A Study on Radio Mean Dd-Distance Number of Some Basic Graphs

International Journal of Applied Engineering Research ISSN 0973-4562 Volume 14, Number 8, 2019 (Special Issue) © Research India Publications. http://www.ripublication.com

A Study on Radio Mean Dd-distance Number of Some Basic Graphs

V. VIOLA Assistant Proferssor, Department of Mathematics, St.Jude’sCollege ,Thoothoor, Tamil Nadu ManonmaniamSundaranar University Tirunelveli,

T. NICHOLAS Former Principal, St. Jude’s College , Thoothoor, Tamil Nadu.

Abstract the definition and the upper bound follows from the A Radio Mean Dd-distance labeling of a triangular inequality. connected graph G is an injective map f from the We introduced the concept of radio mean Dd- vertex set V(G) to the ℕ such that for two distinct distance colouring of a graph G. Radio mean Dd- 푓(푢)+ 푓(푣) vertices u and v of G, 퐷퐷푑(u, v) + ⌈ ⌉ ≥ 1 + distance coloring is a function ƒ : V(G) → ℕ such 2 퐷푑 푓(푢)+ 푓(푣) 퐷푑 퐷푑 that 퐷 (u, v) + ⌈ ⌉ ≥ diamDd(G) + 1, where 푑푖푎푚 (G), where 퐷 (u, v) denote the Dd-distance 2 between u and v and 푑푖푎푚퐷푑(G) denotes the Dd- 푑푖푎푚퐷푑(G) is the Dd-distance diameter of G. A Dd- diameter of G. The radio mean Dd-distance number distance radio coloring number of G is the maximum of f, 푟푚푛퐷푑(f) is the maximum label assigned to any color assigned to any vertex of G. It is denoted by vertex of G. The radio mean Dd-distance number of 푟푚푛퐷푑(G). G, 푟푚푛퐷푑(G) is the minimum value of f of G.In this Radio labeling can be regarded as an paper we find the radio mean Dd-distance number of extension of distance-two labeling which is some standard graphs. motivated by the channel assignment problem Keywords: Dd-distance, Radio mean Dd-distance, introduced by W. K. Hale [6]. G. Chartrand et al.[2] Radio Dd-distance number. introduced the concept of radio labeling of graph. Also G. Chartrand et al.[3] gave the upper bound for Subject Classification (2010): 05C78, 05C15 the radio number of path. The exact value for the radio number of path and cycle was given by Liu and Introduction Zhu [10]. However G. Chartrand et al.[2] obtained By a graph G = (V(G), E(G)) we mean a finite different values for them. They found the lower and undirected graph without loops or multiple edges. upper bound for the radio number of cycle. Liu [9] The order and size of G are denoted by p and q gave the lower bound for the radio number of Tree. respectively. The exact value for the radio number of Hypercube The Dd-distance was introduced by A. Anto Kinsley was given by R. Khennoufa and O. Togni [8]. M. M. and P. Siva Ananthi [1]. For a connected graph G, the Rivera et al.[21] gave the radio number of 퐶푛 × 퐶푛, Dd-length of a connected u – v path is defined the Cartesian product of 퐶푛. In [4] C. Fernandez et al. as퐷퐷푑(u, v) = D(u, v) + deg(u) + deg(v). The Dd- found the radio number for Complete graph, Star radius, denoted by 푟퐷푑(G) is the minimum Dd- graph, Complete Bipartite graph, Wheel graph and eccentricity among all vertices of G. That is 푟퐷푑(G) = Gear graph. M. T. Rahim and I. Tomescu [17] min{푒퐷푑(G) : v ∈ V(G)}. Similarly the Dd-diameter, investigated the radio number of Helm graph. The 퐷퐷푑(G) is the maximum Dd-eccentricity among all radio number for the generalized prism graphs were vertices of G. We observe that for any two vertices u, presented by Paul Martinez et al. in [11]. In this v of G, we have d(u, v) ≤ 퐷퐷푑(u, v). The equality paper, we fined the radioDd-distance coloring of holds if and only if u, v are identical. If G is any some basic graphs. connected graph then the Dd-distance is a metric on Main Results the set of vertices of G. We can check easily 푟퐷푑(G) Theorem 2.1. 퐷푑 퐷푑 The radio mean Dd-distance number of a complete ≤ 퐷 (G) ≤ 2푟 (G). The lower bound is clear from Dd graph 퐾푛, rmn (퐾푛) = n.

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Dd Dd Proof. D (푢푖, 푢푗) = D (푣푖, 푣푗) = 4,1 ≤ i, j ≤ n , i ≠ j 퐷푑 Dd Let {푣1, 푣2, . . . , 푣푛} be the vertex set. Then 퐷 (푣푖, D (푣푖, 푣푗) = 5, 1 ≤ i, j ≤ n , i ≠ j. 푣 ) = 3(n – 1), 1 ≤ i, j ≤ n, i ≠ j Dd 푗 So diam (퐵푛,푛) = 2n + 3. 퐷푑 So 푑푖푎푚 (퐾푛) = 3(n – 1). Without loss of generality, Let, f(푥1) < f(푥2) < 퐷푑 The radio mean Dd-distance condition is 퐷 (u, v) + f(푣1) < . . . < f(푣푛) < f(푢1) < . . . < f(푢푛). 푓(푢)+ 푓(푣) ⌈ ⌉ ≥ diamDd(G) + 1 The radio mean Dd-distance condition is DDd(u, v) + 2 푓(푢)+ 푓(푣) Dd Dd 푓(푣1)+푓(푣2) Dd ⌈ ⌉ ≥ diam (G) + 1. Now, D (v1, v2) + ⌈ ⌉ ≥ diam (퐾푛) + 1 2 2 푓(푥 )+푓(푥 ) Now, DDd(푥 , 푥 ) + ⌈ 1 2 ⌉ ≥ diamDd(퐵 ) + 1 f(v1) + f(v2) ≥ 1, which implies f(v1) = 1 and f(v2) = 1 2 2 푛,푛 2. So f(vi) = i, 1 ≤ i ≤ n. f(푥1) + f(푥2) ≥ 1, which implies f(푥1) = n – 1 and Dd Hence rmn (Kn) = n. f( 푥 2) = n. 푓(푥 )+푓(푣 ) ∎ DDd(푥 , 푣 ) + ⌈ 2 1 ⌉ ≥ diamDd(퐵 ) + 1 Theorem 2.2 2 1 2 푛,푛 The radio mean Dd-distance number of a star graph f(푥2) + f(푣1) ≥ 2n + 1, which implies f(푥2) Dd = n and f(푣1) ≥ n + 1. 퐾1,푛, rmn (퐾1,푛) ≤ 3n – 7, n ≥ 2. Proof. Therefore, f(푣1) = n + 1 푓(푣 )+푓(푣 ) DDd(푣 , 푣 ) + ⌈ 1 2 ⌉ ≥ diamDd(퐵 ) + 1 Let V(퐾1,푛) = {푣0,푣1,푣2, . . . ,푣푛} be the vertex set, 1 2 2 푛,푛 where 푣0 is the apex vertex. f(푣1) + f(푣2) ≥ 4n – 1, which implies f(푣1) = n + 1 Let V(퐾1,푛) = {푣0푣푖,1 ≤ i ≤ n} be the edge set. and f(푣2) ≥ 2n. Dd ThenD (vi, vj) = 4, 1 ≤ i, j ≤ n, i ≠ j Therefore, f(푣2) = 2n. Dd Dd D (푣0, 푣푖) = n + 2, 1 ≤ i ≤ n. So diam (퐾1,푛) = So, f(푣푖) = 2n – (i – 2), 1 ≤ i ≤ n. n + 2. Dd 푓(푣푛)+푓(푢1) Dd D (푣푛, 푢1) + ⌈ ⌉ ≥ diam (퐵푛,푛) + 1 Without loss of generality, Let, f(푣 ) < f(푣 ) < . 2 0 1 f(푣 ) + f(푢 ) ≥ 4n – 3, which implies f(푣 ) = 3n – 2 . . < f(푣 ). 푛 1 푛 푛 and f(푢 ) ≥ n – 1. The radio mean Dd-distance condition is DDd(u, v) + 1 Therefore, f(푢 ) = 3n – 1. 푓(푢)+ 푓(푣) Dd 1 ⌈ ⌉ ≥ diam (G) + 1. Dd 푓(푢1)+푓(푢2) Dd 2 D (푢1, 푢2) + ⌈ ⌉ ≥ diam (퐵푛,푛) + 1 Dd 푓(푣0)+푓(푣1) Dd 2 Now, D (푣 , 푣 ) + ⌈ ⌉ ≥ diam (Kn ) + 1 0 1 2 f(푢1) + f(푢2) ≥ 4n – 1, which implies f(푢1) = 3n – 1 f(푣0) + f(푣1) ≥ 1, which implies f(푣0) = n – 3 and and f(푢2) ≥ n. f(푣1) = n – 2 Therefore, f(푢2) = 3n. Dd 푓(푣1)+푓(푣2) Dd So, f(푢 ) = 3n – (i – 2), 1 ≤ i ≤ n. D (푣1, 푣2) + ⌈ ⌉ ≥ diam (Kn ) + 1 푖 2 Hence rmnDd(B ) ≤ 4n – 2, n ≥ 2. f(푣 ) + f(푣 ) ≥ 2n – 3 which implies f(푣 ) = n – 2 푛,푛 1 2 1 ∎ and f(푣 ) = n – 1. 2 ∗ The subdivition of a starK denoted by S(K ) is Dd 푓(푣2)+푓(푣3) Dd 1,푛 1,푛 D (푣2, 푣3) + ⌈ ⌉ ≥ diam (Kn ) + 1 2 the graph obtained from K1,푛 by f(푣2) + f(푣3) ≥ 2n – 3 which implies f(푣2) = n – 1 inserting a vertex an each edge of K1,푛. and f(푣3) ≥ n – 2 Theorem 2.4 Therefore, f(푣3) = n, f(푣푖) = n + i – 3, 0 ≤ The radio mean Dd-distance number of subdivision i ≤ n. of a star graph S(퐾1,푛), Hence rmnDd(퐾 ) ≤ 2n – 3, n ≥ 4. Dd 1,푛 rmn (S(퐾1,푛)) ≤ 4n – 9, n ≥ 5. ∎ Proof. Note.rmnDd(퐾 ) = n + 1, if n = 2, 3 1,푛 Let V(S(퐾1,푛)) = {푣0, 푣1, 푣2, . . . ,푣푛, 푢1, 푢2, . . . ,푢푛} Theorem 2.3 be the vertex set, where 푣0 is the apex The radio mean Dd-distance number of a bistargraph vertex. Let E(S(퐾 )) = {푣 푣 , 푣 푢 , 1 ≤ i ≤ n } be Dd 1,푛 0 푖 푖 푖 퐵푛,푛, rmn (퐵푛,푛) ≤ 4n – 2, n ≥ 2. the edge set. Proof. Dd Dd Then D (푣푖, 푢푖) = 4, 1 ≤ i ≤ n, D (푣0, 푢푖) = n + Let V(퐵푛,푛) = {푣1,푣2, . . . ,푣푛, 푢1, 푢2, . . . ,푢푛, 푥1, 푥2} 3, 1 ≤ i ≤ n, be the vertex set, where 푥 , 푥 Dd Dd Dd 1 2 D (푣푖, 푢푗) = D (푣푖, 푣푗) = D (푢푖, 푢푗) = 6, 1 ≤ i, j ≤ are the central vertices. Let, E(퐵푛,푛) = {푥1푢푖,푥2푣푖, Dd n , i ≠ j. So diam (S(퐾1,푛)) = n + 3. 푥1푥2:1 ≤ i ≤ n } be the edge set. Without loss of generality, Let, f(푣 ) < f(푢 ) < Dd Dd 0 1 Then D (푥1, 푢푖) = D (푥2, 푣푖) = n + 3, 1 ≤ i ≤ n, f(푢 ) < . . . < f(푢 ) < f(푣 ) < . . . < f(푣 ). Dd 2 푛 1 푛 D (푥1, 푥2) = 2n + 3

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The radio mean Dd-distance condition is DDd(u, v) Theorem 2.6 푓(푢)+ 푓(푣) + ⌈ ⌉ ≥ diamDd(G) + 1. The radio mean Dd-distance number of friendship 2 (푡) Dd (푡) 푓(푣 )+푓(푢 ) graph 퐶 ,rmn (퐶 ) ≤ 4t – 5, t ≥ 3. Now, DDd(푣 , 푢 ) + ⌈ 0 1 ⌉ ≥ diamDd(S(퐾 )) 3 3 0 1 2 1,푛 .Proof. + 1 (푡) Let V(퐶3 ) ={푣0, 푣1,푣2, . . . ,푣푡, 푢푡+1, 푢푡+2, . . . , f(푣0) + f(푢1) ≥ 1, which implies f(푣0) = 1 (푡) and f(푢 ) = 2. 푢2푡}and E(퐶3 ) = {푣푖푣푡 +푖, 푣0푣푖 : 1≤ i ≤ n} 1 Dd Dd 푓(푢 )+푓(푢 ) Then D (푣 ,푣 ) = 2t + 4, D (푣 ,푣 ) = 8. So DDd(푢 , 푢 ) + ⌈ 1 2 ⌉ ≥ diamDd(S(퐾 )) + 1 0 1 1 2 1 2 2 1,푛 Dd (푡) diam (퐶3 ) = 2t + 4. f(푢1) + f(푢2) ≥ 2n – 5, which implies f(푢1) = 2 and Without loss of generality, Let, f(푣0) < f(푣1) < f(푢2) ≥ 2n – 7. f(푣2) < . . . < f(푣푡) < f(푣푡+1) < . . . < f(푣2푡). Therefore, f(푢푖) = 2n + i– 9, 2 ≤ i ≤ n. The radio mean Dd-distance condition is DDd(u, v) + Dd 푓(푢푛)+푓(푣1) Dd D (푢 , 푣 ) + ⌈ ⌉ ≥ diam (S(퐾 )) + 1 푓(푢)+ 푓(푣) Dd 푛 1 2 1,푛 ⌈ ⌉ ≥ diam (G) + 1. f( ) + f( ) 2n – 5, which implies f( ) = 3n – 9 2 푢푛 푣1 ≥ 푢푛 Dd 푓(푣0)+푓(푣1) Dd (푡) Now, D (푣0, 푣1) + ⌈ ⌉ ≥ diam (퐶 ) + 1 and f(푣1) = 3n – 8. 2 3 So, f(푣푖) = 3n + (i – 9), 1 ≤ i ≤ n. f(푣0) + f(푣1) ≥ 1, which implies f(푣0) = 2t – 5 and Dd Hence rmn (S(K1,푛)) ≤ 4n – 9, n ≥ 5. f( 푣 1 ) = 2t – 4. Dd 푓(푣1)+푓(푣2) Dd (푡) ∎ D (푣1, 푣2) + ⌈ ⌉ ≥ diam (퐶3 ) + 1 Dd 2 Note.If n = 2, 3, 4 thenrmn (S(K1,푛)) = 2n + 1 f(푣1) + f(푣2) ≥ 4t – 7, which implies f(푣1) = 2t – 4 Theorem 2.5 and f(푣2) = 2t – 3. The radio mean Dd-distance number of path 푃 푛, f(푣 ) = 2t + i – 5, 1 ≤ i ≤ t Dd 푖 rmn (푃 ) ≤ 2n – 4, n ≥ 6 푓(푣 )+푓(푣 ) 푛 DDd(푣 , 푣 ) + ⌈ 푡 푡+1 ⌉ ≥ diamDd(퐶(푡)) + 1 Proof. 푡 푡+1 2 3 Let V(푃 푛) ={푣1,푣2, . . . ,푣푛} be the vertex set and f(푣푡) + f(푣푡+1) ≥ 4t – 7, which implies f(푣푡) = 3t – 5 E(푃 푛) ={푣푖푣푖+1, 1 ≤ i ≤ n − 1} be the and f(푣푡+1) = 3t – 4. Dd Dd edge set .Then D (푣1,푣푛) = D (푣푛, 푣2) = n + 1, Therefore, f(푣푡+푖) = 3t + i – 5, 1≤ i ≤ t. Dd Dd (푡) D (푣푖, 푣푖+1) = 5,1 ≤ i ≤ n. Hence rmn (퐶3 ) ≤ 4t – 5, t≥ 3. Dd So diam (푃 푛) = n + 1. ∎ Without loss of generality, Let, f(푣푛) < f(푣1) < Theorem 2.7 f(푣2) < . . . < f(푣푛−1). The radio mean Dd-distance number of Fan graph Dd Dd The radio mean Dd-distance condition is D (u, v) + 퐹푛,rmn (퐹푛) ≤ 2n – 4, n ≥ 5. 푓(푢)+ 푓(푣) ⌈ ⌉ ≥ diamDd(G) + 1. .Proof. 2 푓(푣 )+푓(푣 ) Let V(퐹푛) ={푣0,푣푖 : 1≤ i ≤ n } and E(퐹푛) = {푣0푣푖, Now, DDd(푣 , 푣 ) + ⌈ 1 푛 ⌉ ≥ diamDd(푃 ) + 1 1 푛 2 푛 푣푖푣푖+1 : 1≤ i ≤ n}. Dd Dd f(푣1) + f(푣푛) ≥ 1, which implies f(푣푛) = n – 4 and Then D (푣0,푣1) = 2n + 2 , D (푣1,푣2) = n + 5. So Dd f(푣1) = n – 3 . diam (퐹푛) = 2n + 2. Dd 푓(푣1)+푓(푣2) Dd Without loss of generality, Let, f(푣 ) < f(푣 ) < D (푣1, 푣2) + ⌈ ⌉ ≥ diam (푃푛) + 1 0 1 2 f(푣 ) < . . . < f(푣 ). f(푣 ) + f(푣 ) ≥ 2n – 5, which implies f(푣 ) = n – 3 2 푛 1 2 1 The radio mean Dd-distance condition is DDd(u, v) + and f(푣 ) = n – 2. 2 푓(푢)+ 푓(푣) Dd Dd 푓(푣2)+푓(푣3) Dd ⌈ ⌉ ≥ diam (G) + 1. D (푣2, 푣3) + ⌈ ⌉ ≥ diam (푃푛) + 1 2 2 푓(푣 )+푓(푣 ) Now, DDd(푣 , 푣 ) + ⌈ 0 1 ⌉ ≥ diamDd(퐹 ) + 1 f(푣2) + f(푣3) ≥ 2n – 7, which implies f(푣2) = n – 2 0 1 2 푛 and f(푣3) = n – 1. f(푣0) + f(푣1) ≥ 1, which implies f(푣0) = n – 4 and Therefore f(푣푖) = n + i – 4, 1 ≤ i ≤ n – 1. f(푣1) = n – 3. Hence, rmnDd(P ) ≤ 2n – 4, n ≥ 6. Dd 푓 ( 푣 1 ) + 푓(푣2) Dd 푛 D (푣1, 푣2) + ⌈ ⌉ ≥ diam (퐹푛) + 1 ∎ 2 Dd f(푣1) + f(푣2) ≥ 2n – 5, which implies f(푣1) = n – 3 Note.If n = 2, 3, 4, 5diam (푃푛) = n + 1 and Dd and f(푣2) ≥ n – 2. rmn (P 푛) = n. (푡) Therefore, f(푣2) = n – 2. ∗ The graph 퐶 denoted the one point union of t 푓(푣 )+푓(푣 ) 푛 DDd(푣 , 푣 ) + ⌈ 2 3 ⌉ ≥ diamDd(퐹 ) + 1 (푡) (푡) 2 3 2 푛 copies of cycle 퐶푛. The graph 퐶3 (or 퐾3 ) is calledfriendship graph. f(푣2) + f(푣3) ≥ 2n – 7, which implies f(푣2) = n – 2 and f(푣3) = n – 1. f(푣푖) = n + i – 4, 1 ≤ i ≤ n.

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Dd Hence rmn (퐹푛) ≤ 2n – 4, n≥ 5. ∎ [12] P. Murtinez, J. OrtiZ, M. Tomova,and C. Wyles, Note. “Radio Numbers For Generalized Prism Graphs, Dd rmn (퐹푛) ≤ n + 1, 2 ≤ n ≤ 4 Kodai Math. J. , 22, 131-139(1999).

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