_1)$ or $(u_1,v_2)$. Hence, D is a dominating set.

Suppose there is a dominating set $S=\{(x,y),(x^{\prime },y^{\prime })\}$ of size 2. If $x=x^{\prime }$, then $(x,y^{″})$ where $y^{″}\notin \{y,y^{\prime }\}$ is not adjacent to any vertices in S. If $y=y^{\prime }$, then $(x^{″},y)$ where $x^{″}\notin \{x,x^{\prime }\}$ is not adjacent to any vertices in S. If $x\ne x^{\prime }$ and $y\ne y^{\prime }$, then $(x,y^{\prime })$ is not adjacent to any vertices in S. This is a contradiction. So, the domination number is 3. □

Let $s_G$ be the number of singleton sets in $\mathcal{P}\left(G\right)$. This parameter plays an important role in the following theorem and serves as a key term in the resulting bounds.

Theorem 12.

Let G and H be graphs, each with at least two vertices. If H is a regular graph, then

$$\gamma \left({(G\square H)}_{\delta }\right)\le \left\{\begin{array}{cc}2k\left(G\right)-s_G\hfill & if\gamma \left(H_{\delta }\right)=1,\hfill \\ 2k\left(G\right)\hfill & if\gamma \left(H_{\delta }\right)=2,\hfill \\ 3k\left(G\right)+(\gamma \left(H_{\delta }\right)-3)s_G\hfill & otherwise.\hfill \end{array}\right.$$

Furthermore, the bounds are sharp for infinitely many graphs.

Proof. 

Let $k=k\left(G\right)$. For a vertex partition $\mathcal{P}\left(G\right)=\{V_{d_1}\left(G\right),\dots ,V_{d_k}\left(G\right)\}$, we order it so that $|V_{d_1}\left(G\right)|\le |V_{d_2}\left(G\right)|\le \cdots \le | V_{d_k}\left(G\right)|$. It follows that $|V_{d_i}\left(G\right)|=1$ for $i\le s_G$ and $|V_{d_i}\left(G\right)|>1$ for $s_G+1\le i\le k$. Since H is regular, we see that $\mathcal{P}(G\square H)=\{V_{d_1}\left(G\right)\times V\left(H\right),\dots ,V_{d_k}\left(G\right)\times V\left(H\right)\}$ is a vertex partition of the vertices of $G\square H$ based on its degree.

Let $D\subseteq V\left(H\right)$ be a minimum dominating set of $H_{\delta }$. For any $i\le s_G$, we have that $V_{d_i}\left(G\right)\times D$ dominates $V_{d_i}\left(G\right)\times V\left(H\right)$ in ${(G\square H)}_{\delta }$. Thus, we need at most $\gamma \left(H_{\delta }\right)·s_G$ vertices to cover ${\bigcup }_{i=1}^{s_G}{V}_{d_i}\left(G\right)\times V\left(H\right)$. Next, we consider each $i\ge s_G+1$. Let $u_i\in V_{d_i}\left(G\right)$.

Case 1. $\gamma \left(H_{\delta }\right)=1$. Then, H is an empty graph. Let $v,w\in V\left(H\right)$ where $v\ne w$. Each of v and w dominates $H_{\delta }$. We have $(u_i,v)$ dominates $\left\{u_i\right\}\times V\left(H\right)$ in ${(G\square H)}_{\delta }$. Let $(x,y)\in V_{d_i}\left(G\right)\times V\left(H\right)$ where $x\ne u_i$.

• If $y\ne v$, then $(x,y)$ is not adjacent to $(u_i,v)$ in $G\square H$.
• If $y=v$, then $(x,y)$ is not adjacent to $(u_i,w)$ in $G\square H$.

We have the vertex subset $\{(u_i,v),(u_i,w)\}$ dominates $V_{d_i}\left(G\right)\times V\left(H\right)$ in ${(G\square H)}_{\delta }$ for $s_G+1\le i\le k$. Thus, we need at most $2(k-s_G)$ vertices to cover ${\bigcup }_{i=s_G+1}^k{V}_{d_i}\left(G\right)\times V\left(H\right)$. Hence, in this case, $\gamma \left({(G\square H)}_{\delta }\right)\le s_G+2(k-s_G)=2k-s_G$.

To show that this bound is sharp, for $k\ge 2$ and $n\ge 3$, we let $G=K_1+K_3+K_4+\cdots +K_{k+1}$ and H be a null graph $\overline{K_n}$. We see that $s_G=1$ and $\gamma \left(H_{\delta }\right)=1$. Note that

$$G\square H=(K_1\square \overline{K_n})+(K_3\square \overline{K_n})+\cdots +(K_{k+1}\square \overline{K_n}).$$

Since $G\square H$ contains no edge between two vertices of different degrees, it follows that ${(G\square H)}_{\delta }={(K_1\square \overline{K_n})}_{\delta }+{(K_3\square \overline{K_n})}_{\delta }+\cdots +{(K_{k+1}\square \overline{K_n})}_{\delta }$. By Lemma 5, we have $\gamma \left({(K_r\square \overline{K_n})}_{\delta }\right)=2$ for $r\ge 3$. We note $\gamma \left({(K_1\square \overline{K_n})}_{\delta }\right)=\gamma \left(K_n\right)=1$. Thus, $\gamma \left({(G\square H)}_{\delta }\right)=1+2(k-1)=2k-1=2k-s_G$.

Case 2. $\gamma \left(H_{\delta }\right)=2$. Let $v,w\in V\left(H\right)$ be such that $v\ne w$ and the vertex subset $\{v,w\}$ dominates $H_{\delta }$. Then, the vertex subset $\{(u_i,v),(u_i,w)\}$ dominates $\left\{u_i\right\}\times V\left(H\right)$ in ${(G\square H)}_{\delta }$. Consider $(x,y)\in V_{d_i}\left(G\right)\times V\left(H\right)$ where $x\ne u_i$.

• If $y\ne v$, then $(x,y)$ is not adjacent to $(u_i,v)$ in $G\square H$.
• If $y\ne w$, then $(x,y)$ is not adjacent to $(u_i,w)$ in $G\square H$.

It follows that the vertex subset $\{(u_i,v),(u_i,w)\}$ dominates $V_{d_i}\left(G\right)\times V\left(H\right)$ in ${(G\square H)}_{\delta }$ for $s_G+1\le i\le k$. Thus, we need at most $2(k-s_G)$ vertices to cover ${\bigcup }_{i=s_G+1}^k{V}_{d_i}\left(G\right)\times V\left(H\right)$. Hence, in this case, $\gamma \left({(G\square H)}_{\delta }\right)\le 2s_G+2(k-s_G)=2k$.

To show that this bound is sharp, for $k\ge 1$ and $n\ge 4$, we let $G=K_1+\cdots +K_k$ and $H=C_n$. We see that $s_G=1$ and $\gamma \left(H_{\delta }\right)=2$. Note that

$$G\square H=(K_1\square C_n)+\cdots +(K_k\square C_n).$$

Since $G\square H$ contains no edge between two vertices of different degrees, it follows that ${(G\square H)}_{\delta }={(K_1\square C_n)}_{\delta }+\cdots +{(K_k\square C_n)}_{\delta }$. By Lemma 6, we have $\gamma \left({(K_r\square C_n)}_{\delta }\right)=2$ for $r\ge 1$. Thus, ${(G\square H)}_{\delta }=2k$.

Case 3. $\gamma \left(H_{\delta }\right)\ge 3$. For each $i\ge s_G+1$, pick any two vertices $a_i,b_i\in V_{d_i}\left(G\right)$ and any two vertices $v_1,v_2\in V\left(H\right)$. Let $(x,y)\in V_{d_i}\left(G\right)\times V\left(H\right)$.

• If $x\ne a_i$ and $y\ne v_1$, then $(x,y)$ is not adjacent to $(a_i,v_1)$ in $G\square H$.
• If $x=a_i$ and $y\ne v_1$, then $(x,y)$ is not adjacent to $(b_i,v_1)$ in $G\square H$.
• If $x\ne a_i$ and $y=v_1$, then $(x,y)$ is not adjacent to $(a_i,v_2)$ in $G\square H$.

Hence, $\{(a_i,v_1),(b_i,v_1),(a_i,v_2)\}$ dominates $V_{d_i}\left(G\right)\times V\left(H\right)$ in ${(G\square H)}_{\delta }$ for $s_G+1\le i\le k$. Thus, we need at most $3(k-s_G)$ vertices to cover ${\bigcup }_{i=s_G+1}^k{V}_{d_i}\left(G\right)\times V\left(H\right)$. Hence, in this case, $\gamma \left({(G\square H)}_{\delta }\right)\le \gamma \left(H_{\delta }\right)·s_G+3(k-s_G)=3k+(\gamma \left(H_{\delta }\right)-3)s_G$.

To show that this bound is sharp, for $k\ge 2$ and $n\ge 3$, we let $G=K_1+K_3+K_4+\cdots +K_{k+1}$ and $H=K_n$. We see that $s_G=1$ and $\gamma \left(H_{\delta }\right)=n$. Note that

$$G\square H=(K_1\square K_n)+(K_3\square K_n)+\dots +(K_{k+1}\square K_n).$$

Since $G\square H$ contains no edge between two vertices of different degrees, it follows that ${(G\square H)}_{\delta }={(K_1\square K_n)}_{\delta }+{(K_3\square K_n)}_{\delta }+\dots +{(K_{k+1}\square K_n)}_{\delta }$. By Lemma 7, we have $\gamma \left({(K_r\square K_n)}_{\delta }\right)=3$ for $r\ge 3$. We note $\gamma \left({(K_1\square K_n)}_{\delta }\right)=\gamma \left(\overline{K_n}\right)=n$. Thus, ${(G\square H)}_{\delta }=n+3(k-1)=3k+(\gamma \left(H_{\delta }\right)-3)s_G$. □

Consequently, we obtain sharp upper bounds when both graphs in the product are regular.

Corollary 2.

Let G and H be regular graphs, each with at least two vertices. Then,

$$\gamma \left({(G\square H)}_{\delta }\right)\le \left\{\begin{array}{cc}2\hfill & ifmin\{\gamma \left(G_{\delta }\right),\gamma \left(H_{\delta }\right)\}\le 2,\hfill \\ 3\hfill & otherwise.\hfill \end{array}\right.$$

Furthermore, the bounds are sharp for infinitely many graphs.

Proof. 

Note that $k\left(G\right)=k\left(H\right)=1$ and $s_G=s_H=0$. The result follows directly from Theorem 12.

Next, we show that the bounds are sharp. For $n\ge 3$, $min\{\gamma \left({\left(K_2\right)}_{\delta }\right),\gamma \left({\left(C_n\right)}_{\delta }\right)\}\le 2$. Lemma 6 implies that $\gamma \left({(K_2\square C_n)}_{\delta }\right)=2$. In addition, $min\{\gamma \left({\left(K_n\right)}_{\delta }\right),\gamma \left({\left(C_3\right)}_{\delta }\right)\}=3$. Lemma 6 implies that $\gamma \left({(K_n\square C_3)}_{\delta }\right)=3$. Hence, both bounds are sharp. □

Finally, the following theorem is another special case of Theorem 12 where one of the graphs in the product is a cycle. In this case, sharp bounds can also be obtained.

Theorem 13.

Let G be a graph. Then,

$$\gamma \left({(G\square C_n)}_{\delta }\right)\le \left\{\begin{array}{cc}3k\left(G\right)\hfill & ifn=3,\hfill \\ 2k\left(G\right)\hfill & ifn\ge 4.\hfill \end{array}\right.$$

Furthermore, the bounds are sharp for infinitely many graphs.

Proof. 

Let $k=k\left(G\right)$. If $\left|V\right(G\left)\right|=1$, then $G=K_1$ and $k=1$. From Lemma 6, we get $\gamma \left({(G\square C_3)}_{\delta }\right)=3$ and $\gamma \left({(G\square C_n)}_{\delta }\right)=2$ for $n\ge 4$.

For $\left|V\right(G\left)\right|\ge 2$, Theorem 12 implies the result. To show that the bounds are sharp, we let $G=K_3+K_4+\cdots +K_{k+2}$. Note that

$$G\square C_n=(K_3\square C_n)+\cdots +(K_{k+2}\square C_n).$$

So, $G\square C_n$ contains no edge between two vertices of different degrees; hence,

$${(G\square C_n)}_{\delta }={(K_3\square C_n)}_{\delta }+\cdots +{(K_{k+2}\square C_n)}_{\delta }.$$

In the case that $n=3$, by Lemma 6, $\gamma \left({(K_r\square C_3)}_{\delta }\right)=3$ for $3\le r\le k+2$. Hence, $\gamma \left({(G\square C_3)}_{\delta }\right)=3k$. In the case that $n\ge 4$, by Lemma 6, $\gamma \left({(K_r\square C_n)}_{\delta }\right)=2$ for $3\le r\le k+2$. Hence, $\gamma \left({(G\square C_n)}_{\delta }\right)=2k$. □

Table 2. The summary of bounds on the Cartesian product of two graphs. Note that, other than the first case, G and H are required to contain at least two vertices.

Condition	G	H	Upper Bound of $\mathit{\gamma }\left({(\mathit{G}\square \mathit{H})}_{\mathit{\delta }}\right)$	Reference
-	any graph	any graph	$min\{\gamma \left(G_{\delta }\right)·|V\left(H\right)|,\gamma \left(H_{\delta }\right)·|V\left(G\right)\left|\right\}$	Lemma 4
$min\{\gamma \left(G_{\delta }\right),\gamma \left(H_{\delta }\right)\}\le 2$	regular	regular	2	Corollary 2
$min\{\gamma \left(G_{\delta }\right),\gamma \left(H_{\delta }\right)\}>2$	regular	regular	3	Corollary 2
$\gamma \left(H_{\delta }\right)=1$	any graph	regular	$2k\left(G\right)-s_G$	Theorem 12
$\gamma \left(H_{\delta }\right)=2$	any graph	regular	$2k\left(G\right)$	Theorem 12
$\gamma \left(H_{\delta }\right)>2$	any graph	regular	$3k\left(G\right)+(\gamma \left(H_{\delta }\right)-3)s_G$	Theorem 12

presents the upper bounds on the domination numbers of the Cartesian products of graphs.

6. Discussion and Conclusions

In this research, we investigate the domination numbers of the $\delta$-complements of graphs. We provide Nordhaus–Gaddum-type bounds on the domination number between a graph and its $\delta$-complement, bounds on the domination number of the $\delta$-complement of joined graphs, and Cartesian products of graphs. The bounds are in terms of the number of distinct degrees, the domination number of the $\delta$-complement of each graph in the product, or the number of degrees with one vertex, and the given bounds are sharp.

Many interesting aspects of the domination numbers of the $\delta$-complements of graphs are yet to be explored, for example, bounds when both graphs in the product are nonregular, bounds on other types of graph products, and bounds on other types of domination numbers of the $\delta$-complement graphs.