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Preprocessing Techniques for Computing Nash Equilibria Vincent Conitzer Duke University Based on: Conitzer and Sandholm. A Generalized Strategy Eliminability Criterion and Computational Methods for Applying It. AAAI-05 Conitzer and Sandholm. A Technique for Reducing Normal-Form Games to Compute a Nash Equilibrium. AAMAS-06

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Computing Nash equilibria in (2-player) normal-form games Computing one Nash equilibrium is PPAD-complete –Daskalakis, Goldberg, & Papadimitriou ECCC05; Chen & Deng FOCS06 Determining whether a Nash equilibrium with a certain property exists is typically NP-complete –Is there an equilibrium with player 1s utility / each players utility / average utility > k? Even hard to approximate –Is there an equilibrium that puts positive / zero probability on pure strategy s? –Etc. –Gilboa & Zemel GEB 89; Conitzer & Sandholm IJCAI03/extended draft All known algorithms take exponential time –Even Lemke-Howson [Savani & von Stengel FOCS04/Econometrica06]

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Preprocessing games We are solving for an equilibrium (optimal equilibrium, maybe) When can we shrink the game and solve the shrunk game instead? If a strategy is (strictly) dominated, can throw it out If the game has only a small nonzero component, can focus on that We will see generalizations of both of these

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A class of hard games Sandholm, Gilpin, Conitzer AAAI05 0, 20, 33, 00, 00, 20, 00, 2 0, 0 0, 30, 23, 00, 2 2, 42, 0 4, 22, 03, 3 2, 0 2, 42, 04, 2 0, 23, 00, 30, 00, 20, 00, 2 0, 0 3, 00, 20, 30, 2 4, 22, 0 3, 32, 02, 4 1/3 0000 0 0 0 0

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Eliminability concepts Dominance: strategy always does worse than some other (mixed) strategy - strong argument - local reasoning - easy to compute - often does not apply Nash equilibrium: strategy does not appear in support of any Nash equilibrium -weaker argument - global reasoning - hard to compute - applies more often 3, 22, 3 3, 2 4, 00, 1.5 0 Is there something in between that combines good aspects of both? Yes! [Conitzer & Sandholm AAAI05] 3, 22, 3 3, 2 2, 02, 1.5

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Definition as game between attacker and defender Stage 1: Defender specifies probabilities on E strategies (e r * must get > 0) 3, 00, 30, 2 sr4sr4 0, 33, 00, 2 sr3sr3 2, 0 2, 2 sr2sr2 2, 0 2, 2 sr1sr1 sc4sc4 sc3sc3 sc2sc2 sc1sc1 0.4 0.3 0.50.4 Stage 2: Attacker chooses one of the E strategies with positive probability to attack and chooses (possibly mixed) attacking strategy 0.50.4 attacked attacking Stage 3: Defender chooses on which (non-E) strategy to place the remainder of the probability –If attacking outperforms attacked, attacker wins attacked attacking 0.50.40.1 e r * = s r 3, E r = {s r 3, s r 4 }, E c = {s c 3, s c 4 } 3, 00, 30, 2 sr4sr4 0, 33, 00, 2 sr3sr3 2, 0 2, 2 sr2sr2 2, 0 2, 2 sr1sr1 sc4sc4 sc3sc3 sc2sc2 sc1sc1 3, 00, 30, 2 sr4sr4 0, 33, 00, 2 sr3sr3 2, 0 2, 2 sr2sr2 2, 0 2, 2 sr1sr1 sc4sc4 sc3sc3 sc2sc2 sc1sc1

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A spectrum of elimination power The larger the E i sets, the more strategies are eliminable If the E i sets include all strategies, then a strategy is eliminable if and only if no Nash equilibrium places positive probability on it If the E i sets are empty (with the exception of e r *) then e r * is eliminable if and only if it is dominated dominance Nash equilibrium larger E i sets

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Alternative definition Stage 1: Defender specifies probabilities on E sets (e r * must get > 0) 0.4 0.3 0.50.4 Stage 2: Attacker chooses one of the E strategies with positive probability to attack Stage 3: Defender distributes the remainder of the probability (not on E) attacked 0.50.4 attacked 0.50.4 Stage 4: Attacker chooses attacking strategy –If attacking outperforms attacked, attacker wins 0.05 attacked 0.50.40.05 attacking e r * = s r 3, E r = {s r 3, s r 4 }, E c = {s c 3, s c 4 }

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Equivalence Theorem. The alternative definition is equivalent to the original one. Proof based on duality (more specifically, Minimax Theorem [von Neumann 1927] )

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Mixed integer programming approach (using alternative definition) Continuous variables: p i (e i ), p i e -i (s i ), binary: b i (e i ) maximize p r (e r *) subject to –for both i, for any e i E i, Σp -i (e -i ) + Σp -i e i (s -i ) = 1 –for both i, for any e i E i, p i (e i ) b i (e i ) –for both i, for any e i E i and any d i S i, Σp -i (e -i )(u i (e i, e -i )-u i (d i, e -i )) + Σp -i e i (s -i )(u i (e i, s -i )-u i (d i, s -i )) (b i (e i )-1)U i U i is the maximum difference between two of player is utilities Number of binary variables = |E r | + |E c | –Exponential only in this!

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Eliminating strategies in the hard game 0, 20, 33, 00, 00, 20, 00, 2 0, 0 0, 30, 23, 00, 2 2, 42, 0 4, 22, 03, 3 2, 0 2, 42, 04, 2 0, 23, 00, 30, 00, 20, 00, 2 0, 0 3, 00, 20, 30, 2 4, 22, 0 3, 32, 02, 4 1/3 0000 0 0 0 0 ErEr EcEc

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Another preprocessing technique for computing a Nash equilibrium [Conitzer & Sandholm AAMAS06] a l, d ml …a 2, d m2 a 1, d m1 ……… a l, d 2l …a 2, d 22 a 1, d 21 a l, d 1l …a 2, d 12 a 1, d 11 c kn, b k …c k2, b k c k1, b k ……… c 2n, b 2 …c 22, b 2 c 21, b 2 c 1n, b 1 …c 12, b 1 c 11, b 1 G π r, π c a l, Σ i p G (s i )d 1l …a 2, Σ i p G (s i )d i2 a 1,Σ i p G (s i )d i1 Σ j p G (t j )c kj, b k … Σ j p G (t j )c 2j, b 2 Σ j p G (t j )c 1j, b 1 G H H

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Required structure on original game O a l, d ml …a 2, d m2 a 1, d m1 smsm ………… a l, d 2l …a 2, d 22 a 1, d 21 s2s2 a l, d 1l …a 2, d 12 a 1, d 11 s1s1 c kn, b k …c k2, b k c k1, b k ukuk ………… c 2n, b 2 …c 22, b 2 c 21, b 2 u2u2 c 1n, b 1 …c 12, b 1 c 11, b 1 u1u1 tntn …t2t2 t1t1 vlvl …v2v2 v1v1 That is: against any fixed v j, all the s i give the row player the same utility a j against any fixed u i, all the t j give the column player the same utility b i H G

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Solve for equilibrium of G (recursively) smsm … s2s2 s1s1 tntn …t2t2 t1t1 Obtain –Equilibrium distributions p G (s i ), p G (t j ) –Players expected payoffs in equilibrium π r, π c G

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Reduced game R π r, π c a l, Σ i p G (s i )d 1l …a 2, Σ i p G (s i )d i2 a 1,Σ i p G (s i )d i1 s Σ j p G (t j )c kj, b k ukuk …… Σ j p G (t j )c 2j, b 2 u2u2 Σ j p G (t j )c 1j, b 1 u1u1 tvlvl …v2v2 v1v1 Expected payoffs when row player plays the equilibrium of G, column player plays v i Expected payoffs when both players play the equilibrium of G Theorem. p R (u i ), p R (s)p G (s i ); p R (v j ), p R (t)p G (t j ) constitutes a Nash equilibrium of original game. H

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Example v1v1 t1t1 t2t2 u1u1 2, 20, 32, 3 s1s1 1, 24, 00, 4 s2s2 1, 40, 44, 0 t1t1 t2t2 s1s1 0, 4 s2s2 4, 0 0.5 v1v1 t u1u1 2, 21, 3 s 2, 2 0.5 0.25

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A more difficult example = the game that we solved before! v 1 = b 2 t 1 = b 1 t 2 = b 3 u 1 = a 2 2, 20, 32, 3 s 1 = a 1 1, 24, 00, 4 s 2 = a 3 1, 40, 44, 0 b1b1 b2b2 b3b3 a1a1 1, 20, 4 a2a2 0, 32, 22, 3 a3a3 0, 41, 44, 0 But how (in general) do we find the correct labeling of the strategies as u i, s i, v j, t j ? Can it be done in polynomial time?

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Lets try to use satisfiability b1b1 b2b2 b3b3 a1a1 4, 01, 20, 4 a2a2 0, 32, 22, 3 a3a3 0, 41, 44, 0 Say that v(σ) = true if we label σ as one of the s i or t j (that is, we put it in G) If a 1, a 2 are both in G, then b 1 must also be in G because a 1, a 2 get different payoffs against b 1 Equivalently, v(a 1 ) and v(a 2 ) v(b 1 ) –or (-v(a 1 ) or -v(a 2 ) or v(b 1 )) Theorem: satisfaction of all such clauses the condition is satisfied

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Clauses for the example b1b1 b2b2 b3b3 a1a1 4, 01, 20, 4 a2a2 0, 32, 22, 3 a3a3 0, 41, 44, 0 v(a 1 ) and v(a 2 ) v(b 1 ) and v(b 2 ) and v(b 3 ) v(a 1 ) and v(a 3 ) v(b 1 ) and v(b 3 ) v(a 2 ) and v(a 3 ) v(b 2 ) and v(b 3 ) v(b 1 ) and v(b 2 ) v(a 1 ) and v(a 2 ) v(b 1 ) and v(b 3 ) v(a 1 ) and v(a 3 ) v(b 2 ) and v(b 3 ) v(a 1 ) and v(a 2 ) and v(a 3 ) Complete characterization of solutions: –Set at most one variable to true for each player (does not reduce game) –Set all variables to true (G = whole game!) –Only nontrivial solution: set v(a 1 ), v(a 3 ), v(b 1 ), v(b 3 ) to true

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Simple algorithm Algorithm to find nontrivial solution: –Start with any two variables for the same agent set to true –Follow the implications –If all variables set to true, start with next pair of variables

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Solving the example with the algorithm (pass 1) b1b1 b2b2 b3b3 a1a1 4, 01, 20, 4 a2a2 0, 32, 22, 3 a3a3 0, 41, 44, 0 v(a 1 ) and v(a 2 ) v(b 1 ) and v(b 2 ) and v(b 3 ) v(a 1 ) and v(a 3 ) v(b 1 ) and v(b 3 ) v(a 2 ) and v(a 3 ) v(b 2 ) and v(b 3 ) v(b 1 ) and v(b 2 ) v(a 1 ) and v(a 2 ) v(b 1 ) and v(b 3 ) v(a 1 ) and v(a 3 ) v(b 2 ) and v(b 3 ) v(a 1 ) and v(a 2 ) and v(a 3 ) Variables set to true: v(a 1 ) v(a 2 ) v(a 3 ) v(b 1 )v(b 2 )v(b 3 )

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Solving the example with the algorithm (pass 2) b1b1 b2b2 b3b3 a1a1 4, 01, 20, 4 a2a2 0, 32, 22, 3 a3a3 0, 41, 44, 0 v(a 1 ) and v(a 2 ) v(b 1 ) and v(b 2 ) and v(b 3 ) v(a 1 ) and v(a 3 ) v(b 1 ) and v(b 3 ) v(a 2 ) and v(a 3 ) v(b 2 ) and v(b 3 ) v(b 1 ) and v(b 2 ) v(a 1 ) and v(a 2 ) v(b 1 ) and v(b 3 ) v(a 1 ) and v(a 3 ) v(b 2 ) and v(b 3 ) v(a 1 ) and v(a 2 ) and v(a 3 ) Variables set to true: v(a 1 ) v(a 3 )v(b 1 )v(b 3 )

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Algorithm complexity Theorem. Requires at most O((#rows+#columns) 4 ) clause applications –That is, quadratic if the game is square Can improve in practice by caching previous results

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Preprocessing the hard game 2, 44, 23, 3 2, 44, 2 3, 32, 4 0, 21.5, 1.50, 2 2, 42, 04, 23, 3 2, 02, 44, 2 2, 03, 32, 4 0, 20, 33, 00, 00, 20, 00, 2 0, 0 0, 30, 23, 00, 2 2, 42, 0 4, 22, 03, 3 2, 0 2, 42, 04, 2 0, 23, 00, 30, 00, 20, 00, 2 0, 0 3, 00, 20, 30, 2 4, 22, 0 3, 32, 02, 4 0, 33, 0 0, 3 3, 00, 0 1.5, 1.5 3, 00, 30, 0 1.5, 1.50, 0 1.5, 1.5 0, 33, 0 0, 3 3, 00, 0 0, 33, 0 0, 30, 0 3, 00, 3 1/2 1/3 1 1 0 0

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Conclusions Generalized strategy eliminability criterion [AAAI05] –Parameterized definition –At one extreme setting, dominance –At other extreme, whether a strategy is in the support of any Nash –Efficiently computable for settings close to dominance Technique for recursively solving subcomponent [AAMAS06] –Subcomponents solution can be used to shrink original game –Efficient algorithm for finding subcomponent Other techniques? Generalization to extensive form? Thank you for your attention!

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Search-based approaches Suppose we know the support X i of each player is mixed strategy in equilibrium Then, we have a simple linear feasibility problem: –for both i, for any s i X i, Σp -i (s -i )u i (s i, s -i ) = u i –for both i, for any s i S i - X i, Σp -i (s -i )u i (s i, s -i ) u i Thus, we can search over supports –This is the basic idea underlying methods in Dickhaut & Kaplan 91; Porter, Nudelman, Shoham AAAI04; Sandholm, Gilpin, Conitzer AAAI05 Dominated strategies can be eliminated –A type of preprocessing –What other preprocessing techniques exist?

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Vincent Conitzer conitzer@cs.duke.edu CPS 296.1 Repeated games Vincent Conitzer conitzer@cs.duke.edu.

Vincent Conitzer conitzer@cs.duke.edu CPS 296.1 Repeated games Vincent Conitzer conitzer@cs.duke.edu.

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