Dominating the Direct Product of Two Graphs through Total Roman Strategies

Given a graph G without isolated vertices, a total Roman dominating function for G is a function f:V(G)→{0,1,2} such that every vertex u with f(u)=0 is adjacent to a vertex v with f(v)=2, and the set of vertices with positive labels induces a graph of minimum degree at least one. The total Roman domination number γtR(G) of G is the smallest possible value of ∑v∈V(G)f(v) among all total Roman dominating functions f. The total Roman domination number of the direct product G×H of the graphs G and H is studied in this work. Specifically, several relationships, in the shape of upper and lower bounds, between γtR(G×H) and some classical domination parameters for the factors are given. Characterizations of the direct product graphs G×H achieving small values (≤7) for γtR(G×H) are presented, and exact values for γtR(G×H) are deduced, while considering various specific direct product classes.


Introduction
The present investigation is devoted to describe a number of contributions to the theory of total Roman dominating functions while dealing with the direct (or tensor or Kronecker) product of two graphs. Studies concerning parameters in relation to domination in graphs are very frequently present in the last recent years. This might probably be caused by the popularity of some classical problems, like for instance Vizing's conjecture [17,18] for domination in Cartesian products 1 . See [3], for a survey and recent results concerning this conjecture. Several other problems concerning domination parameters in product graphs have occupied the mind of a significant number of investigators. Works of that type concerning direct product graphs are [4,8,12,14].
The (total) Roman domination variants are among the most popular topics of domination in graphs. Both versions have had their birth in connection with some defense strategies related to the ancient Roman Empire (see [13,15]). Studies on (total) Roman domination in product graphs have not escaped from the researchers attention. For instance, [5,6,16,19] are aimed to these goals, although no works appear that considers the Roman domination parameters for the case of direct products. We hence continue with giving new contributions to the theory of parameters related to domination in graph products, specifically we center our attention on the total Roman domination version for the case of the direct product of graphs.
In this work, we consider simple graphs without vertices of degree 0. For a map f : V (G) → {0, 1, 2} and a set of vertices S ⊆ V (G), the weight of S under f is f (S) = v∈S f (v). Moreover, the weight of f is ω(f ) = f (V (G)). Since the function f generates three sets V 0 , V 1 , V 2 such that .
A function f = (V 0 , V 1 , V 2 ) is known to be a Roman dominating function on G whenever all vertices v ∈ V 0 have at least one neighbor u ∈ V 2 . In connection with this, the parameter of G called Roman domination number stands for the least weight among all functions that are proved to be Roman dominating on G. This parameter is usually represented as γ R (G). Such concepts in the theory of graphs were formally introduced in [7], motivated in part by some domination strategies which arose from the antique Roman Empire (see for instance [13,15]). A Roman dominating function f = (V 0 , V 1 , V 2 ) is called a total Roman dominating function if V 1 ∪ V 2 induces a graph without vertices of degree 0. The total Roman domination number of G stands for the minimum possible weight among all total Roman dominating functions on G. This parameter is pointed out as γ tR (G). By a γ tR (G)-function we mean a total Roman dominating function whose weight equals precisely γ tR (G). These concepts of total Roman domination were first introduced in [11] by using some more general settings. The concepts were further specifically introduced and firstly well studied in [2].
A set D = {v 1 , . . . , v r } ⊂ V (G) is called a packing set of G, if N[v i ] ∩ N[v j ] = ∅ for every two different integers i, j ∈ {1, . . . , r}. The packing number of G is the cardinality of a largest possible packing set of G. We represent such cardinality as ρ(G). A packing set induces a subgraph of maximum degree 0, i.e., a graph without edges. If we substitute the closed neighborhoods with open neighborhood in the definition above, then the concept of open packing sets arises. Hence, D is considered to be an open packing set whenever N(v i ) ∩ N(v j ) = ∅ for any two distinct i, j ∈ {1, . . . , r}. Similarly, the parameter called open packing number of G is the cardinality of the largest possible open packing set of G. We write this cardinality by using the notation ρ o (G). We recall that any open packing set represents a set of vertices of the graph which induces a graph having the maximum degree equal to one, and clearly, it could have some vertices whose degree equals zero.
A set D ⊆ V (G) is total dominating if all the vertices of the whole graph G have at least a neighbor in the set D. The cardinality of the smallest total dominating set of G is known as the total domination number of G. This cardinality is then represented as γ t (G). A set being total dominating and having cardinality γ t (G) is said to be a γ t (G)-set. The graph G is called an efficient open domination graph, if there is a total dominating set of G which is simultaneously also an open packing.
The direct product (also known as tensor product or Kronecker product) of two graphs G and H is the graph denoted by G × H whose vertex set is given by V (G × H) = V (G) × V (H) and the edge set is the Cartesian product of the vertex sets of the factors. That is, Figure 1 we show the graph P 6 ×P 6 . As usual, we call the map p G : (g, h) → g a projection of G × H onto G and the map p H : is similarly defined. Note that vertices from a G-layer and from an H g -layer form independent sets of G × H.
The direct product is a graph product (see the exhausting monograph on graph products [9]) in categorical sense, as the end vertices of every edge from G × H project to end vertices of edges in both factors. Consequently, one of the most natural products among all graph products is precisely the direct product, but on the other hand, this also makes this product the most elusive one in many perspectives. So, the connectedness of both factors G and H does not imply the connectedness of the product G × H. (Notice that P 6 × P 6 from Figure 1 is not connected.) To achieve this, one of the factors must also be non-bipartite, see Theorem 5.9 in [9]. One reason for this is that layers form independent sets in G × H. On the other side, the open neighborhoods behave "nice", with respect to the factors, while making a direct product based on the fact Two different total Roman dominating functions on P 6 × P 6 are presented on Figure 1.
The maximum degree of a vertex in a graph G is denoted by ∆(G). Clearly, 1 ≤ ∆(G) ≤ |V (G)| − 1 as we consider only simple graphs having no vertices of degree zero. A leaf of G is a vertex v ∈ V (G) with degree δ G (v) = 1 and in contrast, if δ G (v) = |V (G)|−1, then the vertex v is called as universal vertex. For the specific case of the direct product of two graphs G and H, we recall that δ G×H (g, h) = δ G (g)δ H (h) and ∆(G × H) = ∆(G)∆(H) by (1).

General bounds
We start our exposition with some lower and upper bounds for γ tR (G × H) which are mainly depending on ρ(G), ρ(H), γ tR (G) and γ tR (H).  Proof. We consider a function f on G×H defined as follows.
, then a symmetrical argument to the above one produce a similar conclusion.
As a consequence, we deduce f is a total Roman dominating function on the direct product G × H, which leads to Now, in order we deduce the lower bound, a γ tR (G × H)-function f and a ρ(G)-set S = {u 1 , . . . , u ρ(G) } are considered. Hence, for any integer i ∈ {1, . . . , ρ(G)}, we construct a function h i on H as follows. Also, for any vertex Moreover, note that in this case h i (y) = 2 and also that y In such situation, we similarly get h i (y ′ ) ≥ 1 and y ′ ∈ N H (v).
On the other hand, if h i (v) = 0, then for every vertex As a consequence of these arguments, we deduce that h i is a total Roman dominating function on H whose weight is less than or equal to Hence, we have the following.
By the symmetry of the product, we also deduce that γ tR (G × H) ≥ ρ(H)γ tR (G), and this ends the proof for the case of the lower bound.
Since every graph of order at least three contains at least one total Roman dominating function whose weight equals the total Roman domination number and at least one vertex labeled two, the following result is directly deduced from the result above.
Corollary 2. For every graphs G and H without vertices of degree 0 and of orders at least three, Notice that we can avoid the remarks about maximum cardinality of A 2 and B 2 in Theorem 1. However, the bound is better if we take a γ tR (G)-function and a γ tR (H)-function with maximum cardinality of A 2 and B 2 , respectively. The proof of the upper bound from Theorem 1 remains valid for any total Roman dominating functions g and h of graphs G and H without isolated vertices, respectively, as long as we exchange γ tR (G) and γ tR (H) by ω(g) and ω(h), respectively, in the last step of the proof. Therefore, we can improve the upper bound of Theorem 1, we we next show.
Remark 3. For every two graphs G and H without vertices of degree 0, where such minimum value is understood for every total Roman dominating functions g = (A 0 , A 1 , A 2 ) and h = (B 0 , B 1 , B 2 ) on G and H, respectively.
Despite the fact that the bound above represents an advance with respect to the upper bound of Theorem 1, we have no knowledge of one pair of graphs G and H where the bound given in Remark 3 is better than the upper bound of Theorem 1.
Let D G be a γ t (G)-set. Clearly, the function g = (V (G) −D G , ∅, D G ) total Roman dominating for G and the weight of g is ω(g) = 2γ t (G). Remark 3 yields the following connection.
If the graphs G and H represents efficient open domination graphs, then ρ o (H) = γ t (H) and ρ o (G) = γ t (G) (see Observation 1.1 from [10]), and Corollary 4 implies the following.

Corollary 5. If the graphs G and H represents efficient open domination graphs, then
A graph G is known to be a total Roman graph if it satisfies that γ tR (G) = 2γ t (G). In the case of two total Roman graphs we can develop the upper bound of Corollary 4 to the following result.
Corollary 6. If G and H are two total Roman graphs, then The bound given in Theorem 1 can be enhanced by a factor of 2, whenever one factor is bipartite and the other without triangles as shown next.
Theorem 7. If G is a triangle free graph and H is a bipartite graph of order at least two without isolated vertices, then Proof. Let f and S be defined in a similar manner to that of the proof of Theorem 1 for the lower bound. Clearly, for any vertex A similar argument as the one used to prove Theorem 1 gives the stated bound.
By using a similar argument as the one used while proving the lower bound of Theorem 1, but using an open packing instead of a packing, we can also deduce the lower bound in the following result.
Theorem 8. For any graphs G and H without vertices of degree 0 and of orders at least three, Proof. The stated bound is obtained by considering a similar partition of the vertex of G, as the one used while proving the lower bound of Theorem 1, but instead of using only one vertex as the "center" of each set of the partition, we might need to use now two adjacent vertices as the "centers". This is based on the structure of open packing sets. We consider a γ tR (G × H)-function f , and a ρ o (G)-set S = S 0 ∪ S 1 such that S 0 induces a graph without edges and S 1 induces a regular graph of degree 1. Note that S 0 or S 1 could be empty (although not both at the same time). Now, for every u i ∈ S 0 , we construct a function In the same manner, as in the proof of the lower bound of Theorem 1, we deduce that h i is a total Roman dominating function on H, and so, Now, for any pair of adjacent vertices As a consequence of these arguments, we deduce that h ′ i is a total Roman dominating function on H whose weight is less than or equal to f ( Hence, we have the following.
By the symmetry of the product, we also deduce that , which completes the first part of the proof.
The bound of Theorem 8 can be improved if we consider one bipartite factor and the other without triangles as next stated.
Theorem 9. If G is a graph having no triangles and having a ρ o (G)-set which induces a graph with all components isomorphic to K 2 , and H is a bipartite graph without vertices of degree 0 and of order at least two, then Proof. Let f be a γ tR (G × H)-function, and assume S = induces a non connected graph with at least two components. In concordance with this fact, by using similar arguments as those ones in the proofs for the lower bounds of Theorems 1 and 8, we deduce that for every i ∈ {1, . . . , ρ o (G)/2}, we can construct two total Roman dominating functions . Therefore, we obtain that and the proof is completed.
3 Direct product graphs with small γ tR (G × H) We concentrate our attention in this section on the case when γ tR (G × H) is small. We shall characterize all the direct products graphs G × H for which γ tR (G × H) ≤ 7. For this we need the following class of graphs. A graph G is called triangle centered if there exists a triangle C 3 = xyz in G such that every vertex of G is adjacent to at least two vertices of C 3 . We call such C 3 as the central triangle of a triangle centered graph. Notice that any two vertices of a central triangle form a total dominating set of a triangle centered graph G and we have γ t (G) = 2.
Theorem 10. The following assertions holds for any two graphs G and H without vertices of degree 0. Proof. For (i) notice that there must be at least two adjacent vertices (g, h) and and (g ′ , h) have label 0 and no neighbor with label 2, a contradiction. This already shows that γ tR (G × H) ≥ 3. If γ tR (G ×H) = 3, then either |V 1 ∪V 2 | = 2, which is not possible, or |V 1 ∪V 2 | = 3. In later case there are three vertices of label 1 and no vertex of label 2, a contradiction as we have |V (G × H)| ≥ 4. Hence γ tR (G × H) > 3.
To end with (i) suppose that γ tR (G × H) = 5. Let first |V 2 | = 2 where (g, h), (g 1 , h 1 ) ∈ V 2 . If g = g 1 and h = h 1 , then only one vertex from (g, h 1 ) and (g 1 , h) can have label 1 and the other has label 0 and is not adjacent to a vertex of label 2, a contradiction. So, either g = g 1 or h = h 1 , say g = g 1 . In V 1 is only one vertex, say (g 2 , h 2 ), and it must be adjacent to both vertices of V 2 . This means that h 2 = h and h 2 = h 1 . But then (g, h 2 ) posses label 0 and is not adjacent to a vertex of label 2, a contradiction.
So let |V 2 | = 1 where (g, h) ∈ V 2 and (g ′ , h ′ ) ∈ V 1 is adjacent to (g, h). There are only two more vertices in V 1 and these vertices must be (g, h ′ ) and (g ′ , h) because they are not adjacent to (g, h). If there exists any other vertex from the mentioned four, then such a vertex implies the existence of a vertex of label 0 in G h ∪ H g , a contradiction. Hence we have only four vertices and G × H ∼ = K 2 × K 2 . But in this case we have γ tR (G × H) ≤ 4 as there exists a total Roman dominating function with V 1 = V (G) × V (H). This is the final contradiction and γ tR (G × H) = 5.
The implication (⇐) of item (ii) follows from (i) and the total Roman dominating function with V 1 = V (K 2 )×V (K 2 ). For (⇒) of (ii) suppose that at least one of G and H contains more than three vertices. Hence |V (G) × V (H)| ≥ 6 and if all vertices have label 1, then γ tR (G × H) ≥ 6 > 4. Otherwise, if V 0 = ∅, then also V 2 = ∅. Let (g, h) ∈ V 2 and let (g ′ , h ′ ) ∈ V 1 ∪ V 2 be a neighbor of (g, h). If also (g, h ′ ), (g ′ , h) ∈ V 1 ∪ V 2 , then we have γ tR (G × H) > 4. On the other hand, if at least one of (g, h ′ ) and (g ′ , h) has label 0, then there exists a vertex of label 2 different than (g, h) and (g ′ , h ′ ), meaning that γ tR (G × H) > 4 again and (ii) is done.
For (iii) we start with (⇐). We know from (i) and (ii) that γ tR (G × H) ≥ 6 whenever at least one of G and H contains more than two vertices, which is true in all three cases. Suppose first that each G and H have at least two universal vertices g, g ′ and h, h ′ , respectively, and are of order at least three. If we set is a total Roman dominating function with ω(f ) = 6. Assume now that one factor, say H, is K 2 and that G contains at least three vertices together with a universal vertex g. For for an arbitrary neighbor g ′ of g in G. It is easy to check that f 2 is a total Roman dominating function with ω(f 2 ) = 6. The third possibility is that both G and H are triangle centered graphs with central triangles g 1 g 2 g 3 and h 1 h 2 h 3 , respectively. We define V ′′ is a total Roman dominating function. First notice that V 2 induces a triangle in G × H. Let (g, h) ∈ V 0 . By the definition of the central triangle, g and h are adjacent to at least two vertices of {g 1 , g 2 , g 3 } and {h 1 , h 2 , h 3 }, respectively. Hence, there exists i ∈ {1, 2, 3} such that gg i ∈ E(G) and hh i ∈ E(H), and (g, h) is adjacent to (g i , h i ) ∈ V 2 . Therefore, f is a total Roman dominating function on G × H with ω(f 3 ) = 6. In all three cases we have γ tR (G × H) ≤ 6 and by (i) and (ii) the equality γ tR (G × H) = 6 follows.
For the opposite implication (⇒) of (iii) we have γ tR (G × H) = 6 and analyze the different possibilities for the cardinalities of V 1 and V 2 for a γ tR (G × H)-function f = (V 0 , V 1 , V 2 ). We start with |V 1 | = 0 and |V 2 | = 3 and let (g 1 , h 1 ), (g 2 , h 2 ), (g 3 , h 3 ) ∈ V 2 . As V 1 ∪ V 2 induces a graph without isolated vertices one vertex of these mentioned three, say (g 2 , h 2 ), must be adjacent to the other two. Hence g 1 g 2 , g 2 g 3 ∈ E(G) and h 1 h 2 , h 2 h 3 ∈ E(H). If g 1 g 3 / ∈ E(G), then (g 1 , h 2 ) is a vertex of label 0 not adjacent to a vertex from V 2 . Similar, if h 1 h 3 / ∈ E(H), then (g 2 , h 1 ) is a vertex of label 0 not adjacent to a vertex from V 2 . Hence g 1 g 2 g 3 and h 1 h 2 h 3 form a triangle in G and H, respectively. Suppose that there exists g ∈ V (G) that is either adjacent to exactly one vertex of {g 1 , g 2 , g 3 }, say to g 1 , or to no vertex of {g 1 , g 2 , g 3 }. In both cases we obtain (g, h 1 ) must has label 0, and is not adjacent to any vertex of V 1 ∪ V 2 , which is not possible for a total Roman dominating function f . Thus, every vertex g ∈ V (G) must be adjacent to at least two vertices from {g 1 , g 2 , g 3 } and G is triangle centered. Similarly one shows that H is triangle centered and the third option follows.
We continue with |V 1 | = 2 and |V 2 | = 2. Let (g, h) and (g ′ , h ′ ) be vertices of label 2. Assume first that (g, h) and (g ′ , h ′ ) are adjacent. Hence, the vertices (g, h ′ ) and (g ′ , h) are not adjacent to (g, h) nor to (g ′ , h ′ ) and must have label 1. All the other vertices are in V 0 . Moreover, V 0 = ∅ as the converse leads to a contradiction with f being a γ tR (G × H)-function. Every vertex (g, x), x ∈ V (H) − {h, h ′ } has label 0 and is not adjacent to (g, h). Therefore they must be adjacent to (g ′ , h ′ ), which means that h ′ is a universal vertex of H. Similarly, every vertex (g ′ , x), x ∈ V (H) − {h, h ′ } has label 0 and is not adjacent to (g ′ , h ′ ). So they are adjacent to (g, h), and h is a universal vertex of H. By symmetric arguments, also g and g ′ are universal vertices of G. Thus, both G and H have at least two universal vertices. If both have only two vertices, then we have a contradiction with (ii). Therefore we obtained the first possibility.
Let now (g, h) and (g ′ , h ′ ) be nonadjacent. If they are not in the same (G-or H-) layer, then (g, h ′ ) and (g ′ , h) are not adjacent to (g, h) nor to (g ′ , h ′ ) and must have label 1. All the other vertices must be in V 0 . But, this is a contradiction because V 1 ∪ V 2 induces four isolated vertices. Hence, (g, h) and (g ′ , h ′ ) are in the same G-or H-layer, say in H g . So, g = g ′ . If there exists different h 1 , h 2 ∈ V (H) − {h, h ′ }, then (g, h 1 ), (g, h 2 ) ∈ V 1 , since there are no edges between vertices of H g . A contradiction again, due to no existing edges between vertices of 1 and the other vertex (a, b) from V 1 must be adjacent to all three vertices from H g . This is not possible as (a, b) is contained in one of the layers G h , G h ′ or G h 1 . Again we have a vertex from V 1 ∪ V 2 that is not adjacent to any other vertex of V 1 ∪ V 2 , a contradiction. So, H contains only two vertices h and h ′ , which are adjacent and therefore both universal vertices. If both vertices from V 1 belong to the same G-layer, say G h , then (g, h) is not adjacent to any vertex from V 1 ∪ V 2 , which is not possible. So, we may assume that V 1 = {(g 1 , h), (g 2 , h ′ )}. Clearly gg 1 , gg 2 ∈ E(G), so that V 1 ∪ V 2 induces a subgraph without isolated vertices. Also every vertex (g 3 , h) ∈ V 0 must be adjacent to (g, h ′ ), which means that gg 3 ∈ E(G) and g is an universal vertex of G. (Notice also that in the case when g 1 = g 2 , there always exists g 3 ∈ V (G) − {g, g 1 }, because otherwise we have a contradiction with (ii).) This yields the middle case of (iii).
To end with (iii) let |V 1 | = 4 and h). Clearly all vertices from G h ∪ H g must be in V 1 ∪ V 2 , meaning that one of the factors is K 2 and the other contains three vertices, say H ∼ = K 2 . Moreover, g must be a universal vertex of G. So, either G ∼ = C 3 or G ∼ = P 3 , which is the middle case of (iii) and the proof of (iii) is completed.
We continue with (⇐) of (iv). We may assume that G has exactly one universal vertex g, and that H is different from K 2 with a universal vertex h, and that at most one of G and H is triangle centered. Further, let g ′ and h ′ be arbitrary neighbors of g in G and of h in H, respectively. By (i), (ii) and (iii) we know that γ tR (G×H) ≥ 7. If we set is a total Roman dominating function with ω(f ) = 7. Hence, γ tR (G × H) ≤ 7 and the equality follows.
So we can assume that either g 2 = g 3 or g 2 = g 4 , say that g 2 = g 3 . Moreover, also h 2 = h 4 as otherwise (g 2 , h 4 ) has no neighbor of label 2. If h 2 is not adjacent to some vertex h ∈ V (H), then (g 2 , h) is not adjacent to a vertex of label 2, meaning that h 2 is a universal vertex of H. Similarly, we see that g 2 is a universal vertex of G. We have γ tR (G × H) = 6 by (iii) when both G and H have (at least) two universal vertices, or one is K 2 and the other contains a universal vertex, a contradiction. Hence, one of G or H has at most one universal vertex and the other is not K 2 and we are done in this case.
The second possibility is that |V 1 | = 3 and h 4 ) and (g 5 , h 5 ) are adjacent, then (g 4 , h 5 ), (g 5 , h 4 ) ∈ V 1 , say (g 4 , h 5 ) = (g 2 , h 2 ) and (g 5 , h 4 ) = (g 3 , h 3 ). Suppose that g 1 / ∈ {g 4 , g 5 } and h 1 / ∈ {h 4 , h 5 }. All the vertices of G h 4 − {(g 4 , h 4 ), (g 5 , h 4 )} must be in V 0 and adjacent to (g 5 , h 5 ), meaning that g 5 is a universal vertex of G. Similarly, all the vertices of G h 5 − {(g 4 , h 5 ), (g 5 , h 5 )} must be in V 0 and adjacent to (g 4 , h 4 ), meaning that g 4 is a universal vertex of G. This means that G is triangle centered with central triangle g 1 g 4 g 5 . By symmetric arguments H is triangle centered with central triangle h 1 h 4 h 5 . By (iii) we have γ tR (G × H) = 6, a contradiction. So, either h 1 ∈ {h 4 , h 5 } or g 1 ∈ {g 4 , g 5 }, say h 1 = h 4 . By the same arguments as above, we see that g 4 is a universal vertex of G, and that h 4 and h 5 are universal vertices of H. (Notice that g 1 is not adjacent to g 5 , otherwise also g 5 is universal vertex, a contradiction with (iii).) If H ∼ = K 2 , then we have γ tR (G × H) = 6 by (iii), a contradiction. Otherwise H ≇ K 2 and we are done. Now we can assume that (g 4 , h 4 ) and (g 5 , h 5 ) are not adjacent. If g 4 = g 5 and h 4 = h 5 , then, as in the previous paragraph, we can choose the notation such that (g 4 , h 5 ) = (g 2 , h 2 ) and that (g 5 , h 4 ) = (g 3 , h 3 ). Moreover, (g 1 , h 1 ) must be adjacent to all other vertices from V 1 ∪ V 2 in order to avoid isolated vertices of positive label. Vertices (g 5 , h 1 ) and (g 1 , h 4 ) are from V 0 and must have a neighbor in V 2 . The only possibility is that (g 5 , h 1 ) is adjacent to (g 4 , h 4 ) and (g 1 , h 4 ) is adjacent to (g 5 , h 5 ). The mentioned edges imply that g 4 g 5 ∈ E(G) and h 4 h 5 ∈ E(H), a contradiction with the not adjacency of (g 4 , h 4 ) and (g 5 , h 5 ). It remains that (g 4 , h 4 ) and (g 5 , h 5 ) belong to the same layer, say H g 4 , that is g 4 = g 5 . Every vertex from H g 4 − {(g 4 , h 4 ), (g 4 , h 5 )} is not adjacent to a vertex of label 2 and must poses label 1. We need also at least two vertices of label 1 outside of H g 4 to assure non isolated vertices in V 1 ∪ V 2 . This means that |V (H)| ≤ 3. Every vertex from G h 4 − {(g 4 , h 4 )} must be adjacent to (g 4 , h 5 ) and g 4 is a universal vertex of G. If H ∼ = K 2 , then we have a contradiction with (iii). So either H ∼ = P 3 or H ∼ = C 3 , meaning that also H has a universal vertex and the second possibility is done.
The last option is that |V 1 | = 5 and |V 2 | = 1, where V 2 = {(g, h)}. Clearly all vertices from G h ∪ H g must be in V 1 ∪ V 2 and g and h must be universal vertices of G and H, respectively. We either obtain a contradiction with (iii) (when one factor is K 2 ) or obtain that G ∼ = H ∼ = K 1,2 which yields the desired situation and the proof of (iv) is completed.
We conclude this proof with (v). We have γ tR (G × H) ≥ 8 from assertions (i) − (iv). Let is a total Roman dominating function on G × H. Let (g 1 , h 1 ) ∈ V 0 . Clearly, g 1 is neighbor of g or of g ′ , say of g, and h 1 is neighbor of h or h ′ , say h. Therefore (g 1 , h 1 ) is neighbor of (g, h) and f satisfies the conditions to be total Roman dominating for G × H. Hence, the inequality γ tR (G × H) ≤ 8 is obtained, which leads to the claimed equality.
A wheel graph W n , n ≥ 4, is a join of K 1 and C n−1 and a fan graph F n , n ≥ 2, is a join of K 1 and P n−1 . Clearly W n and F n have exactly one universal vertex when n > 4. In particular, W n and F n are triangle centered whenever n ∈ {4, 5}. For a complete graph K n and a maximum matching M of it, the graph K n − M, n ≥ 5, is a triangle centered graph with a universal vertex whenever n is an odd number. By using Theorem 10 we directly obtain the next results (among others).
We end this section with an alternative presentation with respect to Theorem 10, where we consider the number of vertices in V 1 ∪V 2 of a total Roman dominating function. For the minimum cardinality of V 1 ∪V 2 , we need an additional condition that the cardinality of V 2 must be maximum to be able to characterize them.
Theorem 12. Let f = (V 0 , V 1 , V 2 ) be a γ tR (G × H)-function with the largest possible cardinality for V 2 , where G and H are two graphs of order at least three. The next items are equivalent.
(i) Graphs G and H are triangle centered.
For the direction ((ii) ⇒ (iii)), let γ tR (G × H) = 6 where f = (V 0 , V 1 , V 2 ) is a γ tR (G × H)function with maximum cardinality of V 2 . There exist vertices from G × H in V 0 as there are at least nine vertices in G × H. Consequently V 2 = ∅. Let (g, h) ∈ V 2 and let (g ′ , h ′ ) be a neighbor of (g, h) with f (g ′ , h ′ ) > 0. There exists at least one vertex (x, y) from (G h ∪ H g ) − {(g, h)} of label 0, because γ tR (G × H) = 6. Suppose that (g ′′ , h ′′ ) is a neighbor of (x, y) of label 2. Assume first that (g ′ , h ′ ) = (g ′′ , h ′′ ). The vertices (g ′ , h) and (g, h ′ ) are not adjacent to (g ′ , h ′ ) nor to (g, h). If they have label equal to 1, then all the other vertices have label 0 and every vertex is adjacent to (g, h) or to (g ′ , h ′ ). Let g 1 and h 1 be a third vertex of G and H, respectively. Clearly, (g 1 , h ′ ) and (g ′ , h 1 ) are adjacent to (g, h) and with this, we have gg 1 ∈ E(G) and hh 1 ∈ E(H). Similarly, (g, h 1 ) and (g 1 , h) are adjacent to (g ′ , h ′ ), and with this we get g ′ g 1 ∈ E(G) and h ′ h 1 ∈ E(H).
Clearly, f ′ is a total Roman dominating function with |V ′ 2 | > |V 2 |, a contradiction with the choice of f . Therefore, the label of (g ′ , h) and (g, h ′ ) must be 0 and there exists a third vertex (g 2 , h 2 ) of label 2 that is adjacent to (g ′ , h) and (g, h ′ ). From γ tR (G × H) = 6 it follows that |V 1 ∪ V 2 | = 3.
((iii) ⇒ (i)) Let |V 1 ∪ V 2 | = 3 and let (g 1 , h 1 ), (g 2 , h 2 ), (g 3 , h 3 ) ∈ V 1 ∪ V 2 . As V 1 ∪ V 2 induces a graph without isolated vertices, one vertex of these mentioned three, say (g 2 , h 2 ), must be adjacent to the other two. Thus, g 1 g 2 , g 2 , g 1 ∈ E(G) and h 1 h 2 , h 2 , h 3 ∈ E(H). If g 1 g 3 / ∈ E(G), then (g 1 , h 2 ) is a vertex that is labeled with 0 being not neighbor of a vertex belonging to V 2 . Similarly, if h 1 h 3 / ∈ E(H), then (g 2 , h 1 ) is a vertex whose label is equal to 0 being not neighbor of one vertex from V 2 . Hence g 1 g 2 g 3 and h 1 h 2 h 3 form a triangle in G and H, respectively. Suppose there is a vertex g ∈ V (G) which is either neighbor of exactly one vertex of {g 1 , g 2 , g 3 }, say to g 1 , or to no vertex of {g 1 , g 2 , g 3 }. In both cases the vertex (g, h 1 ) has label 0 and is not adjacent to any vertex of V 1 ∪ V 2 , which is not possible since f is a function which is total Roman dominating. Hence, every vertex g ∈ V (G) is adjacent to two or more vertices from {g 1 , g 2 , g 3 } and G is triangle centered. Similarly, one shows that H is triangle centered.

A general lower bound and its consequences on the direct product
The following lower bound for γ tR (G) depends on the order of G and its maximum degree ∆(G) as well as on a γ tR (G)-function. . Moreover, if in addition |V (G)| = ∆(G)|V 2 | + |V 1 |, then the equality γ tR (G) = |V (G)| − (∆(G) − 2)|V 2 | holds.
If we rewrite the Theorem 13 for the direct product G × H, then we have the following. 2 | holds. The lower bound from Theorem 13 is better when |V 2 | is small as possible. Also, one cannot expect that the mentioned bound behave well when there exists a small quantity of vertices with maximum number of neighbors in G. From this point of view, one can expect that Theorem 13 works at its best for regular graphs. To see this, the following known remark is necessary.