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

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.