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CSCI 2670 Introduction to Theory of Computing

CSCI 2670 Introduction to Theory of Computing. December 7, 2005. Agenda. Today Discussion of Cook-Levin Theorem One more NP-completeness proof Course evaluations Tomorrow Return tests Provide pre-final grades Final review. NP-completeness.

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CSCI 2670 Introduction to Theory of Computing

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  1. CSCI 2670Introduction to Theory of Computing December 7, 2005

  2. Agenda • Today • Discussion of Cook-Levin Theorem • One more NP-completeness proof • Course evaluations • Tomorrow • Return tests • Provide pre-final grades • Final review December 7, 2005

  3. NP-completeness • A problem C is NP-complete if finding a polynomial-time solution for C would imply P=NP Definition: A language B is NP-complete if it satisfies two conditions: • B is in NP, and • Every A in NP is polynomial time reducible to B December 7, 2005

  4. Cook-Levin theorem • SAT = {<B>|B is a satisfiable Boolean expression} Theorem: SAT is NP-complete • If SAT can be solved in polynomial time then any problem in NP can be solved in polynomial time December 7, 2005

  5. Proof of Cook-Levin theorem • First show that SAT is in NP • Easy • The show that SAT  P implies P = NP • Proof converts a non-deterministic Turing machine M with input w into a Boolean expression φ • M accepts string w iff φ is satisfiable • <M,w> is converted into φ in polynomial time December 7, 2005

  6. Variables in proof of Cook-Levin Thm • Ti,j,s is true if tape position i contains symbol s at step j of computation • O(p(n)2) variables • Hi,k is true if M’s tape head is at position i at step k of computation • O(p(n)2) variables • Qq,k is true if M is in state q at step k of the computation • O(p(n)) variables December 7, 2005

  7. Boolean expression • Using the variables Ti,j,k, Hi,k, and Qq,k, Cook & Levin developed a Boolean expression that is satisfiable if and only if M accepts w • Expression reflects proper behavior of M • Qstart,0 ensures starts in start state • H0,0 ensures starts with tape head at start of tape • Ti,j,k Ti,j’,k ensures one symbol per tape cell • Etc… December 7, 2005

  8. Purpose of Boolean expression • If the Boolean expression in the proof of the Cook-Levin theorem can evaluate to TRUE then • Starting at the start state with w on the tape, M can take steps based on the transition function and end at the accept state • I.e., M accepts w • If the Boolean expression is unsatisfiable, then M rejects w December 7, 2005

  9. Length of Boolean expression • O((log p(n)) p(n)2) • Since p(n) is polynomial in the length of the input, so is the Boolean expression • If the satisfiability of the expression can be determined in polynomial time, we can determine in polynomial time whether M accepts or rejects w • Every problem in NP would be in P if SAT is in P December 7, 2005

  10. Another NP-complete problem • 3SAT • The Boolean expression must be in a specific form called 3-cnf • Conjunctive normal form • A literal is a Boolean variable or the negation of a Boolean variable • A clause is the disjunction (OR) of literals • A Boolean formula is in cnf if it is the conjunction (AND) of clauses • It is 3-cnf if all the clauses have 3 literals December 7, 2005

  11. Importance of Cook-Levin theorem • Cook & Levin proved that SAT is NP-complete • If SAT can be solved in polynomial time, then any problem in NP can be solved in polynomial time • The showed that any problem in NP can be polynomially reduced to the SAT problem • The proof that 3SAT is NP-complete is similar • Generate a 3-cnf formula from a TM December 7, 2005

  12. NP-completeness proof • CLIQUE is NP-complete • A clique in an undirected graph is a subgraph with every pair of nodes connected by an edge • CLIQUE = {<G,k> | G is an undirected graph with a k-clique} December 7, 2005

  13. Example This graph has a 4-clique December 7, 2005

  14. Proving CLIQUE is NP-complete • Show CLIQUE is in NP • Show some NP-complete problem can be polynomially reduced to CLIQUE December 7, 2005

  15. Is CLIQUE in NP? • Yes • Given a subset V’ of V • Verify |V’|=k • O(k) time • Verify every pair of vertices in |V’| have an edge in E • O(k2 |E|) time December 7, 2005

  16. Reducing 3SAT to CLIQUE • Create a polynomial time function that converts a 3-cnf Boolean formula to a graph • The graph will have a k-clique if and only if the formula is satisfiable • Cliques in the graph correspond to satisfying assignments of truth values to variables in the formula • Structures in the graph mimic behavior of clauses December 7, 2005

  17. 3SAT reduction to CLIQUE • Start with any 3-cnf formula with k clauses φ = (a1b1c1)(a2b2c2)…(a2b2c2) • Create a graph with 3k nodes • One node for each literal in φ • A single literal may have more than one node • A pair of nodes xi and xj has an edge if (1) they appear in different clauses, (2) i ≠ j and (3) xi ≠ xj December 7, 2005

  18. φ = (x  y  y)  (x  y  y)  (x  x  y) x y y x y x y y x Example Q:Is φ satisfiable? A:Yes: x = y = 1 December 7, 2005

  19. Correctness of construction • Need to show the formula is satisfiable iff the graph has a k-clique • If the formula is satisfiable, there is a way to assign values to the variables such that at least one literal is true in each clause • The corresponding nodes will create a k-clique December 7, 2005

  20. Correctness of construction • Need to show the formula is satisfiable iff the graph has a k-clique • By construction, any k-clique will contain one node from each clause and will not contain both x and x for any variable x • Assigning the variable corresponding to each node the value true will result in one literal in each clause being true December 7, 2005

  21. Are we done? • We have shown that CLIQUE is in NP • We have found a reduction from 3SAT to CLIQUE • The 3-cnf formula is satisfiable iff the graph has a k-clique • What’s left? • Demonstrate the reduction is polynomial • |V| = # of literals in φ • |E| ≤ (|V|-3)(|V|-6)(|V|-9)…(3) = O(|V|2) =O( (# of literals in φ)2 ) December 7, 2005

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