CSCI 2670Introduction to Theory of

Computing

December 7, 2005

December 7, 2005

Agenda

• Today– Discussion of Cook-Levin Theorem– One more NP-completeness proof– Course evaluations

• Tomorrow– Return tests– Provide pre-final grades– Final review

December 7, 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

December 7, 2005

Cook-Levin theorem

• SAT = {<B>|B is a satisfiable Boolean expression}

Theorem: SAT is NP-complete– If SAT can be solved in polynomial

time then any problem in NP can be solved in polynomial time

December 7, 2005

Proof of Cook-Levin theorem

• First show that SAT is in NP– Easy

• The show that SAT P implies P = NP– Proof converts a non-deterministic

Turing machine M with input w into a Boolean expression φ

– M accepts string w iff φ is satisfiable– <M,w> is converted into φ in

polynomial time

December 7, 2005

Variables in proof of Cook-Levin Thm

• Ti,j,s is true if tape position i contains symbol s at step j of computation– O(p(n)2) variables

• Hi,k is true if M’s tape head is at position i at step k of computation– O(p(n)2) variables

• Qq,k is true if M is in state q at step k of the computation– O(p(n)) variables

December 7, 2005

Boolean expression

• Using the variables Ti,j,k, Hi,k, and Qq,k, Cook & Levin developed a Boolean expression that is satisfiable if and only if M accepts w– Expression reflects proper behavior of

M• Qstart,0 ensures starts in start state• H0,0 ensures starts with tape head at start of

tape• Ti,j,k Ti,j’,k ensures one symbol per tape

cell• Etc…

December 7, 2005

Purpose of Boolean expression

• If the Boolean expression in the proof of the Cook-Levin theorem can evaluate to TRUE then– Starting at the start state with w on

the tape, M can take steps based on the transition function and end at the accept state

– I.e., M accepts w

• If the Boolean expression is unsatisfiable, then M rejects w

December 7, 2005

Length of Boolean expression

• O((log p(n)) p(n)2)• Since p(n) is polynomial in the

length of the input, so is the Boolean expression

• If the satisfiability of the expression can be determined in polynomial time, we can determine in polynomial time whether M accepts or rejects w– Every problem in NP would be in P if

SAT is in P

December 7, 2005

Another NP-complete problem

• 3SAT – The Boolean expression must be in a

specific form called 3-cnf• Conjunctive normal form

– A literal is a Boolean variable or the negation of a Boolean variable

– A clause is the disjunction (OR) of literals

– A Boolean formula is in cnf if it is the conjunction (AND) of clauses• It is 3-cnf if all the clauses have 3 literals

December 7, 2005

Importance of Cook-Levin theorem

• Cook & Levin proved that SAT is NP-complete– If SAT can be solved in polynomial

time, then any problem in NP can be solved in polynomial time

• The showed that any problem in NP can be polynomially reduced to the SAT problem

• The proof that 3SAT is NP-complete is similar– Generate a 3-cnf formula from a TM

December 7, 2005

NP-completeness proof

• CLIQUE is NP-complete– A clique in an undirected graph is a

subgraph with every pair of nodes connected by an edge

• CLIQUE = {<G,k> | G is an undirected graph with a k-clique}

December 7, 2005

Example

This graph has a 4-clique

December 7, 2005

Proving CLIQUE is NP-complete

1. Show CLIQUE is in NP2. Show some NP-complete problem

can be polynomially reduced to CLIQUE

December 7, 2005

Is CLIQUE in NP?

• Yes• Given a subset V’ of V

– Verify |V’|=k• O(k) time

– Verify every pair of vertices in |V’| have an edge in E• O(k2 |E|) time

December 7, 2005

Reducing 3SAT to CLIQUE

• Create a polynomial time function that converts a 3-cnf Boolean formula to a graph– The graph will have a k-clique if and

only if the formula is satisfiable– Cliques in the graph correspond to

satisfying assignments of truth values to variables in the formula

– Structures in the graph mimic behavior of clauses

December 7, 2005

3SAT reduction to CLIQUE

• Start with any 3-cnf formula with k clauses

φ = (a1b1c1)(a2b2c2)…(a2b2c2)

• Create a graph with 3k nodes– One node for each literal in φ– A single literal may have more than one

node

• A pair of nodes xi and xj has an edge if (1) they appear in different clauses, (2) i ≠ j and (3) xi ≠ xj

December 7, 2005

Example

φ = (x y y) (x y y) (x x y)

x

y

y

x y y

x

x

y

Q: Is φ satisfiable? A: Yes: x = y = 1

December 7, 2005

Correctness of construction

• Need to show the formula is satisfiable iff the graph has a k-clique

• If the formula is satisfiable, there is a way to assign values to the variables such that at least one literal is true in each clause– The corresponding nodes will create a

k-clique

December 7, 2005

Correctness of construction

• Need to show the formula is satisfiable iff the graph has a k-clique

• By construction, any k-clique will contain one node from each clause and will not contain both x and x for any variable x– Assigning the variable corresponding

to each node the value true will result in one literal in each clause being true

December 7, 2005

Are we done?

• We have shown that CLIQUE is in NP• We have found a reduction from

3SAT to CLIQUE– The 3-cnf formula is satisfiable iff the

graph has a k-clique

• What’s left?– Demonstrate the reduction is

polynomial– |V| = # of literals in φ– |E| ≤ (|V|-3)(|V|-6)(|V|-9)…(3) = O(|V|2)

=O( (# of literals in φ)2 )