Extending the Sierpinski Property to all Cases in the Cups and Stones Counting Problem by Numbering the Stones

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1 Journal of Cellular Automata, Vol. 0, pp Reprints available directly from the publisher Photocopying permitted by license only 2014 Old City Publishing, Inc. Published by license under the OCP Science imprint, a member of the Old City Publishing Group. Extending the Sierpinsi Property to all Cases in the Cups and Stones Counting Problem by Numbering the Stones DAVID ETTESTAD 1 AND JOAQUIN CARBONARA 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: April 30, Accepted: May 6, 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. Ettestad and Carbonara (2010 and 2011) noted that this system is a finite Cellular Automaton, showed two interesting non-recursive formulas for S,1 and showed that the shape of the non-zero terms in the reduced matrix for the Cups and Stones Counting Problem (CSCP) with exactly 2 n + 1 cups is equivalent to Sierpinsi s gaset. We call this the Sierpinsi property. In this paper we extend the problem by numbering the stones, thereby revealing several new and interesting properties of the game. In particular we extend a slightly modified version of the Sierpinsi property to the CSCP with any number of cups by defining a home cup and referencing all the other cups to the home cup. 1 INTRODUCTION 1.1 Definition of the CSCP game Combinatorially, the CSCP can be stated as follows. Definition 1 (The Cups and Stones Counting Problem (CSCP)). General idea: Given an initial configuration and a transition rule to go from one configuration to the next, count the number of different configurations possible. 1 JCA 0166 CarbonaraEttestad V3 1

2 2 DAVID ETTESTAD AND JOAQUIN CARBONARA 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.) to obtain new configurations: Redistribute all the stones in the root cup by removing one stone at a time and placing the stones in consecutive cups in a clocwise fashion, starting at the cup adjacent to the root. The last cup to receive a stone on each application of the T.r. becomes the root in the resulting configuration. Note that although the T.r. changes the location of the root, the numbering of the cups doesn t change. Configurations: Configurations consist of a circle of numbered cups, some having stones, and one cup denoted as root. Two configurations will be consider equal if they are identical when ignoring the numbering of the cups. 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 C SC P,σ. 1.2 CSCP is a Cellular Automaton and the CA theory can be used to understand its properties The CSCP can be viewed as a 1D Cellular Automaton (finite and circular). When the problem of finding the number of different configurations of the CSCP was originally posed (see [4]) a number of partial solutions were proposed (see [5]), none of which used ideas from Cellular Automata theory. In this paper we show (see Figures 9 and 11 and Lemmas 11 and 12) that the CSCP is completely described by the 1D CA rule 102, or rule 60 (see [6]) or Pascal s Triangle modulo 2 (which are all in essence the same). Anything we now about those CA can be used to understand the CSCP and vice versa. The CSCP can be viewed at different granularity levels (see Subsection 1.4) which brings a new perspective to help understand such CA. For example, the 1D CA rule 102 is not reversible, but the CSCP at the configuration level of granularity is reversible. 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 [6] for details. To visualize the CSCP 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 JCA 0166 CarbonaraEttestad V3 2

3 EXTENDING THE SIERPINSKI PROPERTY 3 = 3, σ = 1 S 3,1 = = M 3,1 Bolded circle Special cup Root Underlined entry FIGURE 1 Cups and Stones Counting Problem 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 Figure 1. From now on fix σ to be 1, denote M,1 simply by M, S,1 by S and CSCP,1 by CSCP or just CSCP if is clear from the context. The authors derived formulas for S in [2]. Lemma 1. The transition rule in C SC P will generate different configurations until all the stones are in the root cup. The following configuration will be the one with one stone in every cup, i.e. the initial configuration. Proof. The lemma follows since the transition rule is invertible. 1.3 Overview Now that game has been described, we present an overview of our paper: In the rest of section 1 we give some important definitions and lemmas to lay the foundation for rest of the paper. In section 2 we prove an important theorem that says all but possibly the last two stones placed from the root cup go into empty cups. In section 3 we introduce a novel and revealing way of looing at the CSCP s, namely the numbering of all but one of the stones. Then we use the results of section 2 to show that the numbered stones stay in order for all configurations in all CSCP s. This preliminary result is needed to build to the main theorem of this paper. In section 4 we use the ordering property of the numbered stones to show that the path of any particular numbered stone relative to a home cup doesn t depend on the number of cups in the CSCP. JCA 0166 CarbonaraEttestad V3 3

4 4 DAVID ETTESTAD AND JOAQUIN CARBONARA In section 5 we use the results of the previous sections to extend a slightly modified (and for all computational purposes equivalent) Sierpinsi property to CSCP s with any number of cups, which is the main result of this paper. 1.4 Ways to disect the CSCP game From the definition of the game, it is clear that looing at the various configurations in a given CSCP is the most natural way of analyzing the game. For most of this paper we will do just that. However, for certain proofs and to understand the game better it sometimes helps to have other ways to dissect the evolution of the game in time. Here we present two other methods. Another way to dissect the game is to consider the situation after each individual stone drop as a frame, lie frames in a movie. To do this we brea up the transition described in Definition 1 into parts. Each part consists of picing up one stone from the root cup and putting it into another cup. As before the stones will be put into cups in order, starting with the cup immediately after the root cup in a clocwise direction. The situation after each individual stone is dropped is what we call a frame. The initial configuration is also defined to be a frame. More formally, a frame is defined below. Definition 2. A frame is defined as follows: the initial configuration is a frame. Then the distribution of stones obtained by moving just one stone (as part of the CSCP transitions) is also a frame. See Figure 2. Note that all configurations are frames, but some consecutive configurations will have one or more frames in between them. The third way to dissect the game is by cycles and alignment frames as defined below. Definition 3. In a CSCP with cups, frame n is an alignment frame iff mod +1 (n) = 1. The set of frames in between alignment frames is called a cycle. Alignment frames, which are introduced in this paper for the first time in the CSCP context, are critical to many of the proofs in the rest of the paper. These three ways to organize the game for the case of four cups are shown in Figure 2a. Suppose one loos at the CSCP and doesn t consider the individual transitions or configurations, but simply loos at the order that stones are put into cups. From the transition rule it is clear that one stone is put into each cup in order clocwise. In other words, if a stone is put into cup i at a particular time, then the next stone that is placed will be into cup i + 1, regardless of JCA 0166 CarbonaraEttestad V3 4

5 EXTENDING THE SIERPINSKI PROPERTY 5 FIGURE 2 (a) Configurations, frames and cycles/alignment frames (b) Relationship between drop cup, root cup and home cup. whether the stone placement occurs in the same transition or the following one. As an example, if there are 4 total cups (numbered as in definition 1) then the order of the cup numbers of the stone placements will be 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4,... In general, for n > 1, in frame n a stone is dropped into cup [mod (n)]. Let the cup [mod (n)] in frame n 1 be called the drop cup of the frame. See Figure 2b. 1.5 The Home Cup We will now define one of the most important concepts in this paper, the home cup. Definition 4. Consider the CSCP with cups. In addition to having a root, for each frame we now define a home cup as follows. If a frame is alignment frame i or any subsequent frame up to but not including alignment frame i+1, the home cup is cup mod (i). The relative cup number of a cup (in contrast to the cup number as introduced in Definition 1, which sometimes will be referred to as the absolute cup number) is the number of cups it is JCA 0166 CarbonaraEttestad V3 5

6 6 DAVID ETTESTAD AND JOAQUIN CARBONARA clocwise from the home cup. The relative cup number of the home cup is zero. Thus, in the first frame and the entire first cycle, the home cup is cup 1. In the second alignment frame and the entire second cycle, the home cup is cup 2, etc. See Figure 2b. Note that the home cup moves to a new cup in each alignment frame but stays in the same cup throughout any cycle. In particular the home cup moves one cup clocwise at each alignment frame. There is a ind of symmetry between the root and the home cup. Between any two consecutive configurations, the root is the same in all frames. Between any two consecutive alignment frames, the home cup is the same in all frames. 1.6 The Staircase Matrix The home cup will provide a reference point in addition to the original numbering of the cups (which correspond to the column numbers of M ). In fact, we will now define a new matrix M, where we shift M so that the home cup is the first element in each row (but stays in the same column); in such cases, the matrix would loo lie a staircase with steps of different size. The following definition will be useful dealing with this change in the frame of reference. Definition 5. Consider the CSCP problem with cups. Label the cups as in definition We define the staircase matrix M as the matrix of configurations where the first cup in each row is always the home cup. To obtain M from M, we proceed as follows: Say in configuration g the home cup is cup n; then move cups 1 through n 1 by displacing them to the end of row g, so that cups 1 through n 1 occupy columns + 1 to + n 1. See Figure The home cup determines a partition of the matrix M (as well as M ) into sections (call them step sections). Each step section consists of the set of consecutive rows where the home cup is in the same column. Later we will show that all alignment frames are also configurations and therefore rows in the staircase matrix. So we could also define step sections by saying they each consist of an alignment frame and the configurations in the following cycle. Note that M is not a matrix in the traditional sense. M loos lie a staircase, where each indentation occurs at the places where the home cup JCA 0166 CarbonaraEttestad V3 6

7 EXTENDING THE SIERPINSKI PROPERTY 7 M 4 = M4 = Framed entry Home cup Underlined entry Root cup FIGURE 3 Obtaining M from M changes. In addition, the staircase matrix is a new way to visualize the configurations. Figure 3 illustrates the home cup shift for = 4; note that there are four step sections. 1.7 The Fundamental Lemma Part 2 In [3], we presented a fundamental lemma with two parts. The first part only applied to CSCP with number of cups of the form 2 n + 1. The second part applies to all CSCP, and it was proven as Lemma 26 in [3]. We present it below after an important definition. Definition 6. Consider row i in a staircase matrix M. If the root in that row is the nth term in the row and has m stones in it, we define C (i) = n, and (i) = m. R Note that C (i) + R (i) determines the location of the root in row i + 1. Lemma 2 (Fundamental Lemma Part 2). Let M be the staircase matrix for the CSCP, where is any integer and let n = log 2 [ 2]. Then 1. C (i) + R (i) > [C (i) + R (i) = + 1 or C (i) + R (i) = + 2]. 2. M has 2n+1 step sections. 3. The last row of M is (0,, 0, 0, 0,..., 0). Statement 1 of Lemma 2 is very important and we would lie to explain it more in words. First we must note a couple things: 1. Any frame in which the drop cup is the root must be a configuration. This follows from the definition of the root cup in definition 1. JCA 0166 CarbonaraEttestad V3 7

8 8 DAVID ETTESTAD AND JOAQUIN CARBONARA 2. In the last frame of any cycle the drop cup is the home cup. In the following frame (which is an alignment frame), the drop cup is the following cup, which is also the new home cup. Statement 1 of the Fundamental Lemma Part 2 says that at least one of the two frames mentioned in item 2 above must be a configuration. Fundamental Lemma part 2 forms the basis for the following important lemma. Lemma 3. In any alignment frame: 1. The home cup is the root. 2. There is only one stone in the home cup. 3. The frame is a configuration. Proof. We will use induction on the alignment frame number i. For alignment frame 1, (the first configuration), the lemma is clearly true. Now assume the lemma is true for all alignment frames up to and including alignment frame i, which we will also designate as frame j. Thus the home cup in alignment frame i is the root cup, has only one stone in it, and is a configuration. By the transition rule, in the following frame (frame j + 1), the home cup will be empty, the cup immediately after the home cup will be the root, and frame j + 1 will be a configuration. Therefore in frame j + 2 the cup immediately after the home cup will also be empty. In fact, both the home cup and the cup immediately after it will be empty in all of the remaining frames of that cycle (while every other cup gets a stone put into it) until the very last frame of the cycle, when a stone is put into the home cup. As stated above, according to Fundamental Lemma part 2, either this last frame of the cycle or the next frame (an alignment frame) will be a configuration. Let us consider both possibilities: If the last frame of the cycle is a configuration, then this means the home cup is a root and clearly only has one stone in it, since it had just been empty. Therefore, in the next frame the following cup will be the root and have just one stone in it. (Remember it had been empty also.) Since this next frame is immediately after the last frame of a cycle, it is also an alignment frame, and thus the drop cup, in addition to being the root, is also the new home cup. Since the previous root cup only had one stone in it, this frame will also be a configuration. Thus, in this case the alignment frame i + 1 satisfies all three parts of the lemma. On the other hand, if the next alignment frame is a configuration, then the cup after the old home cup is the root. Since this is an alignment frame, this is JCA 0166 CarbonaraEttestad V3 8

9 EXTENDING THE SIERPINSKI PROPERTY 9 simply the new home cup, so the home cup is the root cup. Since this cup had been empty for the last several frames of the previous cycle, it will have only one stone in it in this alignment frame. Again all three parts of the lemma are satisfied. Having just shown that all alignment frames are also configurations, we add the following definition for clarity: Definition 7. The configurations corresponding to alignment frames are called alignment configurations. The fact that in every alignment configuration the home cup is the root cup and only has one stone in it is very critical to the structure of the CSCP games. This fact is what leads to the similarity among the M s for different numbers of cups. In particular it leads to the core structure of the M used in the proof of the theorem in the next section. 2 EMPTY CUPS THEOREM We now introduce a ey theorem that forms the basis for much of the rest of the paper. Theorem 1. Consider a CSCP with cups. In any configuration, if there are p stones in the root cup, the next p-2 cups (clocwise) will be empty. If we thin about the process of placing stones from the root cup into other cups, this theorem says that all except possibly the last two stones are always placed into empty cups! Outline of proof: We will use induction on. The main point of the proof is that the matrix M for any can be broen up into four main sections, three of which are equivalent to the matrices for smaller values of and one of which is nearly all zero s (see comment in Figure 4). By induction the theorem wors in the smaller matrices and it is simple to show the theorem also is true in the section of nearly all zero s. Before we give the actual proof of the theorem, we present an important definition and explain how to divide up M into four parts. In [3] (Theorem 32 and section 5.2) we showed that because of the Fundamental Lemma, any matrix M can be seen to be a combination of matrices of smaller. To see his more clearly we need to define the cores of the matrices. JCA 0166 CarbonaraEttestad V3 9

10 10 DAVID ETTESTAD AND JOAQUIN CARBONARA FIGURE 4 (a) Four parts (T 1, T 2, T 3, T 4 )ofm7 ; (b) Matrix M 5 with the cores shaded. The diagonal shaded entries correspond to home cups that are also root cups; (c) Matrix M3 with the cores shaded. Comment: The shaded areas in T 1 and T 3 correspond to the shaded areas of (b) and the shaded areas of T 2 correspond to the shaded areas of (c). Definition 8. Let (M ) 1,...,(M ) m be the step sections of M. Each of the step sections is a rectangular submatrix in M. We define the core of each step section to be the submatrix of (M ) i obtained from (M ) i by removing the rows where the root is at the home cup, and by eliminating as well the entries corresponding to the home cup. To help visualize what the cores are, see Figures 4b and 4c. We now describe how to divide M up into four parts (T 1, T 2, T 3, T 4 ) for a general. JCA 0166 CarbonaraEttestad V3 10

11 EXTENDING THE SIERPINSKI PROPERTY 11 Consider a CSCP with cups. Let n be the largest integer such that > 2 n + 1. Let = 2 n j. Clearly j < 2 n + 1. Let the first row that has at least 2 n + 1 stones in the root be called the divider row. Then the top parts (T 1 and T 2 ) will contain all the rows above and including the divider row while the bottom parts (T 3 and T 4 ) will contain the rows below the divider row (see Figure 4a). The left parts (T 1 and T 3 ) will contain the first 2 n + 1 entries in each row of M. The right parts (T 2 and T 4 ) will contain the last j entries in each row of M. We are now ready to prove Theorem 1. Proof. We will use induction on the number of cups. By inspection it is true for = 1. Assume it is true for all CSCP with the number of cups equal to 1 through 1. We now show it is true for cups. It is clear that if a root has only one or two stones in it, the theorem automatically holds. Suppose the root has r stones in it, with r > 2. Since the next root will then be r cups past the present root, we need to show that all the cups in between the present root cup and the cup right before the next root cup are empty. In [3] (Corollary 28, statement 3 and Corollary 29, statement 1), it was shown that whenever the home cup is a root, it has exactly one stone in it, so it obeys the theorem automatically. Thus we only need to consider the roots that are not in home cups. In [3] (Corollary 28, statement 1), the rows in T 1 with the root not in the home cup were shown to be identical to the core sections of M2 n +1, so the theorem is true for all roots in T 1 by induction. A possible exception is the root in what corresponds to the last row in each of the core sections of M2 n +1, since the next root will be outside this particular core. However, by Lemma 2 applied to M2 n +1, the next root in M 2 n +1 will be at most two cups past the last cup in the core. This means that the next root in M will be at most two cups past the last cup in that row in T 1, which means that all the cups that must be empty are actually in the core, and therefore by induction these cups are empty. A similar line of thining proves the theorem for T 2. In [3] (Corollary 28, statement 2), the rows in T 2 with the root were shown to be identical to the core sections of M j+1, so the theorem is true for all roots in T 2 by induction. The same possible exception applies to the roots in the last rows of the cores, but again Lemma 2 taes care of this exception. Also from [3] (Corollary 29, statement 2), the cores within part T 3 are identical to the cores within M2 n +1 except for the fact that one particular cup always has j = 2 n 1 more stones in it. Because of [3] (Corollary 29, statement 2 and Lemma 10) this particular cup always had at least one stone in it even in M2 n +1 (and therefore will not change which cups are empty), and JCA 0166 CarbonaraEttestad V3 11

12 12 DAVID ETTESTAD AND JOAQUIN CARBONARA since it is never the root until the very last row of M2 n +1 (and therefore won t change the number of stones in the root), the theorem still holds by induction. Again Lemma 2 taes care of the cases where the root is the last root in each core section. For the very last row of M2 n +1, by Lemma 2 all stones are in the root cup so clearly the next 2 cups are empty. In T 4 it was shown in [3] (Corollary 29, statement 3) that the root always has only one stone in it so the theorem is automatically true in that part. Since we have shown that the roots in all four parts of M follow the theorem, the entire matrix follows the theorem. Then by induction on the theorem holds. 3 STONE ORDERING Theorem 1 is a fundamental theorem that forms the basis for the rest of this paper. Many interesting results follow from it. One ey result is that, if the stones are numbered and we define an ordering of the stones in the cups, then, with the exception of one particular stone, the order of the stones is the same in all configurations. The one particular stone is the one that was in the root cup in the first configuration. It is very special and behaves very differently than all of the other stones. In fact, to help with this distinction, we will name both the special stone and the other stones. The special stone will be called the gardener and the other stones, which are the ones to be numbered, will be called numbered stones or simply n-stones. These distinctions are summarized in the following definition. Definition 9 (Gardener and stone number definition). Consider the initial configuration of a CSCP, with the cups numbered as in definition 1. The stone that is in the root cup (cup 1) is called the gardener. It will remain the gardener in all configurations, regardless of what cup it is in. All the other stones will be called numbered stones or simply n-stones. The stone in cup i (i > 1) in the initial configuration is labeled stone i-1. Each stone maintains its label throughout the CSCP, regardless of what cup it is in within a particular configuration. Thus, in any CSCP with cups there is one gardener and 1 numbered stones. Definition 10 (Matrix and shifted matrix with numbered stones). In complete analogy with the case of stones without numbers, we define a matrix where each entry is the ordered set of stones in a given cup. We will denote the equivalent to the matrix M by N M (the numbered stone matrix) and JCA 0166 CarbonaraEttestad V3 12

13 EXTENDING THE SIERPINSKI PROPERTY {G} {1} {2} {3} {} {G, 1} {2} {3} {} {} {1, 2} {G, 3} {G} {1, 2} {} {3} {} {G, 1, 2} {} {3} {G} {} {1} {2, 3} {G} {1} {2, 3} {} {} {G, 1} {2, 3} {} {} {} {1, 2, 3} {G} {G} {} {1, 2, 3} {} {G} {1, 2, 3} {} {} {} {G, 1, 2, 3} {} {} FIGURE 5 Left: M 4, Right: NM 4 where G represents the gardener stone and integer i represents stone i G p 2 p 4 p 1 p 3 p 5 Home cup FIGURE 6 Ordering of n-stones in a frame. The label p i represents the i th position of an n-stone. Note that the location of the gardener does not affect the order of the n-stones. the equivalent of M by N M example, see Figure 5. (the numbered stone shifted matrix). For an In order to prove that the numbered stones stay in order, we need to define how we determine their order and how the stones are distributed during the transitions between configurations. Definition 11 (Stone order). The order of the stones is determined as follows: Assume all the stones in each cup are staced vertically. All the stones in a given cup are ahead of all the stones of another cup that is further away (clocwise) from the home cup. Within each cup a stone is considered ahead of another stone if it is located above that stone in the same cup. See Figure 6. Note that in the initial configuration all of the numbered stones are in numerical order. Finally we need to describe which stones are placed into which cups during the transition between configurations. Definition 12 (Stone movement). During a transition between consecutive configurations, when the stones are piced up from the root cup and placed JCA 0166 CarbonaraEttestad V3 13

14 14 DAVID ETTESTAD AND JOAQUIN CARBONARA in the following cups, the numbered stone that was located highest in the root cup is placed in the the cup immediately after the root, located on top of the other stones already in the cup. The numbered stone that was located second highest in the root cup is placed on top in the second cup after the root, etc. When all numbered stones are placed in cups, if the gardener is among the stones piced up, it is placed on top in the cup after the cup where the last n-stone was placed. Lemma 4 (Gardener-Root Lemma). In any configuration, the gardener is always in the root cup. Proof. We ll use induction on configuration number n. The lemma is certainly true in the first configuration by definition 1. Now assume it is true for configuration n. By definition 1, the root cup in configuration n + 1 is the cup where the last stone is placed. But by definition 12, the gardener is the last stone placed. So the gardener is in the root cup in configuration n + 1, and, by induction, for all configurations. The gardener-root lemma leads to an interesting fact about the home cup: Lemma 5 (Home cup/gardener Lemma). In any configuration in any CSPC, no n-stone is ever in the home cup. Either the home cup is empty or it has only the gardener in it. Proof. In [3] (Corollary 28, statement 3 and Corollary 29, statement 1) it was shown that the home cup is either empty or is the root with one stone in it. If it is a root, that one stone must be the gardener by Lemma 4. With these preliminaries finished, we can now present our main ordering theorem. Theorem 2. In any configuration of any CSCP with any number of cups, stone i (for any i) is always in position i in the stone ordering. Proof. Consider a CSCP with cups and corresponding matrix M. Let configuration n be the configuration that corresponds to row n of M. We will use induction on the configuration number n. In other words, we ll assume that configuration n obeys the theorem and show this implies that configuration n + 1 obeys the theorem as well. By construction, the numbered stones in configuration 1 obey the theorem for any CSCP. Now assume that configuration n obeys the theorem. JCA 0166 CarbonaraEttestad V3 14

15 EXTENDING THE SIERPINSKI PROPERTY 15 The only numbered stones that are moved between configurations n and n + 1 are the ones in the root cup of configuration n. Suppose there are q numbered stones in that root cup. If q is zero, only the gardener moves and the n-stones clearly stay in the same order. If q is one, the n-stone that was in the root cup, which was ahead of all the n-stones in the cups following the root, will be placed on top in the cup immediately following the root cup. It will still be ahead of all the n-stones in both that cup and all of the other cups following the root cup, and therefore the order of the n-stones will not change. What if q is greater than 1? From definition 12 it is clear that all the n- stones that are taen from the root cup will maintain their order within themselves since the n-stones that were higher in the root cup go into the cups closer to the root and hence closer to the home cup. However, if there were any n-stones already in the cups they went into, the order of the n-stones as a whole could be changed. But by Theorem 1, the q 1 cups following the root cup are empty. (Remember the last stone placed is the gardener.) Only the very last n-stone (the qth n-stone) might go into a cup with other n-stones already in it. In that case the qth n-stone will be placed on top of the stac in that cup and still be ahead of all the n-stones below it, thus maintaining the order of all the n-stones. There is one other possibility we must consider. If some of the n-stones land in or past the home cup, then these n-stones, instead of being behind most of the other n-stones, would actually be ahead of them, since we measure position relative to the home cup. However, by Lemma 5, no n-stone can be in the home cup of configuration n + 1. Therefore, the farthest along an n-stone may be placed is the cup just before the home cup of configuration n + 1. Therefore the last n-stone placed is still behind the home cup and the order of the n-stones is maintained. Corollary 1 (stone drop ordering). Consider a root cup in a configuration in a CSCP containing stones j 1 and j 2 with j 1 < j 2. In the following configuration, if stones j 1 and j 2 are in cups with relative cup numbers i 1 and i 2 (respectively) then i 1 < i 2. Corollary 2 (root stone movement). Consider stone j in a root cup in a configuration in a CSCP. If there are i n-stones above stone j then in the following configuration stone j will have moved i + 1 cups. Proof of both corollaries. These corollaries follow from Definitions 11 and 12 and Theorem 2. JCA 0166 CarbonaraEttestad V3 15

16 16 DAVID ETTESTAD AND JOAQUIN CARBONARA The fact that the numbered stones stay in numerical order in all configurations leads to another interesting fact about the game. Lemma 6. Stone j never moves more than j cups between consecutive configurations. Proof. Stone j will only move if it is in the root cup. If it is in the root cup, by corollary 2, the number of cups an n-stone moves equals one more than the number of n-stones above it in the root cup. Since the n-stones are in order, there can be at most j 1 n-stones above stone j, and therefore stone j can move at most j cups between configurations. The labeling of the stones as the gardener and numbered stones helps one see another aspect of the CSCP game. While the numbered stones stay in order, the gardener hops all over the place, following the root cup. In fact, one could thin of the gardener as the one that actually pics up the stones from the root cup and distributes them into the following cups. After distributing all the stones, the gardener places himself in the cup after the one with the last deposited numbered stone. This is why we chose the name gardener for this stone. 4 THE PATH OF EACH NUMBERED STONE In this section we show that the location of a particular n-stone in a particular alignment configuration does not depend on the number of cups in the CSCP. Definition 13. In M (and N M ), each row corresponds to a configuration of a CSCP. Similarly, alignment row ninm (and N M ) corresponds to alignment configuration n of the CSCP. Definition 14. Consider a CSCP with cups. The path of stone j, where 1 j <, is the ordered list of the relative cup numbers (see definition 4) that the stone occupies in all of the alignment configurations. As an example the path of stone 2 for = 4 is (2,1,2,1). (See the right side of Figure 7.) Note that in the right side of Figure 7 the home cup of each configuration has been left aligned so that it is in the first column, allowing for easier viewing of the paths. Thus, in the left aligned NM4 each column contains cups with the same relative cup number while in the regular NM4 each column contains cups with the same absolute cup number. JCA 0166 CarbonaraEttestad V3 16

17 EXTENDING THE SIERPINSKI PROPERTY 17 {G} {1} {2} {3} {} {G, 1} {2} {3} {} {} {1, 2} {G, 3} {G} {1, 2} {} {3} {} {G, 1, 2} {} {3} {G} {} {1} {2, 3} {G} {1} {2, 3} {} {} {G, 1} {2, 3} {} {} {} {1, 2, 3} {G} {G} {} {1, 2, 3} {} {G} {1, 2, 3} {} {} {} {G, 1, 2, 3} {} {} {G} {1} {2} {3} {} {G, 1} {2} {3} {} {} {1, 2} {G, 3} {G} {1, 2} {} {3} {} {G, 1, 2} {} {3} {G} {} {1} {2, 3} {G} {1} {2, 3} {} {} {G, 1} {2, 3} {} {} {} {1, 2, 3} {G} {G} {} {1, 2, 3} {} {G} {1, 2, 3} {} {} {} {G, 1, 2, 3} {} {} FIGURE 7 LEFT: NM4 ; RIGHT: Left aligned NM 4 with alignment rows shaded. Note that stone 2 has been bolded in the alignment rows to show its path. We now introduce the concept of a home position for each n-stone because it is where each n-stone repeatedly goes bac to. Definition 15. The home position for stone j in an alignment row is j cups past the home cup. Note that in the initial configuration all n-stones are in their home position. Theorem 3 (Stone path theorem). Consider any CSCP that has a stone j and an alignment row n. The path of stone j up to alignment row n is independent of the number of cups in the CSCP. Equivalently, one could say that the relative cup location of stone j in alignment row n is independent of the the number of cups in the CSCP. As an example, suppose in the CSCP with 40 cups, stone 7 is in relative cup number 5 in the fourth alignment row. Then, according to the theorem, in the CSCP with 433 cups, stone 7 is also in relative cup number 5 in the fourth alignment row. Proof. We ll use double induction on stone number j and alignment row number n. For j = 1, we ll show that the path is simply a sequence of 1 s by using induction on alignment row number n. Clearly in the first alignment row (the initial configuration) stone 1 is in relative cup 1. Now assume in alignment row n stone 1 is in relative cup number number 1. By Lemma 3 in any alignment row the home cup is the root cup with only one stone in it, so in the next configuration the root cup will be the next cup (the cup with stone 1 in it). By Corollary 2 and Theorem 2, stone 1 will therefore be in relative cup 2 in the following configuration. Thus when the home cup moves forward one cup in alignment row n + 1, stone 1 will again be in relative cup 1. Therefore, by JCA 0166 CarbonaraEttestad V3 17

18 18 DAVID ETTESTAD AND JOAQUIN CARBONARA induction on n, stone 1 will be in relative cup 1 in all alignment rows. Thus the path of stone 1 is always just a sequence of 1 s. Therefore, for any CSCP that has an alignment row n, the path of stone 1 starts off with n 1 s. This proves the theorem for j = 1. Now assume the theorem is true for stone j. We ll show it s true for stone j + 1. We ll again use induction on alignment row number n. In alignment row 1 (corresponding to the initial configuration) stone j + 1 is always in relative cup j + 1, so it satisfies the theorem there. Now assume the theorem is true for stone j + 1 in alignment row n. We ll show it s true for stone j + 1in alignment row n + 1. The relative cup number of stone j + 1 in alignment row n + 1 is determined by the absolute cup numbers of both the home cup and stone j + 1in alignment row n + 1. We note that the home cup advances one absolute cup number each alignment row by definition regardless of the total number of cups. The position of stone j + 1 in alignment row n + 1 depends on whether its cup becomes a root or not before the next alignment row. We will now show that this does not depend on the total number of cups. By Lemmas 3 and 4, in alignment configuration n the root is the home cup, which has only the gardener in it. Whether or not the root will land in the cup where stone j + 1 is thus depends on the location of all the n-stones between the home cup and stone j + 1. Because of the ordering of the n-stones, this only includes stones 1 to j. But by induction on j, these must all be in the same relative cup regardless of the total number of cups. Thus, whether or not the root lands where stone j + 1 is located is independent of the number of cups. Next we show that, both in the case where stone j + 1 is in the root cup before the next alignment row, and in the case where it is not, that its relative cup in the next alignment row is independent of the total number of cups. If the root does not land in the cup where stone j + 1 is, the absolute cup number of stone j + 1 stays the same, its relative cup number decreases by one in the next alignment row (because the home cup moves forward one cup) and its relative cup number clearly does not depend on the total number of cups. If the root does land in the cup where stone j + 1 is, then by corollary 2 the number of cups stone j + 1 moves forward equals 1 plus the number of n- stones above stone j + 1 in the cup that it s in. However, by Theorem 2, only n-stones with number less than j + 1 can be above stone j + 1 in that cup. By induction on j, that number is independent of the total number of cups. Therefore, the number of cups stone j + 1 moves forward is independent of JCA 0166 CarbonaraEttestad V3 18

19 EXTENDING THE SIERPINSKI PROPERTY 19 the total number of cups. Therefore the relative cup number of stone j + 1in alignment row n + 1 is again independent of the total number of cups. Thus the location of stone j + 1 in alignment row n + 1 is independent of the number of cups in the CSCP. So by induction on alignment row number, stone j + 1 obeys the theorem. Furthermore, by induction on stone number j, all n-stones obey the theorem. It turns out that the path of every stone will start to repeat itself after a while, if there are enough cups. The following lemma gives more specifics to this statement. Lemma 7. The path of stone j will repeat after every 2 log 2 j terms. Proof. Consider a CSCP with = j + 1 cups. Then by Lemma 2, there are 2 log 2 j alignment rows in the CSCP (we have used the fact that log 2 ( 1) =1 + log 2 ( 2) ). By Lemma 2 the last configuration in this CSCP has all the stones in the root cup. Woring bacwards one can easily see that the previous configuration consisted of the gardener in the home cup and all j n-stones in relative cup 1 (and that it was an alignment configuration). Now consider a different CSCP with large enough that there are more than 2 log 2 j alignment rows. By Theorem 3 (Stone Path Theorem) in alignment row 2 log 2 j the first j n-stones would all be in relative cup 1. Then the next configuration would clearly again have the first j n-stones in the root cup. Then by corollary 3 in the next alignment row the first j n-stones would all be in their home positions 1. But this is precisely the arrangement of the first j n-stones in the first alignment row (i.e. the initial configuration). Since by the Theorem 3 (Stone Path Theorem) the higher numbered n-stones don t affect the relative cup numbers of the first j stones in the alignment rows, the paths of the first j n-stones would repeat themselves for next 2 log 2 j alignment rows. A similar argument shows that their paths would continue to repeat every 2 log 2 j alignment rows. This repetition in the paths is the basis for the following definition: Definition 16. Given a stone j in a CSCP, the pattern of stone j is the first 2 log 2 j terms of its path. 1 Note that although Corollary 3 is proven after the lemma we are proving, the proof of the corollary and Lemma 8 do not depend on Lemma 7. JCA 0166 CarbonaraEttestad V3 19

20 20 DAVID ETTESTAD AND JOAQUIN CARBONARA FIGURE 8 Paths and Patterns for the 8 n-stones of the CSCP with 9 Cups Examples of both paths and patterns are given in Figure 8. The data in Figure 8 suggests the following relationship among patterns: Conjecture 16. Given a stone j + 2 n in a CSCP with stones where 2 n j > 0, the pattern of stone j + 2 n is determined by the following two rules: 1. The first 2 n terms of the pattern are the same as the first 2 n terms of the path of stone j, except that 2 n is added to each term. 2. The last 2 n terms are simply 2 n, 2 n 1, 2 n 2,... 3, 2, 1. 5 THE SIERPINSKI PROPERTY FOR ALL CSCP S In [1] we showed that the 1-D Cellular Automata rule 102 could be used to construct the matrix of 0 s and 1 s obtained by replacing non-zero entries by 1 in the matrix of initiator rows for CSCP with = 2 n + 1. In other words the CSCP with = 2 n + 1 can be characterized by CA rule 102 (such Figure is a discrete version of the Sierpinsi Gaset). We now call this property of the CSCP with = 2 n + 1 the Sierpinsi property (SP). We now show, using results of this paper, that a slightly modified SP is true for any CSCP with any value of. First we need an important lemma. Lemma 8 (Stones Above Lemma). In any alignment row, if stone j is i cups from the home cup, then i j and the number of n-stones numbered less than j in that same cup is exactly j-i. Proof. We ll use induction on the alignment row number. In the first alignment row, stone j is alone in cup j, so the lemma holds. Now assume the JCA 0166 CarbonaraEttestad V3 20

21 EXTENDING THE SIERPINSKI PROPERTY 21 lemma holds in alignment row n. Between alignment row n and alignment row n + 1, either the cup with stone j becomes the root or it doesn t. If it does not become the root before alignment row n + 1, then in alignment row n + 1 that cup will have one additional n-stone. Because the new n-stone is added to the top of the cup and the n-stones are in order, that n- stone must be numbered less then j. This maes a total of ( j i) + 1 = j (i 1) n-stones numbered less than j in the cup. Stone j will not move to a new absolute cup and when the home cup moves one cup forward at the next alignment row, stone j will be i 1 cups from the home cup. Letting i = i 1, stone j is then i cups from the home cup and there are j i n- stones numbered less than j in the cup with stone j, so the lemma is satisfied in alignment row n + 1. On the other hand, if the cup that stone j is in does become the root before the next alignment row, by Corollary 2 stone j will move ( j i) + 1 cups forward before the next alignment row. Combining this with the fact that the home cup itself moves one cup forward at the next alignment row, stone j will be in relative cup number j (its home position) in the next alignment row. By Corollary 1 any n-stones numbered less than j in the root cup would have been dropped into a cup closer to the home cup so there will be no lower numbered n-stones in the cup where stone j is, so again the lemma holds for alignment row n + 1. Thus, whether or not the cup with stone j become the root before alignment row n + 1, the lemma holds for stone j in alignment row n + 1, and by induction, for all alignment rows. From the lemma and its proof above, the following corollary follows: Corollary 3. When a cup becomes a root cup between alignment rows, in the next alignment row all of the n-stones that were in that cup are in their home position. NOTE: From the above lemma and corollary, it is clear that, given a stone in an alignment row, only two things can happen to the stone before the next alignment row: 1. The cup that the stone is in does not become the root before the next alignment row and the stone stays in the same absolute cup. Since the home cup always moves one cup forward in the next alignment row, the relative cup number of the stone will decrease by one. 2. The cup that the stone is in does become the root cup before the next alignment row, and the stone ends in its home position in the next alignment row. JCA 0166 CarbonaraEttestad V3 21

22 22 DAVID ETTESTAD AND JOAQUIN CARBONARA It is enlightening to loo at this from the point of view of the home cup. From that perspective, between any two consecutive alignment rows, any n- stone will either move one cup closer to the home cup or it will jump all the way to its home position (i.e. the farthest away from the home cup it can go). Viewing a given n-stone s progression along several consecutive alignment rows, the stone will slowly move one cup at a time towards the home cup, and then every once in a while, it will jump bac to its home position and start all over again. This is illustrated in the path column of Figure 8. From Lemma 8 three more interesting corollaries and a lemma follow readily. Corollary 4. In an alignment row, stone j can never be farther from the home cup than its home position. Corollary 5. In an alignment row, if stone j is in relative cup i, then stones i, i + 1,i+ 2,, j are all in cup j. Proof. The only way to have j i n-stones numbered less than j (without having any n-stones numbered less than i because of corollary 4) is to have all numbered stones from i to j in the cup. Corollary 6. In an alignment row, no n-stone can be in relative cup i unless stone i is in the cup. If we say that stone i is the owner of its home cup (i.e. relative cup i), then Corollary 6 simply says that no n-stone can be in a cup unless the owner is present. Lemma 9. In an alignment row, if there are p n-stones in a cup, there will be exactly p 1 empty cups following it. Proof. Suppose the cup with p n-stones is relative cup i. Then by Corollary 5 these stones must be numbered i, i + 1,, i + p 1. Since the p 1nstones from i + 1toi + p 1 are not home, no other n-stones can be in their home cups. Therefore the following p 1 cups must be empty. On the other hand, because of stone ordering and Corollary 4, the only place stone i + p (if it exists) could possibly be located is relative cup i + p (i.e. its home cup). Whether or not stone i + p exists, there are exactly p 1 empty cups following relative cup i. JCA 0166 CarbonaraEttestad V3 22

23 EXTENDING THE SIERPINSKI PROPERTY 23 Corresponding cells FIGURE 9 Left: M init 33 Right: Ɣ 33 Before we prove the modified Sierpinsi property for all CSCP, we need to specify more clearly what both the original and modified Sierpinsi properties are. 5.1 The Original Sierpinsi Property The original Sierpinsi Property (SP), which we only proved for CSCP s with a number of cups of the form = 2 n + 1, used a matrix which consisted only of the rows of the matrix M where the root was in the first column. See the left side of Figure 9. These rows were called initiator rows in [3]. Let us therefore call this matrix M init. Definition 17. Given a CSCP with = 2 n + 1, let M init be the matrix consisting only of the rows of the matrix M where the root is in the first column. Unfortunately, the structure of CSCP for 2 n + 1 cannot be described using the original Sierpinsi Property. For example, from Figure 3, where = 4, there are only two initiator rows, and they do not follow the original SP. We have represented the CSCP as a matrix, first as a matrix of numbers, and then as a matrix of sets. Both Sierpinsi properties are based on the binary footprint of these matrices. Next we mae this definition more formal. Definition 18. The binary footprint of a matrix with numerical entries is obtained by replacing in it each nonzero entry by a 1. The binary footprint of a matrix with entries that are sets is obtained by replacing in it each non empty set by a 1 and each empty set by a 0. JCA 0166 CarbonaraEttestad V3 23

24 24 DAVID ETTESTAD AND JOAQUIN CARBONARA Definition 19. Given a CSCP with = 2 n + 1, let Ɣ be the binary footprint of the inverted matrix M init with the first column eliminated. We showed in [1] that Ɣ is identical to the 2D cellular automata obtained by rule 102 (see [6]), which is nown to produce a right aligned Sierpinsi gaset. See the right side of Figure 9. The fact that Ɣ follows rule 102 is the original Sierpinsi property. 5.2 The Modified Sierpinsi Property To explain the modified Sierpinsi property, we need to loo at several matrices that are formed by just the alignment rows of CSPC matrices. Definition 20. Given a CSCP, let M align be the matrix consisting only of the alignment rows of the matrix M. Similarly let (M align ) be the matrix consisting only of the alignment rows of the shifted matrix M. Let L(Malign ) be the left aligned (M align ). Finally let be the binary footprint of the inverted L(M align ) with the first column removed. See Figure M align 5 (M align 5 ) L(M align 5 ) 5 FIGURE 10 The alignment row matrices Definition 21. A CSCP matrix has the modified version of the Sierpinsi property if it can be embedded in the CA rule 60 grid formed (using an initial configuration of a 1 and the rest all 0 s) with the top left corner of over the only 1 in the top row of Rule 60. Note that CA rule 60 is the mirror image of CA rule Proving the Modified Sierpinsi Property for all CSCP Now we will show that has the modifed SP for all. First we show that, for the case = 2 n + 1, the original Sierpinsi property implies the modified version as well, and then we show that the modified SP for = 2 n + 1 implies the modified SP for any. Lemma 10. When = 2 n + 1, the original version of the Sierpinsi property implies the modified version. JCA 0166 CarbonaraEttestad V3 24