An Improved Nordhaus–Gaddum-Type Theorem for 2-Rainbow Independent Domination Number

Abstract

For a graph G, its k-rainbow independent domination number, written as ${\gamma }_{\mathrm{rik}}\left(G\right)$, is defined as the cardinality of a minimum set consisting of k vertex-disjoint independent sets $V_1,V_2,\dots ,V_k$ such that every vertex in $V_0=V\left(G\right)\backslash \left({\cup }_{i=1}^k{V}_i\right)$ has a neighbor in $V_i$ for all $i\in \{1,2,\dots ,k\}$. This domination invariant was proposed by Kraner Šumenjak, Rall and Tepeh (in Applied Mathematics and Computation 333(15), 2018: 353–361), which aims to compute the independent domination number of $G\square K_k$ (the generalized prism) via studying the problem of integer labeling on G. They proved a Nordhaus–Gaddum-type theorem: $5\le {\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n+3$ for any n-order graph G with $n\ge 3$, in which $\overline{G}$ denotes the complement of G. This work improves their result and shows that if $G\ncong C_5$, then $5\le {\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n+2$.

1. Introduction

Throughout the paper, only simple graphs are considered. We refer to [1] for undefined notations. For a graph G, the edge set and vertex set of G are denoted by $E\left(G\right)$ and $V\left(G\right)$, respectively. For any $v_1,v_2\in V\left(G\right)$, they are adjacent in G if $v_1$ and $v_2$ are the endpoints of an identical edge of G. A vertex $w\in V\left(G\right)$ is adjacent to a set $W\subseteq V\left(G\right)$ in G if W contains a vertex $w^{\prime }$ s.t. $w{w}^{\prime }\in E\left(G\right)$. $N_G\left(w\right)$ = $\left\{v\right|vw\in E\left(G\right)\}$ is called the open neighborhood of w and $N_G\left[w\right]$ = $N_G\left(w\right)\cup \left\{w\right\}$ is the closed neighborhood of w. $d_G\left(w\right)=\left|N_G\left(w\right)\right|$ denotes the degree of w in G and $\Delta \left(G\right)=max\left\{d_G\left(w\right)\right|w\in V\left(G\right)\}$. A vertex that has degree ℓ and at least ℓ is called an ℓ-vertex and ${\ell }^{+}$-vertex, respectively. For any $W\subseteq V\left(G\right)$, let $N_G\left(W\right)$ = ${\bigcup }_{w\in W}{N}_G\left(w\right)\backslash W$ and $N_G\left[W\right]$ = $N_G\left(W\right)\cup W$. We say that Wdominates a set $W^{\prime }$ if $W^{\prime }\subseteq N_G\left[W\right]$. Moreover, we use the notation $G-W$ to denote the subgraph of G by deleting vertices in W and their incident edges in G, and $G\left[W\right]=G-\left(V\right(G)\backslash W)$ the subgraph of G induced by W. The ℓ-order complete graph and the ℓ-length cycle are denoted by $K_{\ell }$ and $C_{\ell }$, respectively. As usual, for any two natural numbers $p,q$ with $p<q$, $[p,q]$ represents $\{p,p+1,\dots ,q\}$.

Given a graph G and a subset $W\subseteq V\left(G\right)$, we call W a dominating set (abbreviated as DS) of G if W dominates $V\left(G\right)$. An independent set (abbreviated as IS) of a graph is a set of vertices, no two of which are adjacent in the graph. If a DS W of G is an IS, then W is called an independent dominating set (IDS for short) of G. The independent domination number of G, denoted by $i\left(G\right)$, is the cardinality of a minimum IDS of G. Domination and independent domination in graphs have always attracted extensive attention [2,3] and many variants of domination [4] have been introduced increasingly, for the applications in diverse fields, such as electrical networks, computational biology, and land surveying. Recent studies on these variations include (total) roman domination [5,6], strong roman domination [7], semitotal domination [8,9], relating domination [10], just to name a few.

Let $G\square H$ be the Cartesian product of G and H. In order to reduce the problem of determining $i(G\square K_k)$ into the problem of integer labeling on G, Kraner Šumenjak et al. [11] proposed a new variation of domination, called k-rainbow independent dominating function of a graph G (kRiDF for short), which is a function f from $V\left(G\right)$ to $[0,k]$, s.t., for each $i\in [1,k]$, $V_i$ is an IS and every vertex v with $f\left(v\right)=0$ is adjacent to a vertex u with $f\left(u\right)=i$. Alternatively, a kRiDF f of G may be viewed as an ordered partition $(V_0,V_1,\dots ,V_k)$ such that for each $i\in [1,k]$, $V_i$ is an IS and $N_G\left(x\right)\cap V_i\ne \varnothing$ for every $x\in V_0$, where $V_j$, $j\in [0,k]$, denotes the set of vertices assigned value j under f. The weight $w\left(f\right)$ of a kRiDF f is defined as the number of nonzero vertices, i.e., $w\left(f\right)$ = $\left|V\left(G\right)\right|-|V_0|$. The k-rainbow independent domination number of G, denoted by ${\gamma }_{\mathrm{rik}}\left(G\right)$, is the minimum weight of a kRiDF of G. From the definition, we have ${\gamma }_{ri1}\left(G\right)=i\left(G\right)$. A ${\gamma }_{rik}\left(G\right)$-function represents a kRiDF of G which has weight ${\gamma }_{rik}\left(G\right)$.

Let G be a graph and H a subgraph of G. Suppose that g is a kRiDF of H. We say that a kRiDF f of G is extended from g if $f\left(v\right)=g\left(v\right)$ for every $v\in V\left(H\right)$. To prove that a graph G has a kRiDF, we will first find a $k^{\prime }$RiDF g of a subgraph $G^{\prime }$ of G, $k^{\prime }\le k$, and then extend g to a kRiDF f of G. By using this approach, we describe the structure characterization of graphs G with ${\gamma }_{\mathrm{ri}2}\left(G\right)$ = $\left|V\right(G\left)\right|-1$ (Section 2), and then obtain an improved Nordhaus–Gaddum-type theorem with regard to ${\gamma }_{\mathrm{ri}2}$ (Section 3).

2. Structure Characterization of Graphs $\mathit{G}$ s.t., ${\mathbf{\gamma }}_{\mathrm{ri}2}\left(\mathit{G}\right)$ = $\left|\mathit{V}\right(\mathit{G}\left)\right|-\mathbf{1}$

To get the improved Nordhaus–Gaddum-type theorem in terms of ${\gamma }_{\mathrm{ri}2}$, we have to characterize the graphs G s.t., ${\gamma }_{\mathrm{ri}2}\left(G\right)$ = $\left|V\right(G\left)\right|-1$. For this, we need the following special graphs.

A star $S_n$, $n\ge 1$, is a complete bipartite graph $G[X,Y]$ with $\left|X\right|$=1 and $\left|Y\right|=n$, where the vertex in X is called the center of $S_n$ and the vertices in Y are leaves of $S_n$. Let $S_n^{+}$ be the graph obtained from $S_n$ by adding a single edge connecting an arbitrary pair of leaves of $S_n$ [11]. A double star [12] is defined as the union of two vertex-disjoint stars with an edge connecting their centers. Specifically, for two integers $n,m$ such that $n\ge m\ge 0$ the double star, denoted by $S(n,m)$, is the graph with vertex set $\{u_0,u_1,\dots ,u_n,v_0,v_1,\dots ,v_m\}$ and edge set $\{u_0{v}_0,u_0{u}_i,v_0{v}_j|i\in [1,n],j\in [1,m]\}$, where $u_0{v}_0$ is called the bridge of $S(n,m)$ and the subgraphs induced by $\left\{u_i\right|i\in [0,n]\}$ and $\left\{v_j\right|j\in [0,m]\}$ are called the n-star at $u_0$ and m-star at $v_0$, respectively. Observe that $S(n,m)$ is defined on the premise of $n\ge m$. For mathematical convenience, we denote a double star $S(n,m)$ as a vertex-sequence $v_m{v}_{m-1}\dots v_0{u}_0{u}_1\dots u_n$.

We start with a known result, which characterizes graphs G with ${\gamma }_{\mathrm{ri}2}\left(G\right)=n$. For a fixed graph G, its complement is written as $\overline{G}$.

Lemma 1

([11]). Let G be a graph of order n. Then, ${\gamma }_{\mathrm{ri}2}\left(G\right)=n$ iff G only contains components isomorphic to $K_1$ or $K_2$. And, if ${\gamma }_{\mathrm{ri}2}\left(G\right)=n$, then ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=2$.

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

Lemma 2.

Let H be a subgraph of a fixed graph G and $g=(V_0,V_1,\dots ,V_k)$ be a ${\gamma }_{\mathrm{rik}}\left(H\right)$-function. Then g can be extended to a kRiDF of G with weight at most $\left|V\left(G\right)\right|-|V_0|$.

Proof. 

Let $V\left(G\right)\backslash V\left(H\right)=\{x_1,\dots ,x_{\ell }\}$. We will deal with these vertices in the order of $x_1,\dots ,x_{\ell }$ 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 $V_0\cup \left\{x_i\right\}$; otherwise we update $V_j$ by $V_j\cup \left\{x_i\right\}$. After the last one, i.e., $x_{\ell }$ is handled, we obtain a kRiDF of G. Obviously, the weight of the resulting kRiDF of G is not more than $\left|V\left(G\right)\right|-|V_0|$. □

The following theorem clarifies the structure of connected graphs G with ${\gamma }_{\mathrm{ri}2}\left(G\right)$=$\left|V\right(G\left)\right|-1$.

Theorem 1.

Let G be a connected graph with order $n\ge 3$. Then, ${\gamma }_{\mathrm{ri}2}\left(G\right)=n-1$ iff G is isomorphic to one among $S_{n-1}$, $S_{n-1}^{+}$, $S(n-3,1)$ ($n\ge 4$) and $C_5$.

Proof. 

Let $f=(V_0,V_1,V_2)$ be an arbitrary ${\gamma }_{\mathrm{ri}2}\left(G\right)$-function. Observe that $V_0$ does not contain any 1-vertex; one can readily derive that ${\gamma }_{\mathrm{ri}2}\left(G\right)=n-1$ when G is isomorphic to one of $S_{n-1},S_{n-1}^{+}$, $S(n-3,1)$ and $C_5$. Conversely, suppose that ${\gamma }_{\mathrm{ri}2}\left(G\right)=n-1$, that is, $|V_0|=1$. By Lemma 2, G contains no subgraph H that has a 2RiDF of weight at most $\left|V\right(H\left)\right|-2$. Since ${\gamma }_{\mathrm{ri}2}\left(C_4\right)=2=\left|V\left(C_4\right)\right|-2$ and each $C_k$ for $k\ge 6$ contains a subgraph isomorphic to a 6-order path $P_6$ with ${\gamma }_{\mathrm{ri}2}\left(P_6\right)=4=\left|V\left(P_6\right)\right|-2$, G does not contain any subgraph isomorphic to $C_4$ or $C_k$ for $k\ge 6$. 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 $N_G\left(x\right)$. Suppose to the contrary that G contains a $2^{+}$-vertex y such that $y\notin N_G\left(x\right)$. Let $\{x_1,x_2,x_3\}\subseteq N_G\left(x\right)$ and $\{y_1,y_2\}\subseteq N_G\left(y\right)$. Observe that $|\{x_1,x_2,x_3\}\cap \{y_1,y_2\}|\le 1$ and $|N_G\left(y_i\right)\cap \{x_1,x_2,x_3\}|\le 1$ for $i\in [1,2]$; we WLOG assume that $y_2\notin \{x_1,x_2,x_3\}$, $y_2{x}_2\notin E\left(G\right)$ and $y_2{x}_3\notin E\left(G\right)$. Let f be: $f\left(x\right)=f\left(y\right)=0,f\left(x_2\right)=1,f\left(x_3\right)=2$. Notice that either $y_1=x_j$ or $y_1{x}_j\notin E\left(G\right)$ for some $j\in [2,3]$; we further let $f\left(y_1\right)=f\left(x_j\right)$ and $f\left(y_2\right)=[1,2]\backslash \left\{f\left(y_1\right)\right\}$. Clearly, f is a 2RiDF of $G\left[\{x,x_2,x_3,y,y_1,y_2\}\right]$ of weight $\left|\right\{x,x_2,x_3,y,y_1,y_2\left\}\right|-2$, a contradiction.

Observation 2.G contains at most one $3^{+}$-vertex. Suppose that G has two distinct $3^{+}$-vertices, say x and y. By Observation 1, $xy\in E\left(G\right)$. Let $\{y,x_1,x_2\}\subseteq N_G\left(x\right)$ and $\{x,y_1,y_2\}\subseteq N_G\left(y\right)$. Since G contains no subgraph isomorphic to $C_4$, $|\{x_1,x_2\}\cap \{y_1,y_2\}|\le 1$ and there are no edges between $\{x_1,x_2\}$ and $\{y_1,y_2\}$. 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\}$$\to \{0,1,2\}$ such that $f\left(x\right)$ = $f\left(y\right)$ = 0, $f\left(x_2\right)$ = $f\left(y_2\right)$ = 2 and $f\left(x_1\right)$ = $f\left(y_1\right)$ = 1, is a 2RiDF of $G\left[\{x,y,x_1,x_2,y_1,y_2\}\right]$ of weight $\left|\right\{x,y,x_1,$$x_2,y_1,y_2\left\}\right|-2$, a contradiction.

Observation 3.If G contains a $3^{+}$-vertex x, $N_G\left(x\right)$ has not more than two 2-vertices; in particular, when $N_G\left(x\right)$ contains two 2-vertices, in G these two 2-vertices are adjacent. If not, suppose that $N_G\left(x\right)$ contains three 2-vertices, say $x_1,x_2,x_3$. We WLOG assume that $x_3\notin N_G\left(\{x_1,x_2\}\right)$ and let $N_G\left(x_3\right)=\{x,y_3\}$. Let $N_G\left(x_1\right)=\{x,y_1\}$ (possibly $y_1=x_2$, but $y_1\ne y_3$). By Observation 1, $d_G\left(y_3\right)=1$, i.e., $y_1{y}_3\notin E\left(G\right)$. Let f be: $f\left(x\right)=1,f\left(x_1\right)=f\left(x_3\right)=0,f\left(y_1\right)=f\left(y_3\right)=2$. Obviously, f is a 2RiDF of $G\left[\{x,x_1,y_1,x_3,y_3\}\right]$ of weight $\left|\right\{x,x_1,y_1,x_3,y_3\left\}\right|-2$, a contradiction. Now, suppose that $N_G\left(x\right)$ contains two 2-vertices, say $x_1,x_2$. If $x_1{x}_2\notin E\left(G\right)$, let $N_G\left(x_i\right)=\{x,y_i\},$$i\in [1,2]$. Clearly, $y_1\ne y_2$ and $y_1{y}_2\notin E\left(G\right)$. Let f be: $f\left(x\right)=1,f\left(x_1\right)=f\left(x_2\right)=0,f\left(y_1\right)=f\left(y_2\right)=2$. Then, f is a 2RiDF of $G\left[\{x,x_1,y_1,x_2,y_2\}\right]$ of weight $\left|\right\{x,x_1,x_2,y_1,y_2\left\}\right|-2$, 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 $V\left(G\right)\backslash \left\{x\right\}$ 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 $\Delta \left(G\right)=2$, then G is isomorphic to one of $S_2^{+},S_2,S(1,1)$ and $C_5$. □

The theorem below follows from Theorem 1, Lemma 1, and ${\gamma }_{\mathrm{ri}2}\left(G\right)$ = ${\sum }_{i=1}^k{\gamma }_{\mathrm{ri}2}\left(G_i\right)$, where $G_1,\dots ,G_k$ are the components of G.

Theorem 2.

Given a graph G with order $n\ge 3$, ${\gamma }_{\mathrm{ri}2}\left(G\right)=n-1$ iff G has one component $G_1$ isomorphic to one among $S_{n_1-1}$ ($n_1\ge 3$), $S_{n_1-1}^{+}$ ($n_1\ge 3$), $S(n_1-3,1)$ ($n_1\ge 4$) and $C_5$, and other components are isomorphic to $K_1$ or $K_2$, where $n_1=\left|V\left(G_1\right)\right|$.

3. An Improved Nordhaus–Gaddum Type Theorem for ${\mathbf{\gamma }}_{\mathrm{ri}2}\left(\mathit{G}\right)$

This section is devoted to achieve an improved Nordhaus–Gaddum type theorem by showing that ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n+2$ for every graph $G\ncong C_5$ of order $n\ge 2$, which improves a result obtained by Kraner Šumenjak et al., et al [11]. We first present some fundamental lemmas.

Lemma 3.

For an n-order graph G with $n\ge 3$, if G is $S_{n-1},S_{n-1}^{+}$ or $S(n-3,1)$, then ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le 3$.

Proof. 

If $G\cong S_{n-1}$ or $G\cong S_{n-1}^{+}$, let $V\left(G\right)=\{v_0,v_1,\dots ,v_{n-1}\}$ where $v_0$ is the center and $v_1{v}_2\in E\left(G\right)$ when $G\cong S_{n-1}^{+}$. Define a function f such that $f\left(v_1\right)=1,f\left(v_0\right)=f\left(v_2\right)=2$ and $f\left(v\right)=0$ for every $v\in V\left(\overline{G}\right)\backslash \{v_0,v_1,v_2\}$. Since every vertex in $V\left(\overline{G}\right)\backslash \{v_0,v_1,v_2\}$ is a neighbor of $v_1$ and also $v_2$ in $\overline{G}$, it follows that f is a 2RiDF of $\overline{G}$ of weight 3.

If $G\cong S(n-3,1)$, then $n\ge 4$. Let $V\left(G\right)=\{v_1,v_0,u_0,u_1,\dots ,u_{n-3}\}$, where $v_0{u}_0$ is the bridge of G and $E\left(G\right)=\{v_0{v}_1,v_0{u}_0,u_0{u}_i|i\in [1,n-3]\}$. If $n=4$, then both G and $\overline{G}$ are isomorphic to $P_4$, the path of length 3, and the conclusion holds. If $n\ge 5$, then the function f from $V\left(\overline{G}\right)$ to $[0,2]$ such that $f\left(u_2\right)=2,f\left(u_1\right)=f\left(u_0\right)=1$, and $f\left(v\right)=0$ for every $v\in V\left(\overline{G}\right)\backslash \{u_0,u_1,u_2\}$ is a 2RiDF of $\overline{G}$ with weight 3. □

Lemma 4.

For a graph n-order G, if $G\ncong C_5$ and ${\gamma }_{\mathrm{ri}2}\left(G\right)=4$, then ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n-2$.

Proof. 

Clearly, $n\ge 4$. When $n=4$, ${\gamma }_{\mathrm{ri}2}\left(G\right)=4$ implies that ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=2=n-2$ by Lemma 1. Therefore, we assume that $n\ge 5$. Suppose that ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\ge n-1$. If ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=n$, by Lemma 1 we have ${\gamma }_{\mathrm{ri}2}\left(G\right)=2$, a contradiction. Therefore, ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=n-1$. By Theorem 2 $\overline{G}$ has one component isomorphic to $S_{n_1},S_{n_1}^{+}$, $S(n_2,1)$ or $C_5$ where $n_1\ge 2,n_2\ge 1$, and all of the other components of $\overline{G}$ are isomorphic to $K_1$ or $K_2$.

If $\overline{G}$ contains two vertices u and v s.t. $N_{\overline{G}}\left(\{u,v\}\right)=\varnothing$, then in G both u and v are adjacent to every vertex in $V\left(G\right)\backslash \{u,v\}$. 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 ${\gamma }_{\mathrm{ri}2}\left(G\right)\le 2$ and a contradiction. Therefore, $\overline{G}$ contains no $K_2$ components and contains at most one $K_1$ component, implying that $\overline{G}$ contains at most two components. If $\overline{G}$ contains only one component, it follows that $\overline{G}$ is $S_{n-1},S_{n-1}^{+}$ or $S(n-3,1)$ (since $G\ncong C_5$). By Lemma 3 ${\gamma }_{\mathrm{ri}2}\left(G\right)\le 3$ 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 $V\left(G_1\right)=\left\{u\right\}$ and define a function f as follows: let $f\left(u\right)=1$; $f\left(v_0\right)=f\left(v^{\prime }\right)=2$ 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 $G_2$ by the assumption of $n\ge 5$), $f\left(v_0\right)=f\left(u_0\right)=2$ when $G_2\cong S(n-4,1)$ (where $v_0{u}_0$ is the bridge of $G_2$), or $f\left(u_1\right)=f\left(u_2\right)=2$ 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, all vertices with value 0 are adjacent to u and a vertex with value 2. Hence, f is a 2RiDF of G, which has weight 3, a contradiction. □

Lemma 5.

Suppose that G is an n-order graph satisfying that ${\gamma }_{\mathrm{ri}2}\left(G\right)\ge 4$ and ${\gamma }_{\mathrm{ri}2}\left(G\right)$+${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)$ = $n+3$. Let $f=(V_0,V_1,V_2)$ be an arbitrary ${\gamma }_{\mathrm{ri}2}\left(G\right)$-function. We have

(1)

If $|V_0|\ge 2$, then for any $u,v\in V_0$, there does not exist $u_1,u_2,v_1,v_2$ such that $\{u_1,u_2\}\in N_{\overline{G}}\left(u\right)$, $\{v_1,v_2\}\in N_{\overline{G}}\left(v\right)$ and $u_i{v}_i\notin E\left(\overline{G}\right)$ for $i\in [1,2]$, where $u_1\ne u_2,v_1\ne v_2$ but possibly $u_i=v_i$;

(2)

If $u,v$ are two arbitrary different vertices of $V_0$, then $|N_{\overline{G}}\left(\{u,v\}\right)|\ge 3$;

(3)

$|V_i|\ge 2$ for $i\in [0,2]$.

Proof. 

For (1), if the conclusion is false, then let g be: $g\left(u\right)=g\left(v\right)=0$ and $g\left(u_i\right)=g\left(v_i\right)=i$, $i\in [1,2]$. Then, g is a 2RiDF of $\overline{G}\left[\{u,v,u_1,v_1,u_2,v_2\}\right]$ with weight $\left|\right\{u,v,u_1,v_1,u_2,v_2\left\}\right|-2$. Since $V_1$ and $V_2$ are cliques in $\overline{G}$, $V_i$ contains at most two vertices not assigned 0 under every 2RiDF of $\overline{G}$ for $i\in [1,2]$. Hence, we can extend g to a 2RiDF of $\overline{G}$ with weight at most $|V_0|-2+4=|V_0|+2$, according to Lemma 2. This shows that ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le \left|V_0\right|+2$ and ${\gamma }_{\mathrm{ri}2}\left(G\right)$+${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le |V_1|+|V_2|+|V_0|+2=n+2$, a contradiction.

For (2), if $|N_{\overline{G}}\left(\{u,v\}\right)|\le 2$, let f be: $f\left(v\right)=2,f\left(u\right)=1$, and $f\left(x\right)=0$ for $x\in V\left(G\right)\backslash N_{\overline{G}}\left[\{u,v\}\right]$. It is clear that f is a 2RiDF of $G[V\left(G\right)\backslash N_{\overline{G}}\left(\{u,v\}\right)]$ with weight 2. According to Lemma 2, we can extend f to a 2RiDF of G with weight at most 4 (since $|N_{\overline{G}}\left(\{u,v\}\right)|\le 2$). Thus, ${\gamma }_{\mathrm{ri}2}\left(G\right)=4$ and by Lemma 4 ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n-2$, a contradiction.

For (3), if $|V_0|=1$, then ${\gamma }_{\mathrm{ri}2}\left(G\right)$ = $n-1$. By an analogous argument as that in Lemma 4, we can derive that ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n+2$, a contradiction. In the following, we prove that $|V_1|\ge 2$ (the proof of $|V_2|\ge 2$ is similar to that of $|V_2|\ge 2$). Suppose that $|V_1|=1$ and let $V_1=\left\{u\right\}$. 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 4 we assume that $|V_1|+|V_2|\ge 5$. 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: $g\left(v\right)=0,g\left(v_1\right)=1,g\left(v_2\right)=2$. Since $V_2$ is a clique in $\overline{G}$, we can extend g to a 2RiDF of $\overline{G}$ with weight at most $|V_0|-1+3=|V_0|+2$, according to Lemma 2. This shows that ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le \left|V_0\right|+2$ and hence ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n+2$, a contradiction. Therefore, every vertex in $V_0$ has degree at most 1 in $\overline{G}$, which implies that $|N_{\overline{G}}\left(\{x,y\}\right)|\le 2$ for any two vertices $x\in V_0,y\in V_0$ (observe that $|V_0|\ge 2$). This contradicts (2). □

Lemma 6.

Let G be an n-order graph, $n\ge 4$. For any $u\in V\left(G\right)$, if $H=G-u$, the resulting graph by deleting u and its incident edges from G, is connected and ${\gamma }_{\mathrm{ri}2}\left(H\right)=\left|V\left(H\right)\right|-1$, then G has a 2RiDF f satisfying $f\left(u\right)=1$ and $f\left(v\right)=0$ for some $v\in V\left(H\right)$.

Proof. 

Clearly, $\left|V\right(H\left)\right|\ge 3$. If u has no neighbor in $V\left(H\right)$, then let f be: $f\left(v\right)=g\left(v\right)$ for every $v\in V\left(H\right)$, and $f\left(u\right)=1$, where g is a ${\gamma }_{\mathrm{ri}2}\left(H\right)$-function of H. Since ${\gamma }_{\mathrm{ri}2}\left(H\right)=\left|V\left(H\right)\right|-1$, there exists $v\in V\left(H\right)$ satisfying $f\left(v\right)=g\left(v\right)=0$. If u has a neighbor $u_1\in V\left(H\right)$, there exists a $u_2\in V\left(H\right)$ s.t. $u_1{u}_2\in E\left(H\right)$ since H is connected. Let f be: $f\left(u_1\right)=0,f\left(u\right)=1,f\left(u_2\right)=2$. Then, we can extend f to a desired 2RiDF of G according to Lemma 2. □

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

Theorem 3.

Suppose that G is an n-order graph, $n\ge 2$. If $G\ncong C_5$, then ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n+2$.

Proof. 

We are sufficient to handle the situation $n\ge 5$ since cases of $n\le 4$ are trivial. Let $f_0=(V_0,V_1,V_2)$ be a ${\gamma }_{\mathrm{ri}2}\left(G\right)$-function such that $\overline{G}\left[V_0\right]$ contains the maximum number of components isomorphic to $K_2$. Suppose to the contrary that ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)>n+2$. Then, ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=n+3$ since ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le n+3$ [11], that is,

$${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=\left|V_0\right|+3$$

(1)

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 ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)\le 2+n$, a contradiction; if $|V_1\cup V_2|=3$, then ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=n$ and by Lemma 1 ${\gamma }_{\mathrm{ri}2}\left(G\right)=2$, also a contradiction. Therefore, by Lemma 4,

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

(2)

Then, by Lemma 5 (3) we have $|V_i|\ge 2$ for $i\in [0,2]$. In addition, because, by definition, $\overline{G}\left[V_i\right]$ is a clique, $i\in [1,2]$, it follows that for every 2RiDF $g_0=(V_0^{\prime },V_1^{\prime },V_2^{\prime })$ of $\overline{G}$,

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

(3)

Therefore, by Lemma 2 we can extend every ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)$-function to a 2RiDF of $\overline{G}$ with weight at most ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)+4$, i.e., ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)\ge \left|V_0\right|-1$ by Formula (1).

Claim 1.Denote by ℓ 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, $\ell \le 1-{\ell }^{\prime }$ where ${\ell }^{\prime }=\left|V_0\right|-{\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)$$\le 1$. If not, either ℓ is at least 2 or both ℓ and ${\ell }^{\prime }$ are equal to 1. Suppose that $\ell \ge 2$ and take two vertices $v_1$,$v_2$$\in (V_1\cup V_2)$ such that they are adjacent to all vertices of $(V_1\cup V_2)\backslash \{u,v\}$ in $\overline{G}$. Let $g^{\prime }$ be: $g^{\prime }\left(v_1\right)=1,g^{\prime }\left(v_2\right)=2,g^{\prime }\left(x\right)=0$ for $x\in V_1\cup V_2\backslash \{v_1,v_2\}$. Clearly, $g^{\prime }$ is a 2RiDF of $\overline{G}[V_1\cup V_2]$ and by Lemma 2 we can extend $g^{\prime }$ to a 2RiDF of $\overline{G}$, which has weight at most $|V_0|+2$, a contradiction. Now, suppose that $\ell ={\ell }^{\prime }=1$. Then, ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)=\left|V_0\right|-1$, which indicates that $\overline{G}\left[V_0\right]$ contains a component $H^{\prime }$ s.t. ${\gamma }_{\mathrm{ri}2}\left(H^{\prime }\right)=\left|V\left(H^{\prime }\right)\right|-1$. Since $\ell =1$, there is a vertex v, say $v\in V_1$, which is adjacent to every vertex of $V_2$ in $\overline{G}$. By Lemma 6 $\overline{G}[V\left(H^{\prime }\right)\cup \left\{v\right\}]$ has a 2RiDF $g^{\prime }$ s.t. $g^{\prime }\left(x\right)=0$ for some $x\in V\left(H^{\prime }\right)$ and $g^{\prime }\left(v\right)=1$. Observe that in $\overline{G}$v is adjacent to all vertices of $(V_1\cup V_2)\backslash \left\{v\right\}$; by the rule of Lemma 2 we can extend $g^{\prime }$ to a 2RiDF g of $\overline{G}$ under which there is at most one vertex in $V_1\backslash \left\{v\right\}$ (and $V_2$) not assigned value 0. Thus, $w\left(g\right)\le |V_0|-1+3=|V_0|+2$, a contradiction. This completes the proof of Claim 1.

Now, we WLOG assume $|V_1|\ge |V_2|$. Then, $|V_1|\ge 3$ by Formula (2).

Claim 2.$\overline{G}\left[V_0\right]$ does not contain any isolated vertex v s.t. $N_{\overline{G}}\left(v\right)\cap V_1=\varnothing$. Otherwise, define $f^{\prime }$ as: for $x\in V_2$$f^{\prime }\left(x\right)=2$, and $f^{\prime }\left(v\right)=1$. By Claim 1, in $\overline{G}$, $V_1$ has not more than one vertex adjacent to every vertex in $V_2$; say $v^{\prime }$ if such a vertex exists. We further let $f^{\prime }\left(y\right)=0$ for $y\in V_1\cup (V_0\backslash \left\{v\right\})$ (or for $y\in (V_1\backslash \left\{v^{\prime }\right\})\cup (V_0\backslash \left\{v\right\})$ if $v^{\prime }$ exists). Since in G every vertex in $V_1\cup V_0$ (except for $v^{\prime }$) is adjacent to v and also $V_2$, f is a 2RiDF of G of weight at most $|V_2|+2$, a contradiction. This completes the proof of Claim 2.

We proceed by distinguishing two cases: ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)=\left|V_0\right|-1$ and ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)=\left|V_0\right|$.

Case 1.${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)=\left|V_0\right|-1$. In this case, by Claim 1 each vertex of $V_i$ owns a neighbor belonging to $V_j$ in G where $\{i,j\}$= [1,2]; by Theorem 2, $\overline{G}\left[V_0\right]$ has one component H isomorphic to one of $S_{\left|V\right(H\left)\right|-1}$ ($\left|V\right(H\left)\right|\ge 3$), $S_{\left|V\right(H\left)\right|-1}^{+}$ ($\left|V\right(H\left)\right|\ge 3$), $S\left(\right|V\left(H\right)|-3,1)$ ($\left|V\right(H\left)\right|\ge 4$) and $C_5$, and other components of $\overline{G}\left[V_0\right]$ are isomorphic to $K_1$ or $K_2$. Let $u_0\in V\left(H\right)$ be a vertex with $d_H\left(u_0\right)=\Delta \left(H\right)$. Clearly, $d_H\left(u_0\right)\ge 2$. Let $u_1\in N_H\left(u_0\right)$ and $u_2\in N_H\left(u_0\right)$ be two vertices such that every vertex in $V\left(H\right)\backslash \{u_0,u_1,u_2\}$ has degree in H not exceeding $min\{d_H\left(u_1\right),d_H\left(u_2\right)\}$. By the structure of H, for $i\in [1,2]$, we have that $d_H\left(u_i\right)\le 2$ and if $u_i$ has a neighbor $u_i^{\prime }(\notin \{u_0,u_1,u_2\}$) in H, then $u_0{u}_i^{\prime }\notin E\left(H\right)$. Moreover, by Lemma 5 (1), $(N_{\overline{G}}\left(u_1\right)\cap N_{\overline{G}}\left(u_2\right))\backslash \left\{u_0\right\}=\varnothing$, which implies that each vertex of $V_1\cup V_2$ is adjacent to $u_1$ or $u_2$ in G.

Claim 3.$|V_0\backslash V\left(H\right)|\le 1$. Otherwise, let $\{v_1,v_2\}\subseteq (V_0\backslash V\left(H\right))$. Then, $d_{\overline{G}\left[V_0\right]}\left(v_1\right)\le 1$ and $d_{\overline{G}\left[V_0\right]}\left(v_2\right)\le 1$. Suppose that $d_{\overline{G}\left[V_0\right]}\left(v_1\right)=1$ (the case of $d_{\overline{G}\left[V_0\right]}\left(v_2\right)=1$ can be similarly discussed). Let $v_1{v}_1^{\prime }\in E\left(\overline{G}\left[V_0\right]\right)$ and clearly $d_{\overline{G}\left[V_0\right]}\left(v_1^{\prime }\right)=1$. By Lemma 5 (2), a vertex $v_0\in (V_1\cup V_2)$ is adjacent to $\{v_1,v_1^{\prime }\}$ in $\overline{G}$. We WLOG assume that $v_1{v}_0\in E\left(\overline{G}\right)$. According to Lemma 6, $\overline{G}[V\left(H\right)\cup \left\{v_0\right\}]$ admits a 2RiDF $g^{\prime }$ satisfying $g^{\prime }\left(v_0\right)=1$ and $g^{\prime }\left(x\right)=0$ for some $x\in V\left(H\right)$. Further, let $g^{\prime }\left(v_1\right)=0$ and $g^{\prime }\left(v_1^{\prime }\right)=2$. So $g^{\prime }$ is a 2RiDF of $\overline{G}[V\left(H\right)\cup \{v_0,v_1,v_1^{\prime }\}]$, and by Lemma 2 and Formula (3) we can extend $g^{\prime }$ to a 2RiDF of $\overline{G}$ with weight at most $|V_0|-2+4=|V_0|+2$ (since $g^{\prime }\left(v_1\right)=g^{\prime }\left(x\right)=0$), a contradiction. We therefore assume that $d_{\overline{G}\left[V_0\right]}\left(v_1\right)=d_{\overline{G}\left[V_0\right]}\left(v_2\right)=0$. By Lemma 5 (2) we have $|N_{\overline{G}}\left(\{v_1,v_2\}\right)\cap (V_1\cup V_2)|\ge 3$. WLOG, suppose that in $\overline{G}$, $v_1$ has two neighbors belonging to $V_1\cup V_2$, say $v_{11}$ and $v_{12}$. By Lemma 5 (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 $u_1{v}_{11}\notin E\left(\overline{G}\right)$ and $u_2{v}_{12}\notin E\left(\overline{G}\right)$, or $u_1{v}_{12}\notin E\left(\overline{G}\right)$ and $u_2{v}_{11}\notin E\left(\overline{G}\right)$, which contradicts to Lemma 5 (1) again. This completes the proof of Claim 3.

By Claim 3, we see that $\overline{G}\left[V_0\right]$ contains no component isomorphic to $K_2$ and contains at most one $K_1$ component.

Claim 4.$\overline{G}\left[V_0\right]$ contains a $K_1$ component. If not, we have $\overline{G}\left[V_0\right]$ = H.

Claim 4.1.$(N_{\overline{G}}\left(u_1\right)\cup N_{\overline{G}}\left(u_2\right))\cap (V_1\cup V_2)\ne \varnothing$.

Otherwise, for $i\in [1,2]$, $u_i$ is adjacent to every vertex of $V_1\cup V_2$ in G, and by Lemma 5 (2) $d_H\left(u_i\right)=2$ and $u_1{u}_2\notin E\left(\overline{G}\right)$. Set $\left\{u_i^{\prime }\right\}=N_H\left(u_i\right)\backslash \left\{u_0\right\}$, $i\in [1,2]$; then, $u_0{u}_i^{\prime }\notin E\left(\overline{G}\right)$. Let f be: $f\left(u_1\right)=f\left(u_1^{\prime }\right)=1$, $f\left(u_2\right)=f\left(u_2^{\prime }\right)=2$ and $f\left(x\right)=0$ for any x in $V\left(G\right)\backslash \{u_1,u_1^{\prime },u_2,u_2^{\prime }\}$. So, we get a 2RiDF f of G, which has weight 4, a contradiction. So, Claim 4.1 holds.

Claim 4.2.$|V_1|=3$.

Observe that $|V_1|\ge 3$; it is enough by showing that G admits a 2RiDF f s.t. $w\left(f\right)\le |V_2|+3$. When $u_1{u}_2\in E\left(\overline{G}\right)$, let f be: $f\left(u_i\right)=1$ for $i\in [0,2]$, $f\left(x\right)=0$ for $x\in (V_1\cup V_0)\backslash \{u_0,u_1,u_2\}$, and $f\left(y\right)=2$ for $y\in V_2$. By Lemma 5 (1), in $\overline{G}$, $V_1\cup V_0$ contains no vertex adjacent to $u_1$ and also $u_2$. Therefore, f is a 2RiDF of G of weight $|V_2|+3$. Now, suppose that $u_1{u}_2\notin E\left(\overline{G}\right)$. By Lemma 5 (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: $f\left(u_0\right)=f\left(u_1\right)=1$ (or $f\left(u\right)=f\left(u_0\right)=f\left(u_1\right)=1$ if u exists), $f\left(x\right)=0$ for $x\in (V_1\cup (V_0\backslash \{u_0,u_1\}))$ (or $x\in (V_1\cup V_0)\backslash \{u_0,u_1,u\}$) and $f\left(y\right)=2$ for $y\in V_2$. Notice that by Claim 1 every vertex of $V_0\cup V_1$ 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 $(V_0\cup V_1)\backslash \{u,u_0,u_1\}$ is adjacent to $\{u_0,u_1\}$ in G; f is a 2RiDF of G of weight at most $|V_2|+3$. This completes the proof of Claim 4.2.

By Claim 4.2, we have $2\le |V_2|\le 3$. Let $V_1=\{w_1,w_2,w_3\}$ in the following.

Claim 4.3.In $\overline{G}$, for $\{i,j\}=[1,2]$ every vertex in $V_i$ has not more than one neighbor in $V_j$.

If not, let $v\in V_2$ be adjacent to two vertices of $V_1$ in $\overline{G}$, say $w_1,w_2$. By Lemma 5 (1) $u_1$ or $u_2$ is not adjacent to v in $\overline{G}$, say $u_{1}v\notin E\left(\overline{G}\right)$. If $u_2{w}_3\notin E\left(\overline{G}\right)$, define $g^{\prime }$ as: $g^{\prime }\left(u_i\right)=i$ for every $i\in [0,2]$, $g^{\prime }\left(w_1\right)=g^{\prime }\left(w_2\right)=0,g^{\prime }\left(w_3\right)=2$, $g^{\prime }\left(v\right)=1$. If $u_2{w}_3\in E\left(\overline{G}\right)$, then $u_1{w}_3\notin E\left(\overline{G}\right)$ and let $g^{\prime }$ be: $g^{\prime }\left(u_1\right)=g^{\prime }\left(w_3\right)=1,g^{\prime }\left(w_1\right)=g^{\prime }\left(w_2\right)=0,g^{\prime }\left(v\right)=2$; further, let $g^{\prime }\left(u_2\right)=0$ when $u_{2}v\in E\left(\overline{G}\right)$, or let $g^{\prime }\left(u_2\right)=2$ and $g^{\prime }\left(u_0\right)=0$ when $u_{2}v\notin E\left(\overline{G}\right)$. According to Lemma 2, in either case the $g^{\prime }$ defined above can be extended to a 2RiDF g of $\overline{G}$ under which $g\left(w_1\right)=g\left(w_2\right)=0$ and $g\left(u_0\right)=0$ or $g\left(u_2\right)=0$. Therefore, by Formula (3) $w\left(g\right)\le |V_0|-1+3=|V_0|+2$, a contradiction. With a similar discussion, there is also a contradiction if we assume $V_1$ contains a vertex that has two neighbors in $V_2$ in $\overline{G}$. This completes the proof of Claim 4.3.

Now, we consider $|V_2|$. Suppose that $|V_2|=3$ and let $V_2=\{w_4,w_5,w_6\}$. According to Claim 4.1, we WLOG assume that $u_1{w}_1\in E\left(\overline{G}\right)$. This indicates that $u_2{w}_1\notin E\left(\overline{G}\right)$ by Lemma 5 (1). If $u_2$ has a neighbor in $V_2$, say $u_2{w}_4\in E\left(\overline{G}\right)$, then according to Lemma 5 (1), $u_1{w}_4\notin E\left(\overline{G}\right)$, $w_1{w}_4\in E\left(\overline{G}\right)$, 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 $w_1$ has two neighbors in $V_1$ or $V_2$ in $\overline{G}$, respectively. This contradicts to Claim 4.3). Let f be: $f\left(u_1\right)=f\left(w_1\right)=1,f\left(u_2\right)=f\left(w_4\right)=2$ and $f\left(x\right)=0$ for $x\in V\left(G\right)\backslash \{u_1,u_2,w_1,w_4\}$. 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 5 (1) $V_0\backslash \{u_0,u_1,u_2\}$ contains no vertex adjacent to both $u_i$ and $w_i$ for some $i\in [1,2]$. Hence, f is a 2RiDF of $G[V\left(G\right)\backslash \left\{u_0\right\}]$ of weight 4 and we are able to extend f to a 2RiDF of G with weight at most $5<|V_1|+|V_2|$ according to Lemma 2, a contradiction. Therefore, we may assume that $N_{\overline{G}}\left(u_2\right)\cap V_2=\varnothing$. In this case, when $N_{\overline{G}}\left(u_2\right)\cap V_1=\varnothing$, let f be: $f\left(u_2\right)=2$, $f\left(u_0\right)=f\left(u_1\right)=1$. By Lemma 5 (1) $V_1\cup V_2$ has not more than one vertex $w^{\prime }$ adjacent to both $u_0$ and $u_1$ in $\overline{G}$ and $V_0\backslash \left\{u_0\right\}$ has not more than one vertex $u^{\prime }$ adjacent to $u_2$ in $\overline{G}$; for $x\in V\left(G\right)\backslash \{u_0,u_1,u_2,u^{\prime },w^{\prime }\}$ we further let $f\left(x\right)=0$. Then, f is a 2RiDF of $G[V\left(G\right)\backslash \{u^{\prime },w^{\prime }\}]$ of weight 3 and according to Lemma 2 we can extend f to a 2RiDF of G of weight at most $5<|V_1|+|V_2|$, a contradiction. We therefore suppose that $u_2$ has a neighbor in $V_1$ in $\overline{G}$, say $u_2{w}_2\in E\left(\overline{G}\right)$. With the same argument as $N_{\overline{G}}\left(u_2\right)\cap V_2=\varnothing$, we can show that $N_{\overline{G}}\left(u_1\right)\cap V_2=\varnothing$ as well.

Then, if $w_3{u}_1\notin E\left(\overline{G}\right)$ and $w_3{u}_2\notin E\left(\overline{G}\right)$, the function f: $f\left(u_1\right)=f\left(w_1\right)=1,f\left(u_2\right)=f\left(w_4\right)=2$ and $f\left(x\right)=0$ for $x\in V\left(G\right)\backslash \{u_1,u_2,w_1,w_4,u_0\}$, is a 2RiDF of $G[V\left(G\right)\backslash \left\{u_0\right\}]$ with weight 4, and according to Lemma 2, we are able to extend f to a 2RiDF of G with weight at most $5<|V_1|+|V_2|$, a contradiction. Therefore, we suppose that $w_3{u}_1\in E\left(\overline{G}\right)$ by the symmetry. By Lemma 5 (1), it has that $w_3{u}_2\notin E\left(\overline{G}\right)$, and $u_0{w}_1\notin E\left(\overline{G}\right)$ or $u_0{w}_3\notin E\left(\overline{G}\right)$, say $u_0{w}_1\notin E\left(\overline{G}\right)$ by the symmetry. Let f be: $f\left(u_0\right)=f\left(u_1\right)=1,f\left(u_2\right)=f\left(w_2\right)=2$ and $f\left(x\right)=0$ for $x\in V\left(G\right)\backslash \{u_1,u_2,u_0,w_2,w_3\}$. Since in G every vertex in $V\left(G\right)\backslash \{u_1,u_2,u_0,w_2,w_3\}$ has a neighbor in $\{u_0,u_1\}$ and also $\{u_2,w_2\}$, f is a 2RiDF of $G[V\left(G\right)\backslash \left\{w_3\right\}]$ of weight 4 and according to Lemma 2 we can extend f 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 so Claim 4 holds.

By Claim 4, we see that $\overline{G}\left[V_0\right]$ contains one component isomorphic to $K_1$. Let s be the vertex of the $K_1$ component. We first show that $|N_{\overline{G}}\left(s\right)\cap (V_1\cup V_2)|\le 1$. If not, in $\overline{G}$ we assume that s has two neighbors in $V_1\cup V_2$, say $s_1,s_2$. By Lemma 5 (1) for $i,j\in [1,2]$, $s_i$ (resp. $u_j$) can not be adjacent to $u_1$ and $u_2$ (resp. $s_1$ and $s_2$) simultaneously in $\overline{G}$. This implies that either $s_i{u}_i\notin E\left(\overline{G}\right)$, $i\in [1,2]$, or $s_1{u}_2\notin E\left(\overline{G}\right)$ and $s_2{u}_1\notin E\left(\overline{G}\right)$, which violates Lemma 5 (1) as well. Thus, by Claim 2 $|N_{\overline{G}}\left(s\right)\cap (V_1\cup V_2)|=1$ and the vertex $s^{\prime }$ adjacent to s in $\overline{G}$ belongs to $V_1$. Let f be: $f\left(x\right)=1$ for $x\in V_1$, $f\left(s\right)=2$, $f\left(y\right)=0$ for $y\in V_2\cup V\left(H\right))$. Observe that by Claim 1 all vertices in $V_2$ are adjacent to $V_1$ in G. Hence, every vertex in $V_2\cup V\left(H\right)$ is adjacent to s and also $V_1$ in G. Therefore, f is a 2RiDF of G with weight $|V_1|+1<|V_1|+|V_2|$ (since $|V_2|\ge 2$), a contradiction.

The foregoing discussion shows that there exists a contradiction if we assume that ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)=\left|V_0\right|-1$. In what remains, we handle the case when ${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)=\left|V_0\right|$.

Case 2.${\gamma }_{\mathrm{ri}2}\left(\overline{G}\left[V_0\right]\right)=\left|V_0\right|$. Then by Lemma 1 every component of $\overline{G}\left[V_0\right]$ is isomorphic to $K_1$ or $K_2$. Recall that $|V_i|\ge 2$ for $i\in [0,2]$. Take two vertices $u,v$ in $V_0$ s.t. $uv\in E\left(\overline{G}\right)$ if $\overline{G}\left[V_0\right]$ contains a $K_2$ component and $u,v$ are isolated vertices in $\overline{G}\left[V_0\right]$ otherwise. By Lemma 5 (1), we have

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

(4)

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

Case 2.1.$uv\in E\left(\overline{G}\right)$. Then in $\overline{G}$, $V_0\backslash \{u,v\}$ contains no vertex adjacent to $\{u,v\}$.

Claim 5.In $\overline{G}[V_1\cup V_2]$, $V_1\cup V_2$ contains only vertices with degree at most $|V_1|+|V_2|-2$. Suppose that $V_1$ contains a vertex w such that $w{w}^{\prime }\in E\left(\overline{G}\right)$ for every $w^{\prime }\in V_2$. If $uw\in E\left(\overline{G}\right)$ (or $vw\in E\left(\overline{G}\right)$), define a 2RiDF $g^{\prime }$ of $\overline{G}\left[\{u,v,w\}\right]$ as: $g^{\prime }\left(u\right)=0$ (or $g^{\prime }\left(v\right)=0),g^{\prime }\left(w\right)=1$ and $g^{\prime }\left(v\right)=2$ ($g^{\prime }\left(u\right)=2$). According to Lemma 2 we can extend $g^{\prime }$ to a 2RiDF of $\overline{G}$, under which $(V_1\cup V_2)\backslash \left\{w\right\}$ contains at most two vertices not assigned 0. Thus, $w\left(g\right)\le |V_0|-1+3=|V_0|+2$, a contradiction. We therefore assume that $uw\notin E\left(\overline{G}\right)$ and $vw\notin E\left(\overline{G}\right)$. By Lemma 5 (2), there are at least three vertices in $(V_1\cup V_2)$ that are adjacent to u or v. We WLOG assume that $V_1\cup V_2$ contains a vertex $u^{\prime }$ s.t. $u^{\prime }u\in E\left(\overline{G}\right)$. Construct a 2RiDF $g^{\prime }$ of $\overline{G}\left[\{u,v,u^{\prime },w\}\right]$ as follows: $g^{\prime }\left(u\right)=0,g^{\prime }\left(u^{\prime }\right)=2$, and $g^{\prime }\left(v\right)=g^{\prime }\left(w\right)=1$. Then, by Lemma 2 $g^{\prime }$ can be extended to a 2RiDF g of $\overline{G}$, under which $(V_1\cup V_2)\backslash \{w,u^{\prime }\}$ contains at most one vertex not assigned value 0. Therefore, $w\left(g\right)\le |V_0|-1+3=|V_0|+2$, a contradiction. Similarly, we can also obtain a contradiction if we assume that $V_2$ contains a vertex adjacent to every vertex of $V_1$. So, Claim 5 holds.

By Claim 5, for $\{i,j\}$= [1,2], each vertex of $V_i$ is adjacent to a vertex of $V_j$ in G. If $V_1\cap (N_{\overline{G}}\left(u\right)\cap N_{\overline{G}}\left(v\right))=\varnothing$, then in G all vertices of $V_1$ are adjacent to $\{u,v\}$. Let f be: $f\left(x\right)=2$ for $x\in V_2$, $f\left(y\right)=0$ for $y\in V_1\cup (V_0\backslash \{u,v\})$, and $f\left(u\right)=f\left(v\right)=1$. Obviously, f is a 2RiDF of G s.t. $w\left(f\right)$ = $|V_2|+2<|V_1|+|V_2|$, a contradiction. We therefore assume that $V_1$ contains a vertex s s.t. $su\in E\left(\overline{G}\right)$ and $sv\in E\left(\overline{G}\right)$. Then, in $\overline{G}$, by Lemma 5 (1) no vertex in $V_2\cup (V_1\backslash \left\{s\right\})$ is adjacent to u and v simultaneously. Analogously, the function f: $f\left(v\right)=f\left(u\right)=1,f\left(x\right)=2$ for $x\in V_1$, and $f\left(y\right)=0$ for $y\in V_2\cup (V_0\backslash \{u,v\})$ (and $f\left(s\right)=f\left(v\right)=f\left(u\right)=1,f\left(x\right)=2$ for $x\in V_2$, and $f\left(y\right)=0$ for $y\in (V_1\backslash \left\{s\right\})\cup (V_0\backslash \{u,v\})$) 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, in $\overline{G}$, neither u nor v is a neighbor of $s_1$ and $s_2$ simultaneously; otherwise, we, by the symmetry, suppose that $u{s}_1\in E\left(\overline{G}\right)$ and $u{s}_2\in E\left(\overline{G}\right)$. Let $g^{\prime }$ be: $g^{\prime }\left(v\right)=g^{\prime }\left(s_1\right)=g^{\prime }\left(s_2\right)=0,g^{\prime }\left(u\right)=1$, and $g^{\prime }\left(s\right)=2$. Obviously, $g^{\prime }$ is a 2RiDF of $\overline{G}\left[\{u,v,s,s_1,s_2\}\right]$ with weight 2. According to Lemma 2, we can extend $g^{\prime }$ to a 2RiDF of $\overline{G}$ with weight at most $|V_0|-1+|V_2|+1=|V_0|+2$, a contradiction. In addition, in $\overline{G}$, $s_i,$$i\in [1,2]$, is not adjacent to u and v simultaneously according to Lemma 5 (1). Therefore, we may assume, by the symmetry, that $s_{1}v\notin E\left(\overline{G}\right)$ and $s_{2}u\notin E\left(\overline{G}\right)$.

If no edge between $\{u,v\}$ and $V_2$ in $\overline{G}$ exists, then by Lemmas 5 (2), $u{s}_1\in E\left(\overline{G}\right)$ and $v{s}_2\in E\left(\overline{G}\right)$. Then, the function $g^{\prime }$ such that $g^{\prime }\left(s\right)=g^{\prime }\left(s_1\right)=g^{\prime }\left(v\right)=0,g^{\prime }\left(s_2\right)=2$, and $g^{\prime }\left(u\right)=1$ is a 2RiDF of $\overline{G}\left[\{u,v,s,s_1,s_2\}\right]$ with weight 2. According to Lemma 2, we can extend $g^{\prime }$ 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 $\overline{G}$ contains an edge connecting $\{u,v\}$ and $V_2$, say $v{s}_3\in E\left(\overline{G}\right)$ by the symmetry.

If $s_{4}s\in E\left(\overline{G}\right)$, define $g^{\prime }$ as: $g^{\prime }\left(s_3\right)=2,g^{\prime }\left(s_4\right)=0,g^{\prime }\left(s\right)=1,g^{\prime }\left(v\right)=0$. Then, $g^{\prime }$ is a 2RiDF of $\overline{G}\left[\{s,v,s_3,s_4\}\right]$ with weight 2. By Lemma 2 and Formula 3, we are able to extend $g^{\prime }$ to a 2RiDF of $\overline{G}$ of weight at most $|V_0|-1+3=|V_0|+2$, a contradiction. Consequently, we have $s_{4}s\notin E\left(\overline{G}\right)$. Then, the function $g^{\prime }$ such that $g^{\prime }\left(s_3\right)=0,g^{\prime }\left(s_4\right)=g^{\prime }\left(s\right)=2,g^{\prime }\left(v\right)=1,g^{\prime }\left(u\right)=0$ is a 2RiDF of $\overline{G}\left[\{s,u,v,s_3,s_4\}\right]$ with weight 3, and by Lemma 2 and Formula 3 we can extend $g^{\prime }$ to a 2RiDF of $\overline{G}$ with weight at most $|V_0|-1+3=|V_0|+2$. This contradicts the assumption.

Case 2.2.$uv\notin E\left(\overline{G}\right)$. Then, by the selection of $u,v$ and $f_0$, $\overline{G}\left[V_0\right]$ contains only isolated vertices and G does not admit a ${\gamma }_{\mathrm{ri}2}\left(G\right)$-function for which the induced subgraph of $\overline{G}$ by vertices with value 0 contains $K_2$ components.

For every $x\in V_0$, let $U_i^x=N_{\overline{G}}\left(x\right)\cap V_i$ for $i\in [1,2]$. Let $f^{\prime }$ be: $f^{\prime }\left(x\right)=0$ for $x\in ((V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v))\cup (V_0\backslash \{u,v\})$, $f^{\prime }\left(v\right)=2$, and $f^{\prime }\left(u\right)=1$. Apparently, $f^{\prime }$ is a 2RiDF of $G-(U_1^u\cup U_2^u\cup U_1^v\cup U_2^v))$ with weight 2. According to Lemma 2, we can extend $f^{\prime }$ to a 2RiDF of G with weight at most $\left|\right(U_1^u\cup U_2^u\cup U_1^v\cup U_2^v\left)\right)|+2$. To ensure $|(U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)\left)\right|+2\ge |V_1|+|V_2|$, we have

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

(5)

Claim 6.$|(V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)|=2$ and the two vertices in $(V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)$ are adjacent in $\overline{G}$. Define a 2RiDF $g^{\prime }$ of $\overline{G}\left[V_0\right]$ as: $g^{\prime }\left(u\right)=g^{\prime }\left(v\right)=1$. Suppose that $|(V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)|\le 1$. Since $V_1$ and $V_2$ are cliques in $\overline{G}$ and every vertex in $U_1^u\cup U_2^u\cup U_1^v\cup U_2^v$ is adjacent to u or v in $\overline{G}$, by Lemma 2 we are able to extend $g^{\prime }$ to a 2RiDF g of $\overline{G}$ under which at most one vertex in $V_i,$$i\in [1,2]$, is not assigned value 0 (here if $(V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)$ contains a vertex, say w, then let $g\left(w\right)=2$). Clearly, $w\left(g\right)=w\left(g^{\prime }\right)+2\le \left|V_0\right|+2$, a contradiction. Moreover, if $(V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)$ 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 $\overline{G}$ via letting $g^{\prime }\left(x\right)=0$ when x is in $(V_1\cup V_2)\backslash \{w_1,w_2\}$ and $g^{\prime }\left(w_1\right)=g^{\prime }\left(w_2\right)=2$. However, $w\left(g\right)=w\left(g^{\prime }\right)+2\le \left|V_0\right|+2$, a contradiction. This completes the proof of Claim 6.

By Claim 6, $(V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^v\cup U_2^v)$ contains two adjacent vertices in $\overline{G}$, say $w_1,w_2$. If there exists a $z\in (V_0\backslash \{u,v\})$ s.t. $z{w}_1\in E\left(\overline{G}\right)$ (or $z{w}_2\in E\left(\overline{G}\right)$), then set $g^{\prime }$ as: $g^{\prime }\left(z\right)=g^{\prime }\left(u\right)=g^{\prime }\left(v\right)=1$, $g^{\prime }\left(w_1\right)=0$ (or $g^{\prime }\left(w_2\right)=0$), $g^{\prime }\left(w_2\right)=2$ (or $g^{\prime }\left(w_1\right)=2$). Since in $\overline{G}$ every vertex in $(V_1\cup V_2)\backslash \left\{w_2\right\}$ has a neighbor in $\{z,u,v\}$ and every vertex in $V^{\prime }\backslash \left\{w_2\right\}$ is a neighbor of $w_2$, where $w_2\in V^{\prime }$ for some $V^{\prime }\in \{V_1,V_2\}$, we can extend $g^{\prime }$ to a 2RiDF g of $\overline{G}$ according to Lemma 2. Under g, every vertex in $V^{\prime }\backslash \left\{w_2\right\}$ is assigned value 0 and at most one vertex in $\{V_1,V_2\}\backslash V^{\prime }$ is not assigned value 0. Therefore, $w\left(g\right)\le |V_0|+2$, a contradiction. This demonstrates that in $\overline{G}$ no vertex in $V_0$ is adjacent to $\{w_1,w_2\}$. Furthermore, if there is a $z\in V_0\backslash \{u,v\}$, then by Claim 6 we have $(V_1\cup V_2)\backslash (U_1^u\cup U_2^u\cup U_1^z\cup U_2^z)=\{w_1,w_2\}$ and $(V_1\cup V_2)\backslash (U_1^v\cup U_2^v\cup U_1^z\cup U_2^z)=\{w_1,w_2\}$, which implies that $N_{\overline{G}}\left(z\right)=U_1^u\cup U_2^u\cup U_1^v\cup U_2^v$. Set $g^{\prime }$ as: $g^{\prime }\left(z\right)=1,g^{\prime }\left(u\right)=g^{\prime }\left(v\right)=2$ and $g^{\prime }\left(x\right)=0$ for $x\in U_1^u\cup U_2^u\cup U_1^v\cup U_2^v$. Then, $g^{\prime }$ is a 2RiDF of $\overline{G}-(\{w_1,w_2\}\cup (V_0\backslash \{u,v,z\}))$ with weight 3, and we can extend $g^{\prime }$ to a 2RiDF of $\overline{G}$ with weight at most $\left(\right|V_0|+2-3)+3=|V_0|+2$ according to Lemma 2, a contradiction. So far, we have shown that $V_0=\{u,v\}$, that is, ${\gamma }_{\mathrm{ri}2}\left(G\right)=n-2$.

Now, we define a 2RiDF $f^{\prime }$ of $G\left[\{u,v,w_1,w_2\}\right]$ as follows: $f^{\prime }\left(w_1\right)=f^{\prime }\left(w_2\right)=0$, $f^{\prime }\left(u\right)=1$ and $f^{\prime }\left(v\right)=2$. According to Lemma 2, we can extend $f^{\prime }$ to a 2RiDF f of G with weight at most $n-2$. To ensure $w\left(f\right)\ge {\gamma }_{\mathrm{ri}2}\left(G\right)=n-2$, f must be a ${\gamma }_{\mathrm{ri}2}\left(G\right)$-function (since $w\left(f\right)=n-2$). However, $\overline{G}\left[\left\{w_1{w}_2\right\}\right]$ is isomorphic to $K_2$. This contradicts the selection of $f_0$. Eventually, the proof of Theorem 3 is finished. □

Based on the foregoing analysis, we observed that the upper bound $n+2$ can be attained by graphs $S_r(r\ge 2)$, $S_r^{+}(r\ge 2)$, and $S(r,1)$$(r\ge 1)$, while we did not find other graphs that possess this property. So, we propose a problem as follows.

Question 1. Is it enough to determine graphs G with ${\gamma }_{\mathrm{ri}2}\left(G\right)+{\gamma }_{\mathrm{ri}2}\left(\overline{G}\right)=\left|V\left(G\right)\right|+2$ by $S_r(r\ge 2)$, $S_r^{+}(r\ge 2)$, and $S(r,1)(r\ge 1)$?