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

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