ONE-DIMENSIONAL GAME OF LIFE AND ITS GROWTH FUNCTIONS

. We start vith finitely many l’s and possibly s,me O’s in between. Then each entry in the other rows is obtained from the Base 2 sum of the two numbers diagonally above it in the precedlng row. We may formulate the game as follows: Define dl, j recursvely for I, a non-negative integer, and j an arbitrary integer by the rules


I. INTRODUCTION
To explain in words: (I) describes the starting configuration with finitely many l's, and possibly some O's in between.(Picture this as row zero).(If) says that all entries on both sides of the starting configuration are zero.
(Note that considering the O's on both sides contributes nothing).(IIl) says that each entry (in the other rows) is obtained from the Base 2 sum of the two numbers diagonally above it in the preceding row.Now, if we interpret the number of l's in row i as the coefficient a i of a formal power series, then we obtain a growth function: O0 f(x) aixl.It is interesting that there are cases for which this growth i=O M.H. AHMADI function factors into an infinite product of polynomials.But there are cases in which the pattern is less regular.Nevertheless, ve shall show the following.

THEOREM.
In a One-dimensional Game of Life, no matter what the starting configuration is, (finitely many l's and possibly some O's in between), the associated growth function never represents a rational function.First ve look at some examples: EXAMPLE 1. Suppose the starting configuration has one 1.We get the following configuration called "Fundamental Configuration" throughout this paper.The resulting growth function is" f(x) aixl 1 + 2x + 2x 2 + 4x 3 + 2x 4 + 4x 5 + 4x 6 +

Ro
We observe that the configuration above is essentially the configuration obtained by reducing modulo 2 each element of Pascal's triangle of binomial coefficients.To obtain the reduced Pascal's triangle from configuration in example 1 simply remove every second entry of each roy beginning vith the entry 0 folloing the initial 1 of that ro.Each entry removed ill be a zero.
This connection vith Pascal's triangle can be used to explain the fact that, for each j _> 1, ro 2 j has precisely tvo non-zero entries, one at each end.In other vords, for every integer j _> 1, roy 2 j has tvo copies of starting configuration, one at each end.Therefore, a2J 2a O.
Worthy of notice is that Wolfram [3] has shorn: In Pascal's Triangle of binomial coefficients reduced modulo 2, vhen infinitely many toys are included, the limiting pattern exhibits a fractal self-similarity and is characterized by a "fractal dimension" Iog23.DEFINITION I.For any integer j _> I, the set of rovs between roy 2 j'1 and 2 j-I inclusive, is called the jth Stage.Note that rot 2j, j _> I, is the initial row of the jth Stage.
Stage O, is the Oth roy or the initial roy of the configuration.EXAMPLE 2.
[I" do,.i 1,.j--0,1 t.hcn ve get the flloving configuration" Ro Term The resulting grovth function g(x) may be vritten as: g(x) 2 + 4x + 4x 2 + 8x 3 + 4x 4 + 8x 5 + 8x 6 + 16x 7 4x 8 + EXAMPLE 3. Let d O,j 1 for j 0,1,2.This yields the following configuration" ttov Term The grovth function h(x) is: h(x) 3 + 4x + 6x 2 + 6x 3 + 6x 4 + 8x 5 + 12x 6 + lOx 7 + 6x 8 + FACTORIZATION OF THE FUNDAMENTAL GROITH FUNCTION.If ve look at the configurations in examples 1 -3 carefully, ve observe that there is in each a recurrent triangular pattern.This is dramatically illustrated in the more extensive computer printouts of figures I -4, reproduced below.Note that in these figures the zeros are not printed In fundamental configuration (Fig. 1) since in each Stage ve have to copies of the preceding triangle (or Stage), hence e may factor its growth function, f(x), as follovs: o Fig. 2 The starting configuration is "11" that is two l's M.H. AHMAD / Fig. 3 The starting configuration is "111, that is, three l's" o 4 o o Fig. 4 The starting Configuration is "101, that is, one-zero-one" Because in Fig. 1, the triangle formed of stages 0 and gives the first factor 1 + 2x.Then since it duplicates in the next stage (i.e., in Stage 3), and the corresponding term involving x of the initial roy of stage 3 is x 2 (vhile considering the first triangle as a vhole, i,e., equivalent to 1), then this yields to 1 + 2x2, as the second factor of f(x) in the factorized form, etc.
Similarly, the grovth function in example 2 (see Fig. 2) can be factored as: Note that the factorization of f and g exhibit infinitely many zeros of the power series which cluster at every point inside the unit circle in complex plane.So f and g must be non-rational.
In example 3, the pattern is less regular.The "irrebruIarity" first appears in the last row of stage O, and therefore we can not factor the growth function in this way.Nevertheless, we shall show that the associated growth function of a general configuration-is always non-rational.
It is obvious that there is a natural relationship between an arbitrary starting configuration and the "fundamental configuration" since a general configuration is simp] a "mode 2" sum of shifts of the fundamental one.Indeed it follows easily from the definition given below.
DEFINITION.If the starting configuration has k number of 1's and Inumber of O's interspersed among them at the columns Jl, J2, j/then its entries, ci, j, can be obtained by the formula: k+/ ci =t0 j ,j.t(mod 2) tJl 'Jl where k and lure integers with k _> 1 and I_> O.
Nov, this definition includ.ing the fact that new stages on fundamental configuration begin at the rows 2 3 imply, no matter what the starting configuration is, the new stages always .start at the rows n 2J; and the corresponding terms of the growth series are an x2j where a n 2a13 THEOREII.
In a One-dimensional Gme of Life, no matter what the starting con- figuration is, (finitely many l's and possibly some O's in between), the associated growth function is non-rational.PRI}OF.Suppose, on the contrary, there is some case in which the associated growth function f(x) a 0 + alx + a2 x2 + an xn + + is rational.It is shown in [2] that if a power series with integral coefficients represents a rational function, then it can be expressed in the form -P(x), q(x) are polynomials with integral coefficients and q(O) 1. Therefore if f(x) is rational then there are polynomials P(x)= cixX and q(x) bixi, where c i, b i e I and q(o)= 1 (i.e., b 0 1)such that i=O By doing simplc algebra, this in turn impl+e that" aj.c (blai_ 4-b2ai_ 2 + bak_m) ior .< . .n and, (blat_ 1 + b2ai_, 2 4- 4 bmai_m) for > n. ( (Set a 0 ftr < 0.) That is if f is a.tional then there is a linear recurrence relation (2) fo coefficients of the poer series after a certain number of terms.Assuing that such a linear recursion exists, there are integers Yl' y such that," a Ylai_l Y2ai_2 4 ymai.m, for i > n (a O, for i < 0). (3) No, e may choose i large enough so that i q -2 j for some j and q > m Note.that row q represents the first ro of a ne stage.Sty, we have: aq 2aO, aq+l 2a1' aq+m-1 2am-l, aq+m 2am" (4) From (3) nd (4) ,, have" aq/m/1 --ylaq+m-y2qq+m_ 1 4 + Ymaq+l By (3) (By 4) Therefore, a2q 4a O.But this is a contradiction, because roy 2q 2(2 j) 2 j+1 represents the first roy of a hey stage, i.e.Stage (j + 1), and hence a2q 2a 0 and this completes the proof of the theorem. [ Fig. l.The starting configuration has only one 1.