Population protocols model information spreading in networks where pairwise node exchanges are determined by an external random scheduler. Most of the population protocols in the literature assume that the participating $n$ nodes are honest. Such an assumption may not be, however, accurate for large-scale systems of small devices. Hence, in this work, we study population protocols in a setting where up to $f$ nodes can be Byzantine. We examine the majority (binary) consensus problem against different levels of adversary strengths, ranging from the Full adversary that has complete knowledge of all the node states to the Weak adversary that has only knowledge about which exchanges take place. We also take into account Dynamic vs Static node corruption by the adversary. We give lower bounds that require any algorithm solving the majority consensus to have initial difference $d = \Omega(f + 1)$ for the tally between the two proposed values, which holds for both the Full Static and Weak Dynamic adversaries. We then present an algorithm that solves the majority consensus problem and tolerates $f \leq n / c$ Byzantine nodes, for some constant $c>0$, with $d = \Omega(f + \sqrt{n \log n})$ and $O(\log^3 n)$ parallel time steps, using $O(\log n)$ states per node. We also give an alternative algorithm with $d = \Omega(\min\{f \log^2 n + 1,n\})$. Moreover, we combine both algorithms into one using random coins. The only other known previous work on Byzantine-resilient population protocols tolerates up to $f = o(\sqrt n)$ faulty nodes and was analyzed against a Static adversary; hence, our protocols significantly improve fault-tolerance by an $\omega(\sqrt n)$ factor and all of them work correctly against a stronger Dynamic adversary.
翻译:在对称节点交换由外部随机调度来确定的网络中传播的人口协议模式信息。 文献中的大多数人口协议假定参与的节点是诚实的。 但是, 这样的假设可能对于大型小型设备系统来说并不准确。 因此, 在这项工作中, 我们在一个设置中研究人口协议, 其中最多为$的节点可以是Byzantine。 我们检查了多数( 二进制) 共识问题, 对抗不同水平的对手力量, 从完全了解所有节点状态到只知道要进行交流的弱对手。 我们还考虑到动态与静态的节点腐败。 我们给出了较低的范围, 需要任何算法来解决大多数共识的最初差 $ =\ Omega (f+ 1), 两种拟议值之间的高点, 既保留着我们完全的节点, 也保留了ndal dirdal Rwaltal 对手。 然后我们展示了一种算法, 解决了多数的共识问题, 并容忍美元/ 美元比尚诺埃 。