Nash Equilibrium: P or NP? S Kameshwaran Oct 25, 2002

Complexity prelims P NP Set of problems that have algorithms which have polynomial running time (worst case) in the input length NP Given a solution, checking its correctness is possible in polynomial time But finding a solution is not assured in polynomial time

Complexity prelims Polynomial reductions NP-Complete: Two problems X and Y X p Y: If there is a polynomial algorithm that takes an instance X` of X and creates an instance Y` of Y, such that they are equivalent in terms of the solutions NP-Complete: Most difficult problems in NP If any of these problems can be solved in polynomial time then all problems in NP can be solved in polynomial time

Complexity prelims Cook’s Theorem: If Y p X, where Y  NP-Complete SAT is NP-Complete SAT  NP For all X  NP, X p SAT If Y p X, where Y  NP-Complete X is NP-Hard X  NP, then X  NP-Complete

Nash Equilibrium A strategy combination is in NE if each player chooses a strategy that is the best response to the strategies chosen by others Issues: Existence of NE for a game Uniqueness of NE Pareto-optimality

Nash Equilibrium: Dining in Bangalore (TGIF, TGIF) is good for both the players (Pizza Hut, Pizza Hut) is not a admissible NE Man Woman Pizza Hut TGIF 10, 10 0, 0 20, 20

Nash Equilibrium: Battle of the Sexes Both NE are Pareto-optimal and both are admissible Man Woman Prize Fight Ballet 2, 1 0, 0 1, 2

Nash Equilibrium N players Action set of player i: Si={si1, …, sim(i)} Utility to player i: ui: S  R Strategy of player i: pi={pi1, …, pim(i)} pij >= 0, jpij = 1 Utility of using strategy p: ui(p) = s ui(s) j pj(s)

ui(pi, p*-i) <= ui(p*i, p*-i) Nash Equilibrium N players Action set of player i: Si={si1, …, sim(i)} S = i Si Utility to player i: ui: S  R Strategy of player i: pi={pi1, …, pim(i)} pij >= 0, jpij = 1 Utility of using strategy p: ui(p) = s ui(s) j pj(s) Nash Equilibrium: p* is NE if for all i, and for all pi ui(pi, p*-i) <= ui(p*i, p*-i)

Some more notations xij(p) = ui(sij,p-i) zij(p) = xij(p) – ui(p) Other players use mixed strategy profile p-i and i use pure strategy sij zij(p) = xij(p) – ui(p) p* is NE iff z(p*) <= 0 gij(p) = max[zij(p), 0] p* is NE iff gij(p*) = 0

NE as a fixed point of a function Define y: P  P p* is NE iff it is a fixed point of y y is a continuous function of a compact set P to itself, so a fixed point exists (Brouwer’s FPT) Nash’s proof of existence of NE for N-person games

NE as a fixed point of a correspondence Best response correspondence: BR: P  P Point to set function BR(p) = arg maxq[i ui(qi, p-i)] BR(p) = Set of best response strategies to p p* is a NE iff it is a fixed point of best response correspondence, p*BR(p*) BR is non-empty, closed and convex valued, hence by Kakutani’s FPT, fixed point exists

NE as a solution to complementarity problem Find a pair of vectors x, y which are complementary (orthogonal) to each other Used to prove complementary slackness and optimality in mathematical programming p* is NE iff p* and z(p*) are orthogonal Two-person games: Linear complementary problems

NE as a minimum of a function on a polytope Define v: P  R v(p) = i j [gij(p)]2 v is a continuous, differentiable function on polytope P p* is NE iff it is global minimum of v, v(p*) = 0 Other formulations: Stationary point, semi-algebraic set

Two person zero sum games 2 Players For every strategy x of player 1 and strategy y of player 2, payoff to 1+ payoff to 2=0 What player one loses is gained by the other Player 1 wants to minimize his losses Determine the optimal mixed strategy that minimizes his loss Player 2 wants to maximize his profits Determine the optimal mixed strategy that maximizes his profit

Two person zero sum games Minimax theorem: Expected minimum of the maximum loss = Expected maximum of the minimum profit The above can be converted into two linear programs which are dual to each other LP  P Simplex algorithm can solve it efficiently though it is not bounded in polynomial time

Two person general sum games Two person non-zero sum games can be formulated as linear complementary problems (LCP) LCP: Given a matrix M and a vector q Find w and z, such that w – Mz = q w >= 0, z >= 0, and wizi = 0 for all i Solution technique: Lemke-Howson Algorithm

Two person general sum games LCP Can be characterized as LPs (Mangasarian, 1978) The solution to LCP can be obtained by optimizing a suitable LP Finding the suitable objective function is not easy (depends on matrix M) Possible Approach: Characterize M in terms of two person games (M consists of payoff matrices of both players) Infer the complexity of games using M

Two person general sum games Lemke-Howson Algorithm Finds a sample equilibrium Incomplete for finding all equilibria Computational complexity: still unknown Exponential lower bound is shown (Murty, 1978) Can be exponential even on a zero-sum game Approximation theory cannot be used as there is no objective function is used and one cannot know how close is the solution to the optimal

Two person general sum games Reduction of a SAT to symmetric two-person game (Conitzer, Sandholm, 2002) Following are NP-hard (even for symmetric case) Deciding whether more than one NE exists Deciding whether a given strategy is played in NE Deciding whether a given strategy is never played in NE Deciding if a Pareto-optimal NE exists

N person games NE as a min of a function v(p) = i j [gij(p)]2 Constraints: pij >= 0, jpij = 1 Global minima correspond to NE Local minima may exist Algorithms: Continuous differentiable function, so any global optimization technique can be used Penalty function methods Convergence of these algorithms are very slow

N person games Alternate formulation: Non-linear complementarity problem Sequence of approximations by LCP Similar to Newton’s method Scarf’s algorithms Worst case complexity is exponential in N

Bayesian-Nash Equilibrium Games of incomplete information: A player may not the utility of other players Most real world applications induce such games: Auctions, markets, bargaining, etc Modifications: Ti: Set of types a player can belong to Eg: In bargaining, the seller assumes that buyer belongs to type [10,25], which means that buyer may value the good anywhere between 10 and 25 Let q be the joint probability distribution on T = i Ti q and T are known to all players Each player knows his own type ti  Ti

Bayesian Nash Equilibrium N players Action set of player i: Ai={ai1, …, aim(i)} A = i Ai Utility to player i: ui: A  T  R (utility depends on the type of the player) Strategy of player i, i : Ti  Ai If either Ti or Ai is infinite then the strategy space is infinite Strategy is now a function Utility of using strategy  : ui( , ti) = Expected utility over all the possible types for other players given the type of i is ti

Bayesian Nash Equilibrium Complexity for 2 player case Reduction from Set cover (Conitzer, Sandholm, 2002) Deciding whether a BNE exists is NP-complete If randomization is allowed, then BNE always exists for finite games Infinite case: Strategy space is infinite Finding the utility of a player amounts to optimizing over a function space Function spaces are generally infinite dimensional Assuring the existence of a solution is not possible

Bayesian Nash Equilibrium Games arising out of market design Infinite Games Given the action combination from players, utility to each player is calculated by some optimization problem some resource allocation algorithm which moves the goods from sellers to buyers to maximize some criteria Finding the expected utility: What is the complexity of finding the utility of a player who uses a strategy  ? What is the complexity of deciding the existence of BNE in infinite games?

References Complementarity and Fixed Point Problems, Ed: M. L. Balinski and R. W. Cottle, North-Holland Publishing Company, 1978 Computational complexity of complementary pivot methods, K. G. Murty Characterization of LCPs as LPs, O. L. Mangasarian Complexity results about NE, V. Conitzer and T. Sandholm, 2002 Computation of equilibria in finite games, R. D. McKelvey and A. McLennan, 1996

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