Nested Boolean Functions as Models for QBFs
SAT 2013 Uwe Bubeck and Hans Kleine BΓΌning University of Paderborn
July 11, 2013
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
July 11, 2013
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QBF extends propositional logic by allowing universal and existential quantifiers over propositional variables. Semantics of closed QBF: βπ§ Ξ¦ π§ is true if and only if Ξ¦ π§/0 is true or Ξ¦ π§/1 is true. βπ¦ Ξ¦ π¦ is true if and only if Ξ¦ π¦/0 is true and Ξ¦ π¦/1 is true.
βπ¦1βπ§1βπ¦2βπ§2 π¦1 β¨ Β¬π§1 β§ Β¬π¦1 β¨ π§2 β§ π§1 β¨ π¦2 β¨ Β¬π§2 β§ Β¬π¦2 β¨ π§2
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π¦1 π§1 π¦2 π§1 π§2 π§2 π§2 π§2 π¦2 π¦1 = 0 π¦1 = 1 π§1 = 0 π§1 = 1 π¦2 = 0 π¦2 = 1 π¦2 = 0 π¦2 = 1 π§2 = 0 π§2 = 1 π§2 = 1 π§2 = 1
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QBF as a 2-player game: β and β player alternatingly choose assignments for variables in prefix order.
π¦1 π§1 π¦2 π§1 π§2 π§2 π§2 π§2 π¦2 π¦1 = 0 π¦1 = 1 π§1 = 0 π§1 = 1 π¦2 = 0 π¦2 = 1 π¦2 = 0 π¦2 = 1 π§2 = 0 π§2 = 1 π§2 = 1 π§2 = 1
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βπ¦1βπ§1βπ¦2βπ§2 β¦ βπ¦πβπ§π π π¦1, β¦ , π¦π, π§1, β¦ , π§π = true if and only if βπ¦1 β¦ βπ¦π π π¦1, β¦ , π¦π, π
1 π¦1 , β¦ , π π π¦1, β¦ , π¦π
= true for some π
1, β¦ , π π (Skolem functions).
βπ¦1βπ§1βπ¦2βπ§2 β¦ βπ¦πβπ§π π π¦1, β¦ , π¦π, π§1, β¦ , π§π = false
if and only if
βπ§1 β¦ βπ§π π(π1(), π2 y1 , β¦ , ππ(y1, β¦ , π§πβ1), π§1, β¦ , π§π) = false
for some π1, β¦ , ππ (Herbrand functions).
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Extract Skolem model from cube-resolution proof, Herbrand countermodel from clause-resolution proof.
(no polynomial-size propositional encoding if Ξ£2
π β Ξ 2 π)
ο direct polynomial-size encoding by NBFs ο (counter)models parameterized by free variables
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Closed QBF: either true or false Open QBF: valuation depends on the free variables:
QBF with free vars Truth Assignment Closed QBF Evaluation True False Propositional Formula β Ξ¦(π¨1, β¦ , π¨π ) π’: π¨1, β¦ , π¨r β 0,1 Ξ¦ π’(π¨1), β¦ , π’(π¨r)
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How are the models and countermodels for different assignments to the free variables related to each other?
QBF with free vars Truth Assignment Closed QBF Evaluation True False
Multiple models and/or countermodels
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Idea: Replace all quantified variables with functions over the free variables. βπ€πβπ€πβ1 β¦ βπ€2βπ€1 π π€1, β¦ , π€π, π¨1, β¦ , π¨π β π β1(π¨1, β¦ , π¨π ), β¦ , βπ(π¨1, β¦ , π¨π ), π¨1, β¦ , π¨π
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Why bother about models parameterized by free variables? Non-prenex QBF: βπβπ βπβπ π½ π, π, π, π β§ βπ¦βπ§ πΎ π, π, π¦, π§ e.g. precompute partial certificate. Open QBF with free vars π, π.
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A Nested Boolean Function (NBF) [Cook/Soltys 1999] is a sequence of functions πΊ = π
0, β¦ , π π with
0, β¦ , π π’ given as propositional formulas
π π¦π β π π0 π π1 π¦1 π , β¦ , π ππ π¦π π
for previously defined functions π
π0, β¦ , π ππ .
Example: parity of Boolean variables π
0 π1, π2 β Β¬π1 β§ π2 β¨ (π1 β§ Β¬π2)
π
1 π1, π2, π3, π4 β π 0 π 0 π1, π2 , π 0 π3, π4
π
2 π1, β¦ , π16 β π 1 π 1 π1, β¦ , π4 , β¦ , π 1 π13, β¦ , π16
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QBF: Ξ¦ π β βπ¦ π(π¦, π) NBF: πΊ0 π¦, π β π π¦, π πΊ
1 π
β πΊ0(πΊ0 1, π , π) = 1 if x = 1 is a satisfying choice = πΊ0 1, π = 1 Ξ¦ π β πΊ
1(π)
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QBF: Ξ¦ π β βπ¦ π(π¦, π) NBF: πΊ0 π¦, π β π π¦, π πΊ
1 π
β πΊ0(πΊ0 1, π , π) = 0 if x = 1 is not satisfying = πΊ0 0, π
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QBF: Ξ¦ π β βπ§βπ¦ π π¦, π§, π NBF: πΊ0 π¦, π§, π β π π¦, π§, π πΊ
1 π§, π
β πΊ0 πΊ0 1, π§, π , π§, π πΊ2 π β πΊ
1 πΊ 1 0, π , π
= 0 if y = 0 is not satisfying = πΊ
1 0, π = 0
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QBF: Ξ¦ π β βπ§βπ¦ π π¦, π§, π NBF: πΊ0 π¦, π§, π β π π¦, π§, π πΊ
1 π§, π
β πΊ0 πΊ0 1, π§, π , π§, π πΊ2 π β πΊ
1 πΊ 1 0, π , π
= 1 if y = 0 is satisfying = πΊ
1 1, π
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QBF: Ξ¦ π β βπ§βπ¦ π π¦, π§, π NBF: πΊ0 π¦, π§, π β π π¦, π§, π πΊ
1 π§, π
β πΊ0 πΊ0 1, π§, π , π§, π πΊ2 π β πΊ
1 πΊ 1 0, π , π
β Concise representation of QDPLL branching: innermost call of πΊπ is the first branch,
repetition of the first one if it is already conclusive.
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First branch determines which branch is conclusive. β this is our witness, i.e. (counter)model.
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First branch determines which branch is conclusive. β this is our witness, i.e. (counter)model. QBF: Ξ¦ π β βπ§βπ¦ π π¦, π§, π NBF: πΊ0 π¦, π§, π β π π¦, π§, π πΊ
1 π§, π
β πΊ0 πΊ0 1, π§, π , π§, π πΊ2 π β πΊ
1 πΊ 1 0, π , π
βπ§(π) βπ¦(π)
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QBF: Ξ¦ π β βπ§βπ¦ π π¦, π§, π NBF: πΊ0 π¦, π§, π β π π¦, π§, π πΊ
1 π§, π
β πΊ0 πΊ0 1, βπ§(π), π , π§, π πΊ2 π β πΊ
1 πΊ 1 0, π , π
βπ§(π) βπ¦(π)
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In general: Ξ¦ π β π ππ€π β¦ π 1π€1 π π€1, β¦ , π€π, π Complete equivalence model: βπ π β πΊπβ1 0, βπ+1 π , β¦ , β1 π , π , if π π = β πΊπβ1 1, βπ+1 π , β¦ , β1 π , π , if π π = β Clearly polynomial size, which is not possible with a propositional encoding if Ξ£2
π β Ξ 2 π.
Admittedly more difficult to evaluate. But:
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Skolem/Herbrand (counter)models parameterized by free variables.
polynomial space by nested Boolean functions with
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QBF, e.g. Horn, 2-CNF, etc.