A property of the rule 150 elementary cellular automaton
aa r X i v : . [ n li n . C G ] J a n A property of the rule 150 elementary cellularautomaton [2014/01/20]
Yuji KanekoGraduate school of Mathematical Sciences,the University of Tokyo, 3-8-1 Komaba, Tokyo 153-8914, Japan
Abstract
We studied the rule 150 elementary cellular automaton in terms of thedistribution of the spacings of the singular values of the matieces obtainedfrom proper time evolutions patterns. The distribution has strong resem-brance to that of the random matrices which is derived from Painlev´e Vequation. Some analytic results for the relative period of the ECS are alsopresented.
The relations between random matrices and the Painlev´e equations are wellknown. The distribution of largest eigenvalue of random matrices is derivedfrom Painlev´e II [1] equation and the distribution of nearest neighbor spacingsfor the eigenvalues is derived from Painlev´e V [2]. The relations between arandom matrix and a cellular automaton (CA) have also been interested; thedistribution of largest eigenvalues was investigated in the study of fluctuationsof interfaces in which used a probabilistic CA [3]. A question is wheter we mayhave a similar relation in a deterministic CA.An elementary cellular automaton (ECA) is a one-dimensional, two-valued,three-neighborhood deterministic dynamical system [5]. There are 256 kindsof ECAs and each ECA is numvered by a nonnegative integer from 0 to 255according to its time evolution rule. Let u tn ∈ { , } be the value of the n th cellat time step t ∈ Z ≥ . u t +1 n = u tn − + u tn + u tn +1 (mod 2) (1)The ECA determined by (1) is called the rule 150 ECA. We impose a periodicboundary condition: u tn = u tN + n , where N ∈ Z is the system size. We rewrite(1) as u t +1 = A u t (mod 2) (2)where u t := t ( u t , · · · , u tN ) is the state vector and A is the N × N transitionmatrix. 1he rule150 is time reversible under the condition of N ≡ , N to 1 , T ( u ) ∈ N such that u = A T ( u )150 u (3)for any initial state u . We call the minimum value of T ( u ) the fundamentalperiod of u .Figure 1: Tow typical time evolution patterns of the rule150 ECAThe figures 1.1 show typical aspects(1=white, 0=black) of time evolution ofthe rule150. The left figure corresponds to the initial state with single nonzerocell. The time evolution pattern exhibits a fractal-like pattern. On the otherhand, the initial state of the right figure is a disordered state and its timeevolution pattern and its time evolution pattern is somehow chaotic.Wolfram named Class 3 to the set of the rules which output patterns likethe right figure of Fig. 1.1. He mentioned that “Class 3 yields chaotic aperiodicpatterns”.[5] However, since the word “chaotic” is not well defined, the methodto quantify the Wolframs’ Class has been studied from the beginning of hisproposal. In our previous study [4], we propose a method of the quantificationin terms of distribution of eigenvalues of correlation matrices of time evolutionpatterns. We characterize chaotic behavior of Class 3 by comparing output pat-terns with random patterns (random matrix) by using the spacing distributionof singular values. As a result, it became apparent that there is close agree-ment. In particular, we found very good coincidence in the rule 150. We showthe numerical results in Fig. 1.1. In the left figure of Fig. 1.1, horizontal axisis spacing of eigenvalue and vertical axis is relative frequency of that spacing.The numerical result of the rule 150 with N = 1792 is indicated by dots. Therule150’s singular values calculated from t U c U c , here U c is the periodic time evo-lution pattern U c = t ( u , · · · , u T − ). The solid line shows the theoretical resultsat N → ∞ of random matrix (Gaussian Orthogonal Ensemble GOE) which isderived from the JMMS equation[2]. This equation is a kind of Painlev´e V.The right figure is the semilog plot of the same figure. The rule150’s spacingdistribution is closely coincident with the theoretical curve. There is anotherexample of distribution which agrees with that of a random matrix, which is2igure 2: Distribution of spacings of singular values of the ECA rule150 timeevolution pattern.known as the Montgomery-Odlyzko low. This low insists that the distributionof spacings between zeros of the Riemann zeta function closely agrees with thatof eigenvalues of a random matrix [7].The time evolution of ECA is deterministic, so it is non-trivial that the prop-erty of the matrix which comes from the rule150 is closely related to a randommatrix. Thus theoretical investigation of the distribution of spacings of therule150 is interesting. If we do not require high accuracy, we can find similar-ity between a random matrix and the other ECAs of Class 3 than the rule150ECA[4][8]. However, other rules do not exhibit such accuracy of coincidenceand it is also important to find the reason why the rule 150 is special.The main purpose of this paper is to show that the rule150 has specialproperty in calculating the spacing distribution of a “chaotic” ECA. We will alsoelucidate that the periodic time evolution matrix U c represents the behavior ofCA in the long-time limit. The procedure to obtain the spacing distribution of singular values numericallyis as follows. Here ∼ are the processes of unfolding [9]. Obtain periodic pattern U c which corresponds to random initial state, andcalculate eigenvalues of t U c U c numerically. Sort the eigenvalues { x , . . . , x N } ( x i − ≤ x i , ≤ i ≤ N ). Keep the eigenval-ues from x N/ to x N/ and discards the other eigenvalues, because itis known that the property of eigenvalues which locate in the vicinity ofthe edges of eigenvalues are different from the others. Construct the cumulative distribution function for and multiply N/ Approximate by a polynomial function f ( x ) (smoothing). Obtain spacings s i = y i − y i − , here y i = f ( x i ).3 Repeat – for randomly selected initial states. Let L be the number ofspacings of eigenvalues of one matrix U c , s ( j ) i be the i th spacing of j th U c , and M be the number of random samplings. We construct the set S := { s ( j ) i | i = 1 , , . . . , L ; j = 1 , , . . . , M } and finally obtain relativefrequency P ( s ).Note that we decrease the number of usable spacings in some cases, that is, L ≤ [ N/ −
1, where [ · · · ] denotes the Gauss symbol. One of such cases isthe case
T < N . In this case we cannot use trivial zero eigenvalues and theirsspacings. There was no sample which shows
T < N in the numerical calculationof the rule150. Figure 1.1 was obtained with N = 1792 , M = 10000.Now we consider how to estimate the P ( s ) in the limit N → ∞ by themethods given above. We have to evaluate the average of (4). We denote by Q ( s | U c ) the conditional distribution of spacing s for a given U c , and by R ( U c )the occurrence probability of U c . P ( s ) = Z Q ( s | U c ) R ( U c ) dU c (4)The random matrix which corresponds to t U c U c is the Wishart matrix t XX .Here X is a p × q rectangular random matrix ( p ≥ q ). If we fix the ratio Q = p/q and p → ∞ , q → ∞ , then the spacing distribution of eigenvalues of the Wishartmatrix is the same as that of the GOE. When we compare t U c U c with t XX ,the rule150 has good properties (I) ∼ (III) at N = 7 · m . We give the proofs ofthese statements in the next section.(I) The maximum fundamental period is N .In the books by Wolfram [5] [10], there are results of numerical simulationto investigate behavior of T associated with increase of N . The behaviorof the rule 150 is fairly complicated as the other rules contained Class 3.However we can prove that the maximum fundamental period of the rule150 is exactly N at N = 7 · m ( m ∈ Z + ). This is a similar property ofthe Wishart matrix and we should focus on the rules which have similarproperty with it. Note that T is not proportional to N in general. Forexample, the rule 45 has a very long periods which is close to 2 N [10]. Inthis case, it is predicted that Q ′ := T /N will diverge when N → ∞ .(II) Almost all fundamental periods are equal to N .We cannot know the number of U c that has maximum fundamental periodfrom (I). If fundamental periods of many periodic patterns are less than N ,those sample produce trivial zero eigenvalues. To prevent the occurrenceof this situation, the Wishart matrix has condition p ≥ q . But we can’tchoose T in the ECA. It is essential difference between the random matrixand the ECA.For this point, (II) insists almost every periodic patterns have maximumfundamental period. This statement drive R ( U c ) to constant in the (4)when N → ∞ . Thus we expect that N = 7 · m is very useful for calculationof (4). 4III) The rule150 is reversible.This is obvious from [6]. But this property is very important. Fromthis property we can conclude that random sampling for initial statesis equivalent to random sampling for periodic patterns with maximumfundamental period.The number of independent rules of ECA is 88 [5]. The rules belongto Class 3 are the rules 18,22,30,45,60,90,105,106,122,129,146,150. Thereversible rules in Class 3 are rule 45,105,150 [6]. The rule 45 is reversibleat N = 1 (mod 2) and the rule 105, 150 is reversible at N = 1 , u t +1 n = u tn − + u tn + u tn +1 + 1 (mod 2) (5)An important point to compare the distribution function for an ECA tothat of a random matrix is that the time evolution pattern of the ECA does nothave any symmetry such as translational symmetry. It is also important thatthe ECA has reversiblity because any state is not a decaying state and we canuse all the states in an orbit. Hence these features (I) ∼ (III) are of particularinterest. In periodic boundary conditions, the transient matrix A of the rule 150 iswrettn by the shift matrix Λ and identity matrix E l . Here E l is l × l identitymatrix. In case of l = N , we omit index l .Λ = (cid:18) E N − (cid:19) (6) A = E + Λ + Λ − (7) Theorem 3.1
Assume N = T = 2 m (2 k − and N = 1 , . Then ( E + Λ + Λ − ) T = E . Proof 3.1
In general, it holds that ( E + Λ + Λ − ) m = E + Λ m + Λ − m (8) Hence, T + 2 m = 2 k + m and thus ( E + Λ + Λ − ) T +2 m = ( E + Λ m + Λ − m ) T = E + Λ m + k + Λ − m + k = E + Λ m + Λ − m = ( E + Λ + Λ − ) m When N = 1 , , E + Λ + Λ − is an inverse matrix. So we have thetheorem 3.1. N = 7 · m . Theorem 3.2
The total number of initial states which do not have the maximum fundamentalperiod is · n . Proof 3.2
In this proof, “period” dose not mean the fundamental period, and it involvesmultiples of the fundamental period. The patterns that have periods shorterthan T must have period m and/or · m − . We consider three cases.1. The total number of the states which have period m .We count the number of solutions to the following equation for u ∈ ( Z ) T . ( E + Λ + Λ − ) m u = u (9) From (8), ( E + Λ m + Λ − m ) u = u . Let r := 2 m . In case the solutionshave period m , u n − r + u n + r ≡ holds for all n (We denote theindex of u with mod T ). Thus, the solutions satisfy ∀ n, u n ≡ u n +2 r On the other hand, from the relation
GCM(2 r, T ) = r and the periodicboundary condition, the solutions satisfy the following condition. ∀ n, u n = u n + r = u n +2 r = · · · = u n +6 r The number of those state is m .2. The total number of the states which have period · m − . ( E + Λ + Λ − ) · m − u = u (10) From (8), ( E + Λ m − + Λ − m − ) u = u . Let z := 2 m − and Γ := Λ z , weobtain (Γ − + Γ − + Γ − + Γ − + E + Γ + Γ + Γ + Γ ) u = u (11) Hence, we have only to count the number of the solutions to the followingequation. (Γ − + Γ − + Γ − + Γ − + Γ + Γ + Γ + Γ ) u = Specifically, ∀ n, u n − z + u n − z + u n − z + u n − z + u n + z + u n +3 z + u n +4 z + u n +6 z ≡ This equation is closed under 14 variables { u n + kz } k = − . u ′ := t ( u n − z , u n − z , u n − z , · · · , u n +5 z , u n +6 z , u n +7 z ) So we use new notation u ′ and consider the following simultaneous equa-tions. (Λ − + Λ − + Λ − + Λ − + Λ + Λ + Λ + Λ ) u ′ = (12)6 et K be the number of solutions to (12) . Then the number of solutions ofthe original equation (10) is K z . The rank of the matrix corresponding tothe left hand side of (12) is 6. Therefore K = 2 − = 2 . Consequently,the number of state is (2 ) s = 2 · m .3. The total number of the states which have period m − .The G.C.D of m and · m − is z (= 2 m − ) . The states which have period z satisfy ( E + Λ + Λ − ) z u = ( E + Λ z + Λ − z ) u = u ∀ n, u n − z + u n + z ≡ . Let r = 2 n and the following equationholds. ∀ n, u n = u n + r = u n +2 r = · · · = u n +6 r The number of those states is m .Therefore the total number of states which do not have the maximum fun-damental periodic patterns is · m + 2 m − m = 2 · m We can conclude that when N = 7 · m , m → ∞ , almost all states of the rule150 have the maximum fundamental period patterns from the above theorem.But, we have to pay attention to symmetry of the U c when we calculate thedistribution of spacings for eigenvalues. In this paper we use periodic boundaryconditions and our systems has translational symmetry in space. Accordingto our numerical methods, those symmetry changes the spacing distributionqualitatively [4][8]. For example in case of the rule 90, we observe the stateswhich satisfy u T/ = Λ N/ u at N = 7 · m and the spacing distribution ofeigenvalues of t U c U c disagrees with that of GOE. In the next section, we provethat the rule 150 have few states like this when N = 7 · m . Definition 4.1
Let T ′ ( u ) be the minimum positive integer which satisfies Λ S ( u ) u = A T ′ ( u )150 u (13) for a certain positive integer S ( u ) ∈ N and for any initial state u . We call T ′ = T ′ ( u ) the fundamental relative period of u , and name S = S ( u ) theshift number for this T ′ (1 ≤ S ≤ N ) . If T = T ′ , the shift number S = N . We consier only the case N = 7 · m . Theorem 4.1
When m → ∞ , almost all states of rule150 have periodic patterns which havemaximum fundamental relative periods T ′ = 7 ∗ m = T = N . roof 4.1 In this proof “relative period” dose not mean the fundamental relative period,and it allows multiples of the fundamental relative period. The patterns thathave relative periods smaller than T must have relative periods T ′ = 2 m with S = S := 2 m , S := 2 · m , S := 3 · m and/or T ′ = 7 · m − with S = S := N/ .1. The total number of states belongs to T ′ = 2 m , S = S . ( E + Λ + Λ − ) m u = Λ m u Thus we should solve Λ m u = u . Let r := 2 n and the following equationholds. ∀ n, u n = u n + r = u n +2 r = · · · = u n +6 r (14) The number of those states is calculated as m . Here, the states rep-resented by (14), which have period m in space. We can regard thosestates as the systems with smaller system size N = 2 m . In this case, since Λ N/ = Λ − N/ , A N/ = E and T = 2 m − . The number of these states isvery few as is proven in the previous section. Hence we need not considerthis case.2. The total number of states belongs to T ′ = 2 m , S = S . ( E + Λ + Λ − ) m u = Λ · m u (15) The following equation holds for t = 2 · m . ( E + Λ + Λ − ) · m u = Λ · m u (16) From (15) and (16), we obtain (Λ · m + Λ − · m ) u = (Λ · m + Λ − · m ) u (17) Adding u to both sides of (17), we find A · m u = A · m u (18) Equation (18) means that T = 2 m . Hence we need not consider this case.3. The total number of states belongs to T ′ = 2 m , S = 3 · m .For t = 2 m , · m , · m , the following three equations hold. ( E + Λ + Λ − ) · m u = Λ · m u (19)( E + Λ + Λ − ) · m u = Λ m u (20)( E + Λ + Λ − ) · m u = Λ · m u (21) From (19), (20) and (21), we obtain A m u = Λ · m u (22) From (22) and A m u = Λ · m u , we can obtain the solutions which haveperiod m in space. As the case of the T ′ = 2 m , S = 2 m , these states arevery few and we need not consider in the limit m → ∞ . . The total number of states belongs to T ′ = 7 · m − , S = N/ . ( E + Λ + Λ − ) · m − u = Λ · m − u (23) Using the same arguments as in (11), we obtain (Λ − + Λ − + Λ − + Λ − + E + Λ + Λ + Λ + Λ + Λ ) u ′ = (24) The rank of matrix corresponding to the left hand side of (24) is 7. Thenumber of states is · m − .Thus we can evaluate the upper bound of the number of the states which donot have the periodic patterns with maximum fundamental relative period as · m + 2 · m − . Thus the ratio of the number of the states calculated above tothat of all the states is less than · m + 2 · m − · m (25) The above ratio becomes to zero for m → ∞ . Therefore almost all states of therule 150 have the periodic patterns with maximum fundamental relative period T ′ = N = 7 · m . The distribution of the spacings of the singular values derived from the periodicpatterns of a chaotic cellular automaton was shown numerically to closely coin-cide with that of eigenvalues of random matrices. In this paper, we showed thatthe rule 150 ECA with a particular system sizes N = 7 · m has special featureswhich are suitable for compariosn to random matrices. In this case, almost allperiodic patterns are regarded as square matrices and we can use eigenvaluesinstead of singular values for rectangular matrices. This property satisfies theWishart matrix’s condition for Q ≥
1. The rigorous derivation of the rule150’sspacing distribution is a problem we would like to address in the future.
Acknowledgements
The author wishes to thank Professors T.Tokihiro, J.Mada and S.M.Nishigakifor useful comments.
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