English

The minimum number of clique-saturating edges

Combinatorics 2022-01-14 v2

Abstract

Let GG be a KpK_p-free graph. We say ee is a KpK_p-saturating edge of GG if eE(G)e\notin E(G) and G+eG+e contains a copy of KpK_p. Denote by fp(n,e)f_p(n, e) the minimum number of KpK_p-saturating edges that an nn-vertex KpK_p-free graph with ee edges can have. Erd\H{o}s and Tuza conjectured that f4(n,n2/4+1)=(1+o(1))n216.f_4(n,\lfloor n^2/4\rfloor+1)=\left(1 + o(1)\right)\frac{n^2}{16}. Balogh and Liu disproved this by showing f4(n,n2/4+1)=(1+o(1))2n233f_4(n,\lfloor n^2/4\rfloor+1)=(1+o(1))\frac{2n^2}{33}. They believed that a natural generalization of their construction for KpK_p-free graph should also be optimal and made a conjecture that fp+1(n,ex(n,Kp)+1)=(2(p2)2p(4p211p+8)+o(1))n2f_{p+1}(n,ex(n,K_p)+1)=\left(\frac{2(p-2)^2}{p(4p^2-11p+8)}+o(1)\right)n^2 for all integers p3p\ge 3. The main result of this paper is to confirm the above conjecture of Balogh and Liu.

Keywords

Cite

@article{arxiv.2201.03983,
  title  = {The minimum number of clique-saturating edges},
  author = {Jialin He and Fuhong Ma and Jie Ma and Xinyang Ye},
  journal= {arXiv preprint arXiv:2201.03983},
  year   = {2022}
}