Relational problems (those with many possible valid outputs) are different from decision problems, but it is easy to forget just how different. This paper initiates the study of FBQP/qpoly, the class of relational problems solvable in quantum polynomial-time with the help of polynomial-sized quantum advice, along with its analogues for deterministic and randomized computation (FP, FBPP) and advice (/poly, /rpoly). Our first result is that FBQP/qpoly != FBQP/poly, unconditionally, with no oracle -- a striking contrast with what we know about the analogous decision classes. The proof repurposes the separation between quantum and classical one-way communication complexities due to Bar-Yossef, Jayram, and Kerenidis. We discuss how this separation raises the prospect of near-term experiments to demonstrate "quantum information supremacy," a form of quantum supremacy that would not depend on unproved complexity assumptions. Our second result is that FBPP is not contained in FP/poly -- that is, Adleman's Theorem fails for relational problems -- unless PSPACE is contained in NP/poly. Our proof uses IP=PSPACE and time-bounded Kolmogorov complexity. On the other hand, we show that proving FBPP not in FP/poly will be hard, as it implies a superpolynomial circuit lower bound for PromiseBPEXP. We prove the following further results: * Unconditionally, FP != FBPP and FP/poly != FBPP/poly (even when these classes are carefully defined). * FBPP/poly = FBPP/rpoly (and likewise for FBQP). For sampling problems, by contrast, SampBPP/poly != SampBPP/rpoly (and likewise for SampBQP).

翻译：关系型问题（那些具有多个可能有效输出的问题）与决策问题不同，但人们很容易忘记它们究竟有多大的不同。本文首次研究了FBQP/qpoly类，即借助多项式大小的量子建议可在量子多项式时间内解决的关系型问题，以及其在确定性和随机计算（FP、FBPP）及建议（/poly、/rpoly）中的对应类。我们的第一个结果是，FBQP/qpoly ≠ FBQP/poly无条件成立，无需借助预言机——这与我们对相应决策类的认知形成了鲜明对比。该证明借鉴了Bar-Yossef、Jayram和Kerenidis关于量子和经典单向通信复杂度的分离结果。我们讨论了这一分离如何提升了近期实验展示“量子信息霸权”的前景，这是一种不依赖于未经证明的复杂性假设的量子霸权形式。我们的第二个结果是，FBPP不包含在FP/poly中——也就是说，Adleman定理在关系型问题上不成立——除非PSPACE包含在NP/poly中。我们的证明使用了IP = PSPACE和限时Kolmogorov复杂度。另一方面，我们表明证明FBPP不在FP/poly中将很困难，因为这蕴含了PromiseBPEXP的超多项式电路下界。我们证明了以下进一步结果：* 无条件地，FP ≠ FBPP且FP/poly ≠ FBPP/poly（即使这些类被精确定义）。* FBPP/poly = FBPP/rpoly（对FBQP同样成立）。相反，对于采样问题，SampBPP/poly ≠ SampBPP/rpoly（对SampBQP同样成立）。