English

Transversals in Uniform Linear Hypergraphs

Combinatorics 2018-02-07 v1

Abstract

The transversal number τ(H)\tau(H) of a hypergraph HH is the minimum number of vertices that intersect every edge of HH. A linear hypergraph is one in which every two distinct edges intersect in at most one vertex. A kk-uniform hypergraph has all edges of size kk. It is known that τ(H)(n+m)/(k+1)\tau(H) \le (n + m)/(k+1) holds for all kk-uniform, linear hypergraphs HH when k{2,3}k \in \{2,3\} or when k4k \ge 4 and the maximum degree of HH is at most two. It has been conjectured that τ(H)(n+m)/(k+1)\tau(H) \le (n+m)/(k+1) holds for all kk-uniform, linear hypergraphs HH. We disprove the conjecture for large kk, and show that the best possible constant ckc_k in the bound τ(H)ck(n+m)\tau(H) \le c_k (n+m) has order ln(k)/k\ln(k)/k for both linear (which we show in this paper) and non-linear hypergraphs. We show that for those kk where the conjecture holds, it is tight for a large number of densities if there exists an affine plane AG(2,k)AG(2,k) of order k2k \ge 2. We raise the problem to find the smallest value, kmink_{\min}, of kk for which the conjecture fails. We prove a general result, which when applied to a projective plane of order 331331 shows that kmin166k_{\min} \le 166. Even though the conjecture fails for large kk, our main result is that it still holds for k=4k=4, implying that kmin5k_{\min} \ge 5. The case k=4k=4 is much more difficult than the cases k{2,3}k \in \{2,3\}, as the conjecture does not hold for general (non-linear) hypergraphs when k=4k=4. Key to our proof is the completely new technique of the deficiency of a hypergraph introduced in this paper.

Keywords

Cite

@article{arxiv.1802.01825,
  title  = {Transversals in Uniform Linear Hypergraphs},
  author = {Michael A. Henning and Anders Yeo},
  journal= {arXiv preprint arXiv:1802.01825},
  year   = {2018}
}

Comments

105 pages

R2 v1 2026-06-23T00:12:33.861Z