Outer independent global dominating set of trees and unicyclic graphs

Let G be a graph. A set D ⊆ V (G) is a global dominating set of G if D is a dominating set of G and G. γg(G) denotes global domination number of G. A set D ⊆ V (G) is an outer independent global dominating set (OIGDS) of G if D is a global dominating set of G and V (G) − D is an independent set of G. The cardinality of the smallest OIGDS of G, denoted by γ g (G), is called the outer independent global domination number of G. An outer independent global dominating set of cardinality γ g (G) is called a γ oi g -set of G. In this paper we characterize trees T for which γ g (T ) = γ(T ) and trees T for which γ oi g (T ) = γg(T ) and trees T for which γ oi g (T ) = γ (T ) and the unicyclic graphs G for which γ g (G) = γ(G), and the unicyclic graphs G for which γ g (G) = γg(G).


Introduction
The usual graph theory notions not herein, refer to [15].Let G = (V, E) be a graph with vertex set V = V (G) and edge set E = E(G).The order of G is denoted by n(G) = |V |.A unicyclic www.ejgta.orgThe cardinality of the maximum independent set of graph G, denoted by α(G), is called independence number of G.The diameter of connected graph G is defined as diam(G) = max{d(u, v) : u, v ∈ V (G)}.For a vertex u ∈ V (G), the eccentricity of u, defined as (u) = max{d(u, v) : v ∈ V (G)}.The radius of a graph G defined as R(G) = min{ (u) : Let G be a graph and B be a subset of V (G) and u ∈ B. We say that vertex v is a private neighbor of u respected to B if N [v] ∩ B = {u} and we say that u is an isolated vertex respected to B if N (u) ∩ B = ∅ [15].A vertex v ∈ V (G) is called a leaf, if d(v) = 1.We denote the set of leaves of graph G by L(G).A vertex u ∈ V (G) that is adjacent to a leaf is called a support vertex.We denoted the set of support vertices of G by S(G).A set D ⊆ V (G) is a dominating set of G if every vertex of V (G) − D is adjacent to at least one vertex of D. The cardinality of the smallest dominating set of G, denoted by γ(G), is called the domination number of G.A dominating set of cardinality γ(G) is called a γ-set of G [9].For every u ∈ S(G) delete all the leaves from N (u) except one, then the remaining graph is called the pruned of G and denoted by G p .Further about the pruned graphs and its application we refer to the reference [11].A set S ⊆ V (G) is a global dominating set of G if S is a dominating set of G and G.The cardinality of the smallest global dominating set of G, denoted by γ g (G), is called the global domination number of G [3,6].A set D ⊆ V (G) is an outer independent dominating set (OIDS) of G if D is a dominating set of G and V (G) − D is an independent set of G.The cardinality of the smallest OIDS of G, denoted by γ oi (G), is called the outer independent domination number of G.An outer independent dominating set of cardinality γ oi (G) is called a γ oi -set of G [10].Also the global outer connected dominating set of graphs has already been studied in [1].
One of many applications of global domination have been given in [2], which relates to a communication network modeled by a graph G, where subnetworks are defined by some matching M i of cardinality k.The necessity of these subnetworks could be due for reason of security, redundancy or limitation of recipients for different classes of messages.For this practical case, the global domination number represents the minimum number of master stations needed such that a message issued simultaneously from all masters reaches all desired recipients after traveling over only one communication link.We note that Carrington [4] gave two other applications of global dominating sets for graph partitioning commonly used in the implementation of parallel algorithms.If D ⊆ V (G) and G − D is an independent set, then D is a vertex cover of G.For every graph G without isolate vertices we have β(G) = γ oi (G).All connected graphs G with γ(G) = β(G) have been characterized in [12,16].Actually they characterized the connected graph G with equal domination number and outer independent domination number.Therefore we maybe use for a graph G, γ oi (G) instead of β(G).
In this article we are going to define and study outer independent global dominating set of trees and unicycles.

Preliminaries results
Let τ denote the class of trees with n ≥ 2 vertices and either radius one (that is, stars) or radius two having a vertex u with d(u) ≥ 2 and d(v) ≥ 3 for all v ∈ N (u) [3].Let M be the family of trees T with diam(T ) = 4 and C(T ) ⊆ S(T ) and N be the family of trees T with diam(T ) = 4 and C(T ) S(T ).Definition 2.1.Let u and v be two distinct vertices of tree T .We define = k and non of the internal vertices of the path between u and v is a support vertex, and D(u, v) = 0 if at least one of the internal vertices of the path between u, v is a support vertex.
The following results has a straightforward proof and it is left.
2 .The next result can be found in [9].Theorem 2.1.([9] Theorem 1.1) A dominating set A is a minimal dominating set of G if and only if for each vertex u ∈ A, one of the following two conditions holds: (a) u is an isolate vertex of A, (b) u has a private neighbor with respect to A.
If A is a domination set (OIDS) of G, then every leaf or it's support vertex belongs to A, so we have the bellow observation.
Observation 2.3.Let T = P 2 be a tree and A be a γ-set (γ oi -set) of T .Then the set Lemma 2.1.Let T be a tree and γ oi (T ) = γ(T ).Let A be a γ oi -set of T and D(u, v) > 1, u, v ∈ S(T ).Let P = ux 1 x 2 ...x k−1 v be the path between u and v in T .If ab ∈ E(P ), then a / ∈ A or b / ∈ A.
Proof.On the contrary let a, b ∈ A. Since D(u, v) > 1, so a / ∈ S(T ) or b / ∈ S(T ).Without lose of generality let b / ∈ S(T ).A is a minimal dominating set of T too, therefore by Theorem 2.1, b has a private neighbor with respect to A like x.Since b / ∈ S(T ), so x is not a leaf, therefore x has a neighbor like y that y / ∈ A, thus the vertices x, y are two adjacent vertices in V (T ) − A, therefore A is not an OIDS of T , that is a contradiction.
The following result characterizes the tree T with γ oi (T ) = γ(T ) other than point of view of what Stracke do in Proposition 2.1.
Proof.Let γ oi (T ) = γ(T ) and A be a γ oi (T )-set of T and D(u, v) > 1, u, v ∈ S(T ).Let P = ux 1 x 2 ...x k−1 v be the path between u and v in T .By Observation 2.3 without lose of generality we have S(T ) ⊆ A. By Lemma 2.1, if k is odd then, it will be contradiction.Now let D(u, v) = k ≥ 6, and k is even.
By Lemma 2.1 we have {u, ) > 2 and w ∈ N (x 4 ) − A, then by a similar proof we find that w is dominated by A 1 .Therefore A 1 is a dominating set of T .Conversely, let T be a tree and for every vertices u, v ∈ S(T ) we have D(u, v) = k, k ∈ {0, 1, 2, 4}.Let A be a γ oi -set of T .By Observation 2.

OIGDS of trees
We begin this section with a definition.
The cardinality of the smallest OIGDS of G, denoted by γ oi g (G), is called the outer independent global domination number OIGDN of G.An outer independent global dominating set of cardinality γ oi g (G) is called a γ oi g -set of G. Lemma 3.1.[7] For any graph G, if R(G) ≥ 3, then every dominating set of G is a dominating set of G.
If T is a tree and diam(T ) ≥ 5, then γ g (T ) = γ(T ) and γ oi g (T ) = γ oi (T ).Theorem 3.1.Let T be a tree.Then γ oi g (T ) = γ g (T ) if and only if one of the following conditions holds: Proof.If diam(T ) = 0 or 1 or 2, then the proof is clear.If diam(T ) = 3, then let us observe that S(T ) is a γ oi g -set and γ g -set of T .Now let diam(T ) = 4.If T ∈ M , then let us observe that S(T ) is a γ oi g -set and γ g -set of T and if Let T be a tree.Then γ oi g (T ) = γ oi (T ) if and only if T = P 2 and T / ∈ N and T is not a star.
Proof.If T = P 2 or T ∈ N or T is a star, then it is clear that γ oi g (T ) = γ oi (T ) + 1.We show that for other trees T we have γ oi g (T ) = γ oi (T ).For T = P 1 we have γ oi g (T ) = γ oi (T ).If T ∈ M or diam(T ) = 3, then S(T ) is a γ oi g -set and γ oi -set of T , so γ oi g (T ) = γ oi (T ).Now let γ oi g (T ) = γ oi (T ) and diam(T ) ≥ 5. Let A be a γ oi -set of T .Since A is not a γ oi g -set of T , so there exists a vertex x ∈ V (T ) − A such that x is adjacent to all vertices of A. Since T is not a star so V (T ) − A − {x} = ∅.Since every vertex of V (T ) − A − {x} is adjacent to some vertices of A and all the vertices of A are adjacent to x, so diam(T ) = 3 or 4. that is contradiction.Theorem 3.3.Let T be a tree.Then γ oi g (T ) = γ(T ) if and only if one of the following conditions holds.

Unicyclic graphs
Volkmann denoted by c(x) the distance from x to cycle C.Proof.In this proof we denote the vertex in γ-set and the vertex in γ oi g -set by bold circle • and empty square respectively in the Figure 2. We show that any unicyclic graph G that satisfies condition (a) or (b) or (c) has equal γ oi g (G) and γ(G), and any unicyclic graph G that does not satisfy in the conditions (a), (b) and (c), γ oi g (G) = γ(G) In Figure 2 we have the pruned of all unicycles G such that R(G) ≤ 2 and G satisfies condition (2) of Theorem 20.Let U be the family of all unicycles and V be the family of all unicycles satisfy in condition (1) or (2) or (3) of theorem 20 and W be the family of unicycles satisfy in condition (a) or (b) or (c).It is well known that If G satisfies condition (c), then according to the γ-sets and γ oi g -sets of unicycles in Figure 1 that presented in Figure (2) we have     5. Unicyclic graphs G with γ oi g (G) = γ g (G) Observation 5.1.Let G be a graph and A be a global dominating set of G and Therefore by Observation 2.2, A is a global dominating set of G p , too.Observation 5.2.Let G be a graph and A be a global dominating set of G p and A ∩ L(G p ) = ∅.Then A is a global dominating set of G.
Proof.Let u be an arbitrary vertex of G.
By Observation 2.2 A is a global dominating set of G p and so γ g (G p ) ≤ γ g (G).
Corollary 5.1.Let G be a graph and G p has a γ g -set like A such that A ∩ L(G p ) = ∅.Then γ g (G p ) = γ g (G).
Proof.By Observation 5.2, A is a global dominating set of G, so γ g (G) ≤ γ g (G p ).By Theorem 5.1 the result holds.
Observation 5.3.Let G be a graph and A be an OIGDS of G and Observation 5.4.Let G be a graph and A be an OIGDS of G p and A ∩ L(G p ) = ∅.Then A is an OIGDS of G.
Proof.Since A is a global dominating set of G p , so by Observation 5.2, A is a global dominating set of G, too.Now we show that E( G − A ) = ∅.On the contrary, let e ∈ E( G − A ) and e = uv.
Corollary 5.2.Let G be a graph and G p has a γ oi g -set like Proof.By Observation 5.4, A is an OIGDS of G, so γ oi g (G) ≤ γ oi g (G p ).By Theorem 5.2 the result holds.       . . . .  . . . .Proof.In this proof we denote the vertex in γ g -set and the vertex in γ oi g -set by bold circle • and empty square respectively in the Figures 5,6,...,21.Let U be the set of all unicyclic graphs and A, B and C are the set of all unicyclic graphs satisfying conditions (i), (ii) and (iii), respectively.It is clear that A, B and C are disjoint sets.For every G ∈ U we will show that then G p is one of the graphs in Figure 5.The pruned of all unicyclic graphs G with R(G) ≤ 2 which are not in Figure 1 are presented in Figure 5.According to the presented γ g -sets and γ oi g -sets in Figure 5, 4,8,9,11,12,14,17,18,19,20,22,24,25,26,27,31,33,34,36,37,38,39,40      If If G has a vertex u ∈ S(G) such that there is only one leaf, x, adjacent to u, then the set (S(G) ∪ {x}) − {u} is a γ g -set of G, so γ oi g (G) = γ g (G) + 1.If every vertex u ∈ S(G) is adjacent to at least two leaves (Figure 8), then the set S(G) ∪ C(G) is a γ g -set of G, so γ oi g (G) = γ g (G).If G p = U 13 (Figure 10), then the set S(G) ∪ {a} is a γ oi g -set of G.If there exists i ∈ {1, 2, ..., k} such that d G (x i ) = 2 and y i is the leaf adjacent to x i , then the set (S(G) ∪ {y 11).If G p = U 15 (Figure 12), then the set 13).If G p = U 16 , then the set S(G) ∪ C(G) is a γ oi g -set of G.If G has a vertex u ∈ S(G) such that there is only one leaf, x, adjacent to u, then the set (S(G) ∪ {x}) − {u} is a γ g -set of G, so γ oi g (G) = γ g (G) + 1.If every vertex u ∈ S(G) is adjacent to at least two leaves (Figure 14), then the set S .
Outer independent global dominating set | D. A. Mojdeh and M. Alishahi graph is a connected graph with exactly one cycle.The open neighborhood of vertex u is denoted by N (u) = {v ∈ V (G) : uv ∈ E(G)} and the closed neighborhood of vertex u is denoted by N [u] = N (u) ∪ {u}.For A ⊆ V (G), the open neighborhood and closed neighborhood of A are defined as N (A) = u∈A N (u) and N [A] = u∈A N [u].Let u ∈ V (G) and A ⊆ V (G), then d(u, A) = min{d(u, v) : v ∈ A}.A set D ⊆ V (G) of a simple graph G is a vertex cover of G if every edge of G has at least one end in D. The covering number β(G) is the minimum cardinality of a vertex cover in G.A set B ⊆ V (G) is an independent set of G if for every edge ab ∈ E(G) , a / ∈ B or b / ∈ B.

Proposition 2 . 1 .
([14], Corollary 2.7) For any tree T , γ(T ) = β(T ) if and only if T * = T − N [L(T )] = ∅ or each component of T * is an isolated vertex or a star, where the center of these stars are not adjacent to a vertex of S(T ).
3 we can consider S(T ) ⊆ A. Let H = {u ∈ V (T ) : d(u, S(T )) = 2}.By Lemma 2.1, H ⊆ A. Since S(T ) ∪ H is an OIDS of T , so A = S(T ) ∪ H. Now let B be an arbitrary γ-set of T .By Observation 2.3 we can consider S(T ) ⊆ B. For every disjoint vertices c 1 , c 2 ∈ H it is clear that c 1 , c 2 / ∈ N [S], d(c 1 , c 2 ) ≥ 4, so corresponding to every vertex c ∈ H there exists a vertex a c ∈ B − S(T ) that dominates c and does't dominate any vertex of H − {c}.So |B| ≥ |S(T ) ∪ H| = |A|, thus γ oi (T ) = γ(T ).

Theorem 4 . 1 .
[14] Let G be a unicyclic graph,G * = G − N [L(G)],and C the only cycle of G. Then β(G) = γ(G) if and only if one of the following conditions holds: (1) G = C 4 .(2) C is adjacent to an end vertex, and the graph G * = ∅ or each component of G * is an isolated vertex or a star, where the centers of these stars are not adjacent to a vertex of N (L(G)).(3) C = C 4 , c(x) ≥ 3 for all x ∈ L(G), min{(d(a), d(b)} = 2 for all pairs of adjacent vertices a, b ∈ V (C), and all components T 1 , ..., T k of the subgraph

Theorem 4 . 2 .
Let G be a unicyclic graph, G * = G − N [L(G)], and C the only cycle of G. Then γ oi g (G) = γ(G) if and only if one of the following conditions holds: a) G satisfies the condition (3) of Theorem 4.1.b) R(G) ≥ 3 and G satisfies the condition (2) of Theorem 4.1.c) G p is one of the following graphs:

Figure 1 .
Figure 1.Pruned graphs of some unicyclic with equal OIGDN and DN.
Proof.If L(G) = ∅, then G p = G and the result holds.Let L(G) = ∅ and A be a γ g -set of G and u ∈ S(G).If |A u | ≥ 3 and x, y, z ∈ A u , then the set (A∪{u})−{y, z} is a global dominating set of G that is contradiction, so |A u | ∈ {0, 1, 2}.If |A u | = 2 and x, y ∈ A u , then the set (A∪{u})−{y} is a global dominating set of G, too, therefore there exists a global dominating set of G like B, such that |B u | = 0 or 1 for every u ∈ S(G).We can construct G p from G such that B ⊆ V (G p ). Therefore for any

Proof.
If L(G) = ∅, then G p = G and the result holds.Let L(G) = ∅ and A be a γ oi g -set of G and u ∈ S(G).If |A u | ≥ 3 and x, y, z ∈ A u , then the set (A ∪ {u}) − {y, z} is an OIGDS of G that is contradiction, so |A u | ∈ {0, 1, 2}.If |A u | = 2 and x, y ∈ A u , then the set (A ∪ {u}) − {y} is an OIGDS of G, too, therefore there exists an OIGDS of G like B, such that |B u | = 0 or 1 for every

e e H 4 Figure 7 .
Figure 7. Graphs G with G p = U 6 .

e e H 5 Figure 8 .Figure 9 .
Figure 8. Graphs G with G p = U 7 and every support vertex has at least two leaves Figure 10

Figure 14 .
Figure 14.Graphs G with G p = U 16 and every support vertex has at least two leaves.

Figure 16 .
Figure 16.Graphs G with G p = U 23 and every support vertex has at least two leaves.

Figure 17 .
Figure17.Graphs G with G p = U 28 and every support vertex has at least two leaves.