Bootstrap percolation on the high-dimensional Hamming graph

TL;DR

This paper analyzes the critical probability for bootstrap percolation on high-dimensional Hamming graphs, extending previous results from hypercubes to more general Cartesian products of complete graphs.

Contribution

It extends the asymptotic analysis of the critical probability for 2-neighbour bootstrap percolation from hypercubes to Hamming graphs with arbitrary size, for large dimensions.

Findings

Determines the asymptotic critical probability for Hamming graphs.

Extends previous hypercube results to more general graphs.

Provides bounds valid for large dimensions and vertex counts.

Abstract

In the random $r$-neighbour bootstrap percolation process on a graph $G$, a set of initially infected vertices is chosen at random by retaining each vertex of $G$ independently with probability $p\in (0,1)$, and "healthy" vertices get infected in subsequent rounds if they have at least $r$ infected neighbours. A graph $G$ \emph{percolates} if every vertex becomes eventually infected. A central problem in this process is to determine the critical probability $p_c(G,r)$, at which the probability that $G$ percolates passes through one half. In this paper, we study random $2$-neighbour bootstrap percolation on the $n$-dimensional Hamming graph $\square_{i=1}^n K_k$, which is the graph obtained by taking the Cartesian product of $n$ copies of the complete graph $K_k$ on $k$ vertices. We extend a result of Balogh and Bollob\'{a}s [Bootstrap percolation on the hypercube, Probab. Theory Related Fields. 134 (2006), no. 4, 624-648. MR2214907] about the asymptotic value of the critical probability $p_c(Q^n,2)$ for random $2$-neighbour bootstrap percolation on the $n$-dimensional hypercube $Q^n=\square_{i=1}^n K_2$ to the $n$-dimensional Hamming graph $\square_{i=1}^n K_k$, determining the asymptotic value of $p_c\left(\square_{i=1}^n K_k,2\right)$, up to multiplicative constants (when $n \rightarrow \infty$), for arbitrary $k \in \mathbb N$ satisfying $2 \leq k\leq 2^{\sqrt{n}}$.