Total Roman Domination Edge-Critical Graphs

TL;DR

This paper investigates the properties and classifications of graphs that are critical with respect to total Roman domination number, providing characterizations and exploring their relationships with other domination concepts.

Contribution

It introduces the concepts of $oldsymbol{ ext{k-}oldsymbol{ extgamma_{tR}}}$-edge-critical and supercritical graphs, offering characterizations and linking them to domination and total domination critical graphs.

Findings

Characterization of certain classes of $oldsymbol{ extgamma_{tR}}$-edge-critical graphs.

Establishment of connections between $k$-$oldsymbol{ extgamma_{tR}}$-edge-critical graphs and domination critical graphs.

Basic results on properties of $oldsymbol{ extgamma_{tR}}$-edge-critical graphs.

Abstract

A total Roman dominating function on a graph $G$ is a function $% f:V(G)\rightarrow \{0,1,2\}$ such that every vertex $v$ with $f(v)=0$ is adjacent to some vertex $u$ with $f(u)=2$, and the subgraph of $G$ induced by the set of all vertices $w$ such that $f(w)>0$ has no isolated vertices. The weight of $f$ is $\Sigma _{v\in V(G)}f(v)$. The total Roman domination number $\gamma _{tR}(G)$ is the minimum weight of a total Roman dominating function on $G$. A graph $G$ is $k$-$\gamma _{tR}$-edge-critical if $\gamma _{tR}(G+e)<\gamma _{tR}(G)=k$ for every edge $e\in E(\overline{G})\neq \emptyset $, and $k$-$\gamma _{tR}$-edge-supercritical if it is $k$-$\gamma _{tR}$-edge-critical and $\gamma _{tR}(G+e)=\gamma _{tR}(G)-2$ for every edge $e\in E(\overline{G})\neq \emptyset $. We present some basic results on $\gamma_{tR}$-edge-critical graphs and characterize certain classes of $\gamma _{tR}$-edge-critical graphs. In addition, we show that, when $k$ is small, there is a connection between $k$-$\gamma _{tR}$-edge-critical graphs and graphs which are critical with respect to the domination and total domination numbers.

Equations

$${\gamma }_{tR}(S)=\{\begin{array}{ll}n& \text{if y≥k−1}\\ n-k+y+1& \text{if 1≤y<k−1}\\ n-k+2& \text{if y=0.}\end{array}$$

$$|S|-|V_f^1|$$

$$=x^2-2(n-1)x+(n-1{)}^2$$

$$=(n-1-x{)}^2$$

$$>0.$$

$$2\left(\lfloor \frac{l}{2}\rfloor +1\right)=\{\begin{array}{ll}l+2& \text{if }l\text{ is even}\\ l+1& \text{if }l\text{ is odd,}\end{array}$$

Full text

Keywords: Roman domination; total Roman domination; total Roman domination edge-critical graphs

AMS Subject Classification Number 2010: 05C69

1 Introduction

We consider the behaviour of the total Roman domination number of a graph $G$ upon the addition of edges to $G$. A dominating set $S$ in a graph $G$ is a set of vertices such that every vertex in $V(G)-S$ is adjacent to at least one vertex in $S$. The domination number $\gamma(G)$ is the cardinality of a minimum dominating set in $G$. A total dominating set $S$ (abbreviated by TD-set) in a graph $G$ with no isolated vertices is a set of vertices such that every vertex in $V(G)$ is adjacent to at least one vertex in $S$. The total domination number $\gamma_{t}(G)$ (abbreviated by TD-number) is the cardinality of a minimum total dominating set in $G$. For $S\subseteq V(G)$ and a function $f:S\rightarrow\mathbb{R}$, define $f(S)=\Sigma_{s\in S}f(s)$. A Roman dominating function (abbreviated by RD-function) on a graph $G$ is a function $f:V(G)\rightarrow\{0,1,2\}$ such that every vertex $v$ with $f(v)=0$ is adjacent to some vertex $u$ with $f(u)=2$. The weight of $f$, $\omega(f)$, is defined as $f(V(G))$. The Roman domination number $\gamma_{R}(G)$ (abbreviated by RD-number) is defined as $\min\{\omega(f):f\text{ is an RD-function on }G\}$. For an RD-function $f$, let $V_{f}^{i}=\{v\in V(G):f(v)=i\}$ and $V_{f}^{+}=V_{f}^{1}\cup V_{f}^{2}$. Thus, we can uniquely express an RD-function $f$ as $f=(V_{f}^{0},V_{f}^{1},V_{f}^{2})$.

As defined by Ahanger, Henning, Samodivkin and Yero [1], a total Roman dominating function (abbreviated by TRD-function) on a graph $G$ with no isolated vertices is a Roman dominating function with the additional condition that $G[V_{f}^{+}]$ has no isolated vertices. The total Roman domination number $\gamma_{tR}(G)$ (abbreviated by TRD-number) is the minimum weight of a TRD-function on $G$, that is, $\gamma_{tR}(G)=\min\{\omega(f):f\text{ is a TRD-function on }G\}$. A TRD-function $f$ such that $\omega(f)=\gamma_{tR}(G)$ is called a $\gamma_{tR}(G)$-function, or a $\gamma_{tR}$-function if the graph $G$ is clear from the context; $\gamma_{R}$-functions are defined analogously.

The addition of an edge to a graph has the potential to change its total domination or Roman domination number. Van der Merwe, Mynhardt and Haynes [10] studied $\gamma_{t}$-edge-critical graphs, that is, graphs $G$ for which $\gamma_{t}(G+e)<\gamma_{t}(G)$ for each $e\in E(\overline{G})$ and $E(\overline{G})\neq\emptyset$. We consider the same concept for total Roman domination. A graph $G$ is total Roman domination edge-critical, or simply $\gamma_{tR}$-edge-critical, if $\gamma_{tR}(G+e)<\gamma_{tR}(G)$ for every edge $e\in E(\overline{G})$ and $E(\overline{G})\neq\emptyset$. We say that $G$ is $k$-$\gamma_{tR}$-edge-critical if $\gamma_{tR}(G)=k$ and $G$ is $\gamma_{tR}$-edge-critical. If $\gamma_{tR}(G+e)\leq\gamma_{tR}(G)-2$ for every edge $e\in E(\overline{G})$ and $E(\overline{G})\neq\emptyset$, we say that $G$ is $\gamma_{tR}$-edge-supercritical. If $\gamma_{tR}(G+e)=\gamma_{tR}(G)$ for all $e\in E(\overline{G})$, or $E(\overline{G})=\emptyset$, we say that $G$ is stable.

Pushpam and Padmapriea [11] established bounds on the total Roman domination number of a graph in terms of its order and girth. Total Roman domination in trees was studied by Amjadi, Nazari-Moghaddam, Sheikholeslami and Volkmann [3], as well as by Amjadi, Sheikholeslami and Soroudi [4]. The authors of [4] also studied Nordhaus-Gaddum bounds for total Roman domination in [5]. Campanelli and Kuziak [6] considered total Roman domination in the lexicographic product of graphs. We refer the reader to the well-known books [7] and [8] for graph theory concepts not defined here. Frequently used or lesser known concepts are defined where needed.

We begin with some general results regarding the addition of an edge $e\in E(\overline{G})$ to a graph $G$ in Section 2. In Section 3, we characterize $n$-$\gamma_{tR}$-edge-critical graphs of order $n$. We characterize $4$-$\gamma_{tR}$-edge-critical graphs in Section 4, and, after investigating $\gamma_{tR}$-edge-supercritical graphs in Section 5, we present a necessary condition for $5$-$\gamma_{tR}$-edge-critical graphs in Section 6. In Section 7, we determine the total Roman domination number of spiders and characterize $\gamma_{tR}$-edge-critical spiders. As can be expected, every graph $G$ with $\gamma_{tR}(G)=k\geq 4$ is a spanning subgraph of a $k$-$\gamma_{tR}(G)$-edge-critical graph; a short proof is given in Section 8, where we also show that for any $k\geq 4$, there exists a $k$-$\gamma_{tR}$-edge-critical graph of diameter $2$. We conclude in Section 9 with ideas for future research.

2 Adding an edge

We begin with a result from [9] which bounds the effect the addition of an edge can have on the total domination number of a graph and show that the same bounds hold with respect to the total Roman domination number.

Proposition 2.1**.**

\thlabel

Myn1 [9] For a graph $G$ with no isolated vertices, if $uv\in E(\overline{G})$, then $\gamma_{t}(G)-2\leq\gamma_{t}(G+uv)\leq\gamma_{t}(G)$.

An edge $uv\in E(\overline{G})$ is critical if $\gamma_{tR}(G+uv)<\gamma_{tR}(G)$. The following proposition restricts the possible values assigned to the vertices of a critical edge $uv$ by a $\gamma_{tR}(G+uv)$-function $f$, which will be useful in proving subsequent results. For a graph $G$ and a vertex $v\in V(G)$, the open neighbourhood of $v$ in $G$ is $N_{G}(v)=\{u\in V(G):uv\in E(G)\}$, and the closed neighbourhood of $v$ in $G$ is $N_{G}[v]=N_{G}(v)\cup\{v\}$. When $G\neq K_{2}$, the unique neighbour of an end-vertex of $G$ is called a support vertex.

Proposition 2.2**.**

\thlabel

set added edge Given a graph $G$ with no isolated vertices, if $uv\in E(\overline{G})$ is a critical edge and $f$ is a $\gamma_{tR}(G+uv)$-function, then $\{f(u),f(v)\}\in\{\{2,2\},\{2,1\},\{2,0\},\{1,1\}\}$. If, in addition, $\deg(u)=\deg(v)=1$, then there exists a $\gamma_{tR}(G+uv)$-function $f$ such that $f(u)=f(v)=1$.

Proof. Let $G$ be a graph with no isolated vertices, $uv\in E(\overline{G})$ such that $\gamma_{tR}(G+uv)<\gamma_{tR}(G)$, and $f$ a $\gamma_{tR}$-function on $G+uv$. Suppose for a contradiction that $\{f(u),f(v)\}\notin\{\{2,2\},\{2,1\},\{2,0\},\{1,1\}\}$. Then $\{f(u),f(v)\}\in\{\{0,0\},\{0,1\}\}$. Note that, in either case, the edge $uv$ cannot affect whether $u$ and $v$ are dominated, or whether, in the case where (say) $f(v)=1$, $v$ is isolated. Hence $f$ is a TRD-function of $G$, contradicting $\gamma_{tR}(G+uv)<\gamma_{tR}(G)$. Therefore $\{f(u),f(v)\}\in\{\{2,2\},\{2,1\},\{2,0\},\{1,1\}\}$.

Now, suppose in addition that $\deg(u)=\deg(v)=1$, and let $f$ be a $\gamma_{tR}(G+uv)$-function such that $|V_{f}^{2}|$ is as small as possible. Let $w$ and $x$ be the unique neighbours of $u$ and $v$, respectively, noting that possibly $w=x$. Suppose for a contradiction that $f(u)=2$ (without loss of generality). If $f(v)=0$, then $f(w)>0$, otherwise $u$ would be isolated in $G[V_{f}^{+}]$. Thus, regardless of whether $w=x$ or not, consider the function $f^{\prime}:V(G)\rightarrow\{0,1,2\}$ defined by $f^{\prime}(u)=f^{\prime}(v)=1$ and $f^{\prime}(y)=f(y)$ for all other $y\in V(G)$. Otherwise, if $f(v)\geq 1$, then clearly $f(w)=0$. Thus, regardless of whether $w=x$ or not, consider the function $f^{\prime}:V(G)\rightarrow\{0,1,2\}$ defined by $f^{\prime}(u)=f^{\prime}(w)=1$ and $f^{\prime}(y)=f(y)$ for all other $y\in V(G)$. In either case, $f^{\prime}$ is a $\gamma_{tR}$-function on $G+uv$. However, $|V_{f^{\prime}}^{2}|<|V_{f}^{2}|$, contradicting $|V_{f}^{2}|$ being as small as possible. Hence $f(u)\neq 2$, and thus $f(u)=f(v)=1$. $\square$

Proposition 2.3**.**

\thlabel

tR bounds Given a graph $G$ with no isolated vertices, if ${uv\in E(\overline{G})}$, then $\gamma_{tR}(G)-2\leq\gamma_{tR}(G+uv)\leq\gamma_{tR}(G)$.

Proof. Let $G$ be a graph with no isolated vertices. Clearly, adding an edge cannot increase the total Roman domination number, hence the upper bound holds. Now, let $uv\in E(\overline{G})$. Note that when $\gamma_{tR}(G+uv)=\gamma_{tR}(G)$ the lower bound clearly holds. So assume $\gamma_{tR}(G+uv)<\gamma_{tR}(G)$ and let $f$ be a $\gamma_{tR}(G+uv)$-function. By \threfset added edge, $\{f(u),f(v)\}\in\{\{2,2\},\{2,1\},\{2,0\},\{1,1\}\}$.

First assume $\{f(u),f(v)\}\in\{\{2,2\},\{2,1\},\{1,1\}\}$. Then $f$ is a RD-function of $G$, and the only possible isolated vertices in $G[V_{f}^{+}]$ are $u$ and $v$. Consider the function $f^{\prime}:V(G)\rightarrow\{0,1,2\}$ defined as follows: If $u$ is isolated in $G[V_{f}^{+}]$, choose $u^{\prime}\in N_{G}(u)$ and let $f^{\prime}(u^{\prime})=1$. Similarly, if $v$ is isolated in $G[V_{f}^{+}]$, choose $v^{\prime}\in N_{G}(v)$ and let $f^{\prime}(v^{\prime})=1$. Let $f^{\prime}(x)=f(x)$ for all other $x\in V(G)$. Now, assume instead that $f(u)=2$ and $f(v)=0$ (without loss of generality). Since $u$ is not isolated in $G[V_{f}^{+}]$, $f$ is a TRD-function of $G-v$. Consider the function $f^{\prime}:V(G)\rightarrow\{0,1,2\}$ defined as follows: Let $f^{\prime}(v)=1$. Then, if $v$ is isolated in $G[V_{f^{\prime}}^{+}]$, choose $v^{\prime}\in N_{G}(v)$ and let $f^{\prime}(v^{\prime})=1$. Let $f^{\prime}(x)=f(x)$ for all other $x\in V(G)$. In either case, $f^{\prime}$ is a TRD-function of $G$ and $\omega(f^{\prime})\leq\gamma_{tR}(G+uv)+2$. Thus $\gamma_{tR}(G)\leq\gamma_{tR}(G+uv)+2$, and hence the lower bound holds. $\square$

3 $\gamma_{tR}$-Edge-critical graphs with large TRD-numbers

We now investigate the $\gamma_{tR}$-edge-critical graphs $G$ which have the largest TRD-number, namely $|V(G)|$. A subdivided star is a tree obtained from a star on at least three vertices by subdividing each edge exactly once. A double star is a tree obtained from two disjoint non-trivial stars by joining the two central vertices (choosing either central vertex in the case of $K_{2}$). The corona $\mathop{\mathrm{c}or}(G)$ (sometimes denoted by $G\circ K_{1}$) of $G$ is obtained by joining each vertex of $G$ to a new end-vertex.

Connected graphs $G$ for which $\gamma_{tR}(G)=|V(G)|$ were characterized in [1]. There, Ahanger et al. defined $\mathcal{G}$ as the family of connected graphs obtained from a $4$-cycle $v_{1},v_{2},v_{3},v_{4},v_{1}$ by adding $k_{1}+k_{2}\geq 1$ vertex-disjoint paths $P_{2}$, and joining $v_{i}$ to the end of $k_{i}$ such paths, for $i\in\{1,2\}$. Note that possibly $k_{1}=0$ or $k_{2}=0$. Furthermore, they defined $\mathcal{H}$ to be the family of graphs obtained from a double star by subdividing each pendant edge once and the non-pendant edge $r\geq 0$ times. For $r\geq 0$, define $\mathcal{H}_{r}\subseteq\mathcal{H}$ as the family of graphs in $\mathcal{H}$ where the non-pendant edge was subdivided $r$ times.

Proposition 3.1**.**

\thlabel

Hen1 [1] If $G$ is a connected graph of order $n\geq 2$, then $\gamma_{tR}(G)=n$ if and only if one of the following holds.

1. $(i)$

$G$* is a path or a cycle.* 2. $(ii)$

$G$* is the corona of a graph.* 3. $(iii)$

$G$* is a subdivided star.* 4. $(iv)$

$G\in\mathcal{G}\cup\mathcal{H}$.

Using \threfHen1, we characterize connected $n$-$\gamma_{tR}$-edge-critical graphs as follows.

Theorem 3.2**.**

\thlabel

n edge-crit A connected graph $G$ of order $n\geq 4$ is $n$-$\gamma_{tR}$-edge-critical if and only if $G$ is one of the following graphs:

1. $(i)$

$C_{n}$, $n\geq 4$, 2. $(ii)$

$\mathop{\mathrm{c}or}(K_{r})$, $r\geq 3$, 3. $(iii)$

a subdivided star of order $n\geq 7$, 4. $(iv)$

$G\in\mathcal{G}$, 5. $(v)$

$G\in\mathcal{H}-\mathcal{H}_{0}-\mathcal{H}_{2}$.

Proof. Let $G$ be a connected graph of order $n\geq 4$ with $\gamma_{tR}(G)=n$. First, suppose $G$ is any of the graphs listed in $(i)-(v)$ above. Then, for any $e\in E(\overline{G})$, $G+e$ is not one of the graphs listed in \threfHen1. Therefore $\gamma_{tR}(G+e)<n$ for all $e\in E(\overline{G})$, and thus $G$ is $\gamma_{tR}$-edge-critical.

Otherwise, suppose $G$ is not one of the graphs listed in $(i)-(v)$ above. Note that since $\gamma_{tR}(G)=n$, $G$ is still listed in \threfHen1 $(i)-(iv)$. If $G\cong P_{n}:v_{1},...,v_{n}$, $n\geq 4$, then $G+v_{1}v_{n}\cong C_{n}$ and $\gamma_{tR}(G)=\gamma_{tR}(C_{n})=n$. If $G\cong\mathop{\mathrm{c}or}(F)$, where $F$ is not a complete graph of order at least $3$, then $\gamma_{tR}(G)=\gamma_{tR}(G+uv)$ for any $uv\in E(\overline{F})$. If $G$ is a subdivided star of order less than $7$, then $G=P_{5}$. In each of these cases, $G$ is clearly not $\gamma_{tR}$-edge-critical.

Now consider $G\in\mathcal{H}$. Let $w_{1},...,w_{k}$ be the leaves of $G$, $u_{1},...,u_{k}$ be their respective support vertices, and $v_{1},...,v_{m}$ be the path such that $v_{1}$ and $v_{m}$ are the two support vertices in the original double star $S$, labelled so that $w_{1}$ is adjacent, in $S$, to $v_{1}$. Note that $m=r+2$, and therefore $m\geq 2$. If $G\in\mathcal{H}_{0}$, consider the graph $G+v_{2}w_{1}$, and note that $G+v_{2}w_{1}\in\mathcal{G}$. Therefore, by \threfHen1, $\gamma_{tR}(G+v_{2}w_{1})=n$, and thus $G$ is not $\gamma_{tR}$-edge-critical. Similarly, if $G\in\mathcal{H}_{2}$, consider the graph $G+v_{1}v_{4}$, and note that $G+v_{1}v_{4}\in\mathcal{G}$. Therefore, by \threfHen1, $\gamma_{tR}(G+v_{1}v_{4})=n$, and again $G$ is not $\gamma_{tR}$-edge-critical. $\square$

4 $4$-$\gamma_{tR}$-Edge-critical graphs

Before we characterize the graphs $G$ such that $\gamma_{tR}(G)=4$ and $\gamma_{tR}(G+e)=3$ for any $e\in E(\overline{G})$ (that is, the graphs which are $4$-$\gamma_{tR}$-edge-critical), we present the following result from [11] which characterizes the graphs with a total Roman domination number of $3$, the smallest possible TRD-number. Note that while the authors required that $G$ has girth $3$, the result actually holds in general for any graph $G$ on at least $3$ vertices, as we now show. A *universal vertex *of $G$ is a vertex that is adjacent to all other vertices of $G$.

Proposition 4.1**.**

\thlabel

tR=3 For a graph $G$ of order $n\geq 3$ with no isolated vertices, $\gamma_{tR}(G)=3$ if and only if $\Delta(G)=n-1$, that is, $G$ has a universal vertex.

Proof. Suppose $\gamma_{tR}(G)=3$ and let $f=(V_{f}^{0},V_{f}^{1},V_{f}^{2})$ be a $\gamma_{tR}(G)$-function. If $V_{f}^{2}=\emptyset$, then $|V_{f}^{1}|=3$, and thus $n=3$. Since $G$ has no isolated vertices, this implies that $G=K_{3}$ or $P_{3}$, both of which have a universal vertex. Otherwise, assume $|V_{f}^{2}|=1$ and $|V_{f}^{1}|=1$. Pick $u,v\in V(G)$ so that $f(u)=1$ and $f(v)=2$. Since $G[V_{f}^{+}]$ has no isolated vertices, $uv\in E(G)$. Furthermore, since $\gamma_{tR}(G)=3$, $f(x)=0$ for all other $x\in V(G)$. Therefore $N_{G}[v]=V(G)$, and thus $v$ is a universal vertex.

Conversely, suppose $G$ has a universal vertex $v$, and take any $u\in N_{G}(v)$. Consider the TRD-function $f:V(G)\rightarrow\{0,1,2\}$ defined by $f(v)=2$, $f(u)=1$, and $f(x)=0$ for all other $x\in V(G)$. Since $G$ has at least three vertices, $\gamma_{tR}(G)>2$. Therefore, since $\omega(f)=3$, we conclude that $\gamma_{tR}(G)=3$. $\square$

A galaxy is defined as the disjoint union of two or more non-trivial stars. The characterization of $4$-$\gamma_{tR}$-edge-critical graphs follows; note that this class of graphs is exactly the class of $2$-$\gamma$-edge-critical graphs, as characterized by Sumner and Blitch [12].

Theorem 4.2**.**

\thlabel

4-tR edge-crit A graph $G$ with no isolated vertices is $4$-$\gamma_{tR}$-edge-critical if and only if $\overline{G}$ is a galaxy.

Proof. Let $G$ be a graph of order $n$ with no isolated vertices. Suppose first that $G$ is $4$-$\gamma_{tR}$-edge-critical. Then for any $e\in E(\overline{G})$, $\gamma_{tR}(G+e)=3$, and thus \threftR=3 implies that the addition of any edge to $G$ creates a universal vertex. Therefore, for each edge $uv\in E(\overline{G})$, one of $u$ and $v$ has degree $n-2$ in $G$; that is, one of $u$ and $v$ is a leaf in $\overline{G}$. Since each edge of $\overline{G}$ connects a leaf to either a support vertex or another leaf, the components of $\overline{G}$ are non-trivial stars. Moreover, $\overline{G}$ has at least two components, otherwise $G$ has an isolated vertex.

Conversely, suppose $\overline{G}$ is a galaxy. Since $\overline{G}$ has no isolated vertices, $G$ has no universal vertices, and thus, by \threftR=3, $\gamma_{tR}(G)>3$. Let $u$ and $v$ be vertices in different components of $\overline{G}$, and define $f:V(G)\rightarrow\{0,1,2\}$ by $f(u)=f(v)=2$ and $f(x)=0$ for all other $x\in V(G)$. Clearly $f$ is a TRD-function on $G$, and hence $\gamma_{tR}(G)=4$. Since the deletion of any edge in $\overline{G}$ produces an isolated vertex, the addition of any edge to $G$ creates a universal vertex. Therefore, by \threftR=3, $\gamma_{tR}(G+e)=3$ for all $e\in E(\overline{G})$, and hence $G$ is $4$-$\gamma_{tR}$-edge-critical. $\square$

Corollary 4.3**.**

If $G$ is a connected $(n-2)$-regular graph, then $G$ is $4$-$\gamma_{tR}$-edge-critical.

Having characterized $4$-$\gamma_{tR}$-edge-critical graphs, our next result demonstrates the existence of stable graphs with total Roman domination number $4$.

Proposition 4.4**.**

\thlabel

n-3 reg If $G$ is an $(n-3)$-regular graph of order $n\geq 6$, then $\gamma_{tR}(G)=4$. Moreover, $G$ is stable.

Proof. We prove that $\gamma(G)=2$. Since $G$ is $(n-3)$-regular, its complement $\overline{G}$ is $2$-regular. If $\overline{G}$ is disconnected, let $u$ and $v$ be vertices in different components of $\overline{G}$. Otherwise, if $\overline{G}$ is connected, then $\overline{G}\cong C_{n}$, $n\geq 6$, and thus we can choose $u,v\in V(\overline{G})$ such that $d_{\overline{G}}(u,v)\geq 3$. In either case, $N_{\overline{G}}[u]\cap N_{\overline{G}}[v]=\emptyset$. In $G$, $u$ dominates all vertices in $G-N_{\overline{G}}(u)$ and $v$ dominates all vertices in $G-N_{\overline{G}}(v)$. Therefore $\{u,v\}$ dominates $G$, and thus, since $G$ has no universal vertex, $\gamma(G)=2$.

Now, define $f:V(G)\rightarrow\{0,1,2\}$ by $f(u)=f(v)=2$ and $f(y)=0$ for all other $y\in V(G)$. Since $uv\in E(G)$, $f$ is a TRD-function on $G$ and $\omega(f)=4$, so $\gamma_{tR}(G)\leq 4$. Since $G$ has no universal vertex, $\gamma_{tR}(G)>3$ by \threftR=3, and thus $\gamma_{tR}(G)=4$, as required. Furthermore, since the addition of any edge to $G$ does not create a universal vertex, it follows from \threftR=3 that $\gamma_{tR}(G+e)=\gamma_{tR}(G)$ for all $e\in E(\overline{G})$. Therefore $G$ is stable. $\square$

5 $\gamma_{tR}$-Edge-supercritical graphs

We now consider the graphs $G$ which attain the lower bound in \threftR bounds for all $e\in E(\overline{G})$, that is, $\gamma_{tR}$-edge-supercritical graphs. An edge $uv\in E(\overline{G})$ is supercritical if $\gamma_{tR}(G+uv)=\gamma_{tR}(G)-2$. Haynes, Mynhardt and Van der Merwe [9] defined a graph $G$ to be $\gamma_{t}$-edge-supercritical if $\gamma_{t}(G+e)=\gamma_{t}(G)-2$ for all $e\in E(\overline{G})$. We begin with their characterization of $\gamma_{t}$-edge-supercritical graphs.

Proposition 5.1**.**

\thlabel

Myn2 [9] A graph $G$ is $\gamma_{t}$-edge-supercritical if and only if $G$ is the union of two or more non-trivial complete graphs.

The proof of the previous result relies on the fact that, if $u$ and $v$ are vertices of a graph $G$ with $d(u,v)=2$, then $\gamma_{t}(G)-1\leq\gamma_{t}(G+uv)$. However, the analogous result does not hold with respect to the total Roman domination number, as we now show. Consider the graph $G=\mathop{\mathrm{c}or}(K_{3})$. By \threfHen1, $\gamma_{tR}(G)=6$. Consider any two non-adjacent vertices $u$ and $v$ in $G$ such that $\deg(u)=1$ and $\deg(v)=3$. Clearly $uv$ is a supercritical edge with $d(u,v)=2$, and thus $d(u,v)=2$ does not always imply that $\gamma_{tR}(G)-1\leq\gamma_{tR}(G+uv)$.

As a result, the classification of $\gamma_{tR}$-edge-supercritical graphs will be less straightforward than that of $\gamma_{t}$-edge-supercritical graphs. However, it is easy to see that there are no $5$-$\gamma_{tR}$-edge-supercritical graphs, the smallest possible TRD-number of a $\gamma_{tR}$-edge-supercritical graph, and that the disjoint union of two or more complete graphs of order at least $3$ is $\gamma_{tR}$-edge-supercritical.

Proposition 5.2**.**

\thlabel

no 5-super

2. $(i)$

There are no $5$-$\gamma_{tR}$-edge-supercritical graphs. 3. $(ii)$

If $G$ is the disjoint union of $k\geq 2$ complete graphs, each of order at least $3$, then $G$ is $3k$-$\gamma_{tR}$-edge-supercritical.

Proof.

1. $(i)$

Suppose for a contradiction that $G$ is a $5$-$\gamma_{tR}$-edge-supercritical graph. Then $\gamma(G+uv)=3$ for any edge $uv\in E(\overline{G})$. However, as in the proof of \thref4-tR edge-crit, this implies that $\overline{G}$ is a galaxy, that is, $G$ is $4$-$\gamma_{tR}$-edge-critical, a contradiction. 2. $(ii)$

It follows from \threftR=3 that $\gamma_{tR}(G)=3k$. Moreover, joining any two vertices in different components of $G$ results in a graph with TRD-number $3k-2$. $\square$

6 $5$-$\gamma_{tR}$-Edge-critical graphs

We now investigate the graphs which are $5$-$\gamma_{tR}$-edge-critical. We begin with the following results from [1], which bound $\gamma_{tR}(G)$ in terms of $\gamma_{t}(G)$.

Proposition 6.1**.**

\thlabel

Hen2 [1] If $G$ is a graph with no isolated vertices, then $\gamma_{t}(G)\leq\gamma_{tR}(G)\leq 2\gamma_{t}(G)$. Furthermore, $\gamma_{tR}(G)=\gamma_{t}(G)$ if and only if $G$ is the disjoint union of copies of $K_{2}$.

Note that Amjadi et al. [3] characterized the trees which attain the upper bound in \threfHen2.

Proposition 6.2**.**

\thlabel

Hen3 [1] Let $G$ be a connected graph of order $n\geq 3$. Then $\gamma_{tR}(G)=\gamma_{t}(G)+1$ if and only if $\Delta(G)=n-1$, that is, $G$ has a universal vertex.

By \threftR=3, \threfHen3 implies that, if $G$ is a connected graph of order $n\geq 3$, then $\gamma_{tR}(G)=\gamma_{t}(G)+1$ if and only if $\gamma_{tR}(G)=3$. These results lead to the following observation.

Observation 6.3**.**

\thlabel

tR¿t+1 If $G$ is a connected graph of order $n\geq 3$ such that $\Delta(G)\leq n-2$, then $\gamma_{t}(G)+2\leq\gamma_{tR}(G)\leq 2\gamma_{t}(G)$.

We now provide a result characterizing graphs with $\gamma_{tR}\in\{3,4\}$ in terms of their domination and total domination numbers that will be useful in describing $5$-$\gamma_{tR}$-edge-critical graphs.

Proposition 6.4**.**

\thlabel

t=2 iff tR=34 If $G$ is a connected graph of order $n\geq 3$, then $\gamma_{tR}(G)\in\{3,4\}$ if and only if $\gamma_{t}(G)=2$. Moreover, $\gamma(G)=1$ when $\gamma_{tR}(G)=3$, and $\gamma(G)=2$ when $\gamma_{tR}(G)=4$.

Proof. Suppose first that $\gamma_{t}(G)=2$. By \threfHen2, $2\leq\gamma_{tR}(G)\leq 4$. Clearly $\gamma_{tR}(G)\neq 2$, since $n\geq 3$. Therefore $\gamma_{tR}(G)\in\{3,4\}$.

Conversely, suppose $\gamma_{tR}(G)\in\{3,4\}$. First, if $\gamma_{tR}(G)=3$, then \threftR=3 implies that $G$ has a universal vertex. Therefore $\gamma_{t}(G)=2$ and $\gamma(G)=1$. Otherwise, if $\gamma_{tR}(G)=4$, then \threftR=3 implies that $G$ has no universal vertex. Therefore, by \threftR¿t+1, $\gamma_{t}(G)+2\leq 4$, and thus $\gamma_{t}(G)=2$. Furthermore, since $\gamma(G)\leq\gamma_{t}(G)$ and $G$ has no universal vertex, $\gamma(G)=2$. $\square$

Proposition 6.5**.**

\thlabel

5-tR edge-crit For any graph $G$, if $G$ is $5$-$\gamma_{tR}$-edge-critical, then $G$ is either $3$-$\gamma_{t}$-edge-critical or $G=K_{2}\cup K_{n}$ for $n\geq 3$, in which case $G$ is $4$-$\gamma_{t}$-edge-supercritical.

Proof. Suppose $G$ is $5$-$\gamma_{tR}$-edge-critical. By \threft=2 iff tR=34, $\gamma_{t}(G)>2$ and $\gamma_{t}(G+e)=2$ for any $e\in E(\overline{G})$. Therefore, by \threfMyn1, $G$ is either $3$-$\gamma_{t}$-edge-critical or $4$-$\gamma_{t}$-edge-supercritical. If $G$ is $4$-$\gamma_{t}$-edge-supercritical, then by \threfMyn2, $G$ is the union of two or more non-trivial complete graphs. Since $\gamma_{tR}(G)=5$, this implies that $G=K_{2}\cup K_{n}$ for $n\geq 3$. $\square$

Note that if we add the condition that $G$ is not $6$-$\gamma_{tR}$-edge-supercritical, then the above becomes a necessary and sufficient condition. Clearly $G=K_{2}\cup K_{n}$ is $5$-$\gamma_{tR}$-edge-critical for any $n\geq 3$. Otherwise, if $G$ is $3$-$\gamma_{t}$-edge-critical, then by \threft=2 iff tR=34, $\gamma_{tR}(G)>4$ and $\gamma_{tR}(G+e)\in\{3,4\}$ for any $e\in E(\overline{G})$. By \threfHen2, $\gamma_{tR}(G)\leq 6$, and thus, since $G$ is not $6$-$\gamma_{tR}$-edge-supercritical, $\gamma_{tR}(G)=5$. Hence $G$ is $5$-$\gamma_{tR}$-edge-critical, as required.

7 $\gamma_{tR}$-Edge-critical spiders

A *(generalized) spider *$\mathrm{Sp}(l_{1},...,l_{k}),\ l_{i}\geq 1,\,k\geq 2$, is a tree obtained from the star $K_{1,k}$ with centre $u$ and leaves $v_{1},...,v_{k}$ by subdividing the edge $uv_{i}$ exactly $l_{i}-1$ times, $i=1,...,k$. Thus, a spider $\mathrm{Sp}(2,...,2)$ is a subdivided star. The $u-v_{i}$ paths (of length $l_{i}$) are called the legs of the spider, while $u$ is its head. We now investigate the spiders which are $\gamma_{tR}$-edge-critical. Note that when $k=2$, $\mathrm{Sp}(l_{1},...,l_{k})\cong P_{n}$ for $n\geq 3$, which, by \threfn edge-crit, is not $\gamma_{tR}$-edge-critical. We begin with two propositions restricting $\gamma_{tR}$-edge-criticality in general graphs, which will be useful in classifying $\gamma_{tR}$-edge-critical spiders.

Proposition 7.1**.**

\thlabel

end deg 3 For a graph $G$ with no isolated vertices, if $G$ has an end-vertex $w$ with support vertex $x$ such that $G[N(x)-\{w\}]$ is not complete, then $G$ is not $\gamma_{tR}$-edge-critical.

Proof. Suppose $u,v\in N_{G}(x)-\{w\}$ such that $uv\in E(\overline{G})$. We claim that $\gamma_{tR}(G+uv)=\gamma_{tR}(G)$. Suppose for a contradiction that $\gamma_{tR}(G+uv)<\gamma_{tR}(G)$, and consider a $\gamma_{tR}$-function $f=(V_{f}^{0},V_{f}^{1},V_{f}^{2})$ on $G+uv$. Note that, since $w$ is an end-vertex, $f(x)>0$. By \threfset added edge, $\{f(u),f(v)\}\in\{\{2,2\},\{2,1\},\{2,0\},\{1,1\}\}$. Since $ux,vx\in E(G)$ and at least one of $f(u)$ and $f(v)$ is positive, we can assume without loss of generality that $f(x)=2$. In any case, $f$ is also a TRD-function on $G$, contradicting $\gamma_{tR}(G+uv)<\gamma_{tR}(G)$. Therefore $\gamma_{tR}(G+uv)=\gamma_{tR}(G)$ and $G$ is not $\gamma_{tR}$-edge-critical. $\square$

In a tree, the support vertex of a leaf is called a stem. A stem is called weak if it is adjacent to exactly one leaf, and strong if it is adjacent to two or more leaves. A vertex $b$ of a tree such that $\deg(b)\geq 3$ is called a branch vertex. An *endpath in a tree is a path from a branch vertex to a leaf, all of whose internal vertices have degree $2$. * The next result follows immediately from \threfend deg 3.

Corollary 7.2**.**

\thlabel

strong stems If $T$ is a $\gamma_{tR}$-edge-critical tree, then $T$ contains no stems of degree at least $3$, and hence no strong stems.

Proposition 7.3**.**

\thlabel

no long legs For a graph $G$ with no isolated vertices, if $G$ has two endpaths $v_{0},v_{1},...,v_{k}$ and $u_{0},u_{1},...,u_{m}$, where $k,m\geq 3$ and $v_{k}$ and $u_{m}$ are leaves, then $G$ is not $\gamma_{tR}$-edge-critical.

Proof. We claim that $\gamma_{tR}(G+v_{k}u_{m})=\gamma_{tR}(G)$. Suppose for a contradiction that $\gamma_{tR}(G+v_{k}u_{m})<\gamma_{tR}(G)$, and let $f$ be a $\gamma_{tR}$-function on $G+v_{k}u_{m}$. Then, by \threfset added edge, we may assume $f(u_{m})=f(v_{k})=1$. Define $f^{\prime}:V(G)\rightarrow\{0,1,2\}$ as follows: If $f(v_{k-1})=0$, then clearly $f(v_{k-2})=2$ and $f(v_{k-3})\geq 1$, so let $f^{\prime}(v_{k-1})=f^{\prime}(v_{k-2})=1$. Otherwise, let $f^{\prime}(v_{k-1})=f(v_{k-1})$ and $f^{\prime}(v_{k-2})=f(v_{k-2})$. Similarly, if $f(u_{m-1})=0$, then clearly $f(u_{m-2})=2$ and $f(u_{m-3})\geq 1$, so let $f^{\prime}(u_{m-1})=f^{\prime}(u_{m-2})=1$. Otherwise, let $f^{\prime}(u_{m-1})=f(u_{m-1})$ and $f^{\prime}(u_{m-2})=f(u_{m-2})$. Finally, let $f^{\prime}(y)=f(y)$ for all other $y\in V(G)$. Clearly $f^{\prime}$ is a TRD-function on $G$ and $\omega(f^{\prime})=\omega(f)$, contradicting $\gamma_{tR}(G+v_{k}u_{m})<\gamma_{tR}(G)$. Therefore $\gamma_{tR}(G+v_{k}u_{m})=\gamma_{tR}(G)$, and thus $G$ is not $\gamma_{tR}$-edge-critical. $\square$

Let $S$ be a spider with $k\geq 3$ legs. In what follows, let $c$ be the head of $S$, and let the $k$ legs be labelled $c,v_{i1},v_{i2},...,v_{im_{i}}$, where $i\in\{1,2,...,k\}$, in order of increasing length. Let $m=m_{k}$, that is, $m$ is the length of a longest leg of $S$. We begin by determining the TRD-number of spiders.

Proposition 7.4**.**

\thlabel

tR spider If $S$ is a spider of order $n$ with $k\geq 3$ legs such that $S$ has $y$ legs of length $2$, then

[TABLE]

Proof. Suppose $S$ has $x$ legs of length $1$, and consider a $\gamma_{tR}$-function $f$ on $S$ such that $|V_{f}^{2}|$ is as small as possible. First, suppose $y\geq k-1$. If $y=k$, then $S$ is a subdivided star. Otherwise, if $y=k-1$, then $S$ has exactly one leg that is not of length $2$, and thus either $x=1$ or $x=0$. If $x=1$, then $S$ is the corona of a graph (specifically, $S=\mathop{\mathrm{c}or}(K_{1,k-1})$). Otherwise, if $x=0$, then $m=m_{k}\geq 3$, and $S\in\mathcal{H}_{r}$, where $r=m-3$. In any case, by \threfHen1, $\gamma_{tR}(S)=n$.

Assume therefore that $y<k-1$. Then $S$ has at least two legs that are not of length $2$. Therefore $S$ is not one of the graphs listed in \threfHen1, and thus $\gamma_{tR}(S)<n$. Hence there is some vertex $u\in V(S)$ such that $f(u)=2$ and $f(w)=0$ for at least two vertices $w$ adjacent to $u$. Furthermore, since $f$ is a TRD-function, such a vertex $u$ is not isolated in $S[V_{f}^{+}]$, and thus $\deg(u)\geq 3$. Since $c$ is the only vertex in $S$ with degree at least $3$, $f(c)=2$. Therefore $c$ Roman dominates each $v_{i1}$, and we need $f(v_{i1})$ to be positive for at least one $i$ to ensure that $S[V_{f}^{+}]$ has no isolated vertices.

Consider an arbitrary leg $c,v_{i1},v_{i2},...,v_{im_{i}}$ of $S$. If $m_{i}=1$, then $f(v_{i1})\in\{0,1\}$ in order for $f$ to totally Roman dominate $c$ and $v_{i1}$. If $m_{i}=2$, a total weight of $2$ on $v_{i1}$ and $v_{i2}$ is required in order for $f$ to total Roman dominate $\{v_{i1},v_{i2}\}$. Since $|V_{f}^{2}|$ is as small as possible, $f(v_{i1})=f(v_{i2})=1$. Finally, if $m_{i}>2$, by \threfHen1 and since $f(c)=2$, a total weight of at least $m_{i}-1$ on $v_{i1},...,v_{im_{i}}$ is required in order for $f$ to totally Roman dominate $c$ and $\{v_{i1},...,v_{im_{i}}\}$. Moreover, by the choice of $f$, $f(v_{i1})\in\{0,1\}$ and $f(v_{i2})=\cdots=f(v_{im})=1$. Therefore $\omega(f)\geq n-k+y+1$.

Now, if $y>0$, where (say) $m_{j}=2$, then $f(v_{j1})=1$. By minimality and since $c$ is adjacent to $v_{j1}$, $f(v_{i1})=0$ for each $i$ such that $m_{i}\neq 2$. Then $\gamma_{tR}(S)=\omega(f)=n-k+y+1$, as required. Otherwise, if $y=0$, then $f(v_{i1})=1$ for some $i$ to ensure that $c$ is not isolated in $S[V_{f}^{+}]$. By minimality, $f(v_{j1})=0$ for each $j\neq i$. Therefore $\gamma_{tR}(S)=\omega(f)=n-k+2$. $\square$

The characterization of $\gamma_{tR}$-edge-critical spiders follows. Our result also shows that a spider of order $n$ is $\gamma_{tR}$-edge-critical if and only if it is $n$-$\gamma_{tR}$-edge-critical.

Theorem 7.5**.**

\thlabel

edge-crit spider A spider $S=\mathop{\mathrm{S}p}(l_{1},...,l_{k}),$ $k\geq 3$, is $\gamma_{tR}$-edge-critical if and only if $l_{i}=2$ for each $i$, $1\leq i\leq k-1$, and $l_{k}\in\{2,m\}$, where $m=4$ or $m\geq 6$.

Proof. Suppose $S$ has order $n$. If $l_{i}=2$ for each $i=1,...,k$, then $S$ is a subdivided star and, by \threfn edge-crit, $S$ is $n$-$\gamma_{tR}$-edge-critical. Now, suppose $S$ has exactly one leg of length $m\neq 2$. If $m=1$, then by \threfend deg 3, $S$ is not $\gamma_{tR}$-edge-critical. If $m=3$ or $m=5$, then $S\in\mathcal{H}_{r}$ with $r=0$ or $2$, respectively, and thus, by \threfn edge-crit, $S$ is not $\gamma_{tR}$-edge-critical. If $m=4$ or $m\geq 6$, then $S\in\mathcal{H}_{r}$ with $r=m-3$, and therefore, by \threfn edge-crit, $S$ is $n$-$\gamma_{tR}$-edge-critical. Finally, suppose $S$ has at least two legs that are not of length $2$. Again, by \threfend deg 3, if $S$ has a leg of length $1$, $S$ is not $\gamma_{tR}$-edge-critical. Otherwise, assume $S$ has at least two legs of length at least $3$. Then, by \threfno long legs, $S$ is not $\gamma_{tR}$-edge-critical. $\square$

8 $k$-$\gamma_{tR}$-Edge-critical graphs with minimum

diameter

We now consider the minimum diameter possible in a $k$-$\gamma_{tR}$-edge-critical graph, for $k\geq 4$. There are no $\gamma_{tR}$-edge-critical graphs with diameter $1$, as the only graphs with diameter $1$ are non-trivial complete graphs, which are clearly not $\gamma_{tR}$-edge-critical since $E(\overline{G})=\emptyset$. Therefore, the minimum possible diameter for a $\gamma_{tR}$-edge-critical graph is $2$. Asplund, Loizeaux and Van der Merwe [2] constructed families of $3$-$\gamma_{t}$-edge-critical graphs with diameter $2$. We will show that, for any $k\geq 4$, there exists a $k$-$\gamma_{tR}$-edge-critical graph of diameter $2$. We first present the following proposition which shows that every graph $G$ without a dominating vertex is a spanning subgraph of a $\gamma_{tR}(G)$-edge-critical graph with the same total Roman domination number, which will be useful in proving our result.

Proposition 8.1**.**

\thlabel

span For a graph $G$ with no isolated vertices, if ${\gamma_{tR}(G)=k\geq 4}$, then $G$ is a spanning subgraph of a $k$-$\gamma_{tR}(G)$-edge-critical graph.

Proof. Suppose $\gamma_{tR}(G)=k\geq 4$. If $G$ is $k$-$\gamma_{tR}(G)$-edge-critical, then we are done. Otherwise, there is, by definition, some edge $e_{1}\in E(\overline{G})$ such that $\gamma_{tR}(G+e_{1})=\gamma_{tR}(G)$. Let $G_{1}=G+e_{1}$. If $G_{1}$ is $k$-$\gamma_{tR}(G)$-edge-critical, then we are done. Otherwise, there is some edge $e_{2}\in E(\overline{G_{1}})$ such that $\gamma_{tR}(G_{1}+e_{2})=\gamma_{tR}(G_{1})$. Let $G_{2}=G_{1}+e_{2}$. Continuing in this way, we eventually obtain a graph $G_{i}$ such that for all $e\in E(\overline{G_{i}})$, $\gamma_{tR}(G_{i}+e)<\gamma_{tR}(G_{i})$ and $\gamma_{tR}(G_{i})=\gamma_{tR}(G_{i-1})=\cdots=\gamma_{tR}(G_{1})=\gamma_{tR}(G)$. Since $k\geq 4$, $G_{i}$ is not complete and thus $E(G_{i})\neq\emptyset$. Therefore, $G_{i}$ is a $k$-$\gamma_{tR}(G)$-edge-critical graph, of which $G$ is a spanning subgraph. $\square$

Before demonstrating the existence of $k$-$\gamma_{tR}$-edge-critical graphs of diameter $2$ for any $k\geq 4$, we determine the TRD-number of K_{n}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{m}, where $n,m\geq 2$. Consider the vertices of K_{n}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{m} as an $n\times m$ matrix, where vertices $v_{ij}$ and $v_{st}$ are adjacent if and only if $i=s$ or $j=t$. The rows and columns of the matrix form disjoint copies of $K_{m}$ and $K_{n}$, respectively. If $v_{ij}$ and $v_{st}$ are nonadjacent, then $v_{sj}$ is adjacent to both $v_{ij}$ and $v_{st}$, and hence \mathop{\mathrm{d}iam}(K_{n}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{m})=2.

Proposition 8.2**.**

\thlabel

tR(KnXKm) If $m$ and $n$ are integers such that $m\geq n\geq 2$, then \gamma_{tR}(K_{n}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{m})=2n.

Proof. Let G=K_{n}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{m}. To see that $\gamma_{tR}(G)\leq 2n$, consider the TRD-function $g=(V_{g}^{0},V_{g}^{1},V_{g}^{2})$ on $G$ where $V_{g}^{1}=\emptyset$ and $V_{g}^{2}=\{v_{i1}:1\leq i\leq n\}$.

Now, suppose for a contradiction that $\gamma_{tR}(G)\leq 2n-1$ and consider a TRD-function $f=(V_{f}^{0},V_{f}^{1},V_{f}^{2})$ on $G$ with $\omega(f)=2n-1$. Each vertex $v$ dominates one row and one column of $G$, so if $|V_{f}^{2}|=x$ (note that $x\leq n-1$), then at most $x$ rows and at most $x$ columns are dominated by elements of $V_{f}^{2}$. Let $S$ be the set of vertices undominated by $V_{f}^{2}$. Then $|S|\geq(n-x)(m-x)\geq(n-x)^{2}$. Moreover, $|V_{f}^{1}|=(2n-1)-2x$ since $\omega(f)=2n-1$ and $|V_{f}^{2}|=x$.

If $x=n-1$, then $|V_{f}^{1}|=1$. Since $f$ is a TRD-function and $|S|\geq(n-x)^{2}$, $|S|=1$; say $S=\{w\}$. Hence $V_{f}^{1}=\{w\}$. However, $V_{f}^{2}$ does not dominate $w$, and thus $w$ is isolated in $G[V_{f}^{+}]$, which is a contradiction. Therefore, there is no TRD-function on $G$ with weight $2n-1$ when $x=n-1$.

Otherwise, if $x<n-1$, then

[TABLE]

Therefore, the number of vertices undominated by $V_{f}^{2}$ is greater than $|V_{f}^{1}|$, contradicting $f$ being a TRD-function. Thus there is no TRD-function on $G$ with weight $2n-1$ when $x<n-1$. We conclude that $\gamma_{tR}(G)=2n$. $\square$

Theorem 8.3**.**

\thlabel

tR diam 2 If $k\geq 4$, then there exists a $k$-$\gamma_{tR}$-edge-critical graph of diameter $2$.

Proof. First, assume that $k$ is even; say $k=2l$ for some $l\geq 2$. Let G_{l}=K_{l}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{l}. By \threftR(KnXKm), $\gamma_{tR}(G_{l})=2l$, and, by \threfspan, $G_{l}$ is a spanning subgraph of a $k$-$\gamma_{tR}$-edge-critical graph $G_{l}^{\prime}$. Since $k>3$, \threftR=3 implies that $G_{l}^{\prime}$ has no dominating vertex, and hence $2\leq\mathop{\mathrm{d}iam}(G_{l}^{\prime})\leq\mathop{\mathrm{d}iam}(G_{l})=2$.

Now, consider the case where $k$ is odd; say $k=2l+1$ for some $l\geq 2$. Let $G_{l}^{d}$ be the graph formed by taking K_{l+1}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{l+1} and deleting the vertices in the set $\{v_{j1}:\lfloor\frac{l}{2}\rfloor+2\leq j\leq l+1\}$. Similarly to $G_{l}$, $\mathop{\mathrm{d}iam}(G_{l}^{d})=2$. See Figure $1$.

We claim that $\gamma_{tR}(G_{l}^{d})=2l+1$. To see that $\gamma_{tR}(G_{l}^{d})\leq 2l+1$, consider the following TRD-function on $G_{l}^{d}$: If $l$ is even, place two $2$’s in each of the first $\frac{l}{2}-1$ rows, and one $2$ in each of rows $\frac{l}{2}$ and $\frac{l}{2}+1$, such that they span columns $2$ through $l+1$. At this point, every vertex in $G_{l}^{d}$ is dominated. However, the $2$’s in rows $\frac{l}{2}$ and $\frac{l}{2}+1$ are isolated, so place a $1$ in row $\frac{l}{2}$ such that it shares a column with the $2$ in row $\frac{l}{2}+1$. Otherwise, if $l$ is odd, place two $2$’s in each of the first $\frac{l-1}{2}$ rows, and one $2$ in row $\frac{l+1}{2}$, such that they span columns $2$ through $l+1$. Similarly to the even case, every vertex in $G_{l}^{d}$ is now dominated. However, the $2$ in row $\frac{l+1}{2}$ is isolated, so place a $1$ in row $\frac{l-1}{2}$ such that it shares a column with that $2$. In either case, we have a TRD-function on $G_{l}^{d}$ with weight $2l+1$, hence $\gamma_{tR}(G_{l}^{d})\leq 2l+1$.

Now, suppose for a contradiction that $\gamma_{tR}(G_{l}^{d})<2l+1$, and consider a TRD-function $f=(V_{f}^{0},V_{f}^{1},V_{f}^{2})$ on $G_{l}^{d}$ with $\omega(f)=2l$. We claim that $f(v_{j1})=0$ for all $1\leq j\leq\lfloor\frac{l}{2}\rfloor+1$. If $f(v_{j1})=2$ for $x\geq 1$ vertices in column $1$, the undominated vertices in columns $2$ through $l+1$ form the graph K_{l}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{l+1-x}. By \threftR(KnXKm), a TRD-function on K_{l}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{l+1-x} requires a weight of $2\min\{l,l+1-x\}=2(l+1-x)$. However, since $2x+2(l+1-x)>2l$, this is impossible. Therefore $f(v_{j1})\neq 2$ for all $1\leq j\leq\lfloor\frac{l}{2}\rfloor+1$. If $f(v_{j1})=1$ for $x\geq 1$ vertices in column $1$, the undominated vertices in columns $2$ through $l+1$ (that is, those for which $f$ could be assigned a $2$) form the graph K_{l}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{l+1}. Again by \threftR(KnXKm), a TRD-function on K_{l}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{l+1} requires a weight of $2\min\{l,l+1\}=2l$. However, $x+2l>2l$ for $x\geq 1$, so this is also not possible. Therefore, $f(v_{j1})=0$ for all $1\leq j\leq\lfloor\frac{l}{2}\rfloor+1$.

As a result, in order to totally Roman dominate the first column, there must be a $2$ in each of the first $\lfloor\frac{l}{2}\rfloor+1$ rows, none of which can be in the first column. That is, for each $1\leq s\leq\lfloor\frac{l}{2}\rfloor+1$, $f(v_{st})=2$ for some $2\leq t\leq l+1$. Let $S$ be the set of these vertices. Note that, thus far, we have accounted for a total weight of

[TABLE]

which leaves a weight of $l-2$ if $l$ is even and $l-1$ if $l$ is odd to be assigned. That is, a weight of $2(\lceil\frac{l}{2}\rceil-1)$ remains to be accounted for. We now claim that no two vertices in $S$ can be in the same column. If the vertices in $S$ span fewer than $\lfloor\frac{l}{2}\rfloor+1$ columns, then the vertices which are undominated by $S$ induce a graph containing K_{\lceil\frac{l}{2}\rceil}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{\lceil\frac{l}{2}\rceil} as subgraph. If $l=2$, then no weight remains to dominate this vertex, as $2(\lceil\frac{l}{2}\rceil-1)=0$. Otherwise, if $l>2$, \threftR(KnXKm) implies that \gamma_{tR}(K_{\lceil\frac{l}{2}\rceil}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{\lceil\frac{l}{2}\rceil})=2(\lceil\frac{l}{2}\rceil). However, $2(\lceil\frac{l}{2}\rceil)>2(\lceil\frac{l}{2}\rceil-1)$. In either case, this contradicts $f$ being a TRD-function, and thus no vertices of $S$ share a column.

Therefore, the vertices left undominated by $S$ induce a graph T\cong K_{\lceil\frac{l}{2}\rceil}\mathbin{\raisebox{0.85358pt}{\scriptstyle\square}}K_{\lceil\frac{l}{2}\rceil-1}, with $\lceil\frac{l}{2}\rceil$ rows and $\lceil\frac{l}{2}\rceil-1$ columns. Moreover, the vertices in $S$ are all isolated, as none share a row or column. By \threftR(KnXKm), $\gamma_{tR}(T)=2(\lceil\frac{l}{2}\rceil-1)$. Thus the entire remaining weight is required in order to dominate $T$; necessarily, the vertices in $V_{f}^{+}-S$ belong to rows and columns that do not contain vertices in $S$. However, this still leaves the vertices in $S$ isolated, which contradicts $f$ being a TRD-function on $G_{l}^{d}$. Therefore $\gamma_{tR}(G_{l}^{d})\geq 2l+1$ and we conclude that $\gamma_{tR}(G_{l}^{d})=2l+1$. As in the case where $k$ is even, $G_{l}^{d}$ is a spanning subgraph of a $k$-$\gamma_{tR}$-edge-critical graph with diameter $2$. $\square$

9 Future work

We showed in Section 5 that the disjoint union of two of more complete graphs, each having order at least $3$, is $\gamma_{tR}$-edge-supercritical. We also explained that a proof similar to that of \threfMyn2 does not work for total Roman domination. Hence we pose the following question.

Question 1**.**

Are the disjoint unions of two or more complete graphs, each having order at least $3$, the only $\gamma_{tR}$-edge-supercritical graphs?

Note that if this is the case, \thref5-tR edge-crit automatically becomes a necessary and sufficient condition for a graph to be $5$-$\gamma_{tR}$-edge-critical.

Now consider, for a moment, Roman dominating functions, and suppose a graph $G$ has non-adjacent vertices $u$ and $v$ such that $f(u)=f(v)=0$ for every $\gamma_{R}$-function $f$ on $G$. We claim that $\gamma_{R}(G+uv)=\gamma_{R}(G)$. Suppose $\gamma_{R}(G+uv)<\gamma_{R}(G)$ and let $f$ be a $\gamma_{R}$-function on $G+uv$. Similar to \threfset added edge, we may assume without loss of generality that $f(u)=2$ and $f(v)=0$, otherwise $f$ is a RD-function on $G$ such that $\omega(f)<\gamma_{R}(G)$. However, the function $f^{\prime}$ defined by $f^{\prime}(v)=1$ and $f^{\prime}(y)=f(y)$ for all other $y\in V(G)$ is a $\gamma_{R}$-function on $G$ such that $f^{\prime}(v)>0$, contrary to our assumption. The situation for total Roman domination is different.

For a graph $G$, we define $u\in V(G)$ to be a dead vertex if every $\gamma_{tR}$-function $f$ on $G$ has $f(u)=0$. Not only do there exist graphs $G$ containing non-adjacent dead vertices $u$ and $v$ such that $\gamma_{tR}(G+uv)<\gamma_{tR}(G)$, but it is possible to find such a graph $G$ with $\gamma_{tR}(G+uw)<\gamma_{tR}(G)$ for every edge $uw\in E(\overline{G})$, that is, every edge in $E(\overline{G})$ incident with the dead vertex $u$ is critical. We define the graph $D_{n}$ below and show that $D_{n}$ is such a graph.

Let $D_{n}$ be the graph composed of $n\geq 2$ copies of $K_{4}-e$ sharing a single central vertex as follows: Let $c$ be the central vertex, $w_{1},...,w_{n}$ be the degree two vertices, and $u_{1},...,u_{n}$ and $v_{1},...,v_{n}$ be the remaining vertices (where $u_{i}$ and $v_{i}$ are adjacent for each $i$) such that $c,u_{i},w_{i},v_{i},c$ is a $4$-cycle in $D_{n}$ for each $1\leq i\leq n$. See Figure $2$.

Proposition 9.1**.**

\thlabel

tR(D_n) If $n\geq 2$, then $\gamma_{tR}(D_{n})=2n+1$. Moreover, $w_{i}$ is a dead vertex for each $1\leq i\leq n$.

Proof. To see that $\gamma_{tR}(D_{n})\leq 2n+1$, consider the TRD-function $g:V(D_{n})\rightarrow\{0,1,2\}$ on $D_{n}$ defined by $g(c)=1$, $g(u_{i})=2$ for $1\leq i\leq n$, and $g(y)=0$ for all other $y\in V(D_{n})$.

We claim that, if $f$ is a TRD-function on $D_{n}$ with $\omega(f)\leq 2n+1$, then $f(c)=1$. If $f(c)=2$, then the only vertices that remain undominated in $D_{n}$ are $w_{i}$ for $1\leq i\leq n$. However, since $d(w_{i},w_{j})=4$ for all $i\neq j$, a weight of $2n$ is required in order to totally Roman dominate these vertices, contradicting $\omega(f)\leq 2n+1$. If $f(c)=0$, then since $D_{n}-c$ is the disjoint union of $n$ triangles, \threfHen1 implies that a weight of $3n$ is required to totally Roman dominate the remaining vertices, contradicting $\omega(f)\leq 2n+1$. Therefore $f(c)=1$. Since a weight of at least $2n$ is required to totally Roman dominate the remaining disjoint union of $n$ triangles, we conclude that $\gamma_{tR}(D_{n})=2n+1$.

Now, let $f$ be any $\gamma_{tR}$-function on $D_{n}$. Then $\omega(f)=2n+1$ and $f(c)=1$. To dominate each triangle of $D_{n}-c$ with a weight of $2$, $\{f(u_{i}),f(v_{i})\}=\{0,2\}$ and $f(w_{i})=0$ for each $1\leq i\leq n$. Hence each $w_{i}$ is a dead vertex. $\square$

The following result shows that, for $n\geq 3$, every edge in $E(\overline{D_{n}})$ incident with $w_{i}$ is critical.

Proposition 9.2**.**

If $n\geq 3$, $i\in\{1,...,n\}$ and $w_{i}v\in E(\overline{D_{n}})$, then $\gamma_{tR}(D_{n}+w_{i}v)<\gamma_{tR}(D_{n})$.

Proof. Without loss of generality, consider an edge $w_{1}v\in E(\overline{D_{n}})$. Then (without loss of generality) $v\in\{w_{2},u_{2},c\}$. If $v=w_{2}$, define $f:V(D_{n}+w_{1}v)\rightarrow\{0,1,2\}$ by $f(w_{1})=f(w_{2})=1$, $f(c)=f(u_{3})=\cdots=f(u_{n})=2$, and $f(y)=0$ for all other $y\in V(D_{n})$. Otherwise, if $v\in\{u_{2},c\}$, define $f:V(D_{n}+w_{1}v)\rightarrow\{0,1,2\}$ by $f(c)=f(u_{2})=f(u_{3})=\cdots=f(u_{n})=2$ and $f(y)=0$ for all other $y\in V(D_{n})$. In either case, $f$ is a TRD-function on $D_{n}+w_{1}v$ and $\omega(f)=2n$. Therefore, by \threftR(D_n), every edge $w_{i}v\in E(\overline{D_{n}})$ is critical. $\square$

However, for $n\geq 3$, $D_{n}$ is not $\gamma_{tR}$-edge-critical since (for example) $\gamma_{tR}(D_{n}+u_{1}u_{2})=2n+1$. Furthermore, $D_{2}$ is not $\gamma_{tR}$-edge-critical since (for example) $\gamma_{tR}(D_{2}+w_{1}w_{2})=5$. However, adding edges to $D_{n}$ until a $(2n+1)$-$\gamma_{tR}$-edge-critical graph $D_{n}^{\prime}$ is obtained results in $D_{n}^{\prime}$ having no dead vertices. Hence we pose the following question.

Question 2**.**

Do there exist $\gamma_{tR}$-edge-critical graphs containing dead vertices?

We characterized $\gamma_{tR}$-edge-critical spiders in \threfedge-crit spider. Finding other classes of $\gamma_{tR}$-edge-critical trees and, indeed, characterizing $\gamma_{tR}$-edge-critical trees, remain open problems.