Ramsey's theorem and self-complementary graphs

It is provet d that, given any positive integer k, there exists a self-complementary graph with more than 4 .2 4 k vertices which contains no complete subgraph with k+1 vertices . An application of this result to coding theory is mentioned . A graph will be called s-good if it contains neither a complete subgraph with more than s vertices nor an independent set of more than s vertices . A special case of the celebrated Ramsey's theorem [7] asserts that given any positive integer s there is an n = n(s) such that no graph with more than n(s) vertices is s-good . Apart from the trivial n(1) = 1, only two exact values of n(s) are known [4] ; these are n(2) = 5 and n(3) = 17 . Clearly, a graph G is s-good if and only if its complement G is s-good . It does not seem unlikely that for any s, there is an s-good self-complementary graph with n(s) vertices. This is true at least for s = 2 and s = 3 (and in this case, the s-good graphs with n(s) vertices are unique [6] ) . However, it seems quite difficult to prove this conjecture for all s. We shall denote by n * (s) the greatest integer n * such that there is a self-complementary s-good graph with n * vertices; trivially, n* (s)< n(s) . Theorem . n*(st) > (n*(s) 1)n(t) . Proof . Let Go = (Vo , E0 ) be an s-good self-complementary graph with 30 2 V. Chvátal et al., Ramsey's theorem n* (s) vertices, let fo : Vo Vo be an isomorphism between G and G . It is easy to see that the permutation fo has at most one fixed point and no odd cycles of length > 3 . Therefore there is an s-good self-complementary graph G Z = (Vi , Ei ) with n *(s) or n *(s)-1 vertices and a permutation f: Vl Vl setting up an isomorphism between Gl and Gl such that f has cycles of even length only (and no fixed points) . Consequently, Vl can be split into disjoint sets X and Y with f(X) = Y, f(Y) = X. Let G2 = (V2 , EZ ) be a t-good graph with n(t) vertices. We shall consider the graph G = (Vi X V2 , E) where ((u, v), (w, z)) belongs to E if and only if either {u, w) C El or u = w C X, {v, z) C E2 or finally u = w E Y, {v, z) (I E2 . G is self-complementary ; indeed, the mapping F: Vl X V2 Vl X V2 defined by F(u, v) _ (f (u), v) is an isomorphism between G and G. If Z C Vl X V2 spans a complete subgraph in G then at most s vertices in Z have distinct first coordinates (otherwise G l would not be sgood) and at most t vertices in Z have the same first coordinate (otherwise G2 would not be t-good) . Therefore IZI < st and G, being self-complementary, isst-good . Hence n*(st) > I Vi X V2 1 > (n*(s)-1)n(t) and the proof is finished . Corollary . n *(2t) > 4n(t) . Our original interest in this area was stimulated by the notion of the capacity of a graph as defined by Shannon [9] . One defines the product Gl X G2 X . . . X Gk of graphs Gi = (Vi, Ei ), i = 1, 2, . . ., k, as the graph G = (VI X V2 X . . . X Vk , E) where two distinct vertices (u l , u2 , . . ., uk), (vl , v 2) . . ., vk ) of G are adjacent if and only if, for each i = 1, 2, . . ., k, either {ui , v i) C El or else u i = vi . We denote the largest cardinality of an independent set in G by M(G); evidently, (1) g(GI X G2 X . . . X Gk) > g(GI ) g (G2 ) . . . g(Gk ) . Considering noisy channels in information theory, Shannon [ 9 ] was led to the definition of the capacity 0(G) of a graph G, 8(G) = sup (g(Gk )) i lk k V. Chvátal et al., Ramsey's theorem 303 Obviously, O (G) > p(G) . However, one can have 0 (G) > p (G) ; for instance, if G is the pentagon then p(G) = 2, μ(G 2 ) = 5 . It can be shown that p(G l ) = p(G 2 ) = k implies p(G I X G2 )< n(k) and this bound is best possible . Moreover, this inequality generalizes into the case of more graphs Gi with p(Gi ) not necessarily equal . Apparently Hedrlin [ 5 ] was the first to discover this relation between Ramsey numbers and the capacity problems . However, Hedrlin did not publish his result . Unaware of his contribution, Erdös, McEliece and Taylor [3] recently published an independent derivation of the equivalence . If G = (V, E) is a self-complementary graph with m vertices then μ(G2 ) > m. Indeed, if f is an isomorphism between G and G then the set {(u, f(u)) I u C V} is independent in G 2 = G X G. Hence p(G2 ) > m. Consequently, one has