The%20Game%20of%20Nim%20on%20Graphs:%20NimG

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The Game of Nim on Graphs NimG

• By Gwendolyn Stockman
• With Alan Frieze, and Juan Vera

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The Game of Nim

• 2 players
• n piles of disks, with a1,a2, an 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

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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 cant complete their turn

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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

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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)

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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.

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My Theorem (Notation)

• (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.

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My Theorem
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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,

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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

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Sketch of Proof (cont.)

• And, since
• by above, we have
• So,
• The rest of the theorem is proved by induction.

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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.

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Questions???