International Journal of Recent Technology and Engineering (IJRTE) ISSN: 2277-3878, Volume-9 Issue-2, July 2020 Double Domination Number of Some Families of Graph

Manjula. C. Gudgeri , Varsha

Abstract: In a graph G = (V, E) each is said to dominate 2.4. Double domination number: The “double domination every vertex in its closed neighborhood. In a graph G, if S is a number” γx2 (G) is the smallest cardinality of a double subset of V then S is a double of G if every vertex dominating set of G. [4] in V is dominated by at least two vertices in S. The smallest cardinality of a double dominating set is called the double Example: domination number γx2 (G). [4]. In this paper, we computed some relations between double domination number, domination number, number of vertices (n) and maximum degree (∆) of Helm graph, Friendship graph, Ladder graph, Circular Ladder graph, Barbell graph, Gear graph and Firecracker graph.

Keywords: dominating set, domination number, double dominating set, double domination number. We denote n, ∆ Fig.1 respectively by number of vertices, maximum degree of a graph G. Now consider the graph Fig.1. The set S = {a, b, c, e, f, g} forms a double dominating set of G and double I. INTRODUCTION domination number γx2 = 6. Frank Harary and T. W. Haynes defined and studied the concept of double domination in graphs. Domination and III. DEFINITIONS OF SOME FAMILES OF GRAPHS double domination numbers are defined only for graphs with 3.1. “Helm Graph”: It is designed from a with non- isolated vertices. Every graph with no-isolated vertex n vertices by adjoining a pendant edge at each node of the has a double dominating set. In a graph G= (V, E), a subset cycle and denoted by Hn . [8] S ⊆ V is a dominating set of G if every vertex v of V−S has a neighbor in S.The domination number γ (G) is the 3.2. “Friendship Graph”: It is constructed by joining n minimum size of a dominating set of vertices in G. In a copies of the C3 with a common vertex and graph G with vertex set V and edge set E, a subset S of V is represented by Fn. It is a planar undirected graph with 2n+1 a double dominating set of G if every vertex v ∈ V, is in vertices and 3n edges. [7]

V−S and has at least two neighbors in S. The smallest 3.3. Ladder Graph: The “Ladder graph” Ln can be obtained cardinality of a double dominating set of G is known as the as the Cartesian product of two path graphs P2 and Pn ,where double domination number γx2. [6] [4] [8]. Pn is a path graph. The Ladder graph Ln is a planar undirected graph with 2n vertices and 3n−2 edges. [7] II. DEFINITIONS 3.4. “Circular Ladder graph” CLn: It is a connected 2.1. Dominating set: A non-empty set D⊆ V of a graph G is planar and Hamiltonian graph which is constructible by a “dominating set” of G if every vertex in V-D is adjacent to connecting the four 2-degree vertices in a straight way or by some vertex in D. [1] the Cartesian product of a cycle of length n ≥ 3 and an edge. 2.2. Domination number: The “domination number” γ(G) In symbols, CLn=Cn x P2 . It has 2n nodes and 3n edges. is the number of vertices in a smallest dominating set for G. They are the polyhedral graphs of prisms, so they are more [7] commonly called prism graphs. [7] 2.3. Double dominating set: A subset S ⊆ V (G) is a 3.5. “n-Barbell graph” Bn: It is obtained by connecting “double dominating set” for G if S dominates every vertex two copies of a complete graph Kn by a cut edge. It has 2n of G at least twice. [4] vertices and 2n+1 edges. [8]

3.6. “Gear graph” Gn: It is a wheel graph Wn with a vertex added between every pair of adjacency vertices of the n- cycle. It has 2n+1 vertices and 3n edges. [3]

3.7. “Firecracker Graph” F(n,k) : It is framed by the

concatenation of nk-stars by linking one leaf from each. It has order nk and size nk−1. [2]

Revised Manuscript Received on June 22, 2020. Dr.Manjula. C. Gudgeri, Professor, KLEMSSCET, Belagavi, India. E- mail: [email protected] Mrs. Varsha, Research Scholar, KLEMSSCET, Belagavi, India. E- mail: [email protected]

Published By: Retrieval Number: B3307079220/2020©BEIESP Blue Eyes Intelligence Engineering DOI:10.35940/ijrte.B3307.079220 161 & Sciences Publication

Double Domination Number of Some Families of Graph

IV RESULTS ON DOUBLE DOMINATION OF SOME Theorem 4.4: FAMILES OF GRAPHS For any Circular Ladder graph CLn (n ≥ 3) , double Theorem 4.1: domination number γ×2(CLn) = n.

For any Helm graph Hn n ≥ 3 , double domination number Proof: Circular Ladder graph CLn has 2n nodes ∆ , if n = 3 and 3n edges with vertex set {v1, v2, v3,…..,v2n}. Here n γ (H ) = ×2 n ∆ + 1 , if n ≥ 4 number of nodes is enough to dominate all the vertices of the graph CL twice. Let us consider the double dominating Proof: n set of CLn = {v1, v2, v3,…..,vn}. Thus double domination Case (i) for n=3, number is γ x2 (CLn) = { v1, v2, v3,…..,vn}. Hence γ×2 (CLn) = n. The Helm graph Hn has a single universal vertex v connecting all the vertices of the cycle with vertex set {x1, Theorem 4.5: x , x and each node of the cycle adjoining a pendant edge. 2 3} For any Barbell graph B (n ≥ 3) , double domination Let y , y , y be the pendant vertices. Here vertex v and n 1 2 3 number γ (B ) = 4. pendant vertices are enough to dominate all the vertices of ×2 n graph Hn twice. Here, ∆= n+1. Let us consider the double Proof : The Barbell graph Bn has 2n vertices and 2n+1 dominating set of Hn={v, y1, y2, y3}. Thus double edges with vertex set {x1, x2, x3,………, xn, y1,y2, y3, domination number γ×2 (Hn) is {v, y1, y2, y3}. Hence γ×2(Hn) ……..,yn }. Since Bn is obtained by connecting two copies of = ∆. complete graph. So, any two vertices say x1 and x2 dominates all the vertices of single copy of complete graph Case (ii) for n ≥ 4, twice and since it has two copies. The double dominating set

The Helm graph Hn has a universal vertex v connecting the of Bn = {x1, x2, y1, y2}. Thus double domination number is vertices of the cycle having vertex set {x1 , x2 , x3, ..………, γ×2 (Bn) = {x1, x2, y1, y2}. xn } and each node of the cycle adjoining a pendant edge. Hence γ×2 (Bn) = 4. Let {y1, y2, y3, ……..,yn } be the pendant vertices. Here vertex v and pendant vertices will dominate the remaining Theorem 4.6: vertices of graph Hn twice. Here ∆ = n. Let us consider the For any Gear graph Gn (n ≥ 3), double domination number double dominating set of Hn = {v, y1, y2, y3, ……..,yn }. γ×2 (Gn) = ∆. Thus double domination number is γ×2(Hn) = {v, y1, y2, y3, ……..,yn }. Proof : The Gear graph has 2n+1 vertices with vertex set {v, v1, v2, v3,…..,vn, u1, u2, u3,……,un} and 3n edges with ∆= n. Hence γ×2 (Hn) = ∆ + 1. As Gn is constructed from the wheel graph Wn by adding a Theorem 4.2: vertex between every pair of adjacent vertices of the n-

cycle. Here the vertices of the cycle {v1, v2, v3,…..,vn} For any Friendship graph Fn n ≥ 2 , double domination ∆ dominates all the vertices of Gn twice. Let us consider the number γ×2(Fn) = 1 + . 2 double dominating set of Gn ={ v1, v2, v3,…..,vn }.Thus double domination number is γ×2(Gn) = {v1, v2, v3,…..…, Proof: The Friendship graph Fn has 2n+1 vertices ………vn }. Hence γ×2 (Gn) = ∆. and 3n edges. Let the vertex set be {v, v1, v2, v3,…..,v2n} and ∆ ∆= 2n =≫ n= . Here vertex v and n vertices dominates all Theorem 4.7: 2 the vertices of the graph Fn twice. Let us consider the double For any Firecracker graph F(n, k) (n = 2,k ≥ 3), double dominating set of F = {v, v , v , v ,…..,v }. Thus double n 1 3 5 n domination number γ×2 [F(n,k)] = n ∆. domination number is Proof: The Firecracker graph F(n, k) (n = 2,k ≥ 3), has nk ∆ γ (F ) = {v, v , v , v ,…..,v }. Hence γ (F ) = 1 + . vertices and nk−1 edges. The graph F(n, k) consists of ×2 n 1 3 5 n ×2 n 2 (k−2) pendant vertices in each . Let the vertex set of Theorem 4.3: F(n, k) be {v1, v2, v3,………..,vk, u1, u2, u3,………,uk } and

For any Ladder graph Ln (n ≥ 2) , double domination ∆ = (k − 1). Here all pendant vertices and one of the number γ×2(Ln) = n. remaining vertex dominates all the vertices of the graph F(n, k) twice, for single star. Since the graph F(n, k) has two Proof: The Ladder graph Ln has 2n nodes and (3n−2) edges. stars, let us consider the double dominating set of F(n, k) = Let the vertex set be {x1, x2, x3, ………, xn, y1,y2, y3, {v1, v2, v3,…..,vk-2, vk-1, u1, u2, u3,……,uk-2,uk-1 }. Thus ……..,yn } and the edge set be {xi yi , xi xi+1 , yi yi+1} for double domination number is 1≤ i ≤ n. Let us consider the double dominating set of Ln= {x1, x3, x5,,…………, xn-1, y2, y4, y6, ….…..,yn }. Thus γ×2[F(n, k)] = {v1,v2, v3,…....,vk-2, vk-1, u1, u2, u3,.……,uk-2, double domination number is uk-1 }.

γ×2(Ln) = {x1, x3, x5,,……… xn-1, y2, y4, y6, ……..,yn }. Hence γ×2 [F(n, k)] =2(k−1) = n∆. n n Hence γ ( L ) = + = n. ×2 n 2 2

Published By: Retrieval Number: B3307079220/2020©BEIESP Blue Eyes Intelligence Engineering DOI:10.35940/ijrte.B3307.079220 162 & Sciences Publication International Journal of Recent Technology and Engineering (IJRTE) ISSN: 2277-3878, Volume-9 Issue-2, July 2020

Theorem 4.8: Proof : The Helm graph Hn has 2n+1 vertices and 3n edges which consists of single universal vertex v connecting all For any Firecracker graph F(n,k) with odd values of n (n ≥ the vertices of the cycle and each node of the cycle 3, k ≥ 4), double domination number adjoining a pendant edge with vertex set {v, x1 , x2 , x3, 3푛−1 γ [F(n,k)] = nk − . x4,………, x2n } Here, vertex v and pendant vertices x2 2 dominate all the other vertices of graph Hn twice. Let us Proof: The Firecracker graph F(n,k) (n ≥ 3, k ≥ 4), has nk consider the double dominating set of Hn = {v, x1 , x2 , x3, vertices and nk−1 edges. The graph F(n,k) consists of (k−2) x4,………, xn }. Thus double domination number is γ×2 pendant vertices in each star. Let the vertex set of F(n,k) be (Hn) = {v, x1 , x2 , x3, x4,………, xn }. Hence γ×2(Hn) = n+1-- {u1,u2,u3,…,uk, v1,v2,v3,….,vk, w1,w2,w3,….., ------equation (1). Whereas, in a dominating set of Hn all the wk,x1,x2,x3,…….,xk,y1,y2,y3,……..,yk,……….,k1,k2,k3,…… vertices of the cycle are enough to dominate all the vertices …,knk}. Here, all pendant vertices and some of the other of the graph. Let us consider the dominating set of Hn={ x1 remaining vertices will dominate all the vertices of the , x2 , x3, x4,………, xn }. Thus domination number is γ (Hn) graph F(n,k) twice. Let us consider double dominating set ={ x1 , x2 , x3, x4,………, xn }. of F(n,k) as {u1, u2, u3,……,uk-2, uk-1, v1, v2, v3,…….,vk-2, Hence γ (Hn) = n------equation (2). From equations (1) w1, w2, w3,…….,wk-2, wk-1, x1, x2, x3,…….,xk- and (2), γ×2 (Hn) = γ ( Hn) + 1. 2,y1,y2,y3,……..,yk-2,yk-1,…….,k1,k2,k3,…….., knk-2, knk-1}. The double domination number is γx2[F(n,k)] = Theorem 5.2: {u1,u2,u3,…….., uk-2, uk-1, v1, v2, v3,…..…,vk- For any Friendship graph Fn (n ≥ 2), double domination 2,w1,w2,w3,…..…,wk-2,wk-1,x1,x2,x3,…..,xk-2,y1,y2,y3,….... number γ×2 (Fn) = γ (Fn) + n. ….,yk-2, yk-1,…....,k1, k2, k3,…,knk-2, knk-1}.Thus we have Proof: The Friendship graph Fn has 2n+1 vertices γx2[F(n,k)] = nk − 4, nk − 7, nk − 10, nk − 13, … … .. . and 3n edges. Let the vertex set be {v, v1, v2, v3,…..,v2n}. Here, vertex v and n vertices dominates all the vertices of 3푛−1 Hence, γ [F (n,k)]= nk − . the graph F twice. Let us consider the double dominating x2 2 n set of Fn = {v, v1, v3, v5,…..,vn }. The double domination Theorem 4.9: number is γ×2(Fn) = {v, v1, v3, v5,…..,vn}. Hence γ (F ) = n + 1------equation (1). Whereas, in a dominating For any Firecracker graph F(n,k) with even values of n (n ≥ ×2 n set of F common vertex will dominate all the vertices of the 4,k ≥ 4), double domination number n graph. Let us consider the domination set of Fn = {v}. Thus 3푛−2 γ [F (n, k)] = nk − domination number is γ (Fn) = {v}. x2 2 Hence γ (F ) = 1------equation (2). From equations (1) and Proof: The Firecracker graph F(n,k) (n ≥ 4,k ≥ 4), has nk n (2) γ×2 (Fn) = γ (Fn) + n. vertices and nk − 1 edges. The graph F(n,k) consists of (k−2) pendant vertices in each star. Theorem 5.3:

Let the vertex set of F(n,k) be{u1,u2,u3,……,uk, For any Barbell graph Bn (n ≥ 3) , double domination v1,v2,v3,…….,vk,w1,w2,w3,…….,wk,x1,x2,x3,……,xk,y1,y2,y3, number γ×2 (Bn) = 2 γ ( Bn ). ……..,yk,……..,k1,k2,k3,…….,knk}. Here all pendant Proof : The Barbell graph Bn has 2n vertices and 2n+1 vertices along with some of the rest of vertices will edges with vertex set {x1, x2, x3, x4 ,………, x2n}. Since Bn dominate all the vertices of the graph F(n,k) twice. is obtained by connecting two copies of complete graph. In Let us consider double dominating set of F(n,k) as dominating set of Bn only two vertices dominate the {u1,u2,u3,….....,uk-1,v1,v2,v3,…...... ,vk-2,w1,w2,w3,…...,wk-1, remaining vertices of the graph. The dominating set of Bn = x1,x2,x3,…...,xk-2,y1,y2,y3,…...,yk-1,…...k1,k2,k3,…...,knk-1}. {x1, x2}.Thus domination number is γ (Bn) = {x1, x2}. Hence γ (Bn) = 2------equation (1). From theorem 4.5 γ×2 The double domination number is (Bn) = 4 ------equation (2). From equations (1) and (2)

γx2[F(n,k)] = {u1,u2,u3,…….…,uk-1,v1,v2,v3,…….….,vk-2, γ×2 (Bn) = 2 γ (Bn). w1,w2,w3,….,wk-1,x1,x2,x3,……,xk-2,y1,y2,y3,…..,yk-1,……. Theorem 5.4 : …..…, k1,k2,k3, …….,knk-1}. Thus we have For any Firecracker graph F(n , k) (n = 2, k ≥ 3), double γx2 [F(n,k)] = nk − 5, nk − 8, nk − 11, nk − 14, … … . . . domination number γx2 [F (n,k) ] = γ [F (n,k) ] ∆. 3푛−2 Hence,γ [F (n,k)] = n푘 − . x2 2 Proof : F(n , k) (n = 2,k ≥ 3), has nk vertices and nk−1 edges. It consists of (k−2) pendant vertices in each star. Let V. RELATION BETWEEN DOUBLE DOMINATION the vertex set of F (n,k) be {v1,v2,v3,……,vk, AND DOMINATION NUMBER. u1,u2,u3,…….,uk} and ∆ = (k−1). Here two vertices (i.e., one from each star) are enough to dominate all the vertices Theorem 5.1: of the graph. For any Helm graph H (n ≥ 3), double domination number n γ×2 (Hn) = γ (Hn) + 1.

Published By: Retrieval Number: B3307079220/2020©BEIESP Blue Eyes Intelligence Engineering DOI:10.35940/ijrte.B3307.079220 163 & Sciences Publication

Double Domination Number of Some Families of Graph

The dominating set of F(n,k) = {v1 , u1}. Thus domination International Journal of Pure and Applied Mathematics Volume 116 number is γ [ F (n,k) ] = {v , u }. Hence γ [F (n,k) ] = 2 = n- No. 4 2017, 1035-1042. 1 1 3. Ersin Aslan and Alpay Kirlangic, “Computing The Scattering ------equation (1). From theorem 4.7 Number and The Toughness for Gear Graphs”, Bulletin of Society of Mathematicans Banja Luka, ISSN 0354-5792, ISSN 1986-521X(o) γx2 [F (n,k) ] = n ∆------equation (2). From equations (1) Vol. 18(2011), 5-15. and (2) γx2 [F (n,k) ] = γ [F (n,k) ] ∆. 4. F. Harary and T.W.Haynes, “Double domination in graphs”, Ars Combis. 55 April(2000) 201-213. Theorem 5.5: 5. T.W.Haynes, S.Teresa. Hedetniemi and P.J.Slater, “Domination in Graphs”: Advanced Topics (Marcel Dekker, New york, (1998). For any Firecracker graph F(n , k) where n is odd (n ≥ 6. Mustapha. Chellali, Abdelkader Khelladi and Frederic Maffray,” 3, k ≥ 4), the double domination number is EXACT DOUBLE DOMINATION IN GRAPHS”, Discussiones Mathematicae 291 25 (2005 ) , Volume :25, Issue:3 , 3n−1 page 291–302, ISSN: 2083-5892 . 훾 F n, k = kγ F − . x2 2 7. From Wikipedia, the free encyclopedia https://en.wikipedia.org/wiki/. 8. Wolfram Math World , https://mathworld.wolfram.com/. Proof : The Firecracker graph F(n,k) (n ≥ 3, k ≥ 4), has nk vertices and nk−1 edges. The graph F(n , k) consists AUTHORS PROFILE of (k−2) pendant vertices in each star. Let the vertex set Dr. Manjula. C. Gudgeri, M.Sc., Ph.D. Professor, of F(n,k) be {u1 ,u2 ,u3 ,….,uk ,v1 ,v2 ,v3 ,…, vk ,w1 ,w2 ,w3 Department of Mathematics, KLEMSSCET, Belagavi, ,…,wk ,…….,k1 ,k2 ,k3 ,……, k nk}. Here n vertices (i.e., 590008, India [email protected] central vertex of each star) are enough to dominate all the vertices of the graph. The dominating set of F(n,k) is {u1,v1,w1,…..,k1}. Thus the domination number is γ ={u1, Mrs. Varsha, M.Sc. Research Scholar, Department of v1,w1,….k1}. Hence γ [F (n,k)] = n ------equation (1). Mathematics, KLEMSSCET, Belagavi, 590008, India. E- From theorem 4.8 mail address: [email protected] 3푛−1 γ 퐹 푛, 푘 = nk − ------equation (2). From x2 2 equations (1) and (2)

3푛−1 γ 퐹 푛, 푘 = k γ(F) − . x2 2 Theorem 5.6 : Every Firecracker graph F(n,k) 푛 ≥ 4 , 푘 ≥ 4 , where n is even, the double domination number is

3n−2 γ F n, k = kγ F − . x2 2 Proof : The Firecracker graph F (n,k) (n ≥ 4, k ≥ 4), has nk vertices and nk-1 edges. The graph F(n , k) consists of (k- 2) pendant vertices in each star. Let the vertex set of F(n,k) be {u1 ,u2 ,u3 ,….uk , v1 ,v2 ,v3 ,……vk ,w1 w2 ,w3 ,…wk ,…….k1 ,k2 ,k3 ,……k n k}. Here n vertices (i.e., central vertex of each star) are enough to dominate all the vertices of the graph. The dominating set of F(n,k) is {u1,v1,w1,…..k1}. Thus the domination number is γ = {u1, v1,w1,….k1}. Hence γ (F(n,k) = n ------equation (1). From theorem 4.9

3n−2 γ F n, k = nk − ------equation 2 x2 2 From equations (1) and (2)

3n−2 훾 퐹 푛, 푘 = kγ[F] − . 푥2 2

VI. CONCLUSION We have computed relation between double domination number, domination number, number of vertices and maximum degree of some graphs.

REFERENCE: 1. K. Ameenai Bibi and R.Selvakumar, “The Inverse Split and Non-split Domination in Graphs”, International Journal of Computer Applications ( 0975-8887) , Volume 8-No 7.October 2010. 2. Ashaq Ali , Hifza Iqbal , Waqas Nazeer , Shin Min Kang, “On Topological Indices for the line graph of firecracker graph”.

Published By: Retrieval Number: B3307079220/2020©BEIESP Blue Eyes Intelligence Engineering DOI:10.35940/ijrte.B3307.079220 164 & Sciences Publication