NP-complete - PowerPoint Presentation

Slide1

NP-complete examples

CSCI3130 Tutorial 10

Chun-Ho Hung

chhung@cse.cuhk.edu.hk

Department of Computer Science & Engineering

1Slide2

OutlineReview

P, NP, NPCPolynomial-time Reduction2 problems

Double-SATDominating sethttp://en.wikipedia.org/wiki/Dominating_set_problem

2Slide3

ReviewP, NP, NPCPolynomial-time Reduction

3Slide4

PP is the class of all languages that have poly-time

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

4Slide5

NPNP is the class of all languages that have poly-time

verifierA verifier – a Turing Machine, V,

s.t.Given a potential xx ∈ L

 V accepts input <x, s> for some sSolution sV runs in polynomial time

5Slide6

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 forVerifyingGiven an assignment, just evaluate the truth value

Finding a solution?No efficient algorithm has been discovered yet

6

(x

1

∨

x

2

)

∧

(

x

2

∨

x

3

∨

x

4) ∧ (x1)

f

=Slide7

P versus NPEvery language, L in P, L is also in NP

Let Verifier = Poly-time TM that solves LThereforeP is contained in NPNote: L in NP does not

imply that efficient algorithm that decides Ldoes not exist

7

NP

PSlide8

NPCA language C is NP-complete if:C is in NP

Every language L in NP, L poly-time reduces to CWhat is a reduction

…?8Slide9

ReductionThe

direction of the reduction is very importantSaying “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”

9Slide10

Reduction (Con’t)Consider 2 problems:

BFS on unweighted graph

Shortest path on weighted graphAssume 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

resultThese two “yes” instances corresponds to each other10Slide11

Poly-Time ReductionHow to show that a problem B is not easier than a problem A?

Informally, if B can be solved efficiently, we can solve A efficientlyFormally, 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

11Slide12

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), andy is accepted  x is accepted

12Slide13

Poly-Time Reduction (Implication)

Suppose A reduces to BIf B is polynomial time solvable, then A is polynomial time solvableIf A is not polynomial time solvable, then B is not polynomial time solvable

Contrapositive13

x

y

R

TM for

L’

acc

rej

Poly-time TMSlide14

Poly-Time Reduction (Implication)

Suppose A reduces to BSolving B cannot be easier than solving ASuppose A is “difficult” while B is “easy”However, by this reduction, you find a “easy” way to solve AConsequently

, if A is NPC, then B must be NPC14

x

y

R

TM for

L’

acc

rej

Poly-time TMSlide15

Poly-Time Reduction - P versus NPTo 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 CIf 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

15Slide16

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 ≠ NPEncryption algorithms heavily rely on the assumption that P ≠ NP

P = NP or P ≠ NP is still an open problem16Slide17

Relations

NP

P

NP-C

easy

hard

Is there any problem even harder than NP-C?

Yes! e.g.

I-go

17Slide18

MethodologyTo 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-completeShow that L is in NPPoly-time reduce some NPC problem to L

i.e., design

a polynomial-time reduction from some problem we know to be NP-complete

18Slide19

ExampleProving a problem being NPC

19Slide20

Double-SATProblem:

Double-SAT = {<φ> |

φ is a Boolean formula with at least two satisfying assignments

}Goal:Show that Double-SAT is NP-Complete

20Slide21

Double-SAT (Proof Sketch)Steps:

Show that Double-SAT ∈

NPShow that Double-SAT is

not easier than a certain NPC problemFor the NPC problem, we choose

SATi.e., we want to poly-time reduce Double-SAT to SATShow the correspondence of “yes” instance between reduction

21Slide22

Double-SAT - (1) NP

It is trivial to see that Double-SAT ∈ NPGiven 2 assignments

for φ, and verify whether

both of them satisfy φWe can just evaluate the truth value

in poly-time22Slide23

Double-SAT - (2) ReductionReduction:On input

φ(x1, . . . , xn): 1. Introduce a new variable w

2. Output formula φ’(x1, . . . ,

xn, y) = φ(x1, . . . , xn

) ∧ ( w ∨ w ).

23

x

y

x

∈

L

y

∈

L

’

R

TM for

L’

acc

rej

TM accepts

SAT

Double-SATSlide24

Double-SAT - (3) Correspondencex

∈ L 

y ∈ L’

 : Suppose there is an satisfying assignment, X, for φ(x

1, . . . , xn), we can find two satisfying assignments for φ’(x1, . . . , xn,

w):

Assignment 1 = {X, w=

True

}

Assignment 2 = {X, w=

False

}

φ’

(x

1

, . . . ,

x

n

,

w) =

φ(x1, . . . , xn) ∧ ( w ∨ w )24For {xi}, assign X,then this part = TrueNo matter what w is,this part = TrueSlide25

Double-SAT - (3) Correspondencex

∈ L 

y ∈ L’

: We use contrapositivei.e., to show x

∉ L ⇒ y ∉ L’Indeed, if x

∉

L

,

φ

(x

1

, . . . ,

x

n

)=

False

Then, no matter what the value of

y

is

φ’

(x

1, . . . , xn, y)=False25Slide26

Dominating SetProblem:Dominating-set = {<

G, K> | A dominating set of size K for G

exists}Goal:Show that Dominating-set

is NP-Complete26Slide27

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 graphA dominating set D is a set of vertices that covers all

verticesi.e., every vertex of G is either in D or is adjacent to at least one vertex from D

27Slide28

Dominating Set (Example)Size-2 example : {Yellow vertices}

e

28Slide29

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 coverReduction from Vertex-cover to Dominating-set

Show the correspondence of “yes” instances between the reduction29Slide30

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 Di.e., for each vertex, v, in (V-D), whether v is adjacent to some vertex u in D

30Slide31

Dominating Set - (2) ReductionReduction - Graph transformation

Construct a new graph G' by adding new vertices and edges to the graph G

as follows:31

G

G’

<

G,k

>

∈

L

<G’, k>

∈

L

’

T

Vertex-cover

Dominating-setSlide32

Dominating Set - (2) ReductionReduction - 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 followings32

G

G’

<

G,k

>

∈

L

<G’, k>

∈

L

’

T

Vertex-cover

Dominating-setSlide33

[Recap] Vertex coverA vertex

cover, C, is a set of vertices that covers all edgesi.e., each edge is at least adjacent to some node in C

33

{2, 4}, {3, 4}, {1, 2, 3}

are vertex covers

1

2

3

4Slide34

Dominating Set – Graph Transformation Example

v

w

z

u

G

w

v

z

u

vz

wu

vw

zu

vu

G'

34Slide35

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 G35Slide36

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)

36Slide37

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, a

ll

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

∈ CThis node will cover uv in G’ Thus, C is a valid dominating for G’ (of size at most K)37Slide38

Dominating Set - (3) Correspondence

v

w

z

u

w

v

z

u

vz

wu

vw

zu

vu

Vertex-cover

in G

Dominating-set in

G

'

38Slide39

EndAny questions? (There should be some)

39