Preprocessing of Min Ones Problems: A Dichotomy
aa r X i v : . [ c s . CC ] O c t Preprocessing of Min Ones Problems: A Dichotomy
Stefan Kratsch Magnus Wahlstr¨omSeptember 17, 2018
Abstract
A parameterized problem consists of a classical problem and an additional component,the so-called parameter . This point of view allows a formal definition of preprocessing:Given a parameterized instance (
I, k ), a polynomial kernelization computes an equivalentinstance ( I ′ , k ′ ) of size and parameter bounded by a polynomial in k . We give a completeclassification of Min Ones Constraint Satisfaction problems, i.e., Min Ones SAT(Γ), with re-spect to admitting or not admitting a polynomial kernelization (unless N P ⊆ co-
N P /poly).For this we introduce the notion of mergeability. If all relations of the constraint language Γare mergeable, then a new variant of sunflower kernelization applies, based on non-zero-closed cores. We obtain a kernel with O ( k d +1 ) variables and polynomial total size, where d is the maximum arity of a constraint in Γ, comparing nicely with the bound of O ( k d − )vertices for the less general and arguably simpler d - Hitting Set problem. Otherwise, anyrelation in Γ that is not mergeable permits us to construct a log-cost selection formula,i.e., an n -ary selection formula with O (log n ) true local variables. From this we can con-struct our lower bound using recent results by Bodlaender et al. as well as Fortnow andSanthanam, proving that there is no polynomial kernelization, unless N P ⊆ co-
N P /polyand the polynomial hierarchy collapses to the third level.
Preprocessing and data reduction are ubiquitous, especially in the context of combinatoriallyhard problems. Of course, it is a commonplace that there can be no polynomial-time algo-rithm that provably shrinks every instance of an
N P -hard problem, unless P = N P . Still,there does in fact exist a formal notion of efficient preprocessing, coming from the field of pa-rameterized complexity. There, problems are considered with an additional component, theso-called parameter , intended to express the difficulty of a problem instance, e.g., solution size,nesting depth, or treewidth. This way preprocessing can be defined as a polynomial-time map-ping K : ( I, k ) ( I ′ , k ′ ) such that ( I, k ) and ( I ′ , k ′ ) are equivalent and k ′ as well as the sizeof I ′ are bounded by a polynomial in the parameter k ; K is called a polynomial kernelization .Parameterized complexity originated as a multivariate analysis of algorithms, motivated by thehuge difference in (often trivial) n f ( k ) versus f ( k ) n c algorithms, the latter having a much betterscalability. Kernelization is one possible technique to prove fixed-parameter tractability (i.e.,the existence of an f ( k ) n c algorithm). Indeed it is known that a problem is fixed-parametertractable if and only if it admits a kernelization (see [10]). However, this relation does notimply kernelizations with a polynomial size bound; achieving the strongest size bounds or atleast breaking the polynomial barrier is of high interest. Consider for example the list of im-provements for Feedback Vertex Set , from the first polynomial kernel [7], to cubic [4], andnow quadratic [20]; the existence of a linear kernel is still an open problem. Recently a seminalpaper by Bodlaender, Downey, Fellows, and Hermelin [5] provided the first polynomial lowerbounds on the kernelizability of some problems, based on hypotheses in classical complexity.1sing results by Fortnow and Santhanam [13], they showed that so-called compositional param-eterized problems admit no polynomial kernelizations unless
N P ⊆ co-
N P /poly; by Yap [21],this would imply that the polynomial hierarchy collapses. The existence of such lower boundshas sparked high activity in the field (see related work below).Constraint satisfaction problems (CSP) are a fundamental and general problem setting,encompassing a wide range of natural problems, e.g., satisfiability, graph modification, andcovering problems. CSPs are posed as restrictions, called constraints , on the feasible assignmentsto a set of variables. The constraints are applications of relations from a given constraintlanguage Γ to tuples of variables. The complexity of deciding feasibility of a CSP or findingan assignment that optimizes a certain goal varies according to the constraint language. E.g.,consider
Clique as a Max Ones SAT( {¬ x ∨ ¬ y } ) problem, which is hard to approximateand also W[1]-complete when parameterized by the size of the clique. Khanna et al. [14]classified Boolean CSPs according to their approximability, for the questions of optimizingeither the weight of a solution (Min/Max Ones SAT problems) or the number of satisfied orunsatisfied constraints (Min/Max SAT problems). We study the kernelization properties of MinOnes SAT(Γ), parameterized by the number of true variables, and classify these problems intoadmitting or not admitting a polynomial kernelization. We point out that Max SAT(Γ), as asubset of Max SNP (cf. [14]), admits polynomial kernelizations independent of Γ [15]. Related work
In the literature there exists an impressive list of problems that admit polyno-mial kernels (in fact often linear or quadratic); giving stronger and stronger kernels has becomeits own field of interest. We name only a few results for problems that also have a notion ofarity: O ( k d − ) universe size for Hitting Set with set size at most d [1], O ( k d − ) vertices forpacking k vertex disjoint copies of a d -vertex graph [18], and O ( k d ) respectively O ( k d +1 ) baseset size for any problem from MIN F + Π or MAX NP with at most d variables per clause [15].Let us also mention a few lower bound results that are based on the framework of Bodlaen-der et al. [5]. First of all, Bodlaender et al. [6] provided kernelization-preserving reductions,which can be used to extend the applicability of the lower bounds. Using this, Dom et al. [9]gave polynomial lower bounds for a number of problems, among them Steiner Tree and
Connected Vertex Cover . Furthermore they considered problems that have a k f ( d ) kernel,where k is the solution size and d is a secondary parameter (e.g., maximum set size), and showedthat there is no kernel with size polynomial in k + d ; for, e.g., Hitting Set , Set Cover , and
Unique Coverage . Fernau et al. [12] showed that
Leaf Out Branching does not admit apolynomial kernelization, while
Rooted Leaf Out Branching does. They express that thisgives a Turing kernelization for
Leaf Out Branching , by creating one kernel for each choiceof the root. In [16] the present authors show that a certain Min Ones CSP problem does notadmit a polynomial kernel and employ this bound to show that there are H -free edge deletionrespectively edge editing problems that do not admit a polynomial kernel. Our work
We give a complete classification of Min Ones SAT(Γ) problems with respect toadmittance of polynomial kernelizations. Apart from the hardness dichotomy due to Khanna etal. [14], we distinguish constraint languages Γ by being mergeable or containing at least one re-lation that is not mergeable. For the first case, we provide a new polynomial kernelization basedon non-zero-closed cores . For the latter we show that Min Ones SAT(Γ) is either polynomial-time solvable or does not admit a polynomial kernelization, unless
N P ⊆ co-
N P /poly.
Structure of the paper
We introduce some basic notation and the notion of mergeabilityin Sections 2 and 3. Sections 4 and 5 then form the main part of this work, i.e., the generalpolynomial kernelization for Min Ones SAT(Γ) when all relations of Γ are mergeable, and thelower bound for constraint languages that contain at least one relation that is not mergeable.We conclude in Section 6, with a discussion of implications as well as open problems.2
Boolean Constraint Satisfaction Problems A constraint is an application of a relation R to a tuple of variables ( x , . . . , x r ), requiringthat R ( x , . . . , x r ) holds, allowing repeated variables (e.g., R ( x, x, y )). A constraint language is a set Γ of relations; we shall require throughout that every constraint language Γ is finite,and contains only relations over the boolean domain. A formula F over Γ is a conjunction ofconstraints using relations R ∈ Γ, and V ( F ) denotes the set of variables that occur in F . Anassignment to the variables of F satisfies F if every constraint in F holds under the assignment.The weight of an assignment is the number of variables that it sets to true. Fixing a finite set Γwith relations over the boolean domain defines a Min Ones SAT(Γ) problem: Input:
A formula F over a finite constraint language Γ; an integer k . Parameter: k . Task:
Decide whether there is a satisfying assignment for F of weight at most k .As an example, if R ( x, y ) = { (0 , , (1 , , (1 , } , then Min Ones SAT( R ) is the well-knownproblem Vertex Cover . The approximation properties of such problems have been classifiedby Khanna et al. [14]; in particular, we have the following.
Theorem 1 ([14]) . Let Γ be a finite set of relations over the boolean domain. If Γ is zero-valid,Horn, or width-2 affine (i.e., implementable by assignments, ( x = y ) , and ( x = y ) ), then MinOnes SAT( Γ ) is in P ; otherwise it is N P -complete.
SAT(Γ) denotes the problem of deciding whether any satisfying assignment exists; the clas-sical complexity of these problems was classified by Schaefer [19], and the parameterized com-plexity, for the question of finding a satisfying assignment with exactly k true variables, hasbeen classified by Marx [17]. The problem Min Ones SAT(Γ) is fixed-parameter tractable forevery finite Γ, by a simple branching algorithm; see [17].We need to define a number of types of constraints. Let Γ be a finite set of relationsover the boolean domain. We say that Γ implements a relation R if R is the set of satisfyingassignments for a formula over Γ, i.e., R ( x , . . . , x r ) ≡ V i R i ( x i , . . . , x it ) where each R i ∈ Γ (wedo not automatically allow the equality relation unless = ∈ Γ). A positive clause is a disjunctionof (non-negated) variables. A negative clause is a disjunction of negated variables. We saythat a constraint is zero-valid if a tuple of zeros satisfies it. A constraint is
Horn if it can beimplemented by disjunctions containing at most one unnegated variable each, dual Horn if itcan be implemented by disjunctions containing at most one negated variable each, and
IHSB- (Implicative Hitting Set Bounded-) if it can be implemented by assignments, implications, andnegative clauses. These constraint types can also be characterized by closure properties. Fortwo tuples α = ( α , . . . , α r ), β = ( β , . . . , β r ), let α ∧ β = ( α ∧ β , . . . , α r ∧ β r ), and likewisefor α ∨ β , and write α ≤ β if α i ≤ β i for every 1 ≤ i ≤ r (where 0 and 1 are used for false andtrue values, respectively). We then have that a constraint R is Horn if and only if it is closedunder intersection, i.e., if α, β ∈ R , then α ∧ β ∈ R , and a constraint is dual Horn if and onlyif it is closed under disjunction. Likewise, a constraint R is IHSB- if and only if it is closedunder an operation α ∧ ( β ∨ γ ) for tuples α, β, γ in R . See [8] for more on this. A constraintlanguage Γ is zero-valid (one-valid, Horn, dual Horn, IHSB-) if every R ∈ Γ is.
The characterization of the dichotomy of the kernelizability of Min Ones SAT(Γ) given in thispaper centers around a newly introduced property we refer to as mergeability.
Specifically,we will see that for any finite set Γ of relations over the boolean domain, Min Ones SAT(Γ)3dmits a polynomial kernelization if either Min Ones SAT(Γ) is in P or every relation R ∈ Γis mergeable; in every other case, Min Ones SAT(Γ) admits no polynomial kernelization unless
N P ⊆ co-
N P /poly (which would imply that the polynomial hierarchy collapses to the thirdlevel). In this section, we define this property and give some basic results about it.
Definition 1.
Let R be a relation on the boolean domain. Given four (not necessarily distinct)tuples α, β, γ, δ ∈ R , we say that the merge operation applies if α ∧ δ ≤ β ≤ α and β ∧ γ ≤ δ ≤ γ .If so, then applying the merge operation produces the tuple α ∧ ( β ∨ γ ). We say that R is mergeable if for any four tuples α, β, γ, δ ∈ R for which the merge operation applies, we have α ∧ ( β ∨ γ ) ∈ R .We show some basic results about mergeability. First, we show an alternate presentation ofthe property; this perspective will be important in Section 4, when sunflowers are introduced. Proposition 1.
Let R be a relation of arity r on the boolean domain. Partition the positionsof R into two sets, called the core and the petals ; w.l.o.g. assume that positions through c arethe core, and the rest the petals. Let ( α C , α P ) , where α C is a c -ary tuple and α P an ( r − c ) -arytuple, denote the tuple whose first c positions are given by α C , and whose subsequent positionsare given by α P . Consider then the following four tuples. α = ( α C , α P ) β = ( α C , γ = ( γ C , γ P ) δ = ( γ C , If α through δ are in R , then the merge operation applies, giving us ( α C , α P ∧ γ P ) ∈ R. Furthermore, for any four tuples to which the merge operation applies, there is a partitioningof the positions into core and petals such that the tuples can be written in the above form.
It is straight-forward that this property is preserved by implementations.
Proposition 2.
Mergeability is preserved by assignment and identification of variables, i.e.,if R is mergeable, then so is any relation produced from R by these operations. Further, anyrelation implementable by mergeable relations is mergeable. Next, we show what mergeability implies for a zero-valid relation.
Lemma 1.
Any zero-valid relation R which is mergeable is also IHSB-, and can therefore beimplemented using negative clauses and implications.Proof. We show that α ∧ ( β ∨ γ ) ∈ R for all tuples α, β, γ ∈ R . First, for any two tuples α, β ∈ R ,we can apply the merge operation to the tuples α , 0, β , 0, to show that α ∧ β ∈ R . It can thenbe checked that the operation applies to the tuples α , ( α ∧ β ), ( α ∧ γ ), and ( α ∧ β ∧ γ ), andthat this implies α ∧ ( β ∨ γ ) ∈ R . As previously mentioned, this shows that R is IHSB-. Notethat assignments add no expressive power when R is zero-valid.Note that by Proposition 2, this shows that the only zero-valid relations that can be producedfrom a mergeable relation by assigning or identifying variables are IHSB-. However, this stillleaves room for other positive examples; e.g., the constraints ( x + y + z = 1 (mod 2)) and (( x = y ) → z ) are both mergeable. 4 Kernelization
In this section, we show that Min Ones SAT(Γ) admits a polynomial kernelization if all relationsin Γ are mergeable. For the purpose of describing our kernelization we first define a sunflowerof tuples, similarly to the original sunflower definition for sets. We point out that a similarthough more restricted definition for sunflowers of tuples was given by Marx [17]; accordinglythe bounds of our sunflower lemma are considerably smaller.
Definition 2.
Let U be a finite set, let d ∈ N , and let H ⊆ U d . A sunflower ( of tuples ) withcardinality t and core C ⊆ { , . . . , d } in U is a subset consisting of t tuples that have the sameelement at all positions in C and, in the remaining positions, no element occurs in more thanone tuple. The set of remaining positions P = { , . . . , d } \ C is called the petals .As an example, ( x , . . . , x c , y , . . . , y p ), . . . , ( x , . . . , x c , y t , . . . , y tp ) is a sunflower of car-dinality t with core C = { , . . . , c } , if all y ij and y i ′ j ′ are distinct when i = i ′ . Note that,differing from Marx [17] variables in the petal positions may also occur in the core. For setsof tuples H ⊆ U d , we give a variant of Erd˝os’ and Rado’s Sunflower Lemma [11]. The proofis along the same lines as the original, only requiring an additional factor of d ! for picking theshared core positions. Same as the Sunflower Lemma, this immediately gives a polynomial-timealgorithm for finding a sunflower of tuples. Lemma 2.
Let U be a finite set, let d ∈ N , and let H ⊆ U d . If the size of H is greaterthan k d ( d !) , then it contains a sunflower of cardinality k + 1 .Proof. If d = 1, then a sunflower of size k + 1 can be easily found, since any k + 1 tuples ofarity d = 1 form a sunflower with empty core. Now for induction, assume the lemma to be truefor all d ′ ≤ d − X contain { x , . . . , x d } for each tuple ( x , . . . , x d ) ∈ H . Select a maximal pairwisedisjoint subset F ⊆ X . If | F | ≥ k + 1 then its elements correspond to k + 1 tuples that share novariable, i.e., a sunflower with empty core. Otherwise, if | F | ≤ k , then all other sets of X havea non-empty intersection with some element of F . Since the sets correspond to the tuples of H there must be an element of some set in F , say x , that occurs in at least |H| /kd tuples of H , asthere are at most kd such elements. Therefore, there must be a position, p ∈ { , . . . , d } , suchthat x occurs in position p of at least |H| /kd > k d − (( d − tuples of H .Define H ′ by H ′ = { ( x , . . . , x p − , x p +1 , . . . , x d ) | ( x , . . . , x p − , x, x p +1 , . . . , x d ) ∈ H} . Ob-serve that H ′ ⊆ U d − and |H ′ | > k d − (( d − , implying that a sunflower of cardinality k + 1in H ′ can be found in H ′ ; immediately giving a sunflower in H .Our kernelization requires also the notion of a zero-closed position and the related zero-closure of a relation. The following definition introduces these concepts. Definition 3.
Let R be an r -ary relation. The relation R is zero-closed on position i , if forevery tuple ( t , . . . , t i − , t i , t i +1 , . . . , t r ) ∈ R we have ( t , . . . , t i − , , t i +1 , . . . , t r ) ∈ R . A relationis non-zero-closed if it has no zero-closed positions. We define functions ∆, Π, and ∇ : • ∆ P ( R ) is defined to be the zero-closure of R on positions P ⊆ { , . . . , r } , i.e., the smallestsuperset of R that is zero-closed on all positions i ∈ P . • Π( R ) denotes the non-zero-closed core , i.e., the projection of R onto all positions that arenot zero-closed or, equivalently, the relation on the non-zero-closed positions obtained byforcing x i = 0 for all zero-closed positions.5 ∇ C ( R ) denotes the sunflower restriction of R with core C : w.l.o.g. the relation expressedby [ ∇ C ( R )]( x , . . . , x r ) = R ( x , . . . , x r ) ∧ R ( x , . . . , x c , , . . . ,
0) for C = { , . . . , c } . Thecorresponding core relation is the | C | -ary relation given by R ( x , . . . , x c , , . . . , P and ∇ C extend also to constraints: ∇ C ( R ( x , . . . , x r )) = [ ∇ C ( R )]( x , . . . , x r ).Similarly for Π, but variables in zero-closed positions are removed, e.g., when i is the only zero-closed position of R then Π( R ( x , . . . , x r )) = [Π( R )]( x , . . . , x i − , x i +1 , . . . , x r ). Lemma 3.
Let R be a mergeable relation and let C ∪ P be a partition of its positions into coreand petals. There is an implementation of ∇ C ( R ) using ∆ P ( ∇ C ( R )) and implications.Proof. By Prop. 2, ∇ C ( R ) must be mergeable. Further, assigning any set of values to thevariables in the core produces a zero-valid relation on the petals, which is still mergeable. Thusby Lemma 1, this relation on the petals has an implementation using negative clauses andimplications. For any tuple σ in ∇ C ( R ) or ∆ P ( ∇ C ( R )), let its core assignment be the values itassigns to the positions in C . By definition, we have ∇ C ( R ) ⊆ ∆ P ( ∇ C ( R )).Consider now a tuple σ ∈ ∆ P ( ∇ C ( R )) \ ∇ C ( R ). Assume that σ makes core assignment α C ;thus there is a matching α ∈ ∇ C ( R ), α > σ , with an identical core assignment. As in Prop. 1,write α = ( α C , α P ), and consider the constraint on the petals that is formed by core as-signment α C . We see that this constraint must entail some implication ( y i → y j ), where σ assigns y i = 1 , y j = 0 while α P assigns y i = y j = 1. Further, since ∇ C ( R ) is a sunflowerrestriction, we have β = ( α C , ∈ ∇ C ( R ). Now if ( y i → y j ) does not hold in ∇ C ( R ) ingeneral, then let γ = ( γ C , γ P ) ∈ ∇ C ( R ) be a tuple which assigns y i = 1, y j = 0, andlet δ = ( γ C , ∈ ∇ C ( R ). By Prop. 1, we can now apply the merge operation to tuples α through δ , showing ( α C , α P ∧ γ P ) ∈ ∇ C ( R ). But this is a tuple with core assignment α C whichassigns y i = 1, y j = 0, which is a contradiction. Thus ( y i → y j ) holds in ∇ C ( R ) regardless ofcore assignment, and the constraint ( y i → y j ) can be added to our implementation of ∇ C ( R ),removing the tuple σ .Adding all implications between petals which hold in ∇ C ( R ) removes from our implementa-tion all tuples which are in ∆ P ( ∇ C ( R )) but not in ∇ C ( R ), so that the conjunction of ∆ P ( ∇ C ( R ))with all valid implications is an implementation of ∇ C ( R ).The following technical lemma proves that the relation ∆ P ( ∇ C ( R )), required by Lemma 3,is mergeable. Lemma 4.
Let R be a mergeable relation and let C ∪ P be a partition of its positions into coreand petals. Then ∆ P ( ∇ C ( R )) is mergeable.Proof. Recall that ∆ P ( ∇ C ( R )) is the zero-closure on the petal positions of the sunflower re-striction of R with core C . Let R ′ := ∆ P ( ∇ C ( R )); assume by way of contradiction that R ′ isnot mergeable. Then there are four tuples in R ′ such that applying the merge operation on thetuples creates a tuple not in R ′ . Let C ′ ∪ P ′ be the partition of the positions of R ′ into coreand petals that is used in this counterexample. Grouping the positions of R ′ in four groups,written in the order ( C ′ ∩ C, C ′ ∩ P, P ′ ∩ C, P ′ ∩ P ), naming the groups W through Z , thecounterexample can be written as follows.( W , X , Y , Z ) ∈ R ′ (1)( W , X , , ∈ R ′ (2)( W , X , Y , Z ) ∈ R ′ (3)( W , X , , ∈ R ′ (4)( W , X , Y ∧ Y , Z ∧ Z ) / ∈ R ′ (5)6e will derive a contradiction. First, we note that for each equation (1)–(4), there is a corre-sponding tuple in ∇ C ( R ). ( W , X a , Y , Z a ) ∈ ∇ C ( R ) (6)( W , X b , , Z b ) ∈ ∇ C ( R ) (7)( W , X a , Y , Z a ) ∈ ∇ C ( R ) (8)( W , X b , , Z c ) ∈ ∇ C ( R ) (9)Here, X a and X b are supersets of X , and likewise for X , Z , and Z . Z b and Z c are arbitrary.Using that ∇ C ( R ) is a sunflower restriction and mergeable, we can conclude the following.( W , , , ∈ ∇ C ( R ) (10)( W , , , ∈ ∇ C ( R ) (11)( W , X a ∧ X a , Y ∧ Y , Z a ∧ Z a ) ∈ ∇ C ( R ) (12)The first two come from (7) and (9); the third is produced by a merge operation on (6) and (8)using these two. Now, the tuples which match W on the W -variables form a zero-valid relation.By Lemma 1, this relation is closed under an operation ( α ∧ ( β ∨ γ )). Applying this on thetuples of equations (6), (7), and (12) gives us the following conclusion.( W , X a ∧ ( X b ∨ X a ) , Y ∧ Y , Z a ∧ ( Z b ∨ Z a )) ∈ ∇ C ( R ) (13)In particular, this tuple matches (5) on the W - and Y -variables, and is a superset of it on the X - and Z -variables. Since R ′ is the zero-closure of ∇ C ( R ) on the X - and Z -variables, we havea contradiction.Lemmas 3 and 4 are the foundation for a sunflower-based kernelization for Min Ones SAT(Γ).They show that the sunflower restriction ∇ C ( R ) of some mergeable R -constraint can be imple-mented using its mergeable zero-closure on the petal positions as well as implications. How-ever, ∆ P ( ∇ C ( R )) is not necessarily contained in Γ and such a replacement does not give anyimmediate reduction of the instance. Indeed, the arity of ∆ P ( ∇ C ( R )) is the same as that of R .We address this problem by introducing a new measure of difficulty for formulas, namelythe sum of non-zero-closed cores, based on the following definition. Definition 4.
Let F be a formula and let R be a relation. We define Z ( F , R ) as the set of alltuples ( x , . . . , x t ) where [Π( R )]( x , . . . , x t ) is the non-zero-closed core of an R -constraint in F .For some relations, sunflower restriction, zero-closure, and the relation itself are the same, forcertain selections of core and petal positions. See for example the following mergeable relation: R = { (0 , , , , (0 , , , , (0 , , , , (1 , , , , (1 , , , , (1 , , , , (1 , , , } = ∇ { , , } ( R ) = ∆ { } ( ∇ { , , } ( R ))This is also one of the smallest examples, where a sunflower restriction cannot be expressed usingthe core relation (i.e., all tuples for the core such that the petal variables can take value 0),implications, and negative clauses. Here, the core relation is { (0 , , , (0 , , , (1 , , , (1 , , } ,but no implication or negative clause can exclude the tuple (0 , , ,
1) without also excludingother tuples that do occur in the sunflower restriction. Thus there are mergeable relations forwhich a sunflower-based reduction using Lemma 3 does not lead to any simplification, even interms of Z ( F , R ). We overcome this difficulty by searching for sunflowers among the tuplesof Z ( F , R ). Those are leveraged into a replacement of the R -constraints that contributed thesetuples. The following theorem shows this approach in detail.7 heorem 2. Let Γ be a mergeable constraint language with maximum arity d . Let F be aformula over Γ and let k be an integer. In polynomial time one can compute a formula F ′ overa mergeable constraint language Γ ′ ⊇ Γ with maximum arity d , such that every assignment ofweight at most k satisfies F if and only if it satisfies F ′ and, furthermore, |Z ( F ′ , R ) | ∈ O ( k d ) for every non-zero-valid relation that occurs in F ′ .Proof. We begin constructing F ′ , starting from F ′ = F . While |Z ( F ′ , R ) | > k d ( d !) for any non-zero-valid relation R in F ′ , search for a sunflower of cardinality k + 1 in Z ( F ′ , R ), accordingto Lemma 2. Let C denote the core of the sunflower and apply the following replacement.Remove each R -constraint whose non-zero-closed core matches a tuple of the sunflower, andadd its sunflower restriction with core C using an implementation according to Lemma 3.Repeating this step until |Z ( F ′ , R ) | ≤ k d ( d !) for all non-zero-valid relations R in F ′ completesthe construction.Now, to prove correctness, let us consider a single replacement. We denote the tuples ofthe sunflower by ( x , . . . , x c , y i , . . . , y ip ), with i ∈ { , . . . , k + 1 } , i.e., w.l.o.g. with core C = { , . . . , c } and petals P = { c + 1 , . . . , c + p } . Let φ be any satisfying assignment of weightat most k and consider any tuple ( x , . . . , x c , y i , . . . , y ip ) of the sunflower. There must bea constraint R ( x , . . . , x c , y i , . . . , y ip , z , . . . , z t ) whose non-zero-closed core matches the tuple,w.l.o.g. we take the last positions of R to be zero-closed, let Z be those positions. Thus φ must satisfy R ( x , . . . , x c , y i , . . . , y ip , , . . . , z i are in zero-closed positions. Ob-serve that, by maximum weight k , the assignment φ assigns 0 to all variables y i , . . . , y ip for an i ∈ { , . . . , k + 1 } . Thus φ satisfies also R ( x , . . . , x c , , . . . , R ( x , . . . , x c , y i , . . . , y ip , z , . . . , z t ), it satisfies ∇ C ( R ( x , . . . , x c , y i , . . . , y ip , z , . . . , z t ))too. This permits us to replace each R -constraint, whose non-zero-closed core matches a tupleof the sunflower, by an implementation of its sunflower restriction with core C , according toLemma 3. The implementation uses ∆ P ∪ Z ( ∇ C ( R ( x , . . . , x c , y i , . . . , y ip , z , . . . , z t ))) and impli-cations. By Lemma 4 the added constraints ∆ P ∪ Z ( ∇ C ( R ( x , . . . , x c , ., . . . , . ))) are mergeable,implying that all constraints in F ′ are mergeable.To establish that the construction can be performed efficiently, i.e., in time polynomial inthe size of F , we use as a measure of F ′ the sum of |Z ( F ′ , R ) | over all relations R occurringin F ′ . First, let us observe that, initially, this measure is bounded by the size of F sinceeach R -constraint of F ′ contributes at most one tuple to the corresponding set Z ( F ′ , R ) (recallthat we start with F ′ = F ). Consider again the replacement made in each step: All R -constraints matching one of the tuples of the sunflower are replaced by an implementationusing ˆ R = ∆ P ∪ Z ( ∇ C ( R )) and implications. It is crucial to observe that all added constraintscontribute the same tuple to Z ( F ′ , ˆ R ), consisting only of variables with positions in C . Thisis caused by the application of the zero closure ∆ on all positions but those in C . Hencethe k + 1 tuples of the sunflower are removed, as all matching R -constraints are replaced, andonly one new tuple is added to the set Z ( F ′ , ˆ R ). This decreases the measure, implying that themodification step is applied at most a number of times polynomial in the size of F .Finally let us express the fact that each iteration of the replacement can be done effi-ciently. The set Z ( F ′ , R ) can be generated in one pass over the formula and since the arity isbounded by d there is only a constant number of relations. The applications of Lemma 2 tofind a sunflower among the tuples of the sets Z ( F ′ , R ) take time polynomial in |F | , since thesize |Z ( F ′ , R ) | ∈ O ( |F | ). Observe that the size of F ′ is bounded by a polynomial in |F | at alltimes, since there is only a polynomial number of possible constraints of arity at most d on thevariables of F .Now we are able to derive a polynomial kernelization for Min Ones SAT(Γ). For a giveninstance ( F , k ), it first generates an equivalent formula F ′ according to Theorem 2. However, F ′ F , rather, it allows us to remove variables from F based on conclusions drawnfrom F ′ . This approach avoids the obstacle of a possible lack of expressibility from using onlythe language Γ, and requires no additional assumptions or annotations to be made. Theorem 3.
Let Γ be a mergeable constraint language. Then Min Ones SAT( Γ ) admits apolynomial kernelization.Proof. Let ( F , k ) be an instance of Min Ones SAT(Γ) and let d be the maximum arity ofrelations in Γ. According to Theorem 2, we generate a formula F ′ , such that assignments ofweight at most k are satisfying for F if and only if they are satisfying for F ′ . Moreover, for eachnon-zero-valid relation R , we have that |Z ( F ′ , R ) | ∈ O ( k d ). Note that constraints of F ′ havemaximum arity d . We allow the constant 0 to be used for replacing variables; a constructionfor this not using ( x = 0) follows at the end of the proof.First, according to Lemma 1, we replace each zero-valid constraint of F ′ by an implemen-tation through negative clauses and implications. Next, we address variables that occur onlyin zero-closed positions constraints in F ′ . By definition of zero-closed positions it is immediatethat setting such a variable to 0, does not affect the possible assignments for the other variables.By equivalence of F and F ′ with respect to assignments of weight at most k , the same is truefor F . We replace all such variables by the constant 0 in F and F ′ , maintaining the equivalencewith respect to assignments of weight at most k .Now, let X be the set of variables that occur in a non-zero-closed position of some non-zero-valid constraint of F ′ . For each variable x ∈ X count the number of variables that are impliedby x , i.e., that have to take value 1 if x = 1, by implication constraints in F ′ . If the number ofthose variables is at least k , then there is no satisfying assignment of weight at most k for F ′ that assigns 1 to x . By equivalence of F and F ′ with respect to such assignments, we replace alloccurrences of such a variable x by the constant 0, again maintaining the equivalence property.Finally we replace all variables y ∈ V ( F ′ ) \ X , that are not implied by a variable from X in F ′ ,by the constant 0 in F and F ′ . Note that such variables y occur only in zero-closed positionsand in implications. It can be easily verified that this does not affect satisfiability with respectto assignments of weight at most k . For efficiency of this modification consider the fact thatthe number of implications in F ′ is polynomial in the initial size of F , since there are at mosttwo implications per pair of variables of F . This completes the kernelization.Now we prove a bound of O ( k d +1 ) on the number of variables in F . First, we observe thatall remaining variables of F must occur in a non-zero-closed position of some constraint of F ′ .We begin by bounding the number of variables that occur in a non-zero-closed position of somenon-zero-valid R -constraint, i.e., the remaining variables of the set X . Observe that such avariable must occur in the corresponding tuple of Z ( F ′ , R ). Since there is only a constantnumber of relations of arity at most d and since Z ( F ′ , R ) ∈ O ( k d ), this limits the number ofsuch variables by O ( k d ). For all other variables, their non-zero-closed occurrences must be inimplications, since negative clauses are zero-closed on all positions. Thus, these variables mustbe implied by a variable of X . Since each variable implies at most k − O ( k d +1 ). Finally, the total size of F is polynomial for a fix d , since thenumber of variables is polynomial and the arity of the constraints is bounded.To express the 0-constant, we add k + 1 new variables z , . . . , z k +1 . Every constraint withat least one 0 is replaced by k + 1 copies, each time replacing 0 with a different z i . Clearly oneof the z i takes value 0 in any assignment of weight at most k . Hence the original constraintswith constant 0 are enforced. Conversely, given a satisfying assignment of weight at most k forthe formula before making this replacement, we can easily extend it by assigning 0 to each z i .This construction does not affect our upper bound on the number of variables.9 Kernel Lower Bounds
We will now complete the dichotomy by showing that if Min Ones SAT(Γ) is
N P -complete andsome R ∈ Γ is not mergeable, then the problem admits no polynomial kernelization unless
N P⊆ co-
N P /poly. The central concept of our lower bound construction is the following definition.
Definition 5. A log-cost selection formula of arity n is a formula on variable sets X and Y ,with | Y | = n and | X | = n O (1) , such that there is no solution where Y = 0, but for any y i ∈ Y there is a solution where y i = 1, y j = 0 for j = i , and where a fix number w n = O (log n )variables among X are true. Furthermore, there is no solution where fewer than w n variablesamong X are true.We will show that any Γ as described can be used to construct log-cost selection formulas,and then derive a lower bound from this. The next lemma describes our constructions. Lemma 5.
The following types of relations can implement log-cost selection formulas of anyarity.1. A 3-ary relation R such that { (0 , , , (1 , , , (1 , , } ⊆ R and (1 , , / ∈ R , togetherwith relations ( x = 1) and ( x = 0) .2. A 5-ary relation R such that { (1 , , , , , (1 , , , , , (0 , , , , , (0 , , , , } ⊆ R and (1 , , , , , (0 , , , , / ∈ R , together with relations ( x = y ) , ( x = 1) , and ( x = 0) .Proof. Let Y = { y , . . . , y n } be the variables over which a log-cost selection formula is requested.We will create “branching trees” over variables x i,j for 0 ≤ i ≤ log n , 1 ≤ j ≤ i , as variantsof the composition trees used in [16]. Assume that n = 2 h for some integer h ; otherwise pad Y with variables forced to be false, as assumed to be possible in both constructions.The first construction is immediate. Create the variables x i,j and add a constraint ( x , = 1).Further, for all i, j with 0 ≤ i < h and 1 ≤ j ≤ i , add a constraint R ( x i,j , x i +1 , j − , x i +1 , j ).Finally, replace variables x h,j by y j . By the requirements on R , for every internal variable x i,j ,if x i,j = 1 then one of its children x i +1 , j − and x i +1 , j must be true. Thus by transitivity,some variable on each level of the branching tree must be true, making Y = 0 is impossible.Conversely, for any variable y i on the leaf level, there is a solution where exactly the variablesalong the path from the root node to y i are true. Thus w n = h = log n .The second construction uses the same principle, but the construction is somewhat moreinvolved. Create variables x i,j and a constraint ( x , = 1) as before. In addition, introducefor every 0 ≤ i ≤ h − l i , r i and a constraint ( l i = r i ). Now the intention isthat ( l i , r i ) decides whether the path of true variables from the root to a leaf should take a leftor a right turn after level i . Concretely, add for every i, j with 0 ≤ i ≤ h − ≤ j ≤ i a constraint R ( l i , r i , x i,j , x i +1 , j − , x i +1 , j ). Now for every true variable x i,j , it is not allowedthat x i +1 , j − = x i +1 , j = 0, while depending on l i and r i , either ( x i +1 , j − = 1 , x i +1 , j = 0)or ( x i +1 , j − = 0 , x i +1 , j = 1) is allowed. This rules out the case Y = 0, while for each set ofvalues of l i , r i it is allowed to set among variables x i,j exactly the variables along a path fromthe root to a leaf y i to true, and other variables to false. In total, exactly two variables notamong Y are true per level in such an assignment, making w n = 2 h = 2 log n .We now reach the technical part, where we show that any relation which is not mergeablecan be used to construct a relation as in Lemma 5. The constructions are based on the conceptof a witness that some relation R lacks a certain closure property. For instance, if R is notmergeable, then there are four tuples α, β, γ, δ ∈ R to which the merge operation applies, but10uch that α ∧ ( β ∨ γ ) / ∈ R ; these four tuples form a witness that R is not mergeable. Using theknowledge that such witnesses exist, we use the approach of Schaefer [19], identifying variablesaccording to their occurrence in the tuples of the witness, to build relations with the propertieswe need. Lemma 6.
Let Γ be a set of relations such that Min Ones SAT( Γ ) is N P -complete andsome R ∈ Γ is not mergeable. Under a constraint that at most k variables are true, Γ canbe used to force ( x = 0) and ( x = 1) . Furthermore, there is an implementation of ( x = y ) using R , ( x = 0) , and ( x = 1) .Proof. First of all, we show how to force ( x = 1). Since Min Ones SAT(Γ) is N P -complete, itcontains some relation that is not zero-valid; let R ∈ Γ be such a relation. If R is one-valid,then R ( x, . . . , x ) is equivalent to ( x = 1). Else, let r be the arity of R and let I be a maximalset such that R ( x , . . . , x r ) holds for x i = 1 for i ∈ I , x i = 0 else. Identify all x i , i ∈ I , to asingle variable x , and all x i , i / ∈ I , to a single variable y . This forms a new constraint R ′ ( x, y ),where (1 , ∈ R ′ ( x, y ) and (0 , , (1 , / ∈ R ′ ( x, y ). Thus R ′ is either ( x = 1 ∧ y = 0) or ( x = y ).In the former case we are done; in the latter case, constraints x = y i for 1 ≤ i ≤ k + 1 force x = 1and all y i = 0 in any solution with at most k true variables.Now we can use this to force ( x = 0) and ( x = y ). Let α through δ be a witness that R is not mergeable; let σ = α ∧ ( β ∨ γ ) / ∈ R be the produced tuple. Notice that β < σ < α ,meaning that the positions of R are of four types: those where β < σ , those where σ < α , andoptionally positions which are constant among these tuples, i.e. true in β or false in α . Callthese positions C x , C y , C , and C , in the order they were introduced. Place a variable z = 1in all positions C , if any, and variables x and y in all positions C x resp. C y . Now, if there are nopositions C , then this creates a constraint R ′ ( x, y ) such that R ′ ( x, y ) ∧ R ′ ( y, x ) implements ( x = y ) directly. This can be used to force x = 0: create k variables y i and let x = y i for every i . Inany solution with at most k true variables, all these variables are false.Otherwise, if there are positions C , then place the variable y in these positions as well, andapply R ′ ( x, y ) ∧ R ′ ( y, x ) again; the result is either ( x = y ) or ( x = y = 0). Finally, placing avariable z = 0 in positions C lets us implement ( x = y ) as above. Lemma 7.
Let Min Ones SAT( Γ ) be NP-complete, and not mergeable. Then Min Ones SAT( Γ )can express a log-cost selection formula of any arity.Proof. Let R ∈ Γ be a relation that is not mergeable, and let α through δ be a witness of this. ByProp. 1, partition the positions of R into core and petals in a way that agrees with the witness.Group the variables w.r.t. their values in these four tuples into constant variables Z , Z , non-constant core variables C and C , and non-constant petal variables P , P , P (where theindices indicate membership in α and γ , as β and δ are now determined by this). Identifyvariables according to type, and order them in the order of the previous sentence. We nowhave a relation whose arity depends on which variable types that are represented in the witness.In the case that all seven types are present, we have implemented a 7-ary relation R aboutwhich we know the following (the final tuple is produced on the witness tuples by the mergeoperation). (1 , , , , , , ∈ R (1 , , , , , , ∈ R (1 , , , , , , ∈ R (1 , , , , , , ∈ R (1 , , , , , , / ∈ R
11o distinguish the final tuple from the witness tuples, we can observe that variable types P , P ,and one further non-constant variable type must be represented by the witness. The constantpositions can be ignored by putting variables z = 1 and z = 0 in these positions, by Lemma 6.Thus we implement a relation of arity between three and five.First assume that R is dual Horn. Then the tuple β ∨ γ ∈ R , i.e. (1 , , , , , , ∈ R , andthe variable type C or P must occur. Identify C and P if both occur, and set C = 1if this type occurs, implementing a 3-ary relation R ′ which matches R of Lemma 5, with thevariable types being P , P , and ( P = C ) in the order used in Lemma 5. Indeed, { α, β ∨ γ, β } ⊆ R , representing the positive requirement, while there can be no tuple (1 , , ∈ R ′ ,whether C occurs or not. This and Lemma 6 fulfills the conditions of Lemma 5, part 1.Otherwise R is not dual Horn, in which case it is not closed under disjunction. Using awitness for this, we can implement a 2-ary relation R which is either ( x = y ) or ( ¬ x ∨ ¬ y ).Likewise, by N P -completeness we have a relation which is not Horn, which can implement ( x = y ) or ( x ∨ y ). Combining them, we find that we can always implement ( x = y ), and thus arefree to use R of Lemma 5. We implement a relation R ′ as before, again letting the variabletypes appear in the order ( C , C , P , P , P ). We go through the cases of non-empty non-constant variable types, and show that our relation R ′ can implement a relation matching R or R of Lemma 5.1. If R ′ has arity three, with the third variable type being P or C , then we implement arelation matching R with { (1 , , , (1 , , , (0 , , } ⊆ R ′ and (1 , , / ∈ R ′ .2. If R ′ has arity three, and the third type is C , then we implement a relation R ′ with { (1 , , , (1 , , , (0 , , , (0 , , ⊆ R ′ and (1 , , / ∈ R ′ . Use R ′ ( v, x, y ) ∧ R ′ ( w, x, z ) toimplement a relation matching R .3. If the core type C is not present, then identify C with P . This implements a rela-tion R ′ matching R .4. If the core type C is not present, we need two cases. If (0 , , , / ∈ R ′ , then identify C with P to produce a 3-ary relation matching R . Otherwise, force P = 0 to produce a3-ary relation as in case 2.5. If the petal type P is not present, then R ′ ( v, w, x, y ) ∧ R ′ ( w, v, x, z ) implements a relationmatching R .6. If all five types are present, then R ′ ( v, w, x, y, z ) ∧ R ′ ( w, v, x, z, y ) implements a relationmatching R .Thus in every case, we meet the conditions of part 1 or 2 of Lemma 5.We now show our result, using the tools of [6]. We have the following definition. Let Q and Q ′ be parameterized problems. A polynomial time and parameter transformation from Q to Q ′ is a polynomial-time mapping H : Σ ∗ × N → Σ ∗ × N : ( x, k ) ( x ′ , k ′ ) such that ∀ ( x, k ) ∈ Σ ∗ × N : (( x, k ) ∈ Q ⇔ ( x ′ , k ′ ) ∈ Q ′ ) and k ′ ≤ p ( k ) , for some polynomial p .We will provide a polynomial time and parameter transformation to Min Ones SAT(Γ) fromExact Hitting Set( m ), defined as follows. Input:
A hypergraph H consisting of m subsets of a universe U of size n . Parameter: m . Task:
Decide whether there is a set S ⊂ U such that | E ∩ S | = 1 for every E ∈ H .12t was shown in [6] that polynomial time and parameter transformations preserve polynomialkernelizability; thus our lower bound will follow. To establish a lower bound for Exact HittingSet( m ), we need the following notions from [5, 6]. Let Q be a parameterized problem. A composition algorithm for Q is an algorithm that on input ( x , k ) , . . . , ( x t , k ) ⊆ Σ ∗ × N usestime polynomial in P ti =1 | x i | + k and outputs ( y, k ′ ) with k ′ bounded by a polynomial in k andsuch that ( y, k ′ ) ∈ Q if and only if ( x i , k ) ∈ Q for at least one i ∈ { , . . . , t } . The problem Q isthen said to be compositional .The derived classical problem ˜ Q of Q is defined by ˜ Q = { x k | ( x, k ) ∈ Q} , where / ∈ Σis the blank letter and 1 is any letter from Σ.Following Dom et al. [9], we next give our equivalence of a colored version of the problem,which we call Exact CSP( m + n ), defined as follows. Input:
A CSP instance with n variables of arbitrary finite domain, and m con-straints Exactly-One( v i = b i , . . . , v i r = b i r ) of arbitrary arity, where each v i is avariable and b i a value from the respective variable domain. Parameter: m + n . Task:
Decide whether there is an assignment of a value to every variable thatsatisfies each constraint (i.e. for each constraint, exactly one statement v = b istrue).We show that Exact CSP( m + n ) admits no polynomial kernelization; the result follows by atrivial problem reduction from Exact CSP( m + n ) to Exact Hitting Set( m ). The proof followsthe same lines as the lower bound for Unique Coverage in [9, Sec. 4.2], but the constructionis somewhat simplified, and the lower bound somewhat stronger (as Exact Hitting Set( m ) isequivalent to a special case of Unique Coverage). Lemma 8.
Exact Hitting Set ( m ) admits no polynomial kernelization unless N P ⊆ co-
N P /poly.Proof.
First, Exact CSP( m + n ) is N P -complete (even if all domains have cardinality 2, in whichcase it is the Exact Satisfiability problem). Also, the problem can be solved in time O ∗ ( n m ).Decide for each constraint the identity of the variable which will hit it (but not yet its value).Assuming that a variable v is chosen for a particular constraint, for every statement ( v i = b j ) inthe constraint with v = v i , remove the value j from the domain of the variable v i , and restrictthe domain of v to those values which would hit the constraint. Repeat for all constraints,backtracking if necessary; the size of the search tree is at most n m . Thus, we may assume inour composition algorithm that the number of input instances is bounded by n m (or else wesolve all instances in time polynomial in the total input size).Assume, then, that there are t input instances. Let n be the maximum number of variablesand m the maximum number of constraints; for simplicity of the argument, assume that allinput instances have the same numbers of variables and constraints (or else do trivial paddingwith unary-domain variables or trivially true constraints, such as “variable 1 has exactly onevalue”). Number the variables from 1 to n and the constraints from 1 to m in each inputinstance.Now create the composed instance. First collect all the values of variables numbered i intothe domain of a single variable v ′ i , say with values ( j , j ) signifying “value j in the domain ofinput instance j ”. Similarly concatenate all constraints numbered i into a single constraint,over these new domain values. Note that values stemming from different instances are different,so that a constraint is hit only once in an intended solution (where all values come from the same Dom et al. also give a proof for the problem
Bipartite Perfect Code , which is the same underlyingproblem as here, but the parameterization is different. t as its ID, and write this inbinary form. Let l = ⌈ log t ⌉ = O ( m log n ) be the number of bit levels needed. For each pair ofvalues ( i, i + 1), 1 ≤ i < n , and each bit level j , 1 ≤ j ≤ l , add a testing constraint consistingof all values of v i for which the j :th digit of the ID of the originating instance is 1, and allvalues of v i +1 for which the j :th digit of the ID is 0. This makes O ( mn log n ) extra edges. Ifvariables v i and v j take values from different input instances, then their IDs will differ in someposition, which will lead to one of these testing edges being hit twice or not at all. Otherwise,each testing edge is hit exactly once. By transitivity, this forces all variables in the composedinstance to take values from the same input instance. This completes the compositionalityproof, showing that Exact CSP with parameter m + n admits no polynomial kernelization.The result for Exact Hitting Set with parameter m follows by a simple reduction. Let thevertices of the hitting set be the individual variable values, create for each variable an edgecontaining all its values, and retain all constraints as edges. We get an equivalent instancewith m + n edges.We can now show the main result of this section. Theorem 4.
Let Γ be a constraint language which is not mergeable. Then Min Ones SAT( Γ )is either polynomial-time solvable, or does not admit a polynomial kernelization unless N P ⊆ co-
N P /poly.Proof.
By Theorem 1, Min Ones SAT(Γ) is either polynomial-time solvable or
N P -complete;assume that it is
N P -complete. By Lemma 6 we have both constants and the constraint ( x = y ),and by Lemma 7 we can implement log-cost selection formulas. It remains only to describe thepolynomial time and parameter transformation from Exact Hitting Set( m ) to Min Ones SAT(Γ).Let H be a hypergraph. If H contains more than 2 m vertices, then it can be solved intime polynomial in the input length [3]; otherwise, we create a formula F and fix a weight k so that ( F , k ) is positive if and only if H has an exact hitting set. Create one variable y i,j in F for every occurrence of a vertex v i in an edge E j in H . For each edge E ∈ H , createa selection formula over the variables representing the occurrences in E . Finally, for all pairsof occurrences of each vertex v i , add constraints ( y i,j = y i,j ′ ), and fix k = m + P E ∈H w | E | ,where w i is the weight of an i -selection formula. We have an upper bound on the value of k of O ( m log n ) = O ( m ).Now solutions with weight exactly k correspond to exact hitting sets of H . Note that k isthe minimum possible weight of the selection formulas, which is taken if exactly one occurrencein each edge is picked. By the definition of log-cost selection formulas, any solution where morethan one occurrence has been picked (if such a solution is possible at all) will have a total weightwhich is larger than this, if the weight of the y -variables is counted as well, and thus such asolution to F of weight at most k is not possible.As Exact Hitting Set( m ) is N P -complete, it follows from [6] that a polynomial kernelizationfor Min Ones SAT(Γ) would imply the same for Exact Hitting Set( m ), giving our result.Finally, let us remark that Lemma 6 can be adjusted to provide ( x = 1) and ( x = y ) withoutthe use of repeated variables, and that using standard techniques (see [16] and Theorem 3), wecan show that the lower bound still applies under the restriction that constraints contain norepeated variables. Such a restriction can be useful in showing hardness of other problems, e.g.,as in [16]. 14 Conclusions
We presented a dichotomy for Min Ones SAT(Γ) for finite sets of relations Γ, assuming that thepolynomial hierarchy does not collapse. The characterization of the dichotomy is a new conceptwe call mergeability. We showed that Min Ones SAT(Γ) admits a polynomial kernelization if theproblem is in P or if every relation in Γ is mergeable, while in every other case no polynomialkernelization is possible unless N P ⊆ co-
N P /poly, in which case the polynomial hierarchywould collapse to the third level.It might be interesting to compare our kernelization dichotomy to the approximation prop-erties of Min Ones SAT(Γ), as characterized by Khanna et al. [14]. The mergeability propertycuts through the classification of Khanna et al. as follows (we use the terms from B.4 of [14]).For every Γ such that Min Ones SAT(Γ) is known to be in APX, Min Ones SAT(Γ) admitsa polynomial kernelization, while no problem identified as being Min Horn Deletion-completeis mergeable. The remaining classes are cut through (e.g., among the affine relations, therelation ( x + y + z = 1 (mod 2)) is mergeable, while ( x + y + z = 0 (mod 2)) is not). We alsoget kernelizations for some problems where the corresponding SAT problem is N P -complete,e.g., Min Ones Exact Hitting Set for sets of bounded arity, where no approximation is possibleunless P = N P . Acknowledgements
We are thankful to Gustav Nordh and D´aniel Marx for helpful and interesting discussions.
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