ON KOTZIG S NIM. Xin Lu Tan 1 Wharton Statistics Department, University of Pennsylvania, Philadelphia, Pennsylvania xtan@wharton.upenn.
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1 #G0 INTEGERS (0) ON KOTZIG S NIM Xin Lu Tan Wharton Statistics Department, University of Pennsylvania, Philadelphia, Pennsylvania xtan@wharton.upenn.edu Mark Daniel Ward Department of Statistics, Purdue University, West Lafayette, Indiana mdw@purdue.edu Received: //, Revised: //, Accepted: /9/, Published: 9// Abstract In 9, Anton Kotzig introduced Kotzig s Nim, also known as Modular Nim. This impartial, combinatorial game is played by two players who take turns moving around a circular board; each move is chosen from a common set of allowable step sizes. We consider Kotzig s Nim with two allowable step sizes. Although Kotzig s Nim is easy to learn and has been known for many decades, very few theorems have been established. We prove a new primitive theorem about Kotzig s Nim. We also introduce a series of conjectures about periodicities in Kotzig s Nim, based on computational exploration of the game. We share a free database of our computation, to entice others into further explorations of Kotzig s Nim. We hope the present paper sparks a revival of interest in this enticing game, for which the winning strategy (in most cases) remains an enigma. Dedicated to the memory of Philippe Flajolet.. Introduction Guy and Nowakowski s Unsolved Problems in Combinatorial Games [] asks, A (). Extend the analysis of Kotzig s Nim (WW, ). Is the game eventually periodic in terms of the length of the circle for every finite move set? Analyze the misère version of Kotzig s Nim. Kotzig s Nim [], also known as Modular Nim [], is a combinatorial game, i.e., it has two players who each have perfect information about the game and no chance moves. Kotzig s Nim is named after its inventor, Anton Kotzig []. The game is played on a circle with n N locations, numbered clockwise from 0 through n. Thesameset ofpotential movesm = {m,...,m k } N k is availableto eachplayer, and thus the game is impartial. Players alternate moves, always choosing a move graduate student
2 INTEGERS: (0) m M and moving exactly m units in a clockwise direction on the board. A player cannot land on a previously occupied location. For this reason, the player places a mark at the current board location immediately upon arrival, so that it will never be revisited during the remainder of the game. The game ends when a player is completely obstructed, i.e., a player discovers that every set of potential moves in M leads to a previously occupied location; such an unfortunate player loses in the normal style of play. Kotzig s Nim is a special case of the general problem known as Generalized Geography ( which was proved by Thomas J. Schaefer to be PSPACE-complete []. The two most recent discussions of Kotzig s Nim [, ] have a difference in the way that the game begins. In [], the board is assumed to be empty at the start of the game, so Player I begins the game by moving to any location at the start. Without loss of generality (i.e., by a possible circular shift of the labels), Player I begins the game at location 0. Subsequently, Player II moves to a location m i mod n, and Player I moves to (m i + m i ) mod n, etc. In [], location 0 is already labeled at the start, so Player I begins by moving to a location m i mod n at the start. Subsequently, Player II moves to (m i +m i ) mod n, etc. By pretending that Player II visited location 0 at the start in [], the two interpretations of the game are seen to be exactly the same, except that the roles of the two players are reversed. In other words, Player I in [] corresponds precisely to Player II in []. With the normal play rule, P- and N-positions are defined in the standard recursive way: A P-position is any position from which the Previous player can force a win. An N position is any position from which the Next player who moves can force a win. The sets of P- and N-positions are denoted, respectively, by P and N. All terminal positions are P-positions. From every N-position, there is at least one move to a P-position. From every P-position, every move is to an N-position. Since the roles of Players I and II are exactly reversed in [] and [], then the N and P positions are exactly reversed too, i.e., the set of P positions in [] are exactly the set of N positions in []. We adhere to the conventions of [], in particular, to the convention that location 0 is already labeled at the start of the game, Player I moves to a location m i mod n, etc. The notational convention of [] is Γ = (m,...,m k ;n) N or P, if, respectively, the first or second player can force a win from the start of a game of Kotzig s
3 INTEGERS: (0) Nim with move set M = {m,...,m k } and board size n. We simplify this notation slightly (dropping the equals sign), by writing Γ(m,...,m k ;n) N or P.. An Example As an example of play in Kotzig s Nim, we consider the game with move set M = {,} and n =. This example is discussed on page of [], where the solution is given as a table. Instead, we present the solution for this example as a tree. The first player can move from position 0 to either position or position. If the first player moves to position, then the game is finished after a total of four moves. If the first player moves to position, then the game is finished after a total of six moves. In either case, the second player has a winning strategy. The tree associated with the game play for this example is displayed in Figure. For ease of interpretation, we use solid lines for the first player s moves and dashed lines for the second player s moves. We indicate when the losing player s move is forced (but we never call the winning player s moves forced ). 0 I: II: II: I: I: II: I I II: I: forced I: forced 0, taken 0, taken II: II:, taken 0, taken Figure : The tree associated with the Example in Section. It shows the winning strategy for Player II in Kotzig s Nim Γ(,;). This is a reduced version of the full game tree. It indicates the moves that the winning player should take. (It omits the winning player s other possible moves.)
4 INTEGERS: (0). A New Primitive Theorem In [], Corollary, Theorem, and Theorem completely classify (respectively) the games Γ(,;n), Γ(,;n), and almost Γ(,;n) ( almost because the cases n = ± mod are only conjectures in []). We emphasize that m m = in each case. The argument in p. of [], attributed to Nowakowski, also classifies Γ(,;n). Here, m m =. We are unaware of any proofs in which the complete classification of Γ(,;n) is given. Since m m =, this turned out to be aquite challengingtask, but we give a complete classification and proof in Theorem.. This is the main achievement of the present paper. In Section, we pose many new open questions that the reader may wish to analyze. Theorem.. If n {,,,,},or if n mod and n, then Γ(,;n) P; otherwise, Γ(,;n) N. Proof. Considering the value of n modulo, there are five cases that we consider. We provide tree diagrams to directly establish each of the five cases. In the tree diagrams, the solid lines and circles correspond to moves by Player I, and the dashed lines and circles correspond to moves by PlayerII. We use a notation on some of the figures, in a red font, to indicate which positions on the board have been visited. The diamonds correspond to the diamond strategy, in which the player with the winning strategy chooses the opposite type of move chosen by the other player. For example, in Figure, on the right-hand side of the tree, Player II has recently moved from to. If Player I moves to land at, then PlayerII does the opposite type of move, i.e., moves, to land at ; on the other hand, if Player I moves to land at, then Player II does the opposite type of move, i.e., moves, to land at. For more examples of the diamond strategy, see []. Case I. n = k +. If n =, then Player I is immediately stuck, and thus Γ(,;) P. Otherwise, Γ(,;n) N. For k, Figure shows the game tree that describes a winning strategy for the first player. Player I should start with. Each time Player II wants to move, Player I responds by moving. If Player II makes k consecutive moves of, then Player I should respond by consistently moving during the first k responses, and then Player I should move as the kth response. This is the start of the right-hand side of the tree in Figure. Then the diamond strategy will be used k times in a row, and Player II is forced into a loss. On the other hand, if Player II moves exactly j consecutive times for 0 j k, and then switches to moving on his (j+)st move, Player I should respond by consistently moving for the first j + responses. This is the start of the left-
5 INTEGERS: (0) 0 0 (I,) j, 0 j k (I,) k,i, l,l+ for l j I by definition of j j + l,l+ for l k and k, k diamonds in a row j + j + n l or l+; and l+; for l k j + j + I forced n n k j diamonds in a row l or l+; and l+; for j + l k I forced n n n, taken n n (I forced,) j l,l for l j j j,j + taken Figure : Winning strategy for Player I for Γ(,;k + ), where k and n = k+.
6 INTEGERS: (0) hand side of the tree in Figure. Then k j uses of the diamond strategy force Player II into a loss. Case II. n = k +. If n =, then Player I moves and then Player II is stuck, and thus Γ(,;) N. If n =, then Player II has the winning strategy shown in Figure, so Γ(,;) P. Otherwise, for n = k +, with k, we have Γ(,;n) N; Figure shows the game tree that describes a winning strategy for the first player. Player I should start with. If Player II moves, the play proceeds as on the left of the tree in Figure, from position. If PlayerII movesstrictly less than k times in a row(say, j times in a row), then Player I should respond by consistently moving. Then, after Player II switches to move on his (j +)st move, Player I should respond by moving again. After a series of k j diamonds, Player I can force the remainder of the moves. This is the middle of the tree in Figure. If Player II moves at least k times in a row, Player I should respond by consistently moving during his first k responses, and then switch to moving for his kth response. Afterwards, Player I wins by forcing the remainder of the moves. This is the right-hand side of the tree in Figure. Case III. n mod. This case is quite different because, for large n, Player II has a winning strategy. If n =, each player moves once, then Player I is stuck, so Player II wins. If n =,,, then Player I has the winning strategies in Figure a, Figure b, Figure, respectively, so Γ(,;n) N for n =,,. For n mod, with n, Γ(,;n) P; Figures,, 9, 0,, show the winning strategy for Player II. Case IV. n = k +. We have Γ(,;n) N. For k 0, Figure displays the tree corresponding to a winning strategy for player I. Player I should start with. Let j (for 0 j k ) denote the number of times at the start that Player II chooses to make a move of, before switching to a move of on the (j+)st; Player I should always follow with a move of. If, on the other hand, PlayerII makes k consecutive movesof and if PlayerIresponds with a move of each time then Player II will be forced to switch to on Player II s (k +)st move; thus, naturally j = k in this case. Afterwards, Player I should move from j+ to j+, and then should implement k j diamonds in a row. The remaining moves of Player II are forced. This is depicted in Figure. Case V. n = k. If n = or n =, then Player II has the winning strategy shown in Figures a and, so Γ(,;) P and Γ(,;) P. Also, Γ(,;0) N as in Figure b.
7 INTEGERS: (0) 0 I I I forced I I forced I 0, taken 0, taken, taken Figure : Winning strategy for Player II for Γ(,;).
8 INTEGERS: (0) 0 0 I (I,) j, j k (I,) k, I, l +, l + for l j I by definition of j j + l,l for l k; and, I forced j + j + I I 0 0 j + j + k j diamonds (I forced,) k l or l; and l + ; for j + l k n l,l for l k n k diamonds k diamonds I forced I forced l or l + ; and l + ; for l k n l or l; and l + ; for l k n n n I forced I forced n n (I forced,) j l +, l + for l j j +, taken I forced I forced j +,j + taken 9 9 I forced, taken ( if n = ) ( if n = ) k diamonds l or l + ; and l + ; for l k, taken Figure : Winning strategy for Player I for Γ(,;k + ), where k and n = k+.
9 INTEGERS: (0) 9 0 I I 0 I forced I forced I forced 9 I forced 0 I forced I forced I forced 9 0 I forced I forced, taken, taken, taken Figure : Winning strategy for Player I for (a.) Γ(,;); (b.) Γ(,;).
10 INTEGERS: (0) 0 0 I I 9 9 I I I I 0 0 I I I forced I forced I forced I forced I forced I forced I forced I forced I forced I I I forced 0 9, taken I forced I forced I forced I forced I forced 0 I forced I forced I forced, taken I forced, taken, taken I forced I forced 9, taken,9 taken Figure : Winning strategy for Player I for Γ(,;).
11 INTEGERS: (0) 0 0 see next figure I (, I) j, j k (, I) k,, I l +, l + for l j j + l +,l + for l k ; and n, n n I by definition of j forced j + j + n n k diamonds I I l or l + ; and l + ; for l k n j + j + n n k j diamonds forced 0, taken l or l + ; and l + ; for j + l k n forced, I I n,n n forced, I k diamonds n, l or l + ; and l + ; for l k n forced, I, n, 0 taken ( forced,i) j l,l for l j j j +, j + taken Figure : Winning strategy for Player II for Γ(,;k + ), where k and n = k +.
12 INTEGERS: (0) 0 0 see previous figure I 9 9 I I 0 0 k diamonds l or l, and l, for l k; and n or ; and I I forced I see Figure 9 see Figure 0 if n = I (if n > ) (if n > ) I k diamonds 9 (or if n = ) 9 l or l +, and l +, for l k; n k diamonds 0, taken ( forced,i), forced, I I see Figure b l or l +, and l +, for l < k; (for n > ) n,, forced, I,,0, forced, I if =,I; forced, I, forced, I if =,I, taken l or l, and l, for l k; n or ; ; or ; ; or ; k diamonds, ( forced,i),, 0, 0, forced, I forced, I,0 0 if n >, forced, I, if n = 0, taken,9 9, taken k diamonds 0, 0 taken n 0, taken l or l +, and l +, for l k Figure : Winning strategy for Player II for Γ(,;k + ), where k and n = k +.
13 INTEGERS: (0) 9 9 I 0 0 I I ( if n = ) ( if n = ) k diamonds k diamonds l or l, and l, for l k; n or, l or l, and l, for l k; n or, I I I I forced forced, I or, and or, and, forced I forced forced, I, I forced I forced, I, I forced k diamonds, taken n l or l +, and l +, for l k; I forced, I, taken, k diamonds forced, I l or l +, and l +, for l k; n, 0, taken, taken Figure 9: Winning strategy for Player II for Γ(,;k + ), where k and n = k +.
14 INTEGERS: (0) I if n = I if n = I if n forced, I, or, 0 0 forced, I forced, I I I,, ( if n = ) or 0, forced, I, 9 9 see Fig. a k diamonds l or l +, and l +, for l < k; n, forced, I I, or 0, 0 0 forced, I forced, I forced, I k diamonds, 0, forced, I, 9 9 if =,I, taken, if =,I forced, I l or l, and l, for l k; n or, or, or, forced, I, forced, I if =,I; forced,i if =,I 0,0 taken, forced, I forced, I,, 0 0, or 0,, taken forced, I, forced, I 9, 9 0, taken 0 or,,, 9 taken forced, I if n = 0,0 taken 9, 9 (if n > ), forced, I, k diamonds l or l +, and l +, for l k; n 0, taken Figure 0: Winning strategy for Player II for Γ(,;k + ), where k and n = k +.
15 INTEGERS: (0) I ( if n = ) I k diamonds ( if n = ) l or l, and l +, for l k; n or, k diamonds forced, I l or l, and l +, for l k; n or, forced, I, forced, I,, ( forced,i) forced, I,, 0,,9 9 k diamonds k diamonds l or l +, and l +, for l < k; n l or l +, and l +, for l < k; n if n taken in ; 0,n taken if n not in forced, I n, if n taken in ; 0,n taken if n not in forced, I n, forced, I forced, I forced, I, if taken in ;, taken, 0, forced, I forced, I 0,,9 9 forced, I 0 or, 9, 9, taken 0, taken Figure : Winning strategy for Player II for Γ(,;k + ), where k and n = k +; the left-hand side is Figure a; the right side is Figure b.
16 INTEGERS: (0) 0 0 (I,) j, 0 j k l,l+ for l j j + I by definition of j, or forced if j = k j + j + j + j + k j diamonds in a row l or l+; and l+; for j + l k n (I forced,) j l,l for l j j j,j + taken Figure : Winning strategy for Player I for Γ(, ; k + ), where k and n = k +.
17 INTEGERS: (0) Otherwise, for n 0 mod, with n 0, we have Γ(,;n) N; Figures and show the winning strategy for Player I in these cases. Player I should move at the start. If Player II makes a move of, to position : Player I should move, and then the play splits according to the left-hand side of Figure : If Player II makes a move of, then Player I moves also, and a series of k diamonds ends the game. On the other hand, if Player II makes a move of, then Player I moves too, followed by k diamonds. Then Player II moves, Player I moves, and the game ends with k diamonds. If Player II makes a move of, to position, then Player I should move, to position. The rest of the situation is described on the right-hand side of Figure and throughout Figure. We break the game into cases: (A.) Right-hand side of Figure. Suppose Player II moves ; then Player I should move. If Player II moves, then Player I moves, and a series of k diamonds ends the game. On the other hand, if Player II moves, then Player I moves and uses k diamonds to arrive at position n. Then Player II s next three moves are forced, followed by Player I moving the same way each time (, ;, ;, ). Then, after a series of k diamonds, Player II can make just one more move and then Player I forces the end. (B.) Middle of Figure. If Player II moves a total of j (for j k ) times, then PlayerI responds by moving each time. Then (by definition) Player II moves, Player I moves, and then Player I uses a series of k j diamonds. Finally, Player II is forced to move twice, with Player I responding with a move of each time, and then a series of j pairs of moves (Player II moving, Player I moving ) ends the game. (C.) Right-hand side of Figure. If Player II moves a total of k times, Player I should respond by moving during the first k times, and then move on the (k )nd time. Then the play varies, according to whether Player II moves or : (C part.) If Player II moves, to position, then Player I moves, Player II is forced to move, and then Player I moves. Player II is forced to move for k times, followed by Player I moving each time. Then Player II must move again, and Player I moves to end the game. (C part.) If Player II moves, to position 9, then Player I moves, Player II is forced to move, and then Player I moves. Player II is forced to move for k times, followed by Player I moving each time. Then Player II must move twice, with Player I responding by moving each time, and the game ends.
18 INTEGERS: (0) 0 I I 9 I forced I forced 0 I forced I forced 9 I I forced forced I forced I forced I I 0, taken 0, taken,9 taken, taken Figure : (a.) Winning strategy for Player II for Γ(,;); (b.) Winning strategy for Player I for Γ(,;0).
19 INTEGERS: (0) 9 0 I I 9 I I I I 0 0 forced I I I I I forced forced 0, taken 0, taken I I forced, taken, taken forced I I forced forced I I 9, 0 taken 0, taken Figure : Winning strategy for Player II for Γ(,;).
20 INTEGERS: (0) I I I I I I one or more times 0 0 see next figure k diamonds k diamonds I I l or l + ; and l + ; for l k, taken n l or l + ; and l + ; for l k I forced 9 9 k diamonds k diamonds l or l; and l + ; for l k + n l or l + ; and l + ; for l k k diamonds, taken (I forced,), l or l + ; and l + ; for l k,,,, taken I forced,, k diamonds l or l + ; and l + ; for l k I forced 0 0, taken Figure : Winning strategy for Player I for Γ(,;k), where k and n = k.
21 INTEGERS: (0) 0 0 I I see previous figure I (I,) k, I, see previous figure (I,) j, j k l +,l + for l k ; and, I I l +,l + for l j + j I by definition of j j + j + I forced I forced j + j + k j diamonds l or l + and l + ; for j + l k n (I forced,), (I forced,) k (I forced,) k,,, l,l l k n n l, l for l k (I forced,) j I forced (I forced,) l,l for l j + j + 0 n n n,,,, taken j +, j + taken, taken Figure : Winning strategy for Player I for Γ(,;k), where k and n = k.
22 INTEGERS: (0). Computational Database for Conjectures We make a series of primitive conjectures in Section., and a host of other conjectures in Section., both meant to inspire new research about Kotzig s Nim. These conjectures were inspired by combing through the data generated from an exploratory computation and the resulting database. To assist the reader who is also interested in conjectures from exploratory data, we have posted all of the results of our computations on Kotzig s Nim. All of the computations were performed on an -core.0 GHz Apple MacPro, using C++. The database itself is located at and has, entries at the time of submission. Each entry is one line of text, containing three positive integers m,m,n, and a character value N or P that classifies the starting location of the game of Kotzig s Nim. All triples {m,m ;n} with m < m and n are found in the database. Some entries with n > are also given for particular pairs m,m for which m m is relatively small, e.g.,, those pairs m,m corresponding to Conjectures. through.. Note: In the opening sentence of the Concluding Remarks of [], Fraenkel et al. make the conjecture that PlayerI can win Γ(,;n) for all n = ± mod. The data in our database verifies their conjecture for n... New Conjectures about Primitive Games These six primitive conjectures are in the same spirit as Theorem., but we do not (yet) know of a proof for any of these six, and we would be delighted to correspond with readers who are interested in finding a methodology sufficiently robust to handle them. In each conjecture, the move set M = {m,m } is fixed, and with m m small. The initial positions of the games are classified as N- positions or P-positions, depending on the board length n. Conjecture.. If n {,,,,,,,,,,},or if n mod 0 with n, then Γ(,;n) P; otherwise, Γ(,;n) N. (Verified for n 0.) Conjecture.. If n {,,,,,,9,0,,,,,0,,,,,0,,,,,}, or if n is equivalent to,,, or modulo with n, then Γ(,;n) P; otherwise, Γ(,;n) N. (Verified for n.) Conjecture.. If n is odd and n / {9,,} then Γ(,;n) P; otherwise, Γ(,;n) N. (Verified for n.) Conjecture.. If n {,,,,9,,,9,,,,,,,,9},or if n is equivalent to,,,, or modulo 9 with n, then Γ(,;n) P; otherwise, Γ(,;n) N. (Verified for n 09.)
23 INTEGERS: (0) Conjecture.. If n {,,,,,,9,,,9,,,,,0,,,,,,,,}, or if n is equivalent to,,,, or 0 modulo with n, then Γ(,;n) P; otherwise, Γ(,;n) N. (Verified for n.) Conjecture.. If n {,,,,,9,,,,,,0,,,,}, or if n is equivalent to,,,, or 9 modulo with n 9, or if n mod with n, then Γ(,;n) P; otherwise, Γ(,;n) N. (Verified for n.).. Other New Conjectures about Kotzig s Nim Now we present several more conjectures about other types of periodicities in Kotzig s Nim. The conjectures have been verified for all triples m,m,n found in the database generated from computations described in the introduction to Section.... A Conjecture Related to Primitive Games In [] and [], Γ(, ; n) and Γ(, ; n), were already completely classified. Also Γ(,;n) was completely classified except for n ± mod. Indeed, Γ(m,m +;n) was classified for:. n = k(m +m ) in Theorem of [],. n = k(m +m )+0 in Theorem of [],. n = k(m +m )+ in Theorem of []. The cases m = m + and n = k(m +m )± remain open, but we introduce the following new conjecture: Conjecture.. Let m and m = m +. (.) Consider n = k(m +m ). If k =, then Γ{m,m ;k(m +m ) } P when m mod, Γ{m,m ;k(m +m ) } N If k = or k, then otherwise. Γ{m,m ;k(m +m ) } N when m 0 mod, Γ{m,m ;k(m +m ) } P (.) Consider n = k(m +m )+. otherwise.
24 INTEGERS: (0) If k = 0 or k =, then Γ{m,m ;k(m +m )+} N. If k = or k =, then If k, then Γ{m,m ;k(m +m )+} N when m mod, Γ{m,m ;k(m +m )+} P otherwise. Γ{m,m ;k(m +m )+} P when m mod, Γ{m,m ;k(m +m )+} N otherwise. Remark.. Conjecture. holds for m too, with only two exceptions: Γ(,;k ) N for k ; Γ{,;} P. but these relatively small cases were already known from the complete classifications of Γ(,;n), Γ(,;n), Γ(,;n). We explicitly wrote m in Conjecture. to emphasize the new content.... A Conjecture for a Pair of Moves That Differ by In Theorem in [], where m = m +, the games Γ(m,m ;k(m + m ) ) were classified as all being in N. Here we make an analogous new conjecture for m = m +: Conjecture.9. Let m = m +. When m,m are both odd, then Γ{m,m +;k(m +m ) } P, with one type of exception: When m =, m =, and k 0 mod, then for nonnegative a, we have Γ = {,;+a} N. Remark.0. We did not include the even situation in Conjecture.9, because those games are known to be in N. Indeed, if m and m = m + are even, and if n = k(m +m ), then is relatively prime to n. So by Lemma of [], it is equivalent to analyze Γ( m, m+ ;k(m +m ) ), which is in N by Theorem of []. Thus Γ(m,m +;k(m +m ) ) N too.
25 INTEGERS: (0)... (Relatively) Low Hanging Fruit? The following conjectures materialized after the first author spent some time combing through the data. A few special cases of these conjectures are already known, and we indicate some of these overlaps in the remarks below. On the other hand, in most cases, the conjectures below are disjoint from the known results. Conjecture.. For every pair m,m, Γ{m,m ;(m +m )} N. Remark.. When m and m are both odd, then since n is even Conjecture. follows from Theorem of []. Conjecture.. For every pair m,m, Γ{m,m ;m } N. Remark.. Again, when m and m are both odd, then since n is even Conjecture. follows from Theorem of []. Conjecture.. Letm < m. Ifthereisavaluek such that{m,m ;m m } equals {k,k,k}, {k,k,k}, {k,k,k}, or {k,k,k}, then Γ{m,m ;m m } P. Otherwise, Γ{m,m ;m m } N. Remark.. To prove Conjecture., it is safe to assume, without loss of generality, that m and m are relatively prime. For, if g = gcd(m,m ), then we can write m = ga and m = gb where a,b are relatively prime, and then Γ(m,m ;m m ) is equivalent to Γ(a,b;b a) by Lemma of []. Conjecture.. Let m = m +d where d 0 mod. Then Γ{m,m +d;(m +m )} N. Conjecture.. Let = m < m. If m mod, then Γ{,m ;m +} P; otherwise, Γ{,m ;m +} N. Remark.9. Again, when m and m are both odd, then since n is even Conjecture. follows from Theorem of []. In the following two conjectures (and also in Conjecture.), move m takes a player almost all the way around the circular board. Therefore, for large m, it might be more helpful to interpret a move of m spaces clockwise as (equivalently) a move of n m spaces counterclockwise. Conjecture.0. Let = m < m. If m is equivalent to,, or modulo 0, then Γ{,m ;+m } P; otherwise, Γ{,m ;+m } N.
26 INTEGERS: (0) Conjecture.. Let = m < m. If m is equivalent to,, or modulo 0, then Γ{,m ;+m } P; otherwise, Γ{,m ;+m } N. Inthecasewherem iseveninconjecture.,thenγ(,m ;+m )isequivalent to Γ(, m ;+m ) by Lemma of [], so we get the following immediate corollary (that depends on the veracity of Conjecture., of course): Corollary.. Let = m < m. If m mod or m mod, then Γ{,m ;+m } P; otherwise, Γ{,m ;+m } N. In this last conjecture (and also in Conjecture.), one move takes a player approximately halfway around the circular board. Conjecture.. Let = m < m. If m is equivalent to,, or modulo 0, then Γ{,m ;+m } P; otherwise, Γ{,m ;+m } N.. Conclusion Kotzig s Nim is an excellent game for exploring with students. The game is easy to learn and fast to play, but investigations into the structure of the game require students to learn about a variety of computational and mathematical techniques. Many question remain unanswered. The authors hope that the present paper inspires new investigations. We invite others to utilize our database located at and we urge our readers to continue working towards a complete classification of all Kotzig s Nim games with move set of size two, i.e., games of the form Γ(m,m ;n). Acknowledgments The first author is thankful for the generosity of Kathryn and Art Lorenz through the Joseph Ruzicka Undergraduate Fellowship in the College of Science at Purdue University, which made this project possible. The second author s work is supported by the Center for Science of Information (CSoI), an NSF Science and Technology Center, under grant agreement CCF We thank Richard K. Guy and Richard J. Nowakowski for their wonderful archive of unsolved problems [], which piqued our curiosity to study Kotzig s Nim. We also thank an anonymous reviewer for detailed comments that improved the presentation and clarity of the paper. References [] Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy. Winning Ways For Your Mathematical Plays, volume. A K Peters, Natick, MA, nd edition, 00. See pp..
27 INTEGERS: (0) [] Aviezri S. Fraenkel, Alan Jaffray, Anton Kotzig, and Gert Sabidussi. Modular nim. Theoret. Comput. Sci. (99), 9. [] Richard K. Guy and Richard J. Nowakowski. Unsolved problems in combinatorial games. In Michael H. Albert and Richard J. Nowakowski, editors, Games of No Chance. Cambridge, 009. [] Anton Kotzig. O k-posunutiach (on k-translations; in Slovakian). Časop. pro Pěst. Mat. a Fys. (9),. Extended abstract in French, pp.. [] Thomas J. Schaefer. On the complexity of some two-person perfect-information games. J. Comput. and System Sci. (9),.
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