Using the PCP to prove fixed parameter hardness

Transcription

1 Using the PCP to prove fixed parameter hardness Mohammad T. Hajiaghayi Rohit Khandekar Guy Kortsarz Abstract In all FPT-hardness result we are aware off, the authors use the conjectures W [1] FPT or W [2] FPT. Let n be the size of the problem. The above assumptions imply that for some parameter k, and for any increasing computable function t, it is not possible to tell between the values of k or k + 1 for a yes or a no instance in time t(k) poly(n). Getting stronger inapproximability requires gap reduction that increases the k versus k + 1 gap to a much larger one. This is similar to proving strong inapproximability under P NP. A very difficult task. This motivates our paper. We suggest a new method for proving FPT-hardness avoiding the k versus k + 1 difficulty. We start with the standard idea of Gap reductions via the PCP which may give a very strong gap. Of course, in almost all such reduction the optimum is very large and close to the size of the input. However we add a new idea: Gap preserving/increasing Reductions that drastically reduces opt. Thus since opt is small now and the gap is still large, thus we can get stronger inapproximability in opt. For that reason our inapproximability for setcover is simpler to prove than the hardness for setcover [2] in FOCS Further, while [2] were not able to give inapproximability for clique, not being able to increase the k + 1 versus k gap to a larger gap, in this case, our paper is stronger. We give a strong hardness for clique. All our reduction have time super exponential in opt. In all our hardness proofs k = opt(i) for some instance I. This does not hold in previous reductions,. In addition the value of k is known, in all our reductions, again a new property. Assuming the eth and the pgc (to be described later) we prove that setcover admits no log 1+c opt ratio, for c > 1 for any algorithm with running in time exp ( opt (logf opt) ) poly(n) for constant f > 0. Under the eth alone, we prove that setcover admits no log opt ratio, in time exp ( ) opt (logf opt) poly(n) for constant f > 0. Under the eth we prove clique admits no (c, t(opt)), approximation, for any constant c in time exp(exp(opt d )) with d a constant that depends on c. Our method indicate that the subject of gap increasing/preserving reductions that make opt small may be important. We use aas a toy problem, the Minimum Maximal Independent Set, (MMIS) problem. This problem is not that relevant for FPT, as its not monotone. However, we provide an algorithm that increases the gap to an arbitrarily large gap in opt, and reduces opt to an arbitrary small function h(opt), is interesting, and may have further applications. In particular we show that MMIS is hard to approximate within (r(opt), t(opt)) for any functions r and t. This hardness is already known in k [5]. However, their proof needs to reduce k, and does not requires to reduce opt. Reducing opt is much harder. 1 Introduction The paper is self contained and both people from approximation and FPT communities should be able to understand the paper. Maryland University. Supported in part by NSF CAREER award , NSF grant CCF , ONR YIP award N , and DARPA/AFOSR grant FA hajiagha@cs.umd.edu. KCG holdings Inc., USA rkhandekar@gmail.com Computer Science department, Rutgers University, Camden. guyk@crab.rutgers.edu Partially supported by NSF grant number

2 1.1 Basic notation When using reductions from 3-SAT to optimization problems P, the number of variables is always denoted by q, and the number of clauses is always denoted by m. The size of the instance I of P is always denoted by n. A function f : {1, 2,...} {1, 2,...} is proper if it is a computable, non decreasing function, We deal with proper function only and so we do not mention that again. We denote by poly(n) an arbitrary polynomial in n and by polylog(n) an arbitrary polynomial in log n. 2 Background from F P T 2.1 FPT-hardness For a minimization (resp, maximization) problem P, of size n and a parameter k an algorithm is called an (r(k), t(k))-fpt-approximation algorithm for P if for any instance I of P with optimum opt, the algorithm either computes a feasible solution for I, with value at most k r(k) (resp, at least k/r(k)) or it computes a certificate that k < opt (resp, k > opt), in time t(k) poly(n). For maximization problems the property k/r(k) = o(k) is required as well. A problem P is (r, t)-fpt-inapproximable (or, (r, t)-fpt-hard) if the problem does not admit an (r, t)-fpt-approximation algorithm. 2.2 The Exponential Time Hypothesis The following is our main assumption. It is due to Impagliazzo, Paturi and Zane, [7]. and to the Sparsification lemma of Calabro, Impagliazzo and Paturi, [1]. The eth conjecture: The Exponential Time Hypothesis is that the 3-SAT problems admits no 2 o(q+m) exact solution, with with q the number of variables in the instance, and m the number of clauses. 2.3 The W [i] hierarchy A minimization (respectively, maximization) problem belongs to FPT if given an instance I of the problem of size n and a parameter k, there is a function t so that we can tell in time t(k) poly(n) if there is a feasible solution of value k or smaller (k or larger). The W -hierarchy is a collection of computational complexity classes. The formal definition of those classes is not important here. What is important is to understand are the implications of a problem being W [i]-hard for i 1. The following containments are known: FPT = W [0] W [1] W [2] W [3],.... It is widely believed that FPT W [i] for any i 1. Therefore W [i]-hard problems for i 1, are widely believed to be fixed parameter intractable. The clique problem is W [1]-hard and the setcover problem is W [2]-hard. Hence we should not at this time hope for an FPT algorithm for these problems. However, we can hope that there may be FPT-approximation. for these problems, if possible. Our paper shows that even a good FPT-approximation for setcover and clique is unlikely. 2.4 An important conjecture of Fellows The following is a conjecture by Fellows. setcover is (r, t)-fpt-hard for any functions r and t. clique is (r, t)-fpt-hard for any functions r and t. We are not aware of a citation for this conjecture. 2

3 3 Background from approximation algorithms 3.1 Gap reductions Definition 3.1 A yes instance of a 3-SAT formula is a satisfiable formula. A no instance is any instance that is not a yes instance. Throughout the paper, we use reductions (functions) from 3-SAT formula S, to some instance I S of some problem P. In all cases, for any satisfiable formula S, the optimum for I S is the same. We denote this optimum for all formulas opt y. The same holds for unsatisfiable instances. The optimum for I S for any unsatisfiable formula is the same, and is denoted opt y. Definition 3.2 A ρ-gap reduction from 3-SAT to a minimization (respectively maximization) problem P is a reduction for which if opt n /opt y > ρ (respectively opy y /opy n > ρ). The value of opt y is known in all our reduction (in fact it is known in almost all existing gap reductions). Corollary 3.3 If there is a ρ-gap reduction from an instance S of 3-SAT to an instance I S of a problem P, and also a ρ approximation for this problem, so that combined running time of both procedures is 2 o(m+q), then the eth fails. 3.2 Setcover and Clique A clique is a set of vertices all two pairs of which are neighbors. clique Input: an undirected graph G(V, E), of size n, and an integer k representing the optimum value. Question: Find in time t(k) poly(n) if there exists clique of size at least k. In the setcover problem the input is a universe U and a collection S of subset {S 1, S 2,...} of S. A set cover is a subset S of S so that the union of the sets in S contains all of U. setcover Input: U, S of size n and a parameter k representing the optimum value Question: Find in time t(k) poly(n) if there is a setcover of size at most k. 3.3 Gap preserving/increasing reductions that decrease opt We will describe the two stage process, for minimization problems, that give a large gap and small opt. For maximization problems a similar reduction holds. 1. Start with a ρ-gap reduction (hopefully with large ρ) from 3-SAT to an instance I of P. Thus if opt y, opt n are the optima of (all) yes and (all) no instances, and opt n /opt y > ρ holds. 2. Apply a gap-preserving/increasing reduction from I to another instance I of P so that I has a much smaller opt. Namely, the new gap ρ satisfies ρ ρ and the new opt opt << opt. 3. Let opt s (y) and opt s (n) be the new optima for the yes and no instance of I. The letter s stands for small. We assign k = opt s (y), namely, the value of the optimum for a yes instance for I with the smaller opt. The following is immediate for any W [i]-hard problem for i 1. Corollary 3.4 Assuming the eth, P is (r(opt s (y)), t(opt s (y))) hard as long as r(opt s (y)) ρ and t(opt s (y)) 2 o(q+m). 3

4 3.4 The advantage of using the PCP theorem for proving fixed parameter inapproximability Under the W [1] FPT or W [2] FPT assumptions, for any W [1] or W [2] hard problem, it is eth-hard to tell between the values k and k+1, in time t(k) poly(n) for any polynomial poly(n). All FPT-hardness results that we are aware off start in the this way. Extending the gap of k versus k + 1 to a large gap is quite challenging (albeit possible in some cases). In [2], it is declared that its hard to use the PCP to prove fixed parameter hardness, as the value of opt in these gap reductions is usually very large. Our paper shows that this is not the case. Our proofs under our new method are much simpler as we remove the k versus k + 1 difficulty. In particular, our setcover-fpt-hardness is much easier than the one in [2] (FOCS 2016). A further evidence that our method is simpler is that [2] were not able (yet?) to prove any hardness for clique, while we give a constant hardness for any constant c even under time double exponential in opt, solving a problem [2] were not able to do. 3.5 Comparing FPT-hardness and regular inapproximability theory. It is known that FPT-inapproximability in k, implies the same inapproximability in opt. Indeed, we prove prove that a ρ approximation in opt implies inapproximability in opt which is equivalent. Say that there is a ρ approximation with the parameter opt(i). It runs in time t(opt)p 1 (n) for some polynomial P 1 (n). We run the algorithm for at most t(k) poly(n) steps. If a solution is not returned by then, the algorithmreject. Note that if opt k clearly (when we get to value opt) a ρ(opt) opt ρ(k) k value solution is returned, in time t(opt) P 1 (opt) t(opt) k t(k). Otherwise k < opt which is allowed to be return in FPT-approximatin. This inapproximability is not relevant for the usual inapproximability theory. It can be that for every distance I k < opt(i) is returned, which does not provide any insight on inapproximability. To avoid such loop holes, we highly recommend to give inapproximability only with with k = opt(i) for some input I, hence avoiding the possibility of returning k < opt. 4 Related work In [3], the paper studies hardness results for clique and setcover with sub-exponential time in opt. After looking into this issue further, we think that any reduction with time that is not super exponential in opt just translates a hardness results to FPT-hardness results. For this reason, [3] has no journal version, and all our reductions in this paper, have time super exponential in opt. Downey et. al [5], proved that there is no constant additive approximation for the the parameterized setcover problem. The Minimum Maximal independent set (MMIS) problem is given G(V, E) of size n and a parameter k find in time t(k) poly(n) if there is an independent set of size at most k that is also a dominating set. In [5] an (r(k), t(k))-hardness is given for MMIS for any two functions r and t. In [11], it is shown that a problem called weighted monotone/antimonotone circuit satisfiability does not admit an (r, t) approximation for any two functions r, t. The most related paper to our paper proves: Theorem 4.1 [2] FOCS 2016 For any function t, setcover admits no log 1/4 k ratio algorithm that runs in time t(k). The authors manage to increase the gap from k versus k+1 to gap log 1/4 k. A remarkable achievement, albeit this reduction is very complex. 4

5 5 Our results From now on exp(t) means in this paper c t for some constant c > 1. In the next statement we cite the projection game conjecture (pgc) that is defined only later. However, the implications can be understood without knowing the conjecture. Theorem 5.1 Under the eth and pgc conjectures, setcover is (r, t)-fpt-hard ( for r(opt) ) = (log opt) γ, for a constant γ > 1, in time t(opt) = exp(exp((log opt) γ )) poly(n) = exp opt (logf opt) poly(n) and f = γ 1. The time here is much larger than just exponential in opt. Further Ω(log opt) hardness follows trivially from the known Ω(log n) inapproximability (under P NP ) for setcover [15]. We prove a stronger hardness as γ > 1. A stronger version of the pgc that assumes a size m P 1 (log log m) PCP implies the following: Theorem 5.2 Under eth and the above stronger version of pgc there exists constants d 1, d 2 > 0 so that setcover is (r, t)-fpt-hard for r(opt) = opt d 1 and t(opt) = exp(exp(opt d 2 )). We are able to prove hardness for setcover under eth alone: Theorem 5.3 Under eth alone, setcover cannot be approximated within c ( ) log opt for some constant c, in time exp opt (log opt)f poly(n) for some constant f. Comparison to the result of [2]: It seems appropriate to compare [2] only to Theorem 5.3 because in this case, both hardness is based on the same assumption, namely, only based on the eth. Theorem 5.3 gives a opt hardness which is stronger than log 1/4 k hardness of [2]. On the other hand [2] proves their hardness for any running time t(k), while our running time is some specific super exponential in opt running time t(opt). Thus, Theorem 5.3 and [2], are incomparable. In [2] no hardness is given (yet?) for clique. We are able to give strong lower bounds for clique due to the simplicity of the new method. Marx [10] raised the question if there is some function t so that clique can be approximated by 2 in time t(op T ). We prove that to get an approximation of 2 for clique, t must be doubly exponential in opt or larger, partially answering the question of Marx. Theorem 5.4 Under the eth, for any constant c, clique is (c, t)-fpt-hard for t(opt) = exp(exp(opt d )) for some constant d that depends on c. It is also (r, t)-fpt-hard for some super constant function r(opt) and t(opt) = exp(exp(opt/p(opt))) for an arbitrarily slowly increasing function p(opt). For MMIS study show how to increase the gap and reduce opt to a arbitrarily small function of opt. We prove: Theorem 5.5 MMIS is (r, t) hard in opt for any two functions r, t. Recall that [5] does not reduce opt. They reduce k. But k is just a parameter. Reducing opt to opt is harder, as there must exist some I so that opt = opt(i ). Remark: Due to space limitations the proofs of Theorem 5.2 and 5.3 are given in an appendix. 6 Nearly linear PCP and their relation to setcover We present a reduction from 3-SAT to setcover. But this is done in stages. First we reduce from the PCP (or 3-SAT) to a problem called Label-Cover. Then we present a reduction from Label-Cover to Min-Rep. And then a reduction from Min-Rep to setcover. We now present these reductions. 5

6 6.1 The PCP theorem In the PCP theorem we have a prover, a verifier and some alphabet Σ. Given a SAT instance, the prover defines a PCP string, that is a function of the SAT instance. Each letter in the PCP string belongs to Σ. The prover sends the PCP to the verifier. The verifier chooses, at random, a constant number of locations in the string and checks their content. Using some known predicate on the letters read, the verifier decides if to accept or reject the string. The completeness is the probability that the verifier accepts if the SAT instance is satisfiable. The soundness is an upper bound on the probability that the verifier accepts a no instance (namely, makes a mistake). The size of the PCP is called almost linear if the string has size m 1+o(1) with m the number of causes in the SAT instance. The following theorem is due to [13]. Theorem 6.1 There exists a a constant c such that for any ɛ 1/m c there exists an almost linear PCP of size m 2 logα m poly(m) poly(1/ɛ), for some constant α < 1, with completeness 1 and soundness ɛ. The verifier inspect the PCP in two places. We choose ɛ = (log 2 m) 2. Thus we get the following corollary. Theorem 6.2 Given a SAT instance of size m, there exists a polynomial P 1 and a reduction from SAT to a PCP of size m 2 logα m P 1 (log m) for some constant α < 1, with completeness 1 and soundness ɛ = 1/ log 2 m. The verifier checks two entries of the PCP. In [13], the alphabet is exponential in 1/ɛ. We use the following conjecture raised by Moshkovitz [12]. This conjecture is called the Projection Game Conjecture. The Projection game conjecture: Given a SAT instance of size m it can be reduced to a PCP of size no larger than m 2 logα m P 1 (log m) with completeness 1 and soundness ɛ = 1/ log 2 m and α < 1. Furthermore, the alphabet size is bounded by a polynomial in 1/ɛ namely by P 2 (log m) for some polynomial P 2. The verifier inspects two entries of the PCP. 6.2 The Labelcover and Min-Rep problems and their relation to the PCP theorem The PCP game can be posed as a graph theoretic problem called Label Cover introduced by Arora and Lund. We adopt the presentation of [8]. The input for the Label Cover problem consists of a bipartite graph G(A, B, E), with an explicit partitioning of each of A and B into q pairwise disjoint subsets, namely A = q i=1 A i and B = q j=1 B j. The bipartite graph G induces a super-graph H as follows. The super-vertices (i.e., the vertices of H) are the 2q sets A i and B j. A super-edge (an edge in H) connects two super-vertices A i and B j if there exist some a A i and b B j which are adjacent in G. A pair (a, b) covers a super-edge (A i, B j ) if a A i and b B j, and a, b are adjacent in G. Let S q i=1 A i q i=1 B j. We say that S covers the super-edge (A i, B j ) if there exist two vertices a, b S such that the pair (a, b) covers the super-edge (A i, B j ). The goal in the Label-Cover problem is to select a single vertex from every {A i }, {B j }, called a representative, and cover the maximum number of super edges. Relation between the PCP theorem and Label Cover: Since in the pgc the verifier inspects the PCP in two places we get a connection between the PCP and Label Cover as follows. Let the length of the PCP be l. We define a bijection f : {A i } {B j } {1, 2,..., l}. Thus every entry in the PCP corresponds to either an A i or to some B j and vise versa. The sets {A i }, {B j } contain all of Σ. Let 1 e(a i ) l and 1 e(b j ) l be the entry in the PCP that correspond to a given A i and B j, respectively. Let val(a i ), val(b j ) be the content of entries e(a i ) e(b j ). The main difference between Label Cover and a PCP is that in a PCP a unique vertex val(a i ), val(b j ) was already selected from {A i }, {B j }. Indeed val(a i ), val(b j ) is the value in entries e(a i ), e(b j ). In the Labelcover problem A i, B j contain all of Σ. Hence the values of the respective entries of the PCP were not chosen yet. This means that a choice of unique vertex (letter from Σ) for every {A i }, {B j } uniquely defines a PCP string. And 6

7 conversely: a PCP string uniquely defines a choice of a vertex a i A i for all i and a vertex b j B j for all j. Note that by the pgc, A i, B j < P 2 (log(m))) for some polynomial P 2. We now define the edges of the Label Cover instances as a function of the PCP in a way that the completeness and soundness correspond to the number of superedges covered, by a unique choice of vertices from every {A i }, {B j }. There exists a superedge A i, B j only if A, B j is a query namely, the verifier queries the entries e(a i ), e(b j ) for some specific result of the random coins. Given a query A i, B j and a i A i, b j B j, we put an edge in the Label Cover graph between a i, b j, if when the verifier inspects e(a i ) and e(b j ) and finds out that the content of e(a i ) is a i and the content of e(b j ) is b j, the verifier accepts. From now on we use the terms queries and superedges interchangeably. Note that all queries are equality likely to be asked. Therefore we can speak about the fraction of superedges covered and not the probability that a superedge is covered. We note now that the completeness of 1 of the PCP implies that we can choose a single vertex of every A i, B j and cover all the superedges. Indeed if we select the unique vertices {val(a i )}, {val(b j )} the verifier accepts for any query. By definition of the Label Cover graph (val(a i ), val(b j )) is an edge in the Label Cover problem. Completeness 1 implies that the above choice of unique vertices from {A i }, {B j } covers all superedges. Now, say that the soundness is µ. This implies that for any PCP the prover accepts a no instance with probability no larger than µ. By the above discussion this means that for any choice of a unique representatives from all {A i }, {B j } (or equivalently, for any PCP) the fraction of superedges covered is at must µ. Using the pgc, we get the following. The number of supervertices in the Label-Cover instance that corresponds to the pgc is the length of the PCP. and so it is bounded by m 2 logα m P 1 (log m). The number of vertices in the Label Cover instance is bounded by m 2 logα m P 1 (log m) P 2 (log m) because each A i, B j has size at most Σ which is at most P 2 (log m). For simplicity we set P 3 (log m) = P 1 (log m) P 2 (log m) The completeness 1 implies that there is a choice of a unique vertex out of every A i and B j that covers a all superedges. The soundness of ɛ = 1/ log 2 m means that any selection of one vertex per super vertex (namely, for any PCP) the fraction of superedges covered is at most 1/ log 2 m. 6.3 The Min-Rep problem The Min-Rep problem is to select a minimum size S A B that covers all superedges. This problem was defined in [8]. It is shown in [8] that if Label-Cover is p(n) hard to approximate, then Min-Rep is p(n) hard to approximate. Corollary 6.3 The Min-Rep instance derived from the pgc is ɛ = log m-inapproximable (recall that we chose ɛ = log 2 m.) The projection property: This property means that for any superedge (query) (A i, B j ), each vertex in B j is adjacent to exactly one vertex (its projection) in A i. Namely, A i B j is a collection of vertex disjoint stars with heads in A i. 6.4 Reducing Min-Rep to setcover [9]: sketch We now finish the reduction from the PCP to setcover. Theorem 6.4 There exists a constant d < 1 and a reduction from the decision version of SAT to a setcover of with m 2 logα m P 3 (log m) sets and with O(m 5 ) elements so that in a yes instance the optimum equals m 2 logα m P 1 (log m) and in the no instance opt n opt y d log 2 n. for some constant d Proof. (Sketch) Every one of the m 2 logα m P 3 (log m) vertices contained in A i B j become a set in the setcover. We will describe the setcover instance as a bipartite graph between sets and elements. To represent a set that contains an element we join the two by an edge. Note that n < m 2 and thus log n and log m are equal up to constant factors. For every superedge A i, B j we add a ground set of elements 7

8 M ij of size m. For every star in A i B j join the head of the star to a random half, U, of M ij (a set and an element are connected iff this set contains the element). All the leaves of the star are joined M ij U. In a yes instance there exists a unique choice of representative from every supervertex so that all superedges A i, B j are covered. This means that if the unique representatives of A i, B j are a i, b j, then a i is the head of a star and b j is one of the leaves of the star. By definition these two sets contain equal sized disjoint halves of M ij. Thus M ij is covered by 2 sets. Thus opt y is just the number of A i, B j namely opt y = m 2 logα m P 1 (log m). On the other hand recall that in a no instance an average of 1/ ɛ = log m vertices are needed to be taken out of every A i, B j in order to cover all superedges. However, if we use a solution of this size, we already loose log m in the approximation. On the other hand, if we take an average log m/p sets from every A i, B j, for a large enough constant p, then few of the superedges are covered. If (A i, B j ) is not covered, then each set in S A i and S B j, contain a random half of M ij (since this superedge is not covered, and therefore for any star, the head of a star and one of its leaves were not both chosen). This random property means that we still need log 2 m sets of A i B j, in order to cover M ij. Thus opt n /opt y log m/2 in any case. Since in our reduction the setcover has size polynomial in m, this gives d ln n hardness for some constant d. Remark: This construction can be derandomized losing a small constant in the gap [14] 7 FPT-hardness for Set Cover with super-exponential time in opt In this section we prove Theorem 5.1. We start with the setcover instance of Theorem 6.4. For simplicity we assume that all terms that follow are integral. Correcting the proof using and is trivial. The idea is to make the optimum much smaller preserving the gap. We introduce a set s S as s = p i=1 s i for each subcollection {s 1, s 2,..., s p } S of size p = m/ log m. Lemma 7.1 The number of sets in the new instance S = (U, S ) is 2 o(m). The new instance can be constructed in time 2 o(m). Proof. Recall that the number of sets is n = m 2 logα m P 1 ((log(m))). The new instance size is at most ( ) n = p ( m 2 log α m P 1 ((log(m))) m/ log m We use the inequality ( n p) (ne/p) p to upper-bound this by ( e 2 logα m P 1 (log m) 2 log m) m/ log m = 2 O(m/ log 1 α m) = 2 o(m). The last equality holds since 0 < α < 1. It is easy to see that the new instance can be created in 2 o(m) time. The new optimum for a yes instance, k = opt y /(m/ log m) = opt y (s) = opt y log m/m and of a no instance opt n (s) = opt n log m/m. Thus the gap opt y (s)/opt y (s) remains at least d log n. And the optimum k = opt y (s) is k = opt y (m) ( = 2 logα m ) log m P 1 ((log(m))). Define r(opt) = (log k) γ = exp k (logf k) poly(s) and t(opt) = exp(exp((log k) γ )) for any 1 < γ < 1/α, as given in Theorem 5.1. Note that r(opt) = Õ((logα m) γ ) = o((log α m) 1/α ) = o(log m) and t(opt) = 2 o(m). Thus, setcover is (r(opt), t(opt))-fpt-hard for these functions, proving Theorem 5.1. ). 8

9 8 Fixed parameter hardness for clique In this section we present the inapproximability for clique. Note that graph products play a crucial role in the proof. Theorem 8.1 [4] There exist a reduction from SAT of size m to a 3-SAT of size S m poly(m), so that for a yes instance all clauses of the 3-SAT can be satisfied, and for a no instance at most 1/2 of the clauses can be simultaneously satisfied Theorem 8.2 [6] There exists a constant ɛ and linear reduction from 3-SAT to clique, of some size S 1 so that for a yes instance opt y = m and for a no instance opt n (1 ɛ)m and the number of vertices is 7 m Clearly S 1 = 7m = O(m poly(m)) because the reduction is linear. Composing the reductions: Corollary 8.3 There exists a reduction from the decision version of 3-SAT to the optimization version of clique, of size S 1, so that the number of vertices in the clique instance is S 1 = O(m poly(log m)) so that opt y /opt n 1/(1 ɛ) for some constant ɛ. The optimum opt y = m for a yes instance is known. Proof of the theorem 5.4 We create a new clique instance and later set k to be the new optimum value k = opt y (s). Let f(m) be any super constant slowly growing function. Introduce a vertex into the new graph for each subset of size S 1 /(b log log m f(m)) vertices in the old clique instance, with b a constant to be defined. Every such set is called a supervertex. Two supervertices A, B, are connected by an edge, if A B is a clique, and A B =. The last condition, namely, the fact that two sets that are connected must be disjoint is not needed in the setcover reduction, but it is crucial here. The constant b is such that poly(log m) (log m) b for the term poly(log m) In Corollary 8.3. Lemma 8.4 The new instance of the clique problem has size 2 o(m). Proof. The size of the original clique is bounded by S 1 = O( m log b m). We take all subsets of size S 1 /(b log log m f(m)). Using ( n p) (ne/p) p, we get that the number of supervertices is at most O(log b m m) m/((b log log m)f(m)) = 2 O(m)/f(m) = 2 o(m). The last inequality is because f(m) = ω(1). The number of edges in the new clique instance, being at most the square of the number of vertices, is also 2 o(m). Lemma 8.5 The gap between a yes and a no instance remains 1/(1 ɛ) Proof. Let ρ = S 1 /(b log log m f(m)). Clearly, opy y (s) = opt y /ρ and opt n (s) = opt n /ρ. Hence the ratio between opt n (s) and opt y (s) does not change and is at least 1/(1 ɛ). The resulting hardness: We got a constant gap in which the optimum (of a yes instance) is k = opt y (s) = O(log m b ). Thus k d = o(log m) for some constant d. Thus t(k) = exp(exp(k d )) = 2 o(m). Hence the problem admits a (c, exp(exp(k d )))-FPT-hardness Graph products: Graph products with parameter i, take a graph of size n and produce a graph of size n i so that the gap grows to (1/(1 ɛ)) i and k = opt y (s) grows to k = opt y (s) i, Taking a constant graph product, we can make r(k) c for any constant c. The running time will still exp(exp(k d ))) except that d is a constant that depends on c. To get super constant r(k) we use graph products with i = (f(m)) and the size of the graph remains 2 o(m) by Claim 8.4. Thus gives a gap of r(k) = (1/(1 ɛ)) f(m). Clearly r(k) is a super constant function. The increase in the size of the optimum mean that t(k) = exp(exp(k/p(k))) for an arbitrarily slowly increasing function p(k). 9

10 9 Arbitrary redution of opt We start with 3-SAT instance I with m clauses and q variables. We assume that a yes instance admits a satisfying assignment and in the case of a no instance, any assignment will leave at least one clause unsatisfied. We now describe how to build the new graph G(I) = (V (I), E(I)). The building blocks: 1. For every variable x in C, we define two vertices u x and ū x. The choice of a vertex u x represents an assignment True to x and the choice of ū x represents a False assignment to x. 2. For every clause we add a set W (C) of q copies of the clause. Namely, W (C) = {w C 1,..., wc q }. Intuitively, we want to create a setcover-like instance in which variables are sets and clauses are elements and a variable u x covers C if x C and ū x covers C if x C. Supervertices: Similar to our construction for setcover and clique, we define a new graph H(I) with supervertices that are collections of vertices of the type u x, ū x. Let f(q) be any slowly increasing function of q such that f(q) = ω(1) and assume, by adding dummy clauses if needed, that f(q) divides q. The supervertices of V (I), denoted by v S, correspond to subsets S {u x x C} {ū x x C} satisfying the following two conditions: 1. S = q/f(q), 2. S does not contain both u z, ū z for any variable z (i.e., a set S does not contain a contradiction in the truth value assignment). Edges between two supervertices: Introduce an edge between v S1 and v S2 if and only if there exists some variable x so that either u x or ū x belongs to S 1 and either u x or ū x belongs to S 2 Note that the above gives four cases in which v S1, v S2 are connected. Edges between supervertices and W (C) vertices: Introduce edges as follows: 1. If a variable x C, any supervertex that contains the vertex u x is connected to all vertices of W (C). 2. If a variable x C, any supervertex that contains ū x is connected to all vertices of W (C). Example: Say for example C = (x z w). Then any supervertex that contains u x is connected to all the copies of W (C). Also, every supervertex that contains ū z or u w is connected to all the copies of W (C). Lemma 9.1 The total number of vertices in H(I) is 2 o(q) + qm. The instance G(I) can be constructed in time 2 o(q). Proof. The total number of vertices in H(I) of type v S for S A is at most ( q q/f(q)) < (qe/(q/f(q))) q/f(q) < 2 o(q). Here we again use the inequality ( q k) (qe/k) k. The number of vertices per W (C) is q for a total of q m. Building an MMIS of size f(q) for a yes instance: 1. Start with the set X = {u x x for a variable x}. This set contains for every variable its vertex copy that corresponds to a True assignment. 2. Decompose X to f(q) pairwise disjoint sets each containing q/f(q) vertices. Let these sets be S 1, S 2,..., S q/f(q). We want to derive sets so that v Si is a feasible MMIS, which is of course not the case so far (because not all W (C) are covered). 10

11 3. We now modify sets S i to obtain sets T i as follows. Fix a satisfying assignment τ to the variables. We start by setting T i = S i for all i. If τ(x) is False, then for the unique i so that u x T i, remove u x from T i and add ū x to T i. This is done for all variables. The final T i sets are called the assignment sets. Our solution will be I = {v Ti T i is an assignment set}. Lemma 9.2 The set {v Ti } is independent in H(I). Proof. For the vertices v Ti, v Tj with i j to be connected it must be that some x so that either u x or ū x belongs to T i and either u x or ū x belongs to T j. Clearly, this implies that u x S i S j. This is a contradiction to the fact that the sets {S p } are pairwise disjoint. Lemma 9.3 The f(q) vertices {v Ti } defined above form a dominating set in H(I) and so k = f(q) = opt y (s). Proof. We first show each vertex in W (C) is adjacent to some vertex v Ti. Note that τ satisfies all clauses C. One possibility is that τ(x) is True and x C. Thus the unique assignment set T i that contains u x is connected to all the copies W (C) of C. Alternatively, if τ(x) is False and x C, the unique T i that contains ū x is connected to all copies of W (C). We now show that I dominates every supervertex not in I. Let v S be a vertex of H(I) that does not belong to I. Pick an arbitrary variable x so that either u x S, or ū x S. By construction there is some assignment set T i I that contains u x or ū x. In all the fours cases above, by definition, there is an edge between v Ti and v S. Thus we just proved the following corollary. Corollary 9.4 The yes instance admits a solution of size k = opt y (s) = f(q). As f is an arbitrarily small proper function, the reduction in opt is the best possible, making our point. Lemma 9.5 For a no instance the minimum MMIS is of size larger opt y (s) larger than f 1 (opt) = q. Proof. Let S be the optimum MMIS of the no instance. Note that all super vertices chosen by the optimum have to be consistent. Namely, we can not have u x belonging to one set T i in S and ū x to some T j S because this will imply an edge between v Ti and v Tj and a contradiction. In particular, this implies that vertices {v Ti } represent a (maybe partial) truth assignment to the variables. Since we are dealing with a no instance, there must be a clause C that is not satisfied by this partial assignment. This means that none of the vertices that correspond to literals that satisfy C are in any set of S. For example if C = (x z w) then there there is a set that contains ū x, one that contains u z and one that contains w. Because the assignment does not satisfy C. There may be one set related to z, but it contains u z, and there may be a set for w, but it contains ū w. This means that the q copies W (C) must be present in S, since it is a maximal independent set. Thus the size of S is at least opt y (s) = q. Theorem 9.6 Assuming the eth, MMIS problem is (r, t)-hardness, for any r, t. Proof. Since the new optimum for a yes instance is k = opt y (s) = f(q) where f is an arbitrarily slow growing function, for any given functions r and t, we can make sure that r(f(k)) < f(q)/q = k/f 1(k) and t(f(k)) = 2 o(q+m). Note that by Corollary 9.4 and Claim 9.5, the gap between yes and no instances is opt y (s)/opt y (s) k/f 1(k) = q/f(q). If there existed an (r, t)-fpt-approximation for MMIS, we could distinguish between a yes and a no instance of 3-SAT in time 2 o(q+m), contradicting the ETH. 11

12 10 An open problem Our paper suggests that it is interesting to develop a theory on gap increasing/preserving reductions that make opt much smaller. We provided a starting point but clearly this subject is not well understood. One motivation for this challenge is proving FPT-hardness via our method. However, it seems to us that this challenge is of independent interest and such reductions may have other applications. Acknowledgement We thanks D. Marx and M. Cygan for helpful discussions. References [1] C. Calabro, R. Impagliazzo, and R. Paturi. A duality between clause width and clause density for sat. In CCC, pages , [2] Y. Chen and B. Lin. The constant inapproximability of the parameterized dominating set problem. FOCS 2016, to appear, [3] R. H. Chitnis, M. Hajiaghayi, and G. Kortsarz. Fixed-parameter and approximation algorithms: A new look. In Parameterized and Exact Computation - 8th International Symposium, IPEC 2013, Sophia Antipolis, France, September 4-6, 2013, Revised Selected Papers, pages , [4] I. Dinur and S. Safra. The importance of being biased. In STOC, pages 33 42, [5] R. G. Downey, M. R. Fellows, C. McCartin, and F. A. Rosamond. Parameterized approximation of dominating set problems. Inf. Process. Lett., 109(1):68 70, [6] U. Feige, S. Goldwasser, L. Lovász, S. Safra, and M. Szegedy. Interactive proofs and the hardness of approximating cliques. J. ACM, 43(2): , [7] R. Impagliazzo, R. Paturi, and F. Zane. Which problems have strongly exponential complexity? In FOCS, pages , [8] G. Kortsarz. On the hardness of approximating spanners. Algorithmica, 30(3): , [9] C. Lund and M. Yannakakis. On the hardness of approximating minimization problems. J. ACM, 41(5): , [10] D. Marx. Parameterized complexity and approximation algorithms. Comput. J., 51(1):60 78, [11] D. Marx. Completely inapproximable monotone and antimonotone parameterized problems. J. Comput. Syst. Sci., 79(1): , [12] D. Moshkovitz. The projection games conjecture and the NP-hardness of ln n-approximating setcover. In APPROX, pages , [13] D. Moshkovitz and R. Raz. Two-query PCP with subconstant error. J. ACM, 57(5), [14] M. Naor, L. J. Schulman, and A. Srinivasan. Splitters and near-optimal derandomization. In 36th Annual Symposium on Foundations of Computer Science, Milwaukee, Wisconsin, October 1995, pages , [15] R. Raz and S. Safra. A sub-constant error-probability low-degree test, and a sub-constant errorprobability PCP characterization of NP. In STOC, pages ,

13 A Stronger hardness for setcover via a stronger assumption In this section we prove Theorem 5.2 Let P 1 and P 2 be polynomial as in Corollary 6.4. We assume Conjecture A.1 There exists a constant c > 0 and a pcp of size m polylog(m)p 1 (1/ɛ), for any ɛ so that ɛ 1/m c. This following was conjectured to hold by Moshkovitz in a private communication. We now use the above conjecture and show a much stronger FPT inapproximability for setcover. By Corollary 6.4, and the above conjecture we get the following corollary, using ɛ = c / log 2 m for a large enough constant c : Corollary A.2 There exists a a reduction from 3-SAT to setcover so that: 1. The number of sets is σ = m P 1 (log(m)) 2. The number of elements is poly(m). 3. The value of the optimum in yes instance is exactly opt y = m P 2 ((log(m))) and that in the no instance is at least log m opt y. A proof of along the lines of the proof of Theorems 5.1 and 5.4 gives that setcover is (k d, exp(exp(k d 2 )))- FPT-hard. for some constant d 2. B Hardness for Setcover under the ETH alone The following is proved in [13]. Let P 1 (m) and P 2 (m) be two polynomials as in Corollary 6.4. Theorem B.1 There exists a constant c and a pcp of size m 2 logα m poly(1/ɛ), such that the size of the alphabet is at most exp(1/ɛ) and the gap that can be chosen to be 1/ɛ for any ɛ > 1/m c. Proof. We choose ɛ = ln 2 log α m. Note that exp(ɛ) = 2 logα m. Therefore when using the reduction from 3-SAT to setcover described in Corollary 6.4 we get: Corollary B.2 There exists a constant d > 0, and a constant 0 < α < 1 and a reduction from 3-SAT to setcover so that: 1. The number of sets is m 2 2 logα m 2. The number of elements is poly(m). 3. The gap is d log α m. 4. The optimum of a yes instance does not change, namely, is opt y (s) = 2 logα m poly log(m). Note that the optimum does not change because we still select one vertex out of every supervertex. The gap is d log α k for some constant d. k = opt y (s) = 2 logα k. Thus for some constant c, the problem is (c log ( ) k, exp k (logf k) )-fpt hard for the same constant f > 0, that appears in Theorem