🤖 AI Summary
本文重新探讨了Yao的全表模型中的隐式成员问题，通过使用大型语言模型改进了45年前的Ramsey界，特别是将两次查询的上界从塔函数类型降低到多项式类型。
📝 Abstract
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.