Complexity and Exact Values for [k]-Roman and Strong Roman Domination for Specific Graph Families

Abstract

Motivated by the original idea of defending the Roman Empire, all these domination concepts can be interpreted as vertex-labeling schemes that model the allocation of resources to protect a graph against attacks. A Roman dominating function (RDF) is a labeling of the vertices of a graph with labels in $\{0,1,2\}$ such that every vertex labeled 0 is adjacent to at least one vertex labeled 2. The weight of an RDF is the sum of all vertex labels. Vertices labeled 2 are intended to protect their neighbors labeled 0. The Roman domination number is the minimum weight of an RDF on the graph. In 2017, Álvarez et al. introduced strong Roman domination as a variant of Roman domination designed to protect the vertices of a graph against multiple simultaneous attacks. In 2021, Ahangar et al. defined $\left[k\right]$-Roman domination, another model intended to defend a graph against individual attacks on vertices. In this paper, we investigate the computational complexity of the associated decision problems for $\left[k\right]$-Roman domination and strong Roman domination. Furthermore, we determine exact values of these parameters for several graph families under both variants.

1. Introduction

In this study, we present new results on two variants of Roman domination: $\left[k\right]$-Roman domination and strong Roman domination.

In 1999, Stewart described the defensive strategy decreed by Emperor Constantine I “The Great” to defend the Roman Empire [1]. Subsequently, Cockayne et al. formalized the concept of Roman domination based on this defensive strategy [2].

A graph is denoted by $G=\left(V\right(G),E(G\left)\right)$, where $V\left(G\right)$ is the vertex set and $E\left(G\right)$ is the edge set. Two edges are said to be independent if they do not share any endpoint, that is, no vertex is incident to both edges. We consider only simple, undirected, and finite graphs $G=(V,E)$ without loops or multiedges. A Roman dominating function (RDF) on a finite graph $G=(V,E)$ is a function that assigns to each vertex one of the labels $\{0,1,2\}$, such that every vertex labeled 0 is adjacent to at least one vertex labeled 2. Vertices labeled 2 are intended to protect their neighbors labeled 0. The Roman domination number of the graph, ${\gamma }_R\left(G\right)$, is the minimum weight (the sum of all vertex labels) of an RDF. An RDF f of minimum weight is called a ${\gamma }_R\left(G\right)$-function. For a subset $A\subseteq V\left(G\right)$, we denote by $f\left(A\right)$ the sum ${\sum }_{v\in A}f\left(v\right)$.

Since then, various studies have emerged that attempt to improve the effectiveness of this model. These studies propose different strategies based on increasing the number of units, edges, etc., under single attacks [2,3,4] or simultaneous attacks [5,6,7].

We denote by $\left[m\right]=\{1,2,\dots ,m\}$ the set of consecutive integers starting from 1. The open neighborhood (resp. closed neighborhood) of a vertex u in a graph, $N\left(u\right)$ (resp. $N\left[u\right]$), is the set of adjacent vertices to u (resp. the set of adjacent vertices to $u,$ including the vertex itself). That is, $N\left[u\right]=N\left(u\right)\cup \left\{u\right\}$. We denote by $\delta \left(G\right)$ and $\Delta \left(G\right)$ the minimum and maximum degree of G, respectively.

Given a labeling f of the vertices of a graph $G,$ the active neighborhood of u, denoted by $AN\left(u\right)$, is the set $\{w\in N(u):f(w)>0\}$. Similarly, $AN\left[u\right]=AN\left(u\right)\cup \left\{u\right\}.$

A dominating set of vertices in a graph is a subset $D\subseteq V$ such that every vertex in $V\setminus D$ has a neighbor in D. The domination number, $\gamma \left(G\right)$, of a graph is the minimum cardinality of a dominating set in G. A vertex with degree equal to 1 is called a leaf, and a vertex that is adjacent to exactly one leaf (resp. to at least two leaves) is called a weak support vertex (resp. strong support vertex). A path of length n is a sequence of distinct vertices $P_n:u_0{u}_1\dots u_n$ such that $u_i{u}_{i+1}\in E\left(P_n\right)$ for all $i\in \{0,\dots ,n-1\}$. A cycle on n vertices, denoted $C_n$, is a sequence of distinct vertices $v_1,v_2,\dots ,v_n$ together with edges $v_i{v}_{i+1},$ for $i=1,\dots ,n-1$ and $v_n{v}_1,$ forming a closed loop. A tree is a connected and acyclic graph. A tree with exactly two non-leaf vertices is called a bi-star or a double star, $S_{s,t}$, where the non-leaf vertices are adjacent to s and t leaves, respectively, with $1\le s\le t.$

A caterpillar is a tree T such that the subgraph obtained by deleting all leaves of T is a path; this path is called the spine of the caterpillar. A crown graph $C\left(n\right)$ is the graph obtained from a complete bipartite graph $K_{n,n}$ by removing a complete matching, i.e., a set of n pairwise independent edges. Let H be a graph with a distinguished vertex $r\in V\left(H\right)$ (called the root). The rooted product of a graph G and the rooted graph $(H,r)$, denoted by $G{\circ }_{r}H$, is obtained by taking $\left|V\right(G\left)\right|$ vertex-disjoint copies of H and identifying the root r of the i-th copy with the i-th vertex of G. For integers $n\ge 3$ and $t\ge 1$, the t-fold wheel $W_{n,t}$ is the graph obtained from a cycle $C_n$ (the rim) by adding t new vertices (the hubs) which are pairwise nonadjacent, and making each hub adjacent to every rim vertex. A bipartite graph $G=(A\cup B,E)$ is star-convex if there exists a star T with vertex set A such that, for every $b\in B$, the neighborhood $N\left(b\right)$ induces a connected subtree of T.

One of the strategies used to defend a graph against an individual attack on a vertex is to increase the number of protection units, without leaving strong vertices undefended. Consequently, a series of variations appeared, such as double, triple, or quadruple Roman domination [3,4,8]. To generalize this defensive strategy, Ahangar et al. [3] defined $\left[k\right]$-Roman domination in 2021. A function $f:V\to \{0,1,\dots ,k+1\}$ is said to be a $\left[k\right]$-Roman dominating function, and it is denoted by $\left[k\right]$-RDF, if for every vertex u of the graph we have $f\left(N\right[u\left]\right)\ge k+\left|AN\right(u\left)\right|.$ The minimum weight of a $\left[k\right]$-RDF on a graph G is known as the $\left[k\right]$-Roman domination number ($\left[k\right]$-RDN) of G, and it is denoted by ${\gamma }_{\left[kR\right]}\left(G\right)$. We say that a $\left[k\right]$-RDF with weight equal to ${\gamma }_{\left[kR\right]}\left(G\right)$ is a ${\gamma }_{\left[kR\right]}\left(G\right)$-RDF. Any labeling f of the vertices of a graph may be expressed as an ordered vertex partition $f=(V_0,V_1,\dots ,V_{k+1})$ where $V_j=\{v\in V:f\left(v\right)=j\}.$

Figure 1 shows an RDF and a $\left[k\right]$-RDF on a graph.

A natural question arises: How many units are sufficient to defend a graph against multiple simultaneous attacks? Several authors have investigated this problem [6,9]. To address this problem, in 2017, Alvarez et al. [5] introduced strong Roman domination. A strong Roman dominating function f (for short, StRDF) assigns labels from the set $\{0,1,2,\dots ,⌈{\displaystyle \frac{\Delta }{2}}⌉+1\}$ to the vertices of the graph in such a way that for every vertex v with $f\left(v\right)=0$, there exists at least one neighbor u such that $f\left(u\right)\ge 1+⌈{\displaystyle \frac{1}{2}}|N\left(u\right)\cap B_0|⌉,$ where $B_0=\{w\in V:f\left(w\right)=0\}.$ In other words, under simultaneous attacks, every strong vertex could defend, at least, one-half of its undefended neighbors without leaving itself unprotected. We also denote by $B_1=\{w\in V:f\left(w\right)=1\}$ and by $B_2=V\setminus \left(B_0\cup B_1\right).$ The strong Roman domination number (for short, StRDN) of G is the minimum weight of an StRDF on G and is denoted by ${\gamma }_{StR}\left(G\right).$

It can be observed, see Figure 2, that with two more units compared to Roman domination, at least half of the undefended vertices can be protected against multiple simultaneous attacks.

Let f be a Roman-type dominating function on G and let $u,v\in V\left(G\right)$ with $uv\in E\left(G\right)$.

• We say that u is Roman dominated by v if $f\left(v\right)=2$.
• We say that u is strongly Roman dominated by v if $f\left(v\right)\ge 1+⌈{\displaystyle \frac{1}{2}}|N\left(v\right)\cap B_0|⌉,$ where $B_0=\{w\in V:f\left(w\right)=0\}.$
• For $k\ge 1$, we say that u is $\left[k\right]$-Roman dominated by v whenever $f\left(N\right[v\left]\right)\ge k+\left|AN\right(v\left)\right|$.

Next, we recall some results used in the proofs. Let p be a positive integer. A graph G is called a p-graph if it is equipped with a labeling function $\mathcal{\ell }:V\left(G\right)\to \{1,2,\dots ,p\}$ assigning to each vertex a label from a set of size p.

The clique-width of a graph G, denoted by $cw\left(G\right)$, is the minimum number of labels needed to construct G using the following operations: (i) creation of a new vertex with a given label, (ii) disjoint union of two labeled graphs, (iii) joining by edges all the vertices with label i to all vertices with label j, and (iv) relabeling all vertices with label i to label j.

An r-expression is an algebraic expression that constructs a graph using at most r labels and the operations defining clique-width. A graph has clique-width at most r if and only if it can be generated by an r-expression.

Proposition 1

([5]). Given any graph G, ${\gamma }_{\mathit{R}}\left(G\right)\le {\gamma }_{StR}\left(G\right)\le \left(1+⌈{\displaystyle \frac{\Delta }{2}}⌉\right)\gamma \left(G\right).$

Proposition 2

([5]). Let G be a graph of order $n,$ then, ${\gamma }_{StR}\left(G\right)\le n-⌊{\displaystyle \frac{\Delta \left(G\right)}{2}}⌋.$

Proposition 3

([5]). Let G be a graph of order $n,$ then, ${\gamma }_{StR}\left(G\right)\ge ⌈{\displaystyle \frac{n+1}{2}}⌉.$ Furthermore, if n is odd, then equality holds if and only if $\Delta \left(G\right)=n-1$.

Corollary 1

([5]). For $n\ge 1$, ${\gamma }_{StR}\left(K_{1,n}\right)=⌈{\displaystyle \frac{n+1}{2}}⌉.$

The corona product $G\circ H$ is obtained by taking one copy of G and $\left|V\right(G\left)\right|$ disjoint copies of H, connecting each vertex $v_i\in V\left(G\right)$ to all vertices of the i-th copy $H_i$, so that $V(G\circ H)=V\left(G\right)\cup {\bigcup }_{i}V\left(H_i\right)$ and $E(G\circ H)=E\left(G\right)\cup {\bigcup }_{i}E\left(H_i\right)\cup {\bigcup }_i\{v_{i}u:u\in V\left(H_i\right)\}$.

Proposition 4

([10]). For $m\ge 3$, ${\gamma }_{StR}(C_m\circ K_1)\le ⌈{\displaystyle \frac{5n}{6}}⌉$, where $n=2m.$

Theorem 1

([11]). Let G be a connected graph of order n with $\Delta \left(G\right)\le 4.$ Then, ${\gamma }_{StR}\left(G\right)\ge ⌈{\displaystyle \frac{3n}{5}}⌉.$

Remark 1

([11]). Let G be a connected graph of order n with $\Delta \left(G\right)\le 4$ and let $f=(B_0,B_1,B_2)$ be a ${\gamma }_{St\mathit{R}}\left(G\right)$-RDF. Let $B_2^i=\{v\in B_2:|N\left(v\right)\cap B_0|=i\}$, $i=1,2,3,4$, $B_0^1=\{v\in B_0:|N\left(v\right)\cap B_2|=1\}$ and $B_0^2=\{v\in B_0:|N\left(v\right)\cap B_2|\ge 2\}.$ Clearly, $(B_2^1,B_2^2,B_2^3,B_2^4)$ is a partition of $B_2$ and $(B_0^1,B_0^2)$ is a partition of $B_0$. From this, it follows that $n=|B_0|+|B_1|+|B_2|$, $|B_2|={\sum }_{i=1}^4\left|B_2^i\right|$, and $|B_0|=|B_0^1|+|B_0^2|.$ Among all the possible ${\gamma }_{StR}$-functions f of a graph, we choose one that maximizes $|B_1|.$ Consequently, it is satisfied that $B_2^1=\varnothing ,$$B_2^3=\varnothing$ and $N\left(v\right)\cap B_0\subseteq B_0^1,$ for all $v\in B_2^2.$

2. Complexity Results for [k]-Roman Domination

Recent investigations have established that the decision problem for $\left[k\right]$-Roman domination is NP-complete across several graph classes, including bipartite and chordal graphs [12], and star-convex and comb-convex bipartite graphs [13]. In contrast to these hardness results, this section demonstrates that $\left[k\right]$-Roman domination can be solved in linear time when the graph satisfies certain structural conditions.

Courcelle’s Theorem 2 provides a framework for solving optimization problems in linear time by expressing them in Monadic Second-Order Logic (MSOL).

Monadic Second-Order Logic (MSOL) is a powerful extension of first-order logic that, in addition to quantification over individual vertices, allows quantification over sets of vertices (and, in some variants, sets of edges). In the context of graphs, MSOL formulas are interpreted over the structure $\langle V\left(G\right),R\rangle$, where $R(u,v)$ indicates adjacency between vertices u and v. This expressive framework enables the definition of a wide range of graph properties and combinatorial problems by means of logical formulas involving vertex variables, set variables, and standard logical connectives.

LinEMSOL (Linear Extended Monadic Second-Order Logic) is an extension of MSOL that allows, in addition to the usual logical predicates over vertices, edges, and sets, the use of linear objective functions defined in terms of the cardinalities of monadic sets. In this framework, one can express optimization problems of the form $\min \left\{{\sum }_i{a}_i\right|X_i|:\phi (X_1,\dots ,X_m)\}$, where $\phi$ is an MSOL formula and $a_i$ are fixed integers. Courcelle’s Theorem extends to LinEMSOL, ensuring that such optimization problems are solvable in linear time on graph classes of bounded clique-width, provided that an appropriate graph decomposition (such as an r-expression) is given.

Theorem 2

([14]). Let $\mathcal{C}$ be a class of p-graphs with clique-width at most r such that there is a known $O\left(f\right(\left|V\right|,\left|E\right|\left)\right)$ algorithm which, for each p-graph in $\mathcal{C}$, constructs an r-expression defining it. Then every LinEMSOL problem on $\mathcal{C}$ can be solved in time $O\left(f\right(\left|V\right|,\left|E\right|\left)\right)$.

To demonstrate that we can reduce the complexity of the problem, let us express it in LinEMSOL. Let us consider $f=(V_0,V_1,\dots ,V_{k+1})$ where $v\in V_j$ if and only if $f\left(v\right)=j,$ and let $V_j$ be the monadic sets. Then,

$$\mathbf{Partition}\left(\mathbf{V}\right):\equiv \forall v:\left(\underset{i=0}{\overset{k+1}{\bigvee }}{V}_i\left(v\right)\right)\wedge \left(\forall i\ne j:\neg \left(V_i\left(v\right)\wedge V_j\left(v\right)\right)\right),$$

which is satisfied if the sets $V_j$ represent a partition of V.

Each vertex v must satisfy the condition for a $\left[k\right]$-RDF. Therefore, we need to introduce the predicate corresponding to the set $N\left[v\right]$ as follows: $\mathbf{N}[\mathbf{v},\mathbf{u}]:\equiv (u=v)\vee \mathcal{R}(u,v).$ With this notation, the following predicate represents the condition of $\left[k\right]$-RDF,

$$\begin{array}{c}\mathbf{[}\mathbf{k}\mathbf{]}\mathbf{-}\mathbf{RDF}\mathbf{(}\mathbf{V}\mathbf{)}\mathbf{:}\equiv \hfill \\ {\displaystyle \forall v:\sum _{i=1}^{k+1}i·|\{u:\mathbf{N}[\mathbf{v},\mathbf{u}]\wedge V_i\left(u\right)\}|\ge k+|\left\{u:\mathcal{R}(u,v)\wedge \left(V_1\left(u\right)\vee \dots \vee V_{k+1}\left(u\right)\right)\right\}|.}\hfill \end{array}$$

Thus, finding a minimum-weight $\left[k\right]$-RDF in the graph G can be expressed as the following LinEMSOL-type problem:

$$min\left\{\sum _{j=1}^{k+1}j·\left|V_j\right|:\mathbf{Partition}\left(\mathbf{V}\right)\wedge \mathbf{[}\mathbf{k}\mathbf{]}\mathbf{\text{-}}\mathbf{RDF}\right\}.$$

Consequently, the task of identifying a minimum-weight $\left[k\right]$-RDF is successfully reduced to a LinEMSOL optimization problem. This leads to the following theorem.

Theorem 3.

The decision problem associated with $\left[k\right]$-Roman domination, $\left[k\right]$-RDN, can be expressed as a LinEMSOL optimization problem. Consequently, if an r-representation of the graph is included in the problem input, or if an efficient algorithm is available that provides it, then the $\left[k\right]$-RDN problem can be solved in linear time with respect to the graph order.

Corollary 2.

Since $\left[k\right]\text{-}\mathrm{RDN}$ is expressible in LinEMSOL, it can be solved in $f\left(k\right)·n$ time for any graph class with bounded clique-width, including cographs, distance-hereditary graphs, and complete graphs; and bounded treewidth, including trees, series-parallel graphs, and outerplanar graphs.

3. Complexity Results for Strong Roman Domination

Álvarez et al. introduced the concept of strong Roman domination in their seminal paper (see [5]) and showed that the associated decision problem is NP-complete for planar graphs. In this section, we prove that the StRDP is also NP-complete for several classes of graphs, including subclasses of bipartite graphs.

We begin by formulating the associated decision problem for strong Roman domination.

StRDN PROBLEM

   Instance: A graph $G=(V,E)$ and a positive integer k.

   Question: Does G have an StRD function f with $f\left(V\right)\le k$?

To show that the StRDN PROBLEM is NP-complete, we make use of the restricted version of the Exact Cover by 3-Sets (RX3C) problem (see [15,16]).

RESTRICTED EXACT 3-COVER (RX3C) PROBLEM

   Instance: Let X be a finite set with $\left|X\right|=3q$, and let C be a set of 3-element subsets of X such that each $x\in X$ belongs to exactly three sets in C.

   Question: Does there exist a subset $C^{*}\subseteq C$ such that the sets in $C^{*}$ form a partition of X, i.e.,

$$\bigcup _{S\in C^{*}}S=X\mathrm{and}S\cap S^{\prime }=\varnothing \mathrm{for}\mathrm{all}\mathrm{distinct}S,S^{\prime }\in C^{*}?$$

(such a set $C^{*}$ is called an exact cover of X).

Theorem 4.

The StRDN PROBLEM is NP-complete for star-convex bipartite graphs.

Proof. 

It can be readily verified that the StRDN PROBLEM belongs to the complexity class $\mathcal{NP}$. Indeed, one can check in polynomial time whether a function $f:V\to \{0,1,\dots ,⌈{\displaystyle \frac{\Delta }{2}}⌉+1\}$ is an StRDF of weight at most k.

Next, we prove that the StRDN PROBLEM is NP-complete for star-convex bipartite graphs by constructing a polynomial transformation between RX3C and the StRDN PROBLEM.

Let $I=(X,C)$ be an instance of RX3C with $X=\{x_j:j\in \left[3q\right]\}$ and $C=\{c_j:j\in \left[3q\right]\}$.

We construct (see Figure 3) a star-convex bipartite graph $\Gamma \left(I\right)=\Gamma =(A\cup B,E)$ and we give a positive integer k such that I contains an exact cover if and only if $\Gamma$ has an StRD function f with $f\left(V\right)\le k=6q+2.$

Consider the bipartite graph $\Gamma \left(I\right)=(V,E)$ with $V=A\cup B$, where

$$A=\{a,x_i,z_i:i\in \left[3q\right]\},B=\{c_j,a_1,a_2:j\in \left[3q\right]\}.$$

Then $\left|A\right|=6q+1$ and $\left|B\right|=3q+2$. For $i,j\in \left[3q\right]$, we join $x_i$ to $c_j$ if and only if $x_i\in c_j$, and set

$$E=\{\{a,a_1\},\{a,a_2\}\}\cup \{\{a,c_j\},\{c_j,z_j\}:j\in \left[3q\right]\}.$$

Let us point out that $\Gamma \left(I\right)$ is a star-convex bipartite graph with vertex classes given by $A\cup B$, the set of edges E defined above and the star graph $T_{\Gamma }$ with central vertex a and leaves $\{x_i,z_j\}$ associated to the vertex class B. That is, for each vertex $y\in B,$ the subset $N_{\Gamma \left(I\right)}\left(y\right)\subseteq A$ induces a connected subtree in the graph $T_{\Gamma }.$

Assume that $C^{\prime }\subseteq C$ is an exact cover of X in the RX3C instance $I=(X,C)$, with $|C^{\prime }|=q$. We define a function $f:V(\Gamma (I\left)\right)\to \{0,1,\dots ,\lceil \Delta /2\rceil +1\}$ as follows:

$$f\left(v\right)=\left\{\begin{array}{cc}3,\hfill & \mathrm{if}v=c_j\in C^{\prime },\hfill \\ 1,\hfill & \mathrm{if}v=z_i\mathrm{with}{c}_i\notin C^{\prime },\hfill \\ q+2,\hfill & \mathrm{if}v=a,\hfill \\ 0,\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

Let us verify that f is a strong Roman dominating function:

• Each vertex $x_j\in A\setminus \{a,z_i\}$ belongs to exactly one $c_j\in C^{\prime }$. Since $f\left(c_j\right)=3$ and $c_j$ has at most four neighbors with $f=0$, the strong Roman domination condition is satisfied for $x_j$.
• Each vertex $z_i$ either satisfies $f\left(z_i\right)=1$ or has a neighbor $c_i$ with $f\left(c_i\right)=3$, which guarantees that $z_i$ is strongly Roman dominated.
• For the vertex a, we have $f\left(a\right)=q+2=1+{\displaystyle \frac{1}{2}}|B_0\cap N\left(a\right)|$, where $B_0=\{v\in V:f\left(v\right)=0\}$, ensuring that all vertices in $N\left[a\right]$ are strongly Roman dominated.

Finally, the total weight of f is

$$f\left(V(\Gamma \left(I\right))\right)=f\left(a\right)+\sum _{c_j\in C^{\prime }}f\left(c_j\right)+\sum _{z_i:c_i\notin C^{\prime }}f\left(z_i\right)=(q+2)+3q+2q=6q+2.$$

Therefore, f is a strong Roman dominating function of $\Gamma \left(I\right)$ with weight $6q+2$, as required.

Conversely, let us assume that there exists an StRDF $f=(B_0,B_1,B_2)$ with $f\left(V_{\Gamma }\right)=f\left(V\right)\le 6q+2.$

• Claim 1. $f\left(a_i\right)=0$ for $i=1,2$ and $f\left(a\right)\ge 2.$

Since $N\left(a_1\right)=N\left(a_2\right)=\left\{a\right\},$ it can be readily verified that the minimum weight of f is achieved when $f\left(a_i\right)=0$ and $f\left(a\right)\ge 2.$

• Claim 2. $f\left(x_i\right)=0$ for all $i\in \left[3q\right].$

Let $r=|X^{\prime }|=|\{x_i:f\left(x_i\right)\ne 0\}|$. First of all, we note that since $|N\left(c_j\right)\setminus \left\{a\right\}|=4$ for all $j\in \left[3q\right],$ the minimum weight of $f(N\left[c_j\right]\setminus \left\{a\right\})$ is obtained if and only if $f\left(c_j\right)=3$ and $f(N\left(c_j\right)\setminus \left\{a\right\})=0$. Moreover, there must exist a subset $C_{X^{\prime }}\subseteq C$ with $|C_{X^{\prime }}|=⌈{\displaystyle \frac{3q-r}{3}}⌉$ such that $f\left(c\right)=3$ for all $c\in C_{X^{\prime }},$ ensuring that the set $X\setminus X^{\prime }$ is strongly Roman dominated by $C_{X^{\prime }}$. Then

$${\displaystyle \begin{array}{ccc}\hfill f\left(V\right)& \ge & f\left(X^{\prime }\right)+f\left(C_{X^{\prime }}\right)+f\left(\{z_j:f\left(c_j\right)=0\}\right)+f(N\left(c_j\right)\setminus \left\{z_j\right\}:f\left(c_j\right)=0\})+f\left(a\right)\hfill \\ & \ge & {\displaystyle r+3⌈\frac{3q-r}{3}⌉+3q-⌈\frac{3q-r}{3}⌉+⌈\frac{1}{2}(3q-⌈\frac{3q-r}{3}⌉+2)⌉+1}\hfill \\ & \ge & {\displaystyle r+3(q-⌊\frac{r}{3}⌋)+3q-(q-⌊\frac{r}{3}⌋)+⌈\frac{1}{2}\left(3q-(q-⌊\frac{r}{3}⌋)\right)⌉+2}\hfill \\ & \ge & {\displaystyle 6q+2+r-2⌊\frac{r}{3}⌋+⌈\frac{1}{2}⌊\frac{r}{3}⌋⌉}\hfill \end{array}}$$

It can be easily verified that if $r\ge 1$ then $r-2⌊\frac{r}{3}⌋+⌈\frac{1}{2}⌊\frac{r}{3}⌋⌉\ge 1$. As a consequence, r ≥ 1 implies that f(V) ≥ 6q + 3, a contradiction. Therefore, we must have r = 0, and the claim holds.

• Claim 3. $\left|\right\{z_j:f\left(z_j\right)=0\left\}\right|=q,$ and $f\left(a\right)=q+2.$

Let $s=|Z^{\prime }|=\left|\{z_j:f\left(z_j\right)=0\}\right|.$ By Claim 2, we have that $f\left(x_i\right)=0$ for all $1\le i\le 3q,$ which implies that $f\left(c\right)=3$ for any $c\in C_{Z^{\prime }}=\{c_j:z_j\in Z^{\prime }\}$. Moreover, since $\left|X\right|=3q$ then $s\ge q.$

Furthermore, since $f\left(x_i\right)=0$ for all $1\le i\le 3q,$ $c_j$ must be strongly Roman dominated by the vertex a, whenever $f\left(c_j\right)=0.$ Then,

$$\begin{array}{ccc}\hfill 6q+2\ge f\left(V\right)& \ge & f\left(C_{Z^{\prime }}\right)+f\left(\{z_j:f\left(z_j\right)\ne 0\}\right)+f\left(a\right)\hfill \\ \hfill & \ge & {\displaystyle 3s+3q-s+⌈\frac{3q-s+2}{2}⌉+1}\hfill \\ \hfill & \ge & {\displaystyle \frac{9}{2}q+2+\frac{3}{2}s\ge \frac{9}{2}q+2+\frac{3}{2}q=6q+2}\hfill \end{array}$$

and the inequalities become equalities.

Thus, we have $s=q$ and $f\left(a\right)=q+2,$ and there are exactly q vertices $z_j$ with $f\left(z_j\right)=0$, which implies that the q corresponding vertices $c_j$ satisfy $f\left(c_j\right)=3.$ Clearly, $f\left(c_j\right)=0$ otherwise. Moreover, the remaining $2q$ vertices $z_j,$ with $f\left(z_j\right)\ge 1,$ add $2q$ to the total weight $f\left(V\right),$ which implies that $f\left(z_j\right)=1$ for all $z_j$ with $f\left(z_j\right)\ne 0.$

• Claim 4. $C^{\prime }=\{c_j:f\left(c_j\right)=3\}$ is a solution to the instance $I=(X,C)$ of the RX3C problem.

By Claim 3, we have that $\left|\right\{c_j:f\left(c_j\right)=3\left\}\right|=q.$ By Claim 2, $f\left(x_i\right)=0$ for all $i\in \left[3q\right]$ and f is an StRD function. Therefore, each and every vertex $x_i$ must be adjacent to exactly one vertex $c_j,$ which means that $C^{\prime }$ is a solution to $I=(X,C).$

The latter concludes the proof. □

4. Exact Values for [k]-Roman Domination

This section presents some results obtained for the $\left[k\right]$-Roman domination number that will be presented. The first result allows us to bound the labels of a $\left[k\right]$-Roman dominating function asigned to strong and weak support vertices, leaves, and their neighborhoods.

Lemma 1.

Let G be a non-trivial connected graph and let f be a ${\gamma }_{\left[kR\right]}$-RDF with $k\ge 2$.

(i) 

If v is a weak support vertex and u is a leaf adjacent to v, then $k\le f\left(u\right)+f\left(v\right)\le k+1.$ Furthermore, if $f\left(u\right)+f\left(v\right)=k,$ it follows that $f\left(u\right)=k$ and $f\left(v\right)=0.$

(ii) 

If v is a strong support vertex, then $f\left(v\right)=k+1$ and $f\left(u\right)=0$ for all leaves $u\in N\left(v\right)$.

Proof. 

The first inequality in item (i) follows directly from the definition of a $\left[k\right]$-RDF, since $f\left(u\right)+f\left(v\right)=f\left(N\right[u\left]\right)\ge k+\left|AN\right(u\left)\right|\ge k.$ Now, suppose, for the sake of contradiction, that $f\left(u\right)+f\left(v\right)>k+1.$ Consider the function g defined as follows: $g\left(u\right)=0,$$g\left(v\right)=k+1$ and $g\left(z\right)=f\left(z\right)$ for all other vertices. Since u is perfectly dominated by its support vertex v, and f is a $\left[k\right]$-RDF, it follows that g is also a $\left[k\right]$-RDF. Moreover, $w\left(g\right)=w\left(f\right)-\left(f\left(u\right)+f\left(v\right)\right)+k+1<w\left(f\right),$ which contradicts the optimality of f. Therefore, $f\left(u\right)+f\left(v\right)\le k+1$.

If $f\left(u\right)+f\left(v\right)=k$ and we assume that $f\left(u\right)<k$, then v must be an active neighbor of u. Consequently, $f\left(N\right[u\left]\right)=f\left(u\right)+f\left(v\right)\ge k+\left|AN\right(u\left)\right|\ge k+1,$ which is a contradiction. Hence, we must have $f\left(u\right)\ge k$. Since $f\left(u\right)+f\left(v\right)=k$, we deduce that $f\left(u\right)=k$ and $f\left(v\right)=0$.

Now, we prove item (ii). Let $\{u_i:i\in I\}$ be the set of leaves adjacent to the strong support vertex $v.$ Since v is a strong support vertex, $\left|I\right|\ge 2.$ We consider the function g defined by $g\left(u_i\right)=0$ for all $i\in I$, $g\left(v\right)=k+1$, and $g\left(z\right)=f\left(z\right)$ otherwise. If ${\sum }_{i\in I}f\left(u_i\right)=0$, the result follows immediately, since $f\left(v\right)=k+1$ is required to dominate those leaves. Assume, therefore, that at least one leaf has a non-zero label; without loss of generality, let $f\left(u_1\right)>0$. If for any other leaf, say $u_2$, we had $f\left(u_2\right)=0$, it would follow that $f\left(v\right)=k+1.$ Then, the function g described above would be a $\left[k\right]$-RDF with weight $w\left(g\right)=w\left(f\right)-f\left(u_1\right)<w\left(f\right),$ which is impossible. Thus, under these conditions, $f\left(u_i\right)\ne 0$ for every $i\in I$.

Furthermore, it is also impossible that $f\left(v\right)=0$ because v is a strong support vertex; otherwise, $f\left(u_i\right)=k$ for all $i\in I$. In that case, the function g would be a $\left[k\right]$-RDF with weight $w\left(g\right)=w\left(f\right)-\left|I\right|k+k+1\le w\left(f\right)-2k+k+1=w\left(f\right)-k+1<w\left(f\right),$ which is again a contradiction. Therefore, $f\left(v\right)+f\left(u_i\right)\ge k+1$ for all $i\in I$. Assume, without loss of generality, that $f\left(u_1\right)=min\{f\left(u_i\right):i\in I\}.$ Then, the function g satisfies:

$$\begin{array}{cc}\hfill w\left(g\right)=& w\left(f\right)-\left(\sum _{i\in I}f\left(u_i\right)+f\left(v\right)\right)+k+1\hfill \\ \hfill \le & w\left(f\right)-\left(\right|I|-1)f\left(u_1\right)+k+1-\left(f\left(v\right)+f\left(u_1\right)\right)<w\left(f\right),\hfill \end{array}$$

which leads to a contradiction. This concludes the proof. □

Let $q_i\ge 2$ be integers, and let $K_{1,q_i}^i$, for $i\in \left[n\right]$, be n star graphs with central vertices denoted by $c_i$. Let $H=\left(V\right(H),E(H\left)\right)$ be an arbitrary graph of order n, where $V\left(H\right)=\{v_i:i\in \left[n\right]\}.$ We denote by $\mathcal{H}\left(K_{1,q_i}^i\right)$ the graph obtained by identifying each central vertex $c_i$ of $K_{1,q_i}^i$ with the vertex $v_i$ of the graph H.

Lemma 2.

Let $q_i$ be integers and let H be a graph of order n. Then, ${\gamma }_{\left[kR\right]}\left(\mathcal{H}\left(K_{1,q_i}^i\right)\right)=(k+1)n.$

Proof. 

We consider the previously defined graph $\mathcal{H}\left(K_{1,q_i}^i\right).$ This graph is composed entirely of strong support vertices and their corresponding leaves. The result follows directly from the second item of Lemma 1. □

As a consequence of the previous result, one can readily derive the $\left[k\right]$-Roman domination number for bi-stars.

Proposition 5.

Let p and q be two positive integers. Then:

(i) 

${\gamma }_{\left[kR\right]}\left(S_{1,q}\right)=2k+1$, for $q\ge 1$

(ii) 

${\gamma }_{\left[kR\right]}\left(S_{p,q}\right)=2k+2$, for $p,q\ge 2.$

Proof. 

To prove item (i), we denote by u the weak support vertex and by v the strong support vertex of the bistar $S_{1,q}$. Then, the function f defined as $f\left(u\right)=k,$ $f\left(v\right)=k+1,$ and $f\left(z\right)=0$ for all other vertices, is a $\left[k\right]$-RDF; therefore, ${\gamma }_{\left[kR\right]}\left(S_{1,q}\right)\le 2k+1.$ On the other hand, by Lemma 1, we know that for any $\left[k\right]$-RDF f on $S_{1,q},$ we have that $f\left(N\right[u\left]\right)\ge k$ and $f\left(N\right[v\left]\right)\ge k+1$, which proves the equality.

Item (ii) is a straightforward consequence of Lemma 2 by considering $H=P_2$. □

Given a ${\gamma }_{\left[kR\right]}$-function, the following proposition allows us to establish bounds on the sequence of labels for cycles and paths.

Proposition 6.

Let G be a path or cycle with at least four vertices and let f be a ${\gamma }_{\left[kR\right]}$-function. There cannot exist four consecutive vertices in G assigned the label $⌈{\displaystyle \frac{k+1}{2}}⌉$ under $f.$

Proof. 

First, consider the case $k=1$. Suppose there exist four consecutive vertices $\{u_{i-2},$$u_{i-1},$$u_i,$ $u_{i+1}\}$ in G, all assigned the label $1.$ We can then define the function g by setting $g\left(u_{i-2}\right)=1,$$g\left(u_{i-1}\right)=0,$$g\left(u_i\right)=2,$$g\left(u_{i+1}\right)=0$ and $g\left(v\right)=f\left(v\right)$ otherwise. This function is a $\left[k\right]$-RDF with a total weight $w\left(g\right)<w\left(f\right),$ which contradicts the optimality of $f.$

Now, assume $k\ge 2$ and suppose there are four consecutive vertices $\{u_{i-2},u_{i-1},u_i,u_{i+1}\}$ in G such that $f\left(u_j\right)=⌈{\displaystyle \frac{k+1}{2}}⌉$ for every $j\in \{i-2,i-1,i,i+1\}.$ We define the function g as follows: $g\left(u_{i-2}\right)=g\left(u_{i-1}\right)$$=⌈{\displaystyle \frac{k+1}{2}}⌉$, $g\left(u_i\right)=0$, $g\left(u_{i+1}\right)=k,$ and $g\left(v\right)=f\left(v\right)$ for all other vertices. Since $k\ge 2$, we have $⌈{\displaystyle \frac{k+1}{2}}⌉\ge 2$. Therefore, $u_i$ is $\left[k\right]$-dominated by its two neighbors because $g\left(N\left[u_i\right]\right)=g\left(u_{i-1}\right)+g\left(u_i\right)+g\left(u_{i+1}\right)=⌈{\displaystyle \frac{k+1}{2}}⌉+k\ge k+2=k+\left|AN\left(u_i\right)\right|.$ The weight of g satisfies: $w\left(g\right)=w\left(f\right)+k-2⌈{\displaystyle \frac{k+1}{2}}⌉\le w\left(f\right)-1<w\left(f\right),$ yielding a contradiction. □

The following results allow us to establish general bounds for the $\left[k\right]$-Roman domination number, as well as exact values under certain conditions for cycles and paths.

Remark 2.

If k is even, the previous proposition is true for three consecutive vertices. That is, if k is even, there cannot exist three consecutive vertices in G assigned the label $⌈{\displaystyle \frac{k+1}{2}}⌉$ under f. If such vertices $v_{i-1},v_i,v_{i+1}$ existed, by considering the function g such that $g\left(v_i\right)=⌈{\displaystyle \frac{k+1}{2}}⌉-1$ and $g\left(v\right)=f\left(v\right)$ otherwise for $k\ge 6$, or $g\left(v_{i-1}\right)=g\left(v_{i+1}\right)=k$, $g\left(v_i\right)=0$ and $g\left(v\right)=f\left(v\right)$ otherwise for $k\le 4$, we would obtain that g is a $\left[k\right]$-RDF with a smaller weight than f, which would be a contradiction. Furthermore, if k is odd, the previous statement is not true. It suffices to consider for $k=3$ and $n=10$ the label assignment $(2,2,2,0,3,0,2,2,0,3)$.

In the work of Brezovnik and Zerovnik [17], it is proven (Lemma 8 [17]) that if H is a spanning subgraph of the graph G, then ${\gamma }_{\left[kR\right]}\left(G\right)\le {\gamma }_{\left[kR\right]}\left(H\right)$. Taking the previous result into account, the following holds.

Proposition 7.

Let $k\ge 1,$ and let $P_n$ and $C_n$ be the path and the cycle of order n, respectively, with $n\ge 3$. Then, for any $m\ge 3$,

$${\gamma }_{\left[kR\right]}(C_m\square C_n)\le {\gamma }_{\left[kR\right]}(C_m\square P_n),\mathit{and},\mathit{in}\mathit{particular},{\gamma }_{\left[kR\right]}\left(C_n\right)\le {\gamma }_{\left[kR\right]}\left(P_n\right).$$

Under certain conditions regarding the existence of a $\left[k\right]$-RDF on a path, the inequality shown in the previous result can be improved.

Proposition 8.

Let $n,k\ge 4$ be positive integers and let $P_n$ be a path of order n such that there exists a ${\gamma }_{\left[kR\right]}\left(P_n\right)$-RDF, denoted by $f=(V_0,\varnothing ,V_2,\dots ,V_k,\varnothing ).$ Then, ${\gamma }_{\left[kR\right]}\left(C_n\right)\le {\gamma }_{\left[kR\right]}\left(P_n\right)-1.$

Proof. 

Let $P_n:v_1{v}_2\dots v_n$ be a path and let f be a ${\gamma }_{\left[kR\right]}$-RDF on $P_n$ with $V_{k+1}=\varnothing .$ In particular, it follows that $0<f\left(v_1\right), f\left(v_n\right)<k+1.$ Consider the cycle obtained by adding to $P_n$ the edge $v_1{v}_n$, that is, $C_n=P_n+v_1{v}_n.$ Without loss of generality, we may assume that $f\left(v_1\right)\le f\left(v_n\right).$

Let us define the function g on $C_n$ as follows: $g\left(v_2\right)=min\{k+1,f\left(v_2\right)+f\left(v_1\right)-1\},$$g\left(v_1\right)=0$ and $g\left(z\right)=f\left(z\right)$ otherwise. For any vertex other than $v_1$, the function g satisfies the conditions to be a $\left[k\right]$-RDF, since this property holds for the original function f.

On the other hand, since $2\le f\left(v_1\right)\le f\left(v_n\right),$ it follows that $g\left(v_n\right)\ge 2.$ Furthermore, since f is a $\left[k\right]$-RDF, we have that $f\left(v_1\right)+f\left(v_2\right)\ge k+1$. Consequently, it can be verified that:

$$\begin{array}{cc}\hfill g\left(N_{C_n}\left[v_1\right]\right)& =g\left(v_1\right)+g\left(v_2\right)+g\left(v_n\right)=f\left(v_1\right)+f\left(v_2\right)-1+f\left(v_n\right)\hfill \\ & \ge k+1-1+f\left(v_n\right)\ge k+2=k+\left|A{N}_g\left(v_1\right)\right|.\hfill \end{array}$$

Then, g is a $\left[k\right]$-RDF on the cycle $C_n$ with weight ${\gamma }_{\left[kR\right]}\left(P_n\right)-1.$ This concludes the proof. □

To introduce the next result, we recall that an efficient dominating set is a dominating set S such that $\left|N\right[u]\cap S|=1$ for every vertex $u\in V\left(G\right)$. An efficient graph is a graph having an efficient dominating set.

Theorem 5 establishes that for $n\equiv 0\left(\mathrm{mod}3\right),$ the $\left[k\right]$-Roman domination numbers for $C_n$ and $P_n$ are equal; furthermore, we determine their exact value. To this end, we first provide a preliminary lemma below.

Lemma 3.

[18] For any graph G, if S and $S^{\prime }$ are two different efficient sets then $\left|S\right|= |S^{\prime }| =\gamma \left(G\right).$

We also use a result by Brezovnik and Zerovnik [17].

Proposition 9.

[17] Let G be an efficient graph. Then ${\gamma }_{\left[kR\right]}\left(G\right)=(k+1)\gamma \left(G\right).$

As a consequence of the previous results, we derive the following theorem.

Theorem 5.

Let n be a positive integer such that $n\equiv 0\left(\mathrm{mod}3\right).$ Then,

$${\gamma }_{\left[kR\right]}\left(C_n\right)={\gamma }_{\left[kR\right]}\left(P_n\right)=(k+1){\displaystyle \frac{n}{3}}.$$

Proof. 

Let G be a cycle or a path graph and let $V\left(G\right)=\{v_1,\dots ,v_n\}$ be the vertex set of G. It is sufficient to note that the set $\{v_{3j-1}:j=1,\dots ,{\displaystyle \frac{n}{3}}\}$ is an efficient dominating set in $G.$ □

Finally, we extend the concept of reinforcement, defined in [19] as the minimum number of edges that must be added to a graph to decrease its Roman domination number.

Definition 1.

Let $G=(V,E)$ be a graph. The $\left[k\right]$-Roman domination reinforcement number of G, denoted by $r_{\left[kR\right]}\left(G\right)$, is defined as

$$r_{\left[kR\right]}\left(G\right)=min\left\{\right|F|:F\subseteq E\left(\overline{G}\right)\mathrm{and}{\gamma }_{\left[kR\right]}(G+F)<{\gamma }_{\left[kR\right]}\left(G\right)\},$$

where $E\left(\overline{G}\right)$ denotes the edge set of $\overline{G}$, and $G+F$ is obtained by adding to G the edges of F.

Proposition 10.

Let $P_n$ be a path with $n\ge 4$ vertices. Then, $r_{\left[kR\right]}\left(P_n\right)\in \{1,2\}$. Moreover, there exist two edges $e,e^{\prime }\in E\left(\overline{P_n}\right)$ such that,

(i) 

If $n\equiv 0\left(\mathrm{mod}3\right)$, then $r\le 2$ and ${\gamma }_{\left[kR\right]}(P_n+\{e,e^{\prime }\})\le {\gamma }_{\left[kR\right]}\left(P_n\right)-1$

(ii) 

If $n\equiv 1\left(\mathrm{mod}3\right)$, then $r=1$ and ${\gamma }_{\left[kR\right]}(P_n+e)\le {\gamma }_{\left[kR\right]}\left(P_n\right)-k$

(iii) 

If $n\equiv 2\left(\mathrm{mod}3\right)$, then $r=1$ and ${\gamma }_{\left[kR\right]}(P_n+e)\le {\gamma }_{\left[kR\right]}\left(P_n\right)-1.$

Proof. 

Let $P_n:v_1{v}_2\dots v_n$ be a path of order n.

• Case (i). If $n\equiv 0\left(\mathrm{mod}3\right),$ we observe that ${\gamma }_{\left[kR\right]}\left(P_n\right)=(k+1)⌊{\displaystyle \frac{n}{3}}⌋.$ Let us define the function $f:V(P_n+\{v_2{v}_4,v_6{v}_8\})⟶\{0,1,2,\dots ,k+1\},$ as follows $f\left(v_{3i-1}\right)=k+1$ for $1\le i\le {\displaystyle \frac{n}{3}},$ where $i\ne 2$, $f\left(v_5\right)=k$ and $f\left(v\right)=0$ otherwise. This function is a $\left[k\right]$-RDF with weight $(k+1)⌊{\displaystyle \frac{n}{3}}⌋-1.$ Thus, by adding two edges, we decrease the $\left[k\right]$-Roman domination number of $P_n$ by one.
• Case (ii). If $n\equiv 1\left(\mathrm{mod}3\right),$ we have that ${\gamma }_{\left[kR\right]}\left(P_n\right)=(k+1)⌊{\displaystyle \frac{n}{3}}⌋+k.$ Let us define the function $f:V(P_n+\left\{v_1{v}_3\right\})⟶\{0,1,2,\dots ,k+1\}$ in the following way: $f\left(v_{3i}\right)=k+1$ for $1\le i\le {\displaystyle \frac{n-1}{3}}$ and $f\left(v\right)=0$ or all other vertices. This function is a $\left[k\right]$-RDF in $P_n+\left\{v_1{v}_3\right\},$ with weight $(k+1)⌊{\displaystyle \frac{n}{3}}⌋.$ So, by adding one edge, we decrease the $\left[k\right]$-Roman domination number of $P_n$ by $k.$
• Case (iii). If $n\equiv 2\left(\mathrm{mod}3\right),$ then ${\gamma }_{\left[kR\right]}\left(P_n\right)=(k+1)⌊{\displaystyle \frac{n}{3}}⌋+k+1.$ In this case, the function $f:V(P_n+\left\{v_1{v}_4\right\})⟶\{0,1,2,\dots ,k+1\},$ defined as $f\left(v_2\right)=k$, $f\left(v_{3i+1}\right)=k+1$ for $1\le i\le {\displaystyle \frac{n-2}{3}}$ and $f\left(v\right)=0$ otherwise, is a $\left[k\right]$-RDF in $P_n+\left\{v_1{v}_4\right\},$ with weight $(k+1)⌊{\displaystyle \frac{n}{3}}⌋+k.$ Therefore, by adding one edge, we decrease the $\left[k\right]$-Roman domination number of $P_n$ by one, yielding $r_{\left[kR\right]}\left(P_n\right)=1$. □

5. Exact Values for Strong Roman Domination

In this section, we address the problem of obtaining the exact values for some graph families under strong Roman domination. Specifically, the first result determines the exact values for t-fold wheel graphs.

Proposition 11.

Let $m,t$ be positive integers. Then

$${\gamma }_{StR}\left(W_{m,t}\right)=\left\{\begin{array}{cc}\hfill ⌈{\displaystyle \frac{t}{2}}⌉+2,\hfill & m=3,t\ge 2;\hfill \\ \hfill ⌈{\displaystyle \frac{t}{2}}⌉+3,\hfill & m=4,t\ge 2;\hfill \\ \hfill ⌈{\displaystyle \frac{m}{2}}⌉+t,\hfill & m\ge 5,t=2;\mathit{or}m\equiv 0\left(\mathrm{mod}2\right),m\ge 5\mathit{and}t=3;\hfill \\ \hfill ⌈{\displaystyle \frac{m-1}{2}}⌉+⌈{\displaystyle \frac{t+1}{2}}⌉+2,\hfill & m\ge 5,t\ge 3,\mathit{with}m,t\equiv 1\left(\mathrm{mod}2\right);\mathit{or}\hfill \\ & m\ge 6,t\ge 4,\mathit{with}m+t\equiv 1\left(\mathrm{mod}2\right);\hfill \\ \hfill {\displaystyle \frac{t}{2}}+4,\hfill & m=5,t\equiv 0\left(\mathrm{mod}2\right)\mathit{and}t\ge 4;\hfill \\ \hfill {\displaystyle \frac{m}{2}}+{\displaystyle \frac{t}{2}}+2,\hfill & m\ge 6,t\ge 4\mathit{with}m,t\equiv 0\left(\mathrm{mod}2\right).\hfill \end{array}\right.$$

Proof. 

To prove the result, we distinguish the following cases:

• Case (i). Here, all vertices are universal. Since $n=3+t$, we have $n+1=t+4.$ By Proposition 3, it follows that ${\gamma }_{StR}\left(W_{3,t}\right)=⌈{\displaystyle \frac{t+4}{2}}⌉$, for $t\ge 2.$
• Cases (ii) and (iii). We distinguish two subcases:
  • Subcase (ii): $G=W_{4,t}$, for $t\ge 2,$ and Subcase (iii.1): $G=W_{m,t}$ for $m\ge 5$ and $t=2.$
    • In both instances, there are two non-adjacent vertices with maximum degree $\Delta =n-2$. Using the upper bound from Proposition 2, we have:

      $${\gamma }_{StR}\left(G\right)\le n-⌊{\displaystyle \frac{\Delta }{2}}⌋=n-⌊{\displaystyle \frac{n-2}{2}}⌋=n-⌊{\displaystyle \frac{n}{2}}⌋+1=⌈{\displaystyle \frac{n}{2}}⌉+1=⌈{\displaystyle \frac{\Delta }{2}}⌉+2.$$
    • From Proposition 3, we deduce ${\gamma }_{StR}\left(G\right)\ge ⌈{\displaystyle \frac{n+1}{2}}⌉=⌈{\displaystyle \frac{\Delta +3}{2}}⌉=⌈{\displaystyle \frac{\Delta +1}{2}}⌉+1.$
    • Moreover, if n is odd, since $\Delta =n-2<n-1,$ we have that ${\gamma }_{StR}\left(G\right)>⌈{\displaystyle \frac{n+1}{2}}⌉$ and therefore,

      $${\gamma }_{StR}\left(G\right)\ge ⌈{\displaystyle \frac{n+1}{2}}⌉+1={\displaystyle \frac{\Delta +3}{2}}+1=⌈{\displaystyle \frac{\Delta +2}{2}}⌉+1=⌈{\displaystyle \frac{\Delta }{2}}⌉+2.$$
    • On the other hand, if n is even, ${\gamma }_{StR}\left(G\right)\ge ⌈{\displaystyle \frac{n+1}{2}}⌉=⌈{\displaystyle \frac{\Delta +3}{2}}⌉={\displaystyle \frac{\Delta }{2}}+2=⌈{\displaystyle \frac{\Delta }{2}}⌉+2.$
    • Since $\Delta =n-2=t+2$ for Subcase (ii) and $\Delta =m+2-2=m$ for Subcase (iii.1), the result follows.
  • Subcase (iii.2). For $m\ge 5$ with $m\equiv 0\left(\mathrm{mod}2\right),$ and $t=3$.
    • Using Proposition 2 and the fact that $\Delta =m=n-3$ is even, we obtain ${\gamma }_{StR}\left(W_{m,3}\right)\le n-⌊{\displaystyle \frac{\Delta }{2}}⌋=⌈{\displaystyle \frac{n}{2}}⌉+⌊{\displaystyle \frac{n}{2}}⌋-⌊{\displaystyle \frac{\Delta }{2}}⌋=⌈{\displaystyle \frac{m}{2}}⌉+3.$
    • For the lower bound, the assumption that m is even implies that the graph order n is odd. Consequently, by Proposition 3, we obtain the strict inequality ${\gamma }_{StR}\left(W_{m,3}\right)>⌈{\displaystyle \frac{n+1}{2}}⌉.$ Thus, ${\gamma }_{StR}\left(W_{m,3}\right)\ge ⌈{\displaystyle \frac{n+1}{2}}⌉+1=⌈{\displaystyle \frac{m+4}{2}}⌉+1=⌈{\displaystyle \frac{m}{2}}⌉+3.$
• Case (iv). For $m\ge 5$, $t\ge 3$. We distinguish the following subcases:
  • Subcase (iv.1). $m,t\equiv 1\left(\mathrm{mod}2\right)$ with $m\ge 5$ and $t\ge 3$.
    • Here, n is even. We define the function f as follows: if c is a vertex on the cycle, let $f\left(c\right)=1+⌈{\displaystyle \frac{t+1}{2}}⌉$; if s is an interior vertex, let $f\left(s\right)=1+⌈{\displaystyle \frac{m-1}{2}}⌉$; and let $f\left(z\right)=0$ for $z\ne c,s$. The vertex c covers the $t-1$ undefended interior vertices and its two undefended neighbors in the cycle, while s covers the $m-1$ vertices of the cycle. Thus, all vertices are strongly Roman dominated, which ensures that f is an StRDF:

      $$w\left(f\right)=2+⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-1}{2}}⌉\ge {\gamma }_{StR}\left(W_{m,t}\right).$$

      (1)
    • To prove the lower bound, let f be a ${\gamma }_{StR}$-function. Observe that

      $${\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)\ge |B_1|+|B_2|+⌈{\displaystyle \frac{|B_0|}{2}}⌉.$$

      (2)
    • If $|B_2=\left\{s\right\}|=1,$ and $t=3,$ we have:

      $$\begin{array}{cc}\hfill {\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)& \ge |B_1|+f\left(s\right)=t-1+1+⌈{\displaystyle \frac{m}{2}}⌉\hfill \\ & \ge 1+⌈{\displaystyle \frac{t-1}{2}}⌉+1+⌈{\displaystyle \frac{m+1}{2}}⌉=1+⌈{\displaystyle \frac{t-1}{2}}⌉+2\hfill \\ & +⌈{\displaystyle \frac{m-1}{2}}⌉=⌈{\displaystyle \frac{t+1}{2}}⌉+2+⌈{\displaystyle \frac{m-1}{2}}⌉,\hfill \end{array}$$

      attaining the upper bound for $t=3$.
    • If $|B_2=\left\{c\right\}|=1$ with c on the cycle, and $m=5,$ we have:

      $$\begin{array}{cc}\hfill {\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)& \ge |B_1|+f\left(c\right)\ge 1+⌈{\displaystyle \frac{t+2}{2}}⌉+m-3=1+⌈{\displaystyle \frac{t}{2}}⌉+m-2\hfill \\ & \ge 1+⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-2}{2}}⌉+1=⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-1}{2}}⌉+2,\hfill \end{array}$$

      attaining the upper bound for $m=5.$
    • For the remaining cases, we now show by contradiction that there must exist at least two distinct strong vertices $c,s\in B_2:$
      • Suppose $B_2=\varnothing$.
        Then ${\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)\ge \left|B_1\right|=n=m+t.$ However, this contradicts the upper bound, as $2+⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-1}{2}}⌉\ge {\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)=m+t=⌈{\displaystyle \frac{t}{2}}⌉+⌊{\displaystyle \frac{t}{2}}⌋+⌈{\displaystyle \frac{m}{2}}⌉+⌊{\displaystyle \frac{m}{2}}⌋>⌈{\displaystyle \frac{t+1}{2}}⌉+1+⌈{\displaystyle \frac{m-1}{2}}⌉+1$.
      • Suppose $|B_2| =1$ with a cycle vertex $c\in B_2$ and $m\ge 7.$ It follows that ${\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)\ge \left|B_1\right|+f\left(c\right)\ge 1+⌈{\displaystyle \frac{t+2}{2}}⌉+m-3=1+⌈{\displaystyle \frac{t}{2}}⌉+m-2>1+⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-2}{2}}⌉+1=⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-1}{2}}⌉+2$, a contradiction.
      • Suppose $|B_2| =1$ with an interior vertex $s\in B_2$ and $t\ge 5$. Here, ${\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)\ge \left|B_1\right|+f\left(s\right)=t-1+1+⌈{\displaystyle \frac{m}{2}}⌉>1+⌈{\displaystyle \frac{t-1}{2}}⌉+1+⌈{\displaystyle \frac{m+1}{2}}⌉=1+⌈{\displaystyle \frac{t-1}{2}}⌉+2+⌈{\displaystyle \frac{m-1}{2}}⌉=⌈{\displaystyle \frac{t+1}{2}}⌉+2+⌈{\displaystyle \frac{m-1}{2}}⌉$, a contradiction.
      • Suppose $|B_2| \ge 2$ and all strong vertices are interior vertices. So, ${\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)\ge 2(1+⌈{\displaystyle \frac{m}{2}}⌉)-1+t-2=2⌈{\displaystyle \frac{m+1}{2}}⌉+t-1\ge ⌈{\displaystyle \frac{m+1}{2}}⌉+t+2>⌈{\displaystyle \frac{m-1}{2}}⌉+⌈{\displaystyle \frac{t}{2}}⌉+2=⌈{\displaystyle \frac{m-1}{2}}⌉+⌈{\displaystyle \frac{t+1}{2}}⌉+2$, a contradiction.
      • Suppose $|B_2| \ge 2$ and all strong vertices are on the cycle. Then, ${\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)\ge 2(1+⌈{\displaystyle \frac{t+1}{2}}⌉)-1+m-4=2⌈{\displaystyle \frac{t+1}{2}}⌉+m-3\ge ⌈{\displaystyle \frac{t+1}{2}}⌉+m-1\ge ⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-1}{2}}⌉+3>⌈{\displaystyle \frac{t+1}{2}}⌉+⌈{\displaystyle \frac{m-1}{2}}⌉+2$, a contradiction.
    • Therefore, there must be at least two strong vertices $c,s\in B_2$, one on the cycle and one in the interior. Thus, the following holds:

      $$\begin{array}{cc}\hfill {\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)& \ge |B_1|+|B_2|+⌈{\displaystyle \frac{|B_0|}{2}}⌉\ge \left|B_2\right|+⌈{\displaystyle \frac{|B_0|}{2}}⌉\hfill \\ & =⌈{\displaystyle \frac{t+1}{2}}⌉+2+⌈{\displaystyle \frac{m-1}{2}}⌉.\hfill \end{array}$$
  • Subcase (iv.2). $m\equiv 0\left(\mathrm{mod}2\right)$ and $t\equiv 1\left(\mathrm{mod}2\right)$ with $m\ge 6$ and $t\ge 5$.
    • In this case, n is odd. We proceed as in Subcase (iv.1). A similarly constructed function f satisfies Equations (1) and (2).
      If $|B_2|=1$ with vertex $c\in B_2$ on the cycle and $m=6$ or $s\in B_2$ is an interior vertex with $t=5$, the bounds hold. The The remaining cases—specifically, $B_2=\varnothing$, $|B_2|=1$ with vertex $c\in B_2$ on the cycle and $m\ge 8$ or $s\in B_2$ an interior vertex with $t\ge 7;$ and $|B_2| \ge 2$ with all strong vertices either on the cycle or in the interior—lead to a contradiction. We conclude that there are at least two strong vertices $c,s\in B_2$ (one on the cycle, one interior) and the bounds hold.
  • Subcase (iv.3). $m\equiv 1\left(\mathrm{mod}2\right)$ and $t\equiv 0\left(\mathrm{mod}2\right)$ with $m\ge 7$ and $t\ge 4.$
    • Note that n is odd. By defining f as in the previous subcases, inequalities (1) and (2) are verified.
    • If $|B_2| =1$ and either $c\in B_2$ is on the cycle with $m=7$ or $s\in B_2$ is an interior vertex with $t=4,$ the inequalities become equalities. The remaining cases ($B_2=\varnothing$; $|B_2| =1$ with a cycle vertex $c\in B_2$ and $m\ge 9$ or an interior vertex $s\in B_2$ with $t\ge 6;$ and $|B_2| \ge 2$ with all the strong vertices either on the cycle or in the interior) again lead to a contradiction. Therefore, we can assume that there are at least two strong vertices $c,s\in B_2$ (one on the cycle, one interior), completing the proof for this subcase.
• Case (v). $m=5$ and $t\equiv 0\left(\mathrm{mod}2\right)$ with $t\ge 4$.
  • We denote the vertices of the cycle as $c_1,c_2,c_3,c_4,c_5$. We define the function f as follows: let $f\left(c_1\right)=1+⌈{\displaystyle \frac{t+2}{2}}⌉$; $f\left(c_3\right)=f\left(c_4\right)=1$ and $f\left(z\right)=0$, otherwise. The vertex $c_1$ covers all the undefended vertices. Thus, all vertices are strongly Roman dominated, which ensures that f is an StRDF:

    $$w\left(f\right)=⌈{\displaystyle \frac{t+2}{2}}⌉+3=4+{\displaystyle \frac{t}{2}}\ge {\gamma }_{StR}\left(W_{5,t}\right).$$
    To prove the lower bound, let f be a ${\gamma }_{StR}$-function. We know that the lower bound holds, (2).
    If $|B_2|=1$ with a vertex cycle $c\in B_2$, it follows that

    $${\gamma }_{StR}\left(W_{5,t}\right)=w\left(f\right)\ge \left|B_1\right|+f\left(c\right)\ge 1+⌈{\displaystyle \frac{t+2}{2}}⌉+2=4+⌈{\displaystyle \frac{t}{2}}⌉.$$
    Arguing in the same way as in the previous cases, if $B_2=\varnothing$ or $|B_2|=1$ with a strong interior vertex or $|B_2|\ge 2$, we arrive at a contradiction.
• Case (vi). $m,t\equiv 0\left(\mathrm{mod}2\right)$ with $m\ge 6$ and $t\ge 4$.
  • To begin, we define a function f as follows: if $c_1$ is a vertex on the cycle, then $f\left(c_1\right)=1+{\displaystyle \frac{t}{2}}$; if $c_2\in N\left(c_1\right)$, then $f\left(c_2\right)=1;$ if s is an interior vertex, then $f\left(s\right)=1+{\displaystyle \frac{m-2}{2}};$ and $f\left(z\right)=0$ otherwise. Vertex $c_1$ covers the $t-1$ undefended interior vertices and its sole undefended neighbor in the cycle, while s covers the $m-2$ vertices of the cycle. Thus, all vertices are strongly Roman dominated, ensuring f is an StRDF.

    $$w\left(f\right)=1+{\displaystyle \frac{m-2}{2}}+1+1+{\displaystyle \frac{t}{2}}={\displaystyle \frac{m}{2}}+2+{\displaystyle \frac{t}{2}}\ge {\gamma }_{StR}\left(W_{m,t}\right).$$
    For the reverse inequality, we will show by contradiction that we may assume the existence of two strong vertices, one in the interior and the other on the cycle:
    If $B_2=\varnothing ,$ then ${\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)\ge \left|B_1\right|=n=m+t$, which contradicts the upper bound:

    $$\begin{array}{cc}\hfill 2+{\displaystyle \frac{t}{2}}+{\displaystyle \frac{m}{2}}& \ge {\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)=m+t\hfill \\ & ={\displaystyle \frac{t}{2}}+{\displaystyle \frac{t}{2}}+{\displaystyle \frac{m}{2}}+{\displaystyle \frac{m}{2}}\hfill \\ & >{\displaystyle \frac{t}{2}}+1+{\displaystyle \frac{m}{2}}+1.\hfill \end{array}$$
    If $|B_2|=1$ and either $c\in B_2$ is on the cycle with $m=6$ or $s\in B_2$ is an interior vertex with $t=4$, the inequalities of the lower bound (2) become equalities.
  • Proceeding as in the previous cases, if $|B_2| =1$ and either $c\in B_2$ is on the cycle with $m\ge 8$ or $s\in B_2$ is an interior vertex with $t\ge 6$, using the lower bound, (2), we obtain a contradiction with the upper bound.
  • Similarly, if $|B_2| \ge 2$ and all strong vertices are interior vertices or all are on the cycle, following the same arguments, and using the lower and upper bounds, we again arrive at a contradiction.
  • Therefore, we can assume that there are at least two strong vertices $c,s\in B_2$ (one on the cycle, one interior) and therefore,

    $$\begin{array}{cc}\hfill {\gamma }_{StR}\left(W_{m,t}\right)=w\left(f\right)& \ge |B_1|+|B_2|+⌈{\displaystyle \frac{|B_0|}{2}}⌉\ge 1+1+{\displaystyle \frac{t}{2}}+1+{\displaystyle \frac{m-2}{2}}\hfill \\ & ={\displaystyle \frac{t}{2}}+2+{\displaystyle \frac{m}{2}}.\hfill \end{array}$$

This concludes the proof. □

Next, we determine the exact values of the strong Roman domination number for crown graphs $C\left(n\right),$ see Figure 4, whose values depend strictly on the parity of $n.$

Proposition 12.

Let $C\left(n\right)$ be a crown graph with $2n$ vertices, where $n\ge 3$. Then,

$${\gamma }_{StR}\left(C\left(n\right)\right)=\left\{\begin{array}{cc}n+1,& ifn\equiv 1\left(\mathrm{mod}2\right)\\ n+2,& ifn\equiv 0\left(\mathrm{mod}2\right).\end{array}\right.$$

Proof. 

Let $C\left(n\right)$ denote the crown graph with $2n$ vertices. Let A be the independent set containing n vertices $u_1,u_2,\dots ,u_n$, and let B be the subset consisting of the remaining n vertices, denoted by $v_1,v_2,\dots ,v_n$, where each $v_i$ is a neighbor of $u_j,$ for all $j\ne i.$

To obtain an upper bound for ${\gamma }_{StR}\left(C\left(n\right)\right)$, we define a function f by assigning $f\left(u_1\right)=f\left(v_1\right)=1+⌈{\displaystyle \frac{\Delta }{2}}⌉$ and $f\left(z\right)=0$ for the remaining vertices. Thus, the vertex $u_1$ covers the $n-1$ vertices of the set B, and $v_1$ covers the $n-1$ vertices of A under the conditions of strong Roman domination.

Consequently, f is an StRDF and ${\gamma }_{StR}\left(C\left(n\right)\right)\le w\left(f\right)=2+2⌈{\displaystyle \frac{\Delta }{2}}⌉.$ Since $\Delta =n-1,$ the upper bound holds.

Taking into account Proposition 3, we have that

$${\gamma }_{StR}\left(C\left(n\right)\right)\ge ⌈{\displaystyle \frac{2n+1}{2}}⌉=n+1.$$

Therefore, for odd $n\ge 3,$ we obtain the equality.

Let us assume that $n\ge 4$ is an even integer. We now show that $B_2$ contains at least two vertices. If $B_2=\varnothing$ then ${\gamma }_{StR}\left(C\left(n\right)\right)=w\left(f\right)\ge \left|B_1\right|=2n\ge n+4>n+2$, which is a contradiction.

If $|B_2| =1$, say $u_1\in B_2\cap A$, it follows that

$${\gamma }_{StR}\left(C\left(n\right)\right)=w\left(f\right)\ge \left|B_1\right|+f\left(u_1\right)\ge n+1+⌈{\displaystyle \frac{n-1}{2}}⌉\ge n+⌈{\displaystyle \frac{n+1}{2}}⌉\ge n+3>n+2,$$

which is again a contradiction.

Hence, there are at least two strong vertices in $B_2$. Moreover, if $B_2\subseteq A$ or $B_2\subseteq B$ then $w\left(f\right)\ge 2(1+⌈{\displaystyle \frac{n-1}{2}}⌉)+n-2>n+2,$ which is again a contradiction. Therefore, there is at least one vertex in $B_2\cap A$ and one vertex in $B_2\cap B.$ From this, one can straightforwardly deduce that

$${\gamma }_{StR}\left(C\left(n\right)\right)\ge 1+⌈{\displaystyle \frac{n-1}{2}}⌉+1+⌈{\displaystyle \frac{n-1}{2}}⌉=1+{\displaystyle \frac{n}{2}}+1+{\displaystyle \frac{n}{2}}=n+2.$$

□

Proposition 13.

Let $G=C_m\circ K_1$ be the corona product of a cycle and a singleton $K_1$. Then,

$${\gamma }_{StR}(C_m\circ K_1)=\left\{\begin{array}{cc}\hfill {\displaystyle \frac{3m}{2}},\hfill & ifm\equiv 0\left(\mathrm{mod}4\right)\hfill \\ \hfill {\displaystyle \frac{3m+1}{2}},\hfill & ifm\equiv 1\left(\mathrm{mod}2\right)\hfill \\ \hfill {\displaystyle \frac{3m+2}{2}},\hfill & ifm\equiv 2\left(\mathrm{mod}4\right)\hfill \end{array}\right.$$

Proof. 

We construct a labeling on G and show that it is an StRD function, thereby obtaining an upper bound for ${\gamma }_{StR}\left(G\right)$.

For $m=3$ it can be easily verified that ${\gamma }_{StR}(C_3\circ K_1)=5.$ Let $m=4k+r$ with $k\ge 1$ and $r\in \{0,1,2,3\}$. Let $\left\{v_j\right\}$ be the vertices of the cycle and let $\left\{u_j\right\}$ be the pendant vertices, for $j\in \left[m\right],$ where $u_j$ is adjacent to $v_j$. We define the labeling $f:V\left(G\right)\to \{0,1,2\}$ as follows:

• Cycle vertices: For the blocks of four vertices, we define $f\left(v_{4i+1}\right)=f\left(v_{4i+4}\right)=2$ and $f\left(v_{4i+2}\right)=f\left(v_{4i+3}\right)=0$ for all $i\in \{0,1,\dots ,k-1\}$. If $r>0$, the labels of the remaining vertices $v_j$ (for $j>4k$) are defined as:

  $$\left\{\begin{array}{cc}f\left(v_{4k+1}\right)=2\hfill & \mathrm{if}r=1;\hfill \\ f\left(v_{4k+1}\right)=f\left(v_{4k+2}\right)=2\hfill & \mathrm{if}r=2;\hfill \\ f\left(v_{4k+1}\right)=f\left(v_{4k+3}\right)=2,f\left(v_{4k+2}\right)=0\hfill & \mathrm{if}r=3.\hfill \end{array}\right.$$
• Pendant vertices: The labels of the pendant vertices are determined by their neighbors:

  $$f\left(u_j\right)=\left\{\begin{array}{cc}0\hfill & \mathrm{if}f\left(v_j\right)=2,\hfill \\ 1\hfill & \mathrm{if}f\left(v_j\right)=0.\hfill \end{array}\right.$$

Observe that every vertex labeled 0 is adjacent to at least one vertex labeled 2. Furthermore, every vertex labeled 2 has sufficient weight to satisfy the StRDF condition for its, at most, two neighbors labeled 0. Hence, f is an StRDF of G with total weight:

$$w\left(f\right)=\left\{\begin{array}{cc}6k={\displaystyle \frac{3\left(4k\right)}{2}}={\displaystyle \frac{3m}{2}}\hfill & \mathrm{if}r=0;\hfill \\ 6k+2={\displaystyle \frac{3(4k+1)+1}{2}}={\displaystyle \frac{3m+1}{2}}\hfill & \mathrm{if}r=1;\hfill \\ 6k+4={\displaystyle \frac{3(4k+2)+2}{2}}={\displaystyle \frac{3m+2}{2}}\hfill & \mathrm{if}r=2;\hfill \\ 6k+5={\displaystyle \frac{3(4k+3)+1}{2}}={\displaystyle \frac{3m+1}{2}}\hfill & \mathrm{if}r=3.\hfill \end{array}\right.$$

These values match those stated in the proposition. Therefore, the upper bound holds.

Now, let us prove the lower bound. First, consider the case $r=0$. Let f be an StRD function such that $|B_1|$ is maximum. We claim that:

• There is no vertex $v\in V\left(G\right)$ such that $f\left(v\right)=3.$ Otherwise, v must be a vertex of the cycle, say $v=v_j.$ Then, we would have $f\left(v_j\right)=3$ and $f\left(v_{j-1}\right)=f\left(v_{j+1}\right)=f\left(u_j\right)=0.$ We define the following function $g:$$g\left(v_j\right)=2,g\left(u_j\right)=1$ and $g\left(z\right)=f\left(z\right)$, otherwise. The new function g is an StRD function with the same weight as f but $|B_1^g| > |B_1^f|,$ which is a contradiction.
• Every vertex v with a label 2 has exactly two neighbors having a label 0. Otherwise, if v would have only one neighbor $v^{\prime }$ in $B_0$, the function defined by $g\left(v\right)=g\left(v^{\prime }\right)=1$ would be an StRDF of minimum weight with more vertices labeled 1 than $f,$ yielding again a contradiction.
• As a consequence, every vertex with a label 2 is a vertex $v_j$ of the cycle $C_m;$ its leaf $u_j$ has a label 0; one of its neighbors in the cycle, say $v_{j-1}$, must have a label $0;$ and the corresponding leaf $u_{j-1}$ must have a label $1.$

We now apply the following discharging procedure. Consider $s_0\left(v\right)=f\left(v\right)$ as the initial charge for each vertex v of G. We apply the following rules to obtain the new value of the charges $s\left(v\right)$:

• Each vertex $v_j$ of the cycle with a label 0 receives a charge of 1 unit from its leaf with a label $1,$ so $s_0\left(v_j\right)$ increases by 1 and $s\left(u_j\right)=0.$
• Each vertex $v_j$ of the cycle with a label 2 transfers a charge of ${\displaystyle \frac{1}{2}}$ unit to its neighbor $v_{j-1}$ on the cycle having a label 0, hence $s\left(v_j\right)={\displaystyle \frac{3}{2}}$ and $s_0\left(v_{j-1}\right)$ increases by ${\displaystyle \frac{1}{2}}$.
• Each vertex $v_j$ of the cycle with a label 1, if any, receives ${\displaystyle \frac{1}{2}}$ unit from its leaf $u_j,$ which also has a label 1 because f has minimum weight. So, $s\left(v_j\right)={\displaystyle \frac{3}{2}}$ and $s\left(u_j\right)={\displaystyle \frac{1}{2}}.$

In summary, by this procedure, the total charge does not vary, i.e., ${\sum }_{v\in V}s\left(v\right)$$={\sum }_{v\in V}f\left(v\right)$. After applying the rules to all the vertices of G, we have that:

$$\begin{array}{cc}\hfill {\gamma }_{StR}\left(G\right)& ={\displaystyle \sum _{v\in V}}f\left(v\right)={\displaystyle \sum _{v\in V}}s\left(v\right)\hfill \\ & ={\displaystyle \sum _{v\in B_0}}s\left(v\right)+{\displaystyle \sum _{v\in B_1}}s\left(v\right)+{\displaystyle \sum _{v\in B_2}}s\left(v\right)\hfill \\ & \ge {\displaystyle \sum _{v\in B_0\cap V\left(C_m\right)}}\left(f\left(v\right)+1+{\displaystyle \frac{1}{2}}\right)+{\displaystyle \sum _{v\in B_1\cap V\left(C_m\right)}}\left(f\left(v\right)+{\displaystyle \frac{1}{2}}\right)+{\displaystyle \sum _{v\in B_2\cap V\left(C_m\right)}}\left(f\left(v\right)-{\displaystyle \frac{1}{2}}\right)\hfill \\ & =|B_0\cap V\left(C_m\right)|{\displaystyle \frac{3}{2}}+|B_1\cap V\left(C_m\right)|\left(1+{\displaystyle \frac{1}{2}}\right)+|B_2\cap V\left(C_m\right)|\left(2-{\displaystyle \frac{1}{2}}\right)\hfill \\ & ={\displaystyle \frac{3}{2}}\left|V\left(C_m\right)\right|={\displaystyle \frac{3m}{2}}.\hfill \end{array}$$

The proof for the other two cases of congruences is quite similar. □

Remark 3.

Let G be a connected graph, f a ${\gamma }_{StR}$-function, v a support vertex of the graph, and $\{u_1,\dots ,u_p\}$ be the set of its leaves, with $p\ge 2.$ Then,

$$f\left(v\right)+{\displaystyle \sum _{1\le i\le p}}f\left(u_i\right)\ge 1+⌈{\displaystyle \frac{p}{2}}⌉.$$

Now, we establish a specific result for a particular class of caterpillar graphs, which is subsequently extended to a more general setting, see Figure 5.

Proposition 14.

Let C be a caterpillar graph whose central path consists of n vertices $v_i,$ each having $x_i$ leaves, where $1\le i\le n$ and $x_i\ge 2$. Then,

$${\gamma }_{StR}\left(C\right)=\sum _{i=1}^n\left(1+⌈{\displaystyle \frac{x_i}{2}}⌉\right).$$

Proof. 

Let C be the caterpillar graph as stated in the proposition. Let us define a function f by setting $f\left(v_i\right)=\left(1+⌈{\displaystyle \frac{x_i}{2}}⌉\right)$ for the vertices $v_i$ ($1\le i\le n$) on the central path; and $f\left(z\right)=0$, otherwise. This labelling convention ensures that f is an StRDF and, therefore,

$${\gamma }_{StR}\left(C\right)\le \sum _{i=1}^n\left(1+⌈{\displaystyle \frac{x_i}{2}}⌉\right).$$

On the other hand, by Remark 3, since the vertices $v_i$ are strong support vertices, we know that $f\left(N\left[v_i\right]\right)\ge 1+⌈{\displaystyle \frac{x_i}{2}}⌉$, $1\le i\le n.$ Therefore,

$${\gamma }_{StR}\left(C\right)=\sum _{i=1}^{n}f\left(N\left[v_i\right]\right)\ge \sum _{i=1}^n\left(1+⌈{\displaystyle \frac{x_i}{2}}⌉\right),$$

and the result follows. □

To conclude this section, we provide an upper bound for the StRDN of caterpillars. To this end, let C be a caterpillar whose central path consists of n vertices $v_i$. We denote by $x_i$ the number of leaves attached to the vertex $v_i$ where $x_i\ge 2$, and let $P_{r_i}$ be the induced subpaths of length $r_i$ that may occur between the support vertices $v_i$ and $v_{i+1}$.

Proposition 15.

Let C be a caterpillar whose central path consists of n strong support vertices $v_i$ (having $x_i\ge 2$ leaves each); k weak support vertices; and q induced subpaths of length $r_j$ whose vertices are not support vertices of the caterpillar. Then,

$${\gamma }_{StR}\left(C\right))\le \sum _{i=1}^n\left(1+⌈{\displaystyle \frac{x_i}{2}}⌉\right)+2k+\sum _{j=1}^q⌈{\displaystyle \frac{2r_j}{3}}⌉.$$

Proof. 

To obtain this bound, we define a function f on the caterpillar, ensuring that it is an StRDF. For vertices $v_i$ with two or more leaves, we define $f\left(v_i\right)=1+⌈{\displaystyle \frac{x_i}{2}}⌉$; $f\left(z\right)=2,$ for all weak support vertices; and we label each intermediate subpath of length $r_j,$ for $j\in \left[q\right]$, with a ${\gamma }_{StR}$-function $h_j$ having weight $h_j\left(P_{r_j}\right)=⌈{\displaystyle \frac{2r_j}{3}}⌉.$ All vertices are thus covered under strong Roman domination and, therefore, we obtain:

$${\gamma }_{StR}\left(C\right)\le \sum _{i=1}^n\left(1+⌈{\displaystyle \frac{x_i}{2}}⌉\right)+\sum _{l=1}^{k}f\left(u_j\right)+\sum _{j=1}^q{h}_j\left(P_{r_j}\right)=\sum _{i=1}^n\left(1+⌈{\displaystyle \frac{x_i}{2}}⌉\right)+2k+\sum _{j=1}^q⌈{\displaystyle \frac{2r_j}{3}}⌉.$$

□