We study the extremal Forrelation problem, where, provided with oracle access to Boolean functions $f$ and $g$ promised to satisfy either $\textrm{forr}(f,g)=1$ or $\textrm{forr}(f,g)=-1$, one must determine (with high probability) which of the two cases holds while performing as few oracle queries as possible. It is well known that this problem can be solved with \emph{one} quantum query; yet, Girish and Servedio (TQC 2025) recently showed this problem requires $\widetildeΩ(2^{n/4})$ classical queries, and conjectured that the optimal lower bound is $\widetildeΩ(2^{n/2})$. Through a completely different construction, we improve on their result and prove a lower bound of $Ω(2^{0.4999n})$, which matches the conjectured lower bound up to an arbitrarily small constant in the exponent.

翻译：我们研究极值Forrelation问题：给定对布尔函数$f$和$g$的预言机访问，且保证满足$\textrm{forr}(f,g)=1$或$\textrm{forr}(f,g)=-1$中的一种情况，目标是在执行尽可能少的预言机查询的前提下，以高概率判定属于哪一种情况。众所周知，该问题可通过\emph{一次}量子查询解决；然而，Girish与Servedio（TQC 2025）近期证明该问题的经典查询复杂度需达到$\widetildeΩ(2^{n/4})$，并猜想最优下界应为$\widetildeΩ(2^{n/2})$。通过完全不同的构造方法，我们改进了他们的结果，证明了$Ω(2^{0.4999n})$的下界，该结果与猜想下界在指数上的差异仅为任意小常数。