Solving MAXSAT by Solving a Sequence of Simpler SAT Instances

Transcription

1 Solving MAXSAT by Solving a Sequence of Simpler SAT Instances Jessica Davies and Fahiem Bacchus Department of Computer Science University of Toronto [jdavies

2 The MAXSAT Problem An instance of the MAXSAT problem is given by a CNF formula F and a cost wt(c) associated with falsifying each clause C e.g. F = {(a b c, ), (x y, 4)} (a b c, ) is a hard clause, (x y, 4) is soft A truth assignment π has cost equal to the sum of the costs of the clauses it falsifies Goal: find an optimal truth assignment, i.e., a truth assignment of minimum cost mincost(f)

3 Applications Many optimization problems can be easily encoded as MAXSAT Traveling Salesman MaxCut, MaxClique Problems from planning, probabilistic reasoning (MPE) CNF is commonly used for industrial problems in hardware design Applications of MAXSAT include FPGA routing, design debugging

4 How to solve MAXSAT? 1. Branch and Bound 2. Solve a Sequence of SAT Instances

5 1. Branch and Bound Backtracking search through the partial truth assignments Ineffective for very large instances from industrial applications Solvers include MiniMaxSat [Heras et al. 2008], MaxSatz c [Li et al. 2010], [Davies et al. 2010]

6 2. Solving a Sequence of SAT Instances: Existing Approach Each SAT instance in the sequence encodes a MAXSAT Decision Problem Is there a truth assignment of cost at most k?, then k is varied Exploits the good performance of SAT solvers on large industrial instances Solvers that implement variations of this algorithm include MSUnCore [Marques-Silva and Manquinho 2008], WPM2 [Ansótegui et al. 2010]

7 2. Solving a Sequence of SAT Instances: Existing Approach Bottleneck: the SAT solving time The SAT encodings of the decision problems become larger and harder to solve, especially for non-uniform costs Our approach: solve MAXSAT with a sequence of simpler SAT instances

8 Definitions Definition A core is a set of soft clauses that is inconsistent with the hard clauses. Observation (1) Given a collection of cores, K, and a truth assignment π s.t. π = hard(f), let hs(π) be the set of clauses falsified by π. Then hs(π) is a hitting set of K, i.e. hs(π) κ = κ K.

9 Definitions Example K κ 1 = {(x, 1), ( x, 1)} κ 2 = {(y, 1), ( y x, 1), (x, 1)} κ 3 = {(a, 1), ( a b, 1), ( b a, 1)} κ 1, κ 2, and κ 3 are cores π = {x, y, a, b, c} falsifies ( x, 1) κ 1, ( y x, 1) κ 2 and ( b a, 1) κ 3 the set of clauses falsified by π is a hitting set for K

10 Definitions Example K κ 1 = {(x, 1), ( x, 1)} κ 2 = {(y, 1), ( y x, 1), (x, 1)} κ 3 = {(a, 1), ( a b, 1), ( b a, 1)} κ 1, κ 2, and κ 3 are cores π = {x, y, a, b, c} falsifies ( x, 1) κ 1, ( y x, 1) κ 2 and ( b a, 1) κ 3 the set of clauses falsified by π is a hitting set for K

11 Definitions Definition A Minimum Cost Hitting Set (MCHS) of a collection of cores K is a hitting set for K such that all other hitting sets have greater or equal cost. The cost of a hitting set is the sum of the costs of the clauses it contains MCHS(K) denotes the cost of a minimum cost hitting set of K.

12 Definitions Example K κ 1 = {(x, 1), ( x, 1)} κ 2 = {(y, 1), ( y x, 1), (x, 1)} κ 3 = {(a, 1), ( a b, 1), ( b a, 1)} hs min = {(x, 1), ( a b, 1)} is a MCHS of K MCHS(K) = 2 is the cost of hs min hs C = {(x, 1), C} for any C κ 3 is also a MCHS of K

13 A New Approach to Solving MAXSAT Observation (2) Given a collection of cores K, and a truth assignment π, cost(π) MCHS(K). Therefore, mincost(f) MCHS(K). Observation (3) If π = F \ hs then cost(π) cost(hs). Theorem If π = F \ hs and hs is a min cost hitting set of a collection of cores K, then π is an optimal truth assignment for F. Proof. By Observations 2 and 3, mincost(f) MCHS(K) = cost(hs) cost(π) so π is an optimal truth assignment.

14 The MAXHS Algorithm κ 1 κ 2 κ 3 κ 4... κ n κ n+1 F = hs min F \ hs min yes SAT? no π = F \ hs min Theorem π is optimal for F Theorem If π = F \ hs and hs is a min cost hitting set of a collection of cores K, then π is an optimal truth assignment for F.

15 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) What is mincost(f)?

16 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 SAT-Solver(F) returns (UNSAT, κ 1 )

17 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 MCHS({κ 1 }) = 1

18 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 2 SAT-Solver(F \ {(x, 1)}) returns (UNSAT, κ 2 )

19 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 κ 2 MCHS({κ 1, κ 2 }) = 1

20 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 3 SAT-Solver(F \ {( x, 1)}) returns (UNSAT, κ 3 )

21 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 κ 2 κ 3 MCHS({κ 1, κ 2, κ 3 }) = 2

22 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 4 SAT-Solver(F \ {( x, 1), (x y, 1)}) returns (UNSAT, κ 4 )

23 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 κ 2 κ 3 κ 4 MCHS({κ 1, κ 2, κ 3, κ 4 }) = 2

24 MAXHS: Example (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) SAT-Solver(F \ {(x, 1), (x y, 1)}) returns (SAT, π) where π = { x, y, z} is a solution Theorem π is an optimal truth assignment for F

25 MAXHS: Advantages The MAXHS algorithm achieves a separation of SAT from numerical reasoning Each SAT instance is a subset of the original MAXSAT formula F The SAT refutations are easy in practice SAT solving is no longer a bottleneck (small fraction of total time) The MCHS problem is well-studied in the OR community Sophisticated MIP solvers like CPLEX can be used The strengths of both types of solvers are better exploited

26 MAXHS: Plateaus In the example, the cost of the MCHS did not always increase when a new core was added (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 κ 2 κ 3 MCHS({κ 1, κ 2, κ 3 }) = 2

27 MAXHS: Plateaus In the example, the cost of the MCHS did not always increase when a new core was added (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 κ 2 κ 3 κ 4 MCHS({κ 1, κ 2, κ 3, κ 4 }) = MCHS({κ 1, κ 2, κ 3 }) = 2 4 cores, mincost(f) = 2

28 MAXHS: Plateaus In general, the number of cores generated will be larger than the number of clauses that are falsified by the optimal truth assignment Many of the cores do not increase the MCHS cost e.g. 914 cores to prove that at least 287 clauses must be falsified We investigated several techniques to discourage plateaus

29 Diverse Cores First, find as many disjoint cores as possible MCHS problem is trivial during this phase For some MAXSAT instances, very few additional cores will be needed Encourage the SAT solver to return diverse cores Invert the activities of the variables between iterations Delete learnt clauses between iterations

30 (x, 1) Realizable Hitting Sets ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 κ 2 κ 3 If the MCHS hs 1 = {( x, 1), (x y, 1)} is chosen, we saw earlier that another core κ 4 will be found (x, 1) ( x, 1) (x y, 1) ( y, 1) ( x z, 1) ( z y, 1) κ 1 κ 2 κ 3 If the MCHS hs 2 = {(x, 1), (x y, 1)} is chosen instead, the remaining clauses are satisfiable, and the algorithm terminates immediately with the optimal truth assignment

31 Realizable Hitting Sets It is impossible to falsify both clauses in hs 1 = {( x, 1), (x y, 1)} at the same time hs 1 is not realizable Definition A hitting set H is realizable in a MAXSAT problem F if there exists a truth assignment π such that (a) for each clause c H, π = c, and (b) π = hard(f). MAXRHS is like MAXHS except we find the min realizable hitting set It can be shown that this is also sufficient for correctness

32 Realizable Hitting Sets It is not easy to enforce the realizability condition in an Integer Program formulation We developed our own Branch and Bound solver to find realizable hitting sets Not as fast as CPLEX However, MAXRHS can sometimes reduce the number of iterations enough to pay off Average # MAXHS MAXRHS Family Instances mincost # Iter Time (s) # Iter Time (s) ms/sean > pms/bcp-msp > pms/bcp-mtg > pms/bcp-syn >

33 MAXHS Experiments We used Minisat2.0 for the SAT solver and CPLEX for the hitting sets 1034 unsatisfiable Industrial instances from the 2009 MaxSAT Evaluation 116 unsatisfiable Crafted instances All competing MAXSAT solvers that use a Sequence of SAT approach Msuncore, WBO, PM2, WPM1, WPM2, SAT4J and Qmaxsat 1200s timeout # SAT4J WPM1 WPM2 MAXHS Instances # Solved # Solved # Solved # Solved

34 MAXHS: Newly Solved Instances Instance (# Clauses) mincost # Iters Avg. Core Size Total Time 1_1_10_15 (200) _1_10_10 (200) _10_20 (370) _10_14 (357) _10_15 (361) _10_10 (336) ex5.r (1729) ex5.pi (1785) pdc.r (2999) test1.r (898) rot.b (2199) bench1.pi (1287) max1024.r (1924) max1024.pi (1952) prom2.r (3566) prom2.pi (3584) MAXHS solved 16 partial MAXSAT instances from the Industrial category, not solved by any solvers in the 2009 and 2010 Evaluations and unsolved in our experiments after 5400s

35 MAXHS: Newly Solved Instances Instance (# Clauses) mincost # Iters Avg. Core Size Total Time 1_1_10_15 (200) _1_10_10 (200) _10_20 (370) _10_14 (357) _10_15 (361) _10_10 (336) ex5.r (1729) ex5.pi (1785) pdc.r (2999) test1.r (898) rot.b (2199) bench1.pi (1287) max1024.r (1924) max1024.pi (1952) prom2.r (3566) prom2.pi (3584) From the 2008 PMS Industrial bcp-syn family, submitted by Marques-Silva

36 MAXHS: Newly Solved Instances Instance (# Clauses) mincost # Iters Avg. Core Size Total Time 1_1_10_15 (200) _1_10_10 (200) _10_20 (370) _10_14 (357) _10_15 (361) _10_10 (336) ex5.r (1729) ex5.pi (1785) pdc.r (2999) test1.r (898) rot.b (2199) bench1.pi (1287) max1024.r (1924) max1024.pi (1952) prom2.r (3566) prom2.pi (3584) Many clauses must be falsified (e.g. 287) The hitting set problems are very easy (CPLEX takes < 1 sec. for each)

37 Diverse Costs # Distinct SAT4J WPM1 WPM2 WBO MAXHS Costs Iter Time Iter Time Iter Time Iter Time Iter Time We increased the number of distinct costs on a Linux Upgradeability Weighted Partial MAXSAT instance MAXHS and WBO are immune to this kind of cost diversification

38 Summary and Future Work A new approach for solving large industrial MAXSAT instances Separates the SAT solving from the numerical reasoning Allows SAT solvers and MIP solvers to be better exploited Competitive with existing MAXSAT solvers based on a sequence of SAT instances Can solve problems with larger optima, requiring many clauses to be falsified Future Work How to produce more diverse cores to avoid plateaus? Improve the enforcement of the realizability condition Exploit non-minimum hitting sets