A NEW PUZZLE FOR ITERATED COMPLETE GRAPHS OF ANY DIMENSION

Transcription

1 A NEW PUZZLE FOR ITERATED COMPLETE GRAPHS OF ANY DIMENSION ELIZABETH SKUBAK AND NICHOLAS STEVENSON ADVISOR: PAUL CULL OREGON STATE UNIVERSITY ABSTRACT. The Towers of Hanoi puzzle can be use to label a family of graphs in a way that provies easy (finite-state) coing properties. This puzzle can be moifie to have any o number of towers (greater than three) an the rules ajuste so that similar easy labelings can be create on the corresponing o-imensional graphs. However, the puzzle cannot be irectly extene to an even number of towers an retain its essential structure. Are there other puzzles that exist on even-imensional graphs in this family? A puzzle for the 2-imensional case is known, an is even sol commercially as the Spin-Out Puzzle. We give an extension of this puzzle for graphs of any imension 2 m for m 1. We also explore combinations of this extene Spin-Out Puzzle with the Towers of Hanoi puzzles to create puzzles that correspon to the Sierpinksi graphs for any imension. In aition, we present two new labelings which have simple constructions an some easy coing properties. 1. INTRODUCTION The Towers of Hanoi puzzle is an interesting an often-stuie puzzle that also has curious applications to coing theory. In particular, it can be use to label a family of graphs in a way that provies easy (finite-state) coing properties[10]. In Section 2, we give backgroun information on this family, calle Sierpinski graphs or iterate complete graphs, as well as on coes an labels. It has also been proven in previous work (see [16],[14]) that the Towers of Hanoi puzzle can be moifie to have any o number of towers (greater than three) an the rules ajuste so that similar easy labelings can be create on the corresponing o-imensional graphs. We summarize these results in Section 3. However, the puzzle cannot be irectly extene to an even number of towers an retain its essential structure. Are there other puzzles that exist on these associate even-imensional graphs? A puzzle for the 2-imensional case is known, an is even sol commercially as the Spin-Out Puzzle, which we explore in Section 4. In Section 5, we give an extension of this puzzle to the 4-imensional case base on work by Weaver[28], an also exten this to a puzzle for graphs of any imension 2 m for m 1 in Section 6. We then give the recursive labels of the graphs corresponing to these puzzles, an show that there are finite-state machines that can recognize coewors as well as error-correct wors. In Section 7, we explore combinations of the extene Spin-Out Puzzle with the Towers of Hanoi puzzles to create new puzzles that correspon to the iterate complete graphs for all imensions, incluing the remaining even imensions. These are create by writing any imension as q 2 m for q o; the puzzle will have q towers but also pieces Key wors an phrases. graphs, coes, puzzles, Sierpinski graphs, iterate complete graphs, error-correction. This work was one uring the Summer 2008 REU program in Mathematics at Oregon State University

2 2 Skubak, Stevenson that have 2 m orientations on each tower. The rules of the puzzle are essentially a combination of the rules of both puzzles. As esire, they reuce to the rules for the known puzzles when the imension is o or a power of two. Our search for other labelings prouce two, the Corner-Distance an Subgraph labelings, which have intuitive constructions an several esirable characteristics. We stoppe short of consiering puzzles for them. Section 8 covers the labelings an some basic results. CONTENTS 1. Introuction 1 2. Graphs, Labels, an Coes Iterate Complete Graphs Coes on Graphs: Perfect One-Error-Correcting Coes Labelings an Coewors The G-U Construction 4 3. The SF Labeling an Puzzle Construction of the SF Labeling The SF Puzzle 7 4. Dimension 2: The Spin-Out Puzzle Graph an Label Coewor Recognition Error Correction Dimension 4: The Reflection Puzzle Rules Graph an Label Dimension 2 m Rules Graph an Labeling Recursive Labeling Coewor Recognition Error Correction Other Even Dimensions Dimension General Dimensions Conclusions on the Puzzle for General Dimension New Labelings The Corner-Distance Labeling The Subgraph Labeling Conclusions on the New Labelings 38 References

3 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 3 2. GRAPHS, LABELS, AND CODES In this section we give some backgroun on the specific graphs we will be working with, as well as the basic efinitions for coes on graphs. Together this information forms the basis for our investigations Iterate Complete Graphs. Definition 2.1. A (simple) graph G = (V,E) consists of a finite set V (G) (calle vertices) an a set E(G) (calle eges). Elements of E are unorere pairs of elements of V. Two vertices v 1 an v 2 are ajacent (have an ege between them) if (v 1,v 2 ) E. The ajective simple inicates that any two vertices have at most one ege between them, an that no vertex is ajacent to itself. Definition 2.2. The egree of a vertex v is the number of vertices which are ajacent to v. Definition 2.3. The complete graph on vertices, enote K, is the graph such that all the vertices are pairwise ajacent. That is, V (K ) = an E(K ) = {unorere pairs (a,b) : a,b V (K )}. Figure 1 shows some complete graphs. FIGURE 1. The complete graphs K 3, K 5, an K 8. Definition 2.4. An iterate complete graph, also known as a Sierpinski graph[15], on vertices with n iterations, enote K n, can be efine recursively. K1 is the complete graph on vertices. K n is compose of copies of Kn 1 an eges such that exactly one ege connects each K n 1 subgraph to every other K n 1 subgraph an exactly one vertex in each of the K n 1 subgraphs has egree 1. We say that a graph K n has imension. Definition 2.5. A subgraph M of a graph G consists of a subset V(M) V (G) together with the associate eges. In particular, the copies of K n 1 from which K n is constructe are all subgraphs of K n. Iterate complete graphs are easier to explain using examples. The graph K n can simply be thought of as copies of K n 1 connecte in a nice way, or alternatively as the graph K n 1 with each vertex replace by a copy of K. Figure 2 shows the graphs K6 1, K2 6 an K3 6, illustrating how each graph is constructe from the graph of the previous imension. 106

4 4 Skubak, Stevenson FIGURE 2. The iterate complete graphs K 1 6, K2 6 an K3 6. Definition 2.6. A corner vertex, or simply corner, of the graph K n is a vertex with egree 1. A non-corner vertex is simply a vertex that is not a corner. All non-corner vertices of iterate complete graphs have egree Coes on Graphs: Perfect One-Error-Correcting Coes. Definition 2.7. Let G be a graph an let V be the set of vertices of G. Then a coe on G is a subset C V. A coevertex is a vertex c C. A noncoevertex is a vertex v / C. Definition 2.8. A perfect one-error-correcting coe (or P1ECC) on a graph G is a coe such that: (1) No two coevertices are ajacent. (2) Every noncoevertex is ajacent to exactly one coevertex. Examples of P1ECC s can be foun in the left-han graphs in Figures 3, 4, an Labelings an Coewors. Definition 2.9. A labeling on K n is a metho of assigning strings to the vertices of Kn such that this metho gives a bijection between vertices an strings. The string assigne to a vertex will be calle the label of that vertex. Definition In a labeling of G, a coewor is the label of a coevertex. A noncoewor is the label of a noncoevertex. We say that L n is the labeling of Kn. Which labeling we mean will be clear from the context. Definition Let G be a graph. A labeling of G has the Gray coe property if every pair of ajacent vertices has labels which iffer in exactly one position The G-U Construction. Cull an Nelson[10] prove that etermining whether a given graph has a P1ECC is an NP-complete (ifficult) problem. However, they introuce a relatively simple metho for constructing a P1ECC on K3 n for any iteration n. Also, they prove that this coe is unique up to rotation, with strict uniqueness if a specifie corner of K3 n is require to be a coewor. These results were later foun to generalize to higher imensions, an inepenently by Klavzar, Milutinovic, an Petr in [15]. Cull an Nelson s metho has come to be known as the G-U construction. It is the founation of our methos for coewor recognition an error correction on iterate complete graphs. 107

5 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 5 The G-U construction uses two types of coes on K n: G-coes an U-coes. Let Gn enote Kn with the G-coe an let U n enote Kn with the U-coe. Gn an U n are constructe recursively as follows: To construct G 1, esignate one vertex of K1 as the top vertex an rotate it to the top position. Make this vertex a coevertex. Make the other 1 vertices noncoevertices. To construct U 1, esignate one vertex of K1 as the top vertex an rotate it to the top position. Make all vertices noncoevertices. Figure 3 shows G 1 5 an U 5 1. FIGURE 3. G 1 5 an U 1 5. We now show how to construct G n an U n for arbitrary n: To construct G n when n is even: (1) Make copies of G n 1. (2) Connect each pair of copies so that the top vertex of every copy remains unconnecte. (3) Designate the top vertex of some G n 1 as the top vertex of G n. To construct G n when n is o: (1) Create one copy of G n 1 an 1 copies of U n 1. (2) Connect the top vertices of the copies of U n 1 to istinct non-top corner vertices of G n 1. (3) Connect each pair of copies of U n 1 by one ege such that This ege connects a non-top corner vertex in one copy to a non-top corner vertex in the other copy. Exactly one non-top corner vertex of each U n 1 remains unconnecte. (4) Designate the top vertex of G n 1 as the top vertex of G n. To construct U n when n is even: (1) Make one copy of U n 1 an 1 copies of G n 1. (2) Connect the top vertices of the copies of G n 1 to istinct non-top corner vertices of U n 1. (3) Connect each pair of copies of G n 1 by one ege such that This ege connects a non-top corner vertex in one copy to a non-top corner vertex in the other copy. Exactly one non-top corner vertex of each G n 1 remains unconnecte. (4) Designate the top vertex of U n 1 as the top vertex of U n. To construct U n when n is o: (1) Make copies of U n

6 6 Skubak, Stevenson (2) Connect each pair of copies by a vertex such that the top vertex of every copy remains unconnecte. (3) Designate the top vertex of some U n 1 as the top vertex of U n. This is much easier to unerstan via example. Figure 4 shows G 2 5 an U 2 5. Figure 5 shows G3 5 an U 3 5. FIGURE 4. G 2 5 an U 2 5. FIGURE 5. G 3 5 an U THE SF LABELING AND PUZZLE Because labels an puzzles for o imensions have been establishe in previous papers (see [14], [16]), the focus of our paper is iterate complete graphs with even imension. In this section 109

7 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 7 we summarize those results. The SF labeling on the o imension iterate complete graphs has been establishe to have finite-state machines for coewor recognition an error correction. The SF labeling also has the Gray coe property an correspons to a puzzle calle the SF puzzle. In the case = 3, the SF labeling correspons to the Towers of Hanoi labeling given by Cull an Nelson[10]. It has been emonstrate that even imensional iterate complete graphs o not support SF-like labelings Construction of the SF Labeling. Let 3 be an o number. The labeling of K n is constructe recursively from the labeling of K n 1. Label K 1 as follows: the top vertex is labele 0, then the remaining vertices are labele 1,2,...,( 1) going counterclockwise. Figure 6 shows the SF labeling of K FIGURE 6. The SF labeling of K 1 5. The SF labeling of K n is constructe accoring to the following algorithm: Apply the permutation α to each igit in every label of K n 1, where α(z) = +1 2 z (mo ). Now make copies of α(k n 1 ). Rotate the k th copy 2πk raians counterclockwise, then appen k to each wor in this copy. Finally, connect the copies to form K n. Figure 7 shows the SF labeling of K2 7. Figure 8 shows the SF labeling of K The SF Puzzle. The Towers of Hanoi is playe with n isks all of ifferent size. The isks are stacke on three towers so that no larger isk is stacke on top of a smaller one. The goal is to begin with all isks on one tower an move them to another. We will number the towers 0, 1, an 2. A natural way to label the configurations of isks on towers is with ternary strings as follows. Recor the tower number of the smallest isk. To the right of this number, recor the tower number of the next smallest isk. Continue in this way to obtain a string of length n. Now each vertex of K3 n has an SF label that correspons to a configuration of the Towers of Hanoi puzzle. The labels of ajacent vertices represent configurations which are one legal move from each other. Figure 9 shows the SF labele graph K3 3 corresponing to Towers of Hanoi with 3 isks. Now imagine we have an o number 3 of towers numbere 0 through 1. Like the Towers of Hanoi, configurations of n isks on these towers can be represente by base strings of length n. The SF puzzle has the same rules as Towers of Hanoi. In aition, it has the following rules to restrict the possible moves to those represente by the SF labeling on K n. (1) No isk may be move unless all of the isks smaller than it are stacke together on the same tower. 110

8 8 Skubak, Stevenson FIGURE 7. The SF labeling of K 2 7. (2) When a isk is able to move, if the stack of smaller isks is on tower a an the isk to be move is on tower b, then the isk may only move to tower (2a b) mo. Note that the movement of the smallest isk is unaffecte by these aitional rules; it can always move to any tower. Figure 10 shows configurations corresponing to labels 220 an 224 on K5 3. Here the largest isk can only move between towers 0 an 4, an inee there is an ege between these two vertices in K5 3, as shown in Figure

9 A New Puzzle on Iterate Complete Graphs with Dimension 2 n FIGURE 8. The SF labeling of K FIGURE 9. The labele graph K 3 3 corresponing to the Towers of Hanoi with 3 isks. 4. DIMENSION 2: THE SPIN-OUT PUZZLE A goo starting point for our investigation into puzzles for even-imensional iterate complete graphs is the Spin-Out puzzle by ThinkFun[25]. The goal of the game is to remove a rectangle with seven spinners on it from a plastic case. In the traitional starting position, all seven spinners are vertical, an the rectangle can only be remove when all of the spinners are aligne horizontally. Let the spinners be labele 0 to 6 from the left to the right. The n th spinner can only be turne 112

10 10 Skubak, Stevenson FIGURE 10. Configurations corresponing to labels 220 an 224 on K5 3. largest isk may move between towers 0 an 4. The when the spinners 0 through n 2 are horizontal an spinner n 1 is vertical. Note that the leftmost spinner is free to move at anytime. FIGURE 11. A configuration of the Spin Out puzzle. The spinner uner the arc may move, an we may also slie the large rectangle to the right an move the the leftmost spinner. To represent this puzzle by a labeling on a graph, let each spinner be represente by a bit. If the spinner is horizontal, the bit is 0; if it s vertical, 1. Then let each configuration of the puzzle be represente by a string of seven bits, the leftmost bit corresponing to the leftmost spinner an so on. We associate these labels with vertices, an when we create eges between them representing possible moves of the Spin-Out puzzle, we get a Gray labeling on K2 7. Note that the puzzle can be generalize to use any number of spinners, not just 7. This resulting family of puzzles can be represente by the reflecte binary Gray coe on K2 n, which we now escribe Graph an Label. The reflecte binary Gray coe is a well-known labeling scheme on K n 2 with an easily efine recursive construction[26]. For n = 1, let the reflecte binary Gray coe, G 1, be 0-1. That is, we label one vertex 0 an the other vertex 1. To construct G n : (1) Take two copies of G n 1. Appen 0 to the labels in the first copy an 1 to the labels in the secon copy. 113

11 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 11 FIGURE 12. The reflecte binary Gray coe for n = 1, n = 2, an n = 3 with coe vertices circle (2) Reverse the orer of the strings in the secon copy, an connect the new beginning of the secon copy to the en of the first copy. Note that in some papers, these 0 s an 1 s are sometimes prepene rather than appene; we appen for consistency with previous work an for their relationships to the puzzles Coewor Recognition. If we consier the perfect one error-correcting coe on this labeling as efine by the G-U construction (see Section 2.4), we fin the coe vertices as shown circle on the graphs. Now, given a label, we woul like to know whether or not it is a coewor. More specifically, is there a finite-state machine that recognizes coewors? Yes, there is a machine that can be create irectly from the recursive construction of the label an coe. This coewor recognizer for the reflecte binary Gray coe is shown in Figure 13. This machine is noneterministic, since it has more than one start state. If the iteration is even, the machine starts in state G e. If the iteration is o, the machine starts in state G o. If the machine ens in the accepting state, then the wor is a coewor. We can easily convert this machine to a eterministic machine, though we o not show this here (see [21]) for theory on finite-state machines). Later, (Section 6.4, we will prove a more general case of this recognizer an show a eterministic version. Alternatively, if we require that is a coewor, the only P1ECC on the graphs K2 n has the coewors at every thir vertex. Thus this machine only nees to convert the Gray label to its binary position an etermine if that position is a multiple of 3, both of which can be simply one with finite-state machines Error Correction. If we place a wor into the recognizer, we can ecie whether or not it is a coewor. One motivation for perfect one-error-correcting coes, however, is to be able to easily fin, for any wor, the closest coewor. This is calle error correcting. The error corrector for the reflecte binary Gray coe labeling is shown in Figure 14. The machine reas a string bit by bit, following the transition with the given bit on the left part of the label. The machine then changes that bit to the bit on the right sie of the label. The coewor recognizer above must first be use. If the final state is a G state, the error corrector begins at the 0 state (which returns the string unchange, since it is a coewor). If the final state is either of the C 1 states, the machine begins in 1; if the final state is a C 2 state, it begins in 2. The proof that a generalize version of this error corrector inee works can be foun in Section

12 12 Skubak, Stevenson G e 1 all 0 G o C o 1 C e C e 2 C o 2 all FIGURE 13. Finite-state Recognizer for the Spin-Out Puzzle FIGURE 14. Error-corrector for the Spin-Out Puzzle 5. DIMENSION 4: THE REFLECTION PUZZLE The Reflection Puzzle was first escribe by Weaver[28]. We foun a physical representation of this labeling of the iterate complete graphs of imension 4. It bears a close resemblance to the Spin-Out puzzle. In fact, it is essentially two such puzzles stacke on top of each other. Our goal will still be to change each piece to the zero orientation. In this puzzle, however, each piece is mae up of two spinners that can (sometimes) move inepenently, giving four possible orientations, shown in Figure 15. FIGURE 15. Spinner orientations for the Reflection Puzzle We will again use the labels of the pieces to represent puzzle configurations. An example can be foun in Figure Rules. As usual, the leftmost piece may move at any time. Here, however, it may also move to any orientation, allowing 3 possible moves. For a piece j 0 to be able to move, we must have 115

13 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 13 FIGURE 16. An example puzzle orientation for the Reflection Puzzle labele 0320 pieces 0 through j 2 at orientation 0. Also, if piece j 1 is at 0, then no spinners in piece j can move; thus we nee f ( j 1) 0. So suppose piece j is able to move. Which orientation shoul it change to? Our rule is: if a spinner can move, it oes move. Thus for any vertical spinner in piece j 1, the corresponing spinner in j will change orientations. In Figure 16, the secon to last piece may change from 2 to Graph an Label. To represent this puzzle on a graph, we will associate the labels as efine above with vertices an legal moves with eges. As hope, the graph generate is K4 n. Since only one piece is move at a time in the puzzle, this new labeling is a Gray labeling. To create this labeling when n = 1, the top vertex of K4 1 is labele 0 an each other vertex, moving counterclockwise, is labele with the successive integers through 3. To construct this labeling scheme on K4 n for n > 1: (1) Construct 4 copies of the labeling on K4 n 1, an inex these copies from 0 to 3. (2) Reflect copy 1 vertically, an a a 1 to the en of every label in the copy. (3) Reflect copy 2 both vertically an horizontally, an a a 2 to the en of every label in the copy. (4) Reflect copy 3 horizontally, an a a 3 to the en of every label in the copy. (5) Place copy 0 as the top copy an the others in orer counterclockwise. Finally connect each copy to all of the other copies, as shown in Figure 17. As in the Spin-Out puzzle an associate labeling, this labeling has finite-state recognition an error correction, which we prove generally in the next section. 116

14 14 Skubak, Stevenson FIGURE 17. The Reflection-Metho labelings of K 1 4 an K DIMENSION 2 m The puzzle on four imensions suggests an easy extension of Spin-Out to all imensions which are powers of 2. The extene puzzles will retain the sliing aspect of Spin-Out, but the spinners will be replace by pieces which consist of a stack of spinners. When a piece is compose of m spinners, it will have 2 m possible orientations, since each spinner can be in one of two orientations. For n pieces, there will be (2 m ) n = n configurations. The sliing rules will etermine which pieces can change, an new spinning rules will etermine how the pieces can change. Together they will efine which configurations can change to which configurations. Note that all proofs in this section apply to the Spin-Out an Reflection puzzles. At present, these aitional puzzles are mathematical constructions an they may be ifficult to physically construct so that all the mathematical rules are completely enforce. We nee a way to associate orientations of the pieces with numbers 0 through 1. We efine the orientations of our pieces as follows: For a imension = 2 m, each puzzle piece will consist of m spinners stacke one on top of the other. To fin orientation j, write j as a binary number. To set a piece in this orientation, let the 1 s (rightmost) bit represent the top spinner; a 0 bit means that it is horizontal, while a 1 bit means that it is vertical. Similarly, let the 2 s bit represent the spinner just below the top spinner, the 4 s bit the next spinner, etc. Continue in this manner; the (m 1) s bit will represent the bottom spinner. Thus for each j {0,..., 1} we have a istinct orientation an corresponing binary number. Since there are exactly orientations, we know we have efine all possibilities. Example 6.1. Suppose = 8 = 2 3. That is, m = 3, so the pieces are compose of 3 spinners. Then, for example, the 0 = orientation consists of all horizontal spinners, the 7 = orientation has all vertical spinners, an the 3 = orientation has a horizontal spinner on the bottom with two vertical spinners above it. 117

15 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 15 FIGURE 18. Piece orientations for the Dimension 8 Puzzle Note that the 0 th orientation will always consist of all horizontal spinners. Note also that Spin- Out an the Reflection Puzzle both consistently follow this naming scheme. Now, for an iteration n for n 1, we will have n puzzle pieces. We will call the leftmost piece the 0 th piece an continue numbering the pieces from left to right. Thus the rightmost piece is the (n 1) st piece. Given a configuration of our puzzle, we will label it with a string of characters from {0,..., 1}, where each piece 0 through n 1 is represente by the number of the orientation it is in. To avoi confusion, we will write f ( j) to refer to the orientation of piece j. Example 6.2. Continuing from the example above, Figure 19 has the label FIGURE 19. An example configuration for the Dimension 8 Puzzle. The pieces are numbere left to right 0, 1, 2, an Rules. The rules of this puzzle are nearly the same as for the Reflection Puzzle. First, the 0 th piece may always change orientation, an may change to any other orientation. To spin at least one spinner of the j th piece, we must have that pieces 0 through j 2 are 0 an piece j 1 is not 0. In our notation, we nee f (0) through f ( j 2) to be 0 an f ( j 1) 0. If these conitions are satisfie, then we must move as many spinners of the j th piece as possible; that is, any spinner that can switch between its horizontal an vertical positions must o so. For example, in Figure 19, piece 2 is able to change orientations. Since the bottom spinner of piece 1 is horizontal, the bottom spinner of piece 2 cannot move. However, the other two spinners can move, an so they must become horizontal. Thus we may change the orientation of piece 2 from 7 to 4. The goal of the puzzle is, given some initial configuration, to move all the pieces to orientation 0. Since this puzzle can be represente by K n as we prove shortly, we know that the minimum solution path for any two configurations is not longer than 2 m 1, the iameter of the graph. This quantity is the same as for both the Spin-Out puzzle an the SF Puzzles. Another way of solving this puzzle might be to choose two configurations an move from one to the other. Of course, 118

16 16 Skubak, Stevenson puzzles are fun, but the focus of the paper is the puzzle s relationship to the family of iterate complete graphs, which we will now prove, after some preliminary lemmas. It woul be helpful to be able to know which orientation piece j can move to without thinking in terms of spinners. To o this, we efine a special operation. The operator enotes bitwise aition on two numbers; that is, r s means write both r an s as binary numbers an o a bitwise aition (also known as a XOR, or aition without carry). Lemma 6.3 (Orientation Change Function). If f (0) through f ( j 2) are 0 an f ( j 1) 0, then piece j may move to f ( j 1) f ( j). Because f ( j 1) f ( j) f ( j), piece j can make a change to another istinct orientation. Proof. Recall that we efine our orientations to coincie with binary numbers representing the spinners in a piece; a 0 is a horizontal spinner, an a 1 is a vertical spinner. Then consier that movement of the spinners of piece j are completely ictate by the positions of the spinners of piece j 1. In particular, any horizontal spinners in piece j 1 block movement in piece j of the spinner in the same horizontal plane. Therefore, any 0 s in the binary representation of f ( j 1) ictate that no change shoul occur in the corresponing bits in the binary representation of f ( j). Since 0 + x = x in bitwise aition, our formula hols in this case. For the remaining case, consier the vertical spinners in j 1, i.e. the 1 s in the binary representation of f ( j 1). These allow movement of the corresponing spinners in j an accoring to our rule, the spinners must change. Since = 0 an = 1 in bitwise aition, our formula hols for each bit. Thus the result f ( j 1) f ( j) is the orientation to which piece j may move. Finally, note that since f ( j 1) 0, it has at least one bit equal to 1, which causes a change in that bit in f ( j), so f ( j 1) f ( j) f ( j). Note that f ( j 1) f ( j) can equal f ( j 1). In fact, this occurs if an only if f ( j) = 0. Thus the configuration 0...0x0 may move to 0...0xx. Lemma 6.4 (Reversibility of Moves). All moves in this puzzle are reversible. Proof. Clearly all moves by the leftmost piece are reversible since we may move this piece at any time. To consier moves by some other piece, first let x {1,..., 1} an y {0,..., 1}. Also, let W be the empty string or some fixe string consisting of characters from {0,..., 1}. Now suppose the configuration 0...0xyW may move to 0...0xzW (where there may be any fixe nonnegative number of leaing zeros). By Lemma 6.3, z = x y. Then in 0...0xzW, the piece at z may change to x z = x (x y). It is easy to check that bitwise aition (the XOR) is associative. Also, note that any string ae bitwise to itself is simply the zero string since = = 0. Thus x (x y) = (x x) y = 0 y, which simply equals y. Therefore the configuration 0...0xzW may move to 0...0xyW as esire. Thus we know that is associative as well as commutative, an that each number is its own inverse uner. These properties will be of further use to us Graph an Labeling. Now that we have a puzzle, we wish to show that it is in fact the puzzle we were looking for the puzzle that fits the graph K n (for = 2m ). We label each vertex in K n with the label of some puzzle configuration an connect two vertices if the two configurations are one legal move apart. Since we can only move one piece at a time, the labeling scheme is a 119

17 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 17 Gray coe. Also, by Lemma 6.4, the eges are not irecte; we may always uno the last move we mae. The following lemmas will help in our proof. Lemma 6.5 (Corner Labels). The puzzle configurations 0...0k for k = {0,..., 1} have exactly 1 ajacent configurations (where there may be any nonnegative number of leaing zeros). Proof. By a rule of our puzzle, the 0 th piece, at orientation 0, may always change to any other orientation 1,..., 1. This gives us 1 possible moves. To see that there are no others, note that there is no piece where f ( j 1) 0. Since this is require for a piece j 0 to move, no other moves are possible. Lemma 6.6 (Non-Corner Labels). All other configurations (those not of the form 0...0k for k = {0,..., 1}) have exactly moves. Proof. First, note that any configuration not of the form 0...0k for k = {0,..., 1} instea has the form 0...0xyW where there may be any nonnegative number of leaing zeros, x 0, an where W is the empty string or some string consisting of characters from {0,..., 1}. Again, the first piece may always change to any other orientation 1,..., 1, giving us 1 possible moves. Now, we show that there is exactly one more. Since x 0, the piece j such that f ( j) = y is the only piece that has f (0) through f ( j 2) equal to 0 an f ( j 1) 0. Therefore piece j is the only other piece that may move. By Lemma 6.3, piece j may change from orientation y to some other istinct orientation, giving us a total of moves. Lemma 6.7. For a fixe y {0,..., 1}, the function efining the possible change of the character y in the configuration 0...0xy... of omain x {1,..., 1} is a bijection onto {0,..., 1} \ {y}. That is, for any other character z not y, there is some nonzero x such that 0...0xy... may change to 0...0xz.... Proof. First, let the piece with orientation y be piece j. Since x 0, y oes move, an so this function is well-efine. By Lemma 6.3, piece j actually changes, so y is not a vali element of the range; thus the omain an range are the same size. Now we nee only show that this function is injective. So suppose that both x 1 an x 2 change y to z. That is, x 1 y = z = x 2 y. Bitwise aing y shows that x 1 = x 2. Theorem 6.8. The escribe generalize Spin-Out puzzle fits the graph K n. That is, each vertex in K n represents a configuration of pieces, an each ege represents a vali move between two configurations. Proof. We will use inuction on the number of puzzle pieces n, which will correspon to the iteration of the graph K n. First consier the puzzle with one puzzle piece consisting of m spinners. Because this piece is the 0 th piece, it may change to any orientation at any time, an because it is the only piece, these are the only moves we can make. Clearly this correspons to the complete graph K = K 1. Now assume that the puzzle with (n 1) pieces correspons to the graph K n 1. We will consier the puzzle with n pieces. Note that if we may move the j th piece in some puzzle, the ( j +1) st piece (an any pieces farther right) is essentially irrelevant. As seen in the example in Figure 19, the last piece oes not affect what moves are legal by piece 0 through piece 2. Thus we consier the puzzle with n pieces to be a combination of puzzles, each with n 1 pieces, one puzzle for each 120

18 18 Skubak, Stevenson possible orientation of our ae piece. Therefore, one character representing the orientation of the new piece will be appene to each label. The smaller puzzles are each copies of K n 1 by our inuction hypothesis. Then all that remains is to show that these subgraphs connect in the correct manner. Recall that an iterate complete graph of imension has corner vertices (vertices that are egree 1). We claim that exactly one vertex from each of the K n 1 subgraphs will become a corner vertex of K n : the vertex 0...0, which we know is a corner vertex by Lemma 6.5. To see this, note that in K n, the label will be appene with some r {0,..., 1}. Simply using Lemma 6.5 once more shows that the vertex 0...0r is a corner vertex of K n, giving us such corner vertices. Clearly these are the only ones, since any other vertex of K n 1, when appene by some r {0,..., 1}, cannot have the form 0...0r. By Lemma 6.6, these other vertices are not a corner vertices, an we have verifie this claim. Next we show that every other corner of each K n 1 subgraph connects to exactly one ifferent K n 1 subgraph at one of its corner vertices. Each of these other corner vertices in the K n 1 graphs have the form 0...0s for s {1,..., 1} (Lemma 6.5). Thus, in K n, their new labels have the form 0...0st for s {1,..., 1} an t {0,..., 1}. Then 0...0st may move to 0...0s(s t), which is in the (s t) th subgraph; by Lemma 6.3 this is a ifferent subgraph than the t th subgraph. So consier some fixe K n 1 subgraph, i.e. fix t in 0...0st. By Lemma 6.7, over all values of s (we know s 0), t will change to every character except t itself. This implies that this fixe subgraph must be connecte to each of the other 1 subgraphs exactly once. Since t was arbitrary, this is true for each K n 1 subgraph. This finishes our proof Recursive Labeling. Theorem 6.8 tells us how to construct our graph an labeling from the puzzle. We wish, however, to have a way to recursively construct the labelings on the family of iterate complete graphs K n without referring to the puzzle. Unfortunately, this labeling of the family of iterate complete graphs of these imensions is not unique. For at least imension 8, there exist other labelings that still preserve the esire property of having the reflecte binary Gray coe along each iagonal from However, it oes not appear that any puzzle easily associates with any other labeling we have foun. The base case for our labeling is the complete graph K = K 1. Clearly any labeling of the vertices woul correspon to the puzzle, so without loss of generality we will label some top vertex 0 an label in orer counterclockwise, an call this labeling L 1. Now assume we have L n 1. We know that L n is base on copies of Ln 1 by Theorem 6.8; however, if we simply place each copy own, the eges connecting these subgraphs woul not raw the graph as we have neatly epicte it in this paper. Therefore we will permute each subgraph so that the eges are in the esire locations. Clearly without loss of generality, we may place the 0 th subgraph any way we wish, so we will not permute it. Then to create L n : When i = 0, the copy is place in the top (0 th ) position an 0 is appene. For all other i, the permutation Γ i is applie to the last character of each label in the i th copy of L n 1, where Γ i bitwise as i to the last character in a label. That is, Γ i (...x) =...(x i). Then this i th copy is place in the i th position counterclockwise from the top position, an the character i is appene to each label. 121

19 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 19 Finally, for each i, the vertex at position j from the top position is connecte to the i th corner of j th subgraph. If j = i, the vertex is a corner of the entire graph an no ege is rawn. Example 6.9. Figure 20 shows L 1 8 an L2 8, the recursive labeling for the graphs K1 8 an K2 8. As an example of how to permute a subgraph, look at the subgraph immeiately counterclockwise of the top position, the 1 st subgraph. This was labele by applying Γ 1 to L 1 8 an appening 1. Note that 0 1 = 1, 1 1 = 0, 2 1 = 3, etc. FIGURE 20. The labeling for the first an secon iterations for the imension 8 graph, corresponing to the extene Spin-Out puzzles with 1 an 2 pieces respectively. Lemma Each permutation Γ i, for any i {0,..., 1}, is a graph automorphism on L n. That is, applying any Γ i oes not change the structure or any eges of the graph. Proof. First of all, note that Γ i is in fact a permutation on the vertices. Inee, suppose Γ i (...y) = Γ i (...z). (Since Γ i only changes the last character, we know that the rest of the label must be ientical.) But Γ i (...y) =...(y i) an Γ i (...z) =...(z i). Then y i = z i an y = z, so the labels...y an...z are the same. Thus Γ i is injective, an since Γ i maps a set to itself, Γ i is a bijection. Now, we show that for ajacent vertices u,v in L n, we have that the vertices Γ i(u) an Γ i (v) are ajacent in Γ(L n ). There are three cases. (In all cases, the number of leaing zeros may be positive or zero.) (1) If the vertex is labele 0...0x, then it is ajacent only to the 1 vertices y0...0x for y {0,..., 1} (see Lemma 6.5). Call this set of vertices S. Now Γ i (0...0x) = (x i), which, again by Lemma 6.5, is ajacent only to y0... 0(x i) for y {0,..., 1}. The set of these vertices is clearly equal to Γ i (S). 122

20 20 Skubak, Stevenson (2) If the vertex is labele 0...0xz, then it is ajacent to only the vertices in the set S = {y0...0xz : y {0,..., 1}} {0...0x(x z)}. But Γ i (0...0xz) = 0...0x(z i), which is ajacent only to {y0...0x(z i) : y {0,..., 1}} {0...0x(x z i)} = Γ i (S). (3) The remaining possibility is that the vertex is labele 0...0xzW for some nonempty string of characters W. It is only ajacent to the labels corresponing to all changes of the leaing character in the label, an also to the label 0...0x(x z)w. Clearly, applying Γ i (changing the last character in W ) preserves these ajacencies. Thus all eges are preserve. Theorem This recursive labeling is the same labeling as escribe in Theorem 6.8. Proof. Since Lemma 6.10 shows that applying some Γ i to a subgraph oes not change the eges within that subgraph, we only nee show that the new eges rawn between the subgraphs are in fact the correct connections. Note that Γ 0 is the ientity permutation since we o not permute the 0 th subgraph. So fix some subgraph i in L n, an take the jth corner vertex where j i. Recall that Γ i was first applie to the subgraph an then i was appene. Thus, since it is a corner, the label has the form 0...0( j i)i = v. Now we claime in our construction metho that this vertex shoul connect to the i th corner of j th subgraph. Similar to above, the j th subgraph ha Γ j applie an j appene, so the label is of the form 0...0(i j) j. But this configuration may move to 0...0(i j)(i j j) = 0...0(i j)i = v, since is commutative. Thus the two vertices are inee ajacent. Since i, j were arbitrary, all connections between the subgraphs are correct, an we have proven the theorem Coewor Recognition. Using the G-U construction as escribe earlier in the paper, we know exactly what vertices are coe vertices for each graph. Since these sets are efine recursively, we can use them in combination with the recursive labeling to prouce recursive efinitions for the actual coewors themselves. It will be helpful to refer to a labele graph L n with a coe G n or U n associate with it. Thus we efine Cn as the labele graph Ln with the Gn coe scheme, an H n as Ln with the U n coe scheme. (We use this notation because the coe vertices of Cn are the coewors of the P1ECC on L n, while the coe vertices of Hn are a helper set of vertices.) Then when we write Γ i (C n) or of Γ i(h n ), we mean that it permutes the labels unerneath the coe scheme, without moving the coe vertices of the G n or U n graph. We also use the to enote an appening; x i means appen i to x. Lemma For an o iteration n, C n = Cn 1 For an even iteration n, H n = Hn 1 0 S 1 i=1 Γ i(h n 1 ) i. ) i. 0 S 1 i=1 Γ i(c n 1 Proof. This follows immeiately from the G-U construction an the recursive labeling. For o n, the 0 th subgraph of the labele graph C n is forme by placing a copy of Ln 1 an appening 0. The G-U construction shows that the 0 th subgraph is a G graph, hence the C n 1 0. For all other i, 123

21 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 21 the i th subgraph was labele using Γ i (L n 1 ), an the coe vertices chosen using the U graph. Thus the terms Γ i (H n 1 ). The other equation is obtaine similarly. Recall that in the G-U construction, we often have to rotate some G or U scheme so that the top vertex becomes some other corner vertex. This plays a large role in our coewor efinitions, so we formalize that rotation here. Let R i be the permutation on the graphs G n an U n that rotates the entire set 2πi/ counterclockwise; that is it rotates so that the top corner, formerly at the 0 th corner, now lies at the i th corner. See Figure 21 for an example. Note that the function R i only moves the coe vertices, not the labels associate with them, so that if we write R i (Cn ) or R i(h n ), we mean that R i rotates the coe scheme above the labels without moving or changing them. FIGURE 21. An example of the R function Theorem (1) For n o, R i (C n ) = Γ i(c n ). (2) For n even, R i (H n ) = Γ i(h n ). (3) For n even, Γ i (C n ) = Cn. (4) For n o, Γ i (H n ) = Hn. Proof. These four results are so closely linke that the truth of any one of them epens on the truth of at least one other. We prove them using as little epenency as possible, an at the en show that we know enough base cases to establish each result for all iterations. Beginning with (1), take some C 2m+1 an consier what happens when we apply R i. Recall that the G graph is mae up of one smaller G graph an 1 smaller U graphs. First, the i th subgraph, which ha Γ i applie to it in the the recursive labeling an all of whose labels en in i, has become the one G subgraph. This gives us Γ i (C 2m) i. Next, for j i, the j th subgraph, which ha Γ j applie to it in the recursive labeling an all of whose labels en in j, is a U graph. However, this U graph has been rotate so that its top vertex is its i th corner vertex, giving us R i Γ j (H 2m ) j (where the enotes normal function composition). 124

22 22 Skubak, Stevenson Thus in total we have that R i (C 2m+1 ) = Γ i (C 2m) i S j i R i Γ j (H 2m ) j. But suppose (3) hel; then the first element woul be just C 2m i. Supposing (2) hols allows us to rewrite the union as S j i Γ i Γ j (H 2m) j = S j i Γ i j (H 2m) j. Now applying Γ i to both sies, we woul have Γ i R i (C 2m+1 ) = Γ i (C 2m i) Γ i ( [ Γ i j (H 2m ) j) = C2m 0 [ j i j i Γ i j (H 2m ) ( j i). Since j i, we know j i = i j is not 0, but can be any other character in {1,..., 1}. Thus Γ i R i (C 2m+1 ) = C 2m 0 [ Γ i j (H 2m ) (i j) = C2m+1 j i by Lemma Thus, when their omains are restricte to o iterations of C n, we have that Γ i R i = I, so that R i = Γi 1 = Γ i since numbers are their own inverses uner bitwise aition. Therefore we ve shown (1) if (2) an (3) are true. Now the argument for (2) is only notationally ifferent from above an gives that, when restricte to even iterations of H n, we get that Γ i R i (H 2m ) = H2m 1 0 [ j i Γ i j (C 2m 1 ) ( j i) = H 2m by Lemma 6.12 if (1) an (4) are true. Thus we again have that restricte to this omain R i = Γ i. For (3), take some C 2m an consier its construction. In the labeling the top (0 th ) subgraph was not permute, an then the G 2m 1 graph was place on top, simply giving us C 2m 1 0. Now for any other subgraph i, we know Γ i was applie. Also, however, the G 2m 1 graph was rotate, moving its top corner to the i th corner; that is, R i was also applie. But if (1) hols, then over the omain of C 2m 1, we know R i = Γ i = Γ 1 i, so the two permutations cancel, giving us only C 2m 1 i. Thus if we woul now apply any Γ j, all we woul o is change the last characters, which has no effect on the set of coewors S 1 i=0 C2m 1 i (it only changes their orer on the graph). The argument for (4) is the same as for (3); H 2m+1 is constructe with an unpermute 0 th subgraph, giving us H 2m 0. For other subgraphs i, both Γ i an R i were applie, so that if (2) hols, we have the set of coewors S 1 i=0 H2m i, which is unchange uner any Γ j. Finally, we show that we have a sufficient base to prove these claims for all iterations. First, we know that (4) hols for H 1 no Γ j can have an effect here, since it is the empty set. Next, we establish (1) for C 1, whose sole coewor is 0. We know that Γ i sens i to 0, making i the coewor. Similarly, R i rotates the G graph from the top to the i th corner, again leaving i as the coewor. Thus the two functions are the same here. Now, combining these two cases, we are able to prove (2) for H 2. Also, knowing that (1) hols for C 1 establishes (3) for C2. These cases are sufficient to show each of these claims for all n. Corollary For any imension = 2 m, the coewors for the Powers of Two labeling are escribe by the following recursive efinitions: for an even iteration n, 125

23 C n = S 1 i=0 Cn 1 i an H n = Hn 1 0 S 1 i=1 Γ i(c n 1 ) i for an o iteration n, C n = Cn 1 0 S 1 i=1 Γ i(h n 1 ) i an H n = S 1 i i=0 Hn 1 A New Puzzle on Iterate Complete Graphs with Dimension 2 n 23 Proof. The mile two equations are Lemma The other two were prove within parts (3) an (4) of Theorem Γ U 2 0 G e all G o Γ U 1 1 Γ G i Γ G 2 i Γ U i i 1 2 Γ G i i 1 Γ U Γ G 1 U e all 0 FIGURE 22. Noneterministic finite-state Coewor Recognizer for the Dimension 2 m Puzzle U o Corollary A noneterministic finite-state machine (Figure 22) that reas strings right to left an recognizes coewors follows irectly from the recursive efinitions in corollary Since the recognizer has only one accepting state, we can easily reverse the machine to make it eterministic by making it rea the strings in the opposite irection, left to right. We show this eterministic machine in Figure 23. (For theory on finite-state machines, see [21].) We leave out several eges between the various Γ i states so the structure of the machine can be more easily seen; these eges can be easily ae by following those Γ i permutations irectly. All eges involving the G states an U states are shown. Also note that the machines for the labels of imensions 2 an 4 are of this form Error Correction. Theorem The finite-state machine shown in Figure 24 correctly error-corrects labels. The recognizer must first be use; the ening state if the recognizer is the start state for the corrector. Note that for the transitions labele with a Γ permutation, each character follows that arrow an permutes accoring to that Γ. All other eges o not change the characters. 126

24 24 Skubak, Stevenson 0 G e all G o Γ U Γ G 1 1 Γ U i 1 2 Γ G 2 i Γ U i i Γ G i 1 Γ U 1 1 i Γ G 1 U e U o all 0 FIGURE 23. Deterministic finite-state Coewor Recognizer for the Dimension 2 m Puzzle all all G Γ 1 Γ2 Γ 1 Γ i Γ 1 Γ 2 Γ i Γ U i i 1 1 FIGURE 24. finite-state Error Corrector for the Dimension 2 m Puzzle Proof. If we use the noneterministic coewor recognizer in Figure 22, by a parity argument we fin that for any wor, we en in one of + 1 states: the coewor G e state, the 1 states Γ U i, an the U e state. Thus this must be sufficient information to correct. If a wor ens in the G state, clearly we want to make no change, which is clearly reflecte in the corrector. Now recall that this recognizer reas right to left, so that the leftmost bit is the last rea. Then if a wor ens in some Γ U i state, we can retrace one step, analogous to returning one bit to the right. Note that this will always place us in either the G 0 state or a Γ G i state. Each of these states has exactly one 1-step path to the accepting state; thus we nee only change the last bit to the bit labeling this path to make our wor a coewor. This correspons to the 1 mile states in the error-corrector; since the corrector runs left to right, if we begin in one of these states, we change only the leftmost bit to the require character, which is ictate by Γ i. This is clearly the case when the wor is in the same K 1 complete graph as a coewor an only nees change its leftmost bit, or 0th piece. 127