lect22

lect22 - 6.841 Advanced Complexity Theory Apr 29, 2009...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
6.841 Advanced Complexity Theory Apr 29, 2009 Lecture 22 Lecturer: Madhu Sudan Scribe: Jessica Yuan 1 Last Time Last time, we talked about what a propositional proof system is. It’s a way of proving that a formula is a tautology (i.e., it’s always true). Very quickly, we switched from proving formulas that are always true to formulas that are always false. Without loss of generality, we focused on CNF formulas, since we can transform formulas into CNF form with only a linear blow-up. A proof system is something that can determine whether a CNF formula F is either unsatisfiable: there exists some proof π that is efficiently (poly-time) verifiable deterministically satisfiable: there should be no fake proof that can fool us Why do we care about propositional proof systems? 1) P vs. NP. If there is no proof system that can always provide proofs of polynomial size, this implies P 6 = NP. 2) Understand limits of math reasoning 3) SAT-solving. In industrial applications, there are really good algorithms for SAT-solving in linear time, and the best ones are based on a proof system called resolution . We then looked at some examples of proof systems and decided to focus on resolution. This is the proof system where given clauses C x and D ¯ x , we can obtain the clause C D . Additionally, we can use a weakening mechanism, where given clause C , we can obtain the weaker clause C D . This weakening mechanism is not necessary for proving unsatisfiability, but it is often useful. We looked at some examples of formula families, such as Graph Tautology, and decided to focus on Pigeonhole Principle. Pigeonhole Principle ( PHP m n ) lets us show that if there are m pigeons and n holes, where m > n , there must be two pigeons that share a hole. The unsatisfiable formula consists of two parts: Each pigeon must be in a hole: V i [ m ] W j [ n ] x i j No two pigeons share a hole: V i 1 6 = i 2 [ m ] ,j [ n ] ( x i 1 j x i 2 j ) Combined, this gives us the formula: ^ i [ m ] _ j [ n ] x i j ^ i 1 6 = i 2 [ m ] ,j [ n ] ( x i 1 j x i 2 j ) Today, we’ll focus on how hard it is to prove that this formula is unsatisfiable. 22-1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 Resolution Proof Size We’re interested in the size of resolution proofs, i.e. the total number of symbols. We can also count the number of clauses in the proof (i.e. the length of the derivation of the contradiction). This tends to be easier to deal with than size, and because length is linearly related to size, we can get a good idea of the size of a proof given its length. The width of the clauses is also a good tool for studying the length of proofs. The length of a refutation proof π is denoted as L ( π ). The length of the proof showing that a formula F is unsatisfiable is the length of the shortest proof that proves that F is unsatisfiable: L ( F ` 0) = min π : F ` 0 { L ( π ) } Note that if n is the number of variables in F , L ( F ` 0) . 2 n .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/02/2010 for the course CS 6.841 taught by Professor Madhusudan during the Spring '09 term at MIT.

Page1 / 7

lect22 - 6.841 Advanced Complexity Theory Apr 29, 2009...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online