Theory Comput SystDOI 10.1007/s00224-013-9523-7

Complexity of Rational and Irrational Nash Equilibria

Vittorio Bilò · Marios Mavronicolas

© Springer Science+Business Media New York 2013

Abstract We introduce two new natural decision problems, denoted as ∃ RATIONALNASH and ∃ IRRATIONAL NASH, pertinent to the rationality and irrationality, re-spectively, of Nash equilibria for (finite) strategic games. These problems ask, givena strategic game, whether or not it admits (i) a rational Nash equilibrium where allprobabilities are rational numbers, and (ii) an irrational Nash equilibrium where atleast one probability is irrational, respectively. We are interested here in the complex-ities of ∃ RATIONAL NASH and ∃ IRRATIONAL NASH.

Towards this end, we study two other decision problems, denoted as NASH-EQUIVALENCE and NASH-REDUCTION, pertinent to some mutual properties of thesets of Nash equilibria of two given strategic games with the same number of players.The problem NASH-EQUIVALENCE asks whether or not the two sets of Nash equi-libria coincide; we identify a restriction of its complementary problem that witnesses∃ RATIONAL NASH. The problem NASH-REDUCTION asks whether or not there isa so called Nash reduction: a suitable map between corresponding strategy sets ofplayers that yields a Nash equilibrium of the former game from a Nash equilibriumof the latter game; we identify a restriction of NASH-REDUCTION that witnesses∃ IRRATIONAL NASH.

As our main result, we provide two distinct reductions to simultaneously show that(i) NASH-EQUIVALENCE is co-NP-hard and ∃ RATIONAL NASH is NP-hard, and

Part of the work of the first author was done while visiting University of L’Aquila, Italy andUniversity of Cyprus, Cyprus. The second author has been partially supported by research funds atthe University of Cyprus. Part of his work was done while visiting University of L’Aquila, Italy.

V. Bilò (B)Dipartimento di Matematica e Fisica “Ennio De Giorgi”, Università del Salento,Provinciale Lecce-Arnesano, P.O. Box 193, 73100 Lecce, Italye-mail: vittorio.bilo@unisalento.it

M. MavronicolasDepartment of Computer Science, University of Cyprus, Nicosia 1678, Cypruse-mail: mavronic@cs.ucy.ac.cy

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(ii) NASH-REDUCTION and ∃ IRRATIONAL NASH are both NP-hard, respectively.The reductions significantly extend techniques previously employed by Conitzerand Sandholm (Proceedings of the 18th Joint Conference on Artificial Intelligence,pp. 765–771, 2003; Games Econ. Behav. 63(2), 621–641, 2008).

Keywords Complexity theory · Decision problems · Strategic games · Nashequilibria

1 Introduction

1.1 Motivation and Framework

Understanding the complexity of algorithmic problems pertinent to equilibria in (fi-nite) strategic games is one of the most intensively studied topics in AlgorithmicGame Theory today (see, for example, [2–6, 8, 9, 15, 16, 19] and references therein).Much of this research has focused on Nash equilibria [17, 18], the most influentialequilibrium concept in Game Theory. In the wake of the complexity results for searchproblems about Nash equilibria, a series of breakthrough works [3, 8] shows that,even for two-player games, computing an (exact) Nash equilibrium is complete forPPAD [19], a complexity class to capture the computation of discrete fixed points;so also is computing an approximate Nash equilibrium.

In this work, we continue the study of the complexity of decision problems aboutNash equilibria. The celebrated result of John Nash [17, 18] shows that every (finite)game admits a mixed Nash equilibrium; so, it trivializes the decision problem aboutthe existence of (mixed) Nash equilibria, while it simultaneously leaves open thecomplexity of decision problems about the existence of Nash equilibria with certainproperties.

Gilboa and Zemel [13] were the first to present complexity results (more specif-ically, NP-hardness results) about mixed Nash equilibria (and correlated equilib-ria) for games represented in explicit form; they identified some NP-hard decisionproblems about the existence of (mixed) Nash equilibria with certain properties fortwo-player strategic games. (For example, it is NP-hard to decide if a game admitsa Nash equilibrium where each player receives utility above some threshold.)

Much later, Conitzer and Sandholm [5, 6] provided a very notable unifying reduc-tion, henceforth abbreviated as CS-reduction, to show that all decision problems from[13] and many more are NP-hard. The CS-reduction [5, 6] yields a two-player gameout of a CNF formula φ; it is then shown that the game has a Nash equilibrium withcertain properties (in addition to some fixed pure Nash equilibrium) if and only if φ

is satisfiable. Hence, deciding the properties is NP-hard.The CS-reduction uses literals, variables and clauses from the formula φ, together

with a special strategy f , as the strategies of each player. The essence of the CS-reduction is that (i) both players choosing f results in a pure Nash equilibrium, and(ii) a player could otherwise improve (by switching to f ) unless both players onlyrandomize over literals. More important, a Nash equilibrium where both players onlyrandomize over literals is possible (and has certain properties) if and only if φ issatisfiable.

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1.2 Summary of Contribution

In this paper, we shall extend the work from [5, 6, 13] to decision problems pertinentto rationality and irrationality properties of mixed Nash equilibria. Recall that a Nashequilibrium is rational if all involved probabilities are rational and otherwise it isirrational; so, a pure Nash equilibrium is rational but not vice-versa. All two-playergames have only rational Nash equilibria, while there are already known three-playergames with no rational Nash equilibrium (cf. [18]).

We introduce two new natural decision problems, denoted as ∃ RATIONAL NASHand ∃ IRRATIONAL NASH, respectively; these problems ask whether or not thereis a rational (resp., irrational) Nash equilibrium.1 Hence, both problems ∃ RATIO-NAL NASH and ∃ IRRATIONAL NASH trivialize when restricted to two-player gamesbut become non-trivial for games with at least three players. Since the CS-reduction[5, 6] applies to two-player games, it will not be directly applicable to settling thecomplexity of ∃ RATIONAL NASH and ∃ IRRATIONAL NASH.

1.3 Milestones

1.3.1 Plan

To establish the NP-hardness of the problems ∃ RATIONAL NASH and ∃ IRRA-TIONAL NASH we shall use two new suitable decision problems that make no ref-erence to rationality or irrationality properties of Nash equilibria. These problemswill be NASH-EQUIVALENCE and NASH-REDUCTION, respectively, but yet witness∃ RATIONAL NASH and ∃ IRRATIONAL NASH, respectively. Both problems receiveas input a pair of strategic games SG and SG with the same number of players r ≥ 2;they inquire about some mutual properties of their Nash equilibria. We shall usetwo distinct CS-like reductions with each simultaneously showing that both prob-lems NASH-EQUIVALENCE and ∃ RATIONAL NASH (resp., NASH-REDUCTION and∃ IRRATIONAL NASH) are NP-hard.

To the best of our knowledge, these complexity results for ∃ RATIONAL NASHand ∃ IRRATIONAL NASH are the first NP-hardness results for a decision probleminquiring the existence of a combinatorial object involving rational (resp., irrational)numbers; no such NP-hard problems are listed in [12].

1.3.2 NASH-EQUIVALENCE as a Witness to ∃ RATIONAL NASH

The problem NASH-EQUIVALENCE asks whether the sets of Nash equilibria for thegames SG and SG coincide. Fixing SG to some gadget game yields the restrictedproblem NASH-EQUIVALENCE(SG) with a single input SG.

Assume that (i) the set of Nash equilibria for SG is a subset of those for SG and(ii) SG has no rational Nash equilibrium, then SG may have a rational Nash equilib-rium if and only if the set of Nash equilibria for SG and SG do not coincide. So, the

1We were inspired to study these problems by a corresponding question posed by E. Koutsoupias [14] toM. Yannakakis during his Invited Talk at SAGT 2009.

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existence of a rational Nash equilibrium for SG is a witness to the non-equivalence

of SG and SG under the given assumptions. So, if NASH-EQUIVALENCE(SG) is

NP-hard, then so is ∃ RATIONAL NASH. We show that NASH-EQUIVALENCE(SG)

is NP-hard for an arbitrary but fixed strategic game SG (Proposition 2 and Theo-rem 1). Fixing SG to admit no rational Nash equilibrium yields that ∃ RATIONALNASH is NP-hard (Proposition 4 and Theorem 2).

1.3.3 NASH-REDUCTION as a Witness to ∃ IRRATIONAL NASH

The problem NASH-REDUCTION asks whether there is a Nash reduction from SG toSG. Roughly speaking, a Nash reduction consists of a family of surjective functions,one per player, mapping the strategy set of each player in SG to the strategy set of thesame player in SG. Note that any family of surjective functions induces a map frommixed profiles for SG to mixed profiles for SG in the natural way: probabilities todifferent strategies of a player in SG that map to the same strategy (of the player) inSG are added up. However, a Nash reduction must, in addition, preserve at least oneNash equilibrium: for any Nash equilibrium for SG, there must be a Nash equilibriumfor SG that maps to it.

Assume that (i) there is a Nash reduction from SG to SG, and (ii) SG has onlyrational Nash equilibria. Then SG has at least one rational Nash equilibrium. It fol-lows from the contraposition that, if SG has no rational Nash equilibrium, then either(i′) there is no Nash reduction from SG to SG or (ii′) SG has an irrational Nash equi-librium. Hence, the inexistence of an irrational Nash equilibrium for SG is a witnessto the inexistence of a Nash reduction from SG to SG. So, if NASH-REDUCTION(SG)is NP-hard, then so is ∃ IRRATIONAL NASH.

We show that NASH-REDUCTION(SG) is NP-hard for a fixed strategic game SGwhich (a) is constant-sum with sum r · u, (b) has a unique Nash equilibrium whichis (b/i) fully mixed and in which (b/ii) the utility of each player is u (Proposition 6and Theorem 3). Fixing the gadget SG so that, in addition, it admits no rational Nashequilibrium yields that ∃ IRRATIONAL NASH is NP-hard (Proposition 8 and Theo-rem 4).

1.4 The two CS-like Reductions

Both reductions yield a game SG = SG(φ) with an arbitrary number of players r ≥ 2inherited from the gadget game SG; recall that the CS-reduction yields a two-playergame. Hence, the resulting game may or may not have properties, such as having arational Nash equilibrium or having an irrational Nash equilibrium, which two-playergames necessarily have or necessarily do not have, respectively.

1.4.1 The Reduction to NASH-EQUIVALENCE(SG) and ∃ RATIONALNASH

For each player, the special strategy f from the CS-reduction [5, 6] is replaced bythe strategies of the player in the gadget game SG. Two features of the reduction are:(i) If all players choose strategies as in a Nash equilibrium for SG, the result is aNash equilibrium for the resulting game SG(φ). This implies that SG(φ) necessarily

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has Nash equilibria with desirable properties (such as irrationality), as opposed tonecessarily having a pure Nash equilibrium (in the case of the CS-reduction). (ii) Itstill holds that a player could otherwise improve (by switching to a strategy fromthe gadget game) unless all players only randomize over literals. More important, arational Nash equilibrium where all player only randomize over literals is possible ifand only if the formula φ is satisfiable.

1.4.2 The Reduction to NASH-REDUCTION(SG) and ∃ IRRATIONAL NASH

For each player, the special strategy f from the CS-reduction [5, 6] remains. A fea-ture of the reduction is: (i) It still holds that a player could otherwise improve (byswitching to the special strategy f ) unless all players only randomize over literals.

We use a surjective map from (disjoint) sets of literals in φ to single strategiesfrom the gadget game SG. This allows using utilities from the gadget game to defineutilities when all players choose literals. (In the CS-reduction, these utilities wereidentical for all players.)

So, if all players randomize over literals to induce a Nash equilibrium σ for SG,the result is a profile σ for SG that preserves rationality (resp., irrationality) of σ .More important, σ is an (irrational) Nash equilibrium for SG (and, hence, there is areduction from SG to SG) if and only if the formula φ is satisfiable.

1.5 Other Related Work

Etessami and Yannakakis [9] study the dual search problem, over games with at leastthree players, of computing an approximation to a Nash equilibrium within a speci-fied precision: a rational profile that differs from an (irrational) Nash equilibrium byat most ε in every coordinate. Note that this problem is different than computing anε-approximate Nash equilibrium, which may be very far from an actual Nash equi-librium. It is shown [9, Theorem 4] that placing this problem in NP would implythe breakthrough result that the SQUARE-ROOT-SUM [11] problem in also in NP ,which is a long-standing open problem in Complexity Theory. It is also shown [9,Theorem 18] that the same problem is complete for the complexity class FIXP [9].

In a recent work [10], Fiat and Papadimitriou consider the decision problem aboutthe existence of a (mixed) Nash equilibrium for a two-players game whose players arenot expectation-maximizers. The existence result of Nash [17, 18] does not apply tosuch games, so that the decision problem now becomes non-trivial. Fiat and Papadim-itriou use a generalized rock-paper-scissors game as a gadget game in a reductionarguably simpler than the CS-reduction [5, 6]; thereby, they prove [10, Theorem 5]that it is NP-hard to decide if such a two-players game has a Nash equilibrium.

2 Definitions and Preliminaries

2.1 Background from Game Theory

2.1.1 Games, Strategies and Profiles

A game is a triple SG = 〈[r], {Σi}i∈[r], {Ui}i∈[r]〉, where: (i) [r] = {1, . . . , r} is afinite set of players with r ≥ 2, and (ii) for each player i ∈ [r], Σi is the set of

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strategies for player i, and Ui is the utility function Ui : �k∈[r]Σk → R for player i.For each player i ∈ [r], denote Σ−i = �k∈[r]\{i}Σk ; denote Σ = �k∈[r]Σk . For anyinteger r ≥ 2, denote as r-SG the set of r-player games; so, SG = ⋃

r≥2 r-SG is theset of all games.

A profile is a tuple s of r strategies, one for each player; so s ∈ Σ . For a pro-file s, the vector U(s) = 〈U1(s), . . . ,Ur (s)〉 is called the utility vector. The game SGis constant-sum if there is a constant c such that for each profile s,

∑

i∈[r] Ui (s) = c.A partial profile s−i is a tuple of r − 1 strategies, one for each player other than i; sos−i ∈ Σ−i . For a profile s and a strategy ti ∈ Σi , denote as s−i ti the profile obtainedby substituting strategy ti for strategy si in the profile s.

A mixed strategy for player i ∈ [r] is a probability distribution σi on her strategyset Σi : a function σi : Σi → [0,1] such that

∑

s∈Σiσi(s) = 1. Denote as Support(σi)

the set of strategies s ∈ Σi such that σi(s) > 0. The mixed strategy σi : Σi → [0,1]is rational if all values of σi are rational numbers; else, it is irrational.

2.1.2 Mixed Profiles

A mixed profile σ = (σi)i∈[r] is a tuple of mixed strategies, one for each player.A partial mixed profile σ−i is a tuple of r − 1 mixed strategies, one for each playerother than i. For a mixed profile σ and a mixed strategy τi of player i ∈ [r], denoteas σ−i τi the mixed profile obtained by substituting the mixed strategy τi for themixed strategy σi in the mixed profile σ .

A mixed profile is rational if all its mixed strategies are rational; else, it is ir-rational. So, a profile is the degenerate case of a rational mixed profile where allrational probabilities are either 0 or 1.

2.1.3 Nash Equilibria

A mixed profile σ induces a probability measure Pσ on the set of profiles in thenatural way. Say that the profile s is enabled in the mixed profile σ , and write s ∼ σ ,if Pσ (s) > 0; note that for a profile s,

Pσ (s) =∏

k∈[r]σk(sk).

Under the mixed profile σ , the utility of each player becomes a random variable. So,associated with the mixed profile σ is the expected utility for each player i ∈ [r],denoted as Ui (σ ) and defined as the expectation according to Pσ of her utility for aprofile s enabled in the mixed profile σ ; so,

Ui (σ ) = Es∼σ

(

Ui (s))

=∑

s∈Σ(SG)

Pσ (s) · Ui (s)

=∑

s∈Σ(SG)

(

∏

k∈[r]σk(sk)

)

· Ui (s).

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A pure Nash equilibrium, or Nash equilibrium for short, is a profile s ∈ Σ suchthat for each player i ∈ [r] and for each strategy ti ∈ Σi , Ui (s) ≥ Ui (s−i ti ).

A mixed Nash equilibrium is a mixed profile σ such that for each player i ∈ [r]and for each mixed strategy τi , Ui (σ ) ≥ Ui (σ−i τi). The mixed profile σ is a mixedNash equilibrium if for each player i ∈ [r] and for each strategy ti ∈ Σi , Ui (σ ) ≥Ui (σ−i ti ).

For a strategic game SG, denote as NE(SG) the set of Nash equilibria for SG.Nash equilibria shall be classified as pure, mixed rational (or rational for short) andmixed irrational (or irrational for short) in the natural way.

We shall make extensive use of the following characterization of mixed Nash equi-libria.

Lemma 1 A mixed profile σ is a Nash equilibrium if and only if for each playeri ∈ [r], (i) for each pair of strategies si, s

′i ∈ Support(σi), Ui (σ−i si) = Ui (σ−i s′

i ),(ii) for each strategy s′

i ∈ Support(σi), Ui (σ ) = Ui (σ−i s′i ), and (iii) for each pair

of strategies si ∈ Support(σi) and s′i /∈ Support(σi), Ui (σ−i si) ≥ Ui (σ−i s′

i ).

2.2 Background from Complexity Theory

A decision problem Π is identified with a set of positive instances encoded over{0,1}. The class NP is the set of all decision problems Π for which there is apolynomial-time algorithm V such that for every instance w ∈ {0,1}∗, w is a positiveinstance for Π if and only if there is a certificate c ∈ {0,1}p(|w|) (for some polyno-mial p : N → N) such that V accepts the input 〈w, c〉. The class co-NP is the set ofall complements of decision problems in NP .

Say that the decision problem Π ′ polynomially reduces to the decision problemΠ , denoted as Π ′ ≤P Π , if there is a polynomial-time function f : {0,1}∗ → {0,1}∗such that w is a positive instance for Π ′ if and only if f(w) is a positive instance forΠ . The decision problem Π is NP-hard if for every decision problem Π ′ ∈ NP ,Π ′ ≤P Π .

A boolean formula φ is in Conjunctive Normal Form, denoted as CNF, if it isa conjunction of clauses; that is, φ = ∧

1≤i≤k

∨

1≤j≤l �ij , where �ij is a literal: aboolean variable or its negation. Denote the set of clauses in φ as C(φ) = {∨j∈[l] �ij |i ∈ [k]}. Denote as V(φ) and L(φ) the sets of variables and literals, respectively, inthe formula φ, with |V(φ)| = n and |L(φ)| = 2n. The function v : L(φ) → V(φ) givesthe variable corresponding to a literal. When φ is clear from the context, we shall useL, V and C for L(φ), V(φ) and C(φ), respectively.

An assignment is a function α : V(φ) → {0,1}; it is represented by a tuple ofliterals 〈�1, . . . , �n〉, where for each index j ∈ [n], v(�j ) = vj and �j = 1 (under theassignment). The value of φ under the assignment α, denoted as φ(α), is definedin the natural way. The formula φ is satisfiable if there is a satisfying assignment: anassignment α such that φ(α) = 1. The decision problem CNF SAT, identified with theset of satisfiable boolean formulas in CNF, is the archetypical NP-hard problem [7].

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2.3 Decision Problems about Nash Equilibria

A decision problem about Nash equilibria asks whether or not, given a game, it hasa Nash equilibrium with certain properties. We shall consider the particular (class of)properties:

2.3.1 NASH-EQUIVALENCE

A pair of games SG and SG are Nash-equivalent if NE(SG) = NE(SG). A decisionproblem about Nash equilibria arises in the natural way:

NASH-EQUIVALENCE

INSTANCE: Two games SG and SG from r-SG, for some integer r ≥ 2.QUESTION: Are SG and SG Nash-equivalent?

A restriction of NASH-EQUIVALENCE yields a decision problem parameterized bysome fixed game SG from r-SG, for some integer r ≥ 2; the game SG will be calleda gadget game.

NASH-EQUIVALENCE(SG)

INSTANCE: A game SG from r-SG.QUESTION: Are SG and SG Nash-equivalent?

So, NASH-EQUIVALENCE (SG) ≤P NASH-EQUIVALENCE.

2.3.2 NASH-REDUCTION

Consider now a pair of games SG,SG ∈ r-SG. Assume that for each player i ∈ [r],|Σi | ≥ |Σi |. A surjective map from SG to SG is a family of surjective functions H ={hi : Σi → Σi}i∈[r], one for each player i ∈ [r]; so, a surjective map maps strategiesof each player i ∈ [r] in the game SG to strategies of the same player in the game SGin a surjective way.

Note that the surjective map H induces a corresponding surjective map from theset of profiles Σ for the game SG to the set of profiles Σ for the game SG. In turn,it induces a map H from the set of mixed profiles for SG to the set of mixed profilesfor SG as follows. Consider a mixed profile σ for SG. Then, σ maps to the mixedprofile σ = H(σ ), where, for each player i ∈ [r] and each strategy s ∈ Σi ,

σi (s) =∑

s∈Σi |hi (s)=s

σi(s).

A Nash reduction from the game SG to the game SG is a surjective map fromSG to SG such that there is a Nash equilibrium σ for the game SG for which themixed profile H(σ ) is a Nash equilibrium for the game SG. Say that the game SGNash-reduces to the game SG if there is a Nash reduction from SG to SG. This leadsnaturally to the following decision problem about Nash equilibria:

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NASH-REDUCTION

INSTANCE: Two games SG and SG from r-SG, for some integer r ≥ 2.QUESTION: Does SG Nash-reduce to SG?

A restriction of NASH-REDUCTION yields a decision problem parameterized bysome fixed game SG from r-SG, for some integer r ≥ 2; the game SG will be calleda gadget game.

NASH − REDUCTION(SG)

INSTANCE: A game SG from r-SG.QUESTION: Does SG Nash-reduce to SG?

So, NASH − REDUCTION (SG) ≤P NASH − REDUCTION.

2.4 ∃ RATIONAL NASH and ∃ IRRATIONAL NASH

To the best of our knowledge, the following decision problems about Nash equilibriaare new.

∃ RATIONAL NASH

INSTANCE: A game SG.QUESTION: Does SG have a rational Nash equilibrium?

∃ IRRATIONAL NASH

INSTANCE: A game SG.QUESTION: Does SG have an irrational Nash equilibrium?

3 Complexity of NASH-EQUIVALENCE and ∃ RATIONAL NASH

Fix a gadget game SG = 〈[r], {Σi}i∈[r], {Ui}i∈[r]〉, with r ≥ 2. Denote Σ = Σ1 ×. . . × Σr . Denote u = max{Ui (s)}i∈[r],s∈Σ and u = min{Ui (s)}i∈[r],s∈Σ ; so, u and u

are the maximum and minimum utilities, respectively, of a player in the game SG.We shall use π to denote a permutation on [r].

3.1 The Reduction

Given a CNF formula φ, construct the game SG(φ) = 〈[r], {Σi}i∈[r], {Ui}i∈[r]〉, orSG for short, as follows:

– For each player i ∈ [2], called a special player, Σi := Σi ∪ L(φ) ∪ V(φ) ∪ C(φ);for each player i ∈ [r] \ [2], Σi := Σi ∪ {δ}, where δ is a special strategy. So,each strategy in Σi with i ∈ [r] is inherited from the gadget game SG. For eachplayer i ∈ [2], her remaining strategies come from the formula φ; in more detail,

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these strategies are the literals, variables and clauses, respectively, of the formulaφ. Note that the special players are the only players whose strategies are influencedby the formula φ.

– Fix now a profile s from Σ = Σ1 × . . . × Σr . Use s to partition [r] into

P(s) = {

i ∈ [r] | si ∈ Σi

}

and

P(s) = {

i ∈ [r] | si /∈ Σi

};

loosely speaking, P(s) and P(s) are the sets of players choosing and not choosing,respectively, strategies from the game SG. The utility vector U(s) is depicted in thefollowing table, where v ∈ V(φ), �1, �2, � ∈ L(φ) and c ∈ C(φ):

Case Condition on the profile s Utility vector U(s)

(1) s = 〈�1, �2, δ, . . . , δ〉 with �1 �= �2 〈u + 1, . . . , u + 1〉(2) s = 〈�1, �2, δ, . . . , δ〉 with �1 = �2 〈u − 1, . . . , u − 1〉(3) s = 〈v, �, δ, . . . , δ〉 with v(�) �= v 〈u + 2, u − 1, . . . , u − 1〉(4) s = 〈v, �, δ, . . . , δ〉 with v(�) �= v 〈u + 2 − n,u − 1, . . . , u − 1〉(5) s = 〈c, �, δ, . . . , δ〉 with � /∈ c 〈u + 2, u − 1, . . . , u − 1〉(6) s = 〈c, �, δ, . . . , δ〉 with � ∈ c 〈u + 2 − n,u − 1, . . . , u − 1〉(7) For each i ∈ [r], si ∈ Σi U(〈s1, . . . , sr 〉)

(8) P(s) �= ∅ and P(s) �= ∅ Ui (s) ={

u + 1, if i ∈ P(s)u − 1, if i ∈ P(s)

(9) s = π(t), where t falls in one of π(U(t))

the Cases (1) through (8)

(10) None of the above Ui (s) = u − 1 for each i ∈ [r]Clearly, the construction of SG(φ) from φ is carried out in polynomial time. Forbrevity, we shall also write SG, L, V and C for SG(φ), L(φ), V(φ) and C(φ), respec-tively.

From the construction of the utility functions, we observe that for each profile sand for each player i ∈ [r] \ [2], Ui (s) ≤ u + 1. We also observe that

∑

i∈[2]Ui (s) ≤ 2(u + 1)

with

∑

i∈[2]Ui (s) = 2(u + 1)

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if and only if s = 〈�1, �2, δ, . . . , δ〉 with �1 �= �2. This implies that for any property Psatisfied by at least one pure profile enabled in a mixed profile σ ,

Es∼σ

(

∑

i∈[2]Ui (s) | s satisfies P

)

≤ 2 (u + 1).

For a fixed player i ∈ [r], denote

σi(Σi) =∑

s∈Σi

σi(σ )

and

σi(L) =∑

�∈L

σi(�);

so, σi(Σi) and σi(L) are the probability masses put on strategies from Σi and onliterals from L, respectively. We prove:

Proposition 1 The following conditions are equivalent: (1) φ is satisfiable, (2)NE(SG) �= NE(SG), and (3) SG admits a rational Nash equilibrium in NE(SG) \NE(SG).

Proof We first prove that Condition (1) implies Conditions (2) and (3).

Lemma 2 A satisfying assignment for φ induces a rational Nash equilibrium σ ∈NE(SG) \NE(SG).

Proof Consider a satisfying assignment γ = 〈�1, . . . , �n〉 for φ. The assignment γ in-duces the rational mixed profile σ in which (i) for each player i ∈ [2], for each literal�j with j ∈ [n], σi(�

j ) = 1n

, and (ii) for each player i ∈ [r] \ [2], σi(δ) = 1; so, play-ers 1 and 2 play uniformly on the set of satisfying literals while the rest of the playersplay δ with probability 1. Thus, there are n2 pure profiles of type 〈�1, �2, δ, . . . , δ〉enabled in σ , with �1, �2 ∈ {�1, . . . , �n}, and each occurs with probability 1

n2 .Note that, since both special players are playing literals induced by the same as-

signment, for any pair of indices j, j ′ ∈ [n], it holds �j �= �j ′ . Hence, all the n2

pure profiles enabled in σ fall into Case (1). It follows that for each player i ∈ [r],Ui (σ ) = u + 1. Clearly, none of the non-special players i ∈ [r] \ [2] has an incentiveto deviate from σ , since, by the definition of the utility vectors, there is no profile ssuch that Ui (s) > u+ 1. Hence, in order to prove that σ is a Nash equilibrium, we areleft to consider a player i ∈ [2]. By Case (9) in the definition of the utility vectors, itsuffices to consider player 1. Recall that, by the construction of σ , for each j ∈ [n],σ2(�

j ) = 1n

. There are four cases.

– Player 1 switches to a literal � ∈ L. By Cases (1) and (2), U1(σ−1 �) ≤ u + 1.– Player 1 switches to a variable v ∈ V. Note that there is exactly one literal �j with

j ∈ [n] such that v(�j ) = v. Hence, by Cases (3) and (4) in the definition of the

Theory Comput Syst

utility vectors,

U1(σ−1 v)

=∑

j∈[n]U1

(⟨

v, �j , δ, . . . , δ⟩) · σ2

(

�j)

= 1

n·(

∑

j∈[n]|v(�j )�=v

U1(⟨

v, �j , δ, . . . , δ⟩) +

∑

j∈[n]|v(�j )=v

U1(⟨

v, �j , δ, . . . , δ⟩)

)

= 1

n· ((u + 2) · (n − 1) + (u + 2 − n) · 1

)

= u + 1.

– Player 1 switches to a clause c ∈ C. Since � = 〈�1, . . . , �n〉 is a satisfying assign-ment for φ, there is at least one literal �j with j ∈ [n] such that �j ∈ c. Hence, byCases (5) and (6) in the definition of the utility vectors,

U1(σ−1 c)

=∑

j∈[n]U1

(⟨

c, �j , δ, . . . , δ⟩) · σ2

(

�j)

= 1

n·(

∑

j∈[n]|�j /∈c

U1(⟨

v, �j , δ, . . . , δ⟩) +

∑

j∈[n]|�j ∈c

U1(⟨

v, �j , δ, . . . , δ⟩)

)

= 1

n· ((u + 2) · (n − 1) + (u + 2 − n) · 1

)

= u + 1.

– Player 1 switches to a strategy s ∈ Σ1. By Cases (8) and (10), U1(σ−1 s) ≤ u+1.

So, player 1 cannot improve her utility U1(σ ) by switching to a different strategy. Itfollows that σ is a rational Nash equilibrium in NE(SG) \NE(SG), as needed. �

We continue to prove:

Lemma 3 NE(SG) ⊆ NE(SG).

Proof Fix any Nash equilibrium σ ∈ NE(SG). We shall prove that σ ∈ NE(SG).Since σ is a Nash equilibrium for SG, each player i ∈ [r] cannot improve her

utility by switching to a strategy from Σi . So, we only need to prove that (i) eachplayer i ∈ [2] cannot improve her utility by switching to a strategy from V∪L∪C, and(ii) each player i ∈ [r] \ [2] cannot improve her utility by switching to the strategy δ.

If either a player i ∈ [2] switches to a strategy s′ from V ∪ L ∪ C or a playeri ∈ [r]\[2] switches to δ, then, for each profile s enabled in σ−i s′, it holds P(s) �= ∅(since i ∈ P(s)) and P(s) �= ∅ (since each player k ∈ [r]\{i} is playing only strategiesin Σk).

Theory Comput Syst

By Case (8), Ui (σ−i s′) = u − 1 < u. Hence, no player can improve and σ ∈NE(SG). �

Lemma 4 Consider a Nash equilibrium σ ∈ NE(SG) \ NE(SG). Then, there is noplayer i ∈ [r] such that Support(σi) ⊆ Σi (so, no player can randomize only overstrategies from Σi ).

Proof Assume, by way of contradiction, that there is a player i ∈ [r] such thatSupport(σi) ⊆ Σi .

Since σ /∈ NE(SG), there must be a player k ∈ [r] such that Support(σk) in-cludes strategies outside Σk (otherwise, it would be Support(σk) ⊆ Σk and thenσ ∈ NE(SG) yielding a contradiction). Then, there is a profile s enabled in σ suchthat P(s) �= ∅ (since i ∈ P(s)) and P(s) �= ∅ (since k ∈ P(s)).

By Case (8), Uk(s) = u − 1 and, since σ is a Nash equilibrium, Lemma 1 impliesthat Uk(σ ) = Uk(σ−k sk) = u − 1.

If player k switches to a strategy s ∈ Σk , then, for each profile s enabled in σ−k s,it holds k ∈ P(s). Moreover, for each profile s enabled in σ−k s, we have eitherUk(s) ≥ u or Uk(s) = u + 1 depending on whether P(s) is empty (Case (7)) or not(Case (8)).

Hence, Uk(σ−k s) ≥ u > Uk(σ ). A contradiction to the assumption that σ is aNash equilibrium. �

Lemma 5 Consider a Nash equilibrium σ ∈ NE(SG) \ NE(SG). Then, for eachplayer i ∈ [2], Support(σi) ⊆ L ∪ Σi .

Proof Assume, by way of contradiction, that there is a player i ∈ [2] such thatSupport(σi) ∩ (V ∪ C) �= ∅. So, there is a strategy s ∈ V ∪ C such that σi(s) > 0.

Look at player [2] \ {i} = i′. By Lemma 4, Support(σ ′i ) �

Σi′ which impliesSupport(σ ′

i ) ∩ (V ∪ C ∪ L) �= ∅.So, there is a profile s′ enabled in σ in which (i) none of the two special players

plays a strategy from SG, (ii) one of the two special players, in particular player i,plays a strategy in V ∪ C. Since player i is not playing a literal in σ , it holds

∑

i∈[2]Ui

(

s′) < 2(u + 1).

Note that, by definition of the utility vectors in SG, in a profile s, it holds

∑

i∈[2]Ui (s) ≤ 2(u + 1).

It follows that

Es∼σ

(

∑

i∈[2]Ui (s)|s1 /∈ Σ1 ∧ s2 /∈ Σ2

)

< 2(u + 1),

since s′1 /∈ Σ1 and s′

2 /∈ Σ2.

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By linearity of expectation, one of the two special players, say player 1, has

Es∼σ

(

U1(s)|s1 /∈ Σ1 ∧ s2 /∈ Σ2)

< u + 1.

Clearly,

U1(σ ) = Es∼σ

(

U1(s))

= Es∼σ

(

U1(s) | s1 /∈ Σ1 ∧ s2 /∈ Σ2) · σ1(Σ1) · σ2(Σ2)

+Es∼σ

(

U1(s) | s1 /∈ Σ1 ∧ s2 ∈ Σ2) · σ1(Σ1) · σ2(Σ2)

+Es∼σ

(

U1(s) | s1 ∈ Σ1 ∧ s2 ∈ Σ2) · σ1(Σ1) · σ2(Σ2)

+Es∼σ

(

U1(s) | s1 ∈ Σ1 ∧ s2 /∈ Σ2) · σ1(Σ1) · σ2(Σ2)

< (u + 1) · σ1(Σ1) · σ2(Σ2) + (u − 1) · σ1(Σ1) · σ2(Σ2)

+ α · σ1(Σ1) · σ2(Σ2) + (u + 1) · σ1(Σ1) · σ2(Σ2)

≤ (u + 1) · σ2(Σ2) + α · σ2(Σ2),

where α := Es∼σ (U1(s) | s1 ∈ Σ1 ∧ s2 ∈ Σ2).By the Law of Conditional Expectation,

Es∼σ

(

U1(s) | s1 ∈ Σ1 ∧ s2 ∈ Σ2)

=∑

s∈Σ1

σ1(s) ·Es∼σ

(

U1(s) | s1 ∈ Σ1 ∧ s2 ∈ Σ2 ∧ s1 = s)

.

Hence, there is a strategy s ∈ Σ1 such that

Es∼σ

(

U1(s) | s1 = s ∧ s2 ∈ Σ2) ≥ α.

Now player 1 switches to s. By the Law of Conditional Expectation,

U1(σ−1 s) = Es∼(σ−1s)

(

U1(s))

= Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 /∈ Σ2) · σ2(Σ2)

+Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 ∈ Σ2) · σ2(Σ2)

≥ α · σ2(Σ2) +Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 /∈ Σ2) · σ2(Σ2).

Since in any profile s enabled in σ−1 s such that s1 = s and s2 /∈ Σ2, it holds thatplayer 1 belongs to P(s) and player 2 belongs to P(s), we have that U1(s) = u + 1.Hence,

Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 /∈ Σ2) = u + 1.

It follows that

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U1(σ−1 s) ≥ α · σ2(Σ2) + (u + 1) · σ2(Σ2)

> U1(σ ),

a contradiction. �

We continue to strengthen the necessary condition from Lemma 5 on the Nashequilibria for the game SG (if any) that are different from those for the game SG. Weprove:

Lemma 6 Consider a Nash equilibrium σ ∈ NE(SG) \ NE(SG). Then, for eachplayer i ∈ [2], Support(σi) ⊆ L.

Proof By Lemma 5, we have that for each i ∈ [2], Support(σi) ⊆ L∪ Σi . Assume, byway of contradiction, that there is a player i ∈ [2] such that σi(Σi) > 0. By Lemma 4,we have σi(Σi) < 1. It follows that σi(L) > 0.

Without loss of generality, take i = 2 and consider the expected utility of player 1.We have

U1(σ ) = Es∼σ

(

U1(s))

= Es∼σ

(

U1(s) | s1 ∈ L ∧ s2 ∈ L) · σ1(L) · σ2(L)

+Es∼σ

(

U1(s) | s1 ∈ L ∧ s2 /∈ L) · σ1(L) · σ2(L)

+Es∼σ

(

U1(s) | s1 /∈ L ∧ s2 /∈ L) · σ1(L) · σ2(L)

+Es∼σ

(

U1(s) | s1 /∈ L ∧ s2 ∈ L) · σ1(L) · σ2(L)

≤ (u + 1) · σ1(L) · σ2(L) + (u − 1) · σ1(L) · σ2(L)

+ α · σ1(L) · σ2(L) + (u + 1) · σ1(L) · σ2(L)

< (u + 1) · σ2(L) + α · σ2(L)

where α := Es∼σ (U1(s) | s1 /∈ L ∧ s2 /∈ L), with α ≥ u and the last inequality follows

from the fact that σ1(L) · σ2(L) > 0.By the Law of Conditional Expectation,

Es∼σ

(

U1(s) | s1 /∈ L ∧ s2 /∈ L)

=∑

s∈Σ1

σ1(s) ·Es∼σ

(

U1(s) | s1 ∈ Σ1 ∧ s2 ∈ Σ2 ∧ s1 = s)

.

This implies that there exists s ∈ Σ1 such that

Es∼σ

(

U1(s) | s1 = s ∧ s2 ∈ Σ2) ≥ α.

Now player 1 switches to s. By the Law of Conditional Expectation,

Theory Comput Syst

U1(σ−1 s) = Es∼(σ−1s)

(

U1(s))

= Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 /∈ Σ2) · σ2(Σ2)

+Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 ∈ Σ2) · σ2(Σ2)

≥ α · σ2(Σ2) +Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 /∈ Σ2) · σ2(Σ2).

Since in any profile s enabled in σ−1 s and such that s1 = s, s2 /∈ Σ2, it holds thatplayer 1 belongs to P(s) and player 2 belongs to P(s), U1(s) = u + 1. This implies

Es∼(σ−1s)

(

U1(s) | s1 = s ∧ s2 /∈ Σ2) = u + 1.

Hence, we have

U1(σ−1 s) ≥ α · σ2(Σ2) + (u + 1) · σ2(Σ2) > U1(σ ),

a contradiction. �

Lemma 7 Consider a Nash equilibrium σ ∈ NE(SG) \ NE(SG). Then, for eachplayer i ∈ [2], for each literal � ∈ L, σi(�) + σi(�) ≥ 1

n.

Proof Assume, by way of contradiction, that there is a player i ∈ [2] and a literal� ∈ L such that σi(�)+σi(�) < 1

n. Since the two players are symmetric, take (without

loss of generality) that i = 2.Consider player 1. By Lemma 6 and the definitions of the utility vector (Cases (1)

and (2)), U1(σ ) ≤ u + 1. However, when player 1 switches to the variable v(�),

U1(

σ−1 v(�))

= (u + 2 − n) · (σ2(�) + σ2(�)) + (u + 2) · (1 − (

σ2(�) + σ2(�))

(by Lemma 6)

= (

u + 2 − n − (u + 2)) · (σ2(�) + σ2(�)

) + u + 2

> −n · 1

n+ u + 2

= u + 1,

a contradiction to the fact that σ is a Nash equilibrium. �

Lemma 6 implies that for each player i ∈ [2], ∑

�∈L(σi(�) + σi(�)) = 1. Hence,by Lemma 7:

Corollary 1 Consider a Nash equilibrium σ ∈ NE(SG) \ NE(SG). Then, for eachplayer i ∈ [2] and for each literal � ∈ L, σi(�) + σi(�) = 1

n.

We continue with some additional technical claims:

Lemma 8 Consider a Nash equilibrium σ ∈ NE(SG) \ NE(SG). Then, for eachplayer i ∈ [r] \ [2], σi(δ) = 1.

Theory Comput Syst

Proof By Lemma 6, the two special players are playing only literals. Assume, by wayof contradiction, that there is a player i ∈ [r] \ [2] with σi(δ) < 1, so that σi(Σi) > 0.By the Law of Conditional Expectation,

U1(σ ) = Es∼σ

(

U1(s))

= Es∼σ

(

U1(s) | si = δ) · σi(δ) +Es∼σ

(

U1(s) | si �= δ) · σi(Σi)

≤ (u + 1) · σi(δ) + (u − 1) · σi(Σi)

< u + 1.

If player 1 switches to any strategy s ∈ Σ1, we have

U1(σ−1 s) = u + 1,

since in every profile s enabled in σ−1 s, it holds that U1(s) = u+1. This contradictsthe fact that σ is a Nash equilibrium. �

We now prove:

Lemma 9 Consider a Nash equilibrium σ ∈ NE(SG) \ NE(SG). Then, for eachliteral � ∈ L, σ1(�) · σ2(�) = 0.

Proof Assume, by way of contradiction, that there is a literal � ∈ L such that σ1(�) ·σ2(�) �= 0. Then, there is a profile s enabled in σ such that one special player plays� and the other special player plays �. By Case (2), both special players get a utilityequal to u − 1, so that

∑

i∈[2] Ui (s) < 2(u + 1). Hence,

Es∼σ

(

∑

i∈[2]Ui (s) | s1 ∈ L ∧ s2 ∈ L

)

< 2(u + 1).

By linearity of expectation, there is a player i ∈ [2] with

Es∼σ

(

Ui (s) | s1 ∈ L ∧ s2 ∈ L)

< u + 1.

By Lemma 6, both special players are playing only literals; hence

Ui (σ ) = Es∼σ

(

Ui (s) | s1 ∈ L ∧ s2 ∈ L)

< u + 1.

If player i switches to any strategy s ∈ Σ1, we have

U1(σ−1 s) = u + 1,

since in every profile s enabled in σ−1 s, it holds U1(s) = u + 1. This contradictsthe fact that σ is a Nash equilibrium. �

To complete the proof of Proposition 1, we are left to prove that each of the Con-ditions (3) and (2) implies Condition (1). We prove:

Theory Comput Syst

Lemma 10 Any Nash equilibrium σ ∈ NE(SG) \ NE(SG) is rational and inducesa satisfying assignment for φ.

Proof Fix a Nash equilibrium σ ∈ NE(SG)\NE(SG). By Corollary 1, σ is rational.By Lemma 9, σ induces an assignment for φ in the following natural way: if bothplayers put some positive probability on a literal �, then set � := 1; else set � := 0.Assume, by way of contradiction, that φ is not satisfied by this assignment. Consideran unsatisfied clause c. So, every literal � occurring in c is 0 by the assignment.By Lemma 9, all profiles enabled in σ fall in Case (1). So, both special players getu + 1. If player 1 switches to c, then all profiles enabled in σ−i c fall in Case (5).So, U1(σ−i c) = u + 2, a contradiction. �

Lemma 10 establishes that Condition (3) implies Condition (1). Assume that Con-dition (2) holds. By Lemma 3, there is a Nash equilibrium σ ∈ NE(SG) \NE(SG).By Lemma 10 σ induces a satisfying assignment for φ, so that Condition (1) holds.The equivalence of Conditions (1), (2) and (3) now follows. �

Proposition 1 now yields:

Proposition 2 Fix an arbitrary game SG. Then, CNF SAT ≤P

NASH-EQUIVALENCE (SG).

Consider now the gadget game SG1 = 〈[3], {Σi}i∈[3], {Ui}i∈[3]〉, where (i) for eachplayer i ∈ [3], Σi = {0,1}, and (ii) the utility vectors are depicted in the followingtable:

Profile s Utility vector U(s)

〈0,0,0〉 〈1,3,3〉〈0,0,1〉 〈0,1,2〉〈0,1,0〉 〈3,2,2〉〈0,1,1〉 〈2,3,1〉〈1,0,0〉 〈2,3,1〉〈1,0,1〉 〈2,0,3〉〈1,1,0〉 〈1,0,3〉〈1,1,1〉 〈0,1,2〉

We prove:

Proposition 3 The gadget game SG1 has a unique Nash equilibrium σ , and σ isirrational.

Proof Note first that there is no pure Nash equilibrium as the following table indi-cates:

Theory Comput Syst

Profile Player enabled to improve by switching strategy

〈0,0,0〉 1

〈0,0,1〉 3

〈0,1,0〉 2

〈0,1,1〉 3

〈1,0,0〉 3

〈1,0,1〉 2

〈1,1,0〉 1

〈1,1,1〉 3

Consider now a mixed profile σ . Since there are only two strategies, we shall repre-sent the mixed strategy of player i ∈ [3] in the profile σ with the single probability σi

that player i chooses strategy 0; thus, player i chooses strategy 1 with probability σi .Consequently, the mixed profile σ will be represented by the vector of probabilities〈σ1, σ2, σ3〉. So, player i ∈ [3] is mixed for the profile 〈σ1, σ2, σ3〉 if 0 < σi, σi < 1.

Assume that σ is a Nash equilibrium. Then, Lemma 1 implies that for each mixedplayer i ∈ [3], Ui (σ−i ,0) = Ui (σ−i ,1). Hence, we obtain:

(C/1) Assume that player 1 is mixed. Then,

1 · σ2 · σ3 + 3 · σ2 · σ3 + 2 · σ2 · σ3 = 2 · σ2 · σ3 + 2 · σ2 · σ3 + 1 · σ2 · σ3,

which simplifies to

σ2 σ3 − 4σ2 + 2 = 0.

(C/2) Assume now that player 2 is mixed. Then,

3 · σ1 · σ3 + 1 · σ1 · σ3 + 3 · σ1 · σ3 = 2 · σ1 · σ3 + 3 · σ1 · σ3 + 1 · σ1 · σ3,

which simplifies to

−σ1 σ3 − σ1 + 4σ3 − 1 = 0.

(C/3) Assume finally that player 3 is mixed. Then,

3 · σ1 · σ2 + 2 · σ1 · σ2 + 1 · σ1 · σ2 + 3 · σ2 · σ3

= 2 · σ1 · σ2 + 1 · σ1 · σ2 + 3 · σ1 · σ22 · σ1 · σ2,

which simplifies to

3σ1 σ2 − 3σ2 + 1 = 0.

We shall now examine all possible cases for the mixed Nash equilibrium σ . For allbut one cases, we shall use Conditions (C/1), (C/2) and (C/3) to derive a contradiction;in the remaining case, we shall establish that σ is an irrational profile.

Theory Comput Syst

– Only player 1 is mixed. Then, there is a choice of σ2, σ3 ∈ {0,1}. such that Condi-

tion (C/1) holds. By inspection, a contradiction follows.– Only player 2 is mixed. Then, there is a choice of σ1, σ3 ∈ {0,1} such that Condi-

tion (C/2) holds. By inspection, a contradiction follows.– Only player 3 is mixed. Then, there is a choice of σ1, σ2 ∈ {0,1} such that Condi-

tion (C/3) holds. By inspection, a contradiction follows.– Only players 2 and 3 are mixed. Then, there is a choice of σ1 ∈ {0,1} such that

both Conditions (C/2) and (C/3) hold. Note that Condition (C/3) does not holdfor σ1 = 1. Hence, σ1 = 0. By Lemma 1 (Conditions (2) and (3)), it follows thatU1(σ−1 0) ≥ U1(σ−1 1), which simplifies to σ2 σ3 + 2 − 4σ2 ≤ 0. Hence, by

Condition (C/2), σ3 = 14 ; by Condition (C/3), σ2 = 1

3 . So, σ2 σ3 + 2 − 4σ2 = 34 > 0.

By inspection, a contradiction follows.– Only players 1 and 3 are mixed. Then, there is a choice of σ2 ∈ {0,1} such that

both Conditions (C/1) and (C/3) hold. By inspection, a contradiction follows.– Only players 1 and 2 are mixed. Then, there is a choice of σ3 ∈ {0,1} such that

both Conditions (C/1) and (C/2) hold. By inspection, a contradiction follows.– All players are mixed. By Conditions (C/1), (C/2) and (C/3), it follows that σ1 =

25−√409

12 , σ2 = 13+√409

60 and σ3 = 21−√409

2 , so that σ is irrational.

So, SG1 has a single (mixed) Nash equilibrium, which is irrational. �

Using SG1 as the gadget game in the reduction, Condition (3) of Proposition 1 be-comes:

(3′) SG admits a rational Nash equilibrium.

So, the equivalence of Conditions (1) and (3) implies:

Proposition 4 CNF SAT ≤P ∃ RATIONAL NASH.

Since CNF SAT is NP-hard, Propositions 2 and 4 immediately imply:

Theorem 1 NASH-EQUIVALENCE(SG) is co-NP-hard.

Theorem 2 ∃ RATIONAL NASH is NP-hard.

4 Complexity of NASH-REDUCTION and ∃ IRRATIONAL NASH

Fix a gadget game SG = 〈[r], {Σi}i∈[r], {Ui}i∈[r]〉 with the following properties:

1. SG is constant-sum with constant c = ur ,

Theory Comput Syst

2. SG has a unique fully mixed Nash equilibrium σ such that Ui (σ ) = u for eachplayer i ∈ [r].

Call SG uniformly constant-sum.For each player i ∈ [r] in the game SG, set Σi := {1, . . . ,mi}; so, for each player

i ∈ [r], for each strategy j ∈ [mi], σi (j) is the probability that player i choosesstrategy j in a mixed profile σ . Set m := maxi∈[r] mi . For a player i ∈ [r], denotes(i) = (i + 1)mod r ; so, s(i) is the successor modulo r of player i ∈ [r].

4.1 The Reduction

Consider a boolean formula φ in CNF with |V(φ)| = n ≥ m variables. Set V(φ) :={1,2, . . . , n}; so, the boolean variables in φ are numbered 1,2, . . . , n. For each vari-able k ∈ [n], �k and �k are the two associated literals. For each player i ∈ [r] and foreach variable k ∈ [n], define

gi (�k) = gi (�k) ={

k if k ∈ [mi − 1]mi if k ≥ mi.

Intuitively, for each player i ∈ [r], the function gi maps each literal to a correspond-ing group. In particular, both literals �k and �k associated with variable k ≤ m − 1are mapped to group k, while all literals �k and �k associated with variable k ≥ m

are mapped to group m. Moreover, for each player i ∈ [r] and for each j ∈ [mi]define

ni(j) ={

1 if j ∈ [mi − 1]n − mi + 1 if j = mi.

Intuitively, for each player i ∈ [r], the function ni relates a group with the number ofvariables whose associated literals belong to the given group.

Construct the strategic game SG(φ) = 〈[r], {Σi}i∈[r], {Ui}i∈[r]〉 as follows.

– For each player i ∈ [r], Σi := V(φ) ∪ L(φ) ∪ C(φ) ∪ {f }.– Fix a profile s from Σ = Σ1 × . . . × Σr . Denote f = 〈f, . . . , f 〉. Denote

Pf (s) = {

i ∈ [r] | si = f};

so, |Pf (s)| = r if and only if s = f. The utility vector U(s) is depicted in the fol-

lowing table, where zi = ni(gi (ss(i)))

σs(i)(gi (ss(i))).

Theory Comput Syst

Case Condition on the profile s Utility vector U(s)

(1)s = 〈�1, . . . , �r 〉 with(i) �i ∈ L(φ) for each i ∈ [r],&(ii) �i �= �j for each distinct pair i, j ∈ [r]

U(〈sg(�1), . . . , sg(�r )〉)

(2)(i) si = v ∈ V(φ) for some i ∈ [r],(ii) sj ∈ L(φ) for each j ∈ [r] \ {i},&(iii) v(ss(i)) �= v

Uj (s) ={

u + 1, if j = i

u − 2, if j �= i

(3)(i) si = v ∈ V(φ) for some i ∈ [r],(ii) sj ∈ L(φ) for each j ∈ [r] \ {i}&(iii) v(ss(i)) = v

Uj (s) ={

u + 1 − zi, if j = i

u − 2, if j �= i

(4)(i) si = c ∈ C(φ) for some i ∈ [r],(ii) sj ∈ L(φ) for each j ∈ [r] \ {i}&(iii) ss(i) /∈ c

Uj (s) ={

u + 1, if j = i

u − 2, if j �= i

(5)(i) si = c ∈ C(φ) for some i ∈ [r],(ii) sj ∈ L(φ) for each j ∈ [r] \ {i}&(iii) ss(i) ∈ c

Ui (s) ={

u + 1 − zi, if j = i

u − 2, if j �= i

(6) |Pf (s)| = r Ui (s) = u − 1 for each i ∈ [r](7) 0 < |Pf (s)| < r Ui (s) =

{

u, if si = f

u − 2, if si �= f

(8) None of the above Ui (s) = u − 1 for each i ∈ [r]

Clearly, the construction of SG(φ) from φ is carried out in polynomial time. Forbrevity, we shall also write SG, L, V and C for SG(φ), L(φ), V(φ) and C(φ). Weobserve:

– For each profile s,

∑

i∈[r]Ui (s) ≤ r · u

with∑

i∈[r] Ui (s) = r · u if and only if s = 〈�1, . . . , �r 〉 with (i) �i ∈ L for each

i ∈ [r], and (ii) �i �= �j for each distinct pair i, j ∈ [r]. To see this, note that byCase (1) in the definition of the utility vectors

∑

i∈[r]Ui

(〈�1, . . . , �r 〉) =

∑

i∈[r]Ui

(〈sg(�1), . . . , sg(�r )〉) = r · u.

Hence, for any property P satisfied by at least one profile enabled in a mixed profileσ , it holds that

Es∼σ

(

∑

i∈[r]Ui (s) | s satisfies P

)

≤ r · u.

Theory Comput Syst

Let H be the surjective map from SG to SG defined as follows: for each player i ∈ [r],

hi (s) :={

gi (�), if s = �,

∗, otherwise,

where ∗ means a don’t care situation, i.e., the properties that each function hi has tosatisfy for our purposes are independent of the particular choice of the values ∗. Weprove:

Proposition 5 Consider a gadget game SG satisfying Properties (P1) and (P2).Then, the following conditions are equivalent:

(1) φ is satisfiable,(2) SG admits a Nash equilibrium σ �= f,(3) SG Nash-reduces to SG.

Proof We first prove that Condition (1) implies both Conditions (2) and (3).

Lemma 11 A satisfying assignment for φ induces a Nash equilibrium σ �= f withH(σ ) = σ .

Proof Consider a satisfying assignment γ = 〈�1, . . . , �n〉 for φ. The assignment γ

induces the mixed profile σ in which for each player i ∈ [r] and for each literal �j

with j ∈ [n],

σi

(

�j) := σi (gi (�

j ))

ni(gi (�j ));

so, each player plays on the set of satisfying literals. Clearly, σ �= f. �

We prove some technical claims:

Claim 1 For each i ∈ [r], Ui (σ ) = u.

Proof Clearly,

Ui (σ ) =∑

s∼σ

P(s) · Ui (s)

=∑

s∼σ

∏

k∈[r]

σk(gk(sk))

nk(gk(sk))·Ui

(⟨

g1(s1), . . . ,gr (sr )⟩)

=∑

s∼σ

∑

s∼σ :g(s)=s

∏

k∈[r]

σk(gk(sk))

nk(gk(sk))·Ui

(⟨

g1(s1), . . . ,gr (sr )⟩)

=∑

s∼σ

∑

s∼σ :g(s)=s

∏

k∈[r]

σk (sk)

nk(sk)·Ui (s)

Theory Comput Syst

=∑

s∼σ

∑

s∼σ :g(s)=s

∏

k∈[r] σk (sk)∏

k∈[r] nk(sk)·Ui (s)

=∑

s∼σ

∏

k∈[r]nk(sk)

∏

k∈[r] σk (sk)∏

k∈[r] nk(sk)·Ui (s)

=∑

s∼σ

σ (s) ·Ui (s)

= Ui (σ )

= u,

as needed. �

Claim 2 In σ , no player i ∈ [r] improves by switching.

Proof Fix a player i ∈ [r]. We shall consider all possible switches of player i.

(1) Player i switches to f : Then, for any profile s enabled in σ−i f , either Case(6) or (7) apply, which results to Ui (s) ≤ u. So, Ui (σ−i f ) ≤ u and player i

does not improve.(2) Player i switches to v ∈ V: In this case, since γ is an assignment, player s(i)

plays a literal � such that v(�) = v with probability σs(i)(gs(i)(�))

ns(i)(gs(i)(�)). Then,

Ui (σ−i v)

= Es−i∼σ−i

(

Ui (s−i v))

= Es−i∼σ−i

(

Ui (s−i v) | v(ss(i)) = v) · σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

+Es−i∼σ−i

(

Ui (s−i v) | v(ss(i)) �= v) ·

(

1 − σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

)

=(

u + 1 − ns(i)(gs(i)(�))

σs(i)(gs(i)(�))

)

· σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

(

by Case (3))

+ (u + 1) ·(

1 − σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

)

(

by Case (2))

= u,

and player i does not improve.(3) Player i switches to c ∈ C: Fix a literal � belonging to c. Since γ is a satisfying

assignment, player s(i) plays literal � with probability σs(i)(gs(i)(�))

ns(i)(gs(i)(�)). Then,

Ui (σ−i c)

= Es−i∼σ−i

(

Ui (s−i c))

Theory Comput Syst

= Es−i∼σ−i

(

Ui (s−i c) | v(ss(i)) = �) · σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

+Es−i∼σ−i

(

Ui (s−i c) | v(ss(i)) �= �) ·

(

1 − σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

)

≤(

u + 1 − ns(i)(gs(i)(�))

σs(i)(gs(i)(�))

)

· σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

(

by Case (5))

+ (u + 1) ·(

1 − σs(i)(gs(i)(�))

ns(i)(gs(i)(�))

)

(

by Case (4))

= u,

and player i does not improve.(4) Player i switches to � ∈ L. Then,

Ui (σ−i �)

=∑

s−i∼σ−i

P(s−i ) · Ui (s−i �)

=∑

s−i∼σ−i

∏

k∈[r]\{i}

σk(gk(sk))

nk(gk(sk))·Ui

(

g(s−i �))

=∑

s−i∼σ−i

∑

s−i∼σ−i :g(s−i�)=s−igi (�)

∏

k∈[r]\{i}

σk(gk(sk))

nk(gk(sk))·Ui

(

g(s−i �))

=∑

s−i∼σ−i

∏

k∈[r]\{i}nk(sk)

∏

k∈[r]\{i}

σk (sk)

nk(sk)·Ui

(

s−i gi (�))

=∑

s−i∼σ−i

∏

k∈[r]\{i}nk(sk)

∏

k∈[r]\{i} σk (sk)∏

k∈[r]\{i} nk(sk)·Ui

(

s−i gi (�))

=∑

s−i∼σ−i

σ−i (s−i ) ·Ui

(

s−i gi (�))

=Ui

(

σ−i gi (�))

.

So, Ui (σ−i �) ≤ Ui (σ ) if and only if Ui (σ−i gi (�)) ≤ Ui (σ ), which holdsbecause σ is a Nash equilibrium for SG. Hence, player i does not improve.

Since for each player i ∈ [r], Σi = L ∪ V ∪ C ∪ {f }, it follows that in σ no playeri ∈ [r] improves by switching to some other (mixed) strategy. �

We continue to prove:

Claim 3 H(σ ) = σ .

Theory Comput Syst

Proof For each i ∈ [r] and for each j ∈ [mi], it holds

H(σ )i(j) =∑

�∈L:hi (�)=j

σi(�)

=∑

�∈L:gi (�)=j

σi(�)

={

σi(�j ) if j ∈ [mi − 1],

∑nk=mi

σi(�k) if j = mi

=⎧

⎨

⎩

σi (gi (�j ))

ni (gi (�j ))

if j ∈ [mi − 1],∑n

k=mi

σi (gi (�k))

ni (gi (�k))

if j = mi

={

σi (j) if j ∈ [mi − 1],∑n

k=mi

σi (mi)n−mi+1 if j = mi

={

σi (j) if j ∈ [mi − 1],σi(mi) if j = mi

= σi (j). �

Claims 1, 2 and 3 establish the claim. �

We now prove:

Lemma 12 The profile f = 〈f, . . . , f 〉 is a Nash equilibrium for SG. Moreover, ifthere is a Nash equilibrium σ for which there is a player i ∈ [r] such that σi(f ) = 1,then σ = f.

Proof By Case (6) in the definition of the utility vectors, Ui (f) = u−1 for each playeri ∈ [r]. If player i ∈ [r] switches to a strategy s �= f , then by Case (7) in the definitionof the utility vectors, it follows that Ui (f−i s) = u − 2 < u − 1. Since player i waschosen arbitrarily, this implies that f is a Nash equilibrium.

For the second claim, assume, by way of contradiction, that there is a mixed Nashequilibrium σ other than f for which there is a player i ∈ [r] such that σi(f ) = 1.Since σ �= f, it follows that there is a player i′ ∈ [r] such that σi′(L ∪ V ∪ C) > 0. Thisimplies that σi′(f ) < 1. So,

Ui′(σ )

= Es∼σ

(

Ui′(s))

= Es∼σ

(

Ui′(s) | si′ = f ∧ sk = f ∀k �= i′) · σ (f)

+Es∼σ

(

Ui′(s) | si′ = f ∧ ∃k �= i′ : sk �= f) · σi′(f )

(

1 −∏

k∈[r],k �=i′σk(f )

)

Theory Comput Syst

+Es∼σ

(

Ui′(s) | si′ �= f) · (1 − σi′(f )

)

= (u − 1) · σ (f)(

by Case (6))

+ u · σi′(f )

(

1 −∏

k∈[r],k �=i′σk(f )

)

(

by Case (7))

+ (u − 2) · (1 − σi′(f )) (

by Case (7))

= −σ (f) + u − 2 + 2σi′(f )

= −σi′(f )∏

k∈[r],k �=i′σk(f ) + u − 2 + 2σi′(f )

= σi′(f )

(

2 −∏

k∈[r],k �=i′σk(f )

)

+ u − 2

< 2 −∏

k∈[r],k �=i′σk(f ) + u − 2

= u −∏

k∈[r],k �=i′σk(f ).

If player i′ switches to f she gets,

Ui′(σ−i′ f )

= Es−i′∼σ−i′(

Ui′(s−i′ f ))

= Es−i′∼σ−i′(

Ui′(s−i′ f ) | sk = f ∀k ∈ [r] \ {

i′}) ·

∏

k∈[r],k �=i′σk(f )

+Es−i′∼σ−i′(

Ui′(s−i′ f ) | ∃k ∈ [r] \ {

i′} : sk �= f

) ·(

1 −∏

k∈[r],k �=i′σk(f )

)

= (u − 1) ·∏

k∈[r],k �=i′σk(f )

(

by Case (6))

+ u ·(

1 −∏

k∈[r],k �=i′σk(f )

)

(

by Case (7))

= u −∏

k∈[r],k �=i′σk(f ),

a contradiction. �

The second claim in Lemma 12 states that f is the only Nash equilibrium for whichsome player chooses strategy f with probability 1.

Lemma 13 In a Nash equilibrium σ �= f, σi(f ) = 0 for each i ∈ [r].

Theory Comput Syst

Proof We shall use the following claim which is easily shown by induction on r .

Claim 4 For any r-tuple 〈x1, . . . , xr 〉 with 0 ≤ xi ≤ 1, it holds∏

i∈[r]xi ≤ 1 −

∏

i∈[r](1 − xi).

Assume, by way of contradiction, that there is a Nash equilibrium σ �= f with σi(f ) >

0 for some i ∈ [r]. Call i′ ∈ [r] the player minimizing Es∼σ (Ui′(s) | sk �= f ∀k ∈ [r]).By definition of the players’ utilities in SG,

Es∼σ

(

Ui′(s) | sk �= f ∀k ∈ [r]) ≤ u.

We have

Ui′(σ )

= Es∼σ

(

Ui′(s))

= Es∼σ

(

Ui′(s) | sk �= f ∀k ∈ [r]) ·∏

k∈[r]

(

1 − σk(f ))

+Es∼σ

(

Ui′(s) | si′ �= f ∧ ∃k ∈ [r] : sk = f)

· (1 − σi′(f ))

(

1 −∏

k∈[r],k �=i′

(

1 − σk(f ))

)

+Es∼σ

(

Ui′(s) | si′ = f ∧ ∃k ∈ [r] : sk �= f) · σi′(f )

(

1 −∏

k∈[r],k �=i′σk(f )

)

+Es∼σ

(

Ui′(s) | ∀k ∈ [r] : sk = f) · σ (f)

≤ u ·∏

k∈[r]

(

1 − σk(f )) + (u − 2) · (1 − σi′(f )

)

(

1 −∏

k∈[r],k �=i′

(

1 − σk(f ))

)

+ u · σi′(f )

(

1 −∏

k∈[r],k �=i′σk(f )

)

+ (u − 1) · σ (f)

= u − 2(

1 − σi′(f ))

(

1 −∏

k∈[r],k �=i′

(

1 − σk(f ))

)

− σ (f)

≤ u − 2(

1 − σi′(f )) ·

∏

k∈[r],k �=i′σk(f ) − σ (f) (by Lemma 4)

= u − 2 + σ (f)

< u − 1.

Now player i′ switches to f . We have

Ui′(σ−i′ f )

Theory Comput Syst

= Es−i′∼σ−i′(

Ui′(s−i′ f ))

= Es−i′∼σ−i′(

Ui′(s−i′ f ) | sk = f ∀k ∈ [r] \ {

i′}) ·

∏

k∈[r],k �=i′σk(f )

+Es−i′∼σ−i′(

Ui′(s−i′ f ) | ∃k ∈ [r] \ {

i′} : sk �= f

) ·(

1 −∏

k∈[r],k �=i′σk(f )

)

= (u − 1) ·∏

k∈[r],k �=i′σk(f )

(

by Case (6))

+ u ·(

1 −∏

k∈[r],k �=i′σk(f )

)

(

by Case (7))

= u −∏

k∈[r],k �=i′σk(f )

≥ u − 1,

a contradiction. �

Lemma 14 In a Nash equilibrium σ �= f, for each i ∈ [r] it holds: (i) σi(L) = 1, and(ii) Ui (σ ) = u.

Proof Assume that there is a player i′ ∈ [r] such that Ui′(σ ) < u and consider whathappens when player i′ switches to f . Since, by Lemma 13, none of the players isputting positive probability on f in σ , by Case (7), for any pure profile s enabled inσ−i′ f , it holds Ui′(s) = u. Hence, Ui′(σ−i′ f ) = u, contradicting the fact thatσ is a Nash equilibrium. Thus, to prove the claim, we only need to prove that wheneither (i) or (ii) is violated, there is a player i′ ∈ [r] with Ui′(σ ) < u.

For condition (i), assume, by way of contradiction, that there is a player i ∈ [r]such that σi(L) < 1. So, there is a profile s enabled in σ that does not fall into Case(1). For this profile s, it holds

∑

k∈[r]Uk(s) < r · u.

Recall that, by definition of the players’ utilities in SG, it holds∑

k∈[r] Uk (s) ≤ r · ufor any profile s; hence, it follows that

∑

k∈[r]Uk(σ ) < r · u.

Hence, by linearity of expectation, there is a player i′ ∈ [r] such that Ui′(σ ) < u.For condition (ii), assume, by way of contradiction, that there exists a player

i ∈ [r] such that Ui (σ ) �= u. Since, by definition of the utility functions in SG, itholds

∑

k∈[r] Uk(σ ) ≤ r · u, Ui (σ ) �= u implies that there is a player i′ ∈ [r] such thatUi′(σ ) < u. �

Theory Comput Syst

Lemma 15 In a Nash equilibrium σ �= f, for each i ∈ [r] and for each literal � ∈ L,it holds σs(i)(�) + σs(i)(�) ≥ σs(i)(gs(i)(�))

ns(i)(gs(i)(�)).

Proof Assume, by way of contradiction, that there is a player i ∈ [r] and a literal

� ∈ L such that σs(i)(�) + σs(i)(�) <σs(i)(gs(i)(�))

ns(i)(gs(i)(�)).

Now player i switches to the variable v = v(�). We have

Ui (σ−i v) = Es−i∼σ−i

(

Ui (s−i v))

= Es−i∼σ−i

(

Ui (s−i v) | v(ss(i)) = v) · (σs(i)(�) + σs(i)(�)

)

+Es−i∼σ−i

(

Ui (s−i v) | v(ss(i)) �= v) · (1 − σs(i)(�) − σs(i)(�)

)

=(

u + 1 − ns(i)(gs(i)(�))

σs(i)(gs(i)(�))

)

· (σs(i)(�) + σs(i)(�)) (

by Case (3))

+ (u + 1) · (1 − σs(i)(�) − σs(i)(�)) (

by Case (2))

= (u + 1) − ns(i)(gs(i)(�))

σs(i)(gs(i)(�))· (σs(i)(�) + σs(i)(�)

)

> u + 1 − 1

= u,

a contradiction to Lemma 14. �

Lemma 14 implies that in a Nash equilibrium σ �= f, it holds∑

�∈L(σi(�) +σi(�)) = 1 for each player i ∈ [r]. So Lemma 15, implies the following corollary.

Corollary 2 In any Nash equilibrium σ �= f, for any player i ∈ [r] and for any literal� ∈ L, it holds σi(�) + σi(�) = σi (gi (�))

ni (gi (�)).

Proof Assume, by way of contradiction, that there are a player i ∈ [r] and a literal� ∈ L such that σi(�) + σi(�) >

σi(gi (�))ni (gi (�))

. Then,

∑

�∈L

(

σi(�) + σi(�))

>∑

�∈L

σi (gi (�))

ni(gi (�))

=∑

�:gi (�)∈[mi−1]σi

(

gi (�)) +

∑

�:gi (�)=mi

σi(gi (�))

n − mi + 1

=∑

�∈L

σi

(

gi (�))

= 1,

a contradiction. �

Lemma 16 Consider a Nash equilibrium σ �= f. Then, for each pair of players i, i′ ∈[r] and for each literal � ∈ L, it holds that σi(�) · σi(�) = 0.

Theory Comput Syst

Proof Assume, by way of contradiction, that there are a literal � ∈ L and a pair ofplayers i, i′ ∈ [r] such that σi(�) · σi(�) �= 0. So, a pure profile s such that si = � andsi′ = � is enabled in σ . For such a profile, Case (8) applies which gives Uk(s) = u−1for each k ∈ [r]. This implies that

∑

k∈[r]Uk(σ ) < r · u.

By linearity of expectation, there is a player k ∈ [r] such that Uk(σ ) < u, whichcontradicts Lemma 14. �

To complete the proof of Proposition 5, it remains to prove that each of Condi-tions (2) and (3) implies Condition (1).

Lemma 17 (A) If SG Nash-reduces to SG, then there is a Nash equilibrium σ �= f.(B) Any Nash equilibrium σ �= f induces a satisfying assignment.

Proof Let H be the surjective mapping defined in the reduction from SG to SG. If fis the single Nash equilibrium for SG, then H(f) will be the single Nash equilibriumfor SG (since SG satisfies Property (P1)). By definition of surjective mappings, anyprofile s for SG maps to a profile H(s) for SG. Hence, H(f) is a Nash equilibrium:a contradiction to Property (P2). It follows that if either Condition (2) or (3) holds,then there is a Nash equilibrium σ �= f.

Now note that for each literal � ∈ L and for each player i ∈ [r], σi (gi (�)) > 0because σ is a fully mixed Nash equilibrium for SG. Hence, Lemma 15 implies thateither σi(�) > 0 or σi(�) > 0, but not both (by Lemma 16). Hence, σ induces anassignment γ defined as follows: if σi(�) > 0, then � := 1; else � := 0.

We now show that γ satisfies φ. Assume, by way of contradiction, that φ(γ ) isfalse. Then, there is an unsatisfied clause c. So, every literal in clause c is 0 in γ .Now player i switches to c. All profiles s enabled in σ−i c fall into Case (4), whichimplies Ui (s) = u+ 1. Hence, U1(σ−i c) = u + 1. A contradiction to Lemma 14. �

Lemma 17 establishes that Condition (3) implies Condition (1). Assume that Con-dition (2) holds. Lemma 17 (Condition (B)) establishes that Condition (2) impliesCondition (1). Now the equivalence of Conditions (1), (2) and (3) follows. �

The equivalence of Conditions (1) and (2) in Proposition 5 yields:

Proposition 6 Consider a gadget game SG satisfying Properties (P1) and (P2).Then, CNF SAT ≤P NASH − REDUCTION(SG).

We shall now exhibit a gadget game satisfying Properties (P1) and (P2). Consider thegadget game SG2 = 〈[3], {Σi}i∈[3], {Ui}i∈[3]〉, where for each player i ∈ [3], Σi ={0,1}. The utilities are depicted in the following table.

Theory Comput Syst

Profile s Utility vector U(s)

〈0,0,0〉 〈2,4,−3〉〈0,0,1〉 〈−1,5,−1〉〈0,1,0〉 〈3,−4,4〉〈0,1,1〉 〈2,0,1〉〈1,0,0〉 〈−3,6,0〉〈1,0,1〉 〈4,−6,5〉〈1,1,0〉 〈−1,−1,5〉〈1,1,1〉 〈 3863−173

√471

74 , 6877+107√

4711850 , −1323+57

√471

25 〉

Note that SG2 is constant-sum, with constant equal to r . We prove:

Proposition 7 The gadget game SG2 has a single Nash equilibrium σ , which isirrational and fully mixed, and where for each player i ∈ [3], Ui (σ ) = 1.

Proof Note first that there is no pure Nash equilibrium as the following table indi-cates:

Profile Player enabled to improve by switching strategy

〈0,0,0〉 3〈0,0,1〉 1〈0,1,0〉 2〈0,1,1〉 2〈1,0,0〉 1〈1,0,1〉 2〈1,1,0〉 1〈1,1,1〉 3

We now turn to a mixed profile σ . Since there are only two strategies, we shall rep-resent the mixed strategy of player i ∈ [3] in the profile σ with the single probabilityσi that player i chooses strategy 0; thus, player i chooses strategy 1 with probabil-ity σi . Hence, the mixed profile σ will be represented by the vector of probabilities〈σ1, σ2, σ3〉. So, player i ∈ [3] is mixed for the profile 〈σ1, σ2, σ3〉 if 0 < σi, σi < 1.Assume that σ is a Nash equilibrium. Then, Lemma 1 implies that for each mixedplayer i ∈ [3], Ui (σ−i 0) = Ui (σ−i 1). Hence, we obtain:

(C/1) Assume that player 1 is mixed for the Nash equilibrium σ . Then,

2 · σ2 · σ3 − 1 · σ2 · σ3 + 3 · σ2 · σ3 + 2 · σ2 · σ3

= −3 · σ2 · σ3 + 4 · σ2 · σ3 − 1 · σ2 · σ3 + 3863 − 173√

471

74· σ2 · σ3,

Theory Comput Syst

which simplifies to

(

9 − 3863 − 173√

471

74

)

σ2 σ3 +(

−8 + 3863 − 173√

471

74

)

σ2

+(

2 + 3863 − 173√

471

74

)

σ3 = −2 + 3863 − 173√

471

74.

(C/2) Assume now that player 2 is mixed for the Nash equilibrium σ . Then,

4 · σ1 · σ3 + 5 · σ1 · σ3 + 6 · σ1 · σ3 − 6 · σ1 · σ3

= −4 · σ1 · σ3 + 4 · σ1 · σ3 − 1 · σ1 · σ3 + 6877 + 107√

471

1850· σ1 · σ3,

which simplifies to

(

−10 − 6877 + 107√

471

1850

)

σ1 σ3 +(

11 + 6877 + 107√

471

1850

)

σ1

+(

13 + 6877 + 107√

471

1850

)

σ3 = 6 + 6877 + 107√

471

1850.

(C/3) Assume finally that player 3 is mixed for the Nash equilibrium σ . Then,

−3 · σ1 · σ2 + 4 · σ1 · σ2 + 5 · σ1 · σ2

= −1 · σ1 · σ2 + 1 · σ1 · σ2 + 5 · σ1 · σ2 + −1323 + 57√

471

25· σ1 · σ2,

which simplifies to

(

−5 − 1323 − 57√

471

25

)

σ1 σ2 +(

2 + 1323 − 57√

471

25

)

σ1

+(

10 + 1323 − 57√

471

25

)

σ2 = 5 + 1323 − 57√

471

25.

We shall now examine all possible cases for the mixed Nash equilibrium σ . For allbut one cases, we shall use Conditions (C/1), (C/2) and (C/3) to derive a contradiction;in the remaining case, we shall establish that σ is an irrational profile such that foreach player i ∈ [3], Ui (σ ) = 1.

– Only player 1 is mixed. Then, there is a choice of σ2, σ3 ∈ {0,1} such that Con-

dition (C/1) holds. By inspection, Condition (C/1) has no solution with σ2, σ3 ∈{0,1}. A contradiction.

– Only player 2 is mixed. Then, there is a choice of σ1, σ3 ∈ {0,1} such that Con-

dition (C/2) holds. By inspection, Condition (C/2) has no solution with σ1, σ3 ∈{0,1}. A contradiction.

Theory Comput Syst

– Only player 3 is mixed. Then, there is a choice of σ1, σ2 ∈ {0,1} such that Con-

dition (C/3) holds. By inspection, Condition (C/3) has no solution with σ1, σ2 ∈{0,1}. A contradiction.

– Only players 2 and 3 are mixed. Then, there is a choice of σ1 ∈ {0,1} such that

both Conditions (C/2) and (C/3) hold. Note that for σ1 = 1, Condition (C/2)does not hold for any σ3 ∈ (0,1). Hence, σ1 = 0; so, 0 /∈ Support(σ1) while 1 ∈Support(σ1). Then, Conditions (C/2) and (C/3) imply that σ3 = 42516+107

√471

73443 and

σ2 = 29897−285√

47137762 , respectively. Since 0 /∈ Support(σ1) while 1 ∈ Support(σ1),

Lemma 1 (Condition (ii)) implies that U1(σ−1 0) ≤ U1(σ−1 1), or

2 · σ2 · σ3 − 1 · σ2 · σ3 + 3 · σ2 · σ3 + 2 · σ2 · σ3

≤ −3 · σ2 · σ3 + 4 · σ2 · σ3 − 1 · σ2 · σ3 + 3863 − 173√

471

74· σ2 · σ3,

which simplifies to(

9 − 3863 − 173√

471

74

)

σ2 σ3 +(

−8 + 3863 − 173√

471

74

)

σ2

+(

2 + 3863 − 173√

471

74

)

σ3 ≤ −2 + 3863 − 173√

471

74.

The latter does not hold for σ2 = 29897−285√

47137762 and σ3 = 42516+107

√471

73443 . A con-tradiction.

– Only players 1 and 3 are mixed. Then, there is some choice of σ2 ∈ {0,1} such thatboth Conditions (C/1) and (C/3) hold. By inspection, for any σ2 ∈ {0,1}, Conditions(C/1) and (C/3) have no solutions with σ1 ∈ (0,1) and σ3 ∈ (0,1), respectively.A contradiction.

– Only players 1 and 2 are mixed. Then, there is some choice of σ3 ∈ {0,1} such

that both Conditions (C/1) and (C/2) hold. Note that for σ3 = 1, Condition (C/2)does not hold for any σ1 ∈ (0,1). Hence, σ3 = 0; so, 0 /∈ Support(σ3) while

1 ∈ Support(σ3). Then, Conditions (C/1) and (C/2) imply that σ2 = 26281−1038√

47145907

and σ1 = 42516+107√

47173443 , respectively. Since 0 /∈ Support(σ3) and 1 ∈ Support(σ3),

Lemma 1 (Condition (ii)) implies that U1(σ−1 0) ≤ U1(σ−1 1), or

−3 · σ1 · σ2 + 4 · σ1 · σ2 + 5 · σ1 · σ2

≤ −1 · σ1 · σ2 + 1 · σ1 · σ2 + 5 · σ1 · σ2 + −1323 + 57√

471

25· σ2 · σ3,

which simplifies to(

−5 − 1323 − 57√

471

25

)

σ1 σ2 +(

2 + 1323 − 57√

471

25

)

σ1

+(

10 + 1323 − 57√

471

25

)

σ2 ≥ 5 + 1323 − 57√

471

25.

Theory Comput Syst

The latter is falsified for σ1 = 42516+107√

47173443 and σ2 = 26281−1038

√471

45907 . A con-tradiction.

– All players are mixed. By Conditions (C/1), (C/2) and (C/3), it follows that σ1 =4+√

47165 , σ2 = 102−√

471129 and σ3 = 87−√

471169 . So, σ is irrational. We shall now verify

that for each player i ∈ [3], Ui (σ ) = 1.

– For U1(σ ), note that σ2 = 169σ3+15129 . Since 0 ∈ Support(σ1), Lemma 1 implies

that

U1(σ ) = U1(σ−1 0)

= 2σ2 σ3 − σ2 σ3 + 3σ2 σ3 + 2σ2 σ3

= 2σ2 σ3 − σ2 + σ2 σ3 + 3σ3 − 3σ2 σ3 + 2 + 2σ2 σ3 − 2σ2 − 2σ3

= 2σ2 σ3 − 3σ2 + σ3 + 2

=(

2169σ3 + 15

129σ3 − 3

169σ3 + 15

129+ σ3 + 1

)

+ 1

= 2 · 169

129

(

σ 23 − 174

169σ3 + 42

169

)

+ 1.

But, 87−√471

169 + 87+√471

169 = 2·87169 and 87−√

471169 · 87+√

471169 = (87)2−471

(169)2 = 42169 . This

implies that σ 23 − 2·87

169 σ3 + 42169 = 0. Hence, U1(σ ) = 1.

– For U2(σ ), note that σ1 = −169σ3+9165 . Since 0 ∈ Support(σ2), Lemma 1 implies

that

U2(σ )

=U2(σ−2 0)

= 4σ1σ3 + 5σ1σ3 + 6σ1σ3 − 6σ1σ3

= 4σ1σ3 + 5σ1 − 5σ1σ3 + 6σ3 − 6σ1σ3 − 6 + 6σ1 + 6σ3 − 6σ1σ3

= −13σ1σ3 + 11σ1 + 12σ3 − 6

=(

−13 · −169σ3 + 91

65· σ3 + 11 · −169σ3 + 91

65+ 12σ3 − 7

)

+ 1

= 13 · 169

65σ 2

3 − 13 · 91

65σ3 − 11 · 169

65σ3 + 11 · 91

65+ 12σ3 − 7 + 1

= 13 · 169

65·(

σ 23 − 174

169σ3 + 42

169

)

+ 1.

Recall that σ 23 − 174

169 σ3 + 42169 = 0. Hence, U2(σ ) = 1.

– For U3(σ ), note that σ1 = −129σ2+10665 . Since 0 ∈ Support(σ3), Lemma 1 implies

that

Theory Comput Syst

U3(σ ) = U3(σ−3 0)

= −3σ1σ2 + 4σ1σ2 + 5σ1σ2

= −3σ1σ2 + 4σ1(1 − σ2) + 5(1 − σ1)(1 − σ2)

= −3σ1σ2 + 4σ1 − 4σ1σ2 + 5 − 5σ1 − 5σ2 + 5σ1σ2

= −2σ1σ2 − σ1 − 5σ2 + 5

=(

−2 · −129σ2 + 106

65σ2 − −129σ2 + 106

65− 5σ2 + 4

)

+ 1

=(

2 · 129

65σ 2

2 − 2 · 106

65σ2 + 129

65σ2 − 106

65− 5σ2 + 4

)

+ 1

= 2 · 129

65

(

σ 22 − 204

129σ2 + 77

129

)

+ 1.

But 102−√471

129 + 102−√471

129 = 2·102129 and 102−√

471129 · 102−√

471129 = (102)2−471

(129)2 = 77129 .

This implies that σ 22 − 204

129 σ2 + 77129 = 0. Hence, U3(σ ) = 1.

So, SG2 has a single (mixed) Nash equilibrium σ , which is irrational and where, foreach player i ∈ [3], Ui (σ ) = 1. �

Using SG2 as the gadget game in the reduction, Condition (3) in Proposition 5implies that SG has an irrational Nash equilibrium. Hence, Condition (2) is equivalentto the Condition (2’) SG admits an irrational Nash equilibrium. So, the equivalenceof Conditions (1) and (2) implies:

Proposition 8 CNF SAT ≤P ∃ IRRATIONAL NASH.

Since CNF SAT is NP-hard, Propositions 6 and 8 immediately imply:

Theorem 3 NASH − REDUCTION(SG) is NP-hard.

Theorem 4 ∃ IRRATIONAL NASH is NP-hard.

5 Epilogue

Our work initiates the study of the complexity of decision problems about the ratio-nality and irrationality of Nash equilibria. It remains open to determine the extentof strategic games for which NP-hardness holds for ∃ RATIONAL NASH and ∃ IR-RATIONAL NASH. For example, what happens if we restrict to win-lose games [1]:games where all utilities are 0 or 1? What is the smallest value for the utilities thatsuffices for NP-hardness? A wide avenue for further research concerns the extent ofthe decision problems and the games for which NP-hardness holds.

What is the relation between ∃ IRRATIONAL NASH and the SQUARE-ROOT-SUMproblem [11]? Is ∃ IRRATIONAL NASH SQUARE-ROOT-SUM-hard (cf. [9])?

Theory Comput Syst

Irrational Nash equilibria are a manifestation of a familiar phenomenon in scienceand engineering, where the quantities of interest are solutions to nonlinear equations,so that they can be irrational. Are there other manifestations where deciding irra-tionality is NP-hard?

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