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

2. The Game of Nim 2 players n piles of disks, with a 1, a 2, … a n amounts of disks on each pile, respectively Players take turns decreasing the number of disks on each pile to any strictly smaller, non-negative integer A player loses when there are no disks left to be removed

3. Proposed Versions of NimG 2 players 1 piece is moved along an undirected graph, and discs are removed If discs on vertices:  Could move then remove discs  Could remove discs then move If discs on edges:  Remove discs as you go along an edge Players take turns decreasing the number of disks on each pile to any strictly smaller, non-negative integer. How to win:  If you remove the last disk  The other player can’t complete their turn

4. Grundy Numbers Used to represent wining and losing positions Given an acyclic diagraph H = (V,E) Define: Recursively define: 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.) The Grundy Numbers are calculated in reverse order, starting from a winning position Write a program to calculate the Grundy Numbers for all possible positions in the game tree for nimG (which is an acyclic diagraph)

6. Previous Work: Nim on Graphs Edge version – each edge assigned a non- negative integer Undirected Graphs including:  Bipartite Graphs  Trees  Cycles In A Nim game played on Graphs II (Fukuyama) it was proven that the Grundy Numbers of Nim on Trees and Nim on Cycles can be found completely.

7. (a,b,c):0 means: Where is the piece being moved Note that my Theorem is for the vertex version of NimG, with removing disks then moving. My Theorem (Notation) v0v0 v1v1 v2v2 a disks b disks c disks

8. My Theorem

9. Sketch of Proof Note that, by definition: For, all possible moves from result in: and so, thus, Similarly for, For, all possible moves from result in: and both and so, thus,

10. Sketch of Proof (cont.) If, and then so, Note that the same thing happens for and If,, and then not only is but, So, It can be shown that since for all non- negative integers

11. Sketch of Proof (cont.) And, since by above, we have So, The rest of the theorem is proved by induction.

12. Next Steps Prove or Disprove:  Both vertex versions of NimG are special cases of the edge version, in that they can be transformed into trees. Further examine interesting cases with the bounds on Grundy Numbers  For the remove then move vertex version of NimG, path of 4 vertices, leads to much higher Grundy Numbers, than were found for the 3 vertex version.

13. Questions???