Total Roman Domination Number of Rooted Product Graphs

Let G be a graph with no isolated vertex and f : V(G) → {0, 1, 2} a function. If f satisfies that every vertex in the set {v ∈ V(G) : f (v) = 0} is adjacent to at least one vertex in the set {v ∈ V(G) : f (v) = 2}, and if the subgraph induced by the set {v ∈ V(G) : f (v) ≥ 1} has no isolated vertex, then we say that f is a total Roman dominating function on G. The minimum weight ω( f ) = ∑v∈V(G) f (v) among all total Roman dominating functions f on G is the total Roman domination number of G. In this article we study this parameter for the rooted product graphs. Specifically, we obtain closed formulas and tight bounds for the total Roman domination number of rooted product graphs in terms of domination invariants of the factor graphs involved in this product.


Introduction
The study of domination-related parameters in product graphs is one of the most important and attractive areas of domination theory in graphs. Among the principal attractions of this area, Vizing's Conjecture [1] is possibly the most popular open problem for the theory of domination in graphs. This conjecture asserts that the domination number of the Cartesian product of two graphs is at least equal to the product of their domination numbers. In recent years, several kinds of domination-related parameters in product graphs have been studied. For instance, we cite the following works: total domination [2] and total Roman domination [3] in direct product graphs; Roman domination in lexicographic product graphs [4]; domination-related parameters like classical domination, Roman domination, independence domination, connected domination, convex domination and super domination in rooted product graphs [5]; and Roman domination in Cartesian and strong product graphs [6]. In this article, we study a very well-known variant of domination (total Roman domination) for the case of rooted product graphs.
We consider G = (V(G), E(G)) as a simple graph. Given a vertex v of a graph G, N(v) will denote the open neighborhood of v in G, i.e., the set of vertices of G adjacent to v. The closed neighborhood, denoted by N [v], equals N(v) ∪ {v}. Given a set S ⊆ V(G), its open neighborhood is the set N(S) = ∪ v∈S N(v), and its closed neighborhood is the set N[S] = N(S) ∪ S. As usual, the graph obtained from G by removing all the vertices in S ⊆ V(G) will be denoted by G − S. If S = {v}, for some vertex v, then we simple write G − v. γ R (G) ≤ γ tR (G) ≤ 2γ t (G) (see [17]) and γ tR (G) ≤ γ R (G) + γ(G) (see [19]). Further results on total Roman domination can be found for example, in [19][20][21][22].
The complexity of the total Roman domination number was studied in [18]. The authors showed that the decision problem related to total Roman domination number is NP-hard even when restricted to bipartite graphs and chordal graphs. This suggests finding closed formulas or giving tight bounds on this domination invariant for special families of graphs. These studies are attractive, for instance, for the product graphs, and as previously shown, several studies in this regard have been carried out.
Our goal with this paper is to make some contributions to the study of total Roman domination number for the case of rooted product graphs. In that regard, in the next section we state the intervals in which this parameter can be found in a rooted product graph. Furthermore, we obtain closed formulas and tight bounds for this parameter.

Main Results in Rooted Product Graphs
Given a graph G of order n and a graph H with root v ∈ V(H), the rooted product graph G • v H is defined as the graph obtained from G and H by taking one copy of G and n copies of H and identifying the ith-vertex of G with vertex v in the ith-copy of H for every i ∈ {1, . . . , n} [23]. Figure 1 shows an example of rooted product graph P 4 • v H, where P 4 is the path graph of order four. For every x ∈ V(G), H x will denote the copy of H in G • v H containing x. The restriction of any γ tR (G • v H)-function f to V(H x ) will be denoted by f x and the restriction to V(H x ) \ {x} will be denoted by f − x .
Now, we state some tools, which will be very useful throughout the article.

Lemma 1.
Let H be a graph without isolated vertices and v ∈ V(H) \ S(H). Then Moreover, if γ tR (H − v) = γ tR (H) − 2, then the following conditions hold.
Following is a theorem bounding γ tR (H − v), for a particular vertex v ∈ V(H). Before stating this, we shall need the next remark.

Remark 1.
Let H be a graph without isolated vertices. If v ∈ L s (H), then there exists a γ tR (H)-function f such that f (v) = 0.
, which is a contradiction. Hence, u ∈ V 2 , which implies that f (N(u) ∩ L(H)) ≤ 1. So, and without loss of generality, we can assume that f (v) = 0 as |N(u) ∩ L(H)| ≥ 2. Therefore, the proof is complete. Theorem 1. Let H be a graph without isolated vertices. If v ∈ L s (H), then Proof. By Remark 1 there exists a γ tR (H)-function f such that f (v) = 0. Hence, f restricted to let g be a γ tR (H − v)-function and let u be the support vertex associated to v in H. Since g(u) > 0, we have that the function g , defined by g (v) = 1 and g (x) = g(x) whenever x ∈ V(H) \ {v}, is a TRDF on H. So, γ tR (H) ≤ ω(g ) = ω(g) + 1 = γ tR (H − v) + 1, which completes the proof.
Next, we expose a result for the Roman domination number of rooted product graphs given by Kuziak et al. in [5], which will be used later in the paper. Theorem 2. [5] Let G be a graph of order n ≥ 2. Then for any graph H with root v and at least two vertices, We continue this section with a useful lemma.
The following two statements hold for any vertex x ∈ V(G).
which completes the proof of (i). Now, we proceed to prove (ii). Assume that ω( we define the function f as follows: f (w) = min{ f (w) + 1, 2} and In such a case, we define the function f as follows: f (w) = 2 and f (u) = f (u) whenever u ∈ V(H x ) \ {w}. As in the case above, f is a TRDF on H x such that ω( f ) ≤ γ tR (H) − 1, which is again a contradiction. Thus, N(x) ∩ V(H x ) ⊆ V 0 , and the proof is complete.
Theorem 3. Let G and H be two graphs without isolated vertices. If |V(G)| = n and v ∈ V(H), then the following statements hold.
Proof. First, we proceed to prove (i). Notice that, from any γ tR (H)-function, we can construct a TRDF on G • v H of weight nγ tR (H). Thus γ tR (G • v H) ≤ nγ tR (H), which completes the proof of (i).
Finally, we proceed to prove (ii). Assume that v ∈ V(H) \ S(H). Observe that, from any γ tR (G)-function and any γ tR (H − v)-function we can construct a TRDF on G • v H of weight at most γ tR (G) + nγ tR (H − v), which completes the proof.
The next theorem, which we can consider as one of the main results of this paper, states the intervals in which the total Roman domination number of a rooted product graph can be found.

Theorem 4.
Let G and H be two graphs without isolated vertices. If |V(G)| = n and v ∈ V(H), then one of the following conditions holds.
H)-function and we consider the sets A f , B f and C f defined above. Now, we analyze the following cases.
Case 3. B f = ∅ and C f = ∅. By definition we obtain that This implies that A f ∩ V 2 dominates C f , and as a consequence, A f is a dominating set of G. Hence, |A f | ≥ γ(G). Combining the inequalities above, we deduce that On the other hand, by Proposition 1 we have that . Thus, condition (iii) follows.
Case 4. B f = ∅ and C f = ∅. By Proposition 1 and Theorem 3- Finally, let us define a set S ⊆ V(G) as follows. If x ∈ A f then we choose one vertex u x ∈ N(x) ∩ V(G) and set x, u x ∈ S. For the other vertices, if x ∈ B f then we set x ∈ S. Now, we will prove that S is a total dominating set of G. By definition of S, if x ∈ A f , then x is adjacent to some vertex y ∈ S. Now, by Lemma which implies that z ∈ S, as required. Therefore, S is a total dominating set of G of cardinality at most 2|A f | + |B f | and, as a consequence, So, condition (iii) follows, which completes the proof.
The bounds given in the theorem above are tight. To see this, we consider the following examples where H 1 is the graph shown in Figure 2 (we always assume that G is a graph of order n without isolated vertices). Recall that P n is the path graph of order n.

•
If Theorem 5 gives some conditions to achieve the equality γ tR (G • v H) = γ t (G) + n(γ tR (H) − 2). In this case we can take H, for example, as the graph given in Figure 3. Next, we analyse some particular cases. First, we consider the case in which Theorem 5. Let G ∼ = ∪P 2 be a graph of order n without isolated vertices. If H is a graph such that Proof. Let g be a γ tR (H − v)-function. By Lemma 1-(i), we may assume (without loss of generality) that g (N(v)) = 0. From g , we define the following function g on H. If x ∈ V(H − v) then g(x) = g (x), and g(v) = 0. Moreover, let h be a γ tR (H − N[v])-function and we consider the Let D be a γ t (G)-set. From D, g and h, we define the following function f on G • v H. For every z ∈ D, the restriction of f to V(H z ) is induced from h. Moreover, if z ∈ V(G) \ D, then the restriction of f to V(H z ) is induced from g. By the construction of g and h, it is straightforward to see that f is a TRDF on G • v H. Thus, We only need to prove that γ tR (G which is a contradiction because γ t (G) < n (it is easy to see that γ t (G) = n if and only if G ∼ = ∪P 2 ). Therefore, C f = ∅ for any γ tR (G • v H)-function f . Hence, and by analogy to Cases 3 and 4 in the proof of Theorem 4, we deduce that γ tR (G • v H) ≥ γ t (G) + n(γ tR (H) − 2), which completes the proof.
Notice that the premises given for the graph H in the previous theorem lead to the existence of a γ tR (H)-function f satisfying f (v) = 2. Since Lemma 1 does not guarantee the existence of a graph that satisfies such conditions, in Figure 3 we show a graph that does satisfy them. Theorem 6. Let G be a graph of order n without isolated vertices and H a graph such that

Proof. By Theorem 3-(ii) and the equality γ tR (H
then we can obtain a γ tR (H x )-function g with g(x) = 2 by giving label 1 to some neighbor of x in H x , which is a contradiction as H -function with f x (x) = 2, which contradicts the fact that g(v) ≤ 1 for every γ tR (H)-function g). Now, we define a function h(V 0 , V 1 , V 2 ) on G.
, then we choose one vertex not previously labeled u ∈ N(x) ∩ V(G) (if it exists) and set h(u) = 1. (iv) We set h(x) = f (x) for any other vertex x ∈ V(G) not previously labeled.
We only need to prove that V 1 ∪ V 2 is a total dominating set of G. By definition of h, it is clear to see that every vertex in Thus, we conclude that V 1 ∪ V 2 is a total dominating set of G. So, h is a TRDF on G, and as a consequence, Therefore, the proof is complete. Now, we consider the cases in which either v is a support vertex of H or γ tR (H − v) ≥ γ tR (H). Before doing this, we shall need the following useful lemma. Proof. Let h be a function on G • v H defined from f as follows. For every z ∈ V(G), the restriction of h to V(H z ) is induced from f x . Notice that h is a TRDF on G • v H, which implies that γ tR (G • v H) ≤ nω( f x ), as desired. Theorem 7. Let G and H be two graphs without isolated vertices and |V(G)| = n. If v ∈ S(H) or If ω( f ) = nγ tR (H), then we are done. In such a sense, we suppose that ω( f ) < nγ tR (H). This implies that there exists a vertex which is a contradiction. Hence, f (x) > 0. So, by Lemma 2-(ii) we deduce that x / ∈ C f , which implies that x ∈ B f . Moreover, since f (x) > 0, by Lemma 3 we have that H). This implies that f (z) > 0 for every vertex z ∈ V(G). So, by  we (H). So, C f = ∅ by contrapositive of Proposition 1. Hence, as above, we deduce that γ tR (G . Therefore, in both cases, the lower bound obtained leads to the equality γ tR (G • v H) = n(γ tR (H) − 1), as desired.
Furthermore, if γ tR (H) = γ R (H), then by Theorem 2 we have that n(γ tR (H H). Since γ(G) ≥ 1, the result above leads to Observe that if H is a nontrivial star graph or a double star graph (in both cases we consider the root v ∈ S(H)), then for any graph G of order n without isolated vertices we have that γ tR (G • v H) = n(γ tR (H) − 1) and γ tR (G • v H) = nγ tR (H), respectively. These two particular cases belong to families of graphs that are analyzed in Theorems 9 and 12.
We continue with a result in which v ∈ S(H) or γ tR (H − v) ≥ γ tR (H) − 1.

Theorem 8. Let G and H be two graphs without isolated vertices and |V
)-function and let us consider the function h on H defined as follows.
In Figure 4 we show a graph H which satisfies the premises of Theorem 8. In this case, v ∈ S(H) and for any graph G of order n with no isolated vertex, we obtain that γ tR (G • v H) = n(γ tR (H) − 1) = 7n.  Next, we proceed to discuss a particular case when v ∈ S(H).
for any x ∈ V(G). As x and y are adjacent support vertices, it follows that f (x) = f (y) = 2. Hence, f x is a TRDF on H x , and as a consequence, Case 2. g(v) = 1 for every γ tR (H)-function g.
Now, we analyse other cases for the vertex v ∈ V(H).

Theorem 10.
Let G be a graph without isolated vertices of order n. Let H be a graph such that With the assumptions given, Proposition 1 leads to C f = ∅. Now, we analyse two cases. First, if B f = ∅, then by analogy to Case 1 in the proof of Note that the function g, defined by g(u) = 1 for some vertex u ∈ N(x) ∩ V(H x ) and g(w) = f (w) for every w ∈ V(H x ) \ {u}, is a TRDF on H x of weight γ tR (H). So g is a γ tR (H)-function and satisfies that g(v) > 0, which is again a contradiction. Hence, B f = ∅, and we are done.
Next, we consider the case in which the root v is a strong leaf vertex of H. Theorem 11. Let G and H be two graphs without isolated vertices and |V(G)| = n. If v ∈ L s (H), then In such a sense, we consider the following two cases.
Hence f (u x ) > 0, and so f x is a TRDF on H x of weight ω( f x ) = γ tR (H) − 1, which is a contradiction. Therefore, B f = ∅, and so γ tR (G • v H) = nγ tR (H).
Hence, f (x) = 0, and as a consequence, f (u x ) = 1 (otherwise, if f (u x ) = 2, then, as above, f x is a TRDF on H x of weight γ tR (H) − 1, which is a contradiction). Therefore, B f ⊆ V 0 and A f ∩ V 2 dominates B f . This implies that f restricted to V(G) is an RDF on G, and so, γ R (G) ≤ f (V(G)). Now, we suppose that there exists a vertex z ∈ A f ∩ V 2 such that ω( f z ) = γ tR (H). Let u z ∈ N(z) ∩ V(H z ). Since u z ∈ S(G • v H), it follows that f (u z ) > 0. This implies that the function f z is a γ tR (H z )-function because H z ∼ = H. Now, we observe that the function g, defined by g(z) = 1 and g( , which completes the proof. The following theorem states the total Roman domination number of the rooted product graph G • v H, when the root v is a universal vertex of H. Proof. Let f be a function defined as follows: f (x) = 2 if x ∈ V(G) and f (x) = 0 whenever x ∈ V(G • v H) \ V(G). It is clear to see that f is a TRDF on G • v H. Hence, γ tR (G • v H) ≤ ω( f ) = 2n. Since γ tR (H) = 3, we have that either v ∈ S(H) or γ tR (H − v) ≥ γ tR (H). Thus, Theorem 7 leads to the desired result.
Any corona graph G G can be expressed as a rooted product graph G • v H, where H ∼ = K 1 + G and V(K 1 ) = {v}. Therefore, the following result follows as an immediate consequence of the theorem above.
Theorem 13. Let G be a graph of order n without isolated vertices and G any graph. Then γ tR (G G ) = 2n.

Conclusions and Open Problems
In this paper, we have studied the total Roman domination number of rooted product graphs G • v H. In particular, we were able to give the intervals to which γ tR (G • v H) belongs. In addition, we have obtained closed formulas for this parameter when certain conditions are imposed on the factor graphs G and H involved in this product.
Next, we propose some open problems, which we consider to be interesting.
• Characterize the graphs G and H (and the root v) that satisfy the following equalities. Given a graph G of order n and a family of n graphs H = {H 1 , . . . , H n } with a set of root vertices W = {w 1 , w 2 , . . . , w n }, respectively, the generalized rooted product graph G • W H is defined as the graph obtained from G and H, by identifying the vertex v i of G with the root vertex w i ∈ W of H i for every 1 ≤ i ≤ n. We propose to study the total Roman domination number of generalized rooted product graphs. Funding: This research received no external funding.