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 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

Normal Form Games Row player: Column player: finite set of players {1, …, n} each player has k actions (pure strategies): 1, …, k payoffs of the ith player: P i : {1, …, k} n → R

Nash Equilibrium Row player: Column player: 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

Pure vs. Mixed Strategies 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

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

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

2 players, k actions Representation: two k x k matrices Checking for pure NE: easy –at most k 2 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…

n players, 2 actions 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):

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)

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

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 S is reducible to T if: –f, g easy to compute –g(T(f(x))) is in S(x) f g

Equivalences: GP’05 r-player game G NE of G deg 3 graphical game G’ NE of G’ f g d 2 -player game G’ NE of G’ deg d graphical game G NE of G f 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=payoff 1 +payoff 2 –payoff 1 : BLUE tries to guess which vertex RED chose; RED pays a penalty if BLUE guesses correctly –payoff 2 : if all neighbors of a chosen vertex are also chosen, it gets same payoff as in GG, else 0 d-Graphical Game GG → d 2 -Player Game G

r-Player Game G → 3-Graphical Game GG S i : space of pure strategies of player i S - i = S 1 * … S i-1 *S i+1 *.. * S r x i j : the probability that ith player uses j th strategy x s : x 1 s1 * x 2 s2 … * x r sr (for s in S -i ) u i js : utility of the i th player when he plays j and others play according to s -p NE: 0 ≤ x i j ≤ 1  j x i j =1  s in S u i js x s >  s in S u i j’s x s implies x i j’ = 0 -p

r-Player Game G → 3-Graphical Game GG Vertex V ij for any pair (player=i, action=j) Want: Pr[V ij plays 1] = Pr [i plays j in G]=x i j Idea: graphical games can do math! –Enforce constraints from the previous slide… v1v1 v2v2 v3v3 u Set payoffs to u, v 3 so that p[v 3 ]=p[v 1 ] * p[v 2 ] Need gadgets for +, *, c, =, min, max, …

Equivalences: GP’05 r-player game G NE of G deg 3 graphical game G’ NE of G’ f g d 2 -player game G’ NE of G’ deg d graphical game G NE of G f g

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

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

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

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 circuits S, T “end of the line” f g

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)

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: (s 1, …, s n ) → (s 1 +  1, …, s n +  n ), where  i is a shift in the direction of best response to (s 1, …, s i-1, s i+1, …, s n ) –NE is a point where noone wants to deviate, i.e., a fixpoint

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… Sperner’s Lemma A B C

Opposite Direction: 3D-BROUWER Input: –3D unit cube divided into 2 3n cubelets –c ijk is the center of K ijk –  (c ijk )=c ijk +  ijk,  ijk is in {  0,  1,  2,  3 }, where  1 =( , 0, 0),  2 =(0, , 0)  3 =(0, 0,  ),  0 =(- , - , -  ) –circuit C: {0, 1} 3n → {0, 1, 2, 3} selects  ijk Output: –a panchromatic cubelet, i.e., one that has all of  0,  1,  2,  3 among its 8 neighbors

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

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

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

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?

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?