Minimum lethal sets in grids and tori under 3-neighbour bootstrap percolation

Description

Let $\mathcal{r} ≥ 1$ be any non negative integer and let $G = (V, E)$ be any undirected graph in which a subset $D ⊆ V$ of vertices are initially infected. We consider the following process in which, at every step, each non-infected vertex with at least $\mathcal{r}$ infected neighbours becomes and an infected vertex never becomes non-infected. The problem consists in determining the minimum size $s_r (G)$ of an initially infected vertices set $D$ that eventually infects the whole graph $G$. Note that $s_1 (G)$ = 1 for any connected graph $G$. This problem is closely related to cellular automata, to percolation problems and to the Game of Life studied by John Conway. Note that $s_1(G) = 1$ for any connected graph $G$. The case when $G$ is the $n × n$ grid $G_{n×n}$ and $\mathcal{r} = 2$ is well known and appears in many puzzles books, in particular due to the elegant proof that shows that $s_2(G_{n×n})$ = $n$ for all $n$ ∈ $\mathbb{N}$. We study the cases of square grids $G_{n×n}$ and tori $T_{n×n}$ when $\mathcal{r}$ ∈ {3, 4}. We show that $s_3(G_{n×n})$ = $\lceil\frac{n^2+2n+4}{3}\rceil$ for every $n$ even and that $\lceil\frac{n^2 +2n}{3}\rceil$ ≤ $s_3(G_ {n×n})$ ≤ $\lceil\frac{n^2 +2n}{3}\rceil$ + 1 for any $n$ odd. When $n$ is odd, we show that both bounds are reached, namely $s_3(G_{n×n})$ = $\lceil\frac{n^2 +2n}{3}\rceil$ if $n$ ≡ 5 (mod 6) or $n$ = 2$^p$ − 1 for any $p$ ∈ $\mathbb{N}^*$, and $s_3(G_{n×n})$ = $\lceil\frac{n^2 +2n}{3}\rceil$ + 1 if $n$ ∈ {9, 13}. Finally, for all $n$ ∈ $\mathbb{N}$, we give the exact expression of $s_4(G_{n×n})$ and of $s_r(T_{n×n})$ when $\mathcal{r}$ ∈ {3, 4}.

Additional details