Parameterized Complexity of MaxSat Above Average
Robert Crowston, Gregory Gutin, Mark Jones, Venkatesh Raman, Saket Saurabh
aa r X i v : . [ c s . CC ] D ec Parameterized Complexity of MaxSat Above Average ∗ Robert Crowston † Gregory Gutin † Mark Jones † Venkatesh Raman ‡ Saket Saurabh ‡ This paper is dedicated to the memory of Alan Turing
Abstract In MaxSat , we are given a CNF formula F with n variables and m clauses andasked to find a truth assignment satisfying the maximum number of clauses.Let r , . . . , r m be the number of literals in the clauses of F . Then asat( F ) = P mi =1 (1 − − r i ) is the expected number of clauses satisfied by a random truthassignment (the truth values to the variables are distributed uniformly and in-dependently). It is well-known that, in polynomial time, one can find a truthassignment satisfying at least asat( F ) clauses. In the parameterized problem MaxSat-AA , we are to decide whether there is a truth assignment satisfy-ing at least asat( F ) + k clauses, where k is the (nonnegative) parameter. Weprove that MaxSat-AA is para-NP-complete and thus,
MaxSat-AA is notfixed-parameter tractable unless P=NP. This is in sharp contrast to the sim-ilar problem
MaxLin2-AA which was recently proved to be fixed-parametertractable by Crowston et al. (FSTTCS 2011).In fact, we consider a more refined version of
MaxSat-AA , Max- r ( n ) -Sat-AA , where r j ≤ r ( n ) for each j . Alon et al. (SODA 2010) proved thatif r = r ( n ) is a constant, then Max- r -Sat-AA is fixed-parameter tractable.We prove that Max- r ( n ) -Sat-AA is para-NP-complete for r ( n ) = ⌈ log n ⌉ . Wealso prove that assuming the exponential time hypothesis,
Max- r ( n ) -Sat-AA is not in XP already for any r ( n ) ≥ log log n + φ ( n ), where φ ( n ) is any un-bounded strictly increasing function. This lower bound on r ( n ) cannot bedecreased much further as we prove that Max- r ( n ) -Sat-AA is (i) in XP forany r ( n ) ≤ log log n − log log log n and (ii) fixed-parameter tractable for any r ( n ) ≤ log log n − log log log n − φ ( n ), where φ ( n ) is any unbounded strictlyincreasing function. The proof uses some results on MaxLin2-AA . Satisfiability is a well-known fundamental problem in Computer Science. Its opti-mization version (finding the maximum number of clauses that can be satisfied by atruth assignment) and its generalizations (constraint satisfaction problems) are well ∗ A preliminary version of this paper will appear in the proceedings of LATIN 2012 † Royal Holloway, University of London, Egham, Surrey, UK ‡ The Institute of Mathematical Sciences, Chennai 600 113, India
MaxSat .In parameterized complexity, one identifies a natural parameter k in the input andalgorithms are designed and analyzed to confine the combinatorial explosion to thisparameter, while keeping the rest of the running time to be polynomial in the sizeof the input. More specifically, the notion of feasibility is fixed-parameter tractability where one is interested in an algorithm whose running time is O ( f ( k ) n c ), where f is an arbitrary (typically exponential) function of k , n is the input size and c is aconstant independent of k . When the values of k are relatively small, fixed-parametertractability implies that the problem under consideration is tractable, in a sense.The class of fixed-parameter tractable problems will be denoted by FPT. When theavailable running time is replaced by the much more powerful n O ( f ( k )) , we obtain theclass XP, where each problem is polynomial-time solvable for any fixed value of k. It is well-known that FPT is a proper subset of XP. More details on parameterizedalgorithms and complexity are given at the end of this section.A well-studied parameter in most optimization problems is the size of the solu-tion. In particular, for
MaxSat , the natural parameterized question is whether agiven boolean formula in CNF has an assignment satisfying at least k clauses. Usingthe (folklore) observation that every CNF formula on m clauses has an assignment sat-isfying at least m/ m/ m/
2, for the maximum number of clauses, means that theproblem is interesting only when k > m/
2, i.e., when the values of k are relativelylarge. Hence Mahajan and Raman introduced and showed fixed-parameter tractable,a more natural parameterized question, namely whether the given CNF formula hasan assignment satisfying at least m/ k clauses.This idea of parameterizing above a (tight) lower bound has been followed upin many directions subsequently. For MaxSat alone, better (larger than m/
2) lowerbounds for certain classes of instances (formulas with no pair of unit clauses in conflict,for example) have been proved and the problems parameterized above these boundshave been shown to be fixed-parameter tractable [4, 13]. When every clause has r literals, the expected number of clauses that can be satisfied by a (uniformly)random assignment can easily seen to be (1 − / r ) m , and Alon et al. [1] proved thatchecking whether k more than this many clauses can be satisfied in such an r -CNFformula is fixed-parameter tractable. This problem is known to be Max- r -Sat-AA .The problem Max- r ( n ) -Sat-AA we consider in this paper, is a refinement of thisproblem, where r need not be a constant.More specifically, the problem MaxSat-AA , we address is the following. MaxSat-AA
Instance:
A CNF formula F with clauses c , . . . , c m , and variables x , . . . , x n ,and a nonnegative integer k . Clause c i has r i literals, i = 1 , . . . , m . Parameter: k . Question:
Decide whether there is a truth assignment satisfying at leastasat( F ) + k clauses, where asat( F ) = P mi =1 (1 − − r i ) . In this paper, AA is an abbreviation for Above Average
Max- r ( n ) -Sat-AA is a refinement of MaxSat-AA in which each clausehas at most r ( n ) literals.Note that asat( F ) is the average number of satisfied clauses. Indeed, if we as-sign true or false to each x j with probability 1 / c i being satisfied is 1 − − r i , and by linearity ofexpectation, asat( F ) is the expected number of satisfied clauses. (Since our dis-tribution is uniform, asat( F ) is indeed the average number of satisfied clauses.)Let sat( F ) denote the maximum number of clauses satisfied by a truth assignment.For Boolean variables y , . . . , y t , the complete set of clauses on y , . . . , y t is the set { ( z ∨ . . . ∨ z t ) : z i ∈ { y i , ¯ y i } , i ∈ [ t ] } . Any formula F consisting of one or more com-plete sets of clauses shows that the lower bound sat( F ) ≥ asat( F ) on sat( F ) is tight.Using the derandomization method of conditional expectations (see, e.g., [2]), it iseasy to obtain a polynomial time algorithm which finds a truth assignment satisfyingat least asat( F ) clauses. Thus, the question asked in MaxSat-AA is whether we canfind a better truth assignment efficiently from the parameterized complexity point ofview.
New Results.
Solving an open problem of [6] we show that
MaxSat-AA is notfixed-parameter tractable unless P=NP. More specifically, we show this for
Max- r ( n ) -Sat-AA for r ( n ) = ⌈ log n ⌉ . Also, we prove that unless the exponential time hypothesis(ETH) is false,
Max- r ( n ) -Sat-AA is not even in XP for any r ( n ) ≥ log log n + φ ( n ),where φ ( n ) is any unbounded strictly increasing function . These two results areproved in Section 2.These results are in sharp contrast to the related problems MaxLin2-AA (seeSection 3 for a definition of this problem) and
Max- r -Sat-AA which are known tobe fixed-parameter tractable. Also this is one of the very few problems in the ‘aboveguarantee’ parameterization world, which is known to be hard. See [19] for a fewother hard above guarantee problems.Then, complementing our hardness results, we show that the lower bound aboveon r ( n ) cannot be decreased much further as we prove that Max- r ( n ) -Sat-AA isin XP for any r ( n ) ≤ log log n − log log log n and fixed-parameter tractable for any r ( n ) ≤ log log n − log log log n − φ ( n ), where φ ( n ) is any unbounded strictly increasingfunction. This result generalizes the one of Alon et al. [1] and is proved in Section 3.The problem we study is one of the few problems in the ‘above guarantee param-eterization’ where we parameterize above an instance-specific bound, as opposed toa generic bound, see [19] for a discussion on this issue. Another example of such pa-rameterizations is the problem Vertex Cover parameterized above the maximummatching of the given graph. See [7, 21] for recent results on this problem.We complete this paper in Section 4 by stating an open problem on permutationconstraint satisfaction problems parameterized above average.
Basics on Parameterized Complexity.
A parameterized problem Π can beconsidered as a set of pairs (
I, k ) where I is the problem instance and k (usually anonnegative integer) is the parameter . Π is called fixed-parameter tractable (fpt) ifmembership of ( I, k ) in Π can be decided by an algorithm of runtime O ( f ( k ) | I | c ),where | I | is the size of I , f ( k ) is an arbitrary function of the parameter k only, and c A function f is strictly increasing if for every pair x ′ , x ′′ of values of the argument with x ′ < x ′′ ,we have f ( x ′ ) < f ( x ′′ ) .
3s a constant independent from k and I . Such an algorithm is called an fpt algorithm.Let Π and Π ′ be parameterized problems with parameters k and k ′ , respectively. An fpt-reduction R from Π to Π ′ is a many-to-one transformation from Π to Π ′ , such that(i) ( I, k ) ∈ Π if and only if ( I ′ , k ′ ) ∈ Π ′ with k ′ ≤ g ( k ) for a fixed function g , and (ii) R is of complexity O ( f ( k ) | I | c ).Π belongs to XP if membership of ( I, k ) in Π can be decided by an algorithm ofruntime | I | O ( f ( k )) , where f ( k ) is an arbitrary function of the parameter k only. It iswell-known that FPT is a proper subset of XP [8, 9, 20].Π is in para-NP if membership of ( I, k ) in Π can be decided in nondeterministictime O ( f ( k ) | I | c ), where | I | is the size of I , f ( k ) is an arbitrary function of the pa-rameter k only, and c is a constant independent from k and I . Here, nondeterministictime means that we can use nondeterministic Turing machine. A parameterized prob-lem Π ′ is para-NP-complete if it is in para-NP and for any parameterized problem Πin para-NP there is an fpt-reduction from Π to Π ′ . It is well-known that a param-eterized problem Π belonging to para-NP is para-NP-complete if we can reduce anNP-complete problem to the subproblem of Π when the parameter is equal to someconstant [9].For example, consider the k -Colorability problem, where given a graph G anda positive integer k ( k is the parameter), we are to decide whether G is k -colorable.Since the (unparameterized) Colorability problem is in NP, k -Colorability isin para-NP. k -Colorability is para-NP-complete since 3 -Colorability is NP-complete.For further background and terminology on parameterized algorithms and com-plexity we refer the reader to the monographs [8, 9, 20].For an integer n , [ n ] stands for { , . . . , n } . In this section we give our hardness results. For our results we need the followingproblem as a starting point for our reductions.
Linear-3-Sat
Instance:
A 3-CNF formula F with clauses c , . . . , c m , and variables x , . . . , x n such that m ≤ cn for some fixed constant c . That is, num-ber of clauses in F is linear in the number of variables. Question:
Decide whether there is a truth assignment satisfying F .It is well known that Linear-3-Sat is NP-complete. For example, the well-knowntheorem of Tovey [22] states that the problem is NP-complete even when theinput consists of 3-CNF formula with every variable contained in at most four clauses.
Theorem 1.
Max- r ( n ) -Sat-AA is para-NP-complete for r ( n ) = ⌈ log n ⌉ .Proof. Max- r ( n ) -Sat-AA is in para-NP as given a truth assignment for an instanceΦ of Max- r ( n ) -Sat-AA , we can decide, in polynomial time, whether the assignmentsatisfies at least asat(Φ) + k clauses. To complete our proof of para-NP-completeness,we give a reduction from Linear-3-SAT to Max- r ( n ) -Sat-AA with k = 2.4onsider a formula F with n variables, x , . . . , x n , and m distinct clauses c , . . . , c m . Since F is an input to Linear-3-Sat , we may assume that m ≤ cn forsome positive constant c .We form a Max- r ( n ) -Sat-AA instance F ′ with n ′ = 2 cn variables, the existingvariables x , . . . , x n , together with new variables y , . . . , y n ′ − n , and m ′ = 2 ⌈ log n ′ ⌉ +1 clauses. The set of clauses of F ′ consists of three sets, C , C and C , describedbelow: • C is the complete set of clauses on variables y , . . . , y ⌈ log n ′ ⌉ without the clauseconsisting of all negative literals, ¯ c = ( ¯ y ∨ ¯ y ∨ . . . ∨ ¯ y ⌈ log n ′ ⌉ ). • C = { c i ∨ ¯ y ∨ . . . ∨ ¯ y ⌈ log n ′ ⌉ : i ∈ [ m ] } . • C is a set of m ′ − | C | − | C | clauses on the variables y ⌈ log n ′ ⌉ +1 , . . . , y n ′ − n , oflength ⌈ log n ′ ⌉ such that each variable appears in at least one clause and everyclause consists of only positive literals.We claim that F is satisfiable if and only if F ′ is a Yes -instance of
Max- r ( n ) -Sat-AA for k = 2, thus completing the proof. Since, in F ′ , each clause is of length ⌈ log n ′ ⌉ , we have asat( F ′ ) = (1 − /m ′ ) m ′ = m ′ − . Thus,
Max- r ( n ) -Sat-AA for k = 2 asks whether all the clauses of F ′ can be satisfied.Suppose F is satisfied by a truth assignment x . Extend this assignment to thevariables of F ′ by assigning all y i to be true . Since F is satisfied, all the clauses in C are satisfied. Every clause in C and C contains at least one positive literal, andso is satisfied. Hence F ′ is satisfied.If F ′ is satisfied, then y , . . . , y ⌈ log n ′ ⌉ must all be set to true (otherwise, there isa clause in C that is not satisfied). As a result, the set C of clauses can be simplifiedto the formula F , and thus F must be satisfied.It is not hard to prove Theorem 1 without starting from Linear-3-SAT . We use
Linear-3-SAT to ensure that n ′ = O ( n ) which is necessary in the proof of the nexttheorem. Hereafter, we will assume the Exponential Time Hypothesis (ETH), whichis stated below.Exponential Time Hypothesis (ETH) [16]: There is a positive real s suchthat 3-SAT cannot be solved in time 2 sn n O (1) . Here n is the number ofvariables.Using the sparsification lemma [16, Corollary 1], one may assume that in the inputformula F to , every variable occurs in at most p clauses for some positiveconstant p . For completeness we sketch the proof here. Lemma 1 ([16]) . (Sparsification Lemma)
For every ε > and positive integer r ,there is a constant C so that any r -CNF formula F with n variables, can be expressedas F = ∨ ti =1 Y i , where t ≤ εn and each Y i is an r -CNF formula with at most Cn clauses. Moreover, this disjunction can be computed by an algorithm running in time εn n O (1) . Proposition 1 ([16]) . Assuming the ETH, there is a positive real s ′ such that Linear3-SAT cannot be solved in time O (2 s ′ n ) . roof. Suppose the proposition does not hold. That is, for all positive real s ′ Linear3-SAT can be solved in time O (2 s ′ n ). Now consider a 3-CNF formula F . Using thealgorithm for Linear 3-SAT we will show that for every positive real c , 3-SAT canbe solved in time 2 cn n O (1) , contradicting the ETH. Let ε ′ < c/ . Using SparsificationLemma, we produce linear 3-CNF formulas Y , . . . , Y t , t ≤ ε ′ n , in time 2 ε ′ n n O (1) . Wecan solve all of them in time 2 ε ′ n n O (1) = 2 cn n O (1) and so obtain a solution for F intime O (2 cn ), a contradiction.Thus, it shows that Linear-3-SAT cannot be solved in time 2 o ( n ) unless ETHfails. Using the ETH, we strengthen Theorem 1. Theorem 2.
Assuming the ETH,
Max- r ( n ) -Sat-AA is not in XP for any r ( n ) ≥ log log n + φ ( n ) , where φ ( n ) is any unbounded strictly increasing function of n .Proof. Let φ ( n ) be an unbounded strictly increasing function of n . Note that if thetheorem holds for some unbounded strictly increasing function ψ ( n ) of n , it holdsalso for any strictly increasing function ψ ′ ( n ) such that ψ ′ ( n ) ≥ ψ ( n ) for every n ≥ . Thus, we may assume that φ ( n ) ≤ log log n .Consider a instance I with a linear number of clauses, and reduce it toa Max- r ( n ′ ) -Sat-AA instance with r ( n ′ ) = ⌈ log n ′ ⌉ as in the proof of Theorem 1.Note that I has O ( n ′ ) variables. Let F ′ be the formula of the Max- r ( n ′ ) -Sat-AA instance and let n be the maximum integer such that log n ′ ≥ log log n + φ ( n ). Add n − n ′ new variables to F ′ together with a pair of contradicting unit clauses, ( x ), (¯ x )for each new variable. Let F denote the resulting formula. The total number n ofvariables in F is such that r ( n ) = ⌈ log n ′ ⌉ ≥ log log n + φ ( n ). Note that n ≤ n ′ / φ ( n ) ≤ n ′ / φ ( n ′ ) = 2 o ( n ′ ) , and hence it takes 2 o ( n ′ ) time to construct F .Observe that for any value of k , F ′ is a Yes -instance of
Max- r ( n ′ ) -Sat-AA ifand only if F is a Yes -instance of
Max- r ( n ) -Sat-AA . We established in the proof ofTheorem 1 that F ′ is a Yes -instance of
Max- r ( n ′ ) -Sat-AA for k = 2 if and only ifthe instance is satisfiable. Thus, F is a Yes -instance of
Max- r ( n ) -Sat-AA for k = 2 if and only if the instance I is satisfiable. Therefore, if there wasan XP algorithm for Max- r ( n ) -Sat-AA , then for k = 2 it would have running time n O (1) = 2 o ( n ′ ) , contradicting the ETH. To prove the main result of this section, Theorem 4, we reduce
Max- r ( n ) -Sat-AA to Max- r ( n ) -Lin2-AA defined below.In the problem MaxLin2 , we are given a system S consisting of m equations in n variables, where each equation is of the form Q i ∈ I x i = b , b ∈ {− , } and eachvariable x i may only take a value from {− , } . Each equation is assigned a positiveintegral weight and we wish to find an assignment of values to the variables in orderto maximize the total weight of satisfied equations.Let W be the sum of the weights of all equations in S and let sat( S ) be themaximum total weight of equations that can be satisfied simultaneously. Note that6 / S ). Indeed, consider choosing assignments tothe variables uniformly at random. Then W/ /
2) and so is a lowerbound; to see the tightness consider a system consisting of pairs of equations of theform Q i ∈ I x i = − , Q i ∈ I x i = 1 of weight 1, for some non-empty I ⊆ [ n ]. This leadsto the following parameterized problem: Max- r ( n ) -Lin2-AA Instance:
A system S of equations Q i ∈ I j x i = b j , where b j ∈ {− , } , | I j | ≤ r ( n ), x i ∈ {− , } , and j ∈ [ m ], in which Equation j is assigneda positive integral weight w j , and a nonnegative integer k . Let W = P mj =1 w j . Parameter: k . Question:
Decide whether sat( S ) ≥ W/ k .The excess for x = ( x , . . . , x n ) ∈ {− , } n over S is ε S ( x ) = 12 m X j =1 c j Y i ∈ I j x i , where c j = w j b j . Observe that ε S ( x ) is the difference between the total weight ofequations satisfied by x and W/
2. Thus, the answer to
MaxLin2-AA is Yes if andonly if ε S ( x ) ≥ k for some x .Consider two reduction rules for MaxLin2 introduced in [14].
Reduction Rule 1.
Let A be the matrix over F corresponding to the set of equationsin S , such that a ji = 1 if variable x i appears in equation e j , and otherwise. Let t = rank A and suppose columns a i , . . . , a i t of A are linearly independent. Then deleteall variables not in { x i , . . . , x i t } from the equations of S . Reduction Rule 2.
If we have, for a subset I of [ n ] , an equation Q i ∈ I x i = b ′ I withweight w ′ I , and an equation Q i ∈ I x i = b ′′ I with weight w ′′ I , then we replace this pair byone of these equations with weight w ′ I + w ′′ I if b ′ I = b ′′ I and, otherwise, by the equationwhose weight is bigger, modifying its new weight to be the difference of the two oldones. If the resulting weight is 0, we delete the equation from the system. The two reduction rules are of interest due to the following:
Lemma 2. [14] Let S ′ be obtained from S by Rule 1 or 2. Then the maximum excessof S ′ is equal to the maximum excess of S . Moreover, S ′ can be obtained from S intime polynomial in n and m . Using techniques from linear algebra, the authors of [6] showed the following:
Theorem 3.
Let J be an instance of Max- r ( n ) -Lin2-AA in which system S isreduced with respect to Rules 1 and 2. If n ≥ (2 k − r ( n ) + 1 then the answer to J is Yes . Let I be an instance of MaxSat-AA given by a CNF formula F with clauses c , . . . , c m , and variables x , . . . , x n . It will be convenient for us to denote true and7 alse by − x ∈ {− , } n , the excess ε I ( x ) for x is the number of clauses satisfied by x minus asat( x ). Thus, the answerto I is Yes if and only if there is an assignment x with ε I ( x ) ≥ k. Max- r ( n ) -Sat-AA is related to Max- r ( n ) -Lin2-AA as follows. Results similarto Lemma 3 have been proved in [1, 5]. Lemma 3.
Let I be an instance of Max- r ( n ) -SAT-AA with n variables, m clausesand parameter k . Then in time r ( n ) m O (1) we can produce an instance J of Max- r ( n ) -Lin2-AA with parameter k r ( n ) − such that I is a Yes -instance if and only if J is a Yes -instance and J is reduced by Rule 2. Moreover, for any truth assignment x ∈ {− , } n , ε J ( x ) = ε I ( x ) · r ( n ) − .Proof. Let I be an instance of Max- r ( n ) -SAT-AA with clauses c , . . . , c m and vari-ables x , . . . , x n . For a clause c j , var( c j ) will denote the set of variables in c j and r j the number of literals in c j . For every j ∈ [ m ], let h j ( x ) = 2 r ( n ) − r j [1 − Y x i ∈ var( c j ) (1 + d ij x i )] , where d ij = 1 if x i is in c j and d ij = − x i is in c j .Let H ( x ) = P mj =1 h j ( x ) . We will prove that for a truth assignment x ∈ {− , } n ,we have H ( x ) = 2 r ( n ) ε I ( x ) . (1)Let q j = 1 if c j is satisfied by x and q j = 0, otherwise. Observe that h j ( x ) / (2 r ( n ) − r j )equals 1 − r j if q j = 0 and 1, otherwise. Thus, H ( x ) = P mj =1 [2 r ( n ) − r j q j + (2 r ( n ) − r j − r ( n ) )(1 − q j )]= 2 r ( n ) [ P mj =1 q j − P mj =1 (1 − − r j )]= 2 r ( n ) ε I ( x ) . It follows from (1) that the answer to I is Yes if and only if there is a truthassignment x such that H ( x ) ≥ k r ( n ) . (2)Algebraic simplification of H ( x ) will lead us to the following: H ( x ) = X S ∈F c S Y i ∈ S x i , (3)where F = {∅ 6 = S ⊆ [ n ] : c S = 0 , | S | ≤ r ( n ) } . The simplification can be done intime 2 r ( n ) m O (1) .Observe that by replacing each term c S Q i ∈ S x i with the equation Q i ∈ S x i = 1 if c S ≥ Q i ∈ S x i = − c S <
0, with weight | c S | , the sum P S ∈F c S Q i ∈ S x i canbe viewed as twice the excess of an instance J of Max- r ( n ) -Lin2-AA . Let k r ( n ) − be the parameter of J . Then, by (2), I and J are equivalent.Note that the algebraic simplification of H ( x ) ensures that J is reduced by Rule2. This completes the proof. 8t is important to note that the resulting instance J of Max- r ( n ) -Lin2-AA is notnecessarily reduced under Rule 1 and, thus, reduction of J by Rule 1 may result inless than n variables.From Theorem 3 and Lemma 3 we have the following fixed-parameter-tractabilityresult for Max- r ( n ) -SAT-AA . Theorem 4.
Max- r ( n ) -Sat-AA is (i) in XP for r ( n ) ≤ log log n − log log log n and (ii) fixed-parameter tractable for r ( n ) ≤ log log n − log log log n − φ ( n ) , for anyunbounded strictly increasing function φ ( n ) .Proof. We start by proving Part (ii). Let φ ( n ) be an unbounded strictly increasingfunction of positive integral argument. Note that φ ( n ) can be extended to a continuouspositive strictly increasing function φ ( t ) of real argument t ≥
1. Thus, φ ( t ) has aninverse function φ − ( t ) . We may assume that φ ( n ) ≥ n ≥ n large enough.Let r ( n ) ≤ log log n − log log log n − φ ( n ) and consider an instance I of Max- r ( n ) -Sat-AA . Note that 2 r ( n ) ≤ n . Therefore by Lemma 3, in polynomial time we canreduce I into an instance J of Max- r ( n ) -Lin2-AA , such that I is a Yes -instance ifand only if J is a Yes -instance with parameter k · r ( n ) − . Consider the Max- r ( n ) -Lin2-AA instance J ′ with n ′ variables formed by reducing J by Rule 1. If n ′ ≤ log n , J ′ may be solved in polynomial time by trying all 2 n ′ ≤ n assignments to the variablesof J ′ . Thus, we may assume that n ′ > log n. If n ′ ≥ ( k r ( n ) − r ( n ) + 1, then by Theorem 3 and Lemma 3, I is a Yes -instance.Thus, it remains to consider the case log n < n ′ ≤ ( k r ( n ) − r ( n ). We havelog n ≤ ( k r ( n ) − r ( n ) and so log n ≤ k (log log n ) · log n/ (2 φ ( n ) log log n ) . This simplifies to φ ( n ) ≤ log k and so n ≤ φ − (log k ) . Hence, I can be solved in time2 φ − (log k ) m O (1) by trying all possible assignments to variables of J ′ .Now we will prove Part (i). Let r ( n ) ≤ log log n − log log log n and consider aninstance I of Max- r ( n ) -Sat-AA . As in Part (ii) proof, we reduce I into an instance J of Max- r ( n ) -Lin2-AA , such that I is a Yes -instance if and only if J is a Yes -instance with parameter k · r ( n ) − . Consider the Max- r ( n ) -Lin2-AA instance J ′ with n ′ variables formed by reducing J by Rule 1. If n ′ ≤ k log n , J ′ may be solved in XPtime by trying all 2 n ′ ≤ n k assignments to the variables of J ′ . Thus, we may assumethat n ′ > k log n. If n ′ ≥ ( k r ( n ) − r ( n ) + 1, then by Theorem 3 and Lemma 3, I isa Yes -instance. Thus, it remains to consider the case k log n < n ′ ≤ ( k r ( n ) − r ( n ).We have k log n ≤ kr ( n )2 r ( n ) and so log log n ≤ log log n − log log log n, a contradiction.Thus, this case is impossible and we can solve I in XP time. It would be interesting to close the gap between the inequalities of Theorems 2 and4. Apart from
MaxLin-AA and
MaxSat-AA mentioned above, there are someother constraint satisfaction problems parameterized above a tight lower bound whosecomplexity has been established in the last few years. One example is r -Linear-Ordering-AA for r = 2 and 3. Let r ≥ r -Linear-Ordering ,9iven a positive integer n and a multiset C of r -tuples of distinct elements from [ n ],we wish to find a permutation π : [ n ] → [ n ] which maximizes that the number of satisfied r -tuples, i.e., r -tuples ( i , i , . . . , i r ) such that π ( i ) < π ( i ) < · · · < π ( i r ) . Let m stand for the number of r -tuples in C .Let τ : [ n ] → [ n ] be a random permutation (chosen uniformly from the set ofall permutations). Observe that the probability that an r -tuple is satisfied by τ is1 /r !. Thus, by linearity of expectation, the expected number of r -tuples satisfiedby τ is m/r !. Using conditional expectation derandomization method [2], it is notdifficult to obtain a polynomial time algorithm for finding a permutation π whichsatisfies at least m/r ! r -tuples. Thus, we can easily obtain an 1 /r !-approximationalgorithm for r -Linear Ordering . It is remarkable that for any positive ε thereexists no polynomial (1 /r ! + ε )-approximation algorithm provided the Unique GamesConjecture (UGC) of Khot [17] holds. This result was proved by Guruswami et al. [11] for r = 2, Charikar et al. [3] for r = 3 and, finally, by Guruswami et al. [10] forany r .Observe that every permutation π satisfies exactly one r -tuple in the set { ( i , i , . . . , i r ) : { i , i , . . . , i r } = [ r ] } and, thus, m/r ! is a tight lower bound on the maximum numberof r -tuples that can be satisfied by a permutation π . It is natural to ask what is theparameterized complexity of the following problem r -Linear-Ordering-AA : decidewhether there is a permutation π which satisfies at least m/r ! + k r -tuples, where k ≥ et al. [14] and [12] proved that r -Linear-Ordering-AA is fixed-parameter tractable for r = 2 and r = 3, respectively. The complexityof r -Linear-Ordering-AA for r ≥ r -Linear-Ordering-AA is fixed-parameter tractable for some r , then all permu-tation constraint satisfaction problems of arity r parameterized above average arefixed-parameter tractable too (see [12] for the definition of a permutation constraintsatisfaction problem of arity r and a proof of the above-mentioned fact). Acknowledgment
This research was partially supported by an International Jointgrant of Royal Society.
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