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MaxSAT Resolution and Subcube Sums Yuval Filmus Meena Mahajan - - PowerPoint PPT Presentation
MaxSAT Resolution and Subcube Sums Yuval Filmus Meena Mahajan - - PowerPoint PPT Presentation
MaxSAT Resolution and Subcube Sums Yuval Filmus Meena Mahajan Gaurav Sood Marc Vinyals SAT 2020 The MaxSAT problem Input: F , a CNF formula ( n variables, m clauses). Output: MaxSAT ( F ) = x { 0 , 1 } n { number of clauses
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A ◮ is A ∨ B
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A ◮ is A ∨ B
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A ◮ is A ∨ B ◮ are fragments of x ∨ A and x ∨ B.
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A ◮ is A ∨ B ◮ are fragments of x ∨ A and x ∨ B.
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A ◮ is A ∨ B ◮ are fragments of x ∨ A and x ∨ B.
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A ◮ is A ∨ B ◮ are fragments of x ∨ A and x ∨ B.
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MaxSAT Resolution (subcube view)
Boolean hypercube {0, 1}n x = 0 x = 1 x ∨ A x ∨ B x ∨ B x ∨ A ◮ is A ∨ B ◮ are fragments of x ∨ A and x ∨ B.
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The MaxSAT resolution rule [BonetLevyManya]
x ∨ a1 ∨ . . . ∨ as (x ∨ A) x ∨ b1 ∨ . . . ∨ bt (x ∨ B) A ∨ B (the “standard resolvent”) (weakenings of x ∨ A) x ∨ A ∨ b1 x ∨ A ∨ b1 ∨ b2 . . . x ∨ A ∨ b1 ∨ . . . ∨ bt−1 ∨ bt (weakenings of x ∨ B) x ∨ B ∨ a1 x ∨ B ∨ a1 ∨ a2 . . . x ∨ B ∨ a1 ∨ . . . ∨ as−1 ∨ as
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Using the MaxSAT resolution rule for MaxSAT
Goal: Given a CNF formula F, certify that MaxSAT(F) k. A MaxSAT Resolution proof: ◮ Maintain a multiset of clauses C. ◮ Initially, C = F. ◮ At each step, pick two clauses C1, C2 ∈ C, apply the MaxSAT resolution rule to them, and replace them in C by the consequents. ◮ If C contains ’k’ copies of the empty clause , then MaxSAT(F) k. ◮ Size of the refutation: number of steps.
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Using the MaxSAT resolution rule for MaxSAT (cont’d)
MaxRes is a sound and complete proof system for MaxSAT. ◮ Soundness: If F ⊢MaxRes , . . . ,
- k times
, G, then MaxSAT(F) k. ◮ Completeness: If MaxSAT(F) = k then F ⊢MaxRes , . . . ,
- k times
, G.
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Proof system for SAT: MaxRes vs Resolution
◮ Using MaxRes just to certify unsatisfiability? Stop the derivation as soon as a single is derived. ◮ Resolution simulates MaxRes. (by definition of MaxSAT resolution) ◮ In MaxRes, clauses are deleted after use (though some weakenings are added back). So makes sense to compare MaxRes with fragments of Resolution which restrict reuse. ◮ MaxRes simulates read-once resolution. Easy to see. But not very interesting. ◮ Does MaxRes simulate tree-like resolution? We don’t know yet. Adding a MaxSAT-appropriate weakening rule suffices.
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The MaxSAT weakening rule
C C ∨ x and C ∨ x where x is a variable not in C While applying this rule also, we delete the antecedent and add the consequents to the multiset.
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Our Results - I
◮ MaxResW (MaxRes with the weakening rule) simulates TreeRes. ◮ There is a family of unsatisfiable formulas that is
◮ easy in MaxResW, ◮ easy even in MaxRes, ◮ hard for TreeRes. (Pebbling contradictions on Pyramid graphs, composed with OR2, are hard for TreeRes. [BenSassonWigderson] We add some hint clauses to make it easy to refute in MaxRes; we show via pebbling that despite hints it remains hard for TreeRes.)
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Relations between proof systems
TreeRes MaxRes MaxResW Res ◮ A B denotes that A simulates B and B does not simulate A. ◮ A B denotes that A simulates B. ◮ A B denotes that A does not simulate B.
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Lower bounds for MaxRes?
◮ Resolution simulates MaxResW. So, Res lower bounds translate to MaxResW. ◮ Question: Is MaxResW as strong as Res? Probably not – MaxResW maintains a stronger invariant at each step. ◮ To establish a separation, need lower bound techniques specific to MaxRes. ◮ Technique based on the stronger invariant maintained by MaxResW.
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MaxResW Invariant
Let violF(x): number of clauses of F falsified by assignment x. Invariant maintained: ◮ MaxResW step F ⊢ G. violF(x) = violG(x) for all x ∈ {0, 1}n ◮ MaxResW refutation F ⊢MaxRes , G. violG(x) = violF(x) − 1 for all x ∈ {0, 1}n
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The SubCube Sums proof system
Abstracting the MaxResW invariant, we define SubCubeSums: a new static proof system for SAT. ◮ F: an unsatisfiable CNF formula. ◮ A SubCubeSums refutation: a multiset of clauses G satisfying violG(x) = violF(x) − 1 for all x ∈ {0, 1}n (This implies ∀x, violF(x) ≥ 1; hence F unsat. ) ◮ Size of the refutation: number of clauses in G (counted with multiplicity). ◮ Not a proof system in Cook-Reckhow sense; however, verification possible in randomized polynomial time. Short MaxResW refutation = ⇒ small SubCubeSums refutation.
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Our Results - II
◮ A family of formulas requiring large size in SubCubeSums.
(Tseitin contradictions on expander graphs. Lower bound based on how viol behaves, sizes of viol−1(i). Intrinsically different from lower bound for Res.)
Does not separate MaxResW from Res.
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Our Results - III
◮ SubCubeSums can be recast as a special case of Sherali-Adams. ◮ A family of formulas easy in SubCubeSums but hard in Res.
(Subset Cardinality Formulas; known to be hard for Res.) (PigeonHolePrinciple Formulas; known to be hard for Res. Easy for SubCubeSums – implicit in [LarrosaRollon-SAT20]. We give a direct proof.)
◮ A Lifting Technique: F requires large width in SubCubeSums ⇓ F ◦ XOR requires large size in SubCubeSums.
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Relations among various proof systems
TreeRes MaxRes MaxResW Res SubCubeSums Sherali–Adams ◮ A B denotes that A simulates B and B does not simulate A. ◮ A B denotes that A simulates B. ◮ A B denotes that A does not simulate B.
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