Byzantine Lattice Agreement in Synchronous Systems

TL;DR

This paper introduces three algorithms for Byzantine lattice agreement in synchronous systems, improving round complexity and message efficiency while tolerating up to one-third Byzantine failures.

Contribution

It presents three novel algorithms with different round and message complexities for Byzantine lattice agreement in synchronous systems, expanding the known solutions.

Findings

First algorithm: min {3h(X)+6, 6√f+6} rounds, O(n^2 min{h(X), √f}) messages.

Second algorithm: 3 log n + 3 rounds, O(n^2 log n) messages.

Third algorithm: 4 log f + 3 rounds, O(n^2 log f) messages.

Abstract

In this paper, we study the Byzantine lattice agreement problem in synchronous systems. The lattice agreement problem in crash failure model has been studied both in synchronous and asynchronous systems, which leads to the current best upper bound of $O(\log f)$ rounds in both systems. However, very few algorithmic results are known for the lattice agreement problem in Byzantine failure model. The paper [Nowak et al., DISC, 2019] first gives an algorithm for a variant of the lattice agreement problem on cycle-free lattices that tolerates up to $f < n/(h(X) + 1)$ Byzantine faults, where $n$ is the number of processes and $h(X)$ is the height of the input lattice $X$. The recent preprint by Di et al. studies this problem with a slightly modified validity condition in asynchronous systems. They present a $O(f)$ rounds algorithm by using the reliable broadcast primitive as a first step and following the similar algorithmic framework as the algorithms in crash failure model. In this paper, we propose three algorithms for the Byzantine lattice agreement problem in synchronous systems. The first algorithm takes $\min \{3h(X) + 6,6\sqrt{f} + 6\})$ rounds and $O(n^2 \min\{h(X), \sqrt{f}\})$ messages, where $h(X)$ is the height of the input lattice $X$, $n$ is the total number of processes. The second algorithm runs in $3\log n + 3$ rounds and takes $O(n^2 \log n)$ messages. The third algorithm takes $4 \log f + 3$ rounds and takes $O(n^2 \log f)$ messages. All algorithms can tolerate up to $f < \frac{n}{3}$ Byzantine failures.