International Journal of Mathematics Trends and Technology (IJMTT) – Volume 65 Issue 5 - May 2019

Open Support of a Graph under Multiplication

1S. Balamurugan , 2M. Anitha , 3C. Karnan , 4P. Palanikumar 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 The Optimus Public School, Erode - 638 455 4Department of Mathematics Mannar Thirumalai Naicker College, Madurai 625 004, Tamilnadu, India.

Abstract In this paper, the open support of a vertex 푣 under multiplication and open support of a graph 퐺 under multiplication is defined and studied. Also, we find the value of open support of some namely graphs like Dutch windmill graph, Butterfly graph and Ladder graph. Moreover, we generalized the value of open support under multiplication for any given graph 퐺.

AMS Subject Code: 05C07

Keywords: Vertex Degree, Open Support under multiplication of a Vertex, Open Support under multiplication 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 are called r-regular graphs. (푚) 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 n+2(n-1) edges. The Ladder graph obtained as the cartesian product of two graphs one of which has only one edge: 퐿푛,1 = 푃푛 × 푃1 . An open support of a vertex 푣 under multiplication is defined by 푢∈푁(푣) ‍푑푒푔(푢) and is denoted by 푚푢푙푡(푣). An open support of a graph 퐺 under multiplication is defined by 푢∈푉(퐺) ‍푚푢푙푡(푢) and it is denoted by 푚푢푙푡(퐺).

II. DEFINITIONS

Definition 2.1.Let G=(V,E) be a graph. An open support of a vertex 푣 under multiplication is defined by 푢∈푁(푣) ‍푑푒푔(푢) and is denoted by 푚푢푙푡(푣).

Definition 2.2.Let G=(V,E) be a graph. An open support of a graph 퐺 under multiplication is defined by 푢∈푉(퐺) ‍푚푢푙푡(푢) and it is denoted by 푚푢푙푡(퐺).

III. RESULTS

푛−2 Proposition 3.1.For a Path 푃푛 , (푛 ≥ 2), 푚푢푙푡(푃푛 ) = 4 . Proof: Let 퐺 be a path on 푛 vertices and let 푉(퐺) = {푢1, 푢2, … , 푢푛 } where 푑푒푔(푢1) = 푑푒푔(푢푛 ) = 1 and 푑푒푔(푢푖 ) = 2for 푖 = 2,3, … , 푛 − 1.

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If 푛 = 2, 3, or 4, then clearly 푚푢푙푡(퐺) = 1, 4 and 16 respectively. Let 푛 ≥ 5. Then 푚푢푙푡(푢1) = 푚푢푙푡(푢푛 ) = 2,

푚푢푙푡(푢2) = 푚푢푙푡(푢푛−1) = 2 and

푚푢푙푡(푢푖 ) = Π푣∈푁(푢푖)푑푒푔(푣) = 4. Therefore 2 2 (푛−4) 푛−2 푚푢푙푡(퐺) = Π푢∈푉(퐺)(2 ) × (2 ) × (4 ) = 4 .

Theorem 3.2 For any 푟-regular connected graph 퐺 of order 푛 ≥ 2, 푚푢푙푡(퐺) = 푟푟푛 . Proof: Let 퐺 be a 푟-regular connected graph on 푛 vertices and let 푉(퐺) = {푢1, 푢2, … , 푢푛 } where 푑푒푔(푢푖 ) = 푟 for all 푖. Let 푛 ≥ 2. Then 푚푢푙푡(푢 ) = Π 푑푒푔(푣) = 푟 × 푟 × … × 푟 (r − times) = 푟푟 . 푖 푣∈푁(푢푖) 푟 Thus,푚푢푙푡(푢푖 ) = 푟 and hence 푚푢푙푡(퐺) = 푟푟 × 푟푟 × … × 푟푟 = 푟푟푛 .

푛 Corollary 3.3 For a Cycle 퐶푛 , (푛 ≥ 3), 푚푢푙푡(퐶푛 ) = 4 .

푛(푛−1) Corollary 3.4 For a complete graph 퐾푛 , (푛 ≥ 2), 푚푢푙푡(퐾푛 ) = (푛 − 1) .

Corollary 3.5 For a Petersen graph 푃, 푚푢푙푡(푃) = 330.

푚푛 Proposition 3.6.For a complete bipartite graph 퐾푚,푛 , (푚, 푛 ≥ 1),푚푢푙푡(퐾푚,푛 ) = (푚푛) . Proof: Let 퐺 = 퐾푚 ,푛 be a complete bipartite graph with bipartition (푉1, 푉2) where 푉1 = {푢1, 푢2, … , 푢푚 } and 푉2 = {푣1, 푣2,… , 푣푛 }. Then 푑푒푔(푢푖 ) = 푛 and 푑푒푔(푣푗 ) = 푚 for all 푖, 푗. Thus

푚푢푙푡(푢 ) = Π 푑푒푔(푣) = Π 푚 = 푚푛 fori = 1,2, … , m. 푖 푣∈푁(푢푖) 푣∈푉2 Similarly, 푚 푚푢푙푡(푣푗 ) = 푛 for 푗 = 1,2, … , 푛. Therefore 푚푢푙푡(퐺) = 푚푛푚 × 푛푚푛 = (푚푛)푚푛 .

8 6(푛−2) Proposition 3.7.Let 퐺 = 퐿2푛 , (푛 ≥ 4), be a Ladder graph.Then 푚푢푙푡(퐺) = 2 × 3 . Proof: Let 퐺 = 퐿2푛 , 푛 ≥ 4. Let 푉(퐺) = {푢1, 푢2, … , 푢푛 , 푣1,푣2, … , 푣푛 } where 푑푒푔(푢1) = 푑푒푔(푢푛 ) = 푑푒푔(푣1) = 푑푒푔(푣푛 ) = 2 and 푑푒푔(푢푖 ) = 푑푒푔(푣푖 ) = 3 for 푖 = 2,3, … , 푛 − 1. Then 푚푢푙푡(푢1) = 푚푢푙푡(푣1) = 푚푢푙푡(푢푛 ) = 푚푢푙푡(푣푛 ) = 6,

푚푢푙푡(푢2) = 푚푢푙푡(푣2) = 푚푢푙푡(푢푛−1) = 푚푢푙푡(푣푛−1) = 18 and 푚푢푙푡(푢 ) = Π 푑푒푔(푣) = 3 × 3 × 3 = 27. 푖 푣∈푁(푢푖) Similarly, 푚푢푙푡(푣푖 ) = 27. Therefore,

푚푢푙푡(퐺) = Π푢∈푉(퐺)푚푢푙푡(푢) = 64 × (18)4 × (27)2푛−8 = 24 × 34 × 24 × 38 × 36(푛−4) = 28 × 312+6(푛−4) = 28 × 36(푛−2).

4 3(푛−3) 푛−1 Proposition 3.8.For a Fan 퐹푛, (푛 ≥ 4),푚푢푙푡(퐹푛 ) = 2 × 3 × (푛 − 1) . Proof: Let 퐺 = 퐹푛 (푛 ≥ 4). Let 푉(퐺) = {푢, 푢1, 푢2, … , 푢푛−1} where 푑푒푔(푢) = 푛 − 1, 푑푒푔(푢1) = 푑푒푔(푢푛−1) = 2 and 푑푒푔(푢푖 ) = 3 for 푖 = 2,3, … , 푛 − 2. Then

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푚푢푙푡(푢) = 22 × 3푛−3,

푚푢푙푡(푢1) = 푚푢푙푡(푢푛−1) = 3 × (푛 − 1),

푚푢푙푡(푢2) = 푚푢푙푡(푢푛−2) = 6 × (푛 − 1) and

푚푢푙푡(푢푖 ) = Π푣∈푁(푢푖)푑푒푔(푣) = 9 × (푛 − 1) for 푖 = 2,3, … , 푛 − 3. Therefore

푚푢푙푡(퐺) = Π푣∈푉(퐺)푚푢푙푡(푣) = 22 × 3푛−3 × 32 × (푛 − 1)2 × 62 × (푛 − 1)2 × 9푛−5 × (푛 − 1)푛−5

= 24 × 3푛+1 × (푛 − 1)4 × 32푛−10 × (푛 − 1)푛−5 = 푚푢푙푡(퐺) = 24 × 33(푛−3) × (푛 − 1)푛−1.

4 4 2(푛−4) Theorem 3.9.Let 퐺 = 퐿2푛 . Then 푚푢푙푡(퐺) = 6 × 18 × 27 . 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

푚푢푙푡(푣1) = Π푣∈푁(푣1)푑푒푔(푣) = 푑푒푔(푢1) × 푑푒푔(푣2) 푚푢푙푡(푣1) = 6 Similarly, 푚푢푙푡(푣푛 ) = 푚푢푙푡(푢1) = 푚푢푙푡(푢푛 ) = 6.

푚푢푙푡(푣2) = Π푣∈푁(푣2)푑푒푔(푣) = 푑푒푔(푢2) × 푑푒푔(푣1) × 푑푒푔(푣3) 푚푢푙푡(푣2) = 18 Similarly, 푚푢푙푡(푢2) = 푚푢푙푡(푢푛−1) = 푚푢푙푡(푣푛−1) = 18. For each 푖 = 3,4, … , 푛 − 2,

푚푢푙푡(푣푖 ) = Π푣∈푁(푣푖)푑푒푔(푣) = 푑푒푔(푢푖 ) × 푑푒푔(푣푖−1) × 푑푒푔(푣푖+1) 푚푢푙푡 푣푖 = 27, 푓표푟푎푙푙푖 = 3,4, … , 푛 − 2. Similarly, 푚푢푙푡 푢푖 = 27, 푓표푟푎푙푙푖 = 3,4, … , 푛 − 2. Now, 푚푢푙푡(퐺) = Π푣∈푉(퐺)푚푢푙푡(푣) = Π푖=1,푛 푚푢푙푡(푣푖 ) × Π푖=1,푛 푚푢푙푡(푢푖 ) × Π푖=2,푛−1푚푢푙푡(푣푖 ) × Π푖=2,푛−1푚푢푙푡(푢푖 ) 푛−2 푛−2 × Π푖=3 푚푢푙푡(푣푖 ) × Π푖=3 푚푢푙푡(푢푖 ) 푛−2 푛−2 = 푖=1,푛 ‍6 × Π푖=1,푛 6 × Π푖=2,푛−118 × Π푖=2,푛−118 × Π푖=3 27 × Π푖=3 27 푚푢푙푡(퐺) = 64 × 184 × 272(푛−4).

Theorem 3.10.Let G=(m,m,...,m) be a caterpillar graph.Then 푚푢푙푡(퐺) = (푚 + 1)2(푚+1)(푚 + 2)푚푛 +2(푚+푛−2) Proof. 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 푚푢푙푡(푣1) = Π푖=1푑푒푔(푢1푖 ) × 푑푒푔(푣2) 푚 = Π푖=11(푚 + 2) = 푚 + 2 Similarly, 푚푢푙푡(푣푛 ) = 푚 + 2;

푛 푚푢푙푡(푣2) = Π푖=1푑푒푔(푢2푖 ) × 푑푒푔(푣1) × 푑푒푔(푣3) 푚 = Π푖=11(푚 + 1)(푚 + 2) = (푚 + 1)(푚 + 2) Similarly, 푚푢푙푡(푣푛−1) = (푚 + 1)(푚 + 2);

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For 푖 = 3,4, … , 푛 − 2

푚푢푙푡(푣푖 ) = Π푣∈푁(푣푖)푑푒푔(푣) 푛 = 푑푒푔(푣푖−1) × Π푗 =1푑푒푔(푢푖푗 ) × 푑푒푔(푣푖+1) 푛 = (푚 + 2)Π푗 =11 × (푚 + 2) = (푚 + 2)2 For 푗 = 1,2, … , 푚 푚푢푙푡(푢1푗 ) = 푑푒푔(푣1) = 푚 + 1 Similarly, 푚푢푙푡(푢푛푗 ) = 푚 + 1; For 푖 = 2,3, … , 푛 − 1 and 푗 = 1,2, … , 푚 푠푢푝푝(푢푖푗 ) = 푑푒푔(푣푖 ) = 푚 + 2 Now, 푚푢푙푡(퐺) = Π푣∈푉(퐺)푚푢푙푡(푣) 푛 푛 푚 = Π푖=1푚푢푙푡(푣푖 ) × Π푖=1Π푗 =1푚푢푙푡(푢푖푗 ) 푛−2 = 푚푢푙푡(푣1) × 푚푢푙푡(푣2) × Π푖=3 푚푢푙푡(푣푖 ) × 푚푢푙푡(푣푛−1) × 푚푢푙푡(푣푛 ) 푚 푚 푛−1 푚 × Π푗 =1푚푢푙푡(푢1푗 ) × Π푗 =1푚푢푙푡(푢푛푗 ) × Π푖=2 Π푗=1푚푢푙푡(푢푖푗 ) 2 2 2 푛−2 2 = (푚 + 2) (푚 + 1) (푚 + 2) Π푖=3 (푚 + 2) 푚 2 푛−1 푚 Π푗=1(푚 + 1) Π푖=2 Π푗 =1(푚 + 2) = (푚 + 1)2(푚 + 2)4 (푚 + 2)2 (푛−4)(푚 + 1)2푚 (푚 + 2)푚 (푛−2) = (푚 + 1)2(푚+1)(푚 + 2)4+2푛−8+푚푛 −2푚 푚푢푙푡(퐺) = (푚 + 1)2(푚+1)(푚 + 2)푚푛 +2(푚+푛−2).

Theorem 3.11 Let G be a butterfly graph. Then 푚푢푙푡(퐺) = 216 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; 푚푢푙푡(푥) = Π푣∈푁(푥)푑푒푔(푣) = 푑푒푔(푣1) × 푑푒푔(푣2) × 푑푒푔(푣3) × 푑푒푔(푣4) = 16 For 푖 = 1,2,3,4

푚푢푙푡(푣푖 ) = Π푣∈푁(푣푖)푑푒푔(푣) = 2 × 4 = 8 Now, 푚푢푙푡(퐺) = Π푣∈푉(퐺)푚푢푙푡(푣) = 푚푢푙푡(푣1) × 푚푢푙푡(푣2) × 푚푢푙푡(푣3) × 푚푢푙푡(푣4) × 푚푢푙푡(푥) = 216

(푚) (2푛−1) 푚 Theorem 3.12.Let 퐺 = 퐷(푛) be a Dutch windmill graph. Then mult(G)= 푚2 (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; 푚푢푙푡(푥) = Π푣∈푁(푥)푑푒푔(푣) 푚 푖 푖 = Π푖=1(푑푒푔(푣1) × 푑푒푔(푣푛−1)) 푚 = Π푖=1(4) = 4푚 For 푖 = 1,2, … , 푚, 푖 푚푢푙푡(푣 ) = Π 푖 푑푒푔 (푣) 1 푣∈푁(푣1) = 2푚 × 2 Similarly, 푖 푚푢푙푡(푣푛−1) = 4푚 For 푗 = 2,3, … , 푛 − 2, 푖 푚푢푙푡 푣푗 = Π 푖 푑푒푔 (푣) 푣∈푁(푣푗 ) = 2 × 2 = 4 Now,

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푚푢푙푡 (퐺) = Π푣∈푉(퐺) 푚푢푙푡 (푣). 푚 푛−2 푖 푖 푖 = 푚푢푙푡 푥 × Π푖=1{Π푗 =2 푚푢푙푡 푣푗 × 푚푢푙푡 푣1 × 푚푢푙푡(푣푛−1)} 푚 푚 푛−2 = 4 × Π푖=1{Π푗 =2 4 × 8푚} 푚 푚 푛−3 = 4 × Π푖=1{4 × 8푚} = 4푚 × 4푛−3 × 8푚 푚 = 8푚 × 4푛−2 푚 = 푚 × 23 × 22푛−4 푚 = 푚 × 22푛−1 푚

푛 2 Theorem 3.13.Let 퐺 = 퐾푚 (푎1, 푎2, … , 푎푚 ).Then 푚푢푙푡(퐺) = 푛 . Proof. Let 퐺 = 퐾푚 (푎1, 푎2, … , 푎푚 ) be a multistar graph of order 푚 + 푎1 + 푎2 + ⋯ + 푎푚 . Let 푉(퐺) = {푣1, 푣2, … , 푣푛 , 푢1, 푢2, … , 푢푛 } such that 푑푒푔(푣푖 ) = 푛and 푑푒푔(푢푖 ) = 1 for all 푖 = 1,2, … , 푛.

푚푢푙푡(푢푖 ) = Π푣∈푁(푢푖)푑푒푔(푣) 푚푢푙푡(푢푖 ) = 푛

푚푢푙푡(푣푖 ) = Π푣∈푁(푣푖)푑푒푔(푣) 푚푢푙푡(푣푖 ) = 푛 × 푛 × … × 푛[(푛 − 1)푡푖푚푒푠] 푛−1 푚푢푙푡(푣푖 ) = 푛 Now, 푚푢푙푡(퐺) = Π푣∈푉(퐺)푚푢푙푡(푣) 푛 푛 = Π푖=1푚푢푙푡(푣푖 ) × Π푖=1푚푢푙푡(푢푖 ) 푛 푛−1 푛 = Π푖=1푛 × Π푖=1푛 = 푛푛(푛−1) × 푛푛 2 푚푢푙푡(퐺) = 푛푛 .

REFERENCES

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