ON HARMONIOUS COLORING OF TOTAL GRAPHS OF C(Cn), C(K1,n) AND C(Pn)

In this paper, we present the structural properties of total graph of central graph of cycles Cn, star graphs K1,n and paths Pn denoted by T (C(Cn)), T (C(K1,n)) and T (C(Pn)) respectively. We mainly focus our discussion on the harmonious chromatic number of T (C(Cn)), T (C(K1,n)) and T (C(Pn)).


Introduction
The central graph [6,7] C(G) of a graph G is formed by adding an extra vertex on each edge of G, and then joining each pair of vertices of the original graph which were previously non-adjacent.
A harmonious coloring [2,4,5] of a simple graph G is proper vertex coloring such that each pair of colors appears together on at most one edge.The harmonious chromatic number χ H (G) is the least number of colors in such a coloring.
The total graph of G has vertex set V (G) ∪ E(G), and edges joining all elements of this vertex set which are adjacent or incident in G.The terminology and notations used in this paper can be found in [1,3].The purpose of this paper serves to find the harmonious chromatic number for the total graph of central graph of C n , K 1,n and P n .

Harmonious Coloring on Total Graph of Central Graph of Cycle C n
In T (C(C n )), there are n vertices of degree 4, 2n vertices of degree (n + 1), Ã n 2 !vertices of degree 2(n − 1).
Theorem 2.1.The harmonious chromatic number of total graph of central graph of cycles C n , Each v i is incident with the edges e i , e 0 i−1 , e ij (i 6 = j) and (2 ≤ i ≤ n).Also v 1 is incident with e 1 , e 0 n ,e 13 ,e 1n ,• • • , e 1(n−1) .i.e., Total number of incident edges with v i is (n − 1)∀i = 1, 2, • • • , n.By the definition of total graph the edges incident with v i together with the vertex v i induces a clique of n vertices in n shares exactly (n − 3) vertices with the remaining cliques.Also the vertices v i 's are adjacent with u i 's (1 ≤ i ≤ n) and u i 's are adjacent with v i+1 for every i = 1, 2, Therefore in each clique, the harmonious coloring is performed by distinct n colors.K (1) n is assigned n colors, where as since K  Assign 5 distinct colors to the vertices of K 5 .One of these 5 colors can be assigned to the vertices of K n+2 are assigned by Therefore by induction hypothesis χ If n is even, we prove that,χ For n = 6, the total graph of C(C 6 ) has harmonious chromatic number 26.

Central Graph of Cycle C 6
Figure 6 The harmonious coloring of T (C(C 6 )) is as follows.Assign 6 distinct colors to the vertices of K (1) 6 .One of these 6 colors can be assigned to the vertices of K In T (C(K 1,n )) there are n vertices of degree 4, 2n vertices of degree 2n vertices of degree (n + 2). Therefore • The number of edges, q T (C(K 1,n )) = n 3 + 3n 2 + 10n 2 .
Theorem 3.1.The harmonious chromatic number of total graph of central graph of star graph By the definition of central graph of K 1,n , we denote the vertices of subdivision by By the definition of central graph the subgraph induced by the vertex set and let e ij be the edge of C(K 1,n ), connecting the vertex v i and v j (i < j).

Central Graph of Star
Figure 10 The vertices v, e 0 1 , e 0 2 , • • • , e 0 n induce a clique of order (n + 1) in its total graph.The vertices u i adjacent to e 0 1 and e i (1 . Now the vertices v i and e 0 i together with vertices of S i induce a clique of order (n + 1), (1 a clique of order n.Now we prove that harmonious chromatic number of this graph by induction method.

Case (i)
If n is even If n = 2, then C( K 1,2 ) has 5 vertices.The harmonious coloring for the total graph of C(K 1,2 ), is shown below.

Total Graph of Central Graph of Star Graph K 1,2
Figure 11 From the Fig. 11, 9 is the minimum number of colors as the 5 triangles shares atleast one vertex in common.Therefore χ H (T (C(K 1,2 ))) = 9 = n 2 + 3n + 8 2 , when n = 2. Therefore the result is true if n = 2. Assume that the result is true for any even integer n and we prove the same for n + 2. i.e., χ H (T (C(K 1,n ))) = n 2 + 3n + 8 2 .Let v n+1 , v n+2 be two non adjacent vertices introduced in K 1,n which are adjacent to v. Let u n+1 and u n+2 be the vertices of subdivision in its centralization.Clearly by the structure given in Fig. 10, the total graph C(K 1,n+2 ) has the following structural property.(i) There are (n+3) cliques n+3 of order (n + 3).(ii) There is a clique of order (n + 2).(iii) Each K (i) n+3 has exactly one vertex common with K (j) n+3 where (2 ≤ i ≤ n + 3), (2 ≤ j ≤ n + 3) and i 6 = j.By induction hypothesis the minimum number of colors for )) are e n+1 , e n+2 , u n+1 , u n+2 , e 0 n+1 , e 0 n+2 , v n+1 , v n+2 .Therefore the total number of vertices in T 2 + 8 = 2n + 1 + 8 = 2n + 9. Now we find the minimal harmonious coloring in T (C(K 1,n+2 )) as below.By induction hypothesis T (C(K 1,n )) has an harmonious coloring with the minimum number of n 2 + 3n + 8 2 colors.With this same color assigned to the vertices of T (C(K 1,n+2 )), we assign some new colors to the remaining vertices as below.The vertices u n+1 and u n+2 are assign the same color as in u i (1 ≤ i ≤ n).Then all the remaining vertices are assigned 2n + 5 colors.Therefore
Theorem 4.1.The harmonious chromatic number of total graph of central graph of path P n , χ H (T (C(P n ))) = 4n−3, where n denotes the number of vertices in P n .
On the process of centralization of P n , let u i be the vertex of subdivision of the edges . By the definition of central graph the non adjacent vertices v i and v j of P n , are adjacent in C(P n ) by the edge e ij .

Path P n
Central Graph of Path P n Figure 12 Figure 13 Therefore n shares exactly (n − 3) vertices with the remaining cliques.Also the vertices v i 's are adjacent with u i 's for i Total Graph of Central Graph of Path P n Figure 14 In each of the above said clique, harmonious coloring is performed by distinct n colors.K  Assume that the result is valid for any n, χ H (T (C(P n ))) = 4n − 3. Now consider P n+1 by introducing a vertex v n+1 and consider the incident edges of v n+1 in C(P n+2 ).These edges together with the vertex v n+1 induces one more clique of order n in T (C(P n+1 )) and it shares n − 4 vertices with other cliques.Therefore we need 4 colors for the harmonious coloring of T (C(P n )).Therefore χ H (T (C(P n+1 )) = 4n − 3 + 4 = 4n + 1 = 4(n + 1) − 3. Hence by induction hypothesis χ H (T (C(P n+1 ))) = 4n − 3. 2

n
, it needs distinct (n − 1) colors which are distinct from the set of colors assigned to K

( 2 )
n , it needs only (n − 2) colors and so on.Now we use induction.Case (i)If n is odd, If n = 5, T (C(C 5 )) has the harmonious coloring as follows.Central Graph of Cycle C 5 Figure 3
By the definition of total graph, V (T (C(P n ))) = V (C(P n )) ∪ E(C(P n )).In C(P n ), each v i is incident with the (n − 1) edges e i , e 0 i−1 , e ij for ( 2 ≤ i ≤ n − 1) , (i 6 = j), v 1 is incident with the (n − 1) edges e 1 , e 13 , e 14 , • • • e 1n and v n is incident with (n−1) edges e 0 n−1 , e 2n , e 3n , • • • e (n−2)n , e 1n .The edges incident with v i together with vertex v i induces a clique of n vertices in T (C(P n )

( 1 )
n is assigned n colors, where as since K (2) n shares one vertex with K (1) n , it needs distinct (n − 1) colors which are distinct from the set of colors assigned to K

( 2 )
n , it needs only (n − 2) colors and so on.Now we use induction on n.For n = 1, the results follows obviously.For n = 2, C(P 2 ) is P 3 and T (C(P 2 )) and its harmonious coloring is as follows.Total Graph of Central Graph of Path P 2Figure15