meljun cortes automata theory 23

8/13/2019 MELJUN CORTES Automata Theory 23

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CSC 3130: Automata theory and formal languages

NP-complete problems

Fall 2008ELJUN P. CORTES MBA MPA BSCS ACS

MELJUN CORTES


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Polynomial-time reductions

Language Lpolynomial-time reduces to Lif

there exists a polynomial-time computable map R

that takes an instance xof Linto instanceyof L

s.t.xL if and only if yL

L L(IS) (CLIQUE)

x y(G,k) (G,k)

xL yL(Ghas IS of size k) (Ghas clique of size k)

R


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The Cook-Levin Theorem

Every LNPreduces to SAT

SAT= {f:f is a satisfiable Boolean formula}

(x1x2) (x2x3x4) (x1)f=

is satisfiable

x1 = T x2 = Fx3 = T x4 = T

(x1x2) (x1) (x2)f=

is not satisfiable


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NP-hardness

Language Lis NP-hard if every LinNPreducesto L

Intuitively, NP-hard means harder than all of NP

The Cook-Levin Theorem says

SATis

NP-hard

P

SAT

NP


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NP-complete

Lis NP-completeif Lis NP-hard and LNP

Intuitively, NP-complete means hardest in NP

Recall that SAT NP, so SATis NP-complete

If SATP, then P = NP

P

SAT

NP


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Proof of Cook-Levin Theorem

To prove it, we have to describe a reduction R:

Every LNPreduces to SAT

w Boolean formulaf

wL fis satisfiable

R


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All we know about L: It has a poly-time NTMM Lets look at computation tableauofMon input w

M w w1 w2q0

qacc

S-th configuration

symbol at time T

S

T

SinceMis nondeterministic,there may be many possible

tableaus


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w1 w2q0

qacc

S

T

n= length of input w

heightof tableau isp(n)

for some polynomialp

widthis at mostp(n)

k possible tableau symbols

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

tableau contains u

if notfalse,

u

1 Sp(n)

1 Tp(n)

1 u k


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We will design a formulafsuch that:

w Boolean formulaf

wL fis satisfiable

R

variables off :

assignment to xT, S, u

way to fill up the tableau

satisfying assignment accepting computation tableau

fis satisfiable

xT, S, u

Maccepts w


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We want to construct (in time poly(n)) a formulaf:

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

tableau contains u

if notfalse,

1 Sp(n)1 Tp(n)

1 u k

f(x1, 1, 1, ..., xp(n),p(n), k) =

true, if the xs represent a valid

accepting tableau

if notfalse,


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w1 w2q0

qacc

S

T

u

f=fcellf0fmovefacc

fcell: Every cell contains

exactly one symbol

The first row isq0w1w2...wk...

f0:

fmove: The moves between rowsfollow the transitions ofM

facc: qaccappears somewhere in the last row


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Desired meaning

Implementation:

fcell: Every cell contains exactly one symbol

fcell=fcell1, 1... fcellp(n),p(n) where

fcellT, S = (xT, S, 1... xT, S, k)

(xT, S, 1xT, S, 2)(xT, S, 1xT, S, 3)

...

(xT, S, k-1xT, S, k)

at least one symbol

no two symbols

or: Exactly one of xS, T, 1... xS, T, k is true


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Desired meaning

Implementation:

f0: The first row is q0w1w2...wk...

facc: qaccappears somewhere in the last row

f0 = x1, 1, q0x1, 1, w1x1, 1, w2 ... x1,p(n),

facc = xp(n), 1, qaccxp(n), 2, qacc... xp(n),p(n), qacc


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Valid and invalid windows

6c3a0t0

0c6a0p00

6t3t0u0 0t6t0u00

valid window

invalid window

6t3t0u0

0t6t0q3 0

valid window

6t3q3u0

0t6a0q7 0

valid if d(q3, u) = (q7, a, R)

6q3t0u0 0k6t0q0 0

invalid window

6c3a0t0

0b6a0t00

valid window


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Desired meaning

Implementation:

fmove: The moves between rowsfollow transitions ofM

a aq0

qacc

fmove=fmove1, 1... fmovep(n)-3,p(n)-3

b

q3b a b

b q7c b

fmove2, 2

fmoveT, S =over all

(xT, S, a1xT, S+1, a2xT, S+2, a3xT+1, S, a4xT+1, S+1, a5xT+2, S+1, a6)

valid windows

a1 a2 a3

a4 a5 a6


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Other NP-complete problems

CLIQUE= {(G, k): Gis a graph with a clique of

kvertices

}IS= {(G, k): G is a graph with an independent set ofk vertices}

VC= {(G, k): G is a graph with a vertex cover ofk vertices}

CLIQUE, ISand VC are NP-complete

SAT

NP

CLIQUE IS VC


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Proving NP-hardness

To show Lis NP-hard, it is enough to reducefrom some Lwe already know is NP-hard

For now we can take L= SAT

To show Lis NP-complete,we also need to argue thatLis in NP

This is usually the easy part

roadmap:

SAT

3SAT

IS

CLIQUE VC


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3SAT

SAT= {f:f is a satisfiable Boolean formula}3SAT= {f:f is a satisfiable Boolean formula in

conjunctive normal form withat most 3 distinct literals per clause}

CNF: AND of ORs of literals

literal: xior xi

(conjunctive normal form)

3CNF: CNF with 3 lit/clause

(x1x2) (x2x3x4) (x1)clauseliterals

(x2(x1x2))(x1(x1x2))

gates


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NP-hardness of 3SAT

Theorem

Proof:We describe a reduction Rfrom SAT

Boolean formulaf 3CNF formulaf

f is satisfiable

R

fis satisfiable

3SATis NP-hard


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Reducing SATto 3SAT

Example:f=(x2(x1

x2))

(x1

(x1

x2))

We give extra variables to every gate (wire)

x3

x6 x7

x8 x9

x4 x5

x10

AND

OR NOT

AND

NOT OR

AND

NOT

x2 x1 x2 x1 x1x2

x7= x4x5x4x5 x7T T T T

T T F F

T F T F

T F F T

F T T F

F T F T

F F T F

F F F T

(x4x5x7)(x4x5x7)

(x4x5x7)

(x4x5x7)

(x4x5x7)(x4x5x7)(x4x5x7)(x4x5x7)


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Reducing SATto 3SAT

Step 1: Add variable xn+jfor each gate Gjinf

Step 2: Write 3CNFfjfor each gate Gj,j= {1, ...,t}

Step 3: Output

Boolean formulaf 3CNF formulaf R

f =fn+1fn+2... ftxn+t

requires thatoutput offis true


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Reducing SATto 3SAT

Every satisfying assignment off extends uniquely

to a satisfying assignment off

Conversely, every satisfying assignment off

must contain a satisfying assignment off

Boolean formulaf 3CNF formulaf R

f is satisfiablefis satisfiable


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Independent set

Theorem

IS= {(G, k): G is a graph with an independentset ofk vertices}

1 2

3 4

An independent setis

a subset of vertices so

that no pair is

connected{1, 2}, {1, 3}, {4} are

independent sets

ISis NP-hard

SAT

3SAT

IS

CLIQUE VC


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Reducing 3SATto IS

Proof: We describe a reduction from 3SATto IS

3SAT= {f:f is a satisfiable Boolean formula in 3CNF}

IS= {(G, k): G is a graph with an independent set ofk vertices}

3CNF formulaf (G,k)

Ghas an independentset of size k

R

fis satisfiable


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Reducing 3SATto IS

Example:f=(x1x2)

(x2

x3

x4)

(x1)

TT

TF

FT

FF

TTT

TTF

TFT

TFF

FTT

FTF

FFTFFF

T

F

Put an edge for every inconsistency

all interconnected

x2x3x4x1x2

x1

etc.


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Reducing 3SATto IS

Example:f=(x1x2)

(x2

x3

x4)

(x1)

TT

TF

FF

TTT

TTF

TFT

TFF

FTT

FFTFFF

T

x2x3x4x1x2

x1

G

satisfying assignment off

IS of size 3 in G

x1 = T x2 = Fx3 = T x4 = T

edges = inconsistencies

any IS of size 3 in G

satisfying assignment off


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Reducing 3SATto IS

Ghas a vertexfor every clause Cioffand everysatisfying assignment of Ci

G has an edgebetween any two vertices that

represent inconsistent assignments

kis the number of clauses inf

3CNF formulaf (G,k)R


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Reducing 3SATto IS

Every satisfying assignment offgives an

independent set of size kin G

Conversely, from every IS of size kin Gwe can

extract a consistent and satisfying assignmentoff

3CNF formulaf (G,k)

Ghas an IS of size k

R

fis satisfiable


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Vertex cover

SAT

3SAT

IS

CLIQUE VC

Theorem

VCis NP-hard

A vertex cover is a set

of vertices that

touches (covers)all

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

are vertex covers

1 2

3 4

VC= {(G, k): G is a graph with a vertex cover of sizek}


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Reducing IStoVC

Proof: We describe a reduction from IStoVC

Example

(G,k)

Ghas a VC of size k

R(G,k)

Ghas an IS of size k

1 2

3 4

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

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

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

{1, 2, 3, 4}

independent setsvertex covers


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Reducing IStoVC

Claim

Proof

Sis an independent set of Gif and only

ifSis a vertex cover of G

Sis an independent set of G

no edge has both endpoints in S

every edge has an endpoint in S

Sis a vertex cover of G


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Reducing IStoVC

Set G = G, k= nk(n= number of vertices)

by previous Claim.

(G,k)R(G,k)

Ghas a VC of size kGhas an IS of size k


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The ubiquity of NP-complete problems

We saw a few examples of NP-completeproblems, but there are many more

A surprising fact of life is that most CS problems

are either in Por NP-complete

A 1979 book by Garey and Johnson

lists 100+ NP-complete problems

meljun cortes automata theory 23

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