Download presentation

Presentation is loading. Please wait.

Published byPeter Fereday Modified over 2 years ago

1
Nash Equilibria In Graphical Games On Trees Edith Elkind Leslie Ann Goldberg Paul Goldberg

2
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

3
Matrix (normal form) Games 20 01 1 0 0 3 Row player: Column player: 0 1 0101 0 1 finite set of players {1, …, n} each player has k actions (pure strategies): 1, …, k payoffs of the i th player: P i : {1, …, k} n → R

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

5
Pure vs. Mixed Strategies 1 1 1 1 Row player: Column player: H T HTHT H T NE in pure strategies may not exist! –“matching pennies” Mixed strategy: a probability distribution over actions –50% tail, 50% head

6
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?

7
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!)

8
n-player 2-action games representation: payoffs to each player for every action profile (vector in {0, 1} n ): n2 n numbers graphical games: –players are vertices of a graph –V’s payoff depends on actions of W in N (V) U V –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):

9
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

10
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 n 3 time, or a single NE in n 2 time

11
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)

12
Warm-up: 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) mixed NE: r=1/4, c=2/3

13
Potential best response: v is a PBR to w iff when W plays w, there is a NE for T V in which V plays v. upstream pass: construct PBR V (w) from PBR U1 (v), PBR U2 (v) and PBR U3 (v) downstream pass: root selects its strategy based on the children’s PBR’s; propagates to leaves Algorithm for Trees (KLS’01) TVTV W V U1U1 U2U2 U3U3 v w

14
KLS algorithm: running time For bounded-degree trees, constructs all PBR (and then find a NE) in exp time FPTAS for an -NE: –superimpose PBR with a -grid –there exists a grid point -close to PBR – -NE ( = poly( ) ): no one can gain more than by deviating

15
Computing PBR: Example Payoffs to V: –P 000 = 1, P 001 = -9, P 100 = 9, P 101 = -1, P u1w = 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) UVW.51 1 u v v 1 1.1.9 w (v, u) → (f(u), v) PBR U (v) PBR V (w)

16
Trees: too many segments vvw u t v v1v1 v2v2 v1v1 v2v2 v1v1 v2v2 KLS (NIPS’01): can “trim” PBR Incorrect! W V TU (v,t), (v,u) → (f(u,t), v) u2u2 u1u1 t2t2 t1t1

17
Solutions? Solution 1 (for paths): algorithm of UAI’01 paper, careful analysis –the number of segments/rectangles in each PBR is O(n 2 ) –running time O(n 3 ) Solution 2 (for paths): can pick a subset of each PBR consisting of O(n) segments –O(n 2 ) running time

18
O(n 3 ) algorithm f(u) =(au+b)/(cu+d) u*: cu*+d = 0 [v 1, v 2 ] x {u} => {f(u)} x [v 1, v 2 ] {v} x [u 1, u 2 ] => [f(u 1 ), f(u 2 )] x {v} if u* not in [u 1, u 2 ] [0, f(u 2 )] U [f(u 1 ), 1] x {v} if u 1 ≤ u*≤ u 2 PBR V (w) vs PBR U (v): new segments at v=1 and v=0, some segments break into two --- double in size? no: count the event points u v v w (v, u) → (f(u), v) PBR U (v) PBR V (w) u*

19
Extension to trees? V0V0 V1V1 V2V2 V n-1 VnVn U1U1 T1T1 U n-1 T2T2 U2U2 T n-1 TnTn UnUn

20
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

21
Good Nash Equilibria 20 01 1 0 0 3 Row player: Column player: 0 1 0101 0 1 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…

22
What is a good NE? maximize sum of player’s payoffs guarantee to each player a payoff of at least t i (almost) equal payoffs any combination of those…. Can we use PBR data structure to compute those?

23
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 , deg( ) = n, there is a GG with int payoffs on a path of length O(n) in which in the best NE player 1 plays

24
Approximation Can we use the FPTAS of KLS’01? –superimpose PBR with a -grid Observation: there is a grid point -close to best NE –look for best point on the grid close to PBR dynamic programming – -NE ( = poly( ) ): no one can gain more than by deviating

25
True Nash -NE is not always appropriate –what if players are not willing to lose ? Can we find a (true) NE that is -close to the best (true) NE? Idea: –add borders of rectangles in PBR to the grid –only consider grid points in PBR

26
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?

27
Conclusions Nash equilibria in graphical games on trees complexity still unknown…

Similar presentations

OK

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

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

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

E paper display ppt on ipad Cathode ray tube display ppt on tv Ppt on role of it in indian railways Download ppt on earthquake in india Ppt on electricity for class 10th exam Ppt on formation of a company in india Ppt on sexually transmitted diseases Skill based pay ppt online Ppt on diode family video Free ppt on environmental issues