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arXiv:2611.12795 · v1

A short proof of the Aaronson-Ambainis Conjecture via the completion framework

通过补全框架对Aaronson-Ambainis猜想的简短证明

首次提交 2026-11-10 02:01
版本更新 2026-11-10 02:01
雷达观测 2026-11-10 15:22
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摘要

We give a short proof of the Aaronson-Ambainis conjecture. If a partial Boolean function ff is defined on a δδ-fraction of all nn-bit strings and have quantum query complexity Q(f)Q(f), then, for every 0<ε<10<ε<1, a deterministic algorithm computes ff correctly on at least a 1−ε1-ε fraction of its domain using O(Q(f)6/(εδ)2)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.

作者备注:7 pages

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作者报告利用补全框架给出了Aaronson-Ambainis猜想的一个简短证明,并建立了基于经典对手测度的更一般界。

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v1 · 2026-11-10 02:01 · 最新观测版本

A short proof of the Aaronson-Ambainis Conjecture via the completion framework

We give a short proof of the Aaronson-Ambainis conjecture. If a partial Boolean function ff is defined on a δδ-fraction of all nn-bit strings and have quantum query complexity Q(f)Q(f), then, for every 0<ε<10<ε<1, a deterministic algorithm computes ff correctly on at least a 1−ε1-ε fraction of its domain using O(Q(f)6/(εδ)2)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.

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