The convex domination subdivision number of a graph

Communications in Combinatorics and Optimization Vol. 1 No. 1, 2016 pp.43-56 DOI: 10.22049/CCO.2016.13544 CCO Commun. Comb. Optim. The convex domination subdivision number of a graph M. Dettlaff1 , S. Kosari2 , M. Lemańska1 , S.M. Sheikholeslami2 1 Faculty of Applied Physics and Mathematics Gdańsk University of Technology, ul. Narutowicza 11/12 80-233, Gdańsk, Poland mdettlaff;[email protected] 2 Department of Mathematics Azarbaijan Shahid Madani University, Tabriz, I.R. Iran [email protected] Received: 22 June 2016; Accepted: 8 August 2016; Available Online: 9 August 2016. Communicated by Peter Dankelmann Abstract: Let G = (V, E) be a simple graph. A set D ⊆ V is a dominating set of G if every vertex in V \ D has at least one neighbor in D. The distance dG (u, v) between two vertices u and v is the length of a shortest (u, v)-path in G. An (u, v)-path of length dG (u, v) is called an (u, v)-geodesic. A set X ⊆ V is convex in G if vertices from all (a, b)-geodesics belong to X for any two vertices a, b ∈ X. A set X is a convex dominating set if it is convex and dominating set. The convex domination number γcon (G) of a graph G equals the minimum cardinality of a convex dominating set in G. The convex domination subdivision number sdγcon (G) is the minimum number of edges that must be subdivided (each edge in G can be subdivided at most once) in order to increase the convex domination number. In this paper we initiate the study of convex domination subdivision number and we establish upper bounds for it. Keywords: convex dominating set, convex domination number, convex domination subdivision number. 2010 Mathematics Subject Classification: 05C69 1. Introduction Throughout this paper, G is a simple connected graph with vertex set V (G) and edge set E(G) (briefly V and E). For every vertex v ∈ V (G), the open neighborhood of v is the set N (v) = {u ∈ V (G) | uv ∈ E(G)} and the closed c 2016 Azarbaijan Shahid Madani University. All rights reserved. 44 The convex domination subdivision number of a graph neighborhood of v is the set N [v] = N (v) ∪ {v}. The open neighborhood of a set S ⊆ V is the set N (S) = ∪v∈S N (v), and the closed neighborhood of S is the set N [S] = N (S) ∪ S. The degree of a vertex v is degG (v) = |NG (v)|. A leaf is a vertex of degree one and a universal vertex is a vertex of degree |V (G)| − 1. We denote the number of leaves in a graph G by ℓ(G). The minimum and maximum degree of G are denoted by δ(G) and ∆(G), respectively. The private neighborhood of a vertex u with respect to a set D ⊆ V , where u ∈ D, is the set P NG [u, D] = NG [u] − NG [D − {u}]. If v ∈ P NG [u, D], then we say that v is a private neighbor of u with respect to D. For a set S of vertices of G we denote by G[S] the subgraph induced by S in G. The distance dG (u, v) between two vertices u and v in a connected graph G is the length of a shortest (u, v)–path in G. A (u, v)–path of length dG (u, v) is called (u, v)–geodesic. The greatest distance between any pair of vertices u, v in G is the diameter of G, denoted by diam(G). The girth of a graph G, denoted by g(G), is the length of its shortest cycle. The girth of a graph with no cycle is defined ∞. The edge-connectivity κ′ (G) of G is the minimum number of edges whose removal results in a disconnected graph. Clearly for every graph G, κ′ (G) ≤ δ(G). Consult [14] for the notation and terminology which are not defined here. A set A ⊂ V (G) is a dominating set of G if NG [A] = V . The domination number γ(G) is the minimum cardinality of a dominating set of G, and a dominating set of minimum cardinality is called a γ(G)–set. A set X is weakly convex in G if for any two vertices a, b ∈ X there exists an (a, b)–geodesic such that all of its vertices belong to X. A set X ⊆ V is a weakly convex dominating set if it is weakly convex and dominating. The weakly convex domination number γwcon (G) of a graph G equals the minimum cardinality of a weakly convex dominating set in G. A set X ⊂ V (G) is convex in G if vertices from all (a, b)-geodesics belong to X for any two vertices a, b ∈ X. A set X is a convex dominating set if it is convex and dominating. The convex domination number of a graph G, denoted by γcon (G), equals the minimum cardinality of a convex dominating set in G and a convex dominating set of minimum cardinality is called a γcon (G)–set. The (weakly) convex domination number was first investigated in [15], and since then has been studied by several authors [4, 16, 17]. Let us denote by Guv or Ge the graph obtained from a graph G by subdividing an edge e = uv ∈ E(G). The following result was proved in [7]. Proposition 1. The difference between γcon (G) and γcon (Guv ) and between γcon (Guv ) and γcon (G) can be arbitrarily large. It means that subdividing an edge can arbitrarily increase or decrease the convex domination number. M. Dettlaff et al. 45 The (weakly convex, convex) domination subdivision number sdγ (G) (sdγwcon (G), sdγcon (G)) of a graph G is the minimum number of edges that must be subdivided (where each edge in G can be subdivided at most once) in order to increase the (weakly convex, convex) domination number. (An edge uv ∈ E(G) is subdivided if the edge uv is deleted, but a new vertex x is added, along with two new edges ux and vx. The vertex x is called a subdivision vertex). Since the (weakly convex, convex) domination number of the graph K2 does not change when its only edge is subdivided, we will always assume that when we discuss sdγcon (G) all graphs involved are connected with ∆(G) ≥ 2. The domination subdivision number, defined in Velammal’s thesis [18], has been studied be several authors (see for instance [1, 9, 11, 13]). A similar concept related to connected domination in [10], to Roman domination in [2], to rainbow domination in [5, 8], and to 2–domination in [3]. The purpose of this paper is to initialize the study of the convex domination subdivision number sdγcon (G). Since subdividing an edge may decrease the convex domination number (Proposition 1), it may not be immediately obvious that the convex domination subdivision number is defined for all connected graphs with ∆(G) ≥ 2. We will show this shortly. We make use of the following results in this paper. Proposition 2. [15] If G is a connected graph of order n, then γwcon (G) ≤ γcon (G). Proposition 3. [4] If G 6= Kn and D is a γcon (G)-set, then every cut-vertex belongs to D. 2. Basic properties of convex domination subdivision number In this section, we investigate the basic properties of the convex domination subdivision number of a graph. Theorem 1. Let G be a connected graph on at least three vertices, let ES be a set of edges of G, let H be obtained from G by subdividing the edges in ES , and let S be the set of subdivision vertices. If DH is a convex dominating set of H, but D := DH − S is not a convex dominating set of G, then there exists a cycle of length at most 4 in G through some vertex of D. Proof. We first show that D is a dominating set of G. If v is an arbitrary vertex of G, then either (i) v ∈ D, or (ii) v ∈ NH (w) for some vertex w ∈ DH − S, or (iii) v ∈ NH (w) for some vertex w ∈ S. In case (i) or (ii) it is immediate the v is dominated by D, and in case (iii) w is the subdivision vertex 46 The convex domination subdivision number of a graph of an edge e of H that is incident with v, and that it follows by the convexity of DH that the other end of e is contained in DH and thus is D, so v is also dominated by D. Since D is a dominating set of G, it follows that D is not convex in G. Let a, b ∈ D be two vertices of G such that there exists an (a, b)-geodesic P in G containing vertices of V (G) − D. We assume that a and b have been chosen so that d(a, b) is minimal with this property. Then V (P ) ∩ D = {a, b}. (1) Let P = a, a1 , a2 , . . . , ak , where ak = b. Clearly, k ≥ 2. Now P corresponds to an (a, b)-path PH in H. None of the edges of P , except possibly aa1 and ak−1 b, are in ES , since otherwise P would contain vertices of D in its interior, contradicting (1). Since DH is convex in H, it follows that PH is not an (a, b)geodesic in H. Hence PH is longer than P , so at least one of the edges of P , without loss of generality aa1 , is in ES . Let u be the subdivision vertex of aa1 . Now a1 is dominated in H by some vertex b1 ∈ DH (possibly b1 = b). We claim that b1 6= u. Suppose, to the contrary, that b1 = u. Since DH is convex and since a1 6∈ DH , we conclude that every (u, b)-geodesic in H passing through a. Hence, uPH b is a (u, b)-path in H of length at most k + 1 which is not a ′ ′ (u, b)-geodesic. Let PH be a (u, b)-geodesic in H. Clearly, the length of PH is ′ ′ at most k. Now aPH b corresponds to an (a, b)-path P in G of length at most k − 1 which contradicts d(a, b) = k. Thus b1 6= u. Since b1 , a1 , u, a is a path joining two vertices in DH that contains vertices not in DH , it follows by the convexity of DH that there exists a (b1 , a)-path Q in H of length at most two. The paths a, u, a1 , b1 and Q form a cycle of length at most five in H, which corresponds to a cycle of length at most four in G containing a, as desired. A closer look at the proof of Theorem 1 leads to the next result. Corollary 1. Let G be a connected graph on at least three vertices, let ES be a set of edges of G, let H be obtained from G by subdividing the edges in ES , and let S be the set of subdivision vertices. If DH is a convex dominating set of H, then D := DH − S is a dominating set of G. Theorem 2. For any connected graph G of order n ≥ 3 and size m, sdγcon (G) ≤ m. Proof. Let H be the graph obtained from G by subdividing all edges of G, let T be the set of all subdivision vertices and let DH be a convex dominating set of H. Clearly, H is a bipartite graph with partite sets V (G) and T . It follows that γcon (H) ≥ 2. Since for any two vertices x, y ∈ V (G), every (x, y)-geodesic in H contains at least one subdivision vertex, we conclude that DH ∩T 6= ∅. By M. Dettlaff et al. 47 Corollary 1, D := DH − T is a dominating set of G. Now, let a, b ∈ D be two arbitrary vertices. If P is an (a, b)-geodesic in G, then clearly P corresponds to an (a, b)-geodesic PH in H. Since DH is convex in H, we deduce that V (P ) ⊆ D and so D is convex. Thus D is a convex dominating set of G of size smaller than of γcon (H). This yields sdγcon (G) ≤ m and the proof is completed. A consequence of Theorem 2 is that sdγcon (G) is defined for every connected graph G of order n ≥ 3. Given S, T ⊆ V (G), we write [S, T ] for the set of edges having one end–point in S and the other in T . An edge cut is an edge set of the form [S, S] , where S is a nonempty proper subset of V (G) and S denotes V (G) − S. Theorem 3. For any connected triangle-free graph G of order n ≥ 3, sdγcon (G) ≤ κ′ (G). Proof. Assume ET = [S, S] is an edge cut of G of size κ′ (G), G1 and G2 are the components of G − ET , and H is the graph obtained from G by subdividing the edges of ET . Let T be the set of all subdivision vertices and let DH be a convex dominating set of H and Di = DH ∩ V (Gi ) for i = 1, 2. If DH ∩ T = ∅, then Di 6= ∅ for i = 1, 2, and DH = D1 ∪ D2 . Now for vertices x1 ∈ D1 and x2 ∈ D2 , any (x1 , x2 )-geodesic path intersect T implying that DH ∩T 6= ∅ which leads to a contradiction. Therefore DH ∩ T 6= ∅. By Corollary 1, D := DH − T is a dominating set of G. Now we show that D is convex in G. Assume, to the contrary, that D is not a convex set in G. Let a, b ∈ D be two vertices of G such that there exists an (a, b)-geodesic P in G containing vertices of V (G)−D. We suppose that a and b have been chosen so that d(a, b) is minimal with this property. Then V (P ) ∩ D = {a, b}. (2) Let P = a, a1 , a2 , . . . , ak , where ak = b. Clearly, k ≥ 2. Now P corresponds to an (a, b)-path PH in H. None of the edges of P , except possibly aa1 and ak−1 b, are in ET , since otherwise P would contain vertices of D in its interior, contradicting (2). Since DH is convex in H, we conclude that PH is not an (a, b)-geodesic in H. Hence PH is longer than P , so at least one of the edges of P , without loss of generality aa1 , is in ET . Assume that a ∈ V (G1 ) and a1 ∈ V (G2 ). Let u be the subdivision vertex of aa1 . Now a1 is dominated in H by some vertex b1 ∈ DH (possibly b1 = b). As in the proof of Theorem 1, we have b1 6= u. Since b1 , a1 , u, a is a path joining two vertices in DH that contains vertices not in DH , it follows by the convexity of DH that there exists a (b1 , a)-path Q in H of length at most two. Since ET is an edge-cut of G, we deduce that the (b1 , a)-path Q in H has length two. Let Q = b1 ya. If b1 ∈ D, then b1 ∈ V (G2 ) and y is the subdivision vertex of the edge b1 a and this implies 48 The convex domination subdivision number of a graph that aa1 b1 is a triangle in G, a contradiction. If b1 ∈ T , then y ∈ V (G1 ) and so aa1 y is a triangle in G, a contradiction again. Thus D is a convex set in G and hence D is a convex dominating set of G of size smaller than of γcon (H). This yields sdγcon (G) ≤ κ′ (G) and the proof is completed. A closer look at the proof of Theorem 3 shows that if ET = [S, S] is an edgecut of size one, then b1 , a1 , u, a is the unique (b1 , a)-geodesic in H which is impossible. Hence we obtain the next result. Corollary 2. For any connected graph G of order at least 3 with a cut edge, sdγcon (G) = 1. The next results are immediate consequences of Theorem 3. Corollary 3. For any connected triangle-free graph G of order n ≥ 3, sdγcon (G) ≤ δ(G). Corollary 4. For any connected triangle-free graph G with a cut vertex v, sdγcon (G) ≤ ⌊deg(v)/2⌋. Theorem 4. If G is a connected graph of order n with g(G) ≥ 5, then sdγcon (G) = 1. In particular, for every edge e ∈ E(G), γcon (Ge ) > γcon (G). Proof. Let e = u1 u2 be an arbitrary edge of G. If e is a cut edge, then clearly γcon (Ge ) > γcon (G). Let C = (u1 u2 . . . uk ) be a cycle containing e. Assume Ge is obtained from G by subdividing the edge e with subdivision vertex w and let D be a γcon (Ge )–set. First let {u1 , u2 } ⊆ D. Then we have w ∈ D. Since g(G) ≥ 5, we conclude from Theorem 1 that D − {w} is a convex dominating set of G of size smaller than of γcon (Ge ) as desired. Now, let {u1 , u2 } 6⊆ D. Assume, without loss of generality, that u2 6∈ D. To dominate w, we must have u1 ∈ D. If w ∈ D, then as above D − {w} is a convex dominating set of G of size smaller than of γcon (Ge ), as desired. Suppose that w 6∈ D. To dominate u2 , we must D ∩ NG (u2 ) 6= ∅. Suppose v ∈ D ∩ NG (u2 ). Since g(G) ≥ 5, we deduce that dGe (u1 , v) = 3. Since D is a convex dominating set for Ge , we must have u1 , w, u2 , v ∈ D, a contradiction. It follows that γcon (Ge ) > γcon (G) and hence sdγcon (G) = 1. This completes the proof. Corollary 5. For any connected graph G of order n ≥ 6 with g(G) = 4, sdγcon (G) ≤ ⌊n/2⌋. M. Dettlaff et al. 49 Proof. Let C = (v1 v2 v3 v4 ) be a cycle of G and let without loss of generality that deg(v1 ) = min{deg(vi ) | 1 ≤ i ≤ 4}. Since g(G) = 4, N (v1 ) ∩ N (v2 ) = ∅. 2) ≤ n2 and the result follows It follows that δ(G) ≤ deg(v1 ) ≤ deg(v1 )+deg(v 2 from Corollary 3. It could be of ample interest if one could find the bound for sdγcon (G) posed in the following open problems: Problem 1. Let G be a connected graph of girth four. Is there a constant c such that sdγcon (G) ≤ c. Problem 2. Let G be a connected graph of girth three. Is there a constant c such that sdγcon (G) ≤ c. Let α′ (G) be the maximum number of edges in a matching in G. Proposition 4. α′ (G) < n−1 , 2 Let G be a connected triangle-free graph of order n ≥ 3. If then sdγcon (G) ≤ α′ (G). Proof. Let M = {u1 v1 , . . . , uα′ vα′ } be a maximum matching of G and let X be the independent set of M -unsaturated vertices. Since α′ (G) < n−1 2 , we have |X| ≥ 2. Assume y and z are vertices of X such that deg(y) ≤ deg(z). If yui ∈ E(G), then since the matching M is maximum, zvi ∈ / E(G). Therefore, for all i ∈ {1, 2, . . . , α′ } there are at most two edges between the sets {ui , vi } and {y, z}. So 2 deg(y) ≤ deg(y) + deg(z) ≤ 2α′ and the result follows by Corollary 3. Proposition 5. Let G be a connected graph of order n ≥ 3. If α′ (G) > γcon (G), then sdγcon (G) ≤ α′ (G). Proof. Let M = {u1 v1 , . . . , uα′ vα′ } be a maximum matching of G and let G′ be obtained by subdividing every edge of M . Each convex dominating set of G′ has order at least |M |. Hence γcon (G′ ) > γcon (G) and thus sdγcon (G) ≤ α′ (G). 3. Graphs with small convex domination subdivision number In this section, we consider graphs with small convex domination subdivision number. Proposition 6. Let G be a connected graph of order n ≥ 3. If G satisfies one of the following properties: 50 The convex domination subdivision number of a graph (i) γcon (G) = 1; (ii) γcon (G) = 2 and G contains a γcon (G)–set {a, b} such that N (a) ∩ N (b) = ∅; then sdγcon (G) = 1. Proof. (i) Since n ≥ 3, the graph Ge obtained by subdividing any edge e of G has no universal vertex. Hence γcon (Ge ) > 1 = γcon (G) and so sdγcon (G) = 1. (ii) Let G′ be the graph obtained from G by subdividing the edge ab with subdivision vertex x. Obviously every convex dominating set of G′ contains at least one of a, b, say a, and either two vertices in N (a) ∪ N (b), or x and b. Hence γcon (G′ ) ≥ 3 > γcon (G). Proposition 7. For any connected graph G of order n ≥ 3 with γcon (G) = 2, sdγcon (G) ≤ 2. Proof. Since γcon (G) = 2, ∆(G) ≤ n − 2. Let S = {u, v} be a γcon (G)–set. Assume u′ is a private neighbor of u with respect to S and v ′ is a private neighbor of v with respect to S. Let G′ be the graph obtained from G by subdividing the edges uu′ , vv ′ with subdivision vertices x and y, respectively, and let D be a γcon (G′ )-set. We show that |D| ≥ 3 which implies sdγcon (G) ≤ 2. Suppose to the contrary that |D| ≤ 2. To dominate x, y, we must have |D ∩ {u, u′ }| ≥ 1 and |D ∩ {v, v ′ }| ≥ 1. Since |D| ≤ 2, we have |D ∩ {u, u′ }| = 1 and |D ∩ {v, v ′ }| = 1. Since G[D] is connected and since uv ′ 6∈ E(G) and vu′ 6∈ E(G), we deduce that either D = {u, v} or D = {u′ , v ′ }. In each case, D is not a dominating set of G′ which is a contradiction. Hence γcon (G′ ) = |D| ≥ 3 > γcon (G) and the proof is complete. Proposition 8. Let k ≥ 2 be an integer. For the complete k-partite graph G = Kp1 ,p2 ,...pk with 2 ≤ p1 ≤ p2 ≤ · · · ≤ pk , sdγcon (G) =  1 2 if k = 2 otherwise. Proof. It is clear that any two adjacent vertices form a minimum convex dominating set of G which implies γcon (G) = 2. If k = 2, the result follows from Proposition 6 (ii). Let k ≥ 3 and let V1 , V2 , . . . , Vk be the partite sets of G. By Proposition 7, sdγcon (G) ≤ 2. For any edge e = ab, where a ∈ Vi , b ∈ Vj (i 6= j), the set {a, v} for each v ∈ Vk (k 6∈ {i, j}) forms a minimum convex dominating set of G. It follows that sdγcon (G) ≥ 2. Thus sdγcon (G) = 2 and the proof is complete. M. Dettlaff et al. 51 Proposition 8 shows that the bound in Proposition 7 is sharp. Proposition 9. Let G be a connected graph of order n ≥ 3 with γcon (G) = 3 or 4. If G has a triangle, then sdγcon (G) ≤ 3. Proof. Assume uvw is a triangle in G and let H be the graph obtained from G by subdividing the edges uv, uw, vw by subdivision vertices x, y, z, respectively. Let DH be a γcon (H)-set. To dominate the subdivision vertices, we must have |DH ∩ {u, v, w}| ≥ 2. Assume without loss of generality that u, v ∈ D. Since DH is convex, we must have x ∈ DH . Hence {u, v, x} ⊆ DH . We show that |DH | ≥ 5 which implies sdγcon (G) ≤ 3. Suppose to the contrary that |DH | ≤ 4. To dominate w, we must have DH ∩NH [w] 6= ∅. Assume a ∈ DH ∩NH [w]. Then {u, v, x, a} ⊆ DH . If γcon (G) = 3, then we deuce that sdγcon (G) ≤ 3 as desired. Let γcon (G) = 4. If a = y (the case a = z is similar), then we deduce from dH (a, v) = 3 that w, z ∈ DH which is a contradiction. Assume that a 6∈ {y, z}. If a = w, then we must have y, z ∈ DH which leads to a contradiction again. Hence a 6= w. It follows from |DH | = 4 that au, av ∈ E(G). Hence uva is a triangle in G. It follows that D := DH −{x} = {u, v, a} is a convex dominating set of G contradicting γcon (G) = 4. Thus |DH | ≥ 5 and so sdγcon (G) ≤ 3. This completes the proof. Next we show that the bound in Proposition 9 is sharp when γcon (G) = 4. The following graph was introduced by Haynes et al. in [12]. Let X = {1, 2, . . . , 3(k − 1)} and let Y = {Y ⊂
X : |Y | = k}. Thus, Y . Let Gk be the graph with consists of all k–subsets of X, and so |Y| = 3(k−1) k vertex set X ∪ Y and with edge set constructed as follows: add an edge joining every two distinct vertices of X and for each x ∈ X and Y ∈ Y, add an edge joining x and
Y if and only if x ∈ Y . Then, Gk is a connected graph of order + 3(k − 1). The set X induces a clique in Gk , while the set Y is an n = 3(k−1) k independent set and each vertex of Y has degree k in Gk . Therefore δ(Gk ) = k. Dettlaff et al. [6] proved that γwcon (Gk ) = 2(k − 1). Proposition 10. For any integer k ≥ 3, γcon (Gk ) = 2(k − 1). Proof. It is easy to see that any subset of X of cardinality 2(k − 1) is a convex dominating set of G, and so γcon (Gk ) ≤ 2(k − 1). It follows from Proposition 2 that γcon (Gk ) = γwcon (Gk ) = 2(k − 1) and the proof is complete. Proposition 11. For any integer k ≥ 3, sdγcon (Gk ) ≥ 3. 52 The convex domination subdivision number of a graph Proof. Assume e1 , e2 are two arbitrary edges of Gk and let G′k be the graph obtained from Gk by subdividing the edges e1 , e2 . We show that γcon (G′k ) ≤ γcon (Gk ) = 2(k − 1). Assume ei = ui vi for i = 1, 2. Since every edge of G is incident with at least one vertex of X, we may assume that ui ∈ X for i = 1, 2. If vi ∈ Y for i = 1, 2, then let wi be a neighbor of vi in X − {u1 , u2 }. If v1 ∈ Y and v2 ∈ X (the case v2 ∈ Y and v1 ∈ X is similar), then let w2 = v2 and w1 be a neighbor of v1 in X − {u1 , u2 }. If vi ∈ X for i = 1, 2, then let wi be any vertex of X − {ui , vi | i = 1, 2}. Assume that D = {u1 , u2 , w1 , w2 }. Then |D| ≤ 4. Now extend D to a set D′ of size 2(k − 1) by adding 2(k − 1) − |D| vertices of X − {ui , vi | i = 1, 2}. Clearly D′ is a convex dominating set of G′k , and so γcon (G′k ) ≤ 2(k − 1) = γcon (Gk ). This implies that sdγcon (Gk ) ≥ 3 and the proof is complete. In the case k = 3, Propositions 10 and 11 demonstrate that the bound of Proposition 9 is sharp when γcon (G) = 4. Proposition 12. For every connected triangle-free graph G with γcon (G) = 3, sdγcon (G) ≤ 2. Proof. Let G be triangle-free and let D = {u1 , u2 , u} be a γcon (G)-set. Since G[D] is connected and since G is triangle-free, G[D] is a path. Suppose G[D] = u1 uu2 . It follows from convexity of D that NG (u1 ) ∩ NG (u2 ) = {u}. (3) If ui has no private neighbor with respect to D for some i, then clearly D −{ui } is a convex dominating set of G which is a contradiction. Hence, assume ui has a private neighbor, say vi , with respect to D, for i = 1, 2. It follows that u 6∈ NG (v1 ) ∪ NG (v2 ). (4) Let G′ be the graph obtained from G by subdividing the edges u1 v1 , u2 v2 with vertices x1 , x2 , respectively, and let D′ be a γcon (G′ )-set. We show that |D′ | ≥ 4. Suppose to the contrary that |D′ | ≤ 3. To dominate xi , we must have D′ ∩ {ui , vi } 6= ∅ for i = 1, 2. If {ui , vi } ⊆ D′ for some i, then xi ∈ D′ implying that |D′ | ≥ 4, a contradiction. Let |{ui , vi } ∩ D′ | = 1 for each i. If u1 , u2 ∈ D′ , then clearly u ∈ D′ and so D′ = {u, u1 , u2 }. But then v1 is not dominated by D′ since v1 is a private neighbor of u1 with respect to D in G, a contradiction. If u1 , v2 ∈ D′ (the case u2 , v1 ∈ D′ is similar), then u1 and v2 must have a common neighbor, say w, such that D′ = {u1 , v2 , w}. Now to dominate u2 , we must have wu2 ∈ E(G) which is a contradiction because G is M. Dettlaff et al. 53 triangle-free. Let v1 , v2 ∈ D′ and let D′ = {v1 , v2 , w}. By (4), we have w 6= u. On the other hand, w 6= ui for some i, say i = 1. Since D′ is a dominating set, we must have wu, wu1 ∈ E(G) which leads to a contradiction because G is triangle-free. This completes the proof. Theorem 5. For every connected graph G with γcon (G) = 4, sdγcon (G) ≤ 3. Proof. If G has a triangle, then the result follows by Proposition 9. Henceforth, let G be triangle-free. Let D = {u1 , u2 , u3 , u4 } be a γcon (G)-set such that the size of G[D] is as large as possible. Since the induced subgraph G[D] is connected, we consider three cases. Case 1. G[D] = C4 = (u1 , u2 , u3 , u4 ). Since D is a convex set, we deduce that NG (u1 ) ∩ NG (u3 ) = {u2 , u4 } and NG (u2 ) ∩ NG (u4 ) = {u1 , u3 }. Let G′ be the graph obtained from G by subdividing the edges u1 u2 , u2 u3 , u3 u4 with subdivision vertices x1 , x2 , x3 , respectively. Suppose D1 is a γcon (G′ )-set. To dominate x1 , we must have u1 ∈ D1 or u2 ∈ D1 , to dominate x2 , u2 ∈ D1 or u3 ∈ D1 , and to dominate x3 , u3 ∈ D1 or u4 ∈ D1 . Consider the following subcases. Subcase 1.1. u1 , u3 ∈ D1 (the case u2 , u4 ∈ D1 is similar). Since NG (u1 ) ∩ NG (u3 ) = {u2 , u4 } and P = u1 u4 x3 u3 is a path of length 3 in G′ , we deduce that dG′ (u1 , u3 ) = 3. This implies that u1 , u4 , x3 , u3 ∈ D1 . Now to dominate u2 , we must have NG′ (u2 ) ∩ D1 6= ∅. Since G is triangle-free, we deduce that |D1 | ≥ 5 as desired. Subcase 1.2. u2 , u3 ∈ D1 . Since D1 is a convex set, we have x2 ∈ D1 . If x1 , x3 ∈ D1 , then |D1 | ≥ 5 as desired. Let without loss of generality that x1 6∈ D1 . This implies that u1 6∈ D1 . To dominate u1 , we must have NG′ (u1 ) ∩ D1 6= ∅. Let w ∈ NG′ (u1 ) ∩ D1 . Since G is triangle-free and since NG (u1 ) ∩ NG (u3 ) = {u2 , u4 }, we have dG′ (w, {u2 , x2 , u3 }) ≥ 2. It follows from the convexity of D1 that |D1 | ≥ 5 and we are done. Case 2. G[D] = P4 = u1 u2 u3 u4 . Then u1 u4 6∈ E(G). It follows from the convexity of D that dG (u1 , u4 ) = 3, NG (u1 )∩NG (u3 ) = {u2 } and NG (u2 )∩NG (u4 ) = {u3 }. Suppose G′ is the graph obtained from G by subdividing the edges u1 u2 , u2 u3 , u3 u4 with subdivision vertices x1 , x2 , x3 , respectively. Assume D2 is a γcon (G′ )-set. It now will be shown that |D2 | ≥ 5. To dominate x2 , we must have D2 ∩{u2 , u3 } = 6 ∅. Assume without loss of generality that u2 ∈ D2 . Now to dominate x3 , we must have D2 ∩ {u3 , u4 } = 6 ∅. Consider two subcases. Subcase 2.1. u3 ∈ D2 . Since D2 is a convex set, x2 ∈ D2 . If x1 , x3 ∈ D2 , then |D2 | ≥ 5 and we 54 The convex domination subdivision number of a graph are done. Assume without loss of generality that x1 6∈ D2 . Now to dominate u1 , we must have D2 ∩ NG′ (u1 ) 6= ∅. Let w ∈ D2 ∩ NG′ (u1 ). Therefore {u2 , x2 , u3 , w} ⊆ D2 . Since D2 is a convex set and since G is triangle-free, u4 is not dominated by the set {u2 , x2 , u3 , w} implying that |D2 | ≥ 5 as desired. Subcase 2.2. u4 ∈ D2 . Since G is triangle-free and NG (u2 ) ∩ NG (u4 ) = {u3 }, we have dG′ (u2 , u4 ) ≥ 3. If dG′ (u2 , u4 ) ≥ 4, then it follows from convexity of D2 that |D2 | ≥ 5 and we are done. Let dG′ (u2 , u4 ) = 3 and let Q = u2 w1 w2 u4 is a path with length 3 in G′ . Then {u2 , w1 , w2 , u4 } ⊆ D2 . Since u1 u4 6∈ E(G), dG (u1 , u4 ) = 3 and G is triangle-free, we deduce that u1 is not dominated by {u2 , w1 , w2 , u4 } implying that |D2 | ≥ 5 as desired. Case 3. G[D] = K1,3 . Assume u is the center of G[D] = K1,3 and u1 , u2 , u3 are leaves adjacent to u. If ui has no private neighbor with respect to D for some i, then clearly D − {ui } is a convex dominating set of G which is a contradiction. Henceforth, assume ui has a private neighbor with respect to D, say vi , for each i. Let G′ be the graph obtained from G by subdividing the edges u1 v1 , u2 v2 , u3 v3 with vertices x1 , x2 , x3 , respectively, and let D3 be a γcon (G′ )-set. We show that |D3 | ≥ 5. Assume, to the contrary, that |D3 | ≤ 4. To dominate xi , we must have D3 ∩ {ui , vi } 6= ∅ for each i. If {ui , vi } ⊆ D3 for some i, then xi ∈ D3 implying that |D3 | ≥ 5, a contradiction. Let |{ui , vi } ∩ D3 | = 1 for each i. Now we consider the following subcases. Subcase 3.1. ui , uj ∈ D3 . Assume without loss of generality that u1 , u2 ∈ D3 . Since d(u1 , u2 ) = 2, we must have u ∈ D3 because D3 is a convex set. If u3 ∈ D3 , then D3 = D = {u, u1 , u2 , u3 } and v1 is not dominated by D3 since v1 is a private neighbor of u1 with respect to D, a contradiction. Let v3 ∈ D3 . Then D3 = {u, u1 , u2 , v3 }. Since v3 is a private neighbor of u3 with respect to D, we deduce that dG′ (v3 , {u, u1 , u2 }) ≥ dG (v3 , {u, u1 , u2 }) ≥ 2. Hence, v3 is an isolated vertex in G′ [D3 ] which contradicts the connectedness of G′ [D3 ]. Subcase 3.2. ui , vj , vk ∈ D3 where {j, k} = {1, 2, 3} − {i}. Assume without loss of generality that i = 1. Since v2 is a private neighbor of u2 with respect to D, dG′ (u1 , v2 ) ≥ dG (u1 , v2 ) ≥ 2. First let dG′ (u1 , v2 ) = 2. Assume w ∈ N (u1 )∩N (v2 ). Then D3 = {u1 , w, v1 , v2 } and w must dominate u2 which leads to a contradiction because G is triangle-free. Now let dG′ (u1 , v2 ) ≥ 3. Similarly, we may assume dG′ (u1 , v3 ) ≥ 3. It follows from the convexity of D3 that |D3 | ≥ 5, a contradiction again. Subcase 3.3. v1 , v2 , v3 ∈ D3 . Let D3 = {v1 , v2 , v3 , w}. Then w must be adjacent to ui for each i. Since G′ [D3 ] is connected, we may assume that wv1 ∈ E(G). 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