CSCI 2670
Introduction to Theory of
Computing
November 29, 2005
Agenda
• Today
– More on the class NP
November 29, 2005
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
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
What we know
NP
P
November 29, 2005
coNP
What we don’t know
Are there any problems here?
NP
P
November 29, 2005
coNP
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, and
2. Every A in NP is polynomial time
reducible to B
November 29, 2005
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
• ((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
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 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
Polynomial reductions
Definition: Language A is polynomialtime 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
f
f
November 29, 2005
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
1. 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
BP, then P = NP.
Theorem: If B is NP-complete and B≤PC
for some C in NP, then C is NPcomplete
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
November 29, 2005
Reduction from UHAMPATH
R = “On input <G,a,b>, where G = <V,E> is
a graph with nodes a and b
1. If |V| ≤ 1 reject
2. 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
November 29, 2005
Summary
•
To show a language L is NP-complete
1. Demonstrate L is in NP
2. Find a language C that is known to be
NP-complete
3. Create a function f from C to L
4. Demonstrate that if x is in C then f(x)
is in L
5. Demonstrate that if f(x) is in L then x
is in C
6. Demonstrate f is computable in
polynomial time
November 29, 2005