Pure-Circuit: Tight Inapproximability for PPAD

Argyrios Deligkas, John Fearnley, Alexandros Hollender, Themistoklis Melissourgos

Introduction

The complexity class PPAD has played a central role in determining the computational complexity of many problems arising in game theory and economics [Pap94]. The celebrated results of Daskalakis, Goldberg, and Papadimitriou [DGP09] and Chen, Deng, and Teng [CDT09] established that finding a Nash equilibrium in a strategic form game is PPAD-complete, and subsequent to this breakthrough many other PPAD-completeness results have been shown [CSVY08, CDDT09, VY11, DQS12, Das13, KPR+13, CDO15, CPY17, SSB17, Meh18, Rub18, DFS20, CKK21a, CKK21b, PP21, GH21, FRGH+23, DSZ21, CCPY22].

These celebrated results not only showed that it is PPAD-hard to find an exact equilibrium, but also that finding approximate solutions is PPAD-hard. The result of Daskalakis, Goldberg, and Papadimitriou [DGP09] showed that finding an ε\varepsilon-Nash equilibrium is PPAD-complete when ε\varepsilon is exponentially small, while the result of Chen, Deng, and Teng [CDT09] improved this to show hardness for polynomially small ε\varepsilon. This lower bound is strong enough to rule out the existence of an FPTAS for the problem.

The main open question following these results was whether equilibrium computation problems in PPAD were hard for constant ε\varepsilon, which would also rule out the existence of a PTAS. Here one must be careful, because some problems do in fact admit approximation schemes. For example, in the case of two-player strategic-form games, a quasipolynomial-time approximation scheme is known [LMM03], meaning that the problem cannot be hard for a constant ε\varepsilon unless every problem in PPAD can be solved in quasipolynomial time. But for other types of game such results are not known. This includes polymatrix games, which are nn-player games with succinct representation [Jan68].

In another breakthrough result, Rubinstein [Rub18] developed techniques for showing constant inapproximability within PPAD, by proving that there exists a constant ε\varepsilon such that finding an ε\varepsilon-well-supported Nash equilibrium in a polymatrix game is PPAD-complete. This lower bound is obtained by first showing constant inapproximability for the ε\varepsilon-Generalized-Circuit (ε\varepsilon-GCircuit) problem introduced by Chen, Deng, and Teng [CDT09], and then utilizing the known reduction from GCircuit to polymatrix games [DGP09].

Rubinstein’s lower bound has since been used to show constant inapproximability for other problems. Rubinstein himself showed constant inapproximability for finding Bayesian Nash equilibria, relative approximate Nash equilibria, and approximate Nash equilibria in non-monotone markets [Rub18]. Subsequent work has shown constant inapproximability for finding clearing payments in financial networks with credit default swaps [SSB17], finding equilibria in first-price auctions with subjective priors [FRGH+23], finding throttling equilibria in auction markets [CKK21b], finding equilibria in public goods games on directed networks [PP21], and finding consensus-halving solutions in fair division [FRFGZ18, GHI+22].

Rubinstein’s lower bound is an existential-constant result, meaning that it shows that there exists some constant ε\varepsilon below which the problem becomes PPAD-hard. The fact that such a constant exists is important, since it rules out a PTAS. On the other hand, Rubinstein does not give any concrete lower bound on the size of the constant (understandably so, since this was not the purpose of his work). One could of course deduce such a lower bound by a careful examination of his reduction, but it is clear that this would yield an extremely small constant. Due to this, all of the other results that have utilized Rubinstein’s lower bound are likewise existential-constant results, which rule out PTASs but do not give any concrete lower bounds.

Ultimately, this means that existing work does not rule out an efficient algorithm that finds, say, a 0.0010.001-approximate solution for any of these problems, which would likely be more than enough for most practical needs. Arguably, a player would be more than happy to know that her strategy is an optimal best-response, up to a loss of at most 0.0010.001 in her utility value. Moreover, the existing work gives us no clue as to where the threshold for hardness may actually lie. To address these questions one would need to prove a large-constant inapproximability result, giving hardness for a known substantial constant.

Rubinstein’s lower bound is the ultimate source of all of the recent existential-constant lower bounds, so if one seeks a large-constant lower bound, then Rubinstein’s result is the bottleneck. Attempting to directly strengthen or optimize Rubinstein’s result does not seem like a promising direction. His proof, while ingenious, is very involved, and does not lend itself to easy optimization. Furthermore, it consists of many moving parts, so that even if one was able to optimize each module, the resulting constant would still be very small.

In this paper we introduce the techniques needed to show large-constant inapproximability results for problems in PPAD. Our key technical innovation is the introduction of a new intermediate problem, called Pure-Circuit, which we show to be PPAD-complete.

Then, by reducing onwards from Pure-Circuit, we show large-constant inapproximability results for a variety of problems in PPAD. In this sense, Pure-Circuit now takes on the role that ε\varepsilon-GCircuit has taken in the past, as an important intermediate problem from which all other results of this type are derived.

The Pure-Circuit problem itself can be thought of as a version of ε\varepsilon-GCircuit that is taken to its limits, and also dramatically simplified. In fact, the problem has only two gates (or, in a different formulation, three gates), which should be compared to ε\varepsilon-GCircuit, which has nine distinct gates. Perhaps more importantly, the gates in Pure-Circuit have very weak constraints on their outputs: the gates can be thought of as taking inputs in ,andproducingoutputsin, and producing outputs in, but the gates themselves essentially only care about the values and 11, with all other values being considered to be “bad” or “garbage” values (which we will later simply denote by “⊥\bot”, instead of using values in (0,1)(0,1)). This should be compared to ε\varepsilon-GCircuit gates, where, for example, one must output a value in $thatiswithinthat is within\varepsilon$ of the sum of two inputs.

Combined, these properties make Pure-Circuit a very attractive problem to reduce from when showing a hardness result, since one only has to implement two (or three) gates, and the constraints that one must simulate are very loose, making them easy to implement. We formally introduce Pure-Circuit, and compare it to ε\varepsilon-GCircuit, in Section 2.

Our main result is to show that the Pure-Circuit problem is PPAD-complete. It is worth noting that there is no ε\varepsilon in this result, and in fact the Pure-Circuit problem does not even take a parameter ε\varepsilon in its definition. This is because, in some sense, the Pure-Circuit problem can be viewed as a variant of ε\varepsilon-GCircuit in which we have taken the limit ε→1\varepsilon\to 1. We give further justification of this idea in Section 2, but at a high level, this means that there is no loss of ε\varepsilon in our main hardness result, with the only losses coming when one reduces onwards from Pure-Circuit. The proof of our main result is presented in Section 3, but we present a brief exposition of the main ideas in Section 1.1.

Finally, in Section 4 we present a number of new large-constant hardness results for problems in PPAD, all of which are shown via reductions from Pure-Circuit. We begin by showing that ε\varepsilon-GCircuit is PPAD-hard for all ε<0.1\varepsilon<0.1, giving a direct strengthening of Rubinstein’s lower bound. This also implies large-constant inapproximability results for all of the problems that currently have existential-constant lower bounds proved via GCircuit. However, to determine the constant, one would need to determine the amount of ε\varepsilon that is lost in each of the onward reductions, and these reductions often did not optimize this, since they were proving existential-constant lower bounds.

We argue that the way forward now is providing direct reductions from Pure-Circuit in order to get the best possible hardness results. As evidence of this, we present the first tight inapproximability result for additive approximate equilibria in polymatrix games. Via a direct reduction from Pure-Circuit, we show that finding an ε\varepsilon-well-supported Nash equilibrium in a polymatrix game (PolymatrixWSNE) is PPAD-hard for all ε<1/3\varepsilon<1/3, even when every player only has two actions. This is much stronger than the lower bound of 0.050.05 that we would have obtained by a reduction from our lower bound for GCircuit. It is also a tight result for two-action games: we give a polynomial-time algorithm for finding a 1/31/3-well-supported Nash equilibrium, and so our lower bound completely characterizes the computational complexity of approximate well-supported equilibria in two-action polymatrix games. Similarly, we also provide a tight inapproximability result for computing approximate equilibria in threshold games, a problem introduced by Papadimitriou and Peng [PP21].

We summarize the hardness results obtained through a direct reduction from Pure-Circuit in the table below.

We note that PolymatrixWSNE and ThresholdGameNE have themselves both been used as intermediate problems for showing other constant inapproximability results in PPAD [Rub18, PP21, CKK21b, CCPY22], and thus our lower bounds potentially strengthen those results too. We provide an example in Appendix C, where we show that computing a relative ε\varepsilon-WSNE in a bimatrix game with non-negative payoffs is PPAD-complete for any ε≤1/57\varepsilon\leq 1/57, by using an improved version of a reduction from PolymatrixWSNE due to Rubinstein [Rub18].

What is the intractability threshold for ε\varepsilon-NE and ε\varepsilon-WSNE in polymatrix games? For ε\varepsilon-NE our understanding is far from complete, even in the two-action case, since there is a substantial gap between the 0.0880.088 lower bound and the 1/31/3 upper bound. For ε\varepsilon-WSNE, although the problem is completely resolved for the two-action case, the gap in multi-action polymatrix games is still large, and it seems that improving either the lower bound of 1/31/3, or the trivial upper bound of 11, would require significantly new ideas.

1 Proof Overview for Our Main Result

We begin this proof overview by defining a very weak version of Pure-Circuit. An instance of the problem consists of a Boolean circuit using the standard gates NOT, AND, and OR, but with the following tweak: the circuit is allowed to have cycles. A solution to the problem is an assignment of values to each node of the circuit, so that all gates are satisfied. If we are only allowed to assign values in {0,1}\{0,1\} to the nodes, then it is easy to see that the problem is not a total search problem, i.e., some instances do not have a solution. For example, there is no way to assign consistent values to a cycle of three consecutive NOT gates.

In order to ensure that the problem is total (and can thus be used to prove PPAD-hardness results), we make the value space continuous by extending it to $.WeextendthedefinitionofthelogicalgatesNOT,AND,andORtonon−Booleaninputsinthemostpermissiveway:ifatleastoneinputtothegateisnotapurebit(i.e.,notin. We extend the definition of the logical gates NOT, AND, and OR to non-Boolean inputs in the most permissive way: if at least one input to the gate is not a pure bit (i.e., not in\{0,1\}),thenthegateisallowedtooutputanyvaluein), then the gate is allowed to output any value in.Theattentivereadermightobservethatthisproblemisnowtrivialtosolve:justassignarbitraryvaluesin. The attentive reader might observe that this problem is now trivial to solve: just assign arbitrary values in(0,1)$ to all the gates.

It is thus clear that the definition of the problem needs to be extended, by adding extra gates or by strengthening existing gates, so that the problem becomes PPAD-hard. However, in order to discover the least amount of additional structure needed to make the problem hard, it is instructive to proceed with this definition for now, and attempt to prove hardness.

In order to prove the PPAD-hardness of the problem, we cannot follow Rubinstein’s approach, which goes through the construction of a continuous Brouwer function, because Pure-Circuit only offers very weak gates. Instead, we proceed via a direct reduction from the StrongSperner problem, a discrete problem that is a computational version of Sperner’s Lemma. The problem was shown to be PPAD-hard by Daskalakis, Skoulakis, and Zampetakis [DSZ21] (who called it the HighD-BiSperner problem), and is the “PPAD-analogue” of the StrongTucker problem which was recently used to prove PPA-hardness results [DFHM22]. This approach completely bypasses the continuous aspect of all such existing hardness reductions and enables us to work with the very weak gates that Pure-Circuit offers.

At a high level, our hardness construction works as follows: the Pure-Circuit instance implements the evaluation of the StrongSperner labeling on some input point xx (represented in unary by multiple nodes) and then uses a feedback mechanism to ensure that the circuit is only satisfied if xx is a solution to the StrongSperner instance. The full reduction is presented in Section 3, but we mention here the two main obstacles when trying to implement this idea, and how to overcome them.

The input point x\boldsymbol{x} might not be represented by a valid bitstring. Indeed, since the gates take values in $(andvaluesin(and values in(0,1)essentiallydonotcarryanyinformation),thereisnoguaranteethattheinputessentially do not carry any information), there is no guarantee that the inputxwillberepresentedbybitswill be represented by bits\{0,1\}.ButthentheimplementationoftheStrongSpernerlabeling(whichisgivenasaBooleancircuit)willalsofail.Toresolvethisissue,weintroduceanewgate,thePURIFYgate,which,onanyinput,outputstwovalues,withtheguaranteethatatleastoneofthemisa“pure”bit,i.e.,or. But then the implementation of the StrongSperner labeling (which is given as a Boolean circuit) will also fail. To resolve this issue, we introduce a new gate, the PURIFY gate, which, on any input, outputs two values, with the guarantee that at least one of them is a “pure” bit, i.e., or1.Iftheinputisalreadyapurebit,thenbothoutputsareguaranteedtobecopiesoftheinput.UsingabinarytreeofPURIFYgates,wecannowcreatemanycopiesof. If the input is already a pure bit, then both outputs are guaranteed to be copies of the input. Using a binary tree of PURIFY gates, we can now create many copies ofx$, such that most of them consist only of pure bits, and then use the logical gates to compute the StrongSperner labeling correctly on these good copies.

How to implement the feedback mechanism? Given the outputs of the StrongSperner labeling at all the copies of xx, we now need to provide some kind of feedback to xx, so that xx is forced to change if it is not a solution of StrongSperner. It turns out that this step can be performed if we have access to sorting: given a list of values in $$, sort them from smallest to largest. Unfortunately, this is impossible to achieve with the gates at our disposal, namely standard logical gates and the PURIFY gate. We circumvent this obstacle by observing that: (i) it is sufficient to be able to perform some kind of “weak sorting” (essentially, we only care about pure bits being sorted correctly), and (ii) this weak sorting can be achieved if we make our logical gates robust. For example, the robust version of the AND gate outputs , whenever at least one of its inputs is , irrespective of whether the other input is a pure bit or not.

With these two extensions in hand—namely, the PURIFY gate and the robustness of the logical gates— it is now possible to prove PPAD-hardness of the problem. A very natural question to ask is: Is it really necessary to add both extensions for the problem to be hard? In Appendix A we show that any attempt to weaken the gate-constraints makes the problem polynomial-time solvable. In particular, the introduction of the PURIFY gate is not enough by itself to make the problem PPAD-hard; the robustness of the logical gates is also needed.

The robustness of, say, the AND gate seems like a very natural constraint to impose. It is consistent with the meaning of the logical AND operation, but we also observe in our applications that this “robustness” seems to always be automatically satisfied in all simulations of the AND gate. On the other hand, the PURIFY gate, which might look a bit unnatural or artificial at first, actually corresponds to the simplest possible version of a bit decoder, a crucial tool in all prior works. As mentioned above, we show in Appendix A that these are the minimal gate-constraints that are needed for the problem to be PPAD-hard. In that sense, we argue that Pure-Circuit captures the essence of PPAD-hardness: it consists of the minimal set of ingredients that are needed for a problem to be PPAD-hard.

The attentive reader might have noticed that our gates do not distinguish between different values in (0,1)(0,1). For this reason, it will be more convenient to use a single symbol to denote such values in the definition of Pure-Circuit (Section 2) and in the rest of this paper. As explained in more detail in Section 2, the symbol “⊥\bot” will be used to denote these “garbage” values. In other words, the nodes of the circuit will take values in {0,1,⊥}\{0,1,\bot\} instead of $$.

The Pure-Circuit Problem

In this section we define our new problem Pure-Circuit and state our main result, namely its PPAD-completeness. Before defining Pure-Circuit, we begin by explaining the intuition behind its definition, and how it relates to the Generalized-Circuit (GCircuit) problem.

In the Generalized-Circuit (GCircuit) problem (formally defined in Section 4.1) we are given a circuit and the goal is to assign a value to each node of the circuit so that each gate is computed correctly. Importantly, the circuit is a generalized circuit, meaning that cycles are allowed. If cycles were not allowed, then it would be easy to find values satisfying all gates: just pick arbitrary values for the input gates, and then evaluate the circuit on those inputs.

Every node of GCircuit must be assigned a value in ,andthegatesarearithmeticgates,suchasaddition,subtraction,multiplicationbyaconstant(withoutputtruncatedtoliein, and the gates are arithmetic gates, such as addition, subtraction, multiplication by a constant (with output truncated to lie in), and suitably defined logical gates. Reducing from GCircuit is very useful for obtaining hardness of approximation results, because the problem remains PPAD-hard, even when we allow some error at every gate. In the ε\varepsilon-GCircuit problem, the goal is to assign a value in $toeachnodeofthecircuit,sothateachgateiscomputedcorrectly,uptoanadditiveerrorofto each node of the circuit, so that each gate is computed correctly, up to an additive error of\pm\varepsilon$.

The problem was first defined by Chen et al. [CDT09], who proved that it is PPAD-hard for inverse polynomial ε\varepsilon, and who used it to prove PPAD-hardness of finding Nash equilibria in bimatrix games. Prior to that, Daskalakis et al. [DGP09] had implicitly proved that it is PPAD-hard for inverse exponential ε\varepsilon. Rubinstein’s [Rub18] breakthrough result proved that there exists some constant ε>0\varepsilon>0 such that ε\varepsilon-GCircuit remains PPAD-hard.

In order to get strong inapproximability results, it seems necessary to prove hardness of ε\varepsilon-GCircuit for large, explicit, values of ε\varepsilon. Ideally, we would like to obtain hardness for the largest possible ε\varepsilon. While it is unclear what that value is for GCircuit, in theory, as long as ε<1\varepsilon<1 the output of a gate still carries some information. Namely a gate whose actual output should be cannot take the value 11.

This observation leads us to define a problem to essentially capture the setting ε→1\varepsilon\to 1. In that case, a node carries information only if its value is or 11. Otherwise, its value is irrelevant. As a result, the natural operations to consider in this setting are simple Boolean operations, such as NOT, AND, OR, NAND, and NOR. We only require these gates to output the correct result when their input is relevant, i.e., or 11. For example, the NOT gate should output 11 on input , and output on input 11, but there is no constraint on its output when the input lies in (0,1)(0,1).

Since values in (0,1)(0,1) do not carry any information, and are as such interchangeable (e.g., a value 1/21/2 can be replaced by 1/31/3 without impacting any of the gates), we will instead use the symbol “⊥\bot” to denote any and all values in (0,1)(0,1). In other words, instead of assigning a value in $toeachgate,wewillassignavalueinto each gate, we will assign a value in\{0,1,\bot\},where, where\botisinterpretedasa“garbage”value,i.e.,notcorrespondingtoapurebitvalueoris interpreted as a “garbage” value, i.e., not corresponding to a pure bit value or1.Withthisnewnotation,theupdateddescriptionoftheNOTgatewouldbethatitmustoutput. With this new notation, the updated description of the NOT gate would be that it must output1oninput,itmustoutputoninputon input , it must output on input1,anditcanoutputanything(namely,,, and it can output anything (namely, ,1,or, or\bot)oninput) on input\bot$.

Unfortunately, if we only allow these logical gates, then the problem is trivial to solve: assigning the “garbage” value ⊥\bot (or any value in (0,1)(0,1) if we use the old notation) to every node will satisfy all gates. Thus, we need a gate that makes this impossible.

To achieve this, we introduce the PURIFY gate: a gate with one input and two outputs, which, intuitively, “purifies” its input. When fed with an actual pure bit, the PURIFY gate outputs two copies of the input bit. However, when the input is not a pure bit, the gate still ensures that at least one of its two outputs is a pure bit. In more detail:

If the input is , then both outputs are .

If the input is 11, then both outputs are 11.

If the input is ⊥\bot, then at least one of the outputs is a pure bit, i.e., or 11.

Note that the gate is quite “under-defined”. For example, we do not specify which pure bit the gate should output when the input is ⊥\bot, nor do we specify the output on which this bit appears. This is actually an advantage, because it makes it easier to reduce from the problem, since the less constrained the gates are, the easier it is to simulate them in the target application problem.

The introduction of the PURIFY gate makes the problem non-trivial: if a PURIFY gate appears in the circuit, then assigning the “garbage” value ⊥\bot to all nodes is no longer a solution. However, it turns out that one more modification is needed to make the problem PPAD-hard: we have to make the logical gates robust. For the AND gate, this means the following: if one of its two inputs is , then the output is , no matter what the other input is (even if it is not a pure bit, i.e., if it is ⊥\bot). Similarly, for the OR gate we require that the output be 11 when at least one of the two inputs is 11. Robustness is defined analogously for NAND and NOR.

We show that introducing the PURIFY gate and making the logical gates robust is enough to make the problem PPAD-complete. Next, we define the problem formally and state our main result.

In the definition below, we use the PURIFY and NOR gates, because these two gates are enough for the problem to already be PPAD-complete. However, the problem remains hard for various other combinations of gates and restrictions on the interactions between nodes, as we detail in Corollaries 2.2 and 2.3. In Appendix A we discuss the definition in more detail, and explain why any attempt at relaxing the definition (in particular, removing the robustness) makes the problem polynomial-time solvable.

An instance of Pure-Circuit is given by a vertex set V=[n]V=[n] and a set GG of gate-constraints (or just gates). Each gate g∈Gg\in G is of the form g=(T,u,v,w)g=(T,u,v,w) where u,v,w∈Vu,v,w\in V are distinct nodes and T∈{NOR,PURIFY}T\in\{\textup{{NOR}},\textup{{PURIFY}}\} is the type of the gate, with the following interpretation.

If T=NORT=\textup{{NOR}}, then uu and vv are the inputs of the gate, and ww is its output.

If T=PURIFYT=\textup{{PURIFY}}, then uu is the input of the gate, and vv and ww are its outputs.

We require that each node is the output of exactly one gate.

The following theorem is our main technical result and is proved in Section 3.

The Pure-Circuit problem is PPAD-complete.

The most important part of this statement is of course the PPAD-hardness of Pure-Circuit, but let us briefly discuss the other part, namely the PPAD-membership. This is obtained as a byproduct of our results in Section 4, where we reduce Pure-Circuit to various problems that are known to lie in PPAD. However, there is also a more direct way to prove membership in PPAD, and in particular to establish the existence of a solution, and we briefly sketch it here. Indeed, the Pure-Circuit problem can be reduced to the problem of finding a Brouwer fixed point of a continuous function FF, a problem known to lie in PPAD [Pap94, EY10]. Given an instance of Pure-Circuit with nn nodes, the function F:n→nF:^{n}\to^{n} is constructed by letting x∈nx\in^{n} represent an assignment of values to the nn nodes, and by defining Fi(x)∈F_{i}(x)\in as a continuous function that outputs a valid value for the iith node, given that the other nodes have values according to assignment xx (where any value in (0,1)(0,1) is interpreted as “⊥\bot”). For every type of gate, it is not hard to construct a continuous piecewise-linear function FiF_{i} (or, in the case of PURIFY, two such functions FiF_{i} and FjF_{j}) that satisfies the constraints of that type of gate.

Note that the definition of our logical gates essentially follows Kleene’s strong logic of indeterminacy [Kle52], except that undetermined outputs are not required to take value ⊥\bot, but instead any value in {0,1,⊥}\{0,1,\bot\} can be used. This makes it easier to argue about reductions from Pure-Circuit, since the gadgets implementing the gates have to enforce fewer constraints. As a result of this connection to Kleene’s logic, these gates can in particular be used to implement hazard-free circuits [IKL+19].

1 Alternative Gates and Further Restrictions

In this section, we present various versions of the problem that remain PPAD-complete, in particular versions that use alternative gates and have additional restrictions.

We define the following additional gates.

The Pure-Circuit problem is PPAD-complete, for any of the following choices of gate types:

PURIFY and at least one of {NOR,NAND}\{\textup{{NOR}},\textup{{NAND}}\};

PURIFY, NOT, and at least one of {OR,AND}\{\textup{{OR}},\textup{{AND}}\}.

This follows from Theorem 2.1 by observing that a NOR gate can always be simulated with the given set of gates. Clearly, NOR can be simulated by first using an OR gate and then a NOT gate. Furthermore, OR can be simulated by NOT and AND by applying De Morgan’s laws. Finally, AND can easily be obtained from NOT and NAND, and NOT can be obtained from NAND and PURIFY as follows: first apply a PURIFY gate, and then use its two outputs as the two inputs to a NAND gate. ∎

The hardness result is also robust with respect to restrictions applied to the interaction graph. This graph is constructed on the vertex set V=[n]V=[n] by adding a directed edge from node uu to node vv whenever vv is the output of a gate with input uu. For example, a NOR gate with inputs u,vu,v and output ww yields the two edges (u,w)(u,w) and (v,w)(v,w). On the other hand, a PURIFY gate with input uu and outputs v,wv,w gives the edges (u,v)(u,v) and (u,w)(u,w). Since any given node is the output of at most one gate, it immediately follows that the in-degree of every node is at most 22. However, the out-degree of a node can a priori be arbitrarily large. It is quite easy to show that the problem remains PPAD-complete, even if we severely restrict the interaction graph.

The Pure-Circuit problem remains PPAD-complete, for any choice of gates {PURIFY,X,Y}\{\textup{{PURIFY}},\textup{{X}},\textup{{Y}}\}, where (X,Y)∈{NOT}×{OR,AND,NOR,NAND}(\textup{{X}},\textup{{Y}})\in\{\textup{{NOT}}\}\times\{\textup{{OR}},\textup{{AND}},\textup{{NOR}},\textup{{NAND}}\} or (X,Y)∈{COPY}×{NOR,NAND}(\textup{{X}},\textup{{Y}})\in\{\textup{{COPY}}\}\times\{\textup{{NOR}},\textup{{NAND}}\} and even if we also simultaneously have all of the following restrictions.

Every node is the input of exactly one gate.

In the interaction graph, the total degree of every node is at most 3. More specifically, for every node, the in- and out-degrees, dind_{in} and doutd_{out}, satisfy (din,dout)∈{(1,1),(2,1),(1,2)}(d_{in},d_{out})\in\{(1,1),(2,1),(1,2)\}.

The proof of this corollary is again quite simple, but a bit tedious, so it is relegated to Appendix B.

Using Corollary 2.3, it is also possible to show that Pure-Circuit with only two gates (namely, PURIFY and one of {NOR,NAND}\{\textup{{NOR}},\textup{{NAND}}\}) remains PPAD-complete even if the total degree of every node is at most 4 in the interaction graph. Indeed, NOT gates can be implemented by first using a PURIFY gate and then a NOR/NAND gate. The structural properties of Corollary 2.3 ensure that this yields an interaction graph where the total degree is at most 4 for each node. This can be further reduced to degree 3, if one modifies the definition of Pure-Circuit (Definition 1) so that the two inputs to a NOR/NAND gate are no longer required to be two distinct nodes uu and vv, but can possibly be the same node u=vu=v. However, if the definition is modified in that way, then one must be careful when reducing from Pure-Circuit to make sure to take into account the possibility that u=vu=v when constructing the gadget for a NOR/NAND gate.

PPAD-completeness of Pure-Circuit

This section proves our main technical result, namely that Pure-Circuit is PPAD-complete (Theorem 2.1). We note that membership in PPAD follows immediately from the reduction of the problem to GCircuit in Section 4.1. In order to establish the PPAD-hardness, we present a polynomial-time reduction from a PPAD-complete problem to Pure-Circuit. The canonical PPAD-complete problem is the End-of-Line problem, but, as is usually the case, we do not reduce directly from End-of-Line, but from a problem with topological structure instead, which we introduce next.

We will reduce from the StrongSperner problem, which is based on a variant of Sperner’s lemma [Spe28]. This problem is in essence the same as the HighD-BiSperner problem defined by Daskalakis et al. [DSZ21] and used to prove PPAD-hardness of a problem related to constrained min-max optimization. Furthermore, the corresponding “strong” variant of Tucker’s lemma was used by Deligkas et al. [DFHM22] to provide improved PPA-hardness results for the consensus-halving problem in fair division.

A Boolean circuit computing a labeling λ:[M]N→{−1,+1}N\lambda:[M]^{N}\to\{-1,+1\}^{N} satisfying the following boundary conditions for every i∈[N]i\in[N]:

if xi=1x_{i}=1, then [λ(x)]i=+1[\lambda(x)]_{i}=+1;

if xi=Mx_{i}=M, then [λ(x)]i=−1[\lambda(x)]_{i}=-1.

Note that the requirement that a solution should consist of exactly NN points is without loss of generality. If we find less than NN points that cover all labels, then we can simply re-use the same points multiple times to obtain a list of NN points that cover all labels (there is no requirement on them being distinct). If we find more than NN points that cover all labels, then it is easy to see that we can extract a subset of NN points that still cover all labels in polynomial time [DFHM22, Lemma 3.1].

StrongSperner is PPAD-hard, even when MM is only polynomially large (i.e., given in unary in the input).

This was proven by Daskalakis et al. [DSZ21] by reducing from the SuccinctBrouwer problem, which had been proven PPAD-hard by Rubinstein [Rub16]. The PPAD-hardness can also be proved by a more direct reduction from End-of-Line. Indeed, End-of-Line can be reduced to StrongSperner with N=2N=2 and exponentially large MM by using the techniques of Chen and Deng [CD09]. Then, a snake embedding technique [CDT09, DFHM22] can be used to obtain hardness for the high-dimensional version with small MM, in fact, even for constant MM. For our purposes, the hardness for polynomially large MM is sufficient.

2 Reduction from StrongSperner to Pure-Circuit

Consider an instance λ:[M]N→{−1,+1}N\lambda:[M]^{N}\to\{-1,+1\}^{N} of StrongSperner, where λ\lambda is given as a Boolean circuit and MM is only polynomially large (i.e., given in unary). We will now show how to construct an instance of Pure-Circuit in polynomial time such that from any correct assignment to the nodes, we can extract a solution to the StrongSperner instance in polynomial time. We will make use of the gates PURIFY, AND, OR, NOT, COPY. All these gates can easily be simulated using the two gates PURIFY and NOR, by the arguments in the proof of Corollary 2.2.

We begin the construction of the Pure-Circuit instance by creating nodes ui,1,…,ui,Mu_{i,1},\dots,u_{i,M} for each i∈[N]i\in[N]. We call these nodes the original inputs, and we think of ui,1,…,ui,Mu_{i,1},\dots,u_{i,M} as being the unary representation of an element in [M][M]. Of course, this only makes sense when all these nodes are assigned pure bit values, i.e., or 11. In general, this will not be the case. The rest of the instance can be divided into four parts: the purification stage, the circuit stage, the sorting stage, and the selection stage.

For each (i,j)∈[N]×[M](i,j)\in[N]\times[M], we construct a binary tree of PURIFY gates that is rooted at ui,ju_{i,j} and has leaves ui,j(1),…,ui,j(K)u_{i,j}^{(1)},\dots,u_{i,j}^{(K)}.

There are at least K−NMK-NM good copies, i.e., ∣G∣≥K−NM|G|\geq K-NM.

Observe that in a binary tree of PURIFY gates, if some node has some pure value b∈{0,1}b\in\{0,1\}, then all nodes in the subtree rooted at this node also have value bb. Applying this observation at the root of the tree rooted at ui,ju_{i,j}, we immediately obtain part 2 of the statement.

We assume, without loss of generality, that λ\lambda is given as a Boolean circuit C:({0,1}M)N→{0,1}NC:(\{0,1\}^{M})^{N}\to\{0,1\}^{N} using gates AND, OR, NOT, and, on input z∈({0,1}M)Nz\in(\{0,1\}^{M})^{N}:

The circuit CC outputs λ(z1‾,…,zN‾)∈{−1,+1}N\lambda(\overline{z_{1}},\dots,\overline{z_{N}})\in\{-1,+1\}^{N}, where a −1-1 output is represented by a , and a +1+1 output by a 11. To keep things simple, in the rest of this exposition we will abuse notation and think of λ\lambda as outputting labels in {0,1}N\{0,1\}^{N}.

If the circuit is not originally in this form, then it can be brought in this form in polynomial time.

In the circuit stage, we construct KK separate copies of the circuit CC, using the AND, OR, and NOT gates. For each k∈[K]k\in[K], the kkth copy CkC_{k} takes as input the nodes (ui,j(k))(i,j)∈[N]×[M](u_{i,j}^{(k)})_{(i,j)\in[N]\times[M]} and we denote its output nodes by v1(k),…,vN(k)v_{1}^{(k)},\dots,v_{N}^{(k)}. Since the gates always have correct output when the inputs are pure bits, we immediately obtain the following.

In this stage, for each i∈[N]i\in[N], we would like to have a gadget that takes as input the list of nodes vi(1),…,vi(K)v_{i}^{(1)},\dots,v_{i}^{(K)} (namely, the list of iith outputs of the circuits C1,…,CKC_{1},\dots,C_{K}) and outputs the nodes wi(1),…,wi(K)w_{i}^{(1)},\dots,w_{i}^{(K)}, such that these output nodes are a sorted list of the values of the input nodes (where we think of the values as being ordered 0<⊥<10<\bot<1). Unfortunately, this is not possible given the gates we have at our disposal. However, it turns out that we can do some kind of “weak” sorting by using the robustness of the AND and OR gates (i.e., the fact that AND on input and ss, always outputs , no matter what s∈{0,1,⊥}s\in\{0,1,\bot\} is).

For now assume that we consider values in $(insteadof(instead of\{0,1,\bot\})andthatwehaveaccesstoacomparatorgatethattakestwoinputs) and that we have access to a comparator gate that takes two inputss_{1}andands_{2}andoutputsand outputst_{1}andandt_{2},suchthat, such thatt_{1},t_{2}isthesortedlistis the sorted lists_{1},s_{2}.Formally,wecanwritethisas. Formally, we can write this ast_{1}:=\min\{s_{1},s_{2}\}andandt_{2}:=\max\{s_{1},s_{2}\}.Usingcomparatorgates,itiseasytoconstructacircuitthattakes. Using comparator gates, it is easy to construct a circuit that takesKinputsandoutputstheminsortedorder.Indeed,wecandirectlyimplementasortingnetwork[Knu98],forexample.Evenaverynaiveapproachwillyieldsuchacircuitofpolynomialsize,whichisallweneed.Weimplementthiscircuitwithinputsinputs and outputs them in sorted order. Indeed, we can directly implement a sorting network [Knu98], for example. Even a very naive approach will yield such a circuit of polynomial size, which is all we need. We implement this circuit with inputsv_{i}^{(1)},\dots,v_{i}^{(K)}andoutputsand outputsw_{i}^{(1)},\dots,w_{i}^{(K)}inourPure−Circuitinstance,byreplacingeverycomparatorgatebyANDandORgates.Namely,toimplementacomparatorgatewithinputsin our Pure-Circuit instance, by replacing every comparator gate by AND and OR gates. Namely, to implement a comparator gate with inputss_{1},s_{2}andoutputsand outputst_{1},t_{2},weuseanANDgatewithinputs, we use an AND gate with inputss_{1},s_{2}andoutputand outputt_{1},andanORgatewithinputs, and an OR gate with inputss_{1},s_{2}andoutputand outputt_{2}$. The robustness of the AND and OR gates allows us to prove that this sorting gadget sorts the pure bit values correctly, in the following sense.

We prove the claim by induction. Clearly, all input nodes satisfy the claim. Now consider some node t1t_{1} that is the min⁡\min-output of a comparator gate with inputs s1s_{1} and s2s_{2}, that both satisfy the claim. Recall that this gate will be implemented in the Pure-Circuit by an AND gate with inputs s1,s2s_{1},s_{2} and output t1t_{1}. If the ideal circuit assigns value 1/21/2 to t1t_{1}, then the claim trivially holds for t1t_{1}. If the ideal circuit assigns value 11 to t1t_{1}, then both s1s_{1} and s2s_{2} must have value 11 in the ideal circuit. Since the claim holds for s1s_{1} and s2s_{2}, they also have value 11 in Pure-Circuit, and so the AND gate will ensure that t1t_{1} also has value 11, thus satisfying the claim. Finally, if the ideal circuit assigns value to t1t_{1}, then it must also have assigned value to at least one of s1s_{1} or s2s_{2}. But then, by the claim, Pure-Circuit also assigns value to at least one of s1s_{1} or s2s_{2}, and the robustness of the AND gate ensures that t1t_{1} also has value . The same argument also works with max⁡\max and OR instead. ∎

Note that Lemma 3.4 only guarantees a “weak” type of sorting: some parts of the output list might not be correctly ordered, and the list of output values might not be a permutation of the input values (namely, it can happen that there are more ’s and/or 11’s in the output list than in the input list). However, this “weak” sorting will be enough for our needs as we will see below.

Since the list wi(1),…,wi(K)w_{i}^{(1)},\dots,w_{i}^{(K)}, is now “sorted”, we can select MM nodes from it in such a way that at most one node does not have a pure value. Indeed, this can be achieved by selecting nodes that are sufficiently far apart from each other. We thus select the nodes (wi(j⋅2NM))j∈[M](w_{i}^{(j\cdot 2NM)})_{j\in[M]} and copy their values onto the original input nodes (ui,j)j∈[M](u_{i,j})_{j\in[M]}. Namely, for each j∈[M]j\in[M] we introduce a COPY gate with input wi(j⋅2NM)w_{i}^{(j\cdot 2NM)} and output ui,ju_{i,j}. Recall that K=3NM2≥M⋅2NMK=3NM^{2}\geq M\cdot 2NM, so this is well defined.

This selection procedure ensures that two nice properties hold. First, as mentioned above, at most one of the selected nodes does not have a pure value. This is due to the fact that we select nodes that are far apart from each other in the “sorted” list, and all nodes with non-pure values lie close together in that list. Second, if all good copies agree that the iith output is, say, 11, then all of the selected nodes will have value 11. This is due to the fact that the “sorted” list will contain very few values that are not 11, and these values will lie at the very beginning of the list. Since we do not select any nodes from the beginning of the list (the first node that is selected is at index 2NM2NM), we will thus only select nodes with value 11. A similar argument also applies to the case where all good copies agree that the iith output is . In that case, we use the fact that we do not select any nodes that lie close to the end of the list. We formalize these arguments in the proof of the following lemma.

All good copies are close to each other: ∥u(k)−u(k′)∥∞≤1\|u^{(k)}-u^{(k^{\prime})}\|_{\infty}\leq 1 for all k,k′∈Gk,k^{\prime}\in G.

The description of the reduction is now complete. We have constructed a valid instance of Pure-Circuit in polynomial time. In particular, note that every node is the output of exactly one gate. To complete the proof, it remains to prove that from any solution of the Pure-Circuit instance we can extract a solution to StrongSperner in polynomial time. We do this in the following final lemma.

The points {u(k):k∈G}⊆[M]N\{u^{(k)}:k\in G\}\subseteq[M]^{N} yield a solution to the StrongSperner instance λ\lambda.

Applications

In this section, we derive strong inapproximability lower bounds for PPAD-complete problems, by reducing from Pure-Circuit.

The ε\varepsilon-GCircuit problem was introduced by Chen, Deng, and Teng [CDT09]. In this section, we show that ε\varepsilon-GCircuit is PPAD-hard for all ε<0.1\varepsilon<0.1.

A generalized circuit is a tuple (V,T)(V,T), where VV is a set of nodes, and TT is a set of gates. Each gate t∈Tt\in T is a five-tuple (G,u,v,w,c)(G,u,v,w,c), where GG is a gate type from the set {Gc,G×c,G=,G+,G−,G<,G∨,G∧,G¬}\{G_{c},G_{\times c},G_{=},G_{+},G_{-},G_{<},G_{\lor},G_{\land},G_{\lnot}\}, u,v∈V∪{nil}u,v\in V\cup\{\textsf{nil}\} are input variables, w∈Vw\in V is an output variable, and c∈∪{nil}c\in\cup\{\textsf{nil}\} is a rational constant.

The following requirements must be satisfied for each gate (G,u,v,w,c)∈T(G,u,v,w,c)\in T.

GcG_{c} gates take no input variables and use a constant in $.So. Sou=v=\textsf{nil}andandc\inwheneverwheneverG=G_{c}$.

G×cG_{\times c} gates take one input variable and a constant. So u∈Vu\in V, v=nilv=\textsf{nil}, and c∈c\in whenever G=G×cG=G_{\times c}.

G=G_{=} and G¬G_{\lnot} gates take one input variable and do not use a constant. So u∈Vu\in V, v=c=nilv=c=\textsf{nil}, whenever G∈{G=,G¬}G\in\{G_{=},G_{\lnot}\}.

All other gates take two input variables and do not use a constant. So u∈Vu\in V, v∈Vv\in V, and c=nilc=\textsf{nil} whenever G∉{Gc,G×c,G=,G¬}G\notin\{G_{c},G_{\times c},G_{=},G_{\lnot}\}.

Every variable in VV is the output variable for exactly one gate. More formally, for each variable w∈Vw\in V, there is exactly one gate t∈Tt\in T such that t=(G,u,v,w,c)t=(G,u,v,w,c).

Here the notation a=b±εa=b\pm\varepsilon is used as a shorthand for a∈[b−ε,b+ε]a\in[b-\varepsilon,b+\varepsilon]. We will also make use of gates of type (G>,u,v,w,nil)(G_{>},u,v,w,\textsf{nil}) which enforce the constraint

Gates of type G>G_{>} can be easily built by using a G¬G_{\lnot} gate to negate the output of a G<G_{<} gate.

In the remainder of this section, we prove the following result.

ε\varepsilon-GCircuit is PPAD-hard for every ε<0.1\varepsilon<0.1.

We will reduce from the Pure-Circuit problem that uses the gates NOR and PURIFY, which we showed to be PPAD-hard in Theorem 2.1. We will encode 0 values in the Pure-Circuit problem as values in the range [0,ε][0,\varepsilon] in GCircuit, while 1 values will be encoded as values in the range [1−ε,1][1-\varepsilon,1]. Then, each gate from Pure-Circuit will be simulated by a combination of gates in the GCircuit instance.

A NOR gate (NOR,u,v,w)(\textup{{NOR}},u,v,w) will be simulated by GCircuit gates that compute

which requires us to use G+G_{+}, G<G_{<} and GcG_{c}. We claim that this gate works for any ε<1/9\varepsilon<1/9.

2 Polymatrix Games

In this section, we show a number of results for polymatrix games. After defining this class of games and the equilibrium notions of interest, we provide a simple algorithm that computes a 1/31/3-well-supported Nash equilibrium (WSNE) in polynomial time (Theorem 4.2). We then show that this algorithm is in fact optimal, by proving hardness of finding ε\varepsilon-WSNE for all ε<1/3\varepsilon<1/3 (Theorem 4.3). Next, we also prove hardness of computing ε\varepsilon-Nash equilibria for all ε<273−17≈0.088\varepsilon<2\sqrt{73}-17\approx 0.088 (Theorem 4.4). Both of these hardness results hold for degree 3 bipartite games with at most two strategies per player. Finally, we show that the hardness result for ε\varepsilon-WSNE also holds for win-lose polymatrix games in degree 7 bipartite games with at most two strategies per player (Theorem 4.6).

A polymatrix game is defined by an undirected graph (V,E)(V,E), where each vertex represents a player, and we use n=∣V∣n=|V| to denote the number of players in the game. Each player i∈Vi\in V has a fixed number of actions (also called pure strategies) given by mim_{i}. For each edge (i,j)∈E(i,j)\in E, there is an mi×mjm_{i}\times m_{j} matrix AijA_{ij} giving the payoffs that player ii obtains from their interaction with player jj, and likewise there is an mj×mim_{j}\times m_{i} matrix AjiA_{ji} giving payoffs for player jj’s interaction with player ii.

A strategy profile specifies a mixed strategy for each of the players, and so the set of strategy profiles is given by Σ=Δm1−1×⋯×Δmn−1\Sigma=\Delta^{m_{1}-1}\times\dots\times\Delta^{m_{n}-1}. For each strategy profile s=(s1,s2,…,sn)∈Σ\mathbf{s}=(s_{1},s_{2},\dots,s_{n})\in\Sigma, the payoff to player ii is

In other words, the payoff obtained by player ii is the sum of the payoffs obtained from the interaction of ii with every neighbouring player jj. We denote by s−i\mathbf{s}_{-i} (resp. a−i\mathbf{a}_{-i}) the partial strategy profile (resp. partial action profile) consisting of the strategies (resp. actions) of all players except ii. For any particular pure strategy k∈[mi]k\in[m_{i}], the payoff when player ii plays kk, and all other players play according to s\mathbf{s} is denoted by

where eke_{k} is the mim_{i}-dimensional vector with all entries set to , except for the kk-th that is set to 11.

A pure strategy kk is a best response for player ii in strategy profile s\mathbf{s} if it achieves the maximum payoff over all the pure strategies available to player ii, meaning that

Strategy kk is an ε\varepsilon-best response if it is within ε\varepsilon of being a best response, meaning that

The best response payoff for player ii is the payoff associated with the best response strategies, which we denote as

A strategy profile s\mathbf{s} is a Nash equilibrium if every player achieves their best response payoff, meaning that bri(s−i)=ui(s)\text{br}_{i}(\mathbf{s}_{-i})=u_{i}(\mathbf{s}) for all players ii. The approximation notions that will be of interest to us in this section are the following.

A strategy profile s\mathbf{s} is an ε\varepsilon-Nash equilibrium (ε\varepsilon-NE) if every player’s payoff is within ε\varepsilon of their best response payoff. Formally,

A strategy profile s\mathbf{s} is an ε\varepsilon-well-supported Nash equilibrium (ε\varepsilon-WSNE) if every player only plays strategies that are ε\varepsilon-best responses, meaning that for all ii we have that supp(si)\textup{{supp}}(s_{i}) contains only ε\varepsilon-best response strategies. Formally,

Note that ε\varepsilon-NE and ε\varepsilon-WSNE are both additive notions of approximation, and so values of ε\varepsilon cannot normally be compared across games, since doubling all payoffs in the game would double the value of ε\varepsilon, for example. To deal with this, it is common to normalize the payoffs of the game so that the maximum possible payoff is 1 and the minimum possible payoff is 0, which allows us to meaningfully compare values of ε\varepsilon across games.

We adopt the normalization scheme for polymatrix games given by Deligkas et al. [DFSS17], which is a particularly restrictive scheme, but doing so will allow our lower bounds to be compared to the 0.5+δ0.5+\delta upper bound for finding ε\varepsilon-NEs given in that paper.

Then, if d(i)d(i) denotes the degree of player ii in (V,E)(V,E), we apply the following transformation to each payoff zz in each payoff matrix AijA_{ij}

If Ui=LiU_{i}=L_{i}, then we simply set T(z)=0T(z)=0. This ensures that each player’s maximum possible payoff is 11, and their minimum possible payoff is .

In our hardness reductions, we will directly build games that are already normalized, meaning that Ui=1U_{i}=1 and Li=0L_{i}=0 for all players ii, so no extra normalization step is required.

2.1 A Simple Algorithm for 𝟏/𝟑13\boldsymbol{1/3}-WSNE in Two-Action Polymatrix Games

In this section we present a simple polynomial-time algorithm for computing a 1/31/3-WSNE in a two-action polymatrix game. We begin with a description of the algorithm. The algorithm proceeds in two steps:

In the first step of the algorithm, we check if there exists some player ii such that one of its two actions is always a 1/31/3-best response, no matter what actions the other players pick. More formally, we check if there exists a player ii and an action k∈{0,1}k\in\{0,1\} such that ui(k,s−i)≥max⁡{ui(0,s−i),ui(1,s−i)}−1/3u_{i}(k,\mathbf{s}_{-i})\geq\max\{u_{i}(0,\mathbf{s}_{-i}),u_{i}(1,\mathbf{s}_{-i})\}-1/3 for all strategy profiles s\mathbf{s}. If we find such a player ii, then we fix their strategy to always play action kk, and we remove them from the game, while also updating the payoffs of the other players to reflect the fact that player ii always plays action kk. Note that no matter how we fix the strategies of the remaining players later in the algorithm, player ii is guaranteed to play a 1/31/3-best response action in the original game.

After removing player ii from the game, and updating the payoffs, we again check if there exists some player jj with a 1/31/3-best response action, and if so, fix their strategy and remove them from the game, as above. After at most nn iterations, we are left with a game where none of the remaining players has an action that is always a 1/31/3-best response.

In the second step of the algorithm, we simply fix the strategies of the remaining players so that they mix uniformly between their two actions, i.e., they play action with probability 1/21/2, and action 11 with probability 1/21/2.

We note that the simple approach used by the algorithm is essentially the same as an algorithm given by Liu et al. [LLD21] for computation of exact equilibria in very sparse win-lose polymatrix games. We now prove the following.

The algorithm computes a 1/31/3-WSNE in polynomial time in two-action polymatrix games.

The algorithm clearly runs in polynomial time. To prove its correctness, it suffices to show that for the remaining players (i.e., those that were not removed during Step 1) both their actions are 1/31/3-best-responses, when all the other remaining players mix uniformly.

2.2 Hardness for 𝜺𝜺\boldsymbol{\varepsilon}-WSNE

In this section we prove that computing an ε\varepsilon-WSNE is PPAD-hard for any ε<1/3\varepsilon<1/3, even in two-action polymatrix games. In particular, this shows that the simple algorithm presented in the previous section is optimal.

It is PPAD-hard to find an ε\varepsilon-WSNE in a polymatrix game for all ε<1/3\varepsilon<1/3, even in degree 3 bipartite games with two strategies per player.

We reduce from the variant of Pure-Circuit that uses NOT, AND, and PURIFY gates, and we will specifically reduce from the hardness result given in Corollary 2.3, since it will give us extra properties for the hard polymatrix games.

Given an instance (V,G)(V,G) of Pure-Circuit, we will build a polymatrix game with vertex set VV, meaning each player in the game will simulate a variable from Pure-Circuit. Each player ii will have exactly two strategies, which we will denote as zero and one. The strategy sis_{i} of player ii will encode the value of the variable in the following way.

If sis_{i} places all probability on zero, then this corresponds to setting variable ii to 0.

If sis_{i} places all probability on one, then this corresponds to setting variable ii to 1.

If sis_{i} is a strictly mixed strategy, then this corresponds to setting variable ii to ⊥\bot.

The edges of the game will simulate the gates. The definition of Pure-Circuit requires that each variable vv is the output of exactly one gate gg. An important property of our reduction is that the player representing vv only receives non-zero payoffs from the inputs to the gate gg, and receives zero payoffs from any other gates that use vv as an input. This means that we can reason about the equilibrium condition of each of the gates independently, without worrying about where the values computed by those gates are used.

For a gate g=(NOT,u,v)g=(\textup{{NOT}},u,v), the player vv, who represents the output variable, will play the following bimatrix game against uu, who represents the input variable.

vvuuzeroonezeroone1111 This depiction of a bimatrix game shows the matrices AvuA_{vu} and AuvA_{uv}, with the payoffs for vv being shown in the bottom-left of each cell, and the payoffs for uu being shown in the top-right.

We claim that this gate will work for all ε<1\varepsilon<1.

If uu plays zero as a pure strategy, then one gives payoff 11 to vv and zero gives payoff to vv. So in any ε\varepsilon-WSNE with ε<1\varepsilon<1, only one can be an ε\varepsilon-best response for vv, meaning that vv must play one as a pure strategy, as required by the constraints of the NOT gate.

Using identical reasoning, if uu plays one as a pure strategy, then vv must play zero as a pure strategy in any ε\varepsilon-WSNE.

If uu plays a strictly mixed strategy, then the NOT gate places no constraints on the output, so we need not prove anything about vv’s strategy.

For a gate g=(AND,u,v,w)g=(\textup{{AND}},u,v,w), the player ww, who represents the output variable, will play the following matrix games against uu and vv, who represent the input variables.

wwuuzeroonezeroone12<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mfrac><mn>1</mn><mn>6</mn></mfrac></mrow><annotationencoding="application/x−tex">16</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:2.0074em;vertical−align:−0.686em;"></span><spanclass="mord"><spanclass="mopennulldelimiter"></span><spanclass="mfrac"><spanclass="vlist−tvlist−t2"><spanclass="vlist−r"><spanclass="vlist"style="height:1.3214em;"><spanstyle="top:−2.314em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">6</span></span></span><spanstyle="top:−3.23em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="frac−line"style="border−bottom−width:0.04em;"></span></span><spanstyle="top:−3.677em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">1</span></span></span></span><spanclass="vlist−s">​</span></span><spanclass="vlist−r"><spanclass="vlist"style="height:0.686em;"><span></span></span></span></span></span><spanclass="mclosenulldelimiter"></span></span></span></span></span></span>w\frac{1}{2}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mn>1</mn><mn>6</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{1}{6}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.0074em;vertical-align:-0.686em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3214em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">6</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span>wvvzeroonezeroone12\frac{1}{2}16\frac{1}{6} We claim that this gate will work for all ε<13\varepsilon<\frac{1}{3}.

If both uu and vv play one as a pure strategy, then the payoff to ww for playing zero is , and the payoff to ww for playing one is 1/6+1/6=1/31/6+1/6=1/3. So, if ε<13\varepsilon<\frac{1}{3}, then in all ε\varepsilon-WSNEs the only ε\varepsilon-best response for ww is one, as required by the constraints of the AND gate.

If at least one of uu and vv play zero as a pure strategy, then the payoff to ww for playing zero is at least 1/21/2, while the payoff to ww for playing one is at most 1/61/6. So, if ε<13\varepsilon<\frac{1}{3}, then in all ε\varepsilon-WSNEs the only ε\varepsilon-best response for ww is zero, as required by the constraints of the AND gate.

The AND gate places no other constraints on the variable ww, so we can ignore all other cases, e.g., the case where both uu and vv play strictly mixed strategies.

For a gate g=(PURIFY,u,v,w)g=(\textup{{PURIFY}},u,v,w), the players vv and ww, who represent the output variables, play the following games against uu, who represents the input variable.

vvuuzeroonezeroone13<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mn>1</mn></mrow><annotationencoding="application/x−tex">1</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em;"></span><spanclass="mord">1</span></span></span></span></span>w\frac{1}{3}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1</span></span></span></span></span>wuuzeroonezeroone1113\frac{1}{3} We claim that this gate works for any ε<1/3\varepsilon<1/3. We first consider player vv.

If uu plays zero as a pure strategy, then strategy zero gives payoff 1/31/3 to vv, while strategy one gives payoff , so in any ε\varepsilon-WSNE with ε<1/3\varepsilon<1/3 we have that zero is the only ε\varepsilon-best response for vv.

If uu places at least 1/21/2 probability on one, then strategy zero gives at most 1/61/6 payoff to vv, while strategy one gives at least 1/21/2 payoff to vv. So in any ε\varepsilon-WSNE with ε<1/3\varepsilon<1/3 we have that one is the only ε\varepsilon-best response for vv.

Symmetrically, for player ww we have the following.

If uu places at most 1/21/2 probability on one, then strategy zero gives at least 1/21/2 payoff to ww, while strategy one gives at most 1/61/6 payoff to ww. So in any ε\varepsilon-WSNE with ε<1/3\varepsilon<1/3 we have that zero is the only ε\varepsilon-best response for ww.

If uu plays one as a pure strategy, then strategy one gives payoff 1/31/3 to ww, while strategy zero gives payoff , so in any ε\varepsilon-WSNE with ε<1/3\varepsilon<1/3 we have that one is the only ε\varepsilon-best response for ww.

From these properties, we can verify that the constraints imposed by the PURIFY gate are enforced correctly.

If uu plays zero as a pure strategy, then both vv and ww play zero as a pure strategy.

If uu plays one as a pure strategy, then both vv and ww play one as a pure strategy.

No matter how uu mixes, at least one of vv and ww plays a pure strategy, with vv playing pure strategy zero whenever uu places at most 0.50.5 probability on one, and ww playing pure strategy one whenever uu places at least 0.50.5 probability on one.

As we have argued above, each of the gate constraints from the Pure-Circuit instance are enforced correctly in any ε\varepsilon-WSNE of the polymatrix game with ε<1/3\varepsilon<1/3. Thus, given such an ε\varepsilon-WSNE, we can produce a satisfying assignment to the Pure-Circuit instance using the mapping that we defined at the start of the reduction. All of the games that we have presented are already normalized so that each player’s maximum possible payoff is 1, and minimum possible payoff is 0, so no extra normalization step is necessary.

Since we are reducing from Corollary 2.3, we get some extra properties about the structure of the polymatrix game. Observe that the interaction graph of the polymatrix game is exactly the interaction graph of the Pure-Circuit instance, as defined in Section 2. Thus Corollary 2.3 implies that the polymatrix game is degree three and bipartite. Moreover, by construction, each player has exactly two strategies. ∎

2.3 Hardness for 𝜺𝜺\boldsymbol{\varepsilon}-NE

We now show that computing an ε\varepsilon-NE in polymatrix games is PPAD-complete for any constant ε<273−17≈0.088\varepsilon<2\sqrt{73}-17\approx 0.088 even in degree 33 bipartite games with two strategies per player. It is worth noting that, given our PPAD-hardness result for ε′\varepsilon^{\prime}-WSNE for all ε′<1/3\varepsilon^{\prime}<1/3 in polymatrix games (Theorem 4.3), one can effortlessly get PPAD-hardness for ε\varepsilon-NE for ε\varepsilon smaller than roughly 0.001360.00136 by Lemma 7.4 of [Rub18]. However, by doing a direct reduction from Pure-Circuit we get an upper bound for ε\varepsilon that is more than 64 times greater.

It is PPAD-hard to find an ε\varepsilon-NE in a polymatrix game for all ε<273−17≈0.088\varepsilon<2\sqrt{73}-17\approx 0.088, even in degree 3 bipartite games with two strategies per player.

In the remainder of this section we prove the theorem. Fix any ε<273−17\varepsilon<2\sqrt{73}-17. While for ε\varepsilon-WSNE, we used pure strategies to represent and 11 values in Pure-Circuit, for ε\varepsilon-NE we also have to allow mixed strategies to encode s and 11s. Specifically, given a cutoff δ∈(0,1/2)\delta\in(0,1/2), we map the strategies of the players to values in Pure-Circuit in the following way.

If sis_{i} places probability [1−δ,1][1-\delta,1] on zero, then this corresponds to setting variable ii to 0.

If sis_{i} places probability [1−δ,1][1-\delta,1] on one, then this corresponds to setting variable ii to 1.

Otherwise, this corresponds to setting variable ii to ⊥\bot.

Pick a rational δ∈(0,1/4)\delta\in(0,1/4) that satisfies ε<δ(1−2δ)<273−17\varepsilon<\delta(1-2\delta)<2\sqrt{73}-17. Note that such δ∈(0,1/4)\delta\in(0,1/4) always exists, since 273−17<1/82\sqrt{73}-17<1/8. This choice of δ\delta in particular implies that δ<(1−137−1673)/4=(9−73)/4\delta<(1-\sqrt{137-16\sqrt{73}})/4=(9-\sqrt{73})/4, which we will use later.

For a gate g=(NOT,u,v)g=(\textup{{NOT}},u,v), the player vv, who represents the output variable, will play the same bimatrix game against uu as that in the ε\varepsilon-WSNE case, namely:

vvuuzeroonezeroone1111 We claim that this gadget will correctly simulate the NOT gate for the encoding we use in this reduction. Let uu’s strategy be (1−q,q)(1-q,q) and vv’s strategy be (1−p,p)(1-p,p), where q,p∈q,p\in, for playing zero and one, respectively. We need to show that in any ε\varepsilon-NE:

The expected payoff of vv for playing pure strategy zero is Uv(0)=qU_{v}(0)=q and for pure strategy one it is Uv(1)=1−qU_{v}(1)=1-q. Her expected payoff when playing mixed strategy (1−p,p)(1-p,p) is Uv(p)=(1−p)⋅Uv(0)+p⋅Uv(1)=Uv(0)−p⋅(Uv(0)−Uv(1))=Uv(1)−(1−p)⋅(Uv(1)−Uv(0))U_{v}(p)=(1-p)\cdot U_{v}(0)+p\cdot U_{v}(1)=U_{v}(0)-p\cdot(U_{v}(0)-U_{v}(1))=U_{v}(1)-(1-p)\cdot(U_{v}(1)-U_{v}(0)). Now, in any ε\varepsilon-NE, we must have that Uv(p)≥Uv(1)−εU_{v}(p)\geq U_{v}(1)-\varepsilon and Uv(p)≥Uv(0)−εU_{v}(p)\geq U_{v}(0)-\varepsilon, which implies that

In case (i), we have q≤δq\leq\delta, and thus Uv(1)−Uv(0)=1−2q≥1−2δU_{v}(1)-U_{v}(0)=1-2q\geq 1-2\delta. By (1), it follows that 1−p≤ε/(1−2δ)1-p\leq\varepsilon/(1-2\delta). Since ε<δ(1−2δ)\varepsilon<\delta(1-2\delta), we obtain 1−p≤δ1-p\leq\delta, i.e., p≥1−δp\geq 1-\delta, as desired.

In case (ii), q≥1−δq\geq 1-\delta implies that Uv(0)−Uv(1)=2q−1≥1−2δU_{v}(0)-U_{v}(1)=2q-1\geq 1-2\delta. By (2), we must have p≤ε/(1−2δ)≤δ(1−2δ)/(1−2δ)=δp\leq\varepsilon/(1-2\delta)\leq\delta(1-2\delta)/(1-2\delta)=\delta, since ε<δ(1−2δ)\varepsilon<\delta(1-2\delta).

The following is a crucial observation: when constructing a gadget, we can assume that the input to the gadget is encoded using cutoff δ\delta (and so the gap between encoding of and 11 is least 1−2δ1-2\delta) and that the output of the gadget is encoded using cutoff 1/2−δ1/2-\delta (and so the gap between encoding of and 11 is least 2δ2\delta). The reason for this is that we can use a sequence of consecutive NOT gates to amplify the gap in the encoding of the output from 2δ2\delta up to 1−2δ1-2\delta (or, in other words, to decrease the cutoff from 1/2−δ1/2-\delta down to δ\delta). The following lemma formalizes this observation.

If player v1v_{1}, who is the input to gadget g1g_{1}, satisfies Uv1(0)−Uv1(1)≥2δU_{v_{1}}(0)-U_{v_{1}}(1)\geq 2\delta, then player v2k+1v_{2k+1}, who is the output of gadget g2kg_{2k}, satisfies Uv2k+1(0)−Uv2k+1(1)≥1−2δU_{v_{2k+1}}(0)-U_{v_{2k+1}}(1)\geq 1-2\delta. In particular, player v2k+1v_{2k+1} places probability at least 1−δ1-\delta on zero.

If player v1v_{1} satisfies Uv1(1)−Uv1(0)≥2δU_{v_{1}}(1)-U_{v_{1}}(0)\geq 2\delta, then player v2k+1v_{2k+1} satisfies Uv2k+1(1)−Uv2k+1(0)≥1−2δU_{v_{2k+1}}(1)-U_{v_{2k+1}}(0)\geq 1-2\delta. In particular, player v2k+1v_{2k+1} places probability at least 1−δ1-\delta on one.

Consider any i∈[2k]i\in[2k] and let viv_{i} denote the player who is the input to gadget gig_{i} and vi+1v_{i+1} the player who is the output. We will show that in any ε\varepsilon-NE:

If Uvi(0)−Uvi(1)∈[2δ,1−2δ]U_{v_{i}}(0)-U_{v_{i}}(1)\in[2\delta,1-2\delta], then Uvi+1(1)−Uvi+1(0)≥Uvi(0)−Uvi(1)+C(ε,δ)U_{v_{i+1}}(1)-U_{v_{i+1}}(0)\geq U_{v_{i}}(0)-U_{v_{i}}(1)+C(\varepsilon,\delta).

If Uvi(0)−Uvi(1)≥1−2δU_{v_{i}}(0)-U_{v_{i}}(1)\geq 1-2\delta, then Uvi+1(1)−Uvi+1(0)≥1−2δU_{v_{i+1}}(1)-U_{v_{i+1}}(0)\geq 1-2\delta.

If Uvi(1)−Uvi(0)∈[2δ,1−2δ]U_{v_{i}}(1)-U_{v_{i}}(0)\in[2\delta,1-2\delta], then Uvi+1(0)−Uvi+1(1)≥Uvi(1)−Uvi(0)+C(ε,δ)U_{v_{i+1}}(0)-U_{v_{i+1}}(1)\geq U_{v_{i}}(1)-U_{v_{i}}(0)+C(\varepsilon,\delta).

If Uvi(1)−Uvi(0)≥1−2δU_{v_{i}}(1)-U_{v_{i}}(0)\geq 1-2\delta, then Uvi+1(0)−Uvi+1(1)≥1−2δU_{v_{i+1}}(0)-U_{v_{i+1}}(1)\geq 1-2\delta.

Let us begin by proving statement 1. Let γ:=Uvi(0)−Uvi(1)∈[2δ,1−2δ]\gamma:=U_{v_{i}}(0)-U_{v_{i}}(1)\in[2\delta,1-2\delta] and let pp denote the probability that player viv_{i} plays strategy one. By (2) we obtain that p≤ε/γp\leq\varepsilon/\gamma, and thus Uvi+1(1)−Uvi+1(0)=(1−p)−p=1−2p≥1−2ε/γU_{v_{i+1}}(1)-U_{v_{i+1}}(0)=(1-p)-p=1-2p\geq 1-2\varepsilon/\gamma. We can thus write

where we used γ∈[2δ,1−2δ]\gamma\in[2\delta,1-2\delta] and ε<δ(1−2δ)\varepsilon<\delta(1-2\delta). Statement 3 is proved using the same arguments.

Let us now prove statement 2. Letting γ:=Uvi(0)−Uvi(1)∈[1−2δ,1]\gamma:=U_{v_{i}}(0)-U_{v_{i}}(1)\in[1-2\delta,1], we again obtain that p≤ε/γp\leq\varepsilon/\gamma, and thus

as desired. The proof of statement 4 is completely analogous. ∎

For a gate g=(AND,u,v,w)g=(\textup{{AND}},u,v,w), the player ww, who represents the output variable, will play the following matrix games against uu and vv, who represent the input variables.

wwuuzeroonezeroone12<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mtext> </mtext><mtext> </mtext><mfrac><mrow><mn>1</mn><mo>+</mo><mi>δ</mi></mrow><mrow><mn>6</mn><mo>−</mo><mn>2</mn><mi>δ</mi></mrow></mfrac></mrow><annotationencoding="application/x−tex">  1+δ6−2δ</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:2.1408em;vertical−align:−0.7693em;"></span><spanclass="mspace"style="margin−right:0.1667em;"></span><spanclass="mspace"style="margin−right:0.1667em;"></span><spanclass="mord"><spanclass="mopennulldelimiter"></span><spanclass="mfrac"><spanclass="vlist−tvlist−t2"><spanclass="vlist−r"><spanclass="vlist"style="height:1.3714em;"><spanstyle="top:−2.314em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">6</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">−</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mord">2</span><spanclass="mordmathnormal"style="margin−right:0.0379em;">δ</span></span></span><spanstyle="top:−3.23em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="frac−line"style="border−bottom−width:0.04em;"></span></span><spanstyle="top:−3.677em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">1</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mbin">+</span><spanclass="mspace"style="margin−right:0.2222em;"></span><spanclass="mordmathnormal"style="margin−right:0.0379em;">δ</span></span></span></span><spanclass="vlist−s">​</span></span><spanclass="vlist−r"><spanclass="vlist"style="height:0.7693em;"><span></span></span></span></span></span><spanclass="mclosenulldelimiter"></span></span></span></span></span></span>w\frac{1}{2}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mtext> </mtext><mtext> </mtext><mfrac><mrow><mn>1</mn><mo>+</mo><mi>δ</mi></mrow><mrow><mn>6</mn><mo>−</mo><mn>2</mn><mi>δ</mi></mrow></mfrac></mrow><annotation encoding="application/x-tex">\,\,\frac{1+\delta}{6-2\delta}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.1408em;vertical-align:-0.7693em;"></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mspace" style="margin-right:0.1667em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3714em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">6</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">−</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord">2</span><span class="mord mathnormal" style="margin-right:0.0379em;">δ</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mbin">+</span><span class="mspace" style="margin-right:0.2222em;"></span><span class="mord mathnormal" style="margin-right:0.0379em;">δ</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.7693em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span>wvvzeroonezeroone12\frac{1}{2}  1+δ6−2δ\,\,\frac{1+\delta}{6-2\delta} We claim that this gadget correctly simulates the AND gate (together with a sequence of NOT gates applied to the output). Let player uu play strategy (1−q1,q1)(1-q_{1},q_{1}), and let vv play strategy (1−q2,q2)(1-q_{2},q_{2}), where q1,q2∈q_{1},q_{2}\in. By Lemma 4.5 it suffices to show that

if q1≤δq_{1}\leq\delta or q2≤δq_{2}\leq\delta, then Uw(0)−Uw(1)≥2δU_{w}(0)-U_{w}(1)\geq 2\delta,

if q1≥1−δq_{1}\geq 1-\delta and q2≥1−δq_{2}\geq 1-\delta, then Uw(1)−Uw(0)≥2δU_{w}(1)-U_{w}(0)\geq 2\delta.

where Uw(0)U_{w}(0) and Uw(1)U_{w}(1) are the expected payoff of player ww for strategy zero and one respectively. Observe that Uw(0)=12(1−q1)+12(1−q2)U_{w}(0)=\frac{1}{2}(1-q_{1})+\frac{1}{2}(1-q_{2}) and Uw(1)=1+δ6−2δq1+1+δ6−2δq2U_{w}(1)=\frac{1+\delta}{6-2\delta}q_{1}+\frac{1+\delta}{6-2\delta}q_{2}. Below, we use the fact that (1−3δ)/(3−δ)≥2δ(1-3\delta)/(3-\delta)\geq 2\delta, which follows from δ<(9−73)/4\delta<(9-\sqrt{73})/4.

In case (i), Uw(0)−Uw(1)U_{w}(0)-U_{w}(1) is minimal when q1=δq_{1}=\delta and q2=1q_{2}=1 (and also when q1=1q_{1}=1 and q2=δq_{2}=\delta, which is symmetric). Thus, we necessarily have

and thus the gate is correctly simulated.

For a gate g=(PURIFY,u,v,w)g=(\textup{{PURIFY}},u,v,w), the players vv and ww, who represent the output variables, play the following games against uu, who represents the input variable.

vvuuzeroonezeroone  1+2δ3−2δ<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mn>1</mn></mrow><annotationencoding="application/x−tex">1</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6444em;"></span><spanclass="mord">1</span></span></span></span></span>w\,\,\frac{1+2\delta}{3-2\delta}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mn>1</mn></mrow><annotation encoding="application/x-tex">1</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6444em;"></span><span class="mord">1</span></span></span></span></span>wuuzeroonezeroone11  1+2δ3−2δ\,\,\frac{1+2\delta}{3-2\delta} We claim that this gadget correctly simulates the PURIFY gate (together with a sequence of NOT gates applied to the output). Let player uu play strategy (1−q,q)(1-q,q) where q∈q\in. By Lemma 4.5 it suffices to show that

if q≤δq\leq\delta, then Uv(0)−Uv(1)≥2δU_{v}(0)-U_{v}(1)\geq 2\delta and Uw(0)−Uw(1)≥2δU_{w}(0)-U_{w}(1)\geq 2\delta,

if q≥1−δq\geq 1-\delta, then Uv(1)−Uv(0)≥2δU_{v}(1)-U_{v}(0)\geq 2\delta and Uw(1)−Uw(0)≥2δU_{w}(1)-U_{w}(0)\geq 2\delta,

if q∈(δ,1−δ)q\in(\delta,1-\delta), then ∣Uv(0)−Uv(1)∣≥2δ|U_{v}(0)-U_{v}(1)|\geq 2\delta or ∣Uw(0)−Uw(1)∣≥2δ|U_{w}(0)-U_{w}(1)|\geq 2\delta.

Observe that Uv(0)=1+2δ3−2δ(1−q)U_{v}(0)=\frac{1+2\delta}{3-2\delta}(1-q) and Uv(1)=qU_{v}(1)=q, as well as Uw(0)=1−qU_{w}(0)=1-q and Uw(1)=1+2δ3−2δqU_{w}(1)=\frac{1+2\delta}{3-2\delta}q. Below, we use the fact that (1−2δ)/(3−2δ)≥2δ(1-2\delta)/(3-2\delta)\geq 2\delta, which follows from (1−2δ)/(3−2δ)≥(1−3δ)/(3−δ)≥2δ(1-2\delta)/(3-2\delta)\geq(1-3\delta)/(3-\delta)\geq 2\delta as argued above.

In case (i), we have q≤δq\leq\delta and thus

In case (ii), we have q≥1−δq\geq 1-\delta and thus

For case (iii) we consider two subcases. If q≥1/2q\geq 1/2, then

as above. On the other hand, if q≤1/2q\leq 1/2, then

According to the case analysis above, all gate constraints from the Pure-Circuit instance are enforced correctly in any ε\varepsilon-NE of the polymatrix game. Thus, given an ε\varepsilon-NE solution, we can produce a satisfying assignment to the Pure-Circuit instance using the mapping that we defined at the start of the reduction.

Furthermore, note that the polymatrix instance that we construct is normalized and that it is degree 3, bipartite, and has two strategies per player. In particular, note that the addition of a sequence of NOT gates inside the gadgets for AND gates and PURIFY gates (according to Lemma 4.5) does not increase the degree and does not destroy the bipartiteness of the graph, since the sequence has even length.

2.4 Hardness for Win-Lose Polymatrix games

A polymatrix game is win-lose if for every player i∈[n]i\in[n] it holds that the entries of every payoff matrix AijA_{ij} belong to {0,ai}\{0,a_{i}\}, for some ai∈(0,1]a_{i}\in(0,1]. In other words, player ii either “wins” payoff aia_{i} from the interaction with another player, or “loses” and gets payoff 0. Prior work has shown that computing an ε\varepsilon-WSNE is PPAD-complete for inverse-polynomial ε\varepsilon via a reduction from GCircuit [LLD21]. In this section we will show that hardness holds for all ε<1/3\varepsilon<1/3, by modifying our hardness result for ε\varepsilon-WSNE given earlier.

It is PPAD-hard to find an ε\varepsilon-WSNE in a win-lose polymatrix game for all ε<1/3\varepsilon<1/3, even in degree 7 bipartite games with two strategies per player.

We begin by explaining how to modify the payoff matrices described in Section 4.2.2. Observe that the payoff matrix for NOT gates is already win-lose, so we only need to convert AND gates and PURIFY gates into win-lose payoff matrices. The high level idea for these two gates is to create a number of intermediate players, who all copy the input player’s strategy, and whose payoffs to the output player sum to the same payoffs that we used in the ε\varepsilon-WSNE hardness construction. Figure 1 depicts the interaction graphs for these two gates.

For a gate g=(AND,u,v,w)g=(\textup{{AND}},u,v,w), we will create the AND gadget from Figure 1. The player ww, that represents the output variable, will play against the auxiliary players u1,u2,u3u_{1},u_{2},u_{3}, and v1,v2,v3v_{1},v_{2},v_{3} who in turn will “copy” the values, represent the input variables. The payoff matrices for this gadget will be defined as follows, where X∈{u1,v1}X\in\{u_{1},v_{1}\}, Y∈{u2,u3,v2,v3}Y\in\{u_{2},u_{3},v_{2},v_{3}\}, and when Q=vQ=v, then P∈{v1,v2,v3}P\in\{v_{1},v_{2},v_{3}\} and when Q=uQ=u, then P∈{u1,u2,u3}P\in\{u_{1},u_{2},u_{3}\}.

wwXXzeroonezeroone16<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mfrac><mn>1</mn><mn>6</mn></mfrac></mrow><annotationencoding="application/x−tex">16</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:2.0074em;vertical−align:−0.686em;"></span><spanclass="mord"><spanclass="mopennulldelimiter"></span><spanclass="mfrac"><spanclass="vlist−tvlist−t2"><spanclass="vlist−r"><spanclass="vlist"style="height:1.3214em;"><spanstyle="top:−2.314em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">6</span></span></span><spanstyle="top:−3.23em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="frac−line"style="border−bottom−width:0.04em;"></span></span><spanstyle="top:−3.677em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">1</span></span></span></span><spanclass="vlist−s">​</span></span><spanclass="vlist−r"><spanclass="vlist"style="height:0.686em;"><span></span></span></span></span></span><spanclass="mclosenulldelimiter"></span></span></span></span></span></span>w\frac{1}{6}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mn>1</mn><mn>6</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{1}{6}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.0074em;vertical-align:-0.686em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3214em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">6</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span>wYYzeroonezeroone16<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mi>P</mi></mrow><annotationencoding="application/x−tex">P</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:0.6833em;"></span><spanclass="mordmathnormal"style="margin−right:0.1389em;">P</span></span></span></span></span>Q\frac{1}{6}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>P</mi></mrow><annotation encoding="application/x-tex">P</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.6833em;"></span><span class="mord mathnormal" style="margin-right:0.1389em;">P</span></span></span></span></span>Qzeroonezeroone1111 We claim that this gate will work for all ε<1/3\varepsilon<1/3. Firstly, it is not hard to observe that for the game played between PP and QQ, in every ε\varepsilon-WSNE of the game: if QQ plays zero, then PP plays zero; if QQ plays one, then PP plays one. Hence, we have the following for the gadget.

If both uu and vv play one as a pure strategy, then every auxiliary player u1,u2,u3u_{1},u_{2},u_{3} and v1,v2,v3v_{1},v_{2},v_{3} plays one. Hence, the payoff to ww for playing zero is , and the payoff to ww for playing one is 1/6+1/6=1/31/6+1/6=1/3. So, if ε<1/3\varepsilon<1/3, then in all ε\varepsilon-WSNEs the only ε\varepsilon-best response for ww is one, as required by the constraints of the AND gate.

If at least one of uu and vv play zero as a pure strategy, say that uu plays zero, then the auxiliary players u1,u2,u3u_{1},u_{2},u_{3} will play zero as well. Thus, the payoff to ww for playing zero is at least 1/21/2, while the payoff to ww for playing one is at most 1/61/6. So, if ε<1/3\varepsilon<1/3, then in all ε\varepsilon-WSNEs the only ε\varepsilon-best response for ww is zero, as required by the constraints of the AND gate.

The AND gate places no other constraints on the variable ww, so we can ignore all other cases, e.g., the case where both uu and vv play strictly mixed strategies.

For a gate g=(PURIFY,u,v,w)g=(\textup{{PURIFY}},u,v,w), we will create a PURIFY gadget from Figure 1. Players vv and ww, that represent the output variables, will play against the auxiliary players u1,u2,u3u_{1},u_{2},u_{3}, who in turn will “copy” the values of the input variable that corresponds to player uu. The payoff matrices for this gadget will be defined as follows, where P∈{u1,u2,u3}P\in\{u_{1},u_{2},u_{3}\}, Q∈{v,w}Q\in\{v,w\}, and X∈{u1,u2}X\in\{u_{1},u_{2}\}.

vvXXzeroonezeroone13<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mi>w</mi></mrow><annotationencoding="application/x−tex">w</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:0.4306em;"></span><spanclass="mordmathnormal"style="margin−right:0.0269em;">w</span></span></span></span></span>X\frac{1}{3}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>w</mi></mrow><annotation encoding="application/x-tex">w</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.4306em;"></span><span class="mord mathnormal" style="margin-right:0.0269em;">w</span></span></span></span></span>Xzeroonezeroone13<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mi>Q</mi></mrow><annotationencoding="application/x−tex">Q</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:0.8778em;vertical−align:−0.1944em;"></span><spanclass="mordmathnormal">Q</span></span></span></span></span>u3\frac{1}{3}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mi>Q</mi></mrow><annotation encoding="application/x-tex">Q</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:0.8778em;vertical-align:-0.1944em;"></span><span class="mord mathnormal">Q</span></span></span></span></span>u_{3}zeroonezeroone13<spanclass="katex−display"><spanclass="katex"><spanclass="katex−mathml"><mathxmlns="http://www.w3.org/1998/Math/MathML"display="block"><semantics><mrow><mfrac><mn>1</mn><mn>3</mn></mfrac></mrow><annotationencoding="application/x−tex">13</annotation></semantics></math></span><spanclass="katex−html"aria−hidden="true"><spanclass="base"><spanclass="strut"style="height:2.0074em;vertical−align:−0.686em;"></span><spanclass="mord"><spanclass="mopennulldelimiter"></span><spanclass="mfrac"><spanclass="vlist−tvlist−t2"><spanclass="vlist−r"><spanclass="vlist"style="height:1.3214em;"><spanstyle="top:−2.314em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">3</span></span></span><spanstyle="top:−3.23em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="frac−line"style="border−bottom−width:0.04em;"></span></span><spanstyle="top:−3.677em;"><spanclass="pstrut"style="height:3em;"></span><spanclass="mord"><spanclass="mord">1</span></span></span></span><spanclass="vlist−s">​</span></span><spanclass="vlist−r"><spanclass="vlist"style="height:0.686em;"><span></span></span></span></span></span><spanclass="mclosenulldelimiter"></span></span></span></span></span></span>P\frac{1}{3}<span class="katex-display"><span class="katex"><span class="katex-mathml"><math xmlns="http://www.w3.org/1998/Math/MathML" display="block"><semantics><mrow><mfrac><mn>1</mn><mn>3</mn></mfrac></mrow><annotation encoding="application/x-tex">\frac{1}{3}</annotation></semantics></math></span><span class="katex-html" aria-hidden="true"><span class="base"><span class="strut" style="height:2.0074em;vertical-align:-0.686em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height:1.3214em;"><span style="top:-2.314em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">3</span></span></span><span style="top:-3.23em;"><span class="pstrut" style="height:3em;"></span><span class="frac-line" style="border-bottom-width:0.04em;"></span></span><span style="top:-3.677em;"><span class="pstrut" style="height:3em;"></span><span class="mord"><span class="mord">1</span></span></span></span><span class="vlist-s">​</span></span><span class="vlist-r"><span class="vlist" style="height:0.686em;"><span></span></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span>Puuzeroonezeroone1111 We claim that this gate will work for all ε<1/3\varepsilon<1/3. Similarly to the AND gadget, it is not hard to observe that for the game played between uu and PP, in every ε\varepsilon-WSNE of the game: if uu plays zero, then PP plays zero; if uu plays one, then PP plays one.

If uu plays zero as a pure strategy, then auxiliary players u1,u2,u3u_{1},u_{2},u_{3} play zero as well. Thus, strategy zero gives payoff 1/31/3 to both vv and 11 to ww, while strategy one gives payoff to both vv and ww, so in any ε\varepsilon-WSNE with ε<1/3\varepsilon<1/3 we have that zero is the only ε\varepsilon-best response for both vv and ww.

If uu plays one as a pure strategy, then auxiliary players u1,u2,u3u_{1},u_{2},u_{3} play one as well. Thus, strategy one gives payoff 11 to vv and and payoff 1/31/3 to ww, while strategy zero gives payoff to both vv and ww. So, in any ε\varepsilon-WSNE with ε<1/3\varepsilon<1/3 we have that one is the only ε\varepsilon-best response for vv and ww.

Assume now that uu mixes between pure strategies zero and one. Then, the auxiliary players u1,u2,u3u_{1},u_{2},u_{3} might mix between their pure strategies as well; for i∈{1,2,3}i\in\{1,2,3\} let pip_{i} be the probability player uiu_{i} places on pure strategy zero. Then, we have that the following hold.

The payoff for vv is 13⋅p3\frac{1}{3}\cdot p_{3} for playing zero, and 1−13⋅(p1+p2+p3)1-\frac{1}{3}\cdot(p_{1}+p_{2}+p_{3}) for playing one.

The payoff for ww is 13⋅(p1+p2+p3)\frac{1}{3}\cdot(p_{1}+p_{2}+p_{3}) for playing zero, and 13−13⋅p3\frac{1}{3}-\frac{1}{3}\cdot p_{3} for playing one.

Now assume that in an ε\varepsilon-WSNE, with ε<1/3\varepsilon<1/3, vv mixes between zero and one. Then, it must hold that the difference of the payoffs between the two pure strategies is bounded by 1/3, i.e.,

We claim that in this case ww will have to play zero as a pure strategy. Indeed, observe that the payoff of ww from zero is 13⋅(p1+p2+p3)>23−13⋅p3\frac{1}{3}\cdot(p_{1}+p_{2}+p_{3})>\frac{2}{3}-\frac{1}{3}\cdot p_{3}, where the inequality follows from above. On the other hand, the payoff from one is 13−13⋅p3\frac{1}{3}-\frac{1}{3}\cdot p_{3}. Hence, the difference between the payoffs of the two pure strategies is larger than 1/3. Thus, in any ε\varepsilon-WSNE, with ε<1/3\varepsilon<1/3, if vv plays a mixed strategy, then zero is the only ε\varepsilon-best response for ww.

For the second case, assume that in an ε\varepsilon-WSNE, with ε<1/3\varepsilon<1/3, ww mixes between zero and one. Then, it must hold that the difference of the payoffs between the two pure strategies is bounded by 1/3, i.e.,

We claim that in this case vv will have to play one as a pure strategy. Indeed, observe that the payoff of vv from zero is 13⋅p3<23−13⋅(p1+p2+p3)\frac{1}{3}\cdot p_{3}<\frac{2}{3}-\frac{1}{3}\cdot(p_{1}+p_{2}+p_{3}), where the inequality follows from above. On the other hand, the payoff from one is 1−13⋅(p1+p2+p3)1-\frac{1}{3}\cdot(p_{1}+p_{2}+p_{3}). Hence, the difference between the payoffs of the two pure strategies is larger than 1/3. Thus, in any ε\varepsilon-WSNE, with ε<1/3\varepsilon<1/3, if ww plays a mixed strategy, then one is the only ε\varepsilon-best response for vv.

As we have argued above, each of the gate constraints from the Pure-Circuit instance is enforced correctly in any ε\varepsilon-WSNE of the win-lose polymatrix game with ε<1/3\varepsilon<1/3. Thus, given such an ε\varepsilon-WSNE, we can produce a satisfying assignment to the Pure-Circuit instance using the same mapping as we have used before. In addition, all of the games that we have presented are win-lose games and are already normalized, and in addition every player has two strategies.

Now, observe that the reduction described above does not immediately yield a bipartite graph and in addition the maximum degree of the constructed graph is at most nine (this can happen if the vertex that represents the output of an AND gadget, is the input to a different AND gadget as well). However, it is not difficult to tweak the construction and get these two properties too.

Firstly, we perform the following preprocessing step on the Pure-Circuit instance, which results in having a graph with maximum degree seven in the win-lose game constructed here. We replace every node vv of the Pure-Circuit instance with three nodes v1,v2,v3v_{1},v_{2},v_{3} where: v3v_{3} is the output of a NOT gate with input v2v_{2}; v2v_{2} in turn is the output of a NOT gate with input v1v_{1}; v3v_{3} is the input of the gate(s) that had as input the vertex vv; v1v_{1} is the output of the gate that had as output vv. It is not difficult to see that this transformation does not change the solutions of the original instance. In addition, observe that under this preprocessing step and Corollary 2.3, the maximum degree of the constructed win-lose game is now seven.

Next, in order to get a bipartite graph we modify the gadget for NOT gates as follows. For a gate g=(NOT,u,v)g=(\textup{{NOT}},u,v) we create an auxiliary player u′u^{\prime}, where u′u^{\prime} “copies” the strategy of uu and vv “negates” the strategy of u′u^{\prime}. So, overall vv “negates” the strategy of uu. To achieve this, the game played between u′u^{\prime} and vv corresponds to the game constructed for the NOT gates for general polymatrix games and the game between uu and u′u^{\prime} corresponds to the rightmost game constructed for the AND gate for win-lose polymatrix games. Observe now that the final graph is bipartite: one side contains only the vertices from the preprocessed Pure-Circuit instance and the other side contains only the auxiliary vertices we have created for the gadgets. ∎

3 Threshold Games

Threshold games were introduced by Papadimitriou and Peng [PP21] as an intermediate problem that was used to prove PPAD-hardness for public good games on directed networks. Since then, threshold games have been used to prove PPAD-hardness for throttling equilibria in auction markets [CKK21b] and for the famous Hylland-Zeckhauser scheme [CCPY22].

Let ε∈\varepsilon\in. An instance of the ε\varepsilon-ThresholdGameNE problem consists of a threshold game G(V,E)\mathcal{G}(V,E). The task is to find an ε\varepsilon-approximate equilibrium for G(V,E)\mathcal{G}(V,E).

Papadimitriou and Peng [PP21] proved that there exists a constant ε′>0\varepsilon^{\prime}>0 such that ε′\varepsilon^{\prime}-ThresholdGameNE is PPAD-complete, via a reduction from ε\varepsilon-GCircuit, where ε′<ε/5\varepsilon^{\prime}<\varepsilon/5. Hence, from our hardness result for GCircuit (Theorem 4.1) we obtain that ε\varepsilon-ThresholdGameNE is PPAD-complete for some ε\varepsilon upper bounded by 0.020.02.

Using a direct reduction from Pure-Circuit we can show that the problem is in fact PPAD-complete for any ε<1/6\varepsilon<1/6. Furthermore, this is tight, since we provide a simple algorithm solving the problem in polynomial time for ε=1/6\varepsilon=1/6. We first present the algorithm achieving the upper bound, and then the improved lower bound through a direct reduction from Pure-Circuit.

3.1 Computing a 𝟏/𝟔16\boldsymbol{1/6}-Approximate Equilibrium

The 1/61/6-ThresholdGameNE problem can be solved in polynomial time.

Let G=(V,E)G=(V,E) be the threshold game graph. The algorithm proceeds as follows.

The algorithm clearly runs in polynomial time. It remains to prove that all nodes satisfy the ε\varepsilon-approximate equilibrium condition for ε=1/6\varepsilon=1/6. Note that when we assign a value to a node, all of its original incoming edges are still present in the graph. Furthermore, we only assign a value to a node once. We proceed by considering the nodes in each step separately.

3.2 Hardness of 𝜺𝜺\boldsymbol{\varepsilon}-Approximate Equilibrium for any 𝜺<𝟏/𝟔𝜺16\boldsymbol{\varepsilon<1/6}

ε\varepsilon-ThresholdGameNE is PPAD-complete for every ε<1/6\varepsilon<1/6, even when the in- and out-degree of each node is at most two.

In order to prove that ε\varepsilon-ThresholdGameNE is PPAD-hard for ε<1/6\varepsilon<1/6, we will reduce from Pure-Circuit that uses the gates NOT, NOR, and PURIFY (we include the NOT gate because we will later make use of Corollary 2.3 to argue about the degrees of nodes). We will encode values in the Pure-Circuit problem as values in the range [0,ε][0,\varepsilon] in ThresholdGameNE, while 11 values will be encoded as values in the range [1−ε,1][1-\varepsilon,1]. Then, each gate of Pure-Circuit will be simulated by the corresponding gadget depicted in Figure 2.

A (NOT,u,v)(\textup{{NOT}},u,v) gate will be simulated by the NOT gadget depicted in Figure 2. We argue that the gadget works for any ε<0.25\varepsilon<0.25, and thus in particular for any ε<1/6\varepsilon<1/6.

A (NOR,u,v,w)(\textup{{NOR}},u,v,w) gate will be simulated by the NOR gadget depicted in Figure 2. We claim that the gadget works for any ε<1/6\varepsilon<1/6.

A (PURIFY,u,v,w)(\textup{{PURIFY}},u,v,w) gate will be simulated by the PURIFY gadget depicted in Figure 2. We argue that the gadget works for any ε<1/6\varepsilon<1/6.

The correctness of the reduction follows from the arguments above. Finally, by using Corollary 2.3, the constructed instances of ThresholdGameNE will all satisfy that the in- and out-degree of every node is at most two. Indeed, the new “auxiliary” nodes introduced by the PURIFY gadget all satisfy this, and the contribution of each gadget to the in- and out-degree of “original” nodes is exactly the same as the contribution of the corresponding Pure-Circuit gates in the interaction graph. ∎

Appendix A On the Definition of Pure-Circuit

In this section, we explore different ways to weaken the definition of Pure-Circuit, and show how, in each case, the problem is no longer PPAD-hard.

If we allow all gates, except the PURIFY gate (so only NOT, COPY, OR, AND, NOR, NAND), then the problem becomes polynomial-time solvable. Indeed, it suffices to assign value ⊥\bot to all the nodes.

If we allow all gates, except the ones that perform some kind of negation (so only PURIFY, COPY, OR, AND), then the problem becomes polynomial-time solvable. Indeed, assigning the value 11 to each node (or, alternatively, the value to each node) always yields a solution. More generally, we can make the following observation: for any set of gates that can be implemented by monotone functions, the problem lies in the class PLS, and is thus unlikely to be PPAD-complete. Indeed, as already mentioned in Section 2, we can view any Pure-Circuit instance with nn nodes as a function F:n→nF:^{n}\to^{n}, where each gate is replaced by a continuous function that is consistent with the gate-constraint. Then any fixed point of FF yields a solution to the Pure-Circuit instance. If each of the gates can be replaced by a continuous monotone function, then the problem of finding a fixed point of FF is an instance of Tarski’s fixed point theorem, which is known to lie in PLS [EPRY20].

If we allow all gates (PURIFY, NOT, COPY, OR, AND, NOR, NAND), but we drop the robustness requirement from the logical gates OR, AND, NOR, NAND, then the problem can be solved in polynomial time.

Construct the interaction graph GG of the Pure-Circuit instance, as defined in Section 2.1. In the first stage of the algorithm, as long as there exists a directed cycle in graph GG, we do the following:

Pick an arbitrary directed simple cycle of GG.

Remove all nodes on the simple cycle CC from the graph GG, including all their incident edges.

At the end of this procedure, GG no longer contains any cycles. In the second stage of the algorithm we then repeat the following, until GG is empty:

Pick any source uu of GG (which must exist, since GG contains no cycles).

Let gg be the (unique) gate that has uu as output. Since uu does not have incoming edges in GG, all inputs of gg have already been assigned a value. If uu is the only output of gg, then assign a value to uu that satisfies the gate, and remove uu and its edges from GG. If the gate gg also has another output vv, then gg is a PURIFY gate, and there are two cases:

If vv has not been assigned a value yet, then assign values to both uu and vv such that the gate is satisfied. Remove uu and vv and their edges from GG.

If vv has already been assigned a value, then this happened in the first stage of the algorithm, and both vv and the input to gg were assigned value ⊥\bot (if vv lies on a simple cycle CC, then so does the input of gate gg, because vv has a single incoming edge in the original interaction graph). In that case, we assign value or 11 to uu and gg is satisfied. Remove uu and its edges from GG (vv was already removed in the first stage).

Since a node is removed from GG only when it is assigned a value, all nodes have been assigned a value at the end of the algorithm. We argue that all gates are satisfied. Clearly, any gate that has an output node that is still present in GG after the end of the first stage will be satisfied (by construction of the second stage). Thus, it remains to consider any gate gg such that all its output nodes are removed in the first stage. There are three cases:

gg is a NOT or COPY gate: if the output lies on a simple cycle CC, then so does the input. Both are thus assigned value ⊥\bot and the gate is satisfied.

gg is a (non-robust) OR, AND, NOR, or NAND gate: if the output lies on a simple cycle CC, then so does at least one of its inputs. Thus, at least one input is also assigned value ⊥\bot, and the gate is satisfied. Note that we crucially used the non-robustness of the gate here.

gg is a PURIFY gate: if an output vv of gg lies on a simple cycle CC, then the input uu of gg also lies on CC, and so both are removed at the same time from GG. However, the other output ww of gg cannot lie on that same simple cycle CC, and after uu is removed, ww does not have an incoming edge anymore and will thus not be removed in the first stage. Thus, gg cannot be a PURIFY gate.

It is a natural idea to try to add constant gates in an attempt to make the problem hard for a set of gates for which the problem is not PPAD-hard. By constant gates, we mean a 0-gate which has one output and no input, and which enforces that the value of its output node always be , and a 1-gate defined analogously. No matter which subset of gates S⊆{PURIFY,NOT,COPY,OR,AND,NOR,NAND}S\subseteq\{\textup{{PURIFY}},\textup{{NOT}},\textup{{COPY}},\textup{{OR}},\textup{{AND}},\textup{{NOR}},\textup{{NAND}}\} we use, adding the constant gates 0 and 1 does not change the complexity of the problem. Indeed, it is easy to see that the constants can be “propagated” through the circuit. If a constant is an input to a PURIFY, NOT, or COPY gate, then we can replace the output(s) of that gate by constants that satisfy the gate-constraint. If a constant is an input to an X gate, where X∈{OR,AND,NOR,NAND}\textsf{X}\in\{\textup{{OR}},\textup{{AND}},\textup{{NOR}},\textup{{NAND}}\}, then there are two cases: either we can replace the output by a constant, or we can replace the gate by an X gate with the same input twice (namely, the other input). In order to create a copy of the other input, we can either use the COPY gate, or the NOT gate, or the PURIFY gate. If none of these three gates lies in SS, then Pure-Circuit with gates S∪{0,1}S\cup\{\textsf{0},\textsf{1}\} is polynomial-time solvable, since it is polynomial-time solvable with gates S∪{NOT,0,1}S\cup\{\textup{{NOT}},\textsf{0},\textsf{1}\} (by the propagation argument, and the lack of PURIFY gate).

Appendix B Proof of Corollary 2.3

Let X and Y be as required by the statement of Corollary 2.3. By Corollary 2.2 it immediately follows that Pure-Circuit with gates {PURIFY,X,Y}\{\textup{{PURIFY}},\textup{{X}},\textup{{Y}}\} is PPAD-complete. Now consider an instance of Pure-Circuit with gates {PURIFY,X,Y}\{\textup{{PURIFY}},\textup{{X}},\textup{{Y}}\}. We will explain how to turn it into an instance that satisfies the three conditions of Corollary 2.3. We proceed in three steps, where each step adds more structure to the instance, without destroying any of the structure introduced in a previous step.

Step 3. It remains to enforce that every node is the input of exactly one gate. Currently, this only holds with “at most” instead of “exactly”. We begin by introducing a new node v∗v^{*}. Note that v∗v^{*} is not used as an input of any gate, and it is also the only node that is not used as an output by any gate. Let VsinkV_{sink} be the set of all nodes that are not used as an input by any gate. In particular, v∗∈Vsinkv^{*}\in V_{sink}. Using a binary tree of Y-gates which takes the nodes in VsinkV_{sink} as inputs, we can ensure that the root usinku_{sink} of the binary tree is the only node that is not used as input by any gate. By using additional X-gates at the leaves (when needed) we can ensure that the graph remains bipartite. Note that v∗v^{*} is still the only node that is not used as an output of any gate. We finish the construction by adding an X-gate with input usinku_{sink} and output v∗v^{*}. If that makes the graph non-bipartite, then we can instead introduce a new node vv and gates X(usink,v)\textup{{X}}(u_{sink},v) and X(v,v∗)\textup{{X}}(v,v^{*}) instead. The instance now satisfies all three conditions of the statement. ∎

Appendix C Bimatrix Games

In this section we show lower bounds for finding relative approximate well-supported equilibria in bimatrix games. Unlike the additive approximations we have studied so far, in a relative approximate equilibrium ε\varepsilon is not measuring the difference between the current strategy’s payoff and a deviation’s payoff, but instead, the ratio of that difference over the deviation’s payoff.

Daskalakis [Das13] proved that computing a relative ε\varepsilon-WSNE in bimatrix games with payoffs in $isPPAD−completeforanyconstantis PPAD-complete for any constant\varepsilon\in[0,1).Thisleftopenthequestionofinapproximabilityofbimatrixgameswithnon−negativepayoffsRelative. This left open the question of inapproximability of bimatrix games with non-negative payoffsRelative\varepsilon−WSNE(aswellasrelative-WSNE (as well as relative\varepsilon−NE)arescaleinvariant.Therefore,multiplyingallpayoffsbyapositiveconstantdoesnotaffecttheequilibriaforagiven-NE) are scale invariant. Therefore, multiplying all payoffs by a positive constant does not affect the equilibria for a given\varepsilon..Itisknownthatforbimatrixgameswithpayoffsin.. It is known that for bimatrix games with payoffs inthereisapolynomialtimealgorithmforfindingarelativethere is a polynomial time algorithm for finding a relative1/2−WSNE[FNS07].Thebesthardnessresultforgameswithpayoffsin-WSNE [FNS07]. The best hardness result for games with payoffs inwasgivenbyRubinstein[Rub18],whoshowedthatthereexistsaconstantwas given by Rubinstein [Rub18], who showed that there exists a constant\varepsilonsuchthatrelativesuch that relative\varepsilon$-WSNE is PPAD-hard.

Here we show that the problem is hard for any ε≤1/57\varepsilon\leq 1/57. A bimatrix game is a special case of a polymatrix game (see Section 4.2) where we only have two players connected by an edge. Hence, we use the same notation as that of the aforementioned section. Before presenting this section’s main result, let us define the equilibrium notion that will be studied in this section.

The strategy profile s\mathbf{s} is a relative ε\varepsilon-well-supported Nash equilibrium (relative ε\varepsilon-WSNE) if for every player, each action in the support of her strategy is at least as good as any other action up to a fraction ε\varepsilon of the latter action’s absolute expected utility. In particular, for non-negative utilities we have

Computing a relative ε\varepsilon-WSNE in a bimatrix game with non-negative payoffs is PPAD-complete for any ε≤1/57\varepsilon\leq 1/57.

The remainder of this section is devoted to the proof of this theorem.

We will reduce from the PPAD-complete problem of computing an (additive) ε′\varepsilon^{\prime}-WSNE in a polymatrix game with 2n2n players (nodes) and two actions per player for ε′<1/3\varepsilon^{\prime}<1/3 (as shown in Theorem 4.3) to the problem of computing a relative ε\varepsilon-WSNE in a 2n×2n2n\times 2n bimatrix game for ε=ε′/β\varepsilon=\varepsilon^{\prime}/\beta, where β=18.9\beta=18.9. Our reduction closely follows that of Rubinstein’s [Rub18], but instead of reducing from (additive) ε′\varepsilon^{\prime}-NE in polymatrix games, we reduce from their ε′\varepsilon^{\prime}-WSNE counterpart for which we obtained hardness for significantly higher ε′\varepsilon^{\prime}.

For simplicity, given a strategy of a player in the bimatrix game, we refer to the total probability mass of the two actions corresponding to the polymatrix node i∈[n]i\in[n] as the mass of node ii and we denote it as x(i),y(i)x(i),y(i) for the row and column player, respectively. The actions 0, 1 of node ii and their probability masses are denoted by (i:a)(i:a), and x(i:a)x(i:a), y(i:a)y(i:a), respectively, for a∈{0,1}a\in\{0,1\}. For ease of presentation in what follows, we will refer to the players of the polymatrix game as “nodes” and reserve the word “players” for the bimatrix game’s participants. We denote by NR(i)N_{R}(i) the neighbourhood of node ii that belongs to the RR-side of the polymatrix game (and respectively NC(i)N_{C}(i) for a node ii on the CC-side).

First, we present modified versions of the respective results of [Rub18] tailored to the needs of our reduction. Since their proofs are almost identical to those of the aforementioned paper, we have preserved most of their notation.

In every (x,y)(x,y) relative ε\varepsilon-WSNE, for λ<1−ε2\lambda<\frac{1-\varepsilon}{2}, x(i),y(i)∈[1−ε−2λn,(1−ε−2λ)−1n]x(i),y(i)\in\left[\frac{1-\varepsilon-2\lambda}{n},\frac{(1-\varepsilon-2\lambda)^{-1}}{n}\right].

Let tR=max⁡ix(i)t_{R}=\max_{i}x(i) and iR∗=arg⁡max⁡ix(i)i^{*}_{R}=\arg\max_{i}x(i), and let us call UR((i:a),y)U_{R}((i:a),y) the (expected) payoff that the row player gets from playing action (i:a)(i:a), a∈{0,1}a\in\{0,1\} (respectively, UC(x,(i:a))U_{C}(x,(i:a)) for the column player). We will show that, for any j∈[n]j\in[n], if x(j)<(1−ε−2λ)tRx(j)<(1-\varepsilon-2\lambda)t_{R} then y(j+1)=0y(j+1)=0. For any action (j+1:a)(j+1:a) that the column player picks, her payoff is at most UC(x,(j+1:a))≤x(j)+2⋅λ⋅tRU_{C}(x,(j+1:a))\leq x(j)+2\cdot\lambda\cdot t_{R}. That is due to the imitation game that gives her payoff at most x(j)x(j) and the payoffs induced due to her neighbours k∈N(j+1)k\in N(j+1) in the polymatrix game that give her positive payoffFrom the PPAD-hard ε\varepsilon-WSNE instances we construct for polymatrix games in Section 4.2.2, observe that at most two neighbours (out of three) give positive payoff to the node. Namely, only its parents can give it positive payoff while its children always give it 0 payoff., combined with the fact that x(k)≤tRx(k)\leq t_{R}. Therefore, UC(x,(j+1:a))<(1−ε)tRU_{C}(x,(j+1:a))<(1-\varepsilon)t_{R}, but the column player can guarantee a payoff of at least tRt_{R} by playing action (iR∗+1:a)(i^{*}_{R}+1:a) for any a∈{0,1}a\in\{0,1\}. This would contradict the ε\varepsilon-WSNE condition, therefore y(j+1)=0y(j+1)=0. Similarly, for any j∈[n]j\in[n], if y(j)<(1−ε−2λ)tCy(j)<(1-\varepsilon-2\lambda)t_{C} then x(j)=0x(j)=0, where tC=max⁡iy(i)t_{C}=\max_{i}y(i).

Now we claim that for every i∈[n]i\in[n], x(i),y(i)>0x(i),y(i)>0. For the sake of contradiction assume that x(i)=0x(i)=0 without loss of generality. Then, since λ<(1−ε)/2\lambda<(1-\varepsilon)/2 and tR≥1/n>0t_{R}\geq 1/n>0, we get x(i)<(1−ε−2λ)tRx(i)<(1-\varepsilon-2\lambda)t_{R}, and therefore y(i+1)=0y(i+1)=0. With a similar argument for y(i+1)y(i+1) we deduce from the previous paragraph that x(i+1)=0x(i+1)=0, and this inductively yields that for all i∈[n]i\in[n], x(i)=y(i)=0x(i)=y(i)=0, a contradiction. Therefore, for all i∈[n]i\in[n], x(i),y(i)>0x(i),y(i)>0. This, combined with the above paragraph yields that for all i∈[n]i\in[n], x(i)≥(1−ε−2λ)tR≥(1−ε−2λ)/nx(i)\geq(1-\varepsilon-2\lambda)t_{R}\geq(1-\varepsilon-2\lambda)/n.

The upper bound is deduced by the fact that if tR>(1−ε−2λ)−1/nt_{R}>(1-\varepsilon-2\lambda)^{-1}/{n}. Since there exists a node ii with x(i)≤1/nx(i)\leq 1/n (otherwise ∑j∈[n]x(j)>1\sum_{j\in[n]}x(j)>1), we have that x(i)<(1−ε−2λ)tRx(i)<(1-\varepsilon-2\lambda)t_{R}, which cannot happen as shown in the first paragraph of the proof. Therefore, for any i∈[n]i\in[n], x(i)≤tR≤(1−ε−2λ)−1/nx(i)\leq t_{R}\leq(1-\varepsilon-2\lambda)^{-1}/{n}. A similar argument holds for tCt_{C} which yields the upper bound for y(i)y(i). ∎

In every (x,y)(x,y) relative ε\varepsilon-WSNE, for λ<1−ε2\lambda<\frac{1-\varepsilon}{2}, the expected utilities of both players for playing any action against the strategy of the other player are in [1−ε−2λn,(1+2λ)(1−ε−2λ)−1n]\left[\frac{1-\varepsilon-2\lambda}{n},\frac{(1+2\lambda)(1-\varepsilon-2\lambda)^{-1}}{n}\right].

The lower bound comes from the lower bound of the above lemma and the fact that the row player from any action (i:a)(i:a), a∈{0,1}a\in\{0,1\} against yy, gets at least the payoff of the imitation game, multiplied by y(i:a)+y(i:1−a)=y(i)y(i:a)+y(i:1-a)=y(i). The upper bound comes from the upper bound of the above lemma and the fact that the row player from any action (i:a)(i:a), a∈{0,1}a\in\{0,1\} against yy gets from each entry ((i:a),j)((i:a),j), j∈[n]j\in[n] of the matrix at most payoff 1⋅y(i)1\cdot y(i) from the imitation game, and additional payoff of at most λ⋅y(i)\lambda\cdot y(i) from each of two neighbouring nodes in the polymatrix game (note again that only two out of maximum three neighbours can give her positive payoff). A symmetric argument holds for the column player’s bounds. ∎

Fix β=18.9\beta=18.9. Now we will show that, given an ε′<1/3\varepsilon^{\prime}<1/3, for any (x,y)(x,y) which is a relative ε′/β\varepsilon^{\prime}/\beta-WSNE in the constructed bimatrix game, the marginal distributions of the actions (i:a)(i:a), a∈{0,1}a\in\{0,1\} for each node i∈[n]i\in[n] constitute an (additive) ε′\varepsilon^{\prime}-WSNE of the original polymatrix game. In particular, we will prove that the strategy profile where each node ii in the polymatrix game belonging to the RR-side of the bipartite graph plays (x(i:0)/x(i),x(i:1)/x(i))(x(i:0)/x(i),x(i:1)/x(i)), and similarly each node jj of the CC side of the bipartite graph plays (y(j:0)/y(j),y(j:1)/y(j))(y(j:0)/y(j),y(j:1)/y(j)) is an ε′\varepsilon^{\prime}-WSNE.

For the sake of contradiction, assume that the marginal distributions of the node strategies are not such an ε′\varepsilon^{\prime}-WSNE. Then, given the marginals induced by (x,y)(x,y) there is a node ii on the left side of the bipartite graph (without loss of generality) for whom playing action (i:a)(i:a) for some a∈{0,1}a\in\{0,1\} against yy gives her additive ε′\varepsilon^{\prime} more expected utility than what the action (i:1−a)(i:1-a) gives. This discrepancy, translated in the bimatrix game (where we have multiplied all payoffs of the polymatrix game by λ\lambda) means that the difference in payoff would be at least ε′⋅λ⋅1−ε−2λn\varepsilon^{\prime}\cdot\lambda\cdot\frac{1-\varepsilon-2\lambda}{n} due to Corollary C.3, where ε=ε′/β\varepsilon=\varepsilon^{\prime}/\beta. Then, again by Corollary C.3 which bounds her maximum utility from playing any action (i:a)(i:a) against the mixed strategy yy, we conclude that her relative increase in expected payoff is at least

We now find the value for λ\lambda that maximizes the above expression, namely, λ=−3+17−8ε8\lambda=\frac{-3+\sqrt{17-8\varepsilon}}{8}. For this value of λ\lambda and for every ε=ε′/β≤1/57\varepsilon=\varepsilon^{\prime}/\beta\leq 1/57 we have that

Note that the optimum value of λ\lambda is potentially irrational, and therefore, not suitable as input to the algorithm of our reduction. Nevertheless, the proof goes through for λ=0.1383\lambda=0.1383. The above strict inequality shows that the action that the node deviated to gives the bimatrix player more than relative 1/(1−ε)1/(1-\varepsilon) expected utility, and therefore by definition our initial profile (x,y)(x,y) is not a relative ε\varepsilon-WSNE (for ε=ε′/β\varepsilon=\varepsilon^{\prime}/\beta), a contradiction. This completes the proof of Theorem C.1.

In fact, Theorem C.1 holds for any ε<13β∗\varepsilon<\frac{1}{3\beta^{*}}, where β∗\beta^{*} is the optimum value from the above proof. This β∗\beta^{*} is in (18.86,18.87)(18.86,18.87), but we have picked 18.918.9 for ease of presentation.

Acknowledgements

We thank the anonymous reviewers for comments and suggestions that helped improve the presentation of the paper. We are also very grateful to Steffen Schuldenzucker for feedback on an earlier version of the manuscript, and to Christian Ikenmeyer for pointing out the connection to hazard-free circuits. The second author was supported by EPSRC grant EP/W014750/1 “New Techniques for Resolving Boundary Problems in Total Search”.

References