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The Computational Complexity of Nash Equilibria in Concisely Represented Games?

Grant R. Schoenebeck

?

Salil P. Vadhan?

May 4, 2005

Abstract Games may be represented in many di?erent ways, and di?erent representations of games a?ect the complexity of problems associated with games, such as ?nding a Nash equilibrium. The traditional method of representing a game is to explicitly list all the payo?s, but this incurs an exponential blowup as the number of agents grows. We study two models of concisely represented games: circuit games, where the payo?s are computed by a given boolean circuit, and graph games, where each agent’s payo? is a function of only the strategies played by its neighbors in a given graph. For these two models, we study the complexity of four questions: determining if a given strategy is a Nash equilibrium, ?nding a Nash equilibrium, determining if there exists a pure Nash equilibrium, and determining if there exists a Nash equilibrium in which the payo?s to a player meet some given guarantees. In many cases, we obtain tight results, showing that the problems are complete for various complexity classes.

1

Introduction

In recent years, there has been a surge of interest at the interface between computer science and game theory. On one hand, game theory and its notions of equilibria provide a rich framework for modelling the behavior of sel?sh agents in the kinds of distributed or networked environments that often arise in computer science, and o?er mechanisms to achieve e?cient and desirable global outcomes in spite of the sel?sh behavior. On the other hand, classical game theory ignores computational considerations, and it is unclear how meaningful game-theoretic notions of equilibria are if they are infeasible to compute. Finally, game-theoretic characterizations of complexity classes have proved to be extremely useful even in addressing questions that a priori have nothing to do with games, of particular note being the work on interactive proofs and their applications to cryptography and hardness of approximation [GMR89, GMW91, FGL+ 96, AS98, ALM+ 98]. While the recent work at this interface has been extremely fruitful, some of the most basic questions remain unanswered. In particular, one glaring open question (posed, for example, in [Pap01]) is whether there exists a polynomial-time algorithm to ?nd Nash equilibria in standard, two-player “bimatrix” games. (Recall that a Nash equilibrium speci?es randomized strategies for both players so that neither can increase his/her payo? by deviating from the strategy. The

Many of these results have appeared in the ?rst author’s undergraduate thesis. E-mail: schoeneb@post.harvard.edu. Supported by NSF grant CCR-0133096. ? Division of Engineering and Applied Sciences, Harvard University, 33 Oxford Street, Cambridge, MA, 02138, USA. E-mail: salil@eecs.harvard.edu. Work done in part while a Fellow at the Radcli?e Institute for Advanced Study. Also supported by NSF grant CCR-0133096 and a Sloan Research Fellowship.

? ?

1

fundamental result of Nash [Nas51] is that every game (even with many players) has such an equilibrium.) This two-player Nash equilibrium problem is known to be P-hard [FIKU04], and cannot be NP-hard unless NP = coNP [MP91]. The known algorithms are exponential time, though recently a quasipolynomial-time algorithm has been given for ?nding approximate Nash equilibria [LMM03] with respect to additive error = 1/polylog(n). Given that characterizing the complexity of Nash equilibria problem in two-player games has resisted much e?ort, it is natural to investigate the computational complexity of Nash equilibria in other types of games. In particular, n-player games where each player has only a small (e.g. a constant) number of strategies is potentially easier than two-player games with large strategy spaces. However, in n-player games, the representation of the game becomes an important issue. In particular, explicitly describing an n-player game in which each player has two strategies requires an exponentially long representation (consisting of N = n · 2n payo? values) and complexity of this problem is more natural for games given by some type of concise representation, such as the graph games recently proposed by Kearns, Littman, and Singh [KLS01]. Motivated by the above considerations, we undertake a systematic study of the complexity of Nash equilibria in games given by concise representations. We focus on two types of concise representations. The ?rst are circuit games, where the game is speci?ed by a boolean circuit computing the payo?s. Circuit games were previously studied in the setting of two-player zerosum games, where computing (resp., approximating) the “value” of such a game is shown to be EXP-complete [FKS95] (resp., S2 P-complete [FIKU04]). They are a very general model, capturing essentially any representation in which the payo?s are e?ciently computable. The second are the graph games of Kearns, Littman, and Singh [KLS01], where the game is presented by a graph whose nodes are the players and the payo?s of each player are a function only of the strategies played by each player’s neighbor. (Thus, if the graph is of low degree, the payo? functions can be written very compactly). Kearns et al. showed that if the graph is a tree and each player has only two strategies, then approximate Nash equilibria can be found in polynomial time. Gotlobb, Greco, and Scarcello [GGS03] recently showed that the problem of deciding if a degree-4 graph game has a pure-Nash equilibrium is NP-complete. In these two models (circuit games and graph games), we study 4 problems: 1. IsNash: Given a game G and a randomized strategy pro?le θ, determine if θ is a Nash equilibrium in G, 2. ExistsPureNash: Given a game G, determine if G has a pure (i.e. deterministic) Nash equilibrium, 3. FindNash: Given a game G, ?nd a Nash equilibrium in G, and 4. GuaranteeNash: Given a game G, determine whether G has a Nash equilibrium that achieves certain payo? guarantees for each player. (This problem was previously studied by [GZ89, CS03], who showed it to be NP-complete for two-player, bimatrix games.) We study the above four problems in both circuit games and graphical games, in games where each player has only two possible strategies and in games where the strategy space is unbounded, in nplayer games and in 2-player games, and with respect to approximate Nash equilibria for di?erent levels of approximation (exponentially small error, polynomially small error, and constant error). Our results include: ? A tight characterization of the complexity of all of the problems listed above except for FindNash, by showing them to be complete for various complexity classes. This applies to all 2

of their variants (w.r.t. concise representation, number of players, and level of approximation). For the various forms of FindNash, we give upper and lower bounds that are within one nondeterministic quanti?er of each other. ? A general result showing that n-player circuit games in which each player has 2 strategies are a harder class of games than standard two-player bimatrix games (and more generally, than the graphical games of [KLS01]), in that there is a general reduction from the latter to the former which applies to most of the problems listed above. Independent Results. Several researchers have independently obtained some results related to ours. Speci?cally, Daskalakis and Papadimitriou [DS04] give complexity results on concisely represented graphical games where the graph can be exponentially large (whereas we always consider the graph to be given explicitly), and Alvarez, Gabarro, and Serna [AGS05] give results on ExistsPureNash that are very similar to ours. Organization. We de?ne game theoretic terminology and ?x a representation of strategy pro?les in Section 2. Section 3 contains formal de?nitions of the concise representations and problems that we study. Section 4 looks at relationships between these representations. Sections 5 through 8 contain the main complexity results on IsNash, ExistsPureNash, FindNash, and GuaranteeNash.

2

Background and Conventions

Game Theory A game G = (s, ν) with n agents, or players, consists of a set s = s1 × · · · × sn where si is the strategy space of agent i, and a valuation or payo? function ν = ν1 × . . . × νn where νi : s → R is the valuation function of agent i. Intuitively, to “play” such a game, each agent i picks a strategy si ∈ si , and based on all players’ choices realizes the payo? νi (s1 , . . . , sn ). For us, si will always be ?nite and the range of νi will always be rational. An explicit representation of a game G = (s, ν) is composed of a list of each si and an explicit encoding of each νi . This encoding of ν consists of n · |s| = n · |s1 | · · · |sn | rational numbers. An explicit game with exactly two players is call a bimatrix game because the payo? functions can be represented by two matrices, one specifying the values of ν1 on s = s1 × s2 and the other specifying the values of ν2 . A pure strategy for an agent i is an element of si . A mixed strategy θi , or simply a strategy, for a player i is a random variable whose range is si . The set of all strategies for player i will be denoted Θi . A strategy pro?le is a sequence θ = (θ1 , . . . , θn ), where θi is a strategy for agent i. We will denote the set all strategy pro?les Θ. ν = ν1 × · · · × νn extends to Θ by de?ning ν(θ) = Es←θ [ν(s)]. A pure-strategy pro?le is a strategy pro?le in which each agent plays some pure-strategy with probability 1. A k-uniform strategy pro?le is a strategy pro?le where each agent randomizes uniformly between k, not necessarily unique, pure strategies. The support of a strategy (or of a strategy pro?le) is the set of all pure-strategies (or of all pure-strategy pro?les) played with nonzero probability. We de?ne a function Ri : Θ × Θi → Θ that replaces the ith strategy in a strategy pro?le θ by a di?erent strategy for agent i, so Ri (θ, θi ) = (θ1 , . . . , θi , . . . , θn ). This diverges from conventional notation which writes (θ?i , θi ) instead of Ri (θ, θi ). Given a strategy pro?le θ, we say agent i is in equilibrium if he cannot increase his expected payo? by playing some other strategy (giving what the other n ? 1 agents are playing). Formally agent i is in equilibrium if νi (θ) ≥ νi (Ri (θ, θi )) for all θi ∈ Θi . Because Ri (θ, θi ) is a distribution 3

over Ri (θ, si ) where si ∈ si and νi acts linearly on these distributions, Ri (θ, θi ) will be maximized by playing some optimal si ∈ si with probability 1. Therefore, it su?ces to check that νi (θ) ≥ νi (Ri (θ, si )) for all si ∈ si . For the same reason, agent i is in equilibrium if and only if each strategy in the support of θi is an optimal response. A strategy pro?le θ is a Nash equilibrium [Nas51] if all the players are in equilibrium. Given a strategy pro?le θ, we say player i is in equilibrium if νi (Ri (θ, si )) ≤ νi (θ) + for all si ∈ si . A strategy pro?le θ is an -Nash equilibrium if all the players are in -equilibrium. A pure-strategy Nash equilibrium (respectively, a pure-strategy -Nash equilibrium) is a pure-strategy pro?le which is a Nash equilibrium (respectively, an -Nash equilibrium). Pennies is a 2-player game where s1 = s2 = {0, 1}, and the payo?s are as follows: Player 2 Heads Tails (1, 0) (0, 1) (0, 1) (1, 0)

Player 1

Heads Tails

The ?rst number in each ordered pair is the payo? of player 1 and the second number is the payo? to player 2. Pennies has a unique Nash equilibrium where both agents randomize uniformly between their two strategies. In any -Nash equilibrium of 2-player pennies, each player randomizes between each 1 strategy with probability 2 ± 2 (see Appendix A for details). Complexity Theory A promise-language L is a pair (L+ , L? ) such that L+ ? Σ? , L? ? Σ? , and L+ ∩L? = ?. We call L+ the positive instances, and L? the negative instances. An algorithm decides a promise-language if it accepts all the positive instances and rejects all the negative instances. Nothing is required of the algorithm if it is run on instances outside L+ ∪ L? . Because we consider approximation problems in this paper, which are naturally formulated as promise languages, all complexity classes used in this paper are classes of promise problems. We refer the reader to the recent survey of Goldreich [Gol05] for about the usefulness and subtleties of working with promise problems. A search problem, is speci?ed by a relation R ? Σ? × Σ? where given an x ∈ Σ? we want to either compute y ∈ Σ? such that (x, y) ∈ R or say that no such y exists. When reducing to a search problem via an oracle, it is required that any valid response from the oracle yields a correct answer.

3

Concise Representations and Problems Studied

We now give formal descriptions of the problems which we are studying. First we de?ne the two di?erent representations of games. De?nition 3.1 A circuit game is a game G = (s, ν) speci?ed by integers k1 , . . . , kn and circuits C1 , . . . , Cn such that si ? {0, 1}ki and Ci (s) = νi (s) if si ∈ si for all i or Ci (s) = ⊥ otherwise. In a game G = (s, ν), we write i ∝ j if ?s ∈ s, si ∈ si such that νj (s) = νj (Ri (s, si )). Intuitively, i ∝ j if agent i can ever in?uence the payo? of agent j. De?nition 3.2 [KLS01] A graph game is a game G = (s, ν) speci?ed by a directed graph G = (V, E) where V is the set of agents and E ? {(i, j) : i ∝ j}, the strategy space s, and explicit 4

representations of the function νj for each agent j de?ned on the domain (i,j)∈E si , which encodes the payo?s. A degree-d graph game is a graph game where the in-degree of the graph G is bounded by d. This de?nition was proposed in [KLS01]. We change their de?nition slightly by using directed graphs instead of undirected ones (this only changes the constant degree bounds claimed in our results). Note that any game (with rational payo?s) can be represented as a circuit game or a graph game. However, a degree-d graph game can only represent games where no one agent is in?uenced directly by the strategies of more than d other agents. A circuit game can encode the games where each player has exponentially many pure-strategies in a polynomial amount of space. In addition, unlike in an explicit representation, there is no exponential blow-up as the number of agents increases. A degree-d graph game, where d is constant, also avoids the exponential blow-up as the number of agents increases. For this reason we are interested mostly in bounded-degree graph games. We study two restrictions of games. In the ?rst restriction, we restrict a game to having only two players. In the second restriction, we restrict each agent to having only two strategies. We will refer to games that abide by the former restriction as 2-player, and to games that abide by the latter restriction as boolean. If we want to ?nd a Nash equilibrium, we need an agreed upon manner in which to encode the result, which is a strategy pro?le. We represent a strategy pro?le by enumerating, by agent, each pure strategy in that agent’s support and the probability with which the pure strategy is played. Each probability is given as the quotient of two integers. This representation works well in bimatrix games, because the following proposition guarantees that for any 2-player game there exists Nash equilibrium that can be encoded in reasonable amount of space. Proposition 3.3 Any 2-player game with rational payo?s has a rational Nash equilibrium where the probabilities are of bit length polynomial with respect to the number of strategies and bit-lengths of the payo?s. Furthermore, if we restrict ourselves to Nash equilibria θ where νi (θ) ≥ gi for i ∈ {1, 2} where each guarantee gi is a rational number then either 1) there exists such a θ where the probabilities are of bit length polynomial with respect to the number of strategies and bit-lengths of the payo?s and the bit lengths of the guarantees or 2) no such θ exists. Proof Sketch: If we are given the support of some Nash equilibrium, we can set up a polynomially sized linear program whose solution will be a Nash equilibrium in this representation, and so it is polynomially sized with respect to the encoding of the game. (Note that the support may not be easy to ?nd, so this does not yield a polynomial time algorithm). If we restrict ourselves to Nash equilibria θ satisfying νi (θ) ≥ gi as in the proposition, this merely adds additional constraints to the linear program. This proposition implies that for any bimatrix game there exists a Nash equilibrium that is at most polynomially sized with respect to the encoding of the game, and that for any 2-player circuit game there exists a Nash equilibrium that is at most exponentially sized with respect to the encoding of the game. However, there exist 3-player games with rational payo?s that have no Nash equilibrium with all rational probabilities [NS50]. Therefore, we cannot hope to always ?nd a Nash equilibrium in 5

this representation. Instead we will study -Nash equilibrium when we are not restricted to 2-player games. The following result from [LMM03] states that there is always an -Nash equilibrium that can be represented in a reasonable amount of space. Theorem 3.4 [LMM03] Let θ be a Nash equilibrium for a n-player game G = (s, ν) in which all 2 2 the payo?s are between 0 and 1, and let k ≥ n log(n maxi |si |) . Then there exists a k-uniform -Nash 2 equilibrium θ where |νi (θ) ? νi (θ )| ≤ 2 for 1 ≤ i ≤ n. Recall that a k-uniform strategy pro?le is a strategy pro?le where each agent randomizes uniformly between k, not necessarily unique, pure strategies. The number of bits needed to represent such a strategy pro?le is O(( i min{k, |si |}) · log k). Thus, Theorem 3.4 implies that for any that for any n-player game (g1 , . . . , gn ) = (s, ν) in which all the payo?s are between 0 and 1, there exists an -Nash equilibrium of bit-length poly(n, 1/ , log(maxi |si |)). There also is an -Nash equilibrium of bit-length poly(n, log(1/ ), maxi |si |). We want to study the problems with and without approximation. All the problems that we study will take as an input a parameter related to the bound of approximation. We de?ne four types of approximation: 1a) Exact Fix = 0 in the de?nition of the problem.

1 2

1b) Exp-Approx input 2) Poly-Approx input 3) Const-Approx Fix

≥ 0 as a rational number encoded as the quotient of two integers. > 0 as 1k where = 1/k

> 0 in the de?nition of the problem.

With all problems, we will look at only 3 types of approximation. Either 1a) or 1b) and both 2 and 3. With many of the problems we study, approximating using 1a) and 1b) yield identical problems. Since the notion of -Nash equilibrium is with respect to additive error, the above notions of approximation refer only to games whose payo?s are between 0 and 1 (or are scaled to be such). Therefore we assume that all games have payo?s which are between 0 and 1 unless otherwise explicitly stated. Many times our constructions of games use payo?s which are not between 0 and 1 for ease of presentation. In such a cases the payo?s can be scaled. Now we de?ne the problems which we will examine. De?nition 3.5 For a ?xed representation of games, IsNash is the promise language de?ned as follows: Positive instances: (G, θ, ) such that G is a game given in the speci?ed representation, and θ is strategy pro?le which is a Nash equilibrium for G. Negative instances: (G, θ, ) such that θ is a strategy pro?le for G which is not a -Nash equilibrium.

We use this type of approximation only when we are guaranteed to be dealing with rational Nash equilibrium. This is the case in all games restricted to 2-players and when solving problems relating to pure-strategy Nash equilibrium such as determining if a pure-strategy pro?le is a Nash equilibrium and determining if there exists a pure-strategy Nash equilibrium. 2 We will only consider this in the case where a rational Nash equilibrium is not guaranteed to exist, namely in k-player games for k ≥ 3 for the problems IsNash, FindNash, and GuaranteeNash.

1

6

Notice that when = 0 this is just the language of pairs (G, θ) where θ is a Nash equilibrium of G. The the de?nition of IsNash is only one of several natural variations. Fortunately, the manner in which it is de?ned does not a?ect our results and any reasonable de?nition will su?ce. For example, we could instead de?ne IsNash where: 1. (G, θ, ) a positive instance if θ is an /2-Nash equilibrium of G; negative instances as before. 2. (G, θ, , δ) is a positive instance if θ is an -Nash equilibrium; (G, θ, , δ) is a negative instance if θ is not a + δ-Nash equilibrium. δ is represented in the same way is . Similar modi?cations can be made to De?nitions 3.6, 3.7, and 3.9. The only result a?ected is the reduction in Corollary 4.6. De?nition 3.6 We de?ne the promise language IsPureNash to be the same as IsNash except we require that, in both positive and negative instances, θ is a pure-strategy pro?le. De?nition 3.7 For a ?xed representation of games, ExistsPureNash is the promise language de?ned as follows: Positive instances: Pairs (G, ) such that G is a game in the speci?ed representation in which there exists a pure-strategy Nash equilibrium. Negative instances: (G, ) such that there is no pure-strategy -Nash equilibrium in G. Note that Exact ExistsPureNash is just a language consisting of pairs of games with purestrategy Nash equilibria. De?nition 3.8 For a given a representation of games, the problem FindNash is a search problem where, given a pair (G, ) such that G is a game in a speci?ed representation, a valid solution is any strategy-pro?le that is an -Nash equilibrium in G. As remarked above, when dealing with FindNash in games with more than 2 players, we use Exp-Approx rather than Exact. This error ensures the existence of some Nash equilibrium in our representation of strategy pro?les; there may be no rational Nash equilibrium. De?nition 3.9 For a ?xed representation of games, GuaranteeNash is the promise language de?ned as follows: Positive instances: (G, , (g1 , . . . , gn )) such that G is a game in the speci?ed representation in which there exists a Nash equilibrium θ such that, for every agent i, νi (θ) ≥ gi . Negative instances: (G, , (g1 , . . . , gn )) such that G is a game in the speci?ed representation in which there exist no -Nash equilibrium θ such that, for every agent i νi (θ) ≥ gi ? . When we consider IsNash, FindNash, and GuaranteeNash in k-player games, k > 2, we will not consider Exact, but only the other three types: Exp-Approx, Poly-Approx, and ConstApprox. The reason for this is that no rational Nash equilibrium is guaranteed to exist in these cases, and so we want to allow a small rounding error. With all other problems we will not consider Exp-Approx, but only the remaining three: Exact, Poly-Approx, and Const-Approx. 7

4

Relations between concise games

We study two di?erent concise representations of games: circuit games and degree-d graph games; and two restrictions: two-player games and boolean-strategy games. It does not make since to impose both of these restrictions at the same time, because in two-player, boolean games all the problems studied are trivial. This leaves us with three variations of circuit games: circuit games, 2-player circuit games, and boolean circuit games. Figure 1 shows the hierarchy of circuit games. A line drawn between two types of games indicates that the game type higher in the diagram is at least as hard as the game type lower in the diagram in that we can e?ciently reduce questions about Nash equilibria in the games of the lower type to ones in games of the higher type. However, note that there is not necessarily a reduction from 2-player version of an Exact problem to the Exp-Approx circuit game version of that problem. This also leaves us with three variations of degree-d graph games: degree-d graph games, 2player degree-d graph games, and boolean degree-d graph games. A 2-player degree-d graph game is simply a bimatrix game (if d ≥ 2) so the hierarchy of games is as shown in Figure 1. Again, note that there is not necessarily a reduction from the Exact bimatrix version of a problem to the Exp-Approx graph game version of that problem. It is easy to see that given a bimatrix game, we can always e?ciently construct an equivalent 2-player circuit game. We will presently illustrate a reduction from graph games to boolean strategy circuit games. This gives us the relationship in Figure 1.

Circuit

Circuit

2-player Circuit

Boolean Circuit

2-player Circuit

Boolean Circuit

Circuit Games

Graph Graph

Bimatrix

Boolean Graph

Bimatrix

Boolean Graph

Graph Games

All Games

Figure 1: Relationships between games Theorem 4.1 Given an n-player graph game of arbitrary degree G = (G, s, ν), in logarithmic space, we can create an n -player Boolean circuit game G = (s , ν ) where n ≤ n ≤ n |si | and i=1 logarithmic space function f : Θ → Θ and the polynomial time function g : Θ → Θ 3 with the following properties:

More formally, we specify f and g by constructing, in space O(log(|G|)), a branching program for f and a circuit that computes g.

3

8

1. f and g map pure-strategy pro?les to pure-strategy pro?les. 2. f and g map rational strategy pro?les to rational strategy pro?les. 3. g ? f is the identity map. 4. For each agent i in G there an agent i in G such that for any strategy pro?le θ of G, νi (θ) = νi (f (θ)) and for any strategy pro?le θ of G , νi (θ ) = νi (g(θ )). 5. If θ is an -Nash equilibrium in G then g(θ ) is a log2 k · -Nash equilibrium in G where k = maxi |si |. 6. ? For every θ ∈ Θ, θ is a Nash equilibrium if and only if f (θ) is a Nash equilibrium. ? For every pure-strategy pro?le θ ∈ Θ, θ is an -Nash equilibrium if and only if f (θ) is and -Nash equilibrium. Proof: Construction of G Given a graph game G, to construct G , we create a binary tree ti of depth log |si | for each agent i, with the elements of si at the leaves of the tree. Each internal node in ti represents an agent in G . The strategy space of each of these agents is {left, right}, each corresponding to the choice of a subtree under his node. We denote the player at the root of the tree ti as i. There are n ≤ n |si | players in G , because the number of internal nodes in any tree is i=1 less than the number of leaves. s = {left, right}n . For each i, we can recursively de?ne functions αi : s → si that associate pure strategies of agent i in G with each agent i in ti given a pure-strategy pro?le for G as follows: ? if si = right and the right child of i is a leaf corresponding to a strategy si ∈ si , then αi (s ) = si ? if si = right and the right child of i is another agent j , then αi (s ) = αj (s ). ? If si = left, αi (s ) is similarly de?ned. Notice each agent i in the tree ti is associated with a strategy of si that is a descendant of i . This implies that i is the only player in ti that has the possibility of being associated with every strategy of agent i in G. Let s be a pure-strategy pro?le of G and let s = (s1 , . . . , sn ) be the pure-strategy pro?le of G where si = αi (s ). Then we de?ne the payo? of an agent i in ti to be νi (s ) = νi (Ri (s, αi (s ))). So, the payo? to agent i in tree ti in G is the payo? to agent i, in G, playing αi (s ) when the strategy of each other agent j is de?ned to be αj (s ). By inspection, G can be computed in log space. We note for use below, that αi : s → si induces a map from Θ (i.e. random variables on s) to Θi (i.e. random variables on si ) in the natural way. Construction of f : Θ → Θ Fix θ ∈ Θ. For each agent i in tree ti in G let Li , Ri ? si be the set of leaves in the left and right subtrees under node i respectively. Now let f (θ) = θ where Pr[θi =left] = Pr[θi ∈ Li ]/ Pr[θi ∈ Li ∪ Ri ] = Pr[θi ∈ Li |θi ∈ Li ∪ Ri ]. 9

Note that if i is an agent in ti and some strategy si in the support of θi is a descendant of i , then this uniquely de?nes θi . However, for the other players this value is not de?ned because Pr[θi ∈ Li ∪ Ri ] = 0. We de?ne the strategy of the rest of the players inductively. The payo?s to these players are a?ected only by the mixed strategies associated to the roots of the other trees, i.e. {αj (θ ), i = j}–which is already ?xed–and the strategy to which they are associated. By induction, assume that the strategy to any descendant of a given agent i is already ?xed, now simply de?ne θi to be the pure strategy that maximizes his payo? (we break tie in some ?xed but arbitrary manner so that each of these agents plays a pure strategy). By inspection, this f be computed in polynomial time given G and s, which implies that given G, in log space we can construct a circuit computing f . Construction of g : Θ → Θ Given a strategy pro?le θ for G , we de?ne g(θ ) = (α1 (θ ), . . . , αn (θ )). This can be done in log space because computing the probability that each pure strategy is played only involves multiplying a logarithmic number of numbers together, which is known to be in log space [HAB02]. This only needs to be done a polynomial number of times. Proof of 1 If θ is pure-strategy pro?le, then for each agent i, there exists si ∈ si such that Pr[θi = si ] = 1. So all the agents in ti that have si as a descendant must choose the child whose subtree contains si with probability 1, a pure strategy. The remaining agents merely maximize their payo?s, and so always play a pure strategy (recall that ties are broken in some ?xed but arbitrary manner that guarantees a pure strategy). αi : s → si maps pure-strategy pro?les to pure-strategies, so g(s ) = (α1 (s ), . . . , αn (s )) does as well. Proof of 2 For f we recall that if agent i in tree ti has a descendant in the support of θi , then Pr[f (θ)i =left] = Pr[θi ∈ Li ]/ Pr[θi ∈ Li ∪ Ri ] (Li and Ri are as de?ned in the construction of f ), so it is rational if θ is rational. The remaining agents always play a pure strategy. For g we have Pr[g(θ )i = s] =

s :αi (s )=s Pr[θ

= s ], which is rational if θ is rational.

Proof of 3 Since g(f (θ)) = (α1 (f (θ)), . . . , αn (f (θ))), the claim g ? f = id is equivalent to the following lemma. Lemma 4.2 The random variables αi (f (θ)) and θi are identical. Proof: We need to show that for every si ∈ si , Pr[αi (f (θ)) = si ] = Pr[θi = si ]. Fix si , let i = i0 , i1 , . . . , ik = si be the path from the root i to the leaf si in the tree ti , let dirj ∈ {left, right} indicate whether ij+1 is the right or left child of ij , and let Si be the

10

set of all leaves that are descendants of i . Then

k?1 k?1

Pr[αi (f (θ)) = si ] =

j=0

Pr[f (θ)ij = dirj ] =

j=0

Pr[θi ∈ Sij+1 |θi ∈ Sij ] (by the de?nition of f)

= Pr[θ ∈ Sik |θi ∈ Si0 ] (by Bayes’ Law) = Pr[θi = si ] (because Sik = {si } and Si0 = si )

Proof of 4 We ?rst show that νi (θ ) = νi (g(θ )). Fix some θ ∈ Θ . νi (θ ) = νi (Ri ((α1 (θ ), . . . , αn (θ )), αi (θ ))) (by de?nition of νi ) = νi (α1 (θ ), . . . , αn (θ )) = νi (g(θ )) (by de?nition of g) Finally, to show that νi (f (θ)) = νi (θ), ?x θ ∈ Θ and let θ = f (θ). By what we have just shown νi (θ ) = νi (g(θ )) ? νi (f (θ)) = νi (g(f (θ))) = νi (θ) The last equality comes from the fact that g ? f = id. Proof of 5 Fix some -Nash equilibrium θ ∈ Θ and let θ = g(θ ). We must show that νi (θ) is within log2 k · of the payo? for agent i’s optimal response. To do this we show by induction that νi (Ri (θ, αi (θ )) ≥ νi (Ri (θ, si )) ? d for all si that are descendants of agent i in tree ti , where d is the depth of the subtree with agent i at the root. We induct on d. The result follows by taking i = i, and noting that Ri (θ, αi (θ )) = θ and d ≤ log2 k . We defer the base case and proceed to the inductive step. Consider some agent i in tree ti such that the subtree of i has depth d. i has two strategies, {left, right}. Let Ei = νi (θ ) = ν(Ri (θ, αi (θ )) be the expect payo? of i , and let Opti be the maximum of ν(Ri (θ, si )) over si ∈ si that are descendants of i . Similarly de?ne El , Optl , Er , and Optr for the left subtree and right subtree of i respectively. We know Ei ≥ max{El , Er } ? because otherwise i could do better by playing left or right. By induction El ≥ Optl ? (d ? 1) and Er ≥ Optr ? (d ? 1) . Finally, putting this together, we get that Ei ≥ max{El , Er } ? ≥ max{Optl , Optr } ? (d ? 1) ? = Opti ? d The proof of the base case, d = 0, is the same except that instead of employing the inductive hypothesis, we note that there is only one pure strategy in each subtree and so it must be optimal. Proof of 6 Fix some strategy pro?le θ ∈ Θ and let θ = f (θ). Let θ be a Nash equilibrium and let i be an agent in ti that has a descendant which is a pure strategy in the support of θi . All the strategies in the support of αi (θ ) are also in the support of θi ; but, all the strategies in the support of θi are optimal and therefore agent i cannot do better. All of the remaining agents are in equilibrium because they are playing an optimal strategy by construction. Conversely, 11

if f (θ) is a Nash equilibrium, then g(f (θ)) is also by Part 5 above. But by Part 3 above, g(f (θ)) = θ, and therefore θ is a Nash equilibrium. Let θ be a pure-strategy -Nash equilibrium for G. Fix some agent i, and let si ∈ s be such that Pr[θi = si ] = 1. Then any agent in ti that does not have si as a descendant plays optimally in f (θ). If agent i does have si as a descendant then according to f (θ), agent i should select the subtree containing si with probability 1. Assume without loss of generality this is in the right subtree. If agent i plays right, as directed by f (θ), his payo? will be νi (θ). If he plays left, his payo? will be νi (Ri (θ, si )), where si is the strategy that α assigns to the left child of i . But νi (θ) + ≥ νi (Ri (θ, si )) because θ is an -Nash equilibrium. Now say that f (θ) is a pure-strategy -Nash equilibrium for G where θ ∈ Θ is a pure-strategy pro?le. Fix some agent i, and let si ∈ s be such that Pr[θi = si ] = 1. If si is an optimal response to θ, then agent i is in equilibrium. Otherwise, let si = si be an optimal response to θ. Then let i be the last node on the path from i to si in the tree ti such that i has si as a descendant. By de?nition of f and ν , agent i gets payo? νi (Ri (θ, si )) = νi (θ), but would get payo? νi (Ri (θ, si )) if he switched strategies (because the nodes o? of the path from i to si in the tree ti play optimally). Yet f (θ) is an -Nash equilibrium, and so we conclude that these di?er by less than , and thus agent i is in an -equilibrium in G. Corollary 4.3 There exist boolean games without rational Nash equilibria. Proof: We know that there is some 3-player game G with rational payo?s but no rational Nash equilibrium [NS50]. Applying the reduction in Theorem 4.1 to this game results in a boolean game G . If θ were some rational Nash equilibrium for G , then, by parts 2 and 5 of Theorem 4.1, g(θ ) would be a rational Nash equilibrium for G. Corollary 4.4 With Exp-Approx and Poly-Approx, there is a log space reduction from graph game ExistsPureNash to boolean circuit game ExistsPureNash Proof: Given an instance (G, ) where G is a graph game, we construct the corresponding boolean circuit game G as in Theorem 4.1, and then solve ExistsPureNash for (G , / log2 k). We show that (G, ) is a positive instance of ExistsPureNash if and only if (G , / log2 k) is also. Say that (G, ) is a positive instance of ExistsPureNash. Then G has a pure-strategy Nash equilibrium θ, and, by Parts 1 and 6 of Theorem 4.1, f (θ) will be a pure-strategy Nash equilibrium in G . Now say that (G , / log2 k ) is not a negative instance of ExistsPureNash. Then there exists a pure-strategy pro?le θ that is an / log2 k-Nash equilibrium in G . If follows from Part 5 of Theorem 4.1 that g(θ ) is a pure-strategy -Nash equilibrium in G. We do not mention IsNash or IsPureNash because they are in P for graph games (see Section 5.) Corollary 4.5 With Exp-Approx and Poly-Approx, there is a log space reduction from graph game FindNash to boolean circuit game FindNash. Proof: Given an instance (G, ) of n-player graph game FindNash we transform G into an boolean circuit game G as in the Theorem 4.1. Then we can solve FindNash for (G , / log2 k ), where k is the maximum number of strategies of any agent to obtain an ( / log2 k )-Nash equilibrium θ for G , and return g(θ ) which is guaranteed to be an -Nash equilibrium of G by Part 5 of Theorem 4.1. G and g(θ ) can be computed in log space. 12

Corollary 4.6 With Exp-Approx and Poly-Approx, there is a log space reduction from graph game GuaranteeNash to boolean circuit game GuaranteeNash. Proof: Given an instance (G, , (g1 , . . . , gn )) of graph game GuaranteeNash we transform G into an boolean circuit game G as in the Theorem 4.1. Then we can solve GuaranteeNash for (G , / log2 k , (g1 , . . . , gn , 0, . . . , 0), where k is the maximum number of strategies of any agent. So that we require guarantees for the agents at the roots of the trees, but have no guarantee for the other agents. We show that if (G, , (g1 , . . . , gn )) is a positive instance of GuaranteeNash then so is (G , / log2 k , (g1 , . . . , gn , 0, . . . , 0)). Say that (G, , (g1 , . . . , gn )) is a positive instance of GuaranteeNash. Then there exists some Nash Equilibrium of G, θ, such that νi (θ) ≥ gi for each agent i. But then by Parts 6 and 4 of Theorem 4.1 respectively, f (θ) is a Nash Equilibrium of G and νi (f (θ)) = νi (θ) ≥ gi for each agent i of G and corresponding agent i of G . Say that (G , / log2 k , (g1 , . . . , gn , 0, . . . , 0)) is not a negative instance of GuaranteeNash. Then there exists some ( / log2 k )-Nash equilibrium θ of G such that νi (θ ) > gi ? / log2 k for each agent i at the root of a tree. But then by Parts 5 and 4 of Theorem 4.1 respectively, g(θ ) is an -Nash Equilibrium of G and νi (g(θ)) = νi (θ ) ≥ gi ? / log2 k ≥ gi ? . G can be computed in log space.

5

IsNash and IsPureNash

In this section, we study the problem of determining whether a given strategy pro?le is a Nash equilibrium. Studying this problem will also help in studying the complexity of other problems.

5.1

IsNash

A summary of the results for IsNash is shown in Figure 2. Notice that with Poly-Approx and Const-Approx everything works much as with ExpApprox and Exact, but #P, counting, is replaced by BPP, approximate counting. IsNash is in P for all graph games. When allowing arbitrarily many players in a boolean circuit game, IsNash becomes P#P -complete (via cook reductions). When allowing exponentially many strategies in a 2-player circuit game, it becomes coNP-complete. IsNash for a generic circuit game combines the hardness of these 2 cases and is coNP#P -complete. Proposition 5.1 In all approximation schemes, graph game IsNash is in P. Proof: Given a instance (G, θ, ), where G is a graph game, θ is an -Nash equilibrium if and only νi (θ) + ≥ νi (Ri (θ, si )) for all agents i and for all si ∈ si . But there are only polynomially many of these inequalities, and we can compute νi (θ) and νi (Ri (θ, si )) in polynomial time. Proposition 5.2 In all approximation schemes, 2-player circuit game IsNash is coNP-complete. Furthermore, it remains in coNP for any constant number of players, and it remains hard as long as approximation error < 1. Proof: In a 2-player circuit game, Exact IsNash is in coNP because given a pair (G, θ), we can prove θ, is not a Nash equilibrium by guessing an agent i and a strategy si , such that agent i can do better by playing si . Then we can compute if νi (Ri (θ, si )) > νi (θ) + . This computation is in P because θ is in the input, represented as a list of the probabilities of each strategy in the support of each player. The same remains true if G is restricted to any constant number of agents. 13

IsNash

#P coNP -complete Circuit #P P -complete 2-player Circuit coNPcomplete Graph Boolean Circuit 2-player Circuit coNPcomplete Graph BPP coNP -complete Circuit BPP-complete Boolean Circuit

Bimatrix in P

Boolean Graph

Bimatrix in P

Boolean Graph

Exact or Exp-Approx

Poly-Approx and Const-Approx

Figure 2: Summary of IsNash Results It is coNP-hard because even in a one-player game we can o?er an agent a choice between a payo? of 0 and the output of a circuit C. If the agent settling for a payo? of 0 is a Nash equilibrium, then C is unsatis?able. Notice that in this game, the range of payo?s is 1, and as long as < 1, the hardness result will still hold. In the previous proof, we obtain the hardness result by making one player choose between many di?erent strategies, and thus making him assert something about the evaluation of each strategy. We will continue to use similar tricks except that we will often have to be more clever to get many strategies. Randomness provides one way of doing this. Theorem 5.3 Boolean circuit game Exp-Approx IsNash is P#P -complete via Cook reductions. Proof: We ?rst show that it is P#P -hard. We reduce from MajoritySat which is P#P -complete under Cook reductions. A circuit C belongs to MajoritySat if it evaluates to 1 on at least half of its inputs. Given a circuit C with n inputs (without loss of generality, n is even), we construct an (n + 1)player boolean circuit game. The payo? to agent 1 if he plays 0 is 1 , and if he plays 1 is the output 2 of the circuit, C(s2 , . . . , sn+1 ), where si is the strategy of agent i. The payo?s of the other agents are determined by a game of pennies (for details see Section 2) in which agent i plays against agent i + 1 where i is even. Let = 1/2n+1 , and let θ be a mixed strategy pro?le where Pr[θ1 = 1] = 1, and Pr[θi = 1] = 1 2 for i > 1. We claim that θ is a Nash equilibrium if and only if C ∈MajoritySat. All agents besides agent 1 are in equilibrium, so it is a Nash equilibrium if the ?rst player can do better by 14

m changing his strategy. Currently his expected payo? is 2n where m is the number of satisfying assignments of C. If he changes his strategy to 0, his expected payo? will be 1 . He must change 2 m 1 his strategy only if 2 > 2n + .

Now we show that determining if (G, θ, ) ∈ IsNash is in P#P . θ is an -Nash equilibrium if νi (θ) + ≥ νi (Ri (θ, si )) ? i ? si ∈ {0, 1}. There are only 2n of these equations to check. For any strategy pro?le θ, we can compute νi (θ) as follows:

n

νi (θ) =

s1 ∈supp(θ1 ),··· ,sn ∈supp(θn )

Ci (s1 , s2 , . . . , sn )

j=1

Pr[θj = sj ]

(1)

where Ci is the circuit that computes νi . Computing such sums up to poly(n) bits of accuracy can easily be done in P#P . Remark 5.4 In the same way we can show that, given an input (G, θ, , δ) where and δ are encoded as in Poly-Approx, it is in P#P to di?erentiate between the case when θ is an -Nash equilibrium in G and the case where θ is not a ( + δ)-Nash equilibrium in G. Theorem 5.5 Circuit game Exp-Approx IsNash is coNP#P -complete. We ?rst use a de?nition and a lemma to simplify the reduction: De?nition 5.6 #CircuitSat is the function which, given a circuit C, computes the number of satisfying assignments to C. It is known that #CircuitSat is #P-complete. Lemma 5.7 Any language L ∈ NP#P is recognized by a nondeterministic polynomial-time TM that has all its non-determinism up front, makes only one #CircuitSat oracle query, encodes a guess for the oracle query result in its nondeterminism, and accepts only if the oracle query guess encoded in the nondeterminism is correct. Proof: Let L ∈ coNP#P and let M be a co-nondeterministic polynomial-time TM with access to a #CircuitSat oracle that decides L. Then if M fails to satisfy the statement of the lemma, we build a new TM M that does the following: 1. Use non determinism to: ? Guess non-determinism for M . ? Guess all oracle results for M . ? Guess the oracle query results for M . 2. Simulate M using guessed non-determinism for M and assuming that the guessed oracle results for M are correct. Each time an oracle query is made, record the query and use the previously guessed answer. 3. Use one oracle query (as described below) to check if the actual oracle results correspond correctly with the guessed oracle results. 4. Accept if all of the following occurred: 15

? The simulation of M accepts ? The actual oracle queries results of M correctly correspond with the guessed oracle results of M ? The actual oracle queries results of M correctly corresponds with the guessed oracle results of M Otherwise reject It is straightforward to check that, if M decides L, then M ful?lls the requirements of the Lemma. Now we argue that M has an accepting computation if and only if M does also. Say that a computation is accepted on M . Then the same computation where the oracle answers are correctly guessed will be accepted on M . Now say that an computation is accepted by M . This means that all the oracle answers were correctly guessed, and that the simulation of M accepted; so this same computation will accept on M . Finally, we need to show that step 3 is possible. That is that we can check whether all the queries are correct with only one query. Speci?cally, we need to test if circuits C1 , . . . , Ck with n1 , . . . , nk inputs, respectively, have q1 , . . . , qk satisfying assignments, respectively. For each circuit Ci create a new circuit Ci by adding i?1 nj dummy variables to Ci . Then create a circuit C j=1 which takes as in input an integer i and a bit string X of size k nj , as follows: j=1 1. If the last

k j=i+1 nj

bits of X are not all 0 then C(i, X) = 0,

i j=1 nj

2. Otherwise, C(i, X) = Ci (X) where we use the ?rst

bits of X as an input to Ci .

The circuit C has k (qi · 2n1 +n2 +···+ni?1 ) satisfying assignments where qi is the number of i=1 satisfying assignments of Ci . Note that this number together with the ni ’s uniquely determines the qi ’s. Therefore it is su?cient to check if the number of satisfying assignments of C equals k n1 +n2 +···+ni?1 ). i=1 (qi · 2 Corollary 5.8 Any language L ∈ coNP#P is recognized by a co-nondeterministic polynomial-time TM that has all its non-determinism up front, makes only one #CircuitSat oracle query, encodes a guess for the oracle query result in its nondeterminism, and rejects only if the oracle query guess encoded in the nondeterminism is correct. ? Proof: Say L ∈ coNP#P , then the compliment of L, L, is in NP#P . We can use Lemma 5.7 to ? design a TM M as in the statement of Lemma 5.7 that accepts L. Create a new TM M from M where M runs exactly as M accept switches the output. Then M is a nondeterministic polynomialtime TM that has all its non-determinism up front, makes only one #CircuitSat oracle query, and rejects only if an oracle query guess encoded in the nondeterminism is correct. Proof Theorem 5.5: First we show that given an instance (G, θ, ) it is in coNP#P to determine if θ is a Nash equilibrium. If θ is not a Nash equilibrium, then there exists an agent i with a strategy si such that νi (Ri (θ, si )) > νi (θ). As in the proof of Theorem 5.3 (see Equation 1), we can check this in #P (after nondeterministically guessing i and si ). To prove the hardness result, we ?rst note that by Lemma 5.8 it is su?cient to consider only co-nondeterministic Turing machines that make only one query to an #P-oracle. Our oracle will 16

use the #P-complete problem #Sat, so given an encoding of a circuit, the oracle will return the number of satisfying assignments. Given a coNP#P computation with one oracle query, we create a circuit game with 1 + 2q(|x|) agents where q(|x|) is a polynomial which provides an upper bound on the number of inputs in the queried circuit for input strings of length |x|. Agent 1 can either play a string s1 , that is interpreted as containing the nondeterminism to the computation and an oracle result, or he can play some other strategy ?. The rest of the agents, agent 2 through agent 2q(|x|) + 1, have a strategy space si = {0, 1}. The payo? to agent 1 on the strategy s = (s1 , s2 , . . . , s2q(|x|)+1 ) is 0 if s1 = ?, and otherwise is 1 ? f (s1 ) ? g(s), where f (s1 ) is the polynomial-time function checking if the computation and oracle-response speci?ed by s1 would cause the co-nondeterministic algorithm to accept, and g(s) is a function to be constructed below such that Es2 ,...,s2q(|x|)+1 [g(s)] = 0 if s1 contains the correct oracle query and Es2 ,...,s2q(|x|)+1 [g(s)] ≥ 1 otherwise, where the expectations are taken over s2 , . . . , s2q(|x|)+1 chosen uniformly at random. The rest of the agents, agent 2 through agent 2q(|x|) + 1, receive payo? 1 regardless. This ensures that if agent 1 plays ? and the other agents randomize uniformly, this is a Nash equilibrium if there is no rejecting computation and is not even a 1/2-Nash equilibrium if there is a rejecting computation. If there is a rejecting computation then the ?rst player can just play that computation and his payo? will be 1. If there is no rejecting computation, then either f (s1 ) = 1 or contains an incorrect query result, in which case Es→θ [g(s)] ≥ 1. If either the circuit accepts or his query is incorrect, then the payo? will always be at most 0. Now we construct g(s1 , s2 , . . . , s2q(|x|)+1 ). Let C, a circuit, be the oracle query determined by the nondeterministic choice of s1 , let k be the guessed oracle results, and let S1 = s2 s3 . . . sq(|x|)+1 and S2 = sq(|x|)+2 sq(|x|)+3 . . . s2q(|x|)+1 . For convenience we will write g in the form g(k, C(S1 ), C(S2 )). g(k, 1, 1) = k 2 ? 2n+1 k + 22n g(k, 0, 1) = g(k, 1, 0) = ?2n k + k 2 g(k, 0, 0) = k 2

Now let m be the actual number of satisfying assignments of C. Then, if agent 2 through agent 2q(|x|) + 1 randomize uniformly over their strategies: E[g(k, C(S1 ), C(S2 ))] = (m/2n )2 g(k, 1, 1) + 2(m/2n )(1 ? (m/2n ))g(k, 0, 1) + (1 ? (m/2n ))2 g(k, 0, 0) = 22n (m/2n )2 ? 2n+1 (m/2n )k + k 2 = (m ? k)2 So if m = k then E[g] = 0, but if m = k then E[g] ≥ 1. In the game above, the range of payo?s is not bounded by any constant, so we scale G to make all payments in [0, 1] and adjust accordingly.

Notice that even if we allow just one agent in a boolean circuit game to have arbitrarily many strategies, then the problem becomes coNP#P -complete. We now look at the problem when dealing with Poly-Approx and Const-Approx. Theorem 5.9 With Poly-Approx and Const-Approx, boolean circuit game IsNash is BPPcomplete. Furthermore, this holds for any approximation error < 1. 17

Proof: We start by showing boolean circuit game Poly-Approx IsNash is in BPP. Given an instance (G, θ, ), for each agent i and each strategy si ∈ {0, 1}, we use random sampling of strategies according to θ to distinguish the following two possibilities in probabilistic polynomial time: ? νi (θ) ≥ νi (Ri (θ, si )), OR ? νi (θ) + < νi (Ri (θ, si )) (We will show how we check this in a moment.) If it is a Nash equilibrium then the ?rst case is true for all agents i and all si ∈ {0, 1}. If it is not an -Nash equilibrium, then the second case is true for some agents i and some si ∈ {0, 1}. So, it is enough to be able to distinguish these cases with high probability. Now the ?rst case holds if νi (θ) ? νi (Ri (θ, si )) ≥ 0 and the second case holds if νi (θ) ? νi (Ri (θ, si )) ≤ ? . We can view νi (θ) ? νi (Ri (θ, si )) as a random variable with the range [?1, 1] and so, by a Cherno? bound, averaging a polynomial number of samples (in 1/ ) the chance that the deviation will be more than /2 will be exponentially small, and so the total chance of an error in the 2n computations is < 1 by a union bound. 3 Remark 5.10 In the same way we can show that, given an input (G, θ, i, k, ) where G is a circuit game, θ is a strategy pro?le of G, is encoded as in Poly-Approx, it is in BPP to di?erentiate between the case when νi (θ) ≥ k and νi (θ) < k ? . Remark 5.11 Also, in this way we can show that, given an input (G, θ, , δ) where G is a boolean circuit game, θ is a strategy pro?le of G, and and δ are encoded as in Poly-Approx, it is in BPP to di?erentiate between the case when θ is an -Nash equilibrium in G and the case where θ is not a ( + δ)-Nash equilibrium in G. We now show that boolean circuit game Const-Approx IsNash is BPP-hard. Fix some < 1. Let δ = min{( 1? ) , 1 }. 2 4 We create a reduction as follows: given a language L in BPP there exists an algorithm A(x, r) that decides if x ∈ L using coin tosses r with two-sided error of at most δ. Now create G with |r| + 1 agents. The ?rst player gets paid 1 ? δ if he plays 0, or the output of A(x, s2 s3 . . . sn ) if he plays 1. All the other players have a strategy space of {0, 1} and are paid 1 regardless. The strategy pro?le θ is such that Pr[θ1 = 1] = 1 and Pr[θi = 1] = 1 for i > 1. 2 Each of the players besides the ?rst player are in equilibrium because they always receive their maximum payo?. The ?rst player is in equilibrium if Pr[A(x, s2 s3 . . . sn )] ≥ 1 ? δ which is true if x ∈ L. However, if x ∈ L, then ν1 (θ) = Pr[A(x, s2 s3 . . . sn )] < δ, but ν1 (R1 (θ, 0)) = 1 ? δ. So agent 1 could do better by ν1 (R1 (θ, 0)) ? ν1 (θ) > 1 ? δ ? δ ≥ . Theorem 5.12 With Poly-Approx and Const-Approx, circuit game IsNash is coNPBPP = coMA-complete.4 Furthermore, this holds for any approximation error < 1. ACAPP, the Approximate Circuit Acceptance Probability Problem is the promise-language where positive instances are circuits that accept at least 2/3 of their inputs, and negative instances are circuits that reject at least 2/3 of their inputs. ACAPP is in prBPP and any instances of a BPP problem can be reduced to an instance of ACAPP.

4

Recall that all our complexity classes are promise classes, so this is really prcoNPprBPP .

18

Lemma 5.13 Any language L ∈ NPBPP is recognized by a nondeterministic polynomial-time TM that has all its non-determinism up front, makes only one ACAPP oracle query, encodes an oracle query guess in its nondeterminism, and accepts only if the oracle query guess is correct. Proof: The proof is exactly the same as that for Lemma 5.8 except that we now need to show that we can check arbitrarily many BPP oracle queries with only one query. Because any BPP instance can be reduced to ACAPP we can assume that all oracle calls are to ACAPP. We are given circuits C1 , . . . , Cn and are promised that each circuit Ci either accepts at least 2/3 of their inputs, or accepts at most 1/3 of its inputs. Without loss of generality, we are trying to check that each circuit accepts at least 2/3 of their inputs (simply negate each circuit that accept fewer than 1/3 of its inputs). Using boosting, we can instead verify that circuits C1 , . . . , Cn 1 each reject on fewer than 2n+1 of their inputs). So simply and the Ci circuits together to create a new circuit C , and send C to the BPP oracle. Now if even one of the Ci does not accept 2/3 1 of its inputs, then Ci will accept at most a 2n+1 faction of inputs. So also, C will accept at most 1 a 2n+1 faction of inputs. But if all the Ci accept at least 2/3 of their inputs, then each of the Ci 1 n will accept a least a 1 ? 2n+1 faction of their inputs. So C will accept at least a 1 ? 2n+1 > 2/3 fraction of its inputs. Lemma 5.14 Any language L ∈ coNPBPP is decided by co-nondeterministic TM that only uses one BPP oracle query to ACAPP, has all its nondeterminism up front, encodes an oracle query guess in its nondeterminism, and rejects only if the oracle query guess is correct. Proof: This corollary follows from Lemma 5.13 in exactly the same way as Corollary 5.8 followed from Lemma 5.7. Proof of Theorem 5.12: First we show that circuit game Poly-Approx IsNash is in BPP coNP . Say that we are given an instance (G, θ, ). We must determine if θ is an Nash equilibrium or if it is not even an -Nash equilibrium. To do this, we de?ne a promise language L with the following positive and negative instances: L+ = {((G, θ, ), (i, si )) : si ∈ si , νi (Ri (θ, si )) ≤ νi (θ)} L? = {((G, θ, ), (i, si )) : si ∈ si , νi (Ri (θ, si ) > νi (θ) + } Now if for all pairs (i, si ), ((G, θ, ), (i, si )) ∈ L+ , then θ is a Nash equilibrium of G, but if there exists (i, si ), such that ((G, θ, ), (i, si )) ∈ L? , then θ is not an -Nash equilibrium of G. But L ∈ BPP because, by Remark 5.10, as we saw in the proof of Theorem 5.9, we can just sample νi (θ) ? νi (Ri (θ, si )) = Es←θ [νi (s) ? νi (Ri (s, si ))] to see if it is ≥ 0 or < ? . Remark 5.15 In the same way we can show that, given an input (G, θ, , δ) where G is a circuit game, θ is a strategy pro?le of G, and and δ are encoded as in Poly-Approx, it is in coNPBPP to di?erentiate between the case when θ is an -Nash equilibrium in G and the case where θ is not a ( + δ)-Nash equilibrium in G. Now we show that circuit game Const-Approx IsNash is coNPBPP -hard. The proof is similar to the proof of Theorem 5.5 1 Fix < 1 and let δ = min{ 1? , 4 }. Given a coNPBPP computation with one oracle query, we 2 create a circuit game with q(|x|) + 1 agents, where q is some polynomial which we will de?ne later. Agent 1 can either play a string s1 that is interpreted as containing the nondeterminism to be used 19

in the computation and an oracle answer, or he can play some other strategy ?. The other agents, agent 2 through agent q(|x|) + 1, have strategy space si = {0, 1}. The payo? to agent 1 is δ for ?, and 1?max{f (s1 ), g(s)} otherwise, where f (s1 ) is the polynomial time function that we must check, and Es2 ,...,sq(|x|)+1 [g(s)] > 1 ? δ if the oracle guess is incorrect, and Es2 ,...,sq(|x|)+1 [g(s)] < δ of the oracle guess is correct. The other agents are paid 1 regardless. We claim that if agent 1 plays ? and the other agents randomize uniformly, this is an Nash equilibrium if there is no rejecting computation and is not even a δ-Nash equilibrium if there is a failing computation. In the ?rst case, if the ?rst agent does not play ?, either the computation will accept and his payo? will be 0, or the computation will reject but the guessed oracle results will be incorrect and his expected payo? will be: 1 ? max{f (s1 ), g(s)} = 1 ? max{0, g(s)} = 1 ? E[g(s)] > 1 ? (1 ? δ) = δ So he would earn at least that much by playing ?. In the latter case where there is a failing computation, by playing that and a correct oracle result, agents 1’s payo? will be 1 ? max{f (s1 ), g(s)} > 1 ? δ. And if we compare this to what he would be paid for playing ?, we see that it is greater by[1 ? δ] ? [δ] ≥ . Now we de?ne g(s). Let Cs1 be the circuit corresponding to the oracle query in s1 , and let, (k) Cs1 be the circuit corresponding to running Cs1 k times, and taking the majority vote. We de?ne (k) g(s) = 0 if Cs1 (s2 s3 · · · sq(|x|) ) agrees with the oracle guess in s1 and g(s) = 1 otherwise. Now if the oracle result is correct, then the probability that Cs1 (s2 s3 · · · sq(|x|) ) agrees with it is 1 ? 2?(k) , and if the oracle results is incorrect, the probability that Cs1 (s2 s3 · · · sq(|x|) ) agrees with the oracle results (in s1 ) is 2?(k) , so by correctly picking k, g(s) will have the desired properties. De?ne q(|x|) accordingly. When approximating, it never made a di?erence whether we approximated by a polynomially small amount or by any constant amount less than 1.

(k) (k)

5.2

IsPureNash

In this section we will study a similar problem: IsPureNash. In the case of non-boolean circuit games, IsPureNash is coNP-complete. With the other games examined, IsPureNash is in P. Proposition 5.16 With any approximation scheme, circuit game and 2-player circuit game IsPureNash is coNP-complete. Furthermore, it remains hard for any approximation error < 1. Proof: The proof is the same as that for Proposition 5.2: in the reduction for the hardness result θ is always a pure-strategy pro?le. It is in coNP because it more restricted class of problems than circuit game IsPureNash which is in coNP. Proposition 5.17 With any approximation scheme, Boolean circuit game IsPureNash is Pcomplete, and remains so even for one player and any approximation error < 1. Proof: It is in P because each player has only one alternative strategy, so there are only polynomially many possible deviations, and the payments for each any given strategy can be computed in polynomial time. It is P-hard even in a one-player game because, given a circuit C with no inputs (an instance of CircuitValue which is P-hard), we can o?er an agent a choice between a payo? of 0 and the 20

output of the circuit C. If the agent settling for a payo? of 0 is a Nash equilibrium, then C then must evaluate to 0. Notice that in this game, the range of payo?s is 1, and as long as < 1, the hardness result will still hold. Proposition 5.18 With any approximation scheme, graph game IsPureNash is in P for any kind of graph game. Proof: In all these representation, given a game G there are only players, and each player has only a polynomial number of strategies. To equilibrium, one has to check that for all agents i and strategies si ∈ si , as mentioned there are only polynomially many of these strategies and polynomial time. a polynomial number of check that s is an -Nash νi (s) ≥ νi (Ri (s, si )). But each can be evaluated in

6

Existence of pure-strategy Nash equilibria

We now will use the previous relationships to study the complexity of ExistsPureNash. Figure 3 give a summary of the results.

P2 complete 2-player Circuit

All approximation schemes

Figure 3: Summary of ExistsPureNash Results The hardness of these problem is directly related to the hardness of IsPureNash. We can always solve ExistsPureNash with one more non-deterministic alternation because we can nondeterministically guess a pure-strategy Nash equilibrium, and then check that it is correct. Recall that in the case of non-boolean circuit games, IsPureNash is coNP-complete. With the other games examined, IsPureNash is in P (but is only proven to be P-hard in the case of boolean circuit games; see Subsection 5.2). As shown in Figure 3, with the exception of bimatrix games, this strategy of nondeterministically guessing and then checking is the best that one can do. We ?rst note that ExistsPureNash is an exceedingly easy problem in the bimatrix case because we can enumerate over all the possible pure-strategy pro?les and check whether they are Nash equilibria. ExistsPureNash is interesting because it is a language related to the Nash equilibrium of bimatrix games that is not NP-complete. One particular approach to the complexity of ?nding 21

?

ExistsPureNas h

Circuit

Boolean Circuit NP-complete Graph

in P Bimatrix Boolean Graph

a Nash equilibrium is to turn the problem into a language. Both [GZ89] and [CS03] show that just about any reasonable language that one can create involving Nash equilibrium in bimatrix games is NP-complete; however, ExistsPureNash is a notable exception. If we ask whether this generalizes to concisely represented games, the answer is a resounding No. It seems that the bimatrix case is an exception. In all other cases, ExistsPureNash can be solved with one more alternation than IsPureNash and is complete for that class. Theorem 6.1 Circuit game ExistsPureNash and 2-player circuit game ExistsPureNash are Σ2 P-complete with any of the de?ned notions of approximation. Furthermore, it remains hard as long as approximation error < 1. Proof: Membership in Σ2 P follows by observing that the existence of a pure-strategy Nash equilibrium is equivalent to the following Σ2 P predicate: ?s ∈ s, ? i, si ∈ si νi (Ri (s, si )) ≤ νi (s) + To show it is Σ2 P-hard, we reduce from the Σ2 P-complete problem QCircuitSat2 = {(C, k1 , k2 ) : ?X1 ∈ {0, 1}k1 , ?X2 ∈ {0, 1}k2 C(X1 , X2 ) = 1} where C is a circuit that takes k1 +k2 boolean variables. Given an instance (C, k1 , k2 ) of QCircuitSat2 , create 2-player circuit game G = (s, ν), where si = {0, 1}ki ∪ {0, 1}. Player i has the choice of playing a strategy Xi ∈ {0, 1}ki or a strategy yi ∈ {0, 1}. The payo?s for the ?rst player are as follows: Player 2 X2 y2 C(X1 , X2 ) 1 1 Pennies1 (y1 , y2 )

Player 1

X1 y1

Payo?s of player 1 If both players play an input to C, then player 1 gets paid the results of C on these inputs. If both play a strategy in {0, 1}, the payo? to the ?rst player is the same as that in the game of pennies (1 if y1 = y2 ; 0 if y1 = y2 ). If one player plays an input to C, and the other plays a strategy in {0, 1}, then the ?rst player receives 1. The payo?s of the second player are as follows: Player 2 X2 y2 1 ? C(X1 , X2 ) 0 0 Pennies2 (y1 , y2 ) Payo?s of player 2 Player 2’s payo? is the opposite of the output of C when both players play an input to C. He gets the payo? of the second player of pennies (0 if y1 = y2 ; 1 if y1 = y2 ) when both players play strategies in {0, 1}. Player 2’s payo? is 0 if one player plays an input to C while the other plays a strategy in {0, 1}. Now we show that the above construction indeed gives a reduction from QCircuitSat2 to 2player Circuit Game ExistsPureNash. Suppose that (C, k1 , k2 ) ∈QCircuitSat2 . Then there is 22

Player 1

X1 y1

an X1 is such that ?X2 , C(X1 , X2 ) = 1, and we claim (X1 , X2 ) is a pure-strategy Nash equilibrium where X2 is any input to C. Player 1 receives a payo? of 1 and so cannot do better. Whatever player 2 plays, he will get payo? 0 if he plays an input to C and 0 if he plays a strategy in {0, 1}. So can do no better than playing X2 . Now suppose that (C, k1 , k2 ) ∈QCircuitSat2 , i.e. ?X1 , ?X2 C(X1 , X2 ) = 0. Then we want to show there does not exist a pure-strategy -Nash equilibrium. Because the only payo?s possible are 0 and 1 and we are only considering pure-strategies, if any agent in not in equilibrium, he can do at least 1 better by changing his strategy. If player 1 plays an input X1 to C, then player 2 always has a best response X2 where C(X1 , X2 ) = 0 so that he is paid 1, which is at least better than playing an X2 such that C(X1 , X2 ) = 1 or playing a strategy in {0, 1}. So if there is an pure-strategy -Nash equilibrium where player 1 plays X1 , then player 2 must be playing such an X2 . But in this case, player 1 could do 1 better by playing a strategy in {0, 1}. So no pure-strategy Nash equilibrium where player 1 plays X1 . In the case where player 1 plays a strategy in {0, 1}, player 2’s best response is to play the opposite strategy in {0, 1}. He gets 1 for this strategy and 0 for all others. But then player 1 can do 1 better by ?ipping his strategy in {0, 1}. So no pure-strategy Nash equilibrium exists in this game. Therefore, if ?X1 , ?X2 where C(X1 , X2 ) = 0, there does not exist a pure-strategy -Nash equilibrium for any epsilon < 1. For graph games, it was recently shown by Gottlob, Greco, and Scarcello [GGS03] have shown that ExistsPureNash is NP-complete, even restricted to graphs of degree 4. Below we strengthen their result by showing this also holds for boolean graph games, for graphs of degree 3, and for any approximation error < 1. Theorem 6.2 For boolean circuit games, graph games, and boolean graph games using any of the de?ned notions of approximation ExistsPureNash is NP-complete. Moreover, the hardness result holds even for degree-d boolean graph games for any d ≥ 3 and for any approximation error < 1. Proof: We ?rst show that boolean circuit game Exact ExistsPureNash is in NP. Then, by Theorem 4.1, Exact ExistsPureNash is in NP for graph games as well. Adding approximation only makes the problem easier. Given an instance (G, ) we can guess a pure-strategy pro?le θ. Let s ∈ s such that Pr[θ = s] = 1. Then, for each agent i, in polynomial time we can check that νi (s) ≥ νi (Ri (s, si )) ? for all si ∈ {0, 1}. There are only polynomially many agents, so this takes at most polynomial time. Now we show that ExistsPureNash is also NP-hard, even in degree-3 boolean graph games with Const-Approx for every < 1. We reduce from CircuitSat which is NP-complete. Given a circuit C (without loss of generality every gate in C has total degree ≤ 3; we allow unary gates), we design the following game: For each input of C and for each gate in C, we create player with the strategy space {true, false}. We call these the input agents and gate agents respectively, and call the agent associated with the output gate the judge. We also create two additional agents P1 and P2 with strategy space {0, 1}. We now de?ne the payo?s. Each input agent is rewarded 1 regardless. Each gate agent is rewarded 1 for correctly computing the value of his gate and is rewarded 0 otherwise. If the judge plays true then the payo?s to P1 and P2 are always 1. If the judge plays false then the payo?s to P1 and P2 are the same as the game pennies–P1 acting as the ?rst player, P2 as the second. 23

We claim that pure strategy Nash equilibria only exist when C is satis?able. Say C is satis?able and let the input agents play a satisfying assignment, and let all the gate agents play the correct value of their gate, given the input agents strategies. Because it is a satisfying assignment, the judge plays true, and so every agent–the input agents, the gate agents, P1 , and P2 –receive a payo? of 1, and are thus in a Nash equilibrium. Say C is not satis?able. The judge cannot play true in any Nash equilibrium. For, to all be in equilibrium, the gate agents must play the correct valuation of their gate. Because C is unsatis?able, so no matter what pure-strategies the input agents play, the circuit will evaluate to false, and so in no equilibrium will the judge will play true. But if the judge plays false, then P1 and P2 are playing pennies against each other, and so there is no pure-strategy Nash equilibrium. Because the only payo?s possible are 0 and 1, if any agent is not in equilibrium, he can do at least 1 better by changing his strategy. So there does not exists a pure-strategy -Nash equilibrium for any < 1. Note that the in-degree of each agent is at most 3 (recall that we count the agent himself if he in?uences his own payo?), and that the total degree of each agent is at most 4.

The ?rst thing to notice is that like IsPureNash this problem does not become easier with approximation, even if we approximate as much as possible without the problem becoming trivial. Also, similarly to IsPureNash, any reasonable de?nition of approximation would yield the same results.

7

Finding Nash equilibria

Perhaps the most well-studied of these problems is the complexity of ?nding a Nash equilibria in a game. In the bimatrix case FindNash is known to be P-hard but unlikely to be NP-hard. There is something elusive in categorizing the complexity of ?nding something if we know that it is there. [MP91] studies such problems, including ?nding Nash equilibrium. Recently, [LMM03] showed that if we allow constant error, the bimatrix case FindNash is in quasipolynomial time. The results are summarized in Figure 4. In all types of games, there remains a gap of knowledge of less than one alternation. This comes about because to ?nd a Nash equilibrium we can simply guess a strategy pro?le and then check whether it is a Nash equilibrium. It turns out that in all the types of games, the hardness of FindNash is at least as hard as IsNash (although we do not have a generic reduction between the two). Circuit game and 2-player circuit game Poly-Approx and Const-Approx FindNash are the only cases where the gap in knowledge is less than one alternation. In a circuit game, there may be exponentially many strategies in the support of a Nash equilibrium or the bit length of the probability that a particular strategy is played may be exponentially large. In either case, it would take exponentially long just to write down a Nash equilibrium. In order to avoid this problem, when we are not assured the existence of a polynomially sized Nash equilibrium (or -Nash equilibrium), we will prove hardness results not with FindNash, but with FindNashSimple. FindNashSimple an easier promise language version of FindNash, that always has a short answer. De?nition 7.1 For a ?xed representation of games, FindNashSimple is the promise language de?ned as follows: 24

EXP-hard in

NEXP

FindNash

NP #P

in P

NEXP

Circuit

in P 2-player Circuit

in P 2-player Circuit

Boolean Circuit

Graph

Bimatrix P-hard in P

NP

Boolean Graph

NP

in P

in

with constant error is

~ P

Exact Exp-Approx

Poly-Approx and Const-Approx

Figure 4: Summary of FindNash Results

Positive instances: (G, i, si , k, ) such that G is a game given in the speci?ed representation, and in every -Nash equilibrium θ of G, Pr[θi = si ] ≥ k. Negative instances: (G, i, si , k, ) such that G is a game given in the speci?ed representation, and in every -Nash equilibrium θ of G, Pr[θi = si ] < k. FindNashSimple is easier than FindNash in that a FindNash algorithm can be used to obtain FindNashSimple algorithm of similar complexity, but the converse is not clear. Theorem 7.2 2-player circuit game Exact FindNashSimple is EXP-hard, but can be computed in polynomial time with an NEXP oracle. However, if it is NEXP-hard, it implies that NEXP is closed under complement. In the proof we will reduce from a problem called GameValue. A 2-player game is a zerosum game if ν1 (s) = ?ν2 (s) for all s ∈ s. By the von Neumann min-max theorem, for every 2-player zero-sum game there exists a value ν(G), such that in any Nash equilibrium θ of G, ν1 (θ) = ?ν2 (θ) = ν(G). Moreover, it is know that, given a 2-player circuit game G, it is EXP-hard to decide if ν(G) ≥ 0 [FKS95]. Proof Theorem 7.2: We reduce from 2-player circuit game GameValue. Say we are given such a zero-sum game G = (s, ν) and we want to decide if ν(G) ≥ 0. Without loss of generality, assume the payo?s are between ±1/2. We construct a game G = (s , ν ) as follows: s1 = s1 ∪ {?}, s2 = s2 ∪ {?}, and the payo?s are: 25

?

P EXP-hard

#P NP P -hard

S 2P-hard

2P

BPP-hard Circuit in P Boolean Circuit

BPP NP MA

=P

Graph P-hard in P NP

Bimatrix

Boolean Graph

Player 1

s1 ?

Player 2 s2 1 + ν1 (s1 , s2 ) 0

? 0 1

Payo?s of player 1 in G Player 2 s2 ? ν2 (s1 , s2 ) 1 1 0

Player 1

s1 ?

Payo?s of player 2 in G We claim that if ν(G) ≥ 0 then (G , 2, ?, 1/2) is a positive instance of FindNashSimple, and if ν(G) < 0 then (G , 2, ?, 1/2) is a negative instance of FindNashSimple. Fix θ , a Nash equilibrium for G . Let pi = Pr[θi ∈ si ]. It is straightforward to check that p1 , p2 = 0, 1. Let θ be the strategy pro?le where θi is distributed as θi given that θi ∈ si . This is well de?ned because p1 , p2 = 0. Also, θ is a Nash equilibrium of G because if either player could increase there payo? in G by deviating from θ, they could also increase their payo? in G . We will now relate 1 ? p2 , the probability that agent 2 plays ?, to ν(G). The expected payo? to player 1 is p1 p2 (1 + ν(G)) + (1 ? p1 )(1 ? p2 ), which we can view as a function of p1 . Because agent 1 is in an equilibrium, he must play p1 to maximize this function. This can only happen at the end points (p1 = 0 or 1) or when the derivative with respect to p1 is 0. We have already observed that no Nash equilibrium occurs when p1 = 0 or 1, so one must occur when the derivative is 0. The derivative with respect to p1 is p2 (1 + ν(G)) ? (1 ? p2 ). And so p2 (1 + ν(G)) ? (1 ? p2 ) = 1 1 0 ? p2 = 2+ν(G) ? 1 ? p2 = 1 ? 2+ν(G) . Therefore, if ν(G) ≥ 0 then in any Nash equilibrium θ , Pr[θ2 = ?] ≥ 1 ; but if ν(G) < 0 then in any Nash equilibrium θ , Pr[θ2 = ?] < 1 . 2 2 Now we show that 2-player circuit game Exact FindNashSimple is in NEXP. By Proposition 3.3, a Nash equilibrium that can be encoded in exponential space always exists in a 2-player circuit game. Therefore, the non-determinism can guess a strategy pro?le θ that is at most exponentially long, and then we can check whether it is a Nash equilibrium in EXP. Because the description of the strategy-pro?le that was guessed may be exponential in length, we cannot simply use our result from IsNash to show that we can determine if θ is a Nash equilibrium. However, it is not hard to see that we can verify this in a straight-forward manner by computing, for each agent i, νi (θ) and νi (Ri (θ, si )) for all si ∈ si . If 2-player circuit game FindNashSimple were NEXP-hard under cook reductions, it would also be coNEXP-hard under cook reductions. However, this would imply coNEXP ? NEXP, because in NEXP we could simulate the polynomial-time algorithm with oracle access to FindNashSimple, guessing and verifying FindNashSimple oracle query results as follows: Given an instance (G, i, si , k), nondeterministically guess a Nash equilibrium θ of G, verify that θ is indeed a Nash equilibrium of G, and check whether Pr[θi = si ] ≥ k. This result is analogous to the bimatrix case; everything scales up by an exponential factor. The problem becomes more tedious when we add exponentially small error. The di?culty is that we only know GameValue is hard to solve exactly. Because we introduce an element of approximation, we cannot use the same reduction in a straightforward manner. The reductions 26

from EXP to GameValue used in [FIKU04] and [FKS95] require an error bound that is at least doubly exponentially small. Theorem 7.3 Circuit game Exp-Approx FindNashSimple is EXP-hard, but is in NEXP. However, if it is NEXP-hard, it implies that NEXP is closed under complement. The EXPhardness holds even for circuit games with 6 players. Proof: We ?rst prove that circuit game Exp-Approx FindNashSimple is EXP-hard. We reduce from SuccinctCircuitValue. Given a succinctly represented circuit C, we construct an instance of FindNashSimple based upon a 6-player game G = (s, ν). Let G be the set of gates in C and let N = |G|. We create 3 computing agents: c1 , c2 , and c3 ; and we create 3 enforcing agents: e1 , e2 , and e3 . Each computing agent has the strategy set sci = {g, g : g ∈ G}. Each enforcing agent has the strategy set sei = {g : g ∈ G}. The payo? of ? the enforcing agents and the computing agents are designed so that in any -Nash equilibrium each computing agent must play g or g with probability at least 1/N ? . The payo?s of the computing ? agents are also designed so that each computing agent must play a strategy that corresponds with a correct computation of C. That is, if g evaluates to true, each computing agent must play g with 1 ? probability close to N and g with probability close to 0; and if g evaluates to false, vice versa. 2 + 2 )/ . We de?ne the payo?s of the enforcer agents as follows: Let B = (N sei g g sci g, g ? = g, g ? νe i ?B

B N ?1

We de?ne the payo?s of the computing agents as follows (t will be de?ned momentarily): νc1 (s) = t(sc1 , sc2 , sc3 ) ? νe1 (s) νc2 (s) = t(sc2 , sc3 , sc1 ) ? νe2 (s) νc3 (s) = t(sc3 , sc1 , sc2 ) ? νe3 (s) We now de?ne t(sci , scj , sck ) which will always be either ?N 2 or 0. Let g be the gate such that sci ∈ {g, g }. Let gates g1 and g2 be the two inputs of g. If g1 is a constant, then for this de?nition ? ignore scj and instead use the value of the constant. Do likewise for g2 and sck . Then t(sci , scj , sck ) = ? ?N 2 if scj ∈ {g1 , g1 } (or g1 is a constant) and sck ∈ {g2 , g2 } (or g2 is a constant) and ? ? g(scj , sck ) = sci where g(scj , sck ) is the output of gate g using scj and sck (or the respective constants) as the inputs. ? 0 Otherwise Let = 1/(64N 2 ).

Claim 7.4 In any -Nash equilibrium, θ, for any i ∈ {1, 2, 3} and any g ∈ G, Pr[θci ∈ {g, g }] ≥ ? 1/N ? .

27

Proof of claim: Say not, then for some agent ci and gate g, Pr[θci ∈ {g, g }] = ? 1/N ? p for some p > . We show that in such a case, agent ci can do better by changing his strategy. By playing g, ei will get paid 1 1 B ? p · (?B) + 1 ? +p · > pB N N N ?1 So ei has a strategy to get paid more than pB. But θ is an -Nash equilibrium, so νei (θ) > pB ? . But this means that νci (θ) = t(θci , θcj , θck ) ? νei (θ) ≤ ?νei (θ) < ?pB + because it is always the case that t(θci , θcj , θck ) ≤ 0. If θci is the mixed strategy where agent ci randomizes uniformly over all 2N strategies, νci (Rci (θ, θci )) ≥ ?N 2 . This is because here νei (Rci (θ, θci )) = 0 no matter what θei is and t(sci , scj , sck ) ≥ ?N 2 regardless of the inputs. Because this is a -Nash equilibrium, νci (Rci (θ, θci )) ≤ νci (θ) + , i.e. ?N 2 ≤ (?pB + ) + . Thus p≤ N2 + 2 = . B

When C is correctly evaluated, each gate g ∈ G evaluates to either true of false. If gate g evaluates to true, we call the strategy g correct. If gate g evaluates to false, we call the strategy g ? ? to be the the correct. If a strategy is not correct, we say it is incorrect. For any gate g, de?ne g correct strategy for g and de?ne g ? to be the incorrect strategy for g. In sci there are N correct ? strategies, and N incorrect strategies. Claim 7.5 In any -Nash equilibrium θ, no agent ci plays an incorrect strategy with probability greater than 2 .

Proof of claim: The proof proceeds by induction over the layers of the circuit. We defer the base case (i.e. constant gates). Fix a gate g ∈ G with input gates g1 , g2 . Now ?x a computing agent ci . By induction, assume that g1 and g2 are played incorrectly by each computing agent with probability less than 2 . By Claim 7.4 this implies that each computing agent plays correctly with probability at least 1/N ? 3 . We now show that the claim is true by showing that the expected payo? of ci for playing the correct strategy of gate g is always more than 1 better than his payo? for 2 playing the incorrect strategy for gate g. Therefore, if agent ci is playing an incorrect strategy for gate g with probability ≥ 2 , agent ci could do better by playing the correct strategy for gate g whenever he had previously played the incorrect strategy. Note that strategies played for g, g1 , and g2 that are used in computing payo?s are always independent because t takes its three inputs from three di?erent agents. 28

We ?rst compute νci (Rci (θ, g ? )). First note that t(g ? , scj , sck ) = ?N 2 only if scj = g1 ?? or sck = g2 , and each of these events happens with probability less then 2 by induction. ?? So νci (Rci (θ, g ? )) = t(g ? , θcj , θck ) ? νei (Rci (θ, g ? )) ≥ ?N 2 · Pr[θcj = g1 ] ? N 2 · Pr[θck = g2 ] ? νei (Rci (θ, g ? )) ?? ?? ≥ ?4N 2 ? νei (Rci (θ, g ? )) 1 = ? ? νei (Rci (θ, g ? )). 16

? We next compute νci (Rci (θ, g ? )). First note that t(?? , scj , sck ) = 0 unless scj = g1 ? g ? and sck = g2 and each of these events is independent and happens with probability at least 1/N ? 3 , by induction. So

νci (Rci (θ, g ? )) = t(g ? , θcj , θck ) ? νei (Rci (θ, g ? )) ? ?

? ? ≤ ?N 2 · Pr[θcj = g1 and θck = g2 ] ? νei (Rci (θ, g ? )) ?

≤ ?N 2 · (1/N ? 3 )2 ? νei (Rci (θ, g ? )) = (1 ? 3 N )2 ? νei (Rci (θ, g ? )) < ?9/16 ? νei (Rci (θ, g ? )), where the last inequality holds because < 1/12N . Since νei (Rci (θ, g ? )) = νei (Rci (θ, g ? )), ? ? )) ? ν (R (θ, g ? )) > ?1/16 ? (?9/16) = 1/2. ? we conclude that νci (Rci (θ, g ci ci It remains to analyze the case where g1 , g2 , or both are actually constants in the circuit. The analysis above remains the same. The only fact we used about g1 and g2 is that, by induction, the probability that θcj equals the correct value (gi or gi ) is ≥ 1 ? 3 ? ? and the probability that θcj equals the incorrect value (gi or gi ) is < 2 . Thus, if g1 (resp. g2 ) is a constant input and, as per the de?nition of t, we treat θcj as representing the correct constant value, then the analysis above will still go through. Now we can solve an instance of SuccinctCircuitValue on an instance C by querying FindNashSimple on the instance (G, c1 , o, 2 , ), where o is the output gate, and returning the same answer. By Claim 7.5, if C evaluates to true, in any -Nash equilibrium c1 will play o with prob? ability less than 2 and thus by Claim 7.4 will play o with probability at least 1/N ? 3 . If C evaluates to false, by Claim 7.5, in any -Nash equilibrium c1 will play o with probability less than 2 . Now we show that circuit game FindNashSimple is in NEXP. We can nondeterministically guess an -Nash equilibrium, and by Theorem 3.4 we can always ?nd an -equilibrium that uses at most exponential space. Therefore, the non-determinism can guess a strategy pro?le θ that is at most exponentially long, and then we can check whether θ is an -Nash equilibrium by computing, for each agent i, νi (θ) and νi (Ri (θ, si )) for all si ∈ si . If FindNashSimple were NEXP-hard under cook reductions, it would also be coNEXP-hard under cook reductions. However, then we would get coNEXP ? NEXP, because in NEXP we could simulate the polynomial-time algorithm with oracle access to FindNashSimple, guessing and verifying FindNashSimple oracle query results as follows: Given an instance (G, i, si , k, ), nondeterministically guess an -Nash equilibrium θ of G, verify that θ is indeed an -Nash equilibrium of G, and check whether Pr[θi = si ] ≥ k. 29

Things get easier when we approximate. The main reason is that now we know there exists a Nash equilibrium with a polynomially sized support by Theorem 3.4. Thus we can guess an -Nash equilibrium and using a result like IsNash test that it is such. So here, unlike in the exponential case, the complexity is at most one alternation more than the complexity of the corresponding IsNash problem. De?nition 7.6 A promise language L is in S2 P if there exists polynomial-time computable and polynomially bounded relation R ? Σ? × Σ? × Σ? such that: 1. If x ∈ L+ then ? y such that ? z, R(x, y, z) = 1. 2. If x ∈ L? then ? z such that ? y, R(x, y, z) = 0. Theorem 7.7 Circuit game and 2-player circuit game Poly-Approx and Const-Approx FindNash are S2 P-hard but can be computed by a polynomial-time algorithm with access to a Σ2 P oracle. We ?rst prove a technical lemma that will later free us from messy computations. Lemma 7.8 Let G = (s, ν) be a two-player, zero-sum game. Then if we create a new game G = (s , ν ) where s1 = s1 ∪ {?}, s2 = s2 , and νi (s1 , s2 ) = νi (s1 , s2 ) νi (s1 , s2 ) = α if s1 ∈ s1 if s1 = ?

If ν(G) < α ? 4 , then in any -Nash equilibrium θ of G , Pr[θi ∈ si ] < 1/2. Also, if α + 4 < ν(G), then in any -Nash equilibrium θ of G , Pr[θi ∈ si ] > 1/2. Proof: For the sake of contradiction suppose that ν(G) < α ? 4 and θ is an -Nash equilibrium of G such that p = Pr[θi ∈ si ] ≥ 1 . Let θ1 denote the probability distribution over s1 of θ1 give 2 that θ1 ∈ s1 . θ1 is well de?ned because p > 0. Player 2’s payo? is p(ν2 (R1 (θ , θ1 ))) + (1 ? p)(α) However, player 2 can attain a payo? of p(?ν(G)) + (1 ? p)(α) by playing an optimal strategy in G. Because θ is an -Nash equilibrium, the di?erence of these two values is at most : p(?ν(G)) + (1 ? p)(α) ? p(ν2 (R1 (θ , θ1 ))) + (1 ? p)(α) ≤ /p + ν(G) /p + ν(G) ? ?ν2 (R1 (θ , θ1 )) ≤ ? ν1 (R1 (θ , θ1 )) ≤

In the last step ?ν2 (R1 (θ , θ1 )) = ν1 (R1 (θ , θ1 )) because G is a zero-sum game.

30

Because player 1 can always receive α by playing ?, he receives at least α ? equilibrium. This implies that: α? ≤ ν1 (θ ) = p(ν1 (R1 (θ , θ1 ))) + (1 ? p)(α) ≤ p [ /p + ν(G)] + (1 ? p)(α) < p [ /p + (α ? 4 )] + (1 ? p)(α) ≤ α?

in any -Nash

because ν1 (R1 (θ , θ1 )) ≤ /p + ν(G) because ν(G) < α ? 4

So we have found our contradiction. The other implication follows in a very similar manner. De?nition 7.9 For a ?xed representation of games, PrefixNash is the promise language de?ned as follows: Positive instances: (G, x, 1k , , δ) such that G is a game given in the speci?ed representation, and there exists an -Nash equilibrium θ such that the encoding of θ is of length at most k and begins with the string x. Negative instances: (G, x, 1k , , δ) such that G is a game given in the speci?ed representation, and there exists no ( + δ)-Nash equilibrium θ such that the encoding of θ is of length at most k and begins with the string x. Both and δ are given in the appropriate representation depending on whether we are considering Exact, Exp-Approx, Poly-Approx, or Const-Approx. Proof Theorem 7.7: We ?rst show that circuit game Poly-Approx FindNash can be computed in polynomial time with a Σ2 P oracle. The following claim reduces the problem to showing that PrefixNash is in Σ2 P and that there exists an encoding of an /2-Nash equilibrium that is at most polynomially sized. Claim 7.10 With Exact, with Exp-Approx, and with Poly-Approx, given a circuit game G and such that there exists an /2-Nash equilibrium to G of size k, we can ?nd an -Nash equilibrium in time poly(|(G, ), k) using a PrefixNash oracle.

Proof of claim: Given an instance (G, ) of FindNash, the algorithm uses the PrefixNash oracle to ?nd successive bits of an -Nash equilibrium. The algorithm runs as follows: assuming that the algorithm has already computed the ?rst i bits, x1 x1 . . . xi the algorithm sends (G, x1 x2 . . . xi 0, 1k , 2 + i 2k , 2k ) to the PrefixNash oracle. If it accepts, then it sets xi+1 = 0, if it rejects, then the algorithm sends (G, x1 x2 . . . xi 1, 1k , 2 +i 2k , 2k ) to the PrefixNash oracle. If it accepts, it sets xi+1 = 1. If it rejects, it halts and returns x1 x2 . . . xi as the answer. We ?rst prove correctness. We begin by claiming that for every i where 0 ≤ i ≤ k, the partial solution x1 x2 . . . xi can be extended to an 2 + i 2k -Nash equilibrium encoded by a string of length at most k. The base case, i = 0 is true by the hypotheses of the claim. By induction assume that x1 x2 . . . xi can be extended to an 2 + i 2k -Nash equilibrium encoded by a string of length at most k. Then there are 3 cases: 1) the oracle accepts the ?rst query. In this case, x1 x2 . . . xi 0 can be extended to an 2 + (i + 1) 2k Nash equilibrium encoded by a string of length at most k (otherwise, the oracle would 31

have rejected). 2) The ?rst oracle query rejects, but the second accepts. In the case, x1 x2 . . . xi 1 can be extended to an 2 + (i + 1) 2k -Nash equilibrium encoded by a string of length at most k. 3) The oracle rejects both queries. In this case, the algorithm stops. This completes the proof by induction. However, when the algorithm stops, x1 x2 . . . xi could be extended to an 2 +i 2k -Nash equilibrium encoded by a string of length at most k, but x1 x2 . . . xi 0 and x1 x2 . . . xi 1 cannot. Therefore, x1 x2 . . . xi is the encoding of a Nash equilibrium. By the previous claim, it is an 2 + i 2k -Nash equilibrium. It is straightforward to verify that this algorithm runs in polynomial time. By Theorem 3.4 in every n-player game G, there exists a k-uniform /2-Nash equilibrium, where 2 2 max k = 3n ln(n( /2)2 i {|si |}) . This is polynomially bounded with respect to the encoding of G as a circuit game and | | where is represented as in Poly-Approx. Finally, we show that circuit game Poly-Approx PrefixNash is in Σ2 P. Given an instance (G, x, 1k , , δ) we can guess an encoding of a strategy pro?le θ and then test in coNPBPP that |θ| ≤ k, that the encoding of θ begins with the string x, and whether θ is an -Nash equilibrium or not even an ( +δ)-Nash equilibrium. It is clear that the ?rst two criteria can be tested in coNPBPP . By Remark 5.15 the last criterion can also be checked in coNPBPP . It is straightforward to verify the correctness of this algorithm. We have shown that this version of PrefixNash in Σ2 PBPP . However, Σ2 PBPP = Σ2 P because coNPBPP = coMA ? Σ2 P by [BM88]. Thus an ? coNPBPP -predicate can be replaced by ? Σ2 P-predicate = Σ2 P-predicate. We now show that 2-player circuit game Const-Approx FindNash is S2 P hard. We ?rst follow the proof of [FIKU04] which shows that approximating GameValue in 2-player circuit games is S2 P-hard in order to make a game with value either 1 or -1. Then we employ Lemma 7.8. Recall that if a language is in S2 P then there exists a polynomially balanced and polynomially decidable predicate ? such that x ∈ L+ ? ?y, ?z ?(x, y, z) = 1 and x ∈ L? ? ?z, ?y ?(x, y, z) = 0. Let p(|x|) be a polynomial that bounds the lengths of y and z. Let L be a promise language in S2 P, now construct a game G so that, given an -Nash equilibrium to G , we can determine if a given x is in L+ or L? . Given an x, construct an instance of FindNash (G , ) as follows. First, let G be the 2-player circuit game G = (s, ν) where si = { strings of length ≤ p(|x|)} and ν1 (s1 , s2 ) = ?ν2 (s1 , s2 ) = ?(x, s1 , s2 ) Let < 1. 4 If x ∈ L+ the ?rst player has a strategy s1 such that whatever strategy s2 ∈ s2 player 2 plays, ?(x, s1 , s2 ) evaluates to true. So player 1 has a strategy that guarantees him a payo? of 1. On the other hand, if x ∈ L? the second player has a strategy that guarantees him a payo? of 1. We create a new game G as in Lemma 7.8. G = (s , ν ) where s1 = s1 ∪ {?}, s2 = s2 , and ? νi (s1 , s2 ) = νi (s1 , s2 ) if s1 ∈ si ? ν1 (?, s2 ) = ν2 (?, s2 ) = 0 Then if x ∈ L+ , ν(G) = 1 and so because 0 + 4 < ν(G) by Lemma 7.8 in any -Nash equilibrium θ of G , Pr[θ = ?] < 1/2. However, if x ∈ L? , ν(G) = ?1 and so because ν(G) < 0 ? 4 by Lemma 7.8 in any -Nash equilibrium θ of G , Pr[θ = ?] ≥ 1/2. 32

This hardness result was based o? the hardness of GameValue similarly to the 2-player circuit game FindNash hardness proof which reduced directly from GameValue. The next two hardness results use a di?erent general approach. The hardness of these problems is derived from the hardness of IsNash. We could have obtained the result that Circuit Game FindNash is coMA-hard by using a proof similar to that of Theorem 7.11 below that is based on the hardness of IsNash. However it is known that coMA ? S2 P, so the above is a stronger result. Unlike ExistsPureNash, FindNash is a lot harder in boolean circuit games than in graph games. This is because of the hardness of IsNash in boolean circuit games. Theorem 7.11 Boolean circuit game Exp-Approx FindNash is P#P -hard via cook reductions but can be computed in polynomial time given an NP#P oracle. Proof: We ?rst show that Boolean circuit game Exp-Approx FindNash can be computed in polynomial time given an NP#P oracle. By Claim 7.10, which presents a polynomial time algorithm with a PrefixNash oracle for ?nding a polynomially sized -Nash equilibrium, it is enough to show that PrefixNash for Boolean circuit games is in NP#P and that there exists an encoding of an /2-Nash equilibrium of at most polynomially size. By Theorem 3.4 in every n-player game G, there exists an /2-Nash equilibrium that can be encoded in polynomial space. We now show that circuit game Exp-Approx PrefixNash is in NP#P . Given an instance (G, x, 1k , , δ) we can guess an encoding of a strategy pro?le θ and then test in P#P that |θ| ≤ k, that the encoding of θ begins with the substring x, and whether θ is an -Nash equilibrium or not even an ( + δ)-Nash equilibrium. It is clear that the ?rst two criteria can be tested in P#P . By Remark 5.4 the last criterion can also be checked in P#P . It is straightforward to verify the correctness of this algorithm. The proof of the hardness result is very similar to that of Theorem 5.3. Again, we reduce from MajoritySat which is P#P -complete under Cook reductions. A circuit C belongs to MajoritySat if it evaluates to 1 on at least half of its inputs. Given a circuit C with n inputs , we construct an n+1-player boolean circuit game. The payo?s to agent 1 are as follows: ?

1 2

?

1 n+1 2

for playing 0

? the output of the circuit C(s2 , . . . , sn+1 ), where si is the strategy of agent i, for playing 1 The payo? of the other agents is determined by a game of pennies (for details see Section 2) in n+2 1 . which agent i plays against agent i + 1 where i is even. Let = 2n · 1 2 Now we claim it is possible to determine whether a majority of the inputs satisfy C by checking player 1’s strategy in any -Nash equilibrium. If C belongs to MajoritySat, then Pr[θ1 = 0] < 1/2; If C does not belong to MajoritySat then Pr[θ1 = 0] ≥ 1/2. Say that a majority of the inputs are accepted and let θ be an -Nash equilibrium for G. By Theorem A.1, in pennies to obtain an -Nash equilibria, it is necessary that each player plays each strategy with probability ∈ [1/2 ? 2 , 1/2 + 2 ]. That is, for each i = 2, . . . , n + 1, the random variable θi has statistical distance at most 2 from a uniform random bit. This implies that the joint distribution (θ2 , . . . , θn+1 ) has statistical distance at most 2 · n from Un . Thus, |E[C(θ2 , . . . , θn+1 )] ? E[C(Un )]| ≤ 2 n = (1/2)n+2 . 33

So the payo? to agent 1 for playing 0 is 1 ? 1 and for playing 1 is E[C(s2 , . . . , sn+1 )] ≥ 2 2 n+2 1 n+2 1/2 ? 2 . So by playing s1 = 1, agent 1 expects to do better by at least 1/2 ? 1 ? [1/2 ? 2 1 n+1 1 n+2 ]= 2 > 2 . And so the following claim shows that Pr[θ1 = 0] < 1/2. 2 Claim 7.12 Let θ be an -Nash equilibrium. If there exists a strategy si ∈ si such that νi (Ri (θ, si )) ≥ νi (Ri (θ, si )) + 2 for all si ∈ si , si = si , then Pr[θi = si ] ≥ 1/2. Proof of claim: For the sake of contradiction, assume that θ is an -Nash equilibrium where Pr[θi = si ] < 1/2. Let v = maxsi ∈si ,si =si νi (Ri (θ, si )). νi (θ) < 1 1 1 1 2 νi (Ri (θ, si )) + 2 v ≤ 2 νi (Ri (θ, si )) + 2 (νi (Ri (θ, si )) ? 2 ) = νi (Ri (θ, si )) ? . So by changing his strategy to si , agent i could do better. Therefore θ is not an actual -Nash equilibrium. Now say that C is not a member of MajoritySat and θ is a Nash equilibrium for G. We will show that Pr[θ1 = 0] ≥ 1/2. By the same reasoning as above, in any -Nash equilibrium n+2 |E[C(s2 , . . . , sn+1 )] ? E[C(Un )]| ≤ 1 . 2 n+1 So the payo? to agent 1 for playing 0 is 1 ? 1 and for playing 1 is E[C(s2 , . . . , sn+1 )] ≤ 2 2 1 n+2 1 n+1 1 n . So by playing s1 = 0, agent 1 expects to do better by at least 1/2 ? 2 ? 1/2 ? 2 + 2 1 n 1 n+2 1 n+2 [1/2 ? 2 + 2 ]= 2 > 2 . And so by Claim 7.12, Pr[θ1 = 0] ≥ 1/2. In the previous result, the hardness comes from the hardness of IsNash, so it is not surprising that boolean circuit game FindNash becomes easier when we introduce approximation. Theorem 7.13 Boolean circuit game Poly-Approx and Const-Approx FindNash are BPPhard, but can be computed in polynomial time with an oracle to NPBPP = MA. Proof: We show that Boolean circuit game Poly-Approx FindNash can be computed in polynomial time with an oracle to NPBPP = MA. By Claim 7.10, which gives a polynomial time algorithm with access to a PrefixNash oracle for ?nding a polynomially sized -Nash equilibrium, it is enough to show that PrefixNash is in coNPBPP and that there exists an encoding of an /2-Nash equilibrium that is at most polynomially sized. By Theorem 3.4, in every boolean circuit game G, there exists an /2-Nash equilibrium that can be encoded in polynomial space. We now show that circuit game Poly-Approx PrefixNash is in NPBPP . Given an instance (G, x, 1k , , δ) we can guess an encoding of a strategy pro?le θ and then test in BPP that |θ| ≤ k, that the encoding of θ begins with the substring x, and whether θ is an -Nash equilibrium or not even an ( + δ)-Nash equilibrium. It is clear that the ?rst two criteria can be tested in P. By Remark 5.11 the last criterion can also be checked in BPP. It is straightforward to verify the correctness of this algorithm. We now show that boolean circuit game Const-Approx IsNash is BPP-hard. Given a BPP language L and an instance x, we create a game so that we can tell whether x ∈ L by looking at 1 the ?rst agent’s strategy in any 100 -Nash equilibrium. We create a reduction as follows: given a language L in BPP there exists an algorithm A(x, r) 1 that decides if x ∈ L using coin tosses r with two-sided error of at most 100 . Let n = |r| and let k = log25 100n . 34

n+1

Now create G with n·k+1 agents. Each player has a strategy space of {0, 1}. Let w = w1 w2 . . . wn where wi = XOR(s(i?1)k+2 , s(i?1)k+3 , . . . , si·k+1 ). The ?rst player gets paid: ? 1/2 if he plays 0 ? the output of A(x, w) if he plays 1. All the other players play pennies against each other. So agent i plays pennies with agent i + 1 where i is even. Let = 1/100 We claim that if x ∈ L, then Pr[θ1 = 0] < 1/2 in any -Nash equilibrium, and that if x ∈ L, then Pr[θ1 = 0] ≥ 1/2 in any -Nash equilibrium. Say that x ∈ L and that θ is an -Nash equilibrium for G. By Theorem A.1, in order to be in an -equilibrium, all player but the ?rst, must randomize between their two strategies, playing 0 with probability ∈ [1/2 ? 2 , 1/2 + 2 ]. The bits from the strategies of agents 2 through n · k + 1 are fully independent, and so by the next claim, if we XOR k of them together, the resulting bit is within (4 )k = 1/(100n) of being uniform. Claim 7.14 Let X1 , . . . , Xn be independent random variables where, Xi ∈ {0, 1} and Pr[Xi = 0] ∈ [1/2 ? , 1/2 + ]. Let X = XOR(X1 , . . . , Xn ), then Pr[X = 0] ∈ [1/2 ? (2 )n , 1/2 + (2 )n ]. Proof of claim: First create variables Yi = 2Xi ? 1 (so that Yi ∈ {?1, 1} and if Xi = 0 then Yi = ?1 and if Xi = 1 then Yi = 1). E[Yi ] = 2E[Xi ] ? 1 and so ?2 ≤ E[Yi ] ≤ 2 . Let Y = n Yi . It is straightforward to check that Y = 2X ? 1. i=1 And so (E[Y2]+1) = E[X]. But

n n

|E[Y ]| =

i=1

|E[Yi ]| ≤

i=1 )n 2 (2

|2 | = (2 )n

(2 )n 2 ].

And so Pr[X = 1] = E[X] ∈ [1/2 ?

, 1/2 +

Because each input wi to the circuit is within 1/(100n) of uniform, their joint distribution is within 1/100 of uniform. So 1 |E[A(x, w)] ? E[A(x, Un )]| ≤ 100 where Un is the uniform distribution over strings of length n. So if player 1 plays 0, his payo? is 1/2. But if player 1 plays 1, his payo? is E[A(x, w)] ≥ E[A(x, Un )] ? 1 98 ≥ 100 100

98 Therefore, because agent 1 expects to do better by 100 ? 1 ≥ 2 by playing 1, by Claim 7.12, 2 Pr[θ1 = 0] < 1/2. Say x ∈ L and θ is an -Nash equilibrium of G. Then by the same reasoning as above

|E[A(x, w)] ? E[A(x, Un )]| ≤

1 100

And so the payo? to agent 1 for playing 0 is 1 , but the payo? to player 1 for playing 1 is 2 E[A(x, w)] ≤ E[A(x, Un )] + Therefore, because agent 1 expects to do better by Pr[θ1 = 0] ≥ 1/2. 35

1 2

2 1 ≤ 100 100

2 100

?

> 2 by playing 0, by Claim 7.12,

Finally, we show the complexity for graph games. Theorem 7.15 With any type of approximation, graph game and boolean graph game FindNash can be computed in polynomial time with access to an NP oracle, but neither is NP-hard unless NP = coNP. Furthermore, graph game and boolean graph game FindNash are P-hard, even when restricted to boolean graphs of degree ≥ 3. Proof: We show that graph game Exp-Approx FindNash can be computed in polynomial time with an oracle to NP. By Claim 7.10, which gives a polynomial time algorithm with access to a PrefixNash oracle for ?nding a polynomially sized -Nash equilibrium , it is enough to show that PrefixNash is in NP and that there exists an encoding of a Nash equilibrium that is at most polynomially large. By Theorem 3.4 in every graph game G, there exists an /2-Nash equilibrium that can be encoded in polynomial space. We now show that graph game Exp-Approx PrefixNash is in NP. Given an instance (G, x, 1k , , δ) we can guess an encoding of a strategy pro?le θ and then test in P that |θ| ≤ k, that the encoding of θ begins with the substring x, and whether θ is an -Nash equilibrium or not a ( + δ)-Nash equilibrium. It is clear that the ?rst two criteria can be tested in P. The last criterion can be veri?ed in P by testing that for each agent i and for all si ∈ si that νi (θ) ≥ νi (Ri (θ, si )) ? . There are only polynomially many agents, each agent has only polynomially many strategies, and because the payo? function for agent i is encoded explicitly, νi (θ) can be computed in polynomial time. To show the hardness result, we reduce from CircuitValue. Given a circuit C, we construct a game G with an agent for each gate in C. Each agent has possible strategies {0, 1} and is paid 1 for correctly computing the output of his gate (with respect to the strategies of the agents that correspond to the inputs to his gate), and is paid 0 otherwise. Let = 1/100. We call the strategy of the agent associated with gate g correct if it corresponds with the output of the gate in an evaluation of C. The unique Nash equilibrium of G is where each player plays the correct strategy. Claim 7.16 In any -Nash equilibrium, each player must play the correct strategy with probability ≥1?2 .

Proof of claim: We proceed by induction, but we defer the base case. Assume that the two agents associated with the inputs gates to a particular gate g play the correct pure strategy with probability ≥ 1 ? 2 . Let v be the payo? the agent associated with the gate g with he plays his correct strategy. We know that v ≥ (1?2 )2 because if both of the input agents play their correct strategies then the agent associated with g will receive a payo? of 1 when he plays his correct strategy. If g plays the opposite strategy his payo? will be 1 ? v. Now say that g plays the opposite strategy with probability p. Because he is in an -equilibrium, we know that (1 ? p)v + p(1 ? v) + ≥ v because he can get paid v if he just plays the correct strategy all the time. By simple arithmetic, this implies that p≤ 2v ? 1 ≤ 2(1 ? 2 )2 ?1 (by what we know of v) ≤ 2 (by inspection when ≤ 1 ) 100

36

The base case consists of those agents connected directly to the constant gates. However, if we view the constant gates as agents who always tell the truth, then the previous argument applies. Therefore, in any -Nash equilibrium, each player must play the strategy corresponding with the correct valuation of the circuit with probability ≥ 1 ? 2 . So by looking at the strategy in an -Nash equilibrium of the agent at the output gate, we can correctly deduce the value of the circuit.

8

Existence of Nash equilibria with guaranteed properties

Because FindNash is a search problem where a solution is guaranteed to exist, it is hard to de?ne a nontrivial language from it. It is possible to create languages from FindNash by adding additional constraints on the equilibrium. For example: does there exists a Nash equilibrium where each player is paid at least x amount? does there exists a Nash equilibrium with social welfare x? or does there exists a Nash equilibrium in which player 1 does not play some strategy si ? It turns out that in the bimatrix case, for almost any constraint the language ends up being NP-complete [CS03, GZ89].5 GuaranteeNash is another of such a problem. In our results, each GuaranteeNash problem is complete for the class that was the upper bound for the same instance of FindNash. Figure 5 shows a summary of the results.

GuaranteeNash

NEXPcomplete 2-player Circuit #P NP -complete Boolean Circuit P2 complete 2-player Circuit

Graph

Bimatrix NP-complete

Boolean Graph

Exact or Exp-Approx

Figure 5: Summary of GuaranteeNash Results

Theorem 8.1 Circuit game Exp-Approx GuaranteeNash and 2-player circuit game Exact GuaranteeNash are NEXP-complete.

Note that our results show that ExistsPureNash was an exception to this rule. It was trivial in bimatrix games, but at least NP-hard in every other setting.

5

37

?

Circuit

Circuit

NPBPP MA = complete Boolean Circuit

Graph NPcomplete

Bimatrix

Boolean Graph

except Const-Approx Bimatrix is ~ Const-Approx Bimatrix is in P

Poly-Approx and Const-Approx

Proof: We ?rst show that 2-player circuit game Exact GuaranteeNash is in NEXP. Given instance (G, , (g1 , . . . , gn )), guess a strategy pro?le θ, of at most exponential length, and then check whether θ is a Nash equilibrium that meets the guarantees. The correctness of this algorithm follows from Proposition 3.3 which tells us that if a Nash equilibrium that meets the guarantees exists, then one exists which is at most exponentially large. To check that θ is a Nash equilibrium, we need only check that νi (θ) ≥ νi (Ri (θi , si )) for all agents i and for all si ∈ si . Because there are only 2 agents, and only an exponential number of strategies, there are only exponentially many of these inequalities. To check that θ meets the guarantees, we need only check that νi (θ) ≥ gi for at most polynomially many agents. Therefore, it is enough to show that we can compute νi in EXP. But νi (θ) =

s1 ∈s1 s2 ∈s2

Pr[θ1 = s1 ] · Pr[θ2 = s2 ] · vi (s1 , s2 )

All values that are multiplied or summed have at most exponential bit size, thus ν(θ) can be computed in EXP. We next show that circuit game Exp-Approx GuaranteeNash is in NEXP. Given instance 2 2 max (G, , (g1 , . . . , gn )) we guess a n log(n 2 i |si |) -uniform strategy pro?le θ. We then check whether θ is an -Nash equilibrium that is within of meeting the guarantees. If it is, we accept, otherwise we reject. The correctness of this algorithm follows from Theorem 3.4 which states that if there exists 2 2 max a Nash equilibrium that meets the guarantees, then there will exists a n log(n 2 i |si |) -uniform -Nash equilibrium gets within /2 of the guarantees. To check that θ is an -Nash equilibrium, we need only check that νi (θ) ≥ νi (Ri (θi , si )) + for all agents i and for all si ∈ si . Because there are only a polynomial number of agents, and only an exponential number of strategies, there are only exponentially many of these inequalities. To check that θ meets the guarantees, we need only check that νi (θ) ≥ gi ? for at most polynomially many agents. Therefore, it is enough to show that we can compute νi in EXP. But

n

νi (θ) =

s1 ∈si

···

sn ∈sn i=1

Pr[θi = s1 ] vi (s1 , . . . , sn )

All values that are multiplied or summed have polynomial bit size (because it is a k-uniform strategy pro?le), so the product of n of them is still polynomial. And the sum of exponentially many, is exponential. Thus ν(θ) can be computed in EXP. We now show that 2-player circuit game GuaranteeNash with exponentially small error is NEXP-hard. We use a reduction very similar to the one of Conitzer and Sandholm [CS03] except that instead of reducing from 3SAT, we reduce from the NEXP-complete problem Succinct 3SAT, and we keep track of approximation errors in the reduction. Given a succinct representation of a Boolean formula ? in conjunctive normal form with the set of variables V and the set of clauses C, let N = |V | be the number of variables, and let the set L = {x, x : x ∈ V } be the set of literals. We treat L, the set of literals, as formally distinct ? from V , the set of variables 6 , and de?ne a function v : L → V such that v(x) = v(?) = x. We x construct the 2-player circuit game G = (s, ν) where s1 = s2 = V ∪ C ∪ L so that if ? is satis?able and l1 , . . . , lN are literals that satisfy the formula (exactly one for each variable), then the strategy where each player randomizes uniformly between those N literals is a Nash equilibrium where the

6

So that |V ∪ L| = 3N .

38

expected payo? to each player is N ? 1; however, if ? is not satis?able, then no -Nash equilibrium with payo?s to each player of at least N ? 1 ? exists. De?ne ν1 (s1 , s2 ) = ν2 (s2 , s1 ) as follows: 1. ν1 (l1 , l2 ) = N ? 1 where l1 = ?2 for all l1 , l2 ∈ L l This will ensure each player gets a high payo? for playing the aforementioned strategy. 2. ν1 (l, ? = N ? 4 for all l ∈ L l) This will ensure that each player does not play a literal and its negation. 3. ν1 (v, l) = 0 where v(l) = v, for all v ∈ V , l ∈ L This, along with rule 4, ensures that for each variable v, each agent plays either l or ? with l probability at least 1/N where v(l) = v(? = v. l) 4. ν1 (v, l) = N where v(l) = v, for all v ∈ V , l ∈ L 5. ν1 (l, x) = N ? 4 where l ∈ L, x ∈ V ∪ C This, along with rules 6 and 7, ensures that if both players do not play literals, then the payo?s cannot meet the guarantees. 6. ν1 (v, x) = N ? 4 for all v ∈ V , x ∈ V ∪ C 7. ν1 (c, x) = N ? 4 for all c ∈ C, x ∈ V ∪ C 8. ν1 (c, l) = 0 where l ∈ c for all c ∈ C, l ∈ L This, along with rule 9, ensures that for each clause c, each agent plays a literal in the clause c with probability least 1/N . 9. ν1 (c, l) = N where l ∈ c for all c ∈ C, l ∈ L Let = 1/2N 3 and let the guarantee to each player be N ? 1.

First we show that if ? is satis?able, then there exists a Nash equilibrium with the guarantees. Say that l1 , . . . , lN are literals that satisfy the formula (exactly one for each variable). Then the strategy where each player randomizes uniformly between those N literals is a Nash equilibrium where the expected payo? to each player is N ? 1. The expected payo? to each player is N ? 1 because they will always be playing l1 and l2 where l1 = ?2 and l1 , l2 ∈ L. Secondly, there are l only two rules that pay out more than N ? 1: ν1 (c, l) = N where l ∈ C for all c ∈ C, l ∈ L and ν1 (v, l) = N where v(l) = v, for all v ∈ V , l ∈ L. However, if agent i deviates and plays any clause c, the other player will play a literal in that clause c with probability 1/N because he randomizes between literals in a satisfying assignment. So in this case, agent i’s payo? is at most 1/N · 0 + (N ? 1)/N · N = N ? 1, and so agent i is no better o? than before. Similarly no matter what variable an agent deviates to, his opponent plays a corresponding literal with probability 1/N . Now we show that if ? is not satis?able, in any -Nash equilibrium at least one player fails to receive an expected payo? of N ? 1 ? . Unless both players are playing a literal, the maximum sum of the outcomes is 2N ? 4. We cannot be in this case with probability greater than because otherwise the payo?s will sum to less than 2N ? 2 ? 2 . So both players play elements of L with probability > 1 ? . Now assume that the probability that agent i plays l or ? for some speci?c l is less than l 1/N ? ? 2 (≥ 1/N ? 2 ). Then the expected value for the other player, agent j, to play v(l) is N at least (1/N ? ? 2 ) · 0 + · 0 + (1 ? ? (1/N ? ? 2 ))N = N ? 1 + 2 (the ?rst term is when N N 39

agent i plays l or ? the second is when agent i does not play a literal, and the third term is when l, agent i plays a literal = l, ? So either agent j can do better by changing his strategy or he is l). already receiving N ? 1 + and so the other player does not meet his guarantee (recall the sum of payo?s is at most 2N ? 2). Now we show that for each pair of literal, there is one that is played with probability ≥ 1/N ? 2 ? 1/N 2 while the other is played with probability less than 1/N 2 . If one player plays l and the other one ?l, then the sum of payo?s is 2N ? 8 and so this must happen also with probability ≤ /3, otherwise at least one player will fail to meet his guarantee. Without loss of generality, assume that player 1 plays l more than ? For the sake of contradiction, l. 2 + 2 ) and so plays ? with probability assume player 1 plays l with probability less than 1/N ? (1/N l more than 1/N 2 . (Recall that each player plays either l or ? with probability at least 1/N ? ? 2 ≥ l N 1/N ?2 .) Either the other agent plays l with probability less than 1/N 2 or plays ? with probability l greater than 1/N ? (1/N 2 + 2 ). In either case, the two players play both l and ? with probability l 1 [1/N ? (1/N 2 + 2 )][1/N 2 ] = 1/N 3 ? 1/N 4 ? 2/N 6 ≥ 2N 3 = . Which cannot happen. So player 1 must play l with probability greater than 1/N ? (1/N 2 + 2 ) and by a symmetric argument so must player 2. By the same argument, each must play ? with probability less than 1/N 2 . l So in any -Nash equilibrium that meets the guarantees, we can create a correspondence between literals and truth assignments. We say that a literal is true if it is played more often than its negation. However, if ? is not satis?able, it means that for the corresponding assignment, there exists at least one clause with no satisfying literal. Now by changing his strategy to that clause, agent i will expect to receive a payo? of N whenever the other player, agent j, plays a literal that is not in that clause. Agent j plays a literal with probability > 1 ? , and there only 3 literals in the clause, each of which agent j plays with probability ≤ 1/N 2 . By changing his strategy, agent i will receive at least (1 ? ? 3/N 2 )N > N ? 1 + 2 . So either agent i can do better by changing his strategy or he is already receiving N ? 1 + and so the other player does not meet his guarantee (recall the sum of payo?s is at most 2N ? 2).

Theorem 8.2 Circuit game and 2-player circuit game Poly-Approx and Const-Approx GuaranteeNash are Σ2 P-complete. Proof: We ?rst show that circuit game Poly-Approx GuaranteeNash is in Σ2 PBPP . Given an instance (G, , (g1 , . . . , gn )), we nondeterministically guess a polynomially small strategy pro?le θ. Then we test whether θ is an /2-Nash equilibrium that is within /2 of meeting the guarantees or whether θ is either not an -Nash equilibrium or fails to be within of the guarantees. In the former case, we accept, in the latter case we reject. We now argue the correctness of the algorithm. If (G, , (g1 , . . . , gn )) is a positive instance of GuaranteeNash, then there exists a Nash equilibrium with the guaranteed properties in G. By Theorem 3.4 there exists an /2-Nash equilibrium θ that can be represented in polynomial space where the payo?s of each player are within /2 of the guarantees. So the algorithm will accept upon guessing θ. If (G, , (g1 , . . . , gn )) is a negative instance of GuaranteeNash, then there does not exist any -Nash equilibrium within of meeting the guaranteed properties. So whatever strategy pro?le θ the algorithm guesses, either θ will fail to be an -Nash equilibrium or θ will fail to be within of the guarantees. Therefore the algorithm will always reject θ. It is now left to show that in coNPBPP we can tell wither whether θ is an /2-Nash equilibrium that is within /2 of meeting the guarantees or whether θ is either not an -Nash equilibrium or fails to be within of the guarantees. Note that by Remark 5.15 we can verify whether θ is an 40

/2-Nash equilibrium or not even an -Nash equilibrium in coNPBPP . Also, in BPP we can test whether νi (θ) ? /2 ≥ gi or νi (θ) < gi ? by Remark 5.10. Therefore we can test whether all these properties hold or at least one fails to hold in coNPBPP . Finally, recall from the proof of Theorem 7.7 that Σ2 P = Σ2 PBPP . We now show that 2-player circuit game Const-Approx GuaranteeNash is Σ2 P-hard. We reduce from QCircuitSat2 , which is Σ2 P-complete. QCircuitSat2 = {(C, k1 , k2 ) : ?x ∈ {0, 1}k1 , ?y ∈ {0, 1}k2 C(x, y) = 1} where C is a circuit that takes k1 + k2 boolean variables. Given such an instance (C, k1 , k2 ) create 2-player circuit game G = (s, ν), where si = {0, 1}k1 × {0, 1}k2 ∪ {?}. The payo?s to G will be designed so that if there exists an x0 ∈ {0, 1}k1 such that C(x0 , y) = 1 for all y ∈ {0, 1}k2 , then a Nash equilibrium is for each player to play strategies of the form (x0 , y) (for any y ∈ {0, 1}k2 ) with probability 1. However, if no such x0 exists, the only -Nash equilibrium will be to play ? most of the time. We will only de?ne the payo?s for the ?rst player because the payo?s are symmetric, that is ν1 (s1 , s2 ) = ν2 (s2 , s1 ). 1. x1 = x2 , ν1 ((x1 , y1 ), (x2 , y2 )) = 0 2. ν1 ((x, y1 ), (x, y2 )) = ? 1 ? γ if C(x, y1 ) = C(x, y2 ) = 1 ? 0 if C(x, y1 ) = 1 and C(x, y2 ) = 0, ? 1 if C(x, y1 ) = 0 and C(x, y2 ) = 1, ?

1 2

if C(x, y1 ) = C(x, y2 ) = 0

3. ν1 (?, ?) = γ 4. ν1 ((x1 , y1 ), ?) = 0 5. ν1 (?, (x2 , y2 )) = 1 ? γ

1 1 Let = 100 , γ = 10 , and gi = 1 ? γ. We now show that if (C, k1 , k2 ) ∈ QCircuitSat then there exists a Nash equilibrium that meets the guarantees and if (C, k1 , k2 ) ∈ QCircuitSat then no -Nash equilibrium in which each player is paid within of his guarantees exists. Let (C, k1 , k2 ) ∈QCircuitSat, then there exist some x0 such that for all y, C(x0 , y) = 1. Let θ be the strategy pro?le where both agents play (x0 , 0k2 ) with probability 1. Now the payo? to each agent is 1 ? γ and it is easy to see that this is a Nash equilibrium. Now suppose that (C, k1 , k2 ) ∈QCircuitSat. We must show that no -Nash equilibrium gets within of the guarantees. For the sake of contradiction, assume that such a strategy pro?le θ exists. We ?rst note that for both players to get within of their guarantees, the sum of the payo?s to the agents must be greater than 2 ? 2γ ? 2 ≥ 2 ? 4γ. We claim that Pr[θ1 = (x, y1 ) ∧ θ2 = (x, y2 ) such that C(x, y1 ) = C(x, y2 ) = 1] > 1 ? 4γ. The maximum sum of payo?s for any strategy pro?le is 2 ? 2γ. If both agents do not agree on the x component of the strategy and do not both play a pairs (x, y) which satisfy C, then the maximum sum of their payo?s will be 1. If this happens with probability more than 4γ, the sum of the payo?s will be at most (1 ? 4γ) · (2 ? 2γ) + 4γ · 1 = 2 ? 6γ + 4γ 2 < 2 ? 4γ. So this cannot happen in an -Nash equilibrium that meets the guarantees. However, because (C, k1 , k2 ) ∈QCircuitSat, for any x ∈ {0, 1}k1 there exists some y such that C(x, y) = 0. We claim that if agent 1 unilaterally changes his strategy to θ1 so that every time he had

41

played a strategy (x, y) where C(x, y) = 1 in θ1 he now plays a strategy (x, y ) where C(x, y ) = 0 in θ1 , then ν1 (R1 (θ, θ1 )) > ν1 (θ) + . Agent 1 will always be paid at least as much, and whenever in θ the strategies were such that s1 = (x, y1 ) and s2 = (x, y2 ) where C(x, y1 ) = C(x, y2 ) = 1 the strategies in θ2 will instead be s1 = (x, y1 ) and s2 = (x, y2 ) where C(x, y1 ) = 0 and C(x, y2 ) = 1. And in this case agent 1 will receive γ more than before. However, this happens with probability > 1 ? 4γ. Therefore his payo? will increase by γ ? 4γ 2 > . So there is no -Nash equilibrium where each agent comes within of his guarantees. Theorem 8.3 Boolean circuit game Exp-Approx GuaranteeNash is NP#P -complete. Proof: We ?rst show that boolean circuit game Exp-Approx GuaranteeNash is in NP#P . Given an instance (G, , (g1 , . . . , gn )), we nondeterministically guess a polynomially small strategy pro?le θ. Then we test whether θ is an /2-Nash equilibrium that is within /2 of meeting the guarantees or whether θ is either not an -Nash equilibrium or fails to be within of the guarantees. In the former case, we accept, in the latter case we reject. We now argue the correctness of the algorithm. If (G, , (g1 , . . . , gn )) is a positive instance of GuaranteeNash, then there exists a Nash equilibrium with the guaranteed properties in G. By Theorem 3.4 there exists an /2-Nash equilibrium θ that can be represented in polynomial space where the payo?s of each player are within /2 of the guarantees. So the algorithm will accept upon guessing θ. If (G, , (g1 , . . . , gn )) is a negative instance of GuaranteeNash, then there does not exist any -Nash equilibrium with the guaranteed properties. So whatever strategy pro?le θ the algorithm guesses, either θ will fail to be an -Nash equilibrium or θ will fail to be within of the guarantees. Therefore the algorithm will always reject θ. It is now left to show that in NP#P we can tell whether θ is an /2-Nash equilibrium that is within /2 of meeting the guarantees or whether θ is either not an -Nash equilibrium or fails to be within of the guarantees. We can do this by using a #P oracle to compute ν as in Equation 1 (in proof of Theorem 5.3) to within a polynomial number of bits of accuracy. Therefore in P#P we can test whether νi (θ) + /2 ≥ νi (Ri (θ, si )) or νi (θ) + < νi (Ri (θ, si )) for every agent i and si ∈ {0, 1} and can test whether νi (θ) ≥ gi + /2 or νi (θ) < gi ? for every agent i. We now show that boolean circuit game Exp-Approx GuaranteeNash is NP#P -hard. Say that we have a language L ∈ NP#P . By Corollary 5.7 there exists a non-deterministic TM M that computes L which makes only one query calls to a #CircuitSat oracle, has all its nondeterminism at the beginning, and only accepts computations where the correct oracle query result is encoded in the nondeterminism. Let f (|x|) be a polynomial that bounds the length of a string needed to encode the nondeterminism of M , let g(|x|) (without loss of generality even) be a polynomial that bounds the number of inputs to the circuit queried by M , and let y be a string of bits that encodes the nondeterminism used in M on a particular run. Given an input x we construct a boolean game G with the following agents: f (|x|) agents y1 , . . . , yf (|x|) called y agents, f (|x|) agents y1 , . . . , yf (|x|) called y agents, g(|x|) agents z1 , . . . , zg(|x|) called z agents, and agents J1 , J2 , and J3 called the judges. Let the string y = sy1 sy2 . . . syf (|x|) encode the nondeterminism of M , and let C be the circuit sent to the oracle query using the nondeterminism encoded in y, let k be the oracle query guess encoded by y, let m be the actual number of satisfying assignments of C, and let n be the number of inputs to C. The payo?s are as follows: y agents: agent yi is paid 1 regardless. 42

y agents: agent yi receives payo? 1 if his strategy is the same as yi ’s and 0 otherwise. z agents: The z agents are paid according to a game of pennies (see Section 2). Agent zi plays pennies against agent zi+1 where i is odd. agent J1 : J1 receives payo? agent J2 : J2 receives payo?

1 k+ 2 2n 1 k? 2 2n

if he plays 0 and C(sz1 sz2 . . . szk ) otherwise. if he plays 0 and C(sz1 sz2 . . . szk ) otherwise.

agent J3 : J3 receives payo? 1 if J1 plays 0, J2 plays 1, and M run on input x with the nondeterministic choices encoded by y accepts assuming that the query result encoded by y is correct. Otherwise, J3 receives 0. Let We guarantee that J3 and all the yi be paid 1. We make no guarantees to the other players. = 1/(200 · f (|x|) · g(|x|) · 2n ).

Now we show that if x ∈ L then there exists a Nash equilibrium in G with these guarantees, and if x ∈ L then there exists no -Nash equilibrium in G within of these guarantees. Say x ∈ L. Then there exists a nondeterministic guess y = y1 y2 · · · yf (|x|) such that M accepts x run with the nondeterminism encoded by y, and the query result encoded by y is correct. We claim that the strategy pro?le θ is a Nash equilibrium that meets the guarantees where θ is the strategy pro?le where: syi = syi = yi ; sJ1 = 0, sJ2 = 1, sJ3 = 0, and the z agents randomize uniformly between their two strategies. We ?rst show that each agent is in equilibrium playing θ. The y agents and the y agents are in equilibrium because they all receive payo? 1. The z agents are because they are playing the unique equilibrium strategy of pennies. J1 is in equilibrium because

m he now receives 2n2 and playing 1 yields a payo? of C(sz1 sz2 . . . szk ) which has expectation 2n . However, because y encodes a valid query guess, k = m. Similarly, J2 currently receives payo? m C(sz1 sz2 . . . szk ) which is expected to be 2n = 2k and would receive only 2n2 by changing his n strategy. Finally, J3 ’s payo? is independent of his strategy and so he is also in equilibrium. The y agents all receive their guarantees of 1. J3 also receives his guarantee of 1 because sJ1 = 0, sJ2 = 1, and running M on x with the nondeterminism encoded by y results in an accepting computation. Say x ∈ L, then there exists no -Nash equilibrium within of the guarantees. For the sake of contradiction, assume that an -Nash equilibrium θ exists in which each agent is within of his guarantees. We note that each y agent must play some particular strategy with probability greater than 1 ? (if yi does not, then yi cannot attain a payo? of at least 1 ? ). Let syi be the ? strategy agent yi plays with probability ≥ 1 ? in θ, and let y = sy1 sy2 . . . syf (|x|) . By union bound, ? ? ? ? Pr[θy1 θy2 . . . θyf (|x|) = y ] ≥ 1 ? f (|x|) · . Because < 100f1(|x|) , y is played with probability at least ? ? 99/100. Also, by Theorem A.1, Pr[θzi = 0] ∈ [1/2 ? 2 , 1/2 + 2 ]. E[C(θz1 θz2 . . . θzn )] is within 2 · n ≤ 1/(100 · 2n ) of m/2n . Now because x ∈ L either y encodes a rejecting computation on M , or the query result of y is ? incorrect. In the former case, J3 receives payo? 0 whenever y is played, and so cannot receive more ? than 1/100. In the latter case, k = m. If k < m then agent J1 will receive k+1/2 for playing 0, but 2n k+1/2 m 1 m 1 will receive at least 2n ? 100·2k for playing 1. Because [ 2n ? 100·2k ] ? [ 2n ] > 2 by Claim 7.12 Pr[θJ1 = 1] ≥ 1 and so J3 ’s payo? will be at most 1/2 < 1 ? = gJ3 ? . A symmetric argument 2 handles the case where k > m. k? 1 k+ 1

Theorem 8.4 Boolean circuit game Poly-Approx and Const-Approx GuaranteeNash are NPBPP = MA-complete. 43

Proof: We ?rst show that boolean circuit game Poly-Approx GuaranteeNash is in NPBPP . Given an instance (G, , (g1 , . . . , gn )), we nondeterministically guess a polynomially small strategy pro?le θ. Then we test whether θ is an /2-Nash equilibrium that is within /2 of meeting the guarantees or whether θ is either not an -Nash equilibrium or fails to be within of the guarantees. In the former case, we accept, in the latter case we reject. We now argue the correctness of the algorithm. If (G, , (g1 , . . . , gn )) is a positive instance of GuaranteeNash, then there exists a Nash equilibrium with the guaranteed properties in G. By Theorem 3.4 there exists an /2-Nash equilibrium θ that can be represented in polynomial space where the payo?s of each player are within /2 of the guarantees. So the algorithm will accept upon guessing θ. If (G, , (g1 , . . . , gn )) is a negative instance of GuaranteeNash, then there does not exist any -Nash equilibrium within of meeting the guaranteed properties. So whatever strategy pro?le θ the algorithm guesses, either θ will fail to be an -Nash equilibrium or θ will fail to be within of the guarantees. Therefore the algorithm will always reject θ. It is now left to show that in BPP we can tell whether θ is an /2-Nash equilibrium that is within /2 of meeting the guarantees or whether θ is either not an -Nash equilibrium or fails to be within of the guarantees. Note that by Remark 5.11 we can verify whether θ is an /2-Nash equilibrium or not even an -Nash equilibrium in BPP. By Remark 5.10 in BPP we can determine if νi (θ) ≥ gi ? 2 or νi (θ) ≥ gi ? . Therefore we can test whether all these properties hold or at least one fails to hold using calls to a BPP oracle. We now show that boolean circuit game Const-Approx GuaranteeNash is NPBPP -hard. Say that we have a language L ∈ NPBPP . By Lemma 5.13 there exists a non-deterministic TM M that computes L which makes only one query calls to a BPP oracle for the problem ACAPP, has all its nondeterminism at the beginning, and only accepts computations where the correct oracle query is encoded in the nondeterminism. Let f (|x|) be a polynomial that bounds the length of a string needed to encode the nondeterminism of M , let g(|x|) (without loss of generality even) be a polynomial that bounds the number of inputs to the circuit queried by M , let y be a string of bits that encodes the nondeterminism used in M on a particular run, and let r = log1/(4 ) 100g(|x|) . Given an input x we construct a boolean game G = (s, ν) with the following agents: f (|x|) agents y1 , . . . , yf (|x|) called y agents, f (|x|) agents y1 , . . . , yf (|x|) called y agents, r · g(|x|) agents z1 , . . . , zr·g(|x|) called z agents, and agents J1 and J2 called the judges. Let the string y = sy1 sy2 . . . syf (|x|) encode the nondeterminism of M , and let C be the circuit sent to the oracle query using the nondeterminism encoded in y, let k ∈ {0, 1} be the oracle query guess encoded by y, let m ∈ {0, 1} be the correct query response when C is queried, let n be the number of inputs to C, and let w = w1 w2 . . . wn where wi = XOR(sz(i?1)r+1 , sz(i?1)r+2 , . . . , szi·r ). The payo?s are as follows: y agents: agent yi is paid 1 regardless. y agents: agent yi receives payo? 1 if his strategy is the same as yi ’s and 0 otherwise. z agents: The z agents are paid according to a game of pennies (see Section 2). Agent zi plays pennies against agent zi+1 where i is odd. agent J1 : J1 receives payo?

1 2

if he plays 0 and C(w) if he plays 1.

agent J2 : J2 receives payo? 1 if J1 plays k and M run on input x with the nondeterministic choices encoded by y accepts assuming that the query result encoded by y is correct. Otherwise, J2 receives 0. 44

We guarantee that J2 and all the yi be paid 1. We make no guarantees to the other players. 1 Let = 800·f (|x|)·g(|x|) . Now we show that if x ∈ L then there exists a Nash equilibrium in G with these guarantees, and if x ∈ L then there exists no -Nash equilibrium in G within of these guarantees. Say x ∈ L. Then there exists a nondeterministic guess y = y1 y2 · · · yf (|x|) such that M accepts x run with the nondeterminism encoded by y and the query result encoded by y is correct. We claim that the strategy pro?le θ is a Nash equilibrium that meets the guarantees where θ is the strategy pro?le where: syi = syi = yi ; sJ1 = m, sJ2 = 1, and the z agents randomize uniformly between their two strategies. We ?rst show that in θ each agent is in equilibrium. The y agents and the y agents are in equilibrium because they all receive payo? 1. The z agents are because they are playing the unique equilibrium strategy of pennies. J1 is in equilibrium because if m = 0, 1 then C accepts at most 3 of its inputs. The XOR of uniformly random bits is uniformly random 1 so E[C(w)] = E[C(Un )] ≤ 3 (where Un is the uniform distribution over n-bit strings). And so J1 does better by playing sJ1 = 0 = m. If m = 1 a similar argument works. Finally, J2 ’s payo? is independent of his strategy and so he is also in equilibrium. The y agents all receive their guarantees of 1. J2 also receives his guarantee of 1 because sJ1 = m = k (because the oracle query result encoded by y is correct) and running M on x with the nondeterminism encoded by y results in an accepting computation. Say x ∈ L, then there exists no -Nash equilibrium within of the guarantees. For the sake of contradiction, assume that an -Nash equilibrium θ exists in which each agent is within of his guarantees. We note that each y agent must play some particular strategy with probability greater than 1 ? (if yi does not, then yi cannot attain a payo? of at least 1 ? ). Let syi be the strategy ? agent yi plays with probability ≥ 1 ? in θ, and let y = sy1 sy2 . . . syf (|x|) . By a union bound, ? ? ? ? Pr[θy1 θy2 . . . θyf (|x|) = y ] ≥ 1 ? f (|x|) · . Because < 100f1(|x|) , y is played with probability at least ? ? 99/100. Also Pr[θzi = 0] ∈ [1/2 ? 2 , 1/2 + 2 ] by Theorem A.1. So by Claim 7.14 each bit wi , which is the XOR of r such bits, is within (4 )r ≤ 1/100n of being uniform, and their joint distribution is within 1/100 of uniform. So E[C(w)] is within 1/100 of E[C(Un )]. Now because x ∈ L either y encodes a rejecting computation on M , or the query result of y is ? incorrect. In the former case, J3 receives payo? 0 whenever y is played, and so cannot receive more ? than 1/100. In the latter case, k = m. If k = 0 and m = 1 then agent J1 will receive 1 for playing 2 1 1 1 0, but will receive E[C(w)] ≥ E[C(Un )] ? 100 ≥ 2 ? 100 for playing 1. Because [ 2 ? 100 ] ? [ 1 ] > 2 by 3 3 2 Claim 7.12 Pr[θJ1 = 1] ≥ 1 and so J3 ’s payo? will be at most 1/2 < 1 ? . A symmetric argument 2 handles the case where k = 1 and m = 0. Theorem 8.5 Graph game and boolean graph game GuaranteeNash is NP-complete for all levels of approximation. The results hold even when restricted to degree d graphs, d ≥ 3. Graph game GuaranteeNash is in NP because we can guess a strategy pro?le -uniform strategy pro?le θ and test in polynomial time whether θ is an -Nash 2 equilibrium where each player is payed within of his guarantees. If it is, accept; if not, reject. This algorithm works because by Theorem 3.4, if there exits a Nash equilibrium that meets the 2 2 guarantees, then there exists a n log(n maxi |si |) -uniform -Nash equilibrium that gets within /2 of 2 the guarantees.

n2 log(n2 maxi |si |)

Proof:

To show that it is NP-hard, we reduce from CircuitSat. Given a circuit C we create an instance of boolean graph game GuaranteeNash, (G, , (1, . . . , 1)). We create G with the following 45

agents: 2 input agents x and x for each input x to C and a gate agent g for each gate g in the circuit. Each agent has a strategy space of {0, 1}. Each input agent x is paid 1 regardless. Each input agent, x is paid 1 only if it plays the same strategy as x, the other input agent that represents the same circuit input. Except for the gate agent associated with the output gate, each gate agent g is paid 1 for correctly computing the output of his gate (with respect to the strategies of the agents that correspond to the inputs of his gate), and is paid 0 otherwise. If an input x to the circuit is an input to a gate g, then the agent associated with the gate g receives his payo? according to x’s strategy (not x ’s strategy). The output agent gets paid 1 only if both he correctly computes the output of his gate and that value is 1. Let = 1/100. We now show that if C ∈SAT, then there exists a Nash equilibrium that meets the guarantees, but if C ∈SAT, then there exist no -Nash equilibrium that comes within of meeting the guarantees. Say C has a satisfying assignment. The the strategy pro?le where all input agents play a strategy which corresponds to some satisfying assignment, and the gate agents correctly evaluate their gates is a Nash equilibrium that meets the guarantees. It is a Nash equilibrium because each agent receives a payo? of 1, and so cannot do better. The input agents receive 1 because x and x always play the same strategy. Each gate agent besides the output agent is paid 1 because he correctly computes the output of his gate with respect to the strategies of the two inputs. The output gate agent correctly computes the output of his gate with respect to the strategies of the two inputs; moreover, because this is a satisfying assignment, the output he computes is 1, and so he also receives a payo? of 1. If C has no satisfying assignment, then there exists no -Nash equilibrium that comes within of the guarantees. Every -Nash equilibrium of G which is within of the guarantees corresponds to some evaluation of the circuit. By induction we show that every player in the game (with the exception of the gate agent associated with the output gate) plays a pure strategy that corresponds with some evaluation of the circuit with probability > 1 ? 2 . The base case is the input gates. Every input agent x must play some strategy with probability ≥ 1 ? , otherwise, his strategy will not agree with x ’s strategy with probability ≥ 1 ? no matter what x plays, and so x ’s payo? will be less than 1 ? . Given that the input agents each play some strategy the majority of the time, we can de?ne a correct strategy for gate agent g. We call the strategy of the gate agent g correct if it corresponds with the output of the gate g in an evaluation of C using the strategy that agent x plays the majority of the time as the input to gate x. We claim that in any -Nash equilibrium, each gate agent besides the output agent must play the correct strategy with probability ≥ 1 ? 2 . We proceed by induction. The base case has already been proven. The inductive step is exactly as in the proof of Claim 7.16. Therefore, in any qualifying Nash equilibrium, each player (beside the output gate player) must play a strategy corresponding with the correct valuation of the circuit. But because there is no satisfying assignment, the agent assigned to the output node, will not get a payo? close to 1. For in the correct valuation, his gate evaluates to 0, but if he plays 0, he is paid nothing. So the best he can do is play 1. Because each of the agents corresponding to the input gates play the correct strategy with probability ≥ 1 ? 2 , and the output gate receives nothing when they both play the correct strategy, the most that the output agent can be paid is 4 < 1 ? . Conitzer and Sandholm [CS03] showed that Exact GuaranteeNash is NP-complete in bimatrix games. We observe that the same holds even for Poly-Approx: Theorem 8.6 [CS03] Bimatrix Exact and Poly-Approx GuaranteeNash are NP-complete. 46

Proof: The hardness proof is exactly the same as the proof of Theorem 8.1 except now N is polynomial in the size of the input instead of exponential. It is in NP because we can guess a polynomially-sized strategy pro?le, θ, and then in polynomial time check that it is a Nash equilibrium that satis?es the guarantees. By Proposition 3.3 If such a Nash equilibrium exits, then there exists one of at most polynomial size. ? Theorem 8.7 Bimatrix Const-Approx GuaranteeNash is in P. Proof: Given an instance (G, , (g1 , . . . , gn )) simply look through all the k-uniform strategies, where k = 4 log(4(maxi |si |) for a strategy pro?le that is an -Nash equilibrium where the payo?s to )2 players are within /2 of their guarantees. There are only a quasipolynomial number of k-uniform strategies and checking each strategy takes only polynomial time. If such a strategy is found, accept, otherwise reject. If there is no -Nash equilibrium within of the guarantees, surely the algorithm will not ?nd one. However, if there exists some Nash equilibrium θ that pays o? each player his guaranteed amount, then by Theorem 3.4 there will exist a k-uniform -Nash equilibrium θ that is within /2 of the guarantees, and so the algorithm will ?nd it.

Acknowledgments

We thank Eli Ben-Sasson, Adam Klivans, Ryan O’Donnell, Rocco Servedio, and Amir Shpilka for many discussions about algorithmic aspects of Nash equilibria which informed this work. We thank Mike Kearns, Christos Papadimitriou, David Parkes, and Avi Pfe?er for helpful pointers and advice, and we thank Saurabh Sanghvi for fruitful discussions during the initial stages of this work.

References

[ALM+ 98] Sanjeev Arora, Carsten Lund, Rajeev Motwani, Madhu Sudan, and Mario Szegedy. Proof veri?cation and the hardness of approximation problems. Journal of the ACM, 45(3):501–555, May 1998. [AS98] [BM88] Sanjeev Arora and Shmuel Safra. Probabilistic checking of proofs: A new characterization of NP. Journal of the ACM, 45(1):70–122, January 1998. L?szl? Babai and Shlomo Moran. Arthur-Merlin games: A randomized proof system and a o a hierarchy of complexity classes. Journal of Computer and System Sciences, 36:254– 276, 1988. Vincent Conitzer and Tuoman Sandholm. Complexity results about nash equilibria. In International Joint Conference on Arti?cial Intelligence (IJCAI), 2003.

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Joan Feigenbaum, Daphhe Koller, and Peter Shor. A game-theoretic classi?cation of interactive complexity classes. In Tenth Annual IEEE Conference on Computational Complexity, pages 227–237, 1995. Georg Gottlob, Gianluigi Greco, and Francesco Scarcello. Pure nash equilibria: hard and easy games. In TARK ’03: Proceedings of the 9th conference on Theoretical aspects of rationality and knowledge, pages 215–230, New York, NY, USA, 2003. ACM Press. Sha? Goldwasser, Silvio Micali, and Charles Racko?. The knowledge complexity of interactive proof systems. SIAM Journal on Computing, 18(1):186–208, February 1989.

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[GMW91] Oded Goldreich, Silvio Micali, and Avi Wigderson. Proofs that yield nothing but their validity or all languages in NP have zero-knowledge proof systems. Journal of the ACM, 38(1):691–729, 1991. [Gol05] Oded Goldreich. On promise problems (a survey in memory of Shimon Even [1935– 2004]). Technical Report TR05-018, Electronic Colloquium on Computational Complexity, February 2005. Itzhak Gilboa and Eitan Zemel. Nash and correlated equilibria: Some complexity considerations. Games and Economic Behavior, 1:80–93, 1989. W. Hesse, E. Allender, and D. Barrington. Uniform constant-depth threshold circuits for division and iterated multiplication. Journal of Computer and System Sciences, 65:695–716, 2002. Michael Kearns, Michael L. Littman, and Satinder Singh. Graphical models for game theory. In UAI, pages 253–260, 2001.

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[LMM03] Richard Lipton, Evangelos Markakis, and Aranyak Mehta. Playing large games using simple strategies. In AMC Conference on Electronic Commerce, 2003. [MP91] Nimrod Megiddo and Christos H. Papadimitriou. A note on total functions, existence theorems, and computational complexity. Theoretical Computer Science, 81(1):317–324, 1991. John Nash. Non-cooperative games. Annals of Mathematics, 54(2):286–295, 1951. J. F. Nash and L. S. Shapley. A simple three-person poker game. Contributions to the Theory of Games, 1(24):105–116, 1950. Christos H. Papadimitriou. Algorithms, games, and the internet. In STOC, 2001.

[Nas51] [NS50] [Pap01]

A

Analysis of Pennies

The game of pennies G = (s, ν) involves 2 players. s1 = s2 = {0, 1} and the payo?s are as follows: Player 2 Heads Tails (1, 0) (0, 1) (0, 1) (1, 0)

Player 1

Heads Tails 48

Pennies has a unique Nash equilibrium where both agents randomize uniformly between their two strategies. If the second player does not randomize equally between his strategies, player 1’s best strategy is to play, with probability 1, the strategy that player 2 plays more often. And similarly if player 1 does not randomize equally, player 2 does the opposite of what player 1 plays most often. So the only Nash equilibrium is when both players randomized equally between their two options. The following theorem gives us an idea of what constitutes an -Nash equilibrium in the game of pennies. Theorem A.1 In any -Nash equilibrium of pennies, each player randomizes between each strategy 1 with probability 2 ± 2 . Proof: Say that player 1 plays 1 with probability p and player 2 plays 1 with probability q. Then the payo? to player 1 is pq + (1 ? p)(1 ? q). Now let p = 1 + δ and q = 1 + δ . If agent 1 2 2 plays a pure strategy his payo? will be either q or 1 ? q. In any -Nash equilibrium it must be that pq + (1 ? p)(1 ? q) + ≥ max{q, 1 ? q} ? max{q, 1 ? q} ? [pq + (1 ? p)(1 ? q)] ≤ . Say that δ ≥ 0. Then we get 1 1 1 1 1 + δ ? ( + δ)( + δ ) + ( ? δ)( ? δ ) ≤ ? δ ? 2δδ ≤ ? δ ≤ 2 2 2 2 2 1 ? 2δ Similarly, if δ ≤ 0. Then we get 1 1 1 1 1 ? δ ? ( + δ)( + δ ) + ( ? δ)( ? δ ) ≤ ? ?δ ? 2δδ ≤ ? ?δ ≤ 2 2 2 2 2 1 + 2δ Doing the same thing for agent 2 with δ > 0: 1 1 1 1 1 + δ ? ( + δ)( ? δ ) + ( ? δ)( + δ ) ≤ ? δ + 2δδ ≤ ? δ ≤ 2 2 2 2 2 1 + 2δ And now with δ < 0: 1 1 1 1 1 ? δ ? ( + δ)( ? δ ) + ( ? δ)( + δ ) ≤ ? ?δ + 2δδ ≤ ? ?δ ≤ 2 2 2 2 2 1 ? 2δ Now by substitution and algebraic manipulation, we can see that this conditions require that |δ|, |δ | < 2 .

49

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