Early-Stabilizing Counting

TL;DR

This paper introduces an early-stabilizing counting algorithm for Byzantine fault-tolerant systems that adapts its stabilization time and bit complexity based on the actual number of faults, improving efficiency.

Contribution

It presents a novel early stabilization approach for counting in Byzantine systems, achieving optimal stabilization time and adaptive bit complexity.

Findings

Stabilizes within asymptotically optimal O(f+1) rounds

Message size is O(log^2 n + log C)

Amortized bit complexity is O(n(f log C + log^2 n))

Abstract

Synchronous Counting is the task of reaching agreement on a common round counter in a synchronous system of $n$ nodes with up to $t$ Byzantine faults in a self-stabilizing manner. That is, after transient faults may have arbitrarily corrupted the system state and ceased, the at least $n-t$ non-faulty nodes need to (re-)establish that (i) their local outputs are identical and (ii) increase by $1$ modulo $C$ in each round. An overhead-free reduction from consensus shows that all known lower bounds and impossibilities for consensus carry over to the counting problem. In the other direction, prior work has established that a consensus algorithm $\mathcal{A}$ can be turned into a counting algorithm at small overhead relative to the running time and bit complexity of $\mathcal{A}$, without losing resilience. Taking inspiration from early-stopping consensus protocols, in this work we introduce the concept of early stabilization. That is, if there are $0\le f\le t$ (persistent) faults in an execution, the algorithm should stabilize in a number of rounds that depends on $f$ only. Likewise, we seek to achieve an amortized bit complexity that is adaptive in the number of actual faults $f$. By developing a number of modular building blocks suitable to these goals, we develop a $C$-counting algorithm that stabilizes within asymptotically optimal $O(f+1)$ rounds, has message size $O(\log^2 n + \log C)$, and has amortized bit complexity $O(n(f\log C +\log^2 n))$.