Packing a forest with a graph

We discuss only finite simple and use standard terminology and notation from [1) except as indicated. For any graph we use V( G) and E( G) to denote the vertex set and the edge set of G, respectively. We denote the complement of G by GC. Let G and H be two graphs of order n. We say that there is a packing of G and H if the complement GC contains a subgraph isomorphic to H. In this case, we also say that G and H are packable. There are many papers concerning the of two graphs which have a small number of edges. For example, Sauer and Spencer [6) proved that if IE(G)I ~ n 2 and IE(H)I ~ n 2, then there is a packing of G and H. Bollobas and Eldridge [2) found all the forbidden pairs (G, H) of graphs with .6.(G) < n -1, .6.(H) < n 1, IE(G)I + IE(H)I ~ 2n 3 for which there are no packings of G and H. Slater, Teo and Yap [7) proved that if n 2:: 5, G is a tree, H has n 1 edges and neither G nor H is a star, then there is a packing of G and H. Sauer and Spencer [6] also proved that if 2.6.( G).6.(H) < n, then there is a packing of G and H. For more results, see [1, Chapter 8] and [9]. Bollobas and Eldridge [2] conjectured that if (.6.(G)+l)(.6.(H)+l) ~ n+1, then there is a packing of G and H. This conjecture is still open. Hajnal and Szemeredi [4] proved that if n = 2:: 3 and k 2:: 1) and G is the vertex-disjoint union of k copies of Ie and 0,.(H) ~ k 1, i.e., (.6.(G) + l)(.6.(H) + 1) ~ n, then there is a packing of G and H. The result in the case s = 3 was first obtained by Corradi and Hajnal [3). In this paper, we consider the case that one of G and H is a forest, i.e., a graph with no cycles. To state our result, we define kG to be the vertex-disjoint union of k copies of G for any positive integer k and graph G. For even positive integer n, there is no packing of the two graphs in each of the following three pairs of graphs: ((n/2)K2' KI,n-I), (K(n/2)+1 U H, (n/2)K2) wht;re H is any graph of order n/2 -1 and 'u' means 'vertex-disjoint union', and (Kn/2,nj2, (n/2)K2) with n/2 odd. To see this, we observe that in each pair, the complement of the graph which is not (n/2)K2 does not have a perfect matching. We especially name these three pairs as three forbidden pairs of graphs. We prove the following. .