International Journal of Mathematics Trends and Technology (IJMTT) – Volume 65 Issue 5 - May 2019
A Note on Open Support of a Graph under Addition II 1S. Balamurugan , 2M. Anitha , 3P. Aristotle , 4C. Karnan 1PG Department of Mathematics, Government Arts College, Melur - 625 106, Tamilnadu, India. 2Department of Mathematics Syed Ammal Arts and Science College, Ramanathapuram, Tamilnadu, India. 3Department of Mathematics, SACS MAVMM Engineering College, Madurai, Tamilnadu, India. 4Department of Mathematics The Optimus Public School, Erode - 638 455
Abstract In this paper we defined an open support of a vertex 푣 under addition and open support of a graph G under addition. We calculate the open support for Dutch windmill graph, Butterfly graph and Ladder graph. Also, we generalized the value of open support under addition for any given graph 퐺.
AMS Subject Code: 05C07
Keywords: Vertex Degree, Open Support of a Vertex, Open Support of a Graph
I. INTRODUCTION
In this work we consider finite, undirected, simple graphs 퐺 = (푉, 퐸) with 푛 vertices and 푚 edges. The neighbourhood of a vertex 푣 ∈ 푉(퐺) is the set 푁퐺(푣) of all the vertices adjacent to 푣 in 퐺. For a set 푋 ⊆ 푉(퐺), the open neighbourhood푁퐺(푋) is defined to be ∪푣∈푋 푁퐺(푣) and the closed neighbourhood 푁퐺[푋] = 푁퐺(푋) ∪ 푋. The degree of a vertex 푣 ∈ 푉(퐺) is the number of edges of 퐺 incident with 푣 and is denoted by 푑푒푔퐺(푣) or 푑푒푔(푣). The maximum and the minimum degrees of the vertices of 퐺 are respectively denoted by Δ(퐺) and 훿(퐺). A vertex of a degree 0 in 퐺 is called an isolated vertex and a vertex of degree 1 is called a pendent vertex or an end vertex of 퐺. A vertex of a graph 퐺 is said to be a vertex of full degree if it is adjacent to all other vertices in 퐺. A graph 퐺 is said to be regular of degree 푟 if every vertex of 퐺 has degree 푟. Such graphs (푚) are called r-regular graphs. The Dutch windmill graph퐷푛 , is the graph obtained by taking 푚 copies of the cycle graph 퐶푛 with a vertex in common. The Butterfly graph(also called the bowtie graph and the hourglass graph) is a planar undirected graph with 5 vertices and 6 edges. It can be constructed by joining 2 copies of the cycle graph 퐶3 with a common vertex. It is denoted by 퐵푛 . The ladder graph퐿푛 is a planar undirected graph with 2n vertices and 푛 + 2(푛 − 1) edges. The Ladder graph obtained as the cartesian product of two graphs one of which has only one edge: 퐿푛,1 = 푃푛 × 푃1 .
II. DEFINITIONS
Definition 2.1.Let G=(V,E) be a graph. A open support of a vertex, v under addition is defined by 푢∈푁(푣) 푑푒푔(푢) and it is denoted by 푠푢푝푝(푣).
Definition 2.2.Let G=(V,E) be a graph. A open support of a graph, G under addition is defined by 푣∈푉(퐺) 푠푢푝푝(푣) and it is denoted by 푠푢푝푝(퐺).
III. RESULTS
+ Theorem 3.1 Let 퐺 = 푃푛 be a corona graph. Then 푠푢푝푝(퐺) = 10(푛 − 1). + + Proof: Let 퐺 = 푃푛 be a corona graph. Let 푉(푃푛 ) = {푣1, 푣2,푣3, . . . , 푣푛 , 푢1, 푢2, . . . , 푢푛 } such that 푑푒푔(푣1) = 푑푒푔(푣푛 ) = 2; 푑푒푔(푣푖 ) = 3 for all 푖 = 2,3, … , 푛 − 1; 푑푒푔(푢푗 ) = 1 for all 푗 = 1,2, … , 푛. Now,
푠푢푝푝(푣1) = 푣∈푁(푣1) 푑푒푔(푣) = 푑푒푔(푣2) + 푑푒푔(푢1)
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= 3 + 1 = 4 Similarly, 푠푢푝푝(푣푛 ) = 4;
푠푢푝푝(푣2) = 푣∈푁(푣2) 푑푒푔(푣) = 푑푒푔(푣1) + 푑푒푔(푣3) + 푑푒푔(푢2) = 2 + 3 + 1 = 6 Similarly, 푠푢푝푝(푣푛−1) = 6; For 푖 = 3,4, … , 푛 − 2
푠푢푝푝(푣푖 ) = 푣∈푁(푣푖 ) 푑푒푔(푣) = 푑푒푔(푣푖−1) + 푑푒푔(푣푖+1) + 푑푒푔(푢푖 ) = 3 + 3 + 1 = 7
푠푢푝푝(푢1) = 푣∈푁(푢1) 푑푒푔(푣) = 푑푒푔(푣1) = 2 Similarly, 푠푢푝푝(푢푛 ) = 2; For푖 = 2,3,4, … , 푛 − 1
푠푢푝푝(푢푖 ) = 푣∈푁(푢푖) 푑푒푔(푣) = 푑푒푔(푣푖 ) = 3 Now, 푠푢푝푝(퐺) = 푣∈푉(퐺) 푠푢푝푝(푣) 푛 푛 = 푖=1 푠푢푝푝(푣푖 ) + 푖=1 푠푢푝푝(푢푖 ) 푛−2 = 푠푢푝푝(푣1) + 푠푢푝푝(푣2) + 푖=3 푠푢푝푝(푣푖 ) + 푠푢푝푝(푣푛−1) + 푠푢푝푝(푣푛 ) 푛−1 +푠푢푝푝(푢1) + 푠푢푝푝(푢푛 ) + 푖=2 푠푢푝푝(푢푖 ) 푛−2 푛−1 = 4 + 6 + 푖=3 7 + 6 + 4 + 2 + 푖=2 3 + 2 푠푢푝푝(퐺) = 24 + 7푛 − 28 + 3푛 − 6 = 10푛 − 10 = 10(푛 − 1).
Theorem 3.2.Let G=(m,m,...,m) be a caterpillar graph. Then 푠푢푝푝(퐺) = 5푚푛 + 푚2푛 + 4푛 − 4푚 − 6Proof. Let G=(푚, 푚, … , 푚) be a caterpillar graph. Let 푉(퐺) = {푣1, 푣2,… , 푣푛 , 푢푛 ,푢11 , … , 푢1푚 , … , 푢푛1, … , 푢푛푚 } such that 푑푒푔(푣1) = 푑푒푔(푣푛 ) = 푚 + 1; 푑푒푔(푣푖 ) = 푚 + 2,for all 푖 = 2,3, … , 푛 − 1 and 푑푒푔(푢푖푗 ) = 1 for all 푖 = 1,2, … , 푛 and 푗 = 1,2, … , 푚 푛 푠푢푝푝(푣1) = 푖=1 푑푒푔(푢1푖 ) + 푑푒푔(푣2) 푚 = 푖=1 1 + 푚 + 2 = 2푚 + 2 Similarly, 푠푢푝푝(푣푛 ) = 2푚 + 2;
푛 푠푢푝푝(푣2) = 푖=1 푑푒푔(푢2푖 ) + 푑푒푔(푣1) + 푑푒푔(푣3) 푚 = 푖=1 1 + 푚 + 1 + 푚 + 2 = 3푚 + 3 Similarly, 푠푢푝푝(푣푛−1) = 3푚 + 3; For 푖 = 3,4, … , 푛 − 2
푠푢푝푝(푣푖 ) = 푣∈푁(푣푖 ) 푑푒푔(푣) 푛 = 푑푒푔(푣푖−1) + 푗=1 푑푒푔(푢푖푗 ) + 푑푒푔(푣푖+1) 푛 = 푚 + 2 + 푗=1 1 + 푚 + 2 = 3푚 + 4 For 푗 = 1,2, … , 푚 푠푢푝푝(푢1푗 ) = 푑푒푔(푣1) = 푚 + 1 Similarly, 푠푢푝푝(푢푛푗 ) = 푚 + 1;
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For 푖 = 2,3, … , 푛 − 1 and 푗 = 1,2, … , 푚 푠푢푝푝(푢푖푗 ) = 푑푒푔(푣푖 ) = 푚 + 2 Similarly, 푠푢푝푝(푢푛푗 ) = 푚 + 2; Now, 푠푢푝푝(퐺) = 푣∈푉(퐺) 푠푢푝푝(푣) 푛 푛 푚 = 푖=1 푠푢푝푝(푣푖 ) + 푖=1 푗=1 푠푢푝푝(푢푖푗 ) 푛−2 = 푠푢푝푝(푣1) + 푠푢푝푝(푣2) + 푖=3 푠푢푝푝(푣푖 ) + 푠푢푝푝(푣푛−1) + 푠푢푝푝(푣푛 ) 푚 푚 푛−1 푚 + 푗=1 푠푢푝푝(푢1푗 ) + 푗 =1 푠푢푝푝(푢푛푗 ) + 푖=2 푗=1 푠푢푝푝(푢푖푗 ) 푛−2 = 2푚 + 2 + 3푚 + 3 + 푖=3 3푚 + 4 + 3푚 + 3 + 2푚 + 2 푚 푚 푛−1 푚 + 푗=1 (푚 + 1) + 푗 =1 (푚 + 1) + 푖=2 푗=1 (푚 + 2) = 10푚 + 10 + (3푚 + 4)(푛 − 4) + 푚(푚 + 1) + 푚(푚 + 1) +(푛 − 2)푚(푚 + 2) = 4푛 − 6 + 3푚푛 + 2푚2 + 푚2푛 − 2푚2 + 2푚푛 − 4푚 푠푢푝푝(퐺) = 5푚푛 + 푚2푛 + 4푛 − 4푚 − 6.
Theorem 3.3.Let G be a tadpole graph. Then 푠푢푝푝(퐺) = 4(푚 + 푛) + 2 Proof: Let G be a tadpole graph. Let 푉(퐺) = {푥, 푣1,푣2, … , 푣푚−1, 푢1, … , 푢푛 } such that 푑푒푔(푥) = 3; 푑푒푔(푣푖 ) = 2,for all 푖 = 1,2,3, … , 푚 − 1; 푑푒푔(푢푛 ) = 1 and 푑푒푔(푢푖 ) = 2 for all 푖 = 1,2, … , 푛 − 1 푠푢푝푝(푥) = 푣∈푁(푥) 푑푒푔(푣) = 푑푒푔(푣1) + 푑푒푔(푣푚−1) + 푑푒푔(푢1) = 2 + 2 + 2 = 6
푠푢푝푝(푣1) = 푣∈푁(푣1) 푑푒푔(푣) = 푑푒푔(푥) + 푑푒푔(푣2) = 3 + 2 = 5 Similarly, 푠푢푝푝(푣푚−1) = 5 푎푛푑푠푢푝푝(푢1) = 5; For 푖 = 2,3, … , 푚 − 2
푠푢푝푝(푣푖 ) = 푣∈푁(푣푖 ) 푑푒푔(푣) = 푑푒푔(푣푖−1) + 푑푒푔(푣푖+1) = 2 + 2 = 4 For 푖 = 2,3, … , 푛 − 2
푠푢푝푝(푢푖 ) = 푣∈푁(푢푖) 푑푒푔(푣) = 푑푒푔(푢푖−1) + 푑푒푔(푢푖+1) = 2 + 2 = 4
푠푢푝푝(푢푛−1) = 푣∈푁(푢푛 −1) 푑푒푔(푣) = 푑푒푔(푢푛−2) + 푑푒푔(푢푛 ) = 2 + 1 = 3
푠푢푝푝(푢푛 ) = 푣∈푁(푢푛 ) 푑푒푔(푣) = 푑푒푔(푢푛−1) = 2 Now, 푠푢푝푝(퐺) = 푣∈푉(퐺) 푠푢푝푝(푣) 푚−1 푛 = 푖=1 푠푢푝푝(푣푖 ) + 푠푢푝푝(푥) + 푖=1 푠푢푝푝(푢푖 ) 푚−2 = 푠푢푝푝(푣1) + 푠푢푝푝(푣푚−1) + 푖=2 푠푢푝푝(푣푖 ) + 푠푢푝푝(푥) 푛−2 +푠푢푝푝(푢1) + 푖=2 푠푢푝푝(푢푖 ) + 푠푢푝푝(푢푛−1) + 푠푢푝푝(푢푛 ) 푚−2 푛−2 = 5 + 5 + 푖=2 4 + 6 + 5 + 푖=2 4 + 3 + 2 = 26 + (푚 − 3)4 + 4(푛 − 3) 푠푢푝푝(퐺) = 4푚 + 4푛 + 2.
Theorem 3.4.Let G be a butterfly graph. Then 푠푢푝푝(퐺) = 32 Proof. Let G be a butterfly graph. Let 푉(퐺) = {푥, 푣1, 푣2,푣3,푣4} such that 푑푒푔(푥) = 4; 푑푒푔(푣푖 ) = 2,for all 푖 = 1,2,3,4; 푠푢푝푝(푥) = 푣∈푁(푥) 푑푒푔(푣)
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= 푑푒푔(푣1) + 푑푒푔(푣2) + 푑푒푔(푣3) + 푑푒푔(푣4) = 8 For 푖 = 1,2,3,4
푠푢푝푝(푣푖 ) = 푣∈푁(푣푖 ) 푑푒푔(푣) = 2 + 4 = 6 Now, 푠푢푝푝(퐺) = 푣∈푉(퐺) 푠푢푝푝(푣) = 푠푢푝푝(푣1) + 푠푢푝푝(푣2) + 푠푢푝푝(푣3) + 푠푢푝푝(푣4) + 푠푢푝푝(푥) = 32
(푚) Theorem 3.5 Let 퐺 = 퐷(푛) be a Dutch windmill graph. Then 푠푢푝푝(퐺) = 4푚(푛 + 푚 − 1) (푚 ) 푖 푖 푖 푖 Proof. Let 퐺 = 퐷(푛) be a Dutch windmill graph. Let 푉(퐺) = {푥, 푣1,푣2,푣3, … , 푣푛−1} for 푖 = 1,2, … , 푚 such that 푖 푑푒푔(푥) = 2푚; 푑푒푔(푣푗 ) = 2,for all 푖 = 1,2, … , 푚 and 푗 = 1,2, … , 푛 − 1; 푠푢푝푝(푥) = 푣∈푁(푥) 푑푒푔(푣) 푚 푖 푖 = 푖=1 (푑푒푔(푣1) + 푑푒푔(푣푛−1)) 푚 = 푖=1 (4) = 4푚 For 푖 = 1,2, … , 푚 푖 푠푢푝푝(푣 ) = 푖 푑푒푔(푣) 1 푣∈푁(푣1) = 2푚 + 2 Similarly, 푖 푠푢푝푝(푣푛−1) = 2푚 + 2 For푗 = 2,3, … , 푛 − 2 푖 푠푢푝푝 푣푗 = 푖 deg 푣 푣∈푁 푣푗 = 2 + 2 = 4 Now, 푠푢푝푝(퐺) = 푣∈푉(퐺) 푠푢푝푝(푣) 푚 푛−1 푖 = 푠푢푝푝(푥) + 푖=1 푗=1 푠푢푝푝(푣푗 ) = 4푚 + 2푚(2푚 + 2) + 4푚(푛 − 3) = 4푚2 − 4푚 + 4푚푛 = 4푚(푚 + 푛 − 1)
Theorem 3.6.Let 퐺 = 퐿2푛 . Then 푠푢푝푝(퐺) = 18푛 − 20. Proof. Let 퐺 = 퐿2푛 be a Ladder graph with 2푛 vertices. Let 푉(퐺) = {푣1,푣2,… , 푣푛 ,푢1, 푢2, … , 푢푛 } and 푑푒푔(푣푖 ) = 푑푒푔(푢푖 ) = 2 for all 푖 = 1, 푛; 푑푒푔(푣푖 ) = 푑푒푔(푢푖 ) = 3 for all 푖 = 2,3, … , 푛 − 1. Then
푠푢푝푝(푣1) = 푣∈푁(푣1) 푑푒푔(푣) = 푑푒푔(푢1) + 푑푒푔(푣2) 푠푢푝푝(푣1) = 5 Similarly, 푠푢푝푝(푣푛 ) = 푠푢푝푝(푢1) = 푠푢푝푝(푢푛 ) = 5.
푠푢푝푝(푣2) = 푣∈푁(푣2) 푑푒푔(푣) = 푑푒푔(푢2) + 푑푒푔(푣1) + 푑푒푔(푣3) 푠푢푝푝(푣2) = 8 Similarly, 푠푢푝푝(푢2) = 푠푢푝푝(푢푛−1) = 푠푢푝푝(푣푛−1) = 8. For each 푖 = 3,4, … , 푛 − 2,
푠푢푝푝(푣푖 ) = 푣∈푁(푣푖 ) 푑푒푔(푣) = 푑푒푔(푢푖 ) + 푑푒푔(푣푖−1) + 푑푒푔(푣푖+1) 푠푢푝푝(푣푖 ) = 9,for all 푖 = 3,4, … , 푛 − 2. Similarly, 푠푢푝푝 푢푖 = 9,for all푖 = 3,4, … , 푛 − 2. Now, 푠푢푝푝(퐺) = 푣∈푉(퐺) 푠푢푝푝(푣) = 푖=1,푛 푠푢푝푝 푣푖 + 푖=1,푛 푠푢푝푝 푢푖 + 푖=2,푛−1 푠푢푝푝 푣푖
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푛−2 푛−2 + 푖=2,푛−1 푠푢푝푝 푢푖 + 푖=3 푠푢푝푝 푣푖 + 푖=3 푠푢푝푝 푢푖 푛−2 푛−2 = 푖=1,푛 5 + 푖=1,푛 5 + 푖=2,푛−1 8 + 푖=2,푛−1 8 + 푖=3 9 + 푖=3 9 푠푢푝푝(퐺) = 18푛 − 20.
2 Theorem 3.7.For any graph 퐺, 푠푢푝푝(퐺) = 푣∈푉(퐺) 푑푒푔(푣) . Proof. Let 푣 ∈ 푉(퐺) be an arbitrary vertex of 퐺 such that 푑푒푔(푣) = 푘 and let 푁(푣) = {푣1, 푣2,… , 푣푘 }. Then 푘 푠푢푝푝(푣) = 푖=1 푑푒푔(푣푖 ). Similarly, 푠푢푝푝(푣푖 ) must contain 푑푒푔(푣) as its summand, for each 푖 = 1,2, … , 푘. Hence, if 푑푒푔(푣) = 푘, then 푠푢푝푝(퐺) must contain 푑푒푔(푣) as its summand at exactly 푘 times, since 푠푢푝푝(퐺) = 푣∈푉(퐺) 푠푢푝푝(푣). That is, 푠푢푝푝(퐺) must contain 푘푑푒푔(푣) as its summand. This implies that 푠푢푝푝(퐺) must contain 푑푒푔(푣) 2 as its summand. Since 푣 is arbitrary, 푠푢푝푝(퐺) must contain 푑푒푔(푣) 2 as its summand for 2 all 푣 ∈ 푉(퐺). Hence 푠푢푝푝(퐺) = 푣∈푉(퐺) 푑푒푔(푣) .
Theorem 3.8 Let 퐺 ∘ 퐾1 be a corona product of 퐺 and 퐾1. Then 푠푢푝푝(퐺 ∘ 퐾1) = 푠푢푝푝(퐺) + 4푚 + 2푛, where 푛 and 푚 are the order and size of 퐺 respectively. Proof. Let 퐺 be an any graph of order 푛 and size 푚. Let 퐺 ∘ 퐾1 be a corona product of 퐺 and 퐾1. By Theorem 2 2 (3.7), 푠푢푝푝(퐺 ∘ 퐾1) = 푣∈푉(퐺∘퐾1) 푑푒푔(푣) = 푣∈푉(퐺) 푑푒푔(푣) + 1 + 푛, Since degree of each vertex, 푣 in 퐺 ∘ 퐾1 increases one from its corresponding vertex in 퐺. Therefore 푠푢푝푝(퐺 ∘ 퐾1) = 푠푢푝푝(퐺) + 4푚 + 2푛.
REFERENCES
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