量子优惠券收集器
Quantum Coupon Collector
摘要：
我们研究如何有效地$k$元素集$S\subseteq[n]$可以从一个统一的叠加$|S\rangle$的元素。人们可以把$|S\rangle=\sum_{i\in S}|i\rangle/\sqrt{|S|}$看作是$S$上均匀随机样本的量子版本，就像在“优惠券收集器问题”的经典分析中一样。"我们证明，如果$k$接近$n$，那么我们可以使用比随机样本更少的量子样本来学习$S$。特别地，如果存在$n-k=O(1)$缺失元件，则$|S\rangle$的$O(k)$拷贝就足够了，这与经典试样收集器所需的$Θ(k\log k)$随机样本形成对比。另一方面，如果$n-k=Ω(k)$，则$Ω(k\log k)$量子样本是~必需的。更一般地，我们给出了每个$k$和$n$所需的量子样本数量的严格界限，并且我们给出了有效的量子学习算法。我们还给出了紧边界的模型，我们可以通过$|S\rangle$额外反映。最后，我们将优惠券收集与一个已知的例子相关联，该例子分离了正确和不正确的PAC学习，结果表明在量子情况下没有分离。
英文摘要：
We study how efficiently a $k$-element set $S\subseteq[n]$ can be learned from a uniform superposition $|S\rangle$ of its elements. One can think of $|S\rangle=\sum_{i\in S}|i\rangle/\sqrt{|S|}$ as the quantum version of a uniformly random sample over $S$, as in the classical analysis of the ``coupon collector problem.'' We show that if $k$ is close to $n$, then we can learn $S$ using asymptotically fewer quantum samples than random samples. In particular, if there are $n-k=O(1)$ missing elements then $O(k)$ copies of $|S\rangle$ suffice, in contrast to the $Θ(k\log k)$ random samples needed by a classical coupon collector. On the other hand, if $n-k=Ω(k)$, then $Ω(k\log k)$ quantum samples are~necessary. More generally, we give tight bounds on the number of quantum samples needed for every $k$ and $n$, and we give efficient quantum learning algorithms. We also give tight bounds in the model where we can additionally reflect through $|S\rangle$. Finally, we relate coupon collection to a known example separating proper and improper PAC learning that turns out to show no separation in the quantum case.
机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。
