Domination related parameters in rooted product graphs Dorota Kuziak†, Magdalena Leman´ska‡ and Ismael G. Yero§ † Departament d’Enginyeria Informàtica i Matemàtiques, Universitat Rovira i Virgili, Av. Pa¨ısos Catalans 26, 43007 Tarragona, Spain. 3 1 dorota.kuziak@urv.cat 0 ‡ Department of Technical Physics and Applied Mathematics 2 n Gdan´sk University of Technology, ul. Narutowicza 11/12, 80-233 Gdan´sk, Poland a J magda@mifgate.mif.pg.gda.pl 4 § Departamento de Matemáticas, Escuela Politécnica Superior de Algeciras 1 Universidad de Ca´diz, Av. Ramón Puyol, s/n, 11202 Algeciras, Spain. ] O ismael.gonzalez@uca.es C . h t Abstract a m A set S of vertices of a graph G is a dominating set in G if every vertex outside of [ S is adjacent to at least one vertex belonging to S. A domination parameter of G is 2 relatedtothosesetsofverticesofagraphsatisfyingsomedominationpropertytogether v 4 with other conditions on the vertices of G. Here, we investigate several domination 4 related parameters in rooted product graphs. 6 0 Keywords: Domination; Roman domination; domination related parameters; rooted . 4 product graphs. 0 2 AMS Subject Classification Numbers: 05C12; 05C76. 1 : v i 1 Introduction X r a Domination in graph constitutes a very important area in graph theory [9]. An enormous quantity of researches about domination in graphs have been developed in the last years. Nevertheless, there are still several open problems and incoming researches about that. One interesting question in this area is related to the study of domination related parameters in product graphs. For instance, the Vizing’s conjecture [20, 21], is one of the most popular open problems about domination in product graphs. The Vizing’s conjecture states that the domination number of Cartesian product graphs is greater than or equal to the product of the domination numbers of the factor graphs. Moreover, several kind of domination related parameters have been studied in the last years. Some of the most remarkable examples are the following ones. The domination number of direct product graphs was studied in [3, 11, 16]. The total domination number of direct product graphs was studied in [5]. The upper domination number of Cartesian product graphs was studied in [2]. The independence domination number of Kronecker product graphs was studied in [10]. Several domination 1 related parameters of corona product graphs and the conjunction of two graphs were studied in [8] and [22], respectively. The Roman domination number of lexicographic product graphs was studied in [12]. According to the quantity of works devoted to the study of domination related parameters in product graphs it is noted that not only Vizing’s conjecture is an interesting topic related to domination in product graphs. In this sense, in this paper we pretend to contribute with the study of some domination related parameters for the case of rooted product graphs. We begin by establishing the principal terminology and notation which we will use throughout the article. Hereafter G = (V,E) represents an undirected finite graph without loops and multiple edges with set of vertices V and set of edges E. The order of G is |V| = n(G) and the size |E| = m(G) (If there is no ambiguity we will use only n and m). We denote two adjacent vertices u,v ∈ V by u ∼ v and in this case we say that uv is an edge of G or uv ∈ E. For a nonempty set X ⊆ V and a vertex v ∈ V, N (v) denotes the X set of neighbors that v has in X: N (v) := {u ∈ X : u ∼ v} and the degree of v in X is X denoted by δ (v) = |N (v)|. In the case X = V we will use only N(v), which is also called X X the open neighborhood of a vertex v ∈ V, and δ(v) to denote the degree of v in G. The close neighborhood of a vertex v ∈ V is N[v] = N(v)∪{v}. The minimum and maximum degrees of G are denoted by δ and ∆, respectively. The subgraph induced by S ⊂ V is denoted by hSi and the complement of the set S in V is denoted by S. The distance between two vertices u,v ∈ V of G is denoted by d (u,v) (or d(u,v) if there is no ambiguity). G The set of vertices D ⊂ V is a dominating set of G if for every vertex v ∈ D it is satisfied that N (v) 6= ∅. The minimum cardinality of any dominating set of G is the domination D number of G and it is denoted by γ(G). A set D is a γ(G)-set if it is a dominating set and |D| = γ(G). Throughout the article we follow the terminology and notation of [9]. Given a graph G of order n and a graph H with root vertex v, the rooted product G◦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 vertex u of G with the vertex v in the ith copy of H for every 1 ≤ i ≤ n i [7]. If H or G is the singleton graph, then G ◦H is equal to G or H, respectively. In this sense, to obtain the rooted product G◦H, hereafter we will only consider graphs G and H of orders greater than or equal to two. Figure 1 shows the case of the rooted product graph P ◦ C . Hereafter, we will denote by V = {v ,v ,...,v } the set of vertices of G and by 4 3 1 2 n H = (V ,E ) the ith copy of H in G◦H. i i i Figure 1: The rooted product graph P ◦C . 4 3 It is clear that the value of every parameter of the rooted product graph depends on the root of the graph H. In the present article we present some results related to some domination parameters in rooted product graphs. 2 2 Domination number We begin with the following remark which will be useful into proving next results. Lemma 1. Let G be a graph of order n ≥ 2 and let H be any graph with root v and at least two vertices. If v does not belong to any γ(H)-set or v belongs to every γ(H)-set, then γ(G◦H) = nγ(H). Proof. If A is a dominating set of minimum cardinality in H = (V ,E ), i ∈ {1,...,n}, then i i i i n it is clear that A is a dominating set in G ◦ H. Thus γ(G ◦ H) ≤ nγ(H). Suppose i=1 i v does not belong to any γ(H)-set. Let S be a γ(G◦ H)-set and let S = S ∩V for every i i S i ∈ {1,...,n}. Notice that the set S dominates all the vertices of H except maybe the root i i v which could be dominated by other vertex not in H . i i If v ∈/ S for some j ∈ {1,...,n}, then S is a dominating set in H −v . So γ(H −v ) ≤ j j j j j j j |S |. Moreover, since v does not belong to any γ(H )-set, it is satisfied that γ(H −v ) = j j j j j γ(H ). If |S | < γ(H), then we have that γ(H − v ) ≤ |S | < γ(H ) = γ(H − v ), a j j j j j j j j contradiction. On the other side, if v ∈ S for some l ∈ {1,...,n}, then S is a dominating l l l set in H . So γ(H ) ≤ |S |. Therefore, |S | ≥ γ(H) for every i ∈ {1,...,n} and we obtain that l l l i γ(G◦H) = nγ(H). On the other hand, let us suppose v belongs to every γ(H)-set. Thus, v dominates at least one vertex in H which is not dominated by any other vertex in every γ(H)-set and, as a consequence, γ(H −v) ≥ γ(H). As above, S denotes a γ(G◦H)-set and S = S ∩V for i i every i ∈ {1,...,n}. If v ∈/ S for some j ∈ {1,...,n}, then either S is not a dominating j j j set in H (which is a contradiction) or |S | ≥ γ(H − v ) ≥ γ(H ). On the contrary, if j j j j j v ∈ S , then S is a dominating set in H and |S | ≥ γ(H ). Therefore, we have that j j j j j j n n |S| = |S | ≥ γ(H ) = nγ(H) and the proof is complete. i=1 i i=1 i TheorPem 2. LetPG be a graph of order n ≥ 2. Then for any graph H with root v and at least two vertices, γ(G◦H) ∈ {nγ(H),n(γ(H)−1)+γ(G)}. Proof. It is clear that γ(G◦H) ≤ nγ(H) and also, from Lemma 1, there are rooted product graphs G◦H such that γ(G◦H) = nγ(H). Now, let us suppose that γ(G◦H) < nγ(H). Let V be the set of vertices of G and let V , i ∈ {1,...,n}, be the set of vertices of the i copy H of H in G◦H. Hence, if S is a γ(G◦H)-set, then there exists j ∈ {1,...,n} such i that |S ∩V | < γ(H). Notice that the set S ∩V dominates all the vertices in V excluding j j j v . If |S ∩ V | < γ(H) − 1, then the set (S ∩ V ) ∪ {v } is a dominating set in H and j j j j j |(S ∩V )∪{v }| ≤ |(S ∩V )|+1 < γ(H), which is a contradiction. So, |S ∩V | ≥ γ(H)−1 j j j i for every i ∈ {1,...,n}. Let x be the number of copies H ,H ,...,H of H in which the vertex v of G is not j1 j2 jx ji dominated by S∩V (i.e., v is dominated by a vertex of G belonging to other copy H , with ji ji l l ∈/ {j ,...,j }). On the contrary, let y = n−x be the number of copies H ,H ,...,H of 1 x k1 k2 ky H in which the vertex v of G is dominated by S ∩V or v ∈ S. Note that the y vertices ki ki ki v of G satisfying the above property form a dominating set in G and, as a consequence, ki γ(G) ≤ y. Since n = x + y, we have that x ≤ n − γ(G). Also, notice that if the vertex v of G is dominated by S ∩ V or v ∈ S, then S ∩ V is a dominating set in H . So, ki ki ki ki ki γ(H) ≤ |S ∩V | for every copy H in which the vertex v of G is dominated by S ∩V or ki ki ki ki v ∈ S. Thus we have the following. ki 3 x y γ(G◦H) = |S| = (S ∩V )∪ (S ∩V ) ji ki (cid:12) (cid:12) (cid:12)i[=1 i[=1 (cid:12) (cid:12)x y (cid:12) (cid:12) (cid:12) = (cid:12) |S ∩V |+ |S ∩V |(cid:12) ji ki i=1 i=1 X X ≥ x(γ(H)−1)+yγ(H) = nγ(H)−x ≥ nγ(H)−n+γ(G) = n(γ(H)−1)+γ(G). On the other side, let A be a γ(G ◦ H)-set. Since γ(G ◦ H) < nγ(H), there exists at least one copy H of H such that |A ∩ V | < γ(H), which implies |A ∩ V | ≤ γ(H) − 1. k k k Since A ∩ V dominates all the vertices of H except maybe the root v , we have that if k k k v ∈ A∩V , then A∩V is a dominating set in H, which is a contradiction. So, v ∈/ A∩V . k k k k k Now, as |A∩V | ≥ γ(H)−1 for every i ∈ {1,...,n}, we obtain that |A∩V | = γ(H)−1. So, i k A′ = (A∩V )∪{v } is a γ(H)-set. Let us denote by A′, i ∈ {1,...,n}, the set of vertices of k k i A′ −{v } in each copy H of G◦H. i i Let B be a γ(G)-set and let D = ( n A′) ∪ B. Since A′ dominates the vertices of i=1 i i H −{v } for every i ∈ {1,...,n} and B dominates the vertices of G, we obtain that D is a i i S dominating set in G◦H. Thus n |D| = |A′|+|B| = n(|A′|−1)+|B| = n(γ(H)−1)+γ(G). i i=1 X Therefore, we obtain that γ(G◦H) ≤ n(γ(H)−1)+γ(G) and the result follows. 3 Roman domination number Roman domination number was defined by Stewart in [18] and studied further by some researchers, for instance in [4]. Given a graph G = (V,E), a map f : V → {0,1,2} is a Roman dominating function for G if for every vertex v with f(v) = 0, there exists a vertex u ∈ N(v) such that f(u) = 2. The weight of a Roman dominating function is given by f(V) = f(u). The minimum weight of a Roman dominating function on G is called u∈V the Roman domination number of G and it is denoted by γ (G). A function f is a γ (G)- R R P function in a graph G = (V,E) if it is a Roman dominating function and f(V) = γ (G). R Let f be a Roman dominating function on G and let B , B and B be the sets of 0 1 2 vertices of G induced by f, where B = {v ∈ V : f(v) = i}. Frequently, a Roman i dominating function f is represented by the sets B , B and B , and it is common to denote 0 1 2 f = (B ,B ,B ). It is clear that for any Roman dominating function f on the graph G = 0 1 2 (V,E) of order n we have that f(V) = f(u) = 2|B |+|B | and |B |+|B |+|B | = n. u∈V 2 1 2 1 0 The following lemmas will be useful into proving other results in this section. P Lemma 3. [4] For any graph G, γ(G) ≤ γ (G) ≤ 2γ(G). R Lemma 4. Let G = (V,E) be a graph and let f = (B ,B ,B ) be a γ (G)-function. Then 0 1 2 R for every v ∈ V, 4 (i) If v ∈ B , then γ (G)−1 ≤ γ (G−v) ≤ γ (G). 0 R R R (ii) If v ∈ B , then γ (G−v) = γ (G)−1. 1 R R (iii) If v ∈ B , then γ (G)−1 ≤ γ (G−v) ≤ γ (G)+δ(v)−2. 2 R R R Proof. Let f′ = (A ,A ,A ) be a γ (G−v)-function. By making f′(v) = 1 we have that f′ 0 1 2 R is a Roman dominating function in G. Thus γ (G) ≤ γ (G−v)+1. (1) R R Now, if v ∈ B , then it is clear that γ (G−v) ≤ γ (G) and (i) is proved. 0 R R Moreover, if v ∈ B , then (B ,B −{v},B ) is a Roman dominating function in G−v. 1 0 1 2 Thus γ (G−v) ≤ γ (G)−1. Therefore, by (1) we obtain (ii). R R On the other hand, if v ∈ B , then (B ,B ∪ (N(v) − B ),B − {v}) is a Roman 2 0 1 2 2 dominating function in G−v. Thus γ (G−v) ≤ 2|B −{v}|+|B ∪(N(v)−B )| R 2 1 2 = 2|B |−2+|B |+|N(v)−B | 2 1 2 ≤ γ (G)+δ(v)−2. R Therefore, (iii) is proved. Lemma 5. Let G = (V,E) be a graph. If for every γ (G)-function f = (B ,B ,B ) is R 0 1 2 satisfied that v ∈ B , then 0 γ (G−v) = γ (G). R R Proof. FromLemma 4 (i) we have that γ (G−v) ≤ γ (G). If γ (G−v) < γ (G), then there R R R R existsaγ (G−v)-functionh = (A ,A ,A )suchthath(V−{v}) = γ (G−v) < γ (G),which R 0 1 2 R R leads to h(V −{v}) ≤ γ (G)−1. If h′ is a function in G such that for every u ∈ V, u 6= v, we R have that h′(u) = h(u) and h′(v) = 1, then h′ is a Roman dominating function in G. Thus, γ (G) ≤ h′(V) = h(V −{v})+1 ≤ γ (G). So, γ (G) = h′(V) = h(V −{v})+1 = γ (G) and R R R R we have that h′ is a γ (G)-function such that h′(v) = 1, which is a contradiction. Therefore, R γ (G−v) = γ (G). R R The Roman domination number of rooted product graphs is studied at next. Theorem 6. Let G be a graph of order n ≥ 2. Then for any graph H with root v and at least two vertices, n(γ (H)−1)+γ(G) ≤ γ (G◦H) ≤ nγ (H). R R R Proof. It is clear that γ (G ◦ H) ≤ nγ (H). Let V be the set of vertices of H for every R R i i i ∈ {1,...,n} and let f = (B ,B ,B ) be a γ (G◦H)-function. Now, for every i ∈ {1,...,n} 0 1 2 R (i) and every k ∈ {0,1,2}, let B = B ∩ V . Let j ∈ {1,...,n}. We consider the following k k i cases. (j) (j) (j) (j) (j) Case 1: v ∈ B . If N (v )∩B 6= ∅, then f = (B −{v },B ,B ) is a Roman j 0 Hj j 2 j 0 j 1 2 (j) dominating function in H −v . On the contrary, if N (v )∩B = ∅, then v is adjacent j j Hj j 2 j (k) (j) (j) (j) to some vertex v ∈ B , with k 6= j and, again f = (B − {v },B ,B ) is a Roman k 2 j 0 j 1 2 (j) (j) dominating function in H −v . So, γ (H −v ) ≤ 2|B |+|B |. By Lemma 4 (i) we have j j R j j 2 1 (j) (j) that γ (H −v ) ≥ γ (H)−1. Thus 2|B |+|B | ≥ γ (H)−1. R j j R 2 1 R 5 (j) (j) (j) (j) Case 2: v ∈ B . Hence, it is clear that f = (B ,B − {v },B ) is a Roman j 1 j 0 1 j 2 (j) (j) dominating function in H −v . So, γ (H −v ) ≤ 2|B |+|B |−1. By Lemma 4 (ii) we j j R j j 2 1 (j) (j) have that γ (H −v ) = γ (H)−1. Thus 2|B |+|B | ≥ γ (H). R j j R 2 1 R (j) (j) (j) (j) Case 3: v ∈ B . Thus f = (B ,B ,B ) is a Roman dominating function in H . j 2 j 0 1 2 j (j) (j) So, 2|B |+|B | ≥ γ (H). 2 1 R Now, let V be the set of vertices of G and let A ⊆ V ∩B be the set of vertices of G 0 (l) such that for every vertex v ∈ A is satisfied that N (v )∩B = ∅. So, every vertex v ∈ A l Hl l 2 l (k) is dominated by some vertex in (V −A)∩B , with k 6= l. As a consequence, V −A is a 2 dominating set and γ(G) ≤ n−|A|. Since A ⊆ V ∩B , it is satisfied that |A| equals at most 0 (j) (j) the numbers of copies H of H such that 2|B |+|B | ≥ γ (H)−1 (those copies satisfying j 2 1 R Case 1). Thus we have the following, γ (G◦H) = 2|B |+|B | R 2 1 n (i) (i) = (2|B |+|B |) 2 1 i=1 X n−|A| |A| (i) (i) (i) (i) = (2|B |+|B |)+ (2|B |+|B |) 2 1 2 1 i=1 i=1 X X ≥ (n−|A|)γ (H)+|A|(γ (H)−1) R R = nγ (H)−|A| R ≥ n(γ (H)−1)+γ(G). R Therefore the lower bound is proved. As the following proposition shows, the above bounds are tight. Theorem 7. Let G be a graph of order n ≥ 2 and let H be a graph with root v and at least two vertices. Then, (i) If for every γ (H)-function f = (B ,B ,B ) is satisfied that f(v) = 0, then R 0 1 2 γ (G◦H) = nγ (H). R R (ii) If there exist two γ (H)-functions h = (B ,B ,B ) and h′ = (B′,B′,B′) such that R 0 1 2 0 1 2 h(v) = 1 and h′(v) = 2, then γ (G◦H) = n(γ (H)−1)+γ(G). R R Proof. Letf′ = (B′,B′,B′)beaγ (G◦H)-functionandletV bethesetofverticesofH ,i ∈ 0 1 2 R i i {1,...,n}. Now, forevery i ∈ {1,...,n}, let f = (B(i) = B′∩V ,B(i) = B′∩V ,B(i) = B′∩V ). i 0 0 i 1 1 i 2 2 i From Theorem 6 we have that γ (G ◦ H) ≤ nγ (H). If γ (G ◦ H) < nγ (H), then there R R R R (j) (j) (j) (j) (j) exists j ∈ {1,...,n} such that f (V ) = 2|B |+|B | < γ (H). So f = (B ,B ,B ) is j j 2 1 R j 0 1 2 not a Romandominating function inH . If f′(v ) = 1 or f′(v ) = 2, then every vertex in B(j) j j j 0 (j) (j) (j) (j) is adjacent to a vertex in B and, asa consequence, (B ,B ,B ) is a Romandominating 2 0 1 2 function in H , which is a contradiction. So f′(v ) = 0 and f = (B(j)−{v },B(j),B(j)) is a j j j 0 j 1 2 γ (H −v )-function. Since f(v) = 0 for every γ (H)-function, by Lemma 5 we have that R j j R 6 (j) (j) 2|B |+|B | = γ (H −v) = γ (H) and this is a contradiction. Therefore, γ (G◦H) = 2 1 R R R nγ (H) and (i) is proved. R (i) (i) (i) Toprove(ii),foreveryi ∈ {1,...,n}weconsidertwoγ (H )-functionsh = (A ,A ,A ) R i i 0 1 2 and h′ = (B(i),B(i),B(i)) such that h (v ) = 1 and h′(v) = 2, and let S be a γ(G)-set. Now, i 0 1 2 i i i we define a function g in G◦H in the following way. • For every vertex x belonging to a copy H of H such that the root v ∈ S we make j j g(x) = h′(x) (notice that g(v ) = 2). j • For every vertex y, except the corresponding root, belonging to a copy H of H such l that the root v ∈/ S, we make g(x) = h(x). l • For every root of every copy H satisfying the conditions of the above item we make l g(x) = 0 (note that these vertices are adjacent to a vertex w of G for which g(w) = 2). Since every vertex u ∈ V not in G, with g(u) = 0, is adjacent to a vertex u′ such that j g(u′) = 2 and also, every vertex v of G, with g(v ) = 0, is adjacent to a vertex v ∈ S with l l k g(v ) = 2, we obtain that g is a Roman dominating function in G◦H. Thus k |S| n−|S| (i) (i) (i) (i) γ (G◦H) ≤ (2|B |+|B |)+ (2|A |+|A |−1) R 2 1 2 1 i=1 i=1 X X = |S|γ (H)+(n−|S|)(γ (H)−1) R R = n(γ (H)−1)+|S| R = γ(G)+n(γ (H)−1). R Therefore, (ii) follows by Theorem 6. Ontheother hand, we cansee thattherearerootedproduct graphsforwhich thebounds of Theorem 6 are not achieved. Theorem 8. Let G be a graph of order n ≥ 2 and let H be a graph with root v and at least two vertices. If for every γ (H)-function f is satisfied that f(v) = 1, then R γ (G◦H) = n(γ (H)−1)+γ (G). R R R Proof. Let f = (B ,B ,B ) be a γ (H)-function and let f′ = (B′,B′,B′) be a γ (G)- 0 1 2 R 0 1 2 R function. Now, let us define a function h in G ◦ H such that if u 6= v, then h(u) = f(u). Otherwise, h(u) = f′(u). Since f(v) = 1 for every γ (H)-function, it is satisfied that every R vertex x of G◦H with h(x) = 0 is adjacent to a vertex y in G◦H with h(y) = 2. Thus h is a Roman dominating function in G◦H and we have that n γ (G◦H) ≤ (2|B′|+|B′|)+ (2|B |+|B |−1) R 2 1 2 1 i=1 X = n(γ (H)−1)+γ (G). R R On the other hand, let V , i ∈ {1,...,n}, be the set of vertices of the copy H of H in G◦H i i and let V be the set of vertices of G. Now, let g = (A ,A ,A ) be a γ (G◦H)-function and 0 1 2 R (i) (i) (i) for every i ∈ {1,...,n} let g = (A = A ∩V ,A = A ∩V ,A = A ∩V ). Since the root i 0 0 i 1 1 i 2 2 i v of H satisfies that f(v ) = 1 for every γ (H )-function f, we have the following cases. i i i R i 7 Case 1: If there exists l ∈ {1,...,n} such that g(v ) = 2, then g is a Roman dominating l l (l) (l) function in H , but it is not a γ (H)-function. Thus γ (H ) < 2|A |+|A |, which leads to l R R l 2 1 (l) (l) (l) (l) γ (H ) ≤ 2|A |+|A |−1 = 2|A −{v }|+|A |+1. R l 2 1 2 l 1 Case 2: If there exists j ∈ {1,...,n} such that g(v ) = 1, then g is a Roman dominating j j function in H and g′ = (A(j),A(j)−{v },A(j)) is a Roman dominating function in H −v . j j 0 1 j 2 j j Thus, by Lemma 4 (ii), it is satisfied that (j) (j) γ (H ) = γ (H −v )+1 ≤ 2|A |+|A −{v }|+1. R j R j j 2 1 j Case 3: If there exists i ∈ {1,...,n} such that g (v ) = 0, then we have one of the i i following possibilities: • g is not a Roman dominating function in H . So, v should be adjacent to a vertex i i i v , j 6= i, of G such that g (v ) = 2. Moreover, g′ = (A(i) −{v },A(i),A(i)) is a Roman j j j i 0 i 1 2 dominating function in H − v and by Lemma 4 (ii) it is satisfied that γ (H ) = i i R i (i) (i) γ (H −v )+1 ≤ 2|A |+|A |+1. R i i 2 1 • g is a Roman dominating function in H . Since f(v ) = 1 for every γ (H )-function f, i i i R i we have that g (V ) > γ (H ). Let f be a γ (H )-function. Now, by taking a function i i R i i R i g′ on G◦H, such that if v ∈ V , then g′(v) = f′(v) and if v ∈/ V , then g′(v) = g(v), we i i obtain that g′ is a Roman dominating function for G◦H and the weight of g′ is given by n n g′ V = g V +f (V ) j j i i ! ! j=1 j=1,j6=i [ [ n = g V +γ (H ) j R i ! j=1,j6=i [ n < g V +g (V ) j i i ! j=1,j6=i [ n = g V j ! j=1 [ = γ (G◦H). R and this is a contradiction. As a consequence, we obtainthat if g (v ) = 0, then g isnot a Romandominating functionin i i i H . So, everyvertexv ofGforwhichg(v) = 0isadjacenttoavertexv ,k 6= l, ofGsuchthat i l l k g(v ) = 2 and it is satisfied that the function g′ = (X = A ∩V,X = A ∩V,X = A ∩V) k 0 0 1 1 2 2 is a Roman dominating function in G and γ (G) ≤ 2|X |+|X |. Thus we have the following, R 2 1 8 γ (G◦H) = 2|A |+|A | R 2 1 (i) (i) (i) (i) (i) (i) = (2|A |+|A |)+ (2|A |+|A |)+ (2|A |+|A |) 2 1 2 1 2 1 vXi∈X0 vXi∈X1 vXi∈X2 (i) (i) (i) (i) = (2|A |+|A |)+ (2|A |+|A −{v }|)+ 2 1 2 1 i vXi∈X0 vXi∈X1 (i) (i) + (2|A −{v }|+|A |)+|X |+2|X | 2 i 1 1 2 vXi∈X2 ≥ (γ (H )−1)+ (γ (H )−1)+ (γ (H )−1)+2|X |+|X | R i R i R i 2 1 vXi∈X0 vXi∈X1 vXi∈X2 n ≥ (γ (H )−1)+γ (G) R i R i=1 X = n(γ (H)−1)+γ (G). R R Therefore the result follows. 4 Independent domination number A set of vertices S of a graph G is independent if the subgraph induced by S has no edges. The maximum cardinality of an independent set in G is called the independence number of G and it is denoted by α(G). A set S is a α(G)-set if it is independent and |S| = α(G). A set of vertices D of a graph G is an independent dominating set in G if D is a dominating set and the subgraph hDi induced by D is independent in G [1]. The minimum cardinality of any independent dominating set in G is called the independent domination number of G and it is denoted by i(G). A set D is a i(G)-set if it is an independent dominating set and |D| = i(G). At next we study the independent domination number of rooted product graphs and we begin by studying the independence number. Lemma 9. Let v be any vertex of a graph G. If v belongs to every α(G)-set, then α(G) ≥ α(G−v)+1. Proof. Let S be a α(G−v)-set. Since S is still independent in G, we have α(G) ≥ |S|. If α(G) = |S|, then S is a α(G)-set and v ∈/ S, a contradiction. So, α(G) ≥ α(G−v)+1. Theorem 10. For any graph G of order n ≥ 2 and any graph H with root v and at least two vertices, (i) If there is a α(H)-set not containing the root v, then α(G◦H) = nα(H). (ii) If the root v belongs to every α(H)-set, then α(G◦H) = n(α(H)−1)+α(G). n Proof. Let S , i ∈ {1,...,n}, be a α(H )-set not containing the root v . Hence, S is i i i i=1 i independent in G ◦H. Thus α(G◦ H) ≥ nα(H). If α(G ◦H) > nα(H), then there exists S j ∈ {1,...,n} such that |S | > α(H) and S is independent, a contradiction. Therefore, j j α(G◦H) = nα(H). 9 On the other hand, suppose the root v belongs to every α(H)-set. Let A be a α(H )-set i i n and let B be a α(G)-set. Since v ∈ A for every i ∈ {1,...,n}, by taking A = B∪( A − i i i=1 i {v }) we have that A is independent in G◦H. Thus i S n α(G◦H) ≥ |A| = |B|+ |A −{v }| = n(α(H)−1)+α(G). i i i=1 X Now, let V , i ∈ {1,...,n}, be the set of vertices of the copy H of H in G ◦ H and let V i i be the set of vertices of G. Let X be a α(G ◦ H)-set and let X = X ∩ (V − {v }) for i i i every i ∈ {1,...,n} and let Y = V ∩ X. Notice that Y and X are independent sets. So, i α(H −v ) ≥ |X | and α(G) ≥ |Y| and by Lemma 9 we have that |X | ≤ α(H )−1. Thus i i i i i n n α(G◦H) = |Y|+ |X | ≤ α(G)+ (α(H )−1) = α(G)+n(α(H)−1). i i i=1 i=1 X X Therefore, the proof is complete. Lemma 11. Let G = (V,E) be a graph. Then for every set of vertices A ⊂ V, i(G−A) ≥ i(G)−|A|. Proof. Let us suppose i(G −A) < i(G)−|A|. So, there exists an independent dominating set S ⊂ V − A in G − A such that |S| < i(G) − |A|. Let v ∈ A. If N (v) 6= ∅, then v is S independently dominated by the set S in G. On the contrary, if N (v) = ∅, then the set S S ∪{v} is still independent. So, by adding those vertices which maintain the independence in the set S we obtain a set S′ which is independent and dominating in G and we have that i(G) ≤ |S′| ≤ |S| + |A| < i(G) − |A| + |A| = i(G), which is a contradiction. Therefore, i(G−A) ≥ i(G)−|A|. Lemma 12. If v does not belong to any i(G)-set, then i(G−v) = i(G). Proof. Let S be an i(G)-set. Since v ∈/ S, S is still independent and dominating in G−v. So, i(G − v) ≤ i(G). On the other hand, let A be an i(G − v)-set. Let us suppose that |A| < i(G). So, |A| ≤ i(G) − 1. If N (v) = ∅ in G, then A ∪ {v} is independent and A dominating in G. So, i(G) ≤ |A∪{v}| = |A|+1 ≤ i(G). Thus |A+{v}| = i(G) and this is a contradiction because v does not belong to any i(G)-set. On the contrary, if N (v) 6= ∅, A then A is independent and dominating in G, which is a contradiction (|A| < i(G)). So, |A| ≥ i(G). Therefore, i(G−v) = |A| ≥ i(G) and the result follows. Theorem 13. Let G = (V,E) be a graph of order n ≥ 2 and let H be a graph with root v and at least two vertices. Then n(i(H)−1)+i(G) ≤ i(G◦H) ≤ i(H)α(G)+i(H −v)(n−α(G)). Proof. Let S be an i(G ◦ H)-set and let S = S ∩ V , i ∈ {1,...,n}. If v ∈ S for some i i j j ∈ {1,...,n}, then S is an independent dominating set in H . So, |S | ≥ i(H). On the j j j contrary, if v ∈/ S for some k ∈ {1,...,n}, then S independently dominates all vertices k k of H − v. So, S is an independent dominating set in H −v and by Lemma 11 we have k k k 10

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