论文标题

量子优惠券收集器

Quantum Coupon Collector

论文作者

Arunachalam, Srinivasan, Belovs, Aleksandrs, Childs, Andrew M., Kothari, Robin, Rosmanis, Ansis, de Wolf, Ronald

论文摘要

我们研究了$ k $ element Set $ s \ subseteq [n] $可以从其元素的均匀叠加$ | s \ rangle $中学到的。一个人可以想到$ | s \ rangle = \ sum_ {i \ in s} | i \ rangle/\ sqrt {| s |} $是$ s $均匀随机样品的量子版本,就像在$ k $的$ n $上一样,我们可以在$ n $上示出$ n $。比随机样品。特别是,如果有$ n-k = o(1)$丢失的元素,则与$ | s \ rangle $的$ o(k)$副本相比,与$θ(k \ log k)$相反,经典优惠券收集器需要随机样本。另一方面,如果$ n-k =ω(k)$,则需要​​$ω(k \ log k)$量子样本。 更一般而言,我们对每$ k $和$ n $所需的量子样本的数量进行紧密界定,并提供有效的量子学习算法。我们还在模型中给出了紧密的界限,我们可以通过$ | s \ rangle $进行反思。最后,我们将优惠券的收集与一个已知的示例相关联,该示例将适当和不当的PAC学习分开,事实证明在量子情况下没有显示分离。

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.

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