On an extremal subfamily of an extremal family of nearly strongly regular graphs

We continue the classification of the regular simple graphs in which, for some t, any two adjacent vertices have exactly t common neighbors, and the union of their neighbor sets misses exactly two vertices. Previously it was shown that for any such graph with n vertices, if t > 0 then t + 8 ≤ n ≤ 3t + 6. Here we show that there is exactly one such graph on n = 3t+ 6 vertices, for each t = 1, 2, . . . , namely Kt+2,t+2,t+2 minus a two-factor consisting of triangles. Let G be a simple graph. For an edge e ∈ E(G) with end-vertices u, v let t(e) = |N(u) ∩N(v)| and let J(e) = |N(u) ∪N(v)| (with neighborhoods taken in G, of course). Let t(G) = |E(G)|−1 ∑e∈E(G) t(e), and let J(G) = maxe∈E(G) J(e). It is shown in [3] that if G has m edges and n vertices, then 4m ≤ n(J(G) + t(G)), with equality if and only if G is regular and t is a constant function (equivalently, G is regular and J is a constant function; observe that J(e) + t(e) = d(u) + d(v)). This conclusion also holds if J(G) is defined to be an arithmetic mean, and t(G) is a max. Clearly this result generalizes Mantel’s famous theorem, i.e. Turan’s theorem with r = 2 (see [5]). To agree with the notation in [3] and [4], let us denote by ET (n, J, t) the set of extremal graphs for the inequality above, on n vertices with J = J(G) and t = t(G). ∗ Contract grant sponsor: ONR † Contract grant number: N00014-97-1-1067 Australasian Journal of Combinatorics 25(2002), pp.279–284 That is, ET (n, J, t) consists of the regular graphs on n vertices, of degree 2−1(J + t), with each pair of adjacent vertices having exactly t common neighbors. These graphs are “nearly strongly regular”; they are regular, and, in the lingo of strongly regular graphs (see [5]), there is a common value λ of |N(u) ∩ N(v)| for adjacent vertices u and v (namely, λ = t), but there might not be a μ (a common value of |N(u)∩N(v)| for non-adjacent distinct u and v). Since the totality of such graphs include the strongly regular graphs, we despair of ever achieving a complete catalog, indexed by n, J , and t, of such graphs. But some interesting results have been produced by fixing certain values of J = J(n). In [2] it is shown that ⋃ n,t ET (n, n, t) consists of the regular Turàn graphs, i.e., the complete r–partite graphs (for various r) with parts of equal size. In [3] it is shown that ⋃ n,t ET (n, n− 1, t) consists of the complements of the strongly regular graphs with μ = 1. (In general, if G ∈ str(n, k, λ, μ) then Ḡ ∈ ET (n, n−μ, n+μ−2−2k).) In [4] it is shown that G ∈ ET (n, n− 2, 0) if and only if n is even and either G is bipartite and regular (so G = Kn 2 , n 2 minus a one-factor) or G is one of the two non-bipartite graphs given in [4]. (It has since been pointed out that this result for n ≥ 10 is an easy consequence of a famous theorem of Andraśfai, Erdös and Sós [1].) It is also proven in [4] that for t > 0, if ET (n, n − 2, t) is non-empty then t+ 8 ≤ n ≤ 3t+ 6, and that in the case t = 1, the unique graph in ET (9, 7, 1) is the line graph of K3,3. Our aim here is to show that the extreme n = 3t+6 in the result just mentioned is achievable for every t ≥ 1, and that the graph achieving it is unique. Theorem 1 Suppose that t is a positive integer. Then G ∈ ET (3t + 6, 3t + 4, t) if and only if G = Kt+2,t+2,t+2 − F , where F is the set of edges of a 2-factor of Kt+2,t+2,t+2 consisting of triangles. Remark. The graph G described above is the complement of the line graph of K3,t+2. Proof. It is straightforward to verify that Kt+2,t+2,t+2 −F ∈ ET (3t+ 6, 3t+ 4, t). Suppose that G ∈ ET (3t+6, 3t+4, t). G is regular with degree 2−1(3t+4+ t) = 2t+ 2. For adjacent vertices u, v ∈ V (G), let T = T (u, v) = N(u)∩N(v), A = A(u, v) = N(u) \ (T ∪ {v}), B = B(u, v) = N(v) \ (T ∪ {u}), and X = X(u, v) = {x, y} = V (G) \ (N(u) ∪N(v)). Observe that |A| = |B| = t+ 1. Claim 1. There are no edges among the vertices of T , and every vertex of T is adjacent to each of x and y. Proof. Suppose that w ∈ T ; w has t − 1 neighbors in common with u, other than v, and these must be in T ∪ A. Suppose that w is adjacent to s vertices of T . Then w is adjacent to t − 1 − s vertices of A, and, similarly, of B. Since w might be adjacent to one or both of x, y, and is adjacent to both u and v, we have 2t+ 2 = d(w) ≤ s+ 2(t− 1− s) + 2 + 2 = 2t+ 2− s. It follows that s = 0 and that