On Variable-Weighted 2-SAT and Dual Problems

Transcription

1 SAT 2007, Lissabon, Portugal, May 28-31, 2007 On Variable-Weighted 2-SAT and Dual Problems Stefan Porschen joint work with Ewald Speckenmeyer Institut für Informatik Universität zu Köln Germany

2 Introduction Aim and Focus of this Talk Variable-Weighted Satisfiability Optimization Problems: 1. Exact algorithm for SAT on weighted 2-CNF via identification as special mixed Horn formulas Recall: Unweighted 2-SAT case: Linear time solvable via digraph approach (Aspvall et al., 79) 2. Providing poly-time algorithms for SAT, XSAT, NAESAT on weighted monotone CNF + ( 2) (each variable occurs at most twice) Method: Reduction to appropriate graph problems

3 Introduction Recall: Relevant Variants of SAT XSAT: Input: CNF formula C Question/Searched: Truth assignment setting exactly one literal in each clause of C to true, or output: not existing NAESAT: Input: CNF formula C Question/Searched: Truth assignment setting in each clause one literal to false and one literal to true, or output: not existing Schäfer (1978): XSAT, NAESAT NP-complete for the unrestricted CNF case

4 Introduction Precise Problem Definition: Optimum weight SAT Input: C CNF, w : V (C) R Output: model t of smallest/largest weight among all models of C, or nil if C is not satisfiable Weight of model t is defined as w(t) = x t 1 (1) w(x) = x V (C) w(x)t(x) i.e.: Sum of the weights of all variables set to 1 by t MINW-SAT: minimization, MAXW-SAT: maximization case Similarly, define MINW/MAXW-XSAT and MINW/MAXW- NAESAT on basis of XSAT models, resp., NAESAT models.

5 Introduction Optimum weight model spaces For Π {SAT, XSAT, NAESAT}, (C, w) let T Π (C,w): set of all models of (C, w) T Π min (C,w) TΠ (C, w): all minimum weight models of (C,w) Tmax Π (C, w): analogously

6 Reduction Tools for Variable-Weighted Formulas Reducing Maximization Case to Minimization Case Lemma (MAX-MIN-Lemma) Let Π denote a fixed problem of {SAT, XSAT, NAESAT}. Let A be an algorithm for solving MINW-Π on R-variableweighted instances C CNF in O(f( C )) time then A can easily be modified to A solving MAXW-Π for R-weighted members of CNF in O(f( C )) time. Sketch of Proof: One can show: Thus: Perform A on (C, w). T Π min (C,w) = TΠ max(c, w)

7 Reduction Tools for Variable-Weighted Formulas Bijection-Lemma Proposition: For R-variable-weighted formulas (C,w),(C, w ) assume that there is a model space bijection s.t. F : T Π (C) t t := F(t) T Π (C ) w(t) = w (t ) + α with constant α R independent of t and t. Then: Restriction F µ := F T Π µ (C, w) is a bijection between T Π µ (C,w) and TΠ µ (C, w ), We have T Π µ (C, w) = TΠ µ (C, w ), for µ {min,max}.

8 Optimum Weight 2-SAT Variable weighted 2-SAT: NP-hard minimum weight SAT for variable weighted 2-CNF formulas is NP-hard: Because: Reduction from minimum weight vertex cover special class: monotone formulas; each clause forms an edge of a vertex weighted graph (variables = vertices) For example: Let formula P = (a b) (b c) (a c), Then G P : c a b Obvious: each (minimum weight) vertex cover is a (minimum weight) model

9 Optimum Weight 2-SAT Weighted Horn Formulas Horn formula: at most one positive literal in each clause: Well known: SAT for Horn formulas can be solved in linear time. In contrast to 2-CNF case: Lemma: Minimum weight satisfiability for a Horn formula H and positive weight function w : V (H) R +, can be solved in linear time. Observe: Maximization case cannot be reduced to above minimization case as only non-negative weights are allowed.

10 Optimum Weight 2-SAT Special Case of Mixed Horn Formulas Mixed Horn Formula (MHF): Union of a Horn Formula and a 2-CNF formula: M = C H; class: MHF Observe: Minimization problem for variable-weighted MHF NP-hard Known: Can be solved in time O( n ): n variables of non-negative weights [P., S., 2004] Thus Minimization case for 2-SAT can also be solved in time O( n ) time, for R + -weights. Reason: 2-CNF MHF. Wanted: Exact bound for 2-SAT minimization case on arbitrarily R-weighted variables

11 Optimum Weight 2-SAT Helpful Tool: Selected Complementation For formula C, variable set X V (C) let C X be formula obtained from C by complementing variables in X: For example: yields C = {{x, ȳ, z}, { x, y, z}, {x, y, z}, X := {x, z}} C X = {{ x, ȳ, z}, {x, y, z}, { x, y, z}} Lemma For fixed problem in {SAT,XSAT,NAESAT}, let (C, w) with C CNF, w : V (C) R and X V (C) be arbitrary. Then (C X, w X ) with w X defined as w(x) on V (C) \ X, and w X (x) := w(x) for each x X, satisfies: (i) T Π (C) = T Π (C X ), given by t t X, (ii) T Π min (C,w) = TΠ min (CX, w X ). Sketch of Proof: Use the Bijection-Lemma...

12 Optimum Weight 2-SAT Result Use above Lemma to turn variables of negative weights into positive Then perform MHF-algorithm on resulting instance of nonnegative variable weights only. Possible as 2-CNF is invariant w.r.t. complementation. Thus we obtain: Theorem Minimum (resp. maximum) weight 2-SAT can be solved in time O( n ), for n R-weighted variables. Sketch of Proof: The maximization case follows due to the MAX-MIN-Lemma, because there are no restrictions on variable weights.

13 Optimum Weight 2-SAT Variants: XSAT, NAESAT for weighted 2-CNF Observe: XSAT and NAESAT are essentially the same on 2-CNF (except for unit clauses) Known: Minimization and maximization versions of XSAT, NAESAT can be solved in linear time on 2-CNF with variable weights in R [P., 2005].

14 Algorithmic Strategy Let Π {SAT, XSAT, NAESAT} Aim: Solution of MINW-Π, resp., MAXW-Π on R-variable-weighted monotone formulas (C, w), where each variable occurs at most twice (CNF + ( 2)) Algorithmic strategy: (1) Treat minimization case via reduction to appropriate edge-weighted graph problem (2) Deduce maximization case via MAX-MIN-Lemma

15 Intersection Graph for Monotone Formulas Intersection graph of monotone formula C CNF + : vertices: clauses in C edges: c c iff c and c have a variable in common (non-empty intersection) Observe: Unique variables in C are not represented via graph edges

16 Slide modification: Clause Graph Formula C = {c 1, c 2, c 3 } CNF + ( 2), with c 1 = {t, u,x, y, z}, c 2 = {r,u, v,w, x}, c 3 = {r,v, y} t, z, w are unique, all variable weights have value 1 (omitted in the figure) y c x,u c 1 2 v,r c 3 t,z w c 1 x,u c 2 v,r c 3 y Clause graph corresponding to C.

17 Clause Graph Properties Clause graph has at most 2 C vertices and V (C) edges As each variable occurs in at most two clauses: C x V (C) C(x): all clauses containing x. C(x) 2 V (C)

18 XSAT on monotone weighted Note: minimum weight perfect matching in the clause graph is equivalent to minimum weight XSAT-model of (C,w): set exactly the variables to true labeling matching edges: Lemma: A minimum, resp. maximum, XSAT-model of (C, w) with C CNF + ( 2), w : V (C) R, can be computed, respectively, it can be reported that none exists in time O( V (C) 3 ).

19 We even have : Theorem: Above result also valid for non-monotone class CNF( 2): A minimum, respectively, maximum weight XSAT-model of (C,w), with C CNF( 2), w : V (C) R, can be computed, respectively, it can be reported that none exists in O( V (C) 3 ) time. Proof-Sketch: Perform appropriate weighted-xsatequivalent transformations for monotonizing weighted input formula.

20 1. Eliminating pure literals Given pure literal x we are done by Selected Complementation Lemma setting X = {x}, more precisely: Corollary: For (C,w), with C CNF, w : V (C) R, let x V (C) be a variable only occuring negated in C. Then: for (C x, w x ), where w x : V (C) R is defined as w except for w x (x) := w(x), we have Tmin XSAT (C, w) = Tmin XSAT (C x, w x ). Notice: valid also for SAT, NAESAT

21 2. Treating complemented pairs x, x If there is a clause containing more than one complemented pairs: no XSAT model possible: T XSAT = Lemma: Let c C contain exactly one complemented pair x, x. Let C c be formula obtained from C by removing c and assigning all literals to 0 that occur in c := c {x, x} (which can be empty), and by finally removing all duplicate clauses. Let w c be the restriction of w to V (C c ) = V (C) V (c ). Then: There is a bijection providing T XSAT min (C, w) = T XSAT min (C c, w c ), Notice: Can also be adapted to SAT, NAESAT

22 Transformations for Simplification 3. Weighted Simple Resolution Lemma Let C CNF be cp-free formula and w : V (C) R. Assume c i = {x} u, c j = {x} v C Let C ij be obtained from C as follows: (1) c C(x) : c c {x} v, c C(x) : c c {x} u, (2) set all literals in u v to 0, (3) remove all duplicate clauses from the current clause set. Let w ij := V (C ij ) R be defined as follows: for each y V (C ij ) V (u v), set w ij (y) := w(y), and (1 ) if V + (u v) V (u v) = {z}, then y V (u v) {z} set w ij (y) := w(y) and { w(z) + w(x), if z u, z v w ij (z) := w(z) w(x), else (2 ) if V + (u v) V (u v) =, then set y V (v u) : w ij (y) := w(y) and y V + (u v) : w ij (y) := w(y) w(x) and y V (u v) : w ij (y) := w(y) + w(x). Then: V (C ij ) = V (C) {x} V (u v), C ij C 1 and: Tmin XSAT (C, w) = Tmin XSAT (C ij, w ij ).

23 Next Case: Weighted SAT First recall: edge cover in a graph G = (V, E) of no isolated vertices is a subset F E s.t.: each vertex x V is incident to at least one edge in F. (Edges in edge cover need not to be independent (matching))

24 Example: Edge Cover Bold edges form edge cover (here in addition perfect matching).

25 SAT to Edge Cover Observation 1: Input instance (C, w) can be transformed s.t. resulting formula (C, w ) has strictly positive weights. Observation 2: A minimum weight edge cover F in the clause graph of C yields minimum weight SAT-model of (C, w ) via setting exactly all variables to true labeling edges in F. How to compute minimum weight edge cover? First: consider the minimum cardinality case

26 Minimum Cardinality Edge Cover Well known: Cardinality of minimum edge cover F in G closely related to its matching number ν(g) (= cardinality of a maximum matching): F = V ν(g) Reason: All 2ν(G) vertices in a maximum matching M are covered. Remaining vertices are independent of each other and only have neighbours in the set covered by M. So: for each such vertex add exactly one edge to M.

27 Cardinality Case: Illustration G. Maximum Matching.... M Vertices not in M mutually independent: Select one edge of each and add to M. Obvious: The same works if all weight values are equal.

28 Arbitrarily Weighted Case Here: No immediate connection maximum weight matchings minimum weight edge cover. Reason: Objects should be optimized in opposite direction Way out: Appropriate transformation of weight function w : E R + {0} to new weight function ŵ : E R defined for each edge x y as ( ) ŵ(x y) := w(x y) + +min{w(x z) : z N(x)} + +min{w(y z) : z N(y)} N(v) V : set of all neighbours of vertex v in G Observe: If w = const then ŵ = w

29 Advantage Now: Edges of weight smaller than all incident egdes get positive new weight larger edge weights get negative Maximum weight matching: Only edges of positive weight can be joined Thus: Perform a general maximum weight matching algorithm, for arbitrarily edge-weighted graphs, on (G, ŵ) Let M E be its result. Select in (G, w) an edge of least weight for each vertex in V V (M).

30 Example (G, w ) ^ (G, w ) a 8 r a 4 r z 1 x 4 2 u 3 1 y 3 v 3 x 4 z 2 0 max. weight match. u 0 4 y 0 v 0 Transform edge weights, compute max. weight matching, then come back and select edge of least weight for each uncovered vertex.

31 Result Computation of maximum weight matching in arbitrarily edge weighted graph G = (V, E) possible in time O( V 2 E ) Hence Minimum weight edge cover in (G, w) can be computed in time O( V 2 E ). Recall: Clause graph has O( C ) vertices, O( V (C) ) edges, and C 2 V (C). Thus: Theorem A minimum, resp. maximum weight SAT model of (C, w), C CNF + ( 2), w : V (C) R, can be computed in time O( V (C) 3 ). Clear: Such a model always exists (monotone instance).

32 Last Case: NAESAT Unweighted NAESAT: Solved for unrestricted class CNF( 2) in linear time based on Euler tour techniques (P., Speckenmeyer, Randerath, 2003) Unfortunately: Approach does not seem to work in the weighted case.

33 Recall Graph Factor G = (V, E) connected graph, vertex x in G. For edge set M E: M(x) denotes set of all edges in M incident to x V. Consider integer functions f, g : V Z with f g Recall: f-factor in G is set M E s.t. for each x V holds M(x) = f(x) (does not exist, in general) E.g.: f = 1: 1-factor is perfect matching. 2-factor is a cycle.

34 Example 2-Factor x 1 2 x x 3 x 4 5 x Bold edges form a 2-Factor.

35 Recall General Graph Factor More general: [f, g]-factor in G is edge subset M E s.t. M(x) [f(x), g(x)] for each x V. For an edge-weighted graph (G, w) with w : E R an optimum weight [f, g]-factor, is an [f, g]-factor M of optimal weight w(m).

36 NAESAT Reduction to a MINW Factor Problem Note: Minimum weight NAESAT-model of (C, w ) provided by setting to 1 exactly the variables labeling the edges in a minimum weight [1,deg 1]-factor in G C, if existing. Here denotes the degree function. deg : V (G C ) Z

37 Example: [1, deg 1]-factor x 3 x 1 x 4 x 2 x 5 x 6 Bold edges form [1, deg-1]-factor.

38 Result With standard linear programming techniques: Theorem A minimum, resp. maximum, NAESAT-model of (C, w) with C CNF + ( 2), w : V (C) R, can be computed, respectively, it can be reported that none exists in polynomial time.

39 Concluding Remarks and Open Problems Showed: Minimum/maximum weight SAT for 2-CNF formulas of n R-weighted variables can be solved in time O( n ). Open problem: Construct a faster exact algorithm for optimum weight 2-SAT.

40 Concluding Remarks and Open Problems Provided: Poly-time algorithms for SAT, XSAT, NAESAT on weighted monotone class CNF + ( 2) via reduction to graph problems. (For XSAT we modified existing monotonization scheme tackling also the general, i.e., non-monotone case CNF( 2).) Open problem: Construct monotonization schemes for also solving SAT and NAESAT on weighted CNF( 2). Missing link: Pendant to the weighted simple resolution rule for SAT, NAESAT (Note: In the unweighted cases simple resolution works, [P.,S.,R., 2003]).