AI 中文总结
该论文针对d≥2的固定值，证明了τ_d(n)与σ_d(n)均为n^α_d阶，解决了Alon关于格覆盖最小规模的问题，明确了其阶的范围。
AI 中文摘要
对于固定的d≥2，令τ_d(n)为满足以下条件的集合S⊆{0,…,n}^d的最小规模：S中不同点对确定的仿射直线覆盖该网格；令σ_d(n)为类似的最小值，要求每个网格点都必须位于S中两个不同点连接的闭线段上。Alon的著名结果[GAFA,1991]证明，τ_d(n)的阶介于Ω_d(n^α_d)和O_d(n^α_d log n)之间，其中α_d = d(d-1)/(2d-1)，并询问对数项是否必要。我们证明，对于每个固定的d≥2，有c_d n^α_d ≤ τ_d(n) ≤ σ_d(n) ≤ C_d n^α_d，从而以更强的形式解决了Alon的问题。
英文摘要
For fixed $d\geq 2$, let $τ_d(n)$ be the minimum size of a set $S\subseteq\{0,\ldots,n\}^d$ such that the affine lines determined by pairs of distinct points of $S$ cover the grid. Let $σ_d(n)$ be the analogous minimum when every grid point must lie on the closed segment joining two distinct points of $S$. A celebrated result of Alon [GAFA, 1991] proved that $τ_d(n)$ is of order between $Ω_d(n^{α_d})$ and $O_d(n^{α_d}\log n)$, where $α_d=\frac{d(d-1)}{2d-1}$, and asked whether the logarithm term is necessary. We prove that $$c_d n^{α_d}\leqτ_d(n)\leqσ_d(n)\leq C_d n^{α_d}$$ for every fixed $d\geq 2$, thereby resolving Alon's problem in a stronger form.
Comments12 pages
