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Unformatted text preview: CS 577: Introduction to Algorithms 12/05/06 Homework Solution: 5b Instructor: Shuchi Chawla TA: Siddharth Barman Question 1 Note that Not-All-Equal-SAT ( NAE- SAT ) is in NP . Say we have n boolean variables x 1 to x n and m clauses C 1 to C m in CNF giving us boolean formula φ . Given an assignment σ of the n variables it will take us polynomial amount to verify that σ satisfies each and every clause in the not-all- equal form and hence φ ∈ NAE- SAT can be verified in polytime. We prove that NAE- SAT is NP- complete by reducing Circuit- SAT to it. Essentially given a circuit C we wish to determine a boolean function φ such that φ ∈ NAE- SAT iff C is satisfiable. The procedure followed to get such a φ is a slight modification of the reduction from Circuit- SAT to 3- SAT that was done in class. Also in this reduction we will ensure that every clause has exactly 3 variables (and no less). This is not important for this question but will be useful for the next question. • For NOT gate ¬ , we add clauses ( x v ∨ x u ∨ y ) and ( x v ∨ x u ∨ y ). Here an extra boolean variable y is introduced which is common to all clauses. • For OR gate ( x v = x u ∨ x w ) we introduce ( x v ∨ x u ∨ y ), ( x w ∨ x v ∨ y ) and ( x v ∨ x u ∨ x w ) • For AND gate ( x v = x u ∧ x w ) we introduce ( x v ∨ x u ∨ y ), ( x v ∨ x w ∨ y ) and ( x v ∨ x u ∨ x w ) As mentioned above boolean variable y is common through all the clauses. Now if an assignment “satisfies” the given circuit C we can simply set y = 0, this would insure that φ ∈ NAE- SAT . As each clause of φ is satisfiable and there is at least one literal per clause which is 0. Each clause containing y is obviously satisfied because we set y = 0. Each clause that doesn’t contain y (one each in the OR and AND case) can be checked to be satisfied by a case analysis....
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This note was uploaded on 02/17/2011 for the course CS 577 taught by Professor Joseph during the Spring '08 term at University of Wisconsin.
- Spring '08