Past exam of the MSc in advanced computer science course of the University of Oxford / 2023 / Computational game theory / 2

Read the lower-left entry of each cell as player 1's payoff and the upper-right entry as player 2's payoff, as indicated by the source diagram. Against $L$, player 1 strictly prefers $B$ because $2\pi >1$; against $R$, player 1 strictly prefers $T$ because $3>2$. Against $T$, player 2 strictly prefers $R$ because $6>0$; against $B$, player 2 strictly prefers $L$ because $2\sqrt{2}>2$. Thus the pure strategy Nash equilibria are

For a fully mixed strategy Nash equilibrium, write $p=\mathrm{Pr}(T)$ and $q=\mathrm{Pr}(L)$. Player 1's expected payoffs from the two pure strategies areEquality, required when both pure strategies have positive probability, givesPlayer 2's corresponding expected payoffs areso equality givesBoth probabilities lie strictly between zero and one. Hence the unique fully mixed strategy Nash equilibrium iswith complementary probabilities on $B$ and $R$. Its expected payoffs are

These are all three Nash equilibria. There are no equilibria with just one player genuinely mixing: against each opposing pure strategy, the other player's best response is unique, so that player cannot mix. When both players mix, the two linear indifference equations have the unique solution just obtained. This is support enumeration for a bimatrix game with all possible supports checked.

The definition in this part uses a single Boolean valuation, so the relevant Nash equilibria are pure strategy equilibria of a Boolean game. Use its standard convention that the players' control sets form a set partition of the variables.

Membership follows from an existential choice of a strategy profile and a universal check of all unilateral deviations. Assign payoff one to satisfying a player's goal and zero otherwise. A strategy profile $\sigma$ is a Nash equilibrium precisely when, for every player $i$ and every alternative Boolean valuation ${\tau }_i$ of that player's variables,is true: either the goal is already satisfied or no deviation can satisfy it. Thus the requested property isThe strings encoding $\sigma$, $i$ and ${\tau }_i$ have polynomial length, and evaluating the bracketed propositional formulas takes polynomial time. Invalid encodings of a player index can simply be ignored by the test. This is the existential second level of the polynomial hierarchy, ${\Sigma }_2^{\mathrm{P}}$.

For hardness, reduce the canonical complete problem of deciding a two-block quantified Boolean formulaMake a two-player Boolean game: player 1 controls $X$ and has goal ${\gamma }_1=\top$; player 2 controls $Y$ and has goal ${\gamma }_2=\neg \psi$. Set the query formula to $\varphi =\psi$. The construction has size linear in the input formula.

If an assignment $x$ satisfies $\forall Y\psi (x,Y)$, combine it with any assignment $y$. Player 1 already satisfies its goal, and player 2 cannot satisfy $\neg \psi$ by any unilateral deviation. Thus $(x,y)$ is a Nash equilibrium satisfying the query.

Conversely, let $(x,y)$ be a Nash equilibrium satisfying $\psi$. Player 2 currently has payoff zero. If some alternative $y^{\prime }$ made $\psi (x,y^{\prime })$ false, that deviation would satisfy player 2's goal and increase its payoff to one. The Nash equilibrium condition therefore implies $\forall Y\psi (x,Y)$. This proves the required equivalence and the polynomial-time reduction.

Together, membership and hardness prove ${\Sigma }_2^{\mathrm{P}}$-completeness of existential equilibrium checking in a Boolean game. The reduction works with only two players and one constant goal.