CSCI 256

Data Structures and Algorithm Analysis

Lecture 22

Some slides by Kevin Wayne copyright 2005,

Pearson Addison Wesley all rights reserved,

and some by Iker Gondra

Transition point

• Till now we’ve looked at developing efficient algorithms for a wide range of problems

• Now we look to quantify or characterize the range of problems that can’t be solved efficiently

• We’ll use the notion of reduction to help compare the relative difficulty of different problems

The Universe

NP-Complete

NP

P

Millennium Prize Problems

Copyright © 1990, Matt Groening

Classify Problems According to Computational Requirements

• Q: Which problems will we be able to solve in practice?– A working definition. [Cobham 1964, Edmonds 1965, Rabin

1966] Those with polynomial-time algorithms

• Desiderata: Classify problems according to those that can be solved in polynomial-time and those that cannot

• Frustrating news: Huge number of fundamental problems have defied classification for decades – i.e. can’t give a poly time alg and can’t prove that none exists

• This chapter: Show that these fundamental problems are "computationally equivalent" and appear to be different manifestations of one really hard problem

Polynomial-Time Reduction

• Desiderata': Suppose we could solve Y in polynomial-time. What else could we solve in polynomial time?

• Reduction: Problem X polynomial reduces to problem Y if arbitrary instances of problem X can be solved using– Polynomial number of standard computational steps,

plus– Polynomial number of calls to an oracle (a black box)

that solves problem Y

• Notation: X P Ycomputational model supplemented by special pieceof hardware that solves instances of Y in a single step

Lemma

• Suppose X P Y. If Y can be solved in polynomial time, then X can be solved in polynomial time

• Remark – when you are reading the text note that the text talks of Y P X

Lemma

• Suppose X P Y. If X cannot be solved in polynomial time, then Y cannot be solved in polynomial time

Polynomial-Time Reduction

• Purpose: Classify problems according to relative difficulty

• Design algorithms: If X P Y and Y can be solved in polynomial-time, then X can also be solved in polynomial time

• Establish intractability: If X P Y and X cannot be solved in polynomial-time, then Y cannot be solved in polynomial time

• Establish equivalence: If X P Y and Y P X, we use notation X P Yup to cost of reduction

Independent Set

• Independent Set– Goal is to find S of maximum size

1

3

2

6 7

4 5

Decision Problem

• Call the above an optimization version of the problem

• There is another way to formulate the problem – as a decision problem

• A decision problem is a class of problems for which there is a yes or no answer for each instance of the problem.

• We rephrase the Independent Set problem as the following decision problem

Independent Set – Decision version

- Given a graph G = (V, E) and an integer k, does G contain an independent set of size at least k?

– Ex: Instance: For G below, is there an independent set of size 6? Yes

– Ex: Instance: For G below, is there an independent set of size 7? No

independent set

Sample Application: find set of mutually

non-conflicting points

Equivalence between decision and optimization versions

• Given a method to solve the optimization version, we automatically solve each instance of the associated decision problem; and if we have a method to solve each instance of the decision then we can solve the optimization version (for given G, the the largest k for which there is a yes answer gives us the answer to the optimization version for G)

Vertex Cover

• Vertex Cover– Given a graph G = (V, E), a subset S of the vertices is

a vertex cover if every edge in E has at least one endpoint in S. Goal is to find S of minimum size

1

3

2

6 7

4 5

Vertex Cover – Associated decision version – if you can do optimization version you can do decision version and vice versa

• Convert to decision version of the problem– Given a graph G = (V, E) and an integer k, does G contain a

vertex cover of size at most k?

– Ex: Instance: For G, Is there a vertex cover of size 4? Yes– Ex: Instance: For G, Is there a vertex cover of size 3? No

vertex cover

Sample Application: place guards within

an art gallery so that all corridors are visible at any

time

Relative difficulty of the two problems

• We don’t know how to solve the independent set problem or vertex cover problem in polynomial time; but we can show that they are equivalently hard

• First we need the following theorem:

Vertex Cover and Independent Set

• Theorem: S is an independent set iff V S is a vertex cover.

Recall: Given a graph G = (V, E), a subset S of the vertices is independent if there are no edges between vertices in S.

Recall: Given a graph G = (V, E), a subset S of the vertices is a vertex cover if every edge in E has at least one endpoint in S.

Vertex Cover and Independent Set

• Theorem: S is an independent set iff V S is a vertex cover

– Let S be any independent set– Consider an arbitrary edge, (u, v) ε E– S independent u S or v S u V S or v V S– Thus, V S covers (u, v)

– Let V S be any vertex cover – Consider two nodes u S and v S – Observe (u, v) E (V S is a vertex cover so if (u,v) ε E, u V-

S or v V – S so u S or v S -- a contradiction – Thus, no two nodes in S are joined by an edge S indep set

Vertex Cover and Independent Set Reductions to each other:

• Theorem: VERTEX-COVER P INDEPENDENT-SET

If we have a black box to solve the Vertex Cover problem then we can decide if G has an independent set of size at least k by asking the black box whether G has a Vertex Cover of size at most n-k.

Conversely if we have a black box to solve the Independent set problem we can decide if G has a vertex cover problem of size at most k by asking the black box if G has an independent set of size at least n-k

Set Cover

• SET COVER: Given a set U of elements, a collection S1, S2, . . . , Sm of subsets of U, and an integer k, does there exist a collection of k of these sets whose union is equal to U?

• Sample application– m available pieces of software– Set U of n capabilities that we would like our system to have

– The ith piece of software provides the set Si U of capabilities

– Goal: achieve all n capabilities using fewest pieces of softwareU = { 1, 2, 3, 4, 5, 6, 7 }

k = 2S1 = {3, 7} S4 = {2, 4}

S2 = {3, 4, 5, 6} S5 = {5}

S3 = {1} S6 = {1, 2, 6,

7}

• Claim: VERTEX-COVER P SET-COVER

– Pf: Given a VERTEX-COVER instance, i.e., a graph G = (V, E), and an integer, k, we construct a set cover instance whose size equals the size of the vertex cover instance, so G has a vertex cover of size k iff for the set U, and subsets S1, S2, . . . , Sm of U,

there exists a collection of k of these sets whose union is equal to U

Vertex Cover Reduces to Set Cover

• Claim: VERTEX-COVER P SET-COVER

• Construction– Create SET-COVER instance

• k = k, U = { i | ei E), Sv = { i : ei incident to v } (note Sv U)

– Set-cover of size k iff vertex cover of size k

SET COVER

U = { 1, 2, 3, 4, 5, 6, 7 }k = 2Sa = {3, 7} Sb = {2, 4}

Sc = {3, 4, 5, 6} Sd = {5}

Se = {1} Sf= {1, 2, 6, 7}

a

d

b

e

f c

VERTEX COVER

k = 2e1

e2 e3

e5

e4

e6

e7