AI 中文总结
研究有向图中反馈顶点集问题，通过证明\(fvs(G) \leq \frac{2n + m + h}{9}\)，得出平面图、无向三角形平面有向图及最大度为\(6\)的有向图的反馈顶点集上界，改进了相关已知结果并回答了猜想。
AI 中文摘要
对于有向图\(G\)，用\(fvs(G)\)表示从\(G\)中删除使其无环的最小顶点数。我们证明了一个\(n\)个顶点和\(m\)条弧的有向图\(G\)满足\(fvs(G) \leq \frac{2n + m + h}{9}\)，其中\(h\)表示属于一类特殊有向图的\(G\)的连通分量数。此结果有三个推论：一是当\(G\)是平面图时，\(fvs(G) \leq \frac{2n + m}{9}\)，改进了已知的\(\frac{3n}{5}\)的上界；二是应用于无向三角形的平面有向图时，\(fvs(G) \leq \frac{6n - 8}{13}\)，改进了当前\(\frac{n}{2}\)的最佳界；三是当\(G\)的最大度为\(6\)时，\(fvs(G) \leq \frac{4n}{7}\)且此界是紧的，回答了一个猜想。
英文摘要
For an oriented graph $G$, denote by $fvs(G)$ the minimum number of vertices whose deletion from $G$ makes it acyclic. We show that an oriented graph $G$ on $n$ vertices and $m$ arcs satisfies $fvs(G) \le \frac{2n+m+h}{9}$ where $h$ denotes the number of connected components of $G$ that belong to a special class of oriented graphs. This result has three consequences. First, when $G$ is planar, we obtain that $fvs(G) \le \frac{2n+m}{9}$. In particular, this implies that $fvs(G) \le \frac{5n-6}{9}$ for any planar oriented graph $G$, improving the best known upper bound of $\frac{3n}{5}$~[Borodin, Discrete Mathematics, 1979]. Then, applying this inequality to the planar digraphs without directed triangles, we get that $fvs(G) \le \frac{6n-8}{13}$, which improves the current best bound of $\frac{n}{2}$~[Li and Mohar, SIAM Journal on Discrete Mathematics, 2017]. Finally, when $G$ has maximum degree 6, we have $fvs(G) \le \frac{4n}{7}$ and this bound is tight, answering a conjecture of Ai, Gutin, Liu, Yeo and Zhou~[arXiv:2512.01676, 2025].
Comments15 pages, 14 figures
