发表机构
Rutgers University（罗格斯大学）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文重新审视Yao完整表模型中的隐式成员问题，首次将两次探测界从塔型改进为多项式，并证明成员与搜索问题在多项式损失下等价，方法结合Ramsey理论与非Ramsey论证。
AI 中文摘要
我们重新审视Yao在完整表模型[Yao, 1981]中的隐式成员问题，并获得了据我们所知对其45年历史的Ramsey界的第一批定量改进，最显著的是将两次探测界从塔型降低到多项式。在该模型中，一个$n$元集合$S\subseteq\{1,\ldots,m\}$被存储为一个$n$单元表中的排列，查询决定$x\in S$是否成立。令$G_q(n)$为允许对所有$n$元集合进行$q$次探测成员方案的最大宇宙大小。Yao精确确定了单次探测的情况，证明了对于$n>2$，$G_1(n)=2n-2$，但对于$q\ge2$的行为仍然完全开放。Fiat和Naor[1993]为大小为$\exp(n^c)$的宇宙构造了方案，其中$c>0$为某个常数，$q$为足够大的常数。对于第一个自适应情况，$q=2$，我们证明$G_2(n)=O(n^2(\log n)^2)$。对于每个固定的整数$q\ge3$，我们证明$G_q(n)$至多为高度为$q-1$的塔，其顶部为$n^{1+o(1)}$；特别地，$G_3(n)\le\exp(n^{1+o(1)})$。两次探测的证明完全避免了Ramsey理论；对于更大的固定$q$，我们仅使用Ramsey理论使前$q-1$次探测遵循固定模式，然后通过相同的非Ramsey论证处理最后一次探测。有点令人惊讶的是，对于每个固定的$q$，我们还证明隐式成员问题与隐式搜索一样难，直到宇宙大小的多项式损失。隐式搜索必须在$x$存在时返回包含$x$的单元，否则拒绝。对于类似的搜索阈值$H_q(n)$，我们证明对于每个$q,n$，$H_q(n)\le G_q(n)\le n^q(H_q(n)+1)^{q+1}$。因此，对于每个固定的$q$，一个阈值至多为$\exp(n^{O(1)})$当且仅当另一个阈值也如此。证明最初由ChatGPT 5.5 Pro在没有数学提示的情况下生成；成员-搜索等价性是在追求改进的四次探测界时出现的。作者验证并编辑了证明，并对所有内容负责。
英文摘要
We revisit the implicit membership problem in Yao's full-table model [Yao, 1981] and obtain, to our knowledge, the first quantitative improvements to his 45-year-old Ramsey bounds, most notably reducing the two-probe bound from tower-type to polynomial. In this model, an $n$-set $S\subseteq\{1,\ldots,m\}$ is stored as a permutation in an $n$-cell table, and queries decide whether $x\in S$. Let $G_q(n)$ be the largest universe size admitting a $q$-probe membership scheme for all $n$-sets. Yao determined the one-probe case exactly, proving $G_1(n)=2n-2$ for $n>2$, but the behavior for $q\ge2$ remained wide open. Fiat and Naor [1993] constructed schemes for universes of size $\exp(n^c)$ for some constant $c>0$ and sufficiently large constant $q$. For the first adaptive case, $q=2$, we prove $G_2(n)=O(n^2(\log n)^2)$. For every fixed integer $q\ge3$, we show that $G_q(n)$ is at most a tower of height $q-1$ with top $n^{1+o(1)}$; in particular, $G_3(n)\le\exp(n^{1+o(1)})$. The two-probe proof avoids Ramsey theory altogether; for larger fixed $q$, we use Ramsey theory only to make the first $q-1$ probes follow a fixed pattern, and then handle the last probe by the same non-Ramsey argument. Somewhat surprisingly, for each fixed $q$, we also show that implicit membership is as hard as implicit search up to a polynomial loss in universe size. Implicit search must return the cell containing $x$ when present and reject otherwise. For the analogous search threshold $H_q(n)$, we prove $H_q(n)\le G_q(n)\le n^q(H_q(n)+1)^{q+1}$ for every $q,n$. Thus, for every fixed $q$, one threshold is at most $\exp(n^{O(1)})$ if and only if the other is. The proofs were first generated by ChatGPT 5.5 Pro without mathematical hints; the membership-search equivalence emerged while pursuing an improved four-probe bound. The authors have validated and edited the proofs and assume responsibility for all content.
