The Game of Nim on Graphs Nim G

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

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

Proposed Versions of Nim. G 2 players 1 piece is moved along an undirected graph, and discs are removed l If discs on vertices: l l ¡ ¡ l 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. l How to win: l ¡ ¡ If you remove the last disk The other player can’t complete their turn

Grundy Numbers Used to represent wining and losing positions l Given an acyclic diagraph H = (V, E) l Define: l Recursively define: l l A player is in a winning position if at the end of his/her turn the playing piece is on such that

Grundy Numbers (cont. ) l The Grundy Numbers are calculated in reverse order, starting from a winning position l Write a program to calculate the Grundy Numbers for all possible positions in the game tree for nim. G (which is an acyclic diagraph)

Previous Work: Nim on Graphs Edge version – each edge assigned a nonnegative integer l Undirected Graphs including: l ¡ ¡ ¡ l 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.

My Theorem (Notation) l (a, b, c): 0 means: a disks c disks b disks v 0 v 1 v 2 Where is the piece being moved l Note that my Theorem is for the vertex version of Nim. G, with removing disks then moving. l

My Theorem

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

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

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

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

Questions? ? ?