Total Roman {3}-domination in Graphs

Abstract

For a graph $G=(V,E)$ with vertex set $V=V\left(G\right)$ and edge set $E=E\left(G\right)$, a Roman $\left\{3\right\}$-dominating function (R$\left\{3\right\}$-DF) is a function $f:V\left(G\right)\to \{0,1,2,3\}$ having the property that ${\sum }_{u\in N_G\left(v\right)}f\left(u\right)\ge 3$, if $f\left(v\right)=0$, and ${\sum }_{u\in N_G\left(v\right)}f\left(u\right)\ge 2$, if $f\left(v\right)=1$ for any vertex $v\in V\left(G\right)$. The weight of a Roman $\left\{3\right\}$-dominating function f is the sum $f\left(V\right)={\sum }_{v\in V\left(G\right)}f\left(v\right)$ and the minimum weight of a Roman $\left\{3\right\}$-dominating function on G is the Roman $\left\{3\right\}$-domination number of G, denoted by ${\gamma }_{\left\{R3\right\}}\left(G\right)$. Let G be a graph with no isolated vertices. The total Roman $\left\{3\right\}$-dominating function on G is an R$\left\{3\right\}$-DF f on G with the additional property that every vertex $v\in V$ with $f\left(v\right)\ne 0$ has a neighbor w with $f\left(w\right)\ne 0$. The minimum weight of a total Roman $\left\{3\right\}$-dominating function on G, is called the total Roman $\left\{3\right\}$-domination number denoted by ${\gamma }_{t\left\{R3\right\}}\left(G\right)$. We initiate the study of total Roman $\left\{3\right\}$-domination and show its relationship to other domination parameters. We present an upper bound on the total Roman $\left\{3\right\}$-domination number of a connected graph G in terms of the order of G and characterize the graphs attaining this bound. Finally, we investigate the complexity of total Roman $\left\{3\right\}$-domination for bipartite graphs.

1. Introduction

In this paper, we introduce and study a variant of Roman dominating functions, namely, total Roman $\left\{3\right\}$-dominating functions. First we present some necessary terminology and notation. Let $G=(V,E)$ be a graph of order n with vertex set $V=V\left(G\right)$ and edge set $E=E\left(G\right)$. The open neighborhood of a vertex $v\in V\left(G\right)$ is the set $N_G\left(v\right)=N\left(v\right)=\{u:uv\in E\left(G\right)\}$. The closed neighborhood of a vertex $v\in V\left(G\right)$ is $N_G\left[v\right]=N\left[v\right]=N\left(v\right)\cup \left\{v\right\}$. The open neighborhood of a set $S\subseteq V$ is the set $N_G\left(S\right)=N\left(S\right)={\bigcup }_{v\in S}N\left(v\right)$. The closed neighborhood of a set $S\subseteq V$ is the set $N_G\left[S\right]=N\left[S\right]=N\left(S\right)\cup S={\bigcup }_{v\in S}N\left[v\right]$. We denote the degree of v by $d_G\left(v\right)=d\left(v\right)=\left|N\left(v\right)\right|$. By $\mathsf{\Delta }=\mathsf{\Delta }\left(G\right)$ and $\delta =\delta \left(G\right)$, we denote the maximum degree and minimum degree of a graph G, respectively. A vertex of degree one is called a leaf and its neighbor a support vertex. We denote the set of leaves and support vertices of a graph G by $L\left(G\right)$ and $S\left(G\right)$, respectively. We write $K_n$, $P_n$ and $C_n$ for the complete graph, path and cycle of order n, respectively. A tree T is an acyclic connected graph. The corona $H\circ K_1$ of a graph H is the graph constructed from H, where for each vertex $v\in V\left(H\right)$, a new vertex $v^{\prime }$ and a pendant edge $v{v}^{\prime }$ are added. The union of two graphs $G_1$ and $G_2$ ($G_1\cup G_2$) is a graph G such that $V\left(G\right)=V\left(G_1\right)\cup V\left(G_2\right)$ and $E\left(G\right)=E\left(G_1\right)\cup E\left(G_2\right)$.

A set $S\subseteq V$ in a graph G is called a dominating set if $N\left[S\right]=V$. The domination number $\gamma \left(G\right)$ of G is the minimum cardinality of a dominating set in G, and a dominating set of G of cardinality $\gamma \left(G\right)$ is called a $\gamma$-set of G, [1]. A set $S\subseteq V$ in a graph G is called a total dominating set if $N\left(S\right)=V$. The total domination number ${\gamma }_t\left(G\right)$ of G is the minimum cardinality of a total dominating set in G, and a total dominating set of G of cardinality ${\gamma }_t\left(G\right)$ is called a ${\gamma }_t$-set of G, [2].

Given a graph G and a positive integer m, assume that $g:V\left(G\right)\to \{0,1,2,\dots ,m\}$ is a function, and suppose that $(V_0,V_1,V_2,\dots ,V_m)$ is the ordered partition of V induced by g, where $V_i=\{v\in V:g\left(v\right)=i\}$ for $i\in \{0,1,\dots ,m\}$. So we can write $g=(V_0,V_1,V_2,\dots ,V_m)$. A Roman dominating function on graph G is a function $f:V\to \{0,1,2\}$ such that if $v\in V_0$ for some $v\in V$, then there exists a vertex $w\in N\left(v\right)$ such that $w\in V_2$. The weight of a Roman dominating function (RDF) is the sum $w_f={\sum }_{v\in V\left(G\right)}f\left(v\right)$, and the minimum weight of $w_f$ for every Roman dominating function f on G is called the Roman domination number of G, denoted by ${\gamma }_R\left(G\right)$, see also [3].

Let G be a graph with no isolated vertices. The total Roman dominating function (TRDF) on G, is an RDF f on G with the additional property that every vertex $v\in V$ with $f\left(v\right)\ne 0$ has a neighbor w with $f\left(w\right)\ne 0$. The minimum weight of any TRDF on G is called the total Roman domination number of G denoted by ${\gamma }_{tR}\left(G\right)$. A TRDF on G with weight ${\gamma }_{tR}\left(G\right)$ is called a ${\gamma }_{tR}\left(G\right)$-function.

The mathematical concept of Roman domination, is originally defined and discussed by Stewart [4] in 1999, and ReVelle and Rosing [5] in 2000. Recently, Chellali et al. [6] have introduced the Roman $\left\{2\right\}$-dominating function f as follows. A Roman $\left\{2\right\}$-dominating function is a function $f:V\to \{0,1,2\}$ such that for every vertex $v\in V$, with $f\left(v\right)=0,f\left(N\right(v\left)\right)\ge 2$ where $f\left(N\left(v\right)\right)={\sum }_{x\in N\left(v\right)}f\left(x\right)$, that is, either v has a neighbor u with $f\left(u\right)=2$, or has two neighbors $x,y$ with $f\left(x\right)=f\left(y\right)=1$ [7].

In terms of the Roman Empire, this defense strategy requires that every location with no legion has a neighboring location with two legions, or at least two neighboring locations with one legion each.

Note that for a Roman $\left\{2\right\}$-dominating function (R$\left\{2\right\}$-DF) f, and for some vertex v with $f\left(v\right)=1$, it is possible that $f\left(N\right(v\left)\right)=0$. The sum $w_f={\sum }_{v\in V\left(G\right)}f\left(v\right)$ is denoted the weight of a Roman $\left\{2\right\}$-dominating function, and the minimum weight of a Roman $\left\{2\right\}$-dominating function f is the Roman $\left\{2\right\}$-domination number, denoted by ${\gamma }_{\left\{R2\right\}}\left(G\right)$. Roman $\left\{2\right\}$-domination is a generalization of Roman domination that has also studied by Henning and Klostermeyer [8] with the name Italian domination.

The total Roman $\left\{2\right\}$-domination for graphs are defined as follows [9]. Let G be a graph without isolated vertices. Then $f:V\to \{0,1,2\}$ is total Roman$\left\{2\right\}$-dominating function (TR$\left\{2\right\}$-DF) if it is a Roman $\left\{2\right\}$-dominating function and the subgraph induced by the positive weight vertices has no isolated vertex. The minimum weight $w_f={\sum }_{v\in V\left(G\right)}f\left(v\right)$ of a any total Roman$\left\{2\right\}$-dominating function of a graph G is called the total Roman $\left\{2\right\}$-domination number of G and is denoted by ${\gamma }_{t\left\{R2\right\}}\left(G\right)$. Beeler et al. [10] have defined double Roman domination.

A double Roman dominating function (DRDF) on a graph G is a function $f:V\to \{0,1,2,3\}$ such that the following conditions are hold:

(a)

if $f\left(v\right)=0$, then the vertex v must have at least two neighbors in $V_2$ or one neighbor in $V_3$.

(b)

if $f\left(v\right)=1$, then the vertex v must have at least one neighbor in $V_2\cup V_3$.

The weight of a double Roman dominating function is the sum $w_f={\sum }_{v\in V\left(G\right)}f\left(v\right)$, and the minimum weight of $w_f$ for every double Roman dominating function f on G is called the double Roman domination number of G. We denote this number with ${\gamma }_{dR}\left(G\right)$ and a double Roman dominating function of G with weight ${\gamma }_{dR}\left(G\right)$ is called a ${\gamma }_{dR}\left(G\right)$-function of G, see also [11].

Hao et al. [12] have recently defined total double Roman domination. The total double Roman dominating function (TDRDF) on a graph G with no isolated vertex is a DRDF f on G with the additional property that the subgraph of G induced by the set $\{v\in V(G):f(v)\ne 0\}$ has no isolated vertices. The total double Roman domination number ${\gamma }_{tdR}\left(G\right)$ is the minimum weight of a TDRDF on G. A TDRDF on G with weight ${\gamma }_{tdR}\left(G\right)$ is called a ${\gamma }_{tdR}\left(G\right)$-function. Mojdeh et al. [13] have recently defined the Roman $\left\{3\right\}$-dominating function correspondingly to the Roman $\left\{2\right\}$-dominating function of graphs. For a graph G, a Roman $\left\{3\right\}$-dominating function (R$\left\{3\right\}$-DF) is a function $f:V\to \{0,1,2,3\}$ having the property that $f\left(N\right[u\left]\right)\ge 3$ for every vertex $u\in V$ with $f\left(u\right)\in \{0,1\}$. Formally, a Roman $\left\{3\right\}$-dominating function $f:V\to \{0,1,2,3\}$ has the property that for every vertex $v\in V$, with $f\left(v\right)=0$, there exist at least either three vertices in $V_1\cap N\left(v\right)$, or one vertex in $V_1\cap N\left(v\right)$ and one in $V_2\cap N\left(v\right)$, or two vertices in $V_2\cap N\left(v\right)$, or one vertex in $V_3\cap N\left(v\right)$ and for every vertex $v\in V$, with $f\left(v\right)=1$, there exist at least either two vertices in $V_1\cap N\left(v\right)$, or one vertex in $(V_2\cup V_3)\cap N\left(v\right)$. This notion has been defined recently by Mojdeh and Volkmann [13] as Roman $\left\{3\right\}$-domination.

The weight of a Roman $\left\{3\right\}$-dominating function is the sum $w_f=f\left(V\right)={\sum }_{v\in V}f\left(v\right)$, and the minimum weight of a Roman $\left\{3\right\}$-dominating function f is the Roman $\left\{3\right\}$-domination number, denoted by ${\gamma }_{\left\{R3\right\}}\left(G\right)$.

Now we introduce the total Roman $\left\{3\right\}$-domination concept to consider such situation.

Definition 1.

Let G be a graph G with no isolated vertex. The total Roman $\left\{3\right\}$-dominating function (TR$\left\{3\right\}$-DF) on G is an R$\left\{3\right\}$-DF f on G with the additional property that every vertex $v\in V$ with $f\left(v\right)\ne 0$ has a neighbor w with $f\left(w\right)\ne 0$, in the other words, the subgraph of G induced by the set $\{v\in V(G):f(v)\ne 0\}$ has no isolated vertices. The minimum weight of a total Roman $\left\{3\right\}$-dominating function on G is called the total Roman $\left\{3\right\}$-domination number of G denoted by ${\gamma }_{t\left\{R3\right\}}\left(G\right)$. A ${\gamma }_{t\left\{R3\right\}}\left(G\right)$-function is a total Roman $\left\{3\right\}$-dominating function on G with weight ${\gamma }_{t\left\{R3\right\}}\left(G\right)$.

In this paper We study of total Roman $\left\{3\right\}$-domination versus to other domination parameters. We present an upper bound on the total Roman $\left\{3\right\}$-domination number of a connected graph G in terms of the order of G and characterize the graphs attaining this bound. Finally, we investigate the complexity of total Roman $\left\{3\right\}$-domination for bipartite graphs.

2. Total Roman $\left\{3\right\}$-domination of Some Graphs

First we easily see that ${\gamma }_{\left\{R3\right\}}\left(G\right)\le {\gamma }_{t\left\{R3\right\}}\left(G\right)\le {\gamma }_{tdR}\left(G\right)$, because by the definitions every total Roman $\left\{3\right\}$-dominating function is a Roman $\left\{3\right\}$-dominating function and every total double Roman dominating function is a total Roman $\left\{3\right\}$-dominating function.

In [10] we have.

Proposition 1.

([10] Proposition 2) Let G be a graph and $f=(V_0,V_1,V_2)$ a ${\gamma }_R$-function of G. Then ${\gamma }_{dR}\left(G\right)\le 2|V_1|+3|V_2|$. This bound is sharp.

As an immediate result we also have:

Corollary 1.

Let G be a graph and $f=(V_0,V_1,V_2)$ a total Roman $\left\{2\right\}$-dominating function or a Roman dominating function for which the induced subgraph by $V_1\cup V_2$ has no isolated vertex. Then ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 2|V_1|+3|V_2|$. This bound is sharp.

For some special graphs we obtain the total Roman $\left\{3\right\}$-domination numbers.

Observation 1.

Let $n\ge 2$. Then ${\gamma }_{t\left\{R3\right\}}\left(P_n\right)=\left\{\begin{array}{cc}n+2& ifn\equiv 1\left(\mathrm{mod}3\right)\\ n+1& otherwise\end{array},\right.$

Proof. 

Let $P_n=v_1{v}_2\dots v_n$. Since by assigning 2 to the vertices $v_1$ and $v_n$ and value 1 to the other vertices, we have ${\gamma }_{t\left\{R3\right\}}\left(P_n\right)\le n+2$. Since $f\left(v_1\right)+f\left(v_2\right)\ge 3$ and $f\left(v_{n-1}\right)+f\left(v_n\right)\ge 3$, $f\left(v_{i-1}\right)+f\left(v_i\right)+f\left(v_{i+1}\right)\ge 3$ for $4\le i\le n-3$, $f\left(v_{i-1}\right)+f\left(v_i\right)+f\left(v_{i+1}\right)+f\left(v_{i+2}\right)\ge 3$ for $4\le i\le n-4$ and $f\left(v_{i-2}\right)+f\left(v_{i-1}\right)+f\left(v_i\right)+f\left(v_{i+1}\right)+f\left(v_{i+2}\right)\ge 4$ for $5\le i\le n-4$, we observe that ${\gamma }_{t\left\{R3\right\}}\left(P_n\right)\ge n+1$ and ${\gamma }_{t\left\{R3\right\}}\left(P_n\right)\ge n+2$ if $n\equiv 1\left(\mathrm{mod}3\right)$. If $n=3k$, then by assigning 1 to $v_{3t+1}$ and $v_n$, 2 to $v_{3t+2}$, 0 to $v_{3t}$ except $v_n$, we have ${\gamma }_{t\left\{R3\right\}}\left(P_n\right)\ge 3k+1=n+1$. If $n=2+3k$, then by assigning 1 to $v_{3t+1}$, 2 to $v_{3t+2}$, 0 to $v_{3t}$, we have ${\gamma }_{t\left\{R3\right\}}\left(P_n\right)\ge 3k+1=n+1$. Thus the proof is complete. □

In [10], it has been shown that ${\gamma }_{dR}\left(C_n\right)=n$ if $n\equiv 0,2,3,4(mod6)$ and otherwise ${\gamma }_{dR}\left(C_n\right)=n+1$ and since ${\gamma }_{tdR}\left(G\right)\ge {\gamma }_{dR}\left(G\right)$, we deduce that ${\gamma }_{tdR}\left(C_n\right)\ge n$.

Here we show that ${\gamma }_{t\left\{R3\right\}}\left(C_n\right)=n$ for all $n\ge 3$. If we assign weight 1 to every vertex of $C_n$, then it is a total Roman $\left\{3\right\}$-dominating function of $C_n$. Hence ${\gamma }_{t\left\{R3\right\}}\left(C_n\right)\le n$. In [13], we have shown that ${\gamma }_{\left\{R3\right\}}\left(C_n\right)=n$. Since ${\gamma }_{\left\{R3\right\}}\left(C_n\right)\le {\gamma }_{t\left\{R3\right\}}\left(C_n\right)$, we obtain the desired result.

Observation 2.

${\gamma }_{t\left\{R3\right\}}\left(C_n\right)=n$ .

The next result shows another family of graphs G with ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left|V\left(G\right)\right|$. Let $C_n$ be a cycle with vertices $v_1,v_2,\dots ,v_n$ and $P_m$ be a path with vertices $u_1,u_2,\dots ,u_m$ for which $u_1=v_1$ and for some $2\le i\le m$, $u_i\ne v_j$. Let H be a graph obtained from a cycle $C_n$ and k paths like $P_{m_1},P_{m_2},\dots ,P_{m_k}$ ($1\le k\le n$) such that the first vertex of any path $P_{m_i}$ must be $v_i$. Let G be a graph consisting of m graphs like H such that any both of them have at most one common vertex on their cycles. Figure 1 is a sample of graph G is formed of 4 cycles and 15 paths $P_{m_i}$, where $m_i\equiv 1(mod3)$.

Observation 3.

Let G be the graph constructed as above. If $P_{m_i}$ with vertices $u_{1_i},u_{2_i},\dots ,u_{m_i}$ is a path such that $3\mid (m_i-1)$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left|V\left(G\right)\right|$.

Proof. 

Let f be a function that assign value 1 to every vertex of the cycles and if $3\mid (m_i-1)$, we assign value 2 to vertices with indices $3t$, value 1 to vertices with indices $3t+1$, ($1\le t\le \frac{m_i-1}{3}$) and value 0 to the other vertices of the path $P_{m_i}$, except to the common vertex $u_{1_i}=v_i$ of the cycle. Therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left|V\left(G\right)\right|$. □

Let $C_n$ be a cycle and $P_m$ be a path with m vertices and let the first vertex of $P_m$ be the vertex $v_i$ of $C_n$. If $3\mid m_i$ or $3\mid (m_i-2)$, then ${\gamma }_{t\left\{R3\right\}}(C_n\cup P_m)=\left|V(C_n\cup P_m)\right|+1$. Therefore we have the following result.

Corollary 2.

In the graphs constructed above, if there are l paths $P_{m_1},P_{m_2},\dots ,P_{m_l}$ such that $3\mid m_i$ or $3\mid (m_i-2)$ for $1\le m_i\le l$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left|V\left(G\right)\right|+l$.

The Observation 3 and Corollary 2 show that for every nonnegative integer k, there is a graph G such that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left|V\left(G\right)\right|+k$.

Proposition 2.

If G is a connected graph of order $n\ge 2$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)\ge 3$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$ if and only if G has at least two vertices of degree $\mathsf{\Delta }\left(G\right)=n-1$.

Proof. 

If $n=2$, then the statement is clear. Let now $n\ge 3$ and let $f=(V_0,V_1,V_2,V_3)$ be a total Roman $\left\{3\right\}$-dominating function on G of weight ${\gamma }_{\left\{R3\right\}}\left(G\right)$. If $V_0\ne \varnothing$, then ${\sum }_{u\in N\left(v\right)}f\left(u\right)\ge 3$ for a vertex $v\in V_0$ and thus ${\gamma }_{t\left\{R3\right\}}\left(G\right)\ge 3$. If $V_0=\varnothing$, then $f\left(x\right)\ge 1$ for each vertex $x\in V\left(G\right)$ and therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)\ge n\ge 3$.

If G has at least two vertices of degree $\mathsf{\Delta }\left(G\right)=n-1$, then we may assume v and u are two adjacent vertices of maximum degree. Define the function f by $f\left(v\right)=1$, $f\left(u\right)=2$ and $f\left(x\right)=0$ for $x\in V\left(G\right)\backslash \{v,u\}$. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight 3 and hence ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$.

Conversely, assume that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$. Then there are two adjacent vertices $v,u$ with weights 1 and 2 respectively, for which $n-2$ vertices with weight 0 are adjacent to them, or there are three mutuality adjacent vertices $u,v,w$ with weights 1 for which $n-3$ vertices with weight 0 are adjacent to them. Therefore there are at least two vertices of degree $n-1$. □

As an immediate result we have:

Corollary 3.

If G has only one vertex of degree $\mathsf{\Delta }\left(G\right)=n-1$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4$.

In the follow, total Roman $\left\{3\right\}$-domination and total double Roman domination numbers are compared.

Since any partite set of a bipartite graph is an independent set, the weight of total Roman $\left\{3\right\}$-domination number of any partite set is positive. Therefore we have the following.

Proposition 3.

For any complete bipartite graph we have.

• ${\gamma }_{t\left\{R3\right\}}\left(K_{1,n}\right)=4$,
• ${\gamma }_{t\left\{R3\right\}}\left(K_{m,n}\right)=5$ for $m\in \{2,3\}$ and $n\ge 3$.
• ${\gamma }_{t\left\{R3\right\}}\left(K_{m,n}\right)={\gamma }_{dR}\left(K_{m,n}\right)=6$ for $m,n\ge 4$.

Proof. 

In any complete bipartite graph, let $V\left(G\right)=U\cup W$, where U is the small partite set and W is the big partite set.

1. This follows from Corollary 3.

2. We consider two cases.

(i)

Let $U=\{u_1,u_2\}$ and $W=\{w_1,w_2,\dots ,w_n\}$. Let f be a TR$\left\{3\right\}$DF of $K_{2,n}$. If $f\left(W\right)=2$, then $f\left(U\right)\ge 3$. If $f\left(W\right)=3$, then $f\left(U\right)\ge 2$. If $f\left(W\right)\ge 4$, since $f\left(U\right)$ is positive, then $f\left(V\right)\ge 5$. Therefore $f\left(V\right)\ge 5$. Assigning $f\left(u_1\right)=2,f\left(u_2\right)=1$ and $f\left(w_1\right)=2$, shows that ${\gamma }_{t\left\{R3\right\}}\left(K_{2,n}\right)\le 5$.

(ii)

Let $U=\{u_1,u_2,u_3\}$ and $W=\{w_1,w_2,\dots ,w_n\}$. Using sketch of the proof of item 2, ${\gamma }_{t\left\{R3\right\}}\left(K_{3,n}\right)\ge 5$. If we assign value 1 to the vertices $u_1,u_2,u_3$, weight 2 to $w_1$ and 0 to $w_j,$ for $j\ge 2$, then ${\gamma }_{t\left\{R3\right\}}\left(K_{3,n}\right)\le 5$.

3. The function f with $f\left(u_1\right)=3=f\left(w_1\right)$ and $f\left(u_i\right)=0=f\left(u_j\right)$ for $i,j\ne 1$ is a TR$\left\{3\right\}$DF for $K_{m,n}$. Therefore ${\gamma }_{t\left\{R3\right\}}\left(K_{m,n}\right)\le 6$.

Now let f be a ${\gamma }_{t\left\{R3\right\}}$ function of $K_{m,n}$ for $m,n\ge 4$. If $m,n\ge 5$, then it is easy to see that f should be assigned 0 to at least one vertex of each partite set. Therefore every partite set must have weight at least 3. If, without loss of generality, $n=4$, then let $U=\{u_1,u_2,u_3,u_4\}$. If $f\left(u_i\right)\ge 1$ for $1\le i\le 4$, then $f\left(u_i\right)=1$ for $1\le i\le 4$ and thus $f\left(W\right)\ge 2$. So $f\left(V\right)\ge 6$ and therefore ${\gamma }_{t\left\{R3\right\}}\left(K_{m,n}\right)\ge 6$, and the proof is complete. □

One can obtain a similar result for complete r-partite graphs for $r\ge 3$.

Proposition 4.

Let $G=K_{n_1,n_2,\dots ,n_r}$ be the complete r-partite graph with $r\ge 3$ and $n_1\le n_2\le \dots \le n_r$. Then:

• If $n_1=n_2=1$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$.
• If $n_1=1$ and $n_2\ge 2$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4$.
• If $n_1=2$ or $n_1\ge 3$ and $r\ge 4$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4$.
• If $r=3$ and $n_1\ge 3$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=5$.

Proof. 

Let $V={\bigcup }_{i=1}^r{U}_i$ where $U_i$ is the ith partite set with vertices $\{u_{i_1},u_{i_2},\dots ,u_{i_{n_i}}\}$.

1. This follows from Proposition 2.

2. This follows from Corollary 3.

3. Let $n_1\ge 2$. By Proposition 2, we have ${\gamma }_{t\left\{R3\right\}}\left(G\right)\ge 4$. If $n_1=2$, then define $f\left(u_{1_1}\right)=f\left(u_{1_2}\right)=f\left(u_{2_1}\right)=f\left(u_{3_1}\right)=1$ and $f\left(v\right)=0$ otherwise. Then f is a TR$\left\{3\right\}$-DF on G with $f\left(V\right)=4$ and thus ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4$. Now let $n_1\ge 3$ and $r\ge 4$. Then any TR$\left\{3\right\}$-DF f on G with $f\left(u_{1_1}\right)=f\left(u_{2_1}\right)=f\left(u_{3_1}\right)=f\left(u_{4_1}\right)=1$ and $f\left(v\right)=0$ for the other vertices, is a ${\gamma }_{t\left\{R3\right\}}$ function on G. Therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4$.

4. Let $n_1\ge 3$ and $r=3$, and let f be a TR$\left\{3\right\}$-DF function on G. Since two partite sets must have positive weight, we can assume $f\left(U_2\right)\ge 1$. If $f\left(U_2\right)=1$, then $f(U_1\cup U_3)\ge 4$. If $f\left(U_2\right)=2$, then $f(U_1\cup U_3)\ge 3$. If $f\left(U_2\right)=3$, then $f(U_1\cup U_3)\ge 2$. If $f\left(U_2\right)\ge 4$, then $f(U_1\cup U_3)\ge 1$. Thus $f\left(V\right)\ge 5$. Conversely, define $f\left(u_{1_1}\right)=f\left(u_{2_1}\right)=2$ and $f\left(u_{3_1}\right)=1$ and $f\left(v\right)=0$ otherwise. Then f is a TR$\left\{3\right\}$-DF on G with $f\left(V\right)=5$ and so ${\gamma }_{t\left\{R3\right\}}\left(G\right)=5$. □

Theorem 4.

If G is a graph with $\delta \left(G\right)=\delta \ge 2$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le \left|V\left(G\right)\right|+2-\delta$, and this bound is sharp.

Proof. 

Let $V\left(G\right)=\{u_1,u_2,\dots ,u_n\}$, and let v be a vertex of degree $\delta$ with neighbors $\{u_1,u_2,\dots ,u_{\delta }\}$. Let $U=\{v,u_{\delta +2},u_{\delta +3},\dots ,u_n\}\cup \{u_1,u_2\}$. Define the function f by $f\left(x\right)=1$ for $x\in U$ and $f\left(x\right)=0$ for $x\in V\left(G\right)\backslash U$. Then ${\sum }_{x\in N\left(u\right)}f\left(x\right)\ge 2$ for $u\in U$ and ${\sum }_{x\in N\left(u\right)}f\left(x\right)\ge 3$ for $u\in V\left(G\right)\backslash U$. Therefore f is a total Roman $\left\{3\right\}$-dominating function on G of weight $n+2-\delta$ and thus ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le \left|V\left(G\right)\right|+2-\delta$.

According to Observation 2 and Propositions 3 and 8, we note that ${\gamma }_{t\left\{R3\right\}}\left(C_n\right)=n=\left|V\left(C_n\right)\right|+2-\delta \left(C_n\right)$, ${\gamma }_{t\left\{R3\right\}}\left(K_n\right)=3=\left|V\left(K_n\right)\right|+2-\delta \left(K_n\right)$ for $n\ge 3$, ${\gamma }_{t\left\{R3\right\}}\left(K_{3,3}\right)=5=\left|V\left(K_{3,3}\right)\right|+2-\delta \left(K_{3,3}\right)$, ${\gamma }_{t\left\{R3\right\}}\left(K_{4,4}\right)=6=\left|V\left(K_{4,4}\right)\right|+2-\delta \left(K_{4,4}\right)$, ${\gamma }_{t\left\{R3\right\}}\left(K_{3,3,3}\right)=5=\left|V\left(K_{3,3,3}\right)\right|+2-\delta \left(K_{3,3,3}\right)$ and ${\gamma }_{t\left\{R3\right\}}\left(K_{n_1,n_2,\dots ,n_r}\right)=4=\left|V\left(K_{n_1,n_2,\dots ,n_r}\right)\right|+2-\delta \left(K_{n_1,n_2,\dots ,n_r}\right)$ for $r\ge 4$ and $n_1\le n_2\le \dots \le n_r=2$. All these examples demonstrate that the inequality ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le \left|V\left(G\right)\right|+2-\delta$ is sharp. □

Hao et al. defined in [12] the family of graphs $\mathcal{G}$ as follows and have proved Theorem 5 below. Let $\mathcal{G}$ be the family of graphs that can be obtained from a star $S_t=K_{1,t-1}$ of order $t\ge 2$ by adding a pendant edge to each vertex of $V\left(S_t\right)$ and adding any number of edges joining the leaves of $S_t$.

Theorem 5.

[12] For any connected graph G of order $n\ge 2$,

$${\gamma }_{tdR}\left(G\right)\le 2n-\mathsf{\Delta }$$

with equality if and only if $G\in \{P_2,P_3,C_3\}\cup \mathcal{G}$.

This theorem with a little changing may be explored as follows.

Theorem 6.

For any connected graph G of order $n\ge 2$,

$${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 2n-\mathsf{\Delta }$$

with equality if and only if $G\in \{P_2,P_3\}\cup \mathcal{G}$.

3. Total Roman $\left\{3\right\}$-domination and Total Domination

In this section we study the relationship between total domination and total Roman $\left\{3\right\}$-domination of a graph.

In [10] (Proposition 8) the authors proved that, if G is a graph, then $2\gamma \left(G\right)\le {\gamma }_{dR}\left(G\right)\le 3\gamma \left(G\right)$.

If we use the method of the proof of Proposition 8 of [10], then it is easy to show that:

If G is a graph with a ${\gamma }_{\left\{R3\right\}}$-function $f=(V_0,V_2,V_3)$, then $2\gamma \left(G\right)\le {\gamma }_{\left\{R3\right\}}\left(G\right)\le 3\gamma \left(G\right)$.

In [13] Proposition 17 authors proved that:

If G is a graph, then $\gamma \left(G\right)+2\le {\gamma }_{\left\{R3\right\}}\left(G\right)\le 3\gamma \left(G\right)$, and these bounds are sharp. However, we have the following.

Proposition 5.

If G is a graph without isolated vertices, then ${\gamma }_t\left(G\right)+1\le {\gamma }_{t\left\{R3\right\}}\left(G\right)\le 3{\gamma }_t\left(G\right)$.

Proof. 

Let S be a ${\gamma }_t$-set of G. Then $(V_0=V\backslash S,\varnothing ,\varnothing ,V_3=S)$ is a ${\gamma }_{t\left\{R3\right\}}$-function of G. Therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 3{\gamma }_t\left(G\right)$.

For the lower bound, let $f=(V_0,V_1,V_2,V_3)$ be a ${\gamma }_{t\left\{R3\right\}}$-function of G. We distinguish two cases.

Case 1. Let $|V_2|\ge 1$ or $|V_3|\ge 1$. Then ${\gamma }_t\left(G\right)\le |V_1|+|V_2|+|V_3|\le |V_1|+2|V_2|+3|V_3|-1={\gamma }_{t\left\{R3\right\}}\left(G\right)-1$.

Case 2. Let $V_2=V_3=\varnothing$. By the definition, $\delta \left(G\left[V_1\right]\right)\ge 2$. Therefore, for each vertex $v\in V_1$, the subgraph $G[V_1\backslash \left\{v\right\}]$ does not contain an isolated vertex. Consequently, $V_1\backslash \left\{v\right\}$ is total dominating set of G and hence ${\gamma }_t\left(G\right)\le {\gamma }_{t\left\{R3\right\}}\left(G\right)-1$. □

By Proposition 5 the question may arise as whether for any positive integer r, exists a graph G for which ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+r$, where $1\le r\le 2{\gamma }_t\left(G\right)$. For $r=1$ we have. If G is a connected graph of order $n\ge 2$ with at least two vertices of maximum degree $\mathsf{\Delta }\left(G\right)=n-1$, then Proposition 2 implies that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$. Since ${\gamma }_t\left(G\right)=2$ for such graphs, we observe that ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+1$.

Proposition 6.

If G is a graph without isolated vertices, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+1$ if and only if G has at least two vertices of degree $\mathsf{\Delta }=\left|V\right(G\left)\right|-1$, in the other words ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$ and ${\gamma }_t\left(G\right)=2$.

Proof. 

The part “if“ has been proved. Part “only if“: Let G be a graph with ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+1$. Let $f=(V_0,V_1,V_2,V_3)$ be a ${\gamma }_{t\left\{R3\right\}}\left(G\right)$ function. Therefore $V_1\cup V_2\cup V_3$ is a total dominating set for G, and $|V_1|+|V_2|+|V_3|\ge {\gamma }_t\left(G\right)={\gamma }_{t\left\{R3\right\}}\left(G\right)-1=|V_1|+2|V_2|+3V_3|-1$. Therefore $|V_2|+2|V_3|\le 1$ that is $|V_2|\le 1$ and $|V_3|=0$. If $|V_2|=1=|V_1|$ or $|V_2|=0$ and $|V_1|=3$, then G has at least two vertices of degree $\mathsf{\Delta }\left(G\right)=\left|V\right(G\left)\right|-1$. Now we show that there are not any cases for G. On the contrary, we suppose that there are different cases. (1) $|V_2|=1$ and $|V_1|\ge 2$. (2) $|V_2|=0$ and $|V_1|\ge 4$.

Case 1. Let $V_2=\left\{v\right\}$, $|V_1|\ge 2$. Assume first that there exist two vertices $v_1,v_2\in V_1$ which are adjacent to the vertex v. Then $V_2\cup V_1\backslash \left\{v_1\right\}$ is a ${\gamma }_t\left(G\right)$-set of size $|V_1|$ and so ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2+\left|V_1\right|$, a contradiction. Assume next that there exists only one vertex, say $v_1\in V_1$, which is adjacent to v. Then all other vertices of $V_1$ have at least two neighbors in $V_1$. If $v_2\in V_1$ with $v_2\ne v_1$, then we observe that $V_2\cup V_1\backslash \left\{v_2\right\}$ is a ${\gamma }_t\left(G\right)$-set of size $|V_1|$. It follows that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2+\left|V_1\right|$, a contradiction.

Case 2. Let $|V_2|=0$ and $|V_1|\ge 4$. Then there exist two vertices $v_1,v_2$ in which each of them has neighbors in $V_1\backslash \{v_1,v_2\}$ and $G(V_1\backslash \{v_1,v_2\})$ has no isolated vertex. Therefore $V_1\backslash \{v_1,v_2\}$ is a ${\gamma }_t\left(G\right)$-set that is also a contradiction. □

Now we show that for any positive integer n and integer $2\le r\le 2n$, there exists a graph G for which ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=n+r$.

Proposition 7.

Let n and r be positive integers with $2\le r\le 2n$. Then there exists a graph G for which ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=n+r$.

Proof. 

For graph G with ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=n+2$, we consider the following graph. Let H be the graph consisting of a cycle $C_{n+2}$ with $n\ge 3$ and a vertex set $V_0$ of $\left(\genfrac{}{}{0pt}{}{n+2}{3}\right)$ further vertices. Let each vertex of $V_0$ be adjacent to 3 vertices of $V\left(C_{n+2}\right)$ such that the neighborhoods of every two distinct vertices of $V_0$ are different. Let $V_1=V\left(C_{n+2}\right)$. Then ${\gamma }_{t\left\{R3\right\}}\left(H\right)=n+2$ and ${\gamma }_t\left(H\right)=n$ (Figure 2).

For ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=n+k$, where $3\le k\le n-1$. Let $k=3$. For ${\gamma }_t\left(G\right)=4$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=7$, we consider the cycle $C_7$. For ${\gamma }_t\left(G\right)\ge 5$, let H be the above graph where ${\gamma }_t\left(H\right)=n\ge 3$ and ${\gamma }_{t\left\{R3\right\}}\left(H\right)=n+2\ge 5$. Now we consider $G=H\cup K_3$. Then ${\gamma }_t\left(G\right)\ge n\ge 4$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=n+3$.

Let $k=4$. For ${\gamma }_t\left(G\right)=5$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=9$, we consider the cycle $C_9$. For ${\gamma }_t\left(G\right)\ge 6$, consider the graphs $G^{\prime }$ with ${\gamma }_{t\left\{R3\right\}}\left(G^{\prime }\right)={\gamma }_t\left(G^{\prime }\right)+3$ for ${\gamma }_t\left(G^{\prime }\right)\ge 4$. Now we let $G=G^{\prime }\cup K_3$. Then we have ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+4$.

For $5\le k\le n-1$ we use induction on k. Let for any integer $4\le m\le k-1$ there exist graphs $G^{\prime }$ such that ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+m$ for ${\gamma }_t\left(G\right)\ge m+1$. Let $m=k$. For ${\gamma }_t\left(G\right)=k+1$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2k+1$, we consider the cycle $C_{2k+1}$. For graphs G with ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+k$ for ${\gamma }_t\left(G\right)\ge k+2$, using hypothesis of induction, let $G^{\prime }$ be the graphs with ${\gamma }_{t\left\{R3\right\}}\left(G^{\prime }\right)={\gamma }_t\left(G^{\prime }\right)+k-1$ with ${\gamma }_t\left(G^{\prime }\right)\ge k$. Now we let $G=G^{\prime }\cup K_3$. It can be seen ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+k$ for ${\gamma }_t\left(G\right)\ge k+2$.

We now verify the case of ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2{\gamma }_t\left(G\right)+r$ for $0\le r\le {\gamma }_t\left(G\right)$, that is, we wish to show the existence of graphs G, so that ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n+r$ for $0\le r\le n$. Let $r=0$. For even n, let $G=C_{2n}$. Then ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n$.

For odd $n=2k+1$, if $2n\equiv 1\left(mod3\right)$ or $2n\equiv 0\left(mod3\right)$, then we let $G=P_{2n-1}$, and by Observation 1, it can be seen that ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n$.

If $2n\equiv 2\left(mod3\right)$, consider a cycle $C_{2n-1}$ with an additional vertex a that is adjacent to two vertices $v_1$ and $v_2$. Then ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n$.

For $r=1$ and positive even integer n, consider $G=(\frac{n}{2}-1)P_3\cup C_5^{+}$, where $(\frac{n}{2}-1)P_3$ is the union of $\frac{n}{2}-1$ of path $P_3$ and $C_5^{+}$ is the cycle $C_5$ with a chord, then ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n+1$. For $r=1$ and positive odd integer n, consider $G=(\frac{n-1}{2}-1)P_3\cup P_5^{+}$ where $P_5^{+}$ is the path $P_5$ with an additional vertex adjacent to the second or fourth vertex of $P_5$, then ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n+1$.

For $2\le r\le n-1$, we do as follows. Let $r=2$ and so $n\ge 3$. Let $n=3$ and $2n+2=8$. Let $G_1$ be a graph constructed from path $P_5$ with vertices $v_1,v_2,v_3,v_4,v_5$ with additional vertices $u_{12},u_{13},u_{14},u_{52},u_{53},u_{54},u_{24}$ such that the given vertex $u_{i,j}$ is adjacent to vertices $v_i$ and $v_j$ of $P_5$. Then ${\gamma }_t\left(G_1\right)=3$ and ${\gamma }_{t\left\{R3\right\}}\left(G_1\right)=8$.

Let $n=4$ and $2n+2=10$. Then say $G_2=2C_5^{+}$. Let $n=5$ and so $2n+2=12$. Then say $G_3=C_5^{+}\cup P_5^{+}$. For ${\gamma }_t\left(G\right)=k$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2k+2$, where $r+1\le k\le n$, there consider three cases.

• If $k\equiv 0\left(mod3\right)$, then we say $G=\frac{k-3}{3}{P}_5\cup G_1$.
• If $k\equiv 1\left(mod3\right)$, then we say $G=\frac{k-4}{3}{P}_5\cup G_2$.
• If $k\equiv 2\left(mod3\right)$, then we say $G=\frac{k-5}{3}{P}_5\cup G_3$.

It is easy to verifiable, ${\gamma }_t\left(G\right)=k$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2k+2$.

Let $r=3$ and so $n\ge 4$. For graph $G_1^{\prime }$ with ${\gamma }_t\left(G_1^{\prime }\right)=4$ and ${\gamma }_{t\left\{R3\right\}}\left(G_1^{\prime }\right)=11$, we let $G_1^{\prime }=P_4\cup C_5^{+}$. For graph $G_2^{\prime }$ with ${\gamma }_t\left(G_2^{\prime }\right)=5$ and ${\gamma }_{t\left\{R3\right\}}\left(G_2^{\prime }\right)=13$, we let $G_2^{\prime }=P_4\cup P_5^{+}$. And for graph $G_3^{\prime }$ with ${\gamma }_t\left(G_3^{\prime }\right)=6$ and ${\gamma }_{t\left\{R3\right\}}\left(G_3^{\prime }\right)=15$, we let $G_3^{\prime }=3C_5^{+}$. For ${\gamma }_t\left(G\right)=k$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2k+3$, where $r+1\le k\le n$, there consider three cases.

• If $k\equiv 1\left(mod3\right)$, then we say $G=\frac{k-4}{3}{P}_5\cup G_1^{\prime }$.
• If $k\equiv 2\left(mod3\right)$, then we say $G=\frac{k-5}{3}{P}_5\cup G_2^{\prime }$.
• If $k\equiv 0\left(mod3\right)$, then we say $G=\frac{k-6}{3}{P}_5\cup G_3^{\prime }$.

Let $r\ge 4$ and $n\ge r+1$. For graph G with ${\gamma }_t\left(G\right)=k$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2k+r$ where $r+1\le k\le n$, there consider two cases.

Case 1. Let r be an even integer. Then there exists a graph $G^{\prime }$ for which ${\gamma }_t\left(G^{\prime }\right)=k-(r-2)$ and ${\gamma }_{t\left\{R3\right\}}\left(G^{\prime }\right)=2k-2(r-2)+2$. Now let $G=\frac{r-2}{2}{P}_4\cup G^{\prime }$. Then ${\gamma }_t\left(G\right)=r-2+{\gamma }_t\left(G^{\prime }\right)=k$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3(r-2)+{\gamma }_{t\left\{R3\right\}}\left(G^{\prime }\right)=3(r-2)+2k-2(r-2)+2=2k+r$.

Case 2. Let r be an odd integer. Then there exists a graph $G^{″}$ for which ${\gamma }_t\left(G^{″}\right)=k-(r-3)$ and ${\gamma }_{t\left\{R3\right\}}\left(G^{″}\right)=2k-2(r-3)+3$. If we consider $G=\frac{r-3}{2}{P}_4\cup G^{″}$. Then ${\gamma }_t\left(G\right)=r-3+{\gamma }_t\left(G^{″}\right)=k$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3(r-3)+{\gamma }_{t\left\{R3\right\}}\left(G^{″}\right)=2k+r$.

Finally, we want discuss the case of $r=n$, that is we want to find graphs G with ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3n$. For $n=2$ and $3n=6$, let $G=P_4$. For G with ${\gamma }_t\left(G\right)=3$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=9$, let $G=H_1$ be a graph constructed from $P_5$ with vertices $v_1,v_2,v_3,v_4,v_5$ with three additional vertices $w_1,w_2,w_3$ and three pendant edges $v_2{w}_2$, $v_3{w}_3$, $v_4{w}_4$. Then it can be seen that ${\gamma }_t\left(H_1\right)=3$ and ${\gamma }_{t\left\{R3\right\}}\left(H_1\right)=9$.

Let $n\ge 4$. If n is an even, then let $G=\frac{n}{2}{P}_4$ and if n is an odd, then let $G=\frac{n-3}{2}{P}_4\cup H_1$. In both cases ${\gamma }_t\left(G\right)=n$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3n$. □

4. Total Roman $\left\{3\right\}$ and Total Roman $\left\{2\right\}$-domination

In [13] it has been shown that, for a connected graph G with a ${\gamma }_{\left\{R3\right\}}$-function $f=(V_0,V_2,V_3)$, ${\gamma }_{\left\{R3\right\}}\left(G\right)\ge \gamma \left(G\right)+{\gamma }_{\left\{R2\right\}}\left(G\right)$.

In this section we investigate the relation between total Roman $\left\{3\right\}$ and total Roman $\left\{2\right\}$-domination. First we have the following.

Observation 7.

Let G be a graph and $(V_0,V_1,V_2)$ be a ${\gamma }_{t\left\{R2\right\}}$ function of G. Then $(V_0^{\prime }=V_0,V_2^{\prime }=V_1,V_3^{\prime }=V_2)$ is a TR$\left\{3\right\}$-DF function. Conversely, if $(V_0,V_1,V_2,V_3)$ is a ${\gamma }_{t\left\{R3\right\}}$ of G, then $(U_0=V_0,U_1=V_1\cup V_2,U_2=V_3)$ is a TR$\left\{2\right\}$-DF of G.

Proof. 

The proof is straightforward. □

The following results state the relation between ${\gamma }_{t\left\{R3\right\}}$ and ${\gamma }_{t\left\{R2\right\}}$ of graphs G when ${\gamma }_{t\left\{R3\right\}}\left(G\right)$ is small.

Proposition 8.

Let G be a graph. Then:

• ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$ if and only if ${\gamma }_{t\left\{R2\right\}}\left(G\right)=2$.
• If ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4$, then ${\gamma }_{t\left\{R2\right\}}\left(G\right)=3$.
• If ${\gamma }_{t\left\{R2\right\}}\left(G\right)=3$, then $4\le {\gamma }_{t\left\{R3\right\}}\left(G\right)\le 5$.

Proof. 

1. Let ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$. Then there exist two adjacent vertices $v,u$ with label $2,1$ respectively so that each vertex with label 0 is adjacent to them or there exist three mutually adjacent vertices $v,u,w$ with label 1 so that each vertex with label 0 is adjacent to them. In the first case, we change the vertex with label 2 to the label 1 and in the second case we change one of the vertices with label 1 to the label 0. These changing labels give us a ${\gamma }_{t\left\{R2\right\}}\left(G\right)$-function with weight 2. Conversely, let ${\gamma }_{t\left\{R2\right\}}\left(G\right)=2$. Then there exist two vertices with label 1 for which every vertex is adjacent to them. We change one of the labels to 2, and therefore the result holds.

2. Let ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4$. There are three cases.

2.1.

There exist 4 vertices $v,u,w,z$ with label 1 for which the induced subgraph by them is the cycle $C_4$, the graph $K=K_4-e$ or the complete graph $K_4$. In any induced subgraph, there are no two vertices of them for which any vertex with label 0 is adjacent to them. Thus in the case of a TR$\left\{2\right\}$-DF we change one of the labels 1 to the label 0. Therefore ${\gamma }_{t\left\{R2\right\}}\left(G\right)=3$.

2.2.

There exist 2 vertices $v,u$ with label 1 and one vertex w with label 2, for which the induced subgraph by them is the cycle $C_3$, or the path $P_3=v-w-u$. In any of the two cases each vertex with label 0 is adjacent to $v,w$ or $u,w$ or three of them. Now we change the label of w to 1, and we obtain a ${\gamma }_{t\left\{R2\right\}}$-function for G with weight 3.

2.3.

There exist 2 vertices $v,u$ with label 3 and label 1, respectively, for which the induced subgraph by $v,u$ is $K_2$. By this assumption each vertex with label 0 is adjacent to v, but there maybe exist some vertices (none of them) which are adjacent to u. Now we change the label v to 2, and we obtain a ${\gamma }_{t\left\{R2\right\}}$-function for G with weight 3.

3. Let ${\gamma }_{t\left\{R2\right\}}\left(G\right)=3$. There are two cases.

3.1.

There exist 3 vertices $v,u,w$ with label 1 for which the induced subgraph by $v,u,w$ is the cycle $C_3$ or a path $P_3$. If each vertex with label 0 is adjacent to $v,w$ or $u,w$, then by changing the label w to 2, we obtain a ${\gamma }_{t\left\{R3\right\}}$-function for G with weight 4.

If some vertices with label 0 are adjacent to $v,u$, some of them are adjacent to $v,w$ and the other are adjacent $u,w$, then by changing two vertices of $v,u,w$ to label 2, we obtain a ${\gamma }_{t\left\{R3\right\}}$-function for G with weight 5.

3.2.

There exist 2 vertices $v,u$ with label 2 and label 1, respectively, for which the induced subgraph by $v,u$ is $K_2$. By this assumption each vertex with label 0 is adjacent to v, but there maybe exist some vertices (none of them) which are adjacent to u. Now we change the label v to 3, and we obtain a ${\gamma }_{t\left\{R3\right\}}$-function for G with weight 4. Therefore $4\le {\gamma }_{t\left\{R3\right\}}\left(G\right)\le 5$. □

In the following we want to find the relation between total Roman $\left\{3\right\}$-domination, total domination and total Roman $\left\{2\right\}$-domination of graphs.

Observation 8.

Let G be a connected graph with a ${\gamma }_{t\left\{R3\right\}}$-function $f=(V_0,V_2,V_3)$. Then ${\gamma }_{t\left\{R3\right\}}\left(G\right)\ge {\gamma }_t\left(G\right)+{\gamma }_{t\left\{R2\right\}}\left(G\right)$.

Proof. 

Let $(V_0,V_2,V_3)$ be a ${\gamma }_{t\left\{R3\right\}}$-function of G. Then ${\gamma }_t\left(G\right)\le |V_2|+|V_3|$. If we define $g=(V_0^{\prime }=V_0,V_1^{\prime }=V_2,V_2^{\prime }=V_3)$, then g is a total Roman $\left\{2\right\}$-dominating function on G. Therefore ${\gamma }_t\left(G\right)+{\gamma }_{t\left\{R2\right\}}\left(G\right)\le |V_2|+|V_3|+|V_1^{\prime }|+2|V_2^{\prime }|\le 2|V_2|+3|V_3|={\gamma }_{t\left\{R3\right\}}\left(G\right)$. □

In Observation 8 the condition of ${\gamma }_{t\left\{R3\right\}}$-function $f=(V_0,V_2,V_3)$ is necessary. Because there are many graphs for which the result of Observation 8 does not hold. For example, for the complete graphs $K_n$$(n\ge 2)$, cycles $C_n$ and paths $P_n$ for $n\ge 5$, we observe that ${\gamma }_{t\left\{R3\right\}}\left(G\right)<{\gamma }_t\left(G\right)+{\gamma }_{t\left\{R2\right\}}\left(G\right)$. However, in the following we establish, for any integer $n\ge 5$, there is a graph G such that ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_{t\left\{R2\right\}}\left(G\right)+{\gamma }_t\left(G\right)$.

Proposition 9.

For any positive integer $n\ge 5$, there is a graph G for which ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_{t\left\{R2\right\}}\left(G\right)+{\gamma }_t\left(G\right)$.

Proof. 

For $n=5$ let $G=C_5^{+}$. Then ${\gamma }_{t\left\{R2\right\}}\left(G\right)=3$, ${\gamma }_t\left(G\right)=2$ and ${\gamma }_{t\left\{R3\right\}}\left(G\right)=5$. For $n=6$, let G be a bistar of order 6. Then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=6=4+2={\gamma }_{t\left\{R2\right\}}\left(G\right)+{\gamma }_t\left(G\right)$. For $n=7$, let $G=G_1$ in Figure 3. For $n=8$, let $G=G_2$ in Figure 3. For $n=9$, let $G=G_2$ in Figure 3. For $n\ge 10$, by induction we consider the graph $G=C_5^{+}\cup H$ where the graph H (H may be connected or disconnected) for which ${\gamma }_{t\left\{R3\right\}}\left(H\right)=n-5={\gamma }_{t\left\{R2\right\}}\left(H\right)+{\gamma }_t\left(H\right)$. □

Finally, we show that for any positive integer $n\ge 5$, there is a graph G such that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=n$, ${\gamma }_{t\left\{R2\right\}}\left(G\right)=n-1$ and ${\gamma }_t\left(G\right)=n-2$.

For this, let G be the graph constructed in Proposition 3 as graph H for $n\ge 5$. Then ${\gamma }_{t\left\{R3\right\}}\left(H\right)=n$, ${\gamma }_{t\left\{R2\right\}}\left(H\right)=n-1$ and ${\gamma }_t\left(H\right)=n-2$.

5. Large Total Roman $\left\{3\right\}$-domination Number

In this section, we characterize connected graphs G of order n with ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-k$ for $1\le k\le 4$. For this we use the following result.

Theorem 9.

Let G be a connected graph of order $n\ge 2$. Then ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le \left(3n\right)/2$, with equality if and only if G is the corona $H\circ K_1$ where H is a connected graph.

Proof. 

If $n=2$, then the statement is valid. Let now $n\ge 3$. If $\left|L\right(G\left)\right|\le n/2$, then define $f:V\left(G\right)\to \{0,1,2,3\}$ by $f\left(x\right)=2$ for $x\in L\left(G\right)$ and $f\left(x\right)=1$ for $x\in V\left(G\right)\backslash L\left(G\right)$. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight

$$2\left|L\left(G\right)\right|+n-\left|L\left(G\right)\right|=n+\left|L\left(G\right)\right|\le \frac{3n}{2}.$$

If $\left|L\right(G\left)\right|>n/2$, then define $f:V\left(G\right)\to \{0,1,2,3\}$ by $f\left(x\right)=1$ for $x\in L\left(G\right)$ and $f\left(x\right)=2$ for $x\in V\left(G\right)\backslash L\left(G\right)$. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight

$$\left|L\left(G\right)\right|+2(n-|L\left(G\right)\left|\right)=2n-\left|L\left(G\right)\right|<\frac{3n}{2}.$$

If $G=H\circ K_1$ for a is a connected graph H, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left(3n\right)/2$.

Conversely, let ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left(3n\right)/2$. Then the proof above shows that $\left|L\right(G\left)\right|=n/2$. Assume that there exists a vertex $v\in V\left(G\right)$ which is neither a leaf nor a support vertex. Define $f:V\left(G\right)\to \{0,1,2,3\}$ by $f\left(x\right)=1$ for $x\in L\left(G\right)\cup \left\{v\right\}$ and $f\left(x\right)=2$ for $x\in V\left(G\right)\backslash \left(L\right(G)\cup \{v\left\}\right)$. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight

$$\left|L\left(G\right)\right|+2(n-|L\left(G\right)|-1)+1=2n-\left|L\left(G\right)\right|-1=\frac{3n}{2}-1,$$

a contradiction. Thus every vertex is a leaf or a support vertex. Since $\left|L\right(G\left)\right|=n/2$, we deduce that $G=H\circ K_1$ with a connected graph H. □

Corollary 4.

For any connected graph G of order $n\ge 2$, ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-1$ if and only if $G=P_2$.

Proof. 

Let ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-1$. Then Theorem 9 implies $2n-1\le \left(3n\right)/2$ and thus $n=2$. Clearly, the statement is valid for $P_2$. □

Corollary 5.

For any connected graph G of order $n\ge 3$, ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-2$ if and only if $G\in \{P_3,P_4\}$.

Proof. 

If $G\in \{P_3,P_4\}$, then the statement is valid. Conversely, let ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-2$. Then Theorem 9 implies $2n-2\le \left(3n\right)/2$ and thus $n\le 4$, with equality if and only if $G=P_4$. In the remaining case $n=3$, we observe that $G\in \{P_3,C_3\}$ with ${\gamma }_{t\left\{R3\right\}}\left(P_3\right)=4$ and ${\gamma }_{t\left\{R3\right\}}\left(C_3\right)=3$, and therefore $G=P_3$. □

Next we characterize the graphs G with the property that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2\left|V\left(G\right)\right|-3$.

Theorem 10.

For any connected graph G of order $n\ge 3$, ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-3$ if and only if $G\in \{C_3,P_3\circ K_1,C_3\circ K_1\}$.

Proof. 

If $G\in \{C_3,P_3\circ K_1,C_3\circ K_1\}$, then the statement is valid. Conversely, let ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-3$. If $\mathsf{\Delta }\left(G\right)=2$, then $G\in \{P_n,C_n\}$ and we conclude by Observations 1, 2 that $G=C_3$. If $\mathsf{\Delta }\left(G\right)=3$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-3=2n-\mathsf{\Delta }\left(G\right)$ and so by Theorem 6, $G\in \mathcal{G}$. Therefore $G\in \{P_3\circ K_1,C_3\circ K_1\}$. Let $\mathsf{\Delta }\left(G\right)\ge 4$. Then by Theorem 6, ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 2n-\mathsf{\Delta }\left(G\right)=2n-4<2n-3$. Thus $G\in \{C_3,P_3\circ K_1,C_3\circ K_1\}$, and the proof is complete. □

Let $\mathcal{H}$ be the family of connected graphs order 5 with $\mathsf{\Delta }\left(G\right)=3$ which have exactly one leaf or the tree $T_5$ consisting of the path $v_1{v}_2{v}_3{v}_4$ such that $v_2$ is adjacent to a further vertex w.

Let $\mathcal{F}$ be the family of graphs $G=Q\circ K_1$ with a connected graph Q of order 4.

Observation 11.

If $G\in \{\mathcal{F},\mathcal{H}\}$, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-4$.

Proof. 

Clearly, ${\gamma }_{t\left\{R3\right\}}\left(T_5\right)=2n-4=6$. Let $G\in \mathcal{H}$ be of order 5 with exactly one leaf u. If v is the support vertex of u, then $f:V\left(G\right)\to \{0,1,2,3\}$ with $f\left(v\right)=2$ and $f\left(x\right)=1$ for $x\in V\left(G\right)\backslash \left\{v\right\}$ is a TR$\left\{3\right\}$-DF on G and therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)=6=2n-4$.

If $G=Q\circ K_1$ with a connected graph Q of order 4, then we have seen in proof of Theorem 9 that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left(3n\right)/2=2n-4=12$. □

Theorem 12.

For any connected graph G of order $n\ge 4$, we have ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-4$ if and only if $G\in \{C_4,P_5\}\cup \{claw,paw\}\cup \mathcal{H}\cup \mathcal{F}$ where claw is $K_{1,3}$ and paw is obtained from $K_{1,3}$ by adding one edge between two arbitrary distinct vertices.

Proof. 

Let $G\in \{C_4,P_5\}\cup \{claw,paw\}\cup \mathcal{H}\cup \mathcal{F}$. By Observations 1, 2 and 11, we have ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-4$.

Conversely, let ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-4$. According to Theorem 9, we have $2n-4={\gamma }_{t\left\{R3\right\}}\left(G\right)\le \left(3n\right)/2$ and thus $n\le 8$ with equality if and only if G is the corona $H\circ K_1$ with a connected graph H of order 4. Therefore $G\in \mathcal{F}$ if $n=8$. Let now $n\le 7$.

If $\mathsf{\Delta }\left(G\right)=2$, then $G\in \{P_n,C_n\}$ and by Observations 1, 2, we have $n=2n-4$ which implies $n=4$ and $G=C_4$, or $n+1=2n-4$ which implies $n=5$ and $G=P_5$ or $n+2=2n-4$ which implies $G=P_6$. Since ${\gamma }_{t\left\{R3\right\}}\left(C_4\right)=4=2n-4$ and ${\gamma }_{t\left\{R3\right\}}\left(P_5\right)=6=2n-4$ but ${\gamma }_{t\left\{R3\right\}}\left(P_6\right)=7\ne 2n-4$, we deduce that $G\in \{C_4,P_5\}$.

Let now $\mathsf{\Delta }\left(G\right)=3$. Next we discuss the cases $n=4,5,6$ or $n=7$.

If $n=4$, then for only two graphs G, the claw and the paw, we have ${\gamma }_{t\left\{R3\right\}}\left(G\right)=4=2n-4$.

If $n=5$, it is simply verified that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=6=2n-4$ if an only if $G\in \mathcal{H}$.

If $n=6$, then let v be a vertex of degree 3 with the neighbors $u_1,u_2,u_3$, and let $w_1$ and $w_2$ be the remaining vertices. Assume, without loss of generality, that $w_1$ is adjacent to $u_1$.

Case 1: Assume that $w_2$ is adjacent to $u_1$. Then $f:V\left(G\right)\to \{0,1,2,3\}$ with $f\left(v\right)=f\left(u_1\right)=3$ and $f\left(x\right)=0$ for $x\ne v,u_1$ is a TR$\left\{3\right\}$-DF on G and therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 6$.

Case 2: Assume that $w_2$ is adjacent to $w_1$. Then $f:V\left(G\right)\to \{0,1,2,3\}$ with $f\left(v\right)=f\left(w_1\right)=3$, $f\left(u_1\right)=1$ and $f\left(x\right)=0$ for $x\ne v,u_1,w_1$ is a TR$\left\{3\right\}$-DF on G and therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 7$.

Case 3: Assume that $w_2$ is adjacent to $u_2$ or $u_3$, say $u_2$. If there are no further edges, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=9\ne 2n-4$.

Now assume that there are further edges. If $w_2$ is adjacent to $u_3$, then $f:V\left(G\right)\to \{0,1,2,3\}$ with $f\left(u_1\right)=2$ and $f\left(x\right)=1$ for $x\ne u_1$ is a TR$\left\{3\right\}$-DF on G and therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 7$. If $w_1$ is adjacent to $u_2$, then $f:V\left(G\right)\to \{0,1,2,3\}$ with $f\left(v\right)=f\left(u_2\right)=3$ and $f\left(x\right)=0$ for $x\ne v,u_2$ is a TR$\left\{3\right\}$-DF on G and therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 6$. If $u_1$ is adjacent to $u_2$ and there are no further edges, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)=9\ne 2n-4$. If finally, $u_3$ is adjacent to $u_2$ or $u_1$, say $u_2$, then $f:V\left(G\right)\to \{0,1,2,3\}$ with $f\left(u_1\right)=f\left(u_2\right)=3$, $f\left(v\right)=1$ and $f\left(x\right)=0$ for $x\ne v,u_1,u_2$ is a TR$\left\{3\right\}$-DF on G and therefore ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 7$. Thus we see that there is no graph G of order 6 with ${\gamma }_{t\left\{R3\right\}}\left(G\right)=8=2n-4$.

Let now $n=7$. If $\left|L\right(G\left)\right|\le 2$, then define $f:V\left(G\right)\to \{0,1,2,3\}$ by $f\left(x\right)=2$ for $x\in L\left(G\right)$ and $f\left(x\right)=1$ for $x\in V\left(G\right)\backslash L\left(G\right)$. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight $9<10=2n-4$. If $\left|L\right(G\left)\right|\ge 4$, then define $f:V\left(G\right)\to \{0,1,2,3\}$ by $f\left(x\right)=0$ for $x\in L\left(G\right)$ and $f\left(x\right)=3$ for $x\in V\left(G\right)\backslash L\left(G\right)$. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight $9<10=2n-4$.

Finally, assume that $\left|L\right(G\left)\right|=3$. If G has exactly 3 support vertices, then define $f:V\left(G\right)\to \{0,1,2,3\}$ by $f\left(x\right)=1$ for $x\in L\left(G\right)$, $f\left(x\right)=2$ for $x\in S\left(G\right)$ and $f\left(x\right)=0$ for the remaining vertex. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight $9<10=2n-4$. If G has exactly 2 support vertices, then define $f:V\left(G\right)\to \{0,1,2,3\}$ by $f\left(x\right)=0$ for $x\in L\left(G\right)$, $f\left(x\right)=3$ for $x\in S\left(G\right)$ and $f\left(x\right)=1$ for the remaining two vertices. Then f is a total Roman $\left\{3\right\}$-dominating function on G of weight $8<10=2n-4$.

Let $\mathsf{\Delta }\left(G\right)=4$. By Theorem 6, ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2n-4$ if and only if $G\in \mathcal{G}\subseteq \mathcal{F}$.

Let $\mathsf{\Delta }\left(G\right)\ge 5$. Then by Theorem 6 ${\gamma }_{t\left\{R3\right\}}\left(G\right)\le 2n-5<2n-4$. Therefore the proof is complete. □

6. Complexity

In this section, we study the complexity of total Roman $\left\{3\right\}$-domination of graphs. We show that the total Roman $\left\{3\right\}$-domination problem is $NP$-complete for bipartite graphs. Consider the following decision problem.

Total Roman $\left\{3\right\}$-domination problem TR3DP.

Instance: Graph $G=(V,E)$, and a positive integer $k\le \left|V\right|$.

Question: Does G have a total Roman $\left\{3\right\}$-domination of weight at most k?

It is well-known that the Exact-3-Cover (X3C) problem is NP-complete. We show that the NP-completeness of TR3D problem by reducing the Exact-3-Cover (X3C), to TR3D.

EXACT 3-COVER ($X3C$)

Instance: A finite set X with $\left|X\right|=3q$ and a collection C of 3-element subsets of X.

Question: Is there a subcollection $C^{\prime }$ of C such that every element of X appears in exactly one element of $C^{\prime }$?

Theorem 13.

TR3D is $NP$-Complete for bipartite graphs.

Proof. 

It is clear that TR3DP belongs to $\mathcal{N}\mathcal{P}$. Now we show that, how to transform any instance of X3C into an instance G of TR3D so that, the solution one of them is equivalent to the solution of the other one. Let $X=\{x_1,x_2,\dots ,x_{3r}\}$ and $C=\{C_1,C_2,\dots ,C_t\}$ be an arbitrary instance of X3C.

For each $x_i\in X$, we form a graph $G_i$ obtained from a path $P_5:y_{i_1}$-$y_{i_2}$-$y_{i_3}$-$y_{i_4}$-$y_{i_5}$ by adding the edge $y_{i_2}{y}_{i_5}$. For each $C_j\in C$, we form a star $K_{1,5}$ with center $c_j$ for which one leaf is labeled $l_j$. Let $L=\{l_1,l_2,\dots ,l_t\}$. Now to obtain a graph G, we add edges $l_j{y}_{i_1}$ if $y_{i_1}\in C_j$. Set $k=4t+13r$. Let $H=\langle {\bigcup }_{i=1}^{3r}V\left(G_i\right)\rangle$ be the subgraph of G induced by the ${\bigcup }_{i=1}^{3r}V\left(G_i\right)$. Observe that for every total Roman $\left\{3\right\}$-dominating function f on G with $f\left(V\left(G_i\right)\right)\ge 4$, all vertices on each cycle $C_4=y_{i_2}$-$y_{i_3}$-$y_{i_4}$-$y_{i_5}$-$y_{i_2}$ are total Roman $\left\{3\right\}$-dominated. Moreover, since $G_i$ has a total Roman $\left\{3\right\}$-domination number equal to 6, we can assume that $f\left(V\left(G_i\right)\right)\le 6$. More precisely, if $f\left(V\left(G_i\right)\right)=6$, then, without loss of generality, we may assume that $f\left(y_{i_2}\right)=f\left(y_{i_3}\right)=f\left(y_{i_2}\right)=f\left(y_{i_3}\right)=1$ and $f\left(y_{i_1}\right)=2$. If also, $f\left(V\left(G_i\right)\right)\in \{4,5\}$, then obviously at least one vertex of $G_i$ (including $y_{i_1}$) is not total Roman $\left\{3\right\}$-dominated. In this case, we can assume that vertices of $G_i$ are assigned as $f\left(y_{i_2}\right)=f\left(y_{i_3}\right)=f\left(y_{i_4}\right)=f\left(y_{i_5}\right)=1$ so that, only $y_{i_1}$ is not total Roman $\left\{3\right\}$-dominated and $f\left(y_{i_1}\right)\in \{0,1\}$.

Suppose that the instance X, C of $X3C$ has a solution $C^{\prime }$. We build a total Roman $\left\{3\right\}$-dominating function f on G of weight k. For every $C_j$, assign the value 2 to $l_j$ if $C_j\in C^{\prime }$ and 1 to the other $l_j$ if $C_j\notin C^{\prime }$. Assign value 3 to every $c_j$ and value 0 to each leaf adjacent to $c_j$. Finally, for every i, assign 1 to $y_{i_2},y_{i_3},y_{i_4},y_{i_5}$, and 0 to $y_{i_1}$ of $G_i$. Since $C^{\prime }$ exists, $|C^{\prime }|=r$, the number of $l_j$s with weight 2 is r, having disjoint neighborhoods in $\{y_{1_1},y_{2_1},\dots ,y_{{3r}_1}\}$, where every $y_{i_1}$ has one neighbors assigned 1 and one neighbor assigned 2. Also since the number of $l_j$s with weight 1 is $t-r$. Hence, it can be easily seen that f is a TR3-D function with weight $f\left(V\right)=3t+2r+t-r+12r=k$.

Conversely, let $g=(V_0,V_1,V_2,V_3)$ be a total Roman $\left\{3\right\}$-dominating function of G with weight at most k. Obviously, every star needs a weight of at least 4, and so without loss of generality, we may assume that $g\left(c_j\right)=3$ and all the leaves neighbor of $c_j$ are assigned 0. Since $l_j{c}_j\in E\left(G\right)$, it implies that each vertex $l_j$ can be assigned by 1. Moreover, for each i, $g\left(V\left(G_i\right)\right)\in \{4,6\}$, as mentioned above. We can let the vertices of $G_i$ are assigned the values given in the above paragraph depending on whether $g\left(V\left(G_i\right)\right)=4$ or $g\left(V\left(G_i\right)\right)=6$. Let p be the number of $G_i$s having weight 6. Then $g\left(V\right(H\left)\right)=6p+4(3r-p)=12r+2p$. Now, if $g\left(l_j\right)>1$ for some j, then $l_j$ total Roman $\left\{3\right\}$-dominates some vertex $y_{s_1}$, and, in that case, $g\left(l_j\right)=2$ (since $g\left(y_{i_2}\right)=1)$. Let z be the number of $l_j$s assigned 2 and $t-z$ of others be assigned 1. Then $3t+2z+t-z+12r+2p\le k=4t+13r$, implies that $z+2p\le q$. On the other hand, since each $l_j$ has exactly three neighbors in $\{x_{1_1},x_{2_1},\dots ,x_{{3r}_1}\}$, we must have $3z\ge 3r-p$. From these two inequalities, we achieve at $p=0$ and then $z=q$. Consequently, $C^{\prime }=\{C_j:g\left(l_j\right)=2\}$ is an exact cover for C. □

7. Open Problems

In the preceding sections a new model of total Roman domination, total Roman $\left\{R3\right\}$-domination has been introduced. There are the relationships between the total domination, total Roman $\left\{R2\right\}$-domination and total Roman $\left\{R3\right\}$-domination numbers as follows:

If G is a graph without isolated vertices, then ${\gamma }_t\left(G\right)+1\le {\gamma }_{t\left\{R3\right\}}\left(G\right)\le 3{\gamma }_t\left(G\right)$, (Proposition 5).

If G is a graph without isolated vertices, then ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+1$ if and only if G has at least two vertices of degree $\mathsf{\Delta }=\left|V\right(G\left)\right|-1$, in the other words ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3$ and ${\gamma }_t\left(G\right)=2$. (Proposition 6).

For any positive integer $n\ge 5$, there is a graph G of order n in which ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_{t\left\{R2\right\}}\left(G\right)+{\gamma }_t\left(G\right)$, (Proposition 9).

For a family of graphs we have shown that ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left|V\left(G\right)\right|$, (Observation 3).

We have already characterized graphs G in which ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2\left|V\left(G\right)\right|-r$, where $1\le r\le 4$.

Problems

1. Characterize the graphs G for which ${\gamma }_{t\left\{R3\right\}}\left(G\right)=3{\gamma }_t\left(G\right)$.

2. Does there exist any characterization of graphs G for which ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_t\left(G\right)+r$, where $2\le r\le {\gamma }_t\left(G\right)-2$?

3. For positive integers $n\ge 5$, characterize the graphs G for which ${\gamma }_{t\left\{R3\right\}}\left(G\right)={\gamma }_{t\left\{R2\right\}}\left(G\right)+{\gamma }_t\left(G\right)$.

4. Does there exist any characterization of graphs G for which ${\gamma }_{t\left\{R3\right\}}\left(G\right)=\left|V\left(G\right)\right|$?

5. Can one characterize graphs G in which ${\gamma }_{t\left\{R3\right\}}\left(G\right)=2\left|V\left(G\right)\right|-r$ for $5\le r\le \left|V\right(G\left)\right|-1$?

6. Is it possible to construct a polynomial algorithm for computing of ${\gamma }_{t\left\{R3\right\}}\left(T\right)$ for any tree T?