Dynamically two-coloring a finite graph

Let $G=(V,E)$ be a finite graph whose vertices are going to be colored dynamically. More precisely, consider time periods $t \in \left\{0,1,2\ldots,\right\}$ and for each time $t$ and $i \in V$, let $X_{t}(i) \in \left\{-1,+1\right\}$ be the color of vertex $i$ at time $t$. Define $$X_{t+1}(i) = \operatorname{sgn} \sum_{j \in N(i)} {X_{t}(j)},$$ where $N(i)$ is the set of neighbors of $i$ in $G$. If the sum on the right-hand side is $0$ add a self-loop at $i$ and put $X_{t}(i)$ in the sum as well. In other words, the color of $i$ at time $t+1$ is the color that's predominant among its neighbors at time $t$ (and if it is a tie, its color at time $t$ breaks this tie).

I have a complicated proof that for any configuration of initial colors $\left\{X_{0}(i)\right\}_{i \in V}$ and any vertex $i$ it holds that $X_{t+2}(i)=X_{t}(i)$ for all sufficiently large $t$, but I'd like to see if there are any elementary approaches since this sounds to me like a very fun olympiad problem.

Solution 1:

This was established by Gadi Moran in 1995: http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwjE09iPrNvNAhUL8GMKHVViBM8QFggcMAA&url=http%3A%2F%2Fwww.ams.org%2Ftran%2F1995-347-05%2FS0002-9947-1995-1297535-1%2F&usg=AFQjCNHU8Qb-Dakrnpwv0eaSU_42GfshEA&bvm=bv.126130881,d.cGc

and in fact was shown for infinite graphs, when they have bounded degree, and their growth rate (as a function of distance) is subexponential.