The Szemerédi–Petruska conjecture for a few small values

Let H be a 3-uniform hypergraph of order n with clique number ω(H)=k\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\omega (H)=k$$\end{document}. Assume that the union of the k-cliques of H equals its vertex set, the intersection of all maximum cliques of H is empty, but the intersection of all but one k-clique is non-empty. For fixed m=n-k\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$m=n-k$$\end{document}, Szemerédi and Petruska conjectured the sharp bound n6m+22\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$n\hbox {\,\,\char 054\,\,}{m+2\atopwithdelims ()2}$$\end{document}. In this note the conjecture is verified for m=2,3\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$m=2,3$$\end{document} and 4.


Introduction
Let H be a 3-uniform hypergraph of order n with clique number ω(H ) = k. Assume that the union of the k-cliques of H equals its vertex set, the intersection of its maximum cliques is empty, but the intersection of all but one maximum clique is non-empty. For fixed m = n − k Szemerédi and Petruska [8] conjectured the tight bound n m+2 2 .
Here is the construction that shows this bound is sharp for infinitely many k 3. For m 0 let k − 1 = m+1 2 , set n = m + k, and let X , Y be disjoint sets such that |X | = k −1, |Y | = m +1. Let X = {x 1 , . . . , x k−1 } and let { p 1 , . . . , p k−1 } be the set of all 2-element sets of Y . For every i = 1, . . . , k − 1 set N i = (X \ {x i }) ∪ p i . All triples in some N i , i = 1, . . . , k − 1, define a 3-uniform hypergraph H on the n-element vertex set X ∪ Y . It is straightforward to check that {N 1 , . . . , N k−1 } is the family of all maximum cliques of H , in particular ω(H ) = k, furthermore, k−1 i=1 N i = ∅. Notice also that any k − 2 maximum cliques have a common vertex in X ; therefore, H realizes an extremal hypergraph and shows the bound is sharp.
Indeed it had been conjectured this example is the unique extremal example for m 4.
In Sect. 2 we prove the Szemerédi-Petruska conjecture for m = 2, 3, and 4. The proof, based on τ -critical graphs, makes it possible to characterize all extremal hypergraphs of order n = m+2 2 for m 4. It is worth noting that there are two-two non-isomorphic extremal hypergraphs for m = 2 and 3; meanwhile, the extreme is unique for m = 4 and isomorphic to the Szemerédi-Petruska construction described above.
The transversal number of a hypergraph (or a graph), τ (H ), is defined as the minimum number of vertices in a transversal set containing a vertex from every edge of H . A hypergraph (or graph) is τ -critical if it has no isolated vertex and the removal of every edge decreases its transversal number.
The Szemerédi-Petruska conjecture is equivalent with the statement that m+2 2 is the maximum order of a 3-uniform τ -critical hypergraph with transversal number m, see [4] or [9,Problem 18 (a)]. For the maximum order, Tuza 1 obtained the best known bound 3 4 m 2 + m + 1 using the machinery of τ -critical hypergraphs. An alternative approach to solve the conjecture is proposed by Jobson et al. [6], combining a decomposition process introduced by Szemerédi and Petruska [8] with the skew version of Bollobás's theorem [1], and using tools from linear algebra.
It turns out that the Szemerédi-Petruska conjecture has applications in extremal problems concerning convex sets in the plane, see Jobson et al. [5] and [7]. The validity of the Szemerédi-Petruska conjecture for small values carries relevant information pertaining to those combinatorial geometry problems.

Proof of the conjecture for m = 2, 3, and 4
In the proof presented here τ -critical graphs play a key role. In particular, we use the tight bounds (1) and (2) shown below due to Erdős and Gallai [2] and Gyárfás and Lehel [3], respectively. (2) then we may consider H − v thereby reducing both n and m by one. Hence we may assume i=1 N i = V and |V | = n. Furthermore, we assume that 3, since H needs to have at least two cliques, and the claim is obvious for = 2, even if H consists of two disjoint k-cliques.
Szemerédi and Petruska [8, Lemma 4] observed that each N i , i = 1, . . . , , contains a private pair: For an edge for every 1 j . For each edge p j ∈ G we define the weight Let G − p j denote the graph obtained by the removal of edge p j from G (and nothing else). Obviously we have The conditions N i = ∅ and (4) , and the claim follows. Now we break the proof of Proposition 2.2 into three cases.
Thus the bound in Lemma 2.3 becomes

Lemma 2.4 If
3 and w( p ) = 0, then G = G − p satisfies Since a transversal set in G \ p i is also a transversal set in G \ p i , we have τ (G \ p i ) τ (G \ p i ). Then we obtain If p is neither an isolated edge nor a pendant edge, Then by the definition of the edge weights, we have w(G ) = −1 Case 3.1: G contains a K 4 . If G had two more non-isolated vertices, a and b, then ab / ∈ E, since τ (G) = 3. One edge from each of a and b to K 4 would result in an edge of zero weight. Hence |V 0 | = 5 or 4, and G can be obtained starting with a K 5 , and successively removing edges incident with a common vertex, say a. The corresponding configurations are depicted in Fig. 2 (unlabeled edges have weight 1). We eliminate the 'fake' candidate by applying a general observation, the Triples test, as follows.
Notice that the four sets M i \ {c, d, e}, i = 1, 2, 3, 4, are pairwise disjoint, thus every 3-element set f ⊂ N is disjoint from some M i , 1 i 6. Hence, by the Triples test, f ∈ E for every f ⊂ N , | f | = 3. Therefore, N induces a clique of order k + 1, a contradiction.  Case 3.3: G has a triangle, it has no 5-cycle and no K 4 . The τ -critical subgraph in G is K 2 + K 3 . To get all candidates at most three edges and one new vertex can be added to K 2 + K 3 by keeping the transversal number 3 and creating neither a 5-cycle nor a K 4 . These graphs are listed in Fig. 4; all have at most 15 vertices. Case 3.4: G has no cycle. The τ -critical subgraph in G is 3K 2 . To get all candidates from this one at most three edges can be added by keeping the transversal number 3 and creating no cycle. These graphs are listed in Fig. 5; all have at most 15 vertices. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article's Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article's Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/.