TITLE: Are Pandemic chain reactions confluent? (vertex spills weight to neighbors at threshold, once)
QUESTION [6 upvotes]: Are resolutions of chain reactions order-independent in the board game Pandemic?  More formally:
You're given an undirected graph $G = (V, E)$ and a vertex weight $w \colon V \to \{0, \ldots, 3\}$.  Each vertex is also in a state $s \colon V \to \{\textsf{no outbreak}, \textsf{outbreak}\}$.
To increase a vertex $v$:

If $s(v) = \textsf{outbreak}$ then nothing happens.
Else, if $w(v) < 3$ then $w(v) \gets w(v) + 1$.
Else $s(v) \gets \textsf{outbreak}$ and increase all of $v$'s neighbours.

It is not specified which order $v$'s neighbours should be increased in.  Is such a specification superfluous?
I observe that setting $s(v) \gets \textsf{outbreak}$ is essentially the same as removing $v$ from $G$.  I don't know how that helps, though.
Also, if I understand what confluence means, I think my question can be stated: "is the following rewriting system confluent?", where each element is a specification of $w$ and $s$ and a number of due increases for each vertex $v$, and each rewrite is an increase of some $v$ (for which an increase is due; also decrement the number of increases due for $v$).

REPLY [2 votes]: This can be argued directly, by considering the directed acyclic graph (DAG) within $(V, E)$ that encodes all of the resulting updates; a directed edge $(u, v)$ corresponds to an outbreak at $u$ that either incremented $w(v)$ or set $s(v) \leftarrow \text{outbreak}$. In this sense, the in-degree of a node corresponds to the number of cubes added (plus one if an outbreak was also triggered). The claim is that any order of updates will always result in a DAG with the same in-degree at each node (i.e., the same number of cubes added, and the same outbreaks). 
Assume by way of contradiction that two different orderings result in two DAGs: $D_1 = (V, E_1)$ and $D_2 = (V, E_2)$, with in-degree $\deg^-_{D_1}(v) > \deg^-_{D_2}(v)$ for some $v \in V$. Thus there exists some $u \in V$ such that $e := (u, v) \in E_1$ but $e \not\in E_2$. This can only happen if the parent $u$ had an outbreak in $D_1$ but not in $D_2$ (since $\deg^-_{D_1}(v) > \deg^-_{D_2}(v)$ implies that $D_2$ would have updated $v$ from $u$ if it could). Thus $\deg^-_{D_1}(u) > \deg^-_{D_2}(u)$ as well (since $u$ had an outbreak in $D_1$ but not $D_2$). This argument repeats until we hit the root vertex of $D_1$ (since we are traversing up a directed acyclic path in $D_1$). But the in-degree of the root vertex must be $0$ in both DAGs, a contradiction. $\square$
This could also be turned into an inductive proof to be a bit more formal, but I feel this would only obscure the intuition.
Also note this works for the case of multiple roots as well. In this case we create an imaginary vertex connected to each root (with 3 cubes), and then induce an outbreak at the new root.