How to decide consensus? A combinatorial necessary and sufficient condition and a proof that consensus is decidable but NP-hard

TL;DR

This paper introduces an explicit combinatorial condition called the avoiding set condition that precisely characterizes consensus in stochastic matrices, proving that consensus is decidable but NP-hard.

Contribution

The paper presents the avoiding set condition as a necessary and sufficient criterion for consensus, linking it to existing conditions and establishing the problem's computational complexity.

Findings

The avoiding set condition is both necessary and sufficient for consensus.

Consensus checking is algorithmically decidable but NP-hard.

Existing conditions for consensus can be derived from the avoiding set condition.

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.