GRAPHS WHICH HAVE PANCYCLIC COMPLEMENTS

Let p and q denote the number of vertices and edges of a graph G, respectively. Let A(G) denote the maximum degree of G, and G the complement of G. A graph G of order p is said to be pancycllc if G contains a cycle of each length n, 3 < n < p. For a nonnegative integer k, a connected graph G is said to be of rank k if q p i + k. (For k equal to 0 and i these graphs are called trees and unlcycllc graphs, respectively.) In 1975, I posed the following problem: Given k, find the smallest positive integer Pk’ if it exists, such that whenever G is a rank k graph of order p <_ Pk and A(G) < p 2 then G is pancyclic. In this paper it is shown that a result by Schmeichel and Hakiml (2) guarantees that Pk exists. It is further shown that for k 0, i, and 2, Pk" 5, 6, and 7, respectively.

followed.In particular, p and q shall denote the number of vertices and edges of a graph G, respectively.We let A(G) denote the maximum degree of G and G denote the complement of G.
A graph G of order p is called pancyclic if G contains a cycle of each length n, 3 < n < p.For a nonnegative integer k, a connected graph G is said to be of rank k if q p i + k.Here the number k gives the number of independent cycles in G.When k equals 0 or i these graphs are called trees or unicyclic graphs, respectively.
In this paper we explore the following idea: if G is a graph having, in some sense, little cycle structure relative to its order, then perhaps G will have a great deal of cycle structure.As an example, consider the graph shown in Figure i.This graph is a tree, i.e., a connected graph having no cycles.On the other hand note that its complement is pancyclic.

and its pancyclic complement
In 1975, after obtaining the results for k 0, i, and 2 which are presented here, I posed the following problem: Given k, find the smallest positive integer Pk' if it exists, such that whenever G is a graph of rank k of order p > Pk and A(G) < p 2, then G is pancyclic.Recently, J. A. Bondy has pointed out that the existence of Pk is guaranteed by the following result due to Schmeichel and Hakimi [2].
THEOREM.Let G be a graph with p vertices, q edges, and minimum degree 6 > positive integer Pk such that whenever G is a graph of rank k of order P >Pk and A(G) < p-2, then G is pancyclfc.
PROOF.Let G be a graph of rank k with A(G) < p 2. If G has p 2 vertices, then has p vertices, q p 3p + 2 k edges and minimum 2 degree 6 > 2. Depending on the value of 6, the requirements for q given by the theorem would yield the following inequalities: ( Note that each of the above inequalities is true provided that p is large enough.Hence we can choose Pk to be the least positive integer which makes all the above inequalities true.
The above theorem yields an upper bound for Pk; however, in the known cases it does not give us a very good bound.For example, the theorem < I0 whereas we will show that P2 would tell us that P2-7.In the remainder of the paper we show that for k--0, i, and 2, Pk 5, 6, and 7, respectively.
THEOREM 1.If G is a tree of order p > 5 with A(G) < p 2, then G is pancyclic.
PROOF.The proof is by induction on p.If p 5, then G is a path on 5 vertices.Thus G C 5 + e and so clearly G is pancycllc.Now let p > 6, and assume the result holds for all trees of order less than p.
Let G be a tree of order p with A(G) < p 2. If A(G) < p 3, let v be an end vertex of G.If A(G) p 3, then unless G is the graph of Figure i, we may choose v to be an end vertex adjacent with the unique vertex of degree p-3.Now consider G-v, which is a tree of order p 1 with A(G v) < (p i) 2. Hence by the induction hypothesis, G-v has a cycle of each length n, 3 < n < p i. Therefore so does G.
Since degGv p-2 > p i 2 v must be adjacent in G to two consecutive vertices on the (p-l)-cycle in G-'v.Thus this cycle can be extended in to a cycle of length p.Therefore, is pancyclic.Now by induction the proof is completed.
COROLLARY i.If G is a forest of order p > 5 with A(G) < p 2, then G is pancyclic.
PROOF.Note that there exists a tree H with A(H) < p 2 containing G as a spanning subgraph.Since H is pancyclic and H _ G, G is pancyclic.
THEOREM 2. If G is a unicyclic graph of order p > 6 with A(G) < p 2, then G is pancyclic.
PROOF.Let u I, u 2, u n denote the cycle vertices of G.Among the vertices ul, we will choose one of minimum degree in G; call it u.

CASE i
Suppose n > 4 degGv and v is not a cycle vertex.Then G is the graph of Figure 2. If p > 8, we can remove u and proceed as above.If p 6 or 7, we again get special case graphs, which can be shown to have pancycllc complements.Figdre 3 COROLLARY 2. If G is a graph of order p _> 6 with A(G) < p 2 and G contains exactly one cycle, then G is pancycllc.
The five-cycle C 5 is a unicyclic graph on 5 vertices which, does not have a pancycllc complement.This shows that Pl" 6.The graph shown in Figure 4 is a rank 2 graph of order 6 whose complement does not contain a 3-cycle.Hence P2 > 7. Our final result shows that indeed P2 7.
Figure 4 THEOREM 3. If G is a graph of rank 2 of order p > 7 with A(G) < p-2, then G is pancyclic.
PROOF.We consider three cases.
CASE i. G has a cycle with a diagonal.Let u_, u^, u be the n cycle vertices of G, n > 4, and suppose UlUi is the diagonal of the cycle,  and so we argue as before.Lastly we must consider the case where p 7, A(G) 4, and there does not exist a vertex u with degGu 2 and A(G-u) 3. In this case G must be one of the graphs shown in Figure 5, all of which have pancyclic complements.
H. 3. SRAIGH CASE 2. G has the following configuration as a subgraph.
Again let u be a cycle vertex of smallest degree.Then Also if 6(G) p 3, then clearly it is possible to choose u so that A(G-u) p 4. Hence we may argue as before.
CASE 3. G contains the configuration of Figure 6.
Figure 6 Figure 2 cycle can be extended to a p-cycle in G. Therefore, G is pancycllc. this Choose a cycle vertex u which has the smallest degree in G among the cycle vertices Then