EMBEDDING /c-REGULAR GRAPHS IN /c +1-REGULAR GRAPHS

Given a A:-regular graph G of order n, what is the minimum number v(G) of extra vertices required to embed G in a /<+1-regular graph? Clearly v(G) = 0 precisely when the complement G of G has a 1-factor—in particular, when n ^ 2k is even. Suppose G has no 1-factor: if n, k have opposite parity we show that v(G) = 1, while if n, k have the same parity (which must then be even with n < 2k) we show that v(G) = k + 2. We then investigate regular graphs which fail to have a 1-factor and list all such graphs for 2 ^ A: ^ 12. Finally we prove that if G is a /c-regular graph on n vertices which fails to span a A + 1-reguIar graph, then either G has an odd component, or k + 4 ^ n < 3k/2. For k < 16 we list all forty-one A-regular graphs G for which G has odd component, yet which fail to span a k +1 -regular graph. 1. Let G be a /c-regular graph of order n, and let v(G) denote the minimum number of extra vertices required to realise an embedding of G in a k + 1-regular graph. Then u(G) = 0 precisely when the complement G of G has a 1-factor. In general a /c-regular graph G on n vertices spans some k + r-regular graph precisely when its (n — k — l)-regular complement G has an r-factor. Thus, for example, if n and k have opposite parity and G is not a complete graph, then G automatically spans some k: + 2-regular graph. Henceforth we restrict our attention to the case in which r = 1. In Section 2 we use a result of Bollobas and Eldridge [2] (reprinted as [3, Theorem II.4.2]) to show that v(G) = 0 , 1 , or k + 2. In Section 3 we examine the case in which r(G) ^ 1 in more detail and obtain necessary conditions which make it possible to list all /c-regular graphs G of even order for which 2 < k ^ 12 and v(G) ^ 1. The list is surprisingly short. Almost all the graphs listed fail to span a k+ 1-regular graph for the trivial reason that G has an odd component. We prove (Theorem 2) that if G fails to span a k + 1-regular graph and G has no odd component, then either G is connected and k + 4 ^ n < 3k/2 (so that k ^ 12), or G is disconnected and k + 4 ^ n < 4/c/3 (so that k ^ 16). I should like to thank the referee for his constructive criticisms of the original manuscript.