Lower Bounds Against Sparse Symmetric Functions of ACC Circuits: Expanding the Reach of # SAT Algorithms
aa r X i v : . [ c s . CC ] J a n Lower Bounds Against Sparse Symmetric Functions of ACCCircuits: Expanding the Reach of
SAT Algorithms
Nikhil Vyas ∗ MIT Ryan Williams † MITJanuary 23, 2020
Abstract
We continue the program of proving circuit lower bounds via circuit satisfiability algorithms. So far,this program has yielded several concrete results, proving that functions in
Quasi - NP = NTIME [ n (log n ) O (1) ] and NEXP do not have small circuits (in the worst case and/or on average) from various circuit classes C ,by showing that C admits non-trivial satisfiability and/or SAT algorithms which beat exhaustive searchby a minor amount.In this paper, we present a new strong lower bound consequence of non-trivial
SAT algorithm fora circuit class C . Say a symmetric Boolean function f ( x , . . . , x n ) is sparse if it outputs on O (1) values of P i x i . We show that for every sparse f , and for all “typical” C , faster SAT algorithms for C circuits actually imply lower bounds against the circuit class f ◦ C , which may be stronger than C itself.In particular: • SAT algorithms for n k -size C -circuits running in n /n k time (for all k ) imply NEXP does nothave f ◦ C -circuits of polynomial size. • SAT algorithms for n ε -size C -circuits running in n − n ε time (for some ε > ) imply Quasi - NP does not have f ◦ C -circuits of polynomial size.Applying SAT algorithms from the literature, one immediate corollary of our results is that
Quasi - NP does not have EMAJ ◦ ACC ◦ THR circuits of polynomial size, where EMAJ is the “exact majority” func-tion, improving previous lower bounds against
ACC [Williams JACM’14] and ACC ◦ THR [WilliamsSTOC’14], [Murray-Williams STOC’18]. This is the first nontrivial lower bound against such a circuitclass. ∗ [email protected] , Supported by NSF CCF-1909429. † [email protected] , Supported by NSF CCF-1741615 and NSF CCF-1909429. Introduction
Currently, our knowledge of algorithms vastly exceeds our knowledge of lower bounds. Is it possible tobridge this gap, and use the existence of powerful algorithms to give lower bounds for hard functions?Over the last decade, the program of proving lower bounds via algorithms has been positively addressingthis question. A line of work starting with Kabanets and Impagliazzo [KI04] has shown how deterministicsubexponential-time algorithms for polynomial identity testing would imply lower bounds against arithmeticcircuits. Starting around 2010 [Wil13, Wil14], it was shown that even slightly nontrivial algorithms couldimply Boolean circuit lower bounds. For example, a circuit satisfiability algorithm running in O (2 n /n k ) time (for all k ) on n k -size circuits with n inputs would already suffice to yield the (infamously open) lowerbound NEXP P / poly. More generally, a generic connection was found between non-trivial SAT algo-rithms and circuit lower bounds: Theorem 1.1 ([Wil13, Wil14], Informal) . Let C be a circuit class closed under AND, projections, and com-positions. Suppose for all k there is an algorithm A such that, for every C -circuit of n k size, A determinesits satisfiability in O (2 n /n k ) time. Then NEXP does not have polynomial-size C -circuits. To illustrate Theorem 1.1 with two examples, when C is the class of general fan-in 2 circuits, Theo-rem 1.1 says that non-trivial Circuit SAT algorithms imply NEXP P / poly; when C is the class of Booleanformulas, it says non-trivial Formula-SAT algorithms imply NEXP NC . Both are major open questionsin circuit complexity. Theorem 1.1 and related results have been applied to prove several concrete circuitlower bounds: super-polynomial lower bounds for ACC [Wil14], ACC ◦ THR [Wil18a], quadratic lowerbounds for depth-two symmetric and threshold circuits [Tam16, ACW16], and average-case lower boundsas well [COS18, Che19].Recently, the algorithms-to-lower-bounds connection has been extended to show a trade-off between therunning time of the SAT algorithm on large circuits, and the complexity of the hard function in the lowerbound. In particular, it is even possible in principle to obtain circuit lower bounds against NP with thisalgorithmic approach. Theorem 1.2 ([MW18], Informal) . Let C be a class of circuits closed under unbounded AND, ORs of fan-intwo, and negation. Suppose there is an algorithm A and ε > such that, for every C -circuit C of n ε size, A solves satisfiability for C in O (2 n − n ε ) time. Then Quasi - NP does not have polynomial-size C -circuits. In fact, Theorem 1.2 holds even if A only distinguishes between unsatisfiable circuits from those with atleast n − SAT assignments; we call this easier problem GAP-UNSAT.Intuitively, the aforementioned results show that as the circuit satisfiability algorithms improve in run-ning time and scope, they imply stronger lower bounds. In all known results, to prove a lower bound against C , one must design a SAT algorithm for a circuit class that is at least as powerful as C . Inspecting the proofsof the above theorems carefully, it is not hard to show that, even if C did not satisfy the desired closureproperties, it would suffice to give a SAT algorithm for a slightly more powerful class than the lower bound.For example, in Theorem 1.2, a SAT algorithm running in O (2 n − n ε ) time for n ε -size AND of ORs ofthree (possibly negated) C circuits (on n inputs, of n ε size) would still imply C -circuit lower bounds for Quasi - NP . Our key point here is that these proof methods require a SAT algorithm for a potentially more It is not necessary to know precisely what these conditions mean, as we will use different conditions in our paper anyway. Theimportant point is that these conditions hold for most interesting circuit classes that have been studied, such as AC , TC , NC , NC , and general fan-in two circuits. In this paper, we use the notation
Quasi - NP := S k NTIME [ n (log n ) k ] . owerful circuit class than the class for which we can conclude a lower bound. A compelling question iswhether this requirement is an artifact of our proof method, or is it inherent?
Lower bounds for more powerful classes from SAT algorithms?
We feel it is natural to conjecture thata SAT algorithm for a circuit class C implies a lower bound against a class that is more powerful than C ,because checking satisfiability is itself a very powerful ability. Intuitively, a non-trivial SAT algorithm for C on n -input circuits is computing a uniform OR of n C -circuits evaluated on fixed inputs, in o (2 n ) time.(Recall that a “uniform” circuit informally means that any gate of the circuit can be efficiently computed byan algorithm.) If there were an algorithm to decide the outputs of uniform ORs of C -circuits more efficientlythan their actual circuit size, perhaps this implies a lower bound against OR ◦ C circuits.Similarly, a SAT algorithm for C on n -input circuits can be used to compute the output of any circuitof the form f ( C ( x ) , . . . , C ( x n )) where f is a uniform symmetric Boolean function, C is a C -circuitwith n inputs, and x , . . . , x n is an enumeration of all n -bit strings. Should we therefore expect to provelower bounds on symmetric functions of C -circuits, using a SAT algorithm? This question is particularlysignificant because in many of the concrete lower bounds proved via the program [Wil14, Wil18a, MW18],non-trivial
SAT algorithms were actually obtained, not just SAT algorithms. So our question amounts toasking: how strong of a circuit lower bound we can prove, given the SAT algorithms we already have?
Weuse
SYM to denote the class of Boolean symmetric functions.
Conjecture 1 ( SAT Algorithms Imply Symmetric Function Lower Bounds, Informal) . Non-trivial
SATalgorithms for circuit classes C imply size lower bounds against SYM ◦ C circuits. In particular, all state-ments in Theorem 1.1 and Theorem 1.2 hold when the SAT algorithm is replaced by a
SAT algorithm, andthe lower bound consquence for C is replaced by SYM ◦ C . If Conjecture 1 is true, then existing
SAT algorithms would already imply super-polynomial lowerbounds for
SYM ◦ ACC ◦ THR circuits, a class that contains depth-two symmetric circuits (for which nolower bounds greater than n are presently known) [Tam16, ACW16].More intuition for Conjecture 1 can be seen from a recent paper of the second author, who showedhow SAT algorithms for a circuit class C can imply lower bounds on (real-valued) linear combinationsof C -circuits [Wil18c]. For example, known SAT algorithms for
ACC circuits imply Quasi - NP prob-lems cannot be computed via polynomial-size linear combinations of polynomial-size ACC ◦ THR circuits.However, the linear combination representation is rather constrained: the linear combination is requiredto always output or . Applying PCPs of proximity, Chen and Williams [CW19] showed that the lowerbound of [Wil18c] can be extended to “approximate” linear combinations of C -circuits, where the linearcombination does not have to be exactly or , but must be closer to the correct value than to the incorrectone, within an additive constant factor. These results show, in principle, how a SAT algorithm for a circuitclass C can imply lower bounds for a stronger class of representations than C . In this paper, we take a concrete step towards realizing Conjecture 1, by proving it for “sparse” symmetricfunctions. We say a symmetric Boolean function f ( x , . . . , x n ) is k -sparse if f is on at most k values of P i x i . The -sparse symmetric functions are called the exact threshold (ETHR with polynomial weights) or exact majority (EMAJ) functions, which have been studied for years in both circuit complexity (e.g. [Gre00,BTT92, Han07, Han09, HP10]) and structural complexity theory, where the corresponding complexity class(computing an exact majority over all computation paths) is known as C = P [Wag86].3 heorem 1.3. Let C be closed under AND , negation, and suppose the all-ones and parity function are in C . Let f = { f n } be a family of k -sparse symmetric functions for some k = O (1) . • If there is a
SAT algorithm for n k -size C -circuits running in n /n k time (for all k ), then NEXP doesnot have f ◦ C -circuits of polynomial size. • If there is a
SAT algorithm for n ε -size C -circuits running in n − n ε time (for some ε > ), then Quasi - NP does not have f ◦ C -circuits of polynomial size. Applying known
SAT algorithms for AC [ m ] ◦ THR circuits from [Wil18b], we obtain:
Corollary 1.1.
For all constant depths d ≥ and constant moduli m ≥ , Quasi - NP does not havepolynomial-size EMAJ ◦ AC [ m ] ◦ THR circuits.
Here we briefly explain the new ideas that lead to our new circuit lower bounds.As in prior work [Wil18c, CW19], the high-level idea is to show that if (for example)
Quasi - NP haspolynomial-size EMAJ ◦C , and there is a SAT algorithm for C circuits, then we can design a nondeterminis-tic algorithm for verifying GAP Circuit Unsatisfiability (GAP-UNSAT) on generic circuits that beats exhaus-tive search. In GAP-UNSAT, we are given a generic circuit and are promised that it is either unsatisfiable, orat least half of its possible assignments are satisfying, and we need to nondeterministically prove the unsat-isfiable case. (Note this is a much weaker problem than SAT.) As shown in [Wil13, Wil14, MW18], combin-ing a nondeterministic algorithm for GAP-UNSAT with the hypothesis that Quasi - NP has polynomial-sizecircuits, we can derive that nondeterministic time n can be simulated in time o (2 n ) , contradicting thenondeterministic time hierarchy theorem.Our key idea is to use probabilistically checkable proofs (PCPs) in a new way to exploit the power ofa SAT algorithm. First, let’s observe a task that a
SAT algorithm for C can compute on an EMAJ ◦ C circuit. Suppose our EMAJ ◦ C circuit has the form D ( x ) = " t X i =1 C i ( x ) = s , where each C i ( x ) is a Boolean C -circuit on n inputs, s is a threshold value, and our circuit outputs if andonly if the sum of the C i ’s equals s . Consider the expression E ( x ) := t X i =1 C i ( x ) − s ! . (1)Treated as a function, E ( x ) outputs integers; E ( a ) = 0 when D ( a ) = 1 , and otherwise E ( a ) ∈ [1 , ( t + s ) ] .We first claim that the quantity X a ∈{ , } n E ( a ) (2) We are using the standard Iverson bracket notation, where [ P ] is if predicate P is true, and otherwise. SAT algorithm. To see this, using distributivity,we can rewrite (1) as E ( x ) = X i,j ( C i ∧ C j )( x ) − s X i C i ( x ) + s . Assuming C is closed under conjunction, each C i ∧ C j is also a C -circuit, and we can compute X a ∈{ , } n E ( a ) = X i,j X a ∈{ , } n ( C i ∧ C j )( a ) − s X i X a ∈{ , } n C i ( a ) + s · n by making O ( t ) calls to a SAT algorithm. Thus we can compute (2) using a
SAT algorithm.How is computing (2) useful? This is where PCPs come in. We cannot use (2) to directly solve
SATfor D (otherwise as SAT algorithms imply SAT algorithms we could apply existing work [Wil14], andbe done). But we can use (2) to obtain a multiplicative approximation to the number of assignments that falsify D . In particular, each satisfying assignment is counted zero times in (2), and each falsifying as-signment is counted between and (less than) ( t + s ) times. We want to exploit this, and obtain a fasterGAP-UNSAT algorithm. Given a circuit which is a GAP-UNSAT instance, we start by using an efficienthitting set construction [Gol11] to increase the gap of GAP-UNSAT. We obtain a new circuit C ( x ) which iseither UNSAT or has at least n − o (2 n ) satisfying assignments (Section 2.1). Next (Lemma 3.2) we applya PCP of Proximity and an error correcting code to C , yielding a 3-SAT instance over x and extra variables,with constant gap (similar to Chen-Williams [CW19]), and we amplify this gap using standard serial rep-etition. Finally, we apply the FGLSS [FGL +
91] reduction (Lemma 3.6) to the 3-SAT instance, obtainingIndependent Set instances with a large gap between the YES case and NO case. In particular, for all inputs x , when C ( x ) = 1 there is a large independent set in the resulting graph, and when C ( x ) = 0 , there are onlysmall independent sets in the resulting graph (see Lemma 3.1). Returning to the assumption that Quasi - NP has small EMAJ ◦ C circuits, and applying an easy witness lemma [MW18], it follows that the solutions tothe independent set instance can be encoded by EMAJ ◦ C circuits. Because of the large gap between theYES case and NO case, our multiplicative approximation to the number of UNSAT assignments can be usedto distinguish the unsatisfiable case and the “many satisfying assignments” case of GAP-UNSAT, whichfinishes the argument.One interesting bottleneck is that we cannot directly apply serial repetition and the FGLSS reduction inour argument; we need the PCP machinery we use to behave similarly on all inputs x to the original circuit C . This translates to studying the behavior of these reductions with respect to partial assignments . Whilefor these two reductions we are able to prove that they behave “nicely” with respect to partial assignments,it is entirely unclear that this is true for other PCP reductions such alphabet reduction, parallel repetition,and so on.Our approach is very general; to handle k -sparse symmetric functions, we can simply modify the func-tion E accordingly. We assume general familiarity with basic concepts in circuit complexity and computational complexity [AB09].In particular we assume familiarity with AC , ACC , P / poly , NEXP , and so on.
Circuit Notation.
Here we define notation for the relevant circuit classes. By size C ( h ( n )) we denotecircuits from circuit class C with size at most h ( n ) . 5 efinition 2.1. An EMAJ ◦C circuit (a.k.a. “exact majority of C circuit”) has the general form EMAJ ( C ( x ) , C ( x ) , . . . , C t ( x ) , u ) ,where u is a positive integer, x are the input variables, C i ∈ C , and the gate EMAJ ( y , . . . , y t , u ) outputs if and only if exactly u of the y i ’s output 1. Definition 2.2.
A SUM ≥ ◦ C circuit (“positive sum of C circuits”) has the formSUM ≥ ( C ( x ) , C ( x ) , . . . , C t ( x )) = X i ∈ [ t ] C i ( x ) where C i is either a C -circuit or − times a C -circuit and we are promised that P i ∈ [ t ] C i ( x ) ≥ over all x ∈ { , } n .Given a set of circuits { C i } , we say that f : { , } n → { , } is represented by the positive-sum circuitSUM ≥ ( C ( x ) , C ( x ) , . . . , C t ( x )) if for all x , f ( x ) = 1 when P i ∈ [ t ] C i ( x ) > , and f ( x ) = 0 when P i ∈ [ t ] C i ( x ) = 0 . Definition 2.3.
A circuit class C is typical if there is a k > such that the following hold: • Closure under negation.
For every C circuit C , there is a circuit C ′ computing the negation of C where size ( C ′ ) ≤ size ( C ) k . • Closure under AND.
For every C circuits C and C , there is a circuit C ′ computing the AND of C and C where size ( C ′ ) ≤ ( size ( C ) + size ( C )) k . • Contains all-ones.
The function n : { , } n → { , } has a C circuit of size O ( n k ) . The vast majority of circuit classes that are studied ( AC , ACC , TC , NC , P / poly ) are typical. Thenext lemma shows that the negation of an exact-majority of C circuit can be represented as a “positive-sum”of C circuit, if C is typical. Lemma 2.1.
Let C be typical. If a function f has a EMAJ ◦ C circuit D of size s , then ¬ f can be representedby a SUM ≥ ◦ C circuit D ′ of size poly ( s ) . Moreover, a description of the circuit D ′ can be obtained from adescription of D in polynomial time.Proof. Suppose f is computable by the EMAJ ◦ C circuit D = EMAJ ( D , D , . . . , D t , u ) , where u ∈{ , , . . . , t } . Consider the expression E ( x ) := ( SUM ( D , D , . . . , D t ) − u ) . Note that E ( x ) = 0 when D ( x ) = 1 , and E ( x ) > when D ( x ) = 0 . So in order to prove the lemma, itsuffices to show that E can be written as a SUM ≥ ◦ C circuit. Expanding the expression E , E ( x ) = SUM ( D , D , . . . , D t ) − u · SUM ( D , D , . . . , D t ) + u = t X i,j =1 ( D i ∧ D j ) − u X j =1 t X i =1 D i + u . By Definition 2.3 AND ◦ C = C , each D i ∧ D j is a circuit from C of size poly ( s ) . Since the all-onesfunction is in C , the function x u also has a SUM ◦ C circuit of size O ( t ) . Therefore there are circuits D ′ i ∈ C and t ′ ≤ O ( t ) such that by defining D ′ := SUM ≥ ( D ′ , . . . , D ′ t ′ ) we have D ′ ( x ) = E ( x ) for all x . A notable exception (as far as we know) is the class of depth- d exact threshold circuits for a fixed d ≥ , because we do notknow if such classes are closed under negation. Similarly, we do not know if the class of depth- d threshold circuits is typical. (Inthat case, the only non-trivial property to check is closure under AND; we can compute the AND of two threshold circuits with aquasi-polynomial blowup using Beigel-Reingold-Spielman [BRS95], but not with a polynomial blowup.) rror-Correcting Codes. We will need a (standard) construction of binary error correcting codes withconstant rate and constant relative distance.
Theorem 2.1 ([Spi96]) . There are universal constants c ≥ and δ ∈ (0 , such that for all sufficientlylarge n , there are linear functions EN C n : ( F ) n → ( F ) cn such that for all x = y with | x | = | y | = n , theHamming distance between EN C n ( x ) and EN C n ( y ) is at least δn . In what follows, we generally drop the superscript n for notational brevity. Note that each bit of outputENC ni ( x ) (for i = 1 , . . . , cn ) is a parity function on some subset of the input bits. Murray and Williams [MW18] showed that CAPP/GAP-UNSAT algorithms, i.e., algorithms which distin-guish between unsatisfiable circuits and circuits with ≥ n − satisfying assignments are enough to givelower bounds. For our results, it is necessary to strengthen the “gap”, which can be done using knownhitting set constructions. Lemma 2.2 (Corollary C.5 in [Gol11], Hitting Set Construction) . There is a constant ψ > and apoly ( n, log g ) time algorithm S such that, given a (uniform random) string r of n + ψ · log g bits, S outputs t = O (log g ) strings x , x , . . . , x t ∈ { , } n such that for every f : { , } n → { , } with P x f ( x ) ≥ n − , Pr r [ OR ti =1 f ( x i ) = 1] ≥ − /g . We will use the following “algorithms to lower bounds” connections as black box:
Theorem 2.2 ([MW18]) . Suppose for some constant ε ∈ (0 , there is an algorithm A that for all n ε -size circuits C on n inputs, A ( C ) runs in n − n ε time, outputs YES on all unsatisfiable C , and outputsNO on all C that have at least n − satisfying assignments. Then for all k , there is a c ≥ such thatNTIME [2 log ck /ε n ] SIZE [2 log k n ] . Applying Lemma 2.2 to Theorem 2.2, we observe that the circuit lower bound consequence can beobtained from a significantly weaker-looking hypothesis. This weaker hypothesis will be useful for ourlower bound results.
Theorem 2.3.
Suppose for some constant ε ∈ (0 , there is an algorithm A that for all n ε -size circuits C on n inputs, A ( C ) runs in n /g ( n ) ω (1) time, outputs YES on all unsatisfiable C , and outputs NO on all C that have at least n (1 − /g ( n )) satisfying assignments, for g ( n ) = 2 n ε . Then for all k , there is a c ≥ such that NTIME [2 log ck /ε n ] SIZE [2 log k n ] .Proof. Our starting point is Theorem 2.2 ([MW18]): we are given an m -input, m δ -size circuit D ′ that iseither UNSAT or has at least m − satisfying assignments, and we wish to distinguish between the two caseswith a m − m δ -time algorithm. We set δ = ε/ We create a new circuit D with n inputs, where n satisfies n = m + ψ · log g ( n ) , and ψ > is the constant from Lemma 2.2. (Note that, since g ( n ) is time constructible and g ( n ) ≤ o ( n ) ,such an n can be found in subexponential time.) Applying the algorithm from Lemma 2.2, D treats its n bits of input as a string of randomness r , computes t = O (log g ( n )) strings x , x , . . . , x t ∈ { , } m with7 poly ( m, log g ) -size circuit, then outputs the OR of D ′ ( x i ) over all i = 1 , . . . , t . Note the total size of ourcircuit D is poly ( m, log g ) + O (log g ) · size ( D ′ ) = poly ( n ) + O ( n ε ) · m δ < n δ = 2 n ε as ε = 2 δ .Clearly, if D ′ is unsatisfiable, then D is also unsatisfiable. By Lemma 2.2, if D ′ has m − satisfyingassignments, then D has at least n (1 − /g ( n )) satisfying assignments. As size ( D ) ≤ n ε , by our assump-tion we can distinguish the case where D is unsatisfiable from the case where D has at least n (1 − /g ( n )) satisfying assignments, with an algorithm running in time n /g ( n ) ω (1) . This yields an algorithm for distin-guishing the original circuit D ′ on m inputs and m δ size, running in time n /g ( n ) ω (1) = 2 m g ( n ) O (1) /g ( n ) ω (1) = 2 m /g ( n ) ω (1) ≤ m − n ε ≤ m − n δ ≤ m − m δ , since g ( n ) = 2 n ε . By Theorem 2.2, this implies that for all k , there is a c ≥ such that NTIME [2 log ck /δ n ] SIZE [2 log k n ] . As, ε = 2 δ we get that NTIME [2 log ck /ε n ] SIZE [2 log k n ] . But as the constant can beabsorbed in the constant c hence we get that for all k , there is a c ≥ such that NTIME [2 log ck /ε n ] SIZE [2 log k n ] . In Section 3 we give a reduction from Circuit SAT to “Generalized” Independent Set. Section 4 uses thisreduction to prove lower bounds for EMAJ ◦ C assuming C with running time n − n ε .Section 4.1 uses this result to give lower bound for EMAJ ◦ ACC ◦ THR. Section 5 generalizes these resultsto f ◦ C lower bounds where f is a sparse symmetric function. In Section 6 we give lower bounds forEMAJ ◦ C assuming C with running time n /n ω (1) . The goal of this section is to give the main PCP reduction we will use in our new algorithm-to-lower-boundtheorem. First we need a definition of “generalized” independent set instances, where some vertices havealready been “assigned” in or out of the independent set.
Definition 3.1.
Let G = ( V, E ) be a graph. Let π : V → { , , ∗} be a partial Boolean assignment to V . We define G ( π ) to be a graph with the label function π on its vertices (where each vertex gets the label , or , or no label). We construe G ( π ) as an generalized independent set instance , in which any validindependent set (vertex assignment) must be consistent with π : any independent set must contain all verticeslabeled , and no vertices labeled . Lemma 3.1.
Let k be a function of n . Given a circuit D on X with | X | = n bits and of size m > n , there is apoly ( m, O ( k ) ) -time reduction from D to a generalized independent set instance on graph G D = ( V D , E D ) ,with the following properties. • Each vertex v ∈ V D is associated with a set of pairs S v of the form { ( i, b ) } ⊆ [ O ( n )] × { , } . Theset { S v } is produced as part of the reduction. • Each assignment x to X defines a partial assignment π x to V D such that π x ( v ) = ( if ∃ ( i, b ) ∈ S v such that EN C i ( x ) = b ∗ otherwise , where EN C is the error-correcting code from Theorem 2.1. If D ( x ) = 0 , the maximum independent set in G D ( π x ) equals κ for an integer κ , and furthermoregiven x , it can be found in time poly ( n, m, O ( k ) ) . • If D ( x ) = 1 , then the maximum independent set in G D ( π x ) has size at most κ/ k . Intuitively, the use of Lemma 3.1 is that we will start with a “no satisfying assignment” vs “most as-signments are satisfying” GAP-UNSAT instance from Theorem 2.3. Now in the “no satisfying assignment”case for all x the reduced independent set instance G D ( π x ) has a large independent set instance. Countingthe sum of independent sets over x gives a high value. On the other hand in the ‘most assignments are satis-fying” case for most x the reduced independent set instance G D ( π x ) has a small independent set and for avery few x , G D ( π x ) can have a large independent set. Hence in this case counting the sum of independentsets over all x gives a low value. The difference between the high value and low value is big enough thateven a approximate counting of these values as outlined in Section 1.2 is enough to distinguish and hencesolve the GAP-UNSAT instance.The remainder of this section is devoted to the proof of Lemma 3.1.Let us set up some notation for variable assignments to a formula. Let F be a SAT instance on a variableset Z , and let τ : Z → { , , ⋆ } be a partial assignment to Z . Then we define F ( τ ) to be the formulaobtained by setting the variables in F according to τ . Note that we do not perform further reduction ruleson the clauses in F ( τ ) : for each clause in F that becomes false (or true) under τ , there is a clause in F ( τ ) which is always false (true).For every subsequence Y of variables from Z , and every vector y ∈ { , } | Y | , we define F ( Y = y ) tobe the formula F in which the i th variable in Y is assigned y i , and all other variables are left unassigned. Lemma 3.2 (PCPP+ECC, [CW19]) . There is a polynomial-time transformation that, given a circuit D on n inputs of size m ≥ n , outputs a 3-SAT instance F on the variable set Y ∪ Z , where | Y | ≤ poly ( n ) , | Z | ≤ poly ( m ) , and the following hold for all x ∈ { , } n : • If D ( x ) = 0 then F ( Y = ENC ( x )) on variable set Z has a satisfying assignment z x . Furthermore,there is a poly ( m ) -time algorithm that given x outputs z x . • if D ( x ) = 1 then there is no assignment to the Z variables in F ( Y = ENC ( x )) satisfying more thana (1 − Ω(1)) -fraction of the clauses.where ENC : { , } n → { , } O ( n ) is the linear encoding function from Theorem 2.1. As it is a linearfunction, the i th bit of output ENC i ( x ) satisfies ENC i ( x ) = ⊕ j ∈ U i x j for some set U i . Serial Repetition [DR06] is a basic operation on CSPs/PCPs, in which a new CSP is created whoseconstraints are ANDs of k uniformly sampled clauses from the original CSP. Serial repetition is usuallydone for the purpose of reducing soundness, i.e., reducing the fraction of satisfiable clauses. We now state aderandomized version of serial repetition. Lemma 3.3 (Serial repetition [DR06]) . Given a 3-SAT instance F on n variables denoted by Y with m clauses we can construct a O ( k ) -SAT formula F ′ on the same n variables with m O ( k ) clauses such that:1. If Y = y satisfies F then y satisfies F ′ .2. If F ( Y = y ) is at most − Ω(1) satisfiable then F ′ ( Y = y ) is at most / k satisfiable. Next we prove a stronger version of derandomized serial repetition with guarantees for partial assign-ments. The proof directly follows from the guarantees of standard Serial Repetition (Lemma 3.3).9 emma 3.4 (Serial repetition with partial assignments) . Let k be a function of n . Given a 3-SAT instance F on n variables denoted by Y, Z with m clauses we can construct a O ( k ) -SAT formula F ′ on the same n variables with m · O ( k ) clauses such that:1. If Y, Z = y, z satisfies F then y, z satisfies F ′ .2. If F ( Y = y ) is at most − Ω(1) satisfiable then F ′ ( Y = y ) is at most / k satisfiable.Proof. We prove that just standard serial repetition from Lemma 3.3 suffices for proving this stronger prop-erty.Property 1 directly follows from Property 1 in Lemma 3.3.Define F y = F ( Y = y ) where we treat any clauses that became FALSE or TRUE under Y = y as normal clauses. Let F ′ y be the O ( k ) -SAT formula obtained by applying serial repetition to f y fromLemma 3.4.In Serial Repetition [DR06] it is clear that clauses in F ′ are just ANDs of clauses in F and which clausesare part of the “AND” is only dependent on their index.Due to this F ′ ( Y = y ) i.e. first applying serial repetition then setting Y = y is equivalent to first setting Y = y and then applying serial repetition i.e. F ′ y .By our assumption F y is at most − Ω(1) satisfiable and hence by Property 2 of Lemma 3.3 F ′ y is atmost / k satisfiable. As F ′ y = F ′ ( Y = y ) we have that F ′ ( Y = y ) is at most / k satisfiable.The FGLSS reduction [FGL +
91] maps a CSP Φ to a graph G Φ such that the MAX-SAT value in Φ isequal to the size of the maximum independent set in G Φ . Lemma 3.5 (FGLSS [FGL + . Let F be a k -SAT instance on variable set Y with | Y | = n and m clauses.There exists a poly ( n, m, O ( k ) ) time reduction graph from F to a graph G F = ( V F , E F ) such that: the sizeof maximum independent set in G F is exactly equal to maximum clauses satisfiable in F . We note that a stronger version of the FGLSS reduction [FGL +
91] holds with guarantees for partialassignments. The proof is very similar to the proof of the standard FGLSS reduction (Lemma 3.5).
Lemma 3.6 (FGLSS with partial assignments) . Let F be a k -SAT instance on variable set Y, Z with | Y | + | Z | = n and m clauses. There exists a poly ( n, m, O ( k ) ) time reduction graph from F to an independent setinstance on graph G F = ( V F , E F ) . Each vertex v ∈ V F is a associated to a set T v of ( i ∈ [ | Y | ] , b ∈ { , } ) pairs. For each partial assignment of the form τ : Y → { , } define a partial assignment π τ to V F suchthat: π τ ( v ) = ( if ∃ ( i, b ) ∈ T v such that τ ( Y i ) = b ∗ otherwise , Then the max independent set in G F ( π τ ) equals the max number of clauses satisfiable in F ( τ ) .Proof. Let w be a clause in F and w i denote the i th variable in w . Let ℓ denote a satisfying assignmentto w . For every w, ℓ pair create a vertex in V F . Let v be the vertex associated with a particular w, ℓ . Let T v = { ( w i , ℓ i } represent the assignment w i = ℓ i for ≤ i ≤ k .Make an edge between vertex u and vertex v if the assignment T u and T v contradict each other. Notethat this means that there is always an edge between two vertices associated to the same clause but differentsatisfying assignments i.e. vertices associated with the same clause form a clique.Let x be a assignment for F satisfying κ clauses. We now give an independent set in G F of size κ . Forevery satisfied clause w and and ℓ the assignment to variables of w in x we choose the vertex w, ℓ in the10ndependent set. As there are κ satisfied clauses we choose κ vertices. These vertices form and independentset as if two of these vertices u, v had an edge between them it would mean that the assignments T u and T v contradict each other. This is not possible as all these assignments are partial assignments of x .Consider S to be an independent set in G F of size κ . We now give an assignment to F which satis-fies κ clauses. Note that from vertices corresponding to the same clauses only 1 vertex can be a part ofindependent set as they all form a clique. Hence vertices associated with κ different clauses must be partof the independent set. For a vertex u associated with w, ℓ the partial assignment T u satisfies w . For twovertices u, v in the independent set the partial assignments from T v and T u do not contradict as otherwisethere would be an edge between u and v . Hence we can join all the partial assignments T v for vertices v in the independent set to get a partial assignment which satisfies κ clauses in F ( τ ) . Hence the maximumindependent set in G F ( π τ ) has size at most the maximum number clauses satisfied in F ( τ ) .We next present the proof of Lemma 3.1 which just follows by combining Lemma 3.2, 3.4, and 3.6sequentially. Proof of Lemma 3.1.
The proof follows by applying Lemma 3.2, 3.4 and 3.6 sequentially.We start from a circuit D with input variables X ( | X | = n ) and size m > n . Lemma 3.2 transformthis into a 3-SAT instance F with poly ( m ) clauses on the variable set Y ∪ Z , where | Y | ≤ poly ( n ) , | Z | ≤ poly ( m ) , and the following hold for all x ∈ { , } n : • If D ( x ) = 0 then F ( Y = ENC ( x )) on variable set Z has a satisfying assignment z x . Furthermore,there is a poly ( m ) -time algorithm that given x outputs z x . • if D ( x ) = 1 then there is no assignment to the Z variables in F ( Y = ENC ( x )) satisfying more thana (1 − Ω(1)) -fraction of the clauses.where ENC : { , } n → { , } O ( n ) is the linear encoding function from Theorem 2.1.Applying Lemma 3.4 on F gives us a O ( k ) -SAT formula F ′ on the same Y ∪ Z variables with poly ( m ) · O ( k ) clauses such that:1. If Y, Z = y, z satisfies F then y, z satisfies F ′ .2. If F ( Y = y ) is at most − Ω(1) satisfiable then F ′ ( Y = y ) is at most / k satisfiable.which implies that: • If D ( x ) = 0 then F ′ ( Y = ENC ( x )) on variable set Z has a satisfying assignment z x . Furthermore,there is a poly ( m ) -time algorithm that given x outputs z x . • if D ( x ) = 1 then there is no assignment to the Z variables in F ′ ( Y = ENC ( x )) satisfying more thana / k -fraction of the clauses.Finally applying Lemma 3.6 to F ′ where we consider partial assignments τ which assign Y to ENC ( x ) for some x . Hence τ ( Y i ) = ENC i ( x ) . As τ is fixed by fixing x we rename π τ to π x . S v is just a renamingof T v . Size of the graph is poly ( n + m, poly ( m ) · O ( k ) , O ( k ) ) = poly ( m, k ) as m > n . (cid:3) Main Result
We now turn to the proof of the main result, Theorem 1.3. We will prove the result for EMAJ ◦ C first, andsketch how to extend to f ◦ C for sparse symmetric f in Section 5. Below we prove EMAJ ◦ C lower boundsfor Quasi - NP when we have n − n ε time algorithms for C circuits of size n ε . For the other partsof Theorem 1.3 (on n /n ω (1) ), see Section 6.We note here that in Theorem 1.3 we mentioned polynomial size lower bounds for EMAJ ◦ C we in factprove quasi-polynomial size lower bounds below. Theorem 4.1.
Suppose C is typical, and the parity function has poly ( n ) -sized C circuits. Then for every k , quasi - NP does not have EMAJ ◦ C = H circuits of size O ( n log k n ) , if for some ǫ ∈ (0 , there is a n − n ε for all circuits from class C of size at most n ε .Proof. Let us assume that for a fixed k > , quasi - NP has H = EMAJ ◦C circuits of size O ( n (log k n ) ) whichimplies that quasi - NP ∈ size ( n O (log k n ) ) for general circuits. By Theorem 2.3, we obtain a contradiction iffor some constant δ ∈ (0 , and g ( n ) = 2 n δ we can give a n /g ( n ) ω (1) time nondeterministic algorithmfor distinguishing between:1. YES case: D has no satisfying assignments.2. NO case: D has at least n (1 − /g ( n )) satisfying assignmentsgiven a generic fan-in 2 circuit D with n inputs and size m ≤ h ( n ) := 2 n δ . Under the hypothesis, we willgive such an algorithm for δ = ε/ .Using Lemma 3.1, we reduce the circuit D to an independent set instance G D (with k = log h ( n ) ) on n = poly ( m, O ( k ) ) = poly ( m, O ( k ) ) = poly ( m, h ( n ) O (1) ) = poly ( h ( n )) vertices. We also find subsets S i for every vertex i ∈ [ n ] . Let π x be the partial assignment which assigns a vertex i to if there exist ( j ′ , b ) ∈ S i such that ENC j ′ ( x ) = b . Note that π x does not assign any vertex to . By Lemma 3.1, G D hasthe following properties:1. If D ( x ) = 0 , then G D ( π x ) has an independent set of size κ . Furthermore, given x we can find thisindependent set in poly ( h ( n )) time.2. If D ( x ) = 1 , then in G D ( π x ) , all independent sets have size at most κ/h ( n ) .This means it suffices for us to distinguish between the following two cases:1. YES case: For all x , G D ( π x ) has an independent set of size κ .2. NO case: For at most n /g ( n ) values of x, G D ( π x ) has an independent set of size ≥ κ/h ( n ) . Guessing a succinct witness circuit:
As guaranteed by Lemma 3.1 given an x such that D ( x ) = 0 we can find the assignment A ( x ) to G D which is consistent with π x and represents an independent set ofsize κ in poly ( h ( n )) time. Let A ( x, i ) denote the assignment to the i th vertex in A ( x ) . Given x and vertex i ∈ [ n ] , in time poly ( h ( n )) we can produce ¬ A ( x, i ) . Claim 1.
Under the hypothesis, there is a h ( n ) o (1) -sized EMAJ ◦ C circuit U of size h ( n ) o (1) with x, i asinput representing ¬ A ( x, i ) . roof. Under the hypothesis, for some constant k , we have quasi - NP ⊆ size H [ n log k n ] . Specifically, for p ( n ) = n log k +1 n we have NTIME [ p ( n )] ⊆ size H [ p ( n ) / log n ] ⊆ size H [ p ( n ) o (1) ] . As h ( n ) = 2 n ε ≫ p ( n ) ,a standard padding argument implies NTIME [ poly ( h ( n ))] ⊆ size H [( poly ( h ( n ))) o (1) ] = size H [ h ( n ) o (1) ] .Since ¬ A ( x, i ) is computable in poly ( h ( n )) time, we have that ¬ A ( x, i ) can be represented by a h ( n ) o (1) -sized H = EMAJ ◦ C circuit.Our nondeterministic algorithm for GAP-UNSAT begins by guessing U guaranteed by Claim 1 whichis supposed to represent ¬ A . Then by the reduction in Lemma 2.1 we can covert U to a SUM ≥ ◦ C circuit R for A ( x, i ) of size poly ( h ( n ) o (1) ) = h ( n ) o (1) . Note that if our guess for U is correct, i.e., U = ¬ A , then R represents A .Let the subcircuits of R be R , R , . . . , R t , so that R ( x ) = P j ∈ [ t ] R j , where R j ∈ C and t ≤ h ( n ) o (1) .The number of inputs to R j is n ′ = | x | + log n = n + O (log h ( n )) , and the size of R j is h ( n ) o (1) .Note that R ( x, i ) = 0 represents that the i th vertex is not in the independent set of G D in a solutioncorresponding to x , while R ( x, i ) > represents that it is in the independent set of G D in a solution corre-sponding to x . For all x and i we have ≤ R ( x, i ) ≤ t ≤ h ( n ) o (1) . Verifying that R encodes valid independent sets: We can verify that the circuit R produces an in-dependent set on all x by checking each edge over all x . To check the edge between vertices i and i we need to verify that at most one of them is in the independent set. Equivalently, for all x we check that R ( x, i ) · R ( x, i ) = 0 . As R ( x, i ) ≥ for all x and i we can just verify X x ∈{ , } n R ( x, i ) · R ( x, i ) = 0 . Since R ( x, i ) = P j ∈ [ t ] R j ( x, i ) it suffices to verify that X x ∈{ , } n X j ,j ∈ [ t ] R j ( x, i ) · R j ( x, i ) = 0 . Let R j ,j ( x, i , i ) = R j ( x, i ) · R j ( x, i ) . Since C is closed under AND (upto polynomial factors) R j ,j also has a poly ( h ( n ) o (1) ) = h ( n ) o (1) sized C circuit. Exchanging the order of summations is suffices for usto verify X j ,j ∈ [ t ] X x ∈{ , } n R j ,j ( x, i , i ) = 0 . For fixed i , i , j , j the number of inputs to R j ,j is | x | = n and its size is h ( n ) o (1) ≤ n ε . Hence, forfixed i , i , j , j we can compute P x R j ,j ( x, i , i ) using the n − n ε . Summing over all j , j pairs only adds another multiplicative factor of t = h ( n ) o (1) . Thisallows us to verify that the edge ( i , i ) is satisfied by R . Checking all edges of G D only adds another mul-tiplicative factor of poly ( h ( n )) . Hence the total running time for verifying that R encodes valid independentsets on all x is still n − n ε poly ( h ( n )) . Verifying consistency of independent set produced by R with π x : As we care about the sizes ofindependent sets in G D ( π x ) over all x we need to check if the assignment by R is consistent with π x . As π x only assigns vertices to , we need to verify that all vertices assigned to in π x are in fact assignedto by the assignment given by R ( x, · ) . From Lemma 3.1, we know that π x assigns a vertex i to if for13ome ( j ′ , b ) ∈ S i , ENC j ′ ( x ) = b . To check this condition we need to verify that R ( x, i ) = 0 if for some ( j ′ , b ) ∈ S i , ENC j ′ ( x ) = b . Equivalently, we cn check ( ENC j ′ ( x ) ⊕ b ) · R ( x, i ) = 0 for all x, i, ( j ′ , b ) ∈ S i .Since ( ENC j ( x ) ⊕ b ) R ( x, i ) ≥ for all possible inputs we can just check that X x ∈{ , } n ( ENC j ′ ( x ) ⊕ b ) · R ( x, i ) = 0 for all i, ( j ′ , b ) ∈ S i . As R ( x, i ) = P j ∈ [ t ] R j ( x, i ) we can equivalently verify that X x ∈{ , } n X j ∈ [ t ] ( ENC j ′ ( x ) ⊕ b ) · R j ( x, i ) = 0 for all i, ( j ′ , b ) ∈ S i . Note that R j ′ ( x, i ) has a h ( n ) o (1) sized C circuit. By our assumption parity has apoly ( n ) -sized C -circuit so ( ENC j ( x ) ⊕ b ) also has a poly ( n ) -sized C circuit. Hence ( ENC j ( x ) ⊕ b ) · R j ′ ( x, i ) has a poly ( n, h ( n ) o (1) ) = h ( n ) o (1) -sized C circuit, since C is closed under AND.For fixed ( i, j, j ′ ) , ( ENC j ′ ( x ) ⊕ b ) · R j ( x, i ) ∈ C has | x | = n inputs and size h ( n ) o (1) < n ε . Hencewe can use our assumed P x ∈{ , } n ( ENC j ′ ( x ) ⊕ b ) · R j ( x, i ) in time n − n ε .Summing over all j ∈ [ t ] introduces another multiplicative factor of h ( n ) o (1) . This allows us to verifythe desired condition for a fixed i, ( j ′ , b ) ∈ S i . To check it for all i, ( j ′ , b ) ∈ S i (recall | S i | = O ( n ) byTheorem 2.1) only introduces another multiplicative factor of poly ( h ( n )) · O ( n ) = poly ( h ( n )) in time.Therefore the total running time for verifying consistency w.r.t. π x is n − n ε poly ( h ( n )) .At this point, we now know that R represents an independent set, and that R is consistent with π x . Weneed to distinguish between:1. YES case: For all x , R ( x, · ) represents an independent set of size κ .2. NO case: For at most n /g ( n ) values of x, R ( x, · ) represents an independent set of size ≥ κ/h ( n ) . Lemma 4.1.
For all x such that R ( x, · ) represents an independent set of size a . we have a ≤ P i ∈ [ n ] R ( x, i ) ≤ at .Proof. For every vertex i in the independent set, ≤ R ( x, i ) ≤ t . For all vertices i not in the independentset, we have R ( x, i ) = 0 . Hence a ≤ P i ∈ [ n ] R ( x, i ) ≤ at . Distinguishing between the YES and NO cases:
To distinguish between the YES and NO cases, wenow compute X x ∈{ , } n X i ∈ [ n ] R ( x, i ) (3)This allows us to distinguish between the YES case and NO case as:1. YES case: We have for at least n (1 − /g ( n )) values of x we have an independent set of size at most κ/h ( n ) . By Lemma 4.1 for such x , P i ∈ [ n ] R ( x, i ) ≤ tκ/h ( n ) . for the rest of n /g ( n ) values of x the independent set could be all the vertices in the graph G D . Hence by Lemma 4.1 for such valuesof x , P i ∈ [ n ] R ( x, i ) ≤ tn = poly ( h ( n )) . Hence X x ∈{ , } n X i ∈ [ n ] R ( x, i ) ≤ (2 n /g ( n )) poly ( h ( n )) + 2 n tκ/h ( n ) o (2 n ) + 2 n tκ/h ( n ) [ As h ( n ) = g ( n ) o (1) ] ≤ o (2 n ) + o (2 n κ ) [ As t = h ( n ) o (1) ] ≤ n κ [ As κ >
2. NO case: We have for all x ∈ { , } n the independent set is at least of size κ . Hence by Lemma 4.1the sum is P x ∈{ , } n P i ∈ [ n ] R ( x, i ) > n κ .All that remains is how to compute (3). As R ( x, i ) = P j ∈ [ t ] R j ( x, i ) , we can compute X x ∈{ , } n X i ∈ [ n ] X j ∈ [ t ] R j ( x, i ) = X j ∈ [ t ] X i ∈ [ n ] X x ∈{ , } n R j ( x, i ) For a fixed i, j , R j ( x, i ) ∈ C , it has | x | = n inputs and size ≤ poly ( h ( n ) o (1) ) = h ( n ) o (1) < n ε . Hencewe can use the assumed P x ∈{ , } n R j ( x, i ) in time n − n ε . Summing over all j ∈ [ t ] , i ∈ [ n ] only introduces another h ( n ) o (1) poly ( h ( n )) = poly ( h ( n )) multiplicative factor. Thus therunning time for distinguishing the two cases is n − n ε poly ( h ( n )) .In total our running time comes to n − n ε poly ( h ( n )) = 2 n − n δ + O ( n δ ) ≤ n − n δ = 2 n /g ( n ) ω (1) as g ( n ) = 2 n δ and ε = 4 δ . By Theorem 2.3, this gives us a contradiction which completes our proof.The above theorem when combined with known ACC ◦ THR gives an quasi - NP lower bound for EMAJ ◦ ACC ◦ THR. ◦ ACC ◦ THR Lower bound
We will apply a known
SAT algorithm for
ACC ◦ THR circuits.
Theorem 4.2 ([Wil18b]) . For every pair of constants d, m , there exists a constant ε ∈ (0 , such that n − n ε time for AC [ m ] ◦ THR circuits of depth d and size n ε . Theorem 4.3.
For constants k, d, m , quasi - NP does not have size ( n log k n ) EMAJ ◦ ACC ◦ THR circuits ofdepth d .Proof. We first note that
ACC ◦ THR is indeed typical and can represent ENC ( x ) by poly ( n ) -sized circuitsas ENC ( x ) : { , } n → { , } O ( n ) is a linear function.By Theorem 4.2 we know that for all constants d there exists some constant ǫ ∈ (0 , such that thereexists a n − n ε for all circuits from class ACC ◦ THR of size ≤ n ε anddepth d .The above properties imply that ACC ◦ THR satisfies the preconditions of Theorem 4.1 and hence forevery pair of constant k, d , quasi - NP does not have size ( n log k n ) EMAJ ◦ ACC ◦ THR circuits of depth d . The above theorem can be rewritten as: For constants k, d, m , there exists a constant e such thatNTIME [ n log e n ] does not have n log k n -size EMAJ ◦ ACC ◦ THR circuits of depth d . Here the constant e depends on d and m . Using a standard trick (as in [MW18]) this dependence can be removed as we showbelow. 15 orollary 4.1. There exists an e such that NTIME [ n log e n ] does not have polynomial size EMAJ ◦ ACC ◦ THRcircuits.Proof.
Assume for contradiction that for all e , there exists constants d, m such that NTIME [ n log e n ] has poly-sized EMAJ ◦ AC [ m ] ◦ THR circuit of depth d . This implies that P has poly-sized EMAJ ◦ AC [ m ] ◦ THRcircuits, which further implies that
CIRCUIT EVALUATION problem has poly-sized EMAJ ◦ AC [ m ] ◦ THRcircuit of a fixed constant depth d and fixed constant m . Hence any circuit of size s has an equivalentpoly ( s ) -sized EMAJ ◦ AC [ m ] ◦ THR circuit of depth d . Combining this with our assumption yields: Forall e , there exists constants d, m such that NTIME [ n log e n ] has poly-sized EMAJ ◦ AC [ m ] ◦ THR circuitof depth d . This contradicts Theorem 4.3 and hence our assumption was wrong, which completes theproof. Our lower bounds extend to circuit classes of the form f ◦C where f denotes a family of symmetric functionsthat only take the value on a small number of slices of the hypercube. Formally, let f : { , } n → { , } be a symmetric function, and let g : { , , . . . , n } → { , } be its “companion” function, where for all x , f ( x ) = g ( P i x i ) (here, x i denotes the i -th bit of x ). For k ∈ { , , . . . , n } , we say that a symmetricfunction f is k -sparse if | g − (1) | = k . For example, the all-zeroes function is -sparse, the all-ones functionis n -sparse, and the EMAJ function is -sparse. Theorem 5.1.
Let k < n/ . Every k -sparse symmetric function f : { , } n → { , } can be representedas an exact majority of n O ( k ) ANDs on k inputs.Proof. Given a k -sparse f and its companion function g , consider the polynomial expression E ( x ) := Y v ∈ g − (1) X i x i − v ! . Then E ( x ) = 0 whenever f ( x ) = 1 , and E ( x ) = 0 otherwise. Expanding E into a sum of products, we canwrite E as a multilinear n -variate polynomial of degree at most k , with integer coefficients of magnitude atmost n O ( k ) (since each v ≤ n ). We can therefore write E as the EMAJORITY of n O ( k ) distinct ANDs onup to k inputs.The above theorem immediately implies that for every k -sparse symmetric function f m , any circuit withan f m at the output gate can be rewritten as a circuit with an EMAJ of fan-in at most m O ( k ) at the outputgate (and ANDs of fan-in up to k below that). Corollary 5.1.
For every fixed k , and every k -sparse symmetric function family f = { f n } , Quasi - NP doesnot have polynomial-size f ◦ ACC ◦ THR circuits.
In this section we prove NEXP Lower Bounds under weaker algorithmic assumptions. The proof followsthe same pattern as the proof of lower bound for quasi - NP in Theorem 4.1.16 .1 NEXP Lower Bounds Theorem 6.1 ([Wil14]) . Suppose for some constant ε ∈ (0 , there is an algorithm A that for all poly ( n ) -size circuits C on n inputs, A ( C ) runs in n /n ω (1) time, outputs YES on all unsatisfiable C , and outputsNO on all C that have at least n − satisfying assignments. Then NTIME [2 n ] P / poly . Theorem 6.2.
Suppose there is an algorithm A that for all poly ( n ) -sized circuits C on n inputs, A ( C ) runs in n /g ( n ) ω (1) time, outputs YES on all unsatisfiable C , and outputs NO on all C that have at least n (1 − /g ( n )) satisfying assignments, for any g ( n ) satisfying g ( n ) = n ω (1) , g ( n ) = 2 o ( n ) . Then for all k ,there is a c ≥ such that NTIME [2 n ] P / poly .Proof. Our starting point is Theorem 6.1 ([Wil14]): we are given an m -input, poly ( m ) -size circuit D ′ thatis either UNSAT or has at least m − satisfying assignments, and we wish to distinguish between the twocases with a m /m ω (1) -time algorithm.We create a new circuit D with n inputs, where n satisfies n = m + ψ · log g ( n ) , and ψ > is the constant from Lemma 2.2. (Note that, since g ( n ) is time constructible and g ( n ) ≤ o ( n ) ,such an n can be found in subexponential time.) Applying the algorithm from Lemma 2.2, D treats its n bits of input as a string of randomness r , computes t = O (log g ( n )) strings x , x , . . . , x t ∈ { , } m witha poly ( m, log g ) -size circuit, then outputs the OR of D ′ ( x i ) over all i = 1 , . . . , t . Note the total size of ourcircuit D is poly ( m, log g ) + O (log g ) · size ( D ′ ) = poly ( m ) = poly ( n ) .Clearly, if D ′ is unsatisfiable, then D is also unsatisfiable. By Lemma 2.2, if D ′ has m − satisfyingassignments, then D has at least n (1 − /g ( n )) satisfying assignments. As size ( D ) ≤ poly ( n ) , by ourassumption we can distinguish the case where D is unsatisfiable from the case where D has at least n (1 − /g ( n )) satisfying assignments, with an algorithm running in time n /g ( n ) ω (1) . This yields an algorithmfor distinguishing the original circuit D ′ on m inputs and poly ( m ) size, running in time n /g ( n ) ω (1) = 2 m g ( n ) O (1) /g ( n ) ω (1) = 2 m /g ( n ) ω (1) ≤ m /g ( m ) ω (1) ≤ m /m ω (1) since n > m, g ( n ) = n ω (1) . By Theorem 6.1, this implies that NTIME [2 n ] P / poly Theorem 6.3.
N T IM E [2 n ] does not have poly ( n ) -sized EMAJ ◦ C = H circuits if1. There exists a n /b ( n ) for all poly ( n ) -sized circuits from class C where b ( n ) = n ω (1) C is t ypical and ( ¬ ) ENC i ( x ) has poly ( n ) -sized C circuits.Proof. Let us assume that NTIME [2 n ] has poly ( n ) -sized H = EMAJ ◦ C circuits which implies thatNTIME [2 n ] ∈ P / poly . By Theorem 6.2, we will get a contradiction if we can give a n /g ( n ) ω (1) timenondeterministic algorithm for distinguishing between:1. YES case: D has no solutions.2. NO case: D has at least n (1 − /g ( n )) solutions.17iven a circuit D with n inputs and size m = poly ( n ) where g ( n ) = n ω (1) . We will take a g ( n ) such that g ( n ) = b ( n ) o (1) .Let h ( n ) be a function such that h ( n ) = g ( n ) o (1) , h ( n ) = n ω (1) . Using Lemma 3.1 we reduce D to independent set instance on G D (with k = log h ( n ) ) over n = poly ( m, O ( k ) ) = poly ( m, h ( n )) = poly ( h ( n )) vertices and edges as h ( n ) = n ω (1) and m = poly ( n ) . We also find S i for every vertex i ∈ [ n ] .By Lemma 3.1, G D has the following properties:1. Let D ( x ) = 0 then for G D ( π x ) there exists an independent set of size κ . Further given x we can findthis assignment in poly ( h ( n )) time.2. Let D ( x ) = 1 then for G D ( π x ) all independent sets have size ≤ κ/h ( n ) .where π x is the partial assignment which assigns a vertex i to if there exist ( j ′ , b ) ∈ S i such thatENC j ′ ( x ) = b . π x does not assign any vertex to .This means we need to distinguish between the following two cases:1. YES case: For all x , G D ( π x ) has an independent set of size κ .2. NO case: For at most n /g ( n ) values of x, G D ( π x ) has an independent set of size ≥ κ/h ( n ) . Guessing a succinct witness circuit:
As given an x such that D ( x ) = 1 we can find the assignment A ( x ) to G D which is consistent with π x and represents an independent set of size κ in poly ( h ( n )) time. Let A ( x, i ) denote the assignment to i th vertex in A ( x ) . Given x and vertex i ∈ [ n ] in time poly ( h ( n )) we canproduce ¬ A ( x, i ) . Claim 2.
There exists a poly ( n ) -sized EMAJ ◦ C circuit U with x, i as input representing A ( x, i ) .Proof. As given x and vertex i ∈ [ n ] in time poly ( h ( n )) we can produce ¬ A ( x, i ) . NTIME [2 n ] haspoly-sized EMAJ . C circuits given x and i ∈ [ n ] we can also produce/represent ¬ A ( x ) i by a poly ( n + O (log h ( n ))) = poly ( n ) EMAJ . C circuit.Our nondeterministic algorithm for GAP-UNSAT begins by guessing U guaranteed by Claim 2 whichis supposed to represent ¬ A . Then by the reduction in Lemma 2.1 we can covert U to a SUM ≥ ◦ C circuit R for A ( x, i ) of size poly ( n ) . Note that if our guess for U is correct i.e. U = ¬ A then R represents A .Let subcircuits of R be R , R , . . . , R t i.e. R ( x ) = P j ∈ [ t ] R j where R j ∈ C and t = poly ( n ) . Thenumber of inputs to R j are n ′ = | x | + log n = n + O (log h ( n )) and the size of R j is poly ( n ) .Note that R ( x, i ) = 0 represents that the i th vertex is not part of the independent set in a solution corre-sponding to x while R ( x, i ) > represents that it is part of the independent set in a solution correspondingto x . For all x, i , ≤ R ( x, i ) ≤ t ≤ poly ( n ) . Verifying that R encodes valid independent sets: We can verify that the circuit produces an indepen-dent set by checking each edge over all x . To check the edge between vertices i and i we need to verifythat most one of them is part of the independent set. Equivalently, for all x , R ( x, i ) · R ( x, i ) = 0 . As R ( x, i ) is always ≥ we can just verify X x R ( x, i ) · R ( x, i ) = 0 . R ( x, i ) = P j ∈ [ t ] R j ( x, i ) it suffices to verify that X x X j ,j ∈ [ t ] R j ( x, i ) · R j ( x, i ) = 0 Let R j ,j ( x, i , i ) = R j ( x, i ) · R j ( x, i ) . By definition 2.3, C · C = AND . C = C we R j ,j has apoly ( n ) sized C circuit. Interchanging the summations we get that we need to verify X j ,j ∈ [ t ] X x R j ,j ( x, i , i ) = 0 For a fixed j , j number of inputs to R j ,j are | x | = n and its size is poly ( n ) . Hence, for a fixed pair of j , j we can compute P x R j ,j ( x, i , i ) using the n /b ( n ) .Going over all j , j pairs only adds another multiplicative factor of t = poly ( n ) . This allows us to verifythat the edge ( i , i ) is satisfied by R .Checking all edges only adds another multiplicative factor of poly ( h ( n )) . Hence the total running timefor verifying that R encodes valid independent sets is still n poly ( h ( n )) /b ( n ) . Verifying consistency of independent set produced by R with π x : As we care about the size ofindependent set in G D ( π x ) while R assigns all vertices in G D we need to check if the assignment by R isconsistent with π x . As π x only assigns vertices to we need to verify that all vertices assigned to in π x arein fact assigned to by the assignment given by R ( x, · ) . From Lemma 3.1 we know that π x assigns a vertex i to if for any ( j ′ , b ) ∈ S i , ENC j ′ ( x ) = b . To check this we need to verify that R ( x, i ) = 0 wheneverfor any ( j ′ , b ) ∈ S i , ENC j ′ ( x ) = b . Equivalently, ( ENC j ′ ( x ) ⊕ b ) R ( x, i ) = 0 for all x, i, ( j ′ , b ) ∈ S i . As ( ENC j ( x ) ⊕ b ) R ( x, i ) ≥ we can just check that X x ( ENC j ′ ( x ) ⊕ b ) R ( x, i ) = 0 for all i, ( j ′ , b ) ∈ S i . As R ( x, i ) = P j ∈ [ t ] R j ( x, i ) we can equivalently verify that X x ∈{ , } n X j ∈ [ t ] ( ENC j ′ ( x ) ⊕ b ) R j ( x, i ) = 0 for all i, ( j ′ , b ) ∈ S i . Note that R j ′ ( x, i ) has a poly ( n ) -sized C circuit. By our assumption ( ENC j ( x ) ⊕ b ) has a poly ( n ) -sized C circuit. Hence ( ENC j ( x ) ⊕ b ) R j ′ ( x, i ) has a poly ( n ) -sized C circuit as we are giventhat C is typical.For fixed ( i, j, j ′ ) , ( ENC j ′ ( x ) ⊕ b ) R j ( x, i ) ∈ C has | x | = n inputs and size poly ( n ) . Hence we can use P x ∈{ , } n ( ENC j ′ ( x ) ⊕ b ) R j ( x, i ) in time n /b ( n ) . Goingover all j ∈ [ t ] adds another multiplicative factor of poly ( n ) . This allows us to verify the condition for afixed i, ( j ′ , b ) ∈ S i .To go all i, ( j ′ , b ) ∈ S i ( | S i | = O ( n ) by Theorem 2.1) only adds another multiplicative factor ofpoly ( h ( n )) · O ( n ) = poly ( h ( n )) in time. The total running time for verifying consistency w.r.t. π x is n poly ( h ( n )) /b ( n ) .As we now know that R represents and independent set and that R is consistent with π x we need todistinguish between:1. YES case: For all x , R ( x, · ) represents an independent set of size κ .19. NO case: For at most n /g ( n ) values of x, R ( x, · ) represents an independent set of size ≥ κ/h ( n ) .This is because we are giving a non-deterministic algorithm, and hence we can assume in the YES case that R = A . Claim 3.
For an x such that R ( x, · ) represents an independent set of size a then a ≤ P i ∈ [ n ] R ( x, i ) ≤ at .Proof. For every vertex i which is part of the independent set we have ≤ R ( x, i ) ≤ t while for all vertices i which are not part of the independent set we have R ( x, i ) = 0 . Hence b ≤ P i ∈ [ n ] R ( x, i ) ≤ bt . Distinguishing between YES and NO cases:
To distinguish between YES and NO cases we compute X x ∈{ , } n X i ∈ [ n ] R ( x, i ) This allows us to distinguish between the YES case and NO case as:1. YES case: We have for at least n (1 − /g ( n )) values of x we have an independent set of size at most κ/h ( n ) . By Lemma 3 for such x , P i ∈ [ n ] R ( x, i ) ≤ tκ/h ( n ) . for the rest of n /g ( n ) values of x theindependent set could be all the vertices in the graph G D . Hence by Lemma 3 for such values of x , P i ∈ [ n ] R ( x, i ) ≤ tn = poly ( h ( n )) . Hence X x ∈{ , } n X i ∈ [ n ] R ( x, i ) ≤ (2 n /g ( n )) poly ( h ( n )) + 2 n tκ/h ( n ) ≤ o (2 n ) + 2 n tκ/h ( n ) [ As h ( n ) = g ( n ) o (1) ] ≤ o (2 n ) + o (2 n κ ) [ As t = h ( n ) o (1) ] ≤ n κ [ As κ >
2. NO case: We have for all x ∈ { , } n the independent set is at least of size κ . Hence by Lemma 3 thesum is P x ∈{ , } n P i ∈ [ n ] R ( x, i ) > n κ .All that remains is how to compute P x ∈{ , } n P i ∈ [ n ] R ( x, i ) . As R ( x, i ) = P j ∈ [ t ] R j ( x, i ) we cancompute X x ∈{ , } n X i ∈ [ n ] X j ∈ [ t ] R j ( x, i ) = X j ∈ [ t ] X i ∈ [ n ] X x ∈{ , } n R j ( x, i ) For a fixed i, j , R j ( x, i ) ∈ C , it has | x | = n inputs and size poly ( n ) . Hence we can use P x ∈{ , } n R j ( x, i ) in time n /b ( n ) . Doing the summation for all j ∈ [ t ] , i ∈ [ n ] add another h ( n ) o (1) poly ( h ( n )) = poly ( h ( n )) multiplicative factor. The running time fordistinguishing YES case and NO case is n poly ( h ( n )) /b ( n ) .In total our running time comes to n poly ( h ( n )) /b ( n ) = 2 n /g ( n ) ω (1) . By Theorem 6.2 this gives us acontradiction which completes our proof. 20 eferences [AB09] Sanjeev Arora and Boaz Barak. Computational Complexity - A Modern Approach . CambridgeUniversity Press, 2009.[ACW16] Josh Alman, Timothy M. Chan, and R. Ryan Williams. Polynomial representations of thresholdfunctions and algorithmic applications. In
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