Some Upper Bounds for Signed Star Domination Number of Graphs. S. Akbari, A. Norouzi-Fard, A. Rezaei, R. Rotabi, S. Sabour.

Some Upper Bounds for Signed Star Domination Number of Graphs S. Akbari, A. Norouzi-Fard, A. Rezaei, R. Rotabi, S. Sabour Abstract Let G be a graph with the vertex set V (G) and edge set E(G). A function f : E(G) { 1, +1} is said to be a signed star dominating function of G if e E G (v) f(e) 1, for every v V (G), where E G (v) = {uv E(G) u V (G)}. The minimum of the values of e E(G) f(e), taken over all signed star dominating functions f on G is called the signed star domination number of G and is denoted by γ ss (G). In this paper we show that if G is a connected graph of order n containing a 2-factor and G K 5, then γ ss (G) 7n 6. Among other results it is shown that if G is connected and α(g) 2, then γ ss (G) n + 1, where α(g) is the independence number of G. 1. Introduction Let G be a graph with the vertex set V (G) and the edge set E(G). The order and size of G denote the number of vertices and the number of edges of G, respectively. We denote the minimum degree of the vertices of G by δ(g). For every subset U V (G), we denote the induce subgraph on U by U. Let N(v) denote the neighbour set of v. The complete graph and the cycle of order n, are denoted by K n and C n, respectively. A Hamilton cycle is a cycle in a graph that visits each vertex exactly once. A walk is a sequence v 0, e 1, v 1,..., e k, v k of vertices and edges such that, for 1 i k, the edge Key Words: Signed Star Domination Number, 2-factor, Independence Number 2010 Mathematics Subject Classification: 05C69, 05C78 1

e i has endpoints v i 1 and v i. A trail is a walk with no repeated edge. An even trail is a trail with even number of edges. An Eulerian trail is a trail in a graph which visits every edge exactly once. Similarly, an Eulerian circuit is an Eulerian trail which starts and ends on the same vertex. A 2-factor is a disjoint union of finitely many cycles that cover all vertices of G. A graph whose each vertex has even degree is called even graph. The independence number for a given graph G is the size of maximum independent set of G and is denoted by α(g). Here κ(g) denotes the vertex connectivity of the graph G. We define the value of the edge e by f(e), for every e E(G) and S G (v) = u N(v) f(uv), for every v V (G). Also ω(g) stands for e E(G) f(e). A function f : E(G) { 1, 1}, is said to be a signed star dominating function of G if S G (v) 1, for every v V (G). The minimum value of ω(g), taken over all signed star dominating functions f on G, is called the signed star domination number of G and is denoted by γ ss (G). The signed star dominating function (SSDF) was introduced by Xu in [?]. Several authors have been studied the signed star dominating function (SSDF) and obtained the following bounds. It was proved in [?], for all graphs G of order n without isolated vertices, γ ss (G) n 2. Furthermore, for every graph G of order n 4, γ ss (G) 2n 4, and this bound is sharp. In this paper first we prove that for every graph G of order n, if δ(g) n 2, then γ ss (G) n + 1. Then we show that for a connected graph G (G K 5 ) if G has a 2-factor, then γ ss (G) 7n 6 and the upper bound is sharp. Finally for a connected graph G of order n, and α(g) 2, we prove that γ ss (G) n + 1. The following interesting theorem was proved in [?]. Theorem 1. For every natural number n, the following holds: n 2 if n 0 (mod 2) γ ss (K n ) = n + 1 if n 1 (mod 4) n if n 3 (mod 4). 2. An upper bound for the signed star domination number 2

of graphs containing a 2-factor In this section we would like to provide an upper bound for the signed star domination number of graphs containing a 2-factor. Before stating the proof of the main result the following lemma is needed. Lemma 1. Let G be a connected graph. Then there exists f : E(G) { 1, +1} such that for every v V (G), S G (v) 1 and ω(g) 1. Furthermore, if there is no function f with the mentioned property and ω(g) 0, then G is an even graph of odd size. Therefore, if G has a vertex of odd degree, then there exists a function f with the desired property and ω(g) 0. Proof. First add a new vertex u to G and join u to all vertices of odd degree in G. Call this graph by G. Clearly, every vertex of G has an even degree. Therefore, G contains an Eulerian circuit C. If G has at least one vertex of odd degree, then one can assume that u is the starting point of C. Otherwise, choose an arbitrary vertex of G as the starting point. Assign alternatively +1, 1 to all edges of C, starting by +1. Now, for every v V (G), S G (v) = 0. After removing u in G we have S G (v) 1, for every v V (G). Two cases can be considered: Case (i) C has an even number of edges. In this case the sum of all values of edges of C is zero. Since S G (u) = 0, after removing u, we find that ω(g) = 0. Case (ii) C has an odd number of edges. Hence, if G is an even graph, then ω(g) = +1. Otherwise, ω(g) = 1, since the value of the first edge of C is started by +1. Hence for each case we have, ω(g) 1 and the proof is complete. Theorem 2. Let G be a graph of order n such that δ(g) n 2. Then γ ss(g) n + 1. Proof. If n 4 the proof of the theorem is straightforward. Thus, assume that n 5. By Theorem 10.1.1 of [?], G has a Hamilton cycle C. 3

Let G = G \ E(C). Let C 1,..., C r be the connected components of G. By Lemma??, there is an edge assignment f : E(G) { 1, +1} such that for every C i, ω(c i ) 1, and for each v V (C i ), S Ci (v) 1 and ω(c i ) has minimum possible value. Now, we extend f to E(G) by assigning +1 to every edge of C. Let T = {C i ω(c i ) = 1}. If T 1, then we are done. Otherwise, We know that δ(g ) n 2 2. Thus the order of each component is at least n 2 1. Then we have: r( n 1) n. 2 Since we have n 5, it is easily seen that: r 2 which implies that T 2, except when n = 6 and G is a union of one Hamilton cycle and a perfect matching which are edge disjoint. In this case if we assign +1 to all edges of the Hamilton cycle and 1 to the perfect matching edges, the Theorem 2 is proved. Thus assume T = 2. By Lemma??, every vertex of C i has even degree and also C i has odd size, for every C i T. Now, let e = uv E(C), u V (C 1 ) and v V (C 2 ). One can change f(e) from +1 to 1. Now, consider two Eulerian circuits for C 1 and C 2 and reassign the values of the edges of these two Eulerian circuits by +1 and 1 consecutively, starting from u and v with value +1. Hence, S G (z) 1, for every z V (G). Moreover, we find that ω(g) n 2 + 2 = n and the proof is complete. Lemma 2. For every connected even graph G, one can remove all edges of finitely many even closed trails( even trail: a trail with even number of edges) such that the remaining graph is a union of some isolated vertices and at most one odd cycle. Proof. If G has an even size, then it is an even closed trail and in this case the lemma is proved. Otherwise, let T be an Eulerian circuit of odd size in G. Traverse T from an arbitrary vertex w. Let C be the first cycle in our traverse. If C is an odd cycle, then we can remove C from G such that the resultant graph is an even closed trail. In this case, we are done. Otherwise, if C is an even cycle, remove C from G. The resultant graph has at most one component which its order is at least 2 and every vertex has an even degree. By induction on E(G) this component can be partitioned into at most one odd cycle and some even closed trails. Hence, the proof is complete. 4

Theorem 3. Let G K 5 be a connected graph of order n. If G has a 2-factor, then γ ss (G) 7n 6. Moreover, the upper bound is sharp. Proof. If V (G) 5, the proof is straightforward. Thus, assume that V (G) 6. Let F be a 2-factor of G. Assign +1 to all edges of F and let G 1 = G \ F. First, consider those components of G 1 which have at least a vertex of odd degree. Let C be one of these components. By Lemma??, one can assign +1 and 1 to the edges of C such that for every v V (C), S C (v) 1 and ω(c) 0. By considering the values of the edges of F we conclude that S G (v) +1, for every v V (C). Now, consider all even connected components of G 1, say Z 1,..., Z k. By Lemma??, one can remove a set of even closed trails of each Z i, say T i, such that each even component of the remaining graph, G 2, is either an odd cycle or an isolated vertex. For every T T i, 1 i k, we consecutively assign +1 and 1 to the edges of T. This implies that ω(t ) = 0. Now, we would like to determine the values of edges of the remaining odd cycles in G 2. Let C 1,..., C l be these odd cycles. For every C i with V (C i ) 5, we apply the following algorithm: If there are four vertices a, b, c and d such that ac, cd, bd E(C i ) and ab / F, then join a to b and remove ac, cd and db. Call the new graph by G 2. We claim that for every edge assignment of G 2, there is an edge assignment for G 2 where ω(g 2 ) = ω(g 2 ) and S G 2 (v) = S G 2 (v), for every v V (G 2 ). To see this, let f be an edge assignment for G 2. Then the following hold: 1. If f(ab) = 1 in G 2, then define f(ac) = f(bd) = 1 and f(cd) = 1 in G 2. 2. If f(ab) = 1 in G 2, then define f(ac) = f(bd) = 1 and f(cd) = 1 in G 2. For every remaining edge e G 2, define the value of e equal to f(e) in G 2. Clearly, ω(g 2 ) = ω(g 2 ) and S G 2 (v) = S G 2 (v) for every v V (G 2 ). Hence the claim is proved. By this algorithm, the length of an odd cycle such as C i is decreased by 2. Continue this 5

algorithm until there is no cycle which can be replaced with a shorter one. At this step, for every C i, if s and t are two vertices on C i with a path of length 3 on C i, then one can see that st F. Now, let C i have length 2m+1(m > 1). Two cases may be considered: Case 1. The cycle C i has been obtained from a cycle of length 2m + 3 (m > 1), say C i. We claim that cycle C i shorten too. could be shorten in such a way that the resultant cycle can be Let {v 1 v 2, v 2 v 3,..., v 2m+3 v 1 } be the edges of C i. Let v 1v 4 be that edge whose adding to C i makes the cycle C i. Since C i cannot be shorten, we conclude that v 5 v 8, v 5 v 2m+3 F and so v 2 v 5 / F. Two vertices adjacent to v 7 in F is contained in C i. Thus, v 2 v 7 / F. Hence, if we add v 2 v 5 instead of v 1 v 4 to obtain a cycle of length 2m + 1, then the resultant cycle can be shorten via the edge v 2 v 7. So the claim is proved. Therefore, we can assume that if we apply the algorithm on C i, then we can reduce C i to a triangle. Case 2. C i is a cycle obtained by the removing of some even closed trails from a connected component of G 1 containing vertices of C i. Let T Ci be the set of these even trails. If V (C i ) = V (G), then we alternatively assign +1 and 1 to the edges of C i starting by +1. By this assignment we obtain an SSDF for G whose ω(g) is ω(f ) + ω(c i ) = n + 1. Hence, V (G) 6 implies that n + 1 7n 6 and in this case the theorem is proved. Thus, we can assume that V (C i ) V (G). If V (C i ) 5, then let U Ci denote the set of all isolated vertices obtaining after removing all edges of the elements of T Ci. There is no edge with one end point in C i and another one in V (G) \ (V (C i ) U Ci ). Hence, U Ci > 0. For every triangle, say C i, define U Ci =. Therefore, V (C i ) 5 implies that the Case 1 does not occur for C i. By the definition of U Ci and U Cj, we have U Ci U Cj =, for every i j. We say that C i and C j are friend if i j and there exists some edge in F, say e, with one end point in V (C i ) U Ci and another one in V (C j ) U Cj. Now, let L be the set of all odd cycles remaining after finishing the algorithm. We provide an algorithm on L. In each step, consider two cycles C i and C j which are friend, and let w i w j F, where w i 6

V (C i ) U Ci and w j V (C j ) U Cj. Obviously, ( T T Ci T ) C i and ( T T Cj T ) C j are odd closed trails. Change the value of w i w j from +1 to 1 and assign alternatively +1 and 1 to all edges of ( T T Ci T ) C i starting from w i with +1. Do the same for ( T T Cj T ) C j starting from w j. Then remove C i and C j from L. Since the value of w i w j is changed from +1 to 1 and the sum of all values of the edges in ω(( T (T Ci T Cj ) T ) C i C j ) is 2, we conclude that for every v U Ci V (C i ) U Cj V (C j ), S G (v) 1. We continue this algorithm until we cannot find any two cycles in L which are friend. Assume that L is the set of remaining cycles of L. Consider a set of vertices S with this property that if V (C i ) 5, then S U Ci = 1, for i = 1,..., L. Now, define W = ( L i=1 V (C i)) S. Let W = V (G) \ W and t be the number of triangles in L. Let h = L t. We note that W contains 3t vertices of triangles. Since for every C i L, with V (C i ) 5, we have W U Ci = 1, W 3t+6h. Since W + W = n, we find that W + 3t + 6h n. (1) Now, we count the number of edges in F with one end point in W and the other one in W. Let q be the number of such edges. Since the degree of each vertex in F is 2, we find that q 2 W. Clearly, there is no edge in F with one endpoint in the vertex set of a triangle in L and another endpoint in W, since otherwise this triangle is a friend of another cycle in L and we could continue the algorithm. Moreover, for every cycle of length at least 5 in L, say C i, the vertex u Ci S U Ci. We claim that two edges in F incident with u Ci have other endpoints in W. To see this we note that by assumption, every edge in F which meet V (C i )( V (C i ) 5), have two endpoints in V (C i ). Furthermore, if other end point of these edges are in W \ V (C i ), then C i is a friend of another cycle in L, a contradiction. Therefore, the claim is proved. This implies that: 2(3t) + 2h q, q 2 W. (2) Thus, 3t + h W and so 6t + 7h n by (??). This yields that t + h n 6. Now, we claim that γ ss (G) n + t + h. 7

Let M be the induced subgraph on all edges of G \ C i L E(C i). Then ω(m) n. Now, it suffices for every cycle in L, such as C i, we assign +1 and 1, alternatively to the edges of C i starting with +1. Thus, ω(c i ) = +1. Finally, we obtain an SSDF for G with the following property: ω(g) n + t + h 7n 6, which implies that γ ss (G) 7n 6. Now, we would like to show that the upper bound is sharp. To see this for every positive integer k consider the cycle C 6k. Let V (C 6k ) = {v 1,..., v 6k }. We partition the vertices of even indices into k subsets, each of them containing 3 vertices as follows: T i = {v 6i+2, v 6i+4, v 6i+6 }, for every i, 0 i k 1. Join 3 vertices of each subset, to obtain k triangles. Call this graph by G. We claim that for every SSDF of G, say f, we have ω(g) 7n 6. Clearly, two edges incident with a vertex of degree 2 (odd indices) have value 1. Obviously, for each vertex v of degree 4, at least 3 edges incident with v should have value +1. Thus for every triangle obtained by T i, at least two edges have value +1. This yields that ω(g) 6k + k = 7k. Thus γ ss (G) = 7n 6. The graph shown in Figure 1 is a graph of order 12 with γ ss (G) = 7 6 12 = 14. v 1. v 2 v 12 v 3 v 11 v 4 v 10 v 5 v 9 v 6 v 8 v 7 8

Figure 1: Graph G with γ ss (G) = 7n 6. 3. An upper bound for signed star domination number of graphs with independence number at most 2 Let G be a graph. In this section we prove that if the independence number of G is at most 2, then the signed star domination number of G is at most n + 1. Theorem 4. Let G be a connected graph of order n and α(g) 2. Then γ ss (G) n + 1. Proof. We divide the proof into two parts: (i) First assume that G has no cut vertex. Thus, κ(g) 2 and so κ(g) α(g). Thus by Proposition 10.1.2 of [?], G is Hamiltonian. Let C be a Hamilton cycle of G. Remove all edges of C from G and call the resultant graph by G. Let C 1,..., C k be the connected components of G where V (C 1 ) V (C 2 ) V (C k ). By Lemma??, there is an edge assignment f : E(G) { 1, +1} such that for every C i, ω(c i ) 1, and for each v V (C i ), S Ci (v) 1 and ω(c i ) has minimum possible value. Now, we extend f to E(G) by assigning +1 to every edge of C. Assume that r is the number of connected components of G which have at least 3 vertices. Now, two cases can be considered: Case 1. r = 1. Thus ω(g) = ω(c) + ω(c 1 ) n + 1. Case 2. r = 2. If ω(c 1 ) 0 or ω(c 2 ) 0, then clearly we have ω(g) n + 1. Now, let ω(c 1 ) = ω(c 2 ) = 1. By Lemma 1, every vertex of C 1 and C 2 has even degree and also C 1 and C 2 have odd size. We claim that there exists an edge e = uv, u V (C 1 ) and v V (C 2 ). Assume that there is no such edge. Since G is connected we note that there exists x V (G)\(V (C 1 ) V (C 2 )). Since α(g) 2, there is no independent set of size 3. But since V (C 1 ) 3, V (C 2 ) 3 and the degree of x in C is 2, we conclude that 9

there are x 1 V (C 1 ) and x 2 V (C 2 ) such that xx 1, xx 2 / E(G). Hence {x, x 1, x 2 } is an independent set, a contradiction and the claim is proved. Now, let e = uv, u V (C 1 ) and v V (C 2 ). One can change f(e) from +1 to 1. Now, consider two Eulerian circuits for C 1 and C 2 and reassign the values of the edges of these two Eulerian circuits by +1 and 1 consecutively, starting from u and v with value +1. Hence, S G (z) 1, for every z V (G). Moreover, ω(g) n 2 + 2 = n. Case 3. r 3. Suppose that U = {u 1, u 2, u 3 } V (C 1 ), V = {v 1, v 2, v 3 } V (C 2 ) and W = {w 1, w 2, w 3 } V (C 3 ). Since α(g) 2, the subgraph u i, v j, w t contains at least one edge in C, for every 1 i, j, t 3. So the induced subgraph on U V W has at least 9 edges in C. Thus V (G) = 9. Now, we claim that the induced subgraph on U is a triangle. To see this, let u i u j / E(G ), for 1 i < j 3. Every vertex in V (G)\U is adjacent to at least one of the vertices u i and u j. Hence, the degree of u i or u j in C is at least 3, a contradiction. Similarly, the induced subgraphs on V and W are triangle. Now, consider an arbitrary edge u i v j C. There is an edge assignment for U V such that S U (u i ) = S V (v j ) = 2, and ω( U ) = ω( V ) = 1. Now, change the value u i v j from +1 to 1. Obviously, ω(g) = 10. So γ ss (G) 9 + 1 = V (G) + 1. (ii) Now, assume that G has at least one cut vertex. Clearly, there are two positive integers r and s such that G is the union of K r and K s and there exists a unique vertex v V (K r ) adjacent to v 1,..., v p V (K s ). Consider two following cases: Case 1. s is odd. By Theorem 1, there exists an SSDF for K s such that for every v V (K s ), S Ks (v) 2 and ω(k s ) s + 1. If r is odd, then by Theorem 1, there exists an SSDF such that ω(k r ) r + 1 and since the degree of v is even, S Kr (v) 2, for every v V (K s ). Let ω(k r ) = r. Now, for i = 1,..., p, define { +1 if i is even f(vv i ) = 1 otherwise. 10

Obviously, ω(g) s + 1 + r a, where { 0 if p is even a = 1 otherwise. Now, let ω(k r ) = r + 1. Then one can find an SSDF for K r such that S Kr (v) 4. Now, let f(vv 1 ) = 1, and for every 2 i p, define { 1 if i is even f(vv i ) = +1 otherwise, ω(g) r + s + 2 a, where a = { 2 if p is even 1 otherwise. Otherwise, if r is even, then by Theorem 1, there exists an SSDF such that ω(k r ) = r 2. Now, for i = 1,..., p define, ω(g) r 2 + s + 1 + a, where f(vv i ) = a = { 1 if i is even +1 otherwise, { 0 if p is even 1 otherwise. Case 2. s is even. Consider an edge assignment for K r such that ω(k r ) = γ ss (K r ). If r 1 (mod 4), then by Theorem 1, γ ss (K r ) = r + 1. In this case, clearly, there exists x V (K r ) such that S Kr (x) = 4. With no loss of generality one may assume that S Kr (v) = 4. Now, we assign +1, 1 to the edges T = {vv 1,..., vv p } according to Table??. For each cell of the table with value l, we assign 1 to l edges of T and +1 to the other edges. For instance, for p = 4t + 1, we assign 1 to exactly 2t, 2t and 2t + 2 edges of T, if γ ss (K r ) = r 2, r and r + 1, respectively. Now, we would like to divide the vertices of the set {w w {v 1, v 2,..., v p } and f(vw) = 1 } 11

γ ss (K r ) r 2 r r + 1 p 4t 2t 2t 2t 4t + 1 2t 2t 2t + 2 4t + 2 2t 2t 2t + 2 4t + 3 2t 2t + 2 2t + 2 Table 1: The number of edges with value 1 in terms of p and γ ss into 2-subsets. Consider a perfect matching M of K s containing the corresponding edges of these 2- subsets. We claim that there exists an SSDF for K s such that the values of all edges in M is 1 and ω(k s ) = s 2. It is well-known that the edges of K s can be decomposed into s 1 perfect matchings [?, p.274-275]. If we assign 1 to all edges of s 2 1 arbitrary perfect matchings including M and +1 to all other edges of all other matchings, a required SSDF is obtained. Hence, the claim is proved. Consider this SSDF for K s and change the values of all edges corresponding to 2-subsets from 1 to +1. Now, we have ω(g) = γ ss (K r ) + (p 2q) + γ ss (K s ) + 2( q 2 ), where q is the number of edges between v and V (K s ) with value 1. By applying this formula to the all cases of Table??, we obtain the following table. γ ss (K r ) p 4t 4t + 1 4t + 2 4t + 3 r r 2 r r + 1 2 + 0 + s 2 + 2t r + 0 + s 2 + 2t (r + 1) + 0 + s 2 + (2t) r 2 + 1 + s 2 + 2t r + 1 + s 2 + 2t (r + 1) + ( 3) + s 2 + (2t + 2) r 2 + 2 + s 2 + 2t r + 2 + s 2 + 2t (r + 1) + ( 2) + s 2 + (2t + 2) r 2 + 3 + s 2 + 2t r + ( 1) + s 2 + (2t + 2) (r + 1) + ( 1) + s 2 + (2t + 2) Table 2: ω(g) in terms of γ ss and p For instance, suppose that p = 4t+2 and γ ss (K r ) = r+1. We have (4t+2) 2(2t+2) = 2 and 2( q 2 ) = 2t + 2. Thus γ ss(g) = (r + 1) + ( 2) + s 2 + (2t + 2) r + s + 1. Clearly, 12

the value of each cell in this table is less than or equal r + s + 1. Hence γ ss (G) r + s + 1, and the proof is complete. References [1] R. Diestel, Graph Theory, Graduate Texts in Mathematics, Vol. 173, Springer-Verlag, 2005. [2] C. Wang, The signed star domination numbers of the cartesian product graphs, Discrete Appl. Math. 155 (2007) 1497-1505. [3] D.B. West, Introduction to Graph Theory, Prentice Hall Upper Saddle River, NJ, Second Edition, 2001. [4] G. Chartrand and L.Lesniak, Graphs and Digraphs, Chapman and Hall/CRC, Third Edition, 2000. [5] B. Xu, On edge domination numbers of graphs, Discrete Math. 294, (2005) 311-316 Saieed Akbari s akbari@sharif.edu Department of Mathematical Sciences P.O. Box 11365-9415, Tehran, Iran. Ashkan Norouzi ashkan.afn@gmail.com Alireza Rezaei rezaei70@gmail.com Rahmtin Rotabi rahmtin.rotabi@gmail.com Sara Sabour sara.sabour@gmail.com Department of Computer Engineering Sharif University of Technology Tehran, Iran. 13