2-Outer-Independent Domination in Graphs

We initiate the study of 2-outer-independent domination in graphs. A 2-outer-independent dominating set of a graph G is a set D of vertices of G such that every vertex of V(G)\D\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$V(G) {\setminus} D$$\end{document} has at least two neighbors in D, and the set V(G)\D\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$V(G) {\setminus} D$$\end{document} is independent. The 2-outer-independent domination number of a graph G is the minimum cardinality of a 2-outer-independent dominating set of G. We show that if a graph has minimum degree at least two, then its 2-outer-independent domination number equals the vertex cover number. Then we investigate the 2-outer-independent domination in graphs with minimum degree one.


Introduction
Let G ¼ ðV; EÞ be a graph. The number of vertices of G we denote by n and the number of edges we denote by m, thus jVðGÞj ¼ n and jEðGÞj ¼ m. By the complement of G, denoted by G, we mean a graph which has the same vertices as G, and two vertices of G are adjacent if and only if they are not adjacent in G. By the neighborhood of a vertex v of G we mean the set N G ðvÞ ¼ fu 2 VðGÞ : uv 2 EðGÞg. The degree of a vertex v, denoted by d G ðvÞ, is the cardinality of its neighborhood. By a pendant vertex we mean a vertex of degree one, while a support vertex is a vertex adjacent to a pendant vertex. The set of pendant vertices of a graph G we denote by LðGÞ. We say that a support vertex is strong (weak, respectively) if it is adjacent to at least two pendant vertices (exactly one pendant vertex, respectively). Let dðGÞ (DðGÞ, respectively) mean the minimum (maximum, respectively) degree among all vertices of G. The path (cycle, respectively) on n vertices we denote by P n (C n , respectively). A wheel W n , where n ! 4, is a graph with n vertices, formed by connecting a vertex to all vertices of a cycle C nÀ1 . The distance between two vertices of a graph is the number of edges in a shortest path connecting them. The eccentricity of a vertex is the greatest distance between it and any other vertex. The diameter of a graph G, denoted by diamðGÞ, is the maximum eccentricity among all vertices of G. By K p;q we denote a complete bipartite graph the partite sets of which have cardinalities p and q. By a star we mean the graph K 1;m where m ! 2. Let uv be an edge of a graph G. By subdividing the edge uv we mean removing it, and adding a new vertex, say x, along with two new edges ux and xv. By a subdivided star we mean a graph obtained from a star by subdividing each one of its edges. Generally, let K t 1 ;t 2 ;...;t k denote the complete multipartite graph with vertex set S 1 [ S 2 [ . . . [ S k , where jS i j ¼ t i for positive integers i t. The corona of a graph G on n vertices, denoted by G K 1 , is the graph on 2n vertices obtained from G by adding a vertex of degree one adjacent to each vertex of G. We say that a subset of VðGÞ is independent if there is no edge between any two vertices of this set. The independence number of a graph G, denoted by aðGÞ, is the maximum cardinality of an independent subset of the set of vertices of G. A vertex cover of a graph G is a set D of vertices of G such that for every edge uv of G, either u 2 D or v 2 D. The vertex cover number of a graph G, denoted by bðGÞ, is the minimum cardinality of a vertex cover of G. It is wellknown that aðGÞ þ bðGÞ ¼ jVðGÞj, for any graph G [1]. The clique number of G, denoted by xðGÞ, is the number of vertices of a greatest complete graph which is a subgraph of G. By G Ã we denote the graph obtained from G by removing all pendant and isolated vertices.
A subset D VðGÞ is a dominating set of G if every vertex of VðGÞnD has a neighbor in D, while it is a 2dominating set of G if every vertex of VðGÞnD has at least two neighbors in D. The domination (2-domination, respectively) number of a graph G, denoted by cðGÞ (c 2 ðGÞ, respectively), is the minimum cardinality of a dominating (2-dominating, respectively) set of G. Note that 2-domination is a type of multiple domination in which each vertex, which is not in the dominating set, is dominated at least k times for a fixed positive integer k. Multiple domination was introduced by Fink and Jacobson [2], and further studied for example in [3][4][5][6][7][8][9]. For a comprehensive survey of domination in graphs, see [10].
A subset D VðGÞ is a 2-outer-independent dominating set, abbreviated 2OIDS, of G if every vertex of VðGÞnD has at least two neighbors in D, and the set VðGÞnD is independent. The 2-outer-independent domination number of G, denoted by c oi 2 ðGÞ, is the minimum cardinality of a 2outer-independent dominating set of G. A 2-outer-independent dominating set of G of minimum cardinality is called a c oi 2 ðGÞ-set. The 2-outer-independent domination number of trees was investigated in [11], where it was proved that it is upper bounded by half of the sum of the number of vertices and the number of pendant vertices.
In a distributed network, some vertices act as resource centers, or servers, while other vertices are clients. If a set D of servers is a dominating set, then every client in VðGÞnD has direct (one hop) access to at least one server. 2-dominating sets represent a higher level of service, since every client has guaranteed access to at least two servers. The outer-independence condition means that the clients are not able to connect with each other directly. This may be useful for example for security, when we allow clients to communicate with each other only through servers.
We initiate the study of 2-outer-independent domination in graphs. We show that if a graph has minimum degree at least two, then its 2-outer-independent domination number equals the vertex cover number. Then we investigate the 2-outer-independent domination in graphs with minimum degree one. We find the 2-outer-independent domination numbers for several classes of graphs. Next we prove some lower and upper bounds on the 2-outer-independent domination number of a graph, and we characterize the extremal graphs. Then we study the influence of removing or adding vertices and edges. We also give Nordhaus-Gaddum type inequalities [12].

Preliminary Results
If G is a disconnected graph with connected components G 1 ; G 2 ; . . .; G k , then we can easily see that c oi We have the following inequalities.
Proposition 1 Let G be a graph. Then: Proof (i) Any 2-outer-independent dominating set of a graph is a 2-dominating set of this graph, and thus c 2 ðGÞ c oi 2 ðGÞ. (ii) Let D be a c oi 2 ðGÞ-set, and let A be a maximum clique in G. Since VðGÞnD is independent, we have jðVðGÞnDÞ \ Aj 1. This implies that jDj ! jAj À 1. We now get c oi 2 ðGÞ ¼ jDj ! jAj À 1 ¼ xðGÞ À 1. (iii) Note that the definition of 2-outer-independent domination implies that every 2OIDS of a graph is a vertex cover of this graph, and thus the result follows. h Note that the bounds of the above proposition are tight. It is easy to see that for every integer n ! 3 we have c oi 2 ðK n Þ ¼ c 2 ðK n Þ þ n À 3, for every integer m ! 2 we have c oi 2 ðK 1;m Þ ¼ xðK 1;m Þ þ m À 2 and c oi 2 ðK 1;m Þ ¼ bðK 1;m Þ þm À 1, while c oi 2 ðK 3 Þ ¼ 2 ¼ bðK 3 Þ. We next prove that if a graph has no pendant or isolated vertices, then its 2-outer-independent domination number and vertex cover number are equal.
Proof Let D be a minimum vertex cover of G, and let x 2 VðGÞnD. Clearly, N G ðxÞ D. Since dðGÞ ! 2, the vertex x is adjacent to at least two vertices of D. There are no edges between any two vertices of VðGÞnD, thus the set VðGÞnD is independent. This implies that D is a 2OIDS of the graph G. Consequently, c oi 2 ðGÞ bðGÞ. On the other hand, by Proposition 1 we have c oi 2 ðGÞ ! bðGÞ. Thus c oi 2 ðGÞ ¼ bðGÞ. h Corollary 3 Let G be a graph. If c oi 2 ðGÞ 6 ¼ bðGÞ, then dðGÞ 2 f0; 1g.
Henceforth, we study only connected graphs G with dðGÞ ¼ 1, that is, connected graphs having at least one pendant vertex. Since a pendant vertex has only one neighbor in the graph, it cannot have two neighbors in the dominating set. Thus we have the following property of pendant vertices.
Observation 4 Every pendant vertex of a graph G belongs to every c oi 2 ðGÞ-set.

Connected Graphs with Minimum Degree One
Throughout this section we consider only connected graphs with minimum degree one. We have the following relation between the 2-outer-independent domination number of a graph and the independence number of the graph obtained from it by removing all pendant vertices.
Lemma 5 For every graph G with n vertices we have c oi 2 ðGÞ ¼ n À aðG Ã Þ.
Proof Let D be any c oi 2 ðGÞ-set. By Observation 4, all pendant vertices belong to the set D. Therefore VðGÞnD VðG Ã Þ. The set VðGÞnD is independent, thus aðG Ã Þ ! jVðGÞnDj ¼ n À c oi 2 ðGÞ. Now let D Ã be any aðG Ã Þset. Let us observe that in the graph G every vertex of D Ã has at least two neighbors in the set VðGÞnD Ã . Thus VðGÞnD Ã is a 2OIDS of G. We now get c oi 2 ðGÞ jVðGÞnD Ã j ¼ n À aðG Ã Þ. This implies that c oi 2 ðGÞ ¼ n À aðG Ã Þ. h It is obvious that for every graph G we have 2 c oi 2 ðGÞ n. We now characterize the graphs attaining these bounds.
Proposition 6 Let G be a graph. We have: If all vertices of G belong to the set D, then the graph G has two vertices. Consequently, G ¼ P 2 . Now let x be a vertex of VðGÞnD. The vertex x has to be dominated twice, thus d G ðxÞ ! 2. Since the set VðGÞnD is independent, the vertex x cannot have more than two neighbors in G. This implies that G is a path P 3 as no other vertices can be dominated twice. Now assume that for some graph G we have c oi 2 ðGÞ ¼ n. If G has at least three vertices, then it has a vertex, say x, of degree at least two. Let us observe that Dnfxg is a 2OIDS of the graph G. This implies that c oi 2 ðGÞ n À 1. Therefore the graph G has exactly two vertices, and consequently, it is a path P 2 . h

Corollary 7
For every graph G with at least three vertices we have c oi 2 ðGÞ n À 1. We now consider graphs G such that 3 c oi 2 ðGÞ n À 1. Theorem 8 Let G be a graph of order n ! 3, and let k be an integer such that 3 k n À 1. We have c oi 2 ðGÞ ¼ k if and only if G can be obtained from a connected graph H of order k with jLðHÞj n À k and aðHÞ ¼ n À k, by attaching n À k vertices to H in a way such that every pendant vertex of H is a support vertex of G.
Proof Assume that c oi 2 ðGÞ ¼ k. Lemma 5 implies that aðG Ã Þ ¼ n À k. Clearly, every vertex of VðGÞnVðG Ã Þ is a pendant vertex in G. Let us also observe that every pendant vertex of G Ã is a support vertex of G. Thus jLðG Ã Þj n À jVðG Ã Þj. Now assume that G is a graph obtained from a connected graph H of order k with jLðHÞj n À k and aðHÞ ¼ n À k, by attaching n À k vertices to H in a way such that every pendant vertex of H is a support vertex of G. Let us observe that G Ã ¼ H. Let D be a maximum independent set of H. Clearly, VðGÞnD is a 2OIDS of G, and therefore c oi 2 ðGÞ n À aðHÞ ¼ k. Suppose that c oi 2 ðGÞ\k. Using Lemma 5 we obtain aðHÞ [ n À k, a contradiction. Thus c oi 2 ðGÞ ¼ k. h

Bounds
We have the following upper bound on the 2-outer-independent domination number of a graph in terms of its vertex cover number and the number of pendant vertices.

Proposition 9
If G is a graph with l pendant vertices, then c oi 2 ðGÞ bðGÞ þ l. Proof Let us observe that vertices of any minimum vertex cover of G together with all pendant vertices of G form a 2OIDS of the graph G. h Let us observe that the bound from the previous proposition is tight. Let l be a positive integer, and let H ¼ C 6 . Let x be a vertex of H, and let G be a graph obtained from H by attaching l new vertices and joining them to the vertex x. It is straightforward to see that bðGÞ ¼ 3, while c oi 2 ðGÞ ¼ 3 þ l. We have the following upper bound on the 2-outer-independent domination number of a graph in terms of its vertex cover number and maximum degree.

Proposition 10
For every graph G we have c oi 2 ðGÞ bðGÞDðGÞ.
Proof Let S be a minimum vertex cover of G. The vertices of S together with all pendant vertices of G form a 2OIDS of the graph G. Every vertex of S is adjacent to at most DðGÞ pendant vertices. Thus c oi 2 ðGÞ bðGÞDðGÞ. h

2-Outer-Independent Domination in Graphs 265
Let us observe that the bound from the previous proposition is tight. For stars K 1;m we have We have the following upper bound on the 2-outer-independent domination number of a graph.

Proposition 11
For every graph G with l pendant vertices we have nDðGÞ þ l DðGÞ þ 1 : Proof By Lemma 5 we have c oi 2 ðGÞ ¼ n À aðG Ã Þ. Since every maximal independent set of a graph is a dominating set of this graph, we have cðG Ã Þ aðG Ã Þ. We now get We have the following upper bound on the 2-outer-independent domination number of a graph in terms of its diameter.

Proposition 12
If G is a graph of diameter d, then c oi 2 ðGÞ n À bd=2c.
We have the following upper bound on the 2-outer-independent domination number of a tree in terms of its independence number and the number of support vertices.
Theorem 13 For every tree T of order at least three with s support vertices we have c oi 2 ðTÞ aðTÞ þ s À 1. Proof Let n mean the number of vertices of the tree T. We proceed by induction on this number. If diamðTÞ ¼ 1, sðK 1;m Þ À 1. Now let us assume that diamðTÞ ¼ 3. Thus T is a double star. We have c oi 2 ðTÞ ¼ n À 1 ¼ n À 2 þ 2 À 1 ¼ aðTÞ þ sðTÞ À1.
Now assume that diamðTÞ ! 4. Thus the order n of the tree T is at least five. We obtain the result by the induction on the number n. Assume that the theorem is true for every tree T 0 of order n 0 \n.
First assume that some support vertex of T, say n, is strong. Let y be a pendant vertex adjacent to x. Let T 0 ¼ T À y. We have s 0 ¼ s. Let D 0 be any c oi 2 ðT 0 Þ-set. Obviously, D 0 [ fyg is a 2OIDS of the tree T. Thus c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1. Let us observe that there exists a maximum independent set of T 0 that contains the vertex x. Let A 0 be such a set. It is easy to see that D 0 [ fyg is an independent set of the tree T. Thus aðTÞ ! aðT 0 Þ þ 1. We now get c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1 aðT 0 Þ þ s 0 ¼ aðT 0 Þ þ s a ðTÞ þ s À 1. Henceforth, we can assume that all support vertices of T are weak.
We now root T at a vertex r of maximum eccentricity diamðTÞ. Let t be a pendant vertex at maximum distance from r, v be the parent of t, u be the parent of v, and w be the parent of u in the rooted tree. By T x let us denote the subtree induced by a vertex x and its descendants in the rooted tree T.
Assume that among the children of u there is a support vertex, say x, different from v. Let T 0 ¼ T À T v . We have s 0 ¼ s À 1. Let us observe that there exists a c oi 2 ðT 0 Þ-set that contains the vertex u. Let D 0 be such a set. It is easy to observe that D 0 [ ftg is a 2OIDS of the tree T. Thus c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1. Now let A 0 be a maximum independent set of T 0 . It is easy to observe that D 0 [ ftg is an independent set of T. Thus aðTÞ ! aðT 0 Þ þ 1. We now get c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1 aðT 0 Þ þ s 0 ¼ aðT 0 Þ þ s aðTÞ þ s À 1 . Now assume that u is adjacent to a pendant vertex, say x. It suffices to consider only the possibility when d T ðuÞ ¼ 3.
Obviously, aðTÞ ! a ðT 0 Þ. Let D 0 be any c oi 2 ðT 0 Þ-set. Obviously, D 0 [ fxg is a 2OIDS of the tree T. Thus c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1. We now get c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1 aðT 0 Þ þ s 0 ¼ aðT 0 Þ þ s À 1 aðTÞþ s À 1. Now assume that d T ðuÞ ¼ 2. Let T 0 ¼ T À T v . We have s 0 s. Let D 0 be any c oi 2 ðT 0 Þ-set. By Observation 4 we have u 2 D 0 . It is easy to observe that D 0 [ ftg is a 2OIDS of the tree T. Thus c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1. Now let A 0 be a maximum independent set of T 0 . It is easy to see that D 0 [ ftg is an independent set of the tree T. Thus aðTÞ ! aðT 0 Þ þ 1. We now get c oi 2 ðTÞ c oi 2 ðT 0 Þ þ 1 aðT 0 Þ þ s 0 aðT 0 Þ þ s aðTÞ þ s À 1. h We have the following bounds on the 2-outer-independent domination number of a graph in terms of its order and size.
Proposition 14 For every graph G we have 2n þ 1 À ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ð2n À 1Þ 2 À 8ðm À 1Þ q 2 c oi 2 ðGÞ 2n þ 1 þ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ð2n À 1Þ 2 À 8ðm À 1Þ q 2 : Proof Let D be a c oi 2 ðGÞ-set. Let t denote the number of edges between the vertices of D and the vertices of VðGÞnD. Obviously, m t þ jEðG½DÞj. Since G has at least one pendant vertex, we have t ðjDj À 1ÞÁ jVðGÞnDj þ 1. Notice that jEðG½DÞj ðjDj À 1ÞðjDj À2Þ=2. Now simple calculations imply the result. h We also have the following lower bound on the 2-outerindependent domination number of a graph in terms of its order and size.

Proposition 15
For every graph G we have c oi 2 ðGÞ ! n À m=2. Proof Let D be a c oi 2 ðGÞ-set. Since every vertex of VðGÞnD has at least two neighbors in D, have m ! 2jVðGÞnDj. h Let us observe that the bound from the previous proposition is tight. For positive integers n we have We have the following necessary condition for that a graph attains the bound from the previous proposition.

Proposition 16
If for a graph G we have c oi 2 ðGÞ ¼ n À m=2, then the graph G is bipartite and it has at least m=2 vertices of degree two.
Proof Let D be a c oi 2 ðGÞ-set. Let t denote the number of edges between the vertices of D and the vertices of VðGÞnD. If some vertex of VðGÞnD has degree at least three, then we get m ! t ! 3 þ 2ðjVðGÞnDj À 1Þ ¼ 2jVðGÞnDj þ 1 ¼ 2ðn À c oi 2 ðGÞÞ þ 1 ¼ m þ 1 [ m, a contradiction. Thus every vertex of VðGÞnD has degree two. We have jVðGÞnDj ¼ n À c oi 2 ðGÞ ¼ m=2. Thus there are at least m=2 vertices of degree two. If the set D is not independent, then we get m [ t ¼ 2jVðGÞnDj ¼ 2ðn Àc oi 2 ðGÞÞ ¼ m, a contradiction. Therefore D is an independent set. Since the set VðGÞnD is also independent, the graph G is bipartite. h It is an open problem to characterize the graphs attaining the bound from Proposition 16.

Problem 17
Characterize graphs G such that c oi 2 ðGÞ ¼ n À m=2. We now study the influence of the removal of a vertex of a graph on its 2-outer-independent domination number.
Proposition 18 Let G be a graph. For every vertex v of G we have c oi 2 ðGÞ À 1 c oi 2 ðG À vÞ c oi 2 ðGÞ þ d G ðvÞ À 1. Proof Let D be a c oi 2 ðGÞ-set. If v 6 2 D, then observe that D is a 2OIDS of the graph G À v. Now assume that v 2 D. Let us observe that D [ N G ðvÞnfvg is a 2OIDS of the graph G À v. Therefore c oi 2 ðG À vÞ jD [ N G ðvÞn fvgj jDnfvgj þjN G ðvÞj ¼ c oi 2 ðGÞ þ d G ðvÞ À 1. Now let D 0 be any c oi 2 ðG À vÞ-set. It is easy to see that D 0 [ fvg is a 2OIDS of the graph G. Thus c oi 2 ðGÞ c oi 2 ðG À vÞ þ 1. h Let us observe that the bounds from the previous proposition are tight. For the lower bound, let G ¼ K n , where n ! 4. We have c oi For the upper bound, let G be subdivided star. The vertex of minimum eccentricity we denote by v. Let m denote its degree. We have þ d G ðvÞ À 1. We now study the influence of the removal of an edge of a graph on its 2-outer-independent domination number.
Proposition 19 Let G be a graph. For every edge e of G we have c oi 2 ðG À eÞ 2 fc oi 2 ðGÞ À 1; c oi 2 ðGÞ; c oi 2 ðGÞ þ 1g: Proof Let D be a c oi 2 ðGÞ-set, and let e ¼ xy be an edge of G. Since the set VðGÞnD is independent, some of the vertices x and y belongs to the set D. Without loss of generality we may assume that x 2 D. If y 2 D, then it is easy to see that D is a 2OIDS of the graph G À e. If y 6 2 D, then D [ fyg is a 2OIDS of G À e. Thus c oi 2 ðG À eÞ c oi 2 ðGÞ þ 1. Now let D 0 be a c oi 2 ðG À eÞ-set. If some of the vertices x and y belongs to the set D 0 , then D 0 is a 2OIDS of the graph G. If none of the vertices x and y belongs to the set D 0 , then it is easy to observe that D 0 [ fxg is a 2OIDS of the graph G. Therefore c oi 2 ðGÞ c oi 2 ðG À eÞ þ 1. h Let us observe that the bounds from the previous proposition are tight. For the lower bound, let xy be an edge of the complete graph K 4 . Let G be a graph obtained from K 4 by adding two vertices x 1 ; y 1 , and joining x to x 1 , and y to y 1 . Then c oi 2 ðG À xyÞ ¼ c oi 2 ðGÞ À 1. For the upper bound, consider a path P 4 , and the central edge of it.
Similarly, we have the following result, which immediately follows from Proposition 19, concerning the influence of adding an edge on the 2-outer-independent domination number of a graph.

Nordhaus-Gaddum Type Inequalities
A Nordhaus-Gaddum type result is a lower or upper bound on the sum or product of a parameter of a graph and its 2-Outer-Independent Domination in Graphs 267 complement. In 1956 Nordhaus and Gaddum [12] proved the following inequalities for the chromatic number of a graph G and its complement: 2 ffiffi ffi n p vðGÞ þ vðGÞ n þ 1 and n vðGÞvðGÞ ðn þ 1Þ 2 =4.
We now give Nordhaus-Gaddum type inequalities for the sum of the 2-outer-independent domination number of a graph and its complement.

Theorem 21
For every graph G we have n À 1 c oi 2 ðGÞ þ c oi 2 ðGÞ 2n. Proof Let D be a c oi 2 ðGÞ-set. Since VðGÞnD is an independent set, the vertices of VðGÞnD form a clique in G. Let D be any c oi 2 ðGÞ-set. Let us observe that at most one vertex of VðGÞnD does not belong to D. Therefore jDj ! jVðGÞnDj À 1. We now get c oi 2 ðGÞþ c oi 2 ðGÞ ¼ jDj þjDj ! jDj þ jVðGÞnDj À 1 ¼ n À 1.
Obviously, c oi 2 ðGÞ n and c oi 2 ðGÞ n. Thus c oi 2 ðGÞ þ c oi 2 ðGÞ 2n. h We now prove that the complete graphs of order at most two, and their complements are the only graphs which attain the upper bound from Theorem 21.

Theorem 22
Let G be a graph. We have c oi . Now assume that for some graph G we have c oi 2 ðGÞ þ c oi 2 ðGÞ ¼ 2n. This implies that c oi 2 ðGÞ ¼ n and c oi 2 ðGÞ ¼ n. By Corollary 7, n 2. Consequently, Corollary 23 If G and G are different from K 1 and K 2 , then c oi 2 ðGÞ þ c oi 2 ðGÞ 2n À 1. We now prove that the path P 3 and its complement are the only graphs which attain the bound from the previous corollary.

Theorem 24
Let G be a graph. We have c oi 2 ðGÞ þ c oi 2 ðGÞ ¼ 2n À 1 if and only if G or G is a path P 3 .
Proof We have c oi 2 ðP 3 Þ þ c oi 2 ðP 3 Þ ¼ 5 ¼ 2n À 1. Now assume that for some graph G we have c oi 2 ðGÞ þ c oi 2 ðGÞ ¼ 2n À 1. This implies that c oi 2 ðGÞ ¼ n À 1 or c oi 2 ðGÞ ¼ n À 1. Without loss of generality we assume that c oi 2 ðGÞ ¼ n À 1. By Theorem 8, the graph G is obtained from a complete graph K r , for some r ! 1, by attaching at least one pendant vertex. We show that n ¼ 3. Suppose that n ! 4. Since dðGÞ ¼ 1, we may assume that x is a pendant vertex of G. Thus x has at least two neighbors in the graph G. Therefore VðGÞnfxg is a 2OIDS of G, and consequently, c oi 2 ðGÞ n À 1. We now get c oi 2 ðGÞ þ c oi 2 ðGÞ 2n À 2, a contradiction. We deduce that n ¼ 3. Consequently, G ¼ P 3 . h We next improve the lower bound from Theorem 21.
Theorem 25 For every graph G with l pendant vertices we have c oi 2 ðGÞ þ c oi 2 ðGÞ ! n þ l À 2. Proof By Theorem 8, the graph G is obtained from a connected graph H with aðHÞ ¼ n À c oi 2 ðGÞ, by attaching n À jVðHÞj pendant vertices to H such that any pendant vertex of H is a support vertex of G. Let X ¼ VðGÞnVðHÞ. By Lemma 5 we have c oi 2 ðGÞ ¼ n À aðHÞ. Let S be a maximum independent set in H. Then clearly VðGÞnS is a c oi 2 ðGÞ-set. Let D be a c oi 2 ðGÞ-set. Clearly, G½X and G½S are complete graphs. Thus jD \ Sj ! jSj À 1, and jD \ Xj ! jXj À 1. We now get c oi 2 ðGÞ þ c oi 2 ðGÞ ! jVðGÞj À jSj þ jSj À 1 þ jXj À 1 ¼ n þ jXj À 2 ¼ n þ l À 2: h We now characterize graphs attaining the lower bound from Theorem 21, that is, graphs G for which c oi 2 ðGÞ þ c oi 2 ðGÞ ¼ n À 1. Since c oi 2 ðGÞ ! 2, we may assume that c oi 2 ðGÞ\n À 2. Theorem 26 Let G be a graph such that c oi 2 ðGÞ\n À 2. Then c oi 2 ðGÞ þ c oi 2 ðGÞ ¼ n À 1 if and only if G is obtained from a connected graph H such that aðHÞ ¼ n À c oi 2 ðGÞ and jLðHÞj 1, by attaching one pendant vertex to H such that if H has a pendant vertex x, then x is a support vertex in G.
Proof Assume that for some graph G we have c oi 2 ðGÞ þ c oi 2 ðGÞ ¼ n À 1. By Theorem 8, the graph G is obtained from a connected graph H with aðHÞ ¼ n À c oi 2 ðGÞ, by attaching n À jVðHÞj pendant vertices to H such that any pendant vertex of H is a support vertex of G. Let jVðGÞnVðHÞj ¼ l. By Theorem 25 we have n À 1 ¼ c oi 2 ðGÞ þ c oi 2 ðGÞ ! n þ l À 2. This implies that l 1, and so l ¼ 1. Now the result follows.
Conversely, let G be obtained from a connected graph H with aðHÞ ¼ n À c oi 2 ðGÞ and jLðHÞj 1, by attaching one pendant vertex (say u) to H such that if H has a pendant vertex x, then x is a support vertex in G. By Theorem 8 we have c oi 2 ðGÞ ¼ n À aðHÞ. Let S be a maximum independent set in H. Since c oi 2 ðGÞ\n À 2, we find that jSj ! 3. Let x; y 2 S. Then ðS À fx; ygÞ [ fug is a 2OIDS for G, and thus c oi 2 ðGÞ þ c oi 2 ðGÞ n À jSj þ jSj À 2 þ 1 ¼ n À 1. By Theorem 25, c oi 2 ðGÞ þ c oi 2 ðGÞ ! n þ l À 2 ¼ n À 1, and thus the result follows. h Similarly we obtain the following result.
Theorem 27 Let k n À 1 be a non-negative integer. If G is a graph of order n, then c oi 2 ðGÞ þ c oi 2 ðGÞ ¼ n þ k if and only if G is obtained from a connected graph H such that aðHÞ ¼ n À c oi 2 ðGÞ and jLðHÞj t, by attaching t pendant vertices to H, where t k þ 2, in a way such that if H has a pendant vertex x, then x is a support vertex in G.