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Unformatted text preview: 6.045: Automata, Computability, and Complexity (GITCS) Class 16 Nancy Lynch Today: More NPCompleteness • Topics: – 3SAT is NPcomplete – Clique and VertexCover are NPcomplete – More examples, overview – Hamiltonian path and Hamiltonian circuit – Traveling Salesman problem – More examples, revisited • Reading: – Sipser Sections 7.47.5 – Garey and Johnson • N e x t : – Sipser Section 10.2 3SAT is NPComplete NPCompleteness • Definition: Language B is NPcomplete if both of the following hold: (a) B ∈ NP, and (b) For any language A ∈ NP, A ≤ p B. • Definition: Language B is NPhard if, for any language A ∈ NP, A ≤ p B. NP P 3SAT is NPComplete • SAT = { < φ >  φ is a satisfiable Boolean formula } • Boolean formula: Constructed from literals using operations, e.g.: φ = x ∧ ( ( y ∧ z ) ∨ ( ¬ y ∧ ¬ z ) ) ∧ ¬ ( x ∧ z ) • A Boolean formula is satisfiable iff there is an assignment of 0s and 1s to the variables that makes the entire formula evaluate to 1 (true). • Theorem: SAT is NPcomplete. • 3SAT: Satisfiable Boolean formulas of a restricted kind conjunctive normal form (CNF) with exactly 3 literals per clause. • Theorem: 3SAT is NPcomplete. • Proof: – 3SAT ∈ NP: Obvious. – 3SAT is NPhard: … 3SAT is NPhard • Clause: Disjunction of literals, e.g., ( ¬ x 1 ∨ x 2 ∨ ¬ x 3 ) • CNF: Conjunction of such clauses • Example: ( ¬ x 1 ∨ x 2 ) ∧ ( x 1 ∨ ¬ x 2 ) ∧ ( x 1 ∨ x 2 ∨ ¬ x 3 ) ∧ ( x 3 ) • 3 C N F : { < φ >  φ is a CNF formula in which each clause has exactly 3 literals } • CNFSAT: { < φ >  φ is a satisfiable CNF formula } • 3 S A T : { < φ >  φ is a satisfiable 3CNF formula } = SAT ∩ 3CNF • Theorem: 3SAT is NPhard. • P r o o f : Show CNFSAT is NPhard, and CNFSAT ≤ p 3SAT. CNFSAT is NPhard • Theorem: CNFSAT is NPhard. • P r o o f : – We won’t show SAT ≤ p CNFSAT. – Instead, modify the proof that SAT is NPhard, so that it shows A ≤ p CNFSAT, for an arbitrary A in NP, instead of just A ≤ p SAT as before. – We’ve almost done this: formula φ w is almost in CNF. – It’s a conjunction φ w = φ cell ∧ φ start ∧ φ accept ∧ φ move . – And each of these is itself in CNF, except φ move . – φ move is: • a conjunction over all (i,j) • of disjunctions over all tiles • of conjunctions of 6 conditions on the 6 cells: x i,j,a1 ∧ x i,j+1,a2 ∧ x i,j+2,a3 ∧ x i+1,j,b1 ∧ x i+1,j+1,b2 ∧ x i+1,j+2,b3 CNFSAT is NPhard • S h o w A ≤ p CNFSAT. • φ w is a conjunction φ w = φ cell ∧ φ start ∧ φ accept ∧ φ move , where each is in CNF, except φ move . • φ move is: – a conjunction ( ∧ ) over all (i,j) – of disjunctions ( ∨ ) over all tiles – of conjunctions ( ∧ ) of 6 conditions on the 6 cells: x i,j,a1 ∧ x i,j+1,a2 ∧ x i,j+2,a3 ∧ x i+1,j,b1 ∧ x i+1,j+1,b2 ∧ x i+1,j+2,b3 • We want just ∧ of ∨ ....
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This note was uploaded on 12/26/2011 for the course ENGINEERIN 18.400J taught by Professor Prof.scottaaronson during the Spring '11 term at MIT.
 Spring '11
 Prof.ScottAaronson

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