chapter 10 complexity of approximation (1) l-reduction

Chapter 10 Complexity of Approximation

(1) L-Reduction

Ding-Zhu Du

Traveling Salesman

• Given n cities with a distance table, find a minimum total-distance tour to visit each city exactly once.

. distance

total tour witha find ,and cities those

between tabledistance a with cities Given

:TSP-Approx-

optr

n

n

r

Definition

Proof:

Given a graph G=(V,E), define a distance table on V as follows:

EvurV

Evuvud

),( if ,||

),( if ,1),(

hard.- is TSP-Approx- ,1any For NPrr

Theorem

solvable. time-polynomial being HC implies solvable

time-polynomial being TSP-Approx- r

Contradiction Argument

• Suppose r-approximation exists. Then we have a polynomial-time algorithm to solve Hamiltonian Cycle as follow:

r-approximation solution < r |V|

if and only if

G has a Hamiltonian cycle

Special Case

• Traveling around a minimum spanning tree is a 2-approximation.

solvable. time-polynomial

is TSP-Approx-2 ,inequality triangular

thesatisfies tabledistance n the Whe Theorem

• Minimum spanning tree + minimum-length perfect matching on odd vertices is 1.5-approximation

solvable. time-polynomial

is TSP-Approx-1.5 ,inequality triangular

thesatisfies tabledistance n the Whe Theorem

Minimum perfect matching on odd verticeshas weight at most 0.5 opt.

Knapsack

.any for Hence

.any for Assume

}.1 ,0{

t.s.

max

2211

2211

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:Knapsack-Approx-

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. such that Choose

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solvable. time-polynomial

is Knapsack -Approx-2 Theorem

Proof.

solvable. time-polynomial

is Knapsack -Approx- ,1any For rr

Theorem

(PTAS). schemeion approximat

timepolynomial hasKnapsack s,other wordIn

• Classify: for i < m, ci < a= cG,

for i > m+1, ci > a.• Sort

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MAX3SAT

clauses. satisfied of # the

maximiz toassignmentan find , 3CNF aGiven F

clauses. satisfied

/least at have toassignmentan find

, 3CNF aGiven :MAX3SAT-Aprrox-

ropt

Fr

Theorem

solvable.

time-polynomial is MAX3SAT-Approx-2

;|

0set else

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) clauses(#) (clauses# if

do to1for

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Theorem

hard.-

is MAX3SAT-Approx- ,1constant someFor

NP

rr

This an important result proved using PCP system.

complete).-(APX

complete-SNP MAX is MAX3SAT

Theorem

Class MAX SNP (APX?)

solvable. time-polynomial

is PR-Approx-such that 1

constant a exists thereif SNP MAX tobelongs

PR problemon minimizatior on maximizatiA

rr

L-reduction

|))(()(|

|)())((|

such that on of solutions feasible to

)(on of solutions feasible from maps L2)(

);())((

such that of

instances to of instances from maps (L1)

such that 0, constants twoand , and

functions computable time-polynomial twoare thereif

. and problemson optimizati woConsider t

xhoptyobjb

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x

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xoptaxhopt

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bagh

PL

x )(xh

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solutions

feasible

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solutions

feasible

xh

VC-b

y.cardinalit its minimize cover to

vertexa find ,most at degree with graph aGiven bG

Theorem .3-VC-VC ,1any For PLbb

.3-VC-VC that trivialisit ,3any For PLbb

. degree with graph aconsider ,4For bGb

1 2

3

45

1 2

3

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ofcover - vertexa is

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ofcover - vertexminimum a is

Gcc

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vSvvSv

)()12(2)()'( GoptbnmGoptGopt

).'(|'|)(|)'(|

Thus,

.|||)('|' that Note

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}.'|{)'(

define ,' of 'cover x each verteFor

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Properties

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(P1)

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PTASPTAS

MAX SNPMAX SNP

PTAS

PTAS ,

p

L

.for ion approximat-)1( is )(then

,for ion approximat-)(1 is If

))((

)))(()((1

)(

)())((1

)(

))((

problems.on minimizati are and Both :1

abyg

y

xhopt

xhoptyobjab

xopt

xoptygobj

xopt

ygobj

case

.for ion approximat-)1( is )(then

,for ion approximat-)(1 is If

)(

))())(((1

))((

))())(((1

)(

)())((1

)(

))((

problem.on maximizati

a is and problemon minimizati a is :2

abyg

y

yobj

yobjxhoptab

xhopt

yobjxhoptab

xopt

xoptygobj

xopt

ygobj

case

.for ion approximat-)1( is )(then

,for ion approximat-)(1 is If

))(())(()(

1

1

)())(()(

1

1

)())((

)(

))((

)(

on.minimizati a is andon maximizati a is :3

)(

abyg

y

xhoptxhoptyobj

ab

xoptygobjxopt

xoptygobjopt

xopt

ygobj

xopt

case

x

.for ion approximat-)1( is )(then

,for ion approximat-)(1 is If

)()())((

1

1

))(()())((

1

1

)())(()(

1

1

)())((

)(

))((

)(

problems.on maximizati are and Both :4

)(

abyg

y

yobjyobjxhopt

abxhopt

yobjxhoptab

xoptygobjxopt

xoptygobjopt

xopt

ygobj

xopt

case

x

MAX SNP-complete (APX-complete)

.

SNP, MAXany for and SNP MAX if

complete-SNP MAX is problemon optimizatiAn

pL

Theorem

PTAS. no has then , if i.e.,

hard,- is -Approx- ,1

, problem complete-SNP MAX

NPP

NPrr

MAX3SAT-3

clauses. satisfied of # themaximize

toassignmentan find times,most threeat

appears bleeach varia that 3CNF aGiven F

complete.-SNP MAX is 3-MAX3SAT

Theorem

3-MAX3SATMAX3SAT pL

VC-4 is MAX SNP-complete

graph. a is where

,)(construct , 3CNFeach For

4-VC of inputs3-SAT3MAX of inputs:

4-VC3-SAT3MAX

G

GFfF

f

pL

Proof.

1x

))(( 143231 xxxxxx

2x 3x 4x1x 2x 3x 4x

1c

13c

11c 12c2c

23c

22c21c

hard.-SNP MAX isCover -Set

Theorem

Proof. Cover-Set3-VC pL

. ofcover set a is }|{

ofcover - vertexa is

Then }.|{

set aconstruct ,each For

3.-VC of instancean be ),(Let

ECvsS

GVC

evEes

Vv

EVG

vC

v

Theorem

. unlessCover -Setfor

ionapproximat-) (ln time-polynomial no is There

PNP

no

).(

unlessCover -Setfor ion approximat- ln

time-polynomial no is there,1For

) (log nOnDTIMENP

n

Theorem

Proved using PCP system

).(

unless MCDSfor ion approximat- ln

time-polynomial no is there,1For

) (log nOnDTIMENP

n

Theorem

MCDS

y.cardinalit its minimize set to

dominating connected a find ,graph aGiven G

1x 2x 3x 4x 5x 6x

1S 2S 3S

}.,,{

},,,,{},,,{

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xxxS

xxxxSxxxS

. unless CLIQUEfor

ionapproximat- time-polynomial no is there,1For

PNP

ns s

subgraph.) complete a is (A y.cardinalit its

maximize toclique a find,graph aGiven

clique

G

CLIQUE

Theorem

Proved with PCP system.

disjunct?- is

,1integer an and matrix binary aGiven

:-coin is problem following theProve

dM

dM

NP

matrix?binary -by-

disjunct- a thereis ,0,, integersGiven

:in is problem following theProve 2

nt

dtdn

p

1

2

Exercises

rows.) of # maximum thedeletingby submatrix

disjunct-2 a find ,matrix aGiven :DS-2-(Min

2. size has pool

every that case specialin hard- is DS-2-Min

:Prove

M

NP

3

hint

DS-2-MinVC pm

• Min-2-DS is MAX SNP-complete in the case that all given pools have size at most 2.

4 Prove that

5. Is TSP with triangular inequality MAX SNP-complete?