1. Nash Equilibrium ( p *, q * ) is a N.E. – no player has any incentive to move: PPAD hard problem [DGP’06; CD’06] Q: Why are they so extensively studied? A: “Games with rational players have predictable outcome”. or: “Rational players will have predictable strategy”. Approximate Nash Equilibrium ( p, q ) is ² - N.E. – no player has more than ² -incentive to move: Computable in time O( n log( n )/ ² 2 ) [Lipton, Markakis,Metha’07] 0.3393- N.E. is feasible in polytime [Tsaknakis, Spirakis’07] PPAD -hard to get ² -exact- N.E. Q: Why should we try to fine ² -N.E. ? A1: “Games with rational players have approx. this outcome”. A2: “Rational players will have approximately this strategy”. NOT THE SAME! A game is ( ², ¢ )- approximation stable if for any ² - N.E. ( p, q ) we have that Dist( ( p, q ), ( p *, q * ) ) < ¢ (for some ( p *, q * ) ) Dist( ( p, q ), ( p *, q * ) ) = max { VD( p, p * ), VD( q, q * ) } Warmup 1. Every game is ( ²,1)- approximation stable. 2. Any ( p, q ) s.t. Dist( ( p, q ), ( p *, q * ) ) < ® is a 3 ® - N.E. : - Any deviation has incentive at most ® - For ( p, q ), the gain is at least (Nash-value - 2 ® ) 3. No game is ( ², ¢ )- approximation stable if ¢ < ². - One player moves ² -mass arbitrarily and we get ² -NE. 4. The Striker/Goalie game is ( ², ¢ )- stable for ¢ = ² + O( ² 2 ). On Nash-Equilibria of Approximation-Stable Games Pranjal Awasthi* Nina Balcan ** Avrim Blum* Or Sheffet* Santosh Vempala** * Carnegie Mellon University**Georgia Tech Main Result [LMM07]: Any game has ² - N.E. with | support | =O((1/ ² 2 ) log( n )) Pf: treat p * and q * as distributions and sample S rows from it. Apply Chernoff to show the expected gain changes by no more than ². Thm: Any ( ², ¢ ) -stable game has ² - N.E. of support ( ¢ (2-o(1)) / ² 2 )log( n ) Claim 1: If ( p *, q * ) is the N.E. in a ( ², ¢ )- stable game, then there is no decomposition p * = x ¢ p 1 + (1- x ) p 2 where x < ¾, and the col-player incentive to deviate q * ! e j is roughly the same whether row -player plays p * or p 1 : Pf: If there was such a decomposition, imagine row-player moves ¢ - mass from p 2 to p 1. col-player still plays q *. Row-player has no incentive to move (same support as in p * ). Col-player can gain on any deviation q * ! e j no more than: Claim 2: Denote S =( ¢ / ² ) 2 log( n ). If p is almost uniform (| p | 2 < 1/ S ) then a decomposition like in Claim 1 exists. Pf: Treat p * as a vector. p 1 Ã Sample each ccordinate w.p. ½, then normalize. Apply Hoeffding and show both properties hold. Pf of Theorem: Greedily separate support( p * ) into H (heavy) and L (light). Halt if: (a) either: p | L is roughly uniform ( 8 i, p ( i ) < p ( L )/ S ) (b) or: p ( L ) < ¢ If (a) happens: Apply Claim 2 to p | L and obtain contradiction. So (b) happens. - | H | < S log(1/ ¢ ): E very coordinate added to H, decreases the mass p ( L ) by (1- 1 / S ). - Apply [LMM07] to subsample L ’ ½ L, using ² ’ = ² / 2 ¢. - We get p with | support( p ) | = | H | + | L ’| = O( S log(1/ ¢ )) TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A AAA A A A (1,0) (0,1) ) (1,0) Additional Result Thm: If ¢ < 2 ² – 6 ² 2, then exists O( ² )-N.E. with support O(1/ ² ). Claim: We cannot decompose p * = x ¢ p 1 + (1- x ) p 2 where x 2 [ ¢, ¢ + ² ]. Pf: If we could, then define p ’ as shifting ¢ -mass from p 1 to p 2. Then ( p ’, q * ) is ¢ -far from ( p *, q * ). Only col-player has incentive to deviate, that incentive > ². Using the two ways to write p ’, we get an upper and lower bound on p ’ Ce j, which give a lower bound on V C. Now define p ” as shifting ¢ -mass from p 2 to p 1. Same idea gives an upper bound on V C. Comparing the two bounds give ¢ > 2 ² – 6 ² 2. Pf of Theorem: No decomposition means p * is ¢ -close to a vector of support O(1/ ² ). The Opposite Direction A game is ( ², ¢ ) -smooth if for any ( p, q ), s.t. Dist( ( p, q ), ( p *, q * ) ) < ¢, we have that ( p, q ) is a ² -N.E. Algorithm for finding ( ² / ¢ )-N.E. : Guess V R, V C and find a feasible solution ( p, q ) for the LP Main idea: ( p *, q * ) is a feasible solutions for this LP : - p ’ = shifting ¢ -mass to some e i, outside the support of p *. - For ( p ’, q * ), the row-player gains at least V R - ². - So ( e i, q * ) must gain the row-player at least V R - ² / ¢. Open Problems For ¢ < 2 ² – poly-time algorithm For ¢ = ( ² /32) 1/2 – We exhibit a game with all ² -N.E. of support  (log( n )) For ¢ = ² 1/4 – Finding ² - N.E. is as hard as finding ² 1/4 -N.E. in a regular game. [Balcan, Braverman’10] What happens for intermediate values of ¢ ?