Extended Hardness Results for Approximate Gröbner Basis Computation
aa r X i v : . [ c s . S C ] M a y Extended Hardness Results forApproximate Gr ¨obner Basis Computation
G. Spencer ∗ October 2015
Abstract
Two models were recently proposed to explore the robust hardness of Gr¨obnerbasis computation. Given a polynomial system, both models allow an algorithm toselectively ignore some of the polynomials: the algorithm is only responsible forreturning a Gr¨obner basis for the ideal generated by the remaining polynomials.For the q -Fractional Gr¨obner Basis Problem the algorithm is allowed to ignorea constant (1 − q ) -fraction of the polynomials (subject to one natural structuralconstraint). Here we prove a new strongest-parameter result: even if the algorithmis allowed to choose a (3 / − ǫ ) -fraction of the polynomials to ignore, and needonly compute a Gr¨obner basis with respect to some lexicographic order for theremaining polynomials, this cannot be accomplished in polynomial time (unless P = NP ). This statement holds even if every polynomial has maximum degree 3.Next, we prove the first robust hardness result for polynomial systems of maximumdegree 2: for the q -Fractional model a (1 / − ǫ ) fraction of the polynomials may beignored without losing provable NP-Hardness. Both theorems hold even if everypolynomial contains at most three distinct variables. Finally, for the Strong c -partial Gr¨obner Basis Problem of De Loera et al. we give conditional results thatdepend on famous (unresolved) conjectures of Khot and Dinur, et al. Keywords: AMS Subject Classification: 68Q17Keywords:
Computational algebra, Gr¨obner basis, Complexity, Hardness of approxi-mation, Satisfiability, Theoretical Computer Science
Gr¨obner basis computation is a classic problem in computational algebra (see [11], [2],or [7] for details). A Gr¨obner basis for the polynomial ideal hFi is simply a specialset of polynomial generators for hFi with the property that the leading terms of theGr¨obner basis elements generate all leading terms of the polynomial ideal hFi . Incontrast, a general set of polynomial generators F for hFi may not have leading terms ∗ Smith College. Northampton, MA 01063. USA. Phone: (607)351-3035 .
Early stages of this work weresupported by the Mathematics Research Communities Program of the AMS. hFi are obtained throughcancellation of leading terms for combinations of polynomials from F .It is well-known that Gr¨obner basis computation with respect to a lexicographicorder is not possible in polynomial-time (unless P = N P ). This is obvious becausepolynomial systems can be used to tidily encode the exact solutions of number of NP-Hard combinatorial optimization problems. Using standard results from eliminationtheory, once a Gr¨obner basis for a lexicographic order is in hand, an optimal solutioncan be quickly computed (see [2]). The natural polynomial system for the MinimumVertex Cover Problem (which is NP-Hard) shows that this NP-Hardness for Gr¨obnerbasis computation holds even when the polynomial system has maximum degree 2.Despite these general hardness results, many algebraists seem to hope that if theproblem of Gr¨obner basis computation is restricted to polynomial systems with a fewnice properties, then the problem may be efficiently solvable (in the traditional polynomial-time sense). Indeed, empirical experience running the most famous algorithm for theproblem (Buchberger’s Algorithm [1]) makes it tempting to think that addressing poly-nomials via some clever ordering, or choosing a variable or term ordering based on theinput, could somehow resolve various difficulties (intermediate polynomials of veryhigh degree, etc). Unfortunately, recent results of De Loera et al. [8] and Rolnick andSpencer [10] use results from combinatorial optimization and complexity to prove thateven approximate solutions to Gr¨obner Basis queries cannot be efficiently obtained forlexicographic orders, even for intensely-restricted polynomial-system inputs. In particular, many combinatorial optimization problems are not only hard to solveexactly in polynomial time , they are hard to solve even approximately in polynomialtime . In what meaningful sense might a Gr¨obner basis query be hard to solve evenapproximately ? To explore this idea, in [8], De Loera et al. introduced the c -PartialGr¨obner Basis Problem, and later Rolnick and Spencer introduced the q -FractionalGr¨obner Basis Problem [10]. Both models allow an algorithm to selectively ignoresome of the polynomials in the input subject to a structural constraint on the ignored set:the algorithm is only responsible for returning a Gr¨obner Basis for the ideal generatedby the remaining polynomials. The precise definitions of these problems will appear inthe following sections. Existing hardness results for the Strong c -Partial Gr¨obner Ba-sis Problem have been based on hardness-of-approximation results for graph coloringproblems (see [8] and [10]). Existing hardness results for the q -Fractional Gr¨obner Ba-sis Problem are based on hardness-of-approximation results for the classic Max-3SATProblem (see [10]). We shall be more explicit about these existing results after pre-cisely defining the models. Our contribution.
In this paper we give a new best result for the q -Fractional Gr¨obnerBasis Problem based on the standard assumption that P = N P . Our reduction uses a2014 hardenss-of-approximation result of H˚astad for the Max Not-2 Problem.At a high level, our reduction is very similar to the original Max-3SAT-based resultin [10]: Approximate Gr¨obner Basis computation is used to construct a partial assign-ment that is perfect on some fraction of the input, and then a random assignment pro- Notably, their results and ours allow the algorithm to select any lexicographic order that is convenient.Questions about robust hardness of Gr¨obner basis computation with respect to other widely-studied termorders remain completely open.
Values of (1 − q ) for which the q -Fractional Gr¨obner Basis Problemis NP-Hard for lexicographic orders. NP-Hardness of the classic Gr¨obner BasisProblem is depicted by the black points on the horizontal axis. The main theoremfrom Rolnick and Spencer [10] is depicted in green. Our results overlap [10]: pa-rameter values covered by our extensions but not by [10] are depicted in orange. Allresults depicted hold even if every polynomial in F contains at most 3 variables. Maximum Degree Bound for F ( f r a c ti ono f F t ha t c anb e i gno r e d ) (1 − q )0 . . . . OX RB a s e d cedure with modest expected quality can be used to supplement the fraction of satisfiedclauses beyond what is allowed by known hardness-of-approximation bounds. To beatthe result based on Max 3SAT, we turn to a different form of logical predicate where amuch stronger hardness-of-approximation bound holds. Unfortunately, switching fromthe disjunctions of 3-SAT to Not-2 predicates sacrifices a key property of the varietiesof polynomial systems constructed in [10]. Significant additional work is required tobeat [10]: the constructed polynomial system is more complex, and the quality deliv-ered by both the Gr¨obner Basis-based and coin-flip-based portions of the assignmentrely on structural properties revealed by careful pre-processing of satisfiable Max Not-2 instances. We prove that even for polynomial systems of maximum degree 3 in whicheach polynomial contains at most 3 variables, the (7 /
10 + ǫ ) -Fractional Gr¨obner BasisProblem with respect to lexicographic orders is NP-Hard for any ǫ > .Resolving the issues associated with Not-2 predicates, we realized that a highlysimilar proof could be based on an older 2001 hardness result of H˚astad for the Max-OXR predicate problem [4]. Our OXR-based result gives a weaker parameter ( q =4 / ǫ ) for the q -Fractional model, but the result holds even for polynomial systemswith maximum total degree 2. This is the first robust hardness result for degree 2(matching the degree bound for ordinary NP-Hardness of the traditional Gr¨obner Basiscomputation). Anecdotally, we have found that algebraists seem quite surprised that a3obust-hardness notion holds for polynomial systems of maximum degree 2. Further,it seems unlikely that a “max-degree 2” result can be shown for the c -partial modelof Robust Hardness of Gr¨obner Basis Computation. Our new contributions for the q -Fractional problem are summarized in Figure 1.Finally, we point out 2 new conditional results for the Strong c -partial Gr¨obner Ba-sis Problem of De Loera et al. In [10], Rolnick and Spencer’s reduction for the c -partialmodel applied an older (non-conditional) NP-Hardness result of Lund and Yanakakis[9] for the ApproxColoring ( q, Q ) Problem. In 2009, Dinur, Mossel and Regev [3]showed that Khot’s ↔ conjecture implies surprising strong conditional resultsfor the ApproxColoring ( q, Q ) Problem. Further, Dinur, Mossel and Regev propose anew label-cover-related conjecture that implies an even stronger conditional result forApproxColoring ( q, Q ) . In the reduction from [10], we substitute Dinur et al.s condi-tional results directly in the place of the NP-Hardness result of Lund and Yanakakis:the paramaters that result for the Strong c -partial problem seem very surprising. Aswith both conditional results of Dinur et al., our conditional results either suggest deeprobust hardness of a classic problem, or, by extending Khot’s conjectures into areaswhere there are many tools (graph coloring and computational algebra respectively),perhaps supply useful directions for possible proofs by contradiction. q -Fractional Gr ¨obner Basis Com-putation Rolnick and Spencer showed that the ( + ǫ ) -Fractional Gr¨obner Basis Problem isNP-Hard for lexicographic orders. That result was proved by reducing from the Max-3SAT Problem for satisfiable instances . An input to the problem consists ofa set of logical clauses over a set of literals, where each clause is a disjunction ofat most 3 literals (or negations of literals). The objective in
Max-SAT problems isto choose a truth assignment for the literals that satisfies the largest fraction of theset of clauses. When we study
Max-SAT for satisfiable instances we are guaranteedthat a truth assignment which satisfies all clauses exists: H˚astad’s celebrated resultfor
Max-3SAT shows that even with such a guarantee of satisfiability it is NP-Hard toproduce a truth assignment satisfying a (7 / ǫ ) fraction of the clauses (for any ǫ > ). Rolnick and Spencer define an polynomial system based on an arbitrary instanceof Max-3SAT for satisfiable instances and show how the ability to efficiently solvethe q -Fractional Gr¨obner Basis Problem well for a lexicographic order can be used toconstruct a solution of quality that violates H˚astad’s bound.First, in Section 2.1, we give a new reduction for the q -Fractional Gr¨obner BasisProblem that is similar in character to that in [10], but rather than focusing on Satis-fiability for clauses which are disjunctions (perhaps one of the best known problemsin complexity theory), we turn to a different form of predicate over binary { , } vari-ables. We reduce from Max Not-2 (“Not two”) for satisfiable instances for predicatesof arity 3. “Not two” predicates accept any sum that is not 2, and reject otherwise. Weapply a recent (5 / ǫ ) hardness-of-approximation result for satisfiable instances for4his problem also due to H˚astad (which holds provided that P = N P ). Unlike in [10],for
Max-Not-2 the natural set of “predicate polynomials” unfortunately yields a varietycontaining many points that can’t be interpreted as solutions to our combinatorial prob-lem. To get a better correspondence, we are forced to include an additional family ofpolynomials in our constructed polynomial system, however, by tailoring the
Max-Not-2 instance in an initial stage, we can bound the relative size of this family. As in [10] weanalyze a random assignment procedure on the ignored literals: the Not-2 predicatesrequire careful case analysis to achieve a high enough rate of expected satisfaction,and again we rely on a property made possible by our initial stage of Not-2-instancetailoring.Next, in Section 2.2, we reuse the high-level structure of our Max-Not-2 proof, butnow leveraging the Max-OXR Problem (yet another form of logical predicate). Herewe use an earlier 2001 hardness result of H˚astad that appears weaker in its inapprox-amability parameter. The form of the OXR predicates is worth the sacrifice: thesepredicates naturally yield polynomial systems of maximum total degree 2. Further,these polynomial systems are still sparse in the sense that every polynomial containsat most 3 variables. Thus, we give the first robust hardness result for Gr¨obner Basiscomputation in polynomial systems of max degree 2: the (4 / ǫ ) -Fractional Gr¨obnerBasis Problem is NP-Hard. ( − ǫ ) Fraction of F May beSelectively Ignored
Recall the definition from [10]. For consistency, we duplicate their notation exactly.For polynomial system F , and subset of variables Y , let F Y denote the subset ofpolynomials from F which contain at least one variable from Y . Definition 1. q -Fractional Gr¨obner Basis Problem . Given as input a set of polynomi-als F over a set of variables X , for specified q ∈ [0 , , output the following: • X ′ ⊆ X , such that |F X ′ | ≤ (1 − q ) |F| . • A Gr¨obner Basis for
F\F X ′ . For q = 1 , this is exactly the traditional Gr¨obner Basis problem for F . As described in[10], the set X ′ corresponds to a set of variables chosen by the algorithm to be ignored :all the polynomials containing variables from X ′ are ignored, and the algorithm needonly compute a Gr¨obner Basis for the remaining set of polynomials F\F X ′ . Theorem 2.
Extending Robust Hardness.
Assume that we are working over a poly-nomial ring K [ x , x , x ...x n ] . For any ǫ > : there is no polynomial-time algorithm A that solves the (7 /
10 + ǫ ) -Fractional Gr¨obner Basis Problem with respect to anylexicographic order (unless P = NP). This statement holds even when F has maximumdegree 3, and each polynomial from F contains at most 3 variables. Before we became aware H˚astad’s 2014 paper, we had proved a conditional version of our main contri-bution here based on an ’09 conditional result of O’Donnell and Wu. O’Donnell and Wu’s version gives thesame inapproxamability factor as [5] but relies on the assumption that Khot’s d − to − Conjecture holdsfor some finite d .
5e prove Theorem 2 by reducing from the
Max-Not-2 Problem for satisfiable in-stances of arity 3 . The input to the Max-Not-2 Problem is a set of logical predicates P over a set of literals L . Specifying arity 3 means that each predicate contains at most3 signed literals (a “signed literal” is just a literal in either negated or positive form),e.g. ( l i , l j , ¬ l k ) where l i , l j , l k ∈ L . For a truth assignment to the literals, a predicateis “satisfied” if the number of its signed literals that are true is not 2. If exactly 2 of itssigned literals are true, then the predicate is not satisfied. For example, the predicate ( l i , l j , ¬ l k ) is satisfied by a truth assignment where l i is true, l j is true, and l k is false(all three of the signed literals in the predicate are true for this truth assignment). Onthe other hand, consider a truth assignment in which l i is true, l j is true, and l k is true:for this truth assignment exactly 2 of the signed literals in the predicate are true, so thepredicate is not satisfied. The objective for the Max-Not-2 Problem is to compute atruth assignment that satisfies the highest possible fraction of predicates in P . To saythat we consider the problem on satisfiable instances means that we receive the inputtogether with a guarantee that there exists some truth assignment for L which satisfiesevery predicate in P .Now consider the recent result of H˚astad: Theorem 3. (H˚astad, ’14) For any δ > , given a satisfiable instance of Max Not-2 ofarity 3, there is no polynomial-time algorithm to find a truth assignment that satisfiesa ( + δ ) -fraction of the predicates (unless P = N P ). As is standard (and will be useful in describing and analyzing our reduction), wecan equivalently take an algebraic view of Max Not-2 and consider each literal l i to bea { , } variable x i (where x i = 1 corresponds to l i true, and x i = 0 corresponds to l i false), and translate each predicate into a sum. If a predicate contains a signed literalin positive form, a positive copy of the corresponding variable is added. If a predicatecontains a signed literal in negated form, a term is added in which the correspondingvariable is subtracted from 1. For example, the predicate ( l i , l j , ¬ l k ) becomes the sum x i + x j + (1 − x k ) . It is easy to check that the original predicate is satisfied exactly when its correspondingsum is not 2 (and hence has total value 0, 1, or 3). The language of our proof will dealwith an input specified in terms of such sums for predicates and variables for literals.Describing an input for the Max-Not-2 Problem of arity 3 we will say each predicate has at most three acceptable totals , and exactly one unacceptable total . Finally, before starting the main proof, we mention that as in [10] our reductionwill use the crucial fact that if we possess a Gr¨obner Basis for a polynomial systemwith respect to a lexicographic order, then a point in the variety can be computed effi-ciently by iteratively eliminating the variables one at a time. In fact, our situation willbe simpler than in [10] because the variety of the polynomial system defined in ourreduction is finite so that all partial solutions extend during the elimination procedure.These classic results in elimination theory are covered in the textbook of Cox, Little If the predicate has fewer than three signed literals, the number of achievable acceptable totals maybebe less than 3.
Proof (Theorem 2):
Suppose, for the sake of contradiction, that the A asserted in Theo-rem 2 does exist with q = (7 / ǫ ) for some fixed ǫ > . Given an arbitrary satisfiableinput ( P , L ) of the Max-Not-2 Problem of arity 3 we compute an assignment of for-bidden quality in polynomial time as follows. Our assignment will be determined overthe course of three stages.
Stage 1. Instance Tailoring.
We remove some predicates and literals from ( P , L ) sothat certain useful properties hold.Iterate through the predicates in P one at a time. If p ∈ P has strictly more thanone signed literal corresponding to a single literal l i , then update ( P , L ) according towhich of the following cases applies:1. If p contains three identical signed literals, then p is trivially satisfied (everyassignment for L satisfies p ). Remove p from P .2. Otherwise, if p contains exactly 2 identical signed literals indexed by i then:(a) If p ’s third signed literal is the other form of l i : every satisfying assignmentfor ( P , L ) must cause p to contain exactly 1 true literal (the only alternativeis 2, which can’t be). Thus, we know unequivocally the value x i musttake in every satisfying assignment. Substitute the forced value of x i intoevery predicate containing a signed form of literal l i . Say x i has beenpermanently fixed. Remove the literal indexed by i from L , and remove p from P .(b) If p has no third signed literal: every satisfying assignment for ( P , L ) mustcause p to contain exactly 0 true literals (the only alternative is 2, whichcan’t be). Thus, we know unequivocally the value x i must take in everysatisfying assignment. Substitute the forced value of x i into every predicatecontaining a signed form of literal l i . Say x i has been permanently fixed.Remove the literal indexed by i from L , and remove p from P .(c) If p ’s third signed literal corresponds to some other index j where x j hasnot yet been fixed, do nothing.(d) If p ’s third signed literal originally corresponded to some other index j where x j has already been permanently fixed to 0 or 1, then do nothing.3. Otherwise, it must be that p contains 2 signed literals corresponding to the sameindex i , but with opposing forms (one negated and one positive). Then:(a) If p ’s third signed literal is also for index i , then it must be identical to oneof p ’s opposing-signed literals: this was already covered in sub-case 2(a)above.(b) If p ’s third signed literal corresponds to some other index j for which x j has not yet been fixed: since p ’s opposing literals have sum 1, we knowunequivocally the value x j must take in every satisfying assignment. Sub-stitute the forced value of x j into every predicate containing a signed form7f literal l j . Say x j has been permanently fixed. Remove the literal indexedby j from L , and remove p from P .(c) If p ’s original third signed literal corresponded to some other index j forwhich x j has already been permanently fixed to 0 or 1: since variables areonly fixed to values we know they must take in every satisfying assignment, x j must have been fixed so that the term in which it appeared in predicate p had value 0. Thus, p will be satisfied by any assignment for x i . Remove p from P .(d) If p never had a third signed literal: p is trivially satisfied (every assignmentfor L satisfies p ). Remove p from P .Call the set of all literals fixed during this procedure L f , and the set of all pred-icates removed from the original P by P r . After executing the above procedure forevery p ∈ P , observe that ( P , L ) has the following property. Property 1.
Any predicate p ′ ∈ P that has multiple occurrences of the same literalmust have a very specific form: either p ′ contains two identical signed literals anda third signed literal corresponding to a different index, or p ′ contains two identicalsigned literals and a third term whose value has been permanently fixed to either 0 or1. These forms correspond to sub-cases 2(c) and 2(d) above: in all other sub-cases, p was removed from P . Further observe that variables were only permanently fixed (and removed from theset of literals) when we could reason unequivocally about the value they must take inevery satisfying assignment. Thus, since the original ( P , L ) was satisfiable, the up-dated ( P , L ) is still satisfiable. Before proceeding, notice also that a predicate wasonly removed from P when we could be certain that it would be satisfied by any as-signment that extends the partial assignment already constructed for permanently-fixedvariables.Next, we make a final update to ( P , L ) . Call any literal l ∈ L which appears in atmost one predicate from P a Loner Literal . Call the set of predicates from P whichcontain at least one Loner Literal by P l . We consider temporarily ignoring predicates in P l until all non-loner literals have been fixed. Consider an arbitrary partial assignmentfor ( x , ..., x |L| ) that fixes every literal variable except for the Loner Literals . If apredicate p ∈ P contains one or more Loner Literals (in either positive or negatedform), then this arbitrary partial assignment may be easily extended to satisfy p : bymanipulating the { , } value of a contained Loner Literal at least 2 distinct total sumsmay be reached for predicate p (this follows from Property 1). At most one of thesetotal sums can be equal to 2, and the other must be some acceptable total for p (suchthat p is satisfied). Since the Loner Literals in p appear in no other predicates (bydefinition), their value may be fixed one-by-one in this way to satisfy all predicatesin P l . Since we can successfully satisfy them at the end, we ignore such predicates If the term in which x j appeared had value 1, then since the sum of the terms for the opposing formsigned literals involving l i is always 1, no choice of x i would satisfy p , and this would contradict thesatisfiability of ( P , L ) . Loner Literal , remove that predicate from P . Nextremove all Loner Literals from L .Notice that these removing the predicates in P l may have caused some additionalliterals to become Loner Literals . Successively remove additional rounds of
Loner -containing predicates and
Loner Literals . Mark each
Loner Literal by the round inwhich it was removed: once we have created a partial assignment for the remainingsystem we will fix the values of the
Loner literals in an order that reverses the order inwhich they were removed from L . Such an ordering ensures that as Loners from eachround are returned to the instance, the argument made in the previous paragraph abouthow to choose their value will always apply.We now have the ( P , L ) that we will argue about for the remainder of the reduc-tion. We make a few observations before starting Stage 2. Property 1 still holds, andas a result of the Loner -removal and
Loner -containing-predicate removal process, wehave:
Property 2.
After the updates in Stage 1, Every literal l ∈ L appears in some form(negated or positive) in at least two predicates from P . Property 2 immediately implies a fact crucial to our subsequent analysis:
Lemma 1.
An instance of Max Not-2 of arity over literals L and predicates P forwhich each literal appears in at least 2 predicates has |P| ≥ ( |P| + |L| ) .Proof of Lemma: If there are |L| literals, and each appears at least twice, then theremust be at least |L| appearances of literals. Each predicate contains at most 3 appear-ances of literals, so at minimum there are |L| / predicates in P : |P| ≥ |L||P||P| + |L| ≥ (cid:16) |L||P| + |L| (cid:17) |P||P| + |L| ≥ (cid:16) − |P||P| + |L| (cid:17) (cid:16) |P||P| + |L| (cid:17) ≥ |P| ≥
25 ( |P| + |L| ) . (cid:3) . Consider the two types of predicates removed from the original instance duringStage 1. Each predicate in P r (those removed in the first part of Stage 1) is alreadyguaranteed to be satisfied by any extension of the partial assignment that has been per-manently fixed on L f . Further, for any partial assignment for literals now remaining in L that leaves the values of Loner Literals (from every round) unassigned, each predi-cate removed for containing a
Loner Literal (in any round) can be efficiently satisfiedby appropriate choices for the
Loner Literals .9ince of the predicates removed from P in Stage 1 can be satisfied in oneof these two ways (in polynomial time), any fraction of the predicates we can satisfyfor the remaining instance ( P , L ) will be a lower bound on the fraction of the originalpredicates that are satisfied by our assignment method. Thus, to get a contradiction, itis sufficient to show that for our remaining instance we can satisfy a ( + ǫ ) fraction ofthe remaining predicates P . We construct such an assignment for the remaining literalsin L over two stages. Stage 2. Polynomial Encoding.
Recall the algebraic view of ( P , L ) . For ease ofexposition, rename the x i so that those that remain in L are indexed consecutivelyfrom 1 to |L| .Define a polynomial system based on ( P , L ) as follows. Create a variable y i cor-responding to the i th literal of L . Denote this new set of variables by Y . These y i willreplicate the x i in the algebraic view of Max Not-2: for each i ∈ { , , ..., |L|} createa polynomial y i (1 − y i ) . This gives a set of |L| “literal polynomials” whose mutualroots are exactly { , } |L| .Next, create a polynomial corresponding to each predicate. In the algebraic viewof Max Not-2, each predicate p corresponds to a sum with at most 3 acceptable totaloutcomes (some subset of { , , } gives the acceptable totals for which the sum cor-responds to a satisfied predicate). These sum expressions have been altered in Stage 1as some of the variable values have been permanently fixed. Each acceptable total forthe sum corresponding predicate p will be used to define a linear term, and the prod-uct of these linear terms will give the polynomial corresponding to predicate p . Eachlinear term is p ’s sum (with the fixed variables from Stage 1 substituted in) minus anacceptable total for the sum. For example, recall that the algebraic view of the Not-2predicate ( l i , l j , ¬ l k ) is the sum x i + x j + (1 − x k ) = 2 If none of x i , x j , x k were fixed in Stage 1, then this yields the following polynomial: (cid:16) y i + y j + (1 − y k ) − (cid:17)(cid:16) y i + y j + (1 − y k ) − (cid:17)(cid:16) y i + y j + (1 − y k ) − (cid:17) Since there are at most 3 acceptable totals for p ’s sum, the polynomial constructed is theproduct of at most 3 linear terms, and hence has degree at most 3. Since each predicatehas at most 3 signed literals, each polynomial will contain at most 3 variables. Noticethat when restricted to the Variety for the set of literal polynomials defined above, 0s for p ’s polynomial correspond exactly to the cases in which p ’s sum takes on an acceptabletotal .All of these observations hold if some of the variables in p ’s sum were already fixedin Stage 1. For example, if in Stage 1, x i , x j were not fixed but x k was fixed to 0, then p ’s sum would be x i + x j + (1 − = 2 and consequently p ’s polynomial constructionwould be: (cid:16) y i + y j + (1 − − (cid:17)(cid:16) y i + y j + (1 − − (cid:17)(cid:16) y i + y j + (1 − − (cid:17) = (cid:16) y i + y j + 1 (cid:17)(cid:16) y i + y j (cid:17)(cid:16) y i + y j − (cid:17) . |P| . Let F denote thesystem of polynomials containing both the literal polynomials and the predicate poly-nomials. Notice that every satisfying assignment for ( P , L ) can be interpreted as apoint in the variety defined by F . In particular, since ( P , L ) is satisfiable, V ( hFi ) isnon-empty.Apply algorithm A to solve the q -Fractional Gr¨obner Basis Problem for F for q =(7 /
10 + ǫ ) for some fixed ǫ > . Let Y ′ denote the variables that A selects to ignore.The set Y ′ was chosen so that |F Y ′ | ≤ (1 − q ) |F| ≤ (3 / − ǫ ) |F|≤ (3 / − ǫ )( |P| + |L| ) ≤ (3 / − ǫ ) (cid:16) |P| (cid:17) ≤ (cid:16) − ǫ (cid:17) |P| The third line follows from the second line due to Lemma 1.Let P D denote the set of predicate polynomials that are in F Y ′ , and P R denote theset of predicate polynomials that are in F\F Y ′ . Clearly |P D | ≤ |F Y ′ | ≤ (cid:16) − ǫ (cid:17) |P| ,so: |P R | = |P| − |P D | ≥ |P| − (cid:16) − ǫ (cid:17) |P| = (cid:16)
14 + 52 ǫ (cid:17) |P| (1)That is, A computes a Gr¨obner Basis with respect to a lexicographic order for theideal generated by the polynomials in F\F Y ′ , and inequality (1) says that at least a ( + ǫ ) fraction of the predicate polynomials must be in F\F Y ′ . The satisfiability of ( P , L ) ensured that V ( hFi ) was non-empty, so V ( hF\F Y ′ i ) is certainly non-empty.Given the Gr¨obner Basis for hF\F Y ′ i with respect to a lexicographic order, a point inthe variety of hF\F Y ′ i can be efficiently computed via successive elimination of thevariables: since the variety is finite (it is a subset of { , } | Y \ Y ′ | ) , all partial solutionsextend, and for each successive variable elimination only 2 options must be checked tofind some y i that works. This point in the variety is a vector y of length | Y \ Y ′ | whichis a mutual zero of all polynomials in F\F Y ′ . Each entry in the vector correspondsto some literal variable in our Max Not-2 instance: if y i is 1 in this vector, assign thecorresponding literal x i to be 1, if y i is 0 in this vector, assign the corresponding literal x i to be 0. The vector y is a mutual zero of polynomials in F\F Y ′ : substituting y into anypredicate polynomial in F\F Y ′ gives 0. By our construction of the predicate poly-nomials, this implies that our partial assignment based on y gives an x that yields an acceptable total for every predicate polynomial in F\F Y ′ . That is, every predicatewhose polynomial is in P R is satisfied by our partial assignment for x . Unlike in [10], because of the inclusion of the literal polynomials, and the fact that by definition
F\F Y ′ contains all the literal polynomials corresponding to variables in Y \ Y ′ , this routine does make an assignmentfor every literal x i corresponding to a y i ∈ ( Y \ Y ′ ). tage 3. Supplemental Random Assignment. The literals corresponding to variablesin Y ′ have not yet been assigned values. For these variables, as in [10], consider arandom independent-fair-coin-flip procedure that assigns x i = 1 with probability ,and x i = 0 with probability . We will argue that regardless of the partial assignmentconstructed for x in Stage 2 (and effectively in Stage 1, through the substitution of fixedvariable values into the predicates), in expectation this random procedure satisfies atleast half of the predicates corresponding to the polynomials in P D . As in [10], such aprocedure can be derandomized via the well-known method of conditional expectationsto obtain a deterministic assignment algorithm with quality that matches the expectedvalue.Let p denote a predicate corresponding to a polynomial in P D ⊆ F Y ′ . From theform of F Y ′ , p contains at least one variable corresponding to a y i ∈ Y ′ . Thus, at leastone of p s signed literals will have truth value determined by the coin-flipping proce-dure. We analyze the probability that p has an acceptable total (and is thus satisfied) atthe end of the random assignment procedure.The key point in the following case analysis is that p has some current achievedsum at the end of Stage 2 (before random coin-flipping-based assignment begins), andfor p to be satisfied p ’s total sum must avoid exactly one unacceptable total (namely2). We’ll call the difference of these two values p ’s forbidden margin . For example,if p enters Stage 3 with no signed literals fixed, p ’s forbidden margin will be 2. If p enters Stage 3 with exactly 2 signed literals fixed to be true, and one not yet fixed, then p ’s forbidden margin will be 0.First, suppose that all signed literals in p correspond to unique literals. The coinflips for variable values are independent, so the possible probability spaces are as fol-lows.Number of Un-fixed Signed Margin: Additional Signed ProbabilityLiterals Entering Stage 3 Literals True After Stage 3 Distribution3 { , , , } ( , , , ) { , , } ( , , ) { , } ( , ) Notice: regardless of the number of un-fixed signed literals in p entering Stage 3,there is no single outcome whose probability is strictly more than . Thus, regardlessof the value of p ’s forbidden margin , the probability that margin is realized is less thanor equal to . Thus, the probability that p is satisfied (equivalently, that p ’s total sum issome acceptable total ) is greater than or equal to .Now suppose that the signed literals in p do not correspond to unique literals. From Property 1 in Stage 1, we have that there are only two possible forms for such apredicate. Again, we argue that in any case, the probability that p reaches an acceptabletotal is at least :1. Suppose that p is two identical signed literals with a third signed literal that hasalready been fixed to 0 or 1 in Stage 1. Since p ∈ P D , the variable correspondingto the two identical signed literals must be un-fixed entering Stage 3. If the fixed12hird signed literal has value 0, then the probability that p avoids a total sum of 2is . If the fixed third signed literal has value 1, then the probability that p avoidsa total sum of 2 is .2. Suppose that p is two identical signed literals with a third term that is a signedliteral corresponding to an unrelated index. Again p has some well defined for-bidden margin , and we compile a table of possible probability spaces.State Entering Stage 3 Margin: Additional Signed ProbabilityLiterals True After Stage 3 Distribution(pair fixed at 0 or 2, single un-fixed) { , } ( , ) (pair un-fixed, single fixed at 0 or 1) { , } ( , ) (pair un-fixed, single un-fixed) { , , , } ( , , , ) Again, regardless of the partial assignment at the end of Stage 2, the probabilityof p ’s forbidden margin being realized in Stage 3 is less than or equal to . Thus,the probability of p being satisfied (by reaching an acceptable total ) is greaterthan or equal to .Thus, for an arbitrary predicate p ∈ P D , the probability it is satisfied at the end ofStage 3 is ≥ / . Since the expectation of the sum is the sum of the expectations, wehave that the expected number of predicates satisfied by the random assignment pro-cedure is at least |P D | / . As in [10], this procedure may be efficiently derandomizedusing the method of conditional expectations.Finally, we have a full assignment for the literals L that satisfies every predicate in P R and at least / of the predicates in P D :Total Predicates We Satisfy ≥ |P R | + |P D | |P R | + ( |P| − |P R | )2= |P R | |P| ≥ (cid:16)
14 + 52 ǫ (cid:17) |P| + |P| ≥ (cid:16)
58 + 54 ǫ (cid:17) |P|≥ (cid:16)
58 + δ (cid:17) |P| for some δ > The transition from the second to the third line applies inequality (1). Since ǫ > ,letting δ = ǫ gives the statement about δ in the final line.This reduction runs in polynomial time, and the final inequality shows that ourmethod exceeds the hardness-of-approximation bound of H˚astad for Max Not-2 (listedas Theorem 3 earlier). Thus we have a contradiction. (cid:3) .13 .2 Extended Hardness Result: Gr¨obner Basis Computation forMaximum Degree 2 is Robustly Hard In this section we prove the first robust hardness result for Gr¨obner Basis computationfor polynomial systems of degree at most 2 (matching the degree bound in the standardNP-hardness result for the tradtional Gr¨obner Basis Problem). Our result is for the q -Fractional Gr¨obner Basis Problem defined by Rolnick and Spencer. Theorem 4.
Robust Hardness For Maximum Degree 2.
Assume that we are workingover a polynomial ring K [ x , x , x ...x n ] . For any ǫ > : there is no polynomial-timealgorithm A that solves the (4 / ǫ ) -Fractional Gr¨obner Basis Problem with respectto any lexicographic order (unless P = NP). This statement holds even when F hasmaximum degree 2, and each polynomial from F contains at most 3 variables. Is robust hardness for maximum-degree-2 polynomial systems unique to the q -Fractional Gr¨obner Basis Model? For additional perspective, consider that existinghardness results for the c -partial Gr¨obner Basis problem of De Leora et al. are basedon graph coloring hardness, and are valid for degree bounds that match the chromaticnumber in the corresponding coloring-hardness result. Since 2-colorings are easy tocompute (when they exist), it seems that a non-trivial result for the c -partial Gr¨obnerProblem in systems of maximum degree 2 would need to take a significantly differentapproach.Our proof of Theorem 4 is closely inspired by our proof of Theorem 2, and willdirectly reuse much of the notation and language introduced there. We rely on a (muchearlier) hardness result due to H˚astad for a problem involving logical predicates of arity3 where the system of predicates is satisfiable. The predicates are now of the followingform: OXR ( q , q , q ) = q ∨ ( q ⊕ q ) (2)Here q , q , q are signed literals which represent positive or negated forms of literalsfrom a set L . This clause is true if at least one of q or ( q ⊕ q ) is true. The secondoption ( q ⊕ q ) is often called an “xor” or “exclusive or.” This exclusive or is truewhen exactly one of q or q is true. Describing (2) above, we will say that q is in the special position of p and that q and q are in the symmetric positions of p . Theorem 5. (H˚astad, ’01) For any δ > , given a satisfiable instance of Max OXR ofarity 3, there is no polynomial-time algorithm to find a truth assignment that satisfiesa ( + δ ) -fraction of the predicates (unless P = N P ). The fraction above is intentionally not given in lowest form: the unreduced fractioncalls attention to the fact that OXR predicates are satisfied by 6 out of 8 of the possibletruth settings for the contained literals.At a structural level, the following proof is very similar to that we gave abovefor Theorem 2, and elimination theory is invoked in the same way. The differencesarise from the form of the logical predicates considered: the pre-processing in Stage1 is slightly different, the polynomial system constructed has predicate polynomials oflower degree, and the form of these polynomials impacts the analysis of the Gr¨obner-Basis-based partial truth assignment, and the performance of the subsequent coin-flip-based part of the truth assignment. 14 roof (Theorem 4 ):
Suppose, for the sake of contradiction, that the A asserted inTheorem 4 does exist with q = (4 / ǫ ) for some fixed ǫ > . Given an arbitrarysatisfiable input ( P , L ) of the Max-OXR Problem of arity 3 we compute an assignmentof forbidden quality in polynomial time as follows. Our assignment will be determinedover the course of three stages.
Stage 1. Instance Tailoring.
Stage 1 removes some predicates and literals from ( P , L ) so that certain useful properties hold.Iterate through the predicates in P one at a time. Consider the signed literals in thesymmetric positions of p : if both of these signed literals correspond to the same literal,then update ( P , L ) according to which of the following cases applies.1. If the signed literals in the symmetric positions of p are identical, then their xormust be false (either both of the symmetric-position signed literals are true, orboth of the symmetric-position signed literals are false). Thus, in every satisfyingassignment the special-position signed literal of p must be true. Let l i denote theliteral corresponding to the signed literal in the special position of p . Substitutethe forced value of l i into every predicate containing a signed form of l i . Say l i has been permanently fixed. Remove l i from L , and remove p from P .2. Otherwise the signed literals in the symmetric positions of p are in opposingforms (one positive, one negated). In this case their xor is true for every possibleassignment. Remove p from P .Call the set of all literals fixed during this procedure L f , and the set of all pred-icates removed from the original P by P r . After executing the above procedure forevery p ∈ P , observe that ( P , L ) now has the following property. Property 1. If p ∈ P , then the two signed literals in the symmetric positions of p correspond to unique literals. Literals were only permanently fixed (and removed) when we could reason un-equivocally about the truth value they must take in every satisfying assignment. Thus,since the original ( P , L ) was satisfiable, the updated ( P , L ) is still satisfiable. A pred-icate was only removed from P when we could be certain that it would be satisfied byany assignment that extends the partial assignment already constructed for L f .As in the previous proof, we make a final update to ( P , L ) to remove literals thatappear in only one predicate (and the predicates that contain such literals). Call anyliteral l ∈ L which appears in at most one predicate from P a Loner Literal . Call theset of predicates from P which contain a Loner Literal by P l . We consider temporarilyignoring predicates in P l until all non-loner literals have been fixed. For p ∈ P l ,suppose that the truth values for all literals in p except one loner-literal l i have alreadybeen fixed. Then: If more than one signed loner literal in p remains unfixed, then fix all but one arbitrarily, then proceed. If l i corresponds to the signed literal in the special position of p , then clearly l i may be set so p ’s special-position signed literal is true (and p is satisfied). Bydefinition, this choice for l i (a loner literal) affects no other predicates. • If l i corresponds to a signed literal in a symmetric position of p , then, regardlessof the truth value of the signed literal in p ’s other symmetric position , thereis a truth assignment for l i that makes p ’s xor true (so that p is satisfied). Bydefinition, this choice for l i affects no other predicates.Since we can successfully satisfy them at the end, we ignore predicates containingloner literals for now: if a predicate contains a Loner Literal , remove that predicatefrom P . Next remove all Loner Literals from L .Removing predicates may cause additional literals to become Loner Literals . Suc-cessively remove additional rounds of
Loner -containing predicates and
Loner Literals .Mark each
Loner Literal by the round in which it was removed: once we have created apartial assignment for the remaining system we will fix the values of the
Loner literals in an order that reverses the order in which they were removed from L such that allloner-containing predicates are satisfied.We now have the ( P , L ) that we will argue about for the remainder of the reduc-tion. As in the previous proof, observe that: Property 2.
After the updates in Stage 1, Every literal l ∈ L appears in some form(negated or positive) in at least two predicates from P . In the proof of Lemma 1, there was no dependence on the form of the predicates(aside from arity), so again, Property 2 implies that |P| ≥ ( |P| + |L| ) .As in the previous proof, since of the predicates removed from P in Stage 1can be polynomial-time satisfied after any partial assignment is fixed for the remaininginstance ( P , L ) , it is sufficient to show that for this remaining instance we can satisfy a ( + ǫ ) fraction of the predicates in polynomial time. We construct such an assignmentfor the remaining literals over two stages. Stage 2. Polynomial Encoding.
Define a polynomial system based on ( P , L ) asfollows. Create a variable y i corresponding to the i th literal of L . Denote this setof variables by Y . Create a polynomial y i (1 − y i ) . This gives a set of |L| “literalpolynomials” whose mutual roots are exactly { , } |L| .Next create a set of predicate polynomials. Consider p ∈ P . We create a polyno-mial corresponding to p as follows: it will be the product of 2 terms. The first termwill correspond to p ’s special position. If the signed literal in the special position of p corresponds to literal l i and appears in positive form, then the first term in p ’s polyno-mial will be ( y i − . If the signed literal in the special position of p corresponds toliteral l i and appears in negated form, then the first term in p ’s polynomial will be ( y i ) .The second term in p ’s polynomial will correspond to p ’s xor. For p s xor to be true,there is a single acceptable sum for variables corresponding to p s symmetric-positionsigned literals. Let l j and l k denote the literals corresponding to the signed literals in From Property 1, this other signed literal is not a form of l i . p . We summarize the construction of the second term of p spolynomial below:Form of p s xor Form of second term in product-defined polynomial for p ( l j ⊕ l k ) ( y j + y k − ¬ l j ⊕ l k ) ((1 − y j ) + y k − l j ⊕ ¬ l k ) ( y j + (1 − y k ) − ¬ l j ⊕ ¬ l k ) ((1 − y j ) + (1 − y k ) − The product of the two described terms gives the polynomial for p . Note that l j and l k must be different literals from Property 2, but that i might match one of j or k .Since the literals in L f already have permanently-fixed truth values, the corresponding0s or 1s are substituted into the predicate polynomials defined above. For example, thepredicate ¬ l i ∨ ( l j ⊕ ¬ l k ) gives polynomial y i ( y j + (1 − y k ) −
1) = y i ( y j − y k ) . For further example, if l i was set to 1 in Stage 1, and l j and l k remain unfixed, then ¬ l i ∨ ( l j ⊕ ¬ l k ) would produce the simple predicate polynomial y j − y k . This constructs a set of predicate polynomials of cardinality |P| . Each predicatepolynomial is the product of two terms each of degree at most 1: each predicate poly-nomial has maximum total degree at most 2. Also, as in the previous reduction, eachpredicate polynomial contains at most 3 variables (corresponding to a limit of threesigned literals per OXRLet F denote the system of polynomials containing both the literal polynomialsand the predicate polynomials. By our construction, every satisfying assignment for ( P , L ) can be interpreted as a point in the variety defined by F . In particular, since ( P , L ) is satisfiable, V ( hFi ) is non-empty.Apply algorithm A to solve the q -Fractional Gr¨obner Basis Problem for F for q =(4 / ǫ ) for some fixed ǫ > . Let Y ′ denote the variables that A selects to ignore.The set Y ′ was chosen so that |F Y ′ | ≤ (1 − q ) |F| ≤ (1 / − ǫ ) |F|≤ (1 / − ǫ )( |P| + |L| ) ≤ (1 / − ǫ ) (cid:16) |P| (cid:17) ≤ (cid:16) − ǫ (cid:17) |P| The third line follows from the second line due to Lemma 1.Let P D denote the set of predicate polynomials that are in F Y ′ , and P R denote theset of predicate polynomials that are in F\F Y ′ . Clearly |P D | ≤ |F Y ′ | ≤ (cid:16) − ǫ (cid:17) |P| ,so: |P R | = |P| − |P D | ≥ |P| − (cid:16) − ǫ (cid:17) |P| = (cid:16)
12 + 52 ǫ (cid:17) |P| (3)17hat is, A computes a Gr¨obner Basis with respect to a lexicographic order for theideal generated by the polynomials in F\F Y ′ , and inequality (3) says that at least a ( + ǫ ) fraction of the predicate polynomials must be in F\F Y ′ . The satisfiability of ( P , L ) ensured that V ( hFi ) was non-empty, so V ( hF\F Y ′ i ) is certainly non-empty.Given the Gr¨obner Basis for hF\F Y ′ i with respect to a lexicographic order, a pointin the variety of hF\F Y ′ i can be efficiently computed via successive elimination of thevariables: since the variety is finite (it is a subset of { , } | Y \ Y ′ | ) , all partial solutionsextend, and for each successive variable elimination only 2 options must be checked tofind some y i that works. This point in the variety is a vector y of length | Y \ Y ′ | whichis a mutual zero of all polynomials in F\F Y ′ . Each entry in the vector correspondsto some literal variable in our Max OXR instance: if y i is 1 in this vector, assign thecorresponding literal l i to be True, if y i is 0 in this vector, assign the correspondingliteral l i to be False. Since y ∈ { , } | Y \ Y ′ | , this routine makes an assignment forevery l i corresponding to a y i ∈ Y \ Y ′ .The vector y is a mutual zero of polynomials in F\F Y ′ : substituting y into anypredicate polynomial in F\F Y ′ gives 0. Consider the factored form of the predicatepolynomials we constructed: such a polynomial evaluates to 0 under our constructedtruth assignment if, and only if, either: • the signed literal corresponding to p s special position is True or • the xor over p s symmetric-position signed literals is True (or both).Thus, since substituting y into any predicate polynomial in F\F Y ′ gives 0, our y -basedpartial truth assignment satisfies every predicate whose polynomial is in P R . Stage 3. Supplemental Random Assignment.
The literals corresponding to variablesin Y ′ have not yet been assigned truth values. For these literals, consider a randomindependent-fair-coin-flip procedure that assigns the literal to be True with probability / , and False with probability / . We argue that regardless of the partial truth as-signment constructed in Stage 2 (and effectively in Stage 1), in expectation this randomprocedure satisfies at least half of the predicates corresponding to the polynomials in P D . Such a procedure can be derandomized via the method of conditional expectationsto obtain a deterministic assignment algorithm with quality that matches the expectedvalue.Let p denote a OXR predicate corresponding to a polynomial in P D ⊆ F Y ′ . Fromthe form of F Y ′ , p contains (some form of) at least one literal corresponding to a y i ∈ Y ′ : before the coin-flip procedure, the truth value of such a literal has not yetbeen decided. Through case analysis, we show that the probability that p is satisfied atthe end of the coin-flip procedure is always at least / . Before the coin-flip procedure,suppose that: • The truth value of the signed literal in p s special position has not yet been de-cided. The coin-flip procedure sets this special-position signed literal to be truewith probability / . Thus, the probability that p is true at the end of the coin-flipprocedure is at least / . 18 Otherwise, the truth value of the signed literal in p s special position has alreadybeen decided. Then either: – Exactly one of the signed literals in a symmetric position of p has not yetbeen decided. Since the signed literal in p ’s other symmetric position has afixed value, the probability that p s xor over the two symmetric positions istrue as a result of the coin-flip procedure is / . Thus, the probability that p is true at the end of the coin-flip procedure is at least / . – Otherwise, both of the signed literals in the symmetric positions of p havenot yet been decided. From Property 2, these symmetric-position signedliterals correspond to unique literals. Thus, the probability that p s xor overthe two symmetric positions is true as a result of the coin-flip procedure is / ( p s xor is satisfied by exactly half of the possible truth assignments).Thus, the probability that p is true at the end of the coin-flip procedure is atleast / .Thus, for an arbitrary predicate p ∈ P D , the probability it is satisfied at the end of Stage3 is ≥ / . Since the expectation of the sum is the sum of the expectations, we have thatthe expected number of predicates from P D that are satisfied by the random assignmentprocedure is at least |P D | / . This procedure may be efficiently derandomized usingthe method of conditional expectations.Finally, we have a full assignment for the literals L that satisfies every predicate in P R and at least / of the predicates in P D :Total Predicates We Satisfy ≥ |P R | + |P D | |P R | + ( |P| − |P R | )2= |P R | |P| ≥ (cid:16)
12 + 52 ǫ (cid:17) |P| + |P| ≥ (cid:16)
68 + 54 ǫ (cid:17) |P|≥ (cid:16)
68 + δ (cid:17) |P| for some δ > The transition from the second to the third line applies inequality (3). Since ǫ > ,letting δ = ǫ gives the statement about δ in the final line.This reduction runs in polynomial time, and the final inequality shows that ourmethod exceeds the hardness-of-approximation bound of H˚astad for Max OXR (listedas Theorem 5 earlier). Thus we have a contradiction. (cid:3) . c -partial Gr ¨obnerBasis Problem In addition to problems which have been proved to be NP-Hard, there are problemswhich are conjectured to be NP-hard. Some of these conjectures have special signifi-19ance because they can be used to prove conditional hardness-of-approximation resultsthat perfectly match the best existing positive algorithmic results for famous combina-torial optimization problems. Khot’s seminal work on the Unique Games Conjecture([6]) included several conjectures on the NP-Hardness of specialized variants of theLabel Cover Problem (for this body of work, Khot was awarded the 2014 Rolf Nevan-linna Prize by the International Mathematical Union). These conjectures have beenclosely scrutinized but little evidence has arisen against them [3]. In this section wegive several results that depend on conditional results of Dinur, et al [3].Given polynomial system F on variable set X , call Y ⊆ X an independent set ofvariables if no pair of variables in Y appear in a common polynomial from F . Recallthe Strong c -partial Gr¨obner Problem as defined by De Loera, et al.: Definition 6. [8] Define the
Strong c -partial Gr¨obner Problem as follows. Given asinput, a set F of polynomials on a set X of variables, output the following: • disjoint X , ..., X b ⊆ X , such that b ≤ c and each X i is an independent set ofvariables, • X ′ ⊆ X where X ′ = X \ ( ∪ bi =1 X i ) (i.e., we have taken away at most c indepen-dent sets of variables), • F ′ ⊆ F such that F ′ consists of all polynomials in F involving only variablesin X ′ , • a Gr¨obner basis for hF ′ i over X ′ (where the monomial order of X is restrictedto a monomial order on X ′ ). Existing hardness guarantees for this problem are parametrized statements: in both[8] and [10] the c for which NP-hardness is proved grows only linearly with the degreebound for F . De Loera et al. give such linear relationship with slope of roughly2/3 that is valid starting from surprisingly low degree bounds (polynomial systems ofdegree at most 3). In contrast, Rolnick and Spencer show that a linear relationshipof arbitrarily high slope becomes valid above some sufficiently-high degree bound. Islinear dependence of c on the degree bound for F necessary for NP-Hardness to hold?The answer may be “no” in quite a dramatic sense. We offer two conditional results: Theorem 7. (Hardness conditioned on Khot) If Khot’s ↔ conjecture is true, thenfor arbitrary constant k , the following problem is NP-Hard: solve the Strong k -partialGr¨obner Problem for some lexicographic order. This statement holds even if F hasmaximum degree 4. Theorem 8. (Hardness conditioned on Dinur et al.) If Dinur et al.s ⊲ < -shaped con-jecture is true, then for arbitrary constant k , the following problem is NP-Hard: solvethe Strong k -partial Gr¨obner Problem for some lexicographic order. This statementholds even if F has maximum degree 3. To prove these conditional results, we use the reduction technique from [10] forthe large-linear-slope case of the Strong c -partial Gr¨obner Basis Problem as a tem-plate. That proof uses a ’94 NP-hardness result of Lund and Yanakakis [9] for a certain20orm of decision problem in coloring known as the Approximate Coloring Problem.A PPROX C OLORING ( q, Q ) is the following decision problem: given a graph G decidebetween the following two alternatives: • χ ( G ) ≤ q , or • χ ( G ) ≥ Q .Lund and Yanakakis showed that for any h > , there exists a k for which A P - PROX C OLORING ( k, hk ) is NP-Hard. Rolnick and Spencer use the ability to solve theStrong c -partial Gr¨obner Basis Problem closely as a black box to efficiently decidebetween these two alternatives. If G is k -colorable, their polynomial-time method out-puts a proper coloring of G by hk − colors. Such a coloring is clearly impossible if χ ( G ) ≥ hk , so the output makes clear which alternative holds.In 2009, Dinur, Mossel and Regev proved in [3] that conjectures about the hardenssof certain Label Cover problems have dramatic implications for the A PPROX C OLOR - ING ( q, Q ) Problem. Substituting the following conditional results due to Dinur et al.in the place of the result of Lund and Yanakakis result in the reduction from [10] im-mediately gives Theorems and respectively. Theorem 9. (Dinur et al. ’09) For any constant
Q > , the ↔ conjecture of Khotimplies that A PPROX C OLORING (4 , Q ) is NP-Hard. This also holds for A PPROX C OLORING ( q, Q ) for any q ≥ . Theorem 10. (Dinur et al. ’09) For any constant
Q > , the ⊲ < conjecture impliesthat A PPROX C OLORING (3 , Q ) is NP-Hard. This also holds for A PPROX C OLORING ( q, Q ) for any q ≥ . We have argued about robust hardness only with respect to lexicographic orders. Em-pirically, other term orders often result in much faster compute times for Buchberger’sAlgorithm. For example, during Buchberger’s Algorithm, lexicographic orders havebeen shown to sometimes produce very-large-degree intermediate polynomials, whileother orders (e.g. revlex, grevlex) have fewer such problems. For other term orders,are there analogs of the hardness results here and in [8], [10]?We conclude with a comment. The family of hardness results our proofs arebased on (and also that [10] is based on) represent each member of the family ofarity-three logical predicate problems which H˚astad identified as the arity-3 cases forwhich the very simple random-coin-flip-based truth assignment achieves an approxi-mation ratio that is tight (even for the case of satisfiable instances). In particular, asimple random assignment matches the hardness results of H˚astad stated in this pa-per as Theorems 3 and 5: deciding truth values by fair coin flips gives a randomized (5 / -approximation algorithm for Max Not-2, a randomized (6 / -approximation al-gorithm for Max OXR, and a randomized (7 / -approximation algorithm for Max 3-SAT. H˚astad proved (6 / -tightness for Max OXR and (7 / -tightness Max 3-SAT in21001 [4], but only completed the family of results in 2014 by proving / is tight forMax Not-2 [5].By viewing this family of satisfiability problems in terms of the strength of theRobust-Gr¨obner-Basis results that their tight hardness-of-approximation guarantees canbe used to produce, we have a new way to understand the relationships between them.Referring back to Figure 1, for example, we see a rather surprising message: though ittook 13 additional years for H˚astad to prove that / is truly tight for Max Not-2 on sat-isfiable instances, our Robust-Gr¨obner-Basis view shows that there is a sense in whichhis 2001 result for Max OXR on satisfiable instances (with weaker parameter / ) isactually stronger. In particular, from an algebraic perspective, our Max-OXR basedresult for maximum degree 2 (our Theorem 4) may seem like a much more surpris-ing statement on Robust Hardness of Gr¨obner-Basis computation (despite the weaker q parameter) than our stronger-parameter result for maximum degree 3 (Theorem 2). References [1] Bruno Buchberger. Bruno buchbergers phd thesis 1965: An algorithm for find-ing the basis elements of the residue class ring of a zero dimensional polynomialideal.
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