Radio D-Distance Number of Some Basic Graph

Radio D-Distance Number of Some Basic Graph

K. John Bosco

Assistant Professor, Department of Mathematics,

St. Judes college Thoothoor, Tamil Nadu Affiliated to Manonmaniam Sundaranar University,

Abishekapatti, Tirunelveli

R. Adlin Queen

Research Scholar (Reg. no:23113232092004), Department of Mathematics,

St. Judes college Thoothoor, Tamil Nadu Affiliated to Manonmaniam Sundaranar University,

Abishekapatti, Tirunelveli

Abstract A Radio d-distance labeling of a connected graph G is an injective map f from the vertex set V(G) to such that for two distinct vertices u and v of

G, (, ) + |() ()| + (), where (, ) denotes the d-distance between u and v and () denotes the d-diameter of G. The Radio d-distance number of f, ()

is the maximum label assigned to any vertex of G. The Radio d-

distance number of G, () is the minimum value f of G. In

this paper we find the radio d-distance number of some basic

graphs.

Keywordsd-distance, Radio d-distance, Radio d- distance number.

1. INTRODUCTION

   By a graph = (V(), ()) we mean a finite undirected graph without loops or multiple edges. Let () and () denotes the vertex set and edge set of . The order and size of are denoted by and respectively.T. Jackuline,

   J. Golden Ebenezer Jebamani and D. Premalatha introduced

   the concept of dd-distance by considering the degrees of various vertices presented in the path, in addition to the length of the path.

   Let u, v be two vertices of a connected graph G. Then the d-length of a u-v path defined as dd(u,v) = d(u, v) + deg(u) + deg(v) + deg(u) deg(v), where d(u, v) is the shortest distance between the vertices u and v

   In this paper, we introduced the concept of radio d-distance labeling of a graph G. Radio d-distance labeling

   is a function from V(G) to satisfying the condition

   (, ) + |() ()| 1 + (), where

   () is the d-distance diameter of G. A

   G is defined as the minimum span of a radio coloring of G and is denoted as rn(G).

   Radio labeling can be regarded as an extension of distance-two labeling which is motivated by the channel assignment problem introduced by W. K. Hale [6]. G. Chartrand et al.[2] introduced the concept of radio labeling of graph. Also G. Chartrand et al.[3] gave the upper bound for the radio number of path. The exact value for the radio number of path and cycle was given by Liu and Zhu [10]. However G. Chartrand et al.[2] obtained different values for them. They found the lower and upper bound for the radio number of cycle. Liu [9] gave the lower bound for the radio number of Tree. The exact value for the radio number of Hypercube was given by R. Khennoufa and O. Togni [8]. In [4] C. Fernandez et al. found the radio number for complete graph, Star graph, Complete Bipartite graph, Wheel graph and Gear graph. In this paper, we fined the radio d-distance labeling of some basic graphs.

2. MAIN RESULTS

Theorem 2.1

The radio d-distance number of the complete

graph, () =

Proof

Let, V() = {1,2, , } be the vertex set

, ( , ) = 2 1 ,

d-distance radio labeling number of G is the maximum label

assigned to any vertex of G. It is denoted by (). Let G

be a connected graph of diameter d and let k an integer such

that 1 . A radio k-coloring of G is an assignment

of colors (positive integers) to the vertices of G such that

(, ) + |() ()| 1 + for every two distinct vertices u, v of G. The radio k-coloring number () of a

It is obvious that the () = 2

The radio d-distance condition is

(, ) + |() ()| 1 + () = 2 + 1

Now, fix (1) = 1

radio k-coloring of G is the maximum color assigned to any

(

) + |( ) ( )| 2 + |1 ( )|

vertex of G. The radio k-chromatic number () is min{ () } over all radio k-colorings of G. A radio

k-coloring of G is a minimal radio k-coloring if

1, 2

1 2 2

2 + 1

() = (. When k = Diam(G), the resulting radio

k-coloring is called radio coloring of G. The radio number of

|1 (2)| 1, which implies (2) = 2

() = , 1

Hence, () = ,

Theorem 2.2

The radio d-distance number of a path

( ) 2 4 + 10, 4

Proof.

Let V() = {1, 2,, , n} be the vertex set and E() = {i+1 ; 1 1} be the edge set

Then,(1, ) = (2, ) = + 2,

(1, 2 ) = (n1, ) = 6, (, +1 ) = 9; 2 2, (2, 1 ) = + 5

It is clear that ( ) = + 5

Without loss of generality (1) < (2) < < ()

We shall check the radio d-distance condition

(, ) + |() ()| 1 + () = + 6

Fix (1) = 1 for (1, 2)

For (1, 2)

(1, 2) + |(1) (2)| 5 + |1 (2)| 2 + 3

|1 (2)| 2 2, which implies (2) = 2 1

For (2, 3)

(2, 3) + |(2) (3)| 5 + |2 1 (3)|

2 + 3

|2 1 (3)| 2 2, which implies (3) = 4 3

() = (2 2) 2 + 3, 1 Hence, (1,) = 22 4 + 3, 3

Theore. 2.4

The radio d-distance number of bistar graph,

(,) = 23 + 82 4 + 2, 2

Proof.

Let V(,) = {1, 2,, , n, 1, 2,1,2,, , n,}

be the vertex set

and E(,) = {1 , 2,12,; 1 } be the edge set

Then, (1, ) = (2, ) = 2 + 4; 1 ,

(1, 2) + |(1) (2)| 6 + |1 (2)| + 6

(

) = ( + 2)2, ( , ) = 6; 1 , ,

|1 (2)| , which implies (2) = + 1

For (2, 3)

1, 2

, (i, ) = (, ) = 5; 1 ,

(2, 3) + |(2) (3)| 9 + | + 1 (3)| + 6

| + 1 (3)| 3, which implies (3) = 2 2

It is clear that (,

) = ( + 2)2 = 2 + 4 + 4

() = ( 1) 3 + 7, 2 1

Hence, () 2 4 + 10, 4

Note. ( )= n if n = 2,3

Without loss of generality, (1) < < () <

(2) < (1) < (1) < < ()

We shall check the radio d-distance condition

(, ) + |() ()| 1 + () = 2 + 4 + 5

Fix, ( ) = 1, For ( )

1 1, 2

Theorem 2.3

The radio d-distance number of a star graph,

(1, ) = 22 4 + 3, n 3

Proof.

Let V(1, ) = {0, 1,2, , } be the vertex set, where 0 be the central vertex and

E(1,) = {0 ; 1 } be the edge set

Then, (0, ) = 2 + 2; 1 , (, ) = 5; 1 , ;

So, (1,) = 2 + 2

Without loss of generality,

(1) < (0) < (2) < < ()

We shall check the radio Gd-distance condition

(, ) + |() ()| 1 + () = 2 + 3

Fix (1) = 1, for (1, 0)

(1, 0) + |(1) (2)| 2n + 2 + |1 (0)|

2 + 3

(1,2) + |(1) (2)| 5 + |1 (2)|

2 + 4 + 5

|1 (2)| 2 + 4, which implies

(2) = 2 + 4 + 1

For (2,3)

(2,3) + |(2) (3)|

5 + |2 + 4 + 1 (3)|

2 + 4 + 5

|2 + 4 + 1 (3)| 2 + 4 which implies

(3) = 22 + 8 + 1

() = 2( 1) + (4 4)+ 1, 1

Therefore, () = 3 + 32 4 + 1

For (n, 2)

(n,2) + |() (2)|

|1 (0)| 1, which implies (0) = 2

2n + 4 + | 3

+ 32

4 + 1 (2)|

2 + 4 + 5

| 3 + 32 4 + 1 (2)| 2 + 2 + 1, which

For (, +1), 1 1

( , ) + |( ) ( )| 10 + |1 ( )|

1 2 1 2 2

implies (2) = 3 + 42 2 + 2

For (2, 1)

3 + 4

|2 (2)| 3 6, which implies (2) = 3 5

(2,1) + |(2) (1)|

For ( , ), 1

2 3

2 + 4n + 4 + | 3 + 42 2 + 2 (1)|

2 + 4 + 5

| 3 + 42 2 + 2 (1)| 1, which implies

(1) = 3 + 42 2 + 1

For (1, 1)

(2, 3) + |(2) (3)| 10 + |3 5 (3)|

3 + 4

|3 5 (3)| 3 6, which implies

(3) = 6 11

() = (3 3) 6 + 7, 1

2

(1,1) + |(1) (1)|

2n + 4 + | 3 + 42 2 + 1 (1)| 2 + 4 + 5

| 3 + 42 2 + 1 (1)| 2 + 2 + 1, which

Therefore, () = 3

For (, ), 1

9 + 7

implies (1) = 3 + 52 + 2

(, 1) + |() (1)|

8 + |32 9 + 7 (1)| 3 + 4

For (1, 2)

(1, 2) + |(1) (2)|

|32 9 + 7 (1)| 3 4 which implies

( ) = 32 6 + 3

5 + |3 + 52 + 2 (2)| 2 + 4 + 5

|3 + 52 + 2 (2)| 2 + 4, which implies

1

For ( )

(2) = 3 + 62 + 4 + 2

, +1

, 1 1

(1, 2) + |(1) (2)|

2

For (2, 3)

7 + |3

6 + 3 (2)| 3 + 4

(2, 3) + |(2) (3)|

|32 6 + 3 (2)| 3 3, which implies

(2) = 32 3

5 + |3 + 62 + 4 + 2 (3)| 2 + 4 + 5

|3 + 62 + 4 + 2 (2)| 2 + 4, which implies

(2) = 3 + 72 + 8 + 2

For (

2, 3)

() = 3 + ( + 4)2 + (4 4) + 2, 1

Hence, (,) = 23 + 82 4 + 2, 2

Theorem 2.5

(2, 3) + |(2) (3)| 7 + |32 3n (3)|

3 + 4

|32 3n (3)| 3 3, which implies

(3) = 32 3

2

The radio d-distance number of a subdivision of a

() = 3

+ (3 9) 3 + 6, 1

star, S(1, ) = 62 12 + 6, 3

Proof

Let V(S(1,)) = {0, 1, 2,, , n, 1,2,, , n,} be the vertex set, where 0 is the central vertex and E(S(1,)) = {0 , ; 1 } be the edge set

Hence, S(1,

) 62 12 + 6, 3

REFERENCES

Then, (0, ) = 3 + 3; 1 ,

(i, ) = 7, (, ) = 10; 1 ,

(, ) = 6; 1

It is clear that (S(1, )) = 3 + 3

Without loss of generality (1) < (0) < (2) < <

() < (1) < < ()

We shall check the radio d-distance condition

(, ) + |() ()| 1 + () = 3 + 4

Fix (1) = 1, for (1, 0) 1

(1, 0) + |(1) (0)| 3n + 3 + |1 (0)|

3 + 4

|1 (0)| 1 which implies (0) = 2

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