Explicit solution of the Cauchy problem for cellular automaton rule 172
PPublished in
J. of Cellular Automata , vol. 12, no. 6, pp. 423–44 (2017)
Explicit solution of the Cauchy problem for cellular automatonrule 172
Henryk Fuk´s
Department of MathematicsBrock University
Email: [email protected]
Abstract
Cellular automata (CA) are fully discrete alternatives to partial differential equations (PDE).For PDEs, one often considers the Cauchy problem, or initial value problem: find the solutionof the PDE satisfying a given initial condition. For many PDEs of the first order in time, it ispossible to find explicit formulae for the solution at the time t > t = 0. Can something similar be achieved for CA? We demonstrate that this is indeed possiblein some cases, using elementary CA rule 172 as an example. We derive an explicit expressionfor the state of a given cell after n iteration of the rule 172, assuming that states of all cellsare known at n = 0. We then show that this expression (“solution of the CA”) can be used toobtain an expected value of a given cell after n iterations, provided that the initial condition isdrawn from a Bernoulli distribution. This can be done for both finite and infinite lattices, thusproviding an interesting test case for investigating finite size effects in CA.
1. Introduction
Cellular automata are often described as fully discrete alternatives to partial differential equations(PDEs). In one dimension, a PDE which is first-order in time can be written as u t ( x, t ) = F ( u, u x , u xx , . . . ) , (1)where u ( x, t ) is the unknown function, and the independent variables t and x are commonly inter-preted as, respectively, time and position in space. Both variables t and x , as well as u ( x, t ), takevalues in the set of real numbers.Cellular automata (CA), on the other hand, are typically written as u ( i, n + 1) = f ( u ( i − r, n ) , u ( i − r + 1 , n ) , . . . , u ( i + r, n )) , (2)where f is called a local function and the integer r is called a radius of the cellular automaton.For CA independent variables n (representing time) and i (representing space) are integers, while u ( i, n ) takes values in a finite set of symbols, usually integers. In the case of binary cellularautomata, which are the main focus of this paper, u ( i, n ) takes values in the set { , } , so that f : { , } r +1 → { , } . 1 a r X i v : . [ n li n . C G ] F e b omparing eqs. (1) and (2) we conclude that discrete time n in CA plays the role of t in PDEs, i in CA plays the role of x in PDEs, and r in CA plays a similar role as the degree of the highestderivative in PDEs. In fact, there are some further analogies, but we will not discuss them here. Wewill only mention that there exist discretization schemes (such as ultradiscretization [1]) which allowto construct CA from PDE while preserving some features of the dynamics, but they are beyondthe scope of this paper. We merely want to indicate here that conceptually, cellular automata areclosely related to PDEs, although in contrast to PDEs, all variables in CA are discrete. Moreover,dependent variable u is bounded in the case of CA – a restriction which is not normally imposedon the dependent variable of a PDEs.For PDEs, the initial value problem (also called the Cauchy problem) is often considered. It isthe problem of finding u ( x, t ) for t > u t ( x, t ) = F ( u, u x , u xx , . . . ) , for x ∈ R , t > ,u ( x,
0) = G ( x ) for x ∈ R , (3)where the function G : R → R represents given initial data. A similar problem can be formulatedfor cellular automata: given u ( x, t + 1) = f ( u ( x − r, t ) , u ( x − r + 1 , t ) , . . . , u ( x + r, t )) ,u ( x,
0) = g ( x ) , (4)find u ( x, t ) for t >
0, where the initial data is represented by the function g : Z → { , } .For the problem (4), it is easy to find the value of u ( x, t ) for any x ∈ Z and any t ∈ N by directiteration of the cellular automaton equation (2). Thus, in the algorithmic sense, problem (4) isalways solvable – all one needs to do is to take the initial data g ( x ) and perform n iterations.In contrast to this, the initial value problem for PDE cannot be solved exactly by direct iteration.In some cases, however, one can obtain exact solution in the sense of a formula for u ( x, t ) involving G ( x ). To give a concrete example, consider the classical Burgers equation, u t = u xx + uu x . (5)If u ( x,
0) = G ( x ), one can show that for t > u ( x, t ) = 2 ∂∂x ln (cid:26) √ πt (cid:90) ∞−∞ exp (cid:20) − ( x − ξ ) t − (cid:90) ξ G ( ξ (cid:48) ) dξ (cid:48) (cid:21) dξ (cid:27) . (6)Can we obtain similar formulae for cellular automata? The answer is affirmative in some cases .These cases usually involve “simple” CA rules. The goal of this paper is to demonstrate how toobtain solution of a CA in one of such “simple” cases, using elementary CA rule 172 as an example.We will also show some applications of the solution. Some ideas presented here appeared in apreliminary form in an earlier conference proceedings paper [2].
2. Basic definitions
Let A = { , } be called a symbol set , and let S = { , } Z be the set of all bisequences over A , tobe called a configuration space . A block or word of length n is an ordered set b b . . . b n − , where n ∈ N , b i ∈ A . Let n ∈ N andlet B n denote the set of all blocks of length n over A and B be the set of all finite blocks over A .2or r ∈ N , a mapping f : { , } r +1 (cid:55)→ { , } will be called a cellular automaton rule of radius r .Alternatively, the function f can be considered as a mapping of B r +1 into B = A = { , } .Corresponding to f (also called a local mapping ) we define a global mapping F : S → S suchthat ( F ( s )) i = f ( s i − r , . . . , s i , . . . , s i + r ) for any s ∈ S .A block evolution operator corresponding to f is a mapping f : B (cid:55)→ B defined as follows. Let r ∈ N be the radius of f , and let a = a a . . . a n − ∈ B n where n ≥ r + 1 >
0. Then f ( a ) = { f ( a i , a i +1 , . . . , a i +2 r ) } n − r − i =0 . (7)Note that if b ∈ B r +1 then f ( b ) = f ( b ). The set of n -step preimages of the block b under the rule f is defined as the set f − n ( b ) = { c ∈ B : f n ( c ) = b } . Note that the notion of block preimages hasbeen, in somewhat different context, studied in many earlier works, including [3, 4, 5, 6].Binary rules of radius 1 are called elementary rules, and they are usually identified by theirWolfram number [7] W ( f ), defined as W ( f ) = (cid:88) x ,x ,x =0 f ( x , x , x )2 (2 x +2 x +2 x ) . (8)We will consider, as an example of a “solvable” CA rule, one of the elementary rules, namelythe rule with Wolfram number 172. Its local function f : { , } → { , } is defined as f ( x , x , x ) = (cid:26) x if x = 0, x if x = 1. (9)It is easy to verify that for the above f we have W ( f ) = 172.The reason for which rule 172 was selected is that its dynamics is simple enough to render it“solvable”, yet it is not entirely trivial. Further explanation regarding the meaning of “non-trivial”will be given in the conclusion section. We shall also add that many properties of rule have beenstudied in the past, usually in the context of other elementary CA. Some of these include placeof rule 172 in various CA classifications, structure of its Garden of Eden configurations, algebraicproperties, and various properties of its global function [8, 9, 10, 11].In what follows, whenever we use the symbol f it will signify the local function defined in eq.(9), while f and F will denote, respectively, the corresponding block evolution operator and theglobal function. To familiarize the reader with the concept of the block evolution operator, let ustake as an example b = 1001010. We can compute f ( b ) by applying f to all consecutive triples ofsymbols, that is, f ( b ) = f (100) f (001) f (010) f (101) f (010) = 00111. If we apply f again to 00111,we will obtain f ( b ) = 011, and yet another application of f yields f ( b ) = 1. It is sometimesconvenient to write consecutive images of b under each other, as follows: The above shows, starting from the top, b , f ( b ), f ( b ), and f ( b ).We shall also note that there is usually more than one block c such that f ( b ) = c . For ex-ample, for rule 172, f (0010) = f (0011) = f (1101) = 01. We can, therefore, write f − (01) = { , , } . Similarly, we can write f − (101) = { , , , , , } , (10)3
101 201 301 401 50 61 70 811 90 101 110 1211 1301 1411 150 1601 170 1811 1901 2011 210 2201 230 2411 25011 2601 27011
Figure 1: Finite state machine producing f − (1) for elementary CA rule 172.because all 6 block on the right hand side of the above (and only these blocks) have the propertythat after applying f to them twice, one obtains 101.Our strategy for constructing the solution of rule 172 will be as follows. First, we will construct n -step preimages of 1 (i.e., sets f − n (1)) for various n . We will then try to find patterns in these setwhich would allow us to give a combinatorial description of them, as set of binary strings satisfyingcertain conditions. Once this is done, we will construct a Boolean function which is an indicatorfunction of f − n (1). Such function will then be used to construct an explicit expression for [ F n ( x )] i for any x ∈ { , } Z , i ∈ Z and n ∈ N .
3. Structure of preimage sets
Suppose now that we have a string b of length 2 n + 1 and we want to find out the necessary andsufficient conditions for f n ( b ) = 1. We will try to “guess” these conditions first, formulate them ina rigorous way, and then prove them.In order to “guess” the conditions, one can generate sets f − n (1) for various values of n andtry to discover obvious patters in them. From author’s experience, a good way to do this is tobuild minimal finite state machines generating words of f − n (1). This can be done using AT&TFSM Library [12, 13], and Figure 1 shows an example of a minimal finite state machine (FSM)generating f − (1) for rule 172. In order to generate a preimage of 1 using this picture, start onthe left (at circled zero) and follow the arrows writing down all encountered edge labels until youreach the final state (doubly circled 27). The string of 15 labels obtained this way will be a possiblepreimage of 1, one of many. Obviously there are as many preimage string as paths joining theinitial state and the final state. Note that circled numbers denote internal states of the FSM, andare irrelevant for our purposes.From Figure 1, it is clear that the first 5 symbols of f − (1) are arbitrary, and then we have twopossibilities:(i) 001 followed by 6 arbitrary symbols, or(i) string of 8 symbols without 00 pair anywhere, followed by 10 or 11 (if it ends with 0) or by01 or 11 (if it ends by 1).This observation can be generalized and summarized as the following proposition. Proposition 3.1
Block b of length n + 1 belongs to f − n (1) if and only if it has the structure 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 , (11) or b = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a . . . a n +1 c c , (12)4 n Figure 2: Example of a spatiotemporal pattern produced by rule 172. where a a . . . a n is a binary string which does not contain any pair of adjacent zeros, and c c = (cid:40) (cid:63), if a n +1 = 0 ,(cid:63) , otherwise . (13)We will sketch the proof of the above, leaving out some tedious details. It will be helpful to inspectspatiotemporal pattern generated by rule 172 first, as shown in Figure 2. Careful inspection of thispattern reveals three facts, each of them easily provable in a rigorous way:(F1) A cluster of two or more zeros keeps its right boundary in the same place for ever.(F2) A cluster of two or more zeros extends its left boundary to the left one unit per time step aslong as the left boundary is preceded by two or more ones. If the left boundary of the clusterof zeros is 01, the cluster does not grow.(F3) Isolated zero moves to the left one step at a time as long as it has at least two ones on theleft. If an isolated zero is preceded by 01, it disappears in the next time step.Suppose now that we have a string b of length 2 n + 1 and we want to find out the necessaryand sufficient conditions for f n ( b ) = 1. From (F1) it is clear that the word 001 will remain in thesame position forever, which means that if 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 , (14)then f n ( b ) = 1. What are the other possibilities for b which would result in f n ( b ) = 1?From (F2) we deduce that if there is no cluster of two or more more zeros somewhere in the last n + 3 bits of b , then there is no possibility of the growth of cluster of zeros producing f n ( b ) = 0.The only way to get zero after n iterations of f in such a case would be having zero at the end of b preceded by 11. This means that in order to avoid this scenario, the last 3 bits of b must be 010,011, 101 or 111, or, in other words, the last three bits must be 01 (cid:63) or 1 (cid:63)
1, as in eq. (13). (cid:50)
4. Solving rule 172
We are now almost ready to construct the solution of rule 172. Suppose that x ∈ { , } Z is aninitial configuration, and that we iterate rule 172 n times. What is the value of the central site5fter n iterations of the rule, that is, the value of [ F n ( x )] ? Obviously it can be either 0 or 1, so letus suppose that it is 1, which means that x − n x − n +1 , . . . x n ∈ f − n (1). By the virtue of Proposition3.1, x − n x − n +1 , . . . x n must take one of the two forms, the fist of them being x − n x − n +1 , . . . x n = (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 . (15)The above means that x − x − x = 001, and it will be true if and only if(1 − x − )(1 − x − ) x = 1 . (16)The second possibility, according to Proposition 3.1, is x − n x − n +1 , . . . x n = (cid:63) (cid:63) . . . (cid:63) (cid:124) (cid:123)(cid:122) (cid:125) n − a a . . . a n +1 c c , (17)where a i a i +1 (cid:54) = 00 for i = 1 , , . . . , n and c , c satisfy condition of eq. (13). This means that a a . . . a n +1 = c − c − . . . c t − , and therefore x i x i +1 (cid:54) = 00 for i = − , − , . . . , t −
3, as well as x t − x t = (cid:40) (cid:63), if x t − = 0 ,(cid:63) , otherwise . (18)The second possibility will be realized if and only if (cid:32) t − (cid:89) i = − (1 − ¯ x i ¯ x i +1 ) (cid:33) (¯ x t − x t − + x t − x t ) = 1 , (19)where we used notation ¯ x i = 1 − x . Combining eqs. (16) and (19) we obtain[ F n ( x )] = ¯ x − ¯ x − x + (cid:32) n − (cid:89) i = − (1 − ¯ x i ¯ x i +1 ) (cid:33) (¯ x n − x n − + x n − x n ) . (20)This is the desired solution expressing the value of the central site after n iterations of rule F starting from an initial configuration x . Of course, we can now obtain analogous expression for anyother site [ F n ( x )] j , by simply translating the above formula from j = 0 to an arbitrary position j . Proposition 4.1
Let F be the global function of elementary CA rule 172 and x ∈ { , } Z . Then,after n ∈ N iterations of F , for any j ∈ Z , [ F n ( x )] j = ¯ x j − ¯ x j − x j + j + n − (cid:89) i = j − (1 − ¯ x i ¯ x i +1 ) (¯ x j + n − x j + n − + x j + n − x j + n ) . (21)The above could be called a solution of rule 172. It is an explicit solution of the Cauchy problem forthis rule, expressing the state of a site at position j after n iterations in terms of initial site values.The formula for the solution is very simple, utilizing only addition, subtraction, and multiplicationof site values. As we will see in subsequent sections, it can be very useful in practice, for examplefor constructing probabilistic solutions of the CA rule and investigating finite size effects.6 . Probabilistic solution for infinite configurations Let us now assume that the initial configuration x is drawn from a Bernoulli distribution. Moreprecisely, let x j = X j for j ∈ Z , where X j are independent and identically distributed randomvariables such that P r ( X j = 1) = q , P r ( X j = 0) = 1 − q , where q ∈ [0 , F n ( x )] j in such circumstances? Denoting the expected value by (cid:104)·(cid:105) , let us first note thatdue to translational invariance, (cid:104) [ F n ( x )] j (cid:105) = (cid:104) [ F n ( x )] (cid:105) . In order to compute (cid:104) [ F n ( x )] (cid:105) , we needto calculate the expected value of the right hand side of eq. (20). The following lemma will beuseful for this purpose. Lemma 5.1
Let q ∈ (0 , and let X i be independent and identically distributed Bernoulli randomvariables for i ∈ { , , . . . , n } such that P r ( X i = 1) = q , P r ( X i = 0) = 1 − q . Then (cid:42) n − (cid:89) i =1 (cid:0) − ¯ X i ¯ X i +1 (cid:1)(cid:43) = qλ − λ (cid:0) α λ n − + α λ n − (cid:1) , (22) where λ , = 12 q ± (cid:112) q (4 − q ) , (23) α , = (cid:16) q − (cid:17) (cid:112) q (4 − q ) ± (cid:18) q − (cid:19) . (24)To prove it, let us first define U n = n − (cid:89) i =1 (cid:0) − ¯ X i ¯ X i +1 (cid:1) and V n = ¯ X n n − (cid:89) i =1 (cid:0) − ¯ X i ¯ X i +1 (cid:1) . (25)We observe that U n = U n − (1 − ¯ X n − ¯ X n ) = U n − − ¯ X n V n − , (26)and V n = ¯ X n U n = ¯ X n U n − (1 − ¯ X n − ¯ X n ) = ¯ X n U n − − ¯ X n V n − , (27)where we used the fact X n is a Boolean variable, thus ¯ X n = ¯ X n . This yields the system of recurrenceequations for U n and V n , U n = U n − − ¯ X n V n − , (28) V n = ¯ X n U n − − ¯ X n V n − . Since ¯ X n is independent of both U n − and V n − , we can write (cid:104) U n (cid:105) = (cid:104) U n − (cid:105) − (cid:104) ¯ X n (cid:105)(cid:104) V n − (cid:105) , (29) (cid:104) V n (cid:105) = (cid:104) ¯ X n (cid:105)(cid:104) U n − (cid:105) − (cid:104) ¯ X n (cid:105)(cid:104) V n − (cid:105) . (30)Now, taking into account that (cid:104) ¯ X n (cid:105) = 1 − q , we obtain (cid:20) (cid:104) U n (cid:105)(cid:104) V n (cid:105) (cid:21) = M (cid:20) (cid:104) U n − (cid:105)(cid:104) V n − (cid:105) (cid:21) , (31)where M = (cid:20) q − − q q − (cid:21) . (32)7his recurrence equation is easy to solve, (cid:20) (cid:104) U n (cid:105)(cid:104) V n (cid:105) (cid:21) = M n − (cid:20) (cid:104) U (cid:105)(cid:104) V (cid:105) (cid:21) . (33)Since (cid:104) U (cid:105) and (cid:104) V (cid:105) can be directly computed, (cid:104) U (cid:105) = (cid:104) − ¯ X ¯ X (cid:105) = 1 − (1 − q ) = 2 q − q , (34) (cid:104) V (cid:105) = (cid:104) ¯ X (1 − ¯ X ¯ X ) (cid:105) = (cid:104) ¯ X − ¯ X ¯ X (cid:105) = 1 − q − (1 − q ) = q − q , (35)we obtain (cid:20) (cid:104) U n (cid:105)(cid:104) V n (cid:105) (cid:21) = M n − (cid:20) q − q q − q (cid:21) . (36)The only thing left is to compute is M n − . This can be done by diagonalizing M , M n − = P (cid:34) λ n − λ n − (cid:35) P − , (37)where λ , are eigenvalues of M , as defined in eq. (23), and P is the matrix of eigenvectors of P , P = (cid:34) − q − λ − q − λ (cid:35) , P − = 1 λ − λ − ( λ − λ − − q λ − ( λ − λ − − q − λ + 1 . (38)The final formula for U n and V n is (cid:20) (cid:104) U n (cid:105)(cid:104) V n (cid:105) (cid:21) = P (cid:34) λ n − λ n − (cid:35) P − (cid:20) q − q q − q (cid:21) . (39)By carrying out the multiplications of matrices on the right hand side, after some algebra, weobtain (cid:104) U n (cid:105) = qλ − λ (cid:16) (cid:0) − − q + q + 2 λ − qλ (cid:1) λ n − + (cid:0) q − q − λ + qλ (cid:1) λ n − (cid:17) , (40)in agreement with eq. (22). (cid:50) Before we take the expected value of both sides of eq. (20), let us first rewrite the last factor,¯ x n − x n − + x n − x n = (1 − x n − ) x n − + x n − x n = x n − + x n − ( x n − x n − ) . (41)Using the above, we obtain from eq. (20), (cid:104) [ F n ( x )] (cid:105) = (cid:104) ¯ x − ¯ x − x (cid:105) + (cid:42)(cid:32) n − (cid:89) i = − (1 − ¯ x i ¯ x i +1 ) (cid:33) x n − (cid:43) + (cid:42)(cid:32) n − (cid:89) i = − (1 − ¯ x i ¯ x i +1 ) (cid:33) x n − ( x n − x n − ) (cid:43) . (42)8sing the fact that the expected value of the product of independent random variables is equal tothe product of their expected values, this yields (cid:104) [ F n ( x )] (cid:105) = (cid:104) ¯ x − (cid:105)(cid:104) ¯ x − (cid:105)(cid:104) x (cid:105) + (cid:42) n − (cid:89) i = − (1 − ¯ x i ¯ x i +1 ) (cid:43) (cid:104) x n − (cid:105) + (cid:42)(cid:32) n − (cid:89) i = − (1 − ¯ x i ¯ x i +1 ) (cid:33) x n − (cid:43) (cid:104) x n − x n − (cid:105) . (43)Because (cid:104) x n − x n − (cid:105) = 0, the last term vanishes, and, using the fact that (cid:104) x i (cid:105) = q and (cid:104) ¯ x i (cid:105) = 1 − q ,we obtain (cid:104) [ F n ( x )] (cid:105) = (1 − q ) q + q (cid:42) n − (cid:89) i = − (1 − ¯ x i ¯ x i +1 ) (cid:43) . (44)Using Lemma 5.1, and remembering that the expected value must be the same for any index j , thefinal result is thus (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + q λ − λ (cid:0) α λ n − + α λ n − (cid:1) . (45)The above could be called a probabilistic solution of CA rule 172 for infinite Bernoulli initialconfiguration. Note that since | λ , | <
1, we havelim n →∞ (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q. (46)When q = 1 /
2, eq. (45) becomes, after simplification, (cid:104) [ F n ( x )] j (cid:105) = 18 + (cid:32)
14 + √ (cid:33) (cid:32)
14 + √ (cid:33) n + (cid:32) − √ (cid:33) (cid:32) − √ (cid:33) n . (47)Since + √ is half of ratio divina (the golden ratio), one recognizes a link to Fibonacci numbersin the above. Indeed, it is easy to show that for q = 1 / (cid:104) [ F n ( x )] j (cid:105) = 18 + F n +3 n +2 , (48)where F n is the n -th Fibonacci number.
6. Probabilistic solution for periodic configuration
Suppose now that the initial condition is periodic with period k , that is, x i = x i + k for all i ∈ Z .Although one could of course consider all finite configurations of a given length and determine theirattractors, it will nevertheless be useful to construct a general formula valid for arbitrary k . Thiscould be, for example, useful if one wants to study the dependence of the speed of convergence tothe steady state as a function of k .We will take i = 0 , . . . k − k . Let us further assume that,as before, x j = X j for j ∈ { , , . . . , k − } , where X j are independent and identically distributedrandom variables such that P r ( X j = 1) = q , P r ( X j = 0) = 1 − q , where q ∈ [0 , n iterations of rule 172, we take expectedvalue of both sides of eq. (21) (remembering that indices are now modulo k ), obtaining (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + (cid:42) j + n − (cid:89) i = j − (1 − ¯ x i ¯ x i +1 ) (¯ x j + n − x j + n − + x j + n − x j + n ) (cid:43) . (49)Observe that when j = − j -independent, so the choiceof j does not matter), the only indices of x occurring on the right hand side of the above will bein the range from 0 to n + 2. This means that for for n ≤ k −
3, we will never actually need touse modulo k operation to bring the index to the principal period range. For this reason, eq. (45)remains valid in the periodic case as long as n ≤ k − n ≥ k . In this case, j + n − (cid:89) i = j − (1 − ¯ x i ¯ x i +1 ) = k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) , (50)because in the product on the left hand side there are only k different factors, and (1 − ¯ x i ¯ x i +1 ) m =(1 − ¯ x i ¯ x i +1 ) for any positive integer m . We will once again take advantage of the translationalsymmetry. Since (cid:104) [ F n ( x )] j (cid:105) should be the same for all j , we will take j = k − n , (cid:104) [ F n ( x )] j (cid:105) = (cid:104) [ F n ( x )] k − n (cid:105) = (1 − q ) q + (cid:42)(cid:32) k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:33) (¯ x k − x k − + x k − x ) (cid:43) . (51)Before we proceed further, let us note that¯ x k − x k − + x k − x = ¯ x k − (1 − ¯ x k − ) + (1 − ¯ x k − )(1 − ¯ x ) = 1 − ¯ x k − ¯ x k − − ¯ x + ¯ x ¯ x k − . (52)Since (1 − ¯ x k − ¯ x k − ) (cid:81) k − i =0 (1 − ¯ x i ¯ x i +1 ) = (cid:81) k − i =0 (1 − ¯ x i ¯ x i +1 ), we obtain (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + (cid:42) k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) + (cid:42) ¯ x (¯ x k − − k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) . (53)We will deal with the two expected values on the right hand side separately. Let us start from the10rst one. (cid:42) k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:42) (1 − ¯ x ¯ x ) k − (cid:89) i =1 (1 − ¯ x i ¯ x i +1 )(1 − ¯ x k − ¯ x ) (cid:43) = (cid:42) (1 − ¯ x ¯ x )(1 − ¯ x k − ¯ x ) k − (cid:89) i =1 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:42) (1 − ¯ x ¯ x k − − ¯ x ¯ x + ¯ x ¯ x ¯ x k − ) k − (cid:89) i =1 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:42) k − (cid:89) i =1 (1 − ¯ x i ¯ x i +1 ) (cid:43) − (cid:104) ¯ x (cid:105) (cid:42) ¯ x k − k − (cid:89) i =1 (1 − ¯ x i ¯ x i +1 ) (cid:43) −(cid:104) ¯ x (cid:105) (cid:42) ¯ x k − (cid:89) i =1 (1 − ¯ x i ¯ x i +1 ) (cid:43) + (cid:104) ¯ x (cid:105) (cid:42) ¯ x ¯ x k − k − (cid:89) i =1 (1 − ¯ x i ¯ x i +1 ) (cid:43) . Recall that U n = n − (cid:89) i =1 (cid:0) − ¯ X i ¯ X i +1 (cid:1) and V n = ¯ X n n − (cid:89) i =1 (cid:0) − ¯ X i ¯ X i +1 (cid:1) , (54)and define U (cid:48) n = ¯ X n − (cid:89) i =1 (cid:0) − ¯ X i ¯ X i +1 (cid:1) and V (cid:48) n = ¯ X ¯ X n n − (cid:89) i =1 (cid:0) − ¯ X i ¯ X i +1 (cid:1) . (55)Now we have (cid:42) k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:104) U k − (cid:105) − (1 − q ) (cid:104) V k − (cid:105) − (1 − q ) (cid:104) U (cid:48) k − (cid:105) + (1 − q ) (cid:104) V (cid:48) k − (cid:105) = (cid:104) U k (cid:105) + q (cid:104) U (cid:48) k − (cid:105) − (cid:104) U (cid:48) k − (cid:105) + (1 − q ) (cid:104) V (cid:48) k − (cid:105) = (cid:104) U k (cid:105) + q (cid:104) U (cid:48) k − (cid:105) − (cid:104) U (cid:48) k (cid:105) , (56)where we used the fact that (cid:104) U (cid:48) n (cid:105) , (cid:104) V (cid:48) n (cid:105) satisfy the same recurrence equations as (cid:104) U n (cid:105) , (cid:104) V n (cid:105) .11he second expected value in eq. (53) can be calculated as follows. (cid:42) ¯ x (¯ x k − − k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:42) ¯ x (¯ x k − − − ¯ x k − ¯ x k − )(1 − ¯ x k − ¯ x ) k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:42) ¯ x (¯ x k − − ¯ x k − ¯ x k − − x k − ) k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:42) ¯ x ¯ x k − k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) + (cid:42) ¯ x (1 − ¯ x k − ¯ x k − ) k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) − (cid:42) ¯ x k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) + (cid:42) ¯ x ¯ x k − k − (cid:89) i =0 (1 − ¯ x i ¯ x i +1 ) (cid:43) = (cid:104) V (cid:48) k − (cid:105) + (cid:104) U (cid:48) k (cid:105) − (cid:104) U (cid:48) k − (cid:105) + (1 − q ) (cid:104) U (cid:48) k − (cid:105) . Combining both expected values computed above we get (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + (cid:104) U k (cid:105) + q (cid:104) U (cid:48) k − (cid:105) − (cid:104) U (cid:48) k (cid:105) + (cid:104) V (cid:48) k − (cid:105) + (cid:104) U (cid:48) k (cid:105)− (cid:104) U (cid:48) k − (cid:105) + (1 − q ) (cid:104) U (cid:48) k − (cid:105) = (1 − q ) q + (cid:104) U k (cid:105) − (cid:104) U (cid:48) k − (cid:105) + (cid:104) V (cid:48) k − (cid:105) . Since (cid:104) V (cid:48) k (cid:105) = (1 − q ) (cid:104) U (cid:48) k − (cid:105) − (1 − q ) (cid:104) V (cid:48) k − (cid:105) , (57)we obtain (cid:104) U (cid:48) k − (cid:105) − (cid:104) V (cid:48) k − (cid:105) = 11 − q (cid:104) V (cid:48) k (cid:105) , (58)and therefore (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + (cid:104) U k (cid:105) − − q (cid:104) V (cid:48) k (cid:105) . (59)Now we need to find (cid:104) V (cid:48) k (cid:105) . As noted before, the recurrence equations for (cid:104) U (cid:48) n (cid:105) , (cid:104) V (cid:48) n (cid:105) are the sameas for (cid:104) U n (cid:105) , (cid:104) V n (cid:105) , that is, as in eq. (28). Only initial conditions are different, (cid:104) U (cid:48) (cid:105) = (cid:104) ¯ X (1 − ¯ X ¯ X ) (cid:105) = (cid:104) ¯ X − ¯ X ¯ X (cid:105) = 1 − q − (1 − q ) = q − q , (60) (cid:104) V (cid:48) (cid:105) = (cid:104) ¯ X ¯ X (1 − ¯ X ¯ X ) (cid:105) = (cid:104) ¯ X ¯ X − ¯ X ¯ X (cid:105) = 0 . (61)The formula (39) thus becomes (cid:20) (cid:104) U (cid:48) n (cid:105)(cid:104) V (cid:48) n (cid:105) (cid:21) = P (cid:34) λ n − λ n − (cid:35) P − (cid:20) q − q (cid:21) , (62)where P and P − are defined in eq. (38). After carrying out matrix multiplication and simplificationthis yields (cid:104) V (cid:48) n (cid:105) = q (1 − q ) λ − λ (cid:0) λ n − − λ n − (cid:1) . (63)12he final result is thus (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + qλ − λ (cid:16) α λ k − + α λ k − (cid:17) − q (1 − q ) λ − λ (cid:16) λ k − − λ k − (cid:17) , (64)which simplifies to (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + qλ − λ (cid:16) ( α + 1 − q ) λ k − + ( α − q ) λ k − (cid:17) , (65)where λ , and α , are defined in eq. (23). Note that the above expression does not depend on n ,which means that (cid:104) [ F n ( x )] j (cid:105) becomes constant when n ≥ k .In fact, one can show that when n = k − n = k −
2, eq. (65) remains valid. We will omitdetails, but the reasoning is very similar as in the case of n ≥ k . The final result for the periodiccase can thus be summarized as follows. (cid:104) [ F n ( x )] j (cid:105) = (1 − q ) q + (cid:40) q λ − λ (cid:0) α λ n − + α λ n − (cid:1) if n ≤ k − , qλ − λ (cid:16) ( α + 1 − q ) λ k − + ( α − q ) λ k − (cid:17) if n ≥ k − . (66)
7. Finite vs. infinite configurations
The case of the periodic initial configuration is often interpreted as a finite configuration of k siteswith periodic boundary conditions. We can say, therefore, that we have obtained probabilisticsolutions for both infinite (eq. 45) and finite (eq. 66) configurations. Let us briefly describedifferences between them. To simplify notation, we will define c n = (cid:104) [ F n ( x )] j (cid:105) , and we will call c n the density of ones after n iterations of the rule.Figure 3 shows plots of the density c n versus n for both finite ( k = 20) and infinite configurationswith q = 0 .
5. One can see that initially they are identical, and at n = k − q becomes closerto 1. This is demonstrated Figure 4 which shows the graph of c ∞ = lim n →∞ c n as a function of q .One can see that the difference between steady states of finite and infinite configurations growsrapidly when q approaches 1. In fact, in the vicinity of q = 1, finite configurations tend to densityapproaching zero, while infinite configurations tend to density approaching 1. This shows thedanger of using finite lattices with periodic boundaries as somewhat “resembling” infinite ones, asit is sometimes done in cellular automata simulations and models. We can conclude from Figure 4that finite size effects can be very significant in CA, even leading to outcomes completely oppositethan those expected for an infinite system.
8. Conclusions
We have demonstrated that a simple CA rule, namely rule 172, can be explicitly solved, meaningthat it is possible to obtain a closed form formula for the state of a given cell after n iterations ofthe rule, as in eq. (21). Such formula is useful for further analysis of properties of the rule. Using13 c n n numericalinfinitefinite Figure 3: Plot of c n as a function of n for q = 1 / k = 20 (dashed line). Dots represent average value of thestate of site i = 0 after n iterations of rule 172, for finite periodic configuration of 20 sites, obtainedby direct numerical iteration of randomly generated initial configuration repeated 10 times. c ∞ q finite, k=20infinite Figure 4: Plot of c ∞ as a function of q for infinite initial configuration (solid line) and periodicinitial configuration with period k = 20 (dashed line).14t, we obtained “probabilistic” solutions, that is, the expected value of a cell after n iterations forboth infinite and finite configurations, assuming that the initial state is drawn from a Bernoullidistribution. This, in turn, allowed us to investigate the role and significance of finite size effectsfor rule 172.It should be stressed that probabilistic solution obtained here is exact, and thus it should notbe confused with approximate methods such as the mean-field theory [14] or local structure theory[15, 16]Although the method presented here is, obviously, not a general one, it can be used for otherrules providing that their dynamics is not overly complicated. While it is difficult to pinpointwhat “not overly complicated” means precisely, some empirical observations can be made. Firstof all, let us notice that in rule 172 if a pair of zeros occurs somewhere in the initial condition, itstays in the same place throughout iterations of the rule. The word 00 is thus a blocking word ofrule 172 (see [17] for precise definition). It is known that all CA rules possessing a blocking wordare almost equicontinuous [17], thus rule 172 has that property too. Existence of a blocking wordseverely limits propagation of information between sites, thus making the rule “simple”, and a goodcandidate for solving using the method outlined in this paper. One should add, however, that therule 172 is only almost equicontinuous, but not equicontinuous. Furthermore, its entropy is positive,as shown in the appendix, thus its dynamics is not entirely trivial. The fact that it can neverthelessbe solved is encouraging, and it seems highly probable that other almost-equicontinuous rules withpositive entropy could be solved in a similar fashion. Acknowledgments
Appendix: entropy of rule 172
Let us recall that A = { , } and that the local function of rule 172 is defined as f ( x , x , x ) = (cid:26) x if x = 0, x if x = 1. (67)Corresponding to f , we define a global mapping F : A Z → A Z such that ( F ( x )) i = f ( x i − , x i , x i +1 )for any x ∈ A Z . We will be interested in the entropy of the dynamical system ( A Z , F ), to bedenoted by h ( A Z , F ). Proposition 8.1
Entropy of ( A Z , F ) , where F is the global function of CA rule 172, is positive. Let Σ { , } be the set of all elements of A Z in which words 00 and 010 do not occur. (Σ { , } , σ )is then a subshift of finite type, where σ is the usual shift map, defined as σ ( x ) i = x i +1 . We willfirst show that rule 172 restricted to Σ { , } is equivalent to shift map, F | Σ { , } = σ. (68)Indeed, consider the value of f ( x i − , x i , x i +1 ) for x ∈ Σ { , } . If x i − = 0, we must have x i = 1(because double zeros are forbidden), and x i +1 = 1 (because isolated ones are forbidden), therefore f (0 , x i , x i +1 ) = f (0 , ,
1) = 1 = x i +1 . If, on the other hand, x i − = 1, then by the definition of f f (1 , x i , x i +1 ) = x i +1 . This means that f ( x i − , x i , x i +1 ) = x i +1 for all x ∈ Σ { , } , as required.Let us now compute the entropy of (Σ { , } , σ ) using the method outlined in [18]. First, weneed to find a Markov shift conjugate to (Σ { , } , σ ). This can be done by defining new symbol set B = { , , } := { a, b, c } . If Σ { ba } denotes the set of points of B Z in which the word ba doesnot occur, then it is easy to show that the subshift (Σ { , } , σ ) is conjugate to (Σ { ba } , σ ), whichis a Markov subshift. Adjacency matrix 3 × { ba } , σ ) is defined by M i,j = 1 if ( i, j ) (cid:54) = ( b, a )and M i,j = 0 if ( i, j ) = ( b, a ). Spectral radius of this matrix is (3 + √ /
2, therefore h (cid:0) Σ { ba } , σ (cid:1) = h (cid:0) Σ { , } , σ (cid:1) = h (cid:0) Σ { , } , F (cid:1) = ln 3 + √ , (69)where we used the fact that the entropy is invariant with respect to conjugacy. This shows that h (cid:0) Σ { , } , F (cid:1) >
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