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

The Cook-Levin Theorem

Fall 2011

NP-completeness

P NP(easy to solve)(easy to verify solution)

SAT

IS

CLIQUEVC

3SAT

reduces to

Cook-Levin theorem:

Everything in NP reduces to SAT

NP-complete

Polynomial-time reductions

• Language L polynomial-time reduces to L ’ if

there exists a polynomial-time computable map R that takes an instance x of L into instance y of L ’ s.t.

x ∈ L if and only if y ∈ L ’

L L ’(CLIQUE) (IS)

x y(G, k) (G ’, k’)

x ∈ L y ∈ L ’(G has clique of size k) (G ’ has IS of size k ’)

R

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)

The Cook-Levin Theorem

• To show this, we need a reduction R such that:

Every L ∈ NP reduces to SAT

z Boolean formula f

z ∈ L f is satisfiable

R

NP-completeness of SAT

• All we know: L has a polynomial-time verifier V– Look at the computation tableau of V on input 〈 z, s 〉

Vz | s

z2z1

qacc …

S-th configuration symbol at time T

S

T

…q0

ba


n = length of z

height of tableau is O(nc) for some constant c

width is at most nc

k possible tableau symbols

xT, S, u = true,if the (T, S) cell of

tableau contains uif notfalse,

u

1 ≤ S ≤ nc

1 ≤ T ≤ nc

1 ≤ u ≤ k

S

T

z2z1 …q0

qacc …ba


• We will design a formula f such that:

z Boolean formula f


R

assignment to xT, S, u way to fill up the tableau

satisfying assignment accepting computation tableau

variables of f :xT, S, u

f is satisfiable V accepts 〈 z, s 〉 for some s


• We want to construct (in time O(n2c)) a formula f :

xT, S, u = true,if the (T, S) cell of

tableau contains uif notfalse,

1 ≤ S ≤ nc

1 ≤ T ≤ nc

1 ≤ u ≤ k

f(x1, 1, 1, ..., xnc, nc, k) = true,if the xs represent a legal

and accepting tableauif notfalse,


f = fcell ∧ f0 ∧ fmove ∧ facc

fcell:“Every cell contains exactly one symbol”

“The first row is q0z|s☐...☐” for some s

f0:

fmove:“The moves between rows follow the transitions of V ”

facc: “qacc appears somewhere in the last row”

z2z1

qacc …

…q0

ba


fcell: “Every cell contains exactly one symbol”

fcell = fcell 1, 1 ∧ ... ∧ fcell nc, nc where

fcell T, S = (xT, S, 1 ∨ ... ∨ xT, S, k) ∧ (xT, S, 1 ∧ xT, S, 2) ∧ (xT, S, 1 ∧ xT, S, 3) ∧ ... ∧ (xT, S, k-1 ∧ xT, S, k)

at least one symbol

no cell containstwo or more symbols

“Exactly one of xS, T, 1 ∨ ... ∨ xS, T, k is true”


f0:“The first row is q0z1z2...zn|s1s2...sm

f0 = x1, 1, q0 ∧ x1, 2, z1 ∧ ... ∧ x1, n+1, zn ∧ x1, n+2, |

for some (possible solution) s1s2...sm”


facc: “qacc appears somewhere in the last row”

facc = xnc, 1, qacc ∨ xnc, 2, qacc ∨ ... ∨ xnc, nc, qacc

Legal and illegal windows

… 6a3b0x0 …… 0a6b0x0 …0

legal windows

… 6#3b0a0 …… 0#6b0q5 …0

… 6a3q2a0 …… 0q5a6x0 …0

… 6q2a0b0 …… 0a6b0q2 …0

illegal windows

… 6a3a0☐0 …… 0x6a0☐0 …0

q2

q5

a/xL

… 6a3q2a0 …… 0q5a6b0 …0

… 6q2q2a0 …… 0q2q2x3 …0

… 6#3q2a0 …… 0q5#6x0 …0


fmove:“The moves between rows follow transitions of V ”

fmove = fmove 1, 1 ∧ ... ∧ fmove nc-

3, nc-3

a2a1 a3

a6a4 a5

fmove 2, 2 LW = list of all legal windows

= {(a1, a2, a3, a4, a5, a6): is legal}a1a2a3a4a5a6

z2z1q0 …

qacc …ba

fmove T, S = (xT, S, a1 ∧ xT, S+1, a2 ∧ xT, S+2, a3

∧ xT+1, S, a4 ∧ xT+1, S+1, a5 ∧ xT+2, S+1,

a6)

∨(a1, a2, a3, a4, a5, a6) ∈ LW


z Boolean formula f


R

Let V be a poly-time verifier TM for L.

On input z,Construct the formulas fcell, f0, fmove, and facc

Output the formula f = fcell ∧ f0 ∧ fmove ∧ facc.

R :=

✔

V accepts z if and only if f is satisfiable.Then R is computable in time O(n2c) and

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


• 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

✔


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

CLIQUEIND SET

VERTEX COVER

SAT 3SAT

Direction 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


• 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

SUBGRAPH

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


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

SUBGRAPH

Coloring

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


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


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

• 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


〈〉 ∈ 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


〈〉 ∈ 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

csci 3130: formal languages and automata theory

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