One-dimensional number-conserving cellular automata
OOne-dimensional number-conservingcellular automata
Markus Redeker ∗ February 11, 2020
A number-conserving cellular automaton is a simplified model for a sys-tem of interacting particles. This paper contains two related constructionsby which one can find all one-dimensional number-conserving cellular au-tomata with one kind of particle.The output of both methods is a “flow function”, which describes themovement of the particles. In the first method, one puts increasinglystronger restrictions on the particle flow until a single flow function isspecified. There are no dead ends, every choice of restriction steps endswith a flow.The second method uses the fact that the flow functions can be orderedand then form a lattice. This method consists of a recipe for the slowest flowthat enforces a given minimal particle speed in one given neighbourhood.All other flow functions are then maxima of sets of these flows.Other questions, like that about the nature of non-deterministic number-conserving rules, are treated briefly at the end.
Cellular automata are microscopic worlds: extremely simple spaces in which timepasses and events occur. Sometimes they are used to simulate aspects of this world.It is therefore an interesting question to ask which of these micro-worlds can be in-terpreted as containing particles that can move, collide or stick to each other, but areneither destructed nor created.In this paper I provide a new answer to this question, valid for one-dimensional au-tomata with only one kind of particles. The new idea is that such cellular automata arenow constructed instead of testing whether a given system satisfies the requirements.The cellular automata are constructed in terms of flow functions . This are functionsthat describe how many particles cross the boundary between two cells, depending on ∗ [email protected] a r X i v : . [ n li n . C G ] F e b he neighbourhood of this boundary. By specifying the flow functions stepwise withincreasing accuracy, one can find all number-conserving automata.Flow functions did already occur in the literature. They were described by Hattoriand Takesue [6, Th. 2.2] and by Pivato [8, Prop. 12] but not used for the constructionof transition rules for cellular automata. Imai and Alhazov [7] have already the condi-tions (12) for cellular automata of radius and 1; their Proposition 3 gives a necessaryand significant condition for number-conservation in radius cellular automata – it isthe case of (cid:96) = 1 for Theorem 1 below, so to speak. Background
The following paragraphs contain references to papers that are some-what related to this work. The list is by no means complete. For a little bit of historyof number conservation see Bhattacharjee et al. [1, Sec. 4.6] or the introduction of therelated paper by Wolnik et al. [10].The most important predecessor of the current work is the paper of Hattori andTakesue [6], who found a simple way to test whether a one-dimensional cellular au-tomaton is number-conserving. This method works only a posteriori , but it providesthe base for most later papers about number conservation.Boccara and Fukś [2] describe the behaviour of number-conserving cellular automatawith the help of “motion representation” diagrams. This are diagrams that show themotion of individual particles in finite regions of the automaton. The authors find aset of equations for the transition rule of a number-conserving automaton and solvethem. They also show that number-conserving rules can be identified by verifying thatthe number of particles is conserved on all circular cell configurations of a specifiedsize.Fukś [5] shows that in one-dimensional number-conserving automata, one can as-sign permanent identities to the “particles” and describe the evolution of the cellularautomaton in terms of particle movements.Pivato [8] derives several characterisations of conservation laws in the more generalcontext of cellular automata on finite groups.Durand, Formenti and Róka [4] collect different definition of number conservationand show that they are equivalent. They also derive conditions for number conservationin cellular automata of any dimension.
Overview
The rest of this article, after the introduction and a section with definitions,consists of the following major parts.In Section 3, the flow function for a number-conserving cellular automaton is definedand the flow conditions are derived: conditions on the flow functions that are satisfiedif and only if the functions belong to a number-conserving cellular automaton.In Section 4, all solutions to the flow conditions are found. Examples and a diagramnotation for the flow function follow in the next section.In Section 6, a lattice structure for the set of flow functions is found. This leads tothe notion of a set of “minimal” flow functions from which all other flows can be builtwith the help of lattice operations. A recipe for minimal flow functions is found andexamples for minimal flows are shown. 2wo sections, one about related topics and one with open questions, follow at theend. A one-dimensional cellular automaton is a discrete dynamical system; its states areconfigurations of simpler objects, the cells . The cells are arranged in an infinite line –they are indexed by Z – and the state of each cell is an element of a finite set Σ . In an number-conserving cellular automaton, we picture each cell as a container fora certain number of particles; the number of particles it contains is part of its state.Since the number of states is finite, there is a maximal number C of particles that acell may contain, the capacity of the cellular automaton. We therefore have for eachcell state α ∈ Σ a number α ∈ { , . . . , C } , the particle content of α . The expression c α = C − α (1)stands for the complement of the particle content: It is the maximal number of particlesone can still put into a cell of state α without exceeding its capacity.We leave it open whether there are different states with the same particle content,but for simplicity we will assume that for every n ∈ { , . . . , C } , there is a state α with α = n .Often we will take the numbers { , . . . , C } directly as cell states. Such a state setis called a minimal set of states. Even for then, we will keep the distinction between α and α , to make it clearer whether we are speaking of cells or of the number ofparticles in them. We will need to speak about finite and double infinite sequences of cell states. Finitesequences are also called strings , while the infinite sequences are configurations .A string is an element of Σ ∗ = (cid:83) k (cid:62) Σ k . If a ∈ Σ k is a string, then k is its length ,and we write | a | for the length of a . For the string of length 0 we write (cid:15) .Often we write a string a ∈ Σ k as a product of cell states, in the form a = a . . . a k − . (2)The number of particles in a and its complement are then a = k − (cid:88) i =0 a i and c a = | a | C − a . (3)We will also need substrings of a . They are specified by start point and length, inthe form a m : n = a m . . . a m + n − . (4)3 configuration of a cellular automaton is a doubly infinite sequence of cell states,in the form a = . . . a − a − a − a a a a . . . . (5)The conventions for substrings and particle content are the same for configurationsas for strings, except that a is only defined when the number of the a i that have anon-zero content is finite. Such a configuration is said to have a finite particle content . The configuration of a cellular automaton changes over time. This behaviour – the evolution of the automaton – is specified by the global transition rule ˆ ϕ , which mapsthe configuration at time t to the configuration at time t + 1 . I will now define ˆ ϕ in aslightly more general form than usual.As a function, ˆ ϕ is determined by two integers r and r and a local transition rule ϕ : Σ r + r +1 → Σ , subject only to the condition that r + r (cid:62) . The numbers r and r are the left and right radius of ϕ . (Almost always, we will also require that r and r are nonnegative numbers. The only exceptions are the shift rules in theexample below.)Now we can express the value ˆ ϕ ( a ) for an arbitrary configuration a by the conditionthat ˆ ϕ ( a ) x = ϕ ( a x − r , . . . a x + r ) for all x ∈ Z . (6)If we want to use the colon form for substrings that is defined in (3), we can writethis rule also in the form ˆ ϕ ( a ) x = ϕ ( a x − r : r + r +1 ) . This form, which emphasises thelength of the local neighbourhood but is less symmetric, will be used soon.In the conventional setup of working with cellular automata, one only considers thesymmetric case with r = r = r , where r is called the radius of the transition rule. When we create a new transition rule ϕ v by replacing the radii r and r of ϕ with r (cid:48) = r + v and r (cid:48) = r − v , the behaviour of the new transition rule ˆ ϕ v does not differsignificantly from that of ˆ ϕ , except for a horizontal shift that occurs at each time step.It has then namely the dynamics ˆ ϕ v ( a ) x = ϕ ( a x − r − v : r + r +1 ) , or equivalently ˆ ϕ v ( a ) x + v = ϕ ( a x − r : r + r +1 ) for all x ∈ Z . (7)At every time step, the rule ˆ ϕ v moves therefore the new cell states v positions moreto the right than ˆ ϕ . We can say that ˆ ϕ v is ˆ ϕ as seen by an observer who moves withspeed v to the left.All rules ϕ v can be treated in essentially the same way, independent of the value of v . We therefore introduce a new parameter (cid:96) = r + r to characterise ϕ . The number (cid:96) is, anticipatingly, called the flow length of ϕ . At this point, a parameter with a value of r + r + 1 might look more natural, but we will soonsee that r + r occurs more often in our calculations. hift rules The simplest example for rules with the same dynamics are the shift rules ˆ σ k , which exist for all k ∈ Z . The rule ˆ σ k moves in each time step the states of allcells by k positions to the right.They have r = k , r = − k (therefore (cid:96) = 0 ) and a local transition rule of the form σ k : Σ → Σ . Since all shift rules have the same dynamics, all σ k must be the samefunction: It is the identity. All ˆ σ k are trivially number-conserving, independent of theparticle contents of the cell states.This is not the only representation of the shift rules as cellular automata. We willlater see another representation for the rules ˆ σ k with k (cid:62) . A formal definition for number conservation is still missing. We use this one:A transition function ˆ ϕ is number-conserving if ϕ ( a ) = a for everyconfiguration a with finite particle content.Now consider the following setup: . . . x − x − u . . . u r − (cid:12)(cid:12) u r . . . u (cid:96) − y (cid:96) y (cid:96) +1 . . . . (8)It shows the cells of a configuration, divided into three regions: There is an infiniteregion x at the left, an infinite region y at the right, and at the centre a sequence of (cid:96) cells, u . The vertical bar represents another subdivision, that into the left and the right side of the configuration. Note that u consists of r cells left of the division and r cells at its right.Now we apply a number-conserving transition rule to this configuration and lookat the amount by which the number of particles changes. Let L be the change of thenumber of particles at the left side, and R that at the right side. A positive value of R stands therefore for a particle flow to the right. Since the particle number is conserved,we have R = − L .The interaction in a cellular automaton is local, therefore L can only depend on x and u , and R only on u and y . But R = − L , therefore they both can only dependon u . So we can find a function f : Σ (cid:96) → Z (9)with f ( u ) = R = − L . This is the function which will allow us to describe the essentialproperties of a number-conserving cellular automaton. We call f the flow function of ϕ .Next we try to reconstruct ϕ from f . To do this, we consider the following setupwith (cid:96) + 1 named cells and two boundaries, . . . w . . . w r − (cid:12)(cid:12) w r (cid:12)(cid:12) w r +1 . . . w (cid:96) . . . . (10)We are interested in the state of the cell at the centre in the next time step. Initially,it contains w r particles. One step later, there are ϕ ( w ) particles at this place.5n the other hand, during this transition, f ( w (cid:96) ) particles must have entered the cellregion through the left boundary, and f ( w (cid:96) ) of them must have left it to the right.The number of particles at the centre at the next time step must therefore be ϕ ( w ) = f ( w (cid:96) ) + w r − f ( w (cid:96) ) . (11)With this equation, applied to all neighbourhoods w ∈ Σ (cid:96) +1 , we can partially recon-struct ϕ from f . If the state set Σ is minimal, the transition function can even bereconstructed uniquely from the values of ϕ ( w ) . Otherwise, there are several differ-ent transition functions for the same flow function. (How they are related would bethe subject of another paper.)Since ϕ can be derived from f , it will now be enough to consider f alone. Wetherefore need to find all functions f : Σ (cid:96) → Z for which there is a valid transition rule ϕ . These are exactly those functions f for which the right side of (11) is neither toosmall nor too large, or, more precisely, (cid:54) f ( w (cid:96) ) + w r − f ( w (cid:96) ) (cid:54) C for all w ∈ Σ (cid:96) +1 . (12)These inequalities are the flow conditions . A number-conserving cellular automatonfor f exists if and only if they are satisfied. Great flows have little flows next to them to guide ’em,And little flows have lesser flows, but not ad infinitum.
We will now restrict our work to rules with r = (cid:96) and r = 0 , so that particles onlymove to the right. This will make induction arguments on the neighbourhood sizesimpler.The flow conditions have then the form (cid:54) f ( w (cid:96) ) + w (cid:96) − f ( w (cid:96) ) (cid:54) C for all w ∈ Σ (cid:96) +1 . (13)To find solutions for them, we define some functions related to f , the half-flows . Thereare two families of them, the lower and the upper half-flows , given by the equations f k ( v ) = min { f ( uv ) : u ∈ Σ (cid:96) − k } , (14a) f k ( v ) = max { f ( uv ) : u ∈ Σ (cid:96) − k } , (14b)with (cid:54) k (cid:54) (cid:96) and v ∈ Σ k . (Note that f and f are in fact constants.) Much moreuseful are however the inductive forms of these definitions, f (cid:96) ( w ) = f ( w ) , f k ( v ) = min { f k +1 ( αv ) : α ∈ Σ } , (15a) f (cid:96) ( w ) = f ( w ) , f k ( v ) = max { f k +1 ( αv ) : α ∈ Σ } , (15b)with w ∈ Σ (cid:96) , α ∈ Σ and v ∈ Σ k for (cid:54) k (cid:54) (cid:96) .6 .1 Properties of the half-flows The most important consequence of (15) is the following lemma which, shows how thehalf-flows are related among each other.It especially shows that for (cid:54) k (cid:54) (cid:96) , the flow conditions (13) split into pairs ofinequalities, (cid:54) f k ( w k ) + w k − f k ( w k ) , (16a) f k ( w k ) + w k − f k ( w k ) (cid:54) C. (16b)In the lemma, these inequalities are contained in (17): We can get them by removingthe central terms of (17a) and (17b) and rearranging the remaining inequalities. Lemma 1 (Interaction of half-flows) . Let (cid:54) k < (cid:96) and w ∈ Σ k +1 . If f is the flowfunction of a number-conserving cellular automaton, then f k ( w k ) (cid:54) f k +1 ( w ) (cid:54) f k ( w k ) + w k , (17a) f k ( w k ) − c w k (cid:54) f k +1 ( w ) (cid:54) f k ( w k ) . (17b) Proof.
The two pairs of inequalities can be proved independently of each other, so webegin with (17a).Its proof is a finite induction from k = (cid:96) down to k = 1 . The induction step consistsof showing that that from f k ( w k ) (cid:54) f k ( w k ) + w k for all w ∈ Σ k +1 (18)always follows f k − ( w (cid:48) k − ) (cid:54) f k ( w (cid:48) ) (cid:54) f k − ( w (cid:48) k − ) + w (cid:48) k − for all w (cid:48) ∈ Σ k . (19)The induction can begin because for k = (cid:96) , inequality (18) is equivalent to the left sideof the flow condition (13).To prove the induction step, note that the left side of (18) is independent of thechoice of w . At the right side of (18), we can therefore replace the term f k ( w k ) withthe smallest possible value it can take when we vary w and keep w k − fixed. Theresult is is f k − ( w k − ) , and we get f k ( w k ) (cid:54) f k − ( w k − ) + w k . This is alreadythe right inequality of (19); we only need to write w k as w (cid:48) . The left inequality, f k − ( w (cid:48) k − ) (cid:54) f k ( w (cid:48) ) , follows directly from the inductive definition (15a) of f k .The proof of (17b) is similar. We only need to replace f with f , β with − c β and revert the order of the terms in the inequalities.The next lemma is about the behaviour of f k ( v ) and f k ( v ) for a single value of v . Lemma 2 (Bounds on the half-flows) . Let f be the flow function of a number-conserving cellular automaton. Then (cid:54) f k ( v ) (cid:54) v, (20a) (cid:54) f k ( v ) (cid:54) ( (cid:96) − k ) C + v (20b)7 nd f k ( v ) (cid:54) f k ( v ) (cid:54) f k ( v ) + ( (cid:96) − k ) C (21) are valid for (cid:54) k (cid:54) (cid:96) and v ∈ Σ k .Proof. The first two inequalities follow directly from the definitions of f k ( v ) and f k ( v ) in (14). For both half-flows, we have to consider all flows f ( uv ) with u ∈ Σ (cid:96) − k , i. e.the following setup: . . . u . . . u (cid:96) − k v . . . v k (cid:12)(cid:12) . . . (22)Only the particles in u and v may cross the boundary, and they may only move tothe right. This means that all possible values for f ( uv ) are non-negative, and thesame is true for f k ( v ) and f k ( v ) . This proves the two lower bounds in (20). For theupper bound on f k ( v ) we note that the set of all f ( uv ) includes the case with u = 0 .Then at most v particles may cross to the right, which means that also f k ( v ) (cid:54) v .On the other hand it is also possible that all cells in u contain C particles and that u = ( l − k ) C . Then ( (cid:96) − k ) C + v particles may cross the boundary, which explainsthe upper bound for f k ( v ) .The left side of (21) is clear; the right side can be proved with another setup: . . . v . . . v k (cid:12)(cid:12) w . . . w (cid:96) − k (cid:12)(cid:12) . . . (23)Here we assume that we know f ( vw ) and try to find upper and lower bounds forthe number of particles that cross the left boundary. The highest possible number ofparticles is f ( vw ) + c w , because then the w region will be completely filled up in thenext time step. The smallest particle flow is f ( vw ) − w , because then the w regionwill be empty. This means that f k ( v ) (cid:54) f ( vw ) + c w and f k ( v ) (cid:62) f ( vw ) − w .When we subtract these inequalities, we get f k ( v ) − f k ( v ) (cid:54) c w + w = ( (cid:96) − k ) C ,from which the right side of (21) follows. These two lemmas provide us with enough information to solve the flow conditions.We do it step by step, by constructing a sequence of half-flows that converges to aflow.
Theorem 1 (Flow construction) . All solutions of the flow conditions (13) can befound by constructing a sequence of half-flows f , f , f , f , . . . , f (cid:96) , f (cid:96) , (24) that satisfies the conditions of Lemma 1 and 2. Then f (cid:96) = f (cid:96) , and the function f = f (cid:96) = f (cid:96) is the constructed flow.This especially means that it is always possible, given f k and f k , to find functions f k +1 and f k +1 that satisfy the conditions and that all half-flow sequences lead to aflow. roof. In Lemma 1 and 2 we have already seen that for every flow there is a sequenceof half-flows that satisfies the required inequalities. Therefore, all possible flows canbe reached by the construction of the theorem.It remains to show that the construction can always be completed. We will nowfollow it step by step and show that each step is always possible.The construction starts with the choice of two constants f and f that satisfy theconditions of Lemma 2. For k = 0 , they reduce to f = 0 and (cid:54) f (cid:54) (cid:96)C, (25)so the first step of the construction is always possible.Now assume that k < (cid:96) , that the half-flows f k and f k are already constructedand that we want to construct f k +1 and f k +1 . We must then find solutions for theinequalities of Lemma 1 and 2. I will now show them again, in the order in which wewill try to find a solution for them.The result is the following system of inequalities. For any v ∈ Σ k +1 it contains allconditions on f k +1 ( v ) and f k +1 ( v ) . f k ( v k ) (cid:54) f k +1 ( v ) (cid:54) f k ( v k ) + v k , (26a) (cid:54) f k +1 ( v ) (cid:54) v, (26b) f k ( v k ) − c v k (cid:54) f k +1 ( v ) (cid:54) f k ( v k ) (26c) (cid:54) f k +1 ( v ) (cid:54) v + ( (cid:96) − k − C (26d) f k +1 ( v ) (cid:54) f k +1 ( v ) (cid:54) f k +1 ( v ) + ( (cid:96) − k − C, (26e)There are | Σ | k +1 of such systems of inequalities, but each pair of f k +1 ( v ) and f k +1 ( v ) occurs in only one them. The values of f k +1 ( v ) and f k +1 ( v ) can therefore be chosenindependently for each v .First we note that each of the requirements in (26) can be satisfied when taken atits own. The only cases in which this is not obvious are (26a) and (26c). But whenwe remove the central term in both of them, a half-flow condition (16) for f k or f k remains, which is true by induction. The upper and lower bounds on f k +1 and f k +1 in each of the two requirements are therefore consistent with each other and we canfind solutions for them.Next we verify that all requirements in (26) except the right inequality in (26e) canbe satisfied together. We do that by choosing f k +1 ( v ) = f k ( v k ) , f k +1 ( v ) = min { f k ( v k ) , v + ( (cid:96) − k − C } (27)as our candidate for a solution. This is the “lazy” solution in which f k +1 ( v ) is as smalland f k +1 ( v ) as large as possible. The first two requirements in (26) are then clearlysatisfied. For the next two, we need only to verify the left inequality of (26c). It is9quivalent to the two inequalities f k ( v k ) − v ck (cid:54) f k ( v k ) , f k ( v k ) − v ck (cid:54) v + ( (cid:96) − k − C . (28)The left inequality is part of (26c) and true by induction. For the right inequality,we add on both sides c v k and get f k ( v k ) (cid:54) v k + ( (cid:96) − k ) C . This is the versionof (26d) with v k instead of v and again true by induction.What remains is the right inequality in (26e). It may be violated by the solutioncandidate (27). If that is the case, we move the candidates for f k +1 ( v ) and f k +1 ( v ) stepwise towards each other until we have either found a full solution or cannot con-tinue. If we cannot continue, f k +1 ( v ) cannot be made larger or f k +1 ( v ) smaller. Thatis, one of the left inequalities of (26a) and (26b) and one of the right inequalitiesof (26c) and (26d) must become equalities, or f k +1 ( v ) = f k +1 ( v ) .Three cases can be excluded. If f k +1 ( v ) = f k +1 ( v ) , then all of (26e) is satisfied andwe have a solution to the full system of inequalities. If f k +1 ( v ) = 0 , then f k +1 ( v ) = 0 too and we have again a solution. And f k +1 ( v ) = v cannot occur without f k +1 ( v ) = f k ( v k ) + v k because by induction, f k ( v k ) (cid:54) v k and therefore f k ( v k ) + v k (cid:54) v .So we must have f k +1 ( v ) = f k ( v k ) + v k and f k +1 ( v ) = f k ( v k ) − c v k . Butthen f k +1 ( v ) = f k ( v k ) − c v k (cid:54) f k ( v k ) + ( (cid:96) − k ) C − C + v k = f k +1 ( v ) + ( (cid:96) − k − C .
So the right inequality of (26e) can always be satisfied.In other words, the step from f k and f k to f k +1 and f k +1 is always possible, and f (cid:96) and f (cid:96) can always be constructed. That they are equal follows from (21). Loose and tight half-flow systems
Note that in the construction of Theorem 1, thehalf-flows are used with fewer restrictions than in their original definition (14). Nothingin the new definition requires that the half-flows are maxima or minima of flows. Wewill soon see, in the context of Figure 2, that such less restrained flow systems actuallyoccur.Therefore we will call systems of half-flows that obey (14), tight half-flow systems,while those that do not, loose flow systems.
We will now, as an illustration for Theorem 1, construct all number-conserving ele-mentary cellular automata. 10his would lead to a quite voluminous computation if it were done directly. So Iwill at first introduce a diagram notation for all the flows and half-flows of a number-conserving cellular automaton. With these box diagrams , the computation can beshown in a reasonable amount of space.
Before we can begin to work with elementary cellular automata, I will illustrate themethod by applying it to a simpler example. We will “construct” the flows for theshift rule ˆ σ with the methods of Theorem 1. We will do it for the minimal state setwith capacity C = 1 , i. e. the set Σ = { , } , so that every cell contains at most oneparticle. The diagrams for the computations are shown in Figure 1.We can represent every shift rule σ (cid:96) by a flow function f : Σ (cid:96) → { , . . . (cid:96) } , with f ( w ) = w for all w . Such a function, with (cid:96) = 3 and C = 1 , appears at the bottomof Figure 1. In it, the dots are values of the flow function, the numbers at the left are thepossible strengths of the flows, and the strings at the bottoms are the neighbourhoods.(They are ordered lexicographically by their mirror-images.) The rightmost dot thenexpresses the fact that f (111) = 3 , and the dot left of it that f (011) = 2 .The boxes, or rather their upper and lower edges, represent the half-flows. Theinnermost box at the right, for example, stands for the values of f (11) and f (11) .It is contained in the box for f (1) and f (1) , which itself is contained in the box for f and f .The other three box diagrams in Figure 1 describe the construction of this flowfunction according to Theorem 1. It construction begins with the topmost diagramand must obey the following rules. In them, we will call a box that represents f k ( v ) and f k ( v ) for some v a k -box . • All k -boxes except the outermost must be drawn directly inside a ( k + 1) -box.This ensures that the conditions f k ( v k ) (cid:54) f k +1 ( v ) and f k +1 ( v ) (cid:54) f k ( v k ) of (26a) and (26c) are satisfied. • The maximal height of a k -box is (cid:96) − k . This ensures that (26e) is satisfied. • The upper edge of a k -box is not higher than the highest particle content of theneighbourhoods that it contains.This ensures (26d) because the k -box for f k ( v ) contains all neighbourhoods ofthe form uv with u ∈ Σ (cid:96) − k , and their highest particle content is therefore ( (cid:96) − k ) C + v , as required in the inequality. • Finally we need to make sure that the conditions f k +1 ( v ) (cid:54) f k ( v k ) + v k and f k ( v k ) − c v k (cid:54) f k +1 ( v ) of (26a) and (26c) are satisfied. This is donewith help of the thin lines with arrows that appear in the first three diagrams.There are two kinds of them, one with an upwards-pointing arrow and one witha downwards-pointing array. We will call them the up-arrow and down-arrow lines. 11
123 0 10123 00 10 01 010123 000 100 010 110 001 101 011 1110123 000 100 010 110 001 101 011 111
Figure 1: Box diagrams for the half-flows of σ .The two up-arrow lines in each innermost ( k + 1) -box in a diagram mark theminimal heights that the upper edges of the k -boxes may have that will be drawninside the ( k + 1) -box in the next step. The down-arrow lines mark the maximalheights of their lower edges in a similar way.To find out where to draw them, let us again write v as uα , with u ∈ Σ k and α ∈ Σ . Then the down-arrow line must be drawn at height f k ( u ) + α and theup-arrow line at f k ( u ) − c α . Both lines belong to the ( k + 1) -box for ( uα ) k ,since the names of the k -box grow at the left when k grows.In practice the easiest method is to keep u fixed and then draw the pairs of thinlines for each α . All line pairs have then the same distance from each other, onlytheir heights vary according to α .This is the way in which the diagrams in Figure 1 were constructed. As an example12or the most difficult part, namely the construction of the thin lines, we will now lookat the second diagram in detail. According to the recipe, we will first set u to (thatis, to the string which consists of one empty cell) and construct the thin lines for themaximal and minimal flows of the neighbourhoods and .The inner right box is the box for , and we see from it that f (1) = 0 and f (1) = 2 .When we now set α = 0 , we see that the down-line must be placed at f (1) + 0 = 0 ,and the up-line at f (1) − . And it must be placed at the left of the left innerbox, at the place where the 2-box for will be constructed in the next step. Similarly,when we set α = 1 , we see that the thin lines for must be placed at heights 1 and2, and at the left of the inner right 1-box.In the diagrams of Figure 1, the pairs of thin lines always have the maximal alloweddistance, so there is no choice at the following step. This means that the flow functionfor ˆ σ is the only one with f = 3 – an observation that can be easily extended to all ˆ σ (cid:96) with (cid:96) (cid:62) . Now we can begin with the construction of all number-conserving elementary cellularautomata. Elementary cellular automata, or ECA [9], are simply the one-dimensionalcellular automata with radius 1 and state set
Σ = { , } . They therefore have atransition rule ϕ : Σ → Σ and a flow function of width (cid:96) = 2 for capacity C = 1 . Wemust now find all number-conserving flows.This is done in Figure 2. The figure must be read from left to right and from top tobottom. The first column contains all possible box diagrams for the first step of theconstruction. To the right of each diagram and right-below of it are its refinements,the diagrams for all the possible next steps of the construction. So the second boxdiagram in the first column has four refinements, and the second of them, two.The rightmost column contains the code numbers of the ECA that belong to theflows in column 3. The code numbers were given by Wolfram [9]. If a rule is a shiftrule ˆ σ k , the name of the rule is also given. Since we have here silently switched back tothe symmetrical cellular neighbourhoods with r = r = 1 , the shift speed is differentfrom that what we would have seen in the context of Figure 1.Some rule names are put in braces. They belong to rules that occur more thanonce in the diagram, constructed in different ways. For an unknown reason it is onlyRule 204, the identity function, that occurs more than once. But there is only oneconstruction of Rule 204 that is tight; it is the construction that is not put in braces.One new phenomenon appears here: It can happen that the up-arrow line is belowthe down-arrow line. This happens the first time in the construction of Rule 184. Itis the reason why the boundaries on f and f are drawn with an arrow to distinguishthem, and not just as thin lines.The result of this calculation is that the only number-conserving ECA rules are theshift rules ˆ σ − , ˆ σ , ˆ σ , the so-called “traffic rule” 184, and Rule 226, which is Rule 184with left and right exchanged. 13
12 0 1 00 10 01 11 00 10 01 11
Rule 240 ( ˆ σ )
01 0 1 00 10 01 11 00 10 01 11 (Rule 204)
00 10 01 11 00 10 01 11
Rule 184
00 10 01 11 (Rule 204)
00 10 01 11 00 10 01 11 (Rule 204)
00 10 01 11
Rule 226
00 10 01 11 00 10 01 11
Rule 204 ( ˆ σ ) Rule 170 ( ˆ σ − )Figure 2: Construction of all number-conserving ECA rules. We have now found all flow functions with rule width (cid:96) = 2 and for a minimal stateset Σ of capacity C = 1 . Similar computations for larger (cid:96) or Σ would soon becomeunwieldy. I will therefore now describe another method to get an overview over allflows for given (cid:96) and Σ . It is based on a lattice structure on the set of flows.The same method will also provide an alternate way to construct flow functions.14 .1 Sets of flows To keep the following arguments short (or at least not too long), we will introducesome notation. We will write F ( (cid:96), Σ) for the set of flows of width (cid:96) and state set Σ ,i. e. the set of functions f : Σ (cid:96) → { , . . . , (cid:96)C } , with C = max { α : α ∈ Σ } that satisfythe flow conditions (13).We will also write H ( (cid:96), Σ) for the set of half-flows related to (cid:96) and Σ . It is the setof function families ( f k , f k ) (cid:54) k (cid:54) (cid:96) that satisfy the conditions of Lemma 1 and 2. For asequence ( f k , f k ) k , we will use the single symbol f .If Σ is a minimal state set of capacity C , we will also write these sets as F ( (cid:96), C ) and H ( (cid:96), C ) . When the intended meaning is clear, we may also write F or H . Our next step is the introduction of a partial order on F and H . We define the orderpointwise: For two flows f , g ∈ F , the relation f (cid:54) g shall be true if ∀ v ∈ Σ (cid:96) : f ( v ) (cid:54) g ( v ) . (29)For systems of half-flows f , g ∈ H , conditions similar to (29) must be satisfied for allof their half-flows to make f (cid:54) g true.The minimum f ∧ g and maximum f ∨ g of flow functions f , g ∈ F is definedpointwise too, in the form ∀ v ∈ Σ (cid:96) : ( f ∧ g )( v ) = min { f ( v ) , g ( v ) } , (30a) ∀ v ∈ Σ (cid:96) : ( f ∨ g )( v ) = max { f ( v ) , g ( v ) } . (30b)For systems of half-flows, f ∧ g and f ∧ h are defined in the same way – here theformulas of (30) are applied to all pairs of half-flows in f and g with the same type.The minimum and maximum of a set S of flows (or half-flow systems) is written (cid:86) S and (cid:87) S . For S ⊆ F we have then ∀ v ∈ Σ (cid:96) : ( (cid:94) S )( v ) = min { f ( v ) : f ∈ S } , (31a) ∀ v ∈ Σ (cid:96) : ( (cid:95) S )( v ) = max { f ( v ) : f ∈ S } , (31b)and again the formulas for half-flows are similar ut would be more complex.The following theorem shows that these definitions are useful. Theorem 2.
With the operations ∧ and ∨ as defined above, the sets F ( (cid:96), Σ) and H ( (cid:96), Σ) each form a distributive lattice.Proof. We will prove the theorem for F and then sketch a similar argument for H .First we note that the set { , . . . , C } Σ (cid:96) of functions from Σ (cid:96) to { , . . . , C } is adistributive lattice. This is because { , . . . , C } , as a linear order, is distributive, and { , . . . , C } Σ (cid:96) , as a product of distributive lattices, is so too [3, Proposition 4.8].15t is therefore enough to show that F is closed under minimum and maximum; thenit is a sublattice of the full function space and also distributive [3, Section 4.7].To do this, let f , g ∈ F be two flows and h = f ∧ g . We have then to verify thatthe following inequalities are true for all v ∈ Σ (cid:96) and w ∈ Σ (cid:96) +1 : h ( v ) (cid:62) , (32a) h ( v ) (cid:54) v, (32b) h ( w (cid:96) ) − h ( w (cid:96) ) (cid:54) w (cid:96) , (32c) h ( w (cid:96) ) − h ( w (cid:96) ) (cid:54) c w (cid:96) . (32d)The first two inequalities control the upper and lower bounds on h and are clearlysatisfied. The other two represent the flow conditions (13) for h . We first look at (32c).The two flow values at its left side each stand either for a value of f or of g . If bothbelong to the same flow, say f , then (32c) is actually the same inequality for thatflow and therefore true. The only interesting case is therefore that in which the two h values stem from different flows. Without loss of generality we may assume that thefirst one is from f and the second from g , so that we need to find an upper bound for f ( w (cid:96) ) − g ( w (cid:96) ) . But since h ( w (cid:96) ) = f ( w (cid:96) ) , we must have f ( w (cid:96) ) (cid:54) g ( w (cid:96) ) andcan calculate f ( w (cid:96) ) − g ( w (cid:96) ) (cid:54) g ( w (cid:96) ) − g ( w (cid:96) ) (cid:54) w (cid:96) , (33)so (32c) must be true. Inequality (32d) has the same form and can be proved in thesame way. This proves that f ∧ g ∈ F .When we set instead h = f ∨ g , the proof of (32c) is a bit different. Now we concludeinstead from h ( w (cid:96) ) = g ( w (cid:96) ) that g ( w (cid:96) ) (cid:62) f ( w (cid:96) ) , and (33) becomes f ( w (cid:96) ) − g ( w (cid:96) ) (cid:54) f ( w (cid:96) ) − f ( w (cid:96) ) (cid:54) w (cid:96) . (34)The rest of the argument is the same, and we have now proved that F is a distributivelattice.The space H of half-flows is a subset not of a single function space but the product ofseveral such spaces, one for each half-flow function. This product is still a distributivelattice. Therefore it again is enough to check whether it is closed under ∧ and ∨ .There are a lot more inequalities to consider, but they all can be brought to one of thethree forms h k ( w ) (cid:54) K, (35a) h k ( w ) (cid:62) K, (35b) h j ( v ) − h k ( w ) (cid:54) K . (35c)In these inequalities, the h terms stand either for h or h , with possibly different choicesin the same inequality, K is a constant that does not depend on the half-flow functions,and v ∈ Σ j and w ∈ Σ k . These inequalities have the same form as those for F , thereforethe same arguments can be used, and H too is a distributive lattice.16 .3 Minimal flows as building blocks for all the flows Now, with the lattice structures of F and H , we can use a subset of all flows as buildingblocks for the rest. These are the minimal flows m ( v, k ) = (cid:94) { f ∈ F : f ( v ) (cid:62) k } (36)with v ∈ Σ (cid:96) and k ∈ { , . . . , v } . So m ( v, k ) is the smallest flow that at the neigh-bourhood v has at least the strength k .Every flow can then be represented as a maximum of minimal flows, f = (cid:95) { m ( v, f ( v )) : v ∈ Σ (cid:96) } . (37)To see why this is so, we note that m ( v, f ( v )) is a flow that is less than or equal than f and aggrees with f when evaluated at v . This is true because for m ( v, f ( v )) , theright side of (36) becomes (cid:86) { g ∈ F : g ( v ) (cid:62) f ( v ) } . This expression is the minimumof a set of flows which contains f , therefore m ( v, f ( v )) (cid:54) f . And it is the minimum ofa set of functions whose values at v are greater or equal to f ( v ) and which contains f ,therefore m ( v, f ( v ))( v ) = f ( v ) .The first fact proves that the right side of (37) is less than or equal to f ; the secondfact, that for each neighbourhood v , the right side of (37) has a value that is not lessthan f ( v ) . So the right side of (37) must be f .Besides from being building blocks, the minimal flows are also interesting in theirown right. They answer the questions: If I require that f ( v ) = k , which influence hasthis on other neighbourhoods? Does pushing the particles forward in one place setparticles in other place in motion? We will therefore now construct the minimal flows. Theorem 3.
Let f = m ( a, k ) be a minimal flow in F ( (cid:96), Σ) and b ∈ Σ (cid:96) . Then f ( b ) isthe smallest non-negative number which satisfies the inequality f ( b ) (cid:62) k − min {| u | , | u (cid:48) |} C − w − c w (cid:48) , (38) for all u , v , w , u (cid:48) , w (cid:48) ∈ Σ ∗ with a = uvw and u (cid:48) vw (cid:48) = b .Proof. According to Theorem 1, m ( a, k ) can be constructed from half-flows. We takenow the minimum of all half-flow sequences that lead to m ( a, k ) , i. e. the system M ( a, k ) = (cid:94) { f ∈ H : f (cid:96) ( a ) = f (cid:96) ( a ) (cid:62) k } . (39)It consists of a sequence of half-flows that leads to m ( a, k ) “greedily”: M ( a, k ) is asequence ( f i , f i ) (cid:54) i (cid:54) (cid:96) of half-flow pairs in which every value f ( v ) and f ( v ) is thesmallest possible for which the construction sequence can still end in m ( a, k ) . We willnow construct such a sequence.To do this, we will use the inequalities of Lemma 1 and 2, but in a much morecompressed form. In a first simplification, we write them as f ( w ) (cid:54) f ( vw ) (cid:54) f ( v ) + c w, (40a) f ( v ) − c w (cid:54) f ( vw ) (cid:54) f ( w ) , (40b) f ( v ) (cid:54) f ( v ) (cid:54) f ( v ) + ( (cid:96) − | v | ) C, (40c)17nd they are valid for all v , w ∈ Σ ∗ with | vw | (cid:54) (cid:96) .In two aspects, these inequalities differ from their representation in Lemma 1 and 2:(a) The indices on f and f are dropped, since they can be derived from their arguments,and writing them would make the formulas only more complicated. (b) In the first twolines, the formulas have been iterated and the variables renamed . From the inequality f k +1 ( w ) (cid:54) f k ( w k )+ w k in (17a), valid for w ∈ Σ k +1 , we get by induction f k + n ( w ) (cid:54) f k ( w k ) + w k : n for w ∈ Σ k + n , and then, after renaming w k and w k : n to v and w ,the right side of (40a). The other derivations are similar.You may also have noticed that conditions (20) of Lemma 2 have disappeared. Theyare less important. Their left parts state that all half-flows must be non-negative,which we will keep in mind, while their right parts contain upper bounds to the half-flows. We will not need them because in our construction all half-flows are as small aspossible.In a second compression step, we now express the inequalities (40) with arrows. Wewrite an inequality f ( u ) + n (cid:54) f ( v ) , (41)as an arrow u n −−→ v, (42)and do the same for all other combinations of upper and lower bars. The number n on an arrow is called its strength , and it is omitted when n = 0 .An arrow as in (42) can be interpreted as “if f ( u ) (cid:62) x , then f ( v ) (cid:62) x + n ”. Sowhen we have a chain of arrows, like u m −−→ v n −−→ w we can add their strengths andget a new arrow, in this case u m + n −−−−→ w . Our goal are arrows a n −−→ b and similaronce because they lead to lower bounds on f ( b ) . (Note that, since | a | = | b | = (cid:96) ,we have f ( a ) = f ( a ) and f ( b ) = f ( b ) , so that upper and lower bars on a and b areinterchangeable.)When we now translate the inequalities of (40) into arrows, they become w −→ vw, vw − w −−−−→ v, (43a) v − c w −−−−−→ vw, vw −→ w, (43b) v −→ v, v ( | v |− (cid:96) ) C −−−−−−−→ v . (43c)In principle, we must consider all arrow chains from a to b . But most of them canbe replaced by stronger chains, and only two remain. The following arguments showhow this is done.1. We can ignore all chains in which two arrows of the same form occur in sequence.
An example for such a chain is vww (cid:48) − w (cid:48) −−−−−→ vw − w −−−−→ v , in which the leftarrow of (43a) occurs twice. It can be replaced with the equally strong ar-row vww (cid:48) − ww (cid:48) −−−−−−→ v . The same can be done with the other arrows in (43a)and (43b), which are the only arrows to which this rule applies.18. In the next reduction step, we consider arrow chains that consists of arrows ofthe same “type”. Two arrows have the same type if they either lead from anupper half-flow to an upper half-flow or from a lower to an lower half-flow. Anychain of same-type arrows can be brought to one of the forms vw − w −−−−−→ v −→ uv, (44a) uv −→ v − c w −−−−−→ vw (44b) without loss of strength. In other words, we can assume that a shortening arrow always occurs before alengthening one.To prove this, we first show that if a lengthening arrow is followed by a shorteningarrow, they can be rearranged without loss of strength. For lower half-flows, sucha chain must have the form (a) u −→ uxv − xv −−−−−→ v , or (b) uv −→ uvw − w −−−−→ vw . Form (a) occurs when the first arrow adds cell states that the second takesaway, while (b) occurs when there is a common substring v in the first and thirdhalf-flow of the chain.But in (a), we can remove x and get a stronger arrow chain. This chain is thespecial case of (b) with v = (cid:15) , so that we only need to consider (b). And (b) canbe replaced with (44a), so that we have proved our assertion for chains of twoarrows.Chains of more than two arrows can now be rearranged so that there is first achain of shortening and then one of lengthening arrows. But arrows of the sameform can be condensed to a single arrow, as we have seen before. So we end upagain with (44a).Chains of less than two arrows can be extended by adding “empty” arrows, sayby setting u = (cid:15) . Therefore all sequences of arrows between lower half-flows canbe brought into the form (44a). The proof for upper half-flows is similar.3. Another simplification concerns the type-changing arrows in (43c). We can as-sume that two type-changing arrows never occur directly in sequence . For if theyoccur, as in v −→ v ( | v |− (cid:96) ) C −−−−−−−→ v , we can remove them both and get an arrowchain that is at least as strong.4. Next we consider a chain of three arrows with a type-changing arrow in itscentre. As we now can conclude, they can only have the following two forms(with uv = u (cid:48) v (cid:48) or vu = v (cid:48) u (cid:48) , respectively): v −→ uv −→ u (cid:48) v (cid:48) −→ v (cid:48) , (45a) v − c u −−−−−→ vu ( | vu |− (cid:96) ) C −−−−−−−→ v (cid:48) u (cid:48) − u (cid:48) −−−−−→ v (cid:48) . (45b)But for them we can assume that either u = (cid:15) or u (cid:48) = (cid:15) and simplify the arrowchains accordingly. (In other words, do not take away what you just have added .)19he proof consists of four cases. For for each arrow chain, one must distinguishbetween | u | (cid:62) | u (cid:48) | and | u | (cid:54) | u (cid:48) | . We will look only at one case, namely thatin which | u | (cid:54) | u (cid:48) | is true in (45b). There we can write u (cid:48) = xu for a suitable x ∈ Σ ∗ , v − c u −−−−−→ vu ( | vu |− (cid:96) ) C −−−−−−−→ v (cid:48) xu − xu −−−−−→ v (cid:48) , (46)and then remove u to get a stronger chain, v −→ v ( | v |− (cid:96) ) C −−−−−−−→ v (cid:48) x − x −−−−→ v (cid:48) . (47)The other cases are similar and always lead to a new chain that is at least asstrong as the original one.5. Now we can construct the two chains that lead to inequality (38) of the theorem: a = uvw − w −−−−→ uv −→ uv −→ v − c w (cid:48) −−−−−−→ vw (cid:48) ( | vw (cid:48) |− (cid:96) ) C −−−−−−−−→ vw (cid:48) −→ u (cid:48) vw (cid:48) = b, (48a) a = uvw −→ vw ( | vw |− (cid:96) ) C −−−−−−−−→ vw − w −−−−→ v −→ u (cid:48) v −→ u (cid:48) v − c w (cid:48) −−−−−−→ u (cid:48) vw (cid:48) = b . (48b)We must collect the arrow chains for all possible u , v , w , u (cid:48) , v (cid:48) ∈ Σ ∗ to get allrequirements on f ( b ) for a given f ( a ) .The chains arise naturally once we note that a has maximal length and thattherefore the first arrow must necessarily be shortening. After that, there is onlyone possible successor for each arrow, until the arrow chain ends in b .The first chain has the weight − w − c w (cid:48) − | u (cid:48) | C and the second, − w − c w (cid:48) − | u | C , from which we can see that (38) is the right formula.6. Our argument is not yet complete. The arrow chains of (48) consist of a singlecycle of shortening and lengthening arrows, from uvw to u (cid:48) v (cid:48) w (cid:48) . What if wecreated arrow chains of more than one such cycle? Would we then get moreconditions on f ( b ) ?To resolve this question, we first need shorter expressions for the cycles. Wewrite the chains in (48) as single arrows, as uvw −| u (cid:48) | C − w − c w (cid:48) −−−−−−−−−−−−−→ u (cid:48) v (cid:48) w (cid:48) , (49a) uvw −| u | C − w − c w (cid:48) −−−−−−−−−−−−−→ u (cid:48) v (cid:48) w (cid:48) . (49b)The two cycles have essentially the same form, so it will be enough to consideronly the first one.When we now connect two arrows of the form (49a), the result can always bewritten as uxyzw −| u (cid:48) | C − w − c w (cid:48) −−−−−−−−−−−−−→ u (cid:48) xyzw (cid:48) −| u (cid:48)(cid:48) | C − zw − c w (cid:48)(cid:48) −−−−−−−−−−−−−−−→ u (cid:48)(cid:48) yw (cid:48)(cid:48) . (50)20his is because the action of each arrow can be understood as taking away cellstates from both sides of the string at its left and then adding others, resultingin the string at its right. (In (49), the regions u and w are removed and u (cid:48) and w (cid:48) then added.) A region in the centre is left unchanged. With two arrows, theunchanged region of the first arrow might be shortened by the second. In (50)we therefore have assumed, that xyz is the unchanged region of the first and y that of the second arrow.The strength of the arrow chain in (50) is −| u (cid:48) w (cid:48) u (cid:48)(cid:48) | C − zw − c w (cid:48)(cid:48) . But wecan achieve the same result with a single arrow, uxyzw −| u (cid:48)(cid:48) | C − zw − c w (cid:48)(cid:48) −−−−−−−−−−−−−−−→ u (cid:48)(cid:48) yw (cid:48)(cid:48) . (51)Its strength differs from that of (50) by −| u (cid:48) w (cid:48) | C , which is never a positivenumber. This means that we can replace (50) with (51) in a chain of arrows and,by induction, that a single cycle (49) is enough.7. Finally, what about shorter chains? One could omit the type-changing arrowsand get the chains of the form a = vw − w −−−−→ v −→ u (cid:48) v = b, (52a) a = uv −→ v − c w (cid:48) −−−−−−→ vw (cid:48) = b, (52b)which look as if they could be stronger that those in (48). But in fact they arejust special cases of these chains. One can e. g. see that (52a) is just (48b) with u = w (cid:48) = (cid:15) . Recall that a = a and b = b and note that, since | vw | = (cid:96) , thearrow vw ( | vw |− (cid:96) ) C −−−−−−−−→ vw in (48b) has strength 0. In the same way one can seethat (52b) is a special case of (48a).We have now shown that all relations between f ( b ) and f ( a ) derive from the arrowchains in (48). Therefore the conditions in (38) define m ( a, k ) . With Theorem 3, we can now find examples for minimal number-conserving automata.As before with the elementary cellular automata, we use the minimal state set
Σ = { , } . So we have C = 1 , and every cell may contain at most one particle. Even withthese restrictions, we can find cellular automata with an interesting behaviour. The influence of the particle density
An example is m (0110 , in Figure 3. In thefigure, cells in state 0 are displayed in white, and cells in state 1 in black. Time runsupward, and the line at the bottom is a random initial configuration. This proof is a bit long. Another way to prove this theorem – possibly shorter but less natural –is to derive the inequalities (38) only as necessary conditions and then to verify that the function f defined by them satisfies f ( a ) (cid:62) k and the flow conditions. If the conditions (38) were notsufficient, f would be smaller than all the flows g with g ( a ) (cid:62) k and therefore not be a flow. ensity = 0 . . . . Figure 3: Rule m (0110 , at different densities.In an ordinary cellular automaton, one can expect that the number of ones and zeroesin a random initial configuration has no great influence on the patterns that arise aftera few generation. If necessary, the fraction of cells in state 1 changes over time. Withnumber-conserving automata this is no longer true. Figure 3 therefore contains fourevolutions of m (0110 , with different densities: the density is the fraction of cells instate 1 in the initial configuration – which then stays constant during the evolution ofthe cellular automaton.Especially in the two low density evolutions (with density = 0 . and 0.2) one cansee that particles move with three possible speeds: Isolated particles move with speed1, blocks of two particles, like 0110, move with speed 3, and blocks of three particlesmove 5 steps over two generations (from 0111000000 via 0001101000 to 0000001110)and therefore have a speed of 2.5. For very low densities like 0.1, the evolution isinitially dominated by non-interacting single particles, but we can already see howthey are collected by the faster structures and integrated into their particle stream. Inthe density 0.2 image we can see how the fast structures interact: When a middle speedparticle group interacts with a fast one, a short “traffic jam” of high particle density22 (11010 ,
2) at density 0.25 m (01110 ,
2) at density 0.55 m (01100 ,
2) at density 0.65 m (01100 ,
1) at density 0.65
Figure 4: Some minimal rules.occurs, but then the two particle groups separate again. (Note that the middle-speedgroups before and after the interaction consist of different particles.) With density 0.5,the traffic jams are more common, and also highly regular structures between them.With the moderately high density of 0.6, all interesting behaviour stops early, and theautomaton looks like σ .So m (0110 , already establishes a vaguely traffic-like behaviour, except that thespeed of a particle also depends on the location of the particles next to it. Other phenomena
Figure 4 illustrates other phenomena that occur with minimalnumber-conserving cellular automata.Often, especially at low densities, nothing or almost nothing happens. The evolutionof rule m (11010 , at the top left of the figure is an example. Most of the time, almostall particles are arranged in a pattern in that no non-zero flow arises. From time totime, a disturbance moves over the cells.Rule m (01110 , , at the top right, consists too of times of inactivity and mov-23ng high-densitiy regions. The high-density regions have however their own intricatestructure.The evolution of rule m (01100 , , at the bottom left, has large regions of the highestpossible denity, and the disturbances in them sometimes look like “anti-particles”:periodic structures of emptiness between particles.At the bottom right, the related rule m (01100 , shows a differenent pattern ofalmost regular high-density regions. The disturbances in them show a complex, tree-like pattern. So far, the possibility that two or more states have the same particle content hasrarely been addressed. Such non-minimal state sets have however an important appli-cation: They enable the construction of non-deterministic number-conserving cellularautomata.In a non-deterministic cellular automaton, the value of the transition function ϕ isa set of states, and the next state of a cell with may be any element of the value of ϕ ,when applied to its neighbourhood. Instead of (6), we have ˆ ϕ ( a ) x ∈ ϕ ( a x − r , . . . a x + r ) for all x ∈ Z . (53)If we then have constructed a flow function f for a non-minimal state set, we canconstruct a non-deterministic number-conserving rule with the help of a relaxed formof (11), namely ϕ ( w ) ⊆ { α ∈ Σ : α = f ( w (cid:96) ) + w r − f ( w (cid:96) ) } . (54)Such a function is clearly number-conserving, since all possible choices for the nextstate of a cell have the same particle content.The condition that for each w ∈ Σ (cid:96) +1 , all elements of ϕ ( w ) must have the sameparticle content is also necessary: If there are α , β ∈ ϕ ( w ) with α (cid:54) = β , we canuse a configuration a which contains w as a substring to construct a counterexample.Among the possible successor configurations of a in the next time step, there mustbe two that only differ at one cell, which is in one configuration in state α and inthe other one in state β . Since α (cid:54) = β , number conservation cannot hold for bothconfigurations. We now return to the case where r and r are arbitrary nonnegative numbers. Theflow functions for such two-sided neighbourhoods are related in a very simple way tothe one-sided neighbourhoods that we have so far investigated.24et ˆ ϕ be a global transition rule with arbitrary r and r and let ˆ ϕ (cid:48) be the transitionrule with a one-sided neighbourhood that is related to it. Let f and f (cid:48) be their flowfunctions. Then we can write ˆ ϕ = ˆ ϕ (cid:48) ◦ ˆ σ − r . (55)This is because we can get the effect of ˆ ϕ by first moving the content of all cells by r positions to the left and then applying ˆ ϕ (cid:48) . The corresponding equation for flows is f ( uv ) = f (cid:48) ( uv ) − v, (56)for all u ∈ Σ r and v ∈ Σ r : The shift ˆ σ − r produces an additional flow of − v particles over the boundary between u and v .As a corollary of (56), we see that the new two-sided flows obey the same partialorder as the one-sided do. If g is another flow with radii r and r and f (cid:48) is its one-sidedequivalent, then f (cid:54) g iff f (cid:48) (cid:54) g (cid:48) , (57)as we easily can conclude from (56) and the definition of the partial order in (29).The theory of minimal flows for two-sided neighbourhoods is therefore isomorphicto that for one-sided neighbourhoods. As a good notation for two-sided minimial flowsI would propose m ( u, v ; k ) = m ( uv, k ) − v . (58) The set of all number-conserving one-dimensional cellular automata has, as we haveseen, an intricate structure. It leads to many open questions, of which I will list a few,with comments:1. How many number-conserving automata are there for a given state set Σ andflow length (cid:96) ?2. How many flow functions are there for given Σ and (cid:96) ?The answers to these questions are the same if Σ is a minimal state set. Forboth questions, an explicit formula as answer is probably very complex. Anasymptotic formula could be easier to find and might provide more insight.Boccara and Fukś [2] have already found that for Σ = { , } , there are 5 rulesfor (cid:96) = 2 , 22 rules for (cid:96) = 3 and 428 for (cid:96) = 4 .3. If two number-conserving automata have the same flow function, how is theirbehaviour related?An answer to this question would provide insight into the relation between cel-lular automata and their flow functions. It would also be helpful for the betterunderstanding of non-deterministic number-conserving automata.25
12 0000 1000 0100 1100 0010 1010 0110 1110 0001 1001 0101 1101 0011 1011 0111 1111012 0000 1000 0100 1100 0010 1010 0110 1110 0001 1001 0101 1101 0011 1011 0111 1111
Figure 5: Two incomparable flows, m (0101 , and m (0011 , . m (0101 , m (0011 , Figure 6: The flows m (0101 , and m (0011 , at density 0.4.4. If the behaviour of the automata with flow functions f and g is known, what canbe said about those with flows f ∧ g and f ∨ g ?Ideally, the lattice structure of the flows would provide information about thecellular automata. This problem should first be investigated for minimal statesets, since otherwise it would require an answer to the previous question.5. Given f k and f k , what can be said about ˆ ϕ ?The pairs ( f k , f k ) provide a classification of number-conserving cellular au-tomata, and we should expect that they group automata with similar behaviour.But what does “similar” mean in this context?An argument similar to that for Figure 1 gives a partial result: f = (cid:96)C enforcesthat ˆ ϕ is the shift function ˆ σ (cid:96) . 26. What kind of lattices are the flow sets F ( (cid:96), Σ) ?We can see in Figure 2 that F (2 , { , } ) is a linear order. But there are larger val-ues of (cid:96) for which F ( (cid:96), { , } ) does contain incomparable elements. One exampleis shown in Figure 5 and 6.7. What does the theory for more than one kind of particle look like?There are two types of multi-particle automata. In automata of the first type,each cell contains several containers, each for one kind of particle, and the parti-cles only move between “their” containers. This kind of automaton has a theorythat is a straightforward extension of the one-particle theory.In automata of the second type, there is only one container in each cell andseveral kinds of particles that must fit together into such a container. Here thetheory will be more complex.8. What about particles in higher dimensions?The derivation of Theorem 1 only works in one dimension. For higher-dimensionalcellular automata therefore new ideas are needed.9. Are there practical or theoretical applications for this theory?With practical applications I mean e. g. simulations of physical systems. A the-oretical application could be the construction of a universal number-conservingcellular automaton or something similar.10. Can the theory be simplified?In the proofs and calculations of this paper, a small number of types of inequal-ities are used over and over again. Is there a theory with which the repetitionscan be compressed into a few lemmas at the beginning, such that the actualproofs take only (say) two pages? This would also be helpful for multi-particleand higher-dimensional systems. Acknowledgement
I would like to thank Barbara Wolnik for helpful comments onan earlier version of this text.
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