arXivDaily arXiv每日学术速递 周一至周五更新
arXiv周末暂无论文更新,休息一下吧,周末愉快~~
arXiv 2608.09728math.CO

具有大最小出度的规定阶有向子图

Prescribed-order subdigraphs with large minimum out-degree

Bin Chen, Lanchao Wang

AI总结:

该研究改进了有向图规定阶有向子图最小出度亏格的下界,将其从Ω(log s)提升至Ω(√s),缩小了与上界的差距,解决了Steiner针对竞赛图的相关问题。

AI中文摘要:

Alon 将 d(s) 定义为最大整数 d,使得每个有 2n 个顶点、最小出度至少为 s 的有向图,都包含一个有 n 个顶点、最小出度至少为 d 的有向子图。他证明了 s/2 - d(s) = O(√(s log s)),并进一步询问该亏格是否可被绝对常数界定。Steiner 通过构造合适的竞赛图否定了该问题,证明 s/2 - d(s) = Ω(log s)。我们采用不同构造方法,表明该亏格至少按平方根尺度增长,而非仅对数级,将 Steiner 给出的最佳下界从 Ω(log s) 改进至 Ω(√s),使上下界间仅差 √(log s) 因子,这也完全解决了 Steiner 针对竞赛图宿主提出的问题。更一般地,在 Alon 考虑的更广泛场景中,只要规定子图包含宿主有向图任意固定正比例的顶点(而非恰好一半),我们的构造均适用。

英文摘要:

Alon introduced $d(s)$ as the largest integer $d$ such that every digraph on $2n$ vertices with minimum out-degree at least $s$ contains a subdigraph on $n$ vertices with minimum out-degree at least $d$. He proved that $s/2-d(s)=O(\sqrt{s\log s})$, and further asked whether this deficit can be bounded by an absolute constant. Steiner answered this question in the negative by constructing suitable tournaments, and showed that $s/2-d(s)=Ω(\log s)$. Using a different construction, we show that the deficit grows at least on the square-root scale, rather than merely logarithmically, improving the best known lower bound due to Steiner from $Ω(\log s)$ to $Ω(\sqrt{s})$ and leaving only a factor of $\sqrt{\log s}$ between the lower and upper bounds. This also completely settles a question raised by Steiner for tournament hosts. More generally, in the broader setting considered by Alon, our construction applies whenever the prescribed subdigraphs contain any fixed positive proportion of the vertices of the host digraph rather than specifically one half.

补充信息

↑