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DOMINATOR CHROMATIC NUMBER of SOME GRAPHS the Dominator International Journal of Pure and Applied Mathematics Volume 119 No. 7 2018, 787-795 ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version) url: http://www.ijpam.eu Special Issue ijpam.eu DOMINATOR CHROMATIC NUMBER OF SOME GRAPHS T.Manjula1 and R.Rajeswari2 1Research Scholar Sathyabama University Chennai, India 2Professor Department of Mathematics Sathyabama University Chennai, India ABSTRACT The dominator coloring of a graph G is defined as a proper coloring of G where each vertex of the graph dominates every vertex of atleast one color class. The dominator chromatic number is the least number of color classes used in a dominator coloring of a graph G.Thedominator chromatic number and domination number ofdragon graph andlollipop graph are obtained and a relationship between the chromatic number, domination number and dominator chromatic number is expressed in this paper. Keywords: Coloring, Domination, Dominator coloring, Dragon graph, Lollipop graph 1. INTRODUCTION Let G be a graph with vertex set V and edge set E. A dominating set S is a subset of the vertex set V such that every vertexin the graph either belongs to S or has a neighbor in S. If no proper subset of S is a dominating set, thenS is called a minimum dominating set. The domination number denoted by γ(G) is the order of a minimum dominating set. Dominating sets are useful in routing problems, scheduling problems, assignment problems, computer communication networks, land surveying, assignment problems, coding theory etc.. A proper vertex coloring of a graph G is allocation of colors to the vertices such that two neighboring vertices belong to the different color class. The chromatic number, denoted by χ(G) is the minimum number of colors required among all proper colorings of G. Graph coloring are useful tools in modeling many practical problems such as scheduling, allocation of resources, assignment problems, pattern matching, circuit testing etc.. Graph coloring and domination have numerous applications to today's networks. The area obtained by combining graph coloring and domination is called dominator coloring.A dominator coloring of graph G is an assignment of colors to the vertices of G such that two neighboring vertices are assigned different colors and every vertex dominates all vertices of at least one color class. The minimum 787 International Journal of Pure and Applied Mathematics Special Issue number of color classes required for a dominator coloring of G is called dominator chromatic number, denoted by χd(G). The concept of dominator coloring was introducedin 2006 [1]. The relation between dominator coloring, proper coloring and domination number of different classes of graphs were shown in [2], [3], [5]. The dominator coloring ofprism graph, quadrilateral snake, triangle snake and barbell graph, proper interval graphs, block graphswere also studied in various papers [6], [7], [8]. The algorithmic aspects of dominator colorings in graphs have been discussed by Arumugam S etal. in [4]. A (m, n) dragon graph denoted by D(m, n)is the graph obtained by combining a cycle graph 퐶푚 to a path graph 푃푛 with a bridge. It has (m + n) vertices and ( 푚 + 푛)edges. It is also called as a tadpole graph. The (m, 1)- dragon graph is sometimes known as the m-pan graph. The particular cases of (3, 1) and (4, 1) dragon graphs are known as the paw graph and banner graph respectively. A (m, n) lollipop graph denoted by L(m, n) is obtained by connecting a complete graph 퐾푚 to a path graph 푃푛 with a bridge. It has (m + n) vertices and 푚 푚−1 [ + 푛]edges. 2 In this paper the domination number and dominator chromatic number of dragon graph andlollipop graph are obtained and a relation between the dominator chromatic number, domination number and chromatic number is expressed. 2. DOMINATOR CHROMATIC NUMBER OF DRAGON GRAPH Theorem 2.1:The dominator chromatic number of dragon graphD(m, n) when m n−1 3 + if m = 4, 1 ≤ n ≤ 5 2 n−1 ≥ 3 and ∀푛 is given by 휒 퐷 푚, 푛 = 4 + if m = 4, n ≥ 6 푑 3 m n−1 + + 2 otherwise 3 3 (1) Proof: Consider a dragon graph D(m, n) obtained by combining a cycle graph 퐶푚 to a path graph 푃푛 with a bridge. It has(m + n) vertices and (푚 + 푛)edges. The vertex set V and the edge set E are given by 푉 = 푣푖 /1 ≤ 푖 ≤ 푚 ∪ 푢푗 /1 ≤ 푗 ≤ 푛 E = 푣푖푣푖+1/ 1 ≤ 푖 ≤ 푚 − 1 ∪ 푣1푣푛 ∪ 푢1푣1 ∪ 푢푖 푢푖+1/ 1 ≤ 푖 ≤ 푛 − 1 . The procedure for dominator coloring of vertices is explained below Case 1 : When m = 4 and 1 ≤ n ≤ 5 788 International Journal of Pure and Applied Mathematics Special Issue Let the vertex 푣1 and 푣3 be given color 1 and 3 respectively. The vertices 푣2 n and 푣 are given color 2. Let the vertices 푢 , for1 ≤ 푗 ≤ be assigned 4 2푗 −1 2 n−1 color 3 and the vertices 푢 , 1 ≤ 푗 ≤ be assigned color 3+j respectively. 2푗 2 Then the vertex 푣1 and adjacent vertices of 푣1 dominate color class 1. The vertex푣3 dominates color class 2. For 2≤ 푗 ≤ 5, the vertices 푢푗 dominate color j−1 class 3 + . 2 Since every adjacent vertex is given different color it is a proper coloring of vertices and every vertex of the graph dominates all vertices of atleast one color class. Therefore this is a dominator coloring of vertices with dominator chromatic number n−1 휒 퐷(푚, 푛) = 3 + , for m = 4 , 1 ≤ n ≤ 5. d 2 Figure 1: Dominator chromatic number of dragon graph D(4, 5) is 5 . i.e., 휒 퐷(4, 5) = 5 2 푣2 3 1 3 4 3 5 3 푢 푢 푣3 푣1 푢1 푢2 3 푢4 5 푣 2 4 Case 2 : When m = 4 and n ≥ 6 The vertex 푣1 and 푣3 are given color 1 and 3 respectively. The vertices 푣2 and 푣4 are given color 2. Let the end vertex 푢푛 of the path 푃푛 be assigned the color 3, n when 푛 ≡ 1(푚표푑 3). Otherwise it is given color + 4 . Then the remaining 3 uncolored vertices 푢푗 of the path 푃푛 for 1 ≤ 푗 ≤ 푛 − 1are assigned color 3 when j j≡ 1 푚표푑 3 or color 4 when j≡ 2(푚표푑 3) or color + 4when j≡ 3 0(푚표푑 3) Then the vertices 푣1 and adjacent vertices of 푣1 dominate color class 1. The vertex푣3 dominates color class 2. The vertices 푢푗 for 2≤ 푗 ≤ 푛, dominate color j−1 class + 4 . 3 Since every adjacent vertex is given different color it is a proper coloring of vertices and every vertex of the graph dominates all vertices of atleast one color class. Therefore this is a dominator coloring of vertices with dominator chromatic number 789 International Journal of Pure and Applied Mathematics Special Issue n−1 휒 퐷(푚, 푛) = 4 + , when m = 4 and n ≥ 6. 3 Figure 2: Dominator chromatic number of dragon graph D(4, 9) is 7 . i.e., 휒 퐷(4, 9) = 7 2 푣 2 1 3 3 4 5 3 4 6 3 4 7 푣 푣1 푢1 푢2 푢3 푢4 푢5 푢6 푢7 푢8 푢9 3 푣4 2 d Case 3 : When m ≠ 4 and ∀푛 Let the vertex 푣1 be given color 1. Then the vertices 푣푖 for 2 ≤ 푖 ≤ 푚 of the cycle graph 퐶푚 are assigned the color 3 when i ≡ 0 (mod 3) or color 2 when i+8 i ≡ 2 (mod 3)or color when i ≡ 1 (mod 3). Let the end vertex 푢 of the 3 푛 path 푃푛 be assigned the color 2 when n ≡ 1(mod 3). Otherwise it is given color m+6 n + . After the vertex 푢 , the remaining uncolored vertices푢 of the 3 3 푛 푗 path 푃푛 for 1 ≤ 푗 ≤ 푛 − 1are assigned color 2 when j ≡ 1 ( mod 3) or color 3 m+6 j when j ≡ 2 ( mod 3) or color + when j ≡ 0 (mod 3). 3 3 Then the vertex 푣1 and adjacent vertices of 푣1( i.e.,푣2, 푣푛 and 푢1) dominate i color class 1. For 3 ≤ 푖 ≤ 푚 − 1 the vertices 푣 dominate color class + 3. 푖 3 m+8 j−2 For 2≤ 푗 ≤ 푛 – 1 the vertices 푢 dominate color class + . 푗 3 3 Since every adjacent vertex is given different color it is a proper coloring of vertices and every vertex of the graph dominates all vertices of atleast one color class. Therefore this is a dominator coloring of vertices. Thusthe dominator chromatic number of dragon graph is given by m n−1 휒 퐷(푚, 푛) = + + 2 for any n and m ≠ 4. 3 3 Figure 3: Dominator chromatic number of dragon graph D(6,9) is 7. i.e., 휒 퐷(6, 9) = 7 790 International Journal of Pure and Applied Mathematics Special Issue 2 3 푣2 푣3 1 2 3 5 2 3 6 2 3 7 4 푣 푢1 푢2 푢3 푢4 푢5 푢 푢 푣4 1 6 7 푢8 푢9 2 3 푣5 푣6 Proposition 2.2:The chromatic number dragon graph D(m,n) which is obtained by joining a cycle graph 퐶푚 to a path graph 푃푛 by a bridge is given by 2 푖푓 푚 푖푠 푒푣푒푛 휒 퐷(푚, 푛) = (2) 3 푖푓 푚 푖푠 표푑푑 Lemma2.3:The domination number of dragon graph D(m, n), m ≥ 3 and ∀푛, is given by 푛 2 + 푖푓 푚 = 4 푎푛푑 ∀푛 3 훾 퐷 푚, 푛 = 푚 푛−1 (3) + 표푡ℎ푒푟푤푖푠푒 3 3 Proof : Case 1 : When m=4 If n 푚표푑 3 ≡ 0, then S= {푣3} ∪ {푢푛 } ∪ 푢푖 /1 ≤ 푖 ≤ 푛 푎푛푑 푖 푚표푑 3 ≡ 1 .
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