Notes on k-rainbow independent domination in graphs

TL;DR

This paper studies the $k$-rainbow independent domination number in graphs, improves bounds on its sum with the complement, and proves the NP-completeness of related decision problems, addressing open questions in the field.

Contribution

It refines the Nordhaus-Gaddum-type bounds for the $k$-rainbow independent domination number and establishes NP-completeness for the associated decision problem.

Findings

Improved bounds: $5 \,\leq\, \gamma_{rik}(G)+\gamma_{rik}(\overline{G}) \leq n+2$ for non-5-cycle graphs.

NP-completeness: Deciding the existence of a $k$-rainbow independent dominating function of a given weight is NP-complete.

Responds to open questions by umenjak et al.

Abstract

The $k$-rainbow independent domination number of a graph $G$, denoted $\gamma_{\rm rik}(G)$, is the cardinality of a smallest set consisting of two vertex-disjoint independent sets $V_1$ and $V_2$ for which every vertex in $V(G)\setminus (V_1\cup V_2)$ has neighbors in both $V_1$ and $V_2$. This domination invariant was proposed by {\v{S}}umenjak, Rall and Tepeh in (Applied Mathematics and Computation 333(15), 2018: 353-361), which allows to reduce the problem of computing the independent domination number of the generalized prism $G {\Box} K_k$ to an integer labeling problem on $G$. They proved a Nordhaus-Gaddum-type theorem: $5\leq \gamma_{\rm rik}(G)+\gamma_{\rm rik}(\overline{G})\leq n+3$ for every graph $G$ of order $n\geq 3$, where $\overline{G}$ is the complement of $G$. In this paper, we improve this result by showing that if $G$ is not isomorphic to the 5-cycle, then $5\leq \gamma_{\rm rik}(G)+\gamma_{\rm rik}(\overline{G})\leq n+2$. Moreover, we show that the problem of deciding whether a graph has a $k$-rainbow independent dominating function of a given weight is $\mathcal{NP}$-complete. Our results respond some open questions proposed by \v{S}umenjak, et al.

Equations

$${\gamma }_{ri2}(\overline{G})=|V_0|+3$$

$$|V_1|+|V_2|\ge 5$$

$$|(V_1^{\prime }\cup V_2^{\prime })\cap V_i|\le 2,i=1,2$$

$$|(N_{\overline{G}}(u)\cap N_{\overline{G}}(v))\cap (V_1\cup V_2)|\le 1$$

$$|(V_1\cup V_2)\setminus (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)|\le 2$$

Full text

1 Introduction

All graphs considered in this paper are simple and for notation and terminology not defined here we follow the book [1]. Let $G$ be a graph with vertex set $V(G)$ and edge set $E(G)$. Two vertices are adjacent in $G$ if they are the endpoints of an edge of $G$. We say that a vertex $u\in V(G)$ is adjacent to a set $U\subseteq G$ in $G$ if $U$ contains a vertex adjacent to $u$ in $G$. For any $v\in V(G)$, $N_{G}(v)$=$\{u|uv\in E(G)\}$ is called the open neighborhood of $v$ in $G$ and $N_{G}[v]$=$N_{G}(v)\cup\{v\}$ is the closed neighborhood of $v$ in $G$. Let $d_{G}(v)=|N_{G}(v)|$ denote the degree of $v$ in $G$ and $\Delta(G)=\max\{d_{G}(v)|v\in V(G)\}$. A vertex of degree $k$ and at least $k$ is called a $k$-vertex and $k^{+}$-vertex, respectively. For any $S\subseteq V(G)$, let $N_{G}(S)$=$\bigcup_{v\in S}N_{G}(v)\setminus S$ and $N_{G}[S]$=$N_{G}(S)\cup S$. We say that $S$ dominates a set $S^{\prime}$ if $S^{\prime}\subseteq N_{G}[S]$. Moreover, we use the notation $G-S$ to denote the subgraph of $G$ obtained by deleting vertices of $S$ and their incident edges in $G$, and $G[S]=G-(V(G)\setminus S)$ subgraph of $G$ induced by $S$. The complete graph with $n$ vertices and the cycle of length $n$ are denoted by $K_{n}$ and $C_{n}$, respectively. For two integers $i,j$, $i<j$, we will make use the notation $[i,j]$ to denote the set $\{i,i+1,\ldots,j\}$.

Given a graph $G$ and a subset $D\subseteq V(G)$, we call $D$ a dominating set of $G$ if $D$ dominates $V(G)$. An independent set of a graph is a set of vertices no two of which are adjacent in the graph. If a dominating set $D$ of $G$ is an independent set, then $D$ is called an independent dominating set (IDS for short) of $G$. The independent domination number of $G$, denoted by $i(G)$, is the cardinality of a smallest independent dominating set of $G$. Domination and independent domination in graphs have always attracted extensive attention [2, 3], and many variants of domination [2] have been introduced increasingly, for the applications in diverse fields, such as electrical networks, computational biology, land surveying, etc. Recent studies on these variations include strong roman domination [4], sum of domination number [4], semitotal domination[5, 6], relating domination [7], just to name a few.

Let $G\Box H$ be the cartesian product of $G$ and $H$. To reduce the problem of determining $i(G\Box K_{k})$ to an integer labeling problem on $G$ itself, Šumenjak et al. [8] recently proposed a new variation of domination, called $k$-rainbow independent dominating function of a graph $G$ ($k$RiDF for short), which is a function $f$: $V(G)\rightarrow[0,k]$ such that $V_{i}$ is an independent set and every vertex $v$ with $f(v)=0$ is adjacent to a vertex $u$ with $f(u)=i$, for all $i\in[1,k]$. Alternatively, a $k$RiDF $f$ of a graph $G$ may be viewed as an ordered partition $(V_{0},V_{1},\ldots,V_{k})$ such that $V_{j}$ is an independent set for $j=0,1,\ldots,k$ and $N_{G}(x)\cap V_{i}\neq\emptyset$ for every $x\in V_{0}$ and each $i\in[1,k]$, where $V_{j}$ denotes the set of vertices assigned value $j$ under $f$. The weight $w(f)$ of a $k$RiDF $f$ is defined as the number of nonzero vertices, i.e., $w(f)$=$|V(G)|-|V_{0}|$. The $k$-rainbow independent domination number of $G$, denoted by $\gamma_{\rm rik}(G)$, is the minimum weight of a $k$RiDF of $G$. From the definition, we have $\gamma_{ri1}(G)=i(G)$. A $\gamma_{rik}(G)$-function is a $k$RiDF of $G$ with weight $\gamma_{rik}(G)$.

Let $G$ be a graph and $H$ a subgraph of $G$. Suppose that $g$ is a $k$RiDF of $H$. We say that a $k$RiDF $f$ of $G$ is extended from $g$ if $f(v)=g(v)$ for every $v\in V(H)$. In what follows, to prove that a graph $G$ has a $k$RiDF, we will first find a $k^{\prime}$RiDF $g$ of a subgraph $G^{\prime}$ of $G$, $k^{\prime}\leq k$ and then extend $g$ to a $k$RiDF $f$ of $G$. As for the remaining part of this paper, Section 2 is dedicated to characterizing graphs $G$ with $\gamma_{\rm ri2}(G)$=$|V(G)|-1$, based on which we in Section 3 show an improved Nordhaus-Gaddum-type theorem on the sum of 2-rainbow independent domination number of $G$ and its complement. In Section 4 we are devoted to the proof of $\mathcal{NP}$-completeness of the $k$-rainbow independent domination problem, and in the last section we give a conclusion of this paper.

2 Graphs $G$ with $\gamma_{\rm ri2}(G)$=$|V(G)|-1$

To get the improved Nordhaus-Gaddum-type theorem on the sum of 2-rainbow independent domination number of $G$ and of $\overline{G}$, we have to characterize the graphs $G$ such that $\gamma_{\rm ri2}(G)$=$|V(G)|-1$. For this, we need the following special graphs.

A star $S_{n}$, $n\geq 1$, is a complete bipartite graph $G[X,Y]$ with $|X|$=1 and $|Y|=n$, where the vertex in $X$ is called the center of $S_{n}$ and the vertices in $Y$ are leaves of $S_{n}$. The graph obtained from $S_{n}$ by adding a single edge is denoted $S_{n}^{+}$. A double star [9] is defined as the union of two vertex-disjoint stars with an edge connecting their centers. Specifically, for two integers $n,m$ such that $n\geq m\geq 0$ the double star, denoted by $S(n,m)$, is the graph with vertex set $\{u_{0},u_{1},\ldots,u_{n},v_{0},v_{1},\ldots,v_{m}\}$ and edge set $\{u_{0}v_{0},u_{0}u_{i},v_{0}v_{j}|1\leq i\leq n,1\leq j\leq m\}$, where $u_{0}v_{0}$ is called the bridge of $S(n,m)$ and the subgraphs induced by $\{u_{i}|i=0,1,\ldots,n\}$ and $\{v_{j}|j=0,1,\ldots,m\}$ are called the $n$-star at $u_{0}$ and $m$-star at $v_{0}$. Observe that $S(n,m)$ is defined on the premise of $n\geq m$. For mathematical convenience, we denote a double star $S(n,m)$ as a vertex-sequence $v_{m}v_{m-1}\ldots v_{0}u_{0}u_{1}\ldots u_{n}$.

We start with a known result which characterizes graphs $G$ with $\gamma_{\rm ri2}(G)=n$.

Lemma 2.1**.**

[8]** For any graph $G$ of order $n$, $\gamma_{\rm ri2}(G)=n$ if and only if every connected component of $G$ is isomorphic either to $K_{1}$ or $K_{2}$. In addition, if $\gamma_{\rm ri2}(G)=n$, then $\gamma_{\rm ri2}(\overline{G})=2$, where $\overline{G}$ is the complement of $G$.

The following conclusion is simple but will be used throughout this paper.

Lemma 2.2**.**

Let $G$ be a graph and $H$ a subgraph of $G$. Suppose that $g=(V_{0},V_{1},\ldots,V_{k})$ is a $\gamma_{\rm rik}(H)$-function. Then $g$ can be extended to a kRiDF of $G$ with weight at most $|V(G)|-|V_{0}|$.

Proof  Let $V(G)\setminus V(H)=\{x_{1},\ldots,x_{\ell}\}$. We will deal with these vertices in the order of $x_{1},\ldots,x_{\ell}$ by the following rule: for each $x_{i}$, $i\in[1,\ell]$, let $j\in[1,k]$ be the smallest one such that $x_{i}$ is not adjacent to $V_{j}$ in $G$. If such $j$ does not exist, we update $V_{0}$ by $V_{0}\cup\{x_{i}\}$; otherwise we update $V_{j}$ by $V_{j}\cup\{x_{i}\}$. After the last one, i.e., $x_{\ell}$ is handled, we obtain a $k$RiDF of $G$. Obviously, the weight of the resulting kRiDF of $G$ is at most $|V(G)|-|V_{0}|$. ∎

The following theorem clarifies the structure of connected graphs $G$ with $\gamma_{\rm ri2}(G)=|V(G)|-1$.

Theorem 2.3**.**

Let $G$ be a connected graph of order $n\geq 3$. Then, $\gamma_{\rm ri2}(G)=n-1$ if and only if $G$ is isomorphic to one among $S_{n-1}$, $S_{n-1}^{+}$, $S(n-3,1)$ ($n\geq 4$) and $C_{5}$.

Proof  Let $f=(V_{0},V_{1},V_{2})$ be an arbitrary $\gamma_{\rm ri2}(G)$-function. Observe that $V_{0}$ does not contain any 1-vertex; one can readily derive that $\gamma_{\rm ri2}(G)=n-1$ when $G$ is isomorphic to one of $S_{n-1},S_{n-1}^{+}$, $S(n-3,1)$ and $C_{5}$. Conversely, suppose that $\gamma_{\rm ri2}(G)=n-1$, i.e. $|V_{0}|=1$. By Lemma 2.2, $G$ contains no subgraph $H$ that has a 2RiDF of weight at most $|V(H)|-2$. Since $\gamma_{\rm ri2}(C_{4})=2=|V(C_{4})|-2$, $G$ contains no subgraph isomorphic to $C_{4}$. This also shows that every two vertices of $G$ share at most one neighbor in $G$.

Observation 1. If $G$ contains a $3^{+}$-vertex $x$, then every $2^{+}$-vertex of $G$ belongs to $N_{G}(x)$. Suppose to the contrary that $G$ contains a $2^{+}$-vertex $y$ such that $y\notin N_{G}(x)$. Let $\{x_{1},x_{2},x_{3}\}\subseteq N_{G}(x)$ and $\{y_{1},y_{2}\}\subseteq N_{G}(y)$. Observe that $|\{x_{1},x_{2},x_{3}\}\cap\{y_{1},y_{2}\}|\leq 1$ and $|N_{G}(y_{i})\cap\{x_{1},x_{2},x_{3}\}|\leq 1$ for $i=1,2$; we without loss of generality assume that $y_{2}\notin\{x_{1},x_{2},x_{3}\}$, $y_{2}x_{2}\notin E(G)$ and $y_{2}x_{3}\notin E(G)$. Let $f$ be: $f(x)=f(y)=0,f(x_{2})=1,f(x_{3})=2$. Notice that either $y_{1}=x_{j}$ or $y_{1}x_{j}\notin E(G)$ for some $j\in[2,3]$; we further let $f(y_{1})=f(x_{j})$ and $f(y_{2})=[1,2]\setminus\{f(y_{1})\}$. Clearly, $f$ is a 2RiDF of $G[\{x,x_{2},x_{3},y,y_{1},y_{2}\}]$ of weight $|\{x,x_{2},x_{3},y,y_{1},y_{2}\}|-2$, a contradiction.

Observation 2. $G$ contains at most one $3^{+}$-vertex. Suppose to the contrary that $G$ has two distinct $3^{+}$-vertices, say $x$ and $y$. By Observation 1, $xy\in E(G)$. Let $\{y,x_{1},x_{2}\}\subseteq N_{G}(x)$ and $\{x,y_{1},y_{2}\}\subseteq N_{G}(y)$. Since $G$ contains no subgraph isomorphic to $C_{4}$, $|\{x_{1},x_{2}\}\cap\{y_{1},y_{2}\}|\leq 1$ and there are no edges between $\{x_{1},x_{2}\}$ and $\{y_{1},y_{2}\}$. We assume that $x_{2}\notin\{y_{1},y_{2}\}$ and $y_{2}\notin\{x_{1},x_{2}\}$. Then, the function $f$: $\{x,x_{1},x_{2},y,y_{1},y_{2}\}\rightarrow\{0,1,2\}$ such that $f(x)$=$f(y)$=[math], $f(x_{2})$=$f(y_{2})$=$2$ and $f(x_{1})$=$f(y_{1})$=$1$, is a 2RiDF of $G[\{x,y,x_{1},x_{2},y_{1},y_{2}\}]$ of weight $|\{x,y,x_{1},x_{2},y_{1},y_{2}\}|-2$, a contradiction.

Observation 3. If $G$ contains a $3^{+}$-vertex $x$, then $N_{G}(x)$ contains at most two 2-vertices; in particular, when $N_{G}(x)$ contains two 2-vertices, these two 2-vertices are adjacent in $G$. If not, suppose that $N_{G}(x)$ contains three 2-vertices, say $x_{1},x_{2},x_{3}$. Without loss of generality, we assume that $x_{3}\notin N_{G}(\{x_{1},x_{2}\})$ and let $N_{G}(x_{3})=\{x,y_{3}\}$. Let $N_{G}(x_{1})=\{x,y_{1}\}$ (possibly $y_{1}=x_{2}$, but $y_{1}\neq y_{3}$). By Observation 1, $d_{G}(y_{3})=1$, i.e., $y_{1}y_{3}\notin E(G)$. Let $f$ be: $f(x_{1})=f(x_{3})=0,f(x)=1,f(y_{1})=f(y_{3})=2$. Obviously, $f$ is a 2RiDF of $G[\{x,x_{1},y_{1},x_{3},y_{3}\}]$ of weight $|\{x,x_{1},y_{1},x_{3},y_{3}\}|-2$, a contradiction. Now, suppose that $N_{G}(x)$ contains two 2-vertices, say $x_{1},x_{2}$. If $x_{1}x_{2}\notin E(G)$, let $N_{G}(x_{i})=\{x,y_{i}\},i=1,2$. Clearly, $y_{1}\neq y_{2}$ and $y_{1}y_{2}\notin E(G)$. Let $f$ be: $f(x_{1})=f(x_{2})=0,f(x)=1,f(y_{1})=f(y_{2})=2$. Then, $f$ is a 2RiDF of $G[\{x,x_{1},y_{1},x_{2},y_{2}\}]$ of weight $|\{x,x_{1},x_{2},y_{1},y_{2}\}|-2$, a contradiction.

By the above three observations and the assumption that $G$ is connected, we see that if $G$ contains a $3^{+}$-vertex $x$, then $V(G)\setminus\{x\}$ contains either only 1-vertices ($G\cong S_{n-1}$), or one 2-vertex and $n-2$ 1-vertices ($G\cong S(n-3,1)$), or two adjacent 2-vertices and $n-3$ 1-vertices ($G\cong S_{n-1}^{+}$); if $\Delta(G)=2$, then it is easy to see that $G$ is isomorphic to one of $S^{+}_{2},S_{2},S(1,1)$ and $C_{5}$. ∎

According to Lemma 2.1, Theorem 2.3 and the fact that $\gamma_{\rm ri2}(G)$=$\sum_{i=1}^{k}\gamma_{\rm ri2}(G_{i})$, where $G_{1},\ldots,G_{k}$ are the components of $G$, we have the following theorem.

Theorem 2.4**.**

Let $G$ be a graph of order $n\geq 3$. Then, $\gamma_{\rm ri2}(G)=n-1$ if and only if $G$ has one component $G_{1}$ isomorphic to one among $S_{n_{1}-1}$ ($n_{1}\geq 3$), $S_{n_{1}-1}^{+}$ ($n_{1}\geq 3$), $S(n_{1}-3,1)$ ($n_{1}\geq 4$) and $C_{5}$, and other components are isomorphic to $K_{1}$ or $K_{2}$, where $n_{1}=|V(G_{1})|$.

3 An improved Nordhaus-Gaddum type theorem for $\gamma_{\rm ri2}(G)$

This section is devoted to achieve an improved Nordhaus-Gaddum type theorem by showing that $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})\leq n+2$ for every graph $G\not\cong C_{5}$ of order $n\geq 2$, which improves a result obtained by Šumenjak, et al [8]. Before doing so, we need to establish some simple lemmas.

Lemma 3.5**.**

Let $G$ be a graph of order $n\geq 3$. If $G$ is isomorphic to $S_{n-1},S_{n-1}^{+}$ or $S(n-3,1)$, then $\gamma_{\rm ri2}(\overline{G})\leq 3$.

Proof  If $G\cong S_{n-1}$ or $G\cong S_{n-1}^{+}$, let $V(G)=\{v_{0},v_{1},\ldots,v_{n-1}\}$ where $v_{0}$ is the center and $v_{1}v_{2}\in E(G)$ when $G\cong S_{n-1}^{+}$. Define a function $f$ such that $f(v_{1})=1,f(v_{0})=f(v_{2})=2$ and $f(v)=0$ for every $v\in V(\overline{G})\setminus\{v_{0},v_{1},v_{2}\}$. Since every vertex in $V(\overline{G})\setminus\{v_{0},v_{1},v_{2}\}$ is adjacent to both $v_{1}$ and $v_{2}$ in $\overline{G}$, it follows that $f$ is a 2RiDF of $\overline{G}$ of weight 3.

If $G\cong S(n-3,1)$, then $n\geq 4$. Let $G=v_{1}v_{0}u_{0}u_{1}\ldots u_{n-3}$. If $n=4$, then both $G$ and $\overline{G}$ are isomorphic to $P_{4}$, the path of length 3, and the conclusion holds. If $n\geq 5$, then the function $f$: $V(\overline{G})\rightarrow\{0,1,2\}$ such that $f(u_{1})=f(u_{0})=1,f(u_{2})=2$ and $f(v)=0$ for every $v\in V(\overline{G})\setminus\{v_{0},v_{1},v_{2}\}$ is a 2RiDF of $\overline{G}$ with weight 3.∎

Lemma 3.6**.**

Let $G$ be a graph of order $n$. If $G\not\cong C_{5}$ and $\gamma_{\rm ri2}(G)=4$, then $\gamma_{\rm ri2}(\overline{G})\leq n-2$.

Proof  Clearly, $n\geq 4$. When $n=4$ and $n=5$, the conclusion is easy to prove and we assume that $n\geq 6$. Suppose, to the contrary, that $\gamma_{\rm ri2}(\overline{G})\geq n-1$. If $\gamma_{\rm ri2}(\overline{G})=n$, then $\gamma_{\rm ri2}(G)=2$ by Lemma 2.1, a contradiction. Therefore, $\gamma_{\rm ri2}(\overline{G})=n-1$. By Theorem 2.4 $\overline{G}$ has one component isomorphic to $S_{n_{1}},S_{n_{1}}^{+}$, $S(n_{2},1)$ or $C_{5}$ where $n_{1}\geq 2,n_{2}\geq 1$, and all of the other components of $\overline{G}$ are isomorphic to $K_{1}$ or $K_{2}$.

If $\overline{G}$ contains two vertices $u,v$ such that $N_{\overline{G}}\{u,v\}=\emptyset$, then each of $u$ and $v$ is adjacent to all vertices of $V(G)\setminus\{u,v\}$ in $G$. We can obtain a 2RiDF of $G$ by assigning 1 to $u$, 2 to $v$, and 0 to the remained vertices of $G$. This indicates that $\gamma_{\rm ri2}(G)\leq 2$ and a contradiction. Therefore, $\overline{G}$ contains no $K_{2}$ components and contains at most one $K_{1}$ component, which implies that $\overline{G}$ has at most two components. If $\overline{G}$ contains only one component, then $\overline{G}$ is isomorphic to $S_{n-1},S_{n-1}^{+}$ or $S(n-3,1)$ (since $G\not\cong C_{5}$). By Lemma 3.5 $\gamma_{\rm ri2}(G)\leq 3$ and a contradiction. Therefore, $\overline{G}$ has two components, denoted by $G_{1}$ and $G_{2}$, where $G_{1}\cong K_{1}$ and $G_{2}$ is isomorphic to $S_{n-2},S_{n-2}^{+}$, $S(n-4,1)$ or $C_{5}$. Let $V(G_{1})=\{u\}$ and define a function $f$ as follows: let $f(u)=1$; $f(v_{0})=f(v^{\prime})=2$ when $G_{2}\cong S_{n-2}$ or $G_{2}\cong S_{n-2}^{+}$ (where $v_{0}$ is the center of $G_{2}$ and $v^{\prime}$ is a 1-vertex of $G_{2}$. Since $n\geq 6$ such $v^{\prime}$ does exist), $f(v_{0})=f(u_{0})=2$ when $G_{2}\cong S(n-4,1)$ (where $v_{0}u_{0}$ is the bridge of $G_{2}$), or $f(u_{1})=f(u_{2})=2$ when $G_{2}\cong C_{5}$ (where $C_{5}=u_{1}u_{2}u_{3}u_{4}u_{5}u_{1}$); and all of the other remained vertices are assigned value 0. Clearly, every vertex assigned value 0 is adjacent to $u$ and a vertex assigned value 2. Hence, $f$ is a 2RiDF of $G$ with weight 3, and a contradiction. ∎

Lemma 3.7**.**

Suppose that $G$ is a graph of order $n$ such that $\gamma_{\rm ri2}(G)\geq 4$ and $\gamma_{\rm ri2}(G)$+$\gamma_{\rm ri2}(\overline{G})$=$n+3$. Let $f=(V_{0},V_{1},V_{2})$ be an arbitrary $\gamma_{\rm ri2}(G)$-function. Then,

1. (1)

If $|V_{0}|\geq 2$, then for any two vertices $u,v\in V_{0}$, there are no vertices $u_{1},u_{2},v_{1},v_{2}$ such that $\{u_{1},u_{2}\}\in N_{\overline{G}}(u)$, $\{v_{1},v_{2}\}\in N_{\overline{G}}(v)$ and $u_{i}v_{i}\notin E(\overline{G})$ for $i=1,2$, where $u_{1}\neq u_{2},v_{1}\neq v_{2}$ but possibly $u_{i}=v_{i}$; 2. (2)

Suppose that $u,v$ be two arbitrary vertices of $V_{0}$. Then, $|N_{\overline{G}}(\{u,v\})|\geq 3$. 3. (3)

$|V_{i}|\geq 2$* for $i=0,1,2$.*

Proof  For (1), if the conclusion were false, let $g$ be: $g(u_{i})=g(v_{i})=i$ for $i=1,2$ and $g(u)=g(v)=0$. Then, $g$ is a 2RiDF of $\overline{G}[\{u,v,u_{1},v_{1},u_{2},v_{2}\}]$ with weight $|\{u,v,u_{1},v_{1},u_{2},v_{2}\}|-2$. Since $V_{1}$ and $V_{2}$ are cliques in $\overline{G}$, $V_{i}$, for $i=1,2$, contains at most two vertices not assigned 0 under every 2RiDF of $\overline{G}$. Hence, by Lemma 2.2 $g$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $|V_{0}|-2+4=|V_{0}|+2$. This shows that $\gamma_{\rm ri2}(\overline{G})\leq|V_{0}|+2$ and $\gamma_{\rm ri2}(G)$+$\gamma_{\rm ri2}(\overline{G})\leq|V_{1}|+|V_{2}|+|V_{0}|+2=n+2$, a contradiction.

For (2), if $|N_{\overline{G}}(\{u,v\})|\leq 2$, let $f$ be: $f(u)=1,f(v)=2$ and $f(x)=0$ for $x\in V(G)\setminus N_{\overline{G}}[\{u,v\}]$. Clearly, $f$ is a 2RiDF of $G[V(G)\setminus N_{\overline{G}}(\{u,v\})]$ with weight 2. By Lemma 2.2, $f$ can be extended to a 2RiDF of $G$ with weight at most $4$, since $|N_{\overline{G}}(\{u,v\})|\leq 2$. Thus, $\gamma_{\rm ri2}(G)=4$ and by Lemma 3.6 $\gamma_{\rm ri2}(\overline{G})\leq n-2$, a contradiction.

For (3), if $|V_{0}|=1$, then $\gamma_{\rm ri2}(G)$=$n-1$. By an analogous argument as that in Lemma 3.6, we can derive that $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})\leq n+2$, a contradiction. In the following, we prove that $|V_{1}|\geq 2$ (the proof of $|V_{2}|\geq 2$ is similar to that of $|V_{2}|\geq 2$). Suppose that $|V_{1}|=1$ and let $V_{1}=\{u\}$. Then, every vertex of $V_{0}$ is adjacent to $u$ in $G$, i.e., $u$ is not adjacent to $V_{0}$ in $\overline{G}$. By Lemma 3.6 we assume that $|V_{1}|+|V_{2}|\geq 5$. If $V_{0}$ contains a vertex $v$ with two neighbors $v_{1},v_{2}$ in $\overline{G}$, then $u\notin\{v_{1},v_{2}\}$. Let $g$ be: $g(v)=0,g(v_{1})=1,g(v_{2})=2$. Since $V_{2}$ is a clique in $\overline{G}$, by Lemma 2.2 $g$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $|V_{0}|-1+3=|V_{0}|+2$. This shows that $\gamma_{\rm ri2}(\overline{G})\leq|V_{0}|+2$ and hence $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})\leq n+2$, a contradiction. Therefore, every vertex in $V_{0}$ has degree at most 1 in $\overline{G}$, which implies that $|N_{\overline{G}}(\{x,y\})|\leq 2$ for any two vertices $x\in V_{0},y\in V_{0}$ (observe that $|V_{0}|\geq 2$). This contradicts (2).

Lemma 3.8**.**

Let $G$ be a graph of order $n\geq 4$ and $u\in V(G)$. If $H=G-u$, the resulting graph obtained from $G$ by deleting $u$ and its incident edges, is connected and $\gamma_{\rm ri2}(H)=|V(H)|-1$, then $G$ has a 2RiDF $f$ such that $f(u)=1$ and $f(v)=0$ for some $v\in V(H)$.

Proof  Clearly, $|V(H)|\geq 3$. If $u$ is not adjacent to $V(H)$, then let $f$ be: $f(u)=1$ and $f(v)=g(v)$ for every $v\in V(H)$ where $g$ is a $\gamma_{\rm ri2}(H)$-function of $H$. Since $\gamma_{\rm ri2}(H)=|V(H)|-1$, there is a $v\in V(H)$ satisfying $f(v)=g(v)=0$. If $u$ is adjacent to a vertex $u_{1}\in V(H)$, then there is a vertex $u_{2}\in V(H)$ adjacent to $u_{1}$ since $H$ is connected. Let $f$ be: $f(u_{1})=0,f(u)=1,f(u_{2})=2$. Then, by Lemma 2.2 $f$ can be extended to a desired 2RiDF of $G$. ∎

Now, we turn to the proof of our main result.

Theorem 3.9**.**

Let $G$ be a graph of order $n(\geq 2)$. If $G\not\cong C_{5}$, then $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})\leq n+2$.

Proof  We may assume that $n\geq 5$ as the statement holds trivially when $n=2,3,4$. Let $f_{0}=(V_{0},V_{1},V_{2})$ be a $\gamma_{\rm ri2}(G)$-function such that $\overline{G}[V_{0}]$ contains the maximum number of components isomorphic to $K_{2}$. Suppose to the contrary that $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})>n+2$. Then, $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})=n+3$ since $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})\leq n+3$ [8], that is,

[TABLE]

Formula (1) indicates that every 2RiDF of $\overline{G}$ has weight at least $|V_{0}|+3$. We will complete our proof by constructing a 2RiDF of $\overline{G}$ of weight at most $|V_{0}|+2$ or a 2RiDF of $G$ of weight less than $|V_{1}|+|V_{2}|$.

If $|V_{1}\cup V_{2}|=2$, then $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})\leq 2+n$, a contradiction; if $|V_{1}\cup V_{2}|=3$, then $\gamma_{\rm ri2}(\overline{G})=n$ and by Lemma 2.1 $\gamma_{\rm ri2}(G)=2$, also a contradiction. Therefore, by Lemma 3.6,

[TABLE]

Then, by Lemma 3.7 (3) we have $|V_{i}|\geq 2$ for $i=0,1,2$. In addition, because, by definition, $\overline{G}[V_{i}]$ is a clique, $i=1,2$, it follows that for every 2RiDF $g_{0}=(V^{\prime}_{0},V^{\prime}_{1},V^{\prime}_{2})$ of $\overline{G}$,

[TABLE]

Therefore, by Lemma 2.2 every $\gamma_{\rm ri2}(\overline{G}[V_{0}])$-function can be extended to a 2RiDF of $\overline{G}$ with weight at most $\gamma_{\rm ri2}(\overline{G}[V_{0}])+4$, i.e., $\gamma_{\rm ri2}(\overline{G}[V_{0}])\geq|V_{0}|-1$ by Formula (1).

Claim 1. Let $\ell$ be the number of vertices in $V_{1}\cup V_{2}$ which have degree $|V_{1}|+|V_{2}|-1$ in $\overline{G}[V_{1}\cup V_{2}]$. Then, $\ell\leq 1-\ell^{\prime}$ where $\ell^{\prime}=|V_{0}|-\gamma_{\rm ri2}(\overline{G}[V_{0}])$ $\leq 1$. If not, either $\ell\geq 2$ or $\ell=\ell^{\prime}=1$. Suppose that $\ell\geq 2$ and let $v_{1}$ and $v_{2}$ be two vertices of $V_{1}\cup V_{2}$ that are adjacent to all vertices of $(V_{1}\cup V_{2})\setminus\{u,v\}$ in $\overline{G}$. Let $g^{\prime}$ be: $g^{\prime}(v_{1})=1,g^{\prime}(v_{2})=2,g^{\prime}(x)=0$ for $x\in V_{1}\cup V_{2}\setminus\{v_{1},v_{2}\}$. Clearly, $g^{\prime}$ is a 2RiDF of $\overline{G}[V_{1}\cup V_{2}]$ and by Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $|V_{0}|+2$, a contradiction. Now, suppose that $\ell=\ell^{\prime}=1$. Then, $\gamma_{\rm ri2}(\overline{G}[V_{0}])=|V_{0}|-1$, which indicates that $\overline{G}[V_{0}]$ has a component $H^{\prime}$ such that $\gamma_{\rm ri2}(H^{\prime})=|V(H^{\prime})|-1$. Since $\ell=1$, there is a vertex $v$, say $v\in V_{1}$, which is adjacent to all vertices of $V_{2}$ in $\overline{G}$. By Lemma 3.8 $\overline{G}[V(H^{\prime})\cup\{v\}]$ has a 2RiDF $g^{\prime}$ for which $g^{\prime}(v)=1$ and $g^{\prime}(x)=0$ for some $x\in V(H^{\prime})$. Observe that every vertex in $(V_{1}\cup V_{2})\setminus\{v\}$ is adjacent to $v$ in $\overline{G}$; by the rule of Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF $g$ of $\overline{G}$ under which there is at most one vertex in $V_{1}\setminus\{v\}$ (and $V_{2}$) not assigned value 0. Thus, $w(g)\leq|V_{0}|-1+3=|V_{0}|+2$, a contradiction. This completes the proof of Claim 1. $\blacksquare$

In the following, without loss of generality we assume $|V_{1}|\geq|V_{2}|$. Then, $|V_{1}|\geq 3$ by Formula (2).

Claim 2. $\overline{G}[V_{0}]$ contains no isolated vertex $v$ such that $N_{\overline{G}}(v)\cap V_{1}=\emptyset$. Otherwise, let $f^{\prime}$ be: $f^{\prime}(v)=1,f^{\prime}(x)=2$ for $x\in V_{2}$. By Claim 1, $V_{1}$ has at most one vertex adjacent to all vertices of $V_{2}$ in $\overline{G}$; say $v^{\prime}$ if such a vertex exists. We further let $f(y)=0$ for $y\in V_{1}\cup(V_{0}\setminus\{v\})$ (or for $y\in(V_{1}\setminus\{v^{\prime}\})\cup(V_{0}\setminus\{v\})$ if $v^{\prime}$ exists). Since every vertex in $V_{1}\cup V_{0}$ (except for $v^{\prime}$) is adjacent to both $v$ and $V_{2}$ in $G$, $f$ is a 2RiDF of $G$ of weight at most $|V_{2}|+2$, a contradiction. $\blacksquare$

We proceed by distinguishing two cases: $\gamma_{\rm ri2}(\overline{G}[V_{0}])=|V_{0}|-1$ and $\gamma_{\rm ri2}(\overline{G}[V_{0}])=|V_{0}|$.

Case 1. $\gamma_{\rm ri2}(\overline{G}[V_{0}])=|V_{0}|-1$. In this case, by Claim 1 every vertex in $V_{i}$ has a neighbor in $V_{j}$ in $G$ where $\{i,j\}$=[1,2]; by Theorem 2.4, $\overline{G}[V_{0}]$ has one component $H$ isomorphic to one of $S_{|V(H)|-1}$ ($|V(H)|\geq 3$), $S_{|V(H)|-1}^{+}$ ($|V(H)|\geq 3$), $S(|V(H)|-3,1)$ ($|V(H)|\geq 4$) and $C_{5}$, and other components of $\overline{G}[V_{0}]$ are isomorphic to $K_{1}$ or $K_{2}$. Let $u_{0}\in V(H)$ be a vertex with $d_{H}(u_{0})=\Delta(H)$. Clearly, $d_{H}(u_{0})\geq 2$. Let $u_{1}\in N_{H}(u_{0})$ and $u_{2}\in N_{H}(u_{0})$ be two vertices such that every vertex in $V(H)\setminus\{u_{0},u_{1},u_{2}\}$ has degree in $H$ not exceeding $\min\{d_{H}(u_{1}),d_{H}(u_{2})\}$. By the structure of $H$, for $i=1,2$, we have that $d_{H}(u_{i})\leq 2$ and if $u_{i}$ has a neighbor $u^{\prime}_{i}(\neq u_{0}$) in $H$, then $u_{0}u^{\prime}_{i}\notin E(H)$. Moreover, by Lemma 3.7 (1), $(N_{\overline{G}}(u_{1})\cap N_{\overline{G}}(u_{2}))\setminus\{u_{0}\}=\emptyset$, which implies that every vertex in $V_{1}\cup V_{2}$ is adjacent to $u_{1}$ or $u_{2}$ in $G$.

Claim 3. $|V_{0}\setminus V(H)|\leq 1$. Otherwise, let $\{v_{1},v_{2}\}\subseteq(V_{0}\setminus V(H))$. Then, $d_{\overline{G}[V_{0}]}(v_{1})\leq 1$ and $d_{\overline{G}[V_{0}]}(v_{2})\leq 1$. Suppose that $d_{\overline{G}[V_{0}]}(v_{1})=1$ (the case of $d_{\overline{G}[V_{0}]}(v_{2})=1$ can be similarly discussed). Let $v_{1}v^{\prime}_{1}\in E(\overline{G}[V_{0}])$ and clearly $d_{\overline{G}[V_{0}]}(v^{\prime}_{1})=1$. By Lemma 3.7 (2), there exists a vertex $v_{0}\in(V_{1}\cup V_{2})$ that is adjacent to $\{v_{1},v^{\prime}_{1}\}$ in $\overline{G}$. Without loss of generality, we assume that $v_{1}v_{0}\in E(\overline{G})$. By Lemma 3.8, $\overline{G}[V(H)\cup\{v_{0}\}]$ has a 2RiDF $g^{\prime}$ such that $g^{\prime}(v_{0})=1$ and $g^{\prime}(x)=0$ for some $x\in V(H)$. Further, let $g^{\prime}(v_{1})=0$ and $g^{\prime}(v^{\prime}_{1})=2$. Then, $g^{\prime}$ is a 2RiDF of $\overline{G}[V(H)\cup\{v_{0},v_{1},v^{\prime}_{1}\}]$, and by Lemma 2.2 and Formula (3) $g^{\prime}$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $|V_{0}|-2+4=|V_{0}|+2$ (since $g^{\prime}(v_{1})=g^{\prime}(x)=0$), a contradiction. We therefore assume that $d_{\overline{G}[V_{0}]}(v_{1})=d_{\overline{G}[V_{0}]}(v_{2})=0$. By Lemma 3.7 (2) we have $|N_{\overline{G}}\{v_{1},v_{2}\}\cap(V_{1}\cup V_{2})|\geq 3$. Without loss of generality, we may assume that $v_{1}$ is adjacent to two vertices of $V_{1}\cup V_{2}$ in $\overline{G}$, say $v_{11}$ and $v_{12}$. By Lemma 3.7 (1), $u_{i}$ is not adjacent to both $v_{11}$ and $v_{12}$, and $v_{1j}$ is not adjacent to both $u_{1}$ and $u_{2}$ in $\overline{G}$, where $i\in[1,2]$ and $j\in[1,2]$. Thus, it follows that $u_{1}v_{11}\notin E(\overline{G})$ and $u_{2}v_{12}\notin E(\overline{G})$, or $u_{1}v_{12}\notin E(\overline{G})$ and $u_{2}v_{11}\notin E(\overline{G})$, which contradicts to Lemma 3.7 (1) again. $\blacksquare$

By Claim 3, we see that $\overline{G}[V_{0}]$ contains no component isomorphic to $K_{2}$ and contains at most one $K_{1}$ component.

Claim 4. $\overline{G}[V_{0}]$ contains a $K_{1}$ component. If not, we have $\overline{G}[V_{0}]$=$H$.

Claim 4.1. $(N_{\overline{G}}(u_{1})\cup N_{\overline{G}}(u_{2}))\cap(V_{1}\cup V_{2})\neq\emptyset$. Otherwise, both $u_{1}$ and $u_{2}$ are adjacent to all vertices of $V_{1}\cup V_{2}$ in $G$, and by Lemma 3.7 (2) $d_{H}(u_{i})=2$ for $i=1,2$ and $u_{1}u_{2}\notin E(\overline{G})$. Let $\{u^{\prime}_{i}\}=N_{H}(u_{i})\setminus\{u_{0}\},i=1,2$; then, $u_{0}u^{\prime}_{i}\notin E(\overline{G})$. Let $f$ be: $f(u_{1})=f(u^{\prime}_{1})=1$, $f(u_{2})=f(u^{\prime}_{2})=2$ and $f(x)=0$ for $x\in V(G)\setminus\{u_{1},u^{\prime}_{1},u_{2},u^{\prime}_{2}\}$. Then, $f$ is a 2RiDF of $G$ with weight 4, a contradiction. $\blacksquare$

Claim 4.2. $|V_{1}|=3$. Observe that $|V_{1}|\geq 3$; it is enough to show that $G$ has a 2RiDF $f$ with $w(f)\leq|V_{2}|+3$. When $u_{1}u_{2}\in E(\overline{G})$, let $f$ be: $f(u_{0})=f(u_{1})=f(u_{2})=1,f(x)=0$ for $x\in(V_{1}\cup V_{0})\setminus\{u_{0},u_{1},u_{2}\}$ and $f(y)=2$ for $y\in V_{2}$. By Lemma 3.7 (1), $V_{1}\cup V_{0}$ contains no vertex adjacent to both $u_{1}$ and $u_{2}$ in $\overline{G}$. Therefore, $f$ is a 2RiDF of $G$ of weight $|V_{2}|+3$. Now, suppose that $u_{1}u_{2}\notin E(\overline{G})$. By Lemma 3.7 (1), $V_{1}$ contains at most one vertex adjacent to both $u_{0}$ and $u_{1}$ in $\overline{G}$; say $u$ if such a vertex exists. Let $f$ be: $f(u_{0})=f(u_{1})=1$ (or $f(u_{0})=f(u_{1})=f(u)=1$ if $u$ exists), $f(x)=0$ for $x\in(V_{1}\cup(V_{0}\setminus\{u_{0},u_{1}\}))$ (or $x\in(V_{1}\cup V_{0})\setminus\{u_{0},u_{1},u\}$) and $f(y)=2$ for $y\in V_{2}$. Notice that by Claim 1 every vertex in $V_{0}\cup V_{1}$ is adjacent to $V_{2}$ in $G$, and by the structure of $H$ and the selection of $u_{1}$ and $u_{2}$, every vertex of $(V_{0}\cup V_{1})\setminus\{u_{0},u_{1},u\}$ is adjacent to $\{u_{0},u_{1}\}$ in $G$; $f$ is a 2RiDF of $G$ of weight at most $|V_{2}|+3$. $\blacksquare$

By Claim 4.2, we have $2\leq|V_{2}|\leq 3$. Let $V_{1}=\{w_{1},w_{2},w_{3}\}$ in the following.

Claim 4.3. Every vertex of $V_{i}$ is adjacent to at most one vertex of $V_{j}$ in $\overline{G}$ for $\{i,j\}=[1,2]$. If not, suppose that $V_{2}$ contains a vertex $v$ adjacent to two vertices of $V_{1}$ in $\overline{G}$, say $w_{1},w_{2}$. By Lemma 3.7 (1) $v$ is not adjacent to $u_{1}$ or $u_{2}$ in $\overline{G}$, say $u_{1}v\notin E(\overline{G})$. If $u_{2}w_{3}\notin E(\overline{G})$, let $g^{\prime}$ be: $g^{\prime}(u_{i})=i$ for $i=0,1,2$, $g^{\prime}(w_{1})=g^{\prime}(w_{2})=0,g^{\prime}(w_{3})=2$, $g^{\prime}(v)=1$. If $u_{2}w_{3}\in E(\overline{G})$, then $u_{1}w_{3}\notin E(\overline{G})$ and let $g^{\prime}$ be: $g^{\prime}(u_{1})=g^{\prime}(w_{3})=1,g^{\prime}(w_{1})=g^{\prime}(w_{2})=0,g^{\prime}(v)=2$; further, let $g^{\prime}(u_{2})=0$ when $u_{2}v\in E(\overline{G})$, or let $g^{\prime}(u_{2})=2$ and $g^{\prime}(u_{0})=0$ when $u_{2}v\notin E(\overline{G})$. By Lemma 2.2 in either case we can extended the $g^{\prime}$ defined above to a 2RiDF $g$ of $\overline{G}$ under which $g(w_{1})=g(w_{2})=0$ and $g(u_{0})=0$ or $g(u_{2})=0$. Therefore, by Formula (3) $w(g)\leq|V_{0}|-1+3=|V_{0}|+2$, a contradiction. With a similar argument, we can also get a contradiction if we assume $V_{1}$ contains a vertex adjacent to two vertices of $V_{2}$ in $\overline{G}$. $\blacksquare$

Now, we consider the value of $|V_{2}|$. Suppose that $|V_{2}|=3$ and let $V_{2}=\{w_{4},w_{5},w_{6}\}$. By Claim 4.1, we may assume, without loss of generality, that $u_{1}w_{1}\in E(\overline{G})$. This indicates that $u_{2}w_{1}\notin E(\overline{G})$ by Lemma 3.7 (1). If $u_{2}$ is adjacent to $V_{2}$, say $u_{2}w_{4}\in E(\overline{G})$, then by Lemma 3.7 (1), $u_{1}w_{4}\notin E(\overline{G})$, $w_{1}w_{4}\in E(\overline{G})$, and $u_{1}$ (resp. $u_{2}$) is not adjacent to $\{w_{2},w_{3}\}$ (resp. $\{w_{5},w_{6}\}$) in $\overline{G}$ (otherwise $w_{4}$ or $w_{1}$ is adjacent to two vertices of $V_{1}$ or $V_{2}$ in $\overline{G}$, respectively. This contradicts to Claim 4.3). Let $f$ be: $f(u_{1})=f(w_{1})=1,f(u_{2})=f(w_{4})=2$ and $f(x)=0$ for $x\in V(G)\setminus\{u_{1},u_{2},w_{1},w_{4}\}$. Observe that $w_{1}$ (resp. $w_{4}$) is not adjacent to $\{w_{5},w_{6}\}$ (resp. $\{w_{2},w_{3}\}$) in $\overline{G}$ and by Lemma 3.7 (1) $V_{0}\setminus\{u_{0},u_{1},u_{2}\}$ contains no vertex adjacent to both $u_{i}$ and $w_{i}$ for some $i\in[1,2]$. Hence, $f$ is a 2RiDF of $G[V(G)\setminus\{u_{0}\}]$ of weight 4 and by Lemma 2.2 $f$ can be extended to a 2RiDF of $G$ with weight at most $5<|V_{1}|+|V_{2}|$, a contradiction. Therefore, we may assume that $N_{\overline{G}}(u_{2})\cap V_{2}=\emptyset$. In this case, when $N_{\overline{G}}(u_{2})\cap V_{1}=\emptyset$, let $f$ be: $f(u_{2})=2$, $f(u_{0})=f(u_{1})=1$. By Lemma 3.7 (1) $V_{1}\cup V_{2}$ contains at most one vertex $w^{\prime}$ adjacent to both $u_{0}$ and $u_{1}$ in $\overline{G}$ and $V_{0}\setminus\{u_{0}\}$ contains at most one vertex $u^{\prime}$ adjacent to $u_{2}$ in $\overline{G}$; we further let $f(x)=0$ for $x\in V(G)\setminus\{u_{0},u_{1},u_{2},u^{\prime},w^{\prime}\}$. Then, $f$ is a 2RiDF of $G[V(G)\setminus\{u^{\prime},w^{\prime}\}]$ of weight 3 and by Lemma 2.2 $f$ can be extended to a 2RiDF of $G$ of weight at most $5<|V_{1}|+|V_{2}|$, a contradiction. We therefore suppose that $u_{2}$ is adjacent to $V_{1}$ in $\overline{G}$, say $u_{2}w_{2}\in E(\overline{G})$. Then, with the same argument as $N_{\overline{G}}(u_{2})\cap V_{2}=\emptyset$, we can show that $N_{\overline{G}}(u_{1})\cap V_{2}=\emptyset$ as well.

Then, if $w_{3}u_{1}\notin E(\overline{G})$ and $w_{3}u_{2}\notin E(\overline{G})$, the function $f$: $f(u_{1})=f(w_{1})=1,f(u_{2})=f(w_{4})=2$ and $f(x)=0$ for $x\in V(G)\setminus\{u_{1},u_{2},w_{1},w_{4},u_{0}\}$ is a 2RiDF of $G[V(G)\setminus\{u_{0}\}]$ with weight 4, and by Lemma 2.2 $f$ can be extended to a 2RiDF of $G$ with weight at most $5<|V_{1}|+|V_{2}|$, a contradiction. Therefore, we suppose that $w_{3}u_{1}\in E(\overline{G})$ by the symmetry. By Lemma 3.7 (1), it has that $w_{3}u_{2}\notin E(\overline{G})$, and $u_{0}w_{1}\notin E(\overline{G})$ or $u_{0}w_{3}\notin E(\overline{G})$, say $u_{0}w_{1}\notin E(\overline{G})$ by the symmetry. Let $f$ be: $f(u_{0})=f(u_{1})=1,f(u_{2})=f(w_{2})=2$ and $f(x)=0$ for $x\in V(G)\setminus\{u_{1},u_{2},u_{0},w_{2},w_{3}\}$. Since every vertex in $V(G)\setminus\{u_{1},u_{2},u_{0},w_{2},w_{3}\}$ is adjacent to both $\{u_{0},u_{1}\}$ and $\{u_{2},w_{2}\}$ in $G$, $f$ is a 2RiDF of $G[V(G)\setminus\{w_{3}\}]$ of weight 4 and by Lemma 2.2 $f$ can be extended to a 2RiDF of $G$ of weight at most $5<|V_{1}|+|V_{2}|$, and a contradiction.

A similar line of thought leads to a contradiction if we assume that $|V_{2}|=2$ and proves Claim 4. $\blacksquare$

By Claim 4, we see that $\overline{G}[V_{0}]$ contains one component isomorphic to $K_{1}$. Let $s$ be the vertex of the $K_{1}$ component. We first show that $|N_{\overline{G}}(s)\cap(V_{1}\cup V_{2})|\leq 1$. If not, we assume that $s$ is adjacent to two vertices of $V_{1}\cup V_{2}$ in $\overline{G}$, say $s_{1},s_{2}$. By Lemma 3.7 (1) $s_{i}$ (resp. $u_{j}$) is not adjacent to both $u_{1}$ and $u_{2}$ (resp. $s_{1}$ and $s_{2}$) in $\overline{G}$ for every $i,j\in[1,2]$. This implies that either $s_{i}u_{i}\notin E(\overline{G})$ for $i=1,2$ or $s_{1}u_{2}\notin E(\overline{G})$ and $s_{2}u_{1}\notin E(\overline{G})$, which contradicts to Lemma 3.7 (1) as well. Thus, by Claim 2 $|N_{\overline{G}}(s)\cap(V_{1}\cup V_{2})|=1$ and the vertex $s^{\prime}$ adjacent to $s$ in $\overline{G}$ belongs to $V_{1}$. Let $f$ be: $f(s)=2$, $f(x)=1$ for $x\in V_{1}$, $f(y)=0$ for $y\in V_{2}\cup V(H))$. Observe that by Claim 1 every vertex in $V_{2}$ is adjacent to $V_{1}$ in $G$ and hence every vertex in $V_{2}\cup V(H)$ is adjacent to both $V_{1}$ and $s$ in $G$; $f$ is a 2RiDF of $G$ with weight $|V_{1}|+1<|V_{1}|+|V_{2}|$ (since $|V_{2}|\geq 2$), a contradiction.

The foregoing discussion shows that there exists a contradiction if we assume that $\gamma_{\rm ri2}(\overline{G}[V_{0}])=|V_{0}|-1$. In the following, we consider the case of $\gamma_{\rm ri2}(\overline{G}[V_{0}])=|V_{0}|$.

Case 2. $\gamma_{\rm ri2}(\overline{G}[V_{0}])=|V_{0}|$. Then by Lemma 2.1 each component of $\overline{G}[V_{0}]$ is isomorphic to $K_{1}$ or $K_{2}$. Recall that $|V_{i}|\geq 2$ for $i=0,1,2$. Let $u,v$ be two vertices of $V_{0}$ such that $uv\in E(\overline{G})$ if $\overline{G}[V_{0}]$ contains a $K_{2}$ component and $u,v$ are isolated vertices in $\overline{G}[V_{0}]$ otherwise. By Lemma 3.7 (1), we have

[TABLE]

We deal with two subcases in terms of the adjacency property of $u$ and $v$.

Case 2.1. $uv\in E(\overline{G})$. Then every vertex in $V_{0}\setminus\{u,v\}$ is not adjacent to $\{u,v\}$ in $\overline{G}$.

Claim 5. Every vertex in $V_{1}\cup V_{2}$ has degree at most $|V_{1}|+|V_{2}|-2$ in $\overline{G}[V_{1}\cup V_{2}]$. Suppose that $V_{1}$ contains a vertex $w$ adjacent to all vertices of $V_{2}$ in $\overline{G}$. If $uw\in E(\overline{G})$ (or $vw\in E(\overline{G})$), then by Lemma 2.2 the 2RiDF $g^{\prime}$ of $\overline{G}[\{u,v,w\}]$ such that $g^{\prime}(u)=0$ (or $g^{\prime}(v)=0),g^{\prime}(w)=1$ and $g^{\prime}(v)=2$ ($g^{\prime}(u)=2$) can be extended to a 2RiDF of $\overline{G}$, under which $(V_{1}\cup V_{2})\setminus\{w\}$ contains at most two vertices not assigned 0. Thus, $w(g)\leq|V_{0}|-1+3=|V_{0}|+2$, a contradiction. We therefore assume that $uw\notin E(\overline{G})$ and $vw\notin E(\overline{G})$. By Lemma 3.7 (2), $V_{1}\cup V_{2}$ contains at least three vertices adjacent to $u$ or $v$. Without loss of generality, we may suppose that there is a vertex $u^{\prime}\in V_{1}\cup V_{2}$ that is adjacent to $u$ in $\overline{G}$. Define a 2RiDF $g^{\prime}$ of $\overline{G}[\{u,v,u^{\prime},w\}]$ as follows: $g^{\prime}(u^{\prime})=2,g^{\prime}(u)=0$ and $g^{\prime}(v)=g^{\prime}(w)=1$. Then, by Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF $g$ of $\overline{G}$, under which $(V_{1}\cup V_{2})\setminus\{w,u^{\prime}\}$ contains at most one vertex not assigned value 0. Therefore, $w(g)\leq|V_{0}|-1+3=|V_{0}|+2$, a contradiction. With a similar argument, we can also obtain a contradiction if we assume that $V_{2}$ contains a vertex adjacent to all vertices of $V_{1}$. This completes the proof of Claim 5. $\blacksquare$

By Claim 5, every vertex in $V_{i}$ has a neighbor in $V_{j}$ in $G$ for $\{i,j\}$=[1,2]. If $V_{1}\cap(N_{\overline{G}}(u)\cap N_{\overline{G}}(v))=\emptyset$, then every vertex in $V_{1}$ is adjacent to $u$ or $v$ in $G$. Let $f$ be: $f(u)=f(v)=1,f(x)=2$ for $x\in V_{2}$ and $f(y)=0$ for $y\in V_{1}\cup(V_{0}\setminus\{u,v\})$. Clearly, $f$ is a 2RiDF of $G$ with weight $|V_{2}|+2<|V_{1}|+|V_{2}|$, a contradiction. We therefore assume that $V_{1}$ contains a vertex $s$ such that $su\in E(\overline{G})$ and $sv\in E(\overline{G})$. Then, by Lemma 3.7 (1) $V_{2}\cup(V_{1}\setminus\{s\})$ contains no vertex adjacent to both $u$ and $v$ in $\overline{G}$. Analogously, the function $f$ such that $f(u)=f(v)=1,f(x)=2$ for $x\in V_{1}$ and $f(y)=0$ for $y\in V_{2}\cup(V_{0}\setminus\{u,v\})$ (and $f(u)=f(v)=f(s)=1,f(x)=2$ for $x\in V_{2}$ and $f(y)=0$ for $y\in(V_{1}\setminus\{s\})\cup(V_{0}\setminus\{u,v\})$) is a 2RiDF of $G$ with weight $|V_{1}|+2$ (and $|V_{2}|+3$). This implies that $|V_{1}|=3$ and $|V_{2}|=2$. Let $V_{1}=\{s,s_{1},s_{2}\}$ and $V_{2}=\{s_{3},s_{4}\}$. Then, $\{u,v\}$ contains no vertex adjacent to both $s_{1}$ and $s_{2}$ in $\overline{G}$; otherwise, we, by the symmetry, suppose that $us_{1}\in E(\overline{G})$ and $us_{2}\in E(\overline{G})$. Then, the function $g^{\prime}(u)=1,g^{\prime}(s)=2,g^{\prime}(v)=g^{\prime}(s_{1})=g^{\prime}(s_{2})=0$ is a 2RiDF of $\overline{G}[\{u,v,s,s_{1},s_{2}\}]$ with weight 2, and by Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $|V_{0}|-1+|V_{2}|+1=|V_{0}|+2$, a contradiction. In addition, by Lemma 3.7 (1) $s_{i},i=1,2$, is not adjacent to both $u$ and $v$ in $\overline{G}$. Therefore, we may assume, by the symmetry, that $s_{1}v\notin E(\overline{G})$ and $s_{2}u\notin E(\overline{G})$.

Suppose that there are no edges between $\{u,v\}$ and $V_{2}$ in $\overline{G}$. By Lemmas 3.7 (2), $us_{1}\in E(\overline{G})$ and $vs_{2}\in E(\overline{G})$. Then, the function $g^{\prime}$ such that $g^{\prime}(u)=1,g^{\prime}(s_{2})=2,g^{\prime}(s)=g^{\prime}(s_{1})=g^{\prime}(v)=0$ is a 2RiDF of $\overline{G}[\{u,v,s,s_{1},s_{2}\}]$ with weight 2. By Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $|V_{2}|+1+|V_{0}|-1=|V_{0}|+2$, a contradiction. We therefore assume that there is an edge between $\{u,v\}$ and $V_{2}$ in $\overline{G}$, say $vs_{3}\in E(\overline{G})$ by the symmetry.

If $s_{4}s\in E(\overline{G})$, then the function $g^{\prime}$ such that $g^{\prime}(s_{3})=2,g^{\prime}(s_{4})=0,g^{\prime}(s)=1,g^{\prime}(v)=0$ is a 2RiDF of $\overline{G}[\{s,v,s_{3},s_{4}\}]$ with weight 2, and by Lemma 2.2 and Formula 3 $g^{\prime}$ can be extended to a 2RiDF of $\overline{G}$ of weight at most $|V_{0}|-1+3=|V_{0}|+2$, a contradiction. Consequently, we have $s_{4}s\notin E(\overline{G})$. Then, the function $g^{\prime}$ such that $g^{\prime}(s_{3})=0,g^{\prime}(s_{4})=g^{\prime}(s)=2,g^{\prime}(v)=1,g^{\prime}(u)=0$ is a 2RiDF of $\overline{G}[\{s,u,v,s_{3},s_{4}\}]$ with weight 3, and by Lemma 2.2 and Formula 3 $g^{\prime}$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $|V_{0}|-1+3=|V_{0}|+2$, a contradiction.

Case 2.2. $uv\notin E(\overline{G})$. Then, by the selection of $u,v$ and $f_{0}$, $\overline{G}[V_{0}]$ contains only isolated vertices and $G$ contains no $\gamma_{\rm ri2}(G)$-function for which the subgraph of $\overline{G}$ induced by the set of vertices assigned value 0 contains $K_{2}$ components.

For every $x\in V_{0}$, let $U_{i}^{x}=N_{\overline{G}}(x)\cap V_{i}$ for $i=1,2$. Let $f^{\prime}$ be: $f^{\prime}(u)=1,f^{\prime}(v)=2$ and $f^{\prime}(x)=0$ for $x\in((V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2}))\cup(V_{0}\setminus\{u,v\})$. Obviously, $f^{\prime}$ is a 2RiDF of $G-(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2}))$ with weight 2. By Lemma 2.2 $f^{\prime}$ can be extended to a 2RiDF of $G$ with weight at most $|(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2}))|+2$. To ensure $|(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2}))|+2\geq|V_{1}|+|V_{2}|$, we have

[TABLE]

Claim 6. $|(V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2})|=2$ and the two vertices in $(V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2})$ are adjacent in $\overline{G}$. Let $g^{\prime}$ be a 2RiDF of $\overline{G}[V_{0}]$ such that $g^{\prime}(u)=g^{\prime}(v)=1$. Suppose that $|(V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2})|\leq 1$. Since $V_{1}$ and $V_{2}$ are cliques in $\overline{G}$ and every vertex in $U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2}$ is adjacent to $u$ or $v$ in $\overline{G}$, by Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF $g$ of $\overline{G}$ under which at most one vertex in $V_{i},i=1,2$, is not assigned value 0 (here if $(V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2})$ contains a vertex, say $w$, then let $g(w)=2$). Clearly, $w(g)=w(g^{\prime})+2\leq|V_{0}|+2$, a contradiction. Moreover, if $(V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2})$ contains two nonadjacent vertices in $\overline{G}$, say $w_{1},w_{2}$, then $w_{1}$ and $w_{2}$ are not in the same set $V_{i}$ for some $i\in[1,2]$. Therefore, we can extend $g^{\prime}$ to a 2RiDF $g$ of $\overline{G}$ by letting $g^{\prime}(w_{1})=g^{\prime}(w_{2})=2$ and $g^{\prime}(x)=0$ for $x\in(V_{1}\cup V_{2})\setminus\{w_{1},w_{2}\}$. But $w(g)=w(g^{\prime})+2\leq|V_{0}|+2$, a contradiction. $\blacksquare$

By Claim 6, $(V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2})$ contains two adjacent vertices in $\overline{G}$, say $w_{1},w_{2}$. If $V_{0}\setminus\{u,v\}$ contains a vertex $z$ that is adjacent to $w_{1}$ (or $w_{2}$) in $\overline{G}$, then let $g^{\prime}$ be: $g^{\prime}(u)=g^{\prime}(v)=g^{\prime}(z)=1$, $g^{\prime}(w_{1})=0$ (or $g^{\prime}(w_{2})=0$), $g^{\prime}(w_{2})=2$ (or $g^{\prime}(w_{1})=2$). Since every vertex $(V_{1}\cup V_{2})\setminus\{w_{2}\}$ is adjacent to $\{z,u,v\}$ in $\overline{G}$ and every vertex in $V^{\prime}\setminus\{w_{2}\}$ is adjacent to $w_{2}$ where $w_{2}\in V^{\prime}$ for some $V^{\prime}\in\{V_{1},V_{2}\}$, by Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF $g$ of $\overline{G}$ under which every vertex in $V^{\prime}\setminus\{w_{2}\}$ is assigned value 0 and at most one vertex in $\{V_{1},V_{2}\}\setminus V^{\prime}$ is not assigned value 0. Therefore, $w(g)\leq|V_{0}|+2$, a contradiction. This shows that every vertex in $V_{0}$ is not adjacent to $\{w_{1},w_{2}\}$ in $\overline{G}$. Furthermore, if there exists a vertex $z\in V_{0}\setminus\{u,v\}$, then by Claim 6 we have $(V_{1}\cup V_{2})\setminus(U^{u}_{1}\cup U^{u}_{2}\cup U^{z}_{1}\cup U^{z}_{2})=\{w_{1},w_{2}\}$ and $(V_{1}\cup V_{2})\setminus(U^{v}_{1}\cup U^{v}_{2}\cup U^{z}_{1}\cup U^{z}_{2})=\{w_{1},w_{2}\}$, which implies that $N_{\overline{G}}(z)=U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2}$. Then, the function $g^{\prime}$ such that $g^{\prime}(z)=1,g^{\prime}(u)=g^{\prime}(v)=2$ and $g^{\prime}(x)=0$ for $x\in U^{u}_{1}\cup U^{u}_{2}\cup U^{v}_{1}\cup U^{v}_{2}$ is a 2RiDF of $\overline{G}-(\{w_{1},w_{2}\}\cup(V_{0}\setminus\{u,v,z\}))$ with weight 3, and by Lemma 2.2 $g^{\prime}$ can be extended to a 2RiDF of $\overline{G}$ with weight at most $(|V_{0}|+2-3)+3=|V_{0}|+2$, a contradiction. So far, we have shown that $V_{0}=\{u,v\}$, i.e., $\gamma_{\rm ri2}(G)=n-2$.

Now, let $f^{\prime}$ be: $f^{\prime}(u)=1,f^{\prime}(v)=2$ and $f^{\prime}(w_{1})=f^{\prime}(w_{2})=0$. Clearly, $f^{\prime}$ is a 2RiDF of $G[\{u,v,w_{1},w_{2}\}]$. Then, by Lemma 2.2 $f^{\prime}$ can be extended to a 2RiDF $f$ of $G$ with weight at most $n-2$. To ensure $w(f)\geq\gamma_{\rm ri2}(G)=n-2$, we have $w(f)=n-2$, i.e., $f$ is a $\gamma_{\rm ri2}(G)$-function. We see that the subgraph of $\overline{G}$ induced by $\{w_{1}w_{2}\}$ is isomorphic to $K_{2}$. But this contradicts the selection of $f_{0}$. Eventually, we complete the proof of Theorem 3.9. ∎

4 The $\mathcal{NP}$-completeness

In this section, we study the $\mathcal{NP}$-completeness of the $k$-rainbow independent domination problem. To prove a given problem $P$ to be $\mathcal{NP}$-complete, we have to show that $P\in\mathcal{NP}$ and find a known $\mathcal{NP}$-complete problem that can be reduced to $P$ in polynomial time. Here, by establishing an equivalence relation between the domination problem and $k$-rainbow independent domination problem, we can show that the $k$-rainbow independent domination problem is $\mathcal{NP}$-complete when restricted to bipartite graphs. Three problems involved in our proof are described as follows:

The independent domination problem (IDP) [10].

Input: A graph $G$ and a positive integer $k$;

Property: $G$ has an IDS with at most $k$ vertices.

The domination problem (DP) [11].

Input: A graph $G$ and a positive integer $k$;

Property: $G$ has an dominating set with at most $k$ vertices.

The $k$-rainbow independent domination problem ($k$RiDP).

Input: A graph $G$ and two positive integers $k$ and $k^{\prime}$;

Property: $G$ has a $k$RiDF with weight at most $k^{\prime}$.

The operation of identifying two vertices $x$ and $y$ of a graph $G$ is to replace these vertices by a single vertex incident to all the edges which were incident in $G$ to either $x$ or $y$.

Theorem 4.10**.**

The $k$-rainbow independent domination problem is $\mathcal{NP}$-complete for bipartite graphs.

Proof  The $k$RiDP is a member of $\mathcal{NP}$, since we can check in polynomial time that a function from vertex set to $\{0,1,\ldots,k\}$ has weight at most $k^{\prime}$ and is a $k$RiDF.

When $k=1$, the $k$RiDP is equivalent to the IDP which is $\mathcal{NP}$-complete when $G$ is restricted to bipartite graphs [3]. Therefore, we assume that $k\geq 2$. To show $\mathcal{NP}$-hardness, we give a reduction from the domination problem (DP) for bipartite graphs, which is $\mathcal{NP}$-complete [12]. Given a bipartite $G$ with a bipartition $(X,Y)$ where $X=\{x_{1},x_{2},\ldots,x_{m}\}$ and $Y=\{y_{1},y_{2},\ldots,y_{n}\}$, we construct a new graph $G^{\prime}$ by adding $m+n$ copies of star $S_{k-1}$, denoted by $S_{k-1}(x_{i})$ and $S_{k-1}(y_{j})$ for $i\in[1,m]$ and $j\in[1,n]$, and identifying $w$ and the center of $S_{k-1}(w)$ for all $w\in\{x_{i},y_{j}|i=1,\ldots,m,j=1,\dots,n\}$ (see Figure 1, in which we omit the edges between $X$ and $Y$). Clearly, $G^{\prime}$ is also a bipartite graph. We claim that $G^{\prime}$ has a $k$RiDF with weight $(k-1)(m+n)+\ell$ if and only if $G$ has a dominating set of size $\ell$.

Given a $k$RiDF $f=(V_{0},V_{1},\ldots,V_{k})$ of $G^{\prime}$ with weight $(k-1)(m+n)+\ell$, let $D=(D^{\prime}=V_{1}\cup\ldots\cup V_{k})\cap(X\cup Y)$. Observe that all leaves of $S_{k-1}(x_{i})$ and $S_{k-1}(y_{j})$ for $i\in[1,m]$ and $j\in[1,n]$ belong to $D^{\prime}$; therefore, $|D|=\ell$. Since $f$ is a $k$RiDF, it follows that every vertex in $V_{0}$ is adjacent to at least one vertex in $D$. Notice that $X\cup Y=V_{0}\cup D$; we see that $D$ is a dominating set of $G$.

Now, we assume that $G$ has a dominating set $D$ where $|D|=\ell$. Let $D_{1}=D\cap X$ and $D_{2}=D\cap Y$. We define a function $f$: $V(G^{\prime})\rightarrow[0,k]$ as follows: $f(v)=1$ for every $v\in D_{1}$, $f(v)=2$ for every $v\in D_{2}$ and $f(v)=0$ for every $v\in(X\cup Y)\setminus D$. Since $G$ is bipartite and $D$ is a dominating set of $G$, every vertex $v\in(X\cup Y)\setminus D$ is adjacent to either $D_{1}$ or $D_{2}$ in $G$. If $v$ is adjacent to $D_{i}$ for some $i\in[1,2]$ in $G$, then we assign $[1,k]\setminus\{i\}$ to the $k-1$ leaves of $S_{k-1}(v)$ such that every leaf receives an unique number of $[1,k]\setminus\{i\}$. Finally, for every $u\in D_{i}$ for $i=1,2$, we assign $[1,k]\setminus\{i\}$ to the $k-1$ leaves of $S_{k-1}(u)$ such that every leaf receives an unique number of $[1,k]\setminus\{i\}$. Clearly, $w(f)=|D|+(m+n)(k-1)$, $V_{i}$ is an independent set and every vertex in $V_{0}$ is adjacent to a vertex in $V_{i}$ for all $i\in[1,k]$. Therefore, $f$ is a $k$RiDF with weight $|\ell|+(m+n)(k-1)$. ∎

5 Conclusion

In this paper, we respond some questions proposed by Šumenjak et al. [8], by proving an improved Nordhaus-Gaddum type inequality on $k$-rainbow independent domination number and showing that the problem of deciding whether a graph has a $k$-rainbow independent dominating function of a given weight is $\mathcal{NP}$-complete. In the study, we proved that when $G$ satisfies $\gamma_{\rm ri2}(G)=|V(G)|-1$ and $G\not\cong C_{5}$, it follows that $G$ is isomorphic to $S_{n}(n\geq 2)$, $S_{n}^{+}(n\geq 2)$ or $S(n,1)(n\geq 1)$, and $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})=|V(G)|+2$. Additionally, we observe that $\gamma_{\rm ri2}(S(n,m)+\gamma_{\rm ri2}(\overline{S(n,m)})=|V(S(n,m))|+1$ when $m\geq 2$. Therefore, a question that arises is whether $S_{n}(n\geq 2)$, $S_{n}^{+}(n\geq 2)$ and $S(n,1)(n\geq 1)$ are enough for determining graphs $G$ with the property of $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})=|V(G)|+2$. We formulate this more generally as follows:

Question 5.1**.**

How to characterize graphs $G$ with $\gamma_{\rm ri2}(G)+\gamma_{\rm ri2}(\overline{G})=|V(G)|+2?$

Acknowledgments

This work was supported by the National Natural Science Foundation of China (61872101, 61672051, 61309015, 61702075), the China Postdoctoral Science Foundation under grant (2017M611223).