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
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 ais
• Example

1
1
1
0
0
1
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 cant complete their turn

7
Example of Vertex NimG
7
15
9
6
6
5
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
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
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
NimG on a Path of Length 3 (cont.)

• Will prove that given a,b??? such that agtb 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

a1
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

a1
a2
2
2
1
16
NimG on a Path of Length 3 (cont.)

• So given a,b??? such that agtb then
• And
• Further suppose we have a,b,c,e ???0 such that
• then

1
2
3
4
b1
0
17
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)lt A(v) add node
  (G,A) u as a child to (G,A)v where
• Repeat once for each node added to T

18
Example Create T

• Player P1 starts with setup (G,A)v

12
4
9
3
2
1
6
5
19
Example Create T

• Start with setup (G,A)a

12
4
8
3
2
1
6
5
20
Example Create T

• Start with setup (G,A)b

12
4
8
3
2
1
5
5
21
Example Create T

• Start with setup (G,A)b

12
4
8
3
2
1
5
5
22
Example Create T

• Start with setup (G,A)e

12
4
8
3
2
1
5
4
23
Example Create T

• Start with setup (G,A)e

12
4
8
3
1
1
5
4
24
Example Create T
12
4
9
3
1
1
5
4
25
Labeling T

• If it is player P1s 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
  LV(T)?P1,P2, using a depth first labeling
  starting with the nroot 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
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
Why P1 cant 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
12
11
10
8
P2
4
9
3
8
6
5
2
P1
2
1
6
5
P2
P1
28
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
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
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???