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Fall 2009. NP-complete examples. Xiao Linfu. CSC3130 Tutorial 11. [email protected] Department of Computer Science & Engineering. Outline. Review of P, NP, NP-C 2 problem s Double-SAT Dominating set http://en.wikipedia.org/wiki/Dominating_set_problem. Relations. hard. NP-C.

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### NP-complete examples

Xiao Linfu

CSC3130 Tutorial 11

Department of Computer Science & Engineering

Outline

- Review of P, NP, NP-C
- 2 problems
- Double-SAT
- Dominating sethttp://en.wikipedia.org/wiki/Dominating_set_problem

How to show that a problem R is not easier than a problem Q?

Informally, if R can be solved efficiently, we can solve Q efficiently.

- Formally, we say Q polynomially reduces to R if:
- Given an instance q of problem Q
- There is a polynomial time transformation to an instance f(q) of R
- q is a “yes” instance if and only iff(q) is a “yes” instance

Then, if R is polynomial time solvable, then Q is polynomial time solvable.

If Q is not polynomial time solvable, then R is not polynomial time solvable.

Methodology

- To show L is in NP, you can either (i) show that solutions for L can be verified in polynomial-time, or (ii) describe a nondeterministic polynomial-time TM for L.
- To show L is NP-complete, you have to design a polynomial-time reduction from some problem we know to be NP-complete

- The direction of the reduction is very important
- Saying “A is easier than B” and “B is easier than A” mean different things

- What we have? We know SAT, Vertex Cover problems are NP-Complete!

Double-SAT

- Definition:
- Double-SAT = {<φ> |φ is a Boolean formula with at least two satisfying assignments}

- Show that Double-SAT is NP-Complete.
- (1) First, it is easy to see that Double-SAT ∈ NP.
- non-deterministically guess 2 assignments for φand verify whetherboth satisfy φ.

- (2) Then we show Double-SAT is not easier than SAT.
- Reduction from SAT to Double-SAT

- (1) First, it is easy to see that Double-SAT ∈ NP.

Double-SAT

- Reduction:
- On input φ(x1, . . . , xn):
- 1. Introduce a new variable y.
- 2. Output formula
φ’(x1, . . . , xn, y) = φ(x1, . . . , xn) ∧(y∨ y ).

Dominating set

- Definition: input G=(V,E), K
- Let G=(V,E) be an undirected graph. A dominating set D is a set of vertices in G such that every vertex of G is either in D or is adjacent to at least one vertex from D. The problem is to determine whether there is a dominating set of size K for G.

Dominating set

- Show that Dominating set is NP-Complete.
- (1) First, it is easy to see that Dominating set ∈ NP.
- Given a vertex set D of size K, we check whether (V-D) are adjacent to D.

- (2) Then we show Dominating set is not easier than Vertex cover.
- Reduction from Vertex cover to Dominating set

- (1) First, it is easy to see that Dominating set ∈ NP.

Dominating set

- Reduction
- (1) Graph transformation - Construct a new graph G' by adding new vertices and edges to the graph G as follows: For each edge (v, w) of G, add a vertex vw and the edges (v, vw) and (w, vw) to G' . Furthermore, remove all vertices with no incident edges; such vertices would always have to go in a dominating set but are not needed in a vertex cover of G.

Dominating set

- Reduction
- (1) Graph transformation
- (2) a dominating set of size K in G’ a vertex cover of size K in G

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