Application of an Extremal Result of Erdős and Gallai to the ( n , k , t ) Problem

An extremal result about vertex covers, attributed by Hajnal [4] to Erdős and Gallai [2], is applied to prove the following: If n, k, and t are integers satisfying n ≥ k ≥ t ≥ 3 and k ≤ 2t − 2, and G is a graph with the minimum number of edges among graphs on n vertices with the property that every induced subgraph on k vertices contains a complete subgraph on t vertices, then every component of G is complete.


Introduction
All graphs here will be finite, non-null, and simple.A vertex cover of a graph G is a set S ⊂ V (G) that contains at least one endpoint of every edge in G.The vertex cover number of G is the minimum size of a vertex cover of G, and is denoted by β(G).This parameter is monotone -that is, β(H) ≤ β(G) for all subgraphs H of G. Deleting a vertex or an edge of G causes the vertex cover number to go down by at most 1.An edge or vertex of G whose removal causes such a decrease is said to be β-critical (or vertex-cover critical) for G.The graph G itself is said to be β-critical or vertex-cover critical if β(H) < β(G) for every proper subgraph H of G.It is easy to see that G is β-critical if and only if G has no isolated vertices and every edge of G is β-critical for G.In particular, this means that if β(G) > 0, then G has a vertex-cover critical subgraph H with β(H) = β(G).
S ⊂ V (G) is a vertex cover if and only if V (G) \ S is independent: from this it is easy to see that, if α(G) is the vertex independence number of G, the size of a largest independent (mutually non-adjacent) set of vertices, then α(G) + β(G) = |V (G)|.Therefore, a graph G is β-critical if and only if G has no isolated vertices and, for each e ∈ E(G), α(G−e) = α(G)+1.With this in mind, it is easy to verify that the following are β-critical; (i) K n for n ≥ 2; (ii) odd cycles; and (iii) matchings.
In conformity with the notation by which G + H denotes the disjoint union of G and H, a matching with s edges will be denoted If G is bipartite, the Kőnig-Egerváry Theorem ( [1], [6]) says that β(G) is the maximum number of edges in a matching in G. Therefore, a non-empty bipartite graph is vertex-cover critical if and only if it is a matching.
The extremal result of Erdős and Gallai [2] referred to in the title of this paper concerns the function f defined for s = 1, 2, . . .as The result is that f (s) = 2s.We have not been able to obtain a copy of [2]; we found an attribution to [2] of this result in [4], where Hajnal provides a short proof, suggesting that the proof in [2], a 23-page paper, is not very short.Later in [4], Hajnal, apparently without realizing it, provides an even shorter proof that supplies a stronger conclusion: not only is it true that f (s) = 2s, but, also, sK 2 is the only β-critical graph on 2s vertices with vertex cover number s.We will give this proof here, in a form that the reader will not find in [4].Hajnal there is driving toward a dual form of the result, based on the fact that S is a vertex cover of G if and only if V (G) \ S induces a complete graph in G, the complement of G.
Here is our translation of his second proof.The word "cover" will mean vertex cover.If Proof The proof will be by induction on |I|.Since G has no isolated vertices, the conclusion holds when |I| = 1.
Suppose |I| > 1, and suppose that |I| ≥ |N G (I)| + 1.We will deduce a contradiction.Let v ∈ I and But C covers all edges of G with neither end in I ∪ N G (I), and N G (I) covers each edge of G with at least one end in 2 Application to the (n, k, t) Problem Suppose n ≥ k ≥ t are positive integers.An (n, k, t)-graph is a graph on n vertices such that every induced subgraph of order k contains a clique of order t.The (n, k, t) problem is to determine, for each triple (n, k, t), all the minimum (n, k, t)-graphs -that is, the (n, k, t)graphs with the fewest edges.When t = 1 the only such graph is the graph with n isolated vertices, and when t = 2, the problem can be seen as a complementary version of Turán's Theorem [7]; hence the unique minimum (n, k, 2)-graphs are T n,k−1 , where T n,r denotes the Turán graph on n vertices with r parts.Other easy cases include k = t ≥ 2 and n = k, where the unique extremal graphs are K n and (n − t)K 1 + K t , respectively [5].
The (n, k, t) conjecture is that whenever n ≥ k ≥ t, some minimum (n, k, t)-graph has complete components.The strong (n, k, t) conjecture is that every minimum (n, k, t)-graph has complete components.If the strong (n, k, t) conjecture holds then the (n, k, t) problem is essentially solved in [5] -the extremal graphs are all aK 1 +T n−a,b for particular non-negative integers a,b -although there is room for improvement in the determination of a and b given in [5].
Theorem 2.1 (Erdős and Stone [3]) Suppose F is a family of graphs containing no empty graph, and let Let χ(F) = min{χ(H) : H ∈ F}, and suppose that χ(F) > 2. Let r = χ(F) − 1.Then Explanation: The name F was chosen to connote forbidden subgraphs.Clearly no graph with chromatic number r = χ(F) − 1 can contain a subgraph from F, and clearly the Turán graph T n,r is the graph on n vertices of that chromatic number with the most edges, if n ≥ r.Therefore, |E(T n,r )| ≤ g(n), for n ≥ r.The Erdős-Stone Theorem asserts that if F contains no bipartite graph, then, asymptotically, |E(T n,r )| ∼ g(n).
In the original Erdős-Stone Theorem, F was a singleton; but the more general theorem follows easily from the original, by the following argument.Given F, let H ⊂ F be such that χ(H) = χ(F) > 2, and set F = {H}.Let g be defined with reference to F as g was defined with reference to F. Clearly, g (n) ≥ g(n) for all n, so, for To apply the Erdős-Stone Theorem to the (n, k, t) problem, we define an (n, k, t)-graph to be the complement of an (n, k, t)-graph.In other words, an (n, k, t)-graph is a simple graph on n vertices such that every subgraph H of order k has vertex independence number α(H) ≥ t. (Notice the absence of the word "induced" in this description.)Clearly the (n, k, t) problem is equivalent to the problem of describing the (n, k, t)-graphs with the most edges.
Fix k > t > 2. For n ≥ k, an (n, k, t)-graph is a graph on n vertices with no subgraph from On the other hand, there exists a complete multipartite graph H with k t−1 ≥ 2 parts on k vertices with maximum part size t−1.Clearly H ∈ F and χ(H) → 1 as n → ∞.Therefore, the minimum number of edges in an (n, k, t)-graph, for k and t satisfying k > t > 2 and k > 2t − 2, is asymptotically equivalent, as n → ∞, to |E(T n,r )|, where r = k t−1 − 1.This conclusion by no means proves that T n,r is a minimum (n, k, t)-graph for all n sufficiently large, which is a good thing, because that conclusion would be false.For example, if t = 3, k = 6, so k t−1 = 3, by applying the main result of [5] it can be seen that for all n ≥ 8 the unique (n, 6, 3)-graph with the fewest edges among those with all components complete is K 1 + T n−1,2 .In this case, and in many others, T n,r is an (n, k, t)-graph with number of edges (asymptotically as n → ∞) close to smallest, but not smallest, among (n, k, t)-graphs.
However, the application of the Erdős-Stone Theorem to the (n, k, t) problem is intriguing.For those sharing our prejudices, the asymptotic result reinforces a belief in the truth of the (n, k, t) conjecture.It also points out the following, a nice result that we neglected to include in [5].
. For all sufficiently large n, the unique (n, k, t)-graph with the fewest number of edges among those with every component complete is aK 1 + T n−a,r .
Proof By Corollary 1 of [5], for n ≥ k + r − 1 an (n, k, t)-graph having only complete components and with as few edges as possible will be one of (k Therefore, every subgraph of G on k or fewer vertices has vertex cover number less than k − t + 1. However, if G is not an (n, k, t)-graph then G has an induced subgraph H on k vertices with clique number ω(H) ≤ t − 1.Then H is a subgraph of G of order k with α(H) = ω(H) ≤ t − 1; we have that β(H) = k − α(H) ≥ k − t + 1. Hence we can find a β-critical subgraph X of H with β(X) = k − t + 1.
Theorem 2.5 Suppose that k > t > 2. If k ≤ 2t − 2, then for every n > k the unique (n, k, t)-graph with the fewest edges is (k − t)K 1 + K n−k+t .
Proof Suppose that k ≤ 2t − 2, n > k, and G is an (n, k, t)-graph with the minimum number of edges possible.Then G is an (n, k, t)-graph with the maximum number of edges possible.By Lemma 2.4, G has no β-critical subgraph X on k or fewer vertices such that β(X) = k − t + 1.As Theorem 1.2 gives f (k − t + 1) = 2(k − t + 1) ≤ k, it follows that G has no β-critical subgraph X with β(X) = k − t + 1, because such an X could have no more than f (k − t + 1) ≤ k vertices.
Therefore, β(G) ≤ k − t.By Lemma 2.3, G can have no edges than does K k−t ∨ K n−k+t , and, if G has as many edges as that graph, then G = K k−t ∨ K n−k+t .Since K k−t ∨ K n−k+t is an (n, k, t)-graph, it follows that G = K k−t ∨ K n−k+t , so G = K k−t + K n−k+t = (k − t)K 1 + K n−k+t .

1 − 1 .
Since, for each fixed pair (s, b) with s ≥ 0 and b ≥ 0, |E(T n−s,b )| ∼ n 2 2b , for n sufficiently large the choice of must be b = r.The application of Theorem 1.2 to the (n, k, t) problem concerns values of k and t such that k t−1 ≤ 2, the values about which the Erdős-Stone Theorem has nothing to say.The join of two graphs G and H, denoted G ∨ H, is the graph obtained from the disjoint union of G and H by adding a complete bipartite graph between (G) and V (H).

Lemma 2. 3
Suppose that n > s ≥ 1 are integers.The unique graph of order n with vertex cover number s with the most edges is K s ∨ K n−s .Proof Suppose |V (G)| = n and β(G) = s, and let S ⊂ V (G) be a minimum vertex cover.Then V (G) \ S is an independent set of vertices; clearly G can have no more edges than the copy of K s ∨ K n−s obtained by putting in all S-S edges and all S-(V (G) \ S) edges.On the other hand, G = K s ∨ K n−s has order n and vertex cover number n − α(G) = n − (n − s) = s.Lemma 2.4 Let n > k > t > 2 be integers, and let G be a graph on n vertices.G is an (n, k, t)-graph if and only