English

Every 3-connected $\{K_{1,4},K_{1,4}+e\}$-free split graph of order at least 13 is Hamilton-connected

Combinatorics 2026-03-16 v1

Abstract

A graph GG is {F1,F2,,Fk}\{F_{1}, F_{2},\dots,F_{k}\}-free if GG contains no induced subgraph isomorphic to any FiF_{i} (1ik)(1\leq i \leq k). A connected graph GG is a split graph if its vertex set can be partitioned into a clique and an independent set. Ryj\'{a}\v{c}ek et al. [J. Comb. Theory, Ser. B 134 (2019) 239--263] conjectured that every 44-connected {K1,4,K1,4+e}\{K_{1,4},K_{1,4}+e\}-free graph with minimum degree at least 6 is Hamiltonian and they confirmed the case with connectivity at least 5, where K1,4+eK_{1,4}+e is the graph obtained from K1,4K_{1,4} by adding a new edge. In this paper, we show that every 3-connected {K1,4,K1,4+e}\{K_{1,4},K_{1,4}+e\}-free split graph of order at least 1313 is Hamilton-connected. It implies that Ryj\'{a}\v{c}ek et al.'s conjecture holds for split graphs of order at least 1313.

Keywords

Cite

@article{arxiv.2603.12770,
  title  = {Every 3-connected $\{K_{1,4},K_{1,4}+e\}$-free split graph of order at least 13 is Hamilton-connected},
  author = {Tao Tian and Fengming Dong},
  journal= {arXiv preprint arXiv:2603.12770},
  year   = {2026}
}