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Global security in claw-free cubic graphs
Discrete Applied Mathematics (
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Global security in claw-free cubic graphs Katarzyna Jesse-Józefczyk ∗ , Elżbieta Sidorowicz University of Zielona Góra, Faculty of Mathematics, Computer Science and Econometrics, ul. prof. Z. Szafrana 4a, 65-516 Zielona Góra, Poland
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Article history: Received 18 April 2013 Received in revised form 3 April 2014 Accepted 21 May 2014 Available online xxxx Keywords: Secure set Dominating set Cubic graphs
abstract A secure set in a graph G = (V , E ) is a set of vertices S ⊆ V such that for any subset X ⊆ S, |N [X ] ∩ S | ≥ |N (X ) − S |. A global secure set SD ⊆ V is a secure set that is also a dominating set, i.e., N [SD] = V . In this paper we investigate global secure sets that contain exactly half of the vertices of the graph. In particular we show that every hamiltonian claw-free cubic graph has such a global secure set. Moreover, we prove that in any claw-free cubic graph there is a global secure set that contains at most 5/9 of the vertices of the graph. © 2014 Elsevier B.V. All rights reserved.
1. Introduction In this paper we study the minimum cardinality of global secure sets in claw-free cubic graphs. First we give necessary definitions. Let G = (V , E ) be a graph. Throughout this paper, we consider finite and undirected graphs without loops or multiple edges. An open neighbourhood of a vertex v is the set N (v) = {x ∈ V : v x ∈ E }, whereas the closed neighbourhood of v is the set N [x] = N (x) ∪ {x}. Similarly, an open (closed) neighbourhood of a set X ⊆ V is the set N (X ) = v∈X N (v) (N [X ] = N (X ) ∪ X ). The degree of a vertex v is the number of its neighbours and is denoted by d(v). Cubic graphs are graphs in which every vertex is of degree 3. If d(v) = 1, then we say that v is a pendant vertex. A claw is a connected graph with 1 vertex of degree 3 and 3 pendant vertices. A graph is claw-free if it does not contain a claw as an induced subgraph. We say that a graph G contains a graph H if there is a subgraph (not necessarily induced) of G isomorphic to H. Let B ⊆ V . Then, by ⟨B⟩ we denote the graph induced by the vertices of B. The length of the path is the number of its edges. A hamiltonian cycle is a cycle that passes all vertices of a graph. For all undefined concepts we refer the reader to [2]. Secure sets were introduced by Brigham et al. in [1]. Definition 1 ([1]). Let G = (V , E ) be a graph. For any S = {s1 , s2 , . . . , sk } ⊆ V , an attack on S is any k mutually disjoint sets A = {A1 , A2 , . . . , Ak } for which Ai ⊆ N [si ] − S, 1 ≤ i ≤ k. A defence of S is any k mutually disjoint sets D = {D1 , D2 , . . . , Dk } for which Di ⊆ N [si ] ∩ S, 1 ≤ i ≤ k. Attack A is defendable if there exists a defence D such that |Di | ≥ |Ai | for 1 ≤ i ≤ k. The set S is secure if and only if every attack on S is defendable. Let S = {s1 , s2 , . . . , sk } be a secure set of G = (V , E ). Following the authors of [1] we say that the members of S are the defenders and the vertices of N [S ] − S are the attackers. Let A = {A1 , A2 , . . . , Ak } be an attack on S and D = {D1 , D2 , . . . , Dk } be a defence against A. We say that a vertex v ∈ N [si ] − S attacks si ∈ S if v ∈ Ai . Similarly, we say that a vertex sj ∈ S defends si ∈ S if sj ∈ Di . An attack on S is maximal if every vertex of N [S ] − S attacks a vertex of S. Clearly, to determine whether a set of vertices is secure, it is enough to find a defence for every maximal attack. Brigham et al. also proved that
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Corresponding author. E-mail addresses: [email protected] (K. Jesse-Józefczyk), [email protected] (E. Sidorowicz).
http://dx.doi.org/10.1016/j.dam.2014.05.027 0166-218X/© 2014 Elsevier B.V. All rights reserved.
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the set S is secure if and only if for every X ⊆ S, |N [X ] ∩ S | ≥ |N [X ] − S | [1]. A short proof of this result and an interesting generalization of the definition of secure sets was recently given in [6] by Isaak, Johnson and Petrie. A set D ⊆ V is a dominating set if N [D] = V . If a secure set is also a dominating set, then we say that it is a global secure set. The minimum cardinality of a dominating set and a global secure set is denoted by γ (G) and γs (G), respectively. These numbers are called a domination number and a global security number. For any connected graph G of order n, γ (G) ≤ ⌊n/2⌋ and γs (G) ≥ ⌈n/2⌉. The graphs having domination number equal to half their order are completely characterized. Obtaining such characterization for global secure sets seems a hard task. Probably, it is even impossible. Our supposition follows from the fact that graphs for which it was proven that their global security number equals half their order do not share any specific properties. For more details search [4,7] and [8]. In this paper we prove that each hamiltonian claw-free cubic graph of order n has the global security number equal to n/2. Theorem 1. If a graph G of order n is a hamiltonian claw-free cubic graph, then γs (G) = n/2. We believe that the global security number of non-hamiltonian claw-free cubic graphs will have the same value. Conjecture 1. If a graph G of order n is a claw-free cubic graph, then γs (G) = n/2. As a support of this conjecture we show that any claw-free cubic graph has a global secure set with at most 5/9 of the vertices of the graph. Theorem 2. If G is a claw-free cubic of order n, then γs (G) ≤ 5n/9. The paper is organized as follows. In Section 2, we present some properties of secure sets and global secure sets containing half of vertices of a graph. In Section 3 we consider the structure of global secure sets that contain half of vertices of a clawfree cubic graph. Our two main results, i.e., Theorems 1 and 2 we prove in Sections 4 and 5, respectively. In Section 4, we show that any hamiltonian claw-free cubic graph of order n has a global secure set of cardinality n/2. Whereas in Section 5, we give an upper bound on the global security number of claw-free cubic graphs. We show that any claw-free cubic graph has a global secure set with at most 5/9 of the vertices of the graph. Throughout this paper we use the following drawing convention. In every figure the vertices of a graph that are coloured black belong to a secure set. The white vertices are outside the secure set and the status of grey vertices is not precised and is either irrelevant or is going to be determined. 2. Preliminaries In the forthcoming sections we use the following properties of secure sets. Proposition 1 ([1]). If S1 and S2 are vertex disjoint secure sets in the same graph, then S1 ∪ S2 is a secure set. Observation 1. Let S be a secure set of G. If we add a new edge e that both of its end-vertices are either in or outside S, then S is still the secure set of graph G + e. For more results about properties of secure sets we refer the reader to [1,3,5] and [9]. Next, we present some properties of global secure sets that contain exactly half of the vertices of a graph. Proposition 2. Let G be a graph and S be a global secure set of G such that |V (G)| = 2|S |. Then, (i) in any maximal attack, every vertex must participate in a defence of S, (ii) for every vertex v ∈ S, N [N (v)] ∩ (V − S ) ̸= ∅. Proof. (i) Since S is a dominating set, every vertex of V − S has a neighbour in S. Thus, in any maximal attack there are |V (G)|/2 attackers. If there is a vertex of S that do not participate in a defence, then the number of the attackers exceeds the number of defenders and we have a contradiction with the security of S. (ii) Suppose conversely that there exists a vertex v ∈ S such that N [N (v)] ∩ (V − S ) = ∅. By the definition of a secure set, every vertex of S can defend either itself or one of its neighbours. Since neither v nor any of its neighbours has a neighbour in V − S, v does not participate in any defence, which contradicts (i). Lemma 1. Let G be a graph and S be a global secure set of G such that |V (G)| = 2|S |, and let S1 , . . . , Sk be vertex sets of components of ⟨S ⟩. Then, for any i, j ∈ {1, . . . , k} there does not exist a vertex x ̸∈ S such that x ∈ N (Si ) ∩ N (Sj ). Proof. Suppose conversely that there exists a vertex x ̸∈ S and components of ⟨S ⟩ such that x ∈ N (Si ) ∩ N (Sj ). Consider a maximal attack A on S such that x attacks a vertex y ∈ Si . Let W denote the set of the attackers of Sj . Since A is maximal and by Proposition 2 every vertex of Sj participates in a defence of S, |W | = |Sj |. Now let us modify the attack A in such a way that x attacks a vertex z ∈ Sj , let us denote the obtained attack by A′ . Now the vertices of Sj are attacked by |W | + 1 attackers. Since |W | = |Sj |, the attack A′ cannot be repelled, which contradicts the security of S. Corollary 1. Let G be a graph, S be a global secure set of G such that |V (G)| = 2|S |, and let S ′ be a vertex set of any component of ⟨S ⟩. Then, |S ′ | = |N [S ′ ] − S |.
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Fig. 1. A diamond, a kite and crystal.
Fig. 2. A kite M contained in S ′ .
3. Properties of global secure sets of claw-free cubic graphs In this section we give lemmas describing properties of a global secure set that contains exactly half of vertices of some claw-free cubic graph. Observe that if a cubic graph is claw-free then every its vertex belongs to some triangle. Lemma 2. Let G be a cubic graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component of ⟨S ⟩. Then |E ( S ′ )| ≤ |S ′ |. Proof. Since G is a cubic graph, |N [S ′ ] − S | ≤ 3|S ′ | − 2|E ( S ′ )|. Thus, by Corollary 1, we have |S ′ | ≤ 3|S ′ | − 2|E ( S ′ )|.
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From Lemma 2 it follows that S is either a tree or it has a unique cycle. If G is claw-free and S is a tree, then S has to be a path. For the purpose of the forthcoming lemma we introduce some definitions. We say that K3 (C3 ) is a triangle. A diamond (see Fig. 1) is the graph K4 minus one edge. Furthermore, a bridge of a diamond is an edge between two vertices of degree 3. A kite is a graph obtained by identifying an end-vertex of a path of length 2 with a vertex of a triangle (see Fig. 1). Two vertex-disjoint triangles connected with two independent edges we call a crystal (see Fig. 1). Lemma 3. Let G be a claw-free cubic graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component of ⟨S ⟩. Then,
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(a) S ′ is not a path of length greater than 5, (b) (c) (d) (e) (f) (g)
S contains no cycle of length greater than 3, S ′ does not contain vertices that induce a triangle and belong to a diamond in G, S ′ does not contain a kite, 2 ≤ |S ′ | ≤ 6, if |S ′ | = 2 and S ′ = {x, y}, then e = xy ∈ E and e is a bridge of a diamond or S ′ is in a component of G isomorphic to K4 , S ′ does not contain exactly two non-adjacent vertices of the same diamond.
Proof. (a) Suppose conversely that there is a component S ′ of ⟨S ⟩ that is a path v1 , . . . , vk of length greater than 5. Consider the maximum number of neighbours of S ′ in N [S ′ ] − S. Let us denote this number by p. Since G is claw-free, only v1 and v2 can have two neighbours that are not adjacent to any other vertex of S ′ , and any other vertex vi of S ′ has a neighbour y ∈ V (G) − S such that yvi−1 ∈ E (G) or yvi+1 ∈ E (G), where i ∈ {2, . . . , k − 1}. Furthermore, p is maximal if the vertices v2 , . . . , vk−1 can be divided into pairs and at most one triplet of consecutive vertices that have a common neighbour outside S ′ . Hence p ≤ 4 + (|S ′ | − 2)/2. However, if |S ′ | > 6, then p < |S ′ |. Thus, by Corollary 1, |S ′| ≤ 6, a contradiction. (b) If S ′ is a cycle, then, by Corollary 1, it has length equal exactly to 3. Suppose that S ′ contains a proper subgraph that ′ ′ is a cycle of length greater than 3. Thus, there is a vertex v ∈ S having three neighbours in S . Since G is claw-free, v belongs to a triangle and hence S ′ has two cycles, a contradiction. (c) Let a, b, c , d be the vertices of a diamond D such that a, b, c induce a triangle and bc is a bridge of D. Moreover, let x be a neighbour of a,and y be a neighbour of d such that x, y ̸∈ D (it is possible that x = y or x = d, y = a). Suppose that a, b, c ∈ S. Since S ′ has a unique cycle, then d ̸∈ S. Thus, in any attack on S the vertex c can repel the attack of d and the vertex a can defend itself if x ̸∈ S or help in a defence of x otherwise. Thus, b does not have to participate in any defence of S, which contradicts Proposition 2. (d) Suppose that there is a component of ⟨S ⟩ that contains a kite M. Let S ′ be a vertex set of such component. Furthermore, let V (M ) = {a, b, c , d, e}, T be a triangle of M with V (T ) = {a, b, c } and cd, de ∈ E (M ), see Fig. 2. Since S ′ has a unique cycle, it follows that bd, ae, eb ̸∈ E (⟨S ⟩). Since G is claw-free, d and e must have a common neighbour, say f , and f ̸∈ S. Hence, there are at most 4 vertices in W = N (V (M ))−V (M ). Moreover, c has no neighbours in W , N (d)∩W = {f }, |N (a)∩W | = 1, |N (b)∩ W | = 1, |N (e) ∩ W | = 2 and f ∈ N (e) ∩ W . Observe that for any attack on S we can organize a defence in such a way that vertex d fights against f , depending on the status of neighbours of a, b, e, the vertices a, b, e fight against or help in a defence
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Fig. 3. A part of a component of ⟨S ⟩ that contains exactly two non-adjacent vertices of the same diamond D.
of their neighbours (i.e., we bond vertices a, b and e with their neighbours in W ). It follows, that in any attack on S vertex cdoes not to participate in a defence which have contradicts the assumption that S is a global secure set of G of cardinality n. (e) If S ′ is a path, then by (a) |S ′ | ≤ 6. If S ′ contains a unique cycle, then (b) and (d) imply that |S ′ | ≤ 6. (f) Clearly, if xy is not a bridge of a diamond and a component of G containing S ′ is not K4 , then |N [S ′ ] − S | > |S ′ | which implies that S is not secure. (g) Suppose conversely that there is a component of ⟨S ⟩ with a vertex set S ′ that contains exactly two non-adjacent vertices of the same diamond D. We denote those vertices by c and e and the remaining vertices of D by d and f . Observe, that there is a maximal attack A in which d and f attack e. Let us consider this attack. By Proposition 2, c must participate in a defence of S. Thus, it has a neighbour b ∈ S ′ that is adjacent to a vertex g ̸∈ S. It follows that in any defence against A, c can defend b from g. However, since G is claw-free and S ′ is connected, b is also adjacent to a vertex a ∈ S such that {b, a, g } induces a triangle, see Fig. 3. Moreover, b must participate in a defence of S. Since c defends b against g, it follows that b must defend a. Thus, a must have one more neighbour outside S which gives us the contradiction to the fact that S ′ is a component of ⟨S ⟩. The following lemma is a consequence of Lemma 3(b)–(d) and Corollary 1. ′ Lemma 4. Let G be a claw-free cubic graph of order 2n with a global secure set S on n vertices. Let S be a vertex set of a component of ⟨S ⟩. If S ′ has a unique cycle, then S ′ is either a triangle or a triangle plus one, two or three pendant vertices, and each vertex in N (S ′ ) − S has exactly one neighbour in S.
Lemma 3(a) and Lemma 3(e) imply the next result. Lemma 5. Let graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component G be a claw-free cubic ′ ′ of ⟨S ⟩. If S has no cycle, then S is a path of length 1, 2, 3, 4, or 5. The next lemma gives us one more property of components of global secure sets in claw-free cubic graphs. Lemma 6. Let G be a claw-free cubic graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component of ⟨S ⟩ and S ′ be a path. If a vertex y ̸∈ S ′ has two neighbours in S ′ that do not belong to the same triangle or diamond, then S ′ ′ is a path on 3 vertices and all the vertices of S belong to a cycle of length four of a subgraph of G that is isomorphic to a crystal. Proof. Consider a component of ⟨S ⟩ with vertex set S ′ such that there exists a vertex y ̸∈ S that has two neighbours in S ′ that do not belong to the same triangle or diamond. By Lemma 3(e) and (f), 2 < |S ′ | ≤ 6. Let x and b be the neighbours of y in S ′ that do not belong to the same triangle in G. Without loss of generality we can assume that x is a neighbour of y that belongs to the same triangle T1 , where V (T1 ) = {x, y, z }. Moreover, let T2 be a triangle in G such that V (T2 ) = {b, c , d} and c ∈ S. By Lemma 4 d ̸∈ S. If |S ′ | = 3 then xc ∈ E (G) and we obtain the desired result. Hence let us assume that |S ′ | > 3. Suppose that z ̸∈ S, then there is a path in S ′ of length 2, 3 or 4 that connects c and x. Suppose that there is a path of length 2. Let x0 be an interior vertex of this path. Since G is claw-free, x0 is adjacent either to d or to z. Thus, |N [S ′ ] − S | = 3. This implies that |N [S ′ ] − S | < |S ′ | = 4, which contradicts Corollary 1. If the path connecting c and x has length 3, let x0 , x1 be its interior vertices. In this case either the vertices x0 , x1 have a common neighbour outside S ′ or one of them is adjacent to d and the other one to z. Thus, either |N [S ′ ] − S | = 4 or |N [S ′ ] − S | = 3. In both cases |S ′ | > |N [S ′ ] − S |, which contradicts Corollary 1. Suppose that the path has length 4, let x0 , x1 , x2 be the interior vertices of this path. Hence, two vertices of {x0 , x1 , x2 } have a common neighbour outside S ′ , let e denote this vertex, and the third vertex is adjacent to d, z or e. This implies that |N [S ′ ] − S | = 4 and again we obtain the contradiction with Corollary 1. Now, assume that z ∈ S. If xc ∈ E (G) or zc ∈ E (G), then if |S ′ | = 4, then |N [S ′ ] − S | ≤ 3; otherwise |S ′ | = 5 or |S ′ | = 6 and |N [S ′ ] − S | ≤ 4. Again we obtain a contradiction to Corollary 1. Hence, xc ̸∈ E (G) and zc ̸∈ E (G). It follows that there is a path in S ′ of length 2 or 3 that connects c and x or z. Since G is claw-free in each case, similarly as above, we obtain that |S ′ | > |N [S ′ ] − S | which is in contradiction to Corollary 1 and finishes the proof. From the above lemmas it follows that each component of a graph induced by a minimal global secure set with their neighbourhood outside the minimal global secure set form one of the structures presented in Fig. 4. From Lemma 2 it follows that the component has a unique cycle or is a path. If it has a unique cycle, then, by Lemma 4, the component and its neighbourhood outside the minimal global secure set must form one of the structures presented in figure (c), (f), (i), or (l). By Lemma 3(f), if the component has two vertices, then this component with its neighbourhood outside the minimal global secure set form the structure presented in figure (a). If the component is a path on three vertices, then, by Corollary 1, it has three vertices in its neighbourhood outside the minimal global secure set. Thus, the vertices of the component and their neighbours must form the structure (b) or (d). It follows from Lemma 5 that all the remaining components are paths with
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Fig. 4. Components of a graph induced by a global secure set of cardinality equal to half of the order of a claw-free cubic graph (black vertices). The components are presented with their neighbourhood outside the secure set (white vertices). In the figure we do not present all the edges that may join the white vertices.
Fig. 5. A path of triangles obtained from 4 triangles.
4, 5 or 6 vertices. All such paths must satisfy Corollary 1. Moreover, taking into consideration Lemma 6 we obtain that these paths together with their neighbourhoods outside the minimal global secure set must form one of the structures presented in figures (e), (g), (h), (j) and (k). To obtain a minimal global secure set of cardinality equal half of the order of any claw-free cubic graph we must cover its vertices with some of the structures (a)–(l). 4. Minimum global secure sets in hamiltonian claw-free cubic graphs A path of triangles (a cycle of triangles) is a graph obtained by adding k − 1 edges (k edges) between k vertex-disjoint triangles in such a way that the obtained graph is connected and every triangle has at least one vertex of degree 2. An example of a path of triangles is presented in Fig. 5. By tG we denote the number of triangles in a graph G. When G is clear from the context, we omit the subscript G. We know that a cubic graph has an even number of vertices. Furthermore, in any {claw, diamond}-free cubic graph, except K4 , every vertex belongs to exactly one triangle. Hence, if G is a {claw, diamond}-free cubic graph, then t ≡ 0(mod 4) or t ≡ 2(mod 4). Lemma 7. Let G be a path of triangles of order n. If the number of triangles in G is a multiple of four, then γs (G) = n/2 and there exists a minimal global secure set that contains only vertices of degree 3. Proof. We divide the triangles of G into k disjoint groups of 4 consecutive triangles. In each group we choose, as it is shown in Fig. 6, 6 vertices that form a secure set and dominate all the triangles of the group. The so-obtained set P is a minimal global secure set of G of cardinality n/2 that contains only the vertices of degree 3. Lemma 8. Let G be a {claw, diamond}-free cubic graph of order n. Let G′ be a graph obtain from G by contracting all triangles. If G′ is bipartite, then γs (G) = n/2. Proof. Let (A, B) be a bipartition of G. Since G′ is 3-regular, |A| = |B|. Let S be the set of vertices of triangles in A. Then, S is a global secure set and |S | = n/2.
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Fig. 6. Four consecutive triangles with a secure set that dominates each vertex of those triangles.
Fig. 7. Indexed triangles and vertices of a hamiltonian {claw, diamond}-free cubic graph. The edges between vertices of {y1 , . . . , yt } are not presented.
Fig. 8. Illustration of subcase 1.2.1.
In the next lemma we determine the global security number of hamiltonian {claw, diamond}-free cubic graphs. Observe, that every hamiltonian {claw, diamond}-free cubic graph is obtained from a cycle of triangles by adding edges between vertices of degree 2. In the proof we assume that the triangles and vertices of the graph are indexed as it is shown in Fig. 7. Moreover, for i, j ∈ {1, . . . , t }, where i ≤ j let strij = j − i − 1 and stlij = t − j + i − 1. We say that for two triangles Ti and Tj , strij is the number of separating triangles on the right and stlij is the number of separating triangles on the left side of Ti . Furthermore, if strij ≥ r and stlij ≥ r, then we say that the triangles Ti and Tj are separated by at least r triangles. Lemma 9. If G is a hamiltonian {claw, diamond}-free cubic graph of order n, then γs (G) = n/2. Proof. Let the triangles and vertices of our graph be indexed as in Fig. 7. Case 1. Let t ≥ 10. If t ≡ 0(mod 4) then, by Lemma 7, γs (G) = n/2. Hence t ≡ 2(mod 4). Subcase 1.1. For any edge yi yj ∈ E (G) (i < j) we have i ̸≡ j(mod 2). Let G′ be a graph obtained from G by contracting all triangles. Thus, G′ is bipartite and by Lemma 8, G has a minimum global secure set of cardinality n/2. Subcase 1.2. There exists an edge yi yj ∈ E (G)(i < j) such that i ≡ j(mod 2). Subcase 1.2.1. The triangles Ti and Tj are separated by at least 3 triangles (see Fig. 8). Since i ≡ j(mod 2) and t − 2 is an even number, strij and stlij are odd. Moreover, since t − 2 ≡ 0(mod 4), without loss of generality we may assume that strij ≡ 3(mod 4) and stlij ≡ 1(mod 4). It follows that strij = 3 + 4f and stlij = 5 + 4e, where f , e ∈ {0, 1, 2, . . .}. For the convenience of the notation let us change the indices of the triangles of G in such a way that Ti is from now on T1 . Let P1 = x2 , z2 , x3 , z3 , x4 , yj , zj , xj+1 , zj+1 , xj+2 , zt −2 , xt −1 , zt −1 , xt , zt . Then, P1 is a secure set which dominates only the vertices of the triangles T1 , T2 , T3 , T4 , Tj , Tj+1 , Tj+2 , Tt , Tt −1 , Tt −2 . If there are still vertices of G that are not dominated, then they belong to 4f and 4e consecutive triangles. By Lemma 7, there is a secure set P2 that dominates these vertices and contains exactly half of them but it contains no vertex yi for i ∈ {1, . . . , t }. Thus, by Proposition 1 and Observation 1, P1 ∪ P2 form a global secure set of cardinality n/2. Subcase 1.2.2. strij = 1 or stlij = 1 Again let us change the indices of the triangles of G in such a way that the triangle Ti is from now on T1 . Thus, str1j = 1 or stl1j = 1 which implies that j = 3. Let Tr be a triangle such that y2 yr ∈ E (G). Assume that r ≡ 0(mod 2). Since we are not in Subcase 1.2.1, we have str2r = 1 or stl2r = 1. Without loss of generality we can assume that r = 4. Let P1 denote the set {y1 , z1 , x3 , y3 , y4 , z4 , x5 , z5 , x6 }. By Lemma 7 we can choose a secure set P2
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Fig. 9. Path of triangles and diamonds of type 1.
Fig. 10. Path of triangles and diamonds of type 2.
of cardinality (n − 18)/2 that dominates the vertices of the triangles T7 , . . . , Tt . Now P1 ∪ P2 is a global secure set of G and |P1 ∪ P2 | = n/2. Now suppose that r ̸≡ 0(mod 2). Observe that str3r + stl1,r = t − 4, so str3r + stl1,r ≡ 2(mod 4). Suppose that str3r ≡ 1(mod 4) and stl1,r ≡ 1(mod 4). Let P1 = {y1 , z1 , x3 , y3 } and P2 = {xr , yr , zr , xr +1 , zr −1 }. Then, P1 is a secure set which dominates the triangles T1 and T3 and P2 is a secure set which dominates the triangles T2 , Tr , Tr −1 and Tr +1 . By Lemma 7, there exists a secure set P3 such that |P3 | = (n − 18)/2, and P3 dominates the vertices that are not dominated by P1 and P2 . The union of the sets P1 , P2 and P3 is a global secure set of cardinality n/2. Hence suppose that str3r ≡ 3(mod 4) and stl1,r ≡ 3(mod 4). Let P1 = {zt −1 , xt , zt , x1 , y1 }, P2 = {z2 , y2 , yr , zr , xr +1 } and P3 = {x4 , z4 , x5 , z5 , x6 }. Then, P1 ∪ P2 ∪ P3 is a secure set which dominates the triangles Tt −1 , Tt , T1 , T2 , T3 , T4 , T5 , T6 , Tr , Tr +1 . By Lemma 7, there is a secure set P4 which dominates the remaining vertices and contains exactly half of them. As previously P1 ∪ P2 ∪ P3 ∪ P4 is a global secure set of cardinality n/2. Case 2. Let t < 10. Since every cubic graph has even order, t ∈ {2, 4, 6, 8}. If t = 4 or t = 8, then, by Lemma 7, γs (G) = n/2. If t = 2, then the vertices of any triangle of G form a global secure set of cardinality n/2. Thus let t = 6. Assume that an edge yi yj ∈ E (G)(i < j) iff i ̸≡ j(mod 2). Then, G′ obtained from G by contracting all triangles is bipartite and, by Lemma 8, G has a minimum global secure set of cardinality n/2. Hence, there is an edge yi yj ∈ E (G)(i < j) such that i ≡ j(mod 2). Without loss of generality we can assume that y1 y3 ∈ E. Let P1 = {y1 , z1 , x3 , y3 }. If y2 y5 ∈ E, then let P2 = {z4 , x5 , y5 , z5 , x6 }; otherwise if y2 y4 ∈ E, then P2 = {y4 , z4 , x5 , z5 , x6 }; otherwise y2 y6 ∈ E and P2 = {z4 , x5 , z5 , x6 , y6 }. In every case P1 ∪ P2 is a global secure set of G of cardinality n/2. This observation finishes the proof. A graph is a path of triangles and diamonds of type 1 if it can be obtained from k vertex-disjoint triangles and p vertex-disjoint diamonds by adding k + p − 1 edges between them in such a way that the obtained graph is connected, its maximum degree equals 3 and each triangle has at least one vertex of degree 2 (see Fig. 9). The path of triangles and diamonds of type 2 is obtained from a path of type 1 that starts with a triangle by adding a pendant vertex to one of the vertices of degree 2 of the first triangle of the path. Such path is presented in Fig. 10. Lemma 10. Let P be a path of triangles and diamonds of order n such that 1. 2. 3. 4.
n is even, P is of type 1 or type 2, P ends with a diamond, there is no two consecutive diamonds in P.
Then, P has a global secure set S such that |S | = n/2 and S contains only the vertices of degree three. Proof. We prove the lemma by induction on the number of vertices. Since n is even, the number of triangle of P is even if P is the path of type 1 and odd if P is the path of type 2. It is easy to see that the lemma is true for the path of type 1 without triangles and for the path of type 2 with one triangle. Let D1 , . . . , Dk be diamonds of G. Let P1 be a path of triangles leading D1 and P2 be a path of triangles between D1 and D2 . Similarly as P2 , we define paths P3 , . . . , Pk . Let l(Pi ) denote the number of triangles in Pi . If the path of type 1 starts with a diamond, then l(P1 ) = 0. In each forthcoming case we find two sets S1 and S2 such that S1 ∪ S2 is a global secure set of P of cardinality n/2 that contains only the vertices of degree 3. Case 1. P is of type 1. Subcase 1.1. l(P1 ) = 0 Then P starts with a diamond D1 . We choose two vertices of degree three of D1 to be the secure set S1 of D1 . By the induction hypothesis, P ′ = P \ D1 (P ′ is the path of type 1) has the global secure set S2 having |V (P ′ )|/2 vertices of degree 3. Hence S1 ∪ S2 is the global secure set of P we wanted. Subcase 1.2. l(P1 ) ≥ 4 Let T = T1 ∪ T2 ∪ T3 ∪ T4 . We choose the secure set S1 of T as in Fig. 11. By the induction hypothesis, P ′ = P \ T (P ′ is the path of type 1) has the global secure set S2 that contains only the vertices of degree 3 and |S2 | = |V (P ′ )|/2. Thus, S1 ∪ S2 is the required global secure set of P.
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Fig. 11. Illustration of subcase 1.2. Black vertices belong to S1 .
Fig. 12. Illustration of subcase 1.3. Black vertices belong to S1 .
Fig. 13. Illustration of subcase 1.4. Black vertices belong to S1 .
Fig. 14. Illustration of subcase 1.5. Black vertices belong to S1 .
Fig. 15. Illustration of subcase 2.1. Black vertices belong to S1 .
Fig. 16. Illustration of subcase 2.2. Black vertices belong to S1 .
Subcase 1.3. l(P1 ) = 2 We choose the secure set S1 of T = T1 ∪ T2 ∪ D1 as in Fig. 12. By the induction hypothesis, P ′ = P \ T (P ′ is the path of type 1) there is a global secure set S2 of P ′ such that |S2 | = |V (P ′ )|/2 and for all v ∈ S2 , d(v) = 3. Now S1 ∪ S2 is our global secure set of P. Subcase 1.4. l(P1 ) = 1 Since the number of triangles is even, l(P2 ) ≥ 1. We choose the secure set S1 of T = T1 ∪ T2 ∪ D1 as in Fig. 13. Since we delete two triangles P ′ = P \ T has even number of triangles, so P ′ is the path of type 1. Thus, by the induction hypothesis, P ′ has the global secure set S2 containing only the vertices of degree 3 and |S2 | = |V (P ′ )|/2. Hence S1 ∪ S2 is the required global secure set of P. Subcase 1.5. l(P1 ) = 3 Since the number of triangles is even, l(P2 ) ≥ 1. Let u, v, w be the vertices of D1 , see Fig. 14. We choose the secure set S1 of T = P1 ∪ {u, v, w} as in Fig. 14. Let P ′ = P \ T . P ′ has odd number of triangles and starts with a pendant edge joined to a triangle, so P ′ is of type 2. Again, by the induction hypothesis, P ′ has a global secure set S2 such that |S2 | = |V (P ′ )|/2 and every vertex in S2 is of degree 3. It is easy to see that S1 ∪ S2 is the global secure set of P we search for. Case 2. P is of type 2 Subcase 2.1. l(P1 ) ≥ 3. We choose the secure set S1 of T = T1 ∪ T2 ∪ T3 as in Fig. 15. P ′ = P \ T is a path of type 1 with less vertices. Thus, applying the induction hypothesis to P ′ , there exists a global secure set S2 of P ′ that contains only the vertices of degree 3 and |S2 | = |V (P ′ )|/2. Again S1 ∪ S2 is the global secure set of P we wanted to obtain. Subcase 2.2. l(P1 ) = 1 We choose the secure set S1 of T = T1 ∪ D1 as in Fig. 16. P ′ = P \ T is the path of type 1 with less vertices. Thus, by induction hypothesis, we can choose a global secure set S2 of P ′ such that every vertex of S2 is of degree 3 and |S2 | = |V (P ′ )|/2. Now, S1 ∪ S2 is the required global secure set of P. Subcase 2.3. l(P1 ) = 2 Let u, v, w be the vertices of D1 as it is presented in Fig. 17. We choose the secure set S1 of T = P1 ∪ {u, v, w} as in Fig. 17. P ′ = P \ T is the path of type 2 with less vertices. By the induction hypothesis, P ′ has the global secure set S2 with |V (P ′ )|/2 vertices of degree 3 and S1 ∪ S2 is the global secure set of P that satisfies our requirements.
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Fig. 17. Illustration of subcase 2.3. Black vertices belong to S1 .
Fig. 18. Illustration of Case 1. The edge vw and ut may belong to E (G).
Fig. 19. Illustration of Case 2 when N (y) ∩ S ′ = {x}. The edge vw and ut may belong to E (G).
Fig. 20. Illustration of Case 2 when {x, w} ⊆ N (y) ∩ S ′ . The vertices u and t may be adjacent.
Lemma 11. Let G be hamiltonian claw-free cubic graph of order n with no two adjacent diamonds. Then, γs (G) = n/2. Proof. If G has no diamonds then, by Lemma 9, γs (G) = n/2. Thus, we may assume that G has at least one diamond. It is easy to see that G has a spanning subgraph P isomorphic to the path of type 1 that satisfies the assumptions of Lemma 10. Hence, P has the global secure set S containing |V (P )|/2 vertices such that each vertex of this set is of degree 3. Since edges E (G) \ E (P ) join the vertices that have degree less than 3 in P, by Observation 1, if we add the edges of E (G) \ E (P ), then the set S is still secure. Lemma 12. Let G be a claw-free cubic graph with an induced subgraph isomorphic to two adjacent diamonds. Let G′ be a graph obtained from G by removing these two diamonds and joining the two appearing vertices of degree 2 by an edge. If γs (G′ ) = |V (G′ )|/2, then γs (G) = |V (G)|/2. Proof. If every vertex of G belongs to a diamond, then the vertices of the bridges of the diamonds form a global secure set of cardinality n/2. Thus, suppose that this is not the case. Let D1 , D2 be two adjacent diamonds of G such that D1 is also adjacent to a triangle. Let G′ be a graph obtained from G by removing D1 and D2 and joining the two appearing vertices x, y of degree 2 by an edge e. Let S ′ be a global secure set of G′ of cardinality |V (G′ )|/2. We show that S ′ can be extended to a global secure set S of G such that |S | = |V (G)|/2. Since G′ ̸= K4 , we may assume that N (x) ∩ N (y) = ∅. Let u, v, y be neighbours of x and w, t , x be neighbours of y in G′ . Without loss of generality we can assume that if there are edges joining a vertex of {u, v} with a vertex of {w, t }, then v is adjacent to w or u is adjacent to t. Consider the vertices x, y and their neighbourhood: Case 1 (see Fig. 18). x, y ̸∈ S ′ . Since S ′ is a dominating set, at least one neighbour of x and at least one neighbour of y is in S ′ . Assume that v, w ∈ S ′ . Observe, that the vertices {v, u, w, t } have in G the same neighbourhood as in G′ . Hence, the set S = S ′ ∪ {b1 , c1 , b2 , c2 } is secure.
Case 2 (see Figs. 19 and 20). x ∈ S ′ and y ̸∈ S ′ . Since, by Lemma 3(e), any component of S ′ has at least two vertices, x has ′ ′ ′ ′ a neighbour in S . Let v ∈ S . If x is the only neighbour of y that belongs to S , then S = S ∪ {d1 , a2 , c2 , d2 } (see Fig. 19). If N (y) ∩ S ′ > 1 then, without loss of generality, we can assume that w ∈ S ′ . By Lemma 6, wv ∈ E (G) and u, t ̸∈ S ′ . In this case let S = S ′ ∪ {a1 , b1 , b2 , c2 } (see Fig. 20).
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Fig. 21. Illustration of Case 3.
Fig. 22. Illustration of Case 4. The vertices u and t may be adjacent.
Fig. 23. Illustration of Case 5 when vw ∈ E (G). The vertices u and t may be adjacent.
Fig. 24. Illustration of Case 5 when N ({u, v}) ∩ S ′ = {x}. The vertices u and t may be adjacent.
Case 3 (see Fig. 21). x, y ∈ S ′ and all neighbours of x are in S ′ . Since, by Lemma 3(d), there is no kite in S ′ , x is the only neighbour of y that belongs to S ′ . Moreover, vw, ut ̸∈ E (G). Then, let S = S ′ ∪ {a1 , a2 , c2 , d2 }. Case 4 (see Fig. 22). x, y ∈ S ′ and both of the vertices have exactly one additional neighbour in S ′ . In this case, by Corollary 1, vw ̸∈ E (G). Observe, that for any attack on S ′ we can organize a defence in such a way that x do not participate in a defence of y, and vice versa. Let S = S ′ ∪ {a1 , c1 , c2 , d2 }. Case 5 (see Figs. 23 and 24). x, y ∈ S ′ and x has two neighbours outside S ′ . By Lemma 3(f), y must have additional neighbour in S ′ . Let w ∈ S ′ . Since we are not in Case 3, we may assume that t ̸∈ S ′ . Let S ′′ be a component of S ′ that contains x and y. If |N (u, v) ∩ S ′ | > 1, then, by Lemma 6, |S ′′ | = 3 and vw ∈ E (G), see Fig. 23. Let S = (S ′ − {x}) ∪ {v, b1 , c1 , c2 , d2 }. If N ({u, v}) ∩ S ′ = {x} (see Fig. 24), then in any maximal attack on S ′ , y must defend x, and w defends itself or y against t. In this case let S = S ′ ∪ {a1 , c1 , d1 , d2 }. Thus, in each case the indicated set S is a global secure set of G and |S | = |V (G)|/2.
Now, we are ready to prove the main theorem of this section. Theorem 1. If a graph G of order n is a hamiltonian claw-free cubic graph, then γs (G) = n/2. Proof. Suppose on the contrary that there is a hamiltonian claw-free cubic graph that has no global secure set that contains half of the vertices of the graph. Let G be such a graph with the minimum number of vertices. Let |V (G)| = n. If G has no two adjacent diamonds then either by Lemma 9 or by Lemma 11 we have γs (G) = n/2. Suppose that G has two adjacent diamonds. If all the vertices of G belongs to these diamonds (i.e., n = 8), then the vertices of the bridges of these diamonds form a global secure set of cardinality n/2. Otherwise, let G′ be a graph obtained from G by removing these two diamonds and joining the two appearing vertices of degree 2 by an edge. G′ is still a hamiltonian claw-free cubic graph and has less vertices than G, so by our assumption G′ has a global secure set of cardinality |V (G′ )|/2. Thus, by Lemma 12 the graph G has a global secure set of cardinality n/2, a contradiction. 5. Upper bound on the global security number of claw-free cubic graphs For the purpose of the forthcoming proof we introduce some special types of subgraphs. A (K3 , D)-subgraph G′ of a claw-free cubic graph is a subgraph such that each vertex of G′ belongs either to K3 or a diamond. A K3 -subgraph is a
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Fig. 25. Secure blocks B1 , . . . , B12 and their global super secure sets.
(K3 , D)-subgraph that contains no diamonds. A (K3 , D)-tree is a (K3 , D)-subgraph that is not K4 and has no other induced cycles than K3 . Similarly, a K3 -tree is a (K3 , D)-tree that contains no diamonds. Let G′ be a (K3 , D)-subgraph of a claw-free cubic graph G. A vertex set S ⊆ V (G′ ) is a super secure set in G′ if S is secure in G′ and in G. We say that S is a global super secure set of G′ if S is super secure and dominating in G′ . The following lemma is an easy consequence of the given definitions.
Lemma 13. Let G be a claw-free cubic graph and G1 , G2 , . . . , Gp be its vertex-disjoint (K3 , D)-subgraphs such that V (G) = V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp ). If S1 , S2 , . . . , Sp are global super secure sets in G1 , G2 , . . . , Gp , respectively, then S1 ∪ S2 ∪ · · · ∪ Sp is a global secure set in G. A secure block B is a (K3 , D)-subgraph of a claw-free cubic graph containing a global super secure set with at most 5|V (B)|/9 vertices. Lemma 14. Each graph depicted in Fig. 25 is a secure block. Moreover, the black vertices of each graph form the global super secure set of the block. Let β be a set of secure blocks presented in Fig. 25. Remark 1. Each K3 -tree with k triangles for k = 3, 4, 5 is a secure block. Lemma 15. Any K3 -tree with exactly 7 triangles, except the graph depicted in Fig. 26, can be partitioned into two secure blocks belonging to β .
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Fig. 26. A K3 -tree with exactly 7 triangles.
Consider K3 -tree T7 presented in Fig. 26. Since it contains no global super secure set with at most 5|V (T7 )|/9 vertices, it is not a secure block. Theorem 2. If G is a claw-free cubic of order n, then γs (G) ≤ 5n/9. Proof. First we show that G can be partitioned into vertex-disjoint (K3 , D)-subgraphs G1 , G2 , . . . , Gp such that V (G) = V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp ) (i.e., G1 ∪ G2 ∪ · · · ∪ Gp is a spanning subgraph of G) and each Gi , for i = 1, . . . , p, is a secure block that belongs to β . Next, from Lemma 13, we conclude that S1 ∪ S2 ∪ · · · ∪ Sp is a global secure set of G1 ∪ G2 ∪ · · · ∪ Gp , where S1 , S2 , . . . , Sp are global super secure sets of G1 , G2 , . . . , Gp , respectively. Finally, from the definition of the block and the global super secure set it will follow that γs (G) = 5n/9. Suppose on the contrary that there is no partition of G into secure blocks belonging to β such that all vertices of G are covered. Thus, let G1 , G2 , . . . , Gp be vertex-disjoint (K3 , D)-subgraphs of G such that Gi ∈ β , for i = 1, . . . , p, and V (G) \ (V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp )) is minimum. We will say that G1 , G2 , . . . , Gp is a block covering of G. Consider a subgraph G′ induced by V (G) \ (V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp )). It contains no diamonds, since diamond belongs to β . It also has no connected subgraph containing more than two triangles. Thus, each component of G′ is either a triangle or contains exactly two triangles. Let t be the number of triangles in G′ . Furthermore, from all block coverings that do not cover t triangles of G, let G1 , G2 , . . . , Gp be a block covering with the minimum number of secure blocks isomorphic to B6 . First, observe the following. Claim 1. If we add a triangle to B1 and join it by an edge with a vertex of degree 2 of B1 , then we obtain a K3 -subgraph containing B7 as a spanning subgraph. Claim 2. Let i ∈ {5, 7, 8, 9}. If we add a triangle to Bi and join it by an edge with a vertex of degree 2 of Bi , then we obtain a secure block from β . Claim 3. Let i ∈ {3, 4, 10, 11}. If we add to Bi a triangle and join it by an edge with a vertex of degree 2 of Bi , then we obtain a secure block from β or we can divide the obtained graph into two secure blocks from β . Lemma 15 implies the next claim. Claim 4. If we add a triangle to B12 and join it by an edge with a vertex of degree 2 distinct from x and y, then we obtain a graph that can be divided into two secure blocks that belong to β . Let T be a triangle that is not covered, i.e., T ⊆ G′ . If a vertex of T is adjacent to a vertex of degree 2 of Bi for i ∈ {1, 3, 4, 5, 7, 8, 9, 10, 11}, then by Claim 1, Claim 2 or Claim 3 we can add T to Bi and cover T ∪ Bi by secure blocks. Hence, we obtain a block covering with a smaller number of non-covered triangles, a contradiction. If a vertex of T is adjacent to a vertex of degree 2 of B12 distinct from x and y, then, by Claim 4, we have a block covering with a smaller number of noncovered triangles, a contradiction. If a vertex of T is adjacent to either a vertex z or w or u of B6 , then we divide T ∪ B6 into two secure blocks: one isomorphic to B7 (containing T and two triangles of B6 ) and the other isomorphic to B2 (the diamond of B6 ) and the third triangle of B6 remains non-covered. Thus, we obtain a block covering with the same number of noncovered triangles however with less secure blocks isomorphic to B6 , a contradiction. From these arguments we conclude that any vertex of T is adjacent to x or y of Bi for i ∈ {2, 6, 12} or to a vertex of a non-covered triangle. Furthermore, observe the following. Claim 5. If we delete from B6 or B12 a triangle containing x and y, then we obtain a secure block from β . If T has two vertices adjacent to two distinct secure blocks D1 , D2 isomorphic to B2 , then T ∪ D1 ∪ D2 contains a spanning subgraph Gp+1 that is B3 and hence G1 , G2 , . . . , Gp , Gp+1 is a block covering with a smaller number of non-covered triangles, a contradiction. Suppose that T has exactly one vertex adjacent to a secure block D isomorphic to B2 . Let T ′ be a triangle adjacent to the other vertex of T . If T ′ is not covered (i.e., T ′ ∈ G′ ), then T ∪ T ′ ∪ D is the secure block B4 or contains a spanning subgraph that is the secure block B4 and hence we can cover both T and T ′ . Otherwise, T ′ is a triangle either in B6 or in B12 containing x and y. Thus, by Claim 5 we can delete T ′ from this secure block and T ∪ T ′ ∪ D contains a spanning subgraph that can be a new secure block. Finally suppose that each vertex of T is adjacent to a triangle. If two vertices of T
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are adjacent to one triangle T ′ , then T ∪ T ′ contains a spanning subgraph that is B1 . If each vertex of T is adjacent to a distinct triangle, say T1 , T2 , T3 , then T ∪ T1 ∪ T2 ∪ T3 contains a spanning subgraph that is B9 . Hence, in both cases, using Claim 5, we obtain a block covering with a smaller number of non-covered triangles, a contradiction. Corollary 2. If G is a claw-free cubic graph of order n < 18, then γs (G) = n/2. Acknowledgements The authors wish to thank the referees for their suggestions that helped to improve this paper. The research of Katarzyna Jesse-Józefczyk was funded by the National Science Centre on the basis of the Decision No. DEC-2011/01/N/ST6/00922. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
R.C. Brigham, R.D. Dutton, S.T. Hedetniemi, Security in graphs, Discrete Appl. Math. 155 (2007) 1708–1714. G. Chartrand, L. Lesniak, Graphs & Digraphs, fourth ed., Chapman & Hall/CRC, New York, 2005. R.D. Dutton, On graph’s security number, Discrete Math. 309 (2009) 4443–4447. R. Dutton, Y.Y. Ho, Global secure sets of grid-like graphs, Discrete Appl. Math. 159 (2011) 490–496. R.D. Dutton, R. Lee, R.C. Brigham, Bounds on graph’s security number, Discrete Appl. Math. 156 (2008) 695–704. G. Isaak, P. Johnson, C. Petrie, Integer and fractional security in graphs, Discrete Appl. Math. 160 (2012) 2060–2062. K. Jesse-Józefczyk, The possible cardinalities of global secure sets in cographs, Theoret. Comput. Sci. 414 (1) (2012) 38–46. K. Jesse-Józefczyk, Monotonicity and expansion of global secure sets, Discrete Math. 312 (2012) 3451–3456. K. Kozawa, Y. Otachi, K. Yamazaki, Security number of grid-like graphs, Discrete Appl. Math. 157 (2009) 2555–2561.
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Global security in claw-free cubic graphs Katarzyna Jesse-Józefczyk ∗ , Elżbieta Sidorowicz University of Zielona Góra, Faculty of Mathematics, Computer Science and Econometrics, ul. prof. Z. Szafrana 4a, 65-516 Zielona Góra, Poland
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Article history: Received 18 April 2013 Received in revised form 3 April 2014 Accepted 21 May 2014 Available online xxxx Keywords: Secure set Dominating set Cubic graphs
abstract A secure set in a graph G = (V , E ) is a set of vertices S ⊆ V such that for any subset X ⊆ S, |N [X ] ∩ S | ≥ |N (X ) − S |. A global secure set SD ⊆ V is a secure set that is also a dominating set, i.e., N [SD] = V . In this paper we investigate global secure sets that contain exactly half of the vertices of the graph. In particular we show that every hamiltonian claw-free cubic graph has such a global secure set. Moreover, we prove that in any claw-free cubic graph there is a global secure set that contains at most 5/9 of the vertices of the graph. © 2014 Elsevier B.V. All rights reserved.
1. Introduction In this paper we study the minimum cardinality of global secure sets in claw-free cubic graphs. First we give necessary definitions. Let G = (V , E ) be a graph. Throughout this paper, we consider finite and undirected graphs without loops or multiple edges. An open neighbourhood of a vertex v is the set N (v) = {x ∈ V : v x ∈ E }, whereas the closed neighbourhood of v is the set N [x] = N (x) ∪ {x}. Similarly, an open (closed) neighbourhood of a set X ⊆ V is the set N (X ) = v∈X N (v) (N [X ] = N (X ) ∪ X ). The degree of a vertex v is the number of its neighbours and is denoted by d(v). Cubic graphs are graphs in which every vertex is of degree 3. If d(v) = 1, then we say that v is a pendant vertex. A claw is a connected graph with 1 vertex of degree 3 and 3 pendant vertices. A graph is claw-free if it does not contain a claw as an induced subgraph. We say that a graph G contains a graph H if there is a subgraph (not necessarily induced) of G isomorphic to H. Let B ⊆ V . Then, by ⟨B⟩ we denote the graph induced by the vertices of B. The length of the path is the number of its edges. A hamiltonian cycle is a cycle that passes all vertices of a graph. For all undefined concepts we refer the reader to [2]. Secure sets were introduced by Brigham et al. in [1]. Definition 1 ([1]). Let G = (V , E ) be a graph. For any S = {s1 , s2 , . . . , sk } ⊆ V , an attack on S is any k mutually disjoint sets A = {A1 , A2 , . . . , Ak } for which Ai ⊆ N [si ] − S, 1 ≤ i ≤ k. A defence of S is any k mutually disjoint sets D = {D1 , D2 , . . . , Dk } for which Di ⊆ N [si ] ∩ S, 1 ≤ i ≤ k. Attack A is defendable if there exists a defence D such that |Di | ≥ |Ai | for 1 ≤ i ≤ k. The set S is secure if and only if every attack on S is defendable. Let S = {s1 , s2 , . . . , sk } be a secure set of G = (V , E ). Following the authors of [1] we say that the members of S are the defenders and the vertices of N [S ] − S are the attackers. Let A = {A1 , A2 , . . . , Ak } be an attack on S and D = {D1 , D2 , . . . , Dk } be a defence against A. We say that a vertex v ∈ N [si ] − S attacks si ∈ S if v ∈ Ai . Similarly, we say that a vertex sj ∈ S defends si ∈ S if sj ∈ Di . An attack on S is maximal if every vertex of N [S ] − S attacks a vertex of S. Clearly, to determine whether a set of vertices is secure, it is enough to find a defence for every maximal attack. Brigham et al. also proved that
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Corresponding author. E-mail addresses: [email protected] (K. Jesse-Józefczyk), [email protected] (E. Sidorowicz).
http://dx.doi.org/10.1016/j.dam.2014.05.027 0166-218X/© 2014 Elsevier B.V. All rights reserved.
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the set S is secure if and only if for every X ⊆ S, |N [X ] ∩ S | ≥ |N [X ] − S | [1]. A short proof of this result and an interesting generalization of the definition of secure sets was recently given in [6] by Isaak, Johnson and Petrie. A set D ⊆ V is a dominating set if N [D] = V . If a secure set is also a dominating set, then we say that it is a global secure set. The minimum cardinality of a dominating set and a global secure set is denoted by γ (G) and γs (G), respectively. These numbers are called a domination number and a global security number. For any connected graph G of order n, γ (G) ≤ ⌊n/2⌋ and γs (G) ≥ ⌈n/2⌉. The graphs having domination number equal to half their order are completely characterized. Obtaining such characterization for global secure sets seems a hard task. Probably, it is even impossible. Our supposition follows from the fact that graphs for which it was proven that their global security number equals half their order do not share any specific properties. For more details search [4,7] and [8]. In this paper we prove that each hamiltonian claw-free cubic graph of order n has the global security number equal to n/2. Theorem 1. If a graph G of order n is a hamiltonian claw-free cubic graph, then γs (G) = n/2. We believe that the global security number of non-hamiltonian claw-free cubic graphs will have the same value. Conjecture 1. If a graph G of order n is a claw-free cubic graph, then γs (G) = n/2. As a support of this conjecture we show that any claw-free cubic graph has a global secure set with at most 5/9 of the vertices of the graph. Theorem 2. If G is a claw-free cubic of order n, then γs (G) ≤ 5n/9. The paper is organized as follows. In Section 2, we present some properties of secure sets and global secure sets containing half of vertices of a graph. In Section 3 we consider the structure of global secure sets that contain half of vertices of a clawfree cubic graph. Our two main results, i.e., Theorems 1 and 2 we prove in Sections 4 and 5, respectively. In Section 4, we show that any hamiltonian claw-free cubic graph of order n has a global secure set of cardinality n/2. Whereas in Section 5, we give an upper bound on the global security number of claw-free cubic graphs. We show that any claw-free cubic graph has a global secure set with at most 5/9 of the vertices of the graph. Throughout this paper we use the following drawing convention. In every figure the vertices of a graph that are coloured black belong to a secure set. The white vertices are outside the secure set and the status of grey vertices is not precised and is either irrelevant or is going to be determined. 2. Preliminaries In the forthcoming sections we use the following properties of secure sets. Proposition 1 ([1]). If S1 and S2 are vertex disjoint secure sets in the same graph, then S1 ∪ S2 is a secure set. Observation 1. Let S be a secure set of G. If we add a new edge e that both of its end-vertices are either in or outside S, then S is still the secure set of graph G + e. For more results about properties of secure sets we refer the reader to [1,3,5] and [9]. Next, we present some properties of global secure sets that contain exactly half of the vertices of a graph. Proposition 2. Let G be a graph and S be a global secure set of G such that |V (G)| = 2|S |. Then, (i) in any maximal attack, every vertex must participate in a defence of S, (ii) for every vertex v ∈ S, N [N (v)] ∩ (V − S ) ̸= ∅. Proof. (i) Since S is a dominating set, every vertex of V − S has a neighbour in S. Thus, in any maximal attack there are |V (G)|/2 attackers. If there is a vertex of S that do not participate in a defence, then the number of the attackers exceeds the number of defenders and we have a contradiction with the security of S. (ii) Suppose conversely that there exists a vertex v ∈ S such that N [N (v)] ∩ (V − S ) = ∅. By the definition of a secure set, every vertex of S can defend either itself or one of its neighbours. Since neither v nor any of its neighbours has a neighbour in V − S, v does not participate in any defence, which contradicts (i). Lemma 1. Let G be a graph and S be a global secure set of G such that |V (G)| = 2|S |, and let S1 , . . . , Sk be vertex sets of components of ⟨S ⟩. Then, for any i, j ∈ {1, . . . , k} there does not exist a vertex x ̸∈ S such that x ∈ N (Si ) ∩ N (Sj ). Proof. Suppose conversely that there exists a vertex x ̸∈ S and components of ⟨S ⟩ such that x ∈ N (Si ) ∩ N (Sj ). Consider a maximal attack A on S such that x attacks a vertex y ∈ Si . Let W denote the set of the attackers of Sj . Since A is maximal and by Proposition 2 every vertex of Sj participates in a defence of S, |W | = |Sj |. Now let us modify the attack A in such a way that x attacks a vertex z ∈ Sj , let us denote the obtained attack by A′ . Now the vertices of Sj are attacked by |W | + 1 attackers. Since |W | = |Sj |, the attack A′ cannot be repelled, which contradicts the security of S. Corollary 1. Let G be a graph, S be a global secure set of G such that |V (G)| = 2|S |, and let S ′ be a vertex set of any component of ⟨S ⟩. Then, |S ′ | = |N [S ′ ] − S |.
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Fig. 1. A diamond, a kite and crystal.
Fig. 2. A kite M contained in S ′ .
3. Properties of global secure sets of claw-free cubic graphs In this section we give lemmas describing properties of a global secure set that contains exactly half of vertices of some claw-free cubic graph. Observe that if a cubic graph is claw-free then every its vertex belongs to some triangle. Lemma 2. Let G be a cubic graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component of ⟨S ⟩. Then |E ( S ′ )| ≤ |S ′ |. Proof. Since G is a cubic graph, |N [S ′ ] − S | ≤ 3|S ′ | − 2|E ( S ′ )|. Thus, by Corollary 1, we have |S ′ | ≤ 3|S ′ | − 2|E ( S ′ )|.
′
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From Lemma 2 it follows that S is either a tree or it has a unique cycle. If G is claw-free and S is a tree, then S has to be a path. For the purpose of the forthcoming lemma we introduce some definitions. We say that K3 (C3 ) is a triangle. A diamond (see Fig. 1) is the graph K4 minus one edge. Furthermore, a bridge of a diamond is an edge between two vertices of degree 3. A kite is a graph obtained by identifying an end-vertex of a path of length 2 with a vertex of a triangle (see Fig. 1). Two vertex-disjoint triangles connected with two independent edges we call a crystal (see Fig. 1). Lemma 3. Let G be a claw-free cubic graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component of ⟨S ⟩. Then,
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(a) S ′ is not a path of length greater than 5, (b) (c) (d) (e) (f) (g)
S contains no cycle of length greater than 3, S ′ does not contain vertices that induce a triangle and belong to a diamond in G, S ′ does not contain a kite, 2 ≤ |S ′ | ≤ 6, if |S ′ | = 2 and S ′ = {x, y}, then e = xy ∈ E and e is a bridge of a diamond or S ′ is in a component of G isomorphic to K4 , S ′ does not contain exactly two non-adjacent vertices of the same diamond.
Proof. (a) Suppose conversely that there is a component S ′ of ⟨S ⟩ that is a path v1 , . . . , vk of length greater than 5. Consider the maximum number of neighbours of S ′ in N [S ′ ] − S. Let us denote this number by p. Since G is claw-free, only v1 and v2 can have two neighbours that are not adjacent to any other vertex of S ′ , and any other vertex vi of S ′ has a neighbour y ∈ V (G) − S such that yvi−1 ∈ E (G) or yvi+1 ∈ E (G), where i ∈ {2, . . . , k − 1}. Furthermore, p is maximal if the vertices v2 , . . . , vk−1 can be divided into pairs and at most one triplet of consecutive vertices that have a common neighbour outside S ′ . Hence p ≤ 4 + (|S ′ | − 2)/2. However, if |S ′ | > 6, then p < |S ′ |. Thus, by Corollary 1, |S ′| ≤ 6, a contradiction. (b) If S ′ is a cycle, then, by Corollary 1, it has length equal exactly to 3. Suppose that S ′ contains a proper subgraph that ′ ′ is a cycle of length greater than 3. Thus, there is a vertex v ∈ S having three neighbours in S . Since G is claw-free, v belongs to a triangle and hence S ′ has two cycles, a contradiction. (c) Let a, b, c , d be the vertices of a diamond D such that a, b, c induce a triangle and bc is a bridge of D. Moreover, let x be a neighbour of a,and y be a neighbour of d such that x, y ̸∈ D (it is possible that x = y or x = d, y = a). Suppose that a, b, c ∈ S. Since S ′ has a unique cycle, then d ̸∈ S. Thus, in any attack on S the vertex c can repel the attack of d and the vertex a can defend itself if x ̸∈ S or help in a defence of x otherwise. Thus, b does not have to participate in any defence of S, which contradicts Proposition 2. (d) Suppose that there is a component of ⟨S ⟩ that contains a kite M. Let S ′ be a vertex set of such component. Furthermore, let V (M ) = {a, b, c , d, e}, T be a triangle of M with V (T ) = {a, b, c } and cd, de ∈ E (M ), see Fig. 2. Since S ′ has a unique cycle, it follows that bd, ae, eb ̸∈ E (⟨S ⟩). Since G is claw-free, d and e must have a common neighbour, say f , and f ̸∈ S. Hence, there are at most 4 vertices in W = N (V (M ))−V (M ). Moreover, c has no neighbours in W , N (d)∩W = {f }, |N (a)∩W | = 1, |N (b)∩ W | = 1, |N (e) ∩ W | = 2 and f ∈ N (e) ∩ W . Observe that for any attack on S we can organize a defence in such a way that vertex d fights against f , depending on the status of neighbours of a, b, e, the vertices a, b, e fight against or help in a defence
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Fig. 3. A part of a component of ⟨S ⟩ that contains exactly two non-adjacent vertices of the same diamond D.
of their neighbours (i.e., we bond vertices a, b and e with their neighbours in W ). It follows, that in any attack on S vertex cdoes not to participate in a defence which have contradicts the assumption that S is a global secure set of G of cardinality n. (e) If S ′ is a path, then by (a) |S ′ | ≤ 6. If S ′ contains a unique cycle, then (b) and (d) imply that |S ′ | ≤ 6. (f) Clearly, if xy is not a bridge of a diamond and a component of G containing S ′ is not K4 , then |N [S ′ ] − S | > |S ′ | which implies that S is not secure. (g) Suppose conversely that there is a component of ⟨S ⟩ with a vertex set S ′ that contains exactly two non-adjacent vertices of the same diamond D. We denote those vertices by c and e and the remaining vertices of D by d and f . Observe, that there is a maximal attack A in which d and f attack e. Let us consider this attack. By Proposition 2, c must participate in a defence of S. Thus, it has a neighbour b ∈ S ′ that is adjacent to a vertex g ̸∈ S. It follows that in any defence against A, c can defend b from g. However, since G is claw-free and S ′ is connected, b is also adjacent to a vertex a ∈ S such that {b, a, g } induces a triangle, see Fig. 3. Moreover, b must participate in a defence of S. Since c defends b against g, it follows that b must defend a. Thus, a must have one more neighbour outside S which gives us the contradiction to the fact that S ′ is a component of ⟨S ⟩. The following lemma is a consequence of Lemma 3(b)–(d) and Corollary 1. ′ Lemma 4. Let G be a claw-free cubic graph of order 2n with a global secure set S on n vertices. Let S be a vertex set of a component of ⟨S ⟩. If S ′ has a unique cycle, then S ′ is either a triangle or a triangle plus one, two or three pendant vertices, and each vertex in N (S ′ ) − S has exactly one neighbour in S.
Lemma 3(a) and Lemma 3(e) imply the next result. Lemma 5. Let graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component G be a claw-free cubic ′ ′ of ⟨S ⟩. If S has no cycle, then S is a path of length 1, 2, 3, 4, or 5. The next lemma gives us one more property of components of global secure sets in claw-free cubic graphs. Lemma 6. Let G be a claw-free cubic graph of order 2n with a global secure set S on n vertices. Let S ′ be a vertex set of a component of ⟨S ⟩ and S ′ be a path. If a vertex y ̸∈ S ′ has two neighbours in S ′ that do not belong to the same triangle or diamond, then S ′ ′ is a path on 3 vertices and all the vertices of S belong to a cycle of length four of a subgraph of G that is isomorphic to a crystal. Proof. Consider a component of ⟨S ⟩ with vertex set S ′ such that there exists a vertex y ̸∈ S that has two neighbours in S ′ that do not belong to the same triangle or diamond. By Lemma 3(e) and (f), 2 < |S ′ | ≤ 6. Let x and b be the neighbours of y in S ′ that do not belong to the same triangle in G. Without loss of generality we can assume that x is a neighbour of y that belongs to the same triangle T1 , where V (T1 ) = {x, y, z }. Moreover, let T2 be a triangle in G such that V (T2 ) = {b, c , d} and c ∈ S. By Lemma 4 d ̸∈ S. If |S ′ | = 3 then xc ∈ E (G) and we obtain the desired result. Hence let us assume that |S ′ | > 3. Suppose that z ̸∈ S, then there is a path in S ′ of length 2, 3 or 4 that connects c and x. Suppose that there is a path of length 2. Let x0 be an interior vertex of this path. Since G is claw-free, x0 is adjacent either to d or to z. Thus, |N [S ′ ] − S | = 3. This implies that |N [S ′ ] − S | < |S ′ | = 4, which contradicts Corollary 1. If the path connecting c and x has length 3, let x0 , x1 be its interior vertices. In this case either the vertices x0 , x1 have a common neighbour outside S ′ or one of them is adjacent to d and the other one to z. Thus, either |N [S ′ ] − S | = 4 or |N [S ′ ] − S | = 3. In both cases |S ′ | > |N [S ′ ] − S |, which contradicts Corollary 1. Suppose that the path has length 4, let x0 , x1 , x2 be the interior vertices of this path. Hence, two vertices of {x0 , x1 , x2 } have a common neighbour outside S ′ , let e denote this vertex, and the third vertex is adjacent to d, z or e. This implies that |N [S ′ ] − S | = 4 and again we obtain the contradiction with Corollary 1. Now, assume that z ∈ S. If xc ∈ E (G) or zc ∈ E (G), then if |S ′ | = 4, then |N [S ′ ] − S | ≤ 3; otherwise |S ′ | = 5 or |S ′ | = 6 and |N [S ′ ] − S | ≤ 4. Again we obtain a contradiction to Corollary 1. Hence, xc ̸∈ E (G) and zc ̸∈ E (G). It follows that there is a path in S ′ of length 2 or 3 that connects c and x or z. Since G is claw-free in each case, similarly as above, we obtain that |S ′ | > |N [S ′ ] − S | which is in contradiction to Corollary 1 and finishes the proof. From the above lemmas it follows that each component of a graph induced by a minimal global secure set with their neighbourhood outside the minimal global secure set form one of the structures presented in Fig. 4. From Lemma 2 it follows that the component has a unique cycle or is a path. If it has a unique cycle, then, by Lemma 4, the component and its neighbourhood outside the minimal global secure set must form one of the structures presented in figure (c), (f), (i), or (l). By Lemma 3(f), if the component has two vertices, then this component with its neighbourhood outside the minimal global secure set form the structure presented in figure (a). If the component is a path on three vertices, then, by Corollary 1, it has three vertices in its neighbourhood outside the minimal global secure set. Thus, the vertices of the component and their neighbours must form the structure (b) or (d). It follows from Lemma 5 that all the remaining components are paths with
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Fig. 4. Components of a graph induced by a global secure set of cardinality equal to half of the order of a claw-free cubic graph (black vertices). The components are presented with their neighbourhood outside the secure set (white vertices). In the figure we do not present all the edges that may join the white vertices.
Fig. 5. A path of triangles obtained from 4 triangles.
4, 5 or 6 vertices. All such paths must satisfy Corollary 1. Moreover, taking into consideration Lemma 6 we obtain that these paths together with their neighbourhoods outside the minimal global secure set must form one of the structures presented in figures (e), (g), (h), (j) and (k). To obtain a minimal global secure set of cardinality equal half of the order of any claw-free cubic graph we must cover its vertices with some of the structures (a)–(l). 4. Minimum global secure sets in hamiltonian claw-free cubic graphs A path of triangles (a cycle of triangles) is a graph obtained by adding k − 1 edges (k edges) between k vertex-disjoint triangles in such a way that the obtained graph is connected and every triangle has at least one vertex of degree 2. An example of a path of triangles is presented in Fig. 5. By tG we denote the number of triangles in a graph G. When G is clear from the context, we omit the subscript G. We know that a cubic graph has an even number of vertices. Furthermore, in any {claw, diamond}-free cubic graph, except K4 , every vertex belongs to exactly one triangle. Hence, if G is a {claw, diamond}-free cubic graph, then t ≡ 0(mod 4) or t ≡ 2(mod 4). Lemma 7. Let G be a path of triangles of order n. If the number of triangles in G is a multiple of four, then γs (G) = n/2 and there exists a minimal global secure set that contains only vertices of degree 3. Proof. We divide the triangles of G into k disjoint groups of 4 consecutive triangles. In each group we choose, as it is shown in Fig. 6, 6 vertices that form a secure set and dominate all the triangles of the group. The so-obtained set P is a minimal global secure set of G of cardinality n/2 that contains only the vertices of degree 3. Lemma 8. Let G be a {claw, diamond}-free cubic graph of order n. Let G′ be a graph obtain from G by contracting all triangles. If G′ is bipartite, then γs (G) = n/2. Proof. Let (A, B) be a bipartition of G. Since G′ is 3-regular, |A| = |B|. Let S be the set of vertices of triangles in A. Then, S is a global secure set and |S | = n/2.
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Fig. 6. Four consecutive triangles with a secure set that dominates each vertex of those triangles.
Fig. 7. Indexed triangles and vertices of a hamiltonian {claw, diamond}-free cubic graph. The edges between vertices of {y1 , . . . , yt } are not presented.
Fig. 8. Illustration of subcase 1.2.1.
In the next lemma we determine the global security number of hamiltonian {claw, diamond}-free cubic graphs. Observe, that every hamiltonian {claw, diamond}-free cubic graph is obtained from a cycle of triangles by adding edges between vertices of degree 2. In the proof we assume that the triangles and vertices of the graph are indexed as it is shown in Fig. 7. Moreover, for i, j ∈ {1, . . . , t }, where i ≤ j let strij = j − i − 1 and stlij = t − j + i − 1. We say that for two triangles Ti and Tj , strij is the number of separating triangles on the right and stlij is the number of separating triangles on the left side of Ti . Furthermore, if strij ≥ r and stlij ≥ r, then we say that the triangles Ti and Tj are separated by at least r triangles. Lemma 9. If G is a hamiltonian {claw, diamond}-free cubic graph of order n, then γs (G) = n/2. Proof. Let the triangles and vertices of our graph be indexed as in Fig. 7. Case 1. Let t ≥ 10. If t ≡ 0(mod 4) then, by Lemma 7, γs (G) = n/2. Hence t ≡ 2(mod 4). Subcase 1.1. For any edge yi yj ∈ E (G) (i < j) we have i ̸≡ j(mod 2). Let G′ be a graph obtained from G by contracting all triangles. Thus, G′ is bipartite and by Lemma 8, G has a minimum global secure set of cardinality n/2. Subcase 1.2. There exists an edge yi yj ∈ E (G)(i < j) such that i ≡ j(mod 2). Subcase 1.2.1. The triangles Ti and Tj are separated by at least 3 triangles (see Fig. 8). Since i ≡ j(mod 2) and t − 2 is an even number, strij and stlij are odd. Moreover, since t − 2 ≡ 0(mod 4), without loss of generality we may assume that strij ≡ 3(mod 4) and stlij ≡ 1(mod 4). It follows that strij = 3 + 4f and stlij = 5 + 4e, where f , e ∈ {0, 1, 2, . . .}. For the convenience of the notation let us change the indices of the triangles of G in such a way that Ti is from now on T1 . Let P1 = x2 , z2 , x3 , z3 , x4 , yj , zj , xj+1 , zj+1 , xj+2 , zt −2 , xt −1 , zt −1 , xt , zt . Then, P1 is a secure set which dominates only the vertices of the triangles T1 , T2 , T3 , T4 , Tj , Tj+1 , Tj+2 , Tt , Tt −1 , Tt −2 . If there are still vertices of G that are not dominated, then they belong to 4f and 4e consecutive triangles. By Lemma 7, there is a secure set P2 that dominates these vertices and contains exactly half of them but it contains no vertex yi for i ∈ {1, . . . , t }. Thus, by Proposition 1 and Observation 1, P1 ∪ P2 form a global secure set of cardinality n/2. Subcase 1.2.2. strij = 1 or stlij = 1 Again let us change the indices of the triangles of G in such a way that the triangle Ti is from now on T1 . Thus, str1j = 1 or stl1j = 1 which implies that j = 3. Let Tr be a triangle such that y2 yr ∈ E (G). Assume that r ≡ 0(mod 2). Since we are not in Subcase 1.2.1, we have str2r = 1 or stl2r = 1. Without loss of generality we can assume that r = 4. Let P1 denote the set {y1 , z1 , x3 , y3 , y4 , z4 , x5 , z5 , x6 }. By Lemma 7 we can choose a secure set P2
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Fig. 9. Path of triangles and diamonds of type 1.
Fig. 10. Path of triangles and diamonds of type 2.
of cardinality (n − 18)/2 that dominates the vertices of the triangles T7 , . . . , Tt . Now P1 ∪ P2 is a global secure set of G and |P1 ∪ P2 | = n/2. Now suppose that r ̸≡ 0(mod 2). Observe that str3r + stl1,r = t − 4, so str3r + stl1,r ≡ 2(mod 4). Suppose that str3r ≡ 1(mod 4) and stl1,r ≡ 1(mod 4). Let P1 = {y1 , z1 , x3 , y3 } and P2 = {xr , yr , zr , xr +1 , zr −1 }. Then, P1 is a secure set which dominates the triangles T1 and T3 and P2 is a secure set which dominates the triangles T2 , Tr , Tr −1 and Tr +1 . By Lemma 7, there exists a secure set P3 such that |P3 | = (n − 18)/2, and P3 dominates the vertices that are not dominated by P1 and P2 . The union of the sets P1 , P2 and P3 is a global secure set of cardinality n/2. Hence suppose that str3r ≡ 3(mod 4) and stl1,r ≡ 3(mod 4). Let P1 = {zt −1 , xt , zt , x1 , y1 }, P2 = {z2 , y2 , yr , zr , xr +1 } and P3 = {x4 , z4 , x5 , z5 , x6 }. Then, P1 ∪ P2 ∪ P3 is a secure set which dominates the triangles Tt −1 , Tt , T1 , T2 , T3 , T4 , T5 , T6 , Tr , Tr +1 . By Lemma 7, there is a secure set P4 which dominates the remaining vertices and contains exactly half of them. As previously P1 ∪ P2 ∪ P3 ∪ P4 is a global secure set of cardinality n/2. Case 2. Let t < 10. Since every cubic graph has even order, t ∈ {2, 4, 6, 8}. If t = 4 or t = 8, then, by Lemma 7, γs (G) = n/2. If t = 2, then the vertices of any triangle of G form a global secure set of cardinality n/2. Thus let t = 6. Assume that an edge yi yj ∈ E (G)(i < j) iff i ̸≡ j(mod 2). Then, G′ obtained from G by contracting all triangles is bipartite and, by Lemma 8, G has a minimum global secure set of cardinality n/2. Hence, there is an edge yi yj ∈ E (G)(i < j) such that i ≡ j(mod 2). Without loss of generality we can assume that y1 y3 ∈ E. Let P1 = {y1 , z1 , x3 , y3 }. If y2 y5 ∈ E, then let P2 = {z4 , x5 , y5 , z5 , x6 }; otherwise if y2 y4 ∈ E, then P2 = {y4 , z4 , x5 , z5 , x6 }; otherwise y2 y6 ∈ E and P2 = {z4 , x5 , z5 , x6 , y6 }. In every case P1 ∪ P2 is a global secure set of G of cardinality n/2. This observation finishes the proof. A graph is a path of triangles and diamonds of type 1 if it can be obtained from k vertex-disjoint triangles and p vertex-disjoint diamonds by adding k + p − 1 edges between them in such a way that the obtained graph is connected, its maximum degree equals 3 and each triangle has at least one vertex of degree 2 (see Fig. 9). The path of triangles and diamonds of type 2 is obtained from a path of type 1 that starts with a triangle by adding a pendant vertex to one of the vertices of degree 2 of the first triangle of the path. Such path is presented in Fig. 10. Lemma 10. Let P be a path of triangles and diamonds of order n such that 1. 2. 3. 4.
n is even, P is of type 1 or type 2, P ends with a diamond, there is no two consecutive diamonds in P.
Then, P has a global secure set S such that |S | = n/2 and S contains only the vertices of degree three. Proof. We prove the lemma by induction on the number of vertices. Since n is even, the number of triangle of P is even if P is the path of type 1 and odd if P is the path of type 2. It is easy to see that the lemma is true for the path of type 1 without triangles and for the path of type 2 with one triangle. Let D1 , . . . , Dk be diamonds of G. Let P1 be a path of triangles leading D1 and P2 be a path of triangles between D1 and D2 . Similarly as P2 , we define paths P3 , . . . , Pk . Let l(Pi ) denote the number of triangles in Pi . If the path of type 1 starts with a diamond, then l(P1 ) = 0. In each forthcoming case we find two sets S1 and S2 such that S1 ∪ S2 is a global secure set of P of cardinality n/2 that contains only the vertices of degree 3. Case 1. P is of type 1. Subcase 1.1. l(P1 ) = 0 Then P starts with a diamond D1 . We choose two vertices of degree three of D1 to be the secure set S1 of D1 . By the induction hypothesis, P ′ = P \ D1 (P ′ is the path of type 1) has the global secure set S2 having |V (P ′ )|/2 vertices of degree 3. Hence S1 ∪ S2 is the global secure set of P we wanted. Subcase 1.2. l(P1 ) ≥ 4 Let T = T1 ∪ T2 ∪ T3 ∪ T4 . We choose the secure set S1 of T as in Fig. 11. By the induction hypothesis, P ′ = P \ T (P ′ is the path of type 1) has the global secure set S2 that contains only the vertices of degree 3 and |S2 | = |V (P ′ )|/2. Thus, S1 ∪ S2 is the required global secure set of P.
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Fig. 11. Illustration of subcase 1.2. Black vertices belong to S1 .
Fig. 12. Illustration of subcase 1.3. Black vertices belong to S1 .
Fig. 13. Illustration of subcase 1.4. Black vertices belong to S1 .
Fig. 14. Illustration of subcase 1.5. Black vertices belong to S1 .
Fig. 15. Illustration of subcase 2.1. Black vertices belong to S1 .
Fig. 16. Illustration of subcase 2.2. Black vertices belong to S1 .
Subcase 1.3. l(P1 ) = 2 We choose the secure set S1 of T = T1 ∪ T2 ∪ D1 as in Fig. 12. By the induction hypothesis, P ′ = P \ T (P ′ is the path of type 1) there is a global secure set S2 of P ′ such that |S2 | = |V (P ′ )|/2 and for all v ∈ S2 , d(v) = 3. Now S1 ∪ S2 is our global secure set of P. Subcase 1.4. l(P1 ) = 1 Since the number of triangles is even, l(P2 ) ≥ 1. We choose the secure set S1 of T = T1 ∪ T2 ∪ D1 as in Fig. 13. Since we delete two triangles P ′ = P \ T has even number of triangles, so P ′ is the path of type 1. Thus, by the induction hypothesis, P ′ has the global secure set S2 containing only the vertices of degree 3 and |S2 | = |V (P ′ )|/2. Hence S1 ∪ S2 is the required global secure set of P. Subcase 1.5. l(P1 ) = 3 Since the number of triangles is even, l(P2 ) ≥ 1. Let u, v, w be the vertices of D1 , see Fig. 14. We choose the secure set S1 of T = P1 ∪ {u, v, w} as in Fig. 14. Let P ′ = P \ T . P ′ has odd number of triangles and starts with a pendant edge joined to a triangle, so P ′ is of type 2. Again, by the induction hypothesis, P ′ has a global secure set S2 such that |S2 | = |V (P ′ )|/2 and every vertex in S2 is of degree 3. It is easy to see that S1 ∪ S2 is the global secure set of P we search for. Case 2. P is of type 2 Subcase 2.1. l(P1 ) ≥ 3. We choose the secure set S1 of T = T1 ∪ T2 ∪ T3 as in Fig. 15. P ′ = P \ T is a path of type 1 with less vertices. Thus, applying the induction hypothesis to P ′ , there exists a global secure set S2 of P ′ that contains only the vertices of degree 3 and |S2 | = |V (P ′ )|/2. Again S1 ∪ S2 is the global secure set of P we wanted to obtain. Subcase 2.2. l(P1 ) = 1 We choose the secure set S1 of T = T1 ∪ D1 as in Fig. 16. P ′ = P \ T is the path of type 1 with less vertices. Thus, by induction hypothesis, we can choose a global secure set S2 of P ′ such that every vertex of S2 is of degree 3 and |S2 | = |V (P ′ )|/2. Now, S1 ∪ S2 is the required global secure set of P. Subcase 2.3. l(P1 ) = 2 Let u, v, w be the vertices of D1 as it is presented in Fig. 17. We choose the secure set S1 of T = P1 ∪ {u, v, w} as in Fig. 17. P ′ = P \ T is the path of type 2 with less vertices. By the induction hypothesis, P ′ has the global secure set S2 with |V (P ′ )|/2 vertices of degree 3 and S1 ∪ S2 is the global secure set of P that satisfies our requirements.
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Fig. 17. Illustration of subcase 2.3. Black vertices belong to S1 .
Fig. 18. Illustration of Case 1. The edge vw and ut may belong to E (G).
Fig. 19. Illustration of Case 2 when N (y) ∩ S ′ = {x}. The edge vw and ut may belong to E (G).
Fig. 20. Illustration of Case 2 when {x, w} ⊆ N (y) ∩ S ′ . The vertices u and t may be adjacent.
Lemma 11. Let G be hamiltonian claw-free cubic graph of order n with no two adjacent diamonds. Then, γs (G) = n/2. Proof. If G has no diamonds then, by Lemma 9, γs (G) = n/2. Thus, we may assume that G has at least one diamond. It is easy to see that G has a spanning subgraph P isomorphic to the path of type 1 that satisfies the assumptions of Lemma 10. Hence, P has the global secure set S containing |V (P )|/2 vertices such that each vertex of this set is of degree 3. Since edges E (G) \ E (P ) join the vertices that have degree less than 3 in P, by Observation 1, if we add the edges of E (G) \ E (P ), then the set S is still secure. Lemma 12. Let G be a claw-free cubic graph with an induced subgraph isomorphic to two adjacent diamonds. Let G′ be a graph obtained from G by removing these two diamonds and joining the two appearing vertices of degree 2 by an edge. If γs (G′ ) = |V (G′ )|/2, then γs (G) = |V (G)|/2. Proof. If every vertex of G belongs to a diamond, then the vertices of the bridges of the diamonds form a global secure set of cardinality n/2. Thus, suppose that this is not the case. Let D1 , D2 be two adjacent diamonds of G such that D1 is also adjacent to a triangle. Let G′ be a graph obtained from G by removing D1 and D2 and joining the two appearing vertices x, y of degree 2 by an edge e. Let S ′ be a global secure set of G′ of cardinality |V (G′ )|/2. We show that S ′ can be extended to a global secure set S of G such that |S | = |V (G)|/2. Since G′ ̸= K4 , we may assume that N (x) ∩ N (y) = ∅. Let u, v, y be neighbours of x and w, t , x be neighbours of y in G′ . Without loss of generality we can assume that if there are edges joining a vertex of {u, v} with a vertex of {w, t }, then v is adjacent to w or u is adjacent to t. Consider the vertices x, y and their neighbourhood: Case 1 (see Fig. 18). x, y ̸∈ S ′ . Since S ′ is a dominating set, at least one neighbour of x and at least one neighbour of y is in S ′ . Assume that v, w ∈ S ′ . Observe, that the vertices {v, u, w, t } have in G the same neighbourhood as in G′ . Hence, the set S = S ′ ∪ {b1 , c1 , b2 , c2 } is secure.
Case 2 (see Figs. 19 and 20). x ∈ S ′ and y ̸∈ S ′ . Since, by Lemma 3(e), any component of S ′ has at least two vertices, x has ′ ′ ′ ′ a neighbour in S . Let v ∈ S . If x is the only neighbour of y that belongs to S , then S = S ∪ {d1 , a2 , c2 , d2 } (see Fig. 19). If N (y) ∩ S ′ > 1 then, without loss of generality, we can assume that w ∈ S ′ . By Lemma 6, wv ∈ E (G) and u, t ̸∈ S ′ . In this case let S = S ′ ∪ {a1 , b1 , b2 , c2 } (see Fig. 20).
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Fig. 21. Illustration of Case 3.
Fig. 22. Illustration of Case 4. The vertices u and t may be adjacent.
Fig. 23. Illustration of Case 5 when vw ∈ E (G). The vertices u and t may be adjacent.
Fig. 24. Illustration of Case 5 when N ({u, v}) ∩ S ′ = {x}. The vertices u and t may be adjacent.
Case 3 (see Fig. 21). x, y ∈ S ′ and all neighbours of x are in S ′ . Since, by Lemma 3(d), there is no kite in S ′ , x is the only neighbour of y that belongs to S ′ . Moreover, vw, ut ̸∈ E (G). Then, let S = S ′ ∪ {a1 , a2 , c2 , d2 }. Case 4 (see Fig. 22). x, y ∈ S ′ and both of the vertices have exactly one additional neighbour in S ′ . In this case, by Corollary 1, vw ̸∈ E (G). Observe, that for any attack on S ′ we can organize a defence in such a way that x do not participate in a defence of y, and vice versa. Let S = S ′ ∪ {a1 , c1 , c2 , d2 }. Case 5 (see Figs. 23 and 24). x, y ∈ S ′ and x has two neighbours outside S ′ . By Lemma 3(f), y must have additional neighbour in S ′ . Let w ∈ S ′ . Since we are not in Case 3, we may assume that t ̸∈ S ′ . Let S ′′ be a component of S ′ that contains x and y. If |N (u, v) ∩ S ′ | > 1, then, by Lemma 6, |S ′′ | = 3 and vw ∈ E (G), see Fig. 23. Let S = (S ′ − {x}) ∪ {v, b1 , c1 , c2 , d2 }. If N ({u, v}) ∩ S ′ = {x} (see Fig. 24), then in any maximal attack on S ′ , y must defend x, and w defends itself or y against t. In this case let S = S ′ ∪ {a1 , c1 , d1 , d2 }. Thus, in each case the indicated set S is a global secure set of G and |S | = |V (G)|/2.
Now, we are ready to prove the main theorem of this section. Theorem 1. If a graph G of order n is a hamiltonian claw-free cubic graph, then γs (G) = n/2. Proof. Suppose on the contrary that there is a hamiltonian claw-free cubic graph that has no global secure set that contains half of the vertices of the graph. Let G be such a graph with the minimum number of vertices. Let |V (G)| = n. If G has no two adjacent diamonds then either by Lemma 9 or by Lemma 11 we have γs (G) = n/2. Suppose that G has two adjacent diamonds. If all the vertices of G belongs to these diamonds (i.e., n = 8), then the vertices of the bridges of these diamonds form a global secure set of cardinality n/2. Otherwise, let G′ be a graph obtained from G by removing these two diamonds and joining the two appearing vertices of degree 2 by an edge. G′ is still a hamiltonian claw-free cubic graph and has less vertices than G, so by our assumption G′ has a global secure set of cardinality |V (G′ )|/2. Thus, by Lemma 12 the graph G has a global secure set of cardinality n/2, a contradiction. 5. Upper bound on the global security number of claw-free cubic graphs For the purpose of the forthcoming proof we introduce some special types of subgraphs. A (K3 , D)-subgraph G′ of a claw-free cubic graph is a subgraph such that each vertex of G′ belongs either to K3 or a diamond. A K3 -subgraph is a
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Fig. 25. Secure blocks B1 , . . . , B12 and their global super secure sets.
(K3 , D)-subgraph that contains no diamonds. A (K3 , D)-tree is a (K3 , D)-subgraph that is not K4 and has no other induced cycles than K3 . Similarly, a K3 -tree is a (K3 , D)-tree that contains no diamonds. Let G′ be a (K3 , D)-subgraph of a claw-free cubic graph G. A vertex set S ⊆ V (G′ ) is a super secure set in G′ if S is secure in G′ and in G. We say that S is a global super secure set of G′ if S is super secure and dominating in G′ . The following lemma is an easy consequence of the given definitions.
Lemma 13. Let G be a claw-free cubic graph and G1 , G2 , . . . , Gp be its vertex-disjoint (K3 , D)-subgraphs such that V (G) = V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp ). If S1 , S2 , . . . , Sp are global super secure sets in G1 , G2 , . . . , Gp , respectively, then S1 ∪ S2 ∪ · · · ∪ Sp is a global secure set in G. A secure block B is a (K3 , D)-subgraph of a claw-free cubic graph containing a global super secure set with at most 5|V (B)|/9 vertices. Lemma 14. Each graph depicted in Fig. 25 is a secure block. Moreover, the black vertices of each graph form the global super secure set of the block. Let β be a set of secure blocks presented in Fig. 25. Remark 1. Each K3 -tree with k triangles for k = 3, 4, 5 is a secure block. Lemma 15. Any K3 -tree with exactly 7 triangles, except the graph depicted in Fig. 26, can be partitioned into two secure blocks belonging to β .
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Fig. 26. A K3 -tree with exactly 7 triangles.
Consider K3 -tree T7 presented in Fig. 26. Since it contains no global super secure set with at most 5|V (T7 )|/9 vertices, it is not a secure block. Theorem 2. If G is a claw-free cubic of order n, then γs (G) ≤ 5n/9. Proof. First we show that G can be partitioned into vertex-disjoint (K3 , D)-subgraphs G1 , G2 , . . . , Gp such that V (G) = V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp ) (i.e., G1 ∪ G2 ∪ · · · ∪ Gp is a spanning subgraph of G) and each Gi , for i = 1, . . . , p, is a secure block that belongs to β . Next, from Lemma 13, we conclude that S1 ∪ S2 ∪ · · · ∪ Sp is a global secure set of G1 ∪ G2 ∪ · · · ∪ Gp , where S1 , S2 , . . . , Sp are global super secure sets of G1 , G2 , . . . , Gp , respectively. Finally, from the definition of the block and the global super secure set it will follow that γs (G) = 5n/9. Suppose on the contrary that there is no partition of G into secure blocks belonging to β such that all vertices of G are covered. Thus, let G1 , G2 , . . . , Gp be vertex-disjoint (K3 , D)-subgraphs of G such that Gi ∈ β , for i = 1, . . . , p, and V (G) \ (V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp )) is minimum. We will say that G1 , G2 , . . . , Gp is a block covering of G. Consider a subgraph G′ induced by V (G) \ (V (G1 ) ∪ V (G2 ) ∪ · · · ∪ V (Gp )). It contains no diamonds, since diamond belongs to β . It also has no connected subgraph containing more than two triangles. Thus, each component of G′ is either a triangle or contains exactly two triangles. Let t be the number of triangles in G′ . Furthermore, from all block coverings that do not cover t triangles of G, let G1 , G2 , . . . , Gp be a block covering with the minimum number of secure blocks isomorphic to B6 . First, observe the following. Claim 1. If we add a triangle to B1 and join it by an edge with a vertex of degree 2 of B1 , then we obtain a K3 -subgraph containing B7 as a spanning subgraph. Claim 2. Let i ∈ {5, 7, 8, 9}. If we add a triangle to Bi and join it by an edge with a vertex of degree 2 of Bi , then we obtain a secure block from β . Claim 3. Let i ∈ {3, 4, 10, 11}. If we add to Bi a triangle and join it by an edge with a vertex of degree 2 of Bi , then we obtain a secure block from β or we can divide the obtained graph into two secure blocks from β . Lemma 15 implies the next claim. Claim 4. If we add a triangle to B12 and join it by an edge with a vertex of degree 2 distinct from x and y, then we obtain a graph that can be divided into two secure blocks that belong to β . Let T be a triangle that is not covered, i.e., T ⊆ G′ . If a vertex of T is adjacent to a vertex of degree 2 of Bi for i ∈ {1, 3, 4, 5, 7, 8, 9, 10, 11}, then by Claim 1, Claim 2 or Claim 3 we can add T to Bi and cover T ∪ Bi by secure blocks. Hence, we obtain a block covering with a smaller number of non-covered triangles, a contradiction. If a vertex of T is adjacent to a vertex of degree 2 of B12 distinct from x and y, then, by Claim 4, we have a block covering with a smaller number of noncovered triangles, a contradiction. If a vertex of T is adjacent to either a vertex z or w or u of B6 , then we divide T ∪ B6 into two secure blocks: one isomorphic to B7 (containing T and two triangles of B6 ) and the other isomorphic to B2 (the diamond of B6 ) and the third triangle of B6 remains non-covered. Thus, we obtain a block covering with the same number of noncovered triangles however with less secure blocks isomorphic to B6 , a contradiction. From these arguments we conclude that any vertex of T is adjacent to x or y of Bi for i ∈ {2, 6, 12} or to a vertex of a non-covered triangle. Furthermore, observe the following. Claim 5. If we delete from B6 or B12 a triangle containing x and y, then we obtain a secure block from β . If T has two vertices adjacent to two distinct secure blocks D1 , D2 isomorphic to B2 , then T ∪ D1 ∪ D2 contains a spanning subgraph Gp+1 that is B3 and hence G1 , G2 , . . . , Gp , Gp+1 is a block covering with a smaller number of non-covered triangles, a contradiction. Suppose that T has exactly one vertex adjacent to a secure block D isomorphic to B2 . Let T ′ be a triangle adjacent to the other vertex of T . If T ′ is not covered (i.e., T ′ ∈ G′ ), then T ∪ T ′ ∪ D is the secure block B4 or contains a spanning subgraph that is the secure block B4 and hence we can cover both T and T ′ . Otherwise, T ′ is a triangle either in B6 or in B12 containing x and y. Thus, by Claim 5 we can delete T ′ from this secure block and T ∪ T ′ ∪ D contains a spanning subgraph that can be a new secure block. Finally suppose that each vertex of T is adjacent to a triangle. If two vertices of T
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are adjacent to one triangle T ′ , then T ∪ T ′ contains a spanning subgraph that is B1 . If each vertex of T is adjacent to a distinct triangle, say T1 , T2 , T3 , then T ∪ T1 ∪ T2 ∪ T3 contains a spanning subgraph that is B9 . Hence, in both cases, using Claim 5, we obtain a block covering with a smaller number of non-covered triangles, a contradiction. Corollary 2. If G is a claw-free cubic graph of order n < 18, then γs (G) = n/2. Acknowledgements The authors wish to thank the referees for their suggestions that helped to improve this paper. The research of Katarzyna Jesse-Józefczyk was funded by the National Science Centre on the basis of the Decision No. DEC-2011/01/N/ST6/00922. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
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