An example of a deterministic cellular automaton exhibiting linear-exponential convergence to the steady state
AAn example of a deterministic cellular automatonexhibiting linear-exponential convergence to the steadystate
Henryk Fuk´sandJoel Midgley-Volpato
Department of Mathematics and Statistics, Brock UniversitySt. Catharines, Ontario L2S 3A1, CanadaIn a recent paper [1] we presented an example of a 3-state cellularautomaton which exhibits behaviour analogous to degenerate hyperbolicityoften observed in finite-dimensional dynamical systems. We also calculateddensities of 0, 1 and 2 after n iterations of this rule, using finite statemachines to conjecture patterns present in preimage sets. Here we re-derive the same formulae in a rigorous way, without resorting to any semi-empirical methods. This is done by analyzing the behaviour of continuousclusters of symbols and by considering their interactions. The general question of finding the iterates of the Bernoulli measureunder a given cellular automaton (CA) has been subject of many recentstudies, including, among others, [2, 3, 4, 5, 6, 7, 8] and [9]. A more specificquestion of this type is sometimes called the density response problem : If theprobability of occurence of a certain state in the initial configuration drawnfrom a Bernoulli distribution is given, what is the probability of occurenceof this state after n iterations of the CA rule?Of course, one could ask a similar question about the probability ofoccurence of longer blocks of symbols after n iterations of the rule. Dueto the complexity of CA dynamics, it is clear that questions of this typeare rather hopeless if one wants to know the answer for an arbitrary rule.In spite of this, it may still be possible to find the answer if the rule issufficiently simple.One of the methods which can be used to do this is studying the struc-ture of preimages of short blocks and detecting patterns present in them.This approach has been successfully used for a number of deterministic CArules, such as elementary rules 172, 142, 130 (references [10], [11], and [12] (1) a r X i v : . [ n li n . C G ] A p r respectively), and several others. It has also been used for a special class ofprobabilistic CA known as single-transition α -asynchronous rules [13].Cellular automata are infinitely-dimensional dynamical systems, yet abehaviour similar to hyperbolicity in finite-dimensional systems has beenobserved in many of them. In particular, in some binary cellular automatain one dimension, known as asymptotic emulators of identity , if the initialconfiguration is drawn from a Bernoulli distribution, the expected propor-tion of ones (or zeros) usually tends to its stationary value exponentiallyfast [14]. This type of behaviour is quite common in many other dynamicalsystems. For example, in a linear continuous-time dynamical system givenby ˙ x = A x , if x : R → R n and A is a real n × n matrix with all eigenvaluesdistinct and having negative real parts, x ( t ) tends to zero exponentially fastas t → ∞ . Exponential convergence is also observed in nonlinear systems˙ x = f ( x ) (where f : R n → R n ) in the vicinity of a hyperbolic fixed point,as long as the Jacobian matrix of f evaluated at the fixed point has onlydistinct eigenvalues with negative real parts.If, on the other hand, the matrix A in ˙ x = A x has degenerate (re-peated) eigenvalues, the convergence to the fixed point can be polynomial-exponential, that is, of the form P ( t ) e − bt , where P ( t ) is a polynomial and b >
0. Finite dimensional discrete-time dynamical systems can exhibit anal-ogous behaviour. Consider, for example, the linear system (cid:20) x n +1 y n +1 (cid:21) = (cid:20) − (cid:21) (cid:20) x n y n (cid:21) . (1)The matrix on the right hand side has a degenerate (double) eigenvalue ,and therefore the convergence to the fixed point (0 ,
0) is expected to bepolynomial-exponential (linear-exponential in this case). Indeed, if we ex-plicitly solve the above equation for x n and y n , we obtain (cid:20) x n y n (cid:21) = (cid:18) (cid:19) n (cid:20) − n n − n n (cid:21) (cid:20) x y (cid:21) , (2)and we can clearly see the aforementioned linear-exponential convergence.Very recently, a probabilistic CA has been discovered [15] where the den-sity of ones converges to its stationary value in a linear-exponential fashion,just like in the above example of a degenerate hyperbolic fixed point ina finite-dimensional dynamical systems. This probabilistic CA could beviewed as a simple model for diffusion of innovations, spread of rumors,or a similar process involving transport of information between neighbours.More precisely, it consists of an infinite one-dimensional lattice where eachsite is occupied by an individual who has already adopted the innovation(state 1) or who has not adopted it yet (state 0). Once the individual adopts the innovation, he remains in state 1 forever. Individuals in state 0can change their states to 1 (adopt the innovation) with probabilities de-pending on the state of nearest neighbours. This process can be formallydescribed as a radius 1 binary probabilistic CA with the following transitionprobabilities, w (1 | , w (1 | p, w (1 | , w (1 | , (3) w (1 | q, w (1 | r, w (1 | , w (1 | , where p, q, r are fixed parameters of the model, p, q, r ∈ [0 , w ( d | abc ) one means the probability that site in state b withneighbours a and c changes its state to d in one time step. One can showthat for a certain choice of parameters p, q and r , the expected value of thedensity of ones converges to its steady state in a linear-exponential fashion.Even more recently, we found a deterministic rule with three stateswhich exhibits the same kind of behaviour [1]. This rule, to be defined insec. 2, will be the subject of our subsequent discussion. In [1] we studiedthe structure of preimages of 0, 1 and 2 under the action of this rule, and byemploying finite state machines, we found some patterns in the preimagessets, which in turn allowed us to derive explicit expressions for densities of 0,1 and 2 after n iterations. The finite state machines used in the derivationwere constructed semi-empirically, and no proof of their correctness wasgiven. In the current paper we wish to fill this gap and present a moreformal derivation of the aforementioned expressions, without resorting tofinite state machines.
1. Basic definition
We will start from some basic definitions. For A = { , , } , a finitesequence of elements of A , b = b b . . . , b n , will be called a block (or word )of length n . The set of all blocks of all possible lengths will be denoted by A (cid:63) .Let f : A → A be a local function of a nearest-neighbour cellularautomaton. A block evolution operator corresponding to f is a mapping f : A (cid:63) (cid:55)→ A (cid:63) defined as follows. Let a = a a . . . a n ∈ A n where n ≥ f ( a ) is a block of length n − f ( a ) = f ( a , a , a ) f ( a , a , a ) . . . f ( a n − , a n − , a n ) . (4)If f ( b ) = a , than we will say that b is a preimage of a , and write b ∈ f − ( a ).Similarly, if f n ( b ) = a , than we will say that b is an n -step preimage of a ,and write b ∈ f − n ( a ). Let the density polynomial associated with a string b = b b . . . b n bedefined as Ψ b ( p, q, r ) = p ( b ) q ( b ) r ( b ) , (5)where i ( b ) is the number of occurrences of symbol i in b . If A is a set ofstrings, we define the density polynomial associated with A asΨ A ( p, q, r ) = (cid:88) a ∈ A Ψ a ( p, q, r ) . (6)One can easily show (in a manner similar as done in [14]) that if onestarts with a bi-infinite string of symbols drawn from a Bernoulli distribu-tion where probabilities of 0 , p, q and r , then theexpected proportion of sites in state k after n iterations of rule f is givenby Ψ f − n ( k ) ( p, q, r ). This quantity will be called density of symbols k after n iterations of f .
2. The local rule and its properties
Let us now describe the CA rule which will be the subject of this con-tribution. While studying properties of various 3-state CA rules, we cameacross an interesting specimen of a nearest-neighbour (radius 1) rule with alocal function defined as follows, f ( x , x , x ) = (cid:40) x for x = x > x ,x otherwise , (7)where x , x , x ∈ { , , } . The origins of this rule have been describedin [1]. Here we will only note that it can be equivalently defined as f ( x , x , x ) = x , x , x ) = (1 , ,
0) or ( x , x , x ) = (2 , , , x , x , x ) = (2 , , ,x otherwise , (8)which makes it clear that it differs from the identity rule only on threeneighbourhood configurations, (1 , , , ,
0) and (2 , , f ( x , x , x ) ≤ x , meaning that the state of a given cell cannot increase.This implies that f ( (cid:63), , (cid:63) ) = 0, where (cid:63) denotes an arbitrary symbol fromthe set { , , } . Zero is thus a quiescent state for this CA. it Fig. 1. Sample spatio-temporal pattern generated by 3-state rule 140. White,lighter gray and darker gray cells (blue in color version) correspond, respectively,to 0, 1 and 2.
We also have f (0 , , (cid:63) ) = 1 and f (2 , , (cid:63) ) = 1, which implies that a site instate 1 located at the beginning of a continuous cluster of 1’s of any length(even of length 1, meaning isolated 1) remain in state 1 forever. The same istrue for 2: since f (0 , , (cid:63) ) = 2 and f (1 , , (cid:63) ) = 2, a site in state 2 located atthe beginning of a continuous cluster of 2’s stays in state 2 forever. This canbe observed in Figure 1. In fact, words such as 01 or 02 remain unchangedwhen the rule is iterated, and no information can propagate through a pairsites which are in states 01 or 02. We note in passing that in the CA theorysuch words are called blocking words , and the rules with blocking words areknown to be almost equicontinuous [16].Further inspection of Figure 1 reveals that a continuous cluster of zerosgrows to the left if it is preceded by a cluster of 2’s longer than 1, and italso grows to the left if is preceded by a cluster of 1’s longer than 1.
3. Structure of preimages of 1
We want to find all strings b of length 2 n + 1 such that f n ( b ) = 1. Weknow from the definition of the rule that information can propagate onlyfrom the right to the left, thus the first n − b are arbitrary. Wewill represent n -step preimages of 1 in the form b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a a c c . . . c n , where the allowed values of a , a , a (to be called a prefix ) and c c . . . c n (tobe called a postfix ) need to be determined. The central site of the preimagestring will be, as in the above, denoted by a bold symbol. Since our CA rulehas three states, there are 3 = 27 possible values for the prefix a a a . Notall of them are possible, however. Prefixes 00 , 01 , 02 , 10 , 11 , 12 , , 21 and 22 , which can be represented as (cid:63) (cid:63) , cannot occur in anypreimage of 1. This is because f ( (cid:63), , (cid:63) ) = 0, meaning that if the central siteis in state 0, it will remain in state 0 forever, and consequently f n ( b ) = 0for any string b containing one of the above prefixes.Moreover, prefixes of the type (cid:63) or (cid:63) are not allowed either. Thisis because for the central site to become 1, as required, the transition f (2 , ,
1) = 1 would have to happen somewhere along the way, and forthis the central 2 would have to get 2 as the left neighbour. Since statescan only decrease, not increase, and the left neighbour of the central site is0 or 1, this is not possible.By excluding 9 prefixes of the type (cid:63) (cid:63) and 6 prefixes of the type (cid:63) or (cid:63) we are left with 12 possibilities, 00 , 01 , 02 , 10 , 11 , 12 , 20 , 21 ,22 , 02 , 12 , 22 . All of them are allowed in preimages of 1, providingthat an appropriate suffix is added. In what follows we will find conditionswhich these suffices need to satisfy. We will divide the possible prefixes intofour different types (the reason for this will soon become clear):1. (cid:63) , 22 (cid:63) , 02 , 12
3. 02 , 12
4. 22 For prefixes of type 1, (cid:63) and 22 , the central site is in the state 1 already,thus we have to make sure that it stays in the same state after n iterationsof the rule. The left neighbour of the central 1 is 1 (for (cid:63) ) or it willbecome 1 after one iteration (for 22 ), thus we could potentially be in adanger of the transition f (1 , ,
0) = 0. This could happen only if the central1 belongs to a continuous cluster of ones which is followed by 0 – in sucha case, the cluster of ones will shrink one symbol per time step and thetransition f (1 , ,
0) = 0 will eventually change the central 1 into 0. Wehave, therefore, two choices in avoiding this scenario: either the central 1belongs to a cluster of ones followed by 2, which prevents its shrinking dueto the fact that f (1 , ,
2) = 1, or it belongs to a cluster of ones which extendsall the way to the right. In other words, the postfix for type 2 must be ofthe form c c . . . c n = 1 i (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − − i or c c . . . c n = 1 n , where i ∈ { , , . . . n − } . In the above and in what follows, 1 n denotes thesymbol 1 repeated n times. We will use this convention in the rest of thepaper. For type 2, (cid:63) , 02 , and 12 , note that the central site could become 0only by utilizing the transition f (1 , ,
0) = 0, and for this the left neighbourof the central 1 would have to be 1. This is clearly impossible for (cid:63) as theleft neighbour 0 will always remain 0. In the case of 02 and 12 , the leftneighbour is 2, and it could potentially become 0 by transition f (2 , ,
0) = 0.For this, however, we would need the second left neighbour of the central1 to be in state 2, but this is impossible because site values never increase.Thus all prefixes of type 2 belong to preimages of 1, regardless of the suffix.For type 3, 02 , 12 , the central site is in state 2, and its left neighbouris guaranteed to be in state 2 forever. Consequently, the central 2 mustchange to 1 at some iteration via the f (2 , ,
1) = 1 transition, and in orderfor this to happen, we need to have the right neighbour of the central site instate 1 at some point of time. This can happen when the central 2 belongsto a continuous cluster of 2’s followed by 1, meaning that the postfix mustbe of the form c c . . . c n = 2 i (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − − i , where i ∈ { , , . . . n − } .Type 4 with the prefix 22 is the most complicated one. Similarly asbefore, the central site must change from 2 to 1, and this can only happenvia the transition f (2 , ,
1) = 1, but we do not have a guarantee that theleft neighbour of the central 2 remains in state 2 forever, as it was the thecase of type 3. Nevertheless, the necessary condition for the suffix is thesame as for type 3, meaning that the central 2 must belong to a continuouscluster of 2’s followed by 1. This in not sufficient, however, because whatfollows is also important. In order to understand this clearly, consider twostrings of length 13 iterated 6 times:
The first of these strings has prefix 222, and the second one has prefix 122(thus belonging to type 3). In both of them the central 2 belongs to acontinuous cluster of 2’s followed by 1, but the first one does not produce 1after 6 iterations. This is because the zero which follows propagates to theleft and eventually makes the central site to change to 0 via the transition f (1 , ,
0) = 0. Any string containing 222 as a prefix must, therefore, satisfy an additional property preventing zeros to propagate to the left. This canbe done by a making the last part of the suffix to have the same structureas prefix of type 3, so that the entire suffix takes the form c c . . . c n = 2 i d d . . . d n − − i , (9)where i ∈ { , , . . . n − } and where d d . . . d n − − i = 1 j (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − − i − j or d d . . . d n − − i = 1 n − − i , (10)with j ∈ { , , . . . , n − − i } .Finally, for prefix 222 there is one more possibility not covered by theabove discussion, namely c c . . . c n = 2 n −
10. Below we summarize allthese findings in a form of a single proposition.
Proposition 3.1
Block b belongs to f − n (1) if and only if it is one of thefollowing four types.Type 1: b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a a i (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − − i or b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a a n , (11) where a a a ∈ { , , , } , i ∈ { , , . . . n − } ;Type 2: b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a a (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n , (12) where where a a a ∈ { , , , , } ;Type 3: b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a a i (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − − i , (13) where where a a a ∈ { , } , i ∈ { , , . . . n − } Type 4a: b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a a i c c . . . c n − − i , (14) where a a a = 222 , i ∈ { , , . . . n − } and where c c . . . c n − − i = 1 j (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − − i − j or c c . . . c n − − i = 1 n − − i (15) with j ∈ { , , . . . , n − − i } . Type 4b: b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a a n − , (16) where a a a = 222 .
4. Density polynomials for preimages of 1
Let us now denote the set of strings of type 1 by T , type 2 by T etc.,and let us define λ = p + q + r . Density polynomial for T will be given byΨ T ( p, q, r ) = n − (cid:88) i =0 λ n − ( pq + q + rq + r q ) q i rλ n − i − + λ n − ( pq + q + rq + r q ) q n , (17)which simplifies toΨ T ( p, q, r ) = λ n − r ( λq + r q ) n − (cid:88) i =0 λ − i q i + λ n − ( λq + r q ) q n . (18)By performing summation of the partial geometric sequence in the above,one obtainsΨ T ( p, q, r ) = λ n − r ( λq + r q ) λ n − q n p + r + λ n − ( λq + r q ) q n , (19)which further simplifies toΨ T ( p, q, r ) = r ( λq + r q ) λ n − p + r + λ n − ( λq + r q ) pq n p + r . (20)Similar calculations (omitted here) yieldΨ T ( p, q, r ) = ( λpq + q r + pqr ) λ n − , (21)Ψ T ( p, q, r ) = λ n − r q ( λ n − r n ) . (22)The type 4a is the most complicated. Let us first compute the densitypolynomial for the set of strings of the form c c . . . c k = 1 j (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) k − − j or c c . . . c k = 1 k (23) with j ∈ { , , . . . , k − } . The density polynomial for the above, to bedenoted by h k ( p, q, r ), is given by h k ( p, q, r ) = k − (cid:88) j =0 q j rλ k − − j + q n = r λ k − q k p + r + q k = rλ k p + r + pq k p + r . (24)Having this result, we can write the density polynomial for the entire set T a ,Ψ T a ( p, q, r ) = λ n − r n − (cid:88) i =0 r i qh n − − i ( p, q, r )= λ n − r n − (cid:88) i =0 r i q rλ n − − i p + r + λ n − r n − (cid:88) i =0 r i q pq n − − i p + r = λ n − qr p + r n − (cid:88) i =0 r i λ n − − i + λ n − pqr p + r n − (cid:88) i =0 r i q n − − i = λ n − qr ( λ n − r n )( p + r )( p + q ) + λ n − pqr p + r n − (cid:88) i =0 r i q n − − i . (25)When q (cid:54) = r , we thus obtainΨ T a ( p, q, r ) = λ n − qr ( λ n − r n )( p + r )( p + q ) + λ n − pqr ( q n − r n )( p + r )( q − r ) . (26)When q = r , the last sum becomes (cid:80) n − i =0 r i q n − − i = (cid:80) n − i =0 q n − = q n − n ,therefore Ψ T a ( p, q, q ) = λ n − q ( λ n − q n )( p + q ) + λ n − pq p + q nq n . (27)Finally, type 4b is straightforward,Ψ T b ( p, q, r ) = λ n − r n +1 pq. (28)We are now ready to compute the density polynomial of preimages of 1,by summing density polynomials for T , T , T , T a and T b . This yields, for q (cid:54) = r ,Ψ f − n (1) ( p, q, r ) = r ( λq + r q ) λ n − p + r + λ n − ( λq + r q ) pq n p + r + ( λpq + q r + pqr ) λ n − + λ n − r q ( λ n − r n )+ λ n − qr ( λ n − r n )( p + r )( p + q ) + λ n − pqr ( q n − r n )( p + r )( q − r ) + λ n − r n +1 pq. (29) Collecting together terms for ( qλ ) n , ( rλ ) n , and λ n we obtain, after somealgebra,Ψ f − n (1) ( p, q, r ) = pq (cid:0) − pr + pq + q (cid:1) ( qλ ) n λ ( p + r ) ( q − r )+ qr (cid:0) − p r + p q + pq − pqr + r − q r (cid:1) ( rλ ) n λ ( p + q ) ( q − r )+ q (cid:0) p + p q + 2 p r + pr + 3 pqr + r + r q + q r (cid:1) λ n λ ( p + r ) ( p + q ) , (30)which is the same formula as derived in [1].Similarly, for q = r , we obtainΨ f − n (1) ( p, q, q ) = q ( λq + q ) λ n − p + q + λ n − ( λq + q ) pq n p + q + ( λpq + q + pq ) λ n − + λ n − q ( λ n − q n ) λ n − q ( λ n − q n )( p + q ) + λ n − pq p + q nq n + λ n − q n +1 pq. (31)After simplification and reordering of terms this yieldsΨ f − n (1) ( p, q, q ) = pq ( n + 1) ( qλ ) n λ ( q + p ) + q (cid:0) p + 4 p q + pq − q (cid:1) ( qλ ) n ( q + p ) λ + (cid:0) p + 3 p q + 4 pq + 3 q (cid:1) qλ n λ ( q + p ) , (32)which, again, agrees with the result “guessed” in [1] using finite state ma-chines.
5. Preimages of 2 and their density polynomials
From the definition of the rule we know that a site can be in state 2 onlyif it was in that state at the beginning, that is, sites in state 0 or 1 cannotchange to 2. Moreover, f (2 , ,
0) = 0, f (2 , ,
1) = 1, and in all other cases f ( a , , a ) = 2. This means that a site in state 2 remains in that stateforever if it is preceded by 0 or 1. Therefore, any string of the form b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n or b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n will be an n -step preimage of 2, f n ( b ) = 2. What if 2 is preceded by 2? In this case it must be followed by a sufficientnumber of 2’s before the first 0 or 1 appears, as any 0 or 1 at the end of acluster of 2’s shortens such cluster by one on each iteration. Therefore, for n iterations we need n f n ( b ) = 2 to hold inthis case, b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − n . The above observations can be summarized as follows.
Proposition 5.1
Block b belongs to f − n (2) if and only if it is one of thefollowing three types:1. b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n ,2. b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n ,3. b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − n +2 . This yields the density polynomialΨ f − n (2) ( p, q, r ) = ( p + q ) rλ n − + λ n − r n +2 . (33)
6. Preimages of 0 and their density polynomials
Since everything what is not a preimage of 1 or 2 must be a preimageof 0, we haveΨ f − n (0) ( p, q, r ) = λ n +1 − Ψ f − n (1) ( p, q, r ) − Ψ f − n (2) ( p, q, r ) . (34)After simplification, this yields, for r (cid:54) = q ,Ψ f − n (0) ( p, q, r ) = (cid:0) − pr + pq + q (cid:1) pq ( qλ ) n λ ( p + r ) ( r − q )+ pr (cid:0) − r p + q p + q − r − qr (cid:1) ( rλ ) n λ ( p + q ) ( r − q )+ (cid:0) p + 2 p q + 2 p r + 2 r p + 3 qpr + 2 q p + r + 2 qr + q + q r (cid:1) pλ n ( p + q ) ( p + r ) λ , (35) and for r = q ,Ψ f − n (0) ( p, q, q ) = (cid:0) p + 4 p q + 7 q p + 5 q (cid:1) pλ n λ ( p + q ) − pq ( n + 1) ( qλ ) n λ ( p + q ) − q p (cid:0) p + 8 pq + 6 q (cid:1) ( qλ ) n ( p + q ) λ . (36)Again, similarly as in the density polynomial for 1, the linear-exponentialdependence of the form ( n + 1)( qλ ) n is present in the second term.
7. Density of ones
As already stated, density polynomials Ψ f − n ( k ) ( p, q, r ) represent proba-bility of occurence of k after n iterations starting from a Bernoulli distri-bution with probabilities of 0, 1 and 2 equal to, respectively, p , q , and r ,where p + q + r = 1. If one starts with a symmetric Bernoulli distributionwhere r = q , the probability of occurence of 1 after n steps, to be denotedby P n (1), will be given by eq. (32) as long as one substitutes r = q and q = (1 − p ) /
2. This yields, after simplification, P n (1) = P ∞ (1) − ( An + B ) (cid:18) − p (cid:19) n , (37)where A = ( p − p ) (cid:0) p − p (cid:1) , (38) B = ( p − p ) (cid:0) − p − p − p + 1 (cid:1) , (39) P ∞ (1) = (1 − p ) (cid:0) p + 5 p − p + 3 (cid:1) p ) . (40)One can see that for 0 < p < P n (1) tends to P ∞ (1) as n → ∞ , and thatthe convergence is linear-exponential in n . Such “degenerate” convergencetakes place for probability of occurence of 0 as well, as seen in eq. (36).On the other hand, when r (cid:54) = q in the initial Bernoulli distribution, theconvergence is purely exponential, as in eq. (30) and (35).
8. Conclusions
We presented an example of a 3-state rule exhibiting, under certainconditions, linear-exponential convergence to the steady state. This phe-nomenon is remarkably similar to degenerate hyperbolicity in finite dimen-sional dynamical systems. It is not clear, however, what is the origin of this analogy. One can speculate that “simple” CA rules, such as those which areequicontinuous or almost equicontinuous, can be somewhat approximatedby finite-dimensional systems. Local structure theory could possibly be ap-plicable in this case, as it allows to construct finite-dimensional systemsapproximating orbits of Bernoulli measure under the action of a given CA.It is hoped that this contribution inspires further research on this subject. Acknowledgments [1] H. Fuk´s and J. Midgley-Volpato, “An example of degenerate hyperbolicity incellular automaton with 3 states,” in , J. Kari, I. T¨orm¨a, andM. Szabados, eds., vol. 24 of
TUCS Lecture Notes , pp. 47–55. Turku,Finland, 2015. arXiv:1506.06649 .[2] D. A. Lind, “Applications of ergodic theory and sofic systems to cellularautomata,”