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Nash Equilibria In Graphical Games On Trees. Edith Elkind Leslie Ann Goldberg Paul Goldberg. Games and Strategies. Games: strategic interactions between rational entities Solution concepts: what’s going to happen? dominant strategies Nash equilibrium …. Can it be computed?
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Nash Equilibria In Graphical Games On Trees Edith Elkind Leslie Ann Goldberg Paul Goldberg
Games and Strategies • Games: strategic interactions between rational entities • Solution concepts: what’s going to happen? • dominant strategies • Nash equilibrium • …. • Can it be computed? • if your computer cannot find it, the market probably cannot either
0 0 1 1 0 0 1 1 Matrix (normal form) Games • finite set of players {1, …, n} • each player has kactions (pure strategies): 1, …, k • payoffs of the ith player: Pi: {1, …, k}n→ R Row player: Column player:
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 Row player: Column player:
H H T T H H T T Pure vs. Mixed Strategies • NE in pure strategies may not exist! • “matching pennies” • Mixed strategy: a probability distribution over actions • 50% tail, 50% head Row player: Column player:
Existence of NE • Theorem (Nash 1951): any n-player k-action game in normal form has an equilibrium in mixed strategies can we find one in poly-time?
2 players, n actions • Representation: two n x n matrices • Computation: • all known methods are exptime • can it be NP-hard? no: NE always exists • PPAD-hardness: notion of hardness for total search problems • DGP’06: finding NE in 4-player games is PPAD-hard • CD’06: finding NE in 2-player games is PPAD-hard • DGP reduction uses graphical games (the topic of this talk!)
n-player 2-action games • representation: payoffs to each player for every action profile (vector in {0, 1}n): n2nnumbers • graphical games: • players are vertices of a graph • V’s payoff depends on actions of Win N(V)UV • n players, max degree d => n2d+1numbers 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 W’s payoffs (16 cases): T V U
Complexity: what is known • Bounded-degree trees: • Exp-time algorithm/poly-time approximation algorithm to find all NE (Kearns, Littman, Singh, UAI 2001) • ??? poly-time algorithm to find a single NE (Kearns, Littman, Singh, NIPS’2001) • Heuristics for graphs with cycles • General graphs: • PPAD-complete (DGP’06) even if max deg=3
Our Results (1) • 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 • (a representation for) all NE in n3 time, or • a single NE in n2 time
Our Results (2) • There is a graph of pathwidth 2 on which our algorithm runs in exp time • true for all algorithms that use the basic approach of the UAI’01 paper • The problem is PPAD-complete for bounded pathwidth graphs • Open question: what if pathwidth = 1? • generalizes a cool geometry problem (talk to me if you like those, or see the paper)
0 0 1 1 0 0 1 1 2/3 BR(R) 1/4 Warm-up: 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 1 r 1 mixed NE: r=1/4, c=2/3
U2 Algorithm for Trees (KLS’01) • Potential best response:v is a PBR to w iff when W plays w, there is a NE for TV in which V plays v. • upstream pass: construct PBRV(w) from PBRU1(v), PBRU2(v) and PBRU3(v) • downstream pass: root selects its strategy based on the children’s PBR’s; propagates to leaves W v V TV U3 U1 w
KLS algorithm: running time • For bounded-degree trees, constructs all PBR (and then find a NE) in exp time • FPTAS for an e-NE: • superimpose PBR with a d-grid • there exists a grid point d-close to PBR • e-NE ( e = poly(d) ): no one can gain more than e by deviating
U V W Computing PBR: Example • Payoffs to V: • P000= 1, P001= -9, P100 = 9, P101= -1, Pu1w= 0 for u, w =0, 1 • E0 = EP(V) from playing 0: (1-u)(1-w)*1+(1-u)w*(-9)+u(1-w)*9+uw*(-1) = 1+8u-10w • E1 = EP(V) from playing 1: 0 • E0 = E1 iff w = (8u+1)/10 = f(u) v u 1 1 (v, u) → (f(u), v) .5 PBRU(v) PBRV(w) .5 1 w .1 .9 1 v
Trees: too many segments W v u t t2 u2 v1 V u1 v2 t1 v v w v1 v2 v1 v2 T U (v,t), (v,u) → (f(u,t), v) Incorrect! KLS (NIPS’01): can “trim” PBR
Solutions? • Solution 1 (for paths): algorithm of UAI’01 paper, careful analysis • the number of segments/rectangles in each PBR is O(n2) • running time O(n3) • Solution 2 (for paths): can pick a subset of each PBR consisting of O(n) segments • O(n2) running time
O(n3) algorithm • f(u) =(au+b)/(cu+d) u*: cu*+d = 0 [v1, v2] x {u} => {f(u)} x [v1, v2] {v} x [u1, u2] => [f(u1), f(u2)] x {v} if u* not in [u1, u2] [0, f(u2)] U [f(u1), 1] x {v} if u1≤ u*≤ u2 • PBRV(w) vs PBRU(v): new segments at v=1 and v=0, some segments break into two --- double in size? • no: count the event points v u (v, u) → (f(u), v) u* PBRV(w) PBRU(v) w v
Extension to trees? V0 V1 V2 Vn-1 Vn U1 U2 Un-1 Un T1 T2 Tn-1 Tn
Hardness results • pathwidth 2: our algorithm is not poly-time • and neither is any two-pass algorithm that stores subsets of PBR • pathwidth > k: (probably) all algorithms are not poly-time • finding NE in this case is PPAD-hard • idea: modify the construction in DGP’06
0 0 1 1 0 0 1 1 Good Nash Equilibria Row player: Column player: • Nash equilibria: • (0, 0): total payoff is 3 • (1, 1): total payoff is 4 • (1/4, 2/3): total payoff is 17/12 • not all NE are created equal…
What is a good NE? • maximize sum of player’s payoffs • guarantee to each player a payoff of at least ti • (almost) equal payoffs • any combination of those…. Can we use PBR data structure to compute those?
Can we represent it? • any GG with integer payoffs on a tree has a rational NE • Any PBR consists of segments and rectangles with rational coordinates • Yet, total payoff-maximizing NE may be irrational Our result (EGG’07): for any algebraic a, deg(a) = n, there is a GG with int payoffs on a path of length O(n) in which in the best NE player 1 plays a
Approximation • Can we use the FPTAS of KLS’01? • superimpose PBR with a d-grid • Observation: there is a grid point d-close to best NE • look for best point on the grid close to PBR • dynamic programming • e-NE ( e = poly(d) ): no one can gain more than e by deviating
True Nash • e-NE is not always appropriate • what if players are not willing to lose e? • Can we find a (true) NE that is e-close to the best (true) NE? • Idea: • add borders of rectangles in PBR to the grid • only consider grid points in PBR
Bounded Payoff Nash • Similar algorithm works --- FPTAS • Also for other kinds of “good” NE • If all payment bounds are rational, there is a BP NE that is “almost” rational (deg ≤ 2) • Open question: can we compactly represent all bounded payoff NE? • perhaps by incorporating payoff bounds into PBR?
Conclusions Nash equilibria in graphical games on trees complexity still unknown…