Champion Spiders in the Game of Graph Nim

Transcription

1 Champion Spiders in the Game of Graph Nim Allison Nelson, Sydney Ryan, Chao Xu July 13, 2012 Abstract In the game of Graph Nim, players take turns removing one or more edges incident to a chosen vertex in a graph The player that removes the last edge in the graph wins A spider graph is a champion if it has a Sprague-Grundy number equal to the number of edges in the graph We investigate the the Sprague-Grundy numbers of various spider graphs when the number of paths or length of paths increase 1 Introduction Nim is an impartial two-player game traditionally played with multiple piles of sticks or stones Definition An impartial game is a game in which at any given state state of the game, each player has the same set of possible moves [5] On a player s turn, one or more stones are removed from a pile, and the player that removes the last stone (or stones) is the winner [5] We want to study a variation of this game, called Graph Nim In Graph Nim, players remove edges incident to a vertex and the player to remove the last edge(s) is the winner [1] Definition A move in a game of Graph Nim is the removal of one or more edges adjacent to a single vertex In particular, we are interested in patterns and periodic behavior of the Sprague-Grundy number, or nimber, as we vary the graph that we are playing on Definition The minimal excluded value, or mex, of a set of integers is the smallest nonnegative integer not included in the set Definition The followers of a graph G are the set of all graphs that can be reached in one move on G We denote the followers of G as F (G) Using the previous definitions, we can define the Sprague-Grundy number as follows: Definition Let G be a graph We define a function g(g) = mex({g(h) H F (G)) We call g(g) the Sprague-Grundy number, or nimber of G We denote {g(h) H F (G) = gf (G) 1

2 The Sprague-Grundy Theorem allows us to assume that games of Graph Nim actually have a Sprague-Grundy number Theorem 11 (Sprague-Grundy Theorem) Every impartial game under the normal play convention is equivalent to a nimber We have the following theorem for disjoint games Theorem 12 The Sprague-Grundy number of a game consisting of disjoint components is the nim-sum of the Sprague-Grundy numbers of those components [3] Definition We define the nim-sum of two non-negative integers as binary addition with no carries We denote it as [2] We have shown that for certain classes of graphs the nimbers become periodic as we make the graphs larger Definition A spider graph is a graph with one vertex of degree greater than 2 and all other vertices with degree at most 2 We will extend the class of spider graphs to include paths for our purposes For example, the graph in Figure 1 is a spider graph By studying spider graphs, we describe Figure 1: A spider graph methods for determining the nimbers of several different types of spiders used in a game of Graph Nim We are interested in the periodicity of the nimbers as we increase the length and number of paths attached to the spider The Sprague-Grundy number of a game analyzes whether or not the first player has a winning strategy A non-zero nimber ensures that the first player will always win if he plays the correct strategy A nimber of zero means that the first player does not have a winning strategy and will always lose if his opponent plays the correct strategy [2] To find the Sprague-Grundy number of a graph G, we want to find the minimal excluded value out of the Sprague-Grundy numbers of all graphs that can be obtained in one move on G For example, the graph in Figure 2 has the followers in Figure 3 The nimber for the graph in Figure 2 is g(g) = mex({3, 2, 1, 0) = 4 The following result bounds the Sprague-Grundy number for any graph Theorem 13 g(g) e(g) where G is a graph and e(g) is the number of edges in G 2

3 Figure 2: The graph G Figure 3: Followers of G and their S-G Numbers Proof Proof by contradiction Let G be a counterexample with a minimal number of edges A graph with no edges has a nimber of 0, thus it can t be a counterexample, and thus e(g) > 0 This means g(g) > e(g), so there must be a follower G F (G) whose nimber is e(g) e(g) > e(g ), since G is a follower of G This means that g(g ) > e(g ), which makes G a counterexample as well We assumed that G was the minimal counterexample, so this is a contradiction Thus g(g) e(g) From now on we will use e(g) to denote the number of edges in G We can think of a spider as consisting of a central vertex, or a body, and a set of paths adjacent to it, the legs Definition The body of a spider graph is the vertex of degree > 2 Definition The legs of a spider graph are the paths connected to the body There are two distinct types of moves in Graph Nim on spider graphs, we call them stomps and cuts Definition A stomp removes one or more edges adjacent to the body Definition A cut removes edges not adjacent to the body We want to study spider graphs that are somehow related to see if there is any pattern to their Sprague-Grundy numbers To do this, we consider a mapping between spider graphs and integer partitions If a spider graph consists of a 1 legs of length 1, a 2 legs of length 2,, a n legs of length n, we represent this graph by the partition (n,, n, n 1,, n 1,, 1,, 1) {{{{{{ a n times a n 1 times a 1 times 3

4 We have created programs in both Java and Sage to calculate the nimbers of various graphs represented as partitions and output them in lists that we can sort and analyze (See data in Appendix A and programs in Appendix B ) After looking through the data that was gathered, a few simple patterns were observed 2 Sprague-Grundy Numbers of Spider Graphs First, we consider spider graphs with k legs of length 1 Definition A k-star is a graph consisting of a single vertex and k paths of length 1 adjacent to the vertex We denote a k-star as S k Theorem 21 For all k 1, g(s k ) = k where S k is a star graph with k edges [4] Let P k denote a path of length k Denote P k P {{ k by (n)p k The Sprague-Grundy n times numbers for paths are known [1] The Sprague-Grundy numbers become periodic as the length of the path grows larger than 72 We can use the path nimbers to help analyze simple spider graphs Theorem 22 For all a 1 2, g(3 1 1 a 1 ) = 3 + a 1 Proof The proof is by induction on a 1 Base Case: We verify these first few cases computationally a 1 = 2, g( ) = 5 a 1 = 3, g( ) = 6 a 1 = 4, g( ) = 7 a 1 = 5, g( ) = 8 Inductive Step: Assume that g(3 1 1 k ) = 3 + k for all k < a 1 Consider a 1 We generate these followers and nimbers for a 1 by applying a cut: g(2 1 1 a 1 ) = 2 + a 1 g(p 1 ) g(s a1 +1) = 1 (1 + a 1 ) g(s a1 +1) = 1 + a 1 g(p 2 ) g(s a1 ) = 2 a 1 g(p 1 ) g(s a1 ) = 1 a 1 In addition, we generate these followers by applying a stomp: g(2 1 ) g(s i ) = 2 i for 0 i a 1 g(3 1 1 i ) for 0 i < a 1 We need to show that for each possibility of a 1, the simplified value does not equal a but generates every value below it We know that each follower is bounded by its number of edges, so we simply need to show that there is followers that generate every nimber in {0,, a

5 Case 1 a 1 0 mod 4 The followers of the form P 2 S i for 0 i a 1 will produce nimbers {0,, a 1 1, a The follower P 1 S a1 +1 will give the nimber a 1 and according to Theorem 21 S a1 +1 produces the nimber a Since gf (3 1 1 a 1 ) = {0,, a 1 + 2, g(3 1 1 a 1 ) = a Case 2 a 1 1 mod 4 The followers of the form P 2 S i for 0 i a 1 will produce nimbers {0,, a 1 2, a 1 + 1, a The follower P 1 S a1 will give the nimber a 1 1 By induction, a 1 3 gives nimber a 1 Since gf (3 1 1 a 1 ) = {0,, a 1 +2, g(3 1 1 a 1 ) = a 1 +3 Case 3 a 1 2 mod 4 The followers of the form P 2 S i for 0 i a 1 will produce nimbers {0,, a 1 2, a 1, a 1 +1 The follower P 1 S a1 will give the nimber a 1 1 By induction, g(3 1 1 a 1 1 ) = a and g(3 1 1 a 1 4 ) = a 1 1 Since gf (3 1 1 a 1 ) = {0,, a 1 + 2, g(3 1 1 a 1 ) = a Case 4 a 1 3 mod 4 The followers of the form P 2 S i for 0 i a 1 will produce nimbers {0,, a 1 By induction, g(3 1 1 a 1 1 ) = a and g(3 1 1 a 1 2 ) = a Since gf (3 1 1 a 1 ) = {0,, a 1 + 2, g(3 1 1 a 1 ) = a Since we now know something about graphs of the form a 1, let s look at a 1 Eventually, we hope to generalize this to graphs of the form n an 2 a 2 1 a 1 Theorem 23 For all a 1 4, g(4 1 1 a 1 ) = 4 + a 1 Proof The proof is by induction on a 1 Base Case: We computationally verify the first few cases a 1 = 4, g( ) = 8 a 1 = 5, g( ) = 9 a 1 = 6, g( ) = 10 a 1 = 7, g( ) = 11 Inductive Step: Assume that g(4 1 1 k ) = 4 + k for all k < a 1 Consider a 1 The cuts for a 1 are: g(3 1 1 a 1 ) = 3 + a 1 g(2 1 1 a 1 ) = 2 + a 1 g(p 1 ) g(2 1 1 a 1 ) = 1 (2 + a 1 ) g(p 1 ) g(s a1 +1) = 1 (1 + a 1 ) g(p 2 ) g(s a1 +1) = 2 (1 + a 1 ) g(p 2 ) g(s a1 ) = 2 a 1 g(p 3 ) g(s a1 ) = 3 a 1 The stomp moves for a 1 are: g(p 3 ) g(s i ) = 3 i for 0 i a 1 g(4 1 1 i ) for 0 i < a 1 5

6 We need to show that for each possibility of a 1, the simplified value does not equal a but generates every value below it We know that each follower is bounded by its number of edges, so we simply need to show that there is followers that generate every nimber in {0,, a Case 1 a 1 0 mod 4 The followers of the form P 3 S i for 0 i a 1 will produce nimbers of {0,, a 1 1, a The follower P 1 S a1 +1 will give the nimber a 1 and the follower P 2 S a1 produces the nimber a 1 +2 By induction, we have that g(4 1 1 a 1 2 ) gives the nimber a Since gf (4 1 1 a 1 ) = {0,, a 1 + 3, g(4 1 1 a 1 ) = a Case 2 a 1 1 mod 4 The followers of the form P 3 S i for 0 i a 1 will produce nimbers of {0,, a 1 2, a 1 + 1, a The follower P 3 S a1 will give the nimber a 1 1 By induction, g(4 1 1 a1 4 ) = a 1 and g(4 1 1 a1 1 ) = a Since gf (4 1 1 a 1 ) = {0,, a 1 + 3, g(4 1 1 a 1 ) = a Case 3 a 1 2 mod 4 The followers of the form P 3 S i for 0 i a 1 will produce nimbers of {0,, a 1 The follower a 1 will give the nimber a By induction, g(4 1 1 a1 1 ) = a and g(4 1 1 a1 3 ) = a Since gf (4 1 1 a 1 ) = {0,, a 1 + 2, g(4 1 1 a 1 ) = a Case 4 a 1 3 mod 4 The followers of the form P 3 S i for 0 i a 1 will produce nimbers of {0,, a 1 3, a 1 1, a 1, a The follower P 2 S a1 generates the nimber a 1 2 By induction, g(4 1 1 a 1 2 ) = a and g(4 1 1 a 1 1 ) = a Since gf (4 1 1 a 1 ) = {0,, a 1 + 3, g(4 1 1 a 1 ) = a Since for all cases of a 1 we have generated nimbers from {0,, a 1 +3 we can conclude that the nimber for a 1 is a Now we examine graphs of the form a 1 We will find that we get slightly different behavor than we did for a 1 and a 1 Theorem 24 For all a 1 8, a if a 1 0 or 3 mod 4, g(5 1 1 a 1 ) = a if a 1 1 mod 4, a 1 if a 1 2 mod 4 Proof The proof is by induction on a 1 Base Case: a 1 = 8, g( ) = 13 a 1 = 9, g( ) = 11 a 1 = 10, g( ) = 10 a 1 = 11, g( ) = 16 a 1 = 12, g( ) = 17 6

7 Inductive Step: Assume that our hypothesis holds for all values less than a 1 Now consider the graph a 1 We generate the Sprague-Grundy numbers for each of the follower graphs of a 1 The follower graphs formed from cuts on a 1 are: g(4 1 1 a 1 ) = 4 + a 1 g(3 1 1 a 1 ) = 3 + a 1 g(p 1 ) g(3 1 1 a 1 ) = 1 (3 + a 1 ) g(p 1 ) g(2 1 1 a 1 ) = 1 (2 + a 1 ) g(p 2 ) g(2 1 1 a 1 ) = 2 (2 + a 1 ) g(p 2 ) g(s a1 +1) = 2 (1 + a 1 ) g(p 3 ) g(s a1 +1) = 3 (1 + a 1 ) g(p 3 ) g(s a1 ) = 3 a 1 g(p 4 ) g(s a1 ) = 1 a 1 In addition, the followers created by stomps are: Case 1 a 1 0 mod 4 We have: g(p 4 ) g(s i ) = 1 i for 0 i a 1 g(5 1 1 i ) for 0 i < a 1 g(4 1 1 a 1 ) = 4 + a 1 g(3 1 1 a 1 ) = 3 + a 1 g(p 1 ) g(3 1 1 a 1 ) = 1 (3 + a 1 ) = 2 + a 1 g(p 2 ) g(2 1 1 a 1 ) = 2 (2 + a 1 ) = a 1 g(p 4 ) g(s a1 ) = 1 a 1 = a Next we consider P 4 S i for 0 i a 1 If i 1 mod 2, then the expression i 1 = i 1 If i 0 mod 2, then the expression i 1 = i + 1 Since a 1 is even, then the values {0 a 1 1, a are generated Since all followers have at most a edges we have gf (G) = {0,, a Thus, g(5 1 1 a 1 ) = a for a 1 0 mod 4 Case 2 a 1 1 mod 4 7

8 We have: g(4 1 1 a 1 ) = 4 + a 1 g(3 1 1 a 1 ) = 3 + a 1 g(p 1 ) g(3 1 1 a 1 ) = 1 (3 + a 1 ) = 4 + a 1 g(p 1 ) g(2 1 1 a 1 ) = 1 (2 + a 1 ) = 1 + a 1 g(p 2 ) g(2 1 1 a 1 ) = 2 (2 + a 1 ) = a 1 g(p 2 ) g(s a1 +1) = 2 (1 + a 1 ) = a 1 1 g(p 3 ) g(s a1 +1) = 3 (1 + a 1 ) = a 1 1 g(p 3 ) g(s a1 ) = 3 a 1 = a g(p 4 ) g(s a1 ) = 1 a 1 = a 1 1 Next we consider P 4 S i for 0 i a 1 If i 1 mod 2, then the expression i 1 = i 1 If i 0 mod 2, then the expression i 1 = i + 1 Since a 1 is odd, then the values {0 a 1 are generated For a i j, where j {1, 2, 3, we get these values by the inductive hypothesis: g(5 1 1 a 1 1 ) = a since a mod 4 g(5 1 1 a 1 2 ) = a since a mod 4 g(5 1 1 a 1 3 ) = a 1 3 since a mod 4 Since all other followers have at most a edges, we have gf (G) = {0,, a 1 + 1, a 1 + 3, a Thus, g(5 1 1 a 1 ) = a for a 1 1 mod 4 Case 3 a 1 2 mod 4 We have: g(4 1 1 a 1 ) = 4 + a 1 g(3 1 1 a 1 ) = 3 + a 1 g(p 1 ) g(3 1 1 a 1 ) = 1 (3 + a 1 ) = 2 + a 1 g(p 1 ) g(2 1 1 a 1 ) = 1 (2 + a 1 ) = 3 + a 1 g(p 2 ) g(2 1 1 a 1 ) = 2 (2 + a 1 ) = a g(p 2 ) g(s a1 +1) = 2 (1 + a 1 ) = a 1 1 g(p 3 ) g(s a1 +1) = 3 (1 + a 1 ) = a 1 2 g(p 3 ) g(s a1 ) = 3 a 1 = a 1 2 g(p 4 ) g(s a1 ) = 1 a 1 = a Next we consider P 4 S i for 0 i a 1 If i 1 mod 2, then the expression i 1 = i 1 If i 0 mod 2, then the expression i 1 = i + 1 Since a 1 is even, then the values {0 a 1 1, a are generated For a i j, where j {1, 2,, 5, we get these values by the inductive hypothesis: 8

9 g(5 1 1 a 1 1 ) = a since a mod 4 g(5 1 1 a 1 2 ) = a since a mod 4 g(5 1 1 a 1 3 ) = a since a mod 4 g(5 1 1 a 1 4 ) = a 1 4 since a mod 4 g(5 1 1 a 1 5 ) = a 1 3 since a mod 4 Since all other followers have at most a 1 1 edges we have gf (G) = {0,, a 1 1, a 1 + 1, a 1 + 2, a 1 + 3, a Thus, g(5 1 1 a 1 ) = a 1 for a 1 2 mod 4 Case 4 a 1 3 mod 4 We have: g(4 1 1 a 1 ) = 4 + a 1 g(3 1 1 a 1 ) = 3 + a 1 g(p 1 ) g(2 1 1 a 1 ) = 1 (2 + a 1 ) = 1 + a 1 g(p 3 ) g(s a1 ) = 3 a 1 = a 1 3 g(p 4 ) g(s a1 ) = 1 a 1 = a 1 1 Consider P 4 S i for 0 i a 1 If i 1 mod 2, then the expression i 1 = i 1 If i 0 mod 2, then the expression i 1 = i + 1 Since a 1 is odd, then the values {0 a 1 are generated By induction, we have g(5 1 1 a 1 3 ) = a Since all followers have at most a edges we have gf (G) = {0,, a Thus, g(5 1 1 a 1 ) = a for a 1 3 mod 4 3 Champion Spiders Definition A champion spider is a spider graph whose Sprague-Grundy number is equal to its number of edges In other words, g(n an (n 1) a n 1 1 a 1 ) = n(a n )+(n 1)(a n 1 )+ +a 1 where n is the maximum leg length Notice that Theorem 22 and Theorem 23 tell us that spiders of the form a 1 and a 1 are champion Theorem 31 g(2 a 2 1 a 1 ) = 2a 2 + a 1 if a 1 2a 2 2 for a 1, a 2 N Proof Proof by induction on a 2 Base case: a 2 = 0 g(2 0 1 a 1 ) = a 1 for all a 1 by Theorem 21 Inductive Step: Assume the hypothesis is true for all integers less than a 2 Consider 2 a 2 1 a 1, where a 1 2a 2 2 The followers of 2 a 2 1 a 1 are as follows: 9

10 1 2 a a a a j 1 k (a 2 j)p 1 for all j a 2, k a 1 and j + k < a 2 + a 1 Note: { g(2 j 1 k g(2 j 1 k P 1 ) if a 2 j 1 mod 2 (a 2 j)p 1 ) = g(2 j 1 k ) if a 2 j 0 mod 2 Claim: {0,, 2a 2 + a 1 1 gf (2 a 2 1 a 1 ) For any h {0,, 2a 2 + a 1 1, there are three cases: Case 1 h = 2a 2 + a 1 1 = g(2 a a 1+1 ) by induction Case 2 h = 2a 2 + a 1 2 = g(2 a a 1 ) by induction Case 3 h < 2a 2 + a 1 2 Note that we can write h = 2j + k where 0 j a 2 1 and 2j 2 k < a 1 We will show that there exists a follower graph with nimber equal to h for all h in this range 1 If a 2 j is even, then g(2 j 1 k ) = 2j + k = h by hypothesis 2 If a 2 j is odd and h is odd, then k is odd Therefore k > 2j 2 and k 1 2j 2, so g(2 j 1 k 1 P 1 ) = (h 1) 1 = h 3 If a 2 j is odd and h is even k < a 1 thus k + 1 a 1, so g(2 j 1 k+1 P 1 ) = (h + 1) 1 = h This shows {0,, 2n + a 1 1 gf (2 a 2 1 a 1 ), and because g(2 a 2 1 a 1 ) 2a 2 + a 1, we have g(2 a 2 1 a 1 ) = 2a 2 + a 1 Definition The integer interval between a and b is the set of consecutive integers between and including a and b, denoted [ab] Definition The numbers [ak], a 0, are generated by a graph G if there exists a follower of G whose Sprague-Grundy number equals c for all c [ak] We denote k [ab] = {k x x [ab] Similarly, we denote k+[ab] = {k+x x [ab] Lemma 32 x y x y x + y for x, y Z Proof Since is binary addition without carries, x y x + y Assume x y < x y for some x and y Then x = x (y y) = (x y) y < (x y) y (x y) + y = x This is a contradiction Thus, x y x y Lemma 33 For a, b, k N, [a + kb k] k [ab] 10

11 Proof Let x [a + kb k] We can write t k = x Then x k = t and x k t x + k by Lemma 32 Thus t [(a + k) k(b k) + k] = [ab] so x k [ab] Lemma 34 For b, k N, [0b k] k [0b] Proof Let x [0b k] We can write t k = x, so x k = t This implies t [0 k(b k) + k] Since t 0, we must have that t [0b] Thus x k [0b] When working with a union of intervals, it is useful to know whether they all overlap and can be expressed instead as a single interval We now prove a Lemma allowing us to check whether this is the case, and if so, what the final interval is Lemma 35 Suppose b i 1 b i, and c b i b i 1 0 for all i n If b i a i c for all i, then n [a i b i ] = [ min (a i)b n ] 1 i n Example 36 Let a i = 0, 2, 1, 7 and b i = 3, 5, 8, 10 Then, i=1 n [a i b i ] = [03] [25] [18] [710] i=1 Notice that we have that b i 1 b i and 3 b i b i 1 0 For all i 4, b i a i 3 Thus, we can apply the Lemma to express this as a single interval: Proof We have that [03] [25] [18] [710] = [010] n [a i b i ] [ min (a i) max (b i)] = [ min (a i)b n ] 1 i n 1 i n 1 i n i=1 Assume x [ min (a i)b n ] For the sake of obtaining a contradiction, suppose x [a i b i ] 1 i n for all i Then it is in [b j b j+1 ] for some j < n Since a j+1 b j+1 c b j, we know that n n x [a j+1 b j+1 ] This is a contradiction Thus, [a i b i ] = [ min (a i)b n ] i=1 Lemma 37 For all a 1,, a n N, a n 0 i=1 [0a 1 1] if n = 1, 2 gf (n an 1 a 1 [0a 1 3] if n [35] ) [0a 1 7] if n [627] [0a 1 15] if n [28 ] In particular, for all n, gf (n an 1 a 1 ) [0a 1 15] 11

12 Proof The n = 1 case is implies by Theorem 21 For n > 1, consider all followers of the form n 1 k G, 0 k a 1, where G = (a i )P i 1 Notice that g(1 k G) = g(1 k ) g(g) = k g(g) i=2 We know inequalities about the nimber of paths, [1] 3 if 2 n 4 g(p n ) 7 if 5 n 26 8 if 27 n Then So we have 3 if 2 n 4 g(g) 7 if 5 n if 27 n gf (n an 1 a 1 ) {g(1 k G) 0 k a 1 [0a 1 1] if n = 2 [0a 1 3] if n [35] = g(g) [0a 1 ] [0a 1 7] if n [627] [0a 1 15] if n [28 ] 4 Stability Definition Define the operation + between two spiders to be n an 1 a 1 + n bn 1 b 1 = n an+bn 1 a 1+b 1 Note that this is a vertex contraction on the two root vertices and that it is is associative and commutative Definition The spider G is said to be stable if all spiders of the form G + 1 k where k 0 are champion spiders In other words, if a spider is stable then we can add any number of legs of length 1 to the spider and the resulting spider is champion Definition Let f(g) be the smallest natural number such that G + 1 f(g) is stable If no such number exists, it is defined as Example 41 We have f(3 1 ) = 2 That is, when a 1 2, a 1 is champion, and is not a champion Additionally, f( ) = 1 since is champion but is not 12

13 41 Algorithm for finding f(g) In order to find an algorithm to calculate f(g), we need an algorithm to decide if G + 1 a 1 is stable The following theorem provides a method for determining a graph s stability by checking only a finite number of cases This implies that an algorithm to find f(g) exists Theorem 42 Let G be a spider If G+1 a 1 is champion for 0 a 1 d where d = 13+e(G), then G is stable Proof To show that G + 1 a 1 is stable, we must show that G + 1 a 1 is champion for all a 1 0 We induct on a 1 Base case We have that G + 1 a 1 is champion for 0 a 1 d by the inductive hypothesis Inductive Step Assume the hypothesis is true for all d < a 1 < m Consider the case when m = a 1 In order to show G + 1 a 1 is a champion, we have to show gf (G + 1 a 1 ) = [0e(G + 1 a 1 ) 1] We have {g(g + 1 k ) 0 k < a 1 = [e(g)e(g + 1 a 1 1 )] = [e(g)e(g + 1 a 1 ) 1] and by Lemma 37, we know gf (G + 1 a 1 ) [0a 1 15] [0d ] = [014 + e(g) 15] = [0e(G) 1] Therefore, we can see that G + 1 a 1 is champion Remark In the above proof, there is a sharper bound for d if we know more information about the length of longest leg in G In particular, if we let the longest leg in G have length n, we get: e(g) 1 if n [12] 1 + e(g) if n [35] d = 5 + e(g) if n [627] 13 + e(g) if n [28 ] This comes from applying Lemma 37 slightly differently For example, suppose that we know that n [35] Then, by Lemma 37, we know that gf (G + 1 a 1 ) [0a 1 3] [0d + 1 3] In order to get this equal to [0e(G) 1], we need d = e(g) + 1 If we know that n [12], then by Lemma 37, we know that gf (G + 1 a 1 ) [0a 1 1] [0d + 1 1] In order to get this equal to [0e(G) 1], we need d = e(g) 1 If we know that n [627], then by Lemma 37, we know that gf (G+1 a 1 ) [0a 1 7] [0d + 1 7] In order to get this equal to [0e(G) 1], we need d = e(g) + 5 If we know that n [28 ], then by Lemma 37, we know that gf (G + 1 a 1 ) [0a 1 15] [0d ] In order to get this equal to [0e(G) 1], we need d = e(g) + 13 Theorem 22 and Theorem 23 can be proven by showing f(3 1 ) = 2 and f(4 1 ) = 4 These values can be verified by observing k is a champion for k [ ] = [26], and k is a champion for k [ ] = [49] Using Theorem 42, we have an algorithm, Stable(G), to check if G is stable An algorithm for computing f(g) follows by checking if G + 1 k is stable for every k If the algorithm terminates, then we found f(g) Otherwise f(g) = Note that we have no bounds on f(g), and thus there is no way to determine that f(g) = without running the program for an infinite amount of time 13

14 1: procedure stable(g) 2: bound 13 + e(g) 3: for k from 0 bound do 4: if e(g + 1 k ) g(g + 1 k ) then 5: return f alse 6: end if 7: end for 8: return true 9: end procedure 1: procedure f(g) 2: for k from 0 do 3: if stable(g + 1 k ) then 4: return k 5: end if 6: end for 7: end procedure 42 A result about champions, f(g) and f(g ) Theorem 43 For all spiders G, f(g ) f(g) + e(g) + 15 Proof For convenience, let N = e(g) + a If f(g) = then the theorem is trivially true Suppose f(g) is finite, and let c = f(g) + (N 2 a 1 ) + 15 We want to show for any a 1 c, G a 1 is a champion If gf (G a 1 ) = [0N 1], then G a 1 is a champion Consider the following three cases that will generate all the nimbers in the range Case 1 The follower graphs G + 1 a 1+1 and G + 1 a 1 have the nimbers N 1 and N 2 respectively Case 2 Consider the set of graphs X = {G + 1 f(g) P 1, G + 1 f(g)+1 P 1,, G + 1 a 1 P 1 Each one is a follower of G a 1 Note that all of these graphs are champion spiders union a path of length 1 Thus, we get {g(x) x X = 1 [f(g) + e(g)e(g) a i ] = 1 [c 15N 2] [c 14N 3] by Lemma 33 Case 3 Using Lemma 37, we know gf (G a 1 ) [0a 1 15] [0c 15] 14

15 The three sets cover the entire interval [0N 1], therefore G a 1 is a champion for all a 1 c c = f(g) + (N 2 a 1 ) + 15 = f(g) + e(g) + 15 f(g ) Thus, we have a bound on f(g ) in terms of f(g) It would be useful to have a similar theorem for length 3 legs Theorem 44 Let G be a spider If then G a 1 a 1 max(f(g), f(g )) + e(g) + 16 and e(g a 1 ) 2 or 3 (mod 4), is a champion spider Proof Let c = max(f(g), f(g )) If c =, then the theorem is trivially true Suppose c is finite For convenience, let N = e(g a 1 ) Let a 1 max(f(g), f(g )) e(g) and let c = c + (N a 1 3) + 16 If g(g a 1 ) = N, then the graph is a champion We want to show for any a 1 c and N 2 or 3 (mod 4), G a 1 is a champion Let a 1 c We want to show that gf (G a 1 ) = [0N 1] Consider 4 possible cases that will generate all the nimbers in the range Case 1 The follower graphs G a 1 and G + 1 a 1+1 generate [N 2N 1] Case 2 Consider the set of graphs X = {G + 1 c P 2, G + 1 c+1 P 2,, G + 1 a 1 P 2 Each one is a follower of G a 1 Note that these graphs are champion spiders union a path of length 2 Thus, we get {g(x) x X = 2 [c + N 3 a 1 N 3] = 2 [c 16N 3] [c 14N 5] by Lemma 33 Case 3 gf (G a 1 ) [0a 1 15] [0c 15] Case 4 Notice that [N 4N 3] has not yet been generated Consider the following two cases: If N 2 mod 4, { g(g + 1 a1 2 P 2 ) generates N 3 g(g + 1 a 1 3 P 2 ) generates N 4 If N 3 mod 4, { g(g + 1 a 1+1 P 1 ) generates N 3 g(g + 1 a 1 3 P 2 ) generates N 4 In either case, this shows that [N 4N 3] is generated by some followers Thus, we know that g(g a 1 ) are champions for all a 1 c and N 2 or 3 (mod 4) 15

16 5 Periodicity in the Discrepancy Definition The discrepancy of a graph G, d(g), is defined as d(g) = e(g) g(g) Notice that champion spiders are precisely the spiders with discrepancy 0 We get the following property for the discrepancy Let G, G, H and H be graphs If d(g) = d(g ) and d(h) = d(h ), then d(g H) = d(g H ) Definition Let G be a spider Let c(g) be the smallest natural number such that for all x c(g), d(g + 1 x ) = d(g + 1 x+p(g) ) where p(g) is the smallest period of the sequence d(g + 1 x ), x c(g) We can think of c(g) as the number of new length 1 legs that a spider needs to begin showing periodic behavior in the discrepancy Notice that by definition, f(g) = k if and only if c(g) = k, p(g) = 1, and d(g + 1 k ) = 0 We can use the concept of discrepancy to restate the notion in Theorem 24 Theorem 51 Consider the graph 5 1 The following are true: c(5 1 ) = 8 p(5 1 ) = 4 (d( ), d( ), d( ), d( )) = (0, 3, 5, 0) Can we find and prove values for c(g) and p(g)? If so, then Theorem 24 can be proven by verifying that c(5 1 ) = 8 and p(5 1 ) = 4 In this section, we hope to develop a theorem that gives us an effective way to compute c(g) and p(g) Definition Let L be the set of spiders without any legs of length 1 Theorem 52 Let G L and fix c, p N Let m = max({d(g + 1 k ) c k < c + p {e(g) + 13) If the following conditions hold, then c(g) c and p(g) p 1 For all c x c + p + m, 2 max({c(g ) G G, G L) c 3 lcm({p(g ) G G, G L) p d(g + 1 x ) = d(g + 1 x+p ) Notice that we choose m in this theorem so that d(g + 1 k ) m for all c k < c + p Thus, e(g + 1 k ) g(g + 1 k ) m, so we have g(g + 1 k ) e(g + 1 k ) m This theorem allows us to find c(5 1 ) and p(5 1 ) mechanically, and thus prove theorems similar to Theorem 51 16

17 Proof of Theorem 51 It is known that c(4 1 ) = 4, p(4 1 ) = 1 by Theorem 23, c(3 1 ) = 2, p(3 1 ) = 1 by Theorem 22, and c(2 1 ) = 0, p(2 1 ) = 1 by Theorem 31 Let c = 8 and p = 4 m = max({d(g + 1 k ) c k < c + p {e(g) + 13) = max({d(g + 1 c ), d(g + 1 c+1 ),, d(g + 1 c+p 1 ) {18) = max({d(g ), d(g ), d(g ), d(g ) {18) = max({13 g(g ), 14 g(g ), 15 g(g ), 16 g(g ), 18 = max({0, 3, 5, 18) = 18 We need to check if d(5 1 1 k ) has a period of 4 starting at k = 8 In other words, we need to show that c(g) c and p(g) p 1 By the data in Appendix A2, d(5 1 1 k ) = d(5 1 1 k+4 ) for all 8 k = max({c(4 1 ), c(3 1 ), c(2 1 )) = is a multiple of lcm({p(4 1 ), p(3 1 ), p(2 1 )) = 1 By Theorem 52, c(5 1 ) 8 and p(5 1 ) 4 Since d( ) d( ), we know that c(g) > 7, so c(g) = 8 Since d( ) d( ), we know p(g) 2 and since d( ) d( ) we know p(g) 1 and since p(g) 4, we get that p(g) = 4 Proof of Theorem 52 To show that c(g) c and p(g) p, we must show that for all x c, d(g + 1 x ) = d(g + 1 x+p ) The desired relation between the discrepancy can be expressed as a relation between the nimbers d(g + 1 x ) = d(g + 1 x p ) if and only if g(g + 1 x ) = g(g + 1 x p ) + p We will use induction to show that g(g + 1 x ) = g(g + 1 x p ) + p Proof by induction on x Base Case: The theorem holds for c x c + p + m by hypothesis Inductive Step: Assume the hypothesis is true for all c + p + m < x < t Consider x = t This shows x m > c + p Note: in this proof, we will use G to be a subgraph of a spider G with G L and we will use P to denote some finite set of disjoint paths Let X = {G + 1 k P c + p k x, G + 1 k P F (G + 1 x ) and let Y = {G + 1 k p P G + 1 k P X By this definition, we see that X contains graphs that consist of spider graphs that are covered by either the inductive hypothesis or the conditions in the theorem Y contains the set of spiders for which if you add p legs of length 1 to each, you get an element of X We also define X = F (G + 1 x )\X and Y = F (G + 1 x p )\Y Notice that by definition, X F (G + 1 x ) and Y F (G + 1 x p ) We need to show the following: 17

18 1 [0g(G + 1 x k ) + p 1] gf (G + 1 x ) 2 g(g + 1 x p ) + p gf (G + 1 x ) We will show 1 by showing that Y must generate nimbers in [e(g) + cg(g + 1 x p ) 1], and thus X must generate nimbers in [e(g) + c + pg(g + 1 x p ) + p 1] We will show 2 by showing that nothing in X can have the nimber g(g + 1 x p ) + p by contradiction and that the nimber of everything in X is less than g(g + 1 x p ) + p By the property discussed at the beginning of Section 5, we have for every element G + 1 k P X, e(g ) + e(p ) e(g) and or equivalently d(g + 1 k P ) = d(g + 1 k p P ) g(g + 1 k P ) = g(g + 1 k p P ) + p ( ) Proof that [0g(G + 1 x p ) + p 1] gf (G + 1 x ) Using ( ), we know if G + 1 k p P Y, then we get g(g + 1 k p P ) + p gf (G + 1 x ) The nimbers generated by Y give us information about the nimbers generated by X In fact, if {g(h) H Y [ab], then {g(h ) H X [a + pb + p] We know the nimbers generated by Y Y = F (G + 1 x p ) covers [0g(G + 1 x p ) 1] Consider what can be generated by only Y Since for every H Y, e(h) < e(g + 1 c ) = e(g) + c, Y can only generate elements in [0e(G) + c 1] Then Y must generate [e(g) + cg(g + 1 x p ) 1] Thus X must generate [e(g) + c + pg(g + 1 x p ) + p 1] Lemma 37 shows gf (G + 1 x ) [0x 15] [0(e(G) + c + p ) 15] = [0e(G) + c + p 1] Thus [0g(G + 1 x p ) + p 1] gf (G + 1 x ) Proof that g(g + 1 x p ) + p gf (G + 1 x ) We must show that there is no graph H in X or X such that g(h) = g(g + 1 x p ) First, consider the followers in X Let G + 1 k P X First, we will show that no member of X has the nimber g(g + 1 x p ) + p In other words, g(g + 1 k P ) g(g + 1 x p ) + p To find a contradiction, assume this is false Then, there exist G + 1 k P X, such that g(g + 1 x p ) = g(g + 1 k P ) p (1) = g(g + 1 k p P ) (2) gf (G + 1 x p ) (3) where (2) is true by the inductive hypothesis if G = G or by the conditions required by the theorem if G G This is a contradiction, since for any graph H, g(h) gf (H) Thus g(g + 1 x p ) is not generated by a member of X Second, we will show that no member of X can have the nimber g(g + 1 x p ) + p by showing that if H X, then g(h) < g(g + 1 x p ) + p 18

19 We show this bound by showing e(h) < g(g + 1 x p ) + p Note that by the definition of X, H is of the form G + 1 k P, where k < c + p < x m Since e(g + 1 k P ) e(g + 1 k ), we know e(h) e(g + 1 c+p ) (4) = e(g) + c + p (5) < e(g) + x m (6) Recall that g(g + 1 k ) e(g + 1 k ) m Thus, g(g + 1 x p ) e(g + 1 x p ) m = e(g) + x p m Combining (3) and (6), we get g(g + 1 x p ) + p e(g) + x m > e(h) g(h) So, we get g(g + 1 x p ) + p X X = gf (G + 1 x ) Thus, we have g(g + 1 x p ) + p gf (G + 1 x ) and [0g(G + 1 x p ) + p 1] gf (G + 1 x ), which imply that g(g + 1 x ) = g(g + 1 x p ) + p which means so finally, we know c(g) c and p(g) p d(g + 1 x ) = d(g + 1 x+p ) Notice that although this theorem only considered spiders with no legs of length 1, it can be extended to spiders by noting c(g + 1 k ) = max(c(g) k, 0) and p(g + 1 k ) = p(g) where G L Thus we can easily apply this theorem to spiders with any legs 6 Conjectures on Champion Spiders 61 3 a 3 2a 2 1a 1 Conjecture 61 3 a 3 2 a 2 1 a 1 is a champion spider if a 2 > 0 and for all 3 i 2 j 3 a 3 2 a 2, a 1 > f(3 i 2 j ) + 15 where f(3 i 2 j ) is defined in Section 4 Our attempted proof was by induction on a 3 and a 2 By Theorem 31, we know that 2 a 2 1 a 1 is champion for all a 1 2a 2 2 Thus, f(3 0 2 a 2 ) = 2a 2 2 which exists for all a 2 We can finitely compute f(3 a 2 b ), so our base case consisted of f( ) = 2 and f( ) = 4 and assumed that we could build up from there in our inductive step This turns out to be inadequate as a base case, since we don t have any bounds on f(3 a+1 2 b ) in terms of f(3 a 2 b ) Thus in order for our proof to be sufficient, we need to know the value of f(3 a 2 b ) for all a N in our base case, which is impossible without a nice, evaluatable expression for f(3 a 2 b ) Theorem 44 works toward a bound on f(3 a+1 2 b ) Our inductive step was very similar to the proof of Theorem 31 We broke the interval [03a 3 + 2a 2 + 1a 1 1] into the subintervals [0 5 + f( )], [6 + f( ) 3a 3 + 2a 2 + a 1 3], [3a 3 + 2a 2 + a 1 23a 3 + 2a 2 + a 1 1] 19

20 1: procedure c(g + 1 k ) 2: c, p findcp(g) G L in all the functions 3: return max(0, c k) 4: end procedure 5: procedure p(g + 1 k ) 6: c, p findcp(g) 7: return p 8: end procedure 9: procedure findcp(g) 10: c : p : for all G < G and G L do 13: c 0 max(c(g ), c 0 ) 14: p 0 lcm(p(g ), p 0 ) 15: end for 16: for all c c 0, p 0 p do 17: m 0 18: for all c k < c + p do 19: m max(d(g + 1 k ), m) 20: end for 21: m max(m, e(g) + 15) 22: success true 23: for all c x c + p + m do 24: if d(g + 1 x ) d(g + 1 x+p ) then 25: success f alse 26: end if 27: end for 28: if success then 29: return findboundedcp(g, c, p) 30: end if 31: end for 32: end procedure 33: procedure findboundedcp(g,c,p ) 34: for c from 0 to c do 35: If d(g + 1 x ) = d(g + 1 x+p ) for all c x c + p, stop 36: end for 37: for all p from 0 to p do 38: If d(g + 1 x ) = d(g + 1 x+p ) for all c x c + p, stop 39: end for 40: return c, p 41: end procedure 20

21 and specified followers that achieve all of the numbers in each set However, we realized that our proof assumed that a 2 > 1 for all followers, which is not the case When we noticed this error, we began working on proving the following conjecture Conjecture 62 Let a 3 10 For a 1 { 3a 3 7 3a 3 6 if a 3 even if a 3 odd, 3 a 3 1 a 1 is champion We formed this conjecture by looking at values of d(3 a 3 1 a 1 ), which are contained in Appendix A1 After a certain number of 1s, we noticed that the discrepancy became zero and stayed there for larger values of a 1, indicating that these graphs become stable after a point Conjecture 62 can be shown true for a given a 3 by applying a refined version of Theorem 42 This theorem states that we only need to verify if d(3 a 3 1 k ) = 0 for all 3a 3 6 k 6a 3 5 if a 3 is odd, and if d(3 a 3 1 k ) = 0 for all 3a 3 7 k 6a 3 6 if a 3 is even We have done this for all a 3 30 In fact, if we call m the number of ones for which the graphs become stable, we found predictable behavior of d(3 a 3 1 m 1 ), d(3 a 3 1 m 2 ),, d(3 a 3 1 m 5 ) for a 3 12 In particular, we have the following conjecture Conjecture 63 Let a 3 12 For a 1 = { 3a 3 7 3a 3 6 if a 3 even if a 3 odd, 3 a 3 1 a 1 is stable if and only if we get the following: For a 3 even, d(3 a 3 1 3a3 8 ) = 3 d(3 a 3 1 3a3 9 ) = 0 d(3 a 3 1 3a3 10 ) = 4 d(3 a 3 1 3a3 11 ) = 4 d(3 a 3 1 3a3 12 ) = 8 For a 3 odd, d(3 a 3 1 3a3 7 ) = 5 d(3 a 3 1 3a3 8 ) = 0 d(3 a 3 1 3a3 9 ) = 0 d(3 a 3 1 3a3 10 ) = 3 d(3 a 3 1 3a3 11 ) = 5 Example 64 Let a 3 = 15 Then, let a 1 = 3a 3 6 = 39 If we calculate d(3 a 3 1 a3 6 i ) for i [1, 2, 3, 4, 5], we get the desired values 5,0,0,3,5 This means that is champion Alternatively, if we examine a spider of the form 3 i 1 j that we know to be champion that satisfies the requirements on i and j, we can look it up in the table and if the conjecture is true, we will see the forementioned pattern in the discrepancies of smaller graphs It seems like the larger we force a 3 to be, we can predict d(3 a 3 1 m i ) for larger i 21

22 62 4 a 4 3a 3 2a 2 1a 1 Conjecture 65 f(4 a 4 3 a 3 2 a 2 ) is finite We attempted to prove this in the same manner as Conjecture 62, but with induction on a 4, a 3, and a 2 We discovered similar problems: we would need base cases for graphs of the form 4 i 3 j 1 f(3i 2 j) for all i, j N The proof also incorrectly assumes that a 4, a 3, a 2 are always greater than 1 for all followers 7 Conclusion We have developed methods for computing the Sprague-Grundy numbers of certain types of spider graphs Ultimately, our results discuss only a small subset of these graphs We hope in the future to be able to extend our results using the new methods presented in Section 5 In addition, we plan to further work on the conjectures in Section 6 and show that the graphs 4 a 4 3 a 3 2 a 2 1 a 1 and 3 a 3 2 a 2 1 a 1 become stable Additionally, we would like to be able to say something about the length of the predictable pattern in values of the discrepancy as we force a 3 to be larger, which we mention after Conjecture 63 It is possible that this pattern before acheiving stability could also be generalized to other spider graphs Acknowledgements We would like to thank Neil J Calkin and Janine E Janoski for their invaluable guidance and support this summer Without their knowledge we would not have made the progress that we did In addition, we would also like to thank the National Science Foundation for providing funding this summer through grants DMS and DMS and to Clemson University for hosting us Lastly, we would like to thank Jim Brown, Kevin James, and Mohammed Tessema for leading the program this summer References [1] Neil J Calkin, Kevin James, Janine E Janoski, Sarah Leggett, Bryce Richards, Nathan Sitaraman, and Stephanie Thomas Computing strategies for graphical nim 2009 [2] John H Conway On numbers and games AK Peters, Natick, Mass, 2001 [3] John H Conway Winning ways for your mathematical plays AK Peters, 2001 [4] Mikio Kano Edge-removing games of star type Discrete Mathematics, 151(13): , 1996 [5] Dierk Schleicher and Michael Stoll An introduction to Conway s games and numbers Moscow Math Journal, 6(2): ,

23 A Computational Data for g(4 a 43 a 32 a 21 a 1) List of all g(4 a 4 3 a 3 2 a 2 1 a 1 ) where a 4 = 0, a 3 2, a 2 3, a It is in the format [a 4, a 3, a 2, a 1 ] g(4 a 4 3 a 3 2 a 2 1 a 1 ) [0, 0, 0, 0] 0 [0, 0, 0, 1] 1 [0, 0, 0, 2] 2 [0, 0, 0, 3] 3 [0, 0, 0, 4] 4 [0, 0, 0, 5] 5 [0, 0, 0, 6] 6 [0, 0, 0, 7] 7 [0, 0, 0, 8] 8 [0, 0, 0, 9] 9 [0, 0, 0, 10] 10 [0, 0, 0, 11] 11 [0, 0, 0, 12] 12 [0, 0, 0, 13] 13 [0, 0, 0, 14] 14 [0, 0, 0, 15] 15 [0, 0, 0, 16] 16 [0, 0, 0, 17] 17 [0, 0, 0, 18] 18 [0, 0, 0, 19] 19 [0, 0, 0, 20] 20 [0, 0, 0, 21] 21 [0, 0, 0, 22] 22 [0, 0, 0, 23] 23 [0, 0, 0, 24] 24 [0, 0, 0, 25] 25 [0, 0, 0, 26] 26 [0, 0, 0, 27] 27 [0, 0, 0, 28] 28 [0, 0, 0, 29] 29 [0, 0, 0, 30] 30 [0, 0, 0, 31] 31 [0, 0, 0, 32] 32 [0, 0, 0, 33] 33 [0, 0, 0, 34] 34 [0, 0, 0, 35] 35 [0, 0, 0, 36] 36 [0, 0, 0, 37] 37 [0, 0, 0, 38] 38 [0, 0, 0, 39] 39 [0, 0, 0, 40] 40 [0, 0, 0, 41] 41 [0, 0, 0, 42] 42 [0, 0, 0, 43] 43 [0, 0, 0, 44] 44 [0, 0, 0, 45] 45 [0, 0, 0, 46] 46 [0, 0, 0, 47] 47 [0, 0, 0, 48] 48 [0, 0, 0, 49] 49 [0, 0, 0, 50] 50 [0, 0, 0, 51] 51 [0, 0, 0, 52] 52 [0, 0, 0, 53] 53 [0, 0, 0, 54] 54 [0, 0, 0, 55] 55 [0, 0, 0, 56] 56 [0, 0, 0, 57] 57 [0, 0, 0, 58] 58 [0, 0, 0, 59] 59 [0, 0, 0, 60] 60 [0, 0, 0, 61] 61 [0, 0, 0, 62] 62 [0, 0, 0, 63] 63 [0, 0, 0, 64] 64 [0, 0, 0, 65] 65 [0, 0, 0, 66] 66 [0, 0, 0, 67] 67 [0, 0, 0, 68] 68 [0, 0, 0, 69] 69 [0, 0, 0, 70] 70 [0, 0, 0, 71] 71 [0, 0, 0, 72] 72 [0, 0, 0, 73] 73 [0, 0, 0, 74] 74 [0, 0, 0, 75] 75 [0, 0, 0, 76] 76 [0, 0, 0, 77] 77 [0, 0, 0, 78] 78 [0, 0, 0, 79] 79 [0, 0, 0, 80] 80 [0, 0, 0, 81] 81 [0, 0, 0, 82] 82 [0, 0, 0, 83] 83 [0, 0, 0, 84] 84 [0, 0, 0, 85] 85 [0, 0, 0, 86] 86 [0, 0, 0, 87] 87 [0, 0, 0, 88] 88 [0, 0, 0, 89] 89 [0, 0, 0, 90] 90 [0, 0, 0, 91] 91 [0, 0, 0, 92] 92 [0, 0, 0, 93] 93 [0, 0, 0, 94] 94 [0, 0, 0, 95] 95 [0, 0, 0, 96] 96 [0, 0, 0, 97] 97 [0, 0, 0, 98] 98 [0, 0, 0, 99] 99 [0, 0, 0, 100] 100 [0, 0, 1, 0] 2 [0, 0, 1, 1] 3 [0, 0, 1, 2] 4 [0, 0, 1, 3] 5 [0, 0, 1, 4] 6 [0, 0, 1, 5] 7 [0, 0, 1, 6] 8 [0, 0, 1, 7] 9 [0, 0, 1, 8] 10 [0, 0, 1, 9] 11 [0, 0, 1, 10] 12 [0, 0, 1, 11] 13 [0, 0, 1, 12] 14 [0, 0, 1, 13] 15 [0, 0, 1, 14] 16 [0, 0, 1, 15] 17 [0, 0, 1, 16] 18 [0, 0, 1, 17] 19 [0, 0, 1, 18] 20 [0, 0, 1, 19] 21 [0, 0, 1, 20] 22 [0, 0, 1, 21] 23 [0, 0, 1, 22] 24 [0, 0, 1, 23] 25 [0, 0, 1, 24] 26 [0, 0, 1, 25] 27 [0, 0, 1, 26] 28 [0, 0, 1, 27] 29 [0, 0, 1, 28] 30 [0, 0, 1, 29] 31 [0, 0, 1, 30] 32 [0, 0, 1, 31] 33 [0, 0, 1, 32] 34 [0, 0, 1, 33] 35 [0, 0, 1, 34] 36 [0, 0, 1, 35] 37 [0, 0, 1, 36] 38 [0, 0, 1, 37] 39 [0, 0, 1, 38] 40 [0, 0, 1, 39] 41 [0, 0, 1, 40] 42 [0, 0, 1, 41] 43 [0, 0, 1, 42] 44 [0, 0, 1, 43] 45 [0, 0, 1, 44] 46 [0, 0, 1, 45] 47 [0, 0, 1, 46] 48 [0, 0, 1, 47] 49 [0, 0, 1, 48] 50 [0, 0, 1, 49] 51 [0, 0, 1, 50] 52 [0, 0, 1, 51] 53 [0, 0, 1, 52] 54 [0, 0, 1, 53] 55 [0, 0, 1, 54] 56 [0, 0, 1, 55] 57 [0, 0, 1, 56] 58 [0, 0, 1, 57] 59 [0, 0, 1, 58] 60 23

24 [0, 0, 1, 59] 61 [0, 0, 1, 60] 62 [0, 0, 1, 61] 63 [0, 0, 1, 62] 64 [0, 0, 1, 63] 65 [0, 0, 1, 64] 66 [0, 0, 1, 65] 67 [0, 0, 1, 66] 68 [0, 0, 1, 67] 69 [0, 0, 1, 68] 70 [0, 0, 1, 69] 71 [0, 0, 1, 70] 72 [0, 0, 1, 71] 73 [0, 0, 1, 72] 74 [0, 0, 1, 73] 75 [0, 0, 1, 74] 76 [0, 0, 1, 75] 77 [0, 0, 1, 76] 78 [0, 0, 1, 77] 79 [0, 0, 1, 78] 80 [0, 0, 1, 79] 81 [0, 0, 1, 80] 82 [0, 0, 1, 81] 83 [0, 0, 1, 82] 84 [0, 0, 1, 83] 85 [0, 0, 1, 84] 86 [0, 0, 1, 85] 87 [0, 0, 1, 86] 88 [0, 0, 1, 87] 89 [0, 0, 1, 88] 90 [0, 0, 1, 89] 91 [0, 0, 1, 90] 92 [0, 0, 1, 91] 93 [0, 0, 1, 92] 94 [0, 0, 1, 93] 95 [0, 0, 1, 94] 96 [0, 0, 1, 95] 97 [0, 0, 1, 96] 98 [0, 0, 1, 97] 99 [0, 0, 1, 98] 100 [0, 0, 1, 99] 101 [0, 0, 1, 100] 102 [0, 0, 2, 0] 1 [0, 0, 2, 1] 5 [0, 0, 2, 2] 6 [0, 0, 2, 3] 7 [0, 0, 2, 4] 8 [0, 0, 2, 5] 9 [0, 0, 2, 6] 10 [0, 0, 2, 7] 11 [0, 0, 2, 8] 12 [0, 0, 2, 9] 13 [0, 0, 2, 10] 14 [0, 0, 2, 11] 15 [0, 0, 2, 12] 16 [0, 0, 2, 13] 17 [0, 0, 2, 14] 18 [0, 0, 2, 15] 19 [0, 0, 2, 16] 20 [0, 0, 2, 17] 21 [0, 0, 2, 18] 22 [0, 0, 2, 19] 23 [0, 0, 2, 20] 24 [0, 0, 2, 21] 25 [0, 0, 2, 22] 26 [0, 0, 2, 23] 27 [0, 0, 2, 24] 28 [0, 0, 2, 25] 29 [0, 0, 2, 26] 30 [0, 0, 2, 27] 31 [0, 0, 2, 28] 32 [0, 0, 2, 29] 33 [0, 0, 2, 30] 34 [0, 0, 2, 31] 35 [0, 0, 2, 32] 36 [0, 0, 2, 33] 37 [0, 0, 2, 34] 38 [0, 0, 2, 35] 39 [0, 0, 2, 36] 40 [0, 0, 2, 37] 41 [0, 0, 2, 38] 42 [0, 0, 2, 39] 43 [0, 0, 2, 40] 44 [0, 0, 2, 41] 45 [0, 0, 2, 42] 46 [0, 0, 2, 43] 47 [0, 0, 2, 44] 48 [0, 0, 2, 45] 49 [0, 0, 2, 46] 50 [0, 0, 2, 47] 51 [0, 0, 2, 48] 52 [0, 0, 2, 49] 53 [0, 0, 2, 50] 54 [0, 0, 2, 51] 55 [0, 0, 2, 52] 56 [0, 0, 2, 53] 57 [0, 0, 2, 54] 58 [0, 0, 2, 55] 59 [0, 0, 2, 56] 60 [0, 0, 2, 57] 61 [0, 0, 2, 58] 62 [0, 0, 2, 59] 63 [0, 0, 2, 60] 64 [0, 0, 2, 61] 65 [0, 0, 2, 62] 66 [0, 0, 2, 63] 67 [0, 0, 2, 64] 68 [0, 0, 2, 65] 69 [0, 0, 2, 66] 70 [0, 0, 2, 67] 71 [0, 0, 2, 68] 72 [0, 0, 2, 69] 73 [0, 0, 2, 70] 74 [0, 0, 2, 71] 75 [0, 0, 2, 72] 76 [0, 0, 2, 73] 77 [0, 0, 2, 74] 78 [0, 0, 2, 75] 79 [0, 0, 2, 76] 80 [0, 0, 2, 77] 81 [0, 0, 2, 78] 82 [0, 0, 2, 79] 83 [0, 0, 2, 80] 84 [0, 0, 2, 81] 85 [0, 0, 2, 82] 86 [0, 0, 2, 83] 87 [0, 0, 2, 84] 88 [0, 0, 2, 85] 89 [0, 0, 2, 86] 90 [0, 0, 2, 87] 91 [0, 0, 2, 88] 92 [0, 0, 2, 89] 93 [0, 0, 2, 90] 94 [0, 0, 2, 91] 95 [0, 0, 2, 92] 96 [0, 0, 2, 93] 97 [0, 0, 2, 94] 98 [0, 0, 2, 95] 99 [0, 0, 2, 96] 100 [0, 0, 2, 97] 101 [0, 0, 2, 98] 102 [0, 0, 2, 99] 103 [0, 0, 2, 100] 104 [0, 0, 3, 0] 3 [0, 0, 3, 1] 7 [0, 0, 3, 2] 5 [0, 0, 3, 3] 9 [0, 0, 3, 4] 10 [0, 0, 3, 5] 11 [0, 0, 3, 6] 12 [0, 0, 3, 7] 13 [0, 0, 3, 8] 14 [0, 0, 3, 9] 15 [0, 0, 3, 10] 16 [0, 0, 3, 11] 17 [0, 0, 3, 12] 18 [0, 0, 3, 13] 19 [0, 0, 3, 14] 20 [0, 0, 3, 15] 21 [0, 0, 3, 16] 22 [0, 0, 3, 17] 23 [0, 0, 3, 18] 24 [0, 0, 3, 19] 25 [0, 0, 3, 20] 26 [0, 0, 3, 21] 27 [0, 0, 3, 22] 28 [0, 0, 3, 23] 29 [0, 0, 3, 24] 30 [0, 0, 3, 25] 31 [0, 0, 3, 26] 32 [0, 0, 3, 27] 33 [0, 0, 3, 28] 34 [0, 0, 3, 29] 35 [0, 0, 3, 30] 36 [0, 0, 3, 31] 37 [0, 0, 3, 32] 38 [0, 0, 3, 33] 39 [0, 0, 3, 34] 40 [0, 0, 3, 35] 41 [0, 0, 3, 36] 42 24

25 [0, 0, 3, 37] 43 [0, 0, 3, 38] 44 [0, 0, 3, 39] 45 [0, 0, 3, 40] 46 [0, 0, 3, 41] 47 [0, 0, 3, 42] 48 [0, 0, 3, 43] 49 [0, 0, 3, 44] 50 [0, 0, 3, 45] 51 [0, 0, 3, 46] 52 [0, 0, 3, 47] 53 [0, 0, 3, 48] 54 [0, 0, 3, 49] 55 [0, 0, 3, 50] 56 [0, 0, 3, 51] 57 [0, 0, 3, 52] 58 [0, 0, 3, 53] 59 [0, 0, 3, 54] 60 [0, 0, 3, 55] 61 [0, 0, 3, 56] 62 [0, 0, 3, 57] 63 [0, 0, 3, 58] 64 [0, 0, 3, 59] 65 [0, 0, 3, 60] 66 [0, 0, 3, 61] 67 [0, 0, 3, 62] 68 [0, 0, 3, 63] 69 [0, 0, 3, 64] 70 [0, 0, 3, 65] 71 [0, 0, 3, 66] 72 [0, 0, 3, 67] 73 [0, 0, 3, 68] 74 [0, 0, 3, 69] 75 [0, 0, 3, 70] 76 [0, 0, 3, 71] 77 [0, 0, 3, 72] 78 [0, 0, 3, 73] 79 [0, 0, 3, 74] 80 [0, 0, 3, 75] 81 [0, 0, 3, 76] 82 [0, 0, 3, 77] 83 [0, 0, 3, 78] 84 [0, 0, 3, 79] 85 [0, 0, 3, 80] 86 [0, 0, 3, 81] 87 [0, 0, 3, 82] 88 [0, 0, 3, 83] 89 [0, 0, 3, 84] 90 [0, 0, 3, 85] 91 [0, 0, 3, 86] 92 [0, 0, 3, 87] 93 [0, 0, 3, 88] 94 [0, 0, 3, 89] 95 [0, 0, 3, 90] 96 [0, 0, 3, 91] 97 [0, 0, 3, 92] 98 [0, 0, 3, 93] 99 [0, 0, 3, 94] 100 [0, 0, 3, 95] 101 [0, 0, 3, 96] 102 [0, 0, 3, 97] 103 [0, 0, 3, 98] 104 [0, 0, 3, 99] 105 [0, 0, 3, 100] 106 [0, 1, 0, 0] 3 [0, 1, 0, 1] 1 [0, 1, 0, 2] 5 [0, 1, 0, 3] 6 [0, 1, 0, 4] 7 [0, 1, 0, 5] 8 [0, 1, 0, 6] 9 [0, 1, 0, 7] 10 [0, 1, 0, 8] 11 [0, 1, 0, 9] 12 [0, 1, 0, 10] 13 [0, 1, 0, 11] 14 [0, 1, 0, 12] 15 [0, 1, 0, 13] 16 [0, 1, 0, 14] 17 [0, 1, 0, 15] 18 [0, 1, 0, 16] 19 [0, 1, 0, 17] 20 [0, 1, 0, 18] 21 [0, 1, 0, 19] 22 [0, 1, 0, 20] 23 [0, 1, 0, 21] 24 [0, 1, 0, 22] 25 [0, 1, 0, 23] 26 [0, 1, 0, 24] 27 [0, 1, 0, 25] 28 [0, 1, 0, 26] 29 [0, 1, 0, 27] 30 [0, 1, 0, 28] 31 [0, 1, 0, 29] 32 [0, 1, 0, 30] 33 [0, 1, 0, 31] 34 [0, 1, 0, 32] 35 [0, 1, 0, 33] 36 [0, 1, 0, 34] 37 [0, 1, 0, 35] 38 [0, 1, 0, 36] 39 [0, 1, 0, 37] 40 [0, 1, 0, 38] 41 [0, 1, 0, 39] 42 [0, 1, 0, 40] 43 [0, 1, 0, 41] 44 [0, 1, 0, 42] 45 [0, 1, 0, 43] 46 [0, 1, 0, 44] 47 [0, 1, 0, 45] 48 [0, 1, 0, 46] 49 [0, 1, 0, 47] 50 [0, 1, 0, 48] 51 [0, 1, 0, 49] 52 [0, 1, 0, 50] 53 [0, 1, 0, 51] 54 [0, 1, 0, 52] 55 [0, 1, 0, 53] 56 [0, 1, 0, 54] 57 [0, 1, 0, 55] 58 [0, 1, 0, 56] 59 [0, 1, 0, 57] 60 [0, 1, 0, 58] 61 [0, 1, 0, 59] 62 [0, 1, 0, 60] 63 [0, 1, 0, 61] 64 [0, 1, 0, 62] 65 [0, 1, 0, 63] 66 [0, 1, 0, 64] 67 [0, 1, 0, 65] 68 [0, 1, 0, 66] 69 [0, 1, 0, 67] 70 [0, 1, 0, 68] 71 [0, 1, 0, 69] 72 [0, 1, 0, 70] 73 [0, 1, 0, 71] 74 [0, 1, 0, 72] 75 [0, 1, 0, 73] 76 [0, 1, 0, 74] 77 [0, 1, 0, 75] 78 [0, 1, 0, 76] 79 [0, 1, 0, 77] 80 [0, 1, 0, 78] 81 [0, 1, 0, 79] 82 [0, 1, 0, 80] 83 [0, 1, 0, 81] 84 [0, 1, 0, 82] 85 [0, 1, 0, 83] 86 [0, 1, 0, 84] 87 [0, 1, 0, 85] 88 [0, 1, 0, 86] 89 [0, 1, 0, 87] 90 [0, 1, 0, 88] 91 [0, 1, 0, 89] 92 [0, 1, 0, 90] 93 [0, 1, 0, 91] 94 [0, 1, 0, 92] 95 [0, 1, 0, 93] 96 [0, 1, 0, 94] 97 [0, 1, 0, 95] 98 [0, 1, 0, 96] 99 [0, 1, 0, 97] 100 [0, 1, 0, 98] 101 [0, 1, 0, 99] 102 [0, 1, 0, 100] 103 [0, 1, 1, 0] 4 [0, 1, 1, 1] 6 [0, 1, 1, 2] 7 [0, 1, 1, 3] 5 [0, 1, 1, 4] 9 [0, 1, 1, 5] 10 [0, 1, 1, 6] 11 [0, 1, 1, 7] 12 [0, 1, 1, 8] 13 [0, 1, 1, 9] 14 [0, 1, 1, 10] 15 [0, 1, 1, 11] 16 [0, 1, 1, 12] 17 [0, 1, 1, 13] 18 [0, 1, 1, 14] 19 25

26 [0, 1, 1, 15] 20 [0, 1, 1, 16] 21 [0, 1, 1, 17] 22 [0, 1, 1, 18] 23 [0, 1, 1, 19] 24 [0, 1, 1, 20] 25 [0, 1, 1, 21] 26 [0, 1, 1, 22] 27 [0, 1, 1, 23] 28 [0, 1, 1, 24] 29 [0, 1, 1, 25] 30 [0, 1, 1, 26] 31 [0, 1, 1, 27] 32 [0, 1, 1, 28] 33 [0, 1, 1, 29] 34 [0, 1, 1, 30] 35 [0, 1, 1, 31] 36 [0, 1, 1, 32] 37 [0, 1, 1, 33] 38 [0, 1, 1, 34] 39 [0, 1, 1, 35] 40 [0, 1, 1, 36] 41 [0, 1, 1, 37] 42 [0, 1, 1, 38] 43 [0, 1, 1, 39] 44 [0, 1, 1, 40] 45 [0, 1, 1, 41] 46 [0, 1, 1, 42] 47 [0, 1, 1, 43] 48 [0, 1, 1, 44] 49 [0, 1, 1, 45] 50 [0, 1, 1, 46] 51 [0, 1, 1, 47] 52 [0, 1, 1, 48] 53 [0, 1, 1, 49] 54 [0, 1, 1, 50] 55 [0, 1, 1, 51] 56 [0, 1, 1, 52] 57 [0, 1, 1, 53] 58 [0, 1, 1, 54] 59 [0, 1, 1, 55] 60 [0, 1, 1, 56] 61 [0, 1, 1, 57] 62 [0, 1, 1, 58] 63 [0, 1, 1, 59] 64 [0, 1, 1, 60] 65 [0, 1, 1, 61] 66 [0, 1, 1, 62] 67 [0, 1, 1, 63] 68 [0, 1, 1, 64] 69 [0, 1, 1, 65] 70 [0, 1, 1, 66] 71 [0, 1, 1, 67] 72 [0, 1, 1, 68] 73 [0, 1, 1, 69] 74 [0, 1, 1, 70] 75 [0, 1, 1, 71] 76 [0, 1, 1, 72] 77 [0, 1, 1, 73] 78 [0, 1, 1, 74] 79 [0, 1, 1, 75] 80 [0, 1, 1, 76] 81 [0, 1, 1, 77] 82 [0, 1, 1, 78] 83 [0, 1, 1, 79] 84 [0, 1, 1, 80] 85 [0, 1, 1, 81] 86 [0, 1, 1, 82] 87 [0, 1, 1, 83] 88 [0, 1, 1, 84] 89 [0, 1, 1, 85] 90 [0, 1, 1, 86] 91 [0, 1, 1, 87] 92 [0, 1, 1, 88] 93 [0, 1, 1, 89] 94 [0, 1, 1, 90] 95 [0, 1, 1, 91] 96 [0, 1, 1, 92] 97 [0, 1, 1, 93] 98 [0, 1, 1, 94] 99 [0, 1, 1, 95] 100 [0, 1, 1, 96] 101 [0, 1, 1, 97] 102 [0, 1, 1, 98] 103 [0, 1, 1, 99] 104 [0, 1, 1, 100] 105 [0, 1, 2, 0] 7 [0, 1, 2, 1] 8 [0, 1, 2, 2] 9 [0, 1, 2, 3] 10 [0, 1, 2, 4] 11 [0, 1, 2, 5] 12 [0, 1, 2, 6] 13 [0, 1, 2, 7] 14 [0, 1, 2, 8] 15 [0, 1, 2, 9] 16 [0, 1, 2, 10] 17 [0, 1, 2, 11] 18 [0, 1, 2, 12] 19 [0, 1, 2, 13] 20 [0, 1, 2, 14] 21 [0, 1, 2, 15] 22 [0, 1, 2, 16] 23 [0, 1, 2, 17] 24 [0, 1, 2, 18] 25 [0, 1, 2, 19] 26 [0, 1, 2, 20] 27 [0, 1, 2, 21] 28 [0, 1, 2, 22] 29 [0, 1, 2, 23] 30 [0, 1, 2, 24] 31 [0, 1, 2, 25] 32 [0, 1, 2, 26] 33 [0, 1, 2, 27] 34 [0, 1, 2, 28] 35 [0, 1, 2, 29] 36 [0, 1, 2, 30] 37 [0, 1, 2, 31] 38 [0, 1, 2, 32] 39 [0, 1, 2, 33] 40 [0, 1, 2, 34] 41 [0, 1, 2, 35] 42 [0, 1, 2, 36] 43 [0, 1, 2, 37] 44 [0, 1, 2, 38] 45 [0, 1, 2, 39] 46 [0, 1, 2, 40] 47 [0, 1, 2, 41] 48 [0, 1, 2, 42] 49 [0, 1, 2, 43] 50 [0, 1, 2, 44] 51 [0, 1, 2, 45] 52 [0, 1, 2, 46] 53 [0, 1, 2, 47] 54 [0, 1, 2, 48] 55 [0, 1, 2, 49] 56 [0, 1, 2, 50] 57 [0, 1, 2, 51] 58 [0, 1, 2, 52] 59 [0, 1, 2, 53] 60 [0, 1, 2, 54] 61 [0, 1, 2, 55] 62 [0, 1, 2, 56] 63 [0, 1, 2, 57] 64 [0, 1, 2, 58] 65 [0, 1, 2, 59] 66 [0, 1, 2, 60] 67 [0, 1, 2, 61] 68 [0, 1, 2, 62] 69 [0, 1, 2, 63] 70 [0, 1, 2, 64] 71 [0, 1, 2, 65] 72 [0, 1, 2, 66] 73 [0, 1, 2, 67] 74 [0, 1, 2, 68] 75 [0, 1, 2, 69] 76 [0, 1, 2, 70] 77 [0, 1, 2, 71] 78 [0, 1, 2, 72] 79 [0, 1, 2, 73] 80 [0, 1, 2, 74] 81 [0, 1, 2, 75] 82 [0, 1, 2, 76] 83 [0, 1, 2, 77] 84 [0, 1, 2, 78] 85 [0, 1, 2, 79] 86 [0, 1, 2, 80] 87 [0, 1, 2, 81] 88 [0, 1, 2, 82] 89 [0, 1, 2, 83] 90 [0, 1, 2, 84] 91 [0, 1, 2, 85] 92 [0, 1, 2, 86] 93 [0, 1, 2, 87] 94 [0, 1, 2, 88] 95 [0, 1, 2, 89] 96 [0, 1, 2, 90] 97 [0, 1, 2, 91] 98 [0, 1, 2, 92] 99 [0, 1, 2, 93]