A note on the colour class of a self-complementary graph

We prove that any self-complementary graph G contains no proper overfull subgraph H such that (H) = (G) and |V (H)|¡ |V (G)|. Moreover, a self-complementary graph G is overfull if, and only if, it is regular. This is a support for the following conjecture: a self-complementary graph G is Class 2 if; and only if; G is regular. c © 2000 Elsevier Science B.V. All rights reserved. All graphs we consider are simple. V (G) and E(G) denote the vertex and edge set of a graph G; n = |V (G)| is the order and e = |E(G)| is the size of G. The degree of a vertex v in G is denoted by d(v; G). For a subset X of V we denote by GX the subgraph of G induced by the set X . A graph G = (V; E) is said to be self-complementary if it is isomorphic with its complement (cf. [6]) G= (V; E). Any permutation :V → V such that xy ∈ E if, and only if, (x) (y) 6∈ E is called self-complementing permutation of G. Let ′(G) be the least number of colours su cient for colouring properly the edges of a graph G. By a very well-known theorem of Vizing [7] there are two possibilities: either ′(G) = (G) (and G is called Class 1) or ′(G) = (G) + 1 (and G is called Class 2), where (G) is the maximum vertex degree in G. Holyer [3] proved that the problem of determining the class of a graph is NP-complete. A graph G of order n is said to be overfull if |E|¿ bn=2c (G). Since in any colour class we may have at the most bn=2c edges, it is clear that the overfull graphs are Class 2. The following conjecture has been stated in [8]. Conjecture 1 (Wojda and Zwonek [8]). A self-complementary graph G is Class 2 if, and only if, G is regular. E-mail address: [email protected] (A.P. Wojda) 0012-365X/00/$ see front matter c © 2000 Elsevier Science B.V. All rights reserved. PII: S0012 -365X(99)00191 -0 334 A.P. Wojda /Discrete Mathematics 213 (2000) 333–336 If a self-complementary graph G has a complementing permutation that is a cycle, then we say that G is s.c.-cyclic (see [6]). The following result has been proved in [8]. Theorem 1 (Wojda and Zwonek [8]). If G is an s.c.-cyclic self-complementary graph then G is Class 1: Niessen noticed [5] that, since (G)¿(n− 1)=2 for every self-complementary graph G of order n, Conjecture 1 is related to the following conjecture made by Chetwynd and Hilton. Conjecture 2 (Chetwynd and Hilton [1] and Hilton [2]). Let G be a graph of order n with the maximum vertex degree (G)¿n=3. Then G is Class 2 if, and only if, G contains an overfull subgraph H with the maximum vertex degree (H) = (G). The following proposition has been given in [8] without a proof, we prove it here for completeness. Proposition 2 (Wojda and Zwonek [8]). A self-complementary graph G is overfull if; and only if; G is regular. Proof. If a self-complementary graph G of order n is regular then (G) = (G) = (n− 1)=2 and e(G) = n(n− 1)=4¿ (n− 1)2=4 = bn=2c(n− 1)=2, hence G is overfull. Let us suppose now that G is overfull. Then e(G) = n(n − 1)=4¿ bn=2c (G) = ((n− 1)=2) (G) and therefore, (G)¡n=2. Since G is self-complementary, we have (G) + (G) = n− 1 and, by consequence, (G) = (G) = (n− 1)=2. In this paper we prove. Theorem 3. A self-complementary graph G does not contain any overfull subgraph H with (H) = (G) and |V (H)|¡ |V (G)|. Since for any non trivial self-complementary graph G of order n we have n¿4, (G) + (G) = n − 1 and e(G) = n(n − 1)=4, Theorem 3 is a consequence of the following lemma. Lemma 4. Let G= (V; E) be a graph of order n¿4; such that e(G)6n(n− 1)=4 and (G)¿n − (G) − 1. Then G does not contain any overfull subgraph H such that (H) = (G) and |V (H)|¡n. Proof. Suppose that H is an overfull subgraph of a self-complementary graph G of order n such that (H) = (G) and |V (H)|¡n. It is clear that the subgraph F of G induced in G by the vertex set of H is also an overfull subgraph of G and (F)= (G), A.P. Wojda /Discrete Mathematics 213 (2000) 333–336 335 so without loss of generality we may assume that H is an induced subgraph of G, say H = GV−W , with W ⊂V and |W | = l¿ 0. Since H is overfull with = (H)= (G) we have e(H)¿ b(n−l)=2c. It is a very well known fact that every overfull graph has odd order, thus b(n−l)=2c=(n−l−1)=2 and e(H)¿ n− l− 1 2 : (1) For a vertex w from W and a subset X of the vertex set V , let us denote by e(w; X ) the number of edges which have w for one end vertex, while the second end vertex is in the subset X . Thus the number p of edges of G which are not the edges of H satis es p= ∑ w∈W e(w; V −W ) + 1 2 ∑ w∈W e(w;W ) = ∑ w∈W (e(w; V −W ) + e(w;W )) − 1 2 ∑ w∈W e(w;W ) = ∑ w∈W d(w;G) − 1 2 ∑ w∈W d(w;GW )¿l(n− − 1) − 12 l(l− 1) = l 2 (2n− 2 − l− 1) and e(H)6 n(n− 1) 4 − p6n(n− 1) 4 − 1 2 (2n− 2 − l− 1): (2) Since (H) = we have n− l¿ + 1, thus l6n− − 1: (3) We conclude from (1) and (2) that n− l− 1 2 ¡e(H)6 n(n− 1) 4 − l 2 (2n− 2 − l− 1); hence f(l) = 2l + l(6 − 4n+ 2) + (n− 1)(n− 2 )¿ 0: Since 16l6n− −1; f(l)¿ 0 implies max{f(l): 16l6n− −1}=max{f(1); f(n− −1)}¿ 0. Therefore, f(1)=(n−4)(n−2 −1)¿ 0 or f(n− −1)=−(n−2 )(n−2 − 1)¿ 0: In each case one gets very easily a contradiction, and the lemma follows. For further reading The following reference is also of interest to the reader: [4].