ELSEVIER Information Processing Letters 61 (1997) 169-172

Information

Esing

Ergodicity of linear cellular automata over Z,

Tadakazu Sato Department of Information and Computer Science, Toyo University, 2100, Kujirai, Kawagoe, Saitamu, Japan

Received 25 September 1995 Communicated by K. keda

Keywords: Linear cellular automaton; Local map; Parallel map; Theory of computation

1. Introduction

The study of parallel maps is centrally located in cellular automata theory and has been discussed from various points of view [ 1,4-9,l I, 121.

In [12], Willson regarded the set of all configura- tions as a probability space and introduced measure- preserving parallel maps and gave a sufficient condi- tion for a parallel map to be ergodic.

In this paper, we shall restrict our attention to linear cellular automata over Z, and give the neces- sary and sufficient conditions for a parallel map to be ergodic. As a result, we show that the problem of whether or not a linear parallel map is ergodic is decidable.

2. Preliminaries

In this paper, we discuss for simplicity linear cellular automata with two-dimensional cell spaces, however, we can easily generalize the results ob- tained here to any higher-dimensional cell spaces.

A linear cellular automaton over E, is given by a quadruplet (Z’. Z,, f, N), where h is the set of all integers, Z2 is the set of all pairs of integers called a two-dimensional cell space. For a positive integer m, Z,,, is the set of integers modulo m and denotes a state set of each cell. f is a linear map of Hz into Z,,, called a linear local map over Z, with scope-n,

that is, fix ,,.. ., x,) = Cajxj. N is a finite subset of Z2 and let N={v i, . . . , u,,}. N represents the neigh- bour frame at each cell.

A configuration over Z, is a map of Z2 into Zm and the set of all configurations over Z, is denoted by C(Z,). For a linear local map f with a neighbour frame N, we define a map fr : C(Z,) + C(Z,) as follows: for U,W E C(Z,), fXu)= w ++ W(T) = Caju(r+uj)forall r~2~,wherer+u,,...,r+u, are called the neighbourhood of the cell r E Z2. f% is called a linear parallel map. For s E E2, we define a shift map us : C(Z,) + C(Z,) as follows: for u,wEC(ZJ. oS(u)=w * W(T)-U(T+S) for all r E Z2. The linear parallel map f has a repre- sentation f= = Cajaul.

Now let us give some definitions with respect to local maps which are needed in the following.

Definition 1. (1) f is injectiue on L, if f, is

injective on C(E,). (2) f is surjectioe on Zm if fx is surjective on

C(iz,). (3) f_ has finite order on C(H,) if some power

of f_ equals an identity map on C(H,). (4) f has finite order on Z, if some power of fr

equals a shift transformation on C(E,).

@020-019Q/97/$17.00 8 1997 Elsevier Science B.V. A11 rights reserved. PI/ SOO20-0190(96)00206-2

170 T. &to /Injhwrion Processing Letters 61 (1997) 149-l 72

(5) f has infinite order on Z, if f is injective on Z, but does not have finite order on Z,.

For a local map f = Cajxj with a neighbour frame N, let F( X, Y) denote the polynomial repre-

sentation of fi (see [5]). That is

fi = &lid - F( X, Y) = CaiX”jYYj,

where the summation is taken over all vectors uj = (xj, yj) in N.

The importance of the polynomial representation is that if F( X, Y) and G( X, Y > are the polynomial representations of f_ and g, respectively, then that of the composition of fm and g, equals the product of F( X, Y) and G(X, Y >. For example, if fi has

finite order on C(H,), then F(X, Y >” = 1 for some positive integer n.

Now let us consider a measure on C(Z,) in the

following way [121. The following subset of C@,>

{xEC(ZIm) I x(r,) =a ,,..., x(rs) =a,}

denoted by D,,,( r, , . . . , r,) or simply D,,, is called a

cylinder set over C@,> and let its measure denoted

by p[D,,,(r,,..., r,)] be 1 /m’. Then the measure p on the set of all cylinder sets is uniquely extended to the r-algebra O(Z,) generated by the set of all cylinder sets. We use again the same symbol p as the measure on O(Z,) for each m. Clearly, (C(Z,>, O(Z,), ~1 becomes a probability space.

Let m = II;=, p,5 be the factorization in prime numbers. Since Z, = Zbl, $ . . . $ Z’I:“,, we have that D,( r,, . . . , r,) is also decomposed into Di;(r ,,.. ., r,) d . . . @ Di$r,,. . . , r,), where

D;(r,,...,r,)

=(xEC(.Z~)Ix(r,)=n ,,,..., x(rs)=asj)

and a, I

is the jth component of ~1,. Then we have

3

p[D,,,(r,, . . . . r,)] = l/m”= 1/

j= 1

A parallel map f_ is measurable if for any D E @(Z,), f;](D) E @(iI,). fm is measure-preserving if fr is measurable and for any DE O(Z,), &L(D)) = p(D). A set D E @(Z,) is invariant under fm if E * (0) = D. Let fm be a measure-pre- serving parallel map. Then fm is ergodic if for every

invariant set D E @(Z,,,), we have p(D) = 0 or p(D) = 1. fi is mixing if for any B,D E @(Z,), it holds

lim p(Ln(D> ~B)=P(D)P(B). n-t=

3. Ergodic parallel maps

In this section, we discuss the nature of ergodic parallel maps which must be measure-preserving by definition. Since the parallel maps are continuous [4,9], they are measurable. Furthermore, it is already known that measure-preservability and surjectivity are equivalent concepts with respect to the parallel maps [8]. (It follows directly from the result obtained in [6].)

Ergodicity of a linear parallel map is not an invariant property under shift transformations on C(Z,). The reason for it is that different from injectivity or surjectivity of ones, ergodicity of a linear parallel map can not be determined by its local map only but also depends on its neighbour frame.

Definition 2. A finite subset N of Z2 is rectangular if N has the following form for some element (s, t> E z*:

N={(s+i, t+j)IO<i~m,,Odj~m,)

where M, and m, are positive integers.

A cylinder set D,,,(r,, . . . , r,> is rectangular if the

set {r,,..., rs} is rectangular.

Remark 3. (1) Any cylinder set can be expressed as a union of disjoint rectangular cylinder sets.

(2) Without loss of generality, we can assume that any parallel map has a rectangular neighbour frame.

(3) Let D be a rectangular cylinder set. If f_ is surjective, then L’(D) is a union of recutangular cylinder sets.

T. Sato/lnformation Processing Letters 61 (1997) 169-172 171

Let m = II;_, p,‘j be the factorization in prime numbers. Then it induces naturally the decomposi- tion of f to (f,, . . . , fk>.

Lemma 4. If each component (&I, of fm is mixing, then fm is also mixing.

Proof. We give the proof in the case when k = 2. From [2, Theorem 1.21 and the above Remark 1.

It suffices to show that f, is mixing for all rectangu- lar cylinder sets.

Let D= (D,, D2) and B=(B,, B2) be arbitrary rectangular cylinder sets. From Eq. (1) and measure theory of product spaces 131, we have

=

Lemma 5 [ 111. Let p be a prime and r be a positive integer. For each t (1 ,( t < r>, we have

[A(X,Y)+P’B(X,Y)]“~-’

=A( X, Y)““-’ (mod p’),

where A(X, Y) and B( X, Y ) are arbitrary polyno- mials ouer Z.

From Lemmas 4 and 5, we have the following theorem.

Theorem 6. The following statements are all equiva- lent.

(1) fm is ergodic. (2) For any j (1 < j,C k), <fj), is ergodic. (3) For any j (1 G j G k), <f& is surjective on

C(Z:), but does not havefinite order on C@k). (4j F or some positive integer n, fi is mixing. (5) For some positive integer n, fu is ergodic. (6) For any j (1 Q j < k), F(X, Y) (mod Pi) is

not constant.

Proof. (1) -P (2) Let fS be an ergodic parallel map. By definition, f_ is measure-preserving. Then it is surjective on C(Z,> [12]. (The Ergodic Theorem also shows that f_ must be surjective on C(Z,).> Suppose that (fj), is non-ergodic for some j. Then by definition, there exists a measurable set Bj E O(Zk) such that 0 < p(Bi) < 1 and (&‘(Bj) = Bj. Since f is surjective on Z,, each fi is also surjec- tive on Z;,. Hence put

B=(C,,...,Cj-I, Bj, Cj+l,...,Ck) E O(Z,),

where we simply write C(Z?,) by Cj. Then we have

K’(B) = ((f,)~‘(Cl),...,(f,),*(Bj),

. . ..(f&‘(Ck)) =B.

However, since each p(Ci> = 1 (1 < j G k), we have

P(B) = P(C)) . . . cL(Cj-l)p(Bj)

xp(cj+ I) ’ ’ ’ E.ctck)

=P(Bj).

Hence BE @I@,) (0 < p(B) < 1) is an invariant subset under f=. This contradicts the hypothesis that fi is ergodic.

(2) + (3) It suffices to give the proof only for jth component fj of f. The first part follows from the equivalent concepts between measure-preservability and surjectivity [8]. So we give a proof of the second part. Suppose that (&I, has order n. Consider a cylinder set D E O(Z,) such that 0 < p(D) < l/n. Then the set U zZb<fi)z”(D> is an invariant subset under (fix. This is a contradiction.

(3) + (4) Let f = (f,, . . . , fk) and let Fj( X, Y > denote the polynomial representation of fi. Consider the decomposition of Tj!X, Y > to cr(X, Y) + p(X, Y> wher e a ny coefficient of a( X, Y > can not be divided by pi and any one of p(X, Y > is a multiple of pj. From Lemma 5, we can find a positive integer nj such that F$ X, Y jnl = (Y( X, Y 1”~. By hypothesis, (J.), is surjective on C(Zl;:> but does not have finite order on C(Z?,>. We therefore con- clude that cr(X, Y> is not constant. Then it follows directly from Willson’s result [12] that if all coeffi- cients of f (assuming that F(X, Y) is not constant) are invertible elements of Z,, then f, is mixing. From this, the corresponding local map of a( X, Y 1 is mixing. Hence, each (fj)zJ is mixing. Let n be the Ieast common multipIe of (n,,. . . , nj,. . . , nk). Then

172 T. Sate / Informaiion Processing Letters 61 (I 997) 169-172

each (fj)Z is mixing. From Lemma 4, (fJ” is mix- ing.

(4) + (5) It is obvious. (5) + (1) Suppose that fi is non-ergodic. Then

there exists a measurable set B E @(Z,,,) such that 0 < p(B) < 1 and K’(B) = B. Hence for any posi- tive integer n, we have

E”(B)= ... =L2( B) =f;‘( B) = B.

This shows that ft is non-ergodic. (3) -+ (6) It is known that fj is surjective on Z’?,

iff fj =f (mod p?j> has at least one invertible ele- ment of H’j [lOi Note that in H’j., the elements divisible bp pi are nilpotent andP’the others are invertible elements. (An element a E “2 is nilpotent if uR= 0 for some positive integer n.) Suppose that Fj( X, Y > = F( X, Y > (mod pi) is constant. Then we have the following two cases.

Case 1. All coefficients of Fj( X, Y) are nilpotent elements of “2. This contradicts to the surjectivity of fi.

Case 2. The coefficient u,,~ is an invertible ele- ment and the others are nilpotent. This shows from Lemma 5 that (fj), has finite order. This is a contra- diction.

(6) + (3) It is obvious. So the proof is completed. q

Corollary. There exists an algorithm for deciding wheter or not a linear parallel map is ergodic.

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