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Article

# Total Roman {3}-domination in Graphs

by
Zehui Shao
1,
Doost Ali Mojdeh
2,* and
Lutz Volkmann
3
1
Institute of Computing Science and Technology, Guangzhou University, Guangzhou 510006, China
2
Department of Mathematics, University of Mazandaran, Babolsar 47416-95447, Iran
3
Lehrstuhl II für Mathematik, RWTH Aachen University, 52056 Aachen, Germany
*
Author to whom correspondence should be addressed.
Symmetry 2020, 12(2), 268; https://doi.org/10.3390/sym12020268
Submission received: 4 January 2020 / Revised: 2 February 2020 / Accepted: 3 February 2020 / Published: 9 February 2020

## Abstract

:
For a graph $G = ( V , E )$ with vertex set $V = V ( G )$ and edge set $E = E ( G )$, a Roman ${ 3 }$-dominating function (R${ 3 }$-DF) is a function $f : V ( G ) → { 0 , 1 , 2 , 3 }$ having the property that $∑ u ∈ N G ( v ) f ( u ) ≥ 3$, if $f ( v ) = 0$, and $∑ u ∈ N G ( v ) f ( u ) ≥ 2$, if $f ( v ) = 1$ for any vertex $v ∈ V ( G )$. The weight of a Roman ${ 3 }$-dominating function f is the sum $f ( V ) = ∑ v ∈ V ( G ) f ( v )$ and the minimum weight of a Roman ${ 3 }$-dominating function on G is the Roman ${ 3 }$-domination number of G, denoted by $γ { R 3 } ( G )$. Let G be a graph with no isolated vertices. The total Roman ${ 3 }$-dominating function on G is an R${ 3 }$-DF f on G with the additional property that every vertex $v ∈ V$ with $f ( v ) ≠ 0$ has a neighbor w with $f ( w ) ≠ 0$. The minimum weight of a total Roman ${ 3 }$-dominating function on G, is called the total Roman ${ 3 }$-domination number denoted by $γ t { R 3 } ( G )$. We initiate the study of total Roman ${ 3 }$-domination and show its relationship to other domination parameters. We present an upper bound on the total Roman ${ 3 }$-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 ${ 3 }$-domination for bipartite graphs.

## 1. Introduction

In this paper, we introduce and study a variant of Roman dominating functions, namely, total Roman ${ 3 }$-dominating functions. First we present some necessary terminology and notation. Let $G = ( V , E )$ be a graph of order n with vertex set $V = V ( G )$ and edge set $E = E ( G )$. The open neighborhood of a vertex $v ∈ V ( G )$ is the set $N G ( v ) = N ( v ) = { u : u v ∈ E ( G ) }$. The closed neighborhood of a vertex $v ∈ V ( G )$ is $N G [ v ] = N [ v ] = N ( v ) ∪ { v }$. The open neighborhood of a set $S ⊆ V$ is the set $N G ( S ) = N ( S ) = ⋃ v ∈ S N ( v )$. The closed neighborhood of a set $S ⊆ V$ is the set $N G [ S ] = N [ S ] = N ( S ) ∪ S = ⋃ v ∈ S N [ v ]$. We denote the degree of v by $d G ( v ) = d ( v ) = | N ( v ) |$. By $Δ = Δ ( G )$ and $δ = δ ( G )$, 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 $L ( G )$ and $S ( G )$, 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 $H ∘ K 1$ of a graph H is the graph constructed from H, where for each vertex $v ∈ V ( H )$, a new vertex $v ′$ and a pendant edge $v v ′$ are added. The union of two graphs $G 1$ and $G 2$ ($G 1 ∪ G 2$) is a graph G such that $V ( G ) = V ( G 1 ) ∪ V ( G 2 )$ and $E ( G ) = E ( G 1 ) ∪ E ( G 2 )$.
A set $S ⊆ V$ in a graph G is called a dominating set if $N [ S ] = V$. The domination number $γ ( G )$ of G is the minimum cardinality of a dominating set in G, and a dominating set of G of cardinality $γ ( G )$ is called a $γ$-set of G, [1]. A set $S ⊆ V$ in a graph G is called a total dominating set if $N ( S ) = V$. The total domination number $γ t ( G )$ of G is the minimum cardinality of a total dominating set in G, and a total dominating set of G of cardinality $γ t ( G )$ is called a $γ t$-set of G, [2].
Given a graph G and a positive integer m, assume that $g : V ( G ) → { 0 , 1 , 2 , … , m }$ is a function, and suppose that $( V 0 , V 1 , V 2 , … , V m )$ is the ordered partition of V induced by g, where $V i = { v ∈ V : g ( v ) = i }$ for $i ∈ { 0 , 1 , … , m }$. So we can write $g = ( V 0 , V 1 , V 2 , … , V m )$. A Roman dominating function on graph G is a function $f : V → { 0 , 1 , 2 }$ such that if $v ∈ V 0$ for some $v ∈ V$, then there exists a vertex $w ∈ N ( v )$ such that $w ∈ V 2$. The weight of a Roman dominating function (RDF) is the sum $w f = ∑ v ∈ V ( G ) f ( v )$, 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 $γ R ( G )$, 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 $v ∈ V$ with $f ( v ) ≠ 0$ has a neighbor w with $f ( w ) ≠ 0$. The minimum weight of any TRDF on G is called the total Roman domination number of G denoted by $γ t R ( G )$. A TRDF on G with weight $γ t R ( G )$ is called a $γ t R ( G )$-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 ${ 2 }$-dominating function f as follows. A Roman ${ 2 }$-dominating function is a function $f : V → { 0 , 1 , 2 }$ such that for every vertex $v ∈ V$, with $f ( v ) = 0 , f ( N ( v ) ) ≥ 2$ where $f ( N ( v ) ) = ∑ x ∈ N ( v ) f ( x )$, that is, either v has a neighbor u with $f ( u ) = 2$, or has two neighbors $x , y$ with $f ( x ) = f ( y ) = 1$ [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 ${ 2 }$-dominating function (R${ 2 }$-DF) f, and for some vertex v with $f ( v ) = 1$, it is possible that $f ( N ( v ) ) = 0$. The sum $w f = ∑ v ∈ V ( G ) f ( v )$ is denoted the weight of a Roman ${ 2 }$-dominating function, and the minimum weight of a Roman ${ 2 }$-dominating function f is the Roman ${ 2 }$-domination number, denoted by $γ { R 2 } ( G )$. Roman ${ 2 }$-domination is a generalization of Roman domination that has also studied by Henning and Klostermeyer [8] with the name Italian domination.
The total Roman ${ 2 }$-domination for graphs are defined as follows [9]. Let G be a graph without isolated vertices. Then $f : V → { 0 , 1 , 2 }$ is total Roman${ 2 }$-dominating function (TR${ 2 }$-DF) if it is a Roman ${ 2 }$-dominating function and the subgraph induced by the positive weight vertices has no isolated vertex. The minimum weight $w f = ∑ v ∈ V ( G ) f ( v )$ of a any total Roman${ 2 }$-dominating function of a graph G is called the total Roman ${ 2 }$-domination number of G and is denoted by $γ t { R 2 } ( G )$. Beeler et al. [10] have defined double Roman domination.
A double Roman dominating function (DRDF) on a graph G is a function $f : V → { 0 , 1 , 2 , 3 }$ such that the following conditions are hold:
(a)
if $f ( v ) = 0$, then the vertex v must have at least two neighbors in $V 2$ or one neighbor in $V 3$.
(b)
if $f ( v ) = 1$, then the vertex v must have at least one neighbor in $V 2 ∪ V 3$.
The weight of a double Roman dominating function is the sum $w f = ∑ v ∈ V ( G ) f ( v )$, 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 $γ d R ( G )$ and a double Roman dominating function of G with weight $γ d R ( G )$ is called a $γ d R ( G )$-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 ${ v ∈ V ( G ) : f ( v ) ≠ 0 }$ has no isolated vertices. The total double Roman domination number $γ t d R ( G )$ is the minimum weight of a TDRDF on G. A TDRDF on G with weight $γ t d R ( G )$ is called a $γ t d R ( G )$-function. Mojdeh et al. [13] have recently defined the Roman ${ 3 }$-dominating function correspondingly to the Roman ${ 2 }$-dominating function of graphs. For a graph G, a Roman ${ 3 }$-dominating function (R${ 3 }$-DF) is a function $f : V → { 0 , 1 , 2 , 3 }$ having the property that $f ( N [ u ] ) ≥ 3$ for every vertex $u ∈ V$ with $f ( u ) ∈ { 0 , 1 }$. Formally, a Roman ${ 3 }$-dominating function $f : V → { 0 , 1 , 2 , 3 }$ has the property that for every vertex $v ∈ V$, with $f ( v ) = 0$, there exist at least either three vertices in $V 1 ∩ N ( v )$, or one vertex in $V 1 ∩ N ( v )$ and one in $V 2 ∩ N ( v )$, or two vertices in $V 2 ∩ N ( v )$, or one vertex in $V 3 ∩ N ( v )$ and for every vertex $v ∈ V$, with $f ( v ) = 1$, there exist at least either two vertices in $V 1 ∩ N ( v )$, or one vertex in $( V 2 ∪ V 3 ) ∩ N ( v )$. This notion has been defined recently by Mojdeh and Volkmann [13] as Roman ${ 3 }$-domination.
The weight of a Roman ${ 3 }$-dominating function is the sum $w f = f ( V ) = ∑ v ∈ V f ( v )$, and the minimum weight of a Roman ${ 3 }$-dominating function f is the Roman ${ 3 }$-domination number, denoted by $γ { R 3 } ( G )$.
Now we introduce the total Roman ${ 3 }$-domination concept to consider such situation.
Definition 1.
Let G be a graph G with no isolated vertex. The total Roman ${ 3 }$-dominating function (TR${ 3 }$-DF) on G is an R${ 3 }$-DF f on G with the additional property that every vertex $v ∈ V$ with $f ( v ) ≠ 0$ has a neighbor w with $f ( w ) ≠ 0$, in the other words, the subgraph of G induced by the set ${ v ∈ V ( G ) : f ( v ) ≠ 0 }$ has no isolated vertices. The minimum weight of a total Roman ${ 3 }$-dominating function on G is called the total Roman ${ 3 }$-domination number of G denoted by $γ t { R 3 } ( G )$. A $γ t { R 3 } ( G )$-function is a total Roman ${ 3 }$-dominating function on G with weight $γ t { R 3 } ( G )$.
In this paper We study of total Roman ${ 3 }$-domination versus to other domination parameters. We present an upper bound on the total Roman ${ 3 }$-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 ${ 3 }$-domination for bipartite graphs.

## 2. Total Roman ${ 3 }$-domination of Some Graphs

First we easily see that $γ { R 3 } ( G ) ≤ γ t { R 3 } ( G ) ≤ γ t d R ( G )$, because by the definitions every total Roman ${ 3 }$-dominating function is a Roman ${ 3 }$-dominating function and every total double Roman dominating function is a total Roman ${ 3 }$-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 $γ R$-function of G. Then $γ d R ( G ) ≤ 2 | V 1 | + 3 | V 2 |$. 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 ${ 2 }$-dominating function or a Roman dominating function for which the induced subgraph by $V 1 ∪ V 2$ has no isolated vertex. Then $γ t { R 3 } ( G ) ≤ 2 | V 1 | + 3 | V 2 |$. This bound is sharp.
For some special graphs we obtain the total Roman ${ 3 }$-domination numbers.
Observation 1.
Let $n ≥ 2$. Then $γ t { R 3 } ( P n ) = n + 2 i f n ≡ 1 ( mod 3 ) n + 1 o t h e r w i s e ,$
Proof.
Let $P n = v 1 v 2 … v n$. Since by assigning 2 to the vertices $v 1$ and $v n$ and value 1 to the other vertices, we have $γ t { R 3 } ( P n ) ≤ n + 2$. Since $f ( v 1 ) + f ( v 2 ) ≥ 3$ and $f ( v n − 1 ) + f ( v n ) ≥ 3$, $f ( v i − 1 ) + f ( v i ) + f ( v i + 1 ) ≥ 3$ for $4 ≤ i ≤ n − 3$, $f ( v i − 1 ) + f ( v i ) + f ( v i + 1 ) + f ( v i + 2 ) ≥ 3$ for $4 ≤ i ≤ n − 4$ and $f ( v i − 2 ) + f ( v i − 1 ) + f ( v i ) + f ( v i + 1 ) + f ( v i + 2 ) ≥ 4$ for $5 ≤ i ≤ n − 4$, we observe that $γ t { R 3 } ( P n ) ≥ n + 1$ and $γ t { R 3 } ( P n ) ≥ n + 2$ if $n ≡ 1 ( mod 3 )$. If $n = 3 k$, then by assigning 1 to $v 3 t + 1$ and $v n$, 2 to $v 3 t + 2$, 0 to $v 3 t$ except $v n$, we have $γ t { R 3 } ( P n ) ≥ 3 k + 1 = n + 1$. If $n = 2 + 3 k$, then by assigning 1 to $v 3 t + 1$, 2 to $v 3 t + 2$, 0 to $v 3 t$, we have $γ t { R 3 } ( P n ) ≥ 3 k + 1 = n + 1$. Thus the proof is complete. □
In [10], it has been shown that $γ d R ( C n ) = n$ if $n ≡ 0 , 2 , 3 , 4 ( mod 6 )$ and otherwise $γ d R ( C n ) = n + 1$ and since $γ t d R ( G ) ≥ γ d R ( G )$, we deduce that $γ t d R ( C n ) ≥ n$.
Here we show that $γ t { R 3 } ( C n ) = n$ for all $n ≥ 3$. If we assign weight 1 to every vertex of $C n$, then it is a total Roman ${ 3 }$-dominating function of $C n$. Hence $γ t { R 3 } ( C n ) ≤ n$. In [13], we have shown that $γ { R 3 } ( C n ) = n$. Since $γ { R 3 } ( C n ) ≤ γ t { R 3 } ( C n )$, we obtain the desired result.
Observation 2.
$γ t { R 3 } ( C n ) = n$ .
The next result shows another family of graphs G with $γ t { R 3 } ( G ) = | V ( G ) |$. Let $C n$ be a cycle with vertices $v 1 , v 2 , … , v n$ and $P m$ be a path with vertices $u 1 , u 2 , … , u m$ for which $u 1 = v 1$ and for some $2 ≤ i ≤ m$, $u i ≠ v j$. Let H be a graph obtained from a cycle $C n$ and k paths like $P m 1 , P m 2 , … , P m k$ ($1 ≤ k ≤ n$) 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 $m i ≡ 1 ( mod 3 )$.
Observation 3.
Let G be the graph constructed as above. If $P m i$ with vertices $u 1 i , u 2 i , … , u m i$ is a path such that $3 ∣ ( m i − 1 )$, then $γ t { R 3 } ( G ) = | V ( G ) |$.
Proof.
Let f be a function that assign value 1 to every vertex of the cycles and if $3 ∣ ( m i − 1 )$, we assign value 2 to vertices with indices $3 t$, value 1 to vertices with indices $3 t + 1$, ($1 ≤ t ≤ m i − 1 3$) 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 $γ t { R 3 } ( G ) = | V ( G ) |$. □
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 $3 ∣ m i$ or $3 ∣ ( m i − 2 )$, then $γ t { R 3 } ( C n ∪ P m ) = | V ( C n ∪ P m ) | + 1$. Therefore we have the following result.
Corollary 2.
In the graphs constructed above, if there are l paths $P m 1 , P m 2 , … , P m l$ such that $3 ∣ m i$ or $3 ∣ ( m i − 2 )$ for $1 ≤ m i ≤ l$, then $γ t { R 3 } ( G ) = | V ( G ) | + l$.
The Observation 3 and Corollary 2 show that for every nonnegative integer k, there is a graph G such that $γ t { R 3 } ( G ) = | V ( G ) | + k$.
Proposition 2.
If G is a connected graph of order $n ≥ 2$, then $γ t { R 3 } ( G ) ≥ 3$ and $γ t { R 3 } ( G ) = 3$ if and only if G has at least two vertices of degree $Δ ( G ) = n − 1$.
Proof.
If $n = 2$, then the statement is clear. Let now $n ≥ 3$ and let $f = ( V 0 , V 1 , V 2 , V 3 )$ be a total Roman ${ 3 }$-dominating function on G of weight $γ { R 3 } ( G )$. If $V 0 ≠ ∅$, then $∑ u ∈ N ( v ) f ( u ) ≥ 3$ for a vertex $v ∈ V 0$ and thus $γ t { R 3 } ( G ) ≥ 3$. If $V 0 = ∅$, then $f ( x ) ≥ 1$ for each vertex $x ∈ V ( G )$ and therefore $γ t { R 3 } ( G ) ≥ n ≥ 3$.
If G has at least two vertices of degree $Δ ( G ) = n − 1$, then we may assume v and u are two adjacent vertices of maximum degree. Define the function f by $f ( v ) = 1$, $f ( u ) = 2$ and $f ( x ) = 0$ for $x ∈ V ( G ) \ { v , u }$. Then f is a total Roman ${ 3 }$-dominating function on G of weight 3 and hence $γ t { R 3 } ( G ) = 3$.
Conversely, assume that $γ t { R 3 } ( G ) = 3$. 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 $Δ ( G ) = n − 1$, then $γ t { R 3 } ( G ) = 4$.
In the follow, total Roman ${ 3 }$-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 ${ 3 }$-domination number of any partite set is positive. Therefore we have the following.
Proposition 3.
For any complete bipartite graph we have.
• $γ t { R 3 } ( K 1 , n ) = 4$,
• $γ t { R 3 } ( K m , n ) = 5$ for $m ∈ { 2 , 3 }$ and $n ≥ 3$.
• $γ t { R 3 } ( K m , n ) = γ d R ( K m , n ) = 6$ for $m , n ≥ 4$.
Proof.
In any complete bipartite graph, let $V ( G ) = U ∪ W$, 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 $W = { w 1 , w 2 , … , w n }$. Let f be a TR${ 3 }$DF of $K 2 , n$. If $f ( W ) = 2$, then $f ( U ) ≥ 3$. If $f ( W ) = 3$, then $f ( U ) ≥ 2$. If $f ( W ) ≥ 4$, since $f ( U )$ is positive, then $f ( V ) ≥ 5$. Therefore $f ( V ) ≥ 5$. Assigning $f ( u 1 ) = 2 , f ( u 2 ) = 1$ and $f ( w 1 ) = 2$, shows that $γ t { R 3 } ( K 2 , n ) ≤ 5$.
(ii)
Let $U = { u 1 , u 2 , u 3 }$ and $W = { w 1 , w 2 , … , w n }$. Using sketch of the proof of item 2, $γ t { R 3 } ( K 3 , n ) ≥ 5$. 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 $j ≥ 2$, then $γ t { R 3 } ( K 3 , n ) ≤ 5$.
3. The function f with $f ( u 1 ) = 3 = f ( w 1 )$ and $f ( u i ) = 0 = f ( u j )$ for $i , j ≠ 1$ is a TR${ 3 }$DF for $K m , n$. Therefore $γ t { R 3 } ( K m , n ) ≤ 6$.
Now let f be a $γ t { R 3 }$ function of $K m , n$ for $m , n ≥ 4$. If $m , n ≥ 5$, 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 $f ( u i ) ≥ 1$ for $1 ≤ i ≤ 4$, then $f ( u i ) = 1$ for $1 ≤ i ≤ 4$ and thus $f ( W ) ≥ 2$. So $f ( V ) ≥ 6$ and therefore $γ t { R 3 } ( K m , n ) ≥ 6$, and the proof is complete. □
One can obtain a similar result for complete r-partite graphs for $r ≥ 3$.
Proposition 4.
Let $G = K n 1 , n 2 , … , n r$ be the complete r-partite graph with $r ≥ 3$ and $n 1 ≤ n 2 ≤ … ≤ n r$. Then:
• If $n 1 = n 2 = 1$, then $γ t { R 3 } ( G ) = 3$.
• If $n 1 = 1$ and $n 2 ≥ 2$, then $γ t { R 3 } ( G ) = 4$.
• If $n 1 = 2$ or $n 1 ≥ 3$ and $r ≥ 4$, then $γ t { R 3 } ( G ) = 4$.
• If $r = 3$ and $n 1 ≥ 3$, then $γ t { R 3 } ( G ) = 5$.
Proof.
Let $V = ⋃ i = 1 r U i$ where $U i$ is the ith partite set with vertices ${ u i 1 , u i 2 , … , u i n i }$.
1. This follows from Proposition 2.
2. This follows from Corollary 3.
3. Let $n 1 ≥ 2$. By Proposition 2, we have $γ t { R 3 } ( G ) ≥ 4$. If $n 1 = 2$, then define $f ( u 1 1 ) = f ( u 1 2 ) = f ( u 2 1 ) = f ( u 3 1 ) = 1$ and $f ( v ) = 0$ otherwise. Then f is a TR${ 3 }$-DF on G with $f ( V ) = 4$ and thus $γ t { R 3 } ( G ) = 4$. Now let $n 1 ≥ 3$ and $r ≥ 4$. Then any TR${ 3 }$-DF f on G with $f ( u 1 1 ) = f ( u 2 1 ) = f ( u 3 1 ) = f ( u 4 1 ) = 1$ and $f ( v ) = 0$ for the other vertices, is a $γ t { R 3 }$ function on G. Therefore $γ t { R 3 } ( G ) = 4$.
4. Let $n 1 ≥ 3$ and $r = 3$, and let f be a TR${ 3 }$-DF function on G. Since two partite sets must have positive weight, we can assume $f ( U 2 ) ≥ 1$. If $f ( U 2 ) = 1$, then $f ( U 1 ∪ U 3 ) ≥ 4$. If $f ( U 2 ) = 2$, then $f ( U 1 ∪ U 3 ) ≥ 3$. If $f ( U 2 ) = 3$, then $f ( U 1 ∪ U 3 ) ≥ 2$. If $f ( U 2 ) ≥ 4$, then $f ( U 1 ∪ U 3 ) ≥ 1$. Thus $f ( V ) ≥ 5$. Conversely, define $f ( u 1 1 ) = f ( u 2 1 ) = 2$ and $f ( u 3 1 ) = 1$ and $f ( v ) = 0$ otherwise. Then f is a TR${ 3 }$-DF on G with $f ( V ) = 5$ and so $γ t { R 3 } ( G ) = 5$. □
Theorem 4.
If G is a graph with $δ ( G ) = δ ≥ 2$, then $γ t { R 3 } ( G ) ≤ | V ( G ) | + 2 − δ$, and this bound is sharp.
Proof.
Let $V ( G ) = { u 1 , u 2 , … , u n }$, and let v be a vertex of degree $δ$ with neighbors ${ u 1 , u 2 , … , u δ }$. Let $U = { v , u δ + 2 , u δ + 3 , … , u n } ∪ { u 1 , u 2 }$. Define the function f by $f ( x ) = 1$ for $x ∈ U$ and $f ( x ) = 0$ for $x ∈ V ( G ) \ U$. Then $∑ x ∈ N ( u ) f ( x ) ≥ 2$ for $u ∈ U$ and $∑ x ∈ N ( u ) f ( x ) ≥ 3$ for $u ∈ V ( G ) \ U$. Therefore f is a total Roman ${ 3 }$-dominating function on G of weight $n + 2 − δ$ and thus $γ t { R 3 } ( G ) ≤ | V ( G ) | + 2 − δ$.
According to Observation 2 and Propositions 3 and 8, we note that $γ t { R 3 } ( C n ) = n = | V ( C n ) | + 2 − δ ( C n )$, $γ t { R 3 } ( K n ) = 3 = | V ( K n ) | + 2 − δ ( K n )$ for $n ≥ 3$, $γ t { R 3 } ( K 3 , 3 ) = 5 = | V ( K 3 , 3 ) | + 2 − δ ( K 3 , 3 )$, $γ t { R 3 } ( K 4 , 4 ) = 6 = | V ( K 4 , 4 ) | + 2 − δ ( K 4 , 4 )$, $γ t { R 3 } ( K 3 , 3 , 3 ) = 5 = | V ( K 3 , 3 , 3 ) | + 2 − δ ( K 3 , 3 , 3 )$ and $γ t { R 3 } ( K n 1 , n 2 , … , n r ) = 4 = | V ( K n 1 , n 2 , … , n r ) | + 2 − δ ( K n 1 , n 2 , … , n r )$ for $r ≥ 4$ and $n 1 ≤ n 2 ≤ … ≤ n r = 2$. All these examples demonstrate that the inequality $γ t { R 3 } ( G ) ≤ | V ( G ) | + 2 − δ$ is sharp. □
Hao et al. defined in [12] the family of graphs $G$ as follows and have proved Theorem 5 below. Let $G$ be the family of graphs that can be obtained from a star $S t = K 1 , t − 1$ of order $t ≥ 2$ by adding a pendant edge to each vertex of $V ( S t )$ and adding any number of edges joining the leaves of $S t$.
Theorem 5.
[12] For any connected graph G of order $n ≥ 2$,
$γ t d R ( G ) ≤ 2 n − Δ$
with equality if and only if $G ∈ { P 2 , P 3 , C 3 } ∪ G$.
This theorem with a little changing may be explored as follows.
Theorem 6.
For any connected graph G of order $n ≥ 2$,
$γ t { R 3 } ( G ) ≤ 2 n − Δ$
with equality if and only if $G ∈ { P 2 , P 3 } ∪ G$.

## 3. Total Roman ${ 3 }$-domination and Total Domination

In this section we study the relationship between total domination and total Roman ${ 3 }$-domination of a graph.
In [10] (Proposition 8) the authors proved that, if G is a graph, then $2 γ ( G ) ≤ γ d R ( G ) ≤ 3 γ ( G )$.
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 $γ { R 3 }$-function $f = ( V 0 , V 2 , V 3 )$, then $2 γ ( G ) ≤ γ { R 3 } ( G ) ≤ 3 γ ( G )$.
In [13] Proposition 17 authors proved that:
If G is a graph, then $γ ( G ) + 2 ≤ γ { R 3 } ( G ) ≤ 3 γ ( G )$, and these bounds are sharp. However, we have the following.
Proposition 5.
If G is a graph without isolated vertices, then $γ t ( G ) + 1 ≤ γ t { R 3 } ( G ) ≤ 3 γ t ( G )$.
Proof.
Let S be a $γ t$-set of G. Then $( V 0 = V \ S , ∅ , ∅ , V 3 = S )$ is a $γ t { R 3 }$-function of G. Therefore $γ t { R 3 } ( G ) ≤ 3 γ t ( G )$.
For the lower bound, let $f = ( V 0 , V 1 , V 2 , V 3 )$ be a $γ t { R 3 }$-function of G. We distinguish two cases.
Case 1. Let $| V 2 | ≥ 1$ or $| V 3 | ≥ 1$. Then $γ t ( G ) ≤ | V 1 | + | V 2 | + | V 3 | ≤ | V 1 | + 2 | V 2 | + 3 | V 3 | − 1 = γ t { R 3 } ( G ) − 1$.
Case 2. Let $V 2 = V 3 = ∅$. By the definition, $δ ( G [ V 1 ] ) ≥ 2$. Therefore, for each vertex $v ∈ V 1$, the subgraph $G [ V 1 \ { v } ]$ does not contain an isolated vertex. Consequently, $V 1 \ { v }$ is total dominating set of G and hence $γ t ( G ) ≤ γ t { R 3 } ( G ) − 1$. □
By Proposition 5 the question may arise as whether for any positive integer r, exists a graph G for which $γ t { R 3 } ( G ) = γ t ( G ) + r$, where $1 ≤ r ≤ 2 γ t ( G )$. For $r = 1$ we have. If G is a connected graph of order $n ≥ 2$ with at least two vertices of maximum degree $Δ ( G ) = n − 1$, then Proposition 2 implies that $γ t { R 3 } ( G ) = 3$. Since $γ t ( G ) = 2$ for such graphs, we observe that $γ t { R 3 } ( G ) = γ t ( G ) + 1$.
Proposition 6.
If G is a graph without isolated vertices, then $γ t { R 3 } ( G ) = γ t ( G ) + 1$ if and only if G has at least two vertices of degree $Δ = | V ( G ) | − 1$, in the other words $γ t { R 3 } ( G ) = 3$ and $γ t ( G ) = 2$.
Proof.
The part “if“ has been proved. Part “only if“: Let G be a graph with $γ t { R 3 } ( G ) = γ t ( G ) + 1$. Let $f = ( V 0 , V 1 , V 2 , V 3 )$ be a $γ t { R 3 } ( G )$ function. Therefore $V 1 ∪ V 2 ∪ V 3$ is a total dominating set for G, and $| V 1 | + | V 2 | + | V 3 | ≥ γ t ( G ) = γ t { R 3 } ( G ) − 1 = | V 1 | + 2 | V 2 | + 3 V 3 | − 1$. Therefore $| V 2 | + 2 | V 3 | ≤ 1$ that is $| V 2 | ≤ 1$ 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 $Δ ( G ) = | V ( G ) | − 1$. 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 $| V 1 | ≥ 2$. (2) $| V 2 | = 0$ and $| V 1 | ≥ 4$.
Case 1. Let $V 2 = { v }$, $| V 1 | ≥ 2$. Assume first that there exist two vertices $v 1 , v 2 ∈ V 1$ which are adjacent to the vertex v. Then $V 2 ∪ V 1 \ { v 1 }$ is a $γ t ( G )$-set of size $| V 1 |$ and so $γ t { R 3 } ( G ) = 2 + | V 1 |$, a contradiction. Assume next that there exists only one vertex, say $v 1 ∈ V 1$, which is adjacent to v. Then all other vertices of $V 1$ have at least two neighbors in $V 1$. If $v 2 ∈ V 1$ with $v 2 ≠ v 1$, then we observe that $V 2 ∪ V 1 \ { v 2 }$ is a $γ t ( G )$-set of size $| V 1 |$. It follows that $γ t { R 3 } ( G ) = 2 + | V 1 |$, a contradiction.
Case 2. Let $| V 2 | = 0$ and $| V 1 | ≥ 4$. Then there exist two vertices $v 1 , v 2$ in which each of them has neighbors in $V 1 \ { v 1 , v 2 }$ and $G ( V 1 \ { v 1 , v 2 } )$ has no isolated vertex. Therefore $V 1 \ { v 1 , v 2 }$ is a $γ t ( G )$-set that is also a contradiction. □
Now we show that for any positive integer n and integer $2 ≤ r ≤ 2 n$, there exists a graph G for which $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = n + r$.
Proposition 7.
Let n and r be positive integers with $2 ≤ r ≤ 2 n$. Then there exists a graph G for which $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = n + r$.
Proof.
For graph G with $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = n + 2$, we consider the following graph. Let H be the graph consisting of a cycle $C n + 2$ with $n ≥ 3$ and a vertex set $V 0$ of $n + 2 3$ further vertices. Let each vertex of $V 0$ be adjacent to 3 vertices of $V ( C n + 2 )$ such that the neighborhoods of every two distinct vertices of $V 0$ are different. Let $V 1 = V ( C n + 2 )$. Then $γ t { R 3 } ( H ) = n + 2$ and $γ t ( H ) = n$ (Figure 2).
For $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = n + k$, where $3 ≤ k ≤ n − 1$. Let $k = 3$. For $γ t ( G ) = 4$ and $γ t { R 3 } ( G ) = 7$, we consider the cycle $C 7$. For $γ t ( G ) ≥ 5$, let H be the above graph where $γ t ( H ) = n ≥ 3$ and $γ t { R 3 } ( H ) = n + 2 ≥ 5$. Now we consider $G = H ∪ K 3$. Then $γ t ( G ) ≥ n ≥ 4$ and $γ t { R 3 } ( G ) = n + 3$.
Let $k = 4$. For $γ t ( G ) = 5$ and $γ t { R 3 } ( G ) = 9$, we consider the cycle $C 9$. For $γ t ( G ) ≥ 6$, consider the graphs $G ′$ with $γ t { R 3 } ( G ′ ) = γ t ( G ′ ) + 3$ for $γ t ( G ′ ) ≥ 4$. Now we let $G = G ′ ∪ K 3$. Then we have $γ t { R 3 } ( G ) = γ t ( G ) + 4$.
For $5 ≤ k ≤ n − 1$ we use induction on k. Let for any integer $4 ≤ m ≤ k − 1$ there exist graphs $G ′$ such that $γ t { R 3 } ( G ) = γ t ( G ) + m$ for $γ t ( G ) ≥ m + 1$. Let $m = k$. For $γ t ( G ) = k + 1$ and $γ t { R 3 } ( G ) = 2 k + 1$, we consider the cycle $C 2 k + 1$. For graphs G with $γ t { R 3 } ( G ) = γ t ( G ) + k$ for $γ t ( G ) ≥ k + 2$, using hypothesis of induction, let $G ′$ be the graphs with $γ t { R 3 } ( G ′ ) = γ t ( G ′ ) + k − 1$ with $γ t ( G ′ ) ≥ k$. Now we let $G = G ′ ∪ K 3$. It can be seen $γ t { R 3 } ( G ) = γ t ( G ) + k$ for $γ t ( G ) ≥ k + 2$.
We now verify the case of $γ t { R 3 } ( G ) = 2 γ t ( G ) + r$ for $0 ≤ r ≤ γ t ( G )$, that is, we wish to show the existence of graphs G, so that $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 2 n + r$ for $0 ≤ r ≤ n$. Let $r = 0$. For even n, let $G = C 2 n$. Then $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 2 n$.
For odd $n = 2 k + 1$, if $2 n ≡ 1 ( m o d 3 )$ or $2 n ≡ 0 ( m o d 3 )$, then we let $G = P 2 n − 1$, and by Observation 1, it can be seen that $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 2 n$.
If $2 n ≡ 2 ( m o d 3 )$, consider a cycle $C 2 n − 1$ with an additional vertex a that is adjacent to two vertices $v 1$ and $v 2$. Then $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 2 n$.
For $r = 1$ and positive even integer n, consider $G = ( n 2 − 1 ) P 3 ∪ C 5 +$, where $( n 2 − 1 ) P 3$ is the union of $n 2 − 1$ of path $P 3$ and $C 5 +$ is the cycle $C 5$ with a chord, then $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 2 n + 1$. For $r = 1$ and positive odd integer n, consider $G = ( n − 1 2 − 1 ) P 3 ∪ P 5 +$ where $P 5 +$ is the path $P 5$ with an additional vertex adjacent to the second or fourth vertex of $P 5$, then $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 2 n + 1$.
For $2 ≤ r ≤ n − 1$, we do as follows. Let $r = 2$ and so $n ≥ 3$. Let $n = 3$ and $2 n + 2 = 8$. 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 $γ t ( G 1 ) = 3$ and $γ t { R 3 } ( G 1 ) = 8$.
Let $n = 4$ and $2 n + 2 = 10$. Then say $G 2 = 2 C 5 +$. Let $n = 5$ and so $2 n + 2 = 12$. Then say $G 3 = C 5 + ∪ P 5 +$. For $γ t ( G ) = k$ and $γ t { R 3 } ( G ) = 2 k + 2$, where $r + 1 ≤ k ≤ n$, there consider three cases.
• If $k ≡ 0 ( m o d 3 )$, then we say $G = k − 3 3 P 5 ∪ G 1$.
• If $k ≡ 1 ( m o d 3 )$, then we say $G = k − 4 3 P 5 ∪ G 2$.
• If $k ≡ 2 ( m o d 3 )$, then we say $G = k − 5 3 P 5 ∪ G 3$.
It is easy to verifiable, $γ t ( G ) = k$ and $γ t { R 3 } ( G ) = 2 k + 2$.
Let $r = 3$ and so $n ≥ 4$. For graph $G 1 ′$ with $γ t ( G 1 ′ ) = 4$ and $γ t { R 3 } ( G 1 ′ ) = 11$, we let $G 1 ′ = P 4 ∪ C 5 +$. For graph $G 2 ′$ with $γ t ( G 2 ′ ) = 5$ and $γ t { R 3 } ( G 2 ′ ) = 13$, we let $G 2 ′ = P 4 ∪ P 5 +$. And for graph $G 3 ′$ with $γ t ( G 3 ′ ) = 6$ and $γ t { R 3 } ( G 3 ′ ) = 15$, we let $G 3 ′ = 3 C 5 +$. For $γ t ( G ) = k$ and $γ t { R 3 } ( G ) = 2 k + 3$, where $r + 1 ≤ k ≤ n$, there consider three cases.
• If $k ≡ 1 ( m o d 3 )$, then we say $G = k − 4 3 P 5 ∪ G 1 ′$.
• If $k ≡ 2 ( m o d 3 )$, then we say $G = k − 5 3 P 5 ∪ G 2 ′$.
• If $k ≡ 0 ( m o d 3 )$, then we say $G = k − 6 3 P 5 ∪ G 3 ′$.
Let $r ≥ 4$ and $n ≥ r + 1$. For graph G with $γ t ( G ) = k$ and $γ t { R 3 } ( G ) = 2 k + r$ where $r + 1 ≤ k ≤ n$, there consider two cases.
Case 1. Let r be an even integer. Then there exists a graph $G ′$ for which $γ t ( G ′ ) = k − ( r − 2 )$ and $γ t { R 3 } ( G ′ ) = 2 k − 2 ( r − 2 ) + 2$. Now let $G = r − 2 2 P 4 ∪ G ′$. Then $γ t ( G ) = r − 2 + γ t ( G ′ ) = k$ and $γ t { R 3 } ( G ) = 3 ( r − 2 ) + γ t { R 3 } ( G ′ ) = 3 ( r − 2 ) + 2 k − 2 ( r − 2 ) + 2 = 2 k + r$.
Case 2. Let r be an odd integer. Then there exists a graph $G ″$ for which $γ t ( G ″ ) = k − ( r − 3 )$ and $γ t { R 3 } ( G ″ ) = 2 k − 2 ( r − 3 ) + 3$. If we consider $G = r − 3 2 P 4 ∪ G ″$. Then $γ t ( G ) = r − 3 + γ t ( G ″ ) = k$ and $γ t { R 3 } ( G ) = 3 ( r − 3 ) + γ t { R 3 } ( G ″ ) = 2 k + r$.
Finally, we want discuss the case of $r = n$, that is we want to find graphs G with $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 3 n$. For $n = 2$ and $3 n = 6$, let $G = P 4$. For G with $γ t ( G ) = 3$ and $γ t { R 3 } ( G ) = 9$, 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 $γ t ( H 1 ) = 3$ and $γ t { R 3 } ( H 1 ) = 9$.
Let $n ≥ 4$. If n is an even, then let $G = n 2 P 4$ and if n is an odd, then let $G = n − 3 2 P 4 ∪ H 1$. In both cases $γ t ( G ) = n$ and $γ t { R 3 } ( G ) = 3 n$. □

## 4. Total Roman ${ 3 }$ and Total Roman ${ 2 }$-domination

In [13] it has been shown that, for a connected graph G with a $γ { R 3 }$-function $f = ( V 0 , V 2 , V 3 )$, $γ { R 3 } ( G ) ≥ γ ( G ) + γ { R 2 } ( G )$.
In this section we investigate the relation between total Roman ${ 3 }$ and total Roman ${ 2 }$-domination. First we have the following.
Observation 7.
Let G be a graph and $( V 0 , V 1 , V 2 )$ be a $γ t { R 2 }$ function of G. Then $( V 0 ′ = V 0 , V 2 ′ = V 1 , V 3 ′ = V 2 )$ is a TR${ 3 }$-DF function. Conversely, if $( V 0 , V 1 , V 2 , V 3 )$ is a $γ t { R 3 }$ of G, then $( U 0 = V 0 , U 1 = V 1 ∪ V 2 , U 2 = V 3 )$ is a TR${ 2 }$-DF of G.
Proof.
The proof is straightforward. □
The following results state the relation between $γ t { R 3 }$ and $γ t { R 2 }$ of graphs G when $γ t { R 3 } ( G )$ is small.
Proposition 8.
Let G be a graph. Then:
• $γ t { R 3 } ( G ) = 3$ if and only if $γ t { R 2 } ( G ) = 2$.
• If $γ t { R 3 } ( G ) = 4$, then $γ t { R 2 } ( G ) = 3$.
• If $γ t { R 2 } ( G ) = 3$, then $4 ≤ γ t { R 3 } ( G ) ≤ 5$.
Proof.
1. Let $γ t { R 3 } ( G ) = 3$. 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 $γ t { R 2 } ( G )$-function with weight 2. Conversely, let $γ t { R 2 } ( G ) = 2$. 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 $γ t { R 3 } ( G ) = 4$. 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${ 2 }$-DF we change one of the labels 1 to the label 0. Therefore $γ t { R 2 } ( G ) = 3$.
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 $γ t { R 2 }$-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 $γ t { R 2 }$-function for G with weight 3.
3. Let $γ t { R 2 } ( G ) = 3$. 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 $γ t { R 3 }$-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 $γ t { R 3 }$-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 $γ t { R 3 }$-function for G with weight 4. Therefore $4 ≤ γ t { R 3 } ( G ) ≤ 5$. □
In the following we want to find the relation between total Roman ${ 3 }$-domination, total domination and total Roman ${ 2 }$-domination of graphs.
Observation 8.
Let G be a connected graph with a $γ t { R 3 }$-function $f = ( V 0 , V 2 , V 3 )$. Then $γ t { R 3 } ( G ) ≥ γ t ( G ) + γ t { R 2 } ( G )$.
Proof.
Let $( V 0 , V 2 , V 3 )$ be a $γ t { R 3 }$-function of G. Then $γ t ( G ) ≤ | V 2 | + | V 3 |$. If we define $g = ( V 0 ′ = V 0 , V 1 ′ = V 2 , V 2 ′ = V 3 )$, then g is a total Roman ${ 2 }$-dominating function on G. Therefore $γ t ( G ) + γ t { R 2 } ( G ) ≤ | V 2 | + | V 3 | + | V 1 ′ | + 2 | V 2 ′ | ≤ 2 | V 2 | + 3 | V 3 | = γ t { R 3 } ( G )$. □
In Observation 8 the condition of $γ t { R 3 }$-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 ≥ 2 )$, cycles $C n$ and paths $P n$ for $n ≥ 5$, we observe that $γ t { R 3 } ( G ) < γ t ( G ) + γ t { R 2 } ( G )$. However, in the following we establish, for any integer $n ≥ 5$, there is a graph G such that $γ t { R 3 } ( G ) = γ t { R 2 } ( G ) + γ t ( G )$.
Proposition 9.
For any positive integer $n ≥ 5$, there is a graph G for which $γ t { R 3 } ( G ) = γ t { R 2 } ( G ) + γ t ( G )$.
Proof.
For $n = 5$ let $G = C 5 +$. Then $γ t { R 2 } ( G ) = 3$, $γ t ( G ) = 2$ and $γ t { R 3 } ( G ) = 5$. For $n = 6$, let G be a bistar of order 6. Then $γ t { R 3 } ( G ) = 6 = 4 + 2 = γ t { R 2 } ( G ) + γ t ( G )$. 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 $n ≥ 10$, by induction we consider the graph $G = C 5 + ∪ H$ where the graph H (H may be connected or disconnected) for which $γ t { R 3 } ( H ) = n − 5 = γ t { R 2 } ( H ) + γ t ( H )$. □
Finally, we show that for any positive integer $n ≥ 5$, there is a graph G such that $γ t { R 3 } ( G ) = n$, $γ t { R 2 } ( G ) = n − 1$ and $γ t ( G ) = n − 2$.
For this, let G be the graph constructed in Proposition 3 as graph H for $n ≥ 5$. Then $γ t { R 3 } ( H ) = n$, $γ t { R 2 } ( H ) = n − 1$ and $γ t ( H ) = n − 2$.

## 5. Large Total Roman ${ 3 }$-domination Number

In this section, we characterize connected graphs G of order n with $γ t { R 3 } ( G ) = 2 n − k$ for $1 ≤ k ≤ 4$. For this we use the following result.
Theorem 9.
Let G be a connected graph of order $n ≥ 2$. Then $γ t { R 3 } ( G ) ≤ ( 3 n ) / 2$, with equality if and only if G is the corona $H ∘ K 1$ where H is a connected graph.
Proof.
If $n = 2$, then the statement is valid. Let now $n ≥ 3$. If $| L ( G ) | ≤ n / 2$, then define $f : V ( G ) → { 0 , 1 , 2 , 3 }$ by $f ( x ) = 2$ for $x ∈ L ( G )$ and $f ( x ) = 1$ for $x ∈ V ( G ) \ L ( G )$. Then f is a total Roman ${ 3 }$-dominating function on G of weight
$2 | L ( G ) | + n − | L ( G ) | = n + | L ( G ) | ≤ 3 n 2 .$
If $| L ( G ) | > n / 2$, then define $f : V ( G ) → { 0 , 1 , 2 , 3 }$ by $f ( x ) = 1$ for $x ∈ L ( G )$ and $f ( x ) = 2$ for $x ∈ V ( G ) \ L ( G )$. Then f is a total Roman ${ 3 }$-dominating function on G of weight
$| L ( G ) | + 2 ( n − | L ( G ) | ) = 2 n − | L ( G ) | < 3 n 2 .$
If $G = H ∘ K 1$ for a is a connected graph H, then $γ t { R 3 } ( G ) = ( 3 n ) / 2$.
Conversely, let $γ t { R 3 } ( G ) = ( 3 n ) / 2$. Then the proof above shows that $| L ( G ) | = n / 2$. Assume that there exists a vertex $v ∈ V ( G )$ which is neither a leaf nor a support vertex. Define $f : V ( G ) → { 0 , 1 , 2 , 3 }$ by $f ( x ) = 1$ for $x ∈ L ( G ) ∪ { v }$ and $f ( x ) = 2$ for $x ∈ V ( G ) \ ( L ( G ) ∪ { v } )$. Then f is a total Roman ${ 3 }$-dominating function on G of weight
$| L ( G ) | + 2 ( n − | L ( G ) | − 1 ) + 1 = 2 n − | L ( G ) | − 1 = 3 n 2 − 1 ,$
a contradiction. Thus every vertex is a leaf or a support vertex. Since $| L ( G ) | = n / 2$, we deduce that $G = H ∘ K 1$ with a connected graph H. □
Corollary 4.
For any connected graph G of order $n ≥ 2$, $γ t { R 3 } ( G ) = 2 n − 1$ if and only if $G = P 2$.
Proof.
Let $γ t { R 3 } ( G ) = 2 n − 1$. Then Theorem 9 implies $2 n − 1 ≤ ( 3 n ) / 2$ and thus $n = 2$. Clearly, the statement is valid for $P 2$. □
Corollary 5.
For any connected graph G of order $n ≥ 3$, $γ t { R 3 } ( G ) = 2 n − 2$ if and only if $G ∈ { P 3 , P 4 }$.
Proof.
If $G ∈ { P 3 , P 4 }$, then the statement is valid. Conversely, let $γ t { R 3 } ( G ) = 2 n − 2$. Then Theorem 9 implies $2 n − 2 ≤ ( 3 n ) / 2$ and thus $n ≤ 4$, with equality if and only if $G = P 4$. In the remaining case $n = 3$, we observe that $G ∈ { P 3 , C 3 }$ with $γ t { R 3 } ( P 3 ) = 4$ and $γ t { R 3 } ( C 3 ) = 3$, and therefore $G = P 3$. □
Next we characterize the graphs G with the property that $γ t { R 3 } ( G ) = 2 | V ( G ) | − 3$.
Theorem 10.
For any connected graph G of order $n ≥ 3$, $γ t { R 3 } ( G ) = 2 n − 3$ if and only if $G ∈ { C 3 , P 3 ∘ K 1 , C 3 ∘ K 1 }$.
Proof.
If $G ∈ { C 3 , P 3 ∘ K 1 , C 3 ∘ K 1 }$, then the statement is valid. Conversely, let $γ t { R 3 } ( G ) = 2 n − 3$. If $Δ ( G ) = 2$, then $G ∈ { P n , C n }$ and we conclude by Observations 1, 2 that $G = C 3$. If $Δ ( G ) = 3$, then $γ t { R 3 } ( G ) = 2 n − 3 = 2 n − Δ ( G )$ and so by Theorem 6, $G ∈ G$. Therefore $G ∈ { P 3 ∘ K 1 , C 3 ∘ K 1 }$. Let $Δ ( G ) ≥ 4$. Then by Theorem 6, $γ t { R 3 } ( G ) ≤ 2 n − Δ ( G ) = 2 n − 4 < 2 n − 3$. Thus $G ∈ { C 3 , P 3 ∘ K 1 , C 3 ∘ K 1 }$, and the proof is complete. □
Let $H$ be the family of connected graphs order 5 with $Δ ( G ) = 3$ 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 $F$ be the family of graphs $G = Q ∘ K 1$ with a connected graph Q of order 4.
Observation 11.
If $G ∈ { F , H }$, then $γ t { R 3 } ( G ) = 2 n − 4$.
Proof.
Clearly, $γ t { R 3 } ( T 5 ) = 2 n − 4 = 6$. Let $G ∈ H$ be of order 5 with exactly one leaf u. If v is the support vertex of u, then $f : V ( G ) → { 0 , 1 , 2 , 3 }$ with $f ( v ) = 2$ and $f ( x ) = 1$ for $x ∈ V ( G ) \ { v }$ is a TR${ 3 }$-DF on G and therefore $γ t { R 3 } ( G ) = 6 = 2 n − 4$.
If $G = Q ∘ K 1$ with a connected graph Q of order 4, then we have seen in proof of Theorem 9 that $γ t { R 3 } ( G ) = ( 3 n ) / 2 = 2 n − 4 = 12$. □
Theorem 12.
For any connected graph G of order $n ≥ 4$, we have $γ t { R 3 } ( G ) = 2 n − 4$ if and only if $G ∈ { C 4 , P 5 } ∪ { c l a w , p a w } ∪ H ∪ F$ 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 $G ∈ { C 4 , P 5 } ∪ { c l a w , p a w } ∪ H ∪ F$. By Observations 1, 2 and 11, we have $γ t { R 3 } ( G ) = 2 n − 4$.
Conversely, let $γ t { R 3 } ( G ) = 2 n − 4$. According to Theorem 9, we have $2 n − 4 = γ t { R 3 } ( G ) ≤ ( 3 n ) / 2$ and thus $n ≤ 8$ with equality if and only if G is the corona $H ∘ K 1$ with a connected graph H of order 4. Therefore $G ∈ F$ if $n = 8$. Let now $n ≤ 7$.
If $Δ ( G ) = 2$, then $G ∈ { P n , C n }$ and by Observations 1, 2, we have $n = 2 n − 4$ which implies $n = 4$ and $G = C 4$, or $n + 1 = 2 n − 4$ which implies $n = 5$ and $G = P 5$ or $n + 2 = 2 n − 4$ which implies $G = P 6$. Since $γ t { R 3 } ( C 4 ) = 4 = 2 n − 4$ and $γ t { R 3 } ( P 5 ) = 6 = 2 n − 4$ but $γ t { R 3 } ( P 6 ) = 7 ≠ 2 n − 4$, we deduce that $G ∈ { C 4 , P 5 }$.
Let now $Δ ( G ) = 3$. 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 $γ t { R 3 } ( G ) = 4 = 2 n − 4$.
If $n = 5$, it is simply verified that $γ t { R 3 } ( G ) = 6 = 2 n − 4$ if an only if $G ∈ H$.
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 $f : V ( G ) → { 0 , 1 , 2 , 3 }$ with $f ( v ) = f ( u 1 ) = 3$ and $f ( x ) = 0$ for $x ≠ v , u 1$ is a TR${ 3 }$-DF on G and therefore $γ t { R 3 } ( G ) ≤ 6$.
Case 2: Assume that $w 2$ is adjacent to $w 1$. Then $f : V ( G ) → { 0 , 1 , 2 , 3 }$ with $f ( v ) = f ( w 1 ) = 3$, $f ( u 1 ) = 1$ and $f ( x ) = 0$ for $x ≠ v , u 1 , w 1$ is a TR${ 3 }$-DF on G and therefore $γ t { R 3 } ( G ) ≤ 7$.
Case 3: Assume that $w 2$ is adjacent to $u 2$ or $u 3$, say $u 2$. If there are no further edges, then $γ t { R 3 } ( G ) = 9 ≠ 2 n − 4$.
Now assume that there are further edges. If $w 2$ is adjacent to $u 3$, then $f : V ( G ) → { 0 , 1 , 2 , 3 }$ with $f ( u 1 ) = 2$ and $f ( x ) = 1$ for $x ≠ u 1$ is a TR${ 3 }$-DF on G and therefore $γ t { R 3 } ( G ) ≤ 7$. If $w 1$ is adjacent to $u 2$, then $f : V ( G ) → { 0 , 1 , 2 , 3 }$ with $f ( v ) = f ( u 2 ) = 3$ and $f ( x ) = 0$ for $x ≠ v , u 2$ is a TR${ 3 }$-DF on G and therefore $γ t { R 3 } ( G ) ≤ 6$. If $u 1$ is adjacent to $u 2$ and there are no further edges, then $γ t { R 3 } ( G ) = 9 ≠ 2 n − 4$. If finally, $u 3$ is adjacent to $u 2$ or $u 1$, say $u 2$, then $f : V ( G ) → { 0 , 1 , 2 , 3 }$ with $f ( u 1 ) = f ( u 2 ) = 3$, $f ( v ) = 1$ and $f ( x ) = 0$ for $x ≠ v , u 1 , u 2$ is a TR${ 3 }$-DF on G and therefore $γ t { R 3 } ( G ) ≤ 7$. Thus we see that there is no graph G of order 6 with $γ t { R 3 } ( G ) = 8 = 2 n − 4$.
Let now $n = 7$. If $| L ( G ) | ≤ 2$, then define $f : V ( G ) → { 0 , 1 , 2 , 3 }$ by $f ( x ) = 2$ for $x ∈ L ( G )$ and $f ( x ) = 1$ for $x ∈ V ( G ) \ L ( G )$. Then f is a total Roman ${ 3 }$-dominating function on G of weight $9 < 10 = 2 n − 4$. If $| L ( G ) | ≥ 4$, then define $f : V ( G ) → { 0 , 1 , 2 , 3 }$ by $f ( x ) = 0$ for $x ∈ L ( G )$ and $f ( x ) = 3$ for $x ∈ V ( G ) \ L ( G )$. Then f is a total Roman ${ 3 }$-dominating function on G of weight $9 < 10 = 2 n − 4$.
Finally, assume that $| L ( G ) | = 3$. If G has exactly 3 support vertices, then define $f : V ( G ) → { 0 , 1 , 2 , 3 }$ by $f ( x ) = 1$ for $x ∈ L ( G )$, $f ( x ) = 2$ for $x ∈ S ( G )$ and $f ( x ) = 0$ for the remaining vertex. Then f is a total Roman ${ 3 }$-dominating function on G of weight $9 < 10 = 2 n − 4$. If G has exactly 2 support vertices, then define $f : V ( G ) → { 0 , 1 , 2 , 3 }$ by $f ( x ) = 0$ for $x ∈ L ( G )$, $f ( x ) = 3$ for $x ∈ S ( G )$ and $f ( x ) = 1$ for the remaining two vertices. Then f is a total Roman ${ 3 }$-dominating function on G of weight $8 < 10 = 2 n − 4$.
Let $Δ ( G ) = 4$. By Theorem 6, $γ t { R 3 } ( G ) = 2 n − 4$ if and only if $G ∈ G ⊆ F$.
Let $Δ ( G ) ≥ 5$. Then by Theorem 6 $γ t { R 3 } ( G ) ≤ 2 n − 5 < 2 n − 4$. Therefore the proof is complete. □

## 6. Complexity

In this section, we study the complexity of total Roman ${ 3 }$-domination of graphs. We show that the total Roman ${ 3 }$-domination problem is $N P$-complete for bipartite graphs. Consider the following decision problem.
Total Roman ${ 3 }$-domination problem TR3DP.
Instance: Graph $G = ( V , E )$, and a positive integer $k ≤ | V |$.
Question: Does G have a total Roman ${ 3 }$-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 ($X 3 C$)
Instance: A finite set X with $| X | = 3 q$ and a collection C of 3-element subsets of X.
Question: Is there a subcollection $C ′$ of C such that every element of X appears in exactly one element of $C ′$?
Theorem 13.
TR3D is $N P$-Complete for bipartite graphs.
Proof.
It is clear that TR3DP belongs to $N P$. 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 $X = { x 1 , x 2 , … , x 3 r }$ and $C = { C 1 , C 2 , … , C t }$ be an arbitrary instance of X3C.
For each $x i ∈ X$, 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 $C j ∈ C$, we form a star $K 1 , 5$ with center $c j$ for which one leaf is labeled $l j$. Let $L = { l 1 , l 2 , … , l t }$. Now to obtain a graph G, we add edges $l j y i 1$ if $y i 1 ∈ C j$. Set $k = 4 t + 13 r$. Let $H = 〈 ⋃ i = 1 3 r V ( G i ) 〉$ be the subgraph of G induced by the $⋃ i = 1 3 r V ( G i )$. Observe that for every total Roman ${ 3 }$-dominating function f on G with $f ( V ( G i ) ) ≥ 4$, 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 ${ 3 }$-dominated. Moreover, since $G i$ has a total Roman ${ 3 }$-domination number equal to 6, we can assume that $f ( V ( G i ) ) ≤ 6$. More precisely, if $f ( V ( G i ) ) = 6$, then, without loss of generality, we may assume that $f ( y i 2 ) = f ( y i 3 ) = f ( y i 2 ) = f ( y i 3 ) = 1$ and $f ( y i 1 ) = 2$. If also, $f ( V ( G i ) ) ∈ { 4 , 5 }$, then obviously at least one vertex of $G i$ (including $y i 1$) is not total Roman ${ 3 }$-dominated. In this case, we can assume that vertices of $G i$ are assigned as $f ( y i 2 ) = f ( y i 3 ) = f ( y i 4 ) = f ( y i 5 ) = 1$ so that, only $y i 1$ is not total Roman ${ 3 }$-dominated and $f ( y i 1 ) ∈ { 0 , 1 }$.
Suppose that the instance X, C of $X 3 C$ has a solution $C ′$. We build a total Roman ${ 3 }$-dominating function f on G of weight k. For every $C j$, assign the value 2 to $l j$ if $C j ∈ C ′$ and 1 to the other $l j$ if $C j ∉ C ′$. 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 $C ′$ exists, $| C ′ | = r$, the number of $l j$s with weight 2 is r, having disjoint neighborhoods in ${ y 1 1 , y 2 1 , … , y 3 r 1 }$, 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 $f ( V ) = 3 t + 2 r + t − r + 12 r = k$.
Conversely, let $g = ( V 0 , V 1 , V 2 , V 3 )$ be a total Roman ${ 3 }$-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 $g ( c j ) = 3$ and all the leaves neighbor of $c j$ are assigned 0. Since $l j c j ∈ E ( G )$, it implies that each vertex $l j$ can be assigned by 1. Moreover, for each i, $g ( V ( G i ) ) ∈ { 4 , 6 }$, as mentioned above. We can let the vertices of $G i$ are assigned the values given in the above paragraph depending on whether $g ( V ( G i ) ) = 4$ or $g ( V ( G i ) ) = 6$. Let p be the number of $G i$s having weight 6. Then $g ( V ( H ) ) = 6 p + 4 ( 3 r − p ) = 12 r + 2 p$. Now, if $g ( l j ) > 1$ for some j, then $l j$ total Roman ${ 3 }$-dominates some vertex $y s 1$, and, in that case, $g ( l j ) = 2$ (since $g ( y i 2 ) = 1 )$. Let z be the number of $l j$s assigned 2 and $t − z$ of others be assigned 1. Then $3 t + 2 z + t − z + 12 r + 2 p ≤ k = 4 t + 13 r$, implies that $z + 2 p ≤ q$. On the other hand, since each $l j$ has exactly three neighbors in ${ x 1 1 , x 2 1 , … , x 3 r 1 }$, we must have $3 z ≥ 3 r − p$. From these two inequalities, we achieve at $p = 0$ and then $z = q$. Consequently, $C ′ = { C j : g ( l j ) = 2 }$ is an exact cover for C. □

## 7. Open Problems

In the preceding sections a new model of total Roman domination, total Roman ${ R 3 }$-domination has been introduced. There are the relationships between the total domination, total Roman ${ R 2 }$-domination and total Roman ${ R 3 }$-domination numbers as follows:
If G is a graph without isolated vertices, then $γ t ( G ) + 1 ≤ γ t { R 3 } ( G ) ≤ 3 γ t ( G )$, (Proposition 5).
If G is a graph without isolated vertices, then $γ t { R 3 } ( G ) = γ t ( G ) + 1$ if and only if G has at least two vertices of degree $Δ = | V ( G ) | − 1$, in the other words $γ t { R 3 } ( G ) = 3$ and $γ t ( G ) = 2$. (Proposition 6).
For any positive integer $n ≥ 5$, there is a graph G of order n in which $γ t { R 3 } ( G ) = γ t { R 2 } ( G ) + γ t ( G )$, (Proposition 9).
For a family of graphs we have shown that $γ t { R 3 } ( G ) = | V ( G ) |$, (Observation 3).
We have already characterized graphs G in which $γ t { R 3 } ( G ) = 2 | V ( G ) | − r$, where $1 ≤ r ≤ 4$.
Problems
1. Characterize the graphs G for which $γ t { R 3 } ( G ) = 3 γ t ( G )$.
2. Does there exist any characterization of graphs G for which $γ t { R 3 } ( G ) = γ t ( G ) + r$, where $2 ≤ r ≤ γ t ( G ) − 2$?
3. For positive integers $n ≥ 5$, characterize the graphs G for which $γ t { R 3 } ( G ) = γ t { R 2 } ( G ) + γ t ( G )$.
4. Does there exist any characterization of graphs G for which $γ t { R 3 } ( G ) = | V ( G ) |$?
5. Can one characterize graphs G in which $γ t { R 3 } ( G ) = 2 | V ( G ) | − r$ for $5 ≤ r ≤ | V ( G ) | − 1$?
6. Is it possible to construct a polynomial algorithm for computing of $γ t { R 3 } ( T )$ for any tree T?

## Author Contributions

Conceptualization, D.A.M., Z.S., and L.V.; writing, D.A.M.; revising, L.V., Z.S., and D.A.M. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

## Acknowledgments

The first author’s research work was supported by the National Key R & D Program of China (Grant No. 2019YFA0706402) and the Natural Science Foundation of Guangdong Province under Grant 2018A0303130115, and the second author’s research work has been supported by a research grant from the University of Mazandaran. The authors thank the three referees of this paper for valuable suggestions and useful comments.

## Conflicts of Interest

The authors declare no conflict of interest.

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Figure 1. A sample of graph G.
Figure 1. A sample of graph G.
Figure 2. A graph H with $n = 3$.
Figure 2. A graph H with $n = 3$.
Figure 3. Examples.
Figure 3. Examples.

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MDPI and ACS Style

Shao, Z.; Mojdeh, D.A.; Volkmann, L. Total Roman {3}-domination in Graphs. Symmetry 2020, 12, 268. https://doi.org/10.3390/sym12020268

AMA Style

Shao Z, Mojdeh DA, Volkmann L. Total Roman {3}-domination in Graphs. Symmetry. 2020; 12(2):268. https://doi.org/10.3390/sym12020268

Chicago/Turabian Style

Shao, Zehui, Doost Ali Mojdeh, and Lutz Volkmann. 2020. "Total Roman {3}-domination in Graphs" Symmetry 12, no. 2: 268. https://doi.org/10.3390/sym12020268

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