CSCI 2670 Introduction to Theory of Computing

1. CSCI 2670Introduction to Theory of Computing November 29, 2005

2. Agenda • Today • More on the class NP November 29, 2005

3. The class NP Definition: A verifier for a language A is an algorithm V, where A={w|V accepts <w,c> for some string c} The string c is called a certificate of membership in A. Definition: NP is the class of languages that have polynomial-time verifiers. November 29, 2005

4. Who wants $1,000,000? • In May, 2000, the Clay Mathematics Institute named seven open problems in mathematics the Millennium Problems • Anyone who solves any of these problems will receive $1,000,000 • Proving whether or not P equals NP is one of these problems November 29, 2005

5. NP P coNP What we know November 29, 2005

6. Are there any problems here? What we don’t know NP P coNP November 29, 2005

7. Solving NP problems • The best-known methods for solving problems in NP that are not known to be in P take exponential time • Brute force search November 29, 2005

8. 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 November 29, 2005

9. An NP-complete problem • A formula is Boolean if each of its variables can be assigned the values TRUE (1) or FALSE (0) • A Boolean formula is satisfiable if there is some assignment of values that results in the formula evaluating to TRUE SAT: Is a given Boolean formula satisfiable? • SAT is NP-complete • We will go over the proof next week November 29, 2005

10. Examples • (x  y)  ( x  y) • Satisfiable – e.g., x = y = 1 • ((xy)  (xz))  ((xy)  (yz)) • Satisfiable – e.g., x = 0, y = z = 1 • ((xy)  (xz))  ((xy)  (yz)) • Unsatisfiable November 29, 2005

11. Proving a problem is NP-complete • A problem C is NP-complete if finding a polynomial-time solution for C would imply P=NP • If a polynomial-time solution is found for C, then that solution can be used to find a polynomial-time solution for any other problem in NP • What does this remind you of? • Reductions! November 29, 2005

12. Reductions and NP-completeness • If we can prove an NP-complete problem C can be polynomially reduced to a problem A, then we’ve shown A is NP-complete • A polynomial-time solution to A would provide a polynomial-time solution to C, which would imply P=NP November 29, 2005

13. Polynomial functions Definition: A function f:Σ*Σ* is a polynomial time computable function if some polynomial time Turing machine M exists that halts with just f(w) on its tape, when started on any input w. November 29, 2005

14. f f Polynomial reductions Definition: Language A is polynomial-time reducible to language B, written A ≤P B, if a polynomial time computable function f:Σ*Σ* exists, where for every w w  A iff f(w)  B November 29, 2005

15. Reductions & NP-completeness Theorem: If A ≤P B and BP, then AP. Proof: Let M be the polynomial time algorithm that decides B and let f be the polynomial reduction from A to B. Consider the TM N N = “On input w • Compute f(w) • Run M on f(w) and output M’s result” Then N decides A in polynomial time. November 29, 2005

16. Implications of NP-completeness Theorem: If B is NP-complete and BP, then P = NP. Theorem: If B is NP-complete and B≤PC for some C in NP, then C is NP-complete November 29, 2005

17. Showing a problem in NP-complete • Two steps to proving a problem L is NP-complete • Show the problem is in NP • Demonstrate there is a polynomial time verifier for the problem • Show some NP-complete problem can be polynomially reduced to L November 29, 2005

18. NP-completeness proof • LPATH = {<G, a, b, k> | G is a graph with nodes a and b and a simple path of length k from a to b} • A simple path has no repeated nodes • We will use the fact that the following language is NP-complete • UHAMPATH = {<G,a,b> | G is a graph with a Hamiltonian path from a to b} • A Hamiltonian path visits each node exactly once November 29, 2005

19. LPATH is NP-complete • First show LPATH is in NP • Can we verify a solution to LPATH in polynomial time? • Yes • Check the certificate is a simple length-k path from a to b November 29, 2005

20. Reduction from UHAMPATH R = “On input <G,a,b>, where G = <V,E> is a graph with nodes a and b • If |V| ≤ 1 reject • Check if <G,a,b,|V|-1> is in LPATH” November 29, 2005

21. Two questions • Need to demonstrate that <Ga,b> is in UHAMPATH iff <G,a,b,|V|-1> is in LPATH • This is clear from the definitions of UHAMPATH and LPATH • Need to demonstrate that the reduction is in polynomial time • The reduction takes O(|V|) time (to evaluate |V| November 29, 2005

22. Insight • We know (because I said so) that UHAMPATH is NP-complete • Therefore, a polynomial time solution to UHAMPATH would imply P = NP • We showed that we can convert UHAMPATH to LPATH in polynomial time • Therefore, a polynomial solution to LPATH would provide a polynomial solution to UHAMPATH • LPATH must be NP-complete November 29, 2005

23. Summary • To show a language L is NP-complete • Demonstrate L is in NP • Find a language C that is known to be NP-complete • Create a function f from C to L • Demonstrate that if x is in C then f(x) is in L • Demonstrate that if f(x) is in L then x is in C • Demonstrate f is computable in polynomial time November 29, 2005