Non-isolating Bondage in Graphs

Bulletin of the Malaysian Mathematical Sciences Society, Dec 2015

A dominating set of a graph \(G = (V,E)\) is a set D of vertices of G such that every vertex of \(V(G){\setminus }D\) has a neighbor in D. The domination number of a graph G, denoted by \(\gamma (G)\), is the minimum cardinality of a dominating set of G. The non-isolating bondage number of G, denoted by \(b'(G)\), is the minimum cardinality among all sets of edges \(E' \subseteq E\) such that \(\delta (G-E') \ge 1\) and \(\gamma (G-E') > \gamma (G)\). If for every \(E' \subseteq E\) we have \(\gamma (G-E') = \gamma (G)\) or \(\delta (G-E') = 0\), then we define \(b'(G) = 0\), and we say that G is a \(\gamma \)-non-isolatingly strongly stable graph. First we discuss various properties of non-isolating bondage in graphs. We find the non-isolating bondage numbers for several classes of graphs. Next we show that for every non-negative integer, there exists a tree having such non-isolating bondage number. Finally, we characterize all \(\gamma \)-non-isolatingly strongly stable trees.

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Non-isolating Bondage in Graphs

Bull. Malays. Math. Sci. Soc. Non-isolating Bondage in Graphs Marcin Krzywkowski 0 1 Mathematics Subject Classification 0 1 0 Faculty of Electronics, Telecommunications and Informatics, Gdansk University of Technology , Gdan ́sk , Poland 1 Department of Pure and Applied Mathematics, University of Johannesburg , Johannesburg , South Africa A dominating set of a graph G = (V , E ) is a set D of vertices of G such that every vertex of V (G) D has a neighbor in D. The domination number of a graph G, denoted by γ (G), is the minimum cardinality of a dominating set of G. The nonisolating bondage number of G, denoted by b (G), is the minimum cardinality among all sets of edges E ⊆ E such that δ(G − E ) ≥ 1 and γ (G − E ) > γ (G). If for every E ⊆ E we have γ (G − E ) = γ (G) or δ(G − E ) = 0, then we define b (G) = 0, and we say that G is a γ -non-isolatingly strongly stable graph. First we discuss various properties of non-isolating bondage in graphs. We find the nonisolating bondage numbers for several classes of graphs. Next we show that for every non-negative integer, there exists a tree having such non-isolating bondage number. Finally, we characterize all γ -non-isolatingly strongly stable trees. Communicated by Xueliang Li. Research partially supported by the Polish National Science Centre Grant 2011/02/A/ST6/00201. Marcin Krzywkowski-Research fellow of the Claude Leon Foundation. Domination; Bondage; Non-isolating bondage; Graph; Tree - 05C05 · 05C69 1 Introduction Let G = (V , E ) be a graph. By the neighborhood of a vertex v of G, we mean the set NG (v) = {u ∈ V (G) : uv ∈ E (G)}. The degree of a vertex v, denoted by dG (v), is the cardinality of its neighborhood. Let δ(G) mean the minimum degree among all vertices of G. By a leaf we mean a vertex of degree one, while a support vertex is a vertex adjacent to a leaf. We say that a support vertex is strong (weak, respectively) if it is adjacent to at least two leaves (exactly one leaf, respectively). 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 diam(G), is the maximum eccentricity among all vertices of G. We denote the path (cycle, respectively) on n vertices by Pn (Cn, respectively). A wheel Wn, where n ≥ 4, is a graph with n vertices, formed by connecting a vertex to all vertices of a cycle Cn−1. Let T be a tree, and let v be a vertex of T . We say that v is adjacent to a path Pn if there is a neighbor of v, say x , of degree two such that the tree resulting from T by removing the edge vx , and which contains the vertex x , is a path Pn . Let K p,q denote a complete bipartite graph the partite sets of which have cardinalities p and q. By a star we mean a connected graph in which exactly one vertex has degree greater than one. A subset D ⊆ V (G) is a dominating set, abbreviated DS, of G if every vertex of V (G)\D has a neighbor in D. The domination number of a graph G, denoted by γ (G), is the minimum cardinality of a dominating set of G. For a comprehensive survey of domination in graphs, see for example [ 5 ]. The bondage number b(G) of a graph G is the minimum cardinality among all sets of edges E ⊆ E such that γ (G − E ) > γ (G). The concept of bondage in graphs was introduced in [ 2 ] and further studied for example in [ 1,3,4,6–9 ]. We define the non-isolating bondage number of a graph G, denoted by b (G), to be the minimum cardinality among all sets of edges E ⊆ E such that δ(G − E ) ≥ 1 and γ (G − E ) > γ (G). Thus b (G) is the minimum number of edges of G that have to be removed in order to obtain a graph with no isolated vertices, and with the domination number greater than that of G. If for every E ⊆ E we have γ (G − E ) = γ (G) or δ(G − E ) = 0, then we define b (G) = 0, and we say that G is a γ -non-isolatingly strongly stable graph. First we discuss various properties of non-isolating bondage in graphs. We find the non-isolating bondage numbers for several classes of graphs. Next we show that for every non-negative integer, there exists a tree having such non-isolating bondage number. Finally, we characterize all γ -non-isolatingly strongly stable trees. 2 Results We begin with the following well known observations. For every graph G of diameter at least two there exists a γ (G)-set that contains all support vertices. If H is a subgraph of G such that V (H ) = V (G), then γ (H ) ≥ γ (G). If n is a positive integer, then γ ( Pn) = (n + 2)/3 . For every integer n ≥ 3 we have γ (Cn) = (n + 2)/3 . Observation 1 If n is a positive integer, then γ (Kn) = 1. Observation 2 For every integer n ≥ 4 we have γ (Wn) = 1. Observation 3 Let p and q be positive integers such that p ≤ q. Then γ (K p,q ) = 1 if p = 1; 2 otherwise. First we calculate the non-isolating bondage numbers of paths. Lemma 4 For any positive integer n we have ⎧ 0 if n = 1, 2, 3, 4, 5, 7; ⎪⎪ b ( Pn) = ⎨ 1 if n ≥ 6 and n = 3k + 1; ⎪⎪⎩ 2 if n ≥ 10 and n = 3k + 1. Proof Let us observe that if a path has at most five or exactly seven vertices, then removing any edges does not increase the domination number, or gives an isolated vertex. Assume that n = 6 or n ≥ 8. First assume that n = 3k. We have γ ( Pn) = (n + 2)/3 = (3k + 2)/3 = k. We also have γ ( Pn−2) + γ ( P2) = n/3 + 1 = k +1 > γ ( Pn). Thus b ( Pn) = 1 if n = 3k and n ≥ 6. Now assume that n = 3k +2. We have γ ( Pn) = (n +2)/3 = (3k +4)/3 = k +1. We also have γ ( Pn−4)+γ ( P4) = n/3 + 2 = k + 2 > γ ( Pn). Thus b ( Pn) = 1 if n = 3k + 2 and n ≥ 8. Now assume that n = 3k + 1. We have γ ( Pn) = (n + 2)/3 = (3k + 3)/3 = k + 1. Let us observe that removing any edge does not increase the domination number. We have γ ( Pn−6) + γ ( P4) + γ ( P2) = (n − 4)/3 + 3 = (3k − 3)/3 + 3 = k + 2 > γ ( Pn). Therefore b ( Pn) = 2 if n = 3k + 1 and n ≥ 10. We now investigate the non-isolating bondage in cycles. Lemma 5 For every integer n ≥ 3 we have b (Cn) = 0 if b ( Pn) = 0; b ( Pn) + 1 if b ( Pn) = 0. Proof We have γ ( Pn) = γ (Cn). Clearly, Cn − e = Pn. This implies that b (Cn) = 0 if b ( Pn) = 0, while b (Cn) = b ( Pn) + 1 if b ( Pn) = 0. We now find the non-isolating bondage numbers of complete graphs. Proposition 6 If n is a positive integer, then b (Kn) = 0 (n + 1)/2 for n = 1, 2, 3; for n ≥ 4. Proof Obviously, b (K1) = 0 and b (K2) = 0. We have K3 − e = C3 and b (C3) = 0. This implies that b (K3) = 0. Now assume that n ≥ 4. By Observation 1 we have γ (Kn) = 1. Let us observe that the domination number of a graph equals one if and only if the graph has a universal vertex. Given a complete graph, we increase the domination number if and only if for every vertex we remove at least one incident edge. If n is even, then we remove n/2 = (n + 1)/2 edges. If n is odd, then we remove (n − 1)/2 + 1 = (n + 1)/2 = (n + 1)/2 edges. We now calculate the non-isolating bondage numbers of wheels. Proposition 7 For integers n ≥ 4 we have b (Wn) = 2 if n = 4; 1 if n ≥ 5. Proof Since W4 = K4, using Proposition 6 we get b (W4) = b (K4) = 5/2 = 2. Now assume that n ≥ 5. By Observation 2 we have γ (Wn) = 1. The domination number of a graph equals one if and only if it has a universal vertex. Removing an edge of Wn incident to the vertex of maximum degree gives a graph without universal vertices. Therefore b (Wn) = 1 for n ≥ 5. We now investigate the non-isolating bondage in complete bipartite graphs. Proposition 8 Let p and q be positive integers such that p ≤ q. Then ⎧ 0 b (K p,q ) = ⎨⎪⎪ 4 ⎪⎪⎩ p if p = 1, 2; if p = 3; otherwise. Proof Let E (K p,q ) = {ai b j : 1 ≤ i ≤ p and 1 ≤ j ≤ q}. If p = 1, then obviously b (K p,q ) = 0 as removing any edge produces an isolated vertex. Now assume that p ≥ 2. By Observation 3 we have γ (K p,q ) = 2. Let E be a subset of the set of edges of K2,q such that δ(K2,q − E ) ≥ 1. Each vertex bi is adjacent to a1 or a2 in the graph K2,q − E . Observe that the vertices a1 and a2 form a dominating set of K2,q − E . Therefore b (K2,q ) = 0. Now assume that p = 3. It is not very difficult to verify that removing any three edges does not increase the domination number while not producing an isolated vertex. We have γ (K3,q − a1b2 − a1b3 −a2b1 − a3b1) = 3 > 2 = γ (K3,q ). Therefore b (K3,q ) = 4. Now assume that p ≥ 4. If we remove at most p − 1 edges, then there are vertices ai and b j which have degrees q and p, respectively. It is easy to observe that the vertices ai and b j still form a dominating set. Let us observe that γ (K p,q − a1b1 − a2b1 − a3b2 −a4b2 − a5b2 − · · · − a pb2) = 3 > 2 = γ (K p,q ). Therefore b (K p,q ) = p if p ≥ 4. The authors of [ 2 ] proved that the bondage number of any tree is either one or two. Theorem 9 ([ 2 ]) For every tree T we have b(T ) ∈ {1, 2}. Let us observe that for every non-negative integer there exists a tree with such non-isolating bondage number. We have b ( P4) = 0. For positive integers k, consider trees Tk of the form presented in Fig. 1. It is not difficult to verify that b (Tk ) = k. Hartnell and Rall [ 3 ] characterized all trees with bondage number equal to two. We characterize all trees with the non-isolating bondage number equal to zero, that is, all γ -non-isolatingly strongly stable trees. We now show that joining two γ -non-isolatingly strongly stable trees gives us also a γ -non-isolatingly strongly stable tree. Lemma 10 Let T1 and T2 be vertex-disjoint γ -non-isolatingly strongly stable trees. Let x be a support vertex of T1 and let y be a leaf of T2. Let T be a tree obtained by joining the vertices x and y. If γ (T ) = γ (T1) + γ (T2), then the tree T is also γ -non-isolatingly strongly stable. Proof Let E1 be a subset of the set of edges of T such that δ(T − E1) ≥ 1. If x y ∈ E1, then we get γ (T − E1) = γ (T1− E1∩ E (T1))+γ (T2− E1∩ E (T2)) = γ (T1)+γ (T2) = γ (T ). Now assume that x y ∈/ E1. Let z be the neighbor of y other than x . If yz ∈/ E1, then let E2 = E1 ∪ {x y}. Similarly as earlier we get γ (T − E2) = γ (T ). We have γ (T − E1) ≤ γ (T − E2), and consequently, γ (T − E1) = γ (T ). Now assume that yz ∈ E1. Let E3 = E1 ∪ {x y}\{yz}. Similarly as earlier we get γ (T − E3) = γ (T ). Let D2 be a γ (T − E3)-set that contains the vertices x and z. It is easy to observe that D2 is also a DS of the graph T − E1. Therefore γ (T − E1) ≤ γ (T − E3). This implies that γ (T − E1) = γ (T ). We now conclude that b (T ) = 0. We next show that a subtree of a γ -non-isolatingly strongly stable tree is also γ -non-isolatingly strongly stable. Lemma 11 Let T be a γ -non-isolatingly strongly stable tree. Assume that T is a subtree of T such that T − T has no isolated vertices. Then b (T ) = 0. Proof If T consists of a single vertex, then obviously b (T ) = 0. Thus assume that T = K1. Let E1 be the minimum subset of E (T ) such that T is a component of T − E1. Now let E be a subset of E (T ) such that δ(T − E ) ≥ 1. Notice that δ(T − E1 − E ) ≥ 1. The assumption b (T ) = 0 implies that γ (T − E1) = γ (T ) and γ (T − E1 − E ) = γ (T ). We have T − E1 − E = T − E ∪ (T − T ) and T − E1 = T ∪ (T − T ). We now get γ (T − E ) = γ (T − E1 − E ) − γ (T − T ) = γ (T ) − γ (T − E1) + γ (T ) = γ (T ). This implies that b (T ) = 0. For the purpose of characterizing all γ -non-isolatingly strongly stable trees, we introduce a family T of trees T = Tk that can be obtained as follows. Let T1 ∈ { P1, P2}. If k is a positive integer, then Tk+1 can be obtained recursively from Tk by one of the following operations. • Operation O1: Attach a vertex by joining it to any support vertex of Tk . • Operation O2: Attach a path P2 by joining one of its vertices to a vertex of Tk , which is adjacent to a path P1 or P4, or is not a leaf and is adjacent to a support vertex. • Operation O3: Attach a path P3 by joining one of its leaves to a vertex of Tk adjacent to a path P1 or P3. • Operation O4: Attach a path P5 by joining one of its leaves to any support vertex of Tk . We now prove that every tree of the family T is γ -non-isolatingly strongly stable. Lemma 12 If T ∈ T , then b (T ) = 0. Proof We use induction on the number k of operations performed to construct the tree T . If T = P1, then obviously b (T ) = 0. If T = P2, then b (T ) = 0 as removing the edge gives isolated vertices. Let k be a positive integer. Assume that the result is true for every tree T = Tk of the family T constructed by k − 1 operations. Let T = Tk+1 be a tree of the family T constructed by k operations. First assume that T is obtained from T by Operation O1. Let x be the attached vertex, and let y be its neighbor. Let z be a leaf adjacent to y and different from x . Let D be a γ (T )-set that contains all support vertices. The set D is minimal, thus x ∈/ D. Obviously, D is a DS of the tree T . Therefore γ (T ) ≤ γ (T ). Now let E be a subset of the set of edges of T such that δ(T − E ) ≥ 1. Since both x and z are leaves of T , we have x y ∈/ E and yz ∈/ E . The assumption b (T ) = 0 implies that γ (T − E ) = γ (T ). Let us observe that there exists a γ (T − E )-set that contains the vertex y. Let D be such a set. It is easy to see that D is a DS of the graph T − E . Thus γ (T − E ) ≤ γ (T − E ). We now get γ (T − E ) ≤ γ (T − E ) = γ (T ) ≤ γ (T ). On the other hand, we have γ (T − E ) ≥ γ (T ). This implies that γ (T − E ) = γ (T ), and consequently, b (T ) = 0. Now assume that T is obtained from T by Operation O2. The vertex to which is attached P2 we denote by x . Let v1v2 be the attached path. Let v1 be joined to x . If x is adjacent to a leaf or a support vertex, say a, then let D be a γ (T )-set that contains all support vertices. We have v2 ∈/ D as the set D is minimal. It is easy to observe that D\{v1} is a DS of the tree T . If x is adjacent to a path P4, then we denote it by abcd. Let a and x be adjacent. Let us observe that there exists a γ (T )-set that contains the vertices v1, c, and x . Let D be such a set. It is easy to observe that D\{v1} is a DS of the tree T . We conclude that γ (T ) ≤ γ (T ) − 1. Now let E be a subset of the set of edges of T such that δ(T − E ) ≥ 1. Since v2 is a leaf of T , we have v1v2 ∈/ E . If x v1 ∈ E , then δ(T − (E ∩ E (T ))) ≥ 1. We get γ (T − E ) = γ ( P2 ∪ T − (E \{x v1})) = γ (T − (E ∩ E (T ))) + γ ( P2) = γ (T ) + 1 ≤ γ (T ). Now assume that x v1 ∈/ E . By Tx (Tx , respectively), we denote the component of T − E (T − E , respectively) which contains the vertex x . If δ(T −(E ∩ E (T ))) ≥ 1, then let Dx be any γ (Tx )-set. It is easy to see that Dx ∪ {v1} is a DS of the tree Tx . Thus γ (Tx ) ≤ γ (Tx ) + 1. We now get γ (T − E ) = γ (T − E − Tx ) + γ (Tx ) ≤ γ (T − E − Tx ) + γ (Tx ) + 1 = γ (T − E − Tx ) + γ (Tx ) + 1 = γ (T − E ) + 1 = γ (T ) + 1 ≤ γ (T ). Now assume that δ(T − (E ∩ E (T ))) = 0. This implies that x is the only isolated vertex of T − (E ∩ E (T )), and so x is not adjacent to any leaf in the trees T and T . Consequently, Tx consists only of the vertex x , and Tx is a path P3. Let us observe that δ(T − (E \{x a})) ≥ 1. Let Ta be the component of T − E , which contains the vertex a. Now let Ta be a tree obtained from Ta by attaching a vertex to the vertex a. We now get γ (T − E ) = γ (T − E − Tx ) + γ ( P3) = γ (T − E − Tx ) + 1 = γ (T − E −Tx −Ta )+γ (Ta )+1 ≤ γ (T − E −Tx −Ta )+γ (Ta )+1 = γ ((T − E −Tx −Ta )∪Ta )+1 = γ (T −(E \{x a}))+1 = γ (T − E )+1 = γ (T )+1 ≤ γ (T ). We conclude that γ (T − E ) = γ (T ), and consequently, b (T ) = 0. Now assume that T is obtained from T by Operation O3. The vertex to which is attached P3 we denote by x . If x is a support vertex, then using Lemma 10, for T1 = T and T2 = P3, we get b (T ) = 0. Now assume that x is adjacent to a path P3, say abc. Let a and x be adjacent. The attached path we denote by v1v2v3. Let v1 be joined to x . Let us observe that there exists a γ (T )-set that contains all support vertices and does not contain the vertex v1. Let D be such a set. We have v3 ∈/ D as the set D is minimal. Observe that D\{v2} is a DS of the tree T . Therefore γ (T ) ≤ γ (T ) − 1. Now let E be a subset of the set of edges of T such that δ(T − E ) ≥ 1. We have v2v3 ∈/ E as the vertex v3 is a leaf. If x v1 ∈ E , then v1v2 ∈/ E ; otherwise we get an isolated vertex. Let us observe that δ(T − (E ∩ E (T ))) ≥ 1. We get γ (T − E ) = γ ( P3 ∪ T − (E \{x v1})) = γ (T − (E ∩ E (T ))) + γ ( P3) = γ (T ) + 1 ≤ γ (T ). Now assume that x v1 ∈/ E . Because of the similarity between the paths abc and v1v2v3 adjacent to the vertex x , it suffices to consider only the possibility when x a ∈/ E . Let us observe that δ(T − (E ∩ E (T ))) ≥ 1. By Tx (Tx , respectively), we denote the component of T − E (T − (E ∩ E (T )), respectively) which contains the vertex x . If v1v2 ∈/ E , then let Dx be any γ (Tx )-set. It is easy to see that Dx ∪ {v2} is a DS of the tree Tx . Thus γ (Tx ) ≤ γ (Tx )+1. We now get γ (T − E ) = γ (T − E − Tx )+γ (Tx ) ≤ γ (T − E − Tx ) + γ (Tx ) + 1 = γ (T − E − Tx ) + γ (Tx ) + 1 = γ (T − E ) + 1 = γ (T ) + 1 ≤ γ (T ). Now assume that v1v2 ∈ E . Because of the similarity between the paths abc and v1v2v3, it suffices to consider only the possibility when ab ∈ E . Let Dx be a γ (Tx )-set that contains all support vertices (so x ∈ Dx ). It is easy to see that Dx is a DS of the tree Tx . Thus γ (Tx ) ≤ γ (Tx ). We get γ (T − E ) = γ (T − E −Tx )+γ (Tx ) ≤ γ (T − E −Tx )+γ (Tx ) = γ (T − E −Tx )+γ (Tx ) = γ (T − E ) = γ (T ) ≤ γ (T ). We now conclude that γ (T − E ) = γ (T ), and consequently, b (T ) = 0. Now assume that T is obtained from T by Operation O4. By Lemma 4 we have b ( P5) = 0. Using Lemma 10, for T1 = T and T2 = P5, we get b (T ) = 0. We now prove that if a tree is γ -non-isolatingly strongly stable, then it belongs to the family T . Lemma 13 Let T be a tree. If b (T ) = 0, then T ∈ T . Proof If diam(T ) ∈ {0, 1}, then T ∈ { P1, P2} ⊆ T . If diam(T ) = 2, then T is a star. The tree T can be obtained from P2 by an appropriate number of Operations O1. Thus T ∈ T . Now assume that diam(T ) ≥ 3. Thus the order n of the tree T is at least four. We obtain the result by the induction on the number n. Assume that the lemma is true for every tree T of order n < n. First assume that some support vertex of T , say x , is strong. Let y be a leaf adjacent to x . Let T = T − y. Let D be a γ (T )-set that contains all support vertices. It is easy to see that D is a DS of the tree T . Thus γ (T ) ≤ γ (T ). Now let E be a subset of the set of edges of T such that δ(T − E ) ≥ 1. Since b (T ) = 0, we have γ (T − E ) = γ (T ). Let us observe that there exists a γ (T − E )-set that contains the vertex x . Let D be such a set. The set D is minimal, thus y ∈/ D. Obviously, D is a DS of the graph T − E . Therefore γ (T − E ) ≤ γ (T − E ). We now get γ (T − E ) ≤ γ (T − E ) = γ (T ) ≤ γ (T ). On the other hand, we have γ (T − E ) ≥ γ (T ). This implies that γ (T − E ) = γ (T ), and consequently, b (T ) = 0. By the inductive hypothesis, we have T ∈ T . The tree T can be obtained from T by Operation O1. Thus T ∈ T . Henceforth, we assume that every support vertex of T is weak. We now root T at a vertex r of maximum eccentricity diam(T ). Let t be a leaf at maximum distance from r , v be the parent of t , and u be the parent of v in the rooted tree. If diam(T ) ≥ 4, then let w be the parent of u. If diam(T ) ≥ 5, then let d be the parent of w. If diam(T ) ≥ 6, then let e be the parent of d. By Tx we denote the subtree induced by a vertex x and its descendants in the rooted tree T . Assume that dT (u) ≥ 3. Thus some child of u is a leaf or a support vertex other than v. Let T = T − Tv. By Lemma 11 we have b (T ) = 0. By the inductive hypothesis, we have T ∈ T . The tree T can be obtained from T by Operation O2. Thus T ∈ T . Now assume that dT (u) = 2. Assume that dT (w) ≥ 3. First assume that there is a child of w other than u, say k, such that the distance of w to the most distant vertex of Tk is three. It suffices to consider only the possibility when Tk is a path P3, say klm. Let T = T − Tu . By Lemma 11 we have b (T ) = 0. By the inductive hypothesis, we have T ∈ T . The tree T can be obtained from T by Operation O3. Thus T ∈ T . Now assume that some child of w is a leaf. Let T = T − Tu . By Lemma 11 we have b (T ) = 0. By the inductive hypothesis, we have T ∈ T . The tree T can be obtained from T by Operation O3. Thus T ∈ T . Thus there is a child of w, say k, such that the distance of w to the most distant vertex of Tk is two. Consequently, k is a support vertex of degree two. Due to the earlier analysis of the children of the vertex u, it suffices to consider only the possibility when dT (w) = 3. Let T = T − Tw. It is easy to observe that D ∪ {v, k} is a DS of the tree T . Thus γ (T ) ≤ γ (T ) + 2. We have δ(T − dw − uv − wk) ≥ 1. We now get γ (T − dw − uv − wk) = γ (T ∪ P2 ∪ P2 ∪ P2) = γ (T ) + 3γ ( P2) = γ (T ) + 3 ≥ γ (T ) + 1 > γ (T ). This implies that b (T ) = 0, a contradiction. If dT (w) = 1, then T = P4. Let T = P2 ∈ T . The tree T can be obtained from T by Operation O2. Thus T ∈ T . Now assume that dT (w) = 2. First assume that there is a child of d other than w, say k, such that the distance of d to the most distant vertex of Tk is four or one. It suffices to consider only the possibilities when Tk is a path P4, or k is a leaf. Let T = T − Tw. Let us observe that there exists a γ (T )-set that contains the vertex d. Let D be such a set. It is easy to observe that D ∪ {v} is a DS of the tree T . Thus γ (T ) ≤ γ (T ) + 1. We have δ(T − dw − uv) ≥ 1. We now get γ (T − dw − uv) = γ (T ∪ P2 ∪ P2) = γ (T ) + 2γ ( P2) = γ (T ) + 2 ≥ γ (T ) + 1 > γ (T ). This implies that b (T ) = 0, a contradiction. Now assume that there is a child of d, say k, such that the distance of d to the most distant vertex of Tk is three. It suffices to consider only the possibility when Tk is a path P3, say klm. Let T = T − Tl . Due to the similarity of T to the tree T from the previous case when d is adjacent to a leaf, we conclude that b (T ) = 0. On the other hand, by Lemma 11 we have b (T ) = 0, a contradiction. Now assume that there is a child of d, say k, such that the distance of d to the most distant vertex of Tk is two. Thus k is a support vertex of degree two. Let T = T − Tk . By Lemma 11 we have b (T ) = 0. By the inductive hypothesis, we have T ∈ T . The tree T can be obtained from T by Operation O2. Thus T ∈ T . If dT (d) = 1, then T = P5. Let T = P2 ∈ T . The tree T can be obtained from T by Operation O3. Thus T ∈ T . Now assume that dT (d) = 2. First assume that e is adjacent to a leaf, say k. Let T = T − Td . By Lemma 11 we have b (T ) = 0. By the inductive hypothesis, we have T ∈ T . The tree T can be obtained from T by Operation O4. Thus T ∈ T . Now assume that e is not adjacent to any leaf. Let E be the set of edges incident with e excluding ed. Let G = T − Td − e. Let D be any γ (G )-set. It is easy to observe that D ∪ {d, v} is a DS of the tree T . Thus γ (T ) ≤ γ (G ) + 2. We have δ(T − (E ∪ {dw, uv})) ≥ 1. We now get γ (T − (E ∪ {dw, uv})) = γ (G ∪ P2 ∪ P2 ∪ P2) = γ (G ) + 3γ ( P2) = γ (G ) + 3 ≥ γ (T ) + 1 > γ (T ). This implies that b (T ) = 0, a contradiction. As an immediate consequence of Lemmas 12 and 13, we have the following characterization of all γ -non-isolatingly strongly stable trees. Theorem 14 Let T be a tree. Then b (T ) = 0 if and only if T ∈ T . Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. 1. Domke , G. , Laskar , R.: The bondage and reinforcement numbers of γ f for some graphs . Discret . Math. 167 ( 168 ), 249 - 259 ( 1997 ) 2. Fink , J. , Jacobson , M. , Kinch , L. , Roberts , J.: The bondage number of a graph . 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Marcin Krzywkowski. Non-isolating Bondage in Graphs, Bulletin of the Malaysian Mathematical Sciences Society, 2016, 219-227, DOI: 10.1007/s40840-015-0290-2