Decision vs. Optimization So far languages L f defined by boolean - - PowerPoint PPT Presentation

Search Problems and coNP

Nabil Mustafa Computability and Complexity

Decision vs. Optimization

So far languages Lf defined by boolean function f x ∈ Lf ⇐ ⇒ f (x) = 1 Given G = (V , E), is G k-colorable? Given G = (V , E), does G have independent set of size k? Given a 3-CNF formula φ, is φ satisfiable? What about the corresponding search questions Given G = (V , E), compute a k-coloring of G Given G = (V , E), compute an independent set of size k in G Given 3-CNF formula, compute a satisfying assignment

Example

Let A = {a1, . . . , an} be a set of n distinct integers Poly-time function: IsItSorted(π(A)) = 1 iff π(A) is sorted Then, using IsItSorted(A), can one actually sort n numbers? First try: Given n numbers A = {a1, . . . , an} For each permutation π′ of A, check IsItSorted(π′(A)) Total calls to IsItSorted(A): n! A better try: Start with the sorted set A1 = a1 Check IsItSorted(A1 ◦ a2) and IsItSorted(a2 ◦ A1) Exactly one of the above has answer 1, say latter one Then set A2 = a2 ◦ a1 Adding similarly, the i-th element: i calls to IsItSorted Total calls: n

General Reduction from Search to Decision

Claim Assume P = NP . Then for every NP language L, there exists a poly-time TM that, on input x ∈ L, outputs a certificate for x. Proof. Recall the proof of the Cook-Levin theorem Take L and do the reduction to CNF satisfiability of φ . The possible transition rules followed during the computation

• f x translated into boolean variables Ri

Now assume we can find a satisfying assignment for φ Ri

This gives us the certificate for x ... assuming satisfiability

Computing Certificate for CNF Satisfiability

Claim Assume P = NP . Then there exists a poly-time TM to find a satisfying assignment for any formula φ. Proof. Lets assume the variables are x1, . . . , xn and formula φ Repeat the following for i = 1 . . . n: If SAT(φ(xi = 1, xi+1, . . . , xn)) = 1, set xi = 1 Else If SAT(φ(xi = 0, xi+1, . . . , xn)) = 1, set xi = 0 Else return failure. Number of calls to SAT: 2n

Example

Independent Set problem Given G = (V , E) and an integer k, compute an independent set

• f size k in G.

Black box: Given G = (V , E), does G have an independent set of size k? Solution: Find and remove useless nodes one by one Pick v and ask: Does G \ v have independent set of size k? If yes then v is useless so remove it from G If no, then v is in independent set. Keep v Repeat from step 1 until the remaining graph has no edges.

Finding independent set

Concrete Example Compute independent set of size k = 2 in this graph. Answer: Yes

Languages With No Short Certificate

L ∈ NP iff for each x ∈ L, ∃ poly-sized certificate u certifying the existence of x ∈ L. Are there languages with no short certificate of acceptance? SAT: all formulae φ which have no satisfying assignment. Is SAT ∈ NP ? I.e., φ ∈ SAT have a short certificate? How do you certify that φ has no satisfying assignment? Possibility: list all assignments and show no satisfying one Certificate size: Ω(2n) On the other hand, if φ / ∈ SAT, the certificate? Just the condition of class NP Now define the set of languages where the non-existence of x in the language has a short certificate.

The Class coNP

Definition of coNP A language L ∈ coNP iff for every x / ∈ L, ∃u of size O(p(|x|)) and a poly-time TM M such that M(x, u) = 0. Definition coNP = {L : L ∈ NP } L ∈ coNP = ⇒ x / ∈ L has a short certificate Claim: SAT ∈ coNP

Examples

No Hamiltonian Cycle Instance: A graph G Question: Does G have no Hamiltonian cycle? Tautology Instance: A CNF-formula f Question: Is f satisfied by all assignments?

coNP is not complement of NP

Claim: coNP = {L : L / ∈ NP }

• Incorrect. A language can be in both NP and coNP !

Claim: Any language in P is in NP and coNP Any language in P is in NP : Trivial. Any language in P is in coNP : L ∈ P = ⇒ TM M: M(x) = 1 ↔ x ∈ L Claim: L ∈ P M′(x) = 1 − M(x). Then M′(x) = 1 ↔ x ∈ L Hence L ∈ P and so L ∈ coNP Corollary: coP = P .

coNP complete languages

Definition A language L is coNP complete iff

Claim A language L is coNP complete iff L is NP complete. Proof. L is NP complete = ⇒ L is coNP complete Show: Deciding x ∈ L′ can be reduced to g(x) ∈ L Deciding x ∈ L′ can be reduced to x / ∈ L′ Deciding x / ∈ L′ can be reduced to g(x) / ∈ L Deciding g(x) / ∈ L can be reduced to g(x) ∈ L

coNP complete languages

Claim The language Tautology is coNP complete. Claim The language No Hamiltonian Path is coNP complete. Claim The language No Satisfiability is coNP complete.

A Claim for coNP

Claim If L is coNP complete, and L ∈ NP , then coNP = NP Proof. L ∈ coNP and L ∈ NP L ∈ NP : x ∈ L has a short certificate L ∈ coNP : x / ∈ L has a short certificate L is coNP complete L is NP complete L′ ∈ NP = ⇒ L′ ∈ coNP : x / ∈ L′ has a short certificate x / ∈ L′ ⇐ ⇒ g(x) / ∈ L ⇐ ⇒ g(x) ∈ L L′ ∈ coNP = ⇒ L′ ∈ NP : x ∈ L′ has a short certificate x ∈ L′ ⇐ ⇒ g(x) ∈ L ⇐ ⇒ g(x) ∈ L

Complexity Classes Scenery

Decidable coNP P

NP Undecidable