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Unformatted text preview: CS103 HO#56 SlidesMore on Complexity 6/1/11 1 CS103 Mathematical Foundations of Computing 6/1/11 FINAL EXAM: Monday, June 6, 12:15 – 3:15 Braun Auditorium More on Complexity DOUBLESAT = { Φ  Φ is a Boolean formula with two satisfying assignments} Show that DOUBLESAT is NPcomplete. 1. DOUBLESAT NP since ... 2. We can show SAT ≤ p DOUBLESAT with the TM F that computes a polynomial time reduction f: F = "On input Φ , where Φ is a Boolean formula with variables x 1 , x 2 , ... , x m : ... " So Φ SAT if and only if f( Φ ) DOUBLESAT. More on Complexity DOUBLESAT = { Φ  Φ is a Boolean formula with two satisfying assignments} Show that DOUBLESAT is NPcomplete. 1. DOUBLESAT NP since two assignments can be verified in polynomial time. 2. We can show SAT ≤ p DOUBLESAT with the TM F that computes a polynomial time reduction f: F = "On input Φ , where Φ is a Boolean formula with variables x 1 , x 2 , ... , x m : 1. Let Φ ' be Φ (x ¬x) where x is a new variable. 2. Output Φ ' ." So Φ SAT if and only if f( Φ ) DOUBLESAT. More on Complexity DOUBLESAT = { Φ  Φ is a Boolean formula with two satisfying assignments} Show that DOUBLESAT is NPcomplete. 1. DOUBLESAT NP since two assignments can be verified in polynomial time. 2. We can show SAT ≤ p DOUBLESAT with the TM F that computes a polynomial time reduction f: F = "On input Φ , where Φ is a Boolean formula with variables x 1 , x 2 , ... , x m : 1. Let Φ ' be Φ (x ¬x) where x is a new variable. 2. Output Φ ' ." If Φ SAT, then Φ ' has at least two satisfying assignments: the original satisfying assignment with x = T and x = F. If Φ ' DOUBLESAT, then Φ is satisfiable since x does not appear in Φ . So Φ SAT if and only if f( Φ ) DOUBLESAT. More on Complexity The CircuitSatisfiability Problem A Boolean circuit is a collection of AND, OR, and NOT gates connected by wires, with no loops....
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 Spring '09

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