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**Department of Computer Science & Engineering**

NP-complete examples CSCI3130 Tutorial 10 Chun-Ho Hung Department of Computer Science & Engineering

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**Outline Review 2 problems P, NP, NPC Polynomial-time Reduction**

Double-SAT Dominating set

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Review P, NP, NPC Polynomial-time Reduction

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**P P is the class of all languages that have poly-time algorithm**

e.g., Shortest path on a directed graph, Sorting

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**NP NP is the class of all languages that have poly-time verifier**

A verifier – a Turing Machine, V, s.t. Given a potential x x ∈ L V accepts input <x, s> for some s Solution s V runs in polynomial time

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NP (con’t) While verifying a solution of a NP problem is easy (in poly-time), finding a solution could be more difficult An 3SAT instance - Find a satisfying assignment for Verifying Given an assignment, just evaluate the truth value Finding a solution? No efficient algorithm has been discovered yet f = (x1∨x2 ) ∧ (x2 ∨x3 ∨x4) ∧ (x1)

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**P versus NP Every language, L in P, L is also in NP Therefore**

Let Verifier = Poly-time TM that solves L Therefore P is contained in NP Note: L in NP does not imply that efficient algorithm that decides L does not exist NP P

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**NPC A language C is NP-complete if: What is a reduction…? C is in NP**

Every language L in NP, L poly-time reduces to C What is a reduction…?

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**Reduction The direction of the reduction is very important**

Saying “A is easier than B” and “B is easier than A” mean different things “A (polynomially) reduces to B” means “B is not easier than A”

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**Reduction (Con’t) Consider 2 problems:**

BFS on unweighted graph Shortest path on weighted graph Assume we have a TM, V, which solves 2) We can reduce 1) to 2): Given an instance of 1), convert it into an instance of 2): Copy the graph, add weight=1 to every edge in 2) Run this instance on V, output result These two “yes” instances corresponds to each other

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Poly-Time Reduction How to show that a problem B is not easier than a problem A? Informally, if B can be solved efficiently, we can solve A efficiently Formally, we say A polynomially reduces to B if: Given an instance a of problem, x There is a polynomial time transformation to an instance of B, y = f(x) x is a “yes” instance if and only if y is a “yes” instance

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**Poly-Time Reduction (Con’t)**

Suppose A poly-time reduces to B Then there exists a poly-time TM, R, s.t., Given an instance of A, x, transforms it to an instance of B, y = f(x), and y is accepted x is accepted

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**Poly-Time Reduction (Implication)**

Suppose A reduces to B If B is polynomial time solvable, then A is polynomial time solvable If A is not polynomial time solvable, then B is not polynomial time solvable Contrapositive acc Poly-time TM R TM for L’ x y rej

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**Poly-Time Reduction (Implication)**

Suppose A reduces to B Solving B cannot be easier than solving A Suppose A is “difficult” while B is “easy” However, by this reduction, you find a “easy” way to solve A Consequently, if A is NPC, then B must be NPC acc Poly-time TM R TM for L’ x y rej

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**Poly-Time Reduction - P versus NP**

To show P = NP, one could try to show that a NPC problem, C, can be solved in polynomial time. Why? Every problem in NP poly-time reduces to C If C can be solved in poly-time, so does each problem in NP Then NP = P!! But this is not that easy and it is counter-intuitive (to most people) too

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P versus NP (Again) Most believe that P ≠ NP, because intuitively searching for a solution is more difficult than verifying a solution What does P = NP imply? Know how to verify a solution in poly-time Know how to find a solution in poly-time (?!) Indeed we prefer P ≠ NP Encryption algorithms heavily rely on the assumption that P ≠ NP P = NP or P ≠ NP is still an open problem

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**Relations hard easy Is there any problem even harder than NP-C?**

Yes! e.g. I-go P easy

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**Methodology To show L is in NP, you can either**

Show that solutions for L can be verified in polynomial-time, or Describe a nondeterministic polynomial-time TM for L (Come back to this if we have enough time) To show L is NP-complete Show that L is in NP Poly-time reduce some NPC problem to L i.e., design a polynomial-time reduction from some problem we know to be NP-complete

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Example Proving a problem being NPC

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**Double-SAT Problem: Goal:**

Double-SAT = {<φ> | φ is a Boolean formula with at least two satisfying assignments} Goal: Show that Double-SAT is NP-Complete

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**Double-SAT (Proof Sketch)**

Steps: Show that Double-SAT ∈ NP Show that Double-SAT is not easier than a certain NPC problem For the NPC problem, we choose SAT i.e., we want to poly-time reduce Double-SAT to SAT Show the correspondence of “yes” instance between reduction

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**Double-SAT - (1) NP It is trivial to see that Double-SAT ∈ NP**

Given 2 assignments for φ, and verify whether both of them satisfy φ We can just evaluate the truth value in poly-time

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**Double-SAT - (2) Reduction**

On input φ(x1, , xn): 1. Introduce a new variable w 2. Output formula φ’(x1, , xn, y) = φ(x1, , xn) ∧ ( w ∨ w ). acc R TM for L’ x y rej SAT Double-SAT x ∈ L y ∈ L’ TM accepts

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**Double-SAT - (3) Correspondence**

x ∈ L y ∈ L’ : Suppose there is an satisfying assignment, X, for φ(x1, , xn), we can find two satisfying assignments for φ’(x1, , xn, w): Assignment 1 = {X, w=True} Assignment 2 = {X, w=False} φ’(x1, , xn, w) = φ(x1, , xn) ∧ ( w ∨ w ) For {xi}, assign X, then this part = True No matter what w is, this part = True

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**Double-SAT - (3) Correspondence**

x ∈ L y ∈ L’ : We use contrapositive i.e., to show x ∉ L ⇒ y ∉ L’ Indeed, if x ∉ L, φ(x1, , xn)=False Then, no matter what the value of y is φ’(x1, , xn, y)=False

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**Dominating Set Problem: Goal:**

Dominating-set = {<G, K> | A dominating set of size K for G exists} Goal: Show that Dominating-set is NP-Complete

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**Dominating Set (Definition)**

Problem: Dominating-set = {<G, K> | A dominating set of size (at most) K for G exists} Let G=(V,E) be an undirected graph A dominating set D is a set of vertices that covers all vertices i.e., every vertex of G is either in D or is adjacent to at least one vertex from D

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**Dominating Set (Example)**

Size-2 example : {Yellow vertices} e

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**Dominating Set (Proof Sketch)**

Steps: Show that Dominating-set ∈ NP. Show that Dominating-set is not easier than a NPC problem We choose this NPC problem to be Vertex cover Reduction from Vertex-cover to Dominating-set Show the correspondence of “yes” instances between the reduction

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**Dominating Set - (1) NP It is trivial to see that Dominating-set ∈ NP**

Given a vertex set D of size K, we check whether (V-D) are adjacent to D i.e., for each vertex, v, in (V-D), whether v is adjacent to some vertex u in D

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**Dominating Set - (2) Reduction**

Reduction - Graph transformation Construct a new graph G' by adding new vertices and edges to the graph G as follows: T G G’ <G,k> ∈ L Vertex-cover Dominating-set <G’, k> ∈ L’

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**Dominating Set - (2) Reduction**

Reduction - Graph transformation (Con’t) 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 We skip the discussion of this subtle part in the followings T G G’ <G,k> ∈ L Vertex-cover Dominating-set <G’, k> ∈ L’

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[Recap] Vertex cover A vertex cover, C, is a set of vertices that covers all edges i.e., each edge is at least adjacent to some node in C 1 2 3 4 {2, 4}, {3, 4}, {1, 2, 3} are vertex covers

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**Dominating Set – Graph Transformation Example**

w v z u vz wu vw zu vu G' v w z u G

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**Dominating Set - (3) Correspondence**

A dominating set of size K in G’ A vertex cover of size K in G Let D be a dominating set of size K in G’ Case 1): D contains only vertices from G Then, all new vertices have an edge to a vertex in D D covers all edges D is a valid vertex cover of G

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**Dominating Set - (3) Correspondence**

A dominating set of size K in G’ A vertex cover of size K in G Let D be a dominating set of size K in G’ Case 2): D contains some new vertices (vertex in the form of uv) (We show how to construct a vertex cover using only old vertices, otherwise we cannot obtain a vertex cover for G) For each new vertex uv, replace it by u (or v) If u ∈ D, this node is not needed Then the edge u-v in G will be covered After new edges are removed, it is a valid vertex cover of G (of size at most K)

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**Dominating Set - (3) Correspondence**

A dominating set of size K in G’ A vertex cover of size K in G Let C be a vertex cover of size K in G For an old vertex, v ∈ G’ : By the definition of VC, all edges incident to v are covered v is also covered For a new vertex, uv ∈ G’ : Edge u-v must be covered, either u or v ∈ C This node will cover uv in G’ Thus, C is a valid dominating for G’ (of size at most K)

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**Dominating Set - (3) Correspondence**

vw v w v w vz wu vu z u z u zu Vertex-cover in G Dominating-set in G'

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End Any questions? (There should be some)

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