量子甲骨文审讯:以几乎一半的价格获得所有信息

W. van Dam
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引用次数: 87

摘要

考虑一个量子计算机与一个域大小为N的二进制oracle相结合。它显示了对oracle的N/2+/spl径向/N调用如何足以以大于95%的概率猜测oracle的整个内容(是一个N位字符串)。这与需要N次调用才能完成相同任务的经典计算机形成了对比。从这个结果可以得出,如果允许很小的错误概率,任何以oracle的N位作为输入的函数都可以使用N/2+/spl径向/N查询来计算。通过使用N/2+O(/spl radial /N) oracle查询,可以使错误概率任意小。文章的第二部分对“近似讯问”进行了探讨。这是当只请求N个oracle位的特定部分时。同样在这种情况下,量子算法优于经典协议。给出了一个例子,其中N/10个查询的量子过程返回一个字符串,其中80%的位是正确的。任何经典协议都需要6N/10个查询来建立这样的正确比率。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Quantum oracle interrogation: getting all information for almost half the price
Consider a quantum computer in combination with a binary oracle of domain size N. It is shown how N/2+/spl radic/N calls to the oracle are sufficient to guess the whole content of the oracle (being an N bit string) with probability greater than 95%. This contrasts the power of classical computers which would require N calls to achieve the same task. From this result it follows that any function with the N bits of the oracle as input can be calculated using N/2+/spl radic/N queries if we allow a small probability of error. It is also shown that this error probability can be made arbitrary small by using N/2+O(/spl radic/N) oracle queries. In the second part of the article 'approximate interrogation' is considered. This is when only a certain fraction of the N oracle bits are requested. Also for this scenario does the quantum algorithm outperform the classical protocols. An example is given where a quantum procedure with N/10 queries returns a string of which 80% of the bits are correct. Any classical protocol would need 6N/10 queries to establish such a correctness ratio.
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