Lecture11

# Lecture11 - CME 305: Discrete Mathematics and Algorithms...

This preview shows pages 1–2. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CME 305: Discrete Mathematics and Algorithms Instructor: Professor Amin Saberi ([email protected]) February 9, 2010 Lecture 11: Computation and Intractability So far we’ve studied several problems for which ’efficient’ algorithms are known, such as max-flow/min- cut, linear programming, bipartite matching, minimum spanning tree. By ’efficient’, we mean solvable in polynomial time. However, it is often the case that if you change the problem statement for an ’easy’ problem slightly, the problem becomes ’hard’, i.e. that the best known algorithms take an exponential number of operations despite many years of study. Some examples of “easy problems”, and their “hard” counterparts: “easy” “hard” shortest path longest path MST min. spanning k-connected graphs, k > 1 2-coloring k-coloring for k > 2 For many of the most fundamental problems in computer science, biology, combinatorics and elsewhere, there is no efficient algorithm known. Moreover, for many problems we don’t know how to prove that no efficient algorithms exist. However, some progress has been made. A large class of these difficult problems have been shown to be “equivalent”. This is the class of NP-completeness and the topic of lecture today. 1 Polynomial Time Reducibility Problem X is polynomially time reducible to problem Y ( X ≤ p Y ) iff for every instance of X there is an algorithm that solves X with polynomially many operations and polynomially many calls to a “black box” that solves a given instance of Y . In other words, if for every instance of problem X we can represent it as a instance of Y (or sequence of such instances) using a polynomial number of computations, then we say that X is harder than Y and we write X ≥ p Y . If Y can be in turn solved through black box calls that solves intances of X , i.e. Y is also harder than X , then X and Y are equivalent and we write X = p Y ....
View Full Document

## This document was uploaded on 06/17/2010.

### Page1 / 3

Lecture11 - CME 305: Discrete Mathematics and Algorithms...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online