Outline for today
Stat155
Game Theory
Lecture 2: Combinatorial Games
Combinatorial games:
Positions, moves, terminal positions, impartial/partisan, progressively
bounded, directed graphs.
Progressively bounded impartial games.
The sets N and P.
Theorem: Someone can win.
Peter Bartlett
Examples
Subtraction
Chomp
August 30, 2016
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Example: Subtraction Game
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Example: Subtraction Game
15 chips
Two players: I and II.
Does Player I have an advantage from having the first move?
Players alternate moves; Player I starts.
If both players play optimally, who will win?
At each move, the player can remove 1 or 2 chips.
What is an optimal strategy?
A player wins when they take the last chip
(so that the other player cannot move).
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Example: Subtraction Game
Example: Subtraction Game
Let x be the number of chips remaining.
Suppose you move next.
Can you guarantee a win?
Write N as the set of positions where the Next player to move can
guarantee a win, provided that they play optimally.
Write P as the set of positions where the other player—the player that
moved Previously—can guarantee a win, provided that they play optimally.
For x = 0?
0 ∈ P.
For x ∈ {1, 2}?
1, 2 ∈ N.
For x = 3?
3 ∈ P.
For x = 4?
4, 5 ∈ N.
For x = 5?
6 ∈ P.
For x = 6?
7, 8 ∈ N.
For x ∈ {7, 8}?
15 ∈ P.
For x = 15?
Player II can always win.
What is Player II’s optimal strategy?
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Definitions: Combinatorial games
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Definitions: Combinatorial games
A combinatorial game has:
Two players, Player I and Player II.
A set of positions X .
For each player, a set of legal moves between positions, that is, a set
of ordered pairs, (current position, next position):
MI , MII ⊂ X × X .
Terminology:
An impartial game has the same set of legal moves for both players:
MI = MII .
A partisan game has different sets of legal moves for the players.
A terminal position for a player has no legal move to another position.
x is terminal for player I if there is no y ∈ X with (x, y ) ∈ MI .
Players alternately choose moves; Player I goes first from some
starting position x0 ∈ X .
Play continues until some player cannot move.
Normal play: the player that cannot move loses the game.
Misère play: the player that cannot move wins the game.
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Definitions: Combinatorial games
Example: Subtraction Game
The subtraction game is an impartial combinatorial game:
A combinatorial game is progressively bounded if, for every starting
position x0 ∈ X , there is finite bound on the number of moves before
the game ends. (That is, if B(x) denotes the maximum number of
moves before the game ends, then B(x) < ∞.)
A strategy for a player is a function that assigns a legal move to each
non-terminal position.
If XNT is the set of non-terminal positions for Player I, then
SI : XNT 7→ X is a strategy for player I if,
for all x ∈ XNT , (x, SI (x)) ∈ MI .
A winning strategy for a player from position x:
guaranteed to result in a win for that player from that position.
Positions X = {0, 1, 2, . . . , 15}.
Moves = {(x, y ) ∈ X × X : y ∈ {x − 1, x − 2}}.
Terminal position: 0.
“Normal play”: the player who moves to 0 wins.
i.e., the player in the terminal position loses.
Progressively bounded (from x ∈ X , there can be no more than x
moves until the terminal position).
A winning strategy for any starting position x ∈ N:
S(x) = 3bx/3c.
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Impartial combinatorial games as directed graphs
Example: Subtraction Game
Positions = nodes.
What is the graph for the subtraction game?
Moves = edges.
Every edge from a node in P leads into N.
Terminal positions = nodes without outgoing edges.
There is an edge from a node in N to a node in P.
“Normal play”: the player who moves to a terminal position wins.
The winning strategy chooses one of these edges.
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Impartial combinatorial games as directed graphs
Impartial combinatorial games and winning strategies
Theorem
What directed graphs correspond to progressively bounded games?
In a progressively bounded impartial
combinatorial game under normal
play, X = N ∪ P.
That is, from any initial position, one
of the players has a winning strategy.
Acyclic graphs
... where all paths from a node have bounded length.
What is B(x)?
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Impartial combinatorial games and winning strategies:
Proof
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Impartial combinatorial games and winning strategies:
Proof
Inductive hypothesis: All x with B(x) ≤ n are in either N or P.
Base case: B(x) = 0 only for terminal positions. But then x ∈ P.
Inductive step: If B(x) = n + 1, then every legal move leads to y with
B(y ) ≤ n, so y ∈ N ∪ P. Consider all the legal next positions y .
Either
1
2
Why did we need the progressively bounded condition?
All of these y are in N, which implies x ∈ P, or
Some legal move leads to a y in P, which implies x ∈ N.
(Recall: B(x) denotes the maximum number of moves before the game
ends.)
Thus, every x is in N ∪ P.
From any initial position, one of the players has a winning strategy.
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Key Ideas: Progressively bounded impartial games
Example: Chomp
Theorem: Every position is in N (positions where the next player has
a winning strategy) or P (positions where the previous player has a
winning strategy).
P: Every move leads to N.
N: Some move leads to P (hence cannot contain terminal positions).
Chomp is an impartial combinatorial game:
Positions X = {non-empty subsets of chocolate block :
left-closed and below-closed}.
Moves = {(x, y ) ∈ X × X : y = x − chomp}.
Terminal position: bottom left square (broccoli).
“Normal play”: the player left with the broccoli loses.
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Example: Chomp
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Example: Chomp
Chomp is progressively bounded (from x ∈ X with |x| blocks
remaining, there can be no more than |x| − 1 moves until the terminal
position).
Hence, there is a winning strategy for one of the players.
Theorem
Every non-terminal rectangle is in N.
Which player?
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Example: Chomp
Example: Chomp
Why?
Because from a rectangle r ∈ X , there is a legal move (r , r 0 ) ∈ M
that we can always choose to skip, that is, for any move (r 0 , s) ∈ M,
we also have (r , s) ∈ M.
(What is this r 0 ?)
2
We showed that, from a non-terminal rectangle, there is a winning
strategy for Player I.
But we didn’t construct a winning strategy.
In particular, we didn’t prove that r 0 is a good or bad move.
Examples:
Why does this imply r ∈ N?
There are two cases:
1
This proof technique is called strategy stealing.
(We’ll encounter it again.)
1
r 0 ∈ P (which implies r ∈ N), and
r 0 ∈ N. In this case, there is an s ∈ P with (r 0 , s) ∈ M. But then we
know that (r , s) ∈ M, also implying r ∈ N.
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Key Ideas: Progressively bounded impartial games
Theorem: Every position is in N (positions where the next player has
a winning strategy) or P (positions where the previous player has a
winning strategy).
P: Every move leads to N.
N: Some move leads to P (hence cannot contain terminal positions).
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2
For a 3 × 2 rectangle, r 0 is a winning move.
(Check!)
For a large-enough square (n × n rectangle with n > 2), r 0 is not a
winning move.
(What is a winning move?)
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