CSCI 2670Introduction to Theory of

Computing

November 29, 2005

November 29, 2005

Agenda• Today

– More on the class NP

November 29, 2005

The class NPDefinition: A verifier for a language

A is an algorithm V, whereA={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

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

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What we know

NPP coNP

November 29, 2005

What we don’t know

NPP coNP

Are there any problems here?

November 29, 2005

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

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:

1. B is in NP, and2. Every A in NP is polynomial time

reducible to B

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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

Examples• (x y) ( x y)

– Satisfiable – e.g., x = y = 1• ((xy) (xz)) ((xy) (yz))

– Satisfiable – e.g., x = 0, y = z = 1• ((xy) (xz)) ((xy) (yz))

– Unsatisfiable

November 29, 2005

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

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

Polynomial functionsDefinition: 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.

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Polynomial reductionsDefinition: 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) Bf

f

November 29, 2005

Reductions & NP-completeness

Theorem: If A ≤P B and BP, then AP.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 w1. Compute f(w)2. Run M on f(w) and output M’s result”Then N decides A in polynomial time.

November 29, 2005

Implications of NP-completeness

Theorem: If B is NP-complete and BP, 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

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

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

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

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Reduction from UHAMPATHR = “On input <G,a,b>, where G =

<V,E> is a graph with nodes a and b

1. If |V| ≤ 1 reject2. Check if <G,a,b,|V|-1> is in

LPATH”

November 29, 2005

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

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

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Summary• To show a language L is NP-

complete1. Demonstrate L is in NP2. Find a language C that is known to be

NP-complete3. Create a function f from C to L4. Demonstrate that if x is in C then f(x)

is in L5. Demonstrate that if f(x) is in L then x

is in C6. Demonstrate f is computable in

polynomial time