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l14-handout

# l14-handout - Lecture 14/15 More NP-completeness Outline...

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Lecture 14/15: More NP-completeness David Dill Department of Computer Science 1 Outline NP-completeness finish CSAT 3SAT Independent Set (IS) Node Cover (NC) 2 From NNF to CNF Assume formula is in NNF now. A recursive function tocnf : nnf formulas cnf formulas can be defined to convert a formula to CNF, which will be a list of clause formulas. tocnf ( α ) returns [[ α ]] if α is a literal. All other formulas are either or formulas. tocnf ( α β ) = append ( tocnf ( α ) , tocnf ( β )) The tricky case is tocnf ( α β ) . How do we combine tocnf ( α ) and tocnf ( β ) in polynomial time? Suppose tocnf ( α ) = [ g 1 , g 2 , . . . , g p ] and tocnf ( β ) = [ h 1 , h 2 , . . . , h q ] where g i and h j are clause formulas. [ g 1 , g 2 , . . . , g p ] represents the logical formula g 1 g 2 . . . g p , and h is also a conjunction. We need a CNF formula that is “equivalent” to ( g 1 g 2 . . . g p ) ( h 1 h 2 . . . h q ) . 3 From NNF to CNF, continued Here is something that doesn’t work: We could use distributive law to distribute over . But that gives exponential blowup in the formula size! Instead, use a trick: add new variables to the clause formulas. Then create a fresh variable y and convert the formulas to ( y g 1 ) ( y g 2 ) . . . ( y g p ) ( ¬ y h 1 ) ( ¬ y h 2 ) . . . ( ¬ y h q ) . Intuition: If y = T , the whole formula is true IFF the h clauses are true. 4

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Correctness of CNF conversion The second transformation preserves satisfiability, but not logical equivalence. Def: φ 1 is equisatisfiable with φ 2 if if φ 1 is satisfiable iff φ 2 is satisfiable. Thm: The logical formula represented by tocnf ( α ) is equisatisfiable to α . Proof sketch: By induction on the structure of propositional formulas. Suppose α is an NNF formula and β = tocnf ( α ) .
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l14-handout - Lecture 14/15 More NP-completeness Outline...

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