By $E$, denote the set of excited states $E=\left\{1,2,\ldots,e\right\}$ and by $R$ the set of refractory states $R=\left\{e+1,e+2,\ldots,e+r\right\}$. By $0$, denote the equilibrium state. The alphabet $A$ is $A=\left\{0,1,\ldots,e,e+1,e+2,\ldots,e+r\right\}$. Let $X=A^{\mathbb{Z}}$ denote the space of all bi-infite configurations. Let $\eta\in X$. By $\eta_n(x)$, denote the state on position $x$ at time $n$. Let $T\colon X\to X$ describe the following dynamics: $$ \eta_{n+1}(x)=\begin{cases}i+1, & \text{ if }\eta_n(x)=i, ~1\leq i\leq e+r-1\\0, & \text{ if }\eta_n(x)=e+r\\0, & \text{ if }\eta_n(x)=0\text{ and }(\eta_n(x-1)\notin E, \eta_n(x+1)\notin E)\\1, & \text{ if }\eta_n(x)=0\text{ and }(\eta_n(x-1)\in E\text{ or }\eta_n(x+1)\in E)\end{cases} $$
My question is if it is possible to compute (or estimate) the topological entropy $h(X,T)$ in case $r\geq e$.
Remark
For $A=\left\{0,1,2\right\}, E=\left\{1\right\}$ and $R=\left\{2\right\}$ it is known that $h(X,T)=2\ln\rho$ where $\rho$ is the largest eigenvalue of $\lambda^3-\lambda^2-1=0$, see "Some Rigorous Results for the Greenberg–Hastings Model" by Steif and Durrett. You can find this paper and the proof of the mentioned result here, pp. 677. There, one essential step in the computation was to determine the set $Y=\bigcap_{n\geq 0}T^nX$ and to use that $h(X,T)=h(Y,T)$. The heart of the idea was that in this special case, each $y\in Y$ has a separating position $n\in\mathbb{Z}\cup\left\{\pm\infty\right\}$ such that to the left of this position, there is a right moving section and, to the right of this position, there is a left moving part.
Hence, my first idea now was to imitate this proof by trying to characterize the set $Y$ for the general setting (without knowing if this might be helpful in the general setting). Up to my attempts, it seems to be that $Y$ consists of those configurations for which there is an $n\in\mathbb{Z}\cup\left\{-\infty,+\infty\right\}$ such that
on position $n$ and to its left
- $0$ has one of $0, 1, 2,\cdots, e$ to its left,
- j has one of $j+1,j+2,\cdots j+e$ to its left for $1\leq j\leq e+r-1$
- $e+r$ has one of $0,1,\cdots e-1$ to its left
to the right of position $n$,
- $0$ has one of $0, 1, 2,\cdots, e$ to its right,
- j has one of $j+1,j+2,\cdots j+e$ to its right for $1\leq j\leq e+r-1$
- $e+r$ has one of $0,1,\cdots e-1$ to its right
In case, we have $E=\left\{1\right\}$ and any number of refractory states, i.e. $R=\left\{2,3,\ldots,e+r\right\}$, this implies - to my opinion - that we can use the same technique to compute $h(X,T)$ as done in the cited paper, since the separating position again separates a right- from a left-moving part. We only have to modify the involved sets and factor mapping. Hence, one should get $h(X,T)=2\ln\rho$ with $\rho$ being the largest positive eigenvalue of the polynomial $\lambda^{e+r+1}-\lambda^{e+r}-1=0$. This covers the situation in the paper.
In case we have more than just one excited state, i.e. $e\geq 2$, the set $Y$ does not longer consist only of such nice configurations, so one should have $h(X,T)\geq 2\ln\rho$. In order to compute $h(X,T)$ explicitly, I guess, one has to compute $h(X,T)$ in a completely new way or maybe an explicit computation is not possible at all. I do not know yet. That's why I am searching for some motivation/ inspiration here. Maybe one shall start with the case $E=\left\{1,2\right\}, R=\left\{3,4\right\}$, i.e. $r=e=2$. Maybe anybody has an idea how to start the computation. I did not find a way yet. Maybe, one can again find some factor map in order to go over to an easier system with same topological entropy.
Thanks in advance for any kind of inspiration!
Edit: I tried to construct a factor map. Let $E=\left\{1,2\right\}, R=\left\{3,4\right\}, A=\left\{0,1,2,3,4\right\}$, i.e. $e=r=2$. By $Y$ denote the set consisting of all configurations as described above. By $Y'$ denote the same set but for $E=\left\{1\right\}, R=\left\{2,3,4\right\}$.
Define the map $U\colon Y\to Y'$ by saying what are the pictures of triples:
$$ 000\mapsto 000\\ 001\mapsto 001\\ 002\mapsto 0012\\ 012\mapsto 012\\ 013\mapsto 0123\\ 023\mapsto 0123\\ 024\mapsto 01234\\ 123\mapsto 123\\ 124\mapsto 1234\\ 134\mapsto 1234\\ 130\mapsto 12340\\ 234\mapsto 234\\ 230\mapsto 234\\ 240\mapsto 2340\\ 241\mapsto 23401\\ 340\mapsto 340\\ 341\mapsto 3401\\ 300\mapsto 340\\ 301\mapsto 3401\\ 302\mapsto 34012\\ 400\mapsto 400\\ 401\mapsto 401\\ 402\mapsto 4012\\ 412\mapsto 4012\\ 413\mapsto 4123 $$ Start at the separating position $n$ of $y\in Y$ and look at the triples to its left resp. its pictures (adding as least linking entries as needed between the pictures of the triples in order to get a configuration in $Y'$) and, similarly, to the triples starting at position $n+1$.
Let "$|$" separate the $n$-th from the $(n+1)$-th position. So, for example,
$$ \ldots 1000420321042003 | 12302413400001\ldots \mapsto \ldots 100043210432104321043|1234012340123400001\ldots $$
As far as I see this is a surjection fulfilling $$ U\circ T=T\circ U. $$
Since $Y'\subset Y$ and using an estimation by Bowen, we shall have $$ h(Y',T)=2\ln\rho\leq h(Y,T)\leqslant h(Y',T)+\sup_{y'\in Y'}h(U^{-1}(\left\{y'\right\}),T)=2\ln\rho + \sup_{y'\in Y'}h(U^{-1}(\left\{y'\right\}),T). $$
So, maybe the question is if we can compute/estimate the supremum.
Unfortunately, determining $U^{-1}(\left\{y'\right\})$ for some $y'\in Y'$ and, hence, $h(U^{-1}(\left\{y'\right\}),T)$ seems to be difficult. Maybe, my suggested map $U$ is simply not good enough and there is a better one. Maybe it is problematic that the pictures of the triples under $U$ have different lengths.
Anyway, I was not yet successful in finding a way to handle this. I think it may be possible also that the supremum is not finite.
Edit 2 I think I found a way to show that the entropy is infinity by approximating the max. number of separated sets. So, from my point of view the thread is through.
I'm just going to focus on proving that for $|A|=5$, the $|E|=1$ system is not isomorphic to $|E|=2$ system. To do this I will count the total number of period $5$ elements, ie points where $T^5(x) = x$, and show that in the $|E|=1$ case there are countably many, whereas in the $|E|=2$ case there are uncountably many.
Observation 1:
Given a period $5$ point, $x$, either all coordinates are $0$, or every coordinate is such that $\{ T^i(x)_n\ :\ 0\leq i \leq 4\} = \{0,1,2,3,4\}$. In the first case we call the coordinate fixed, and in the second varying.
The transformation rule tells us that we are either $0$ and waiting to be awakened, or positive and counting through the relevant values. If a coordinate of a period $5$ point is ever non $0$, then it must be a varying coordinate. A varying coordinate and a fixed coordinate cannot be adjacent, otherwise, the fixed coordinate will be awakened at some time. Therefore, a period $5$ point is composed of all varying coordinates, or all fixed coordinates.
So, we should only worry about periodic points with all varying coordinates.
Observation 2:
Coordinates must be awakened, and if coordinate $n$ awakens coordinate $n+1$, $n+1$ cannot awaken $n$.
So, firstly, we say coordinate $n$ is
All coordinates (of interesting points) must be awakened somehow, by observation 1. So, say $x_n$ is $0$. Either $1\leq x_{n+1} \leq |E|$ or $1\leq x_{n-1} \leq |E|$. So, $n$ is either left, right or bi-awakened.
But, if $n$ is right awakened, then when $x'_{n+1}=0$, $|A|-|E|\leq x'_n < |A|$ and by assumption, this must be a refractory state. So, $n+1$ is not left awakened.
Observation 3:
If we map coordinates to $L, R$ or $B$ according to how they are awakened, we have either all $L$s, all $R$s, or something with an infinite sequence of $L$s, possibly a single $B$, then an infinite string of $R$s.
We call this string the awakening string, and I claim this is relatively clear.
Observation 4:
If $|E|=1$ there are a finite number of points with each of the possible awakening strings.
Let $n$ be the rightmost left-awakened coordinate, and let it be $0$ at time $t$. The $n-1$ coord is $1$ at time $t$ (because it is awakening $n$), the $n-2$ coord must be $2$ (because it awakened $n-1$ at $t-1$), and this sequence of implication, determining the coordinate values to the left of $n$ continues.
And, by symmetry, the same thing happens going to the right of the leftmost, right-awakened coordinate. Which is hard to say, but I think it makes sense.
This implies that there are only a countable number of period $5$ points when $|E|=1$.
Observation 5:
If $|E|=2$, there is a distinct period $5$ point for every bi-infinite sequence of $1$'s and $2$.
Fix the awakening string to be all $L$'s, and then use the string of ones and twos to determine what value awakens the coordinate to the right. That's a slightly hazy description, but I think you should be able to see what I'm saying.
This implies there are an uncountable infinity of period $5$ points when $|E|=2$.
Hope that's some kind of help.