Teachers: Wieslaw Zielonka

Games and Graphs

• $G\stackrel{\scriptscriptstyle\mathrm{def}}{=}(V,E)$
• $V\stackrel{\scriptscriptstyle\mathrm{def}}{=}{V}_A\bigsqcup V_B$
• $A$: Alice, $B$: Bill

$Paths$:

the set of finite paths in $G$

$Path{s}^{\infty }$:

the set of infinite paths

$W_A\subseteq Path{s}^{\infty }$:

the set of infinite paths winning for Alice

$W_B=Path{s}^{\infty }\mathrm{\setminus }{W}_A\subseteq Path{s}^{\infty }$:

the set of infinite paths winning for Bill

Strategy $\sigma$ for Alice:

is a mapping $\sigma :V^{\ast }{V}_A⟶V$ such that $\forall x\stackrel{\scriptscriptstyle\mathrm{def}}{=}{v}_1\cdots v_n\in V^{\ast }{V}_A(v_n,\sigma (x))\in E$

$(\sigma ,\tau )$-strategy profile:

if $\sigma$ (resp. $\tau$) is strategy for Alice (resp. Bill).

A path $p\stackrel{\scriptscriptstyle\mathrm{def}}{=}{v}_1\cdots v_n$ is consistent with the strategy $\sigma$ of Alice:

if for each $i$ such that $v_i\in V_A$, $v_{i+1}=\sigma (v_1\cdots v_i)$

If we fix an initial vertex $v\in V$, then there exists a unique infinite path $p(v,\sigma ,\tau )$ which starts at $v$ and is consistent with $\sigma$ and $\tau$.

$\sigma$ is a winning strategy for Alice for the initial vertex $v$:

if each infinite path $p$ strating at $v$ and consistent with $\sigma$ belongs to $W_A$.

For a given initial vertex $v$, $\sigma ,\tau$, or none of them, may be a winning strategy.

The game is determined:

if for each initial vertex, one of the players has a winning strategy.

Memoryless/positional strategies

The strategy $\sigma$ is memoryless (or positional):

iff for all $x\stackrel{\scriptscriptstyle\mathrm{def}}{=}yv\in V^{\ast }{V}_A$: $\sigma (x)=\sigma (v)$

Ex: If $\sigma$ and $\tau$ are memoryless and the graph is finite: all the plays are “lasso” loops in the graph.

Normally, when considering games on graphs, one looks for a partition:

Moreover, usually: people look for memoryless strategy that are independent on the initial vertex.

$OneSte{p}_A(X)$:

set of vertices such that, strating from one of these vertices, Alice can reach $X$ in exactly one move.

$$\begin{array}{c}OneSte{p}_A(X)\stackrel{\scriptscriptstyle\mathrm{def}}{=}\{v\in V_A\mid \exists w\in X;(v,w)\in E\}\\ \cup \{v\in V_B\mid \forall w\in V,(v,w)\in E⟹w\in X\}\end{array}$$

and:

Then

and if $G$ is finite, there exists $k$ such that $X_{k+1}=X_k$.

For all $v\in V$,

Let $v\in Reac{h}_A(X)$ such that $rank(v)>0$.

If

• $v\in V_A$, then there exists $w$ s.t. $(v,w)\in E$ and $rank(w)<rank(v)$
• $v\in V_B$, then for all $w$ s.t. $(v,w)\in E$, $rank(w)<rank(v)$

Büchi condition

Büchi: Alice wins iff the play visits infinitely often a fixed set $R$ of vertices.

If $X\subseteq Y$:

• $OneSte{p}_A(X)\subseteq OneSte{p}_A(Y)$
• $Reac{h}_A(X)\subseteq Reac{h}_A(Y)$

So the following mapping is clearly monotone: $Y⟼OneSte{p}_A(Reac{h}_A(Y))$

And on top of that:

NB: $Reac{h}_A(Y)\mathrm{\setminus }Y$ because there may exist some $v\in Y$ s.t. $v\notin OneSte{p}_A(Reac{h}_A(Y))$, that is: for which Bill has a strategy to leave $Reac{h}_A(Y)$ in one step.

Let $W$ s.t. $\varphi (W)=W$. Then Alice has a strategy to visit $R$ infinitely often.

(Indeed:

so $Reac{h}_A(W\cap R)\mathrm{\setminus }(W\cap R)=\varnothing$, and $Reac{h}_A(W\cap R)\subseteq W\cap R$)

Let

Then by induction: this sequence is decreasing: $Y_{i+1}\subseteq Y_i\forall i$.

So $\bigcap _i{Y}_i$ is a fixed point of $\varphi$. And it is the greatest fixed point. Indeed, if $W$ is another such fixed point. Then $W\subseteq V$, and by applying $\varphi$ over and over, we get the result.

Path consistent with Bill’s strategy at $Y_i\mathrm{\setminus }{Y}_{i+1}$

• either $\forall i\ge 2$, $v_i\notin R$
• or there exists $j\ge 2$ s.t. $v_j\in Y_i^c$ and all vertices $v_2,\dots ,v_{j-1}$ do not belong to $R$.

Forgetting the first moment, the index corresponds to the maximal number of times that Alice can force the visit to $R$.

Example

Imagine a pie that two players share to eat.

1. Player 1 proposes a proportion $x_1$ of the pie to player 2
2. Player can either accept, in which the proportions gotten are $(1-x_1,x_1)$, or rejects.
3. etc… (as described below)

With proba 1, the game stops.

Nash equilibria: for each $0\le x\le 1$, the proportion $(x,1-x)$ is a Nash equilibrium. Indeed: P1 never accepts less than $x$, and never proposes more than $1-x$. Symmetrically for P2.

Can we find a subgame-perfect equilibrium? (where each time P1 (resp. P2) plays, he/she makes the same proportion).

Backward induction: begin at node $1$ and go backward (along the $\delta$ arrow):

1. Expectation for each player: $(z,1-z)$
2. At $0^{\prime }$, expectation: $(1-\delta )\times (0,0)+\delta (z,1-z)=(\delta z,\delta (1-z))$
3. At $2^{\prime }$: P2 proposes $(z\delta ,1-z\delta )$, so that P1 accepts and P2’s payoff is $1-z\delta >(1-z)\delta$
4. etc…

So now:

Therefore: