The Game of Nim on Graphs: NimG

1. The Game of Nim on Graphs: NimG Gwendolyn Stockman Alan Frieze and Juan Vera

2. The Game of Nim • 2 players • n piles of disks, with a1, a2, … an disks on each pile • Players take turns removing disks from each pile • A player loses when there are no disks left

3. Impartial Games • Geography, Nim, and NimG are impartial games • A game in which the only difference between the two players is which one goes first • Every position of an impartial game has a Grundy Number or Nim-Value

4. x y w z Sprague-Grundy Numbers (Nim-Sums) • Used to represent wining and losing positions • Given a game tree T = (V,E) • Recursively define: with • A player is in a winning position if at the end of his/her turn the playing piece is on such that

5. Grundy Numbers (cont.) • Let a1, a2, … an be the binary representations of the number of disks on each of the n piles • In the game of Nim the Grundy number for a setup is the bit-wise sum (mod 2) of the ai‘s • Example: 1 1 1 1 0 0 1 1 1 1 0 0 0 0 The Grundy number is 0 so this is a winning position!

6. Proposed Versions of NimG • 2 players • 1 piece is moved along an undirected graph with no self-loops, and discs are removed • If discs on vertices: • Could move first and then remove discs • Could remove discs and then move • Possibly allow moves to empty verticies • If discs on edges: • Remove discs as you go along an edge • Players take turns removing disks from piles • How to win: • The other player can’t complete their turn

7. Example of Vertex NimG 7 15 9 6 5 6 5 4 21 21 10 10 8 8 5 13 2

8. Geography • Two players alternately move a piece on a graph until one loses by being unable to make an legal moves • Directed or Undirected • Edge Geography • No edge repeated • Vertex Geography • No vertex repeated • Undirected Vertex Geography • Contained in a version of Vertex NimG with: • 1 chip on each vertex • Not allowed to move to empty vertices • Solvable in polynomial Time • Second player has a winning strategy IFF the graph has a perfect matching (Fraenkel 1993) • Undirected Edge Geography • PSPACE-complete (Fraenkel 1993)

9. Previous Work: Nim on Graphs • In A Nim game played on Graphs I and II (Fukuyama 2003) Edge NimG was examined • Proved that the Grundy Numbers can be found completely on: • Bipartite Graphs • Trees • Cycles • Main Method: matchings on graphs • Contains normal Nim • 2 vertices with n edges with a1, a2, … an disks on each edge • Contains Undirected Edge Geography • 1 disk on each edge

10. andisks a0disks an-1disks a1disks … v0 v1 vn-1 vn The Game: Vertex NimG • The game from now on: • Vertex NimG • Moves to empty vertices allowed • Notation for Vertex NimG on a path of length N • (a0,a1,…,an-1,an):0 represents: • Where is the piece being moved

11. Theorem 1: Vertex NimG on path of length 2

12. 0 0 x 0 x x 0 0 0 x 0 NimG on a Path of Length 3 • Not so simple on longer paths • Grundy Numbers bounded by a function of d, the number of disks on the graph • First note that x  • Easy to see that • is a child for and • Easy to see that • is a child and • Rest omitted for brevity

13. (0,a,b,0):1 (0,0,b,0):2 (0,1,b,0):2 (0,2,b,0):2 (0,b-1,b,0):2 (0,b,b,0):2 0 a b 0 NimG on a Path of Length 3 (cont.) • Will prove that given a,bsuch that a>b then • Because with

14. NimG on a Path of Length 3 (cont.) • Take a and we want to know the value of • Examine the children of that position in the game tree and find a+1 a 1 1

15. NimG on a Path of Length 3 (cont.) • Take a and we want to know the value of • Examine the children of that position in the game tree and find a+1 a+2 2 1 2

16. (0,a,b,0):1 (0,0,b,0):2 (0,1,b,0):2 (0,2,b,0):2 (0,b-1,b,0):2 (0,b,b,0):2 (0,1,b,0):0 NimG on a Path of Length 3 (cont.) • So given a,bsuch that a>b then • And • Further suppose we have a,b,c,e{0}such that then 1 2 3 4 b+1 0

17. u v x y (G, A):v (G, A’):u Vertex NimG on Any Graph • Represent positions of NimG as (G, A):v where G is the graph, A is the amount function for G, and marker is at vertex vV(G) • A reduced game tree is used to find if a winning strategy exists from a given position, and if one does, what it is • Create the reduced game tree, T, where each node is a position, by • making (G,A):v the root • For each uN(v) such that A(u)< A(v) add node (G,A’): u as a child to (G,A):v where • Repeat once for each node added to T

18. v h c b a d v c e a b f v b c a v b a c h Example: Create T • Player P1 starts with setup (G,A):v 12 4 9 3 2 1 6 5

19. v h b b c a d v c d e a b f a d b v b d Example: Create T • Start with setup (G,A’):a 12 4 8 3 2 1 6 5

20. v h c b b a d v c d e a b f d a Example: Create T • Start with setup (G,A’’):b 12 4 8 3 2 1 5 5

21. v h b b c a d v e f c d e a b f b f e v f e a Example: Create T • Start with setup (G,A’’):b 12 4 8 3 2 1 5 5

22. v h c b b a d v e f c d e f a b f e f c b f Example: Create T • Start with setup (G,A’’’):e 12 4 8 3 2 1 5 4

23. v h c b b a d v e f c d e f a b f f c b e Example: Create T • Start with setup (G,A’’’):e 12 4 8 3 1 1 5 4

24. v h c b b a d v e e e f f f c d e f f f a b f e f f f e f Example: Create T 12 4 9 3 1 1 5 4

25. Labeling T • If it is player P1’s turn then label all even levels, including 0, of T, P1, and all odd levels P2 • Label each node either P1 or P2 • Player P1 has a winning strategy from nodes labeled P1 • Player P2 has a winning strategy from nodes labeled P2 • Label all nV(T), with labeling function L:V(T){P1,P2}, using a depth first labeling starting with the n=root of V(T) • Apply the depth first labeling to all children of n in T • Label n with L(n) • The root of T is labeled P1 IFF there is a winning strategy for player P1

26. v b c b a e e e f f f d b w f e f f f d v b b f e a c w a f f f c f e e f b w v d e f f f w a e c b v e Example: Labeling T • Depth First Labeling of T : P1 does NOT have a winning strategy from (G,A):v! n a leaf and on a P2-level n a leaf and on a P2-level or n on a P1-level and at least one child of n labeled P1 or n on a P1-level and at least one child of n labeled P1 P1 or n on a P2-level and all children of n labeled P1 L(n):= n a leaf and on a P1-level n a leaf and on a P1-level P1 or n on a P2-level and at least one child of n labeled P2 or n on a P2-level and at least one child of n labeled P2 P2 or n on a P1-level and all children of n labeled P2 or n on a P1-level and all children of n labeled P2 P2 = A node labeled P2 = A node labeled P1 P1 = A node whose subtree is being labeled P2 = The node currently being examined P1

27. v h d v c f b v e b f f e f c d e f v a f e a b f v v Why P1 can’t win • Assume that players will always follow a winning strategy if one exists • If P1 moves to a, b, or c then P2 will win • What if P1 moves to h? P1 still loses! P1 11 12 10 8 P2 4 9 2 5 8 6 3 P1 2 1 6 5 P2 P1

28. (G, A):v (G, A’):u The Winning Strategy: Case 1 • If player P2 just moved from vertex u to vertex v, creating setup (G, A):v, and (G, A’):u is a child of (G, A): v labeled P1 • Remove r[1, A(v)- A(u)] disks from vertex v and move to vertex u, creating setup with • Replace the sub-tree with root (G, A’):u with the tree of root created as described before (G, A):v

29. (G, A):w (G, A’):x The Winning Strategy: Case 2 • Otherwise, given setup (G, A):v pick any child (G, A’):u of node (G, A):v in T that is labeled P1 • remove r[1, A(v)- A(u)] disks from vertex v and move to vertex u, creating setup with • Replace the sub-tree with root (G, A’):u with the tree of root created as described before • The alterations to T insure that T maintains the property that for any node y=(G, A):w V(T)

30. 1 node root of T  n-1 nodes  n-2 nodes  n-d nodes NimG on any Graph: Analysis • Tree T is exponential in size • What if there is a maximum number of disks allowed on each vertex? • Polynomial – at most nd number of nodes, where d is the maximum number of disks allowed on a vertex and n=V(G) • There does not exist any path starting from the root with a node (G,A):v and a node for any amount functions A and

31. Questions???