models for qbfs

Models for QBFs Uwe Bubeck and Hans Kleine Bning University of - PowerPoint PPT Presentation

SAT 2013 Nested Boolean Functions as Models for QBFs Uwe Bubeck and Hans Kleine Bning University of Paderborn July 11, 2013 Outline Introduction: QBF and (Counter-)Models Free Variables and Models NBF Representation


  1. SAT 2013 Nested Boolean Functions as Models for QBFs Uwe Bubeck and Hans Kleine Büning University of Paderborn July 11, 2013

  2. Outline • Introduction: QBF and (Counter-)Models • Free Variables and Models • NBF Representation • Conclusion Uwe Bubeck NBFs as Models for QBFs 2

  3. Section 1 Introduction: QBF and (Counter-)Models

  4. Quantified Boolean Formulas QBF extends propositional logic by allowing universal and existential quantifiers over propositional variables. Semantics of closed QBF: ∃𝑧 Φ 𝑧 is true if and only if Φ 𝑧/0 is true or Φ 𝑧/1 is true. ∀𝑦 Φ 𝑦 is true if and only if Φ 𝑦/0 is true and Φ 𝑦/1 is true. Uwe Bubeck NBFs as Models for QBFs 4

  5. Tree Models ∀𝑦 1 ∃𝑧 1 ∀𝑦 2 ∃𝑧 2 𝑦 1 ∨ ¬𝑧 1 ∧ ¬𝑦 1 ∨ 𝑧 2 ∧ 𝑧 1 ∨ 𝑦 2 ∨ ¬𝑧 2 ∧ ¬𝑦 2 ∨ 𝑧 2 𝑦 1 𝑦 1 = 0 𝑦 1 = 1 𝑧 1 𝑧 1 𝑧 1 = 0 𝑧 1 = 1 𝑦 2 𝑦 2 𝑦 2 = 0 𝑦 2 = 1 𝑦 2 = 0 𝑦 2 = 1 𝑧 2 𝑧 2 𝑧 2 𝑧 2 𝑧 2 = 0 𝑧 2 = 1 𝑧 2 = 1 𝑧 2 = 1 Uwe Bubeck NBFs as Models for QBFs 9

  6. Function Models 1/2 QBF as a 2-player game: ∃ and ∀ player alternatingly choose assignments for variables in prefix order. 𝑦 1 𝑦 1 = 0 𝑦 1 = 1 𝑧 1 𝑧 1 𝑧 1 = 0 𝑧 1 = 1 𝑦 2 𝑦 2 𝑦 2 = 0 𝑦 2 = 1 𝑦 2 = 0 𝑦 2 = 1 𝑧 2 𝑧 2 𝑧 2 𝑧 2 𝑧 2 = 0 𝑧 2 = 1 𝑧 2 = 1 𝑧 2 = 1 Uwe Bubeck NBFs as Models for QBFs 10

  7. Function Models 2/2 ∀𝑦 1 ∃𝑧 1 ∀𝑦 2 ∃𝑧 2 … ∀𝑦 𝑜 ∃𝑧 𝑜 𝜚 𝑦 1 , … , 𝑦 𝑜 , 𝑧 1 , … , 𝑧 𝑜 = true if and only if ∀𝑦 1 … ∀𝑦 𝑜 𝜚 𝑦 1 , … , 𝑦 𝑜 , 𝑔 1 𝑦 1 , … , 𝑔 𝑜 𝑦 1 , … , 𝑦 𝑜 = true for some 𝑔 1 , … , 𝑔 𝑜 (Skolem functions). ∀𝑦 1 ∃𝑧 1 ∀𝑦 2 ∃𝑧 2 … ∀𝑦 𝑜 ∃𝑧 𝑜 𝜚 𝑦 1 , … , 𝑦 𝑜 , 𝑧 1 , … , 𝑧 𝑜 = false if and only if ∃𝑧 1 … ∃𝑧 𝑜 𝜚(𝑕 1 (), 𝑕 2 y 1 , … , 𝑕 𝑜 (y 1 , … , 𝑧 𝑜−1 ), 𝑧 1 , … , 𝑧 𝑜 ) = false for some 𝑕 1 , … , 𝑕 𝑜 (Herbrand functions) . Uwe Bubeck NBFs as Models for QBFs 11

  8. Motivation • Important applications: solver certificates, explanations, ... • Balabanov and Jiang (2012): Extract Skolem model from cube-resolution proof, Herbrand countermodel from clause-resolution proof. • Problem: compact representation 𝑄 ≠ Π 2 𝑄 ) (no polynomial-size propositional encoding if Σ 2 • Contributions:  direct polynomial-size encoding by NBFs  (counter)models parameterized by free variables Uwe Bubeck NBFs as Models for QBFs 12

  9. Section 2 Free Variables and Models

  10. Semantics of Free Variables Closed QBF: either true or false Open QBF: valuation depends on the free variables: True QBF with Closed Truth Assignment Evaluation free vars QBF False Φ(𝑨 1 , … , 𝑨 𝑠 ) 𝑢: 𝑨 1 , … , 𝑨 r → 0,1 Φ 𝑢(𝑨 1 ), … , 𝑢(𝑨 r ) ≈ Propositional Formula Uwe Bubeck NBFs as Models for QBFs 14

  11. Free Variables and Models True QBF with Closed Truth Assignment Evaluation free vars QBF False Multiple models and/or countermodels How are the models and countermodels for different assignments to the free variables related to each other? Uwe Bubeck NBFs as Models for QBFs 15

  12. Complete Equivalence Models 1/2 Idea: Replace all quantified variables with functions over the free variables. ∀𝑤 𝑜 ∃𝑤 𝑜−1 … ∀𝑤 2 ∃𝑤 1 𝜚 𝑤 1 , … , 𝑤 𝑜 , 𝑨 1 , … , 𝑨 𝑠 ≈ 𝜚 ℎ 1 (𝑨 1 , … , 𝑨 𝑠 ), … , ℎ 𝑜 (𝑨 1 , … , 𝑨 𝑠 ), 𝑨 1 , … , 𝑨 𝑠 Uwe Bubeck NBFs as Models for QBFs 16

  13. Complete Equivalence Models 2/2 Why bother about models parameterized by free variables? Non-prenex QBF: ∀𝑏∃𝑐 ∀𝑑∃𝑒 𝛽 𝑏, 𝑐, 𝑑, 𝑒 ∧ ∀𝑦∃𝑧 𝛾 𝑏, 𝑐, 𝑦, 𝑧 Open QBF with free vars 𝑏, 𝑐 . e.g. precompute partial certificate. Uwe Bubeck NBFs as Models for QBFs 17

  14. Section 3 NBF Representation

  15. Nested Boolean Functions A Nested Boolean Function (NBF) [Cook/Soltys 1999] is a sequence of functions 𝐺 = 𝑔 0 , … , 𝑔 𝑙 with • initial functions 𝑔 0 , … , 𝑔 𝑢 given as propositional formulas 𝑗 𝑦 𝑗 ≔ 𝑔 𝑗 , … , 𝑔 𝑗 • compound functions 𝑔 𝑘 0 𝑔 𝑘 1 𝑦 1 𝑘 𝑠 𝑦 𝑠 for previously defined functions 𝑔 𝑘0 , … , 𝑔 𝑘𝑠 . Example: parity of Boolean variables 𝑔 0 𝑞 1 , 𝑞 2 ≔ ¬𝑞 1 ∧ 𝑞 2 ∨ (𝑞 1 ∧ ¬𝑞 2 ) 𝑔 1 𝑞 1 , 𝑞 2 , 𝑞 3 , 𝑞 4 ≔ 𝑔 0 𝑔 0 𝑞 1 , 𝑞 2 , 𝑔 0 𝑞 3 , 𝑞 4 𝑔 2 𝑞 1 , … , 𝑞 16 ≔ 𝑔 1 𝑔 1 𝑞 1 , … , 𝑞 4 , … , 𝑔 1 𝑞 13 , … , 𝑞 16 Uwe Bubeck NBFs as Models for QBFs 19

  16. Quantifier Encoding in NBF 1/2 Φ 𝒜 ≔ ∃𝑦 𝜚(𝑦, 𝒜) QBF: Φ 𝒜 ≈ 𝐺 1 (𝒜) 𝐺 0 𝑦, 𝒜 ≔ 𝜚 𝑦, 𝒜 NBF: 𝐺 0 (𝐺 0 1, 𝒜 , 𝒜) 𝐺 1 𝒜 ≔ = 1 if x = 1 is a satisfying choice = 𝐺 0 1, 𝒜 = 1 Uwe Bubeck NBFs as Models for QBFs 20

  17. Quantifier Encoding in NBF 1/2 Φ 𝒜 ≔ ∃𝑦 𝜚(𝑦, 𝒜) QBF: 𝐺 0 𝑦, 𝒜 ≔ 𝜚 𝑦, 𝒜 NBF: 𝐺 0 (𝐺 0 1, 𝒜 , 𝒜) 𝐺 1 𝒜 ≔ = 0 if x = 1 is not satisfying = 𝐺 0 0, 𝒜 Uwe Bubeck NBFs as Models for QBFs 21

  18. Quantifier Encoding in NBF 2/2 Φ 𝒜 ≔ ∀𝑧∃𝑦 𝜚 𝑦, 𝑧, 𝒜 QBF: 𝐺 0 𝑦, 𝑧, 𝒜 ≔ 𝜚 𝑦, 𝑧, 𝒜 𝐺 1 𝑧, 𝒜 ≔ 𝐺 0 𝐺 0 1, 𝑧, 𝒜 , 𝑧, 𝒜 NBF: 𝐺 2 𝒜 ≔ 𝐺 1 𝐺 1 0, 𝒜 , 𝒜 = 0 if y = 0 is not satisfying = 𝐺 1 0, 𝒜 = 0 Uwe Bubeck NBFs as Models for QBFs 22

  19. Quantifier Encoding in NBF 2/2 Φ 𝒜 ≔ ∀𝑧∃𝑦 𝜚 𝑦, 𝑧, 𝒜 QBF: 𝐺 0 𝑦, 𝑧, 𝒜 ≔ 𝜚 𝑦, 𝑧, 𝒜 𝐺 1 𝑧, 𝒜 ≔ 𝐺 0 𝐺 0 1, 𝑧, 𝒜 , 𝑧, 𝒜 NBF: 𝐺 2 𝒜 ≔ 𝐺 1 𝐺 1 0, 𝒜 , 𝒜 = 1 if y = 0 is satisfying = 𝐺 1 1, 𝒜 Uwe Bubeck NBFs as Models for QBFs 23

  20. Quantifier Encoding in NBF 2/2 Φ 𝒜 ≔ ∀𝑧∃𝑦 𝜚 𝑦, 𝑧, 𝒜 QBF: 𝐺 0 𝑦, 𝑧, 𝒜 ≔ 𝜚 𝑦, 𝑧, 𝒜 𝐺 1 𝑧, 𝒜 ≔ 𝐺 0 𝐺 0 1, 𝑧, 𝒜 , 𝑧, 𝒜 NBF: 𝐺 2 𝒜 ≔ 𝐺 1 𝐺 1 0, 𝒜 , 𝒜 → Concise representation of QDPLL branching: innermost call of 𝐺 𝑗 is the first branch, outermost call of 𝐺 𝑗 is the second branch, or a repetition of the first one if it is already conclusive. Uwe Bubeck NBFs as Models for QBFs 24

  21. Complete Equiv. Model in NBF 1/2 First branch determines which branch is conclusive. → this is our witness, i.e. (counter)model. Uwe Bubeck NBFs as Models for QBFs 25

  22. Complete Equiv. Model in NBF 1/2 First branch determines which branch is conclusive. → this is our witness, i.e. (counter)model. Φ 𝒜 ≔ ∀𝑧∃𝑦 𝜚 𝑦, 𝑧, 𝒜 QBF: ℎ 𝑦 (𝒜) 𝐺 0 𝑦, 𝑧, 𝒜 ≔ 𝜚 𝑦, 𝑧, 𝒜 𝐺 1 𝑧, 𝒜 ≔ 𝐺 0 𝐺 0 1, 𝑧, 𝒜 , 𝑧, 𝒜 NBF: 𝐺 2 𝒜 ≔ 𝐺 1 𝐺 1 0, 𝒜 , 𝒜 ℎ 𝑧 (𝒜) Uwe Bubeck NBFs as Models for QBFs 27

  23. Complete Equiv. Model in NBF 1/2 Φ 𝒜 ≔ ∀𝑧∃𝑦 𝜚 𝑦, 𝑧, 𝒜 QBF: ℎ 𝑦 (𝒜) 𝐺 0 𝑦, 𝑧, 𝒜 ≔ 𝜚 𝑦, 𝑧, 𝒜 𝐺 1 𝑧, 𝒜 ≔ 𝐺 0 𝐺 0 1, ℎ 𝑧 (𝒜), 𝒜 , 𝑧, 𝒜 NBF: 𝐺 2 𝒜 ≔ 𝐺 1 𝐺 1 0, 𝒜 , 𝒜 ℎ 𝑧 (𝒜) Uwe Bubeck NBFs as Models for QBFs 28

  24. Complete Equiv. Model in NBF 2/2 In general: Φ 𝒜 ≔ 𝑅 𝑜 𝑤 𝑜 … 𝑅 1 𝑤 1 𝜚 𝑤 1 , … , 𝑤 𝑜 , 𝒜 Complete equivalence model: ℎ 𝑗 𝒜 ≔ 𝐺 𝑗−1 0, ℎ 𝑗+1 𝒜 , … , ℎ 1 𝒜 , 𝒜 , if 𝑅 𝑗 = ∀ 𝐺 𝑗−1 1, ℎ 𝑗+1 𝒜 , … , ℎ 1 𝒜 , 𝒜 , if 𝑅 𝑗 = ∃ Clearly polynomial size, which is not possible with a 𝑄 ≠ Π 2 𝑄 . propositional encoding if Σ 2 Admittedly more difficult to evaluate. But: Equiv. model checking PSPACE-hard even if ℎ 𝑗 𝒜 ∈ {0,1} . Uwe Bubeck NBFs as Models for QBFs 29

  25. Section 4 Conclusion

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