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K-consistency and SAT Teague Lasser

K-consistency and SAT Teague Lasser. Petke, J. and Jeavons, P. Local Consistency and SAT-Solvers CP 2010 pp.398-413. Introduction. The relationship between constraint satisfaction problems and boolean satisfiability problems has been an area of active research in the last decade

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K-consistency and SAT Teague Lasser

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  1. K-consistency and SAT Teague Lasser Petke, J. and Jeavons, P. Local Consistency and SAT-Solvers CP 2010 pp.398-413

  2. Introduction • The relationship between constraint satisfaction problems and boolean satisfiability problems has been an area of active research in the last decade • CP solvers attempt to learn new constraints but the belief of the community has been that attempting to enforce stronger consistency (k-consistency) than generalized arc consistency is inefficient • SAT solvers which rely primarily on resolution to produce refutations have shown remarkable

  3. k-Consistency • K-consistency extends arc consistency and path consistency to arbitrarily high levels of consistency • A nearly-optimal algorithm KS for k-consistency for any k proposed in 1989 suggests that if k-consistency is used as a method of relaxing the problem its time complexity is O(n2(a + 1)n) where n is the number of variables and a is a number that relates the number of labelings by a constraint on the domain1 [1] Cooper, M. An Optimal k-Consistency Algorithm Artificial Intelligence 41 pp.89-95

  4. Converting a CSP to a SAT problem • Take a CSP of the form P = {V, D, C} • Introduce a set of boolean variables of the form xvd for each v∈V with domain element d∈Dv • Along the each independent domain, ensure that only one variable can be true for the clause to be satisfied. ∨d∈DvXvd , ¬xvi∨¬xvj for all i,j∈Dv • Encode constraints as clauses of 2 or more boolean variables for each partial assignment that does not satisfy the constraint ∨v∈S¬xvf(v)

  5. Resolution • We can solve clauses by inference across some clauses of the form • C1∨ x and C2∨ ¬x to produce a new clause (the resolvent) of the form C1∨ C2 • If we have a collection of clauses of the form Ci∨¬xi for • i =1, 2,...,r, where each xi is a Boolean variable, and a purely positive clause • x1∨ x2∨···∨ xr, then we can deduce the clause C1∨ C2∨···∨ Cr. • They dubbed this form of inference positive-hyper-resolution and the resultant • clause C1∨ C2∨···∨ Cr the positive-hyper-resolvent.

  6. K-Consistency and Positive-Hyper-Resolution Theorem 1. The k-consistency closure of a CSP instance P is empty if and only if its direct encoding as a set of clauses has a positive-hyper-resolution refutation of width at most k. Lemma 1.1. Let P be a CSP instance, and let Φ be its direct encoding as a set of clauses. If Φ has no positive-hyper-resolution refutation of width k or less, then the k-consistency closure of P is non-empty. Lemma 1.2. Let P be a CSP instance, and let Φ be its direct encoding as a set of clauses. If the k-consistency closure of P is non-empty, then Φ has no positive-hyper-resolution refutation of width k or less.

  7. Proof discussion ∨d∈DvXvd , ¬xvi∨¬xvj for all i, j∈Dv ∨v∈S¬xvf(v) C1VC2Vx1 ¬x1V¬x2 x2VC3VC4 C1VC2vC3VC4

  8. Theorem 2. If a set of non-empty clauses Δ over n Boolean variables has a positive-hyper-resolution refutation of width k and length m, where all derived clauses contain only negative literals, then the expected number of restarts required by a standard randomised SAT-solver to discover that Δ is unsatisfiable is less than mnk(n on k).

  9. Theorem 3. If a set of non-empty clauses Δ over n Boolean variables has a positive-hyper-resolution refutation of width k and length m, where all derived clauses contain only negative literals, then the expected number of restarts required by a standard randomised SAT-solver using the Decision learning scheme to discover that Δ is unsatisfiable is less than m(n on k).

  10. Theorem 4. If the k-consistency closure of a CSP instance P is empty, then the expected number of restarts required by a standard randomised SAT-solver using the Decision learning scheme to discover that the direct encoding of P is unsatisfiable is O(n2kd2k), where n is the number of variables in P and d is the maximum domain size.

  11. A problem with no solution • n = ((d-1)*w + 2)*w over w groups d = {0..d-1} tree width = 2w - 1

  12. On SAT solvers

  13. Acknowledgements • Thank you to Justyna Petke for allowing me to use her modified SAT solver • Thank you to Peter for allowing me to use his laptop

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