Nash Equilibria In Graphical Games On Trees Revisited

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

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

3. Mixed Strategies • pure strategy = action • mixed strategy = probability distribution over actions • pi= Prob [i plays 1] • expected payoff of the ith player for a strategy profile p= (p1, …, pn): EP(i) = E[Pi(a) | Prob[ai=1] = pi]

4. 0 0 1 1 0 0 1 1 Nash Equilibrium • 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? Row player: Column player:

5. 0 0 1 1 0 0 1 1 1 2/3 BR(R) 1/4 Finding NE in 2-player 2-action Games Row player: Column player: BR(C) 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 c 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 r 1 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): n2nnumbers • 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 =>n2d+1 numbers 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): T V U

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 c c = BR(r) r W v=PBRV(w) V T’ U3 U1 U2

10. U V W 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: • P000=1, P001=-9, P100=9, P101=-1, PU1W=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) v u 1 1 (v, u) → (f(u), v) .5 .5 1 w .1 .9 1 v

11. U V W Computing PBR: General Case • EP(V) from playing 0: a1uw+a2u+a3w+a4 • EP(V) from playing 1: b1uw+b2u+b3w+b4 • V is indifferent between 0 and 1 iff w = f(u) = Au+B/Cu+D • PBRV(W)=L0 U f(PBRU(V)) U L1 • For paths, we can show that for any V, PBRV(W) consists of polynomially many segments (rectangles if degenerate)

12. Computing PBR on Trees • Trees: similar algorithm • Indifference function: w = L1(u1, …, un)/L2(u1, …, un) • Potential best response can be exponential in size!