AA Multistage View on 2-Satisfiability
Till Fluschnik
Technische Universität Berlin, Faculty IV, Algorithmics and Computational Complexity, Germanytill.fl[email protected]
Abstract
We study q -SAT in the multistage model, focusing on the linear-time solvable 2 -SAT . Herein, given asequence of q -CNF fomulas and a non-negative integer d , the question is whether there is a sequenceof satisfying truth assignments such that for every two consecutive truth assignments, the numberof variables whose values changed is at most d . We prove that Multistage 2-SAT is NP-hard evenin quite restricted cases. Moreover, we present parameterized algorithms (including kernelization)for
Multistage 2-SAT and prove them to be asymptotically optimal.
Theory of computation, Discrete mathematics
Keywords and phrases satisfiability, temporal problems, symmetric difference, parameterizedcomplexity, problem kernelization
Funding
Till Fluschnik : Supported by DFG, project TORE (NI 369/18).
Acknowledgements
I thank Hendrik Molter and Rolf Niedermeier for their constructive feedbacks. q -Satisfiability ( q -SAT ) is one of the most basic and best studied decision problems incomputer science: It asks whether a given boolean formula in conjunctive normal form, whereeach clause consists of at most q literals, is satisfiable. q -SAT is NP-complete for q ≥
3, while2 -Satisfiability (2 -SAT ) is linear-time solvable [1]. The recently introduced multistagemodel [17, 24] takes a sequence of instances of some decision problem (e.g., modeling oneinstance that evolved over time), and asks whether there is a sequence of solutions to themsuch that, roughly speaking, any two consecutive solutions do not differ too much. Weintroduce q -SAT in the multistage model, defined as follows. Multistage q -SAT (M q SAT)
Input : A set X of variables, a sequence Φ = ( φ , . . . , φ τ ), τ ∈ N , of q -CNF formulas overliterals over X , and an integer d ∈ N . Question : Are there τ truth assignments f , . . . , f τ : X → {⊥ , >} such that (i) for each i ∈ { , . . . , τ } , f i is a satisfying truth assignment for φ i , and (ii) for each i ∈ { , . . . , τ − } , it holds that |{ x ∈ X | f i ( x ) = f i +1 ( x ) }| ≤ d ?Constraint (ii) of M q SAT can also be understood as that the Hamming distance of twoconsecutive truth assignments interpreted as n -dimensional vectors over {⊥ , >} is at most d ,or when considering the sets of variables set true, then the symmetric difference of twoconsecutive sets is at most d .In this work, we focus on M2SAT yet relate most of our results to M q SAT . We study
M2SAT in terms of classic computational complexity and parameterized algorithmics [13]. We identify false and true with ⊥ and > , respectively. a r X i v : . [ c s . CC ] N ov Multistage 2-SAT m para-NP-h.(Thm. 3.4) m para-NP-h.(Thm. 3.4) m = 6 n FPT, no PK(Thm. 4.1) n FPT, no PK(Thm. 4.1) O ( n ) | no 2 o ( n ) † d para-NP-h.(Thm. 3.4) d para-NP-h.(Thm. 3.4) d = 1 | lin. time if d = 0 n − d XP, W[1]-h.(Thm. 5.1) n − d XP, W[1]-h.(Thm. 5.1) n O ( n − d ) | no n o ( n − d ) τ para-NP-h.(Thm. 3.1) τ para-NP-h.(Thm. 3.1) τ = 2 | lin. time if τ = 1 m + d para-NP-h.(Thm. 3.4) m + d para-NP-h.(Thm. 3.4) m = 6 and d = 1 m + n − d FPT, no PK(Thm. 4.3) m + n − d FPT, no PK(Thm. 4.3) no ( n − d ) f ( m ) PK τ + d XP, W[1]-h.(Thm. 3.1 & 6.1) τ + d XP, W[1]-h.(Thm. 3.1 & 6.1) n O ( τ · d ) | no n o ( d ) · f ( τ ) ‡ τ + n − d XP, W[1]-h.(Thm. 3.6) τ + n − d XP, W[1]-h.(Thm. 3.6) no n o ( n − d ) · f ( τ ) m + n FPT, no PK(Thm. 7.1) m + n FPT, no PK(Thm. 7.1) no n f ( m,d ) PK m + τ FPT, PK(Thm. 7.4) m + τ FPT, PK(Thm. 7.4) O ( mτ ) PK | no ( m + τ ) − ε PK ¶ n + τ FPT, PK(Thm. 7.4) n + τ FPT, PK(Thm. 7.4) O ( n τ ) PK | no O ( n − ε τ ) PK § Figure 1
Our results for
Multistage 2-SAT . Each box gives, regarding to a parameterization (toplayer), our parameterized classification (middle layer) with additional details on the correspondingresult (bottom layer). Arrows indicate the parameter hierarchy: An arrow from parameter p to p indicates that p ≤ p . “PK” and “no PK” stand for “polynomial problem kernel” and“no polynomial problem kernel unless NP ⊆ coNP / poly”, respectively. † : unless the ETHbreaks (Thm. 3.4). ‡ : unless the ETH breaks (Thm. 3.1). ¶ : unless NP ⊆ coNP / poly (Thm. 7.6) § : unless NP ⊆ coNP / poly (Thm. 3.1). Motivation.
In theory as well as in practice, it is common to model problems as q -SAT - oreven 2 -SAT -instances. Once being modeled, established solvers specialized on q -SAT areemployed. In some cases, a sequence of problem instances (e.g., modeling a problem instancethat changes over time) is to solve such that any two consecutive solutions are similar in someway (e.g., when costs are inferred for setup changes). Hence, when following the previouslydescribed approach, each problem instance is first modeled as a q -SAT instance such that asequence of q -SAT -instances remains to be solved. Comparably to the single-stage setting,understanding the multistage setting could give raise to a general approach for solvingdifferent (multistage) problems. With M q SAT we introduce the first problem that modelsthe described setup. Note that, though a lot of variants of q -SAT exist, M q SAT is one ofthe very few variants that deal with a sequence of q -SAT -instances [34]. Our Contributions.
Our results for
Multistage 2-SAT are summarized in Figure 1. Weprove
Multistage 2-SAT to be NP-hard, even in fairly restricted cases: (i) if d = 1 and . Fluschnik 3 the maximum number m of clauses in any stage is six, or (ii) if there are only two stages.These results are tight in the sense that M2SAT is linear-time solvable when d = 0 or τ = 1.While NP-hardness for d = 1 implies that there is no ( n + m + τ ) f ( d ) -time algorithm for anyfunction f unless P = NP, where n denotes the number of variables, we prove that whenparameterized by the dual parameter n − d (the minimum number of variables not changingbetween any two consecutive layers), M2SAT is W[1]-hard and solvable in O ∗ ( n O ( n − d ) )time. We prove this algorithm to be tight in the sense that, unless the Exponential TimeHypothesis (ETH) breaks, there is no O ∗ ( n o ( n − d ) )-time algorithm. Further, we prove that M2SAT is solvable in O ∗ (2 O ( n ) ) time but not in O ∗ (2 o ( n ) ) time unless the ETH breaks.Likewise, we prove that M2SAT is solvable in O ∗ ( n O ( τ · d ) ) time but not in O ∗ ( n o ( d ) · f ( τ ) ) timefor any function f unless the ETH breaks. As to efficient and effective data reduction, weprove M2SAT to admit problem kernelizations of size O ( m · τ ) and O ( n τ ), but none ofsize ( n + m ) O (1) , O (( n + m + τ ) − ε ), or O ( n − ε τ ), ε >
0, unless NP ⊆ coNP / poly. Related Work. q -SAT is one of the most famous decision problems with a central role inNP-completeness theory [12, 30], for the (Strong) Exponential Time Hypothesis [28, 29], andin the early theory on kernelization lower bounds [6, 23], for instance. In contrast to q -SAT with q ≥
3, 2 -SAT is proven to be polynomial- [31], even linear-time [1] solvable. Severalapplications of 2 -SAT are known (see, e.g., [11, 18, 25, 33]). In the multistage model, variousproblems from different fields were studied, e.g. graph theory [2, 3, 10, 21, 22, 24], facilitylocation [17], knapsack [5], or committee elections [8]. Also variations to the multistagemodel were studied, e.g. with a global budget [26], an online-version [4], or using differentdistance measures for consecutive stages [8, 22].
We denote by N and N the natural numbers excluding and including zero, respectively.Frequently, we will tacitly make use of the fact that for every n ∈ N , 0 ≤ k ≤ n , it holds truethat 1 + P ki =1 (cid:0) ni (cid:1) = P ki =0 (cid:0) ni (cid:1) ≤ n k ≤ n k . Satisfiability.
Let X denote a set of variables. A literal is a variable that is either positiveor negated (we denote the negation of x by ¬ x ). A clause is a disjunction over literals. Aformula φ is in conjunctive normal form (CNF) if it is of the form V i C i , where C i is a clause.A formula φ is in q -CNF if it is in CNF and each clause consists of at most q literals. Antruth assignment f : X → {⊥ , >} is satisfying for φ (or satisfies φ ) if each clause is satisfied,which is the case if at least one literal in the clause is evaluated to true (a positive variableassigned true, or a negated variable assigned false). For a, b ∈ {⊥ , >} , let a ⊕ b := ⊥ if a = b ,and a ⊕ b := > otherwise. For X ⊂ X , an truth assignment f : X → {⊥ , >} is called partial . We say that we simplify a formula φ given a partial truth assignment f (we denotethe simplified formula by φ [ f ]) if each variable x ∈ X is replaced by f ( x ), and then eachclause containing an evaluated-to-true literal is deleted. Parameterized Algorithmics.
A parameterized problem L is a set of instances ( x, p ) ∈ Σ ∗ × N , where Σ is a finite alphabet and p is referred to as the parameter. A parameterizedproblem L is (i) fixed-parameter tractable (in FPT) if each instance ( x, p ) can be decided The O ∗ -notation suppresses factors polynomial in the input size. Multistage 2-SAT for L in f ( p ) · | x | O (1) time, and (ii) in XP if each instance ( x, p ) can be decided for L in | x | g ( p ) time, where f, g are computable functions only depending on p . If L is W[1]-hard, it ispresumably not in FPT. A problem bikernelization for a parameterized problem L to aparameterized problem L takes any instance ( x, p ) of L and maps it in polynomial time toan equivalent instance ( x , p ) of L (the so-called problem bikernel) such that | x | + p ≤ f ( p )for some computable function f . A problem kernelization is a problem bikernelizationwhere L = L . If f is a polynomial, the problem (bi)kernelization is said to be polynomial. Aparametric transformation from a parameterized problem L to a parameterized problem L maps any instance ( x, p ) of L in f ( p ) · | x | O (1) time to an equivalent instance ( x , p ) of L such that p ≤ g ( p ) for some functions f, g each only depending on p . If there is a parametrictransformation from L to L with L being W[1]-hard, then L is W[1]-hard. If f ( p ) ∈ p O (1) and g ( p ) ∈ O ( p ), then we have a linear parametric transformation [27]. If there is a linearparametric transformation from a problem L to problem L with L admitting a problemkernelization of size h ( p ), then L admits a problem bikernelization of size h ( p ). Preprocessing on
Multistage 2-SAT . Due to the following data reduction, we can safelyassume each stage to admit a satisfying truth assignment. (cid:73)
Reduction Rule 1.
If a stage exists with no satisfying truth assignment, then return no .Reduction Rule 1 is correct and applicable in linear time. Multistage 2-SAT is linear-time solvable if the input consists of only one stage, or if all ornone variables are allowed to change its truth assignment between two consecutive stages. (cid:73)
Observation 1.
Multistage 2-SAT is linear-time solvable if (i) τ = 1 , (ii) d = 0 , or(iii) d = n . Proof.
Let ( X, Φ = ( φ , . . . , φ τ ) , d ) be an instance of M2SAT . Case (i) : τ = 1. Polynomial-time many-one reduction to 2 -SAT with instance ( X, φ ). Case (ii) : d = 0. Polynomial-timemany-one reduction to 2 -SAT with instance ( X, φ ), wheret φ = V τi =1 φ i , Case (iii) : d = n .Solve each of the τ instances ( X, φ ) , . . . , ( X, φ τ ) of 2 -SAT individually (Turing reduction). (cid:74) We will prove that the cases (i) and (ii) in Observation 1 are tight:
Multistage 2-SAT becomes NP-hard if τ ≥ d = 1 (Section 3.2). For the case (iii)in Observation 1 the picture looks different: we prove Multistage 2-SAT to be polynomial-time solvable if n − d ∈ O (1) (Section 5). In this section, we prove that
Multistage 2-SAT becomes NP-hard if τ ≥
2. In fact, weprove the following. (cid:73)
Theorem 3.1.
Multistage 2-SAT is NP -hard, even for two stages, where the variablesappear all negated in one and all positive in the other stage. Moreover, Multistage 2-SAT (i) is W[1] -hard when parameterized by d even if τ = 2 , (ii) admits no n o ( d ) · f ( τ ) -time algorithm for any function f unless the ETH breaks, and (iii) admits no problem kernelization of size O ( n − ε · f ( τ )) for any ε > and function f ,unless NP ⊆ coNP / poly . . Fluschnik 5 We will reduce from the following NP-hard problem:
Weighted 2-SAT
Input : A set of variables X , a 2-CNF φ over X , and an integer k . Question : Is there satisfying truth assignment for φ with at most k variables set true?When parameterized by the number k of set-to-true variables, Weighted 2-SAT is W[1]-complete [15, 20]. Moreover,
Weighted 2-SAT admits no n o ( k ) -time algorithm unless theETH breaks [9] and no problem bikernelization of size O ( n − ε ), ε >
0, unless NP ⊆ coNP / poly [14]. (cid:73) Construction 1.
Let (
X, φ, k ) be an instance of
Weighted 2-SAT , where φ = V mi =1 C i .Construct Φ = ( φ , φ ), where φ := φ and φ := V x ∈ X ( ¬ x ) consists of n size-one clauses,where each variable appears negated in one clause. Finally, set d := k . (cid:5) (cid:73) Lemma 3.2.
Let I = ( X, φ, k ) be an instance of Weighted 2-SAT , and let I = ( X, Φ , d ) be an instance of Multistage 2-SAT obtained from I using Construction 2. Then, I is a yes -instance if and only if I ’ is a yes -instance. Proof. ( ⇒ ) Let f be a satisfying truth assignment for I . We claim that f : X → {⊥} , x
7→ ⊥ , and f := f is a solution to I . Note that f and f satisfy φ and φ , respectively.Moreover, |{ x ∈ X | f ( x ) = f ( x ) }| = |{ x ∈ X | f ( x ) = >}| ≤ d = k .( ⇐ ) Let ( f , f ) be a solution to I . Note that f : X → {⊥} . Since d = k , there areat most k variables set to true by f . Hence, f is a satisfying truth assignment for I withat most k variables set to true, and thus I is yes -instance. (cid:74) Proof of Theorem 3.1.
Construction 1 forms a polynomial-time many-one reduction to aninstance with two stages with d = k . Note that Weighted 2-SAT remains NP-hard if allliterals are positive (e.g., via a reduction from
Vertex Cover ). Hence,
M2SAT is NP-hard,even if the variables appear all negated in one and all positive in the other stage. Moreover,unless the ETH breaks,
M2SAT admits no n o ( d ) · f ( τ ) -time algorithm for any function f sinceno n o ( k ) -time algorithm exists for Weighted 2-SAT [9]. As Construction 1 also forms aparametric transformation,
M2SAT is W[1]-hard when parameterized by d even if τ = 2.Moreover, Construction 1 forms a linear parametric transformation from Weighted 2-SAT parameterized by | X | to M2SAT parameterized by n · f ( τ ) for any function f . Hence, M2SAT admits no problem kernel of size O ( n − ε · f ( τ )) for any ε > f ,unless NP ⊆ coNP / poly. (cid:74)(cid:73) Remark 3.3.
Theorem 3.1(iii) can be generalized to
Multistage q -SAT : Instead from Weighted -SAT , we reduce (in an analogous way) from Weighted q -SAT which admitsno problem bikernelization of size O ( n q − ε ), ε >
0, unless NP ⊆ coNP / poly [14]. Thus,unless NP ⊆ coNP / poly, Multistage q -SAT admits no problem kernel of size O ( n q − ε · f ( τ ))for any ε > f . In this section, we prove that
Multistage 2-SAT becomes NP-hard if d = 1 and themaximum number m of clauses in any stage is six. In fact, we prove the following. (cid:73) Theorem 3.4.
Multistage 2-SAT is NP -hard, even if the number of clauses in eachstage is at most six and d = 1 . Moreover, unless the ETH breaks, Multistage 2-SAT admits no O ∗ (2 o ( n ) ) -time algorithm. Multistage 2-SAT (cid:73)
Construction 2.
Let (
X, φ ) be an instance of , where φ = V mi =1 C i and each clauseconsists of exactly three literals. Let ‘ ji , j ∈ { , , } , denote the literals in C i for each i ∈{ , . . . , m } . Construct instance ( X , Φ , d ) of M2SAT as follows. First, construct X := X ∪ B ,where B := { b , b , b } . Let φ B := ( b ∨ b ) ∧ ( b ∨ b ) ∧ ( b ∨ b ) , and φ ¬ B := ( ¬ b ∨ ¬ b ) ∧ ( ¬ b ∨ ¬ b ) ∧ ( ¬ b ∨ ¬ b ) . Next, construct Φ := ( φ i , . . . , φ m ) as follows. For each i ∈ { , . . . , m } , construct φ i − := φ ¬ B , and φ i := ( ‘ i ∨ b ) ∧ ( ‘ i ∨ b ) ∧ ( ‘ i ∨ b ) ∧ φ B Finally, set d := 1. (cid:5) (cid:73) Observation 2.
In every solution to an instance obtained from Construction 2, in eachodd stage exactly two b j are set to false and in each even stage exactly two b j are set to true. Proof.
Clearly, in every satisfying truth assignment for φ i − = φ ¬ B , i ∈ { , . . . , m } , at leasttwo b j are set to false. In every satisfying truth assignment for φ i , to satisfy φ B , at leasttwo variables from B must be set to true. As d = 1, exactly one of B being set to false in asatisfying truth assignment φ i − can be set to true in a satisfying truth assignment for φ i ,which implies that already one of B must be set to true in φ i − . (cid:74)(cid:73) Lemma 3.5.
Let I = ( X, φ ) be an instance of , and let I = ( X , Φ , d ) be aninstance of Multistage 2-SAT obtained from I using Construction 2. Then, I is a yes -instance if and only if I ’ is a yes -instance. Proof. ( ⇒ ) Let f : X → {⊥ , >} be a satisfying truth assignment for φ . We construct truthassignments f , . . . , f τ : X → {⊥ , >} as follows. Let f i ( x ) = f ( x ) for all i ∈ { , . . . , τ } andall x ∈ X . Next, for each i ∈ { , . . . , m } , f i assigns exactly two variables from B to true suchthat φ i is satisfied. This is possible since at least one clause from φ i is already set to true byone true literal. It remains to show that for each i ∈ { , . . . , m } , there is an truth assignmentof f i − to the variables from B such that exactly two are set to false (in which case φ i − is satisfied), and |{ b ∈ B | f i − ( b ) = f i − ( b ) }| ≤ i = 1, interpret f i − = f i )and |{ b ∈ B | f i − ( b ) = f i ( b ) }| ≤
1. Since | B | = 3, there is a j ∈ { , . . . , } suchthat f i − ( b j ) = f i ( b j ) = > . Set f i − ( b j ) = > and f i − ( b ‘ ) = ⊥ for ‘ ∈ { , . . . , } \ { j } .Observe that |{ b ∈ B \ { b j } | f i − ( b ) = f i − ( b ) }| = |{ b ∈ B \ { b j } | f i − ( b ) = f i ( b ) }| = 1,what we needed to show.( ⇐ ) By Observation 2, between every two consecutive stages, exactly one variable in B changes its true-false value. Hence, each variable from X is assigned the same value in eachstage, i.e., f i ( x ) = f j ( x ) for every x ∈ X and every i, j ∈ { , . . . , τ } . Let f : X → {⊥ , >} bethe truth assignment of the variables in X with f ( x ) := f ( x ) for all x ∈ X . Since in everyeven stage, by Observation 2, exactly one variable from B is set to false, at least one literalmust be set to true. It follows that each clause in the instance is satisfied by f , thatis, f is a satisfying truth assignment for I . Thus, I is a yes -instance. (cid:74) Proof of Theorem 3.4.
Construction 2 forms a polynomial-time many-one reduction toan instance with d = 1, m = 6, and n = | X | + 3. Hence, M2SAT is NP-hard, evenif d = 1 and m = 6, and, unless the ETH breaks, admits no O ∗ (2 o ( n ) )-time algorithm sinceno O ∗ (2 o ( | X | ) )-time algorithm exists for [9]. (cid:74) . Fluschnik 7 k Allowed Changes
In this section, we prove that
Multistage 2-SAT is W[1]-hard when parameterized by thelower bound n − d on the number of unchanged variables between any two consecutive stages. (cid:73) Theorem 3.6.
Multistage 2-SAT is W[1] -hard when parameterized by n − d evenif τ = 2 , and, unless the ETH breaks, admits no O ∗ ( n o ( n − d ) · f ( τ ) ) -time algorithm for anyfunction f . We reduce from the following NP-hard problem:
Multicolored Independent Set (MIS)
Input : An undirected, k -partite graph G = ( V , . . . , V k , E ). Question : Is there an independent set S such that | S ∩ V i | = 1 for all i ∈ { , . . . , k } ? MIS is W[1]-hard with respect to k [19] and unless the ETH breaks, there is no f ( k ) · n o ( k ) -timealgorithm [32]. (cid:73) Construction 3.
Let I = ( G = ( V , . . . , V k , E )) be an instance of MIS and let V := V ] · · · ] V k , n := | V | , and V i = { v i , . . . , v i | V i | } for all i ∈ { , . . . , k } . We constructan instance I = ( X, ( φ , φ ) , d ) with d := n − k as follows. Let X := X ∪ · · · ∪ X k with X i = { x ij | v ij ∈ V i } for all i ∈ { , . . . , k } . Let for all i ∈ { , . . . , k } φ ∗ i := ^ j,j ∈{ ,..., | V i |} , j = j ( ¬ x ij ∨ ¬ x ij ) , and let φ E := ^ { v ij ,v i j }∈ E ( ¬ x ij ∨ ¬ x i j ) . Let φ := ^ x ∈ X ( x ) and φ := φ E ∧ ^ i ∈{ ,...,k } φ ∗ i . This finishes the construction. (cid:5) (cid:73)
Lemma 3.7.
Let I = ( G = ( V , . . . , V k , E )) be an instance of MIS , and let I =( X, ( φ , φ ) , d ) be an instance of Multistage 2-SAT obtained from I using Construction 3.Then, I is a yes -instance if and only if I ’ is a yes -instance. Proof. ( ⇒ ) Let S = { v j , . . . , v kj k } ⊆ V be an independent set with S ∩ V i = { v ij i } forall i ∈ { , . . . , k } . Let X S := { x j , . . . , x kj k } be the variables in X corresponding to thevertices in S . Let f : X → {⊥ , >} , x
7→ > and f : X → {⊥ , >} be defined as f ( x ) = ( > , if x ∈ X S , ⊥ , otherwise . Clearly, f satisfies φ . Further, observe that for each r ∈ { , . . . , k } , f satisfies φ ∗ r sinceall variables from X i except for x ij i is set to ⊥ . Since S is an independent set, and onlyvariables corresponding to vertices from S are set to true by f , f satisfies φ E . It followsthat f satisfies φ , and hence, f = ( f , f ) is a solution for I .( ⇐ ) Let f = ( f , f ) be a solution to I . Let S := { v ij ∈ V | f ( v ij ) = >} . We claimthat S is an independent set in G with | S ∩ V i | = 1 for all i ∈ { , . . . , k } .First, observe that S is an independent set in G : Suppose not, then there are v ij , v i j ∈ S such that { v ij , v i j } ∈ E . By construction, f ( x ij ) = f ( x i j ) = > . Since φ E contains theclause ( ¬ x ij ∨ ¬ x i j ), f does not satisfy φ E (and, thus, φ ), contradicting the fact that f is asolution. It follows that S is an independent set in G . Multistage 2-SAT
It remains to show that | S ∩ V i | = 1 for all i ∈ { , . . . , k } . Observe that for all i ∈{ , . . . , k } , |{ x ∈ X i | f ( x ) = >}| ≤
1. Suppose not, that is, there is some i ∈ { , . . . , k } such that there are x, y ∈ X i with x = y and f ( x ) = f ( y ) = > . By construction, φ ∗ i contains the clause ( ¬ x ∨ ¬ y ), which is evaluated to false under f . This is a contradictionto the fact that f is a solution.Observe that for all i ∈ { , . . . , k } , |{ x ∈ X i | f ( x ) = >}| >
0: By construction of φ , weknow that f ( x ) = > for all x ∈ X . Since d = n − k , there are at least k vertices set to trueby f . If for some i ∈ { , . . . , k } we have that |{ x ∈ X i | f ( x ) = >}| = 0, then, by the pigeon-hole principle, there is an i ∈ { , . . . , k } with i = i such that |{ x ∈ X i | f ( x ) = >}| ≥ | S ∩ V i | = 1 for all i ∈ { , . . . , k } . It followsthat S is a solution to I . (cid:74) Proof of Theorem 3.6.
Construction 3 runs in polynomial time and outputs an equivalentinstance (Lemma 3.7) with two stages and d = n − k . As Construction 1 also forms aparametric transformation, M2SAT is W[1]-hard when parameterized by n − d even if τ = 2.Moreover, unless the ETH breaks, M2SAT admits no n o ( n − d ) · f ( τ ) -time algorithm for anyfunction f since no n o ( k ) -time algorithm exists for MIS . (cid:74) m + n − d In this section, we prove that
Multistage 2-SAT is fixed-parameter tractable regardingthe number of variables (Section 4.1) and regarding the parameter m + n − d , the maximumnumber of clauses over all input formulas and the minimum number of variables not changingbetween any two consecutive stages (Section 4.2). We prove that
Multistage 2-SAT is fixed-parameter tractable regarding the number ofvariables. (cid:73)
Theorem 4.1.
Multistage 2-SAT is solvable in O (min { n n d , n } · τ · ( n + m )) time. Proof.
Let I = ( X, φ, d ) be an instance of
M2SAT with Φ = ( φ , . . . , φ τ ). Construct thedigraph D with vertex set V = V ] · · · ] V τ ] { s, t } and arc set A as follows. Add twodesignated vertices s and t to V . For each i ∈ { , . . . , τ } , for every truth assignment f satisfying φ i , add a vertex v if to V i . Note that there are at most 2 n truth assignments, wherewe can test for each truth assignment whether it is satisfying in O ( n + m ) time. Add thearc ( s, v ) for all v ∈ V and the arc ( v, t ) for all v ∈ V τ . Moreover, for each i ∈ { , . . . , τ − } ,add the arc ( v if , v i +1 g ) if and only if |{ x ∈ X | f ( x ) = g ( x ) }| ≤ d . This finishes theconstruction of D . Note that | V i | ≤ n , and each vertex (except for s ) has outdegree atmost P dj =1 (cid:0) nj (cid:1) ≤ n d . Hence, | A | ∈ O (min { n n d , n } τ ).It is not difficult to see that D admits an s - t path if and only if I is a yes -instance (see,e.g., [8, 21, 22]). Checking whether D admits an s - t path can be done in O ( | V | + | A | ). (cid:74)(cid:73) Remark 4.2.
Theorem 4.1 is asymptotically optimal regarding n unless the ETH breaks (The-orem 3.4). Moreover, Theorem 4.1 is easily adaptable to Multistage q -SAT with q ≥ q ≥
3, the number of truth assignments is 2 n and each is verifiable in linear time. . Fluschnik 9 m + n − d We prove that
Multistage 2-SAT is fixed-parameter tractable regarding the parameter m + n − d . (cid:73) Theorem 4.3.
Multistage 2-SAT is solvable in O (4 m + n − d ) τ ( n + m )) time. To prove Theorem 4.3, we will show that either Theorem 4.1 applies with n ≤ m + n − d )or the following. (cid:73) Lemma 4.4.
Multistage 2-SAT solvable in O ( τ ( n + m )) time if m < d . Proof.
Let I = ( X, φ, d ) be an instance of
M2SAT with Φ = ( φ , . . . , φ τ ) on n variables andeach formula contains at most m clauses. Due to Reduction Rule 1, we can safely assume thateach formula of Φ admits a satisfying truth assignment. Let X i ⊆ X be the set of variablesappearing as literals in φ i for each i ∈ { , . . . , τ } . Note that | X i | ≤ m for each i ∈ { , . . . , τ } .Compute in linear time a satisfying truth assignment f : X → {⊥ , >} for φ . Computefor each i ∈ { , . . . , τ } in linear time a satisfying truth assignment f i : X i → {⊥ , >} for φ i .Next, iteratively for i = 2 , . . . , τ , set for all x ∈ Xf i ( x ) = ( f i ( x ) , if x ∈ X i ,f i − ( x ) , if x ∈ X \ X i . Clearly, truth assignment f i satisfies φ i . Moreover, for all i ∈ { , . . . , τ } it holds that |{ x ∈ X | f i − ( x ) = f i ( x ) }| ≤ | X i | ≤ m < d , and hence ( f , . . . , f τ ) is a solution to I . (cid:74) Proof of Theorem 4.3.
Let I = ( X, φ, d ) be an instance of
M2SAT with Φ = ( φ , . . . , φ τ )on n variables and each formula contains at most m clauses. We distinguish how 2( m + n − d )relates to 2 n − d . Case 1 : m + n − d ) ≥ n − d . Since d ≤ n , it follows that 2( m + n − d ) ≥ n . Dueto Theorem 4.1, we can solve I in O (min { n n d , n } τ ( n + m )) ⊆ O (4 m + n − d ) τ ( n + m ))time. Case 2 : m + n − d ) < n − d . We have that2( m + n − d ) < n − d ⇐⇒ m < d. Due to Lemma 4.4, we can solve I in O ( τ ( n + m )) time. (cid:74)(cid:73) Remark 4.5.
Theorem 4.3 can be adapted for
Multistage q -SAT for every q ≥ qm < d and we check for a satisfying truth assignment foreach stage in O ∗ (2 qm ) time. To adapt the proof of Theorem 4.3, we then relate q ( m + n − d )with qn − ( q − d and either employ the adapted Theorem 4.1 (see Remark 4.2), or theadapted Lemma 4.4. We prove that
Multistage 2-SAT is in XP when parameterized by the lower bound n − d on non-changes between consecutive stages, the parameter “dual” to d . (cid:73) Theorem 5.1.
Multistage 2-SAT is solvable in O ( n n − d )+1 · n − d ) τ ( n + m )) time. Let I = ( X, Φ = ( φ , . . . , φ τ ) , d ) be a fixed yet arbitrary instance with n variables. Two partialtruth assignments f Y : Y → {⊥ , >} and f Z : Z → {⊥ , >} with Y, Z ⊆ X are called compatible if for all x ∈ Y ∩ Z it holds that f Y ( x ) = f Z ( x ). For two compatible assignments f Y , f Z , wedenote by f Y ∪ f Z := ( f Y ( x ) , x ∈ Y,f Z ( x ) , x ∈ Z \ Y. With a similar idea as in the proof of Theorem 4.1, we will construct a directed graph withterminals s and t such that there is an s - t path in G if and only if I is a yes -instance. (cid:73) Construction 4.
Given I , we construct a graph G = ( V, E ) with vertex set V := V → ∪ V → ∪ · · · ∪ V τ − → τ ∪ { s, t } , where for each Y, Z ∈ (cid:0) Xn − d (cid:1) , we have that ( f Y , f Z ) ∈ V i → i +2 if and only if f Y , f Z arecompatible and each of φ i [ f Y ], φ i +1 [ f Y ∪ f Z ], and φ i +2 [ f Z ] is satisfiable, and the followingarcs: (i) ( s, v ) for all v ∈ V → , (ii) ( v, t ) for all v ∈ V τ − → τ , and (iii) (( f Y , f Z ) , ( f Y , f Z )) ∈ V i → i +2 × V j → j +2 if j = i + 1 and f Z = f Y (implying that Z = Y ). (cid:5) (cid:73) Lemma 5.2.
Construction 4 computes a graph of size O ( n n − d )+1 · n − d ) τ ) and can bedone in O ( n n − d )+1 · n − d ) τ ( n + m )) time. Proof.
To construct a set V i → i +2 , we compute each tuple ( f Y , f Y ) in O ( n n − d ) · n − d ) )time, and check whether they are compatible in O ( n + m ) time, and whether each of φ i [ f Y ], φ i +1 [ f Y ∪ f Z ], and φ i +2 [ f Z ] is satisfiable, each in O ( n + m ) time. Since for any f Y , f Z we can check whether f Y = f Z in O ( n ) time, we add the O ( n n − d ) · n − d ) ) many arcsfrom V i → i +2 to V i +1 → i +3 in O ( n n − d )+1 · n − d ) ) time. In total, G can be constructedin O ( n n − d )+1 · n − d ) τ ( n + m )) time. (cid:74)(cid:73) Lemma 5.3.
Let I be an instance of Multistage 2-SAT and let G be the graph obtainedfrom applying Construction 4 to I . Then, I is a yes -instance if and only if G admitsan s - t paths. Proof. ( ⇒ ) Let f = ( f , . . . , f τ ) be a solution to I . For each i ∈ { , . . . , τ − } , since |{ x ∈ X | f i ( x ) = f i +1 ( x ) }| ≤ d , there is a set Y i ⊆ { x ∈ X | f i ( x ) = f i +1 ( x ) } with | Y i | = n − d . Observe that v fi := ( f i | Y i , f i +1 | Y i +1 ) ∈ V i → i +2 : Clearly φ i [ f i | Y i ] and φ i [ f i +2 | Y i +1 ]satisfiable. Note that f i | Y i , f i +1 | Y i +1 are compatible since Y i ∩ Y i +1 ⊆ { x ∈ X | f i ( x ) = f i +1 ( x ) = f i +2 ( x ) } . Moreover, φ i +1 [ f i | Y i ∪ f i +1 | Y i +1 ] is satisfiable since f i +1 ( x ) = f i | Y i ∪ f i +1 | Y i +1 ( x ) for all x ∈ Y i ∪ Y i +1 . It follows that there is an s - t path in G with the arcsequence (( s, v f ) , ( v f , v f ) , . . . , ( v fτ − , t )).( ⇐ ) Let P be an s - t path in G . By construction of G , P contains s , t , and fromeach V i → i +2 exactly one vertex. Moreover, if arc (( f X , f Y ) , ( f X , f Y )) is contained in P ,then f Y = f X . Let ( s, (( f Y i , f Y i +1 )) τ − i =1 , t ) be the sequence of vertices in P . We know thatthere exists an f : X \ Y → {⊥ , >} that satisfies φ [ f Y ], and hence f := f ∪ f Y satisfies φ .Moreover, we know that for all i ∈ { , . . . , τ − } , there exists f i : X \ ( Y i − ∪ Y i ) → {⊥ , >} that satisfies φ i [ f Y i − ∪ f Y i ], and hence f i := f i ∪ f Y i − ∪ f Y i satisfies φ i . Finally, we know thatthere exists an f τ : X \ Y τ − → {⊥ , >} that satisfies φ τ [ f Y τ − ], and hence f τ := f τ ∪ f Y τ − satisfies φ τ .It remains to show that |{ x ∈ X | f i ( x ) = f i +1 ( x ) }| ≤ d for all i ∈ { , . . . , τ − } . Notethat f i ( x ) = f i +1 ( x ) for all x ∈ Y i , and since | Y i | = n − d , the claim follows. (cid:74) . Fluschnik 11 Algorithm 1
XP-algorithm on input instance (
X, φ, d ). foreach X ⊆ X : | X | ≤ τ · d do // n τ · d many foreach f : X → {⊥ , >} do // | X | many φ ∗ ← simplify ( φ , f ); foreach g , g , . . . , g τ : g i ∈ F ( X ) ∀ i ∈ { , . . . , τ } do // τ | X | τ · d many foreach i ∈ { , . . . , τ } do f i ( x ) ← f i − ( x ) ⊕ g i ( x ) ∀ x ∈ X ; φ ∗ i ← simplify ( φ i , f i ); if ( X \ X , ( φ ∗ , . . . , φ ∗ τ ) , is a yes -instance of M2SAT then return yes // decidable in linear time (Observation 1) return no Proof of Theorem 5.1.
Given an instance I = ( X, Φ = ( φ , . . . , φ τ ) , d ) of M2SAT , ap-ply Construction 4 in O ( n n − d )+1 · n − d ) τ ( n + m )) time to obtain graph G with terminals s and t of size O ( n n − d )+1 · n − d ) τ ) (Lemma 5.2). Return, in time linear in the size of G , yes if G admits an s - t path, and no otherwise (Lemma 5.3). (cid:74)(cid:73) Remark 5.4.
Theorem 5.1 is asymptotically optimal regarding n − d unless the ETHbreaks (Theorem 3.6). Moreover, Theorem 5.1 does not generalize to Multistage q -SAT for q ≥
3, as M q SAT is already NP-hard for one stage and hence for any number n − d . In this section, we prove that
Multistage 2-SAT is in XP when parameterized by τ + d . (cid:73) Theorem 6.1.
Multistage 2-SAT is solvable in O ( n τ · d · τ · d +1 · τ · ( n + m )) time. Let I = ( X, Φ = ( φ , . . . , φ τ ) , d ) be a fixed yet arbitrary instance with τ · d < n , asotherwise Theorem 4.1 applies. On a high level, our Algorithm 1 works as follows: (1) Guess q ≤ τ · d variables X ⊆ X that will change over time. (2) Guess an initial truth assignment of the variables in X . (3) For each but the first stage, guess the at most min { q, d } possible variables to change. (4) Set the variables to the guessed true or false values, delete clauses which are set to true. (5)
Return yes if the resulting instance with d = 0 is a yes -instance (linear-time checkable). (6) If the algorithm never (for all possible guesses) returned yes , then return no .For any X ⊆ X , define the set of all truth assignments to variables of X with atmost min {| X | , d } true values by F ( X ) := (cid:8) f : X → {⊥ , >} (cid:12)(cid:12) |{ x ∈ X | f ( x ) = >}| ≤ min {| X | , d } (cid:9) . With the next two lemmas, we prove that Algorithm 1 is correct and runs in XP-timeregarding τ + d . (cid:73) Lemma 6.2.
Algorithm 1 returns yes if and only if the input instance is a yes -instance.
Proof. ( ⇒ ) If Algorithm 1 returns yes , then for some X ⊆ X , and some f , . . . , f τ thatsimplified φ , . . . , φ τ to φ ∗ , . . . , φ ∗ τ , instance I ∗ := ( X \ X , ( φ ∗ , . . . , φ ∗ τ ) ,
0) is a yes -instanceof
M2SAT . Let f ∗ , . . . , f ∗ τ : X \ X → {⊥ , >} be a solution to I ∗ . Let h , . . . , h τ be defined as h i ( x ) := f i ( x ) if x ∈ X , and h i ( x ) := f ∗ i ( x ) otherwise, i.e., if x ∈ X \ X . We claimthat ( h , . . . , h τ ) is a solution to I . Observe that h i satisfies φ i for each i ∈ { , . . . , τ } .Moreover, for each i ∈ { , . . . , τ − } we have |{ x ∈ X | h i ( x ) = h i +1 ( x ) }| = |{ x ∈ X | g i +1 ( x ) = >}| ≤ d .( ⇐ ) Let h = ( h , . . . , h τ ) be a solution to I . Let X ⊆ X with | X | ≤ τ · d the set ofall variables which change at least once over the stages their true-false value. Algorithm 1guesses X in line 1. Let f = ( f , . . . , f τ ) be such that for each i ∈ { , . . . , τ } , f i : X →{⊥ , >} is h i restricted to the variables in X . In line 2, Algorithm 1 guesses f . Since h isa solution to I , we know that |{ x ∈ X | h i ( x ) = h i +1 ( x ) }| = |{ x ∈ X | f i ( x ) = f i +1 ( x ) }| ≤ min {| X | , d } for each i ∈ { , . . . , τ − } . It follows that for each i ∈ { , . . . , τ } there existsa g i ∈ F ( X ) such that f i ( x ) = f i − ( x ) ⊕ g i ( x ). Algorithm 1 guesses g , . . . , g τ in line 4,and finds f in line 6. Let ( φ ∗ , . . . , φ ∗ τ ) be the formulas ( φ , . . . , φ τ ) simplified according to f ,as done by Algorithm 1 in line 3 and line 6. Since h is a solution, f = ( f , . . . , f τ ) wherefor each i ∈ { , . . . , τ } , f i : X \ X → {⊥ , >} is h i restricted to the variables in X \ X , is asolution to ( X \ X , ( φ ∗ , . . . , φ ∗ τ ) , X \ X , ( φ ∗ , . . . , φ ∗ τ ) ,
0) is a yes -instance, andconsequently Algorithm 1 returns yes in line 8. (cid:74)(cid:73)
Lemma 6.3.
Algorithm 1 runs in O ( n τ · d · τ · d +1 τ ( n + m )) time. Proof.
The running time T ( I ) is T ( I ) ≤ (1 + n τ · d ) · T ( I ), where T ( I ) is the worst-caserunning time inside the first for-loop (line 2 to line 8). Analogously, we have T ( I ) ≤ τ · d · T ( I ), and T ( I ) ∈ O ( n + m ) + (1 + ( τ · d ) d ) τ − · T ( I ). Now, T ( I ) ∈ O ( τ ( n + m )),as line 6 can be done in O ( n + m ) time with ( τ −
1) executions of this line, and checking theif-condition for line 8 can be done in O ( τ ( n + m )) time. We arrive at T ( I ) ∈ O ((1 + n τ · d ) · τ · d · (( n + m ) + (1 + τ · d ) τ · d · τ ( n + m ))) ⊆ O ( n τ · d · τ · d +1 · τ ( n + m )) (cid:74) We are set to prove the main result from this section.
Proof of Theorem 6.1.
Let I = ( X, Φ = ( φ , . . . , φ τ ) , d ) be an instance of M2SAT with n variables and at most m clauses in each stage’s formula. If τ · d ≥ n , then, by Theorem 4.1,we know that M2SAT is solvable in O (2 τ · d · τ ( n + m )) time. Otherwise, if τ · d < n , thenAlgorithm 1 runs in O ( n τ · d · τ · d +1 τ ( n + m )) time (Lemma 6.3) and correctly decides I (Lemma 6.2). (cid:74)(cid:73) Remark 6.4.
Theorem 6.1 is asymptotically optimal regarding d unless the ETH breaks (The-orem 3.1). Moreover, Theorem 6.1 is not adaptable to Multistage q -SAT with q ≥ Multistage q -SAT with q ≥ τ + d ∈ O (1). In this section, we study efficient and provably effective data reduction for
Multistage 2-SAT in terms of problem kernelization. We focus on the parameter combinations n + m , n + τ ,and m + τ . We prove that no problem kernelization of size polynomial in n + m existsunless NP ⊆ coNP / poly (Section 7.1), and that a problem kernelization of size quadraticin m + τ and of size cubic in n + τ exists (Section 7.2). Finally, we prove that no problemkernel of size truly subquadratic in m + τ exists unless NP ⊆ coNP / poly (Section 7.2.1). . Fluschnik 13 When parameterized by n + m , efficient and effective data reduction appears unlikely. (cid:73) Theorem 7.1.
Unless NP ⊆ coNP / poly , Multistage 2-SAT admits no problem kernelof size polynomial in n f ( m,d ) , for any function f only depending on m and d . We will prove Theorem 7.1 via an AND-composition [6, 7], that is, we prove that given t instances of Multistage 2-SAT , each with d = 1 and the same number of variablesand stages, we can compute in polynomial time an instance of Multistage 2-SAT suchthat all input instances are yes if and only if the output instance is yes , and the numberof variables and the maximum number of clauses in one stage does not exceed those fromall input instances. Drucker [16] proved that if a parameterized problem admits an AND-composition from an NP-hard problem, then it admits no polynomial problem kernelization,unless NP ⊆ coNP / poly. (cid:73) Construction 5.
Let I , . . . , I t be t instances of M2SAT with d = 1, m = 6, n variables,and τ stages, where I i = ( X i , Φ i , d ) with X i = { x i , . . . , x in } and Φ i = ( φ i , . . . , φ iτ ). Constructthe instance I := ( X, Φ , d ) as follows. Construct the set X = { x , . . . , x n } of variables, andidentify x j with x ij for each i ∈ { , . . . , t } , j ∈ { , . . . , n } . In a nutshell, we constructthe sequence of formulas by chaining up the input instances’ formulas, and add n stagesbetween any two consecutive instances each consisting of the always-true formula ( x ∨ ¬ x )—these ensure a reconfiguration of the last truth assignment to the initial truth assignmentof the subsequent instance. Formally, construct Φ = ( φ , . . . , φ t · ( τ + n ) ) as follows. Forall i ∈ { , . . . , t } , j ∈ { , . . . , τ + n } , set (where S ( i ) := ( i − · ( τ + n )) φ S ( i )+ j := ( φ ij , if 1 ≤ j ≤ τ, ( x ∨ ¬ x ) , if τ + 1 ≤ j ≤ τ + n. Finally, set d = 1. (cid:5) (cid:73) Lemma 7.2.
Let I , . . . , I t be t instances of Multistage 2-SAT with d = 1 , m = 6 , n variables, and τ stages, and let I be the instance obtained from Construction 5. Then,each I i is a yes -instance if and only if I is a yes -instance. Proof. ( ⇐ ) Let ( f , . . . , f t ( τ + n ) ) be a solution to I . It is not difficult to see that, foreach i ∈ { , . . . , t } , the sequence ( f S ( i )+1 , . . . , f S ( i )+ τ ) is a solution to I i .( ⇒ ) For each i ∈ { , . . . , t } , let ( f i , . . . , f iτ ) denote a solution for I i . We construct asolution f = ( f , . . . , f t ( τ + n ) ) for I as follows. For each i ∈ { , . . . , t } and j ∈ { , . . . , τ } ,set f S ( i )+ j := f ij . For each i ∈ { , . . . , t − } , we define f S ( i )+ τ +1 , . . . , f S ( i )+ τ + n iterativelyas follows. For j = 1 , . . . , n , let f S ( i )+ τ + j ( x ) := ( f S ( i )+ τ +( j − ( x ) , if x ∈ X \ { x j } ,f S ( i +1)+1 ( x ) , if x = x j . Observe that for each j ∈ { , . . . , n } , it holds true that |{ x ∈ X | f S ( i )+ τ +( j − ( x ) = f S ( i )+ τ + j ( x ) }| ≤
1, and that f S ( i )+ τ + n = f S ( i +1)+1 . (cid:74) Proof of Theorem 7.1.
Construction 5 forms an AND-composition (Lemma 7.2) from anNP-hard special case of
M2SAT (Theorem 3.4) to
M2SAT when parameterized by n + m ,in fact, mapping m and d to a constant. Thus, due to Drucker [16], M2SAT admits noproblem kernelization of size polynomial in n f ( m,d ) for any function f only depending on m and d . (cid:74) (cid:73) Remark 7.3.
Due to Theorem 4.1,
Multistage 2-SAT yet admits a problem kernel ofsize 2 O ( n ) . We prove problem kernelizations of size polynomial in n + τ and m + τ . (cid:73) Theorem 7.4.
Multistage 2-SAT admits a linear-time computable problem kernelizationof size O ( n τ ) and of size O ( m · τ ) . We employ the following two immediate reduction rules (each is clearly correct and applicablein linear time): (cid:73)
Reduction Rule 2.
In each stage, delete all but one appearances of a clause in the formula. (cid:73)
Reduction Rule 3.
Delete a variable that appears in no stage’s formula as a literal.
Proof of Theorem 7.4.
Observe that there are at most N := 2 n + (cid:0) n (cid:1) ∈ O ( n ) manypairwise different clauses. After exhaustively applying Reduction Rule 2, we have m ≤ N ∈O ( n ). After exhaustively applying Reduction Rule 3, it follows that for each variable, thereis at least one clause, and hence, n ≤ · m · τ . (cid:74)(cid:73) Remark 7.5.
Theorem 7.4 adapts easily to
Multistage q -SAT . Herein, the problem kernelsizes are O ( n q · τ ) and O ( q · m · τ ).Subsequently, we prove that a linear kernel appears unlikely. (cid:73) Theorem 7.6.
Unless NP ⊆ coNP / poly , Multistage 2-SAT admits no problem kernelof size O (( m + n + τ ) − ε ) for any ε > . To prove Theorem 7.6, we show that there is a linear parametric transformation from
VertexCover parameterized by | V | to Multistage 2-SAT parameterized by n + m + τ . (cid:73) Construction 6.
Let I = ( G, k ) with G = ( V, E ) be an instance of
Vertex Cover .Denote the vertices V = { v , . . . , v n } . We construct the instance I = ( X, Φ , d ) of M2SAT with d = k and Φ = ( φ , φ , . . . , φ n ) as follows. Let X = X V ∪ B with X V = { x i | v i ∈ V } and B = { b , . . . , b k } . Let φ := n ^ i =1 ( ¬ x i ) ∧ k ^ j =1 ( ¬ b j ) and φ i := ^ { v i ,v j }∈ E ( x i ∨ x j ) ∧ (V kj =1 ( b j ) if i mod 2 = 0 , V kj =1 ( ¬ b j ) if i mod 2 = 1 , ∀ i ∈ { , . . . , n } . Note that τ + m + | X | ∈ O ( n ), since each vertex degree is at most n − (cid:5) (cid:73) Lemma 7.7.
Let I = ( G, k ) be an instance of Vertex Cover , and let I = ( X , Φ , d ) be the instance of Multistage 2-SAT obtained from I using Construction 6. Then, I is a yes -instance if and only if I is a yes -instance. EFERENCES 15
Proof. ( ⇒ ) Let V ⊆ V be a size-at-most- k vertex cover of G . Let X W := { x i ∈ X V | v i ∈ W } . Define f : X → {⊥ , >} such that f ( x ) = ⊥ for all x ∈ X . Define f , . . . , f n : X →{⊥ , >} and f ∗ : X V → {⊥ , >} as f i ( x ) = f ∗ ( x ) , if x ∈ X V , > , if x ∈ B and i mod 2 = 0 , ⊥ , if x ∈ B and i mod 2 = 1 , where f ∗ ( x ) = ( > , if x ∈ X W , ⊥ , if x ∈ X V \ X W . Observe that |{ x ∈ X | f ( x ) = f ( x ) }| = |{ x ∈ X V | f ∗ ( x ) = >}| = | X W | = | W | ≤ k .Moreover, for each i ∈ { , . . . , n − } , we have that |{ x ∈ X | f i ( x ) = f i +1 ( x ) }| = | B | = k .It is not difficult to see that f i satisfies φ i for each i ∈ { , . . . , τ } . Hence, ( f , f , . . . , f n ) is asolution to I .( ⇐ ) Let f = ( f , f , . . . , f n ) be a solution to I . By construction of φ , it must holdthat f ( x ) = ⊥ for all x ∈ X . Moreover, by construction of φ , we know that f ( x ) = ⊥ for all x ∈ B , and hence X := { x ∈ X V | f ( x ) = f ( x ) } = { x ∈ X V | f ( x ) = >} has | X | ≤ k . Since for each i ∈ { , . . . , n − } , we have that { x ∈ X | f i ( x ) = f i +1 ( x ) } = B by construction, we know that for each i, j ∈ { , . . . , n } it holds true that f i ( x ) = f j ( x )for all x ∈ X V . We claim that W = { v i ∈ V | x i ∈ X } is a size-at-most- k vertex coverof G . We know that | W | = | X | ≤ k . Suppose towards a contradiction that there is anedge { v i , v j } ∈ E disjoint from W . This implies that f i ( x i ) = f i ( x j ) = ⊥ . By construction, φ i contains the clause ( x i ∨ x j ), which is not satisfied by f i . This contradicts the fact that f is a satisfying truth assignment. It follows that W is a size-at-most- k vertex cover of G , andthus, I is a yes -instance. (cid:74) Proof of Theorem 7.6.
Construction 6 is a linear parametric transformation (Lemma 7.7)such that τ + m + | X | ∈ O ( | V | ). Since Vertex Cover admits no problem bikernelizationof size O ( | V | − ε ), ε > (cid:74)(cid:73) Remark 7.8.
Theorem 7.6 can be easily adapted to
Multistage q -SAT when tak-ing q -Hitting Set as source problem [14], ruling out problem kernelizations of size O (( n + m + τ ) q − ε ), ε > ⊆ coNP / poly). While is linear-time solvable, its multistage model
Multistage 2-SAT is intractablein even surprisingly restricted cases. This is also reflected by the fact that several of ourdirect upper bounds are already asymptotically optimal. By our results, the most interestingdifference between
Multistage 2-SAT and
Multistage q -SAT , with q ≥
3, is that theformer is efficiently solvable if the numbers of stages and allowed consecutive changes areconstant, which is not the case for the latter (unless P = NP). Finally, our results showthat exact solutions are far from practical, waving the path for randomized or heuristicapproaches.
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