- Email: [email protected]
Generalization of Achromatic Colouring of Central Graphs
Electronic Notes in Discrete Mathematics 33 (2009) 147–152 www.elsevier.com/locate/endm
Generalization of Achromatic Colouring of Central Graphs K. Thilagavathi
1
N. Roopesh
2
Department of Mathematics, Kongunadu Arts and Science College, Coimbatore - 641 029, India.
Abstract In this paper we discuss the achromatic coloring of graphs, and the achromatic number of central graph of Wheel graphs and Gear graphs and generalize the achromatic colouring of central graphs Keywords: Achromatic coloring, achromatic number, central graph, wheel graphs and gear graphs.
1
Introduction
Let G be a finite undirected graph with no loops and multiple edges. The Central graph [9, 10] C(G) of a graph G is obtained by subdividing each edge of G exactly once and joining all the non adjacent vertices of G. By the definition,PC (G) = p + q.For any (p, q) graph there exists exactly p vertices of degree (p − 1) and q vertices of degree 2 in C(G). An achromatic colouring [5] of a graph is a proper vertex colouring such that each pair of colour classes is adjacent by at least one edge. The largest possible number of colours in an achromatic colouring is called the achromatic number, and is denoted by ψ 1 2
Email:[email protected] Email:[email protected]
1571-0653/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2009.03.021
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Achromatic colouring of wheel graphs
Wheel graphs A wheel graph Wn of order n is a graph that contains a cycle of order n − 1, and for which every graph vertex in the cycle is connected to one other graph vertex (which is known as the hub). The edges of a wheel which includes the hub are called spokes. The wheel Wn can be defined as the graph K1 + Cn−1 . Theorem 2.1 For the wheel graph Wn ,ψ[C(Wn )] = n + 1. Proof. Let Wn be the wheel graph on n verticesv1 , v2 , v3 , ...vn wherevn is the hub. Consider its central graphC(Wn ). let v( i, j) represents the newly introduced vertex in the edge connectingvi andvj ,1 ≤ i, j ≤ n. In C(Wn ) , we can observe that for i = 2, 3, 4, ...n − 2,vi is adjacent with all the vertices exceptvi−1 , vi+1 and vn . Further v1 and vn−1 are not adjacent with v2 , vn−i , vn and vn−2 , v1 , vn respectively. Now assign a proper colouring to these vertices as follows. Consider the colour class C = {c1 , c2 , c3 , ...cn+1 }. For i = 1, 2, 3, ..., n− 1 assign the colour ci+1 to vi and c1 to vn . For i = 2, 3, 4, ...n − 2 we miss the pairs (ci , ci+1 ), (ci−1 , ci ) and (c2 , cn ) and for i = 2, 3, 4, ...n we miss the pair (c1 , ci ) . Thus the colouring of vi,j must be in such a way that it accommodates all these pairs. Thus colour vi,j as follows. Consider the star graph formed by the vertices vn and vi,n , i = 1, 2, 3, ..., n − 1, for i = 2, 3, ..., n − 1,colour the vertex vi,n as ci and v1,n as cn+1 . This will accommodate all the missing pairs except (c1 , cn ) and (ci , cn + 1), i = 3, 4, ..., n . To include these pairs, for i = 2, 3, 4, , ...n − 2 colour vi,j , j = i + 1 as cn+1 and v1,n−1 , v1,2 as c1 and cn respectively. Then the above said colouring is achromatic. To prove the above said colouring is achromatic, we consider a pair (ci , cj ) Case 1 i = 2, 3, 4, ..., n − 1, n, j = i + 1 Then the pair (ci , cj ) is given by the pair(vi i, n, vi ), i = 2, 3, 4, ..., n − 1 and if i = n the required pair is given by (vn−2,n−1 , vn−1 ). Case 2 i = 1, 2, 3, ..., n − 1, n, j = i + 1 Sub case 2.1 i = 1, j = 2, 3, 4, ..., n − 1 The required pair (ci , cj ) is given by (vn , vj,n ). Sub case 2.2 i = 1, j = n, n + 1 Then the pairs (vn−1 , v1,n−1 ) and (vn , v1,n ) respectively will stand for the pair
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(ci , cj ). Sub case 2.3 i = 2, 3, 4, ..., n − 2, j = n, n + 1 Here (vi−1 , vj−1 ) will stand for the pair (ci , cj ). Sub case 2.4 i = 2, 3, 4, ..., n − 2, j = n If i = 2,(ci , cj ) is given by the pair (vi−1 , vi−1,i ) . When i = 3, 4, 5, ..., n − 2, (ci , cj ) is given by the pair (vi−1 , vn−1 ). Sub case 2.5 i = 2, 3, 4, ..., n − 2, n − 1, j = n + 1 If i = 2 ,(ci , cj ) is given by the pair (vi−1 , vi−1,n ) . For the rest of values of i ,(vi−1 , vi−1,i ) stands for the required pair (ci , cj ). Thus the given colouring is achromatic. And by the very construction, it is the maximal colour class. Therefore ψ[c(Wn )] = n + 1. 2
3
Achromatic Colouring of Central graph of Gear graphs
Gear graphs A gear graph sometimes also known as a bipartite wheel graph, is a wheel graph with a graph vertex added between each pair of adjacent graph vertices of the outer cycle. It is denoted by Gn where n denotes the number of vertices of the wheel graph. Theorem 3.1 For any gear graph Gn , ψ[C(Gn )] = 2n − 1. Proof. Let Gn be a gear graph formed from the wheel graph Wn of n vertices. Let the vertex set ofGn be{v1 , v2 , v3 , ..., v2n−1 } where v2n−1 is the hub of the wheel. Consider its central graph C(Gn ) . Let vi,j represents the newly introduced vertex in the edge connecting vi and vj , 1 ≤ i, j ≤ 2n − 1 in C(Gn ) .Note that in C(Gn ), for i = 2, 3, 4, ..., 2n − 3, vi is adjacent with neither vi−1 nor vi+1 , v2n−2 is not adjacent with v − 2n − 3 and v1 , v1 is not adjacent with v2 and v2n−2 . Also v2n−1 is not adjacent with v2m+1 , m = 0, 1, 2, ..., n − 2 (or v2n−1 is not adjacent with v2m , m = 1, 2, 3, ..., n − 1). Now assign a proper vertex colouring as follows:Consider the colour class K = {k1 , k2 , k3 , ...k2n−1 } . For i = 1, 2, 3, ..., 2n−1 , assign the colour ki to vi . Such a colouring will not contribute some pairs, namely (k1 , k2 ), (k1 , k2n−2 ), (ki , ki+1 ), (ki−1 , ki ) , where i = 2, 3, 4, ..., 2n−3, (k2n−2 , k2n−3 ), (k2n−2 , k1 ) and (ki , k2m+1 ) or (ki , k2m ) where i = 2n − 1 and m = 0, 1, 2, ..., n − 2 (or i = 2n − 1 and m = 1, 2, 3, ...n − 1). Thus the colouring to be achromatic, the vertices vi,j should be coloured in such a way that it accommodate all these pairs.Consider the following colour-
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ing. Assign the colour kj+1 to the vertices vi,j wherei = 1, 2, 3, ...2n − 4 and j = i + 1 . Colour the vertices v2n−3,2n−2 and v1,2n−2 as k1 and k2 respectively. Without loss of generality, assume that v1 is not adjacent with v2n−1 . For i = 2m + 1, m = 0, 1, 2, ..., n − 3 colour the vertices vi,2n−1 as ki+2 and for m = n − 2 colour vi,2n−1 ask1 (if v1 is adjacent with v2n−1 , for i = 2m, m = 1, 2, 3, ..., n−2 colour the vertices vi,2n−1 as ki+2 and for m = n−1 colour vi,2n−1 as k1 .) Then the above said colouring is achromatic. For, consider any pair(ki , kj ) . Case 1 i = 1, 2, 3, ..., 2n − 2, j = i + 1 Sub case 1.1 i = 1, j = i + 1 The required pair is given by (v1 , v1,1,2n−2 ) Sub case 1.2 i = 2, 3, 4, ..., 2n − 3, j = i + 1 The required pair (ki , kj ) is given by (vi , vi−1,i ). Sub case 1.3i = 2n − 2, j = i + 1 Then (v2n−2 , v2n−1 ) will stand for the pair (ki , kj ) . Case 2 i = 1, 2, 3, ..., 2n − 1, j = i + 1 Sub case 2.1 i = 1, j = i + 1, j = 2n − 2 and 2n − 1 The required pair is given by (vi , vj ) . If i = 1, j = 2n − 2 the required pair is (v2n−3,2n−2 , v2n−2 ). Sub case 2.2 i + 2, 3, 4, ..., 2n − 2, j = i + 1, 2n − 1 The required pair (ki , kj ) is given by (vi , vj ) . Sub case 2.3 i = 1, 2, 3, ..., 2n − 3, j = 2n − 1 Suppose v1 is not adjacent with v2n−1 , Then the required pair is given by (vi , v2n−1 ) if i is even. Fori = 1 the required pair is (v2n−3,2n−1 , v2n−1 ). And for the rest of the odd values (vi−2,2n−1 , v2n−1 ) will stand for the pair (ki , kj ). If v1 is adjacent with v2n−1 , the required pair is given by (vi , v2n−1 if i is odd. For i = 2 the required pair is (v2n−2,2n−1 , v2n−1 ) . For the rest of the even values of i pair (ki , kj ) is given by (vi−2,2n−1 , v2n−1 ) . Hence the colouring is achromatic. Further by the construction, it is the maximal colouring. There fore ψ[C(Gn )] = 2n − 1 . 2
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Achromatic number of central graphs
Theorem 4.1 Achromatic number of central graph of any simple graph G(V, E) with |V | = n is less than or equal to n + 1. Proof. Consider the graph G(V, E) with V = v1 , v2 , v3 , ..., vn . Let N (vi ) = {v1 , v2 , v3 , ..., vk }, k < n , . Then by the definition of central graph, in C(G), vi is not adjacent with any of the vj s, j = 1, 2, 3, ..., k instead vi is adjacent with all vj , j > k, and vi,j , j = 1, 2, 3, ..., k .Consider the colour class K = {c1 , c2 , c3 , ..., cn } .Assign the colour ci to vi . Then in C(G) we have the pairs of the form (ci , cj ) ,j > k . Let m < k , now to include the pair (ci , cm ) either we should colour any of the new neighbor of vi as cm or colour any of the new neighbor of vm as ci . Now the following cases may arise. Case1:The colouring of vi,j ’ s will accommodate exactly the remaining pairs.Thus further introduction of new colours is not possible. Therefore it is a maximal colouring and ψ[C(G)] = n. Case2:The colouring of vi,j ’ s fails to accommodate any remaining pair. This case implies that the colouring of the graph with these n colours is not achromatic. Thus to get an achromatic colouring we should use less than n colours. Therefore ψ[C(G)] < n. Case3:The colouring of vi,j ’ s will accommodate the remaining pairs and leave some vertices without colour. In this case assign the colour cn + 1 to these vertices. Then one of the following two sub cases may occur. Sub case1:For each i we can find a (ci , cn + 1) pair.Then clearly ψ[C(G)] = n + 1. Note that even though there are so many vi,j remaining to be coloured we cant colour these vi,j with two or more different colours because there is no edge in between any two vi,j s. Sub case2:For some i the pair (ci , cn+1 ) is missing. Then replace the colour cn + 1 with any of the remaining colours selected from the previously considered colour class K so that the colouring is proper.Then ψ[C(G)] < n + 1. Thus in all the cases ψ[C(G)] ≤ n + 1. 2
References [1] R.A. Brualdi and H.J.Rayser, Combinatorial Matrix Theory, Cambridge University press, 1991. [2] J. A. Bondy and U.S.R. Murty (1976), Graph theory with Applications, London: MacMillan .
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[3] Jonathan gross and Jay Yellen (2004), Hand book of Graph theory, CRC Press, New York [4] Keith Edwards (1997), The harmonious chromatic number and the achromatic number, Surveys in combinatorics 1997 by R (Rosmary) Bailey. [5] Frank Harary (1969), Graph Theory, Narosa Publishing home . [6] Frank Harary and Stephen Hedetniemi, The achromatic number of a graph, Journal of Combinatorial Theory,(1970), 154-161 . [7] Frank Harary, Stephen Hedetniemi and Geert Prins, An interpolation theorem for graphical homomorphisms, Portugaliae Mathematica, vol.26-Fasc.4. [8] K. Thilagavathi and N. Roopesh, Achromatic colouring of central graphs and split graphs, Far East Journal of Applied Mathematics, volume 30, issue 3, 359-369. [9] K.Thilagavathi and Vernold Vivin,J., ”Harmonious Colouring of Graphs”, Far East J. Math.Sci.(FJMS), Volume 20 No.2, 189-197. [10] K.Thilagavathi, Vernold Vivin,J and Akbar Ali.M.M, On Harmonious Colouring of Central Graphs” Advances and applications in discrete mathematics, vol. 2, 17-33. [11] Vernold Vivin,J((May 2007),”Harmonious Colouring of Total graphs, n-leaf, Central Graphs and Circumdetic graphs, PhD thesis.
Generalization of Achromatic Colouring of Central Graphs K. Thilagavathi
1
N. Roopesh
2
Department of Mathematics, Kongunadu Arts and Science College, Coimbatore - 641 029, India.
Abstract In this paper we discuss the achromatic coloring of graphs, and the achromatic number of central graph of Wheel graphs and Gear graphs and generalize the achromatic colouring of central graphs Keywords: Achromatic coloring, achromatic number, central graph, wheel graphs and gear graphs.
1
Introduction
Let G be a finite undirected graph with no loops and multiple edges. The Central graph [9, 10] C(G) of a graph G is obtained by subdividing each edge of G exactly once and joining all the non adjacent vertices of G. By the definition,PC (G) = p + q.For any (p, q) graph there exists exactly p vertices of degree (p − 1) and q vertices of degree 2 in C(G). An achromatic colouring [5] of a graph is a proper vertex colouring such that each pair of colour classes is adjacent by at least one edge. The largest possible number of colours in an achromatic colouring is called the achromatic number, and is denoted by ψ 1 2
Email:[email protected] Email:[email protected]
1571-0653/$ – see front matter © 2009 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2009.03.021
148
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K. Thilagavathi, N. Roopesh / Electronic Notes in Discrete Mathematics 33 (2009) 147–152
Achromatic colouring of wheel graphs
Wheel graphs A wheel graph Wn of order n is a graph that contains a cycle of order n − 1, and for which every graph vertex in the cycle is connected to one other graph vertex (which is known as the hub). The edges of a wheel which includes the hub are called spokes. The wheel Wn can be defined as the graph K1 + Cn−1 . Theorem 2.1 For the wheel graph Wn ,ψ[C(Wn )] = n + 1. Proof. Let Wn be the wheel graph on n verticesv1 , v2 , v3 , ...vn wherevn is the hub. Consider its central graphC(Wn ). let v( i, j) represents the newly introduced vertex in the edge connectingvi andvj ,1 ≤ i, j ≤ n. In C(Wn ) , we can observe that for i = 2, 3, 4, ...n − 2,vi is adjacent with all the vertices exceptvi−1 , vi+1 and vn . Further v1 and vn−1 are not adjacent with v2 , vn−i , vn and vn−2 , v1 , vn respectively. Now assign a proper colouring to these vertices as follows. Consider the colour class C = {c1 , c2 , c3 , ...cn+1 }. For i = 1, 2, 3, ..., n− 1 assign the colour ci+1 to vi and c1 to vn . For i = 2, 3, 4, ...n − 2 we miss the pairs (ci , ci+1 ), (ci−1 , ci ) and (c2 , cn ) and for i = 2, 3, 4, ...n we miss the pair (c1 , ci ) . Thus the colouring of vi,j must be in such a way that it accommodates all these pairs. Thus colour vi,j as follows. Consider the star graph formed by the vertices vn and vi,n , i = 1, 2, 3, ..., n − 1, for i = 2, 3, ..., n − 1,colour the vertex vi,n as ci and v1,n as cn+1 . This will accommodate all the missing pairs except (c1 , cn ) and (ci , cn + 1), i = 3, 4, ..., n . To include these pairs, for i = 2, 3, 4, , ...n − 2 colour vi,j , j = i + 1 as cn+1 and v1,n−1 , v1,2 as c1 and cn respectively. Then the above said colouring is achromatic. To prove the above said colouring is achromatic, we consider a pair (ci , cj ) Case 1 i = 2, 3, 4, ..., n − 1, n, j = i + 1 Then the pair (ci , cj ) is given by the pair(vi i, n, vi ), i = 2, 3, 4, ..., n − 1 and if i = n the required pair is given by (vn−2,n−1 , vn−1 ). Case 2 i = 1, 2, 3, ..., n − 1, n, j = i + 1 Sub case 2.1 i = 1, j = 2, 3, 4, ..., n − 1 The required pair (ci , cj ) is given by (vn , vj,n ). Sub case 2.2 i = 1, j = n, n + 1 Then the pairs (vn−1 , v1,n−1 ) and (vn , v1,n ) respectively will stand for the pair
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(ci , cj ). Sub case 2.3 i = 2, 3, 4, ..., n − 2, j = n, n + 1 Here (vi−1 , vj−1 ) will stand for the pair (ci , cj ). Sub case 2.4 i = 2, 3, 4, ..., n − 2, j = n If i = 2,(ci , cj ) is given by the pair (vi−1 , vi−1,i ) . When i = 3, 4, 5, ..., n − 2, (ci , cj ) is given by the pair (vi−1 , vn−1 ). Sub case 2.5 i = 2, 3, 4, ..., n − 2, n − 1, j = n + 1 If i = 2 ,(ci , cj ) is given by the pair (vi−1 , vi−1,n ) . For the rest of values of i ,(vi−1 , vi−1,i ) stands for the required pair (ci , cj ). Thus the given colouring is achromatic. And by the very construction, it is the maximal colour class. Therefore ψ[c(Wn )] = n + 1. 2
3
Achromatic Colouring of Central graph of Gear graphs
Gear graphs A gear graph sometimes also known as a bipartite wheel graph, is a wheel graph with a graph vertex added between each pair of adjacent graph vertices of the outer cycle. It is denoted by Gn where n denotes the number of vertices of the wheel graph. Theorem 3.1 For any gear graph Gn , ψ[C(Gn )] = 2n − 1. Proof. Let Gn be a gear graph formed from the wheel graph Wn of n vertices. Let the vertex set ofGn be{v1 , v2 , v3 , ..., v2n−1 } where v2n−1 is the hub of the wheel. Consider its central graph C(Gn ) . Let vi,j represents the newly introduced vertex in the edge connecting vi and vj , 1 ≤ i, j ≤ 2n − 1 in C(Gn ) .Note that in C(Gn ), for i = 2, 3, 4, ..., 2n − 3, vi is adjacent with neither vi−1 nor vi+1 , v2n−2 is not adjacent with v − 2n − 3 and v1 , v1 is not adjacent with v2 and v2n−2 . Also v2n−1 is not adjacent with v2m+1 , m = 0, 1, 2, ..., n − 2 (or v2n−1 is not adjacent with v2m , m = 1, 2, 3, ..., n − 1). Now assign a proper vertex colouring as follows:Consider the colour class K = {k1 , k2 , k3 , ...k2n−1 } . For i = 1, 2, 3, ..., 2n−1 , assign the colour ki to vi . Such a colouring will not contribute some pairs, namely (k1 , k2 ), (k1 , k2n−2 ), (ki , ki+1 ), (ki−1 , ki ) , where i = 2, 3, 4, ..., 2n−3, (k2n−2 , k2n−3 ), (k2n−2 , k1 ) and (ki , k2m+1 ) or (ki , k2m ) where i = 2n − 1 and m = 0, 1, 2, ..., n − 2 (or i = 2n − 1 and m = 1, 2, 3, ...n − 1). Thus the colouring to be achromatic, the vertices vi,j should be coloured in such a way that it accommodate all these pairs.Consider the following colour-
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ing. Assign the colour kj+1 to the vertices vi,j wherei = 1, 2, 3, ...2n − 4 and j = i + 1 . Colour the vertices v2n−3,2n−2 and v1,2n−2 as k1 and k2 respectively. Without loss of generality, assume that v1 is not adjacent with v2n−1 . For i = 2m + 1, m = 0, 1, 2, ..., n − 3 colour the vertices vi,2n−1 as ki+2 and for m = n − 2 colour vi,2n−1 ask1 (if v1 is adjacent with v2n−1 , for i = 2m, m = 1, 2, 3, ..., n−2 colour the vertices vi,2n−1 as ki+2 and for m = n−1 colour vi,2n−1 as k1 .) Then the above said colouring is achromatic. For, consider any pair(ki , kj ) . Case 1 i = 1, 2, 3, ..., 2n − 2, j = i + 1 Sub case 1.1 i = 1, j = i + 1 The required pair is given by (v1 , v1,1,2n−2 ) Sub case 1.2 i = 2, 3, 4, ..., 2n − 3, j = i + 1 The required pair (ki , kj ) is given by (vi , vi−1,i ). Sub case 1.3i = 2n − 2, j = i + 1 Then (v2n−2 , v2n−1 ) will stand for the pair (ki , kj ) . Case 2 i = 1, 2, 3, ..., 2n − 1, j = i + 1 Sub case 2.1 i = 1, j = i + 1, j = 2n − 2 and 2n − 1 The required pair is given by (vi , vj ) . If i = 1, j = 2n − 2 the required pair is (v2n−3,2n−2 , v2n−2 ). Sub case 2.2 i + 2, 3, 4, ..., 2n − 2, j = i + 1, 2n − 1 The required pair (ki , kj ) is given by (vi , vj ) . Sub case 2.3 i = 1, 2, 3, ..., 2n − 3, j = 2n − 1 Suppose v1 is not adjacent with v2n−1 , Then the required pair is given by (vi , v2n−1 ) if i is even. Fori = 1 the required pair is (v2n−3,2n−1 , v2n−1 ). And for the rest of the odd values (vi−2,2n−1 , v2n−1 ) will stand for the pair (ki , kj ). If v1 is adjacent with v2n−1 , the required pair is given by (vi , v2n−1 if i is odd. For i = 2 the required pair is (v2n−2,2n−1 , v2n−1 ) . For the rest of the even values of i pair (ki , kj ) is given by (vi−2,2n−1 , v2n−1 ) . Hence the colouring is achromatic. Further by the construction, it is the maximal colouring. There fore ψ[C(Gn )] = 2n − 1 . 2
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Achromatic number of central graphs
Theorem 4.1 Achromatic number of central graph of any simple graph G(V, E) with |V | = n is less than or equal to n + 1. Proof. Consider the graph G(V, E) with V = v1 , v2 , v3 , ..., vn . Let N (vi ) = {v1 , v2 , v3 , ..., vk }, k < n , . Then by the definition of central graph, in C(G), vi is not adjacent with any of the vj s, j = 1, 2, 3, ..., k instead vi is adjacent with all vj , j > k, and vi,j , j = 1, 2, 3, ..., k .Consider the colour class K = {c1 , c2 , c3 , ..., cn } .Assign the colour ci to vi . Then in C(G) we have the pairs of the form (ci , cj ) ,j > k . Let m < k , now to include the pair (ci , cm ) either we should colour any of the new neighbor of vi as cm or colour any of the new neighbor of vm as ci . Now the following cases may arise. Case1:The colouring of vi,j ’ s will accommodate exactly the remaining pairs.Thus further introduction of new colours is not possible. Therefore it is a maximal colouring and ψ[C(G)] = n. Case2:The colouring of vi,j ’ s fails to accommodate any remaining pair. This case implies that the colouring of the graph with these n colours is not achromatic. Thus to get an achromatic colouring we should use less than n colours. Therefore ψ[C(G)] < n. Case3:The colouring of vi,j ’ s will accommodate the remaining pairs and leave some vertices without colour. In this case assign the colour cn + 1 to these vertices. Then one of the following two sub cases may occur. Sub case1:For each i we can find a (ci , cn + 1) pair.Then clearly ψ[C(G)] = n + 1. Note that even though there are so many vi,j remaining to be coloured we cant colour these vi,j with two or more different colours because there is no edge in between any two vi,j s. Sub case2:For some i the pair (ci , cn+1 ) is missing. Then replace the colour cn + 1 with any of the remaining colours selected from the previously considered colour class K so that the colouring is proper.Then ψ[C(G)] < n + 1. Thus in all the cases ψ[C(G)] ≤ n + 1. 2
References [1] R.A. Brualdi and H.J.Rayser, Combinatorial Matrix Theory, Cambridge University press, 1991. [2] J. A. Bondy and U.S.R. Murty (1976), Graph theory with Applications, London: MacMillan .
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[3] Jonathan gross and Jay Yellen (2004), Hand book of Graph theory, CRC Press, New York [4] Keith Edwards (1997), The harmonious chromatic number and the achromatic number, Surveys in combinatorics 1997 by R (Rosmary) Bailey. [5] Frank Harary (1969), Graph Theory, Narosa Publishing home . [6] Frank Harary and Stephen Hedetniemi, The achromatic number of a graph, Journal of Combinatorial Theory,(1970), 154-161 . [7] Frank Harary, Stephen Hedetniemi and Geert Prins, An interpolation theorem for graphical homomorphisms, Portugaliae Mathematica, vol.26-Fasc.4. [8] K. Thilagavathi and N. Roopesh, Achromatic colouring of central graphs and split graphs, Far East Journal of Applied Mathematics, volume 30, issue 3, 359-369. [9] K.Thilagavathi and Vernold Vivin,J., ”Harmonious Colouring of Graphs”, Far East J. Math.Sci.(FJMS), Volume 20 No.2, 189-197. [10] K.Thilagavathi, Vernold Vivin,J and Akbar Ali.M.M, On Harmonious Colouring of Central Graphs” Advances and applications in discrete mathematics, vol. 2, 17-33. [11] Vernold Vivin,J((May 2007),”Harmonious Colouring of Total graphs, n-leaf, Central Graphs and Circumdetic graphs, PhD thesis.