IN GRAPHS. Hamideh Aram. Department of Mathematics Gareziaeddin Center, Khoy Branch Islamic Azad University, Khoy, Iran

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1 Discussiones Mathematicae Graph Theory 38 (018) doi: /dmgt.054 ETERNAL m-security BONDAGE NUMBERS IN GRAPHS Hamideh Aram Department of Mathematics Gareziaeddin Center, Khoy Branch Islamic Azad University, Khoy, Iran Maryam Atapour Department of Mathematics Faculty of Basic Sciences University of Bonab, Bonab, I.R. Iran and Seyed Mahmoud Sheikholeslami Department of Mathematics Azarbaijan Shahid Madani University Tabriz, I.R. Iran Abstract An eternal m-secure set of a graph G = (V,E) is a set S 0 V that can defend against any sequence of single-vertex attacks by means of multiple guard shifts along the edges of G. The eternal m-security number σ m (G) is the minimum cardinality of an eternal m-secure set in G. The eternal m- security bondage number b σm (G) of a graph G is the minimum cardinality of a set of edges of G whose removal from G increases the eternal m-security number of G. In this paper, we study properties of the eternal m-security bondage number. In particular, we present some upper bounds on the eternal m-security bondage number in terms of eternal m-security number and edge connectivity number, and we show that the eternal m-security bondage number of trees is at most and we classify all trees attaining this bound. Keywords: eternal m-secure set, eternal m-security number, eternal m- security bondage number. 010 Mathematics Subject Classification: 05C69.

2 99 H. Aram, M. Atapour and S.M. Sheikholeslami 1. Introduction Throughout this paper, G is a simple connected graph with vertex set V = V(G) and edge set E = E(G) and of order n and size m. For every vertex v V, the open neighborhood of v is the set N(v) = {u V(G) : uv E(G)} and the closed neighborhood of v is the set N[v] = N(v) {v}. The degree deg(v) of v is the number of edges incident with v or, equivalently, deg(v) = N(v). The degree sequence of G is (deg(v 1 ),deg(v ),...,deg(v n )), typically written in nondecreasing order. The minimum and maximum degree of vertices in V(G) are denoted by δ(g) and (G), respectively. Let E(A,B) denote the set of all edges with one endpoint in A and the other endpoint in B, e(a,b) be the cardinality of E(A,B), and E u denote the set of edges incident to u. A leaf of a graph G is a vertex of degree 1 and a support vertex of G is a vertex adjacent to a leaf. A support vertex is called strong support vertex if it is adjacent to at least two leaves. The distance between two vertices x and y is denoted by d(x,y) and the diameter of G is denoted by diam(g). A set S of vertices in a graph G is called a dominating set if every vertex in V iseitheranelementofs orisadjacenttoanelementofs. Thedomination number of G, denoted by γ(g), is the minimum cardinality of a dominating set of G. A γ(g)-set is a dominating set of G of size γ(g). For a more thorough treatment of domination parameters and for terminology not presented here see [5,11]. The bondage number b(g) of a graph G is the minimum cardinality of a set of edges of G whose removal from G increases the domination number of G. The bondage number was introduced by Fink et al. [] and was studied by several authors, for example [4,6,8 10]. For more information on this topic we refer the reader to the survey article by Xu [1]. An eternal 1-secure set of a graph G is a set S 0 V that can defend against any sequence of single-vertex attacks by means of single-guard shifts along the edges of G. That is, for any k and any sequence v 1,v,...,v k of vertices, there exists a sequence of guards u 1,u,...,u k with u i S i 1 and either u i = v i or u i v i E, suchthateachsets i = (S i 1 {u i }) {v i }isadominatingset. Itfollows that each S i can be chosen to be an eternal 1-secure set. The eternal 1-security number of G, denoted by σ 1 (G), is the minimum cardinality of an eternal 1-secure set. The eternal 1-security number was introduced by Burger et al. [1] using the notation γ. In order to reduce the number of guards needed in an eternal secure set, Goddard et al. [3] considered allowing more guards to move. Suppose that in responding to each attack, every guard may shift along an incident edge. The eternal m-security number σ m (G) is the minimum number of guards to handle an arbitrary sequence of single attacks using multiple guard shifts. A suitable placement of the guards is called an eternal m-secure set (EmSS). An EmSS of size σ m (G) is called a σ m (G)-set.

3 Eternal m-security Bondage Numbers in Graphs 993 The eternal m-security bondage number b σm (G) of a graph G is the minimum cardinality of a set of edges of G whose removal from G increases the eternal m- security number of G. Since in the study of eternal m-security bondage number the assumption σ m (G) < n is necessary, we always assume that when we discuss b σm (G), all graphs involved satisfy σ m (G) < n, i.e., all graphs are nonempty. An edge set B with σ m (G B) > σ m (G) is called the eternal m-secure bondage set. A b σm (G)-set is an eternal m-secure bondage set of G of size b σm (G). In this paper, we initiate the study of the eternal m-security bondage number in graphs and we establish some bounds on the eternal m-security bondage number in terms of vertex degree, eternal m-security number and edge connectivity number. We also show that the eternal m-security bondage number of trees is at most and we characterize all trees attaining this bound.. Preliminaries and Exact Values The proof of the following four results can be found in [3]. Proposition A. For any graph G, γ(g) σ m (G). A set P V(G) is called a k-packing if d(u,v) > k for each pair of vertices u,v P, u v. The k-packing number α k (G) is the cardinality of a maximum k-packing in G. Note that α 1 (G) = α(g) is the independence number of G. Proposition B. For any graph G, σ m (G) α(g). Proposition C. 1. σ m (K n ) = 1.. σ m (K r,s ) = for r,s 1,r +s σ m (P n ) = n. 4. σ m (C n ) = n 3. Proposition D. For any graph G, σ m (G) (diam(g)+1)/. Next results are immediate consequences of Propositions C and D. Corollary 1. For any graph G, σ m (G) = 1 if and only if G K n. Corollary. For n, we have b σm (K n ) = 1. Corollary 3. For n 5, b σm (C n ) = 1. { 1 if n is even, Corollary 4. For n 3, b σm (P n ) = if n is odd. Proposition E [7]. For any graph G, α (G) γ(g).

4 994 H. Aram, M. Atapour and S.M. Sheikholeslami Corollary 5. For any graph G, α (G) σ m (G). Observation 6. Let G be a graph and H be a spanning subgraph of G such that σ m (H) = σ m (G). If K = E(G)\E(H), then b σm (H) b σm (G) b σm (H)+ K. Proof. Let F be a b σm (H)-set. Then σ m (G) = σ m (H) < σ m (H F) = σ m (G (K F)), which implies that b σm (G) K F = K + F = b σm (H)+ K. Let now T be a b σm (G)-set. Then we have σ m (H) = σ m (G) < σ m (G T) σ m (H T). Thus b σm (H) T = b σm (G) and the proof is complete. Proposition 7. If G contains a vertex adjacent to at least three leaves, then b σm (G) = 1. Proof. Let u be adjacent to the leaves u 1,u,u 3. Consider the graph G obtained from G by deleting the edge uu 1. Let S be a σ m (G )-set which contains u (we may assume that S is a response to an attack on u). Obviously u 1 S and S \{u 1 } is an EmSS of G and so σ m (G) σ m (G ) 1. Hence, b σm (G) = 1. Next we determine the eternal m-security bondage number of complete bipartite graphs. Proposition 8. For m n, b σm (K m,n ) =. Proof. By Proposition C, σ m (K m,n ) =. If m = n =, then clearly b σm (K, ) =. Assume that m 3. It is not hard to see that for any edge e = uv E(G), the set S = {u,v} is an EmSS of K m,n e and so b σm (K m,n ). Now we show that b σm (K m,n ). Suppose that X = {u 1,...,u m } and Y = {v 1,...,v n } be the partite sets of K m,n and let F = {v 1 u 1,v 1 u }. Let S be a σ m (K m,n F)-set which contains u 1. To dominate v 1, we have v 1 S or u i S for some i 3. If v 1 S, then u is not dominated by {u 1,v 1 } and so S 3. Let v 1 / S. Assume without loss of generality that u 3 S. Then u is not dominated by {u 1,u 3 } and this implies that S 3. Hence, b σm (K m,n ) and the proof is complete. 3. Bounds on the Eternal m-security Bondage Number In this section, we present various bounds on the eternal m-security bondage number. We start with an observation. Observation 9. Let G be a connected graph. If σ m (G v) σ m (G) for some vertex v V(G), then b σm (G) deg(v).

5 Eternal m-security Bondage Numbers in Graphs 995 Proof. First, note that σ m (G E v ) σ m (G). If σ m (G E v ) > σ m (G), then we are done. Suppose σ m (G E v ) = σ m (G) and let S be a σ m (G E v )-set. Clearly, v S and S \{v} is an EmSS of G v. It follows that and the proof is complete. σ m (G E v ) 1 σ m (G v) σ m (G), Theorem 10. Let G be a connected graph and uv E(G). Then b σm (G) deg(u)+deg(v) 1 N(u) N(v). Proof. Let X be the set consisting of all edges incident with u and v with exception of the edges E(v,N(u)). Then X = deg(u) + deg(v) 1 N(u) N(v), u is an isolated vertex in G X and v is only adjacent to the vertices of N G (u) N G (v). Let S be a σ m (G X)-set which contains v (we may assume a response to an attack on v). It is easy to verify that S \ {u} is an EmSS of G and hence σ m (G) σ m (G X) 1. This completes the proof. Corollary 11. For any nonempty graph G, b σm (G) δ(g)+ (G) 1. Theorem 1. Let G be a connected graph with degree sequence (d 1,d,...,d n ). Then b σm (G) d α +d α+1 1, where α is the independence number of G. Proof. Let V(G) = {v 1,v,...,v n } and let deg(v i ) = d i for each i. Since the set {v 1,...,v α+1 }isnotindependent, thereisanedgev i v j forsome1 i < j α+1. It follows from Theorem 10 that b σm (G) deg(v i )+deg(v j ) 1 deg(v α )+deg(v α+1 ) 1, and the proof is complete. Theorem 13. Let G be a connected graph and u,v be two vertices of G with d(u,v) =. Then b σm (G) deg(u)+deg(v). Proof. Let w be a common neighbor of u and v and let X be the set consisting of all edges incident with u and v. Then X = deg(u) + deg(v) and u,v are isolated vertices in G X. Let S be a σ m (G X)-set which contains w (we may assume a response to an attack on w). Obviously u,v S and we can easily check that S \{u} is an EmSS of G and so σ m (G) < σ m (G X ). Thus b σm (G) X = deg(u)+deg(v) as desired.

6 996 H. Aram, M. Atapour and S.M. Sheikholeslami Corollary 14. Let G be a connected graph of order n with degree sequence (deg(v 1 ),deg(v ),...,deg(v n )). Then b σm (G) deg(v α )+deg(v α +1). Proof. Clearly, the set {v 1,...,v α +1} is not a -packing. Hence, d(v i,v j ) for some 1 i j α +1 and the result follows by Theorems 10 and 13. Next result is an immediate consequence of Corollaries 5 and 14. Corollary 15. If G is a connected graph with degree sequence (deg(v 1 ),deg(v ),...,deg(v n )), then b σm (G) deg(v σm )+d(v σm+1). Theorem 16. For any connected graph G, b σm (G) (σ m (G) α (G)+1) (G). Proof. By Corollary 5, α (G) σ m (G). Let s = σ m (G) α (G) + 1 and U = {u 1,...,u α } be a maximum -packing in G. Clearly, U V(G). Let T be asubsetofv(g) U ofsizesandletg bethegraphobtainedfromgbyremoving all edges incident to the vertices in T. Obviously, E(G) E(G ) s. Now we have σ m (G ) α (G ) α (G)+s = α (G)+σ m (G) α (G)+1 = σ m (G)+1 > σ m (G) and the proof is complete. The next result is an immediate consequence of Theorem 16. Corollary 17. If σ m (G) = α (G), then b σm (G) (G). The edge connectivity number κ (G) of a connected graph G is the minimum number of edges that have to be removed out of G to decompose G in two components. The inequality κ (G) δ(g) is immediate. Next result is an improvement of Corollary 11. Theorem 18. If G is a nontrivial connected graph, then b σm (G) (G)+κ (G) 1. Proof. Let K be a set of edges such that κ (G) = K and G K is disconnected. Assume that G 1 and G are the components of G K. It is easy to see that σ m (G) σ m (G 1 ) + σ m (G ) = σ m (G K). If σ m (G) < σ m (G 1 ) + σ m (G ), then b σm (G) κ (G) and we are done. Let σ m (G) = σ m (G 1 ) + σ m (G ). We claim that there is a vertex v V(G i ) such that v is incident to an edge of K and σ m (G i E v ) > σ m (G i ) for some i. In this case we have σ m (G K E v ) > σ m (G), which implies that b σm (G) deg(v)+κ (G) 1 (G)+κ (G) 1.

7 Eternal m-security Bondage Numbers in Graphs 997 Assume, to the contrary, that σ m (G 1 E v ) = σ m (G 1 ) for each vertex v V(G 1 ) incident to an edge of K and σ m (G E v ) = σ m (G ) for every vertex v V(G ) incident to an edge of K. Let u 1 u K where u i V(G i ) for i = 1,. Let S i be a σ m (G i E ui )-set for i = 1,. Clearly, u 1 S 1 and u S. It is easy to verify that S = S 1 S \{u 1 } is an eternal m-secure set of G which implies that σ m (G) σ m (G 1 E u )+σ m (G E v ) 1 = σ m (G 1 )+σ m (G ) 1 = σ m (G) 1, a contradiction. This completes the proof. Proposition 19. If σ m (G) =, then b σm (G) δ(g)+1. Proof. Let u V(G) be a vertex of minimum degree. If σ m (G u) σ m (G), then the result follows by Observation 9. Let σ m (G u) σ m (G) 1. Then obviously σ m (G u) = 1 and so G u is a complete graph. By Corollary, we have b σm (G) b σm (G u)+δ(g) = δ(g) Complete Multipartite Graphs In this section we determine the eternal m-security bondage number of complete multipartite graphs yielding that the eternal m-security bondage number can be arbitrary large. Theorem 0. Let t 3 and G = K n1,n,...,n t be the complete t-partite graph with n 1,n,...,n t. Then b σm (G) =. 3(t 1) Proof. Let X 1,X,...,X t be the partite sets of G and let X i = { x i } 1,...,xi n i for 1 i t. Clearly σ m (G) =. Assume { X = x 1 1x j 1,xs 1 x s+1 1 : j t, 1 s t 1 } if t is odd and { X = x 1 1x j 1,xs 1 x s+1 1,x 1 1x t : j t, 1 s t } when t is even. Obviously, X = 3(t 1). It is easy to see that for any eternal. m-securesets ofg X containingx 1 1, wehave S 3andsob σ m (G) 3(t 1) Now we show that b σm (G) 1 and let G = G F. 3(t 1) 3(t 1). Let F be a set of edges of size at most

8 998 H. Aram, M. Atapour and S.M. Sheikholeslami Claim. For each x V(G ), there exists a vertex x such that N G [x] N G [x ] = V(G ). Proof. Assume, to the contrary, that there exists a vertex x V(G ) such that N G [x] N G [v] V(G ) for each v V(G ). Without loss of generality we may assume that x = x 1 1. For i t, let F i = F { x 1 1 xi j : 1 j n i}. Let first F i = for each i t. Since N G [x 1 1 ] N G [v] V(G ) for each v V(G ), we have x 1 j v F for every v V(G )\X 1 and for some j n 1. This implies that F V(G )\X 1 t, a contradiction. Assume that F i for some i t. We consider two cases. Case 1. F i 1 for each i t. Let I {,...,t} be the set of all elements such that F i = 1 for each i I and let J = {,...,t}\i. Without loss of generality, assume that { [ x 1 1] xi 1 : i [ I} ] F. We estimate the number of edges in F as follows. Since N G x 1 1 NG x i 1 V(G ) for i I, there exists a vertex z i such that z i x i 1,zi x 1 1 / E(G ). Obviously, z i / X i ( j J X j). If z i X 1, then E i = { x 1 1 xi 1,xi 1 zi} F, and if z i X l for some l I {i}, then z i = x l 1 and E i = { x 1 [ ] [ 1 xi 1,xi 1 xl 1,x1 1 1} xl F. Since NG x 1 1 NG x j] s V(G ) for j J and 1 s n j, there exists a vertex zs j such that zsx j j s,zsx j 1 1 / E(G ). We note that ( ) (1) zs j X 1 i I X i for j J and 1 s n j. If zs j X 1, then x j szs j F \ ( i I E i), and if z j s X i for some i I, then zs j = x i 1 and xj szs j F \ ( i I E i) again. Since nj, we conclude that F { zsx j j } s : 1 s n j for each j J. By (1) we have { z j s x j } { s : 1 s n j z j s x j s : 1 s n j } = for j j. Hence, we have F i I E i + which is a contradiction. j J ( F 3 I + J 3 ( I + J ) = { }) zsx j j s : 1 s n j 3(t 1) Case. F i for some i t. Let I {,...,t} be the set of all elements i such that F i, J {,...,t} be the set of all elements j such that F j = 1 and R = {,...,t} \ (I J). Without loss of generality, assume that { x 1 1 x i 1,x1 1 xi,x1 1 xj 1 : i I, j J} F. We estimate the number of edges in F as follows. Obviously, i I F [ ] [ i I. Since N G x 1 1 NG x j 1] V(G ) for each j J, there exists a vertex z j such that z j x j 1,zj x 1 1 / E(G ). Obviously, z j / X j ( r R X r). If z j X i, for some i I, then E j = { x 1 1 xj 1,xj 1 zj} F, and if z j X l for some l J {j}, then z j = x l 1 and E j = { x 1 1 xj 1,xj 1 xl 1,x1 1 1} xl F.,

9 Eternal m-security Bondage Numbers in Graphs 999 As in Case 1, we can see that F { } zsx r r s : 1 s n r for each r R, and { z r s x r } { s : 1 s n r z r s x r s : 1 s n r } = for r r. Hence, we have F i I F i + j J E j + ( { }) j J F zsx j j s : 1 s n j I + 3 J + R 3 ( I + J + R ) = 3(t 1), which is a contradiction. Now, for each v V(G ), let x v V(G ) be a vertex such that N G [v] N G [x v ] = V(G ). We show that the set S v = {v,x v } is an EmSS of G. Obviously, S v is a dominating set of G. Consider an attack on a vertex u of V(G ). Then one of v or x v is adjacent to u. Let uv E(G ). If x u is adjacent to x v, then we can shift guards from v and x v to u and x u, respectively. Let x u x v / E(G ). Then x u v,ux v E(G ) and we can shift guards from v and x v to x u and u, respectively. Therefore, σ m (G ) = and this implies that b σm (G) Thus b σm (G) = and the proof is complete. 3(t 1) 3(t 1). 5. Trees In this section, we first prove that for any nontrivial tree T, b σm (T) and then we characterize all trees attaining this bound. Theorem 1. For any tree T of order n, b σm (T). Proof. If diam(t), then T is a star and the result is immediate. Let diam(t) 3. Suppose P := v 1 v v k is a diametral path in T and root T at v k. Obviously, k 4. If deg(v ) =, then b σm (T) by Theorem 10. Let deg(v ) 3. Then v is adjacent to a leaf v other than v 1. Since d(v 1,v ) =, Theorem 13 implies that b σm (T). This completes the proof. Next, we provide a constructive characterization of all trees attaining the bound of Theorem 1. For this purpose, we describe a procedure to build a family T of trees as follows. Let T be the family of trees such that a path P 3 is a tree in T and if T is a tree in T, then the tree T obtained from T by the following four operations which extend the tree T by attaching a tree to a vertex v V(T), called an attacher, is also a tree in T (see Figure 1). Operation T 1. If v V(T), then T 1 adds a path vxy to T. Operation T. If v V(T), then T adds a star K 1,3 with central vertex y and leaves x,w,z and joins x to v.

10 1000 H. Aram, M. Atapour and S.M. Sheikholeslami v x y v x y T 1 : T : z w T 3 : v leaf w x y z T 4 : v leaf x 1 x Figure 1. The four operations. Operation T 3. If v V(T) is a leaf, then T 3 adds a pendant edge vw and a star K 1, with central vertex x and leaves y,z and joins x to v. Operation T 4. If v V(T) is a leaf, then T 4 adds two new stars K 1, centered at x 1 and x, and joins v to x 1 and x. We start with some lemmas. Lemma. Let G be a graph and v V(G). If G is the graph obtained from G by attaching a path vxy, then σ m (G ) = σ m (G) + 1. In particular, b σm (G ) b σm (G). Proof. Clearly, addingxtoanyσ m (G)-setyieldsanEmSSofG andsoσ m (G ) σ m (G)+1. LetnowS beaσ m (G )-setcontainingy (wemayassumearesponseto an attack on y). If x S, then the set (S \{x,y}) {w}, where w N G [v]\s, is anemssofg. Ifx S, thens \{y}isanemssofg. Thusσ m (G) σ m (G ) 1 and so σ m (G ) = σ m (G)+1. Lemma 3. Let G be a graph and v V(G). If G is the graph obtained from G by adding a star K 1,3 with central vertex y and leaves x,w,z and joining x to v, then σ m (G ) = σ m (G)+. In particular, b σm (G ) b σm (G). Proof. Clearly, adding x and y to any σ m (G)-set yields an EmSS of G and so σ m (G ) σ m (G) +. Suppose now S is a σ m (G )-set containing z (we may assume a response to an attack on z). Since S is a dominating set, we must have S {y,w} 1. Ifx S thentheset(s \{x,y,z,w}) {u}, whereu N G [v]\s is an EmSS of G, and if x S then the set S \ {x,y,z,w} is an EmSS of G. Hence σ m (G) σ m (G ) and this implies that σ m (G ) = σ m (G)+. Lemma 4. Let G be a graph and let v V(G). If G is the graph obtained from G by adding a pendant edge vw and a star K 1, with central vertex x

11 Eternal m-security Bondage Numbers in Graphs 1001 and leaves y,z and joining x to v, then σ m (G ) = σ m (G) +. In particular, b σm (G ) b σm (G). Proof. Clearly, adding x,y to any σ m (G)-set containing v yields an EmSS of G and so σ m (G ) σ m (G) +. Assume now that S is a σ m (G )-set. As in the proof of Lemma 3, we may assume that y S and S {x,z} 1. Since S is a dominating set, we must have S {v,w} 1. If S {x,y,z,w} 3, then let S = (S \{x,y,z,w}) {u} where u N G [v]\s, and if S {x,y,z,w} =, then let S = S \{x,y,z,w}. Clearly, S is an EmSS of G and hence σ m (G) σ m (G ). Thus σ m (G ) = σ m (G)+. Lemma 5. Let G be a graph and let v V(G). If G is the graph obtained from G by adding two new stars K 1, centered at x 1,x and joining v to x 1,x, then σ m (G ) = σ m (G)+3. In particular, b σm (G ) b σm (G). Proof. Lety i,z i betheleavesadjacenttox i fori = 1,. Clearly,addingx 1,x,y 1 to any σ m (G)-set containing v yields an EmSS of G and so σ m (G ) σ m (G)+3. Let now S be a σ m (G )-set. As above we may assume that y 1 S, S {x 1,z 1 } 1 and S {x,y,z } 1. It is easy to see that S {x,y,z,v}. If S {x,y,z } =, then let S = (S {x 1,y 1,z 1,x,y,z }) {u} where u N G [v]\s, andif S {x,y,z } = 1, thenlets = S \{x 1,y 1,z 1,x,y,z }. Clearly, S is an EmSS of G and hence σ m (G) σ m (G ) 3. Thus σ m (G ) = σ m (G)+3. Lemma 6. Let T T and u V(T). If T is a tree obtained from T by adding a pendant edge uu, then σ m (T ) = σ m (T). Proof. Let T be a tree obtained from T by adding the pendant edge uu. If S is a σ m (T )-set, then let S = S if u S and S = (S {u }) {w}, where w N T [u]\s, whenu S. Clearly, S isanemssforgandsoσ m (T) σ m (T ). Now we show that σ m (T ) σ m (T). Let P 3 = v 1 v v 3 and let T be obtained from P 3 by successive operations T 1,...,T m, respectively, where T i {T 1,T,T 3,T 4 } for 1 i m, if m 1, and T = P 3 if m = 0. The proof is by induction on m. If m = 0, then clearly the statement is true. Assume m 1 and that the statement holds for all trees which are obtained from P 3 by applying at most m 1 operations. Suppose T m 1 is a tree obtained by applying the first m 1 operations T 1,...,T m 1 and let T be obtained from T m 1 by adding a new part to the attacher v. Assume that T m 1 is obtained from T m 1 by adding a pendant edge uu when u V(T m 1 ). We consider four cases. Case 1. T m = T 1. Then T is obtained from T m 1 by attaching a path vxy to v V(T m 1 ). If u V(T m 1 ), then by the inductive hypothesis, σ m (T m 1 ) = σ m (T m 1 ) and by Lemma we have σ m(t ) = σ m (T). Suppose u {x,y}. Let T = T {y,u }. Then, obviously, T is obtained from T m 1 by adding the

12 100 H. Aram, M. Atapour and S.M. Sheikholeslami pendant edge xv. By the inductive hypothesis, σ m (T ) = σ m (T m 1 ). Let S be a σ m (T )-set containing x. Then S {y} is an EmSS of T and by Lemma we have σ m (T ) σ m (T )+1 = σ m (T m 1 )+1 = σ m (T). Case. T m = T. Then T is obtained from T m 1 by adding a star K 1,3 with central vertex y and leaves x,w,z and joining x to v. If u V(T m 1 ), then the result follows from the induction hypothesis and Lemma 3. Assume that u {x,y,z,w}. Let T = T {y,z,w}. By the induction hypothesis, we have σ m (T ) = σ m (T m 1 ). Let S be a σ m (T )-set containing x. Then the set S {y,z} if u w and the set S {y,w} if u = w, is an EmSS of T and so σ m (T ) σ m (T )+. By Lemma 3, we obtain σ m (T ) σ m (T )+ = σ m (T m 1 )+ = σ m (T). Case 3. T m = T 3. Then T is obtained from T m 1 by attaching a pendant edge vw at v and adding a star K 1, with central vertex x and leaves y,z and joining x to v. If u V(T m 1 ), then we deduce from the induction hypothesis and Lemma 4 that σ m (T ) = σ m (T). If u = x or u = y (the case u = z is similar), then let T = T {u,x,y,z}. Obviously, T is obtained from T m 1 by adding the pendant edge vw at v. By the induction hypothesis, we have σ m (T ) = σ m (T m 1 ). Clearly, adding x,y to any σ m (T )-set yields an EmSS of T and so σ m (T ) σ m (T )+ = σ m (T m 1 )+ = σ m (T). If u = w, then let T = T {u,x,y,z}. Obviously, T is obtained from T m 1 by adding the pendant edge vw at v. By the inductive hypothesis, σ m (T ) = σ m (T m 1 ). Let S be a σ m (T )-set containing w. Then S {v,x} if v / S and S {x,y} if v S, is an EmSS of T and so σ m (T ) σ m (T )+ = σ m (T m 1 )+ = σ m (T). Case 4. T m = T 4. Then T is obtained from T m 1 by adding two stars K 1, with central vertices x 1 and x and joining x 1,x to v V(T m 1 ). Let y i,z i be the leaves adjacent to x i for i = 1,. If u V(T m 1 ), then the result follows from the induction hypothesis and Lemma 5. If u = x 1 (the case u = x is similar), then adding x 1,y 1,x to any σ m (T m 1 )-set containing v yields an EmSS of T and we deduce from Lemma 5 that σ m (T ) σ m (T m 1 )+3 = σ m (T). Assume that u = y 1 (the cases u = z 1,u = y,u = z are similar). Let T = T {x 1,y 1,z 1,u,y,z }. Obviously, T is obtained from T m 1 by adding pendant edge vx at v. By the inductive hypothesis, we have σ m (T ) = σ m (T m 1 ).

13 Eternal m-security Bondage Numbers in Graphs 1003 Clearly, adding x 1,y 1,y to any σ m (T )-set containing x, yields an EmSS of T and this implies that σ m (T ) σ m (T )+3 = σ m (T m 1 )+3 = σ m (T). Hence σ m (T ) σ m (T). Thus σ m (T ) = σ m (T) and the proof is complete. Theorem 7. If T T, then b σm (T) =. Proof. Let T T, e E(T) and T = T e. Clearly σ m (T ) σ m (T). Now we show that σ m (T ) σ m (T). Let P 3 := v 1 v v 3 and let T be obtained from P 3 by successive operations T 1,...,T m, respectively, where T i {T 1,T,T 3,T 4 } for 1 i m if m 1 and T = P 3 if m = 0. The proof is by induction on m. If m = 0, then the statement is true by Corollary 4. Assume m 1 and that the statement holds for all trees obtained from P 3 by applying at most m 1 operations. Suppose T m 1 is a tree obtained by applying the first m 1 operations T 1,...,T m 1. We consider four cases. Case 1. T m = T 1. Then T is obtained from T m 1 by attaching a path vxy at v V(T m 1 ). If e E(T m 1 ), then we deduce from the induction hypothesis and Lemma that σ m (T ) = σ m (T m 1 e)+1 = σ m (T m 1 )+1 = σ m (T). If e = vx, then clearly σ m (T ) = σ m (T m 1 )+1 = σ m (T). Assume that e = xy. Let T = T {y}. Then T is obtained from T m 1 by adding a pendant path vx at v. Clearly σ m (T ) = σ m (T )+1. It follows from Lemmas 6 and that σ m (T ) = σ m (T )+1 = σ m (T m 1 )+1 = σ m (T). Case. T m = T. Then T is obtained from T m 1 by adding a star K 1,3 with central vertex y and leaves x,w,z and joining x to v. If e E(T m 1 ), then by the inductive hypothesis and Lemma 3 we have σ m (T ) = σ m (T m 1 e)+ = σ m (T m 1 )+ = σ m (T). If e = vx, then clearly σ m (T ) = σ m (T m 1 ) + = σ m (T). If e = xy, then let T = T {y,z,w}. By Lemma 6, we have σ m (T ) = σ m (T m 1 ). Clearly σ m (T ) = σ m (T )+ and by Lemma 3 we have σ m (T ) = σ m (T )+ = σ m (T m 1 )+ = σ m (T). Assume that e = yz. Let T = T {z,w}. Then T is obtained from T m 1 by OperationT 1 andsot Tandσ m (T ) = σ m (T m 1 )+1. ByLemma6, wehave

14 1004 H. Aram, M. Atapour and S.M. Sheikholeslami σ m (T +yw) = σ m (T ). Now it is easy to check that σ m (T ) σ m (T +yw)+1 and by Lemma 3 we have σ m (T ) σ m (T )+1 = σ m (T m 1 )+ = σ m (T). Case 3. T m = T 3. Then T is obtained from T m 1 by adding a pendant edge vw at a leaf v V(T m 1 ) and adding a star K 1, with central vertex x and leaves y,z and joining x to v. If e E(T m 1 ), then we conclude from the induction hypothesis and Lemma 4 that σ m (T ) = σ m (T m 1 e)+ = σ m (T m 1 )+ = σ m (T). If e = vw, then let T = T {y,z,w}. Then we have σ m (T ) = σ m (T m 1 ) by Lemma 6. On the other hand, adding y,w to any σ m (T )-set containing x, yields an EmSS of T and we deduce from Lemma 4 that σ m (T ) σ m (T )+ = σ m (T m 1 )+ = σ m (T). If e {xv,xy,xz}, then let T = T {x,y,z}. Then T is obtained from T m 1 by attaching a pendant edge vw. By Lemma 6, we have σ m (T ) = σ m (T m 1 ). On the other hand, adding x,y to any σ m (T )-set yields an EmSS of T and it follows from Lemma 4 that σ m (T ) σ m (T )+ = σ m (T m 1 )+ = σ m (T). Case 4. T m = T 4. Then T is obtained from T m 1 by adding two stars K 1, with central vertices x 1 and x and joining x 1,x to a leaf v. Let y i,z i be the leaves adjacent to x i for i = 1,. If e E(T m 1 ), then by the inductive hypothesis and Lemma 5 we have σ m (T ) = σ m (T m 1 e)+3 = σ m (T m 1 )+3 = σ m (T). If e = x 1 v or e = x 1 y 1, then let T = T {x 1,y 1,z 1,y,z }. Then T is obtained from T m 1 by attaching a pendant edge vx at v. By Lemma 6, we have σ m (T ) = σ m (T m 1 ). On the other hand, adding x 1,y 1,y to any σ m (T )- set containing x yields an EmSS of T and it follows from Lemma 5 that σ m (T ) σ m (T )+3 = σ m (T m 1 )+3 = σ m (T). In the other cases, we can see that σ m (T ) σ m (T) as above. Hence σ m (T ) σ m (T). Thus σ m (T ) = σ m (T) and this implies that b σm (T). Now the result follows from Theorem 1. Now we are ready to prove the main theorem of this section.

15 Eternal m-security Bondage Numbers in Graphs 1005 Theorem 8. Let T be a tree of order n 3. Then b σm (T) = if and only if T T. Proof. According to Theorem 7, we only need to prove the necessity. We proceed by the induction on n. If n = 3, then the result is trivial. Assume that n 4 and the statement holds for all trees T of order less than n. Let T be a tree of order n with b σm (T) =. Since b σm (K 1,n 1 ) = 1, we have diam(t) 3. Suppose P := v 1 v k is a diametral path in T such that deg(v ) is as small as possible and root T at v k. If deg(v ) =, then let T = T {v 1,v }. By Lemma, we have σ m (T) = σ m (T ) + 1 and b σm (T ) =. It follows from the induction hypothesis that T T. Now T can be obtained from T by Operation T 1 and hence T T. Let deg(v ) 3. We conclude from Proposition 7 that deg(v ) = 3. Let w v 1 be a leaf adjacent to v. If deg(v 3 ) =, then let T = T {v 1,v,v 3,w}. By Lemma 3, we have σ m (T) = σ m (T ) + and b σm (T ) =. By the induction hypothesis, we obtain T T. Now T can be obtained from T by Operation T and so T T. Let deg(v 3 ) 3. We consider the following cases. Case 1. There exists a path v 3 xy in T such that x {v,v 4 }. By the choice of diametral path and Proposition 7, we have deg(x) = 3. If v 3 is a support vertex and u is a leaf adjacent to v 3, then it is not hard to see that deleting the edge v 3 u increases the eternal m-security number which leads to a contradiction. Suppose v 3 is not a support vertex. If v 3 is adjacent to a support vertex w other than x,v,v 4, then as above we may assume that deg(w) = 3. It is easy to see that deleting the edge v 3 w increases the eternal m-security number which leads to a contradiction. Hence, deg(v 3 ) = 3. Let T = T {v 1,v,w,x,y,z} where y and z are the leaves adjacent to x. Then σ m (T) = σ m (T )+3 and b σm (T ) = by Lemma 5. We deduce from the induction hypothesis that T T and so T can be obtained from T by Operation T 4. Hence T T. Case. Any neighbor of v 3, except v,v 4, is a leaf. Let u be a leaf adjacent to v 3. If deg(v 3 ) 4, then it is easy to see that deleting the edge v 3 u increases the eternalm-security number and sob σm (T) = 1, acontradiction. Thus deg(v 3 ) = 3. Let T = T {v 1,v,u,w}. By Lemma 4 we have σ m (T) = σ m (T ) + and b σm (T ) =. It follows from the inductive hypothesis that T T. By Operation T 3, T can be obtained from T and so T T. This completes the proof. References [1] A.P. Burger, E.J. Cockayne, W.R. Gröndlingh, C.M. Mynhardt, J.H. van Vuuren and W. Winterbach, Infinite order domination in graphs, J. Combin. Math. Combin. Comput. 50 (004)

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