CSCI 3130: Formal languages and automata theory

CSCI 3130: Formal languages and automata theory

Andrej Bogdanov

http://www.cse.cuhk.edu.hk/~andrejb/csc3130

The Chinese University of Hong Kong

More NP-complete problems

Fall 2010

NP-complete problems

CLIQUE

IND SET

VERTEX COVER

DOM SET MIN COVER

HAM PATH

HAM CYCLE

PARTITION

SUBSET SUM

P

NP

NP-complete

SAT

3SAT

SAT and 3SAT

SAT = { 〈〉 : is a satisfiable Boolean formula}

3SAT = { 〈〉 : is a satisfiable 3CNF formula}

(x1∨x2∨x2 ) ∧ (x2∨x3∨x4) ∧ (x2∨x3∨x5) ∧ (x1∨x5∨x5)

((x1∨x2 ) ∧ (x1∨x2)) ∨ ((x1∨(x2 ∧ x3)) ∧ x3)

Reducing SAT to 3SAT

• Example: =(x2∨(x1∧x2 ))∧(x1∧(x1∨x

2 ))

We add extra variables for every gate (“wire”)

x3

x6 x7

x8 x9

x4 x5

x10

AND

OR NOT

AND

NOT OR

AND

NOT

x2 x1 x2 x1 x1 x2

x7 = x4 ∧ x5x4 x5 x7

T T T TT T F FT F T FT F F TF T T FF T F TF F T FF F F T

(x4∨x5∨x7)(x4∨x5∨x7)

(x4∨x5∨x7)

(x4∨x5∨x7)

(x4∨x5∨x7)∧(x4∨x5∨x7)∧(x4∨x5∨x7)∧(x4∨x5∨x7)

Turning gates into 3CNFs

AND

T T T TT T F FT F T FT F F TF T T FF T F TF F T FF F F T

z = y ∧ xx y z

x y

z

(x∨y∨z)∧(x∨y∨z) ∧(x∨y∨z)∧(x∨y∨z)

OR

T T T TT T F FT F T TT F F FF T T TF T F FF F T FF F F T

z = y ∨ xx y z

x y

z

(x∨y∨z)∧(x∨y∨z) ∧(x∨y∨z)∧(x∨y∨z)

NOT

T T FT F T F T TF F F

z = xx z

x

z

(x∨z)∧(x∨z)

(x∨x∨z)∧(x∨x∨z)

Gj:

j:

Reducing SAT to 3SAT

Boolean formula 3CNF formula ’R

’ = n+1 ∧ n+2 ∧ ... ∧ t ∧ (xn+t ∨xn+t ∨xn+t )

requires thatoutput of f is true

On input , where is a boolean formula

Construct and output the following 3CNF formula ’:

R:=

’ has extra variables xn+1, ..., xn+t one for each gate Gj in

For each gate Gj, construct the formula fj Output

Example

AND

NOT

x1 x2

f = x3

x4

f’ := (x2∨x2∨x3)∧(x2∨x2∨x3)(x1∨x3∨x4)∧(x1∨x3∨x4) ∧(x1∨x3∨x4)∧(x1∨x3∨x4)

∧

(x4∨x4∨x4)∧

AND

NOT

T F

T

T

x1 = T, x2 = F, x3 = T, x4 = T

is a satisfying assignment of f’

NOT gate

AND gate

output

Reducing SAT to 3SAT

• Every satisfying assignment of f extends uniquely to a satisfying assignment of f’

• In the other direction, in every satisfying assignment of f’ the part x1, ..., xn is a satisfying assignment of f

Boolean formula f 3CNF formula f’R

f’ is satisfiablef is satisfiable ✔

How to argue NP-completeness

SUBGRAPH = { 〈 G, H 〉 : G and H are graphs whereH is a subgraph of G}

G H

SUBGRAPH is NP-complete

1 2 3

4 5

6 7

1 2

3

4

Every edge of H must be included in G

1 2 3

4 5

6 7

How to argue NP-completeness

• Step 1: SUBGRAPH is in NP

G H1 2

3

4

What is a solution for SUBGRAPH?

A solution s is a map from vertices of H to vertices of G(1 4)(2 2)(3 5)(4 7)➝ ➝ ➝ ➝

V := On input 〈 G, H, s 〉For every edge {u, v} of H:

Find u’ and v’ such that u➝u’ and v➝v’ in s.

If {u’, v’} is not an edge in G, reject.

Otherwise, accept.

Running time = O(m)m = number edges in H

✔

How to argue NP-completeness

• Step 2: Some NP-complete L reduces to SUBGRAPH

CLIQUEIND SET

VERTEX COVER

SAT 3SAT

DOM SETDirection of reduction is important!

You want to show SUBGRAPH is harder than L

So you assume you can solve SUBGRAPH, and show how to solve L

This requires converting instances of L into instances of SUBGRAPH How to choose L?

Tips on choosing L

• Both CLIQUE and SUBGRAPH are about graph containment, so their complexities may be related

• It usually makes sense to try and reduce from the problem that looks the most related

G H1 2 3

4 5

6 7

1 2

3

4

G

4

CLIQUE SUBGRAPH

How to argue NP-completeness

• Step 3: Reduce CLIQUE to SUBGRAPH

SUBGRAPH = { 〈 G, H 〉 : H is a subgraph of G}

CLIQUE = { 〈 G, k 〉 : G has a k-clique}

R := On input 〈 G, k 〉Construct the k-clique H

Output 〈 G, H 〉

G has a k-clique

H is a subgraph of G

✔

Cover for triangles

• A graph has a k-cover for triangles if there are k vertices that touch all the triangles

Does it have a 2-cover for triangles?

Does it have a 1-cover for triangles?

No, because it has two vertex-disjoint triangles

Yes

TRICOVER = { 〈 G, k 〉 : G has a k-cover for triangles}

TRICOVER is NP-complete

Cover for triangles

• Step 1: TRICOVER is in NP

What is a solution for TRICOVER?

A solution s is a subset of vertices like {2,5}V := On input 〈 G, k, s 〉 , where s is a set of vertices of size k

For every triple {u, v, w} of vertices:

If {u, v}, {u, w}, {v, w}, are all edges in G:

If none of u, v, or w are in s, reject.

Otherwise, accept. Running time = O(n3)

✔

1 2 3

45

6 7

CLIQUEIND SET

VERTEX COVER

SAT 3SAT

DOM SETSUBGRAPH

Cover for triangles

• Step 2: Some NP-complete L reduces to TRICOVER

TRICOVER = { 〈 G, k 〉 : G has a k-cover for triangles}

Some vertex in every triangle is covered

Some endpoint of every edge is covered

VC = { 〈 G, k 〉 : G has a k-vertex cover}

Cover for triangles

• Step 3: Reduce VC to TRICOVER

TRICOVER = { 〈 G’, k’ 〉 : G’ has k’-cover for triangles}

VC = { 〈 G, k 〉 : G has a k-vertex cover (for edges)}

Idea: Replace edges by triangles

G G’

vertex-cover in G cover for triangles in G’

Cover for triangles

• Step 3: Reduce VC to TRICOVER

R := On input 〈 G, k 〉 , where G has n vertices and m edges

Construct the following graph G’:G’ has n + m vertices v1, ..., vn, and uij for every edge {vi, vj} of G

For every edge {vi, vj} of G, include the edges {vi, vj}, {vi, uij}, {vj, uij} in G’

Output 〈 G’, k 〉 .Running time = O(m)m = number of edges in G

Cover for triangles

• Step 4: Argue that the reduction is correct

G G’

〈 G, k 〉 ∈ VC 〈 G’, k 〉 ∈ TRICOVER

G has a k-vertex cover C C is a k-triangle cover of G’

Old triangles from G are covered

New triangles in G’ also covered

✔

Cover for triangles

• Step 4: Argue that the reduction is correct

G G’

〈 G, k 〉 ∈ VC 〈 G’, k 〉 ∈ TRICOVER

G has a k-vertex cover C’ G’ has a k-vertex cover C

Some vertices in C may notcome from G!

But we can move them and stillcover the same triangle

✔

C’ is obtained after moving somevertices of C

Since C’ covers all triangles in G’,it must cover all edges in G

✔

Coloring

CSCI 3230

CSCI 2110

CSCI 3160CSCI 3130

Suppose you need to schedule exams on same day in 3 time slots

Y R B

1030-1230130-330 430-630

CSCI 3190

Some people are enrolled inmultiple classes, so there may be conflicts

Task: Schedule the exams so there are no conflicts

Coloring

Given a graph, we want to colorthe vertices so every edge has endpoints of different colors G

Y R B

This is called a valid 3-coloring

3COL = { 〈 G 〉 : G has a valid 3-coloring}

3COL is NP-complete

Coloring

• Step 1: 3COL is in NP

What is a solution for 3COL?

A solution s is a coloring of vertices like {1B,2R,3Y,4R,5B}

V := On input 〈 G, s 〉For every edge {u, v} of G:

If u and v are assigned same color in s, reject.

Otherwise, accept.

Running time = O(m)

✔

G

1 2

3

4 5

CLIQUEIND SET

VERTEX COVER

SAT 3SAT

DOM SETSUBGRAPH

Coloring

• Step 2: Some NP-complete L reduces to 3COL

3COL = { 〈 G 〉 : G has a valid 3-coloring}

3COL does not look likeany of these problems

In such a case it is a goodidea to start with 3SAT

Coloring

• Step 3: Reduce 3SAT to 3COL

3COL = { 〈 G 〉 : G has a valid 3-coloring}

3SAT = { 〈〉 : is a 3CNF that has a satisfying assignment}

Idea: Assignment of Coloring of G

Part I: 3 special vertices T (true), F (false), and X

T F

X G

Coloring

Part 2: For each variable xi

xi

X

xi

Either xi has color of T and xi has color of F

Or xi has color of F and xi has color of T

Example: (x1∨x1∨x2)∧(x1∨x2∨x3 )

x1 x1 x2 x2 x3 x3

X

T F

T F F T F T

Coloring

• To “encode” the clauses of , we need a gadget

Can be colored if and only if at least one literal is colored TT

x1

x2

x3

x

yz

If x or y is colored T, z can be colored TIf x and y are colored F, z must be colored F

Part 3: For each clause like (x1∨x2∨x3 )

Coloring reduction: An example

Example: (x1∨x1∨x2)∧(x1∨x2∨x3 )

x1 x1 x2 x2 x3 x3

X

T F

T F F T F T

Coloring reduction: An example

Example: (x1∨x1∨x2)∧(x1∨x2∨x3 )

x1 x1 x2 x2 x3 x3

X

T F

T F F T F T

Coloring

• Step 3: Reduce 3SAT to 3COL

R := On input 〈〉 , where is a 3CNF:

Construct the following graph G:Add 3 special vertices T, F, X connected by a triangle.

For every variable xi of , add vertices xi and xi and include the triangle xi, xi, X.

Output 〈 G 〉 .

Running time = O(m + n)n = number of variablesm = number of clauses

For every clause u∨v∨w of , connect u, v, w, and T using the gadget below.

T

u

v

w

Coloring

• Step 4: Argue that the reduction is correct

〈〉 ∈ 3SAT

has a SAT assignment a

✔(x1∨x1∨x2)∧(x1∨x2∨x3 )

x1 x1 x2 x2 x3 x3

X

T F

〈 G 〉 ∈ 3COL

G has a valid 3-coloring

Each literal gets “color” as in a

Each clause gadget contains a true literal, so it can be colored

Coloring

• Step 4: Argue that the reduction is correct

〈〉 ∈ 3SAT

has a SAT assignment

✔(x1∨x1∨x2)∧(x1∨x2∨x3 )

x1 x1 x2 x2 x3 x3

X

T F

〈 G 〉 ∈ 3COL

G has a valid 3-coloring c

Each variable gets value as in G

Since each clause gadget is colored properly,each clause must contain a true literal

xi = true, if node xi has same color as T false, if node xi has same color as F