Burning number and optimal burning sequences

Question

I am reading about graph burning, which is a deterministic discrete-time process on graphs (Bonato, Anthony, A survey of graph burning, Contrib. Discrete Math. 16, No. 1, 185-197 (2021). ZBL1457.05068.)

I have a problem with the definition of the burning number $b(G)$ and the optimal burning sequence. In particular, it is mentioned that if a graph $G = (V(G),E)$ has a burning sequence $(x_1, x_2, \dots x_k)$, then for each pair $i,j$ with $1 \leq i < j \leq k$ and $d(x_i,x_j) \geq j-i$, the set equation $$N_{k-1}[x_1] \cup N_{k-2}[x_2] \cup \dots N_0[x_k] = V(G) $$

holds. Here, $N_i[x] = \{u \in V(G) | d(u,v) \leq i\} \cup \{x\}$.

What I do not understand is the following: what if, in the optimal burning sequence of the graph, no vertex is picked at the last round? Then, the length of the burning sequence may not be equal to the burning number of the graph. For instance, assume we have a complete graph $K_n$. The burning number of $K_n$ is obviously $2$, which is achieved by selecting any vertex of the graph as a source at roung $t=1$. However, here the burning sequence has only one vertex, namely the vertex selected at $t=1$. For this graph, is $k=1$ or $k=2$? Put another way, do we allow an empty vertex to be part of the burning sequence? Otherwise, if $k=1$, then clearly the set equation does not hold. Thus, I am confused about this particular case.

Answer 1

The condition we're given on choosing sources is that for $1 \le i < j \le k$, $d(x_i, x_j) \ge j-i$ because:

...otherwise, if $d(x_i, x_j) = l < j-i$, then $x_j$ will be burned at round $l+i < j$.

This indicates that the criterion for choosing a source on round $j$ is that it has not been burned in a round prior to $j$; it is okay to choose a source that would be burned in round $j$ anyway. To put it in other words, a round of burning happens as follows:

1. In round $i$, you choose a vertex $x_i$ which has not yet burned.
2. Then, vertices burn if they are adjacent to vertices that have burned in previous rounds. (This could cause $x_i$ to burn, and that does not invalidate our choice in step 1.)
3. Then, if $x_i$ did not burn in the previous step, $x_i$ burns anyway.

In the specific case of a complete graph, $k=2$, and $x_1$ and $x_2$ can be any two vertices. In the first round:

1. We choose vertex $x_1$.
2. Nothing burns due to being adjacent to previously burned vertices, because there are no such vertices.
3. But then, $x_1$ burns anyway, because we chose it.

In the second round:

1. We choose vertex $x_2$, distinct from $x_1$.
2. All remaining vertices burn, because they are adjacent to $x_1$.
3. Nothing happens in step 3, because $x_2$ already burned in step 2.

We have $N_1[x_1] \cup N_0[x_2] = V(G)$.