如何达成共识？一个组合充要条件及一致性可判定但NP-难的证明
How to decide consensus? A combinatorial necessary and sufficient condition and a proof that consensus is decidable but NP-hard
摘要：
一个随机阵集合${\cal P}$是一个一致集，如果对于每个其元素属于${\cal P}$的矩阵序列$P(1), P(2), \ldots$和每个初态$x(0)$，由$x(t) = P(t) P(t-1) \cdots P(1) x(0)$定义的状态序列收敛到一个其元素都相同的向量。在本文中，我们引入了一个“避免集条件”的紧集矩阵，并证明了在我们的主要定理，这个明确的组合条件是必要的和充分的共识。我们表明，在文献中提出的共识的条件可以直接从避免集条件。避免集条件很容易用初等算法检查，因此我们的结果也证明了一致性是算法可判定的。避免集合条件的直接验证可能需要多于多项式时间的操作。然而，这可能是任何共识检查算法的情况下，因为我们也证明了在本文中，除非$P=NP$，共识不能在多项式时间内决定。
英文摘要：
A set of stochastic matrices ${\cal P}$ is a consensus set if for every sequence of matrices $P(1), P(2), \ldots$ whose elements belong to ${\cal P}$ and every initial state $x(0)$, the sequence of states defined by $x(t) = P(t) P(t-1) \cdots P(1) x(0)$ converges to a vector whose entries are all identical. In this paper, we introduce an "avoiding set condition" for compact sets of matrices and prove in our main theorem that this explicit combinatorial condition is both necessary and sufficient for consensus. We show that several of the conditions for consensus proposed in the literature can be directly derived from the avoiding set condition. The avoiding set condition is easy to check with an elementary algorithm, and so our result also establishes that consensus is algorithmically decidable. Direct verification of the avoiding set condition may require more than a polynomial time number of operations. This is however likely to be the case for any consensus checking algorithm since we also prove in this paper that unless $P=NP$, consensus cannot be decided in polynomial time.
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。
