We give a short proof of the Aaronson-Ambainis conjecture. If a partial Boolean function $f$ is defined on a $δ$-fraction of all $n$-bit strings and have quantum query complexity $Q(f)$, then, for every $0<ε<1$, a deterministic algorithm computes $f$ correctly on at least a $1-ε$ fraction of its domain using $O( Q(f)^6 / (εδ)^2)$ queries. Our proof uses the completion framework and establishes a more general bound in terms of the classical adversary measure. The only result we use from Turturean's recent proof is a special case of his certificate inequality, for which we provide a self-contained proof. We also derive the stronger variant of the Aaronson-Ambainis Conjecture, the bounded-polynomial influence conjecture, from our main result in the appendix.
$\mathsf{EQP}\subseteq\mathsf{IP}$ Does Not Relativize
Sinan Oral, Supartha Podder
AI 速读 基于摘要
作者构造递归预言机 O 与一元语言 L，使 L 属于 EQP^O 但不属于 IP^O∪PH^O，从而把 BQP 与 IP 的预言机分离强化到精确量子计算。
quant-ph · cs.CC
查看原文摘要
We construct a recursive oracle $O$ and a unary language $L\in\mathsf{EQP}^O\setminus(\mathsf{IP}^O\cup\mathsf{PH}^O)$, strengthening the $\mathsf{BQP}$ versus $\mathsf{IP}$ oracle separation of Bouland et al. to exact quantum computation. The separation is based on a variant of the Forrelation problem in which the normalized correlation of one Boolean function with the Fourier transform of another is promised to be exactly $+1/2$ or $-1/2$. The main technical contribution is a repair method that constructs distributions supported on these exact levels while preserving the indistinguishability needed for the $\mathsf{IP}$ and $\mathsf{PH}$ lower bounds. A uniform polynomial-time quantum algorithm distinguishes the two cases with certainty using only three oracle queries.
For bounded-error randomized query complexity, we resolve the composition conjecture up to logarithmic factors. The general upper bound $R(f\circ g) = O\!\left(R(f)R(g)\log(R(f)+1)\right)$ was already known. Although Meiburg recently disproved constant-factor multiplicative composition for total Boolean functions, the conjecture allowing polylogarithmic losses remained open. We settle this question by proving that, for all (possibly partial) Boolean functions $f$ and $g$, \[ R(f\circ g) = Ω\!\left(\frac{R(f)R(g)}{\log(n+1)}\right), \] where $n$ is the input size for $f$.