Lecture 1: Zerosum perfect information games, Reachability games
Teachers: Wieslaw Zielonka
Game Theory techniques in CS: Introduction
Finite twoplayer zerosum perfect information games
Simplest games: finite ones, finite trees, outcomes on the leaves
$u_i$ utility function (payoff function) for player $i$
Start from the leaves, and backward induction: compute what player 1 earns provided both players act rationally:

player 1 tries to maximises his/her outcome

player 2 tries to minimise player 1’s outcome.
 Couple (profile) of strategies:

(σ^1, σ^2)
$Σ_1, Σ_2$: sets of strategies of player 1 & 2.
 Payoff:

u: Σ_1 × Σ_2 ⟶ ℝ
 Game value:

$v ≝ u(σ^1, σ^2)$
For zerosum games, one talks about optimal strategies.
Theorem (Zermelo): Finite perfect information zerosum games have a game value.
Exs:
 chess is a finite game.

combinatorial game: the Hex game is a finite zerosum game with only two outcomes: win or lose (the first player to have a path from a side of his/her color to the opposing side wins: there’s always a winner).
 the first player (white) to play has a winning strategy
Extensive perfect information games
You may have “chance nodes” in the tree: the child chosen from such nodes is drawn from a fixed probability distribution.
⟶ Value to maximise, for each player: the expectancy of his/her gain
For each stategy profile $(σ^1, \ldots, σ^n)$, one compute the value $\sum\limits_{ i } p_i o_i$, where $p_i$ is the probability of the outcome $o_i$.
Nash equilibria
 Strategy profile $(σ^1, \ldots, σ^n)$ is a Nash equilibrium:

iff for all $τ^i ∈ Σ_i$, u_i(σ^1, \ldots, σ^n) ≥ u_i(σ^1, σ^{i1}, τ^i, σ^{i+1}, σ^n)
NB: it’s in no player’s interest to change his/her strategy.
Each subgame perfect equilibrium is a Nash equilibrium. But the converse is not true.
Games
 Game:

 there are 2 players (Left and Right)
 there are several positions with a fixed starting position
 for each position: a set of available moves specifying the next position
 Left and Right move alternatively
 both players have complete information
 no chance/probabilistic move
 the loser is the one who can’t move anymore at the end
Games as directed acyclic graph (DAG): vertices = positions. Edges in either player’s color, or in a neutral color (if the move can be take by both players).
There is no infinite path (even though the tree may be infinite).
 Winning positions:

all the positions such that the player who plays in this position (the current player) has a winning strategy
 Losing positions:

all the positions such that the player who plays in this position (the current player) has no winning strategy
All the vertices with no outgoing edge are losing (the player can’t play ⟶ he/she loses).
The game of Nim
Typical combinatorial game.
 Operation $\oplus$:

$n \oplus m$ = bitwise XOR between $\underbrace{(n)2}{\rlap{\text{binary representation}}}$ and $(m)_2$
Winning strategy: when $\bigoplus_i n_i ≠ 0$ (where the $n_i$’s are the number of coins on the $i$th stack).
Poker Nim: each player has a personal stack of coins which is finite, and new move: each player can add a certain number of their personal coins to any stack on the board.
⟶ can be reduced to the Nim game: when the opponent is in a losing position and adds personal coins, all the winning player has to do is to remove the coins the opponent just added.
Reachability games
Simplest zerosum games
Let $G = (V, E)$ be a directed graph. Two players: A(lice) and B(ob): $V ≝ V_A ∪ V_B$.
Suppose that for each vertex, there is at least one outgoing edge.
 Reachability game:

we have a set of vertices $R ⊆ V$ and the question is: « Does Alice have a strategy such that someone ends up in $R$? »
Notation:
What does it mean to win in one step?
Similary, one defines $n\text{Step}_A(R)$:
 $0\text{Step}_A(R) ≝ R$
 $1\text{Step}_A(R)$ as above
 $(n+1)\text{Step}_A(R) ≝ n\text{Step}_A(R) ∪ 1\text{Step}_A\Big(n\text{Step}_A(R)\Big)$
Clearly, $(n\text{Step}A(R)){n ∈ ℕ}$ is nondecreasing. There exists $i$ such that

Let $V_A \ni v ∈ Reach_A(R)$. Let $i$ be the smallest index such that $v ∈ i\text{Step}_A(R)$. If $i > 0$, Alice can play a move $(v, w)$ which makes the index decrease stricly. Otherwise, we’re done.

If $v ∈ V \backslash Reach_A(R)$, all Alice’s moves remain in $V \backslash Reach_A(R)$, and similarly, Bob has a strategy to remain in this set.
We’ve calculated the (smallest) fixed point of
each vertex has a finite number of outgoing edges.
NB: if there were an infinite number of outgoing edges for Bob, it may be the case that each one of them are in a $i\text{Step}_A(R)$ (ex: the first one is in $1\text{Step}_A(R)$, the second one in $2\text{Step}_A(R)$, etc…) but there exists no finite $j$ such that $j\text{Step}_A(R)$ contains them all (as there’s an infinite number of them). Same reasoning for the complement.
The first such games were considered: in the 50’s.
Topological games: consider an infinite binary tree (going on the left (resp. right): labelled $0$ (resp. $1$)). $V = (0+1)^\ast$
 Height zero (root): Player 1’s move
 Height one (root): Player 2’s move
 Height two (root): Player 1’s move
 and so on…
$W ⊆ \lbrace 0, 1 \rbrace^ω$. The first player wins if $p ∈ W$.
Distance:
 Open sets for this metric:

the sets of the form $X \lbrace 0, 1 \rbrace^ω$, where $X ⊆ \lbrace 0, 1 \rbrace^\ast$
For open sets $W ≝ X \lbrace 0, 1 \rbrace^ω$: there’s a winning strategy: it’s the reachability set of $X$.
And with the axiom of choice, we can show that there are sets that have no winning strategy.
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