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The Computational Complexity of Finding a Nash Equilibrium

The Computational Complexity of Finding a Nash Equilibrium. Edith Elkind, U. of Warwick. Based On…. Reducibility Among Equilibrium Problems (Goldberg, Papadimitriou): Aug 2005 The Complexity of Computing a Nash Equilibrium (Goldberg, Daskalakis, Papadimitriou): Sep 2005

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The Computational Complexity of Finding a Nash Equilibrium

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  1. The Computational Complexityof Finding a Nash Equilibrium Edith Elkind, U. of Warwick

  2. Based On… • Reducibility Among Equilibrium Problems(Goldberg, Papadimitriou): Aug 2005 • The Complexity of Computing a Nash Equilibrium(Goldberg, Daskalakis, Papadimitriou): Sep 2005 • 3-NASH is PPAD-Complete(Chen, Deng): Nov 2005 • Three-Player Games Are Hard (Daskalakis, Papadimitriou):Nov 2005 • Settling the Complexity of 2-Player Nash-Equilibrium(Chen, Deng): Dec 2005

  3. 0 0 1 1 0 0 1 1 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:

  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 Row player: Column player:

  5. 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:

  6. Existence of NE • Theorem (Nash 1951): any game in normal form has an equilibrium in mixed strategies $1 000 000 question: how to find one?

  7. 0 0 1 1 0 0 1 1 Finding mixed NE in 2 x 2 Games 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 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 NE: r=1/4, c=2/3 Row player: Column player:

  8. 2 players, k actions • Representation: two k x k matrices • Checking for pure NE: easy • at most k2 of them • Checking for mixed NE: • all straightforward methods are exptime • Lemke-Howson algorithm is exptime, too (previous talk) • For 2 players all NE are rational • but not for 3 and more players…

  9. n players, 2 actions • Representation: payoffs to each player for every action profile (vector in {0, 1}n): n2n 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 =>n2d+1numbers 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

  10. Algorithms for NE in Graphical Games • 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): • shown to be incorrect in E., Goldberg, Goldberg, ACM EC’06 • Graphs of max degree 2: • poly-time algorithm (EGG’06)

  11. Is Finding NE NP-hard? • Reminder: a problem P is NP-hard if you can reduce 3-SAT to it: • “yes”-instance 3-SAT→ “yes”-instance of P • “no”-instance 3-SAT→ “no”-instance of P • Problem: each instance of NASH is a “yes”-instance! • every game has a NE • need complexity theory for search problems • Side note: pure Nash for n players, NE of total value > K are NP-hard

  12. g f • S is reducible to T if: • f, g easy to compute • g(T(f(x))) is in S(x) Reducibility Among Search Problems • S associates x in X with a solution set S(x) • Total search problem: for any x, S(x) is not empty S: X Y T: X’ Y’ If T is easy, so is S

  13. r-player game G NE of G g g f f deg 3 graphical game G’ NE of G’ Equivalences: GP’05 deg d graphical game G NE of G d2-player game G’ NE of G’

  14. d-Graphical Game GG → d2-Player Game G • Color the graph of GG: d(u,v) ≤ 2 color(u) ≠ color(v) • Each color is a player of G • RED chooses a red vertex in GG and an action for that vertex in GG • payoff=payoff1+payoff2 • payoff1: BLUE tries to guess which vertex RED chose; RED pays a penalty if BLUE guesses correctly • payoff2: if all neighbors of a chosen vertex are also chosen, it gets same payoff as in GG, else 0

  15. r-Player Game G → 3-Graphical Game GG • Si: space of pure strategies of player i • S- i= S1 * … Si-1*Si+1 *.. * Sr • xij: the probability that ith player uses jth strategy • xs: x1s1 * x2s2 … * xrsr (for s in S-i) • uijs: utility of the ith player when he plays j and others play according to s NE:0 ≤ xij≤ 1 Sjxij=1 Ss in S uijsxs > Ss in S uij’sxs implies xij’= 0 -p -p

  16. r-Player Game G → 3-Graphical Game GG • Vertex Vij for any pair (player=i, action=j) • Want: Pr[Vij plays 1] = Pr [i plays j in G]=xij • Idea: graphical games can do math! • Enforce constraints from the previous slide… v1 v2 v3 Need gadgets for +, *, c, =, min, max, … u Set payoffs to u, v3 so that p[v3]=p[v1] * p[v2]

  17. r-player game G NE of G g g f f deg 3 graphical game G’ NE of G’ Equivalences: GP’05 deg d graphical game G NE of G d2-player game G’ NE of G’

  18. g f 4 X r-player game G NE of G 9-player game G’ NE of G’ Finding NE in a 4-player game is as hard as finding NE in a r-player game for any constant r Combining Reductions: GP’05

  19. Completeness Results? • Can we prove that any total search problem is reducible to r-NASH? • Not really: the class T of all total search problems is a semantic class • not known how to find complete problems for these • Want to pick a large subclass S of T s.t. • S includes some natural problems • there are problems that are complete for S • in particular, r-NASH is complete for S

  20. 00000 11001 01011 01011 END OF THE LINE • Input: Boolean circuits S (Successor), P (Predecessor): • n inputs, n outputs • S(0n) ≠ 0n, P(0n) = 0n • Output: x ≠ 0n s.t. • S(P(x)) ≠ x or P(S(x)) ≠ x Intuition: G=(V, E): • V = Sn; • E = {(x,y) | y=S(x), x=P(y)}

  21. PPAD • PPAD: Polynomial Parity Argument, Directed version • PPAD is the class of all search problems that are reducible to END OF THE LINE search problem solution g f circuits S, T “end of the line”

  22. r-NASH is in PPAD • Proof on Nash’s theorem: • existence of NE reduces to Brouwer’s fixpoint theorem • Brouwer’s fixpoint theorem reduces to Sperner’s lemma • Sperner’s lemma is proven by a parity argument (similar to END OF THE LINE) • Reduction of r-NASH to END OF THE LINE can be extracted from these proofs (Papadimitriou 94)

  23. Brouwer’s Fixpoint Theorem • Brouwer’s Theorem: Any continuous mapping from the simplex to itself has a fixpoint. • Nash  Brouwer proof sketch: • set of all strategy profiles → simplex • mapping: (s1, …, sn) → (s1+d1, …, sn+dn), where di is a shift in the direction of best response to (s1, …, si-1, si+1, …, sn) • NE is a point where noone wants to deviate, i.e., a fixpoint

  24. B C A Sperner’s Lemma • Proper coloring: • vertices on BC are not blue • vertices on AC are not green • vertices on AB are not yellow • Sperner’s Lemma: there exists a trichromatic triangle • Brouwer’s theorem  Sperner’s Lemma: • x is blue if the grad(F) at x points away from A, etc. • trichromatic triangle “has no direction” • repeat at increased resolution…

  25. Opposite Direction: 3D-BROUWER • Input: • 3D unit cube divided into 23n cubelets • cijkis the center of Kijk • f(cijk)=cijk+dijk, dijk is in {d0, d1, d2, d3}, where • d1=(a, 0, 0), d2=(0, a, 0) • d3=(0, 0, a), d0=(-a, -a, -a) • circuit C: {0, 1} 3n→ {0, 1, 2, 3} selects dijk • Output: • a panchromatic cubelet, i.e., one that has all of d0, d1, d2, d3 among its 8 neighbors

  26. 3D-BROUWER is PPAD-complete • Papadimitriou (1994) shows that a more complicated version of 3D-BROUWER is PPAD-complete • This version was proven hard in DGP’05 • Reduction from END OF THE LINE • embed the line L into 3d cube • “protect” L from color 0 using three other colors • color the rest of inner cubelets with 0

  27. r-NASH vs 3D BROUWER • Existence of NE follows from Brouwer’s fixpoint theorem • NE are special cases of Brouwer’s fixpoints • just how special? • Can any fixpoint be represented as a NE of a game? • DGP’05: YES!  4-NASH is PPAD complete • Proof: • 4-NASH  deg 3 Graphical Nash • graphical games can compute fixpoints

  28. 4-NASH to 3-NASH • Daskalakis, Papadimitriou: modify arithmetic gadgets so that the graph is 3-colorable • Chen, Deng: same gadgets, but allow for small error

  29. 2-NASH • Chen, Deng: • avoid graphical games • reduce directly from 3D-BROUWER to 2-NASH using arithmetic gadgets similar to graphical game gadgets • Game over?

  30. Graphical Games: Open Problems • Degree: • deg 3 PPAD-complete (DGP’05b) • deg 2 polynomial time solvable (EGG’06) • Pathwidth: • paths: poly-time • pathwidth 1: maybe algorithm from EGG’06 still works • pathwidth 2: any KLS-style algo is exptime (EGG’06) • pathwidth > r, r constant: PPAD-complete (EGG’06) • Finding NE on trees?

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