Inapproximability Results for Approximate Nash Equilibria
aa r X i v : . [ c s . G T ] A p r Inapproximability Results for Approximate NashEquilibria
Argyrios Deligkas a , John Fearnley b , Rahul Savani b a Technion-Israel Institute of Technology b University of Liverpool
Abstract
We study the problem of finding approximate Nash equilibria that satisfy certainconditions, such as providing good social welfare. In particular, we study theproblem ǫ -NE δ -SW: find an ǫ -approximate Nash equilibrium ( ǫ -NE) that iswithin δ of the best social welfare achievable by an ǫ -NE. Our main result isthat, if the exponential-time hypothesis (ETH) is true, then solving (cid:0) − O( δ ) (cid:1) -NE O( δ )-SW for an n × n bimatrix game requires n e Ω(log n ) time. Building onthis result, we show similar conditional running time lower bounds on a numberof decision problems for approximate Nash equilibria that do not involve socialwelfare, including maximizing or minimizing a certain player’s payoff, or findingapproximate equilibria contained in a given pair of supports. We show quasi-polynomial lower bounds for these problems assuming that ETH holds, wherethese lower bounds apply to ǫ -Nash equilibria for all ǫ < . The hardness ofthese other decision problems has so far only been studied in the context ofexact equilibria. Keywords:
Approximate Nash equilibrium, constrained equilibrium,quasi-polynomial time, lower bound, Exponential Time Hypothesis.
1. Introduction
One of the most fundamental problems in game theory is to find a Nashequilibrium of a game. Often, we are not interested in finding any Nash equi-librium, but instead we want to find one that also satisfies certain constraints.For example, we may want to find a Nash equilibrium that provides high socialwelfare , which is the sum of the players’ payoffs.In this paper we study such problems for bimatrix games , which are two-player strategic-form games. Unfortunately, for bimatrix games, it is knownthat these problems are hard. Finding any Nash equilibrium of a bimatrixgame is
PPAD -complete [1], while finding a constrained Nash equilibrium turns ✩ The authors were supported by EPSRC grant EP/L011018/1. A short version of thispaper appeared at WINE 2016. ut to be even harder. Gilboa and Zemel [2] studied several decision problemsrelated to Nash equilibria. They proved that it is NP -complete to decide whetherthere exist Nash equilibria in bimatrix games with some “desirable” properties,such as high social welfare. Conitzer and Sandholm [3] extended the list of NP -complete problems of [2] and furthermore proved inapproximability resultsfor some of them. Recently, Garg et al. [4] and Bilo and Mavronicolas [5, 6]extended these results to many player games and provided ETR -completenessresults for them.
Approximate equilibria.
Due to the apparent hardness of finding exact Nashequilibria, focus has shifted to approximate equilibria. There are two natu-ral notions of approximate equilibrium, both of which will be studied in thispaper. An ǫ -approximate Nash equilibrium ( ǫ -NE) requires that each playerhas an expected payoff that is within ǫ of their best response payoff. An ǫ -well-supported Nash equilibrium ( ǫ -WSNE) requires that both players only playstrategies whose payoff is within ǫ of the best response payoff. Every ǫ -WSNEis an ǫ -NE but the converse does not hold, so a WSNE is a more restrictivenotion.There has been a long line of work on finding approximate equilibria [7,8, 9, 10, 11, 12, 13]. Since we use an additive notion of approximation, itis common to rescale the game so that the payoffs lie in [0 , . . quasi-polynomial time approximation scheme (QPTAS) forfinding approximate Nash equilibria. The algorithm of Lipton, Markakis, andMehta finds an ǫ -NE in n O ( log nǫ ) time [14]. They proved that there is always an ǫ -NE with support of logarithmic size, and then they use a brute-force searchover all possible candidates to find one. We will refer to their algorithm as theLMM algorithm.A recent breakthrough of Rubinstein implies that we cannot do better than aQPTAS like the LMM algorithm [15]: assuming an exponential time hypothesisfor PPAD (PETH), there is a small constant, ǫ ∗ , such that for ǫ < ǫ ∗ , everyalgorithm for finding an ǫ -NE requires quasi-polynomial time. Briefly, PETHis the conjecture that EndOfTheLine , the canonical
PPAD -complete problem,cannot be solved faster than exponential time.
Constrained approximate Nash equilibria.
While deciding whether agame has an exact Nash equilibrium that satisfies certain constraints is NP -hardfor most interesting constraints, this is not the case for approximate equilibria,because the LMM algorithm can be adapted to provide a QPTAS for them. Thequestion then arises whether one can do better.Let the problem ǫ -NE δ -SW be the problem of finding an ǫ -NE whose socialwelfare is within δ of the best social welfare that can be achieved by an ǫ -NE.Hazan and Krauthgamer [16] and Austrin, Braverman and Chlamtac [17] provedthat there is a small but constant ǫ such that ǫ -NE ǫ -SW is at least as hard as2nding a hidden clique of size O(log n ) in the random graph G n, / . This wasfurther strengthened by Braverman, Ko, and Weinstein [18] who showed a lowerbound based on the exponential-time hypothesis (ETH), which is the conjecturethat any deterministic algorithm for 3SAT requires 2 Ω( n ) time. More precisely,they showed that under ETH there is a small constant ǫ such that any algorithmfor O( ǫ )-NE O( ǫ )-SW requires n poly( ǫ ) log( n ) − o (1) time . We shall refer to thisas the BKW result.It is worth noting that the Rubinstein’s hardness result [15] almost makesthis result redundant. If one is willing to accept that PETH is true, which is astronger conjecture than ETH, then Rubinstein’s result says that for small ǫ werequire quasi-polynomial time to find any ǫ -NE, which obviously implies thatthe same lower bound applies to ǫ -NE δ -SW for any δ . Our results.
Our first result is a lower bound for the problem of finding ǫ -NE δ -SW. We show that, assuming ETH, that there exists a small constant δ such that the problem (cid:16) − g · δ (cid:17) -NE (cid:16) g · δ (cid:17) -SW requires n e Ω(log n ) time , where g = .To understand this result, let us compare it to the BKW result. First,observe that as δ gets smaller, the ǫ in our ǫ -NE gets larger, whereas in theBKW result, ǫ get smaller. Asymptotically, our ǫ approaches 1 /
8. Moreover,since δ ≤
1, our lower bound applies to all ǫ -NE with ǫ ≤ − g ≈ . ǫ -NE δ -SW, we use it to prove lower boundsfor other problems regarding constrained approximate NEs and WSNEs. Ta-ble 1 gives a list of the problems that we consider. For each one, we provide areduction from ǫ -NE δ -SW to that problem. Ultimately, we prove that if ETHis true, then for every ǫ < finding an ǫ -NE with the given property in an n × n bimatrix game requires n e Ω(log n ) time. Techniques.
At a high level, the proof of our first result is similar in spirit tothe proof of the BKW result. They reduce from the problem of approximating While the proof in [18] produces a lower bound for 0 . − O( ǫ ))-SW, this is in agame with maximum payoff O(1 /ǫ ). Therefore, when the payoffs in this game are rescaled to[0 , ǫ -NE ǫ -SW. Although the paper claims that they obtain a n e O(log n ) lower bound, the proof reducesfrom the low error result from [19] (cf. Theorem 36 in [20]), which gives only the weaker lowerbound of n poly( ǫ ) log( n ) − o (1) . Here e Ω(log n ) means Ω( log n (log log n ) c ) for some constant c . roblem description Problem definition Problem 1: Large payoffs u ∈ (0 ,
1] Is there an ǫ -NE ( x , y ) such thatmin( x T R y , x T C y ) ≥ u ?Problem 2: Restricted support S ⊂ [ n ] Is there an ǫ -NE ( x , y ) with supp( x ) ⊆ S ?Problem 3: Two ǫ -NE d ∈ (0 , ǫ -NE with TV distance ≥ d ?Problem 4: Small largest probabil-ity p ∈ (0 ,
1) Is there an ǫ -NE ( x , y ) with max i x i ≤ p ?Problem 5: Small total payoff v ∈ [0 ,
2) Is there an ǫ -NE ( x , y ) such that x T R y + x T C y ≤ v ?Problem 6: Small payoff u ∈ [0 ,
1) Is there an ǫ -NE ( x , y ) such that x T R y ≤ u ?Problem 7: Large total support size k ∈ [ n ] Is there an ǫ -WSNE ( x , y ) such that | supp( x ) | + | supp( y ) | ≥ k ?Problem 8: Large smallest supportsize k ∈ [ n ] Is there an ǫ -WSNE ( x , y ) such thatmin {| supp( x ) | , | supp( y ) |} ≥ k ?Problem 9: Large support size k ∈ [ n ] Is there an ǫ -WSNE ( x , y ) such that | supp( x ) | ≥ k ?Problem 10: Restricted support S R ⊆ [ n ] Is there an ǫ -WSNE ( x , y ) with S R ⊆ supp( x )? Table 1: The decision problems that we consider. All of them take as input abimatrix game (
R, C ) and a quality of approximation ǫ ∈ (0 , ǫ -NE, and Problems 7 - 10 relate to ǫ -WSNE.4he value of a free game . Aaronson, Impagliazzo, and Moshkovitz showed quasi-polynomial lower bounds for this problem assuming ETH [19]. A free gameis played between to players named Merlin and Merlin , and a referee namedArthur. The BKW result creates a bimatrix game that simulates the free game,where the two players take the role of Merlin and Merlin , while Arthur issimulated using a zero-sum game.Our result will also be proved by producing a bimatrix game that simulatesa free game. However, there are a number of key differences that allow usto prove the stronger lower bounds described above. The first key difference isthat we use a different zero-sum game to simulate Arthur. Our zero-sum game isinspired by the one used by Feder, Nazerzadeh, and Saberi [21]. The advantageof this construction is that it is capable of ensuring that play distributions thata very close to uniform in all approximate Nash equilibria, which in turn givesus a very accurate simulation of Arthur.The downside of this zero-sum game is that it requires 2 n rows to forcethe column player to mix close to uniformly over n rows. Arthur is requiredto pick two questions uniformly from a set of possible questions. The freegames provided by Aaronson, Impagliazzo, and Moshkovitz have question setsof linear size, so if we reduce directly from these games, we would end up withan exponentially sized bimatrix game. The conference version of this paper [22]resolved the issue by using a sub-sampling lemma, also proved by Aaronson,Impagliazzo, and Moshkovitz, that produces a free game with logarithmicallysized question sets. This allowed us to produce polynomially sized bimatrixgames, but at the cost of needing randomization to implement the reduction,and so the result depended on the randomized version of the ETH.In this version, we show that we are able to assume only the ETH by usinga stronger result that was discovered by Babichenko, Papadimitriou, and Ru-binstein [23]. Their results imply that approximating the value of a free gamerequires quasipolynomial time even when the size of the question sets is loga-rithmic in the game size. They do not explicitly formulate this result, but it isclearly implied by their techniques. For the sake of completeness, we providean exposition of their ideas in Section 3.The second main difference between our result and the BKW result is thatwe use a different starting point. The BKW result uses the PCP theoremof Moshkovitz and Raz [24], which provides a completeness/soundness gap of1 vs δ for arbitrarily small constant δ in the label cover problem. The useof this powerful PCP theorem is necessary, as their proof relies on the largecompleteness/soundness gap produced by that theorem. This choice of PCPtheorem directly impacts the running time lower bound that they produce, asthe (log n ) − o (1) term in the exponent arises from the blowup of n o (1) fromthe PCP theorem.In contrast to this, our stronger simulation of Arthur allows us to use thePCP theorem of Dinur [25] as our starting point. This PCP theorem onlyinvolves a blowup of n polylog( n ), which directly leads to the improved e Ω(log n )exponent in our lower bound. The improved blowup of the PCP theorem comes5t the cost of providing a completeness/soundness gap of only 1 vs 1 − ǫ for ǫ < , but our simulation is strong enough to deal with this. It is also worthnoting that if a PCP theorem with a constant completeness/soundness gap andlinear blow up is devised in the future, then the exponent in our lower boundwill improve to Ω(log n ).One final point of comparison is the size of the payoffs used in our simulation.The zero-sum games that we use have payoffs in the range ( − , − g · δ bound on the quality of approximation. In contrast to this,the zero-sum games used by the BKW result have payoffs of size O ( ǫ ), whichultimately means that their lower bound only applies to the problem ǫ -NE ǫ -SW. Other related work.
The only positive result for finding ǫ -NE with goodsocial welfare that we are aware of was given by Czumaj, Fasoulakis, and Jur-dzi´nski [26, 27]. In [26], they showed that if there is a polynomial-time algorithmfor finding an ǫ -NE, then for all ǫ ′ > ǫ there is also a polynomial-time algorithmfor finding an ǫ ′ -NE that is within a constant multiplicative approximation ofthe best social welfare. They also give further results for the case where ǫ > .In [27] they derived polynomial-time algorithms that compute ǫ -NE for ǫ ≥ −√ that approximate the quality of plutocratic and egalitarian Nash equilibria tovarious degrees.
2. Preliminaries
Throughout the paper, we use [ n ] to denote the set of integers { , , . . . , n } .An n × n bimatrix game is a pair ( R, C ) of two n × n matrices: R gives payoffsfor the row player and C gives the payoffs for the column player.Each player has n pure strategies. To play the game, both players simul-taneously select a pure strategy: the row player selects a row i ∈ [ n ], and thecolumn player selects a column j ∈ [ n ]. The row player then receives payoff R i,j , and the column player receives payoff C i,j .A mixed strategy is a probability distribution over [ n ]. We denote a mixedstrategy for the row player as a vector x of length n , such that x i is the prob-ability that the row player assigns to pure strategy i . A mixed strategy of thecolumn player is a vector y of length n , with the same interpretation. If x and y are mixed strategies for the row and the column player, respectively, then wecall ( x , y ) a mixed strategy profile . The expected payoff for the row player understrategy profile ( x , y ) is given by x T R y and for the column player by x T C y .We denote the support of a strategy x as supp( x ), which gives the set of purestrategies i such that x i > Nash equilibria.
Let y be a mixed strategy for the column player. The set of pure best responses against y for the row player is the set of pure strategies thatmaximize the payoff against y . More formally, a pure strategy i ∈ [ n ] is a bestresponse against y if, for all pure strategies i ′ ∈ [ n ] we have: P j ∈ [ n ] y j · R i,j ≥ P j ∈ [ n ] y j · R i ′ ,j . Column player best responses are defined analogously.6 mixed strategy profile ( x , y ) is a mixed Nash equilibrium if every purestrategy in supp( x ) is a best response against y , and every pure strategy insupp( y ) is a best response against x . Nash [28] showed that every bimatrixgame has a mixed Nash equilibrium. Observe that in a Nash equilibrium, eachplayer’s expected payoff is equal to their best response payoff. Approximate Equilibria.
There are two commonly studied notions of ap-proximate equilibrium, and we consider both of them in this paper. The firstnotion is that of an ǫ -approximate Nash equilibrium ( ǫ -NE), which weakens therequirement that a player’s expected payoff should be equal to their best re-sponse payoff. Formally, given a strategy profile ( x , y ), we define the regret suffered by the row player to be the difference between the best response payoffand the actual payoff: max i ∈ [ n ] (cid:0) ( R · y ) i (cid:1) − x T · R · y . Regret for the columnplayer is defined analogously. We have that ( x , y ) is an ǫ -NE if and only if bothplayers have regret less than or equal to ǫ .The other notion is that of an ǫ -approximate-well-supported equilibrium ( ǫ -WSNE), which weakens the requirement that players only place probability onbest response strategies. We say that a pure strategy j ∈ [ n ] of the row playeris an ǫ -best-response against y if:max i ∈ [ n ] (cid:0) ( R · y ) i (cid:1) − ( R · y ) j ≤ ǫ. An ǫ -WSNE requires that both players only place probability on ǫ -best-responses.Formally, the row player’s pure strategy regret under ( x , y ) is defined to be:max i ∈ [ n ] (cid:0) ( R · y ) i (cid:1) − min i ∈ supp( x ) (cid:0) ( R · y ) i (cid:1) . Pure strategy regret for the columnplayer is defined analogously. A strategy profile ( x , y ) is an ǫ -WSNE if bothplayers have pure strategy regret less than or equal to ǫ .Since approximate Nash equilibria use an additive notion of approximation,it is standard practice to rescale the input game so that all payoffs lie in therange [0 , , ,
1] at the very end. To avoid confusion,we will refer to an ǫ -approximate Nash equilibrium in this game as an ǫ -UNE,to mark that it is an additive approximation in an unscaled game. Two-prover games.
A two-prover game is defined as follows.
Definition 1 (Two-prover game).
A two-prover game T is defined by a tu-ple ( X, Y, A, B, D , V ) where X and Y are finite sets of questions , A and B arefinite sets of answers , D is a probability distribution defined over X × Y , and V is a verification function of the form V : X × Y × A × B → { , } . The game is a co-operative and played between two players, who are calledMerlin and Merlin , and an adjudicator called Arthur. At the start of thegame, Arthur chooses a question pair ( x, y ) ∈ X × Y randomly according to D .7e then sends x to Merlin and y to Merlin . Crucially, Merlin does notknow the question sent to Merlin and vice versa. Having received x , Merlin then chooses an answer from A and sends it back to Arthur. Merlin similarlypicks an answer from B and returns it to Arthur. Arthur then computes p = V ( x, y, a, b ) and awards payoff p to both players. The size of the game, denoted |T | = | X × Y × A × B | is the total number of entries needed to represent V asa table.A strategy for Merlin is a function a : X → A that gives an answer for everypossible question, and likewise a strategy for Merlin is a function b : Y → B .We define S i to be the set of all strategies for Merlin i . The payoff of the gameunder a pair of strategies ( s , s ) ∈ S × S is denoted as p ( T , s , s ) = E ( x,y ) ∼D [ V ( x, y, s ( x ) , s ( y ))] . The value of the game, denoted ω ( T ), is the maximum expected payoff tothe Merlins when they play optimally: ω ( T ) = max s ∈ S max s ∈ S p ( T , s , s ) . Free games.
A two-prover game is called a free game if the probability distri-bution D is the uniform distribution U over X × Y . In particular, this meansthat there is no correlation between the question sent to Merlin and the ques-tion sent to Merlin . We are interested in the problem of approximating thevalue of a free game within an additive error of δ . FreeGame δ Input: A free game T and a constant δ > p such that | ω ( T ) − p | ≤ δ .
3. Hardness of approximating free games
The exponential time hypothesis (ETH) is the conjecture that any determin-istic algorithm for solving 3SAT requires 2 Ω( n ) time. Aaronson, Impagliazzo,and Moshkovitz have shown that, if ETH holds, then there exists a small con-stant ǫ > ǫ requires quasi-polynomial time. However, their result is not suitablefor our purposes, because it produces a free game in which the question andanswer sets have the same size, and to prove our result, we will require that thequestion sets have logarithmic size when compared to the answer sets.The conference version of this paper [22] solved this issue by using a sub-sampling lemma, also proved by Aaronson, Impagliazzo, and Moshkovitz, whichshows that if we randomly choose logarithmically many questions from the orig-inal game, the value of the resulting sub-game is close the value of the original.8owever, this comes at the cost of needing randomness in the reduction, and soour result depended on the truth of the randomized ETH, which is a strongerconjecture.In this exposition, we will instead use a technique of Babichenko, Papadim-itriou, and Rubinstein [23], which allows us to produce a free game with alogarithmic size question set in a deterministic way. The result that we needis a clear consequence of their ideas, but is not explicitly formulated in theirpaper. For the sake of completeness, in the rest of this section we provide ourown exposition of their ideas.
The PCP theorem.
The starting point of the result will be a 3SAT instance φ .We say that the size of a formula φ is the number of variables and clauses inthe formula. We define SAT( φ ) ∈ [0 ,
1] to be the maximum fraction of clausesthat can be satisfied in φ . The first step is to apply a PCP theorem. Theorem 1 (Dinur’s PCP Theorem [25]).
Given any 3SAT instance φ ofsize n , and a constant ǫ in the range < ǫ < , we can produce in polynomialtime a 3SAT instance ψ where: • The size of ψ is n · polylog( n ) . • Every clause of ψ contains exactly 3 variables and every variable is con-tained in at most d clauses, where d is a constant. • If SAT( φ ) = 1 , then SAT( ψ ) = 1 . • If SAT( φ ) < , then SAT( ψ ) < − ǫ . After applying the PCP theorem given above, we then directly construct afree game. Observe that a 3SAT formula can be viewed as a bipartite graphin which the vertices are variables and clauses, and there is an edge betweena variable x i and a clause C j if and only if x i is appears in C j . In particular,the 3SAT formulas produced by Theorem 1 correspond to bipartite graphs withconstant degree, since each clause has degree at most 3, and each variable hasdegree at most d .The first step is to apply the following lemma, which allows us to partitionthe vertices of this bipartite graph. The lemma and proof are essentially identicalto [23, Lemma 6], although we generalise the formulation slightly, because theoriginal lemma requires that the two sides of the graph have exactly the samenumber of nodes and that the graph is d -regular. Lemma 1 ([23]).
Let ( V, E ) be a bipartite graph with | V | = n , where V = U ∪ W are the two sides of the graph, and where each node has degree at most d . Suppose that U and W both have a constant fraction of the vertices, andhence | U | = c · n and | W | = c = (1 − c ) · n for some constants c < and c < . We can efficiently find a partition S , S , . . . , S √ n of U and a partition T , T , . . . , T √ n of W such that each set has size at most √ n , and for all i and j we have | ( S i × T j ) ∩ E | ≤ d . roof 1. The algorithm is as follows. First we arbitrarily split U into √ n many sets S , S , . . . , S √ n , and so each set S i has size c √ n < √ n . Then weiteratively construct the partition of W into sets T , T , . . . , T √ n in the followingway. We initialize each set T j to be the empty set. In each iteration, we pick avertex of w ∈ W that has not already been assigned to a set. We find a set T j such that | T j | ≤ · √ n , and such that for all i we have | ( S i × T j ) ∩ E | ≤ · d .We assign w to T j and repeat.Obviously, for the algorithm to be correct, we must prove that for each ver-tex w that is considered, there does exist a set T j that satisfies the requiredconstraints. For this, we rely on the following two properties. • The average number of vertices in a set T j is at most c √ n < √ n , and soby Markov’s inequality strictly less than half the sets can have size morethan √ n , and so we lose strictly less than half the sets T j to the sizeconstraint. • Since each vertex has degree at most d , the graph has at most dn edges,and so the average number of edges between each pair of sets S i and T j is dn/ ( √ n · √ n ) = d . Again, using Markov’s inequality we can conclude thatthere are at most / d pairs of sets S i and T j that have more than d edges between them. Hence, even in the worst case, we can lose at most / d sets T j to the edge constraints.So, we lose strictly less than half the sets to the size constraints, and / d ≤ / the sets to the edge constraints. Hence, by the union bound, we have shown thatthere is at least one set T j that satisfies both constraints simultaneously. (cid:3) A free game.
Note that Lemma 1 can be applied to the 3SAT formula thatarises from Dinur’s PCP theorem, because the number of variables and num-ber of constraints are both a constant fraction of the number of nodes in theassociated bipartite graph, and because each vertex has either has degree d ordegree 3. We use this to construct the following free game, which is highlyreminiscent of the clause variable game given by Aaronson, Impagliazzo, andMoshkovitz [19]. Definition 2.
Given a 3SAT formula φ of size n , we define a free game F φ inthe following way.
1. Arthur begins by applying Dinur’s PCP theorem to φ to obtain a formula ψ of size N = n polylog( n ) , and then uses Lemma 1 to split the variables of ψ into sets S , S , . . . , S √ n and the clauses of ψ into sets T , T , . . . , T √ N .
2. Arthur picks an index i uniformly at random from [ √ N ] , and indepen-dently an index j uniformly at random from [ √ N ] . He sends S i to Merlin and T j to Merlin .
3. Merlin responds by giving a truth assignment to every variable in S i , and Merlin responds by giving a truth assignment to every variable that isinvolved with a clause in T j .
10. Arthur awards the Merlins payoff if and only if both of the followingconditions hold. • Merlin returns an assignment that satisfies all clauses in T j . • For every variable v that appears in S i and some clause of T j , theassignment to v given by Merlin agrees with the assignment to v given by Merlin . Note that this condition is always satisfied when S i and T j share no variables. Arthur awards payoff otherwise. If n is the size of φ , then when we write F φ down as a free game ( X, Y, A, B, D , V ),the number of questions in the sets X and Y is p n polylog( n ), and the numberof answers in A and B is 2 √ n polylog( n ) , where the extra polylog( n ) factor arisesdue to the application of the PCP theorem.The following lemma shows that if φ is unsatisfiable, then the value of thisfree game is bounded away from 1. Again, the ideas used to prove this lemma areclearly evident in the work of Babichenko, Papadimitriou, and Rubinstein [23]. Lemma 2 ([23]). If φ is satisfiable then ω ( F φ ) = 1 . If φ is unsatisfiable then ω ( F φ ) ≤ − ǫ/ d . Proof 2.
The case where
SAT( φ ) = 1 is straightforward. Since there exists asatisfying assignment for φ , there also exists a satisfying assignment for ψ . Ifthe two Merlins play according to this satisfying assignment, then they obviouslyachieve an expected payoff of .For the other claim, first observe that we can assume that both Merlins playdeterministic strategies, since the game is co-operative, and therefore nothingcan be gained through randomization. So, let s be a strategy for Merlin . Ob-serve that since S , S , . . . , S √ N partition the variables of ψ , we have that s yields an assignment to the variables of ψ .Let us fix an arbitrary deterministic strategy s for Merlin . We have thatthe payoff to Merlin for an individual question T j can be computed as follows: • For every set S i for which there are no edges between the variables in S i and T j , Merlin gets payoff “for free.” • Otherwise,
Merlin gets payoff only if the assignments to the clausesagree with the assignment implied by s .From this, we can see that when Merlin plays s , Merlin can maximize hispayoff by playing the strategy that agrees everywhere with the assignment chosenby Merlin . So let s denote this strategy.Since φ is unsatisfiable, the PCP theorem tells us that SAT( ψ ) < − ǫ . Thus,there are at least ǫdN clauses that are not satisfied when s is played against s . Since Lemma 1 ensures that the maximum number of edges between twosets is d , there must therefore be at least ǫdN/ d = ǫN/ d pairs of sets thatgive payoff to the Merlins under s and s . Since there are exactly N pairsof sets in total, this means that the expected payoff to the Merlins is bounded by − ǫ/ d . (cid:3) F φ ,rather than the construction originally given in that paper. Theorem 2 ([23]).
Assuming ETH, there is a small constant δ below whichthe problem FreeGame δ cannot be solved faster than N e O(log N ) , even when thequestion sets have size log N . Proof 3.
Lemma 2 implies that if we can approximate the value of F φ withan additive error of less than ǫ/ d , then we can solve the satisfiability problemfor φ .Assume, for the sake of contradiction, that there exists an algorithm that cansolve the FreeGame δ problem in time N o ( log N (log log N ) c ) for some constant c thatwill be fixed later. Observe that the free game F φ has size N = O (2 √ n polylog( n ) ) ,and so the hypothesized algorithm would run in time: exp (cid:18) o (cid:18) log N (log log( N )) c (cid:19)(cid:19) = exp (cid:18) o (cid:18) n polylog( n )(log( √ n polylog( n ))) c (cid:19)(cid:19) . If we set c to be greater than the degree of the polynomial in the polylog( n ) from the numerator, then we can conclude that the running time would be o ( n ) ,which would violate the ETH. (cid:3)
4. Hardness of approximating social welfareOverview.
In this section, we study the following social welfare problem fora bimatrix game G = ( R, C ). The social welfare of a strategy profile ( x , y ) isdenoted by SW( x , y ) and is defined to be x T R y + x T C y . Given an ǫ ≥
0, wedefine the set of all ǫ equilibria as E ǫ = { ( x , y ) : ( x , y ) is an ǫ -NE } . Then, we define the best social welfare achievable by an ǫ -NE in G asBSW( G , ǫ ) = max { SW( x , y ) : ( x , y ) ∈ E ǫ } . Using these definitions we now define the main problem that we consider: ǫ -NE δ -SWInput: A bimatrix game G , and two constants ǫ, δ > ǫ -NE ( x , y ) s.t. SW( x , y ) is within δ of BSW( G , ǫ ).We show a lower bound for this problem by reducing from FreeGame δ . Let F be a free game of size n from the family of free games that were used to prove12heorem 2 (from now on we will drop the subscript φ , since the exact construc-tion of F is not relevant to us.) We have that either ω ( F ) = 1 or ω ( F ) < − δ for some fixed constant δ , and that it is hard to determine which of these is thecase. We will construct a game G such that for ǫ = 1 − g · δ , where g < is afixed constant that we will define at the end of the proof, we have the followingproperties. • ( Completeness ) If ω ( F ) = 1, then the unscaled BSW( G , ǫ ) = 2. • ( Soundness ) If ω ( F ) < − δ , then the unscaled BSW( G , ǫ ) < − g · δ ).This will allow us to prove our lower bound using Theorem 2. We use F to construct a bimatrix game, which we will denote as G through-out the rest of this section. The game is built out of four subgames, which arearranged and defined as follows. ❅❅ I II R − D C D D − D • The game (
R, C ) is built from F in the following way. Each row of thegame corresponds to a pair ( x, a ) ∈ X × A and each column corresponds toa pair ( y, b ) ∈ Y × B . Since all free games are cooperative, the payoff foreach strategy pair ( x, a ) , ( y, b ) is defined to be R ( x,a ) , ( y,b ) = C ( x,a ) , ( y,b ) = V ( x, y, a ( x ) , b ( y )) . • The game ( D , − D ) is a zero-sum game. The game is a slightly modifiedversion of a game devised by Feder, Nazerzadeh, and Saberi [21]. Let H be the set of all functions of the form f : Y → { , } such that f ( y ) = 1for exactly half of the elements y ∈ Y . The game has | Y × B | columnsand | H | rows. For all f ∈ H and all ( y, b ) ∈ Y the payoffs are( D ) f, ( y,b ) = ( g · δ if f ( y ) = 1,0 otherwise. If | Y | is not even, then we can create a new free game in which each question in | Y | appears twice. This will not change the value of the free game. The game ( − D , D ) is built in the same way as the game ( D , − D ), butwith the roles of the players swapped. That is, each column of ( − D , D )corresponds to a function that picks half of the elements of X . • The game (0 ,
0) is a game in which both players have zero matrices.Observe that the size of (
R, C ) is the same as the size of F . The game( D , − D ) has the same number of columns as C , and the number of rows isat most 2 | Y | ≤ O (log |F| ) = |F| O (1) , where we are crucially using the fact thatTheorem 2 allows us to assume that the size of Y is O (log |F| ). By the samereasoning, the number of columns in ( − D , D ) is at most |F| O (1) . Thus, thesize of G is |F| O (1) , and so this reduction is polynomial. To prove completeness, it suffices to show that, if ω ( F ) = 1, then thereexists a (1 − g · δ )-UNE of G that has social welfare 2. To do this, assumethat ω ( F ) = 1, and take a pair of optimal strategies ( s , s ) for F and turnthem into strategies for the players in G . More precisely, the row player willplace probability | X | on each answer chosen by s , and the column player willplace probability | Y | on each answer chosen by s . By construction, this givesboth players payoff 1, and hence the social welfare is 2. The harder part is toshow that this is an approximate equilibrium, and in particular, that neitherplayer can gain by playing a strategy in ( D , − D ) or ( − D , D ). We prove thisin the following lemma. Lemma 3. If ω ( F ) = 1 , then there exists a (1 − g · δ ) -UNE ( x , y ) of G with SW( x , y ) = 2 . Proof 4.
Let ( s , s ) ∈ S × S be a pair of optimal strategies for Merlin and Merlin in F . For each ( x, a ) ∈ X × A and each ( y, b ) ∈ Y × B , we define x ( x, a ) = ( | X | if s ( x ) = a . otherwise. y ( y, b ) = ( | Y | if s ( y ) = b . otherwise.Clearly, by construction, we have that the payoff to the row player under ( x , y ) is equal to p ( F , s , s ) = 1 , and therefore ( x , y ) has social welfare .On the other hand, we must prove that ( x , y ) is a (1 − g · δ ) -UNE. To doso, we will show that neither player has a deviation that increases their payoffby more than (1 − g · δ ) . We will show this for the row player; the proof for thecolumn player is symmetric. There are two types of row to consider. • First suppose that r is a row in the sub-game ( R, C ) . We claim that thepayoff of r is at most . This is because the maximum payoff in R is ,while the maximum payoff in − D is . Since the row player alreadyobtains payoff in ( x , y ) , row r cannot be a profitable deviation. Next suppose that r is a row in the sub-game ( D , − D ) . Since we have P b ∈ B y ( y, b ) = | Y | for every question y , we have that all rows in D havethe same payoff. This payoff is · (cid:18)
41 + 4 · δ (cid:19) = 21 + 4 g · δ = 2 − g · δ g · δ . Since δ ≤ and g ≤ we have
81 + 4 g · δ ≥
81 + 4 g ≥ . Thus, we have shown that the payoff of r is at most − g · δ . Thus therow player’s regret is at most − g · δ . (cid:3) We now suppose that ω ( F ) < − δ/
2, and we will prove that all (1 − g · δ )-UNE provide social welfare at most 2 − g · δ . Throughout this subsection, wewill fix ( x , y ) to be a (1 − g · δ )-UNE of G . We begin by making a simpleobservation about the amount of probability that is placed on the subgame( R, C ). Lemma 4. If SW( x , y ) > − g · δ , then • x places at least (1 − g · δ ) probability on rows in ( R, C ) , and • y places at least (1 − g · δ ) probability on columns in ( R, C ) . Proof 5.
We will prove the lemma for x ; the proof for y is entirely symmetric.For the sake of contradiction, suppose that x places strictly less than (1 − g · δ ) probability on rows in ( R, C ) . Observe that every subgame of G other than ( R, C ) is a zero-sum game. Thus, any probability assigned to these sub-gamescontributes nothing to the social welfare. On the other hand, the payoffs in ( R, C ) are at most . So, even if the column player places all probability oncolumns in C , the social welfare SW( x , y ) will be strictly less than · (1 − g · δ ) + g · δ · − g · δ , a contradiction. (cid:3) So, for the rest of this subsection, we can assume that both x and y place atleast 1 − g · δ probability on the subgame ( R, C ). We will ultimately show that,if this is the case, then both players have payoff at most 1 − · δ + mg · δ for someconstant m that will be derived during the proof. Choosing g = 1 / (2 m + 2)then ensures that both players have payoff at most 1 − g · δ , and therefore thatthe social welfare is at most 2 − g · δ .15 two-prover game. We use ( x , y ) to create a two-prover game. First, wedefine two distributions that capture the marginal probability that a question isplayed by x or y . Formally, we define a distribution x ′ over X and a distribution y ′ over Y such that for all x ∈ X and y ∈ Y we have x ′ ( x ) = P a ∈ A x ( x, a ) , and y ′ ( y ) = P b ∈ B y ( y, b ) . By Lemma 4, we can assume that k x ′ k ≥ − g · δ and k y ′ k ≥ − g · δ .Our two-prover game will have the same question sets, answer sets, andverification function as F , but a different distribution over the question sets.Let T ( x , y ) = ( X, Y, A, B, D , V ), where D is the product of x ′ and y ′ . Note thatwe have cheated slightly here, since D is not actually a probability distribution.If kDk = c <
1, then we can think of this as Arthur having a 1 − c probabilityof not sending any questions to the Merlins and awarding them payoff 0.The strategies x and y can also be used to give a us a strategy for the Merlinsin T ( x , y ) . Without loss of generality, we can assume that for each question x ∈ X there is exactly one answer a ∈ A such that x ( x, a ) >
0, because if there are twoanswers a and a such that x ( x, a ) > x ( x, a ) >
0, then we can shift allprobability onto the answer with (weakly) higher payoff, and (weakly) improvethe payoff to the row player. Since (
R, C ) is cooperative, this can only improvethe payoff of the columns in (
R, C ), and since the row player does not moveprobability between questions, the payoff of the columns in ( − D , D ) does notchange either. Thus, after shifting, we arrive at a (1 − g · δ )-UNE of G whosesocial welfare is at least as good as SW( x , y ). Similarly, we can assume that foreach question y ∈ Y there is exactly one answer b ∈ B such that y ( y, b ) > s x for Merlin in the following way. Foreach question x ∈ X , the strategy s x selects the unique answer a ∈ A suchthat x ( x, a ) >
0. The strategy s y for Merlin is defined symmetrically.We will use T ( x , y ) as an intermediary between G and F by showing thatthe payoff of ( x , y ) in G is close to the payoff of ( s x , s y ) in T ( x , y ) , and that thepayoff of ( s x , s y ) in T ( x , y ) is close to the payoff of ( s x , s y ) in F . Since we havea bound on the payoff of any pair of strategies in F , this will ultimately allowus to bound the payoff to both players when ( x , y ) is played in G . Relating G to T ( x , y ) . For notational convenience, let us define p r ( G , x , y ) and p c ( G , x , y ) to be the payoff to the row player and column player, respectively,when ( x , y ) is played in G . We begin by showing that the difference between p r ( G , x , y ) and p ( T ( x , y ) , s x , s y ) is small. Once again we prove this for the payoffof the row player, but the analogous result also holds for the column player. Lemma 5.
We have | p r ( G , x , y ) − p ( T ( x , y ) , s x , s y ) | ≤ g · δ. Proof 6.
By construction, p ( T ( x , y ) , s x , s y ) is equal to the payoff that the rowplayer obtains from the subgame ( R, C ) , and so we have p ( T ( x , y ) , s x , s y ) ≤ p r ( G , x , y ) . On the other hand, since the row player places at most g · δ probabil-ity on rows not in ( R, C ) , and since these rows have payoff at most g · δ < ,we have p r ( G , x , y ) ≤ p ( T ( x , y ) , s x , s y ) + 4 g · δ . (cid:3) elating T ( x , y ) to F . First we show that if ( x , y ) is indeed a (1 − g · δ )-UNE,then x ′ and y ′ must be close to uniform over the questions. We prove thisfor y ′ , but the proof can equally well be applied to x ′ . The idea is that, if y ′ issufficiently far from uniform, then there is set B ⊆ Y of | Y | / y ′ places significantly more than 0 . D , − D ) that corresponds to B , will have payoff at least 2, while the payoffof ( x , y ) can be at most 1 + 3 g · δ , and so ( x , y ) would not be a (1 − g · δ )-UNE.We formalise this idea in the following lemma. Define u X to be the uniformdistribution over X , and u Y to be the uniform distribution over Y . Lemma 6.
We have k u Y − y ′ k < g · δ and k u X − x ′ k < g · δ . We begin by proving an auxiliary lemma.
Lemma 7. If k u Y − y ′ k ≥ c then there exists a set B ⊆ Y of size | Y | / suchthat X i ∈ B y ′ i >
12 + c − g · δ. Proof 7.
We first define d = y ′ − u Y , and then we partition Y as follows U = { y ∈ Y : d y > | Y | } ,L = { y ∈ Y : d y ≤ | Y | } . Since k y ′ k ≥ − g · δ and k u k = 1 , we have that X y ∈ U d y ≥ c/ − g · δ, X y ∈ L d y ≤ − c/ g · δ. We will prove that there exists a set B ⊆ Y of size | Y | / such that P y ∈ B d y ≥ c/ − g · δ .We have two cases to consider, depending on the size of U . • First suppose that | U | > | Y | / . If this is the case, then there must exista set B ⊆ U with | B | = | U | / and P i ∈ B d i ≥ c/ − g · δ . We can thenadd arbitrary columns from U \ B to B in order to make | B | = | Y | / , andsince d i > for all i ∈ U , this cannot decrease P i ∈ B d i . Thus, we havecompleted the proof for this case. • Now suppose that | U | ≤ | Y | / . If this is the case, then there must exist aset C ⊆ L with | C | = | L | / and P i ∈ C d i ≥ − c + g · δ . So, let C ′ ⊆ C bean arbitrarily chosen subset such that | C ′ | + | U | = | Y | / . This is possiblesince | L | = | Y | − | U | and hence | L | / | Y | / − | U | / , which implies that L | / | U | > | Y | / . Setting B = C ′ ∪ U therefore gives us a set with | B | = | Y | / such that X i ∈ B d i ≥ ( c/ − g · δ ) − ( c/ g · δ )= c/ − · g · δ . So we have completed the proof of this case, and the lemma as a whole. (cid:3)
We can now proceed with the proof of Lemma 6.
Proof 8 (Proof of Lemma 6).
Suppose, for the sake of contradiction thatone of these two properties fails. Without loss of generality, let us assume that k u Y − y ′ k ≥ c . We will show that the row player can gain more than in pay-off by deviating to a new strategy, which will show that ( x , y ) is not a -UNE,contradicting our assumption that it is a (1 − g · δ ) -UNE.By assumption, x places at least − g · δ probability on rows in ( R, C ) . Themaximum payoff in R is , and the maximum payoff in − D is . On the onehand, the rows in D give payoff at most / (2 + g · δ ) ≤ . So the row player’spayoff under ( x , y ) is bounded by (1 − g · δ ) · g · δ ) · g · δ. On the other hand, we can apply Lemma 7 with c = 16 g · δ to find a set B ⊆ Y such that X i ∈ B y ′ ( i ) >
12 + 16 g · δ − g · δ. = 12 + 2 g · δ = 1 + 4 g · δ . So, let r B be the row of D that corresponds to B . This row has payoff g · δ for every entry in B . So, the payoff of row r B must be at least (cid:18) g · δ (cid:19) · (cid:18)
41 + 4 g · δ (cid:19) = 2 . Thus, the row player can deviate to r B and increase his payoff by at least − g · δ ,and ( x , y ) is not a (1 − g · δ ) -UNE. (cid:3) With Lemma 6 at hand, we can now prove that the difference between p ( T ( x , y ) , s x , s y ) and p ( F , s x , s y ) must be small. This is because the questiondistribution D used in T ( x , y ) is a product of two distributions that are close touniform, while the question distribution U used in F is a product of two uniformdistributions. In the following lemma, we show that if we transform D into U ,then we do not change the payoff of ( s x , s y ) very much.18 emma 8. We have | p ( T ( x , y ) , s x , s y ) − p ( F , s x , s y ) | ≤ g · δ. Proof 9.
The distribution used in F is the product of u Y and u X , while thedistribution used in T ( x , y ) is the product of y ’ and x ′ . Furthermore, Lemma 6tells us that k u Y − y ′ k < g · δ and k u X − x ′ k < g · δ . Our approach is totransform u X to x ′ while bounding the amount that p ( F , s x , s y ) changes. Oncewe have this, we can apply the same transformation to u Y and y ′ .Consider the effect of shifting probability from a question x ∈ X to a differ-ent question x ∈ X . Since all entries of V are in { , } , if we shift q probabilityfrom x to x , then p ( F , s x , s y ) can change by at most q . This bound also holdsif we remove probability from x without adding it to x (which we might do since k x k may not be .) Thus, if we shift probability to transform u X into x ′ , thenwe can change p ( F , s x , s y ) by at most g · δ .The same reasoning holds for transforming u Y into y ′ . This means that wecan transform F to T ( x , y ) while changing the payoff of ( s x , s y ) by at most g · δ ,which completes the proof. (cid:3) Completing the soundness proof.
The following lemma uses the boundsderived in Lemmas 5 and 8, along with a suitable setting for g , to bound thepayoff of both players when ( x , y ) is played in G . Lemma 9. If g = , then both players have payoff at most − g · δ when ( x , y ) is played in G . Proof 10.
Lemmas 5 and 8 tell us that | p r ( G , x , y ) − p ( T ( x , y ) , s x , s y ) | ≤ g · δ, | p ( T ( x , y ) , s x , s y ) − p ( F , s x , s y ) | ≤ g · δ. Hence, we have | p r ( G , x , y ) − p ( F , s x , s y ) | ≤ g · δ . However, we know that p ( F , s x , s y ) ≤ − δ/ . So, if we set g = , then we we will have that p r ( G , x , y ) ≤ − · δ + 68138 · δ = 1 − · δ = 1 − g · δ. (cid:3) Hence, we have proved that SW( x , y ) ≤ − g · δ . We can now state the theorem that we have proved in this section. We firstrescale the game so that it lies in [0 , G is g · δ ≤ − g · δ ≥ −
4. To rescale this game, we add 4 to allthe payoffs, and then divide by 8. Let us refer to the scaled game as G s . Observethat an ǫ -UNE in G is a ǫ -NE in G s since adding a constant to all payoffs doesnot change the approximation guarantee, but dividing all payoffs by a constantdoes change the approximation guarantee. So, we have the following theorem.19 heorem 3. If ETH holds, then there exists a constant δ below which the prob-lem ( − g · δ ) -NE ( g · δ ) -SW, where g = , requires n e Ω(log n ) time. Proof 11.
By Lemmas 3 and 9, we have • if ω ( F ) = 1 then there exists a (1 − g · δ ) -UNE of G with social welfare . In the rescaled game this translates to a ( − g · δ ) -NE of G s withsocial welfare + = . • if ω ( F ) < − δ then all (1 − g · δ ) -UNE of G have social welfare at most (1 − g · δ ) + (1 − g · δ ) = 2 − g · δ . After rescaling, we have that all ( − g · δ ) -NE of G s have social welfare social welfare at most − g · δ − g · δ − g · δ By Theorem 2, assuming ETH we require |F| e Ω(log |F| ) time to decide whetherthe value of F is or − δ for some small constant δ . Thus, we also require n e Ω(log | n | ) to solve the problem ( − g · δ ) -NE ( g · δ ) -SW. (cid:3)
5. Hardness results for other decision problems
In this section we study a range of decision problems associated with approx-imate equilibria. Table 1 shows all of the decision problems that we consider.Most are known to be NP -complete for the case of exact Nash equilibria [2, 3].For each problem in Table 1, the input includes a bimatrix game and a qualityof approximation ǫ ∈ (0 , ǫ -NEand ǫ -WSNE. Since ǫ -NE is a weaker solution concept than ǫ -WSNE, i.e., every ǫ -WSNE is an ǫ -NE, the hardness results for ǫ -NE imply the same hardnessfor ǫ -WSNE. We consider problems for ǫ -WNSE only where the correspondingproblem for ǫ -NE is trivial. For example, observe that deciding if there is an ǫ -NE with large support is a trivial problem, since we can always add a tinyamount of probability to each pure strategy without changing our expectedpayoff very much.Our conditional quasi-polynomial lower bounds will hold for all ǫ < , so letus fix ǫ ∗ < for the rest of this section. Using Theorem 3, we compute from ǫ ∗ the parameters n and δ that we require to apply Theorem 3. In particular, set δ ∗ to solve ǫ ∗ = ( − g · δ ∗ ), and choose n ∗ as δ ∗ . Then, for n > n ∗ and δ = δ ∗ wecan apply Theorem 3 to bound the social welfare achievable if ω ( F ) < − δ ∗ as u := 108 − δ ∗ . Theorem 3 implies that in order to decide whether the game G s possesses an ǫ ∗ -NE that yields social welfare strictly greater than u requires n ˜ O (log n ) time, where δ no longer appears in the exponent since we have fixed it as the constant δ ∗ .20roblem 1 asks to decide whether a bimatrix game possesses an ǫ ∗ -NE wherethe expected payoff for each player is at least u , where u is an input to theproblem. When we set u = , the conditional hardness of this problem is animmediate corollary of Theorem 3.For Problems 2 - 9, we use G s to construct a new game G ′ , which adds onerow i and one column j to G s . The payoffs are defined using the constants u and ǫ ∗ , as shown in Figure 1. G ′ = j , + ǫ ∗ G s ...0 , + ǫ ∗ i + ǫ ∗ , · · · + ǫ ∗ , , Figure 1: The game G ′ .In G ′ , the expected payoff for the row player for i is at least + ǫ ∗ irrespec-tive of the column player’s strategy. Similarly, the expected payoff for j is atleast + ǫ ∗ irrespective of the row player’s strategy. This means that: • If G s possesses an ǫ ∗ -NE with social welfare , then G ′ possesses at leastone ǫ ∗ -NE where the players do not play the pure strategies i and j . • If every ǫ ∗ -NE of G s yields social welfare at most u , then in every ǫ ∗ -NE of G ′ , the players place almost all of their probability on i and j respectively.Note that ( i , j ) is a pure exact Nash equilibrium.Problem 2 asks whether a bimatrix game possesses an ǫ -NE where the rowplayer plays with positive probability only strategies in a given set S . Let S R ( S C ) denote the set of pure strategies available to the row (column) player fromthe subgame ( R, C ) of G s . To show the hardness of Problem 2, we will set weset S = S R .Recall that G s is created from F . First, we prove in Lemma 10 that if ω ( F ) = 1,then G ′ possesses an ǫ ∗ -NE such that the answer to Problem 2 is “Yes”. Notethat we actually argue in Lemma 10 about the existence of an ǫ ∗ -WSNE, sincethis stronger claim will be useful when we come to deal with Problems 7 - 9. Lemma 10. If ω ( F ) = 1 , then G ′ possesses an ǫ ∗ -WSNE ( x , y ) such that supp( x ) ⊆ S R . Under ( x , y ) , both players get payoff , so SW( x , y ) = .Moreover, | supp( x ) | = | X | and max i x i ≤ | X | , where X is the question set of Merlin in F . Proof 12.
The proof of Lemma 3 shows that, if ω ( F ) = 1 , then G possesses an ǫ ∗ -WSNE ( x , y ) where the expected payoff for each player is and supp( x ) ⊆ S R . he reason that ( x , y ) is well supported is that all rows in supp( x ) have equalexpected payoff. Moreover, x is a uniform mixture over a pure strategy set ofsize | X | , where X is the question set of Merlin in F . Since G s is obtainedfrom G by adding 4 to the payoffs and dividing by 8, ( x , y ) is as an ǫ ∗ -NE in G s where each player has payoff . To complete the proof we show that ( x , y ) in an ǫ ∗ -NE for G ′ , which is the same as G s apart from the additional pure strategies i and j . Since i and j yield payoff + ǫ ∗ , but not more, the claim holds. (cid:3) Next we prove that if ω ( F ) < − δ ∗ , then the answer to Problem 2 is “No”. Lemma 11. If ω ( F ) < − δ ∗ , then in every ǫ ∗ -NE ( x , y ) of G ′ it holds that x i > − ǫ ∗ − ǫ ∗ and y j > − ǫ ∗ − ǫ ∗ . Proof 13.
Let G s := ( P, Q ) and suppose that ( x , y ) is an ǫ ∗ -NE of G ′ . FromTheorem 3 we know that if ω ( F ) < − δ ∗ , then in any ǫ ∗ -NE of G s we havethat each player gets payoff at most u < . Under ( x , y ) in G ′ the row playergets payoff x T P y < (1 − x i ) · (1 − y j ) ·
58 + x i · (1 − y j )( 58 + ǫ ∗ ) + x i · y j = x i · ((1 − y j ) · ǫ ∗ + y j ) + (1 − y j ) · . From the pure strategy i , the row player gets P i · y = (1 − y j )( 58 + ǫ ∗ ) + y j . In order for ( x , y ) to be an ǫ ∗ -NE it must hold that x T P y ≥ P i y − ǫ ∗ . Usingthe upper bound on x T P y that we just derived, we get: x i > − ǫ ∗ (1 − y j ) · ǫ ∗ + y j . (1) By symmetry, we also have that the column player must play j with probability: y j > − ǫ ∗ (1 − x i ) · ǫ ∗ + x i . (2) Recall that in this section ǫ ∗ is a constant. Observe that the right-hand sideof (2) is increasing in x i , and we can thus use it to replace x i in (2) as follows: y j > − ǫ ∗ (1 − ǫ ∗ (1 − y j ) ǫ ∗ + y j ) ǫ ∗ + 1 − ǫ ∗ (1 − y j ) ǫ ∗ + y j = 1 − ǫ ∗ ǫ ∗ (1 − y j ) ǫ ∗ + y j + 1 − ǫ ∗ (1 − y j ) ǫ ∗ + y j = 1 − (1 − y j ) ǫ ∗ + y j ǫ ∗ ǫ ∗ + (1 − y j ) ǫ ∗ + y j − ǫ ∗ . oting that ( ǫ ∗ + (1 − y j ) ǫ ∗ + y j − ǫ ∗ ) ≥ , by rearranging we get that y j (1 − ǫ ∗ ) + y j (2 ǫ ∗ − > . Then, since ǫ ∗ < , we have − ǫ ∗ > , and we get that y j > − ǫ ∗ − ǫ ∗ = 1 − ǫ ∗ − ǫ ∗ . By symmetry, we have x i > − ǫ ∗ − ǫ ∗ , which completes the proof. (cid:3) Next we recall Problems 3 and 4 and we show that, as for Problem 2, Lem-mas 10 and 11 can also be used to immediately show that there are instancesof these decision problems where the answer is “Yes” if and only if ω ( F ) = 1.Given two probability distributions x and x ′ , the Total Variation (TV) dis-tance between them is max i {| x i − x ′ i |} . We define the TV distance between twostrategy profiles ( x , y ) and ( x ′ , y ′ ) to be the maximum over the TV distance of x and x ′ and the TV distance of y and y ′ . Problem 3 asks whether a bimatrixgame possesses two ǫ -NEs with TV distance at least d . In order to apply Lem-mas 10 and 11, we will set d = 1 − ǫ ∗ − ǫ ∗ . Then an instance G ′ of Problem 3 is“Yes” when ω ( F ) = 1 since the ǫ ∗ -NE ( x , y ) identified in Lemma 10, has TVdistance one from the pure exact Nash equilbrium ( i , j ). Lemma 11 says that,if ω ( F ) < − δ ∗ , every ǫ ∗ -NE ( x , y ) of G ′ has x i > − ǫ ∗ − ǫ ∗ and so all ǫ ∗ -NEare within TV distance 1 − ǫ ∗ − ǫ ∗ of each other.Problem 4 asks to decide whether there exists an ǫ -NE where the row playerdoes not play any pure strategy with probability more than p . For this problem,we set p = | X | , where X is the question set for Merlin . According Lemma 10,if ω ( F ) = 1, then an instance G ′ of Problem 4 is a “Yes”. Lemma 11 says that,if ω ( F ) < − δ ∗ , then for all ǫ ∗ -NE ( x , y ) of G ′ , max i x i ≥ x i > − ǫ ∗ − ǫ ∗ > | X | .Problem 5 asks whether a bimatrix game possesses an ǫ -NE with socialwelfare at most v , and Problem 6 asks whether a bimatrix game possesses an ǫ -NE where the expected payoff of the row player is at most u . We fix v = for Problem 5, and for Problem 6 we fix u = . As we have already explainedin the proof of Lemma 10, if ω ( F ) = 1, then there is an ǫ ∗ -NE for G ′ such thatthe expected payoff for each player is and thus the social welfare is . So, if ω ( F ) = 1, then the answer to Problems 5 and 6 is “Yes”. On the other hand,from the proof of Lemma 11 we know that if ω ( F ) < − δ ∗ , then in any ǫ ∗ -NEof G ′ both players play the strategies i and j with probability at least 1 − ǫ ∗ − ǫ ∗ .So, each player gets payoff at least (1 − ǫ ∗ − ǫ ∗ ) > , since ǫ ∗ < , from theirpure strategies i and j . So, if ω ( F ) < − δ ∗ , then the answer to Problems 5and 6 is “No”.Problems 7 - 9 relate to deciding if there exist approximate well-supported equilibria with large supports (for ǫ -NE these problems would be trivial). Prob-lem 7 asks whether a bimatrix game possesses an ǫ -WSNE ( x , y ) with | supp( x ) | + | supp( y ) | ≥ k . Problem 8 asks whether a bimatrix game possesses an ǫ -WSNE ( x , y ) with min {| supp( x ) | , | supp( y ) |} ≥ k . Problem 9 asks whether23 bimatrix game possesses an ǫ -WSNE ( x , y ) with | supp( x ) | ≥ k . Recall that X and Y are the question sets of Merlin and Merlin respectively that were usedto define F and in turn G s . We will fix k = | X | = | Y | for all three problems.If ω ( F ) = 1, then Lemma 10 says that there exists an ǫ ∗ -WSNE ( x , y ) for G ′ such that | supp( x ) | = | supp( y ) | = k and thus the answer to Problems 7 - 9 is“Yes”. On the other hand, if ω ( F ) < − δ , then we will prove that there is aunique ǫ ∗ -WSNE where the row player plays only the pure strategy i and thecolumn player plays the pure strategy j . Lemma 12. If ω ( F ) < − δ ∗ , then there is a unique ǫ ∗ -WSNE ( x , y ) in G ′ such that x i = 1 and y j = 1 . Proof 14.
We consider only the case that ω ( F ) < − δ ∗ . Then Lemma 11says that in every ǫ ∗ -NE of G ′ the column player plays the pure strategy j withprobability at least − ǫ ∗ − ǫ ∗ . Against j , the row player gets for all pure strategies i = i and for i . Thus, in any ǫ ∗ -NE of G ′ , for every pure strategy i = i , therow player gets at most ǫ ∗ − ǫ ∗ from every pure strategy i , and the row playergets at least − ǫ ∗ − ǫ ∗ from i . So, in every ǫ ∗ -WSNE the row player must playonly the pure strategy i since from every other pure strategy the player suffersregret at least − ǫ ∗ − ǫ ∗ , which is strictly larger than ǫ ∗ for every ǫ ∗ < . Inturn, against i , every pure strategy j = j for the column player yields zero payoffwhile the strategy j yields payoff 1. So, the unique ǫ ∗ -WSNE of G ′ is x i = 1 and y j = 1 . (cid:3) Hence, when ω ( F ) < − δ ∗ the answer to Problems 7 - 9 is “No”. Thus, wehave shown the following: Theorem 4.
Assuming ETH, any algorithm that solves the Problems 2 - 9 forany constant ǫ < requires n ˜Ω(log n ) time. Finally, for Problem 10, we define a new game G ′′ by extending G ′ . We addthe new pure strategies i ′ for the row player and j ′ for the column player. Thepayoffs are shown in Figure 2. Recall that Problem 10 asks whether a bimatrixgame possesses an ǫ -WSNE such that every strategy from a given set S is playedwith positive probability.In order to prove our result we fix S = i ′ . First, we prove that if ω ( F ) = 1then the game G ′′ possesses an ǫ ∗ -WSNE ( x , y ) such that i ′ ∈ supp( x ). Then weprove that if ω ( F ) < − δ , then for any ǫ ∗ -WSNE ( x , y ) it holds that i ′ / ∈ supp( x ). Lemma 13. If ω ( F ) = 1 , then G ′′ possesses an ǫ ∗ -WSNE ( x , y ) such that i ′ ∈ supp( x ) . Proof 15.
Lemma 10 says that if ω ( F ) = 1 , then G ′ possesses an ǫ ∗ -WSNE ( x ′ , y ′ ) that gives payoff for each player, and x ′ is uniform on a set of size | X | .We construct the required ǫ ∗ -WSNE of G ′′ from ( x ′ , y ′ ) as follows. We add i ′ to the support of x ′ so that x is a uniform mixture over supp( x ′ ) ∪ i ′ . For thecolumn player, we extend y ′ by adding zero probability for j ′ . ′′ = j ′ , G ′ ... , i ′ , · · · , , Figure 2: The game G ′′ . Against y , pure strategies in supp( x ′ ) give payoff , pure strategy i in G ′ yields payoff + ǫ ∗ , and i ′ gives payoff . Thus, since ( x ′ , y ′ ) is an ǫ ∗ -WSNE of G ′ , x has pure regret at most ǫ ∗ against y , as required. What remains is to showthat the pure regret of y is no more than ǫ ∗ against x . Recall that, in G ′ , against x ′ , the payoff of each pure strategy in supp( y ′ ) is . Now consider G ′′ . Since,against i ′ , the column player gets for all j ∈ supp( y ) , the column player stillgets against x for all j ∈ supp( y ) . Moreover, against x , the payoff of j ′ is | X || X | +1 · < . Thus, since ( x ′ , y ′ ) is an ǫ ∗ -WSNE of G ′ , we have that ( x , y ) isan ǫ ∗ -WSNE of G ′′ with i ′ ∈ supp( x ) , which completes the proof. (cid:3) Lemma 14. If ω ( F ) < − δ ∗ , then for any ǫ ∗ -WSNE ( x , y ) of G ′′ it holds that i ′ / ∈ supp( x ) . Proof 16.
We prove that the unique ǫ ∗ -WSNE of G ′′ is the pure profile ( i , j ) .Using exactly the same arguments as in the proof of Lemma 11 we can provethat if ω ( F ) < − δ ∗ , then in any ǫ ∗ -NE of G ′′ it holds that x i > − ǫ ∗ − ǫ ∗ and y j > − ǫ ∗ − ǫ ∗ . Then, using exactly the same arguments as in Lemma 12 we canget that the pure strategy j for the column player yields payoff at least − ǫ ∗ − ǫ ∗ while any other pure strategy, including j ′ , yields payoff at most ǫ ∗ − ǫ ∗ . Hence, inany ǫ ∗ -WSNE of G ′′ the column player must play only the pure strategy j . Then,in order to be in an ǫ ∗ -WSNE the row player must play the pure strategy i . Ourclaim follows. (cid:3) The combination of Lemmas 13 and 14 gives the following theorem.
Theorem 5.
Assuming the ETH, any algorithm that solves the Problem 10 forany constant ǫ < requires n ˜Ω(log n ) time. Acknowledgements.
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