On the Reversibility of ECAs with Fully Asynchronous Updating: The Recurrence Point of View

Abstract

The reversibility of classical cellular automata is now a well-studied topic but what is reversibility when the evolution of the system is stochastic? In this context, we study a particular form of reversibility: the possibility of returning infinitely often to the initial condition after a random number of time steps. This corresponds to the recurrence property of the system. We analyse this property for the 256 elementary cellular automata with a finite size and a fully asynchronous updating, that is, we update only one cell, randomly chosen, at each time step. We show that there are 46 recurrent rules which almost surely come back to their initial condition. We analyse the structure of the communication graph of the system and find that the number of the communication classes may have different scaling laws, depending on the active transitions of the rules (those for which the state of the cell is modified when an update occurs).

Keywords

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Appendix

We calculate here an equivalent the number of communication classes of rule 105:ADEH for a ring of size n. For simplicity, let us denote this number by $u_n = {\mathcal C}_{DH}(n)$.

This quantity verifies the following linear recurrence equation:

$$ u_n = u_{n-1} + u_{n-2} + u_{n-4} \text { with } u_0=3 , u_1=2, u_2=2 , u_3=5 . $$

The values $u_0$ and $u_1$ are defined only for convenience and do not correspond to ${\mathcal C}_{DH}(0)$ and ${\mathcal C}_{DH}(1)$, which are excluded form our study. To solve this equation, we need to find the roots of the polynomial

$$ P(x) = x^4 - x^3 - x^2 - 1 = (x+1)(x^3 - 2 x ^2 + x - 1).$$

There are two real roots, $\lambda$ and $-1$, and two complex roots, which are conjugate, and which we denote by ${h}\,$ and ${\bar{h}\,}$. The general solution can thus be written:

$$\begin{aligned} u_n= A \lambda ^n + B {h}\,^n + C {\bar{h}\,}^n + D (-1) ^n , \end{aligned}$$

(1)

where A, B, C, D are four constants which belong to the set of complex numbers.

Looking at the first terms, we have:

$$ \begin{array}{ccccccccc} u_{0} =&{}{}&{} A\,\, +&{}{}&{} B\,\, + &{}{}&{} C &{}{+}\,\, D &{}= {3} \\ u_{1} =&{}{\lambda }&{} A\,\, +&{}{{h}\,}&{} B\,\, + &{}{{\bar{h}\,}}&{} C &{}{-}\,\, D &{}= {2} \\ u_{2} =&{}{\lambda ^2}&{} A\,\, +&{}{{h}\,^2}&{} B\,\, + &{}{{\bar{h}\,}^2}&{} C &{}{-}\,\, D &{}= {2} \\ u_{3} =&{}{\lambda ^3}&{} A\,\, +&{}{{h}\,^3}&{} B\,\, + &{}{{\bar{h}\,}^3}&{} C &{}{-}\,\, D &{}= {5} \\ \end{array} $$

Now let us combine these equations with the evaluation of $q = u_3 - 2 u_2 + u_1 - u_0$. We find that $q = Q(\lambda ) A + Q({h}\,) B + Q({\bar{h}\,}) C - (1+2+1+1) D = 5 - 2\times 2 + 2 - 3 = 0$, where $Q(x)= x^3 - 2 x ^2 + x - 1$. Since by definition $Q(\lambda ) = Q({h}\,) = Q({\bar{h}\,}) = 0$, we find $D= 0$.

In a second step, we evaluate

$$ q'= u_2 - {\bar{h}\,}u_1 = (\lambda ^2 - \lambda {\bar{h}\,}) A + ({h}\,^2 - {h}\,{\bar{h}\,}) B = 2 - 2{\bar{h}\,}$$

and

$$ q''= {h}\,(u_1 - {\bar{h}\,}u_0) = (\lambda - {\bar{h}\,}){h}\,A + {h}\,({h}\,- {\bar{h}\,}) B = {h}\,( 2 - 3 {\bar{h}\,}). $$

The evaluation of $q' - q''$ allows one to get rid of B and we have:

$$\begin{aligned} \big ( \lambda ^2 - \lambda {\bar{h}\,}- h (\lambda - {\bar{h}\,}) \big ) A = 2 - 2{\bar{h}\,}- h (2 - 3{\bar{h}\,}). \end{aligned}$$

(2)

It is now useful to write:

$$\begin{aligned} Q(x)= & {} x^3 - 2 x ^2 + x - 1 \\= & {} (x-\lambda )\,(x-{h}\,)\,(x-{\bar{h}\,}) \\= & {} x^3 - (\lambda +{h}\,+{\bar{h}\,}) \,\, x^2 + (\lambda {h}\,+ \lambda {\bar{h}\,}+ {h}\,{\bar{h}\,}) \,\, x + \lambda {h}\,{\bar{h}\,}, \end{aligned}$$

from which we obtain by identification $\lambda +{h}\,+{\bar{h}\,}= 2$ and $\lambda {h}\,{\bar{h}\,}=1$.

Now, by using the relationships ${h}\,+{\bar{h}\,}= 2 - \lambda$ and ${h}\,{\bar{h}\,}=1/\lambda$ in Eq. 2, we obtain: $(2\lambda ^2 - 2 \lambda + 3) A = (2\lambda ^2 - 2 \lambda + 3)$, which leads to $A = 1$.

The algebraic or numerical estimation of the roots of Q gives: $\lambda \approx 1.75488$ and ${h}\,= 0.12256 + 0.74486 i$, whose module is strictly smaller than 1.

We thus obtain $u_n \sim \lambda ^n$, which was the desired result. $\square$