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\begin{document}

\title{Restrained Roman reinforcement number in graphs}
%\author{M. Hajjari and S.M. Sheikholeslami\vspace{2mm}\\Department of Mathematics\\Azarbaijan Shahid Madani University\\Tabriz, I.R. Iran\\\texttt{m.hajjari260@gmail.com}\\ \texttt{s.m.sheikholeslami@azaruniv.ac.ir}}
\date{}
\maketitle

\begin{abstract}
A restrained Roman dominating function (RRD-function) on a graph $G=(V,E)$
is a function $f$ from $V$ into $\{0,1,2\}$ satisfying: \textrm{(i)
}every vertex $u$ with $f(u)=0$ is adjacent to a vertex $v$ with $f(v)=2$; \textrm{(ii)} the subgraph induced by the
vertices assigned 0 under $f$ has no isolated vertices. The weight of an
RRD-function is the sum of its function value over the whole set of vertices, and
the restrained Roman domination number is the minimum weight of an
RRD-function on $G.$ In this paper, we begin the study of the restrained
Roman reinforcement number $r_{rR}(G)$ of a graph $G$ defined as the
cardinality of a smallest set of edges that we must add to the graph to
decrease its restrained Roman domination number. We start by showing that
the decision problem associated with the restrained Roman reinforcement
problem is NP-hard. Then several properties as well as
some sharp bounds of the restrained Roman reinforcement number  are presented.

\end{abstract}

\textbf{keywords:} Restrained Roman domination, restrained Roman
reinforcement.

\section{Introduction}

For definitions and notations not given here we refer to \cite{hhs}. We
consider simple graphs $G$ with vertex set $V=V(G)$ and edge set $E=E(G)$. The
\textit{order} of $G$ is $n=n(G)=|V|$. The \textit{open neighborhood} of a
vertex $v,$ denoted $N(v)$ (or $N_{G}(v)$ to refer to $G$) is the set $\{u\in
V(G)\mid uv\in E\}$ and its \textit{closed neighborhood} is the set
$N[v]=N_{G}[v]=N(v)\cup\{v\}$. The \textit{degree} of vertex $v\in V$ is
$d(v)=d_{G}(v)=|N(v)|$. The \textit{maximum }and \textit{minimum degree} in
$G$ are denoted by $\Delta=\Delta(G)$ and $\delta=\delta(G)$, respectively. A
vertex of degree one is called a \textit{leaf} and its neighbor is called a
\textit{support vertex}. 
As usual, the \emph{path} (\emph{cycle},  \emph{complete
}$p$\emph{-partite graph}, respectively) of order $n$ is denoted by $P_{n}$
($C_{n}$, $K_{n_{1},n_{2},\ldots,n_{p}}$, respectively). A \emph{star
}of order $n\geq2$ {is the graph }$K_{1,n-1}.$ 
%A tree $T$ is a \emph{double
%star} if it contains exactly two vertices that are not leaves. A double star
%with respectively $p$ and $q$ leaves attached at each support vertex is
%denoted by $DS_{p,q}.$
For a subset $S\subseteq V$, the subgraph induced by $S$ in $G$ is denoted as
$G[S]$. 

A subset $S\subseteq V$ is a \emph{dominating set} of $G$ if every
vertex in $V\setminus S$ has a neighbor in $S$. The \emph{domination number
}$\gamma(G)$ is the minimum cardinality of a dominating set of $G$. In 1990,
Kok and Mynhardt {\cite{KM} }introduced the \emph{reinforcement number} $r(G)$
of a graph $G$ as the minimum numbers of edges that have to be added to the
graph $G$ in order to decrease the domination number. However, it was assumed
that $r(G)=0$ for all graphs $G$ with domination number one. Since then, the
concept of reinforcement has been defined and studied for several other
domination parameters, such as Roman domination \cite{JS}, total Roman
domination \cite{AAC}, Italian domination \cite{HSS}, double Roman domination
\cite{amjadi} and rainbow domination \cite{ADF,SAKS}.

The concept of restrained Roman domination has been introduced in 2015 by Leely Pushpam and Padmapriea \cite{PP} as a new variation of Roman domination. An \textit{restrained Roman dominating function}
(RRD-function, for short) on a graph $G$ is a function $f:V\longrightarrow
\{0,1,2\}$ having the properties that \textrm{(i)
}every vertex $u$ with $f(u)=0$ is adjacent to a vertex $v$ with $f(v)=2$; and 
\textrm{(ii)} the subgraph induced by the
vertices assigned 0 under $f$ has no isolated vertices. 
 The weight of an RRD-function $f$ is the sum $w(f)=\sum_{v\in
V(G)}f(v)$ and the  restrained Roman domination number is the minimum weight of an
RRD-function on $G.$ This parameter has been studied by several authors \cite{AAC1,AAC2}.
Any RRD-function $f$ on $G$ can be simply referred to as $f=(V_{0},V_{1},V_{2})$,
where $V_{i}=\{v\in V(G):f(v)=i\}$ for $i\in\{0,1,2\}$.

In this paper, we are interested in begining the study of the \emph{restrained
Roman reinforcement} number $r_{rR}(G)$ of a graph $G$ defined as the
cardinality of a smallest set of edges $F\subseteq E(\overline{G})$ such that
$\gamma_{rR}(G+F)<\gamma_{rR}(G),$ where $\overline{G}$ denotes the complement
graph of $G.$ In the case that there is no subset of edges $F$ with
$\gamma_{rR}(G+F)<\gamma_{rI}(G)$, we define $r_{rR}(G)=0$. Since for any
nontrivial connected graph $G,$ $\gamma_{rR}(G)\geq2,$ we deduce that
$r_{rR}(G)=0$ for all nontrivial connected graphs with $r_{rR}(G)=2$.
Moreover, we say that a subset $E^{\prime}\subseteq E(\overline{G})$ {is an
}$r_{rR}(G)$-set if $|E^{\prime}|=r_{rR}(G)$ and $\gamma_{rR}(G+E^{\prime
})<\gamma_{rR}(G)$.

In the remainder of this paper, we prove that the decision problem corresponding to
the problem of computing $r_{rR}${$(G)$ {is NP-hard for an arbitrary graph }%
}$G${{. Then various properties of the restrained Roman reinforcement number
are investigated and some sharp bounds on it are presented. }

%We make use of the following results.





We finish this section by observing that any $r_{rR}(G)$-set of a connected graph
$G$ with $\gamma_{rR}(G)\geq3$ can decrease the restrained Roman domination
number of $G$ by {at most two. }

\begin{proposition}
\label{frame}Let $G$ be a connected graph with $\gamma_{rR}(G)\geq3$. If $F$
is {an} $r_{rR}(G)$-set, then
\[
\gamma_{rR}(G)-2\leq\gamma_{rR}(G+F)\leq\gamma_{rR}(G)-1.
\]
{Both }bounds are sharp.
\end{proposition}

%\begin{proof}
\noindent\textbf{Proof.} By assumption, $\gamma_{rR}(G+F)<\gamma_{rR}(G),$
whence the upper bound. {To }show the lower bound, {let us assume that
}$\gamma_{rR}(G+F)\leq\gamma_{rR}(G)-3$. Let $f$ be a $\gamma_{rR}%
(G+F)$-function and let $uv\in F$ {such that }$0\in\{f(u),f(v)\}$. {If such an
edge does not exist, then }$f$ {is an RRD-function of }$G$ {leading to the
contradiction }$\gamma_{rR}(G)\leq\gamma_{rR}(G+F).$ {Hence we suppose that
}$uv$ {exists. Let }$F^{\prime}=F-\{uv\}.$ {Without }loss of generality,
{suppose} that $f(u)=0$. If $f(v)=1$, then $f$ {is an RRD-function of }$G$ {leading to the
contradiction }$\gamma_{rR}(G)\leq\gamma_{rR}(G+F)$ too.
 {Hence }assume that $f(v)\neq1$.

First let $f(v)=2$. If $u$ has a neighbor $w$ in $G+F^{\prime}$ with $f(w)\geq
1$, then the function $g$ defined by $g(w)=2$ and $g(x)=f(x)$ otherwise, is
{an }RRD-function of $G+F^{\prime}$ yielding {as above to the contradiction}
$\gamma_{rR}(G+F^{\prime})<\gamma_{rR}(G)$. Hence we assume that each neighbor
$u$ in $G+F^{\prime}$ is {assigned }0 under $f$. Let $x_{1},\dots,x_{k}$ be the
neighbors of $u$ in $G+F^{\prime}$. If $k=1$ and $x_{1}$ has a neighbor
{assigned }0 other than $u$, then the function $g(u)=1$ and $g(x)=f(x)$
otherwise, is {an} RRD-function of $G+F^{\prime}$ yielding $\gamma
_{rR}(G+F^{\prime})\leq\gamma_{rR}(G+F)+1<\gamma_{rR}(G),$ {a contradiction}. If
$k=1$ and $x_{1}$ has no neighbor {assigned }0 other than $u$, then the
function $g(u)=g(x_{1})=1$ and $g(x)=f(x)$ otherwise, is {an} RRD-function of
$G+F^{\prime}$ {and thus }$\gamma_{rR}(G+F^{\prime})\leq\gamma_{rR}(G+F)+2<\gamma
_{rR}(G),$ a contradiction too. Hence assume that $k\geq2$. If {some }$x_{i}$
has no neighbor {assigned }0 other than $u,$ then the function $g(x_{i})=2$
and $g(x)=f(x)$ otherwise, is an RRD-function of $G+F^{\prime}$ yielding {again
}$\gamma_{rR}(G+F^{\prime})<\gamma_{rR}(G)$. {Hence we assume that for each
}$i,$ $x_{i}$ has {at least two }neighbors {assigned 0 under }$f.$ {In this
case, }the function $g(u)=1$ and $g(x)=f(x)$ otherwise, is {an }RRD-function
of $G+F^{\prime}$ {and thus }$\gamma_{rR}(G+F^{\prime})<\gamma_{rR}(G).$

Finally, assume that $f(v)=0$. Since $F$ is an $r_{rR}(G)$-set, we {can
}suppose, without loss of generality, that all neighbors of $u$ in $G+F^{\prime}$
have positive labels under $f$. {Now, if }$v$ has a neighbor with {weight }0
in $G+F^{\prime}$, then the function $g(u)=1$ and $g(x)=f(x)$ otherwise, is {an}
RRD-function of $G+F^{\prime}$ {while if }$v$ has no neighbor with {weight }0 in
$G+F^{\prime}$, then the function $g(u)=g(v)=1$ and $g(x)=f(x)$ otherwise, is
{an }RRD-function of $G+F^{\prime}.$ {Both situations} yield the contradiction
$\gamma_{rR}(G+F^{\prime})<\gamma_{rR}(G).$ {Consequently, }$\gamma
_{rR}(G+F)\geq\gamma_{rR}(G)-2.$

The cycle $C_{4}$ is a simplest example of a graph for which the lower bound
of Proposition \ref{frame} is attained, while the sharpness of the upper bound
can be seen for the cycle $C_{6}.$ $\hfill\Box$

\begin{figure}[h]
\centering
\begin{tikzpicture}%[scale=.3, transform shape]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%(1)
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\node at (-2.2,1.9) {$z_i$};
%\node at (-1.3,2.0) {$q_i$};
%\node at (-1.3,2.6) {$a_i$};
%\node at (-1.3,3.23) {$b_i$};
%\node at (-1.3,3.93) {$d_i$};
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\node at (-2.0,-0.3) {$x_i$};
%\node at (-1.3,-0.3) {$w_i$};
\node at (-.5,-0.3) {$\overline{x}_i$};
\node at (-1.3,-1) {$K_i$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%(2)
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\node at (2.95,2.8) {$s_1$};
\node at (2.9,.9) {$s_3$};
%\node at (2.95,1.5) {$s_2$};
%\node at (4.5,2) {$s_4$};
%\node at (2.95,-5.4) {$v_1$};
%\node at (4.35,-5.4) {$v_2$};
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%\node at (2.05,-6.4) {$v_5$};
%\node at (2.05,-5.6) {$v_6$};
%\node at  (2.1,1.25) {$t_1$};
%\node at (1,.5) {$t_2$};
%\node at (2.1,-.3) {$t_3$};
%\node at  (4.1,1.25) {$m_1$};
%\node at (5.6,.5) {$m_2$};
%\node at (4.1,-.3) {$m_3$};
\node at (3.2,-1) {$H_1$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%
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\node at (8.9,2.8) {$s_1'$};
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\node at (9.7,1.7) {$s_2'$};
%\node at (6.5,-3.5) {$s_4$};
%\node at (2.95,-5.4) {$v_1$};
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%\node at (2.05,-6.4) {$v_5$};
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%\node at (10.1,-.3) {$m_3'$};
\node at (9.2,-1) {$H_2$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{tikzpicture}
\caption{The graphs $K_{i}$ and $H=H_{1}\cup H_{2}$}%
\label{FAZ}%
\end{figure}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{NP-hardness result}

The aim of this section, is to show that the decision problem associated with
the Restrained Roman reinforcement is NP-hard. Consider the following
decision problem.\newline

\textbf{Restrained  Roman problem (RR-reinforcement)}

\textbf{Instance:} \emph{A nonempty graph $G$ and a positive integer $k$. }

\textbf{Question:} \emph{Is }$r_{rR}(G)\leq k$ ?\newline

\bigskip

We show that the NP-hardness of the RR-reinforcement problem by transforming
the well-known 3-SAT problem to it in polynomial time. Recall that the 3-SAT
problem specified below was proven to be NP- complete in \cite{GJ}.

\bigskip

\smallskip\noindent\textbf{3-SAT problem}

\smallskip\noindent\textbf{Instance: } A collection $\mathcal{C}=\{C_{1}%
,C_{2},\ldots,C_{m}\}$ of clauses over a finite set $U$ of variables such that
$|C_{j}|=3$ for every $j\in\{1,2,\ldots,m\}.$

\smallskip\noindent\textbf{Question: } Is there a truth assignment for $U$
that satisfies all the clauses in $\mathcal{C}$?

\bigskip

\begin{theorem}
The RR-reinforcement problem is NP-complete for arbitrary graphs.
\end{theorem}

%\begin{proof}
\noindent\textbf{ Proof.} Let $X=\{x_{1},x_{2},\ldots,x_{n}\}$ and
$\mathcal{C}=\{C_{1},C_{2},\ldots,C_{m}\}$ be an arbitrary instance of 3-SAT
problem. We will build a graph $G$ and a positive integer $k$ such that $r_{rR}(G)\leq k$ if and only if
$\mathcal{C}$ is satisfiable.

{We associate to each variable }$x_{i}\in X$
{a copy of the graph} $K_{i}$ {as }depicted {in Figure }\ref{FAZ}, for each $i\in\{1,2,\ldots,n\},$. For each
$j\in\{1,2,\ldots,m\}$, {we associate to each }clause $C_{j}=\{u_{j}%
,v_{j},w_{j}\}\in\mathcal{C}$ a  vertex $c_{j}$ {by adding the }edge-set
$E_{j}=\{c_{j}u_{j},c_{j}v_{j},c_{j}w_{j}\}$. Finally, {we }enclose the graph $H$
illustrated in Figure \ref{FAZ} {by connecting vertices} $s_{1},s_{1}^{\prime}$
to every vertex $c_{j}$. Clearly, the resulting graph $G$ is of order
$8n+m+19$ {and size }$11n+5m+27$ {and hence }$G$ {can be build in
polynomial time. }Set $k=1$. {Figure \ref{FAZ1} provides an example of the
resulting graph} when $X=\{x_{1},x_{2},x_{3},x_{4}\}$ and $\mathcal{C}%
=\{C_{1},C_{2},C_{3}\}$, where $C_{1}=\{x_{1},x_{2},\bar{x}_{3}\}$,
$C_{2}=\{\bar{x}_{1},x_{2},x_{4}\}$ {and} $C_{3}=\{\bar{x}_{2},x_{3},x_{4}\}$.

\begin{figure}[h]
\centering
\begin{tikzpicture}%[scale=.3, transform shape]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%(1)
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%\node at (-1.3,2.0) {$q_1$};
%\node at (-1.3,2.6) {$a_1$};
%\node at (-1.3,3.23) {$b_1$};
\node at (-.4,1.9) {$y_1$};
\node at (-2.0,-0.3) {$x_1$};
%\node at (-1.3,-0.3) {$w_1$};
\node at (-.5,-0.3) {$\overline{x}_1$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%(2)
\node [draw, shape=circle,fill=black,scale=0.3] (p2) at (0.9,1.8) {};
\node [draw, shape=circle,fill=white,scale=0.3] (q2) at (1.6,1.8) {};
\node [draw, shape=circle,fill=white,scale=0.3] (a2) at (1.6,2.4) {};
\node [draw, shape=circle,fill=white,scale=0.3] (b2) at (1.6,3) {};
\node [draw, shape=circle,fill=white,scale=0.3] (r2) at (2.3,1.8) {};
\node [draw, shape=circle,fill=white,scale=0.3] (u2) at (0.9,0) {};
\node [draw, shape=circle,fill=white,scale=0.3] (w2) at (1.6,0) {};
\node [draw, shape=circle,fill=black,scale=0.3] (u'2) at (2.3,0) {};
\draw [line width=0.8pt](p2)--(q2)--(r2)--(u'2)--(w2)--(u2)--(p2);
\draw [line width=0.8pt](b2)--(p2)--(a2)--(r2)--(b2)(w2)--(q2);
\node at (0.7,2) {$z_2$};
%\node at (1.6,2) {$q_2$};
%\node at (1.6,2.6) {$a_2$};
%\node at (1.6,3.23) {$b_2$};
\node at (2.5,2) {$y_2$};
\node at (0.7,-0.25) {$x_2$};
%\node at (1.6,-0.250) {$w_2$};
\node at (2.3,-0.25) {$\overline{x}_2$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%(3)
\node [draw, shape=circle,fill=white,scale=0.3] (p3) at (3.8,1.8) {};
\node [draw, shape=circle,fill=white,scale=0.3] (q3) at (4.5,1.8) {};
\node [draw, shape=circle,fill=white,scale=0.3] (a3) at (4.5,2.4) {};
\node [draw, shape=circle,fill=white,scale=0.3] (b3) at (4.5,3) {};
\node [draw, shape=circle,fill=black,scale=0.3] (r3) at (5.2,1.8) {};
\node [draw, shape=circle,fill=black,scale=0.3] (u3) at (3.8,0) {};
\node [draw, shape=circle,fill=white,scale=0.3] (w3) at (4.5,0) {};
\node [draw, shape=circle,fill=white,scale=0.3] (u'3) at (5.2,0){};
\draw  [line width=0.8pt](u3)--(p3)--(q3)--(r3)--(u'3)--(w3)--(u3);
\draw [line width=0.8pt](b3)--(p3)--(a3)--(r3)--(b3)(w3)--(q3);
\node at(3.6,2)  {$z_3$};
%\node at (4.5,2) {$q_3$};
%\node at (4.5,2.6) {$a_3$};
%\node at (4.5,3.23) {$b_3$};
\node at (5.4,2) {$y_3$};
\node at (3.8,-0.25) {$x_3$};
%\node at (4.5,-0.25)  {$w_3$};
\node at (5.2,-0.25) {$\overline{x}_3$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%(4)
\node [draw, shape=circle,fill=white,scale=0.3] (p4) at (6.7,1.8) {};
\node [draw, shape=circle,fill=white,scale=0.3] (q4) at (7.4,1.8) {};
\node [draw, shape=circle,fill=white,scale=0.3] (a4) at (7.4,2.4) {};
\node [draw, shape=circle,fill=white,scale=0.3] (b4) at (7.4,3) {};
\node [draw, shape=circle,fill=black,scale=0.3] (r4) at (8.1,1.8) {};
\node [draw, shape=circle,fill=black,scale=0.3] (u4) at (6.7,0) {};
\node [draw, shape=circle,fill=white,scale=0.3] (w4) at (7.4,0) {};
\node [draw, shape=circle,fill=white,scale=0.3] (u'4) at (8.1,0) {};
\draw  [line width=0.8pt](u4)--(p4)--(q4)--(r4)--(u'4)--(w4)--(u4);
\draw [line width=0.8pt](b4)--(p4)--(a4)--(r4)--(b4)(w4)--(q4);
\node at(6.5,2) {$z_4$};
%\node at (7.4,2) {$q_4$};
%\node at (7.4,2.6) {$a_4$};
%\node at (7.4,3.23) {$b_4$};
\node at (8.3,2) {$y_4$};
\node at (6.8,-0.2) {$x_4$};
%\node at (7.4,-0.2) {$w_4$};
\node at (8.1,-0.2) {$\overline{x}_4$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%
\node [draw, shape=circle,fill=white,scale=0.3] (c1) at (0.8,-2.2){};
\node [draw, shape=circle,fill=white,scale=0.3] (c2) at (3.2,-2.2){};
\node [draw, shape=circle,fill=white,scale=0.3] (c3) at(5.6,-2.2){};
\node at (0.6,-1.8) {$c_1$};
\node at (3.2,-1.9) {$c_2$};
\node at (5.55,-1.85) {$c_3$};
\draw  [line width=0.8pt](u2)--(c1)--(u1)(c1)--(u'3);
\draw  [line width=0.8pt](u'1)--(c2)--(u2)(c2)--(u4);
\draw  [line width=0.8pt](u'2)--(c3)--(u3)(c3)--(u4);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\node [draw, shape=circle,fill=black,scale=0.3] (s1) at (0.2,-3.2){};
\node [draw, shape=circle,fill=black,scale=0.3] (s2) at (0.2,-5){};
\node [draw, shape=circle,fill=white,scale=0.3] (s3) at (0.2,-4){};
\node [draw, shape=circle,fill=white,scale=0.3] (s4) at (1.2,-3.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (t1) at (-1.1,-4.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (t2) at (-2.25,-5){};
\node [draw, shape=circle,fill=white,scale=0.3] (t3) at (-1.1,-5.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (p1) at (1.1,-4.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (p2) at (2.25,-5){};
\node [draw, shape=circle,fill=white,scale=0.3] (p3) at (1.1,-5.5){};
%\node [draw, shape=circle,fill=black,scale=0.3] (v1) at (3.2,-5.6){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v2) at (4.1,-5.6){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v3) at (4.1,-6.4){};
%\node [draw, shape=circle,fill=black,scale=0.3] (v4) at (3.2,-6.4){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v5) at (2.3,-6.4){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v6) at (2.3,-5.6){};
\draw  [line width=0.8pt](c2)--(s1)--(c1)(s1)--(c3);
\draw  [line width=0.8pt](s2)--(t1)--(t2)--(t3)--(s2)--(t2);
\draw  [line width=0.8pt](s1)--(s2)--(s3);
\draw  [line width=0.8pt](s1)--(s3)--(s4)--(s1);
\draw  [line width=0.8pt](s2)--(p1)--(p2)--(p3)--(s2)--(p2);
\node at (0,-3.3) {$s_1$};
\node at (0,-4.7) {$s_3$};
%\node at (0,-4) {$s_2$};
%\node at (1.5,-3.5) {$s_4$};
%\node at (2.95,-5.4) {$v_1$};
%\node at (4.35,-5.4) {$v_2$};
%\node at (4.35,-6.4) {$v_3$};
%\node at (3.2,-6.6) {$v_4$};
%\node at (2.05,-6.4) {$v_5$};
%\node at (2.05,-5.6) {$v_6$};
%\node at  (-1.1,-4.25) {$t_1$};
%\node at (-2.4,-5) {$t_2$};
%\node at (-1.1,-5.7) {$t_3$};
%\node at  (1.1,-4.25) {$m_1$};
%\node at (2.6,-5) {$m_2$};
%\node at (1.1,-5.7) {$m_3$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\node [draw, shape=circle,fill=white,scale=0.3] (s11) at (6.2,-3.2){};
\node [draw, shape=circle,fill=black,scale=0.3] (s21) at (6.2,-5){};
\node [draw, shape=circle,fill=white,scale=0.3] (s31) at (6.5,-4){};
%\node [draw, shape=circle,fill=white,scale=0.3] (s41) at (6.2,-3.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (t11) at (5.1,-4.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (t21) at (4.25,-5){};
\node [draw, shape=circle,fill=white,scale=0.3] (t31) at (5.1,-5.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (p11) at (7,-4.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (p21) at (8,-5){};
\node [draw, shape=circle,fill=white,scale=0.3] (p31) at (7,-5.5){};
%\node [draw, shape=circle,fill=black,scale=0.3] (v1) at (3.2,-5.6){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v2) at (4.1,-5.6){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v3) at (4.1,-6.4){};
%\node [draw, shape=circle,fill=black,scale=0.3] (v4) at (3.2,-6.4){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v5) at (2.3,-6.4){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v6) at (2.3,-5.6){};
\draw  [line width=0.8pt](c2)--(s11)--(c1)--(s11)--(c3);
\draw  [line width=0.8pt](s21)--(t11)--(t21)--(t31)--(s21)--(t21);
\draw  [line width=0.8pt](s21)--(s11)--(s31)--(s21);
\draw  [line width=0.8pt](s21)--(p11)--(p21)--(p31)--(s21)--(p21);
\node at (6.6,-3.3) {$s_1'$};
\node at (6,-4.7) {$s_3'$};
%\node at (6.8,-3.8) {$s_2'$};
%\node at (6.5,-3.5) {$s_4$};
%\node at (2.95,-5.4) {$v_1$};
%\node at (4.35,-5.4) {$v_2$};
%\node at (4.35,-6.4) {$v_3$};
%\node at (3.2,-6.6) {$v_4$};
%\node at (2.05,-6.4) {$v_5$};
%\node at (2.05,-5.6) {$v_6$};
%\node at  (5.5,-4.25) {$t_1'$};
%\node at (4.1,-5) {$t_2'$};
%\node at (5.1,-5.8) {$t_3'$};
%\node at  (7.1,-4.25) {$m_1'$};
%\node at (8.4,-5) {$m_2'$};
%\node at (7.1,-5.7) {$m_3'$};
\end{tikzpicture}
\caption{An instance of the restrained Italian reinforcement number problem
resulting from an instance of 3-SAT. Here $k=1$ and $\gamma_{rR}(G)=22$, where
the black vertex $p$ means there is a RRDF $f$ with $f(p)=2$.}%
\label{FAZ1}%
\end{figure}It is easy to verify that for any $\gamma_{rR}(G)$-function $g$ we must
have $\sum_{v\in V(K_{j})}g(v)\geq4$ for each $j\in\{1,2,\ldots,n\}$ .
Moreover, to restrained Roman dominate all vertices of $V(H)$, we need
$\sum_{i=1}^{m}f(c_{i})+g(V(H))\geq6$. Thus $\gamma_{rR}(G)=w(g)\geq
4n+6$. The equality is attained since one can easily define {an
}RRD-function of $G$ with weight $4n+6$.
% {(you can follow the
%example of the function given on the resulting graph of Figure \ref{FAZ1})}.
%{Consequently, $\gamma_{rI}(G)=4n+6$.}

{In the following, we } show that $\mathcal{C}$ is satisfiable if and only
if $r_{rR}(G)=1$. Let $\mathcal{C}$ be  satisfiable and {
}$t:U\rightarrow\{T,F\}$  a satisfying function for $\mathcal{C}$. We
build a subset $S$ of vertices of $G$ as follows. If $t(x_{i})=T$, then
put the vertices $x_{i}$ and $y_{i}$ in $S$; {while} if $t(x_{i})=F$, then put
the vertices $\overline{x_{i}}$ and $z_{i}$ in $S$. So $|S|=2n$. {Define the function} $h$ {on }$V(G)$ by $h(x)=2$ for every $x\in
S,h(s_{1})=1,h(s_{3})=h(s_{3}^{\prime})=2$ and $h(y)=0$ for the remaining
vertices. It is easy to verify that $h$ is {an }RRD-function of $G+s_{4}s_{3}$
of weight $4n+5<\gamma_{rR}(G)=4n+6$, and hence $r_{rR}(G)=1$.

Conversely, let $r_{rR}(G)=1$. Then, there is an edge $e=uv\in
E(\overline{G})$ for which $\gamma_{rR}(G+e)<4n+6$. Let $g=(V_{0},V_{1}%
,V_{2})$ be a $\gamma_{rR}(G+e)$-function. Since whatever the added edge $e,$
{we have }$g(K_{i})\geq4$, {and thus vertices }$u$ and $v$ {cannot both belong
to }$V(K_{i})$ {(for otherwise }$\gamma_{rR}(G+e)\geq4n+6$). {Consequently, we
can examine }the following {situations}.

{Either }$u,v\in\{c_{i}\mid 1\le i\le m\}$ or $u\in V(K_{i})$ for some $i$
and $v\in\{c_{j}\}\cup V(H)$ {for some }$j$. {In this case, as seen above},
one can check that $\omega(g)\geq4n+6$, a contradiction. {The same can be
seen when }$u\in\{c_{1},c_{2},\ldots,c_{m}\}$ and $v\in V(H).$ {Therefore,
assume that }$u,v\in V(H)$. {Since }$r_{rR}(G)=1$ {and }$g(K_{i})\geq4$, {we must have
}$\sum_{j=1}^{m}g(c_{j})+g(V(H))<6.$ {Since also whatever the added edge }$e,$
{we have }$g(V(H))\geq5,$ {we conclude that }$g(V(H))=5.$ {Indeed, we
have }$g(s_{1}^{\prime})=0,$ $g(s_{1})\leq1$ {and} $\sum_{j=1}^{m}g(c_{j})=0$.
{In addition, we note that }if $\{x_{i},\bar{x_{i}}\}\subseteq V_{2}$ or
$\{x_{i},\bar{x_{i}}\}\cap V_{1}\neq\emptyset$ for some $i$, then
$g(K_{i})\geq5$ {which results in the contradiction }$\gamma_{rR}(G+e)\geq4n+6$.
Thus, $\left\vert \{x_{i},\bar{x_{i}}\}\cap V_{2}\right\vert \leq1$ {and
}$\{x_{i},\bar{x_{i}}\}\cap V_{1}=\emptyset$ for {every }$i\in\{1,\dots,n\}$.
Therefore {each} vertex $c_{j}$ {must have a neighbor in }$\{x_{i},\bar{x_{i}%
}\}$ for some $i$ {which is labeled a 2.} {In this case, define the mapping }$t:U\rightarrow
\{T,F\}$  by
\begin{equation}
t(x_{i})=%
\begin{cases}
T & \text{if \ \ $f(x_{i})=2$ }\\
F & \text{otherwise \ \ }%
\end{cases}
\label{eq2}%
\end{equation}
{for $i\in\{1,\ldots,n\}$}. We shall show that $t$ is a satisfying truth
assignment for $\mathcal{C}$. It is enough to show that every clause in
$\mathcal{C}$ is satisfied by $t$. Consider arbitrary clause $C_{j}%
\in\mathcal{C}$ {for some }$j\in\{1,\dots,m\}$. If $c_{j}$ is dominated by
$x_{i}$, then $g(x_{i})=2$ and so $t(x_{i})=T$. If $c_{j}$ is dominated by
$\bar{x_{i}}$, then $g(\bar{x_{i}})=2$ and hence $t(x_{i})=F$ and $t(\bar{x_{i}%
})=T$. Therefore, in {either case} the clause $C_{j}$ is satisfied. The
arbitrariness of $j$ shows that all clauses in $\mathcal{C}$ are satisfied by
$t$, that is, $\mathcal{C}$ is satisfiable. {This completes the proof of the
}theorem. $\hfill\Box$
%\end{proof}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Exact values}

In this section we determine the restrained Roman reinforcement number of
some classes of graphs including paths, cycles and complete $p$-partite graphs
{for any integer }$p\geq2$. 
%{Here, we recall some results that we will use later. 
As observed in \cite{PP}, for every connected graph }$G$ {of
order }$n\geq2,$ $2\leq\gamma_{rR}(G)\leq n.$ {A characterization of all
connected graphs of order }$n$ {with }$\gamma_{rR}(G)\in\{1,2,3,n\}$ {was
provided by in \cite{PP,AAC2} as follows. } 


{{
\begin{figure}[h]
\centering
\begin{tikzpicture}%[scale=.3, transform shape]
\node [draw, shape=circle,fill=black,scale=0.5] (a1) at  (1,0) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a2) at  (3,0) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a3) at  (3,1) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a4) at  (2,1.5) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a5) at  (1,1) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a6) at  (0,1.6) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a7) at  (0,1.2) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a8) at  (0,0.8) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a9) at  (0,0.4) {};
\draw(a9)--(a5)--(a8)--(a5)--(a7)--(a5)--(a6)--(a5)--(a4)--(a3)--(a2)--(a1)--(a5);
\node at (2,-.9) {The graph $B_4$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\node [draw, shape=circle,fill=black,scale=0.5] (a1) at  (5,0) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a2) at  (7,0) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a3) at  (7,1) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a4) at  (6,1.5) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a5) at  (5,1) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a6) at  (4,1.4) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a7) at  (4,1) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a8) at  (4,.6) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a9) at  (8,1.2) {};
\node [draw, shape=circle,fill=black,scale=0.5] (a10) at  (8,.8) {};
\draw(a8)--(a5)--(a7)--(a5)--(a6)--(a5)--(a4)--(a3)--(a10)--(a3)--(a9)--(a3)--(a2)--(a1)--(a5);
\node at (6,-.9) {The graph $B_{2,3}$};
\node at (1,1.3) {$u_1$};
\node at (5,1.3) {$u_1$};
\node at (2,1.8) {$u_2$};
\node at (6,1.8) {$u_2$};
\node at (7,1.3) {$u_3$};
\node at (3,1.3) {$u_3$};
\node at (1,-.3) {$u_5$};
\node at (5,-.3) {$u_5$};
\node at (7,-.3) {$u_4$};
\node at (3,-.3) {$u_4$};
\end{tikzpicture}
\caption{Graphs $B_4$ and $B_{2,3}$.}
\label{fig-0}
\end{figure}
}

Let $C=(u_1u_2u_3u_4u_5)$ be a cycle of length  5 and let $B_p$ be
 obtained from $C$ by adding $p \geq 1$ pendant edges at
{$u_1$} and  $B_{p,q}$ be obtained from $C$ by adding
$p \geq 1$ pendant edges at  {$u_1$} and $q \geq 1$ pendant
edges at  {$u_3$} (See Fig. \ref{fig-0}).
\begin{prelem}\label{n}
{\em \cite{PP} Let $G$ be a connected graph of order $n\ge 2$. Then \\
(a) $\gamma_{rR}(G) =1$ if and only if $G=K_1$.

\noindent (b) $\gamma_{rR}(G)=2$ if and only if $n=2$ or
$\Delta(G)=n-1$ and $\delta(G)\ge 2$.

\noindent (c) $\gamma_{rR}(G)=n$ if and only if $G\simeq C_4,C_5,B_p,B_{p,q}$ or
$G$ is a tree with ${\rm diam}(G)\le 5$.}
\end{prelem}

\begin{prelem}\label{pp1}
{\em \cite{AAC2} Let $G$ be a connected graph of order $n\ge 4$. Then $\gamma_{rR}(G)=3$ if and only {if $G$ satisfies one of the following conditions}:
\begin{itemize}
  \item[{(i)}] $\Delta(G)=n-1$ and $G$ has exactly one leaf.
  \item[{(ii)}] $\Delta(G)=n-2$ and $G$ has a vertex $u$ of degree $n-2$ such that the induced subgraph $G[N(u)]$ has no isolated vertex.
\end{itemize}
}
\end{prelem}




{The exact values of the restrained Roman domination number have been
established in \cite{PP} for paths, cycles, complete }$p$-partite graphs.

\begin{prelem} \label{o3}
{ (a) $\gamma_{rR} (P_n) = n$ for $1 \leq n \leq 6$ and
$\gamma_{rR} (P_n) =\left\lceil \frac{2n+1}{3} \right\rceil +1$
for $n \geq 7$.\\ (b) $\gamma_{rR}(C_n) = 2\left\lceil
\frac{n}{3}\right\rceil$ when $n \not\equiv 2\;({\rm mod}\;3)$ and
$\gamma_{rR}(C_n) = 2\left\lceil \frac{n}{3}\right\rceil +1$
otherwise.\\ (c) $\gamma_{rR}(K_{m,n})=4$ for $m,n\ge2$.\\ (d) {If }$K_{n_{1},n_{2},\ldots,n_{p}}$ {is }the complete
$p$-partite graph such that $p\geq3$ and $n_{1}\leq n_{2}\leq\ldots\leq n_{p}%
$, {then }$\gamma_{rR}(K_{1,n_{2},\ldots,n_{p}})=2$, $\gamma_{rR}(K_{2,n_{2}%
,\ldots,n_{p}})=3$ and $\gamma_{rR}(K_{n_{1},n_{2},\ldots,n_{p}})=4$ for
$n_{1}\geq3$.}
\end{prelem}


{We now are ready to establish the restrained Roman reinforcement number for
paths, cycles and complete }$p$-partite graphs. { }

\begin{proposition}
\label{pn1}\emph{For $n\geq3$, }$r_{rR}$\emph{$(P_{n})=1$.}
\end{proposition}

%\begin{proof}
\noindent\textbf{ Proof.} Let $P_{n}:=w_{1}w_{2}\ldots w_{n}$. If $n\equiv0\,(\mathrm{{mod}%
\,3})$, then the function $g$ defined by $g(w_{3i+1})=2$ for $0\leq
i\leq(n-3)/3$ and $g(w)=0$ otherwise, is {an} RRD-function of $P_{n}%
+w_{1}w_{n}$ of weight $\frac{2n}{3}.$ If $n\equiv2\,(\mathrm{{mod}\,3)}$,
then the function $g$ defined by $g(w_{n})=2$, $g(w_{3i+1})=2$ for $0\leq
i\leq(n-5)/3$ and $g(x)=0$ otherwise, is {an} RRD-function of $P_{n}%
+w_{1}w_{n-2}$ of weight $\frac{2n+2}{3}.$ Finally, if $n\equiv
1\,(\mathrm{{mod}\,3)}$, then the function $g$ defined by $g(w_{n})=1$,
$g(w_{3i+1})=2$ for $0\leq i\leq(n-4)/3$ and $g(w)=0$ otherwise, is {an}
RRD-function of $P_{n}+w_{1}w_{n-1}$ of weight $\frac{2n+1}{3}.$ {Thus }$r_{rR}(P_{n})=1.\hfill\Box$
%\end{proof}


\begin{proposition}
\label{cn1}\emph{For $n\geq4$,
\[
r_{rR}(C_{n})=\left\{
\begin{array}
[c]{ll}%
2 & \text{\textrm{if }}n\equiv0\,(\mathrm{{mod}\,3})\\
1 & \mathrm{otherwise.}%
\end{array}
\right.
\]
}
\end{proposition}

%\begin{proof}
\noindent\textbf{ Proof.} Assume that $C_{n}:=(w_{1}w_{2}\ldots w_{n})$ be a
cycle on $n$ vertices. If $n\equiv r\pmod 3$ where $r\in\{1,2\}$, then {by a
similar argument to that used in }the proof of Proposition \ref{pn1}, we can
see that $r_{rR}(C_{n})=1$. Hence we assume that $n\equiv0\pmod 3$. First,
{since }the function $g$ defined by $g(w_{n-2})=1$, $g(w_{3i+1})=2$ for $0\leq
i\leq(n-6)/3$ and $g(x)=0$ otherwise, is {\color{red}}an RRD-function of
$C_{n}+\{w_{1}w_{n-1},w_{1}w_{n-3}\}$ of weight $(2n-3)/3=\gamma_{rR}%
(C_{n})-1$ (Proposition \ref{o3}-(b)), {we deduce that }$r_{rR}(C_{n})\leq2$.

Now we prove the inverse inequality. To this end, we {need only to }show that {adding an
arbitrary edge }$e$ {cannot decrease }$\gamma_{rR}(C_{n}).$ {Observe that }for
any edge $e\in\overline{C_{n}}$, { }$\gamma_{rR}(C_{n}+e)\leq\gamma_{rR}%
(C_{n})$. Let $e$ be an arbitrary edge in $\overline{C_{n}}$ and let $f$ be a
$\gamma_{rR}(C_{n}+e)$-function. {Suppose first that there are }three
consecutive vertices $w_{i},w_{i+1},w_{i+2}$ such that $f(w_{i})=f(w_{i+1}%
)=f(w_{i+2})=0$, say {for }$i=1$. Then the edge $e$ must join $w_{2}$ to
{some }vertex {assigned 2, say }$w_{k}$, {with }$k\notin\{1,3\}.$ {Also,
}to dominate $w_{1}$ and $w_{3}$, we must {also }have $f(w_{4})=f(w_{n})=2$.
{Consider the cycles }$C^{\prime}:=(w_{2}w_{3}\ldots w_{k})$ {of order }$k-1$
{and }$C^{\prime\prime}:=(w_{2}w_{k}\ldots w_{n}w_{1})$ {of order }$n-k+3.$
{Let }$k-1\equiv s_{1}\pmod 3${ and }$n-k+3\equiv s_{2}\pmod 3.$ {Notice that
}$s_{1}=0$ and $s_{2}=2;$ $s_{1}=2$ and $s_{2}=0$ or $s_{1}=s_{2}=1.$ {Assume
that }$k-1\equiv0\pmod 3$ {(the case }$n-k+3\equiv0\pmod 3$ {is similar). Then
}$n-k+3\equiv2\pmod 3,$ {and since the restrictions of }$f$ on $V(C^{\prime})$
and $V(C^{\prime\prime})$ {are RRD-functions, we deduce from Proposition
\ref{o3}, that }%

\begin{align*}
\gamma_{rR}(C_{n}+e)  &  =f(V(C^{\prime}))+f(V(C^{\prime\prime}))-2\\
&  \geq\gamma_{rR}(C^{\prime})+\gamma_{rR}(C^{\prime\prime})-2=\frac
{2(k-1)}{3}+\frac{2(n-k+3)+3+2}{3}-2\\
&  =\frac{2n+3}{3}>\gamma_{rR}(C_{n}).{\ }%
\end{align*}
{Assume now that }$s_{1}=s_{2}=1.$ {Then, as above, it follows by Proposition
\ref{o3} that }%

\begin{align*}
\gamma_{rR}(C_{n}+e)  &  =f(V(C^{\prime}))+f(V(C^{\prime\prime}))-2\\
&  \geq\gamma_{rR}(C^{\prime})+\gamma_{rR}(C^{\prime\prime})-2=\frac
{2(k-1)+3+1}{3}+\frac{2(n-k+3)+3+1}{3}-2\\
&  =\frac{2n+6}{3}>\gamma_{rR}(C_{n}).{\ }%
\end{align*}
{Thus in either case we obtain a contradiction.}
%$\gamma_{rI}(C_{n}+e)=\gamma_{rI}(C_{n}).$
{Next }suppose there are three consecutive vertices $w_{i},w_{i+1},w_{i+2}$
such that $f(w_{i})+f(w_{i+1})+f(w_{i+2})=1$, say {for} $i=1$. If $f(w_{2}%
)=1$, then $f(w_{1})=f(w_{3})=0$ and {each of }$w_{1}$ {and }$w_{3}$ must be
adjacent a vertex {assigned }2 {as well as to }a vertex {assigned }0. {This
possible only if }$e=w_{1}w_{3}$ and so $H=(C_{n}+e)-w_{2}$ is a cycle on
$n-1$ vertices, {where the restriction of }$f$ to $H$ {is an RRD-function. It
follows that }$\gamma_{rR}(C_{n}+e)=f(V(H))+1\geq\gamma_{rR}(H)+1,$ {and by
Proposition \ref{o3}, we obtain }$\gamma_{rR}(C_{n}+e)\geq\frac
{2(n-1)+3+2}{3}+1>\gamma_{rR}(C_{n})${ which is  a contradiction} {Hence we
can assume that} $f(w_{2})=0$. Without loss of generality, {let }$f(w_{1})=1$
and $f(w_{3})=0$. To dominate $w_{2}$, the edge $e$ must join $w_{2}$ to a
vertex with label 2, say }$w_{k}$.
{Likewise for }$w_{3}$ we must have $f(w_{4})=2$. {Now, consider the cycles
}$C^{\prime}:=(w_{2}w_{3}\ldots w_{k})$ {of order }$k-1$ {and the path
}$P^{\prime}:=w_{k}\ldots w_{n}w_{1}$ {of order }$n-k+2.$ {Let }$k-1\equiv
s_{1}\pmod 3${ and }$n-k+2\equiv s_{2}\pmod 3.${ Notice that }$s_{1}=0$ and
$s_{2}=1;$ $s_{1}=1$ and $s_{2}=3$ or $s_{1}=s_{2}=2.$ {Notice also that the
restrictions of }$f$ on $V(C^{\prime})$ and $V(P^{\prime})$ {are
RRD-functions, and thus }$\gamma_{rR}(C_{n}+e)=f(V(C^{\prime}))+f(V(P^{\prime
}))-2\geq\gamma_{rR}(C^{\prime})+\gamma_{rR}(P^{\prime})-2.$ {Now using
Propositions \ref{o3}-(a,b), {we get a
contradiction as before.} {Finally, let  $f(w_{i})+f(w_{i+1})+f(w_{i+2})\ge 2$ for each $1\le i\le n$ where the sum in  indices is taken modulo $n$. Then we have $$\gamma_{rR}(C_n+e)=\frac{1}{3}\sum_{i=1}^n(f(w_{i})+f(w_{i+1})+f(w_{i+2}))\ge \frac{2n}{3}=\gamma_{rR}(C_n).$$ Thus $\gamma_{rR}(C_n+e)=\gamma_{rR}(G).$ }
%we deduce as before that for each
%situation }$\gamma_{rI}(C_{n}+e)\geq\gamma_{rI}(C_{n}).${ }
Consequently, $r_{rR}(C_{n})=2$.$\hfill\Box$
%\end{proof}
}

\begin{proposition}
\label{completeb}\emph{For integers $1\leq r\leq s$ with $r+s\ge 3$,
\[
r_{rR}(K_{r,s})=\left\{
\begin{array}
[c]{lll}%
1 & \mathrm{if} & r=1,2,3\hfill.\\
r-2 & \mathrm{if} & r\geq4.\hfill
\end{array}
\right.
\]
}
\end{proposition}

%\begin{proof}
\noindent\textbf{ Proof.} Let $X=\{x_{1},x_{2},\ldots,x_{r}\}$ and
$Y=\{y_{1},y_{2},\ldots,y_{s}\}$ be the {partite sets }of $K_{r,s}$. If $r=1$,
then the function $g$ defined by $g(x_{1})=2$, $g(y_{1})=g(y_{2})=0$ and
$g(x)=1$ otherwise, is {an} RRD-function of $K_{1,s}+y_{1}y_{2}$ of weight
$n-1$ and it follows  from Proposition \ref{n}-(c) that $r_{rR}(K_{1,s})=1$. If
$r=2$, then the function $g$ defined by $g(x_{1})=2$ and $g(x)=0$ otherwise,
is {an} RRD-function of $K_{2,s}+x_{1}x_{2}$ of weight $2$ and we get
from Proposition \ref{n}-(b) that $r_{rR}(K_{2,s})=1$. If $p=3$, then the
function $g$ defined by $g(x_{1})=2,g(x_{2})=1$ and $g(x)=0$ otherwise, is {an
}RRD-function of $K_{3,s}+x_{1}x_{3}$ of weight $3$ and by Proposition
\ref{o3}-(c), we have $r_{rR}(K_{3,s})=1$.

Let $r\geq4$. First we observe that the function $g$ defined by $g(x_{1}%
)=2,g(x_{2})=1$ and $g(x)=0$ otherwise, is {an }RRD-function of $K_{r,s}%
+\{x_{1}x_{i}\mid3\leq i\leq r\}$ of weight $3$ and {thus }by Proposition
\ref{o3}-(c), $r_{rR}(K_{r,s})\leq r-2$. We now show that $r_{rR}%
(K_{r,s})\geq r-2$. Let $F$ be {an} $r_{rR}(K_{r,s})$-set. Then $2\leq
\gamma_{rR}(K_{r,s}+F)\leq3$. By
Propositions \ref{n}-(b) and  \ref{o3} we must have $\Delta(K_{r,s}+F)\geq r+s-2$ and this
implies that $|F|\geq r-2$. Thus $r_{rR}%
(K_{r,s})=r-2$ and the proof is complete. $\hfill\Box$
%Hence assume that $\gamma_{rR}(K_{r,s}+F)=3$. {By
%Proposition \ref{33}, }$K_{p,q}+F\in\Psi\cup(\Theta\setminus\Omega)$. If
%$K_{p,q}+F\in\Psi$, then $\Delta(K_{p,q}+F)\geq p+q-2$ {and thus }$|F|\geq
%p-2$. {Hence suppose }that $K_{p,q}+F\in\Theta\setminus\Omega$ and let $x,y,z$
%be the vertices of $H$ (see the construction of graphs in Family $\Theta$). If
%$|X\cap\{x,y,z\}|=3$ (the case $|Y\cap\{x,y,z\}|=3$ is similar), then each
%vertex in $X-\{x,y,z\}$ must be adjacent to at least to two vertices in
%$\{x,y,z\}$ implying that $|F|\geq2(p-3)\geq p-2$. {Also, if }$|X\cap
%\{x,y,z\}|=2$ (the case $|Y\cap\{x,y,z\}|=2$ is similar), then each vertex in
%$X-\{x,y,z\}$ must be adjacent to at least to one vertex in $\{x,y,z\}$
%implying that $|F|\geq p-2$. {In any case, we conclude that }$r_{rI}%
%(K_{3,q})=p-2$ and the proof is complete. $\hfill\Box$
%\end{proof}


\begin{proposition}
Let $K_{n_{1},n_{2},\ldots,n_{p}}$ be the complete $p$-partite graph such that
$p\geq3$ and $3\leq n_{1}\leq n_{2}\leq\ldots\leq n_{p}$. Then $r_{rR}%
(K_{n_{1},n_{2},\ldots,n_{p}})=n_{1}-2$.
\end{proposition}

\noindent\textbf{ Proof.} Let $G=K_{n_{1},n_{2},\ldots,n_{p}}$ and $X_{1}=\{x_{1},\dots,x_{n_{1}}\}$,
$X_{2}=\{y_{1},\dots,y_{n_{2}}\},\dots,X_{p}$ be the {partite sets }of
$G$. Let $F$ be {an} $r_{rR}(G)$-set. Then $2\le \gamma_{rR}(G+F)\le 3$, {and by }Propositions \ref{n}-(b) and  \ref{o3} we must have $\Delta(G+F)\geq n_{1}+\dots+n_{p}-2$ {implying that }$|F|\geq
n_{1}-2$. On the other hand, the function $g$ defined by $f(x_{1})=2, f(x_2)=1$
and $f(x)=0$ otherwise, is {an} RRD-function of $G+\{x_{i}x_{1}\mid3\leq i\leq n_{1}\}$ {yielding }${r}_{rR}(G)\leq|F|=n_{1}-2$. Consequently, $r_{rR}(G)=n_{1}-2$. $\hfill\Box$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Graphs  {with }small restrained Roman reinforcement
number}

In this section, we study graphs with small restrained Roman reinforcement
number. {We begin with the following lemma.}

\begin{lemma}
\label{nn}If $G$ is a connected graph of order $n\geq3$ with $\gamma
_{rR}(G)=n$, then $r_{rR}(G)=1$.
\end{lemma}

%\begin{proof}
\noindent\textbf{Proof.} By Proposition \ref{n}, $G\simeq C_4,C_5,B_p,B_{p,q}$ or
$G$ is a tree with ${\rm diam}(G)\le 5$.  If $G\in\{C_{4},C_{5}\}$, then
the {desired result }follows from Proposition  \ref{cn1}. If $G\in\{B_p,B_{p,q}\}$, then the function $g$ defined by $g(u_1)=2, g(u_2)=g(u_5)=0$ and $g(x)=1$ otherwise, is an RRD-function of $G+u_2u_5$ and hence $r_{rR}(G)=1$. Hence, we assume that $G$ is a tree with diameter at most 5. Let $v_1v_2\dots v_k\;(k\ge 3)$ be a diametral path in $G$. Define the function $f$ by $f(v_1)=f(v_3)=0, f(v_2)=2$ and $f(x)=1$ for the remaining vertices. Clearly, $f$ is an RRD-function of $G+v_1v_3$ and hence $r_{rR}(G)=1$.
$\hfill\Box$
%\end{proof}

\begin{proposition}
\label{aaa}Let $G$ be a connected graph of order $n\geq4$ with $\gamma
_{rI}(G)\geq3.$ {If }$f=(V_{0},V_{1},V_{2})$ {is }a $\gamma_{rR}(G)$-function
with $V_{1}\neq\emptyset$, {then }
\[
r_{rR}(G)=1.
\]
\end{proposition}
%\begin{proof}
\noindent\textbf{Proof.} Let $f=(V_{0},V_{1},V_{2})$ be a $\gamma_{rR}%
(G)$-function with $V_{1}\neq\emptyset$. If $\gamma_{rR}(G)=n$, then the desired result comes from Lemma \ref{nn}. Hence assume that $\gamma_{rR}(G)<n$. Then $V_0\neq\emptyset$ and so $V_2\neq \emptyset$. Since $G$ is a connected graph and $V_1\neq\emptyset$, there exists a vertex $w\in V_1$ such that $w$ is dominated by $V_0\cup V_2$.  If $w$ is adjacent to a vertex in $V_2$, then the function $g$ defined by $g(w)=0$ and $g(x)=f(x)$ otherwise, is an RRD-function of $G+wz$ where $z\in V_0$, of weight less than $\gamma_{rR}(G)$ and so $r_{rR}(G)= 1$. If $w$ is adjacent to a vertex in $V_0$, then the function $g$ defined by $g(w)=0$ and $g(x)=f(x)$ otherwise, is an RRD-function of $G+wu$ where $u\in V_2$, of weight less than $\gamma_{rR}(G)$ and so $r_{rR}(G)=1$. This completes the proof. $\hfill\Box$


\begin{proposition}
Let $G$ be a connected graph of order $n$ with $\gamma_{rR}(G)\geq3$.
Then $r_{rR}(G)=1$ if and only if $\gamma_{rR}(G)=n$ or $G$ {has a function
}$f=(V_{0},V_{1},V_{2})$ of weight less than $\gamma_{rR}(G)$ such that one of
the following conditions holds.

\begin{itemize}
\item[(i)] $G[V_{0}]$ has at most two isolated vertices and $V_2$ dominates $V_{0}$.

\item[(ii)] $G[V_{0}]$ has no isolated vertices and there is exactly one
vertex $v\in V_{0}$ which is not dominated by $V_2$.
\end{itemize}
\end{proposition}

%\begin{proof}
\noindent\textbf{Proof.} If $\gamma_{rR}(G)=n$, then by Lemma \ref{nn} we have
$r_{rR}(G)=1$. Hence suppose that $\gamma_{rR}(G)<n$, and let $f=(V_{0}%
,V_{1},V_{2})$ be a function on $G$ with weight less than $\gamma_{rR}(G)$
satisfying (i) or (ii). Since $\omega(f)\leq n-2$, we have $|V_{0}|\geq2$. If
$|V_{2}|\geq1$, then let $u\in V_{2}$. {Now, if }(ii) holds, then $V_{2}\neq\emptyset$ and $f$ is {an }RRD-function of $G+uv$. {Assume now that }(i) holds. If $G[V_{0}]$
has two isolated vertices $w,v$, then $f$ is {an }RRD-function of $G+\{wv\}$
and if $G[V_{0}]$ has exactly one isolated vertex, say $w$, then $f$ is {an
}RRD-function of $G+\{wz\}$, where $z$ {is any vertex in }$V_{0}-\{w\}$. Hence
{in either case }$r_{rR}(G)=1$.

Conversely, let $r_{rR}(G)=1$ and suppose that $\{uv\}$ {is an }$r_{rR}%
(G)$-set. If $\gamma_{rR}(G)=n$, then we are done. Hence suppose that
$\gamma_{rR}(G)\leq n-1$ and let $f$ be a $\gamma_{rR}(G+\{uv\})$-function.
{Notice that vertices }$u$ and $v$ {cannot be assigned both positive value
under }$f$ {(otherwise }$f$ {is an RRD-function of }$G$). Without loss of
generality, assume that $f(u)=0$. {If }$f(v)=0$, then $f$ satisfies (i). Hence
assume that $f(v)\geq1$. If $u$ is adjacent to a vertex with label 2 other than $v$, then $f$ {is an
RRD-function of }$G$. Hence $u$ is not dominated by $V_2$ in $G$ and so $f$ satisfies (ii).  This completes the
proof.$\hfill\Box$
%\end{proof}

%
%\begin{proposition}
%\label{v2}Let $G$ be a connected graph of order $n$ with $\gamma
%_{rR}(G)\geq3$ {having }a $\gamma_{rR}(G)$-function $f=(V_{0},V_{1},V_{2})$
%such that $|V_{1}|=1$. Then $r_{rR}(G)\leq1$.
%\end{proposition}
%
%%\begin{proof}
%\noindent\textbf{Proof.} If $\gamma_{rR}(G)=n$, then the desired result comes from Lemma \ref{nn}. Thus suppose that $\gamma_{rR}(G)<n.$ {Then }$V_{0}%
%\neq\emptyset.$ {Let }$f$ be a $\gamma_{rR}(G)$-function such that $|V_{1}%
%|=1$. Let $V_{1}^{f}=\{v\}$ and let $N(v)=\{v_{1},\dots,v_{t}\}$. Notice that
%because of $|V_{1}|=1,$ for each $i$, {either }$f(v_{i})=2$ or $v_{i}$ has a
%neighbor {assigned }2. If $f(v_{i})=2$ for some $i$, then {no edge join }$v$
%to $V_{0}$ {and thus for any } $w\in V_{0}$ the function $g$ defined on
%$G+\{vw\}$ by $g(v)=0$ and $g(x)=f(x)$ otherwise, is {an }RRD-function of
%$G+\{wv\}$ of weight $\omega(f)-1$. {Hence we can assume that no }$v_{i}$ {is
%assigned 2. In that case, for any }$w\in V_{2}$ the function $g$ defined on
%$G+\{vw\}$ by $g(v)=0$ and $g(x)=f(x)$ otherwise, is {an }RRD-function of
%$G+\{wv\}$ of weight $\omega(f)-1$. Therefore, {in either case} we have
%$r_{rR}(G)=1$. $\hfill\Box$
%%\end{proof}


\begin{proposition}
Let $G$ be a connected graph of order $n$ with $\gamma_{rR}(G)\ge3$. If
$\delta(G)=1$, then $r_{rR}(G)\le2$.
\end{proposition}

%\begin{proof}
\noindent\textbf{Proof.} {If }$\gamma_{rR}(G)=n$, then the result comes from Lemma \ref{nn}. Hence we assume that $\gamma_{rR}(G)<n$. Let $f=(V_{0},V_{1},V_{2})$ be a $\gamma_{rR}(G)$-function. {We have  }%
$V_{0}\neq\emptyset$ because $\gamma_{rR}(G)<n$. Thus $V_2\neq\emptyset$. Let $u$ be a support vertex
of $G$ and $u_{1}$ be the leaf neighbor of $u$. By definition we
have $f(u_{1})\geq1$. If $f(u_1)=1$ or $f(u)=1$, then the desired result comes from Proposition \ref{aaa}. Assume that  $f(u_1)=2$ and $f(u)\neq 1$. This implies that $f(u)=0$. Let $w$ be a neighbor of $u$ with label 0. To Roman dominate $w$, there is a vertex $v\neq u_1$ such that $f(v)=2$.  Then the function $g$ defined on $G+\{u_1v,uv\}$ by $g(u_1)=1$ and $g(x)=f(x)$ otherwise, is {an} RRD-function of $G+\{uv\}$
with weight $\omega(f)-1$. {Consequently, }$r_{rR}(G)\leq 2$.$\hfill\Box$
%\end{proof}

%
%\begin{proposition}
%Let $G$ be a connected graph of order $n\geq3$ with $\gamma_{rI}(G)\ge3$. If
%$G$ has at least two leaves, then
%\[
%r_{rI}(G)\le2.
%\]
%This bound is sharp for a double star $DS_{p,q}\;(q\ge p\ge3)$.
%\end{proposition}
%
%%\begin{proof}
%\noindent\textbf{Proof.} If $\gamma_{rI}(G)=n$, then the result {follows from
%}Lemma \ref{nn}. Hence we assume that $\gamma_{rI}(G)<n$, {and let }%
%$f=(V_{0},V_{1},V_{2})$ be a $\gamma_{rI}(G)$-function. {Clearly, }$V_{0}%
%\neq\emptyset$. {Let }$w\in V_{0}$, {and let }$u_{1}$ and $v_{1}$ be two
%leaves of $G$ {whose support vertices are }$u$ and $v,$ respectively (possibly
%$u=v$). By definition $f(u_{1}),f(v_{1})\geq1$. If $f(u)\geq1$ (the case
%$f(v)\geq1$ is similar), then the function $g$ defined on $G+\{u_{1}%
%v_{1},u_{1}w\}$ by $g(u_{1})=0$ and $g(x)=f(x)$ otherwise, is {an
%}RID-function of $G+\{u_{1}v_{1},u_{1}w\}$ of weight $\omega(f)-1$. Hence,
%assume that $f(u)=f(v)=0$. Since $G$ is not a star, we have $(V_{1}\cup
%V_{2})-\{u_{1},v_{1}\}\neq\emptyset$. Let $z\in(V_{1}\cup V_{2})-\{u_{1}%
%,v_{1}\}$. Define the function $g$ on $G+\{u_{1}z,v_{1}z\}$ by $g(z)=\min
%\{2,f(z)+1\}=2$, $g(u_{1})=g(v_{1})=0$ and $g(x)=f(x)$ otherwise. Clearly, $g$
%is {an }RID-function of $G+\{u_{1}z,v_{1}z\}$ with weight $\omega(f)-1$.
%{Therefore, in either case we have} $r_{rI}(G)\leq2$. $\hfill\Box$

\begin{corollary}
For any tree $T$ of order $n\geq3$, $r_{rR}(T)\leq2.$
\end{corollary}

{We know that }$r_{rR}(P_{2})=0$ and $r_{rR}(T)=1$ {if }$T$ {is a  tree of diameter at most five}. {It
would be interesting to characterize all trees }$T$ such that
$r_{rR}(T)=1.$

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\section{Bounds on $r_{rR}(G)$}
In this section, we present  some sharp upper bounds for $r_{rR}(G)$.
Given a set $S \subseteq V $ of vertices in a graph $G$ and a vertex $v \in S$,  the external private neighborhood of $v$ with respect to $S$ in the set $epn(v,S)=\{u\in V-S\mid N(u)\cap S=\{v\}\}$.

\begin{proposition}
\label{aaaaa}Let $G$ be a connected graph of order $n\geq4$ with $\gamma
_{rI}(G)\geq3.$ {If }$f=(V_{0},V_{1},V_{2})$ {is }a $\gamma_{rR}(G)$-function
with $V_{2}\neq\emptyset$, {then }
\[
r_{rR}(G)\leq\min\{|epn(v,V_2)\cap V_0|\;:\; v\in V_{2}\}.
\]
\end{proposition}

%\begin{proof}
\noindent\textbf{Proof.} Let $f=(V_{0},V_{1},V_{2})$ be a $\gamma_{rR}%
(G)$-function with $V_{2}\neq\emptyset$. {If }$|epn(v,V_2)\cap V_0|=0$ {for some vertex
}$v\in V_{2},$ {then reassigning }$v$ {the value 1 instead of 2 provides an
RRD-function of weight less than }$\gamma_{rR}(G)$ {leading to a
contradiction.} Hence $|epn(v,V_2)\cap V_0|\geq1$ {for every }$v\in V_{2}.$ Let $u$ be
a vertex in $V_{2}$ such that $|epn(u,V_2)\cap V_0|=\min\{|epn(v,V_2)\cap V_0|\;:\; v\in V_{2}\}$
and let $epn(v,V_2)\cap V_0=\{u_{1},\dots,u_{\epsilon}\}$. If $|V_2|\ge 2$ and $w\in V_2-\{u\}$, then the function $g$ defined by $g(u)=1$ and $g(x)=f(x)$ otherwise, is {an} RRD-function of $G+\{wx\mid x\in epn(v,V_2)\cap V_0\}$ of weight less than $\gamma_{rR}(G)$ and so $r_{rR}(G)\leq\min\{|epn(v,V_2)\cap V_0|\;:\; v\in V_{2}\}$.
Hence assume that $V_2=\{u\}$. Then $u$ dominates all vertices in $V_0$. Since $\gamma
_{rR}(G)\geq3$, we have $V_1\neq\emptyset$ and the desired result comes from Proposition \ref{aaa}.
 $\hfill\Box$
%\end{proof}


\bigskip

{We observe that }for any $\gamma_{rR}(G)$-function $f=(V_{0},V_{1},V_{2})$,
every vertex $u$ of $V_{2}$ {can have }at most $d_{G}(u)$ neighbors in $V_{0}%
$. {Whence the following corollary. }

\begin{corollary}
\label{V22}Let $G$ be a connected graph with $\gamma_{rI}(G)\ge3$ and
$f=(V_{0},V_{1},V_{2})$ a $\gamma_{rI}(G)$-function {with
$|V_{2}|\ge1$}. Then $r_{rR}(G)\leq\Delta.$
\end{corollary}

%The bounds of Proposition \ref{V2} and Corollary \ref{V22} are sharp for the
%double star $DS_{r,r}$ with $r\ge2$.


\begin{corollary}
Let $G$ be a connected graph with $\gamma_{rR}(G)\geq3$ containing a path
$v_{1}v_{2}v_{3}v_{4}v_{5}$ in which $d_{G}(v_{i})=2$ for $i\in\{2,3,4\}$.
Then $r_{rR}(G)\leq2$.
\end{corollary}

%\begin{proof}
\noindent\textbf{Proof.} If $\gamma_{rR}(G)=n$, then the result is immediate
{from }Lemma \ref{nn}. Hence we assume that $\gamma_{rR}(G)<n$, {and let
}$f=(V_{0},V_{1},V_{2})$ be a $\gamma_{rR}(G)$-function. By Proposition \ref{aaa}, we may assume that $V_1=\emptyset$. Then we must have  $2\in
\{f(v_{2}),f(v_{3}),f(v_{4})\}$ and  {the result }follows from Proposition
\ref{aaaaa}. $\hfill\Box$
%\end{proof}


%\begin{proposition}
%\label{V1} Let $G$ be a connected graph of order $n\geq4$ with $\gamma
%_{rI}(G)\geq3$ and let $f=(V_{0},V_{1},V_{2})$ a $\gamma_{rI}(G)$-function
%with $V_{1}\neq\emptyset$. Then
%\[
%r_{rI}(G)\leq2+\min\{\epsilon(v)\mid v\in V_{1}\}.
%\]
%
%\end{proposition}
%
%%\begin{proof}
%\noindent\textbf{Proof.} {Let }$f=(V_{0},V_{1},V_{2})$ be a $\gamma_{rI}%
%(G)$-function with $V_{1}\neq\emptyset$. If $\gamma_{rI}(G)=n$, then the
%result {follows from} Lemma \ref{nn}. Hence we assume that $\gamma_{rI}(G)<n$.
%{Clearly, }$V_{0}\neq\emptyset$. {Let }$v\in V_{0}$, {and let }$u$ be a vertex
%in $V_{1}$ such that $\epsilon(u)=\min\{\epsilon(w)\mid w\in V_{1}\}.$ {Let
%}$N_{\epsilon}(u)=\{u_{1},\dots,u_{\epsilon}\}$ when $\epsilon(u)\geq1$ {and
%let }$V_{1}-\{u\}=\{z_{1},\dots,z_{k}\}$.
%
%First {assume that }$N(u)\cap V_{0}=\emptyset$. If $V_{2}\neq\emptyset$ and
%$w\in V_{2}$, then {consider }the function $g$ defined by $g(u)=0$ and
%$g(x)=f(x)$ otherwise. {One can check that }$g$ {has weight }$\omega(f)-1$
%{and it is an RID-function on the graph }$G^{\prime}$ {obtained from }$G$ {by
%adding the edge }$vu$ {and possibly the edge }$uw$ {if }$uw\notin E(G).$ Hence
%we can assume that $V_{2}=\emptyset$. {Now, if }$\sum_{x\in N(u)}f(x)\geq2$,
%then the function $g$ {defined on }$G+\{vu\}$ by $g(u)=0$ and $g(x)=f(x)$
%otherwise, is {an } RID-function of weight $\omega(f)-1$. {Moreover, if }%
%$\sum_{x\in N(u)}f(x)=0$, then the function $g$ {defined }on $G+\{z_{1}%
%u,z_{2}u\}$ by $g(u)=0$ and $g(x)=f(x)$ otherwise, is {an} RID-function of
%weight $\omega(f)-1$. {Finally, if }$\sum_{x\in N(u)}f(x)=1$, {then s}ince
%$\gamma_{rI}(G)\geq3$, there is a vertex $z\in V-N[u]$ with $f(z)=1$. {In this
%case, the function }$g$ {defined }on $G+\{vu,zu\}$ by $g(u)=0$ and $g(x)=f(x)$
%otherwise, {is an }RID-function of weight $\omega(f)-1$. {All of the above
%situations lead to }$r_{rI}(G)\leq2,$ {and the result is valid. }
%
%{For the sequel, we can assume that }$N(u)\cap V_{0}\neq\emptyset$. If
%$V_{2}\neq\emptyset$ and $w\in V_{2}$, then the function $g$ {defined }on
%$G_{1}=G+\{wu_{1},\dots,wu_{\epsilon},wu\}$ by $g(u)=0$ and $g(x)=f(x)$
%otherwise, is {an} RID-function of $G_{1}$ of weight $\omega(f)-1$. Hence
%assume that $V_{2}=\emptyset$. Since $\gamma_{rI}(G)\geq3$, we have
%$|V_{1}|\geq3$ and {thus }for any vertex $u_{i}$, there exists a vertex
%$u_{i}^{\prime}\in V_{1}$ such that $u_{i}u_{i}^{\prime}\not \in E(G)$. {Now,
%if }$\sum_{x\in N(u)}f(x)\geq2$, then the function $g$ on $G_{2}%
%=G+\{u_{i}u_{i}^{\prime}\mid1\leq i\leq\epsilon(u)\}$ defined by $g(u)=0$ and
%$g(x)=f(x)$ otherwise, is {an} RID-function of $G_{2}$ of weight $\omega
%(f)-1$. If $\sum_{x\in N(u)}f(x)=0$, then the function $g$ on $G_{3}%
%=G+\{z_{1}u,z_{2}u,u_{i}u_{i}^{\prime}\mid1\leq i\leq\epsilon(u)\}$ by
%$g(u)=0$ and $g(x)=f(x)$ otherwise, is {an} RID-function of $G_{3}$ of weight
%$\omega(f)-1$. {Finally, assume that }$\sum_{x\in N(u)}f(x)=1$. {Without loss
%of generality, assume that }$z_{2}\not \in N[u]$ {and let }$F=\{z_{2}%
%u,u_{i}u_{i}^{\prime}\mid1\leq i\leq\epsilon(u)\}.$ {By defining the function
%}$g$ on $G+F$ as before, {we can see that }$g$ is {an }RID-function of $G+F$
%of weight $\omega(f)-1$. {Consequently, in either case }we get $r_{rI}%
%(G)\leq\epsilon(u)+2=2+\min\{\epsilon(v)\mid v\in V_{1}\}$. This completes the
%proof. $\hfill\Box$
%%\end{proof}
 \begin{figure}[h]
\centering
\begin{tikzpicture}%[scale=.3, transform shape]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%(1)
\node [draw, shape=circle,fill=black,scale=0.3] (u) at (10.5,4.5){};
%\node [draw, shape=circle,fill=black,scale=0.3] (u0) at (12.5,4.5){};
\node [draw, shape=circle,fill=white,scale=0.3] (u1) at (7,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (u2) at (8,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (v1) at (9,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (v2) at (10,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (u11) at (11,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (u21) at (12,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (v11) at (13,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (v21) at (14,2.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1u1) at (7,1.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1u2) at (8,1.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1v1) at (9,1.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1v2) at (10,1.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1u11) at (11,1.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1u21) at (12,1.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1v11) at (13,1.7){};
\node [draw, shape=circle,fill=white,scale=0.3] (1v21) at (14,1.7){};
%\node [draw, shape=circle,fill=white,scale=0.3] (w1) at (8.5,1.7){};
%\node [draw, shape=circle,fill=white,scale=0.3] (w2) at (8.5,.8){};
\node [draw, shape=circle,fill=black,scale=0.3] (v) at (10.5,0.2){};
%\node [draw, shape=circle,fill=black,scale=0.3] (w) at (12.5,0.2){};
%\node [draw, shape=circle,fill=white,scale=0.3] (y5) at (10,1){};
%\node [draw, shape=circle,fill=white,scale=0.3] (y6) at (11,1){};
%\node [draw, shape=circle,fill=black,scale=0.3] (v1) at (3.2,-5.6){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v2) at (4.1,-5.6){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v3) at (4.1,-6.4){};
%\node [draw, shape=circle,fill=black,scale=0.3] (v4) at (3.2,-6.4){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v5) at (2.3,-6.4){};
%\node [draw, shape=circle,fill=white,scale=0.3] (v6) at (2.3,-5.6){};
%\draw  [line width=0.8pt](x1)--(y1)--(x2)--(y4)--(x3)--(y6)--(x4)--(y5)--(y6);
%\draw  [line width=0.8pt](u)--(u1)--(v)--(u2)--(u);
%\draw  [line width=0.8pt](w)--(v11)--(v21)--(w)--(u11)--(u21)--(w);
%\draw  [line width=0.8pt](u0)--(v1)--(v2)--(u0)--(u1)--(u2)--(u0)--(u11)--(u21)--(u)--(v11)--(v21)--(u0);
\draw  [line width=0.8pt](u)--(v1)--(v2)--(u)--(u1)--(u2)--(u)--(u11)--(u21)--(u)--(v11)--(v21)--(u);
\draw  [line width=0.8pt](v)--(1v1)--(1v2)--(v)--(1u1)--(1u2)--(v)--(1u11)--(1u21)--(v)--(1v11)--(1v21)--(v);
%\draw  [line width=0.8pt](w)--(1v1)--(1v2)--(w)--(1u1)--(1u2)--(w)--(1u11)--(1u21)--(w)--(1v11)--(1v21)--(w);
\draw  [line width=0.8pt](u2)--(1u2);
\draw  [line width=0.8pt](v2)--(1v2);
\draw  [line width=0.8pt](u1)--(1u1);
\draw  [line width=0.8pt](v1)--(1v1);
\draw  [line width=0.8pt](u21)--(1u21);
\draw  [line width=0.8pt](v21)--(1v21);
\draw  [line width=0.8pt](u11)--(1u11);
\draw  [line width=0.8pt](v11)--(1v11);
%\draw  [line width=0.8pt](u0)--(u11)--(u21)--(u0)--(v11)--(v21)--(u0);
%\draw  [line width=0.8pt](v1)--(v2);
%\draw  [line width=0.8pt](w1)--(w2);
%\node at (7,3) {$u_1$};
%\node at (7.8,3) {$u_2$};
%\node at (9,2.9) {$v_1$};
%\node at (9.8,2.9) {$v_2$};
\node at (10.5,4.7) {$x$};
%\node at (12.5,4.7) {$y$};
\node at (10.5,-.1) {$y$};
%\node at (12.5,0) {$w$};
%\node at (8.5,.5) {$w_1$};
%\node at (8.5,2) {$w_2$};
%\node at (11,.7) {$y_6$};
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{tikzpicture}
\caption{A graph $G$ of order 18 and $r_{rR}(G)=4$ }%
\label{FAZ23333}%
\end{figure}


Using Propositions \ref{aaa} and \ref{aaaaa}  we {obtain }the next result.

\begin{theorem}
\label{n/2}For any graph $G$ of order $n\geq3$, we
have \ {%
\[
r_{rR}(G)\leq\max\{1,\frac{2n-\gamma_{rR}(G)}{\gamma_{rR}(G)}\}.
\]
Moreover, the bound is sharp.}
\end{theorem}

%\begin{proof}
\noindent\textbf{Proof.}
{If }$\gamma_{rR}(G)=2$, then $r_{rR}(G)=0$ {and
the result is true. }If $\gamma_{rR}(G)=n$, then by Lemma \ref{nn},
$r_{rR}(G)=1$ {and the desired }result {follows. }Hence we assume that
$3\leq\gamma_{rR}(G)<n$, {and let }$f=(V_{0},V_{1},V_{2})$ be a $\gamma
_{rR}(G)$-function. If $V_1\neq \emptyset$, then the result comes from Proposition \ref{aaa}.
Hence we assume that $V_1=\emptyset$. Then $\frac{\gamma_{rR}(G)}{2}=|V_{2}|\geq2$ and clearly  $|epn(u,V_2)\cap V_0|\leq
\frac{2n-\gamma_{rR}(G)}{\gamma_{rR}(G)}$ for some $u\in V_{2}$. Now, the result is immediate by
Proposition \ref{aaa}.


To show the sharpness, consider the graph $G$ illustrated in Figure
\ref{FAZ23333}.
It is easy to see that $\gamma_{rR}(G)=4$ and the function $f$ on $G$ defined
by $f(x)=f(y)=2$ and $f(z)=0$ otherwise, is {the }unique 
$\gamma_{rR}(G)$-function. Then $r_{rR}(G)\leq  \frac{2n-\gamma_{rR}(G)}{\gamma_{rR}(G)}=8$. Now let $F$ be
{an} $r_{rR}(G)$-set. Then $\gamma_{rR}(G+F)\le 3$ and so $\Delta(G+F)\ge n-2$ (see Propositions \ref{n}-(ii) and \ref{o3}). This implies that $|F|\ge 8$.
Consequently, $r_{rR}(G)=8=\frac{2n-\gamma_{rR}(G)}{\gamma_{rR}(G)}$. $\hfill\Box$


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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\end{document} 