1. Nash Equilibria In Graphical Games On Trees Revisited Edith Elkind Leslie Ann Goldberg Paul Goldberg (University of Warwick) (To appear in ACM EC’06)

2. Normal Form Games (with 2 actions per player) 20 01 1 0 0 3 Row player: Column player: 0 1 0101 0 1 finite set of players {1, …, n} each player has 2 actions (pure strategies): 0 and 1 payoffs of the ith player: P i : {0, 1} n → R

3. Mixed Strategies pure strategy = action mixed strategy = probability distribution over actions p i = Prob [i plays 1] expected payoff of the i th player for a strategy profile p = (p 1, …, p n ): EP(i) = E[P i (a) | Prob[a i =1] = p i ]

4. Nash Equilibrium 20 01 1 0 0 3 Row player: Column player: 0 1 0101 0 1 Nash equilibrium: a strategy profile such that noone wants to deviate given other players’ strategies, i.e., each player’s strategy is a best response to others’ strategies: –(0, 0) and (1, 1) are both NE. –any other NE?

5. Finding NE in 2-player 2-action Games 20 01 1 0 0 3 Row player: Column player: 0 1 0101 0 1 Suppose R plays 1 w.p. r EP(C) from playing 0: (1-r)*1 EP(C) from playing 1: r*3 1-r > 3r iff r < ¼ Suppose C plays 1 w.p. c EP(R) from playing 0: (1-c)*2 EP(R) from playing 1: c*1 (1-c)*2 > c iff c < 2/3 1/4 1 r BR(C) c 1 2/3 BR(R) NE: r=1/4, c=2/3

6. NE for n-player 2-action games (poly-time) algorithm for NE in n-player games? representation: payoffs to each player for every action profile (vector in {0, 1} n ): n2 n numbers graphical games: –players are associated with the vertices of a graph; –each player’s payoff depends on his own action and actions of his neighbors –n players, max degree d => n2 d+1 numbers T U V W t=0, u=0, v=0, w=0: 12 t=1, u=0, v=0, w=0: 31 …. t=1, u=1, v=1, w=1: -6 W’s payoffs (16 cases):

7. Related Work Bounded-degree trees: –Exp-time algorithm/poly-time approximation algorithm to find all NE (Kearns, Littmann, Singh, UAI 2001) –??? poly-time algorithm to find a single NE (Kearns, Littmann, Singh, NIPS’2001) General graphs: –can it be NP-hard? no: NE always exists –hardness notion for total functions: PPAD-hardness –NE in graphical games with d ≥ 3 is PPAD-complete (GP, DGP, STOC’06)

8. Our Results Algorithm in NIPS’01 paper is incorrect (does not always output a NE) We fix the NIPS’01 algorithm, but… –our algorithm runs in poly-time on paths with a trick, also on cycles can be used to find all NE (rather than a single one) –there is a graph of pathwidth 2 on which our algorithm (and all algorithms that use the basic approach of the UAI’01 paper) runs in exp time The problem remains PPAD-complete for bounded pathwidth graphs Open question: what if pathwidth = 1?

9. Algorithm for Trees Recall from 2-player case: best response function Potential best response: v is a PBR to w iff when W plays w, there is a NE for T’ in which V plays v. Bottom-up approach: information propagates from the leaves to the root T’ W V U1U1 U2U2 U3U3 v=PBR V (w) c = BR(r) r c

10. Computing PBR: Example Payoffs to U: 0 if U=V, 1 if U≠V –EP(U) from playing 0: v; EP(U) from playing 1: 1-v Payoffs to V: –P 000 =1, P 001 =-9, P 100 =9, P 101 =-1, P U1W =0 for all U, W –EP(V) from playing 0: (1-u)(1-w)*1+(1-u)w*(-9)+u(1-w)*9+uw*(-1) –V is indifferent btw 0 and 1 iff w = (8u+1)/10 = f(u) UVW.51 1 1 1.1.9 u v v w (v, u) → (f(u), v)

11. Computing PBR: General Case EP(V) from playing 0: a 1 uw+a 2 u+a 3 w+a 4 EP(V) from playing 1: b 1 uw+b 2 u+b 3 w+b 4 V is indifferent between 0 and 1 iff w = f(u) = Au+B/Cu+D PBR V (W)=L 0 U f(PBR U (V)) U L 1 For paths, we can show that for any V, PBR V (W) consists of polynomially many segments (rectangles if degenerate) UVW

12. Computing PBR on Trees Trees: similar algorithm Indifference function: w = L 1 (u 1, …, u n )/L 2 (u 1, …, u n ) Potential best response can be exponential in size!