Complex Systems 3 (1989) 615- 633

Particles in Soliton Cellular Automata

Athanassios S. FokasEl ena P. Papadopoulou

Yiannis G. SaridakisDepartment of Math ematics and Computer Scien ce,

andInstitute for Nonlinear St udies,

Clarkson University, Potsdam, NY 13676, USA

Abstract . An initial configurat ion of the filter cellular automatonintroduced by Park, Steiglitz, and Thurston can be thought of as con­sisting of a number of particles. Here we present a complet e analysisof the evolution of single particles. In particular, conditions are givenfor th e existence of periodic particles, and the period is computed interms of the initial data.

1. Introduction

We study a particular class of cellula r automata [1 ] called filt er automat a,recently introduced in [2] and [3]. This cellular automaton cons ists of acollection of zeros or ones, which evolve in time according to t he followingrule:

Let

at : ... Oa~ . .. a: .. . aiO . . . , L < 00 (1.1)

be the state at time t, where a; = 0 or 1 for all i and a~ = 1, at 1 aret he firs t and the last 1 respectively in at . Then the next state is calculated ,sequentially from left to rig ht, as follows:

HI [ 1, sevenai = 0, s odd or zero

1 r

h -=- " H I " twere s - ~ai_j +~ai+j'j = r j =O

(1.2)

(one assumes that a;+l = 0 for i far enough to the left. In the above, r is afixed integer called the radius and r ~ 2).

@ 1989 Complex Systems Publications , Inc.

616 A .S. Fokas, E. Papadopou lou, and Y. Saridakis

As an illustration of the above "parity rule," suppose that r = 4, thena;+l dep ends only on the following sites (note the convention that t increasesdownward s)

i + 1 i +2 i+3 i+ 4

• • • •••

• • • •i -4 i-3 i-2 i-I

An example for r = 4 is shown in figur e 1. It was shown numericallyin [2,3] t hat this filter cellu lar automaton has the remarkable pro pe rty ofexhibit ing solitonic behavior. Namely, if one thinks of a state as a collect ionof "part icles," for most of the initi al st ates considered in t he numerical ex­periments of [2,3], the particles emerging after interaction were identical tothe original ones.

An example of such an int eraction is given in figure 2. Indeed, if one con­siders the state aO to consist of the particles 1001101111 and 1011, then thestate a8 indicat es that these particles interact solito nically (in this exampler = 3).

The aim of our program of study is to develop an analytical approachfor the above cellular au tomat on. Our st arting point is the work of Pap-

at : _... 0a'+': 0 0 0 1 0

oo

o1

1o

1 10000000000000101

o 1 1 0o

Figure 1: Evolution of a configuration for one time step .

t=Ot -1

t=2t=3t=4t=5t =6

t"'7t =8

100 110 11110000010 11100110100100111101100101 100011 1110011111 010110 11011 001

11111010011111011001111001011 110 101 000 11

110110 101 1000 10 1011 1101001 001 0011 011 11 11

1011 0000010011 01111

Figure 2: Solitonic int eraction of two particles.

Particles in Soliton Cellular A utomata 617

a! : •••• 0 [!] 0at+1

: 0 0 0 1 0 1 0o1

[TIo

1 1 0 [Q]o 0 0 0

0000 [TIo 1 0 1 1

o 1 1 [Q]o

Figure 3: Evolution of a configuration using FRT.

at : 0 0 0 0 [TI 0a f+1 : • • • 0 0 0

oo

[TIo

0 [Q] 0 0 000 [TI010 0 00 00

oo

[Q]1

Figure 4: Moving frame of reference.

atheodorou et al. [4], where an equ ivalent bu t more explicit ru le than (1.2)was introduced. This rule, called fast rule theorem (FRT) , consis ts of thefollowing four steps :

1. P lace the first 1 in a box.

2. P ut boxes every T + 1 bit s, unless there exist at least T +1 zeros after agiven box; in this case, put a box at the first ava ilab le 1 afte r t he zeros .

3. Change the boxes to their complements and leave the rest unchanged.

4. Displace everything by T to the left.

As an illustration of the above fast rule consider again the state at offigure 1. The one-step evolution of at using FRT is demonstrated in figure 3.

In what follows we find it convenient to use a moving frame ofreference byomi tting the step (4) of the fast ru le theorem. This affects only the absolutedisplacement of t he particles but not their rela tive displacement (to studyinteractions of particles one needs relative disp lacements only). Using thisconvent ion, figure 3 becomes figure 4.

The FRT indicates that any configuration may be seen as a collect ionof particles; a particle is an integer mult iple of T + 1 consecut ive sit es which

618 A .S. Fobs, E. Papadopoulou, and Y. Saridakis

evolves from t to t +1 independently of the rest of the configuration. Further­more, each particle is a collection of the so-called basic strings (BS), whichare r + 1 consecutive sites starting with a boxed site [5]. If a particle consist sof a single BS, we call it a simple particle.

We use the examples of figures 4 and 2 to illustrate the above notions:

1. The state at of figure 4 consists of the particles A = 1010111110 andB = 11011. The particle A consists of the bas ic strings, A l = 10101and A2 = 11110, while B is a simple particle consist ing of the BSB = 11011. The particles A,B are separated by six zeros.

2. The state a t+l of figure 4 consists of the particles A = 101011101 andB = 10111. The particle A consists of the BS's Al = 10101, A 2 =11010, while B is a simp le particle consisting of the BS B = 10111.

3. The st at e a l of figure 2 consists of a single part icle A = 10011010010011110100, which in turn consists of the BS's A l = 1001, A2 = 1010,A3 = 0100, A4 = 1111, AS = 0100.

An initial configuration consists in general of a number of particles. Thusin orde r t o study such a configuration one needs to develop a theory fort he evolut ion of single particles, as well as a theory for the int eraction ofparticles. In this paper, we present a complete analysis for the evolut ion ofsingle particles.

In more detail, regarding the evolution of a single particle we find thatthere exist three and only three cases:

1. The particle is periodic , and t he period p is uniquely given in terms ofthe initial dat a at .

2. The particle loses one basic string from the left or right ends, and thuscannot be periodic; this occurs if and only if the initial dat a at sa t isfya simple conditio n.

3. The particle splits into at least two part icles; necessary and sufficientcond itions are given for splitting to occur and although these cond it ionsare rather comp licated, they are expressed only in terms of the initialdata at.

As an ap plication of the ab ove results we show that if a particle containstwo or more identi cal noncon secutive basic strings then splitting will occurduring the evolution of this particle.

In order to follow the evolution of particles generated from the splittingof a given particle, one needs to study the interaction of particles . Theinteraction of two simple particles is comp lete ly characterized analyticallyin [7] where it is found th at :

1. If two simp le particles have different speed and they interact , thenthey interact solitonically and although they may interact a number oft imes , they finally separate with the faster particle moving in front ofthe slower one.

Particles in Soliton Cellular Automata 619

2. If two simple particles have the same speed and are close eno ugh sothat they interact, then there exist two cases; either they will inte ractexactly once and in the sequence they will separate traveling indepen­dently of each other or they will form a new periodic configuration byinteracting forever .

The general case of the interaction of two arbitrary particles is consi deredin [7J.

2. Evolution of single particles

Defin it ion 1. Let a be 1 or 0 and let a denote the complement of a, i.e. 0or 1 respectively. Tbe operation EB is called "exclusive or" and is defined bytbe following,

a EB 0 = a, a EB 1 = a. (2.1)

The above operation is extended in a straightforward way to basic strings.For example, let A = 0111, B = 1010, then A EB B = 1101. Furthermore ifA, B I

, .. . ,BL are basic strings

(2.2)

Ev idently, the operation EB is both commutative and associative.

Defin it io n 2. Let A be an arbitrary basic string containing R. 1 'so Ai denotesthe basic string obtained via the following operation: Leave the part of A upto the itb 1 uncbanged and replace the remaining part of A witb zeros.

For example, let A = 10011010. Tben R. = 4 and Al = 10000000, A 2 =10010000, A3 = 10011000, A4 = A.

Defin it ion 3. Let A ,B be two arbitrary basic strings and let d be any int eger1 ::; d ::; r + 1. Tbea A ldB denotes the basic string const ructed as follows:Leave the first d positions of A uncbanged and replace the last r + 1 - dpositions of A by tbe last r + 1 - d positions of B .

For example, let A = 101011, B = 011100, d = 3. Thea AbB = 101100 .

Theorem 1 (Evolu t ion of a Single Particle.) Consider the single parti­cle

(2.3)

consisting of L basic strings, A I , A 2, ••• , AL (0 denotes tbe null BS consist ­

ing of r +1 zeros) .Suppose tbat tbe BS's

(2.4)

620 A.S. Fokas, E. Papadopoulou , and Y. Seiidekis

contain £0,£1, ' " ,£L 1 's respectively.Let {£lO}, {d1

} , • •• , {dL} denote the set of indices indi cating where 1 'soccur in th e above BS's, i.e.,

{dm}~ {d;:',d~, ·· ·,d(m}' 1::;d;:' <d~ < .. . <d'l:n ::;r +1,d~=1,

o::; m < L, (2.5)

where d'[' denotes the posi tion of the i th 1 of the BS Am EB Am+! (AO ==AL+l = 0 ).

Assume that

(a) AL -=f T, where T ~ 10 ·· ·0 is th e trivial BS.

(b) Ai -=f Ai+! for all i = 1,2,·· ·,L-1.

Then

1. Th e particle (2.3) splits for the first time into at least two particles attime:

(a) t = £0 + £1 + ... + £k- l + i, 0 ::; k ::; L, 1 ::; i < £k, £-1 ~ 0 iff,

Ai+11 k Ai = Ak+!1kAk (2.6)di+1 d.:

for some j = k + 1, k + 2, · . . , L ,0,1 , "' , k - 1 (if k 0 thenj = 1,2 ,·· · ,L - 1 and AO= 0).

(b) t = £0 + £1 + . . . + e., 0 ::; k ::; L - 1 iff

Ai+1 1 Ai - Ak+1dk +1 -

1(2.7)

forsomej = k+1,k+2,···,L,O,1,···,k -1 (ifk 0 thenj=1 ,2,· · · ,L -1).

2. Assume that splitting does not occur until time t ~ 1. Th en theparticle (2.3) loses for the first time at time tone BS from the left end,iff t = £0 + £1 + ... + £k, 0 ::; k ::; L - 1, and

(2.8)

Moreover the particle of (2.3) cannot lose BS's from it s right end forall times. Equations (2.8) imply that a particle loses a BS at time t iffat time t - 1 the first BS of this particle is the trivial BS T .

3. If splitt ing does not occur and if th e particle does not lose any BS'suntil t = £0+£1 + . . ' + £k ~ tk, i.e. if conditions (2.6)- (2.8) are violateduntil t = tk, th en the evolution of the particle (2.3) for t ::; tk is givenby

(a) t = £0 + £1 + . . . + £m + i, m ::; k - 1,0 ::; k ::; L , 1 ::; i < £m+!,L 1 ~ 0, AO ~ 0 :

(Am+! EB Am+2 ) i EB Am+1 EB (Am+2 . .. A LAOA I . . . Am) (2.9)

Particles in Soliton Cellular Automata

d1 d? d?+l d' }+ ! j+l j+l" } } }

t = 0: 0... I ... I I .. . I .. . ad? .. . ad? ... ad?t l ., "ad? ... ad? ... ad o.+ >

III W @fJt = i : ." ad?.. I

< 0 >

Figure 5: Schematic representation of FRT at t = i .

(b) t = .eo + .e1+...+ .em m ::; k < L

Am+1 EB (Am+2Am+3.. . ALAoA l . . . Am)

(c) and if k = L then at

t = .eo + .e1 + ... + .eL : Al A2. . . AL

621

(2.10)

(2.11)

4. If equations (2.6)-(2.8) are violated for all 1 ::; t ::; tL th en the part icle(2.3) is periodic with period p,

(2.12)

Proof. 1. Suppose that the first BS, A \ has 1's at positions d~ , .. . ,d~o' Leta~o .. . a~o be t he elements of the jth BS, Aj, at the corresponding positio ns

1 '0

~"'" <flo. Then at t = i <.eo the arbitrary BS Aj changes as follows (using

the FRT ): Replace the elements a~, . . . , a~? by their compleme nt s and leave

the rest unchanged . This is precisely Af EB Aj. Hence

t= O Al .. · ALt = i, i <.eo: AfEB (A1A2.. . ALO) = (AoEBA1)iEBAoEB(A1A2 . ALAO)

or see figure 5. At t = i, the first 1 is at posi tion rfl+l' Therefore sp littingwill occur between Aj and Aj+l iff the site a~?+ 1 is followed by zeros up to

an d including the site aJdt1 . But at t = i t he BS Aj becomes Af EB Aj and BS

1+1

Aj+l becomes Af EB Aj+l. Therefore, the following condit ions must be valid :

Af EB Aj = 0 to the right and excluding posi tion d?+l' and

Af EB Aj+l = 0 up to and includ ing position d?+I'

T herefore, Aj = AI to the right of and excl udi ng pos itio n d?+1 or Aj = 0 tothe right of and excluding position rfl+l' and Aj+l = Af up to and including

622 A .S. Fokas, E. Papadopoulou, and Y. Saridakis

(2.20)

position «; These two conditions can be writ ten in the compact form

Aj+l ldO Aj = A11d9Ao , AO== O. (2.13)'+1 ,

Using the fast rul e theorem (FRT) an d the argumen ts used above, it followsth at the particle (2.3) at t = i o + £1 + ... + i k - 1 + i , 1 ::; i < i k , is given by

(Ak EEl Ak+!)i EEl Ak EEl (Ak+1. .. A1AoAI .. . Ak-1). (2.14)

Hence splitting occurs between the BS's Aj and Aj+! iff:

Ak EEl Aj = 0 to the right of and exclud ing position df+1 and

AkEElAj+! is identical to (AkEElAk+1)i up to and including position din.

Therefore, Aj must be equal to Ak to th e right of and excluding the positiondf+l and Aj+l must be equa l to Ak+1 up to and including position df , Aj+lmust be equal to Ak to th e right of and excluding posit ion df up to andincluding position df+l' These condit ions can be written in th e compactform

Aj +l ldk Aj = Ak+1Idk Ak. (2.15)_+1 I

Finally at t = i o + i 1 + ...+ i k th e particle is (using the FRT)

Ak+1 EEl (Ak+2. . . ALAOAI . . . Ak). (2.16 )

Hence splitting occurs between the BS' s Aj and Aj+! iff:

(a) the part of Ak+1 EEl Aj to the right of d~+! is zero and(b) Ak+l EEl Aj+l is identical to Ak+l EEl Ak+2 up to d1+1. These condi­

tions imply (2.7) .

2. Assume that

t - 1 : ./P jp . .. AL, Al == T. (2.17)

Then the application of the FRT implies that

t : Al EEl (AIA2.. . ALO) Al EEl AIAl EEl jF .. · Al EEl ALAI

= 0 Al EEl (A2 ALo )

= 0 A~ EEl(fF ALo)

Hence, the particle has lost one BS, namely AI . Let us now -assume that att the part icle lost its first BS, and that at

t_1: AIA2 . .. AL, Al:j:.T. (2.18)

Then by applying the FRT we have

t : Al EEl (AI A2. . . ALO), (2.19)

with Al EEl Al :j:. O. But this contradicts the assumption that at t th e particleloses its first BS. It remains to prove that the trivial BS T can app ear as thefirst BS of the particle at t - 1, t ~ 1 if and only if:

(a) t = i o + i 1 + ... + f.k , 0 ~ k ~ L - 1, and

(b) Ak+! ldk 0 = Ak+2 ldk 0 (AL+! = AO = 0).tk lk

Part icl es in Soliton Cellular A utomata 623

To prove (a)- (b) let us assume that the particle has evolved without anypathologies, i.e. without splitting or loss of a BS, unti l time t = Ro + RI +. .. + Rk - b 0 :::; k:::; L , L I ~ O. Then following th e FRT,

(2.21)

t Ro+...+Rk - I + j :(Ak EB Ak+l) j EB Ak EB (Ak+1 Ak+2 . . . AL 0 Al .. . Ak),

1 :::; j < Rk . (2 .22)

Observe now that the first r + 1 positions of the particle at t his stage, thatIS ,

contain Rk - j l's. Thus, for j < Rk - 1, Rk - j > 1, and therefore the firstBS cannot be the trivial. To prove (b) , we first observe that

t = f a +...+Rk - 1 :(A k EB Ak+l)lk_1 EB Ak EB (A k+I Ak+2.. . AL 0 AI . .. Ak).

Thus the first BS, at this stage, is the trivial iff th e BS (Ak EB Ak+1 )lk- I EB(Ak EB Ak+2 ) contains zeros in its first d~k - 1 positions and 1 at the positiondt (in order to avoid splitting). The above condition holds iff

(2.23)

or equivalently,

(2.24)

The bound k :::; L - 1 is obt ained by observing that the particle cannot loselength at t = f o +...+h . To prove this note th at

(2.25)

whence, the first BS is the trivial iff

(2.26)

or, equivalently,

(2.27)

But this contradicts the fact that cfi = 1, and the proof follows.

624 A .S. Fobs, E. Papadopoulou, and Y. Saridakis

To comp lete the proof of 2, it remains to show that the part icle cannotlose lengt h from its right end at all times. Following similar arg uments, withthe ones developed for the "left-end" case , one can show that the particlemay lose its last BS at t > 1, iff at t - 1 the last BS is the trivial one T .Therefore, assuming that the particle has evolved without any pathologiesuntil t ime t - 1, we have

where jP contains 1 at its first sit e and jiLhowever, the FRT, this may only occur if

~ 1 ~2 ~L

t -2 :AA · · ·A

T

(2.28)

10 · . . O. Following,

(2.29)

~1 ~2

where A = T and A has 0 in its first site. But in thi s case we would havesplitt ing, which contradicts the assumption that the particle evolves wit houtany pathologies unt il t - 1. This completes th e proof.

3,4. Assuming that splitt ing or loss of a BS does not occur, the FRTimplies (2.9) , (2.10). In part icular equat ion (2.10) at t = P yields

(2.30)

•Remark 1. For special choices of A~ ... AL it is possible to achieve period­icity with period p less then p, where p = p[m ; p, m integers. For exam pleparticles with period one are characterized in [5}. However, for generic datath e period is p.

In what follows we present several examples illust rating splitting and theloss of BS's.

Example 1 (Splitting of a Particle. ) Given the part icleA1A2A3A4 withAl = 101011, A2 = 101111, A3 = 000111, A4 = 001110, determine if andwhen splitting will occur.

We first construct t he BS's (2.4); Al = 101011, At EEl A2 = 000100,A2 EEl A3 = 101000, A3 EEl A4 = 001001, A4 = 001110. Also £0 = 4, £1 = 1,£2 = 2, £3 = 2, £4 = 3 and eti = 1, ~ = 3, dg = 5, ell = 6, d~ = 4, di = 1,d~ = 3, di = 3, d~ = 6, df = 3, d~ = 4, d~ = 5.

The cond it ions for splitting at time t = i < £0 = 4 depend on the BS's(see equat ions (2.6) )

(2.31)

Part icles in Soliton Cellular A utomata

t =O 101011 10111100011 1001110t=1 0010 110011111001 11101110100000t =2 0000 11000111101111100110101000t=3 000001000101101101100100101010t =4 00000000010010110010010110101100000t=5 000000 10100010000110111100010t =6 00100000000 100111110010010000

Figure 6: Split ting of a particle .

625

At i = 1, the above BS' s are AllrO and A 2b Al, A 3bA2, A413A3, thus therelevant BS's are 100000 and 101011, 000111, 001111. Since 100000 is dif­ferent than the other three BS' s above, no split t ing occurs at t = 1. Ati = 2 the BS's (2.31) are Al l30 and A 21 sAl, A 31sA

2, A 41sA3, i.e., 101000and 101111, 000111, 001111, hence no split ting occurs at t = 2. At i = 3 theBS's (2.31) are AlisO and A216Al, A 316A 2, A4 16A3, i.e. , 101010 and 101111,101111, 001110, hence no splitting occurs at t = 3. For t = 4 = f a we needto investigate (see equation (2.7)),

(2.32)

Since Ai+1 14Ai yields 101111, 000111, 001111, no split ting occurs at t = 4.Similarly for t = 4 + 1 = fa + f 1 we need to investigate

A 2 = 101111 and Ai+l ld2Ai , j = 2, 3,4, 0.1

Since Ai+114Ai yields 001111, 000111, 001110,100000 , no split ting occurs att = 5. The conditions for splitting at t = 4+ 1+i = fo+fl +i, 0 < i < f 2 = 2,depend on

(2.33)

At i = 1, the above BS's are A3lrA 2 and A413A3, ObA\ A l b O, A2bO, thusthe relevant BS 's are 001111, and 001111, 000110, 101000, 101011. Thussplitting will occur at t = 6 between A3 and A4 . This can be verified byfollowing the evolut ion of the particle Al A 2A3A4 numerically (see figur e 6) .

E xample 2 (Lo ss of a B S.) Given th e particle Al A 2A 3A4 wi th A l = 1110,A2 = 1010, A 3 = 1011, A 4 = 0101, det ermine if and when a BS will be lost .

626

t EO

t =lt=2

t=3t =4

A .S. Fokas, E. Papadopo uJou, and Y. Serideki«

11101010101101010110001000111101100000 1001100111100 1110000000100010110 111110

00000001111110100100

Figure 7: Loss of BS from left end.

We first construct the BS's (2.4), A l = 1110, A l EBA2 = 0100 , A2 EBA3 =0001 , A3 EB A4 = 1110, A4 = 0101. Also £0 = 3, £1 = 1, £2 = 1, £3 = 3, £4 = 2and d~o = 3, d}, = 2, di, = 4, 43 = 3, di. = 4.

For loss of a BS at t = £0 = 3 we need (see equation (2.8» A 1 1l00 =°A21dO O. But A1bO = 1110 and A2130 = 0100, thus no loss of a BS occurs

'0at t = 3. Similarly at t = f o +£1 = 4 we require A2 1d' 0 = A3 1d' O. Since

i1 i ]

A 2bO = A3120 = 1000, the particle Al A2A3A4 will lose a BS at t = 4. Thisis verified numerically in figure 7.

Remark 2. The conditions (2.8) of theorem 2.1 can also be written as: Ifdt ::; d~+l then the particle wi1J lose a BS at t = £0 +£1 +...+£k'

Example 3 (Loss ofa BS.) ConsidertheparticleA1A2A3withA1 = 11010,A2 = 01100 , A3 = 01101. Then Al EB A2 = 10010, A2 EB A3 = 00001 and

{~} = {1,2,4},{d1} = {1,3,4}, {d2

} = {5},{d3} = {2,3, 5}.

Since d~o = 4 > d~ = 1 no loss of a BS occurs at t = £0 = 3. However ,d}, = 4 ::; 5 = d~, thus the above particle loses a BS at t = £0 +£1 = 6. Thisis verified numerically in figure 8.

Example 4 (A Periodic Particle.) Consider the particle A l A2A3 whereA l = 10101, A2 = 10111, A3 = 01010. Th e B S's (2.4) are given by AI,Al EB A2 = 00010, A 2 EB A3 = 11101, A3. Th us £0 = 3, £1 = 1, £2 = 4, £3 = 2.It can be shown that equations (2.6)-(2.8) are violated for t ::; 10. Thus theabove particle is periodic with p = £0 + £1 + £2 + £3 = 10. This is verifiednumerically in figure 9, by comparing the states t = 0 and t = 10.

Particles in Soliton Cellular Automata

t=O 10101101110101000000000000000000000t=1 00101001111101010000000000000000000t=2 00001000111111010100000000000000000t=3 00000000101111110101000000000000000t=4 000001110110111000100000000000t=5 0110100111100101000000000t=6 0010101111110101100000000t=7 0000101011111101110000000t=8 0000001010111111110100000t=9 00010101111010101000t=10 00000101011011101010

Figure 8: Loss of BS from left end.

t=O 110100110001101t=1 01010111001110110000t=2 00010101001010111000t=3 000001011010111110 10t=4 00110001110101010000t=5 00010000110111010100t=6 00000000010110010110

Figure 9: Periodic par ticle.

627

628 A.S. Fokas, E. Papadopoulou, and Y. Saridakis

Theorem 1 provides a compl ete description of the evolution of a singleparti cle pro vided that AL I- T and Ai I- Ai+! . In what follows we invest igatewha t happens if these assumptions are violated.

Lemma 1. Consider th e particle (2.3) defined in theorem 1.(1) Suppose that th e assumption (a) of theorem 1 is violated, i.e. AL = T ,

then th e particle loses its last BS at t = 1 and therefore it cannot be periodic.Furthermore, if there exists an index J such that AJ = AJ+I = ... = AL == Tthen the particle loses L - J + 1 BS's from th e right en d at t = 1. Thes e areth e only cases in which loss of BS's from the right end of the part icle mayoccur. (2) Suppose that assumption (b) of theorem 2.1 is violated, namely,assume th at there exist q consecutive identical BS's. Let Am+I be th e firstBS repeated q times in the particle (2.3), i.e.

Furth ermore assum e that neither splitting nor loss of a BS occurs until atleast t = £0 +...+£m - 1. Then the part icle will lose q or q - 1 BS's from theleft end at t = £0 +£1+...+£m == tm depending on whether the con dit ion

(2.34)

is satisfied or violated respectively.

Proof. 1. Since d~ = 1 and df = dtL

= 1 (using AL = T) it follows th atth e first box will be placed at the position d~ of Al and the last box at theposit ion dt

L• By changing the boxes to their comp lements the last BS AL

becomes 0 and the particle now consists of only L - 1 BS's. Similarly forthe case that there exist L - J t rivial BS's at the end of the particle.

2. Since splitting does not occur and the particle does not lose any BS'suntil t = tm - 1 we have

or in terms of BS 's,

t = tm- I : Am EB Am+!lAm EB Am+2 1· · . lAm EB Am+qlAm

EBAm+q+!I· · · IAml·· · IAmEB Am-I.

At t = tm we have

(2.36)

t = tm : Am+! EB Am+2 IAm+! EB Am+3 1·· · IAm+I EB Am+q IAm+ I

EBAm+q+I j. . · IAm+I EB Am. (2.37)

Particles in Soliton Cellular Automata

teO 10100100010001001111t=l 0010 11001100110001111000t=2 000011101110111001011010t=3 011001100110110100101000t =4 001000100010100101101100t=5 000000000000101101001110

Figure 10: Loss of q - 1 BS's.

629

But the first q - 1 BS's are 0 since Am+! = . . . = Am+q. Hence the particleat t = t-; is

(2.38)

that is, it has lost q - 1 BS's. Equat ion (2.38) yields the evolut ion of theparticle

and if the condi tion (2.34) is sat isfied, theorem 1 implies that th e particlewill lose one mor e BS. This completes the proof. •

Example 5 (Identical Consecutive BS 's .) Consider the particleAIA2A3A4As with Al = 1010, A2 = A3 = A4 = 0100 and AS = 1111.In this case m + 1 = 2, i.e. m = 1 and q = 3. Also £0 = 2, £1 = 3, thusth e particle loses BS's at tm = to + t I = 5. Furthermore, equation (2.34) isviolated, hen ce th e particle loses 2 BS's at t = 5 (see figure 10).

Example 6 (Identical Consecutive BS's.) Consider the particleAl A 2 A3A4 with Al = 1100, A 2 = A3 = 0010, A4 = 0011. In ilue casem = ·1, q = 2, £0 = 2, £1 = 3, and the condition (2.34) is satisfied:A2 !30 = A4 130 = 0010. Thus the particle wi11 lose 2 BS's at t = 5 (seefigure 11).

3. Applications

In what follows we will prove that if a particle contains the zero BS, or if itcontains two or more identi cal non consecutive BS's, then splitting will occurduring th e evolut ion of this par ticle. First we prove the following lemma.

630

t=Ot=lt=2t=3t"4t=5

A .S. Fokas, E. Papadopoulo u, and Y. Saridakis

110000 1000 1000 110100101010 10101 11000000011 10111011111 100

011001100 1110 100100000100010001100001 10000000000000100101110

Figure 11: Loss of q BS's.

Lemma 2. Consider the particle

AOB (3.1)

with (i) A ::j:. B, (ii) A ::j:. T ::j:. B, and (iii) A ::j:. 0 ::j:. B. Then splitting willoccur at some time t in the evolu tion of the above particle.

Proof. If splitting does not occur then the evolut ion of the above particleis as follows and the particle is periodic.

where lA, l B denote the number of l 's in the BS 's A and B respectively.We will prove that in the above evolut ion splitting will occur either at

some t < lA or at 2l A < t < 2l A + l B.For split t ing at t = i < lA th e conditions (2.6) must be sati sfied , i.e. ,

(3 .2)

The conditions (2.6) for splitting at t = 2lA +i are

(3.3)

Particles in Soliton Cellular Automata 631

but dA = {dLat, . . . d1A

} and dB = {df,d~, .. . ,dfB} with dt = 1 anddf = 1; otherwise we would have a splitting at t = O. For the rest of theelements in dA , dB we have

(a) d1 < d~ or

(b) at > d~ or

(c) # = df.

If (a) is true then B ldAO = A ldAO i.e. (3.2) is valid and splitting will2 1

occur at t = 2. If (b) is true then A ldBO = B ldBO i.e. (3.3) is true and2 1

splitting will occur at t = 2fA +1. If (c) is true then (a), (b), or (c) will holdfor a: and df and following the arguments above either (3.2) or (3.3) will besatisfied for some index dt, since A =f:. B, therefore splitting necessarily willoccur and that comp letes the proof.

Lemma 2 can be trivially generalized for the general case of a particlewith L BS's containing one 0 BS.

Lemma 3. Consider the particle

ABA (3.4)

with A =I- B, A =f:. T =f:. B, and A =I- 0 =I- B. Then splitting will occur atsome time t in the evolution of (3.4).

Proof. If splitting does not occur then the evolution of the above particleis as follows:

t = 0 A B A

t = fA 0 AEBB 0 A

t = fA + fAffiB 0 AEBB B AEBBt = fA + 2lAffiB 0 A 0 AEBBt = 2lA + 2lAffiB 0 A B A

From lemma 2, however, we have that splitting will occur either at somet > fA or at fA +2fAffiB < t < 2fA +2fAffiB and that completes the proof.•

Remark 3. Splitting in the evolution of a particle does not necessarily implyloss of periodicity.

632 A.S. Fokas, E. Papadopoulou, and Y. Saridakis

Example 7. (A particle containing the zero BS.) (r = 3)

t=O 101000001 100t =l 0010100001001000t=2 0000101001101010t c 3 0010111000101000t=4 0000 110000001010t c 5 010010000010

Splitting occurs at t = 5.

Example 8 (Identical nonconsecutive B S 's.) (r = 3)

t =O 10101100101011 11t =l 0010010000100111 1000t=2 00000110000001011010t=3 0010010000011 1100100

Split ting occurs at t = 3.

Ackn ow le d gments

This work was partially supported by the Office of Naval Research undergrant numb er N00014-88K-0447, National Science Foundation und er grantnumber DMS-8803471, and Air Force Office of Scientific Research under grantnumber 87-0310 and 88-0073.

R eferences

[1] S. Wolfram , Theory and Applications of Cellular A utomata (World Scien­tific, 1986).

[2] J .K. Park, K. Steiglitz, and W.P . Thurston, "Soliton-like behavior in au­tom ata," Physics: D, 19 (1986) 423-432.

[3] K. Steiglitz, I. Kamal, and A. Watson , "Embending computation in one­dimensional auto mata by phase coding solitons," IEEE Trans . on Comp ut­ers, 37 (1988) 138- 145.

[4] T .S. Papatheodorou , M.J. Ablowitz, and Y.G. Saridakis, "A rule for fastcomputation and analysis of soliton automata," Studies in Appl. Mat h., 79(1988) 173-174.

Particles in Soliton Cellular Automata 633

[5] T .S. Papatheodorou and A.S. Fokas, "Evolution theory and characterizationof periodic particles and solitons in cellular automata," Studies in Appl.Math ., 80 (1989) 165-182.

[6] A.S. Fokas, E.P. Pap adopoulou, Y.G. Saridakis, and M.J. Ablowitz, "In­teraction of simple particles in soliton cellular automata," Studies in Appl.Math., 81 (1989) 153-180.

[7] A.S. Fokas, E.P. Papadopoulou, and Y.G . Saridakis, "Soliton Cellular Au­tomata," Clarkson University preprint INS#127 (1989) 153-180.