REVIEW
How to decide consensus? A combinatorial necessary and sufficient condition and a proof that consensus is decidable but NP-hard
Not yet reviewed by Pith; the record is open.
This paper has not been read by Pith yet. Machine review is queued; the pith claim, tier, and objections will appear here once it completes.
SPECIMEN: schema-true, not a live event
T0 review · schema-true
One-sentence machine reading of the paper's core claim.
pith:XXXXXXXX · record.json · timestamp
How to decide consensus? A combinatorial necessary and sufficient condition and a proof that consensus is decidable but NP-hard
read the original abstract
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.
discussion (0)
Sign in with ORCID, Apple, or X to comment. Anyone can read and Pith papers without signing in.