The Cups and Stones Counting Problem, The Sierpinski Gasket, Cellular Automata, Fractals and Pascal s Triangle

Transcription

1 Journal of Cellular Automata, Vol. 6, pp Reprints available directly from the publisher Photocopying permitted by license only 2011 Old City Publishing, Inc. Published by license under the OCP Science imprint, a member of the Old City Publishing Group. The Cups and Stones Counting Problem, The Sierpinsi Gaset, Cellular Automata, Fractals and Pascal s Triangle DAVID ETTESTAD 1,JOAQUIN CARBONARA 2,MICHAEL JANSMA 2, MICHELLE RUA 2 AND KATHERINE SEMBER 2 1 Dept of Physics, State University of New Yor, College at Buffalo, 1300 Elmwood Av, Buffalo NY 14222, USA ettestdj@buffalostate.edu 2 Dept of Mathematics, State University of New Yor, College at Buffalo, 1300 Elmwood Av, Buffalo NY 14222, USA carbonjo@buffalostate.edu Received: May 11, Accepted: May 11, In 1992 Barry Cipra posed an interesting combinatorial counting problem. In essence, it ass for the number S,σ of configurations possible if a circular arrangement of cups, each having σ stones, is modified by applying a particular transition rule that changes the distribution of stones. Carbonara and Green (1998) studied the integer sequence S,1 and presented a recursive formula for it: S = 2S 2 r r j S d+1 + d2 r 2 r+1 where = 2 r d > 2, r 0 and 0 < d 2 r. Ettestad and Carbonara (2010) noted that this system is a finite Cellular Automaton, showed two interesting non-recursive formulas for S, and claimed that the shape of the non-zero terms in the reduced matrix for the Cups and Stones Counting Problem (CSCP) with 2 n + 1 cups is equivalent to the Sierpinsi Gaset. We are proving that claim in this paper. In doing this, we extend the classic definition of the Sierpinsi Gaset to discrete geometric figures. Keywords: Discrete fractals, Discrete geometry 1 INTRODUCTION In 1992 Barry Cipra posed an interesting combinatorial counting problem (see [1]). We call our version of it the Cups and Stones Counting Problem (CSCP). In [2] Ettestad and Carbonara claimed that the shape of the non-zero 421

2 422 J. CARBONARA et al. terms in the reduced matrix for the CSCP with 2 n +1 cups is equivalent to the Sierpinsi Gaset (see [3]). We are proving that in this paper. In doing this, we extend the classic definition of the Sierpinsi Gaset (see [4]) to discrete geometric figures. Combinatorially, the CSCP can be stated as follows. The Cups and Stones Counting Problem (CSCP):Given an initial configuration and a transition rule to go from one configuration to the next, count the number of different configurations possible. Initial configuration: Arrange in a circle cups, numbered in clocwise order each having σ stones with cup number one mared as a special cup (which we call the root). Transition rule (T.r.): Pic up all the stones in the root. Distribute them one cup at a time in a clocwise fashion going around the circle of cups, until all stones that were piced up are reallocated. The last cup to receive a stone on each application of the T.r. becomes the root in the resulting configuration. Note that although T.r. changes the location of the root, the numbering of the cups doesn t change. Notation: We denote by CSCP,σ the ordered list of different configurations in the CSCP where the initial configuration has cups, each having σ stones. Let S,σ be the number of configurations in CSCP,σ. The CSCP can be viewed as a 1D Cellular Automaton (finite and circular). A Cellular Automaton can be defined as a dynamical system with 4 components {L,, N, f }, where L is a lattice of cells (in our case, the lattice is the set of cups), is the set of values allowed in each individual cell (in our case the possible number of stones in each cup), N is the neighborhood of cells affecting the value of each cell at the next time step (in our case the entire set of cups) and f is the transition rule (in our case T.r.) that uses the neighborhood N to determine the value of a cell in the next time step. See [2] and [5] for details. To visualize the CA, we arrange the number of stones in each cup in a matrix M,σ such that M,σ (i, j) is the number of stones in cup j in the i th configuration. Thus each column of M,σ corresponds to a particular cup and each row of M,σ corresponds to a particular configuration in CSCP,σ. Note this means that M,σ has columns and S,σ rows. To eep trac of the location of the root we will underline the entry in each row corresponding to the root. We show an example in Fig. 1. FIGURE 1 Cups and Stones Counting Problem

3 THE CUPS AND STONES COUNTING PROBLEM 423 From now on fix σ to be 1, denote M,1 simply by M, S,1 by S and CSCP,1 by CSCP. In [2] Ettestad and Carbonara proved the following theorems: Theorem 1. Let = N + 1. Let N = w i=1 2ɛ i be the binary expansion of N (ɛ 1 >ɛ 2 >ɛ 3...). Then: S = 2 r+2 3 r r+1 if N = 2 r+1 ( + 1)2 ɛ w 3 ɛ i 2 (ɛ 1 ɛ i ) (i 1) i=1 Theorem 2. Let S be defined as above. Then r I + Otherwise (1) S = 4 t 3 t (n 2)2 t + a(n) (2) where t and n are determined (uniquely) by expressing as 2 t + 1 n with 0 n < 2 t 1 and a(n) is the total number of odd entries in the first n rows of Pascal s triangle. In [6], it was shown that the basic building blocs in this counting problem are the matrices M where = 2 n + 1, and n I. The structure of M was also described in detail. Also in [6], a reduced matrix was defined for those matrices M where = 2 n + 1, consisting only of the rows where the root is in the first column. We shall call these reduced matrices RM in this paper (see Fig. 2). FIGURE 2 M 5 and RM 5

4 424 J. CARBONARA et al. In [2] we claimed that the distribution of non-zeros in RM for = 2 n + 1 is identical to the distribution of ones in the Sierpinsi Gaset. We re proving that in this paper. 2 PROOF OF THE EQUIVALENCE We ll use the notation of [2] and [6]. We will call initiator rows the first row and the rows where a stone has been added to cup 1 since the previous configuration. We will later show that RM consists exactly of all the initiator rows. To show that the pattern in RM is equivalent to the Sierpinsi Gaset (details of what equivalent means will be addressed later), we need to compare it to a formula that is nown to produce the gaset. Cellular automaton Rule 102 (see [5]) is well nown to produce the Sierpinsi Gaset (note that since one is discrete and the other continuous, an explanation is due, and will be provided in the last section of this paper). We use as initial configuration for Rule 102 a row of all 0 s except a single 1 on the right. (See Fig. 4b.) The value of each term w in the following row is determined by the term y directly above it and the terms x and z on the left and right of y respectively. Since each of these is a 0 or 1 there are eight possible combinations of x, y, z and therefore eight cases to the rule that determines the particular cellular automaton. We will write these cases as [x, y, z] w. Rule 102 is given in Fig. 3. Note the result does not depend on the first of the three digits, so we can simplify our notation to: [x,0,0] 0 [x,0,1] 1 [x,1,0] 1 [x,1,1] 0 where x is 0 or 1. FIGURE 3 Notation for Rule 102

5 THE CUPS AND STONES COUNTING PROBLEM 425 FIGURE 4 (a) The left matrix is RM 33 (b) The right matrix is Rule 102. It is helpful to compare RM and Rule 102 side by side, which we do in the Fig. 4. We have assumed =33. Note that, for example, the 4 shaded entries in the left and right matrices correspond to each other. We note three differences between the Sierpinsi Gaset formed by Rule 102 and the Sierpinsi Gaset formed by RM. 1. The Sierpinsi Gaset produced by Rule 102 is inverted from the Sierpinsi Gaset produced by RM. 2. Although it is the non-zero terms in both cases that produce the gaset, Rule 102 has only 0 s and 1 s while RM can have any non-negative integer less than. 3. RM has an extra column of 1 s on the left. We also note that an interesting additional fact is that Rule 102 is not reversible, but CSCP is reversible. We want to show the equivalence of the two matrices, subject to the constraints listed above. We do that with the following theorem: Theorem 3 (The matrix RM is equivalent to the 1D CA Rule 102). Consider the matrix RM with = 2 n + 1 and n I. Replace all integers greater than or equal to one with a one. Consider any sequence of three consecutive integers on any row of the matrix (excluding the first row and the first column). The number above the center digit will be given by Rule 102 applied to the three consecutive integers. Before we prove this theorem, we need some notation and two lemmas. Notation: Let Ɣ CSCP (i.e. Ɣ is one configuration in CSCP ). Let r(ɣ) be the sum of the cup number of the root in Ɣ and the number of stones in the root. Note that r(ɣ) determines where the root is in the next configuration. The first lemma is ey to the entire structure of the M and is proved in [6].

6 426 J. CARBONARA et al. Lemma 1 (Fundamental Lemma part 1 or FLP1). Let Ɣ CSCP, with = 2 n + 1 and n I. If Ɣ is not the last configuration, then 1. The last cup of Ɣ is not the root. 2. If r(ɣ) >, then r(ɣ) = + 1. This leads to the following corollary. Recall that initiator rows of M are the first row and the rows where a stone has been added to cup 1 since the previous configuration. Corollary 1. Let = 2 n + 1. Then, in all the initiator rows in M, cup one is the root and has one stone in it. Proof of Corollary 1: We ll use induction on the location of the initiator row. In the first configuration, cup one is the root and has one stone. Next, assume the l th initiator row has cup one as the root with one stone in it. Then in the configuration immediately following, cup one will clearly have no stones. By definition, the next initiator row will be the next time the number of stones in cup one increases. This happens in the configuration right after the one where r(ɣ) >. But from FLP1, this means r(ɣ) = + 1, so cup one will be the root in the next configuration. This is by definition the (l + 1) th initiator row, and cup one will clearly have exactly one stone in it in addition to being the root. QED Note that, except for the first configuration, if cup 1 is the root, it must have had a stone added to it since the last configuration, and therefore it is an initiator row. This fact along with Corollary 1 clearly show that, if = 2 n +1, RM consists of precisely the initiator rows of M. Corollary 2. Let = 2 n + 1. Then, in all the initiator rows in M after the first, the second column is a zero. Proof of Corollary 2: Since in all initiator rows the first cup has one stone and it is the root by Corollary 1, in the configurations immediately after an initiator row, the second cup will be the root and thereafter become and remain empty, until and including (by FLP1), the next initiator row. QED Notation: Let I i be the ith initiator row in M. Let I i (j) bethejth term in Ii. Lemma 2 (The Zero-Root Lemma). Let = 2 n + 1 and n I. Consider any initiator row I i in M with i = 1. Consider any term I i (j) with j = I i (j) is zero if and only if there is another term above it, in the same column and below the last initiator row, which is a root. 2. If I i(j) > 0 then Ii (j) = Ii 1 (j) + 1. Proof of The Zero-Root Lemma: From Corollary 1 it is clear that in between consecutive initiator rows only two things can happen to any individual cup that is not cup 1:

7 THE CUPS AND STONES COUNTING PROBLEM It gets a stone added and becomes the root, in which case it will be empty in the following initiator row. 2. It gets a stone added but does not become the root, in which case it will have exactly one more stone in the following initiator row than it did in the previous one. These two cases cover both parts of the Lemma, and so it is true. QED Corollary 3. Let = 2 n + 1. Then there are exactly 2 n initiator rows in M. Proof of Corollary 3: The first initiator row is the first row of M, and it is all ones. By FLP1, the last cup is never a root before the last row in M. Therefore by Lemma 2 above, the last cup in initiator row 2 has two stones. Repeating this line of reasoning shows that the last cup in initiator row l has l stones in it. Now by construction, the last row of M must contain a cup which is the root and has all the stones in it. (That is the only way to have the next configuration have one stone in each cup lie the original configuration.) Since the last cup is never the root until the last configuration (by FLP1), it never loses any stones and therefore it must be the cup with all 2 n + 1 stones in it in the last configuration. Since the last cup must also be the root in the last configuration, the following are true: 1. The last cup has one more stone in it than the previous initiator row. 2. The last row is not an initiator row (by Corollary 1). Combining these two points, one can clearly see that in the last initiator row the last cup has 2 n stones and therefore there are 2 n initiator rows. QED Corollary 4. The l th initiator row has l 1 zeros after the first entry. Proof of Corollary 4: From Corollaries 1 and 2 we now that the first two entries in the second initiator row are a 1 (which is the root) and a 0. A simple calculation shows that in the next two configurations the root will be in the second and third cups respectively. By the Zero-Root-Lemma, part 1, the second and third cups in the third initiator row are therefore zeros. By a straight forward application of induction, the l th initiator row has l 1 zeros after the first entry. QED Now we are finally ready to prove Theorem 3 by going through each of the four cases of Rule 102 above to see if they are satisfied by RM. 1. Case [x,0,0] 0: Assume I i(j) = 0 and Ii (j + 1) = 0 where i, j > 1. Since Ii (j) = 0, by Lemma 2 the jth cup was the root in a configuration Ɣ in between I i 1 and I i. Let Ɣ(j) denote the jth cup in configuration Ɣ (which we are assuming is the root). Let us now suppose I i 1 (j) > 0. Then (1) the number of stones in Ɣ(j)

8 428 J. CARBONARA et al. is one more than in the jth cup of the previous configuration, (2) there would be at least two stones in Ɣ(j) and (3) the (j +1)st cup would definitely NOT be the root in the configuration after Ɣ. Then by Lemma 2, I i (j + 1) > 0. Since this is not true, the supposition I i 1 (j) > 0 is not true and hence I i 1 (j) = 0. Therefore for all i > 1, j > 1, [I i(j) = 0 and Ii (j + 1) = 0] Ij 1 (i) = 0. In other words, [x,0,0] Case [x,0,1] 1: Assume I i(j) = 0 and Ii (j + 1) 1 where i, j > 1. Since Ii (j) = 0, by Lemma 2 the jth cup was the root in a configuration Ɣ in between I i 1 and I i. Let Ɣ(j) denote the jth cup in configuration Ɣ (which we are assuming is the root). Let us now suppose I i 1 (j) = 0. Then (1) the number of stones in Ɣ(j) is exactly one, and (2) the (j + 1)st cup would definitely be the root in the configuration after Ɣ. Then by Lemma 2, I i (j + 1) = 0. Since this is not true, the supposition I i 1 (j) = 0 is not true and hence I i 1 (j) 1. Therefore for all i > 1, j > 1, [I i(j) = 0 and Ii (j + 1) 1] I i 1(j) 1. In other words, [x,0,1] 1. For the last two cases we need another lemma (which we will prove later). Lemma 3 (Zero Follower Lemma). Let = 2 n + 1, with n I. Consider any initiator row I i in M. 1. If 1 < j < and I i(j) > 1 then Ii (j + 1) = If 1 < j < and I i(j) = 1 then Ii (j + 1) > 0. The contrapositives of the statements lead to the following corollaries. Corollary 5. If there are two consecutive nonzero integers, not at the beginning, in any initiator row in M, then the left one must be a 1. Corollary 6. If a zero follows any nonzero term, not at the beginning, in any initiator row in M, the nonzero term must be bigger than one. Continuing on with the last two cases of Rule 102: 3. Case [x,1,0] 1 Assume I i(j) 1 and Ii (j+1) = 0 where i, j > 1. By corollary 6, Ii (j) > 1, which by Lemma 2 means I i 1 (j) 1. Therefore for all i > 1, j > 1, [I i(j) 1 and I i (j + 1) = 0] Ii 1 (j) 1. In other words, [x,1,0] Case [x,1,1] 0 Assume I i(j) 1 and Ii (j+1) 1 where i, j > 1. By corollary 5, Ii (j) = 1, which by Lemma 2 means I i 1 (j) = 0. Therefore for all i > 1, j > 1, [I i(j) 1 and Ii (j + 1) 1] Ii 1 (j) = 0. In other words, [x,1,1] 0. We have now showed that, assuming Lemma 3 is proven, that all four cases of Rule 102 are obeyed in RM (going upwards).

9 THE CUPS AND STONES COUNTING PROBLEM 429 Proof of Lemma 3: We ll use induction on n. By inspection it is true for n = 1, 2 and 3. Assume it is true for a general n, that is for RM 2 n +1. We ll show it is also true for RM 2 n +1 where n = n + 1. The matrix RM 2 n+1 +1 (which by Corollary 3 has 2 n+1 rows) can be divided into 4 quadrants delimited by columns 1 to 2 n + 1 and 2 n + 2to2 n and rows 1 to 2 n and 2 n + 1to2 n+1 (see Fig. 5). We ll call these quadrants UL (upper left), UR, LL (lower left) and LR respectively. We ll prove the Lemma for each quadrant separately. Proof for the UL quadrant: We will show that RM 2 n +1 is a submatrix of RM 2 n+1 +1 and is embedded in the upper left corner. Consider the first 2 n + 1 columns of both M 2 n +1 and M 2 n The first row in both is all ones with the root in the first column. The configurations that follow are therefore identical as long as changes, from one configuration to the next, only happen within the first 2 n + 1 columns. This will be true until the root plus the root value is greater than 2 n + 1. The next configuration in M 2 n +1 after that will be the next initiator row with the root in column 1 by FLP1, while in M 2 n+1 +1 the next configuration will therefore have the root in column 2 n + 2. From this configuration on in M 2 n+1 +1 the only changes from a configuration to the next will occur in the last 2 n columns, until the value of the root plus its position is greater than 2 n+1 + 1, in which case the next configuration will be an initiator row for M 2 n+1 +1, with the root in column 1, by FLP1. Therefore, the second initiator rows in both M 2 n +1 and M 2 n+1 +1 are identical for the first 2 n + 1 columns; this means that in the reduced matrices RM 2 n +1 and RM 2 n+1 +1, the first 2 n + 1 columns of the second row are identical. A similar reasoning shows that the first 2 n + 1 columns of the third row are identical. This process continues until the last row of M 2 n +1, where the root is in position 2 n + 1 and equals 2 n + 1. Therefore RM 2 n +1 is embedded in the upper left corner of RM 2 n+1 +1 (see Fig. 5) FIGURE 5 RM 2 n +1 and RM 2 n+1 +1 with n =

10 430 J. CARBONARA et al. Proof for the UR quadrant: Next we want to show that RM 2 n +1 (with the first column removed), is embedded as a submatrix of RM 2 n+1 +1 in the upper right corner. We now that the initiator rows have the root in the first cup and the value of the root is 1 by Corollary 1 of FLP1. Compare columns 2 through 2 n + 1 and columns 2 n + 2 through 2 n+1 + 1ofRM 2 n They start off identical (all ones). In the next configuration (as well as every configuration immediately after an initiator row), the root will be in the second column by Corollary 2. Consider the configurations in M 2 n+1 +1 where the root is at a cup c, with c 2 n +1 and such that c+(value of root) > 2 n +1. The configuration after that will have the root in column 2 n +2 by a simple application of FLP1 to M 2 n +1. The configurations after will have columns 2 n +2to2 n+1 +1 identical to columns 2 to 2 n + 1 in the configurations after the previous initiator row. This will lead to the next initiator row having identical entries in columns 2 through 2 n + 1 and columns 2 n + 2 through 2 n The same argument applies for each initiator row until the root is in position 2 n + 1, which corresponds to the last row of M 2 n +1. Therefore RM 2 n +1 (with the first column removed), is embedded as a submatrix of RM 2 n+1 +1 in the upper right corner (see Fig. 5). By induction, Lemma 3 is true for columns j, with 1 < j < 2 n + 1 and 2 n + 1 < j < 2 n in the upper left and right quadrants. What about for j = 2 n + 1? The first initiator row has all ones, so Lemma 3 is true. For the remaining initiator rows covering the upper quadrants, we now that their entry in column 2 n + 1 is greater than one (see the proof of Corollary 3) and the entry in the next column we now is a 0 from the embedding and Corollary 1. So Lemma 3 is again true. Proof of the LL quadrant: Now consider the lower left quadrant of RM 2 n+1 +1 (i.e. the first 2 n + 1 columns of the last 2 n rows). By Corollary 4 the l th initiator row has l zeros after the first column. Therefore, the LL quadrant has all zeros as entries except in the first column and the Lemma 3 is true in the entire LL quadrant. Proof of the LR quadrant: From the constructions above, the (2 n ) th initiator row consists of a1inthefirst cup, followed by 2 n 1 zeros, followed by 2 n, followed by 2 n 1 zeros, followed by 2 n. It is a simple computational exercise to show that the (2 n + 1) th initiator row consists ofa1inthefirst cup, followed by 2 n zeros, followed by 2 n 1 ones, followed by 2 n + 1 (see rows 8 and 9 of the right side of Fig. 5). We want to show that RM 2 n +1 (with the first column removed and 2 n added to each term in the last column), is embedded as a submatrix of RM 2 n+1 +1 in the bottom right corner. Since the LL quadrant is all zeroes (except in the first column), the column 2 n + 1inM 2 n+1 +1 will be the root somewhere between all consecutive initiator rows in the LR quadrant. Note that the first row of the LR quadrant has all ones except in the last position. The FLP1 applied to M 2 n+1 +1 implies that the last column is never the root until the last row. These three facts and a similar reasoning as used in the UR quadrant, show

11 THE CUPS AND STONES COUNTING PROBLEM 431 that RM 2 n +1 (with the first column removed and 2 n added to each term in the last column), is embedded as a submatrix of RM n+1 in the bottom right corner. By induction, Lemma 3 is true for columns j, with 1 < j < 2 n + 1 and 2 n + 1 < j < 2 n in the bottom left and right quadrants. What about for j = 2 n + 1? Since all terms in the LL quadrant in column 2 n + 1 are zero, Lemma 3 is also true. QED Even though Rule 102 is said to create the Sierpinsi Gaset, we could not find any formal presentation of this statement. Furthermore, Rule 102 is a discrete geometric object, while the Sierpinsi Gaset is a continuous object. Rule 102 is really Pascal s Triangle Mod 2 in disguise (to be shown below) which is also said to create the Sierpinsi Gaset, but again we could not find any formal presentation of this statement in the literature. To mae these ideas more complete (and even to mae sense in the first place) we include in this paper the following section, where we first define, mimicing the usual definition of a (continuous) fractal, a discrete geometric fractal (DGF) and then show how Rule 102 and Pascal s Triangle are DGF s that closely approximate the shape of Sierpinsi Gaset. 3 DISCRETE GEOMETRIC FRACTALS, THE 1D CA RULE 102, PASCAL S TRIANGLE MOD 2 AND THE SIERPINSKI GASKET It is widely quoted that Pascal s Triangle Mod 2 and the 1D CA Rule 102 produce the Sierpinsi Gaset. The original 1915 definition of the Sierpinsi Gaset (see [4]) is of a continuous geometric object, but Pascal s Triangle Mod 2 and the 1D CA Rule 102 are discrete objects. In this section we formalize several important facts in this respect. Fractals are defined in [7] as a rough or fragmented shape that can be split into parts, each of which is (at least approximately) a reduced-size copy of the whole. For example, the main fractal we are concerned with here, the Sierpinsi Gaset, is usually constructed recursively as illustrated below: Note that each term in the sequence illustrated in Fig. 6 is obtained from the previous term by maing the substitution shown in Fig. 7 everywhere possible within the previous term. The general idea in the process above can be adapted to a discrete situation. In order to mae sense of the definition of fractal in the case of Pascal s FIGURE 6 Sequence leading to Sierpinsi Gaset

12 432 J. CARBONARA et al. FIGURE 7 Substitution in the above sequence Triangle Mod 2, which is discrete, below we create a definition of discrete geometric fractal which closely mimics the continuous case. Definition 1. A discrete geometric fractal DGF is defined as lim n SF[n] (using the Hausdorff s measure, see [8]), where SF[n], a discrete geometric figure, is constructed with the following components and procedures: - A discrete geometric shape S, called the recursive figure. - An infinite sequence of discrete geometric shapes with copies of S in it. The n th element of this sequence called the subfractal SF[n]. The first subfractal, called SF[1], is the recursive figure S. - A discrete geometric shape s, called the recursive subfigure, contained and clearly located in S. The subfigure s is a scaled down version of S by a magnification factor m. - A rule R to create SF[n + 1] from SF[n]. The rule R is characterized by the following: i) Locates copies of S in SF[n] which are clearly identified by construction and transforms them into a different discrete shape, following a template that explains how to transform a single S. ii) To find the replacement of each S (i.e. what S is transformed into) the rule R locates each copy of s in S and dictates how many copies of s will be used (which is referred to as the number of parts p) and how will they be used to create the new shape. iii) After the transformation of S, each s is then replaced by S which allows the copies of S to be clearly identified in SF[n + 1]. This recursive process allows each part to be a reduced copy of the whole, as required for a shape to be a fractal. Example 1. Let the recursive figure, S, be FIGURE 8 Recursive figure S which by definition equals SF[1], and the recursive subfigure, s, be each of the sets of three points circled below

13 THE CUPS AND STONES COUNTING PROBLEM 433 FIGURE 9 Recursive subfigures in S which is a scaled down version of S by a magnification factor of 3. We then apply the rule to the subfractal which tells us how many recursive subfigures are used and how we should place them. Here we use four recursive subfigures (the number of parts is four) and place them lie this: FIGURE 10 Intermediate step after applying the rule, part ii Then we change the recursive subfigures bac into the recursive figure. So SF[2] = FIGURE 11 Next subfractal (SF[2]) We can continue this process to ultimately determine the DGF which is the lim SF[n]. n Next we use the definition of DGF to relate the Sierpinsi Gaset to Pascal s triangle matrix Mod 2 (right shifted as needed). Let the recursive subfigure s = 1, the recursive figure S = 0 1 and let the rule be to replace each 1 1 copy of s in S by S. Then we get a sequence of subfractals SF[1], SF[2],... (see Fig. 12). We will show that the discrete geometric fractal obtained as lim n SF[n] is Rule 102 or equivalently a right shifted Pascal s Triangle Mod 2. Note that 0 is considered empty space, and that in our scheme, in going from SF[n] to SF[n+1] we also replace 0 by 0 0. Note that if we 0 0 replace each 1 by a dot in the subfractal above, omit the 0 s and draw them insidea1by1box, the figure of the limit is what is commonly nown as the Sierpinsi Gaset (see Fig. 13).

14 434 J. CARBONARA et al. SF[1]= FIGURE 12 First three subfractals SF[2]= SF[3]= FIGURE 13 Figures obtained from SF[1] to SF[6] replacing each 1 by a dot and ignoring the 0 s The exact identification of the figures above (either having 1 s and 0 s or just points) with the Sierpinsi Gaset follows from an identification of the corners of each triangle (except the three corners in the extreme top right, extreme bottom left and extreme bottom right) in the Sierpinsi Gaset with pairs of 1 s in Pascals Triangle (as in Fig. 14). Theorem 4. The 1D cellular automaton Rule 102 is equivalent to the right shifted Pascal s Triangle Mod 2. They are both discrete geometric fractals. Proof: It is clear that the one-dimensional cellular automata Rule 102 (x 00 0, x 01 1, x 10 1, x 11 0) is addition Mod 2 of the middle and (( right ) numbers. ( On) the( other hand, )) the recursion for Pascal s Triangle is n n 1 n 1 = + which we tae Mod 2. So the right 1 justified Pascal s Triangle Mod 2 and the matrix from Rule 102 with the initial condition (n 1) 0 s and one 1 are equal. Next, we show by induction that the Discrete Geometric Fractal with S = 0 1, s = 1, and the identity Rule where we just replace each s by S in each 1 1 instance of S is in fact Rule 102 (or equivalently Pascal s Triangle Mod 2). First we need two definitions. Define, for any positive integer i, the precursors of a digit in SF[i] to be the number directly above the digit and the number diagonally above and to the right of the digit. We define the daughter digits of a digit in SF[i] to be the four digits in SF[i + 1] that replace it. We now continue with the induction proof.

15 THE CUPS AND STONES COUNTING PROBLEM 435 FIGURE 14 First column: Steps in the construction of the traditional Sierpinsi Gaset. Second: Steps in the construction of the right justified Sierpinsi Gaset. Third: Corners of white triangles in the Sierpinsi Gaset. Fourth: Right justified Pascal s Triangle Mod 2 with pairs of 1 s corresponding to a single corner circled. The base case is true since SF[1] agrees with Rule 102. Now we will show that if SF[n] follows Rule 102, then SF[n + 1] also does. We will do this by considering an arbitrary digit in SF[n] (not in the first row or last column), and then showing that all four of its daughter digits in SF[n + 1] obey rule 102. First, suppose the arbitrary digit in SF[n] is a 1. Its precursors are either 01 or 10. Suppose its precursors are 01 (See Fig. 15a). If we now replace each of these digits with its corresponding daughter digits, we get the result shown in Fig. 15b. Finally, no matter what the digit to the right of the original 1 in SF[n] is, its top left daughter digit is a 0, as shown in Fig. 15c. If we now examine all four daughter digits (shown in bold in Fig. 15) of the original FIGURE 15 Transition from SF[n] to SF[n+1]

16 436 J. CARBONARA et al. FIGURE 16 Other transitions from SF[n] to SF[n+1] arbitrary 1 in SF[n], we see that they all obey rule 102. For instance, the top right bolded digit in Fig. 15c (a 1 ) has precursors 0 and 1 (shown in italics). Now there are three other cases to consider for an arbitrary digit in SF[n] - a 1 with precursors 10 and a 0 with precursors of either 00 or 11 (See Fig. 16a). Following the same steps as above to get the corresponding pattern in SF[n + 1], we get the results shown in Fig. 16c. Again it is clear that all of the daughter digits of the original digit in SF[n] obey rule 102. Since the original digit in SF[n] was arbitrary, we have shown that SF[n + 1] obeys rule 102. QED 4 CONCLUSIONS We have shown that the reduced matrix representation of CSCP when = 2 n + 1 is equivalent to Rule 102 and Pascal s Triangle Mod 2. The Sierpinsi Gaset, which is a fractal and a continuous figure, is commonly identified with Pascal s Triangle Mod 2, which is a discrete figure. We therefore defined a discrete version of fractals (Discrete Geometric Fractals) which have the property that parts of it are identical to the whole (as it is usually required by continuous fractals). Using this definition we showed that Rule 102 (and hence the reduced matrix representation of CSCP when = 2 n + 1) is a Discrete Geometric Fractal (in fact, a discrete version of the Sierpinsi Gaset).

17 THE CUPS AND STONES COUNTING PROBLEM ACKNOWLEDGMENTS The authors J. Carbonara, M. Jansma, M. Rua and K. Sember gratefully acnowledge the support of the CSUMS program of the National Science Foundation grant numbers and REFERENCES [1] B. Cipra. Proposed problem Mathematics Magazine, 65 N1:56, [2] D. Ettestad and J. Carbonara. Formulas for the number of states of an interesting finite cellular automaton and a connection to pascal s triangle. Journal of Cellular Automata, Vol. 5:pp , [3] M Casartelli E Agliari and E Vivo. Metric characterization of cluster dynamics on the sierpinsi gaset. Journal of Statistical Mechanics: Theory and Experiment, 2010, [4] W. Sierpinsi. Sur une courbe dont tout point est un point de ramification. C. R. A. S., 160: , [5] S. Wolfram. A New Kind of Science. Wolfram media Inc, [6] D. Ettestad and J. Carbonara. Fractal properties of the matrix for the cups and stones counting problem. International Journal of Pure and Applied Mathematics, 29 No 1:81 106, [7] B.B. Mandelbrot. The Fractal Geometry of Nature. W.H.Freeeman and Company, [8] Kenneth Falconer. Fractal Geometry. Wiley, 2003.