Theory of Computer Science

Theory of Computer Science

G. R¨oger

Spring Term 2019

University of Basel Computer Science

Exercise meeting 9 — Solutions

Exercise 9.1

Consider propositional formulaϕ=¬(A∨(¬B∧C)).

(a) Specify formulaχ_all as it is used in the polynomial reduction ofSATto3SAT. Solution:

Formula χall over {XA, XB, XC, X_¬B, X_(¬B∧C), X_{(A∨(¬B∧C))}, X¬(A∨(¬B∧C))} is built from the following subformulas:

χ_A= (X_A↔A) χB= (XB ↔B) χC= (XC↔C) χ¬B= (X¬B ↔ ¬XB)

χ_(¬B∧C)= (X_(¬B∧C)↔(X_¬B∧XC))

χ_{(A∨(¬B∧C))}= (X_{(A∨(¬B∧C))}↔(X_A∨X_(¬B∧C))) χ¬(A∨(¬B∧C))= (X¬(A∨(¬B∧C))↔ ¬X_{(A∨(¬B∧C))})

Then

χall=χA∧χB∧χB∧χ¬B∧χ_(¬B∧C)∧χ_{(A∨(¬B∧C))}∧χ¬(A∨(¬B∧C)),

always using the logically equivalent CNF formula that results from replacing the abbre- viation ↔ with the corresponding formula, e.g. χ_{(A∨(¬B∧C))} ≡ (X_{(A∨(¬B∧C))} → (X_A ∨ X_(¬B∧C)))∧((XA ∨X_(¬B∧C)) → X_{(A∨(¬B∧C))}) ≡ (¬X_{(A∨(¬B∧C))} ∨(XA ∨X_(¬B∧C)))∧ (¬(XA∨X_(¬B∧C))∨X_{(A∨(¬B∧C))}).

(b) I ={A7→F, B7→T, C 7→T} is a model ofϕ. Specify the corresponding model ofχall. Solution:

I⁰={XA7→F, XB7→T, XC7→T, X¬B 7→F, X_(¬B∧C)7→F, X_{(A∨(¬B∧C))}7→F, X¬(A∨(¬B∧C))7→T}

Exercise 9.2

The decision problemSAT(satisfiability) is defined as follows:

Given: a propositional logic formula ϕ Question: Isϕsatisfiable?

The general problemGenSAT(model generation) is defined as follows:

Given: a propositional logic formula ϕ

Output: a model forϕor a message that none exists

Show that if there is a polynomial algorithm for SAT then there is a polynomial algorithm for GenSAT.

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Solution:

We specify an algorithm that solvesGenSAT. It uses transformations ψ[v 7→T] andψ[v 7→F]

that replace every occurrence ofv with a small valid (e.g.v∨ ¬v) and unsatisfiable (e.g.v∧ ¬v) formula, respectively. Then

• I⁰ is a model ofψ[v7→T] iffI with I(v) =T and I(v⁰) =I⁰(v⁰) forv⁰6=v is a model ofψ and

• I⁰ is a model ofψ[v7→F] iffI withI(v) =F andI(v⁰) =I⁰(v⁰) forv⁰6=v Moreover, ifψis satisfiable andψ[v7→T] is unsatisfiable thenψ[v7→F] is satisfiable.

The algorithm proceeds as follows:

Call the algorithm forSATon inputϕ. If it is unsatisfiable, output that it has no model.

Otherwise, we successively build a modelI forϕ, starting withϕ⁰:=ϕ, as follows:

While there is still an unassigned variablev, call theSATalgorithm forϕ⁰[v7→T]. If the answer is yes, setI(v) =T and continue withϕ⁰:=ϕ⁰[v7→T], otherwise setI(v) =F and continue with ϕ⁰:=ϕ⁰[v7→F].

Since the number of variables is bound by the size of ϕ, there is only a polynomial number of iterations. The size of the last formulaϕ⁰ is at most k times larger thanϕ for some constantk (k= 2 if we only count variable occurrences to determine the size). Hence, if every call toSATis possible in polynomial time then the overall runtime is polynomial.

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