NP-complete examples CSC3130 Tutorial 11 Xiao Linfu Department of Computer Science & Engineering Fall 2009

Outline Review of P, NP, NP-C 2 problems –Double-SAT –Dominating set

Relations NP P NP-C easy hard Is there any problem even harder than NP-C? Yes! e.g. I-go

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: 1.Given an instance q of problem Q 2.There is a polynomial time transformation to an instance f(q) of R 3.q is a “yes” instance if and only if f(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. Polynomial Time Reduction

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 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 φ and verify whether both satisfy φ. –(2) Then we show Double-SAT is not easier than SAT. Reduction from SAT to Double-SAT

Double-SAT Reduction: –On input φ (x 1,..., x n ): –1. Introduce a new variable y. –2. Output formula φ ’(x 1,..., x n, y) = φ (x 1,..., x n ) ∧ ( 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 - example {yellow vertices} is an example of a dominating set of size 2. e

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

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 – graph transformation vw zu wv zu vzwu vw zu vu G G'

Dominating set Reduction –(1) Graph transformation –(2) a dominating set of size K in G’  a vertex cover of size K in G

Thank you!