关于隐可满足性问题的概率试验的复杂性
On the complexity of probabilistic trials for hidden satisfiability problems
摘要：
解决2SAT所需的最少信息量和时间是多少？当实例已知时，它可以在多项式时间内求解，但这在不知道实例的情况下也是可能的吗？Bei，Chen和Zhang（STOC '13）考虑了一个模型，其中通过向特殊的oracle提出可能的分配来访问输入。这个预言机在遇到提案不满足的某些约束时，只返回约束索引。事实证明，在这个模型中，即使是1SAT也不能在多项式时间内求解，除非P=NP。因此，我们考虑一个模型，在该模型中，输入是通过提出分配给变量的概率分布来访问的。然后，oracle返回该分布最有可能违反的约束的索引。我们表明，这种方式获得的信息是足够的解决1SAT在多项式时间，即使条款可以重复。对于2SAT，只要没有重复子句，我们甚至可以在多项式时间内学习隐藏实例的等价公式，从而也解决了它。我们表明，在这种情况下，1QSAT可以解决在多项式时间恒定精度，和2QSAT可以学习在多项式时间逆多项式精度。
英文摘要：
What is the minimum amount of information and time needed to solve 2SAT? When the instance is known, it can be solved in polynomial time, but is this also possible without knowing the instance? Bei, Chen and Zhang (STOC '13) considered a model where the input is accessed by proposing possible assignments to a special oracle. This oracle, on encountering some constraint unsatisfied by the proposal, returns only the constraint index. It turns out that, in this model, even 1SAT cannot be solved in polynomial time unless P=NP. Hence, we consider a model in which the input is accessed by proposing probability distributions over assignments to the variables. The oracle then returns the index of the constraint that is most likely to be violated by this distribution. We show that the information obtained this way is sufficient to solve 1SAT in polynomial time, even when the clauses can be repeated. For 2SAT, as long as there are no repeated clauses, in polynomial time we can even learn an equivalent formula for the hidden instance and hence also solve it. Furthermore, we extend these results to the quantum regime. We show that in this setting 1QSAT can be solved in polynomial time up to constant precision, and 2QSAT can be learnt in polynomial time up to inverse polynomial precision.
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