Explorations of ternary cellular automata and ternary density classification problems
EExplorations of ternary cellular automata and ternarydensity classification problems
Henryk Fuk´s and Roman Procyk Department of Mathematics and Statistics, Brock UniversitySt. Catharines, Ontario L2S 3A1, Canada Department of Physics, McGill UniversityMontreal, Quebec H3A 2T8, CanadaWhile binary nearest-neighbour cellar automata (CA) have been stud-ied in detail and from many different angles, the same cannot be said aboutternary (three-state) CA rules. We present some results of our explorationsof a small subset of the vast space of ternary rules, namely rules possessingadditive invariants. We first enumerate rules with four different additive in-variants, and then we investigate if any of them could be used to constructa two-rule solution of generalized density classification problem (DCP). Weshow that neither simple nor absolute classification is possible with a pairof ternary rules where the first rule is all-conserving and the second one isreducible to two states. Similar negative result holds for another versionof DCP we propose: symmetric interval-wise DCP. Finally we show an ex-ample of a pair of rules which solve non-symmetric interval-wise DCP forinitial configurations containing at least one zero.PACS numbers: 05.45.-a
1. Introduction
In both the theory and applications of cellular automata, a lot of efforthas been invested in studying binary rules. Given the importance of binarylogic and binary arithmetic in todays computing, this is of course quiteunderstandable.Nevertheless, the next possible arithmetical and logical systems in termsof the number of allowed states, namely the ternary arithmetic and ternarylogic, enjoyed some (if only limited) popularity in the past. To give concreteexamples, the wooden calculating machine [1] built by Thomas Fowler in1840 operated in balanced ternary arithmetic, using three digits -1, 0 and 1.In the early 20th century, Polish mathematician Jan (cid:32)Lukasiewicz inventedand formalized a three-valued logic [2], and many followers developed his (1) a r X i v : . [ n li n . C G ] F e b ideas in subsequent years. In the second half of the 20th century, experi-mental computers based on ternary arithmetic were constructed, beginningwith the the most famous one, the Setun computer developed in 1958 atthe Moscow State University [3]. In 1980’s, ternary CMOS memory chips(ROM) were built at Queens University in Canada. In the most recenttimes, the possibility of quantum computing based on qtrits instead qbitshas been suggested [4].All of the above indicates that ternary arithmetic and ternary logic arestill interesting areas to explore, and that they may have some applicationpotential. This inspired us to probe into the huge space of ternary cellularautomata, with the hope of finding some interesting rules, possibly appli-cable to solving computational problems such as the density classificationproblem.Of course, even if one considers only nearest-neighbour ternary cellularautomata, the number of possible rules is huge, 3 ≈ . · . Thereis no hope to study all of them systematically, as it has been done in thecase of elementary (binary nearest-neighbour) cellular automata, which forma small set of 256 rules. For this reason, we have decided to take a closerlook at only a small subset of ternary rules, namely those which obey simpleconservation laws, hoping to find some which would be useful in constructingsolutions of computational problems similar to classical density classificationproblem (see [5] for review and references).The paper is organized as follows. First we enumerate ternary nearest-neighbour rules with various additive invariants. Then we investigate if anyof the rules found could be used to solve a generalization of the densityclassification problem to 3 states.Let us define some basic concepts first. We will be mostly concerned withthe space S = { , , } L of finite-length configurations of length L . We willimpose periodic boundary conditions on configurations x ∈ S , so that for x = ( x , x , . . . x L − ), the index i in x i is to be always taken modulo L , i.e., i ∈ Z /L . The local function (also called local transition function or localrule ) of a ternary nearest-neighbour cellular automaton will be a function f : { , , } → { , , } , while the corresponding global function F : S → S will be defined as ( F ( x )) i = f ( x i − , x i , x i +1 ) (1)for all i ∈ Z /L .
2. Conservation laws
Interesting classes of ternary rules are those with additive invariants. LetΨ( x ) be some function of x ∈ { , , } . If, for any periodic configuration x of length (period) L we have (cid:80) L − i =0 Ψ( x i ) = (cid:80) L − i =0 Ψ(( F ( x )) i ), then we saythat F conserves Ψ.According to a well-known theorem of Hattori and Takesue [6], Ψ isconserved if and only if for all x , x , x ∈ { , , } we haveΨ( f ( x , x , x )) − Ψ( x ) = Ψ( f ( p, p, x ))+ Ψ( f ( p, x , x )) − Ψ( f ( p, p, x )) − Ψ( f ( p, x , x )) , (2)where p is an arbitrarily chosen element of the alphabet { , , } . For thesake of convenience, we will use p = 0 in all what follows.Eq. (2) can be understood as a sort of discrete form of the continuityequation ∂ρ/∂t = − ∂j/∂x describing transport of some quantity with den-sity ρ , where j is the flux of this quantity. It can be transformed to another,equivalent form, where on the left hand side one has the local change of Ψin one time step, analogous to ∂ρ/∂t , and on the right hand side one hasan expression analogous to the spatial derivative of the flux of Ψ (see [6] forfurther details).Hattori-Takesue theorem makes it quite easy to check if a given ternaryrule f conserves Ψ – all one needs to do is to verify the above condition forall 3 = 27 possible sets of values of x , x , x ∈ { , , } .Examples of some possible choices of Ψ are:(a) Ψ ( x ) = (1 − x )(2 − x ), number of cells in state 0 is conserved;(b) Ψ ( x ) = x (2 − x ), number of cells in state 1 is conserved;(c) Ψ ( x ) = x ( x − s ( x ) = x , sum of cell values is conserved (so-called number-conservingrule ).Note that conservation of any two of the above implies conservation of theremaining two. Rules which conserve all (a)–(d) will be called all-conserving .
3. Enumeration
We will first enumerate ternary rules conserving various Ψs. The follow-ing proposition provides a summary of such enumeration.
Proposition 1
Among nearest-neighbour cellular automata with three states,there is exactly(a) · = 2359296 rules conserving Ψ (the same applies to Ψ and Ψ ); (b) 144 rules conserving Ψ s ;(c) 15 all-conserving rules.Proof: Claim (b) is a known results [7], thus we will prove only (a) and (c). Let f : { , , } → { , , } be a local function of a CA conserving 1’s. Accordingto the Hattori-Takesue theorem, it must satisfyΨ ( f ( x , x , x )) − Ψ ( x ) = Ψ ( f (0 , , x ))+ Ψ ( f (0 , x , x )) − Ψ ( f (0 , , x )) − Ψ ( f (0 , x , x )) (3)for all x , x , x ∈ { , , } , where Ψ ( x ) = x (2 − x ).When values of arguments of f are restricted to only 0’s and 1’s, f must return outputs compatible with one of the five number-conservingelementary cellular automata (ECA) rules with Wolfram numbers 184, 226,170, 240 or 204, where compatibility is defined as follows. We say thatternary rule f is compatible with binary rule g if, for x , x , x ∈ { , } , f ( x , x , x ) = (cid:40) g ( x , x , x ) = 1,0 or 2 if g ( x , x , x ) = 0. (4)In the above, we have “0 or 2” because we only want to preserve the numberof 1’s.We will first enumerate rules which, when their arguments are restrictedto 0’s and 1’s, are compatible with rule 184. Let us define, for x , x , x ∈{ , , } , a x +3 x + x = f ( x , x , x ) . (5)For rules compatible with rule 184 we must have a = f (0 , ,
0) = 0 or 2 ,a = f (0 , ,
1) = 0 or 2 ,a = f (0 , ,
0) = 0 or 2 ,a = f (0 , ,
1) = 1 ,a = f (1 , ,
0) = 1 ,a = f (1 , ,
1) = 1 ,a = f (1 , ,
0) = 0 or 2 ,a = f (1 , ,
1) = 1 . Note that the above implies Ψ ( a ) = Ψ ( a ) = Ψ ( a ) = Ψ ( a ) = 0. Theremaining a i must satisfy 27 conditions obtained from eq. (3). All of them are listed below in two columns, including those which reduce to identities.0 =0 , , Ψ ( a ) =Ψ ( a ) , , , Ψ ( a ) =Ψ ( a ) , Ψ ( a ) =Ψ ( a ) , Ψ ( a ) =Ψ ( a ) , Ψ ( a ) =Ψ ( a ) , , , Ψ ( a ) − ( a ) , − − , , Ψ ( a ) − − ( a ) , Ψ ( a ) − ( a ) + Ψ ( a ) − Ψ ( a ) , Ψ ( a ) − ( a ) + Ψ ( a ) − Ψ ( a ) , Ψ ( a ) − ( a ) + Ψ ( a ) − Ψ ( a ) , Ψ ( a ) = − Ψ ( a ) − Ψ ( a ) , Ψ ( a ) = − Ψ ( a ) − Ψ ( a ) , Ψ ( a ) = − Ψ ( a ) , Ψ ( a ) = − Ψ ( a ) − Ψ ( a ) , Ψ ( a ) =1 − Ψ ( a ) − Ψ ( a ) , Ψ ( a ) =Ψ ( a ) − Ψ ( a ) − Ψ ( a ) , Ψ ( a ) =Ψ ( a ) − Ψ ( a ) , Ψ ( a ) =Ψ ( a ) − Ψ ( a ) , Ψ ( a ) =0 . From the above, if we discard all equations which are identically true andintroduce variables b i = Ψ ( a i ), we obtain b − b ,b − − b ,b − b + b − b ,b − b + b − b ,b − b + b − b ,b = − b − b ,b = − b − b , b = − b ,b = − b − b ,b =1 − b − b ,b = b − b − b ,b = b − b ,b = b − b ,b =0 . This is a linear system of 14 equations with 19 unknowns. We can, therefore,solve this system for b , b , b , . . . b in terms of b , b , b , b , b , obtaining b = 1 + b ,b = b ,b = − b + 1 + b + b ,b = − b + 1 + b + b ,b = − b + 1 + b + b ,b = − b − b ,b = − b − b , b = − b ,b = − b − b ,b = 1 − b − b ,b = b − b − b ,b = b − b ,b = b − b ,b = 0 . Recall that the only allowed values of b i are 0 or 1, thus we must have b = 0. By the same token, we also need b = 0, b = 0, and b = 0. Thisleaves b = 1 ,b = b ,b = − b + 1 ,b = − b + 1 ,b = − b + 1 ,b = 0 ,b = 0 , b = 0 ,b = 0 ,b = 1 ,b = b ,b = 0 ,b = 0 ,b = 0 , meaning that we have only one parameter left, b . Obviously, b ∈ { , } ,thus we obtain two solutions of our original system. The first correspondsto b i = 0 for i ∈ { , , , , , , , , , , , , , } ,b i = 1 for i ∈ { , , , , } , and the second to b i = 0 for i ∈ { , , , , , , , , , , , , , } ,b i = 1 for i ∈ { , , , , } . In both the above solutions, the number of b i ’s with zero values is exactly 14.Each b i = 0 admits two corresponding values of a i , a i = 0 or a i = 2, becausewe defined b i = Ψ ( a i ) = a i (2 − a i ). Moreover, values of 4 parameters a , a , a , and a also admit two possible values, 0 or 2. This means thateach of these two solutions corresponds to 2 possible sets a , a , . . . a satisfying 27 Hattori-Takesue conditions. We thus have total 2 CA rulesconserving 1’s and compatible with rule 184.For rules compatible with ECA 226, the calculations are almost identical,yielding also 2 rules. For rules compatible with ECA 170 or 240, by usingsimilar reasoning as above, we obtain in both cases 2 rules. For rulescompatible with ECA 204, the total number of rules turns out to be 3 · ,again by a similar reasoning (omitted here).The total number of rules conserving 1’s is, therefore, 2 + 2 + 2 +2 + 3 · = 9 · , as claimed in (a).For part (c), we took the advantage of the fact that every all-conservingrule must also be number-conserving. One can, therefore, simply test all 144number-conserving rules for conservation of the number of 0’s (the numbersof 1’s and 2’s will then be automatically conserved, so they do not evenneed to be checked). We carried out his procedure and found that 15 rulesshown in Table 1 are the only all-conserving ternary rules.
4. Classification problems
The following two-rule solution of the density classification problem iswell known [8].
Proposition 2 (H.F. 1997)
Let x be a periodic binary configuration oflength L and density ρ = L (cid:80) L − i =0 x i , and let n = (cid:98) ( L − / (cid:99) , m = (cid:98) ( L − / (cid:99) . Then • F m ( F n ( x )) = 0 L if ρ < / , • F m ( F n ( x )) = 1 L if ρ > / , • F m ( F n ( x )) = . . . . . . if ρ = 1 / . In the above, F and F are global functions of elementary cellular au-tomata 184 and 232, and 0 L (1 L ) denotes a string of length L of all zeros(all ones).We say that the pair of rules (184, 232) classifies densities, or solves the density classification problem (DCP), because iterating rule 184 sufficientnumber of times followed by analogous iteration of rule 232 produces ho-mogeneous string of all zeros if initially we had more zeros than ones, andhomogeneous string of all ones if we had more ones than zeros at the begin-ning. It is worth noting that the above two-rule solution of DCP has beenproposed because single-rule solution of this problem does not exist [9, 10].There are three obvious ways to generalize the density classificationproblem to 3 states (or more). • in the final configuration all sites are to be in state 0 ifthere are more zeros than other symbols in the initial configuration. Ifthere are more non-zero symbols than zeros in the initial configuration,then in the final state all sites are to be in state 1. • simple majority: in the final configuration all sites are to be in state k if symbols k form the majority in the initial configuration. • absolute majority: in the final configuration all sites are to be in state k if symbols k form the absolute majority in the initial configuration.Note that all three are compatible with the binary DCP, meaning that whenthe initial configuration contains only 0’s and 1’s, they reduce to the binaryDCP.Since the binary DCP has no single-rule solution, the same applies to allthree generalized problems as well. But are there any two-rule solutions forthese problems? In 1999, H.F. Chau et al. constructed two-rule solution to n -ary simple majority problem [11]. Their solution, however, uses rules ofrather large neighbourhood size. For ternary rules it requires that the firstrule has neighbourhood radius of at least 45. Despite the fact that Chau’ssolution uses some very interesting ideas, we will not discuss it here, as wewish to focus on nearest-neighbour rules only. Along the same vein, sincewe restrict our attention to ternary rules only, we will not be concerned withsolutions of density classification problems which require very large numberstates, such as, for example, the work of Brice˜no et al. [12].Since in the known two-rule solution of the binary DCP the first ruleconserves the number of zeros (and ones), it is reasonable to expect thatfor ternary rules, the first rule of the solution should also conserve therelevant quantity. For 0-majority problem, the fist rule should thus be Ψ -conserving, and for the simple or absolute majority problem, it should beall-conserving. Since we have enumerated various rules with invariants, wecan search among the relevant rules for a possible candidate for the firstrule of the solution.For the 0-majority problem , we do not even have to search among the9 · Ψ -conserving rules, because the solution is trivial to construct. Let f ( x , x , x ) = f ( φ ( x ) , φ ( x ) , φ ( x )) for all x , x , x ∈ { , , } , (6)where we define φ (0) = 0, φ (1) = 1, φ (2) = 1 and where f is the localrule ECA 184. The pair f and ECA 232 solve 0-majority DCP in the samefashion as rules 184 and 232 solve the binary DCP. Obviously, one couldalso define two other variants of the 0-majority problem (1-majority and2-majority), and construct their two-rule solutions in an analogous way.
5. Simple and absolute density classification
For the remaining two versions of DCP, we have not found any pair ofternary rules solving these versions. We strongly suspect that such a pairdoes not exist, although we were able to prove only a partial non-existenceresult, to be presented below.Let us define, for a given ternary rule f , three functions f | ( x , x , x ) = f ( x , x , x ) ,f | ( x , x , x ) = f (2 x , x , x ) / ,f | ( x , x , x ) = f ( x + 1 , x + 1 , x + 1) − , (7)where x , x , x ∈ { , } . If any of these functions returns only values 0 or1 for all x , x , x ∈ { , } , we will say that the relevant binary reduction of f exists. If all three binary reductions exist, we will call f reducible to twostates . Proposition 3
There exists no pair of nearest-neighbour ternary rules solv-ing the simple (or absolute) majority density classification task such that thefirst rule of the pair is all-conserving and the second rule is reducible to twostates.Proof:
Suppose that there exist such a pair of ternary rules solving thesimple majority density classification, and the first rule f in this pair is all-conserving, while the second one is reducible to two states. Table 1 showsWolfram numbers of binary projections for all 15 all-conserving rules. Onecan clearly see that the first rule, when restricted to binary configurations,behaves as one of the five rules among 204 (identity), 170 (left shift), 240(right shift), 184, and 226. Since identity and shifts do not change thearrangement of symbols, f behaving as rule 204, 170 or 240 on on binaryconfigurations would require that the second rule performed the entire taskof the density classification on its own ( f would do nothing). We assumedreducibility of the second rule, thus its reduction to { , } would have to bea solution of the binary classification problem. This, however is impossible– single-rule solution of the binary classification problem does not exist.Therefore, when we restrict configurations to { , } , the all-conservingrule f must satisfy ∀ x , x , x ∈ { , } : f ( x , x , x ) = f ( x , x , x ) , or ∀ x , x , x ∈ { , } : f ( x , x , x ) = f ( x , x , x ) . (8)This is equivalent to saying that f | = f or f | = f . (9)Exactly the same reasoning applies in the case of reductions to { , } , yield-ing the requirement f | = f or f | = f . (10)For configurations restricted to { , } we obtain f | = f or f | = f . (11)A quick glance at Table 1 convinces us that there is no all-conserving rulesatisfying simultaneously conditions of eq. (9), (10), and (11). This demon-strates that an all-conserving f with the desired properties does not exist.For absolute majority DCP, the proof is identical, as absolute majorityand simple majority are the same when we restrict configurations to twostates only. (cid:3) Wolfram number f f | f | f | Table 1. List of the Wolfram codes of the 15 all-conserving ternary rules (firstcolumn). Wolfram numbers of their binary projections are shown in columns f | , f | , and f | .
6. Interval-wise density classification
In addition to the variants of the DCP described in previous section,there exist yet another possible way to generalize the density classificationproblem which includes the “classical” binary DCP as a special case.Suppose we want to classify finite strings of length L over the alphabetof M symbols A = { , , . . . , M − } . Let ρ ( x ) = (cid:80) L − i =0 x i (cid:80) L − i =0 max A be called density of configuration x = ( x , x , . . . .x L − ), where max A is thelargest element of the alphabet A , so that max A = M −
1. This definitionguarantees that ρ ∈ [0 , ρ ( x ) = 1 L ( M − L − (cid:88) i =0 x i . Let p , p , . . . , p M − be real numbers satisfying 0 < p < p < . . . < p M − <
1. The pair of rules with global functions F and G solves interval-wise density classification problem if there exist integers n, m (possibly dependingon L ) such that for every configuration x = ( x , x , . . . .x L − ) we have G m F n ( x ) = 0 L if ρ ( x ) ∈ [0 , p ) ,G m F n ( x ) = 1 L if ρ ( x ) ∈ ( p , p ) ,G m F n ( x ) = 2 L if ρ ( x ) ∈ ( p , p ) , · · · G m F n ( x ) = ( M − L if ρ ( x ) ∈ ( p M − , . Obviously, when M = 2 and p = 1 /
2, the above reduces to the “classical”binary density classification problem, with known solution by the pair ofECA 184 and 232. Note that we intentionally made the intervals open atpoints p , p , . . . p M − , to be compatible with the standard DCP, where theclassification of strings with equal number of zeros and ones is not required.When p i = 1 /M for i = 1 , , . . . M −
1, we call the problem symmetricinterval-wise density classification problem. For ternary rules ( M = 3),does a two-rule solution of the symmetric interval-wise DCP exist? Again,based on our extensive heuristic search, we suspect that the answer is no,but we were able to prove the non-existence only for rules reducible to twostates.Suppose that such solution indeed existed, consisting of rules with globalfunctions F and G , both reducible to two states. This would mean that forsome m and n we have G m F n ( x ) = 0 L if ρ ( x ) ∈ [0 , / ,G m F n ( x ) = 1 L if ρ ( x ) ∈ (1 / , / ,G m F n ( x ) = 2 L if ρ ( x ) ∈ (2 / , . Now let us suppose that x is binary, consisting only of 0’s and 1’s. In sucha case, the above would reduce to G m F n ( x ) = 0 L if 0 ≤ ρ ( x ) < / ,G m F n ( x ) = 1 L if 1 / < ρ ( x ) < / , or, using the definition of ρ ( x ) = L (cid:80) x i , G m F n ( x ) = 0 L if 0 ≤ L L − (cid:88) i =0 x i < / ,G m F n ( x ) = 1 L if 1 / < L L − (cid:88) i =0 x i < / . This would imply existence of a pair of elementary rules having the propertythat G m F n ( x ) consists of all zeros if zeros occupy 2 / ·
256 = 65536 possible pairs. Thisproves the following.
Proposition 4
There is no pair of nearest-neighbour ternary rules reducibleto two states which would solve the symmetric interval-wise density classi-fication problem.
7. Non-symmetric interval-wise classification
If there is no solution of the symmetric interval-wise problem, can we atleast perform non-symmetric classification with two ternary rules?We performed an intensive heuristic search for such rules, and aftersome tinkering with rule tables, by trial and error we found the followinginteresting pair of ternary rules.
Conjecture 1
Let F be the ternary nearest-neighbour rule with Wolframnumber 6478767664173, and G be the rule with Wolfram number 7580606234490.For any finite ternary string x of length L , containing at least one zero, let ρ = 12 L L − (cid:88) i =0 x i . Then G L F L ( x ) = 0 L if ρ ( x ) ∈ [0 , / ,G L F L ( x ) = 1 L if ρ ( x ) ∈ (2 / , / ,G L F L ( x ) = 2 L if ρ ( x ) ∈ (3 / , . This means that the pair of rules (6478767664173, 7580606234490) “almost”solves the interval-wise DCP with p = 2 / p = 3 /
4. We use the word“almost” because it only works for configurations which contain at least onezero. Configurations which do not satisfy this property can be misclassified- for example, 1 L has density 1 /
2, thus should produce 0 L in the end, yet1 L is a fixed point of both rules 6478767664173 and 7580606234490 .We performed extensive numerical experiments to verify the above con-jecture, and it appears to be valid. Below we provide a sketch of a possibleproof of this result. Figures 1 and 2 show definitions of rules F and G .The first of them (rule 6478767664173) is number-conserving, and its binary We wish to thank anonymous referee for pointing out this fact. f (0 , ,
0) = 0 f (0 , ,
1) = 0 f (0 , ,
0) = 1 f (0 , ,
1) = 2 f (1 , ,
0) = 0 f (1 , ,
1) = 0 f (1 , ,
0) = 0 f (1 , ,
1) = 1 f (0 , ,
0) = 0 f (0 , ,
2) = 1 f (0 , ,
0) = 1 f (0 , ,
2) = 2 f (2 , ,
0) = 0 f (2 , ,
2) = 1 f (2 , ,
0) = 1 f (2 , ,
2) = 2 f (1 , ,
1) = f (1 , ,
2) = f (1 , ,
1) = f (1 , ,
2) = f (2 , ,
1) = f (2 , ,
2) = f (2 , ,
1) = f (2 , ,
2) = f (0 , ,
2) = 2 f (0 , ,
1) = 1 f (1 , ,
2) = 1 f (1 , ,
0) = 1 f (2 , ,
1) = 0 f (2 , ,
0) = 1Fig. 1. Definition of rule 6478767664173. Entries with bold outputs correspond torestriction to configurations consisting only of 1’s and 2’s, on which the above ruleis equivalent to rule 184. f (0 , ,
0) = 0 f (0 , ,
1) = 0 f (0 , ,
0) = 0 f (0 , ,
1) = 0 f (1 , ,
0) = 0 f (1 , ,
1) = 0 f (1 , ,
0) = 1 f (1 , ,
1) = 1 f (0 , ,
0) = 0 f (0 , ,
2) = 0 f (0 , ,
0) = 0 f (0 , ,
2) = 2 f (2 , ,
0) = 0 f (2 , ,
2) = 2 f (2 , ,
0) = 2 f (2 , ,
2) = 2 f (1 , ,
1) = 1 f (1 , ,
2) = 1 f (1 , ,
1) = 1 f (1 , ,
2) = 2 f (2 , ,
1) = 1 f (2 , ,
2) = 2 f (2 , ,
1) = 2 f (2 , ,
2) = 2 f (0 , ,
2) = 0 f (0 , ,
1) = 0 f (1 , ,
2) = 0 f (1 , ,
0) = 2 f (2 , ,
1) = 1 f (2 , ,
0) = 1Fig. 2. Definition of rule 7580606234490. Its binary projections are ECA 192, 232,and 232, and this can be verified by inspecting the first three columns. projection f | (defined as in eq. 7) is ECA 184. This rule plays an analo-gous role as rule 184 in the two-rule solution of the binary DCP, namely, itprepares the configuration for further processing without changing its den-sity ρ ( x ). After sufficiently many iterations (which we simply take to be L ),this rule eliminates certain symbols and substrings, as shown in Table 2.One can see that, for example, when ρ ( x ) < /
3, after L iterations of rule F , the configuration may contain substrings 00 and 11 as well as symbols0, 1, and 2, while substrings 22 are always absent. When ρ ( x ) ∈ (2 / , / ρ ( x ) > / substring ρ ( x ) < / ρ ( x ) ∈ (2 / , / ρ ( x ) > /
400 Yes No No11 Yes Yes No22 No No Yes0 Yes No No1 Yes Yes Yes2 Yes Yes Yes
Table 2. Presence of selected substrings in the final configuration after iteratingrule 6478767664173 L times. The second rule G , with Wolfram number 6478767664173, plays a rolesimilar to rule 232 in the two-rule solution of the binary DCP. Its threebinary projections f | , f | and f | exist, and their Wolfram numbersare, respectively, 192, 232, and 232. This rule grows clusters of 0’s if theyare present, and if not, it just behaves like rule 232, that is, grows clustersof 1’s in the absence of pairs 22, and grows clusters of 2’s in the absence ofpairs 11. A quick look at Table 2 reveals that these three cases will occurafter we iterate rule F on initial configurations with densities, respectively, ρ ( x ) < / ρ ( x ) ∈ (2 / , / ρ ( x ) > /
4. Note that presence of 0’sis required to grow clusters of 0’s, thus the need for additional conditionimposed in the initial configuration (it mus contain at least one zero).The final effect, therefore, of iterations of rule G starting with F L ( x )will be all zeros when ρ ( x ) < /
3, all ones when ρ ( x ) ∈ (2 / , / ρ ( x ) > /
4, exactly as claimed in Conjecture 1. Examples ofthree cases of density classification by the aforementioned rules are shownin Figure 3.Obviously the above is only a sketch of a proof, and it needs furtherelaboration. Our statement about rules 6478767664173 and 7580606234490,therefore, must remain a conjecture for now.
8. Conclusions and future work
We have demonstrated that except the trivial case of 0-majority, thereexist no two-rule solution of various density classification problems (sim-ple majority, absolute majority, symmetric interval-wise) in the domain ofternary nearest-neighbour rules which would be analogous to the known so-lution of DCP by the pair of ECA 184 and 232. By “analogous” we mean asolution consisting of two rules reducible to two states in which the first ruleserves as a pre-processor preserving relationship between densities, thus isall-conserving for simple/absolute majority problem, or number-conservingfor interval-wise problem. it it it ρ ( x ) = 0 . ρ ( x ) = 0 . ρ ( x ) = 0 . Fig. 3. Spatiotemporal patterns illustrating a two-rule solution of non-symmetricinterval-wise DCP by rules 6478767664173 and 7580606234490. White color repre-sents 0’s, gray 1’s, and blue 2’s.
This naturally brings up a question if two-rule solutions exist if onerelaxes the restriction of the first rule possessing additive invariant or therestriction of rules being reducible. While such a possibility cannot be ex-cluded, we seriously doubt it. However, it seems quite possible that extend-ing the neighbourhood size to two nearest neighbours may help to produce atwo-rule solution. We plan to investigate this possibility in the near future.
Acknowledgement:
H.F. acknowledges financial support from the Natural Sciences and En-gineering Research Council of Canada (NSERC) in the form of Discovery Grant. We thankanonymous referees for comments which helped to improve the paper.
REFERENCES [1] P. Vass,
The Power of Three . Boundstone Books, 2015.[2] J. (cid:32)Lukasiewicz, “On three-valued logic,” in
Selected works by Jan(cid:32)Lukasiewicz
New Journal of Physics (2013), no. 5, 053007.[5] P. de Oliveira, “On density determination with cellular automata: results,constructions and directions,” Journal of Cellular Automata (2014),no. 5–6, 357–385.[6] T. Hattori and S. Takesue, “Additive conserved quantities in discrete-timelattice dynamical systems,” Physica D (1991) 295–322.[7] N. Boccara and H. Fuk´s, “Number-conserving cellular automaton rules,” Fundamenta Informaticae (2002) 1–13, arXiv:adap-org/9905004 .[8] H. Fuk´s, “Solution of the density classification problem with two cellularautomata rules,” Phys. Rev. E (1997) 2081R–2084R, arXiv:comp-gas/9703001 .[9] M. Land and R. K. Belew, “No perfect two-state cellular automata fordensity classification exists,” Phys. Rev. Lett. (1995), no. 25, 5148–5150.[10] A. Buˇsi´c, N. Fat`es, J. Mairesse, and I. Marcovici, “Density classification oninfinite lattices and trees,” Electron. J. Probab. (2013) 22 pp.[11] H. F. Chau, L. W. Siu, and K. K. Yan, “One dimensional n-ary densityclassification using two cellular automaton rules,” International Journal ofModern Physics C (1999), no. 05, 883–889.[12] R. Brice˜no, P. M. de Espan´es, A. Osses, and I. Rapaport, “Solving thedensity classification problem with a large diffusion and small amplificationcellular automaton,” Physica D: Nonlinear Phenomena261