FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY

15-453

NP = NTIME(nk)k N

Theorem: L NP if there exists a poly-time Turing machine V(erifier) with

L = { x | y(witness) |y| = poly(|x|) and V(x,y) accepts }

Proof:

(1) If L = { x | y |y| = poly(|x|) and V(x,y) accepts } then L NP

Because we can guess y and then run V

(2) If L NP thenL = { x | y |y| = poly(|x|) and V(x,y) accepts }

Let N be a non-deterministic poly-time TM that decides L and define V(x,y) to accept if y is an accepting computation history of N on x

A language is in NP if and only if there exist polynomial-length certificates for

membership to the language

SAT is in NP because a satisfying assignment is a polynomial-length

certificate that a formula is satisfiable

NP = all the problems for which once you have the answer it is easy (i.e. efficient) to verify

P = NP?

If P = NP…

Cryptography as we know it would not be possible (e.g. RSA)

Mathematicians would be out of a job

AI program become perfect as exhaustive search is efficient

If P = NP…

Writing symphonies is as easy as listening to them.

Being a chef is as easy as eating.

Writing Shakespeare is as easy as recognizing Shakespeare.

Generation is as easy as recognition:

POLY-TIME REDUCIBILITY

A language A is polynomial time reducible to language B, written A P B, if there is a polynomial time computable function

f : Σ* Σ*, where for every w,

w A f(w) B

f is called a polynomial time reduction of A to B

NP Complete Problem: hardest problem in NP

Intuitively, L is harder than L’ if L’ is polynomial reducible to L.

NP Complete Problem: If such a problem has an efficient algorithm (in P), then every other problem also has efficient algorithm.

Theorem (Cook-Levin): SAT is NP-complete

Corollary: SAT P if and only if P = NP

PNP

SAT

Any thing in NP P 3SAT

You can think of -> as “easier than”. SAT is the hardest problem in NP.

NP-Complete Problems

• 3SAT

• k-Clique

• Vertex Cover

• Independent Set

3SAT

• Problem: Given a CNF where each clause has 3 variables, decide whether it is satisfiable or not.

• (x1 x1 x2) (x1 x2 x2) (x1 x2 x2)

3SAT = { | y such that y is a satisfying assigment to and is in 3cnf }

3SAT is NP Complete

How about 2SAT?

PNP

3SAT

SAT

What we want to prove?

How to prove?

1. We can convert (in polynomial time) a given SAT instance S into a 3SAT instance S’ such that

• If S is satisfiable, then S’ is satisfiable.• If S’ is satisfiable, then S is satisfiable.

Key Observation

x1 x2 x3 x4 x5 is satisfiable

if and only if

(x1 x2 z1) ( z1 x3 z2 )( z2 x4 z3) ( z3 x4 x5)

is satisfiable.

Polynomial Time Reduction

Clause in SATx1

x1 x2

x1 x2 x3

x1 x2 x3 x4

x1 x2 x3 x4 x5

Clauses in 3SATx1 x1 x1

x1 x1 x2

x1 x2 x3

(x1 x2 z1) ( z1 x3 x4)

(x1 x2 z1) ( z1 x3 z2)

( z2 x4 z3) ( z3 x4 x5)

CLIQUE

b

a

e

c

d f

g

k-clique = complete subgraph of k nodes

K-CliquesA K-clique is a set of K nodes with all K(K-A K-clique is a set of K nodes with all K(K-1)/2 possible edges between them1)/2 possible edges between them

This graph contains a 4-clique

CLIQUE = { (G,k) | G is an undirected graph with a k-clique }

Theorem: CLIQUE is NP-Complete

(1) CLIQUE NP

(2) 3SAT P CLIQUE

Assume a reasonable encoding of graphs (example: the adjacency matrix is reasonable)

Brute Force Algorithm: Try out all {n choose k} possible locations for the Try out all {n choose k} possible locations for the k cliquek clique

PNP

CLIQUE

3SAT

CLIQUE is NP-Complete

3SAT P CLIQUE

We transform a 3-cnf formula into (G,k) such that

3SAT (G,k) CLIQUE

The transformation can be done in time that is polynomial in the length of

3SAT P CLIQUE

We transform a 3-cnf formula into (G,k) such that

3SAT (G,k) CLIQUE

If has m clauses, we create a graph with m clusters of 3 nodes each, and set k=mEach cluster corresponds to a clause. Each node in a cluster is labeled with a literal from the clause.

We do not connect any nodes in the same cluster

We connect nodes in different clusters whenever they are not contradictory

(x1 x1 x2) (x1 x2 x2) (x1 x2 x2)

x1 x1

x1 x2

x2 x2

x2 x2

x1

k = #clauses

clause

#nodes = 3(# clauses)

(x1 x1 x1) (x1 x1 x2) (x2 x2 x2) (x2 x2 x1)

x1

x1

x2

x2

x1 x1

x2 x2

x1 x2

x2

x1

This graph contains an independent set of size 3

Independent Set

An An independent set independent set is a set of nodes with no is a set of nodes with no edges between themedges between them

Independent Set

G(V,E)

Problem: Given a graph G and k, is there a size k independent set?

Complement of GGiven a graph G, let G*, the complement of G, be the graph such that two nodes are connected in G* if and only if the corresponding nodes are not connected in G

G G*

Key Observation

• For a graph G, vertex set S is an independent set if and only if S is clique in G*.

Let G be an n-node graphIs there size clique in G?

Is there size k independent set in G*?

VERTEX COVER

b

a

e

c

d

b

a

e

c

d

vertex cover = set of nodes that cover all edges

Vertex Cover is NP Complete

• Given a Graph G(V,E), decide if there is k vertex such that every edge is covered by one of them?

K-Indep Set P Vertex Cover

Key Observation

• For a graph G(V,E), S is a independent set if and only if V-S is a vertex cover.

Other NP-Complete Problems

• Travelling Salesman Problem

• Hamiltonian Path

• Max Cut

• Subset Sum

• Integer Programming

• ….

Other Problems in NP

• Graph Isomorphism

• Factoring Number

We don’t know if they are NP Complete or not.