 Open Access
 Authors : K. John Bosco, R. Adlin Queen
 Paper ID : IJERTV12IS070014
 Volume & Issue : Volume 12, Issue 07 (July 2023)
 Published (First Online): 26072023
 ISSN (Online) : 22780181
 Publisher Name : IJERT
 License: This work is licensed under a Creative Commons Attribution 4.0 International License
Radio DDistance 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 ddistance 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 ddistance between u and v and () denotes the ddiameter of G. The Radio ddistance 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 ddistance number of some basic
graphs.
Keywordsddistance, Radio ddistance, Radio d distance number.

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 dddistance 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 dlength of a uv 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 ddistance labeling of a graph G. Radio ddistance labeling
is a function from V(G) to satisfying the condition
(, ) + () () 1 + (), where
() is the ddistance 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 distancetwo 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 ddistance labeling of some basic graphs.

MAIN RESULTS
Theorem 2.1
The radio ddistance number of the complete
graph, () =
Proof
Let, V() = {1,2, , } be the vertex set
, ( , ) = 2 1 ,
ddistance 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 kcoloring 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 kcoloring number () of a
It is obvious that the () = 2
The radio ddistance condition is
(, ) + () () 1 + () = 2 + 1
Now, fix (1) = 1
radio kcoloring of G is the maximum color assigned to any
(
) + ( ) ( ) 2 + 1 ( )
vertex of G. The radio kchromatic number () is min{ () } over all radio kcolorings of G. A radio
kcoloring of G is a minimal radio kcoloring if
1, 2
1 2 2
2 + 1
() = (. When k = Diam(G), the resulting radio
kcoloring is called radio coloring of G. The radio number of
1 (2) 1, which implies (2) = 2
() = , 1
Hence, () = ,
Theorem 2.2
The radio ddistance 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 ddistance 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 ddistance 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 ddistance condition
(, ) + () () 1 + () = 2 + 4 + 5
Fix, ( ) = 1, For ( )
1 1, 2
Theorem 2.3
The radio ddistance 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 Gddistance 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 ddistance 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 ddistance 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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