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Contents lists available atScienceDirect

Theoretical

Computer

Science

www.elsevier.com/locate/tcs

Approximating

MAX

SAT

by

moderately

exponential

and

parameterized

algorithms

Bruno Escoffier

a

,

b

,∗

,

Vangelis Th. Paschos

c

,

d

,

Emeric Tourniaire

c

aSorbonneUniversités,UPMC,LIP6,F-75005,Paris,France bCNRS,UMR7606,LIP6,F-75005,Paris,France

cPSLResearchUniversity,UniversitéParis-Dauphine,LAMSADE,CNRS,UMR7243,France dInstitutUniversitairedeFrance,France

a

r

t

i

c

l

e

i

n

f

o

a

b

s

t

r

a

c

t

Articlehistory:

Received29October2012 Accepted21October2014 Availableonline13November2014 Keywords:

Maximumsatisfiability Exponentialtimealgorithms Approximationalgorithms

We studyapproximationofthe maxsatproblemby moderatelyexponentialalgorithms. Thegeneralgoaloftheissueofmoderatelyexponentialapproximationistocatch-upon polynomialinapproximability,byprovidingalgorithmsachieving,withworst-caserunning timesimportantlysmallerthanthoseneededforexactcomputation,approximationratios unachievableinpolynomialtime. Wedevelopseveralapproximationtechniquesthatcan beappliedto maxsatinordertogetapproximationratiosarbitrarilycloseto 1.

©2014ElsevierB.V.All rights reserved.

1. Introduction

Optimumsatisfiabilityproblemsareofgreatinterestfromboththeoreticalandpracticalpointsofview.Letusonlynote thatseveralsubproblemsof maxsatand minsatareamongthefirstcompleteproblemsformanyapproximabilityclasses

[1,16].Ontheotherhand,inmanyfields(including artificialintelligence,databasesystem,mathematicallogic,

. . .

) several problemscanbeexpressedintermsofversionsof sat[3].

Satisfiabilityproblems haveinparticulardrawnmajor attentioninthefield ofpolynomial timeapproximation aswell asin thefield of parameterizedand exactsolution by exponential time algorithms.Our goalin thispaperis to develop approximation algorithms for maxsat withrunning times which,though being exponential, are much lower than those ofexact algorithms,andwitha better approximationratio thanthe one achievedinpolynomial time.This approachhas alreadybeenconsideredfor maxsatin

[14,20]

,whereinteresting tradeoffsbetweenrunningtimeandapproximationratio are given. It has also been considered for several other well known problems such as minimumsetcover [7,12], min coloring[5,6], maxindependentsetand minvertexcover[8], minbandwidth[13,18],etc.Similarissuesariseinthefield ofFPT algorithms, whereapproximation notions havebeen introduced,forinstance, in[9,15].In thisarticle,we propose severalimprovementsoftheresultsof

[14]

and

[20]

usingvariousalgorithmictechniques.

Givenasetofvariablesandasetofdisjunctiveclauses, maxsatconsistsoffindingatruthassignmentforthevariables thatmaximizesthenumberofsatisfiedclauses.Inwhatfollows,wedenoteby X

= {

x1

,

x2

,

. . . ,

xn

}

thesetofvariablesand

by C

= {

C1

,

C2

, . . . ,

Cm

}

theset ofclauses.Eachclauseconsistsofa disjunctionofliterals,a literalbeingeitheravariable

ResearchpartiallysupportedbytheFrenchAgencyforResearchundertheDEFISprogram“Timevs.OptimalityinDiscreteOptimization”,

ANR-09-EMER-010,andbytheprojectALGONOWoftheresearchfundingprogramTHALIS(co-financedbytheEuropeanSocialFund–ESFandGreeknationalfunds).

*

Correspondingauthor.

E-mailaddresses:bruno.escoffier@upmc.fr(B. Escoffier),paschos@lamsade.dauphine.fr(V.Th. Paschos),emeric.tourniaire@lamsade.dauphine.fr (E. Tourniaire).

http://dx.doi.org/10.1016/j.tcs.2014.10.039 0304-3975/©2014ElsevierB.V.All rights reserved.

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xi or its negation

¬

xi.A

ρ

-approximation algorithm for maxsat (with

ρ

<

1) is an algorithm that finds an assignment satisfyingatleastafraction

ρ

ofthemaximalnumberofsimultaneouslysatisfiedclauses.Thebestknownratioguaranteed by apolynomialtime approximationalgorithmis

α

=

0

.

796 obtainedin

[2]

,whileitisknownthattheproblem(andeven its restrictiontoinstanceswhereevery clausescontainexactlythreeliterals)isnotapproximableinpolynomialtimewith ratio 7

/

8

+



,forany



>

0,unless P

=

NP [19]. Arecentresult

[22]

statesthat forany



>

0,achievinga ratio7

/

8

+



is evenimpossibleto getintime O

(

2m1−

)

forany





>

0 unlesstheexponential timehypothesis fails.1 Thislatterresult

motivatesthestudyofexponentialtimeapproximationalgorithms.

Dealingwithexactsolution,

[10]

givesanexactalgorithmworkingintime O

(

1

.

3247m

)

whichisthebestknownbound

sofarwrt.thenumberofclauses.2Dealingwiththenumbern ofvariables,thetrivialO

(

2n

)

boundhasnotyetbeenbroken down,andthisconstitutesoneofthemainopenproblemsinthefieldofexactexponentialalgorithms.Theparameterized version of maxsatconsists,givenasetofclauses C andanintegerk,offindingatruthassignmentthatsatisfiesatleastk clauses, ortooutputanerrorifnosuchassignmentexists.In

[10]

theauthorsgiveaparameterizedalgorithmfor maxsat runningintime O

(

1

.

3695k

)

.

Usingthesamenotationasin

[10]

,wesaythatavariablex isan

(

i

,

j

)

-variableifitoccurspositivelyinexactlyi clauses

and negatively in exactly j clauses. Forany instance C of maxsat, we will denote by OPT

(

C

)

(or OPT if no ambiguity occurs)anoptimalsetofsatisfiedclauses.Finally,wedenoteby

α

theratioguaranteedbyapolynomialtimeapproximation algorithm.Ingeneral,

ρ

willdenotetheapproximationratioofanalgorithm,and,whendealingwithexponentialcomplexity,

γ

willbethebasisoftheexponentialtermexpressingit.

Inordertofixideas,letusgiveafirstsimplealgorithm,usefultounderstandsomeofourresults.Inparticular,itisone ofthebasicstonesoftheresultsin

[14]

.Itisbaseduponthefollowingtwowellknownreductionrules.

Rule1. Any clause containing an

(

h

,

0

)

- or a

(

0

,

h

)

-literal, h



1, can be removed from the instance. This is correct becausewecansetthisliteralto

TRUE

or

FALSE

andsatisfytheclausesthatcontainit.

Rule2. Any

(

1

,

1

)

-literalcanbe removed too.LetC1

=

x1

x2

∨ · · · ∨

xp andC2

= ¬

x1

x2

∨ · · · ∨

xq betheonly two

clausescontainingthevariablex1.Ifthereexisttwooppositeliterals



and

¬

inresp.C1andC2,thenwecansatisfyboth

clausesbysettingx1 to

TRUE

and



to

FALSE

andthereforewecanremovetheseclauses.Otherwise,wecanreplacethese

clauses by C

=

x2

∨ · · · ∨

xp

x2

∨ · · · ∨

xq. The optimum inthe initial instanceis the optimumin the reducedinstance

plus 1.

Algorithm1. Buildatreeasfollows.Eachnode islabeledwithasub-instance of maxsat.The rootis theinitialinstance. The emptyinstances(instanceswithnoclauses)aretheleaves.Foreachnodewhoselabelisanon-emptysub-instance,if one ofthereductionsaboveapplies,thenthenodehasonechildlabeledwiththeresulting(reduced)sub-instance.Else,a variablex isarbitrarilychosenandthenodehastwochildren:inthefirstone,theinstancehasbeentransformedbysetting

x to

FALSE

(the literalsx havebeenremoved andtheclauses containingthe literal

¬

x aresatisfied); inthesecond one,

x issetto

TRUE

andthecontraryhappens.Finally,forbothchildrentheemptyclauses aremarkedunsatisfied.Thus,each node representsapartialtruthassignment.Anoptimalsolutionisatruthassignment correspondingtoaleafthathasthe largestnumberofsatisfiedclauses.



Toevaluatethecomplexityof

Algorithm 1

,wecountthenumberofleavesinthetree.Notethatifthenumberofleaves isT

(

n

)

,thenthealgorithmobviouslyworksintimeO

(

T

(

n

))

.Inthesequel,inordertosimplifynotationswewilluseT

(

n

)

todenoteboththenumberofleaves(whenweexpressrecurrences)andthecomplexity.Weconsidertwowaystocountthe numberofleaves.Theformerisbymeansofthevariables.Eachnodehastwochildrenforwhichthenumberofremaining variables decreasesby1.Thisleadstoanumberofleaves T

(

n

)



2

×

T

(

n

1

)

andthereforeT

(

n

)

=

O

(

2n

)

.Thesecondis bymeansoftheclauses.Oneachnode,ifthechosen variableisan

(

i

,

j

)

-variable,thenthefirstchildwillhaveitsnumber ofclauses decreased byatleasti andthesecond childby atleast j.Theworst case, usingthetworeduction rulesgiven above,isi

=

1 and j

=

2 (ori

=

2 and j

=

1),thatleadstoT

(

m

)

=

T

(

m

1

)

+

T

(

m

2

)

andthereforeT

(

m

)

=

O

(

1

.

618m

)

. In [14],the authorsshoweda waytotransformanypolynomial time approximationalgorithm (withratio

α

) for max satintoan approximationalgorithmwithratio

ρ

(forany

α

ρ

1) andrunningtime O

(

1

.

618α)(1−α)−1m

)

.Thebasic idea ofthisalgorithmistobuild thesametreeasin

Algorithm 1

uptothefactthatwestop thebranching whenenough clauses aresatisfied.Then the

α

-approximationpolynomialalgorithmisappliedontheresultingsub-instances. Asalready mentioned, the bestvalue of

α

is0.796 [2].Dealing withcomplexity depending on thenumber ofvariables,using local search techniquesHirsch

[20]

devisesforany



>

0 andanyk

2 arandomizedalgorithmthatfindwithhighprobability a

(

1



)

approximation for max-k-sat (restriction of the problem to instances withclauses of size at mostk) in time

O

((

2

c,k

)

n

)

wherec,k

=

2



/(

k

(

1

+



))

.For max-2-sat,thecomplexityisimproveddowntoO

((

2

3



/(

1

+

3



))

n

)

.

The paperisorganized asfollows.InSections

2

and

3

weproposesome improvements(foranyratio

ρ

)oftheresults of [14].

Fig.

1illustrates therelationship approximation ratio-running time ofthedifferentmethods we develop inthese sectionsandcomparethemwiththeresultin

[14]

.Moreprecisely,inSection

2

twofirstresultsarepresented:thefirstone

1 Thishypothesis[21]saysthat maxsatwhereeachclausehasthreeliteralsisnotsolvableinsubexponentialtimewrt.thenumberofvariables. 2 WeusethestandardnotationO(f)todenote f×p(m+n)forsomepolynomialp.

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Fig. 1. Evaluation of complexities for different methods.

Fig. 2. Comparisonbetweenthealgorithmof[20]for max-2-sat (uppercurve),thealgorithmsfor maxsat(intermediatecurve)and max-2-sat (lowercurve) thatwillbegiveninSection4.

usesroughlythesametechniqueasin

[14]

(leadingto

Algorithm 2

in

Fig. 1

)whilethesecondoneusesadifferentapproach consistingofsplittingtheinstancein“small”sub-instances(Algorithm 3in

Fig. 1

).InSection

3

,wefurtherimprovethese resultsforsome ratiosusinganothertechnique consistingofapproximatelypruningasearch tree(Algorithm 5in

Fig. 1

). Notethat

Fig. 1

isdrawn using

α

=

0

.

796 asthe bestpolynomial timeapproximation ratio,buta figureofsimilar shape wouldfollowwithotherpossiblevaluesof

α

.Inthesesections,we alsoshow thatsimilar resultscanbe derived forFPT approximationalgorithmswhere,givenaratio

ρ

andanintegerk,onehaseithertooutputasolutionthatsatisfiesatleast

ρ

k clausesortoassertthatnosolutionsatisfies(atleast)k clauses.

Alltheseresultsdealwithcomplexitydependingonthenumberofclauses.InSection

4

,weconsidercomplexity depend-ingonthenumberofvariablesandimprovetheresultsof

[20]

(see

Fig. 2

) bygivingforanyratio

ρ

1 a

ρ

-approximate algorithmthatis(1)deterministic,(2)validfor maxsat(norestrictionontheclauses length)and(3)withamuchsmaller runningtime(foranyratio

ρ

).WeconcludethearticleinSection

5

wherewealsobrieflydiscussthe minsatproblem.

2. Firstresults

Weprovide inthissection twofirst improvementsoftheresultgivenin

[14]

.The firstone,givenin Section

2.1

,uses thesameideaas

[14]

whilethesecondoneusesacompletelydifferenttechniqueandachieveimprovedrunningtimes(for someapproximationratios)bysplittingtheinitialinstanceinsub-instancesofsmallersize.

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Fig. 3. Forming the q subsets of clauses.

2.1. Usingabetterparameterizedalgorithm

In thissectionwe briefly mentionthat thesametechnique asin

[14]

leads toan improvedresultwhenwe build the searchtreeaccordingtothealgorithmfrom

[10]

insteadofthesearchtreepresentedinSection

1

.Wesoderivethefollowing algorithm,thatisstrictlybetterthantheoneof

[14]

(see

Fig. 1

inSection

1

).

Algorithm2. Buildasearch-treeastheparameterized algorithmof

[10]

does.3 Stop thedevelopment ofthistreeateach

node where atleast

(

m

(

ρ

α

)/(

1

α

))

clausesare satisfied (recallthat

α

isthe bestknown polynomial approximation ratiofor maxsat),orwhentheinstanceisempty.Foreachleafoftheso-prunedtree,applyapolynomial

α

-approximation algorithmtocompletetheassignmentoftheremainingvariables;thus,eachleafofthetreecorrespondstoacompletetruth assignment.Returntheassignmentsatisfyingthelargestnumberofclauses.



Proposition1.Forany

ρ

suchthat

α

ρ

1,Algorithm2achievesapproximationratio

ρ

intimeO

(

1

.

3695m(ρα)/(1−α)

)

.

Proof. Consider first the running time. The parameterized algorithm of [10] buildsa search tree where the worst case recurrence relationis T

(

k

)

2T

(

k

3

)

+

2T

(

k

7

)

,wheretheparameterk isthenumberofsatisfiedclauses,leading toa globalcomplexityofO

(

1

.

3695k

)

.Here,webuildthistreeandstoptheconstructionineachleavewherem

(

ρ

α

)/(

1

α

)

clausesaresatisfied.ThisleadstoarunningtimeofO

(

1

.

3695m(ρα)/(1−α)

)

.

Wenowhandletheapproximationratio.First,ifthenumber

|

OPT

|

ofclausessatisfiedbyanoptimalsolutionOPT isless thanm

(

ρ

α

)/(

1

α

)

,then

Algorithm 2

obviouslyfindsanoptimumsolution.Otherwise,letusconsiderthebranchofthe branchingtreewheretheleafcorrespondstoapartialoptimaltruthassignmentsatisfyingclausesinOPT.Denotebyk0 the

numberof clausessatisfiedin thisleaf (k0



m

(

ρ

α

)/(

1

α

)

),i.e.by thepartialassignment corresponding tothisleaf.

Usingthisassignment,wegetaresultinginstanceinwhichitispossibletosatisfy

|

OPT

| −

k0clauses (becausetheoptimal

assignmentsatisfies

|

OPT

|

clauses).Consequently,the

α

-approximationalgorithmcalledby

Algorithm 2

willsatisfyatleast

α

(

|

OPT

| −

k0

)

moreclauses.So,finally,atleast: k0

+

α



|

OPT

| −

k0



=

k0

(

1

α

)

+

α

|

OPT

| 

m

(

ρ

α

)

+

α

|

OPT

|

 |

OPT

|(

ρ

α

)

+

α

|

OPT

| =

ρ

|

OPT

|

clauseswillbesatisfied.

2

2.2. Splittingtheclauses

In [8],it isshownthat ageneric methodcan give interestingmoderately exponentialapproximation algorithmsif ap-pliedin(maximization)problemssatisfyingsome hereditaryproperty(apropertyissaidtobe hereditaryifforanyset A

satisfying thisproperty, andany B

A, B satisfies this propertytoo). maxsatcan be seen assearchingfora maximum subset of clausessatisfying theproperty “canbe satisfiedby a truthassignment”, andthisproperty isclearly hereditary. Therefore, wecan adaptthe splittingmethodintroduced in

[8]

to transformanyexactalgorithm intoa

ρ

-approximation algorithm,foranyrational

ρ

,andwithrunningtimeexponentiallyincreasingwith

ρ

.

Algorithm3.Let p

,

q betwointegerssuchthat

ρ

=

p

/

q.Splitthesetofclausesintoq pairwisedisjointsubsets A1

,

· · · ,

Aq

ofsizem

/

q (atmost



m

/

q

ifm

/

q isnotaninteger).Then,considertheq subsetsCi

=

Ai

Ai+1

∪ · · · ∪

Ai+p−1 (iftheindex

is larger thanq, take itmodulo q) for i

=

1

,

· · · ,

q (seeFig. 3). On eachsubset, apply some exact algorithm for maxsat. Returnthebesttruthassignmentamongthemassolutionforthewholeinstance.



3 Notethatweuseonlythesearch-treeofthealgorithmof[10](inparticular,nottheinitialkernelizationinit),sothatatleastonebranchofthetree

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Proposition2.Givenanexactalgorithmfor maxsatrunningintimeO

(

γ

m

)

,Algorithm3achievesapproximationratio

ρ

intime

O

(

γ

ρm

)

.

Proof. Algorithm 3callsq timesanexactalgorithm(whoserunningtimeis O

(

γ

m

)

).Thentheboundoftherunningtime

easilyfollowsfromthefactthateachsubsetCi containsatmostp



m

/

q

ρ

m

+

p clauses.

Fortheapproximationratio,notefirstthatifwerestrictaninstancewithasetC ofclausestoanewinstancewithanew set C

C of clauses,thenan optimalsolutionforC satisfiesatleastthesameamount ofclausesin C thanan optimal solutionforC (inotherwords,therestrictionofanysolutionforC toCisfeasibleforC),i.e.,

|

OPT

(

C

)

C

|

 |

OPT

(

C

)

|

.In particular,fori

=

1

,

. . . ,

q,

|

OPT

(

Ci

)

|  |

OPT

(

C

)

Ci

|

.

Now,note that by construction ofthe Ci’s, we easily see that each clause appears inexactly p amongthe q subsets

C1

,

C2

,

. . . ,

Cq,andthisholdsinparticularforanyclauseinOPT.Thus,



iq=1

|

OPT

(

C

)

Ci

| =

p

× |

OPT

(

C

)

|

.Bythediscussion

above,



qi=1

|

OPT

(

Ci

)

|



p

× |

OPT

(

C

)

|

.Since



qi=1

|

OPT

(

Ci

)

| 

q

×

maxqi=1

|

OPT

(

Ci

)

|

,thenmaxqi=1

|

OPT

(

Ci

)

| 

qp

|

OPT

(

C

)

|

.

2

NotethatifweconsideranFPTalgorithmworkingintimeO

(

γ

k

)

,usingitin

Algorithm 3

withparameter

ρ

k (insteadof

anexactone)leadstoa

ρ

approximateFPTalgorithmworkingintime O

(

γ

ρk

)

.Also,itisworthnoticingthat

Algorithm 3

isfasterthan

Algorithm 2

forratioscloseto1(see

Fig. 1

inSection

1

).

3. Approximatepruningofthesearchtree

Informally, the idea of an approximate pruning of the search tree is based upon the fact that, if we seek, say, a 1

/

2-approximation fora maximization problem,then when asearch-tree based algorithmselects aparticular datumd

forinclusioninthesolution,onemayremoveoneotherdatumdfromtheinstance(without,ofcourse,includingitinthe solution).At worst,d ispartofan optimalsolutionandislost byour solution.Thus,globally,thenumberof datainan optimumsolutionisatmosttwotimesthenumberofdatainthebuiltsolution.Ontheotherhand,withtheremovalofd, thesizeofthesurvivinginstanceisreducednotby1(duetotheremovalofd)butby2.

Thismethod can beadapted to maxsat inthe followingway:revisit

Algorithm

1andrecall that its worst casewith respecttom istobranchona

(

1

,

2

)

-literalandtofix1(satisfied)clauseontheone sideand2(satisfied)clauses onthe otherside.Ifwedecidetoalsoremove1moreclause(arbitrarilychosen)intheformercaseand2moreclauses(arbitrarily chosen)inthelatterone,thisleadstoarunningtime T

(

m

)

satisfyingT

(

m

)



T

(

m

2

)

+

T

(

m

4

)

,i.e.,T

(

m

)



O

(

1

.

27m

)

. Sinceinthebrancheswehavesatisfiedatleasts

1 clause(resp.,s

2 clauses)whiletheoptimumsatisfiesatmosts

+

1 clauses(resp.,s

+

2 clauses),wegetanapproximationratio0

.

5.

Thisbasicprinciple isnot sufficienttoget aninteresting resultfor maxsat,butit canbe improvedasfollows.Letus considerthe leftbranch wherethe

(

1

,

2

)

-literal issetto true,satisfyingaclause C1.Instead ofthrowingawayone other

clause,wepicktwoclauses C2 andC3suchthatC2 containsaliteral



andC3containstheliteral

¬

,andweremovethese

twoclauses. Anytruth assignmentsatisfieseitherC2 or C3,meaning thatinthisbranchwewillsatisfy atleast2clauses

(C1 andone among C2 andC3), whileatworst the optimumwill satisfy thesethreeclauses. Inthe other branchwhere

2clauses aresatisfied,wepicktwopairs ofclauses containingopposite literalsandwe removethem.Thistrickimproves both theapproximationratioandthe runningtime:nowwe havean approximationratio2

/

3 (2 clausessatisfiedamong 3clausesremovedinonebranch,4clauses satisfiedamong6clauses removedintheother branch),andtherunningtime satisfiesT

(

m

)



T

(

m

3

)

+

T

(

m

6

)

,i.e., T

(

m

)

=

O

(

1

.

17m

)

.

Inwhatfollows,wegeneralizetheideassketchedaboveinordertoworkforanyratio

ρ

∈ Q

.

Algorithm4.Letp andq betwo integerssuchas pq

=

1ρ21ρ . Webuildthesearchtreeand, onanyofitsnodes,wecount thenumberofsatisfiedclausesfromtheroot(wedonotcountheretheclausesthathavebeenarbitrarilyremoved).Each timewe reachamultipleofq,we pick p pairsofclauseswithoppositeliteralsandweremovethemfromtheremaining sub-instance.Whensuchasub-instanceonanode isempty, wearbitrarilyassignavalue onanystillunassignedvariable. Finally,wereturnthebesttruthassignmentsoconstructed.



Notethatit mightbethecasethatatsome pointitisimpossibletofind p pairsofclauses withopposite literals.But thismeans that (after removing q

<

p pairs) each variable appears onlypositively oronly negatively,andthe remaining instanceiseasilysolvableinlineartime.

Theorem1.Algorithm4runsintimeO

(

1

.

618m(2ρ−1)

)

andsatisfiesatleast

ρ

· |

OPT

|

clauses.

Proof. Considerthe leafwherethevariablesare setlikeinanoptimum solution.In thisleaf,assume thatthenumberof satisfiedclausesis s

×

q

+

s(where s

<

q);again, wedonotcount theclausesthat havebeenarbitrarily removed.Then, thealgorithmhasremoveds

×

2p clausesarbitrarily,amongwhichatleasts

×

p arenecessarilysatisfied.Intheworstcase, the s

×

p other clauses are in OPT; hence,

|

OPT

| 

2sp

+

sq

+

s.So, the approximation ratioof Algorithm 4is at least:

(

sq

+

sp

+

s

)/(

sq

+

2sp

+

s

)

ρ

.

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Table 1

Runningtimesforthealgorithmof[10]andAlgorithm5.

Case [10] Algorithm 5 4.0 a) T(m)=T(m−1)+T(m−5) T(m)=T(m−1−χ)+T(m−5−4χ) 4.0 b) T(m)=T(m−1)+T(m−7)+T(m−10) T(m)=T(m−1−χ)+T(m−7−6χ)+T(m−10−9χ) 4.1 T(m)=2T(m−3) T(m)=2T(m−3−3χ) 4.2 T(m)=T(m−2)+T(m−3) T(m)=T(m−2−2χ)+T(m−3−2χ) 4.3 T(m)=2T(m−6)+T(m−2) T(m)=2T(m−6−6χ)+T(m−2−2χ) 4.4 T(m)=T(m−3)+T(m−2) T(m)=T(m−3−3χ)+T(m−2−2χ) 4.5 Same as for 4.3 4.6 Same as for 4.4 4.7 T(m)=2T(m−5) T(m)=2T(m−5−5χ) 4.8 T(m)=2T(m−5)+2T(m−7) T(m)=2T(m−5−5χ)+2T(m−7−6χ) 4.9 a) Same as for 4.1 4.9 b) Same as for 4.0 a) 4.10 T(m)=T(m−1)+1 T(m)=T(m−1)+1 4.11 a) T(m)=2T(m−4) T(m)=2T(m−4−4χ) 4.11 b) Same as for 4.0 a) 4.12 a) Same as for 4.0 a) 4.12 b) T(m)=2T(m−8)+T(m−1) T(m)=2T(m−8−7χ)+T(m−1−χ)

Wenowestimatetherunningtimeof

Algorithm 4

.Foreachnode i ofthetree,denotebymithenumberofclausesleft

inthesurvivingsub-instanceofthisnode,byzithenumberofsatisfiedclausesfromtherootofthetree(wedonotcount

theclausesthathavebeenarbitrarilyremoved)andsetti

=

mi

− (

2p

/

q

)(

zimod q

)

.

Fortherootofthetree,zi

=

0 andthereforeti

=

m. Leti be anode withtwochildren j (atleastoneclausesatisfied)

and g (atleasttwoclausessatisfied).Letusexaminequantitytjwhenexactlyoneclauseissatisfied.Inthiscase,zj

=

zi

+

1.

Ontheotherhand:i)Ifzj mod q

=

0,then wehavenot reachedthethresholdnecessarytoremovethe2p clauses. Then,

mj

=

mi

1 andtj

=

mj

2p

/

q

(

zjmod q

)

=

mi

1

2p

/

q

((

zimod q

)

+

1

)

=

ti

1

2p

/

q.Ifzjmod q

=

0,thenzimod q

=

q

1 andthethresholdhasbeenreached;so2p clauseshavebeenremoved.Then,mj

=

mi

1

2p,tj

=

mj

=

mi

1

2p

andti

=

mi

2p

/

q

(

q

1

)

=

mi

2p

+

2p

/

q.Finally,tj

=

ti

1

2p

/

q.Therefore,inbothcases i) and ii),tj



ti

1

2p

/

q.

Of course, by a similar argument, ifwe satisfy g clauses, then the quantity ti is reducedby g

(

1

+

2p

/

q

)

. Thisleads to

a running time T

(

t

)



T

(

t

1

2p

/

q

)

+

T

(

t

2

4p

/

q

)

and hence T

(

t

)

=

1

.

618t/(1+2p/q). Since initially t

=

m, we get

T

(

m

)

=

1

.

618m/(1+2p/q).Takingintoaccountthat p

/

q

= (

ρ

1

)/(

1

2

ρ

)

,wegetimmediately1

/(

1

+

2p

/

q

)

=

2

ρ

1.

2

Algorithm 4canbeimprovedifinsteadofusingthesimplebranchingruleinthetree,themoreinvolvedcaseanalysis of

[10]

isused.AsalreadynotedinSection

2.1

,weuseonlythesearch-treeofthealgorithmof

[10]

,thatensures thatat leastonebranchofthetreecorresondstoapartialoptimalassignment.Thisderivesthefollowingalgorithm.

Algorithm5.Letp andq betwointegerssuchas pq

=

1ρ21ρ .Buildthesearch-treeof

[10]

and,oneachnodeofit,countthe numberofsatisfiedclausesfromtheroot.Eachtimeamultipleofq isreached,pickp pairsofclauseswithoppositeliterals andremovethemfromtheresultingsub-instance.Returnthebesttruthassignmentsoconstructed.



Toestimatetherunningtimeof

Algorithm 5

,weusenearly thesameanalysisasin

[10]

.Theonlydifferenceisthat,at each stepofthesearch tree,

[10]

countswithoutdistinctionthesatisfiedandtheunsatisfiedclauses (clausesthat became empty), whereas we have to make a difference in the complexity analysis: a satisfied clause reduces the quantity t by

1

+

2p

/

q in

Algorithm 5

,whileanunsatisfiedclausereducesitbyonly1.

By an exhaustive comparative studybetween the cases analyzed in [10] andAlgorithm 5, it can be shownthat for any

ρ

the worst case is always reached by the case(noted by 4.2 in [10]) T

(

m

)

=

T

(

m

2

)

+

T

(

m

3

)

, that becomes

T

(

m

)

=

T

(

m

2

2

χ

)

+

T

(

m

3

2

χ

)

with

χ

=

2p

/

q intheanalysisof

Algorithm 5

.

Indeed,incase4.2of

[10]

(“thereis

(

2

,

2

)

-literalx that occursatleastonceasa unitclause”), abranchingisdoneon the variablex.Ontheoneside,2clauses aresatisfiedwhile,ontheother side,2aresatisfiedand1becomes emptyand thusitisremovedfromtheinstance.For

[10]

,thisleadstoacomplexity T

(

m

)



T

(

m

2

)

+

T

(

m

3

)

.For

Algorithm 5

,this gives T

(

m

)



T

(

m

2

4p

/

q

)

+

T

(

m

3

4p

/

q

)

.Tosimplifytheseresults,set

χ

=

2p

/

q.Then T

(

m

)



T

(

m

2

2

χ

)

+

T

(

m

3

2

χ

)

,whichleadstoT

(

m

)

=

O

(

γ

m

)

with

γ

thelargestrealsolutionoftheequation

γ

2χ+3

γ

1

=

0.

Fortheother cases,acomparativestudybetweenthealgorithmof

[10]

and

Algorithm 5

issummarizedin

Table 1

.Its thirdcolumngivesequationswhoselargestrealsolutionsaretheworstcaserunningtimesfor

Algorithm 5

.

Dependingontheratio

ρ

weseek,therunningtimeofthealgorithmisgivenbytheworstcaseofallthecasesgivenin

Table 1.However,onecan showthatforany

ρ

theworstcaseisalwaysreachedby thecase4.2.Letusshow anexample (the other casesaresimilar). Consider theequations f4.2

(

X

)

=

X2χ+3

X

1

=

0 and f4.0

(

X

)

=

X4χ+5

X3χ+4

1

=

0.

The largest realsolution ofthe former isalways larger than the largestreal solutionof thelatter one. Indeed,let

χ

be anypositivevalue.Remarkfirst that f4.0

(

X

)

= (

4

χ

+

5

)

X4χ+4

− (

3

χ

+

4

)

X3χ+3

>

0;hence,function f

(7)

Fig. 4. Illustration ofAlgorithm 6.

X

>

1.Whatwenowneedtoshowisthatif f4.2

(

X

)

=

0,then f4.0

(

X

)



0 (thismeansthatthezeroof f4.0 isbeforethat

of f4.2): f4.2

(

X

)

=

0

X2χ+3

=

X

+

1

X3χ+4

=

+2

+

+1

X4χ+5

=

X2χ+3

+

X2χ+2

X4χ+5

=

X2χ+2

+

X

+

1

X4χ+5

X3χ+4

1

=

X2χ+2

+

X

+2

+1

f4.0

(

X

)

=

X2χ+2

+

X

+2

+1

f4.0

(

X

)

=



+1

1



+1

X





0

andtheresultfollows.

Now,theclaimof

Theorem 1

dealingwiththeapproximationratioof

Algorithm 4

identicallyappliesalsofor

Algorithm 5

. Puttingalltheabovetogether,thefollowingtheoremholds.

Theorem2.Forany

ρ

<

1,Algorithm5achievesapproximationratio

ρ

on maxsatwithrunningtimeT

(

m

)

=

O

(

γ

m

)

,where

γ

is

thelargestrealsolutionoftheequationX2α+3

X

1

=

0 and

α

=

2ρ−2 1−2ρ .

Itisvery well-knownthat, inevery maxsat-instance,atleastm

/

2 clausescanbealways greedilysatisfied.Soforany suchformula,

|

OPT

| 

m

/

2.Hence,anyalgorithmwithrunningtimefunctionofm isaparameterizedalgorithmfor maxsat. Thishowevermaylead touninterestingresults(itsrunningtimemaybeworse thanthat oftheparameterizedalgorithm of[10]),butwecanimprovethis. Indeed,wecanshow thatthepruningmethodjustdescribedcanbedirectlyappliedto theparameterizedalgorithmof

[10]

fortheachievementofthefollowingparameterizedapproximationresult.

Proposition3.Forany

ρ

<

1, maxsatisapproximablewithinratio

ρ

intimeO

(

1

.

3695(2ρ−1)k

)

,wherek isthemaximumnumber

ofsatisfiedclausesintheinstance.

4. Splittingthevariables

Inthissection,wepresenttwoalgorithmsthatapproximate maxsatwithinanyapproximationratiosmallerthan1,and withacomputation timedependingonn (thenumberofvariables).Asmentionedintheintroduction,Hirsch

[20]

devises forany



>

0 andanyk

2 arandomizedalgorithmthatfindwithhighprobabilitya

(

1



)

approximationfor max-k-sat intime O

((

2

c,k

)

n

)

wherec,k

=

2



/(

k

(

1

+



))

(notethatc,k

0 whenk

→ ∞

,sothisdoesnotgiveacomplexity cn

withc

<

2 for maxsat).For max-2-sat,thecomplexityisimproveddowntoO

((

2

3



/(

1

+

3



))

n

)

.

Aswewillsee,thefirstalgorithmofthissection(Algorithm 6–see

Fig. 4

foranillustration)improvestheseresults.It buildsseveraltrees.Then,ineachofthem,asfor

Algorithm 2

inSection

2.1

,itcutsthetreeatsomepoint andcompletes variables’assignmentusingapolynomialapproximationalgorithm.

Algorithm6.Letp andq betwointegerssuchthatp

/

q

= (

ρ

α

)/(

1

α

)

.Buildq subsetsX1

,

· · · ,

Xqofvariables,eachone

containing roughly p

/

q

×

n variables,whereeach variableappearsinexactly p subsets(asin

Algorithm 3

inSection 2.2). Foreachsubset Xi,constructacompletesearchtree,consideringonlythevariablesinthesubset(i.e.,thedepthofeachof

(8)

thesetreesisexactly

|

Xi

|



p

/

q

×

n).Foreachoftheleavesofthesetrees,runapolynomialtimealgorithmguaranteeinga

ratio

α

onthesurvivingsub-instance.Returnthebesttruthassignmentamongthosebuilt.



Eachofthetreesbuiltby

Algorithm 6

isabinarytreeandhasdepthroughlypn

/

q (moreprecisely,atmostpn

/

q

+

p).So its runningtimeis O

(

2np/q

)

.Notealsothat,oneachofthesetrees,atleastoneleafisapartialassignmentofanoptimal

(global)truthassignment.Wewillcallsuchaleafanoptimalleaf.

Lemma1.Atleastoneofoptimalleafhasatleast qp

× |

OPT

|

satisfiedclauses(beforeapplyingthepolytimeapproximationalgorithm).

Proof. Remark that every clause Ci in OPT contains atleastone true literal; pick one ofthem fromeach clause Ci and

denotethevariablecorrespondingtothisliteralbyVar

(

Ci

)

.Let,foreachvariablex,C

(

x

)

bethesetofclausesfromOPT for

whichx or

¬

x isthepickedliteral,i.e.,

x

X ,C

(

x

)

= {

Ci

OPT

/

Var

(

Ci

)

=

x

}

.Baseduponthis,OPT

=



xXC

(

x

)

.

Inthetreeobtainedontheset Xi,denoteby

Λ

ithesetofsatisfiedclausesonsomeoptimalleafandset

λ

i

= |Λ

i

|

.Then,



xXiC

(

x

)

⊆ Λ

i and,byconstruction,

i

,

j,C

(

xi

)

C

(

xj

)

= ∅

.Wesohave:

λi





xXi



C

(

x

)



q



i=1

λi



q



i=1



xXi



C

(

x

)



Aseveryx belongstoexactlyp subsetsamongtheq setsXi,itholdsthat: q



i=1

i

| 

p

×



xX



C

(

x

)

 =

p

× |

OPT

|

(1)

From

(1)

,itisimmediatelyderivedthat:

q max i=1

i

| 

1 q q



i=1

i

| 

p q

×



xX



C

(

x

)

 =

p q

× |

OPT

|

thatconcludestheproof.

2

Proposition4.Algorithm6achievesapproximationratio

ρ

.

Proof. By

Lemma 1

,amongalltheoptimalleaves,atleastonesatisfies

λ



pq

×|

OPT

|

clauses.Asanoptimalleafcorresponds toanoptimaltruthassignment,itispossibletocompletethisassignmentintoanoptimal(global)solution.Inotherwords, thereexist

|

OPT

|

− λ

remainingclausesthatbecometrueonthesurvivingsub-instance.Ifthepolynomialalgorithmcalled by

Algorithm 6

achievesapproximationratio

α

,itwill computea solutionthat satisfiesatleast

α

× (|

OPT

− |λ|)

clauses. Hence,thenumberofsatisfiedclauseswillbeatleast:

|λ| +

α

×



|

OPT

− |λ|



=

α

|

OPT

| + (

1

α



α

|

OPT

| + (

1

α

)

p q

|

OPT

|

thatleadstoanapproximationratioof

α

+ (

1

α

)

pq

=

ρ

.

2

Puttingalltheabovetogether,thefollowingtheoremholds.

Theorem3.Algorithm6achievesratio

ρ

intimeO

(

2n(ρα)/(1−α)

)

,forany

ρ

1.

Algorithm 6 is both deterministic and validfor maxsat. Moreover in [20] the best running time is obtainedfor the restrictedproblem max-2-sat forwhicha

ρ

= (

1



)

-approximatesolutionisfoundintime O

((

2

3

(

1

ρ

)/(

4

3

ρ

))

n

)

. Interestingly enough,

(

2

3

(

1

ρ

)/(

4

3

ρ

))

is greater than 2α)/(1−α) (with

α

=

0

.

796) for any

ρ

<

1. Finally,note thatfor max-2-sat onecanuse

Theorem 3

withthebestpolynomialtimeapproximationratioknownforthisproblem,i.e.,

α

=

0

.

931[17] (notethat max-2-sat is notapproximable inpolynomial timewithin ratio0.955 unless P

=

NP [19]). See

Fig. 2inSection

1

foracomparisonofrunningtimes.

Algorithm 6buildsafullsearch treeoneach subsetofvariables.Inparticular,whentheratiosought

ρ

tendsto1,the basis oftheexponentinthecomplexity tendsto2.Then,one mightaskthefollowingquestion:supposethat thereisan exact algorithm solving maxsatin O

(

γ

n

)

(forsome

γ

<

2), is itpossible tofind a

ρ

approximation algorithm intime

O

(

γ

n

(9)

exactone wouldallowtotake advantageofanypossibleimprovementofthe exactsolutionof maxsat,whichisnotthe casein

Algorithm 6

.Notethatfindinganexactalgorithmintime O

(

γ

n

)

forsome

γ

<

2 isafamousopenquestionfor max

sat(cf.thestrongexponentialtimehypothesis

[21]

)aswellasforsomeothercombinatorialproblems.Ithasveryrecently receivedapositiveanswerfortheHamiltoniancycleproblemin

[4]

.

Indeed, we propose in what follows a

ρ

-approximation algorithms working in time O

(

γ

n

ρ

)

with

γ

ρ

<

γ

for any

ρ

∈ ]

α

,

1

[

. Wefirst give a simple solution (Algorithm 7) that we improvelater (Algorithm 8). Algorithm 7moves inthe samespiritas

Algorithm 6

but,insteadofbuildingafullbranchingtreeon Xi,callsanexactalgorithmonthesub-instance

inducedbytheset Xi.

Algorithm7.Let

ρ

∈ Q

and p and q two integers such that p

/

q

=

ρ

. Buildq subsets ofvariables, each one containing

p

/

q

×

n variables(asin

Algorithm 6

).Foreachsubsetofvariables Xi:

a) RemovefromtheinstancethevariablesnotinXi andanyemptyclause.

b) Runtheexactalgorithmontheresultingsub-instance,thusobtainingatruthassignmentforthevariablesin Xi.

c) Completethisassignmentwitharbitrarytruth-valuesforthevariablesnotin Xi.

Among all the truth assignments produced, return theone that satisfiesthe largest numberof clauses in thewhole

in-stance.



In

Algorithm 7

,theexactalgorithmcalledinstepb)runsintime O

(

γ

ρn

)

.Itsapproximationratioistheoneclaimedin Lemma 1.Indeed,theexactalgorithmsatisfiesatleastthesameamountofclausesastheoptimalbranching(fortheglobal instance)woulddo.Moreprecisely, foreach Xi,andforanyx

Xi,theclauses containingx (andinparticulartheclauses

in C

(

x

)

) are not removed fromtheinstance.The optimalbranching wouldthen satisfy atleast



xXi

|

C

(

x

)

|

clauses and, obviously,theexactalgorithmwouldsatisfyevenmore.Hence,thefollowingresultholds.

Proposition5.Algorithm7achievesapproximationratio

ρ

intimeO

(

γ

ρn

)

,whereO

(

γ

n

)

istherunningtimeofanexactalgorithm

for maxsat.

As onecan see,in stepc) ofAlgorithm 7,variables outside Xi, i

=

1

,

. . . ,

q, areassignedarbitrarily, so,atworst their

truthvaluemaysatisfynoadditionalclause.Notethatonemightwanttouseanapproximationalgorithmintheremaining instance asin Algorithm 6; however, the sameanalysis would not work since the exact solution obtained by the exact algorithm on the sub-instance might be completely different from the partial assignment of a global optimal solution. Nevertheless,weareabletoproposeanimprovementbycompletingpartialsolutionsinsuchawaythat,roughlyspeaking, atleasthalfoftheremainingclausesaresatisfied.

Wenowpropose

Algorithm 8

,thatisanimprovementof

Algorithm 7

.Forany

ρ

∈]

α

,

1

]

,itachievesapproximationratio

ρ

andrunsintime O

(

γ

n

ρ

)

.

Algorithm8.Letp

,

q

∈ Q

besuchthat p

/

q

=

2

ρ

1.Buildq subsetsofvariables X1

,

. . . ,

Xq,eachonecontaining p

/

q

×

n

variables(asin

Algorithm 6

).Foreach Xi runthefollowingsteps:

i) assignweight 2toevery clausecontaining onlyvariables in Xi,andweight 1toevery clausecontaining atleastone

variablenotin Xi;

ii) removefromtheinstancethevariablesnotin Xi;removeemptyclauses;

iii) solveexactlythis maxweightedsatresultinginstance,thusobtainingatruthassignmentforthevariablesinXi; iv) completethe assignmentwitha greedyalgorithm: foreach

(

i

,

j

)

-literal,if i

>

j,then theliteral issetto

TRUE

,else

it is setto

FALSE

(and the instanceis modified accordingly)and returnthe best among thetruth-assignments

so-produced.



Lemma2.Ifthereisa maxsat-algorithmworkingintimeO

(

γ

n

)

,thentheinstancesof maxweightedsatinAlgorithm8canbe

solvedwiththesameboundontherunningtime.

Proof. Notethat theonlyweights assignedby Algorithm 8 are1and2.In sucha weighted instance,we canadd anew variablex0andreplaceeachclausec ofweight2bythreenewclauses:c,c

x0 andc

∨ ¬

x0.Thus,ifc issatisfied,thenit

willcountinthenewinstanceasthreesatisfiedclauses.Otherwise,exactlyoneofthethreenewclauseswillbesatisfied. Thus,theso-builtinstanceof maxsatisequivalenttotheinitial maxweightedsat-instancebuiltby

Algorithm 8

.

2

Theorem4.Algorithm8achievesapproximationratio

ρ

intimeO

(

γ

(2ρ−1)n

)

,whereO

(

γ

n

)

istherunningtimeofanexact

algo-rithmfor maxsat.

(10)

Fig. 5. Division of clauses according to a subset Xiof variables.

Fortheapproximationratio,usingthesamenotationasbefore,consideroneparticularliteralineachclausesatisfiedby some optimum solutionOPT,andlet C

(

x

)

be thesubset oftheseclausessuch that thepickedliteral isx or

¬

x.Then,as shownbefore,thereexistsasubset Xisuchthat



xXi

|

C

(

x

)

| 

p

q

|

OPT

|

.Considernowsucha Xi,anddenoteby(see

Fig. 5

):

A thesubset ofclauses containing only variables in X

\

Xi, A+ (resp., A) the subset ofclauses from A that arein the optimum (resp.,arenot intheoptimum); B thesubset ofclauses containingatleastone variablein Xi andone variable

in X

\

Xi; B1+(resp., B2+)thesubsetofclausesfrom B thatareintheoptimumandwhosechosen variableVar

(

c

)

isin Xi

(resp.,notin Xi); B−thesubset ofclausesfrom B thatarenotintheoptimum; C theremaining clauses,i.e.,theclauses thatcontainonlyvariablesin Xi,C+(resp.,C−)thesubsetofclausesfromC thatareintheoptimum(resp.,arenotinthe

optimum). Notethatwhenremovingvariablesin X

\

Xi,clausesin A becomeempty, sotheremainingclauses areexactly

thoseinB

C .Withthesenotations,OPT

=

A+

B1+

B2+

C+and



xXi

|

C

(

x

)

| = |

B1+

| + |

C+

|

.Then,forthechosen Xi:



B1+

 +|

C+

| 

p q

|

OPT

| =

p q



|

A+

| +



B1+

 +

B2+

 +|

C+

|



(2)

Withrespecttostepiii)of

Algorithm 8

,denotebyB1thesubsetofsatisfiedclausesfromB andbyC1 thesubsetofsatisfied

clausesfromC .AsOPT isaparticularsolutionofweight

|

B1

+

|

+

2

|

C+

|

forthisweightedsatproblem,wehave:

|

B1

| +

2

|

C1

| 



B1+

 +

2

|

C+

|

(3)

Thegreedyalgorithminstepiv)willsatisfyatleasthalfoftheremainingclausescontainingatleastoneliteralfrom X

\

Xi,

i.e.,theset

(

B

\

B1

)

A.Finally,thenumberofsatisfiedclausesisatleast:

|

B1

| + |

C1

| +

|

B

| − |

B1

| + |

A

|

2

= |

C1

| +

|

B1

|

2

+

|

B

|

2

+

|

A

|

2 (3)



|

B1+

|

2

+ |

C+

| +

|

B

|

2

+

|

A

|

2





B1+

 +|

C+

| +

|

B + 2

|

2

+

|

A+

|

2

So,theapproximationratioachievedisatleast:

|

B1 +

| + |

C+

| +

|B 2 +|+|A+| 2

|

B1+

| + |

C+

| + |

B2+

| + |

A+

|

=

1 2



1

+

|

B 1 +

| + |

C+

|

|

B1+

| + |

C+

| + |

B2+

| + |

A+

|

(2)



1 2



1

+

p q

=

q

+

p 2q

=

ρ

thatcompletestheproof.

2

For instance, suppose that maxsat is solvable in O

(

1

.

657n

)

, which is the running time to solve Hamiltonian cycle in [4]. Then Algorithm 8 achievesa 0.9-approximationintime O

(

1

.

576n

)

while Algorithm 6achieves thesame ratioin time O

(

1

.

703n

)

.

5. Discussion

We have proposed in this paperseveral algorithms that constitute a kindof “moderately exponential approximation schemata” for maxsat.Theyguarantee approximation ratiosthat are unachievable inpolynomial time unless P

=

NP, or even intime 2m1− undertheexponential time hypothesis.To obtaintheseschemata, severaltechniqueshavebeen used

comingeitherfromthepolynomialapproximationorfromtheexactcomputation.Furthermore,

Algorithm 8

inSection

4

is a kindofpolynomialreductionbetweenexactcomputationandmoderatelyexponential approximationtransformingexact algorithmsrunningon“small”sub-instancesintoapproximationalgorithmsguaranteeinggoodratiosforthewholeinstance. Wethinkthatresearchinmoderatelyexponentialapproximationisaninterestingresearchissueforovercominglimitsposed tothepolynomialapproximationduetothestronginapproximabilityresultsprovedinthelatterparadigm.

Weconcludethispaperwithawordaboutanotherverywellknownoptimumsatisfiabilityproblem,the minsatproblem that, given a set ofvariables anda set ofdisjunctive clauses, consistsof finding a truth assignment that minimizesthe

(11)

numberofsatisfiedclauses.A

ρ

-approximationalgorithmfor minsat(with

ρ

>

1)isanalgorithmthatfindsanassignment satisfyingatmost

ρ

timestheminimalnumberofsimultaneouslysatisfiedclauses.

In

[11]

an approximability-preservingreduction between minvertexcoverand minsatispresented transformingany

ρ

-approximation for the former problem into a

ρ

-approximation for the latter problem. This reduction can be used to translateanyresultonthe minvertexcoverproblemintoaresultonthe minsat,thenumberofverticesinthe minvertex coverinstancebeingthenumberofclausesinthe minsatinstance.Forinstance,theresultsfrom

[8]

for minvertexcover leadtothefollowingparameterizedapproximationresultfor minsat:foreveryinstanceof minsatandforanyr

∈ Q

,ifthere

existsasolutionfor minsatsatisfyingatmostk clauses,itispossibletodeterminewithcomplexityO

(

1

.

28rk

)

a2

r-approximation

ofit.

Wealsonotethatthemethodusedin

Algorithm 6

canbeappliedaswellto minsatwiththefollowingmodificationof thealgorithm.Letp

,

q

∈ Q

besuchthat p

/

q

=

2

ρ

1.Buildq subsetsofvariables,eachonecontaining p

/

q

×

n variables.

Foreachsubset,constructasearchtree,consideringonlythevariablesinthesubset(thedepthofthetreesisp

/

q

×

n).For eachleafofanyoftheso-builttrees,usesomepolynomialalgorithmwithratio

α

onthesurvivingsub-instance.Returnthe bestofthetruthassignmentscomputed.

The complexityof themodificationjustdescribed isthe sameasthatof

Algorithm

6,i.e., O

(

2n(αρ)/(α−1)

)

(thebest knownratiois

α

=

2),andasimilaranalysisderivesanapproximationratio

α

− (

α

1

)

qp

=

ρ

.

References

[1]G.Ausiello,P.Crescenzi,G.Gambosi,V.Kann,A.Marchetti-Spaccamela,M.Protasi,ComplexityandApproximation.CombinatorialOptimization Prob-lemsandTheirApproximabilityProperties,Springer-Verlag,Berlin,1999.

[2]A.Avidor,I.Berkovitch,U.Zwick,ImprovedapproximationalgorithmsforMAXNAE-SATandMAXSAT,in:T.Erlebach,G.Persiano(Eds.),Proc. Work-shoponApproximationandOnlineAlgorithms,WAOA’05,in:LectureNotesinComputerScience,vol. 3879,Springer-Verlag,2006,pp. 27–40. [3]R.Battiti,M.Protasi,Algorithmsandheuristicsformax-sat,in:D.Z.Du,P.M.Pardalos(Eds.),HandbookofCombinatorialOptimization,vol.1,Kluwer

AcademicPublishers,1998,pp. 77–148.

[4]A.Björklund,DeterminantsumsforundirectedHamiltonicity,in:Proc.FOCS’10,IEEEComputerSociety,2010,pp. 173–182. [5]A.Björklund,T.Husfeldt,M.Koivisto,Setpartitioningviainclusion–exclusion,SIAMJ.Comput.39 (2)(2009)546–563.

[6]N.Bourgeois,B.Escoffier,V.Th.Paschos,Efficientapproximationof mincoloringbymoderatelyexponentialalgorithms,Inform.Process.Lett.109 (16) (2009)950–954.

[7]N. Bourgeois, B.Escoffier, V.Th. Paschos,Efficient approximationof minsetcover bymoderately exponential algorithms,Theoret. Comput. Sci. 410 (21–23)(2009)2184–2195.

[8]N.Bourgeois,B.Escoffier,V.Th.Paschos,Approximationof maxindependentset, minvertexcoverandrelatedproblemsbymoderatelyexponential algorithms,DiscreteAppl.Math.159 (17)(2011)1954–1970.

[9]L.Cai,X.Huang,Fixed-parameterapproximation:conceptualframeworkandapproximabilityresults,in:H.L.Bodlaender,M.A.Langston(Eds.),Proc. InternationalWorkshoponParameterizedandExactComputation,IWPEC’06,in:LectureNotesinComputerScience,vol. 4169,Springer-Verlag,2006, pp. 96–108.

[10]J.Chen,I.A.Kanj,Improvedexactalgorithmsfor maxsat,DiscreteAppl.Math.142(2004)17–27.

[11]P.Crescenzi,R.Silvestri,L.Trevisan,Toweightornottoweight:whereisthequestion?,in:Proc.IsraeliSymposiumonTheoryofComputingand Systems,ISTCS’96,IEEE,1996,pp. 68–77.

[12]M.Cygan,L.Kowalik,M.Wykurz,Exponential-timeapproximationofweightedsetcover,Inform.Process.Lett.109 (16)(2009)957–961. [13]M.Cygan,M.Pilipczuk,Exactandapproximatebandwidth,Theoret.Comput.Sci.411 (40–42)(2010)3701–3713.

[14]E.Dantsin,M.Gavrilovich,E.A.Hirsch,B.Konev, maxsatapproximationbeyondthelimitsofpolynomial-timeapproximation,Ann.PureAppl.Logic 113(2002)81–94.

[15]R.G.Downey,M.R.Fellows,C.McCartin,Parameterizedapproximationproblems,in:H.L.Bodlaender,M.A.Langston(Eds.),Proc.InternationalWorkshop onParameterizedandExactComputation,IWPEC’06,in:LectureNotesinComputerScience,vol. 4169,Springer-Verlag,2006,pp. 121–129.

[16]B.Escoffier,V.Th.Paschos,Asurveyonthestructureofapproximationclasses,Comput.Sci.Rev.4 (1)(2010)19–40.

[17]U.Feige,M.X.Goemans,Approximatingthevalueoftwoproverproofsystems,withapplicationstomax2satandmaxdicut,in:Proc.IsraeliSymposium onTheoryofComputingandSystems,ISTCS’95,1995,pp. 182–189.

[18]M.Fürer,S.Gaspers,S.P.Kasiviswanathan,Anexponentialtime2-approximationalgorithmforbandwidth,in:Proc.InternationalWorkshopon Param-eterizedandExactComputation,IWPEC’09,in:LectureNotesinComputerScience,vol. 5917,Springer,2009,pp. 173–184.

[19]J.Håstad,Someoptimalinapproximabilityresults,in:Proc.STOC’97,1997,pp. 1–10.

[20]E.A.Hirsch,Worst-casestudyoflocalsearchforMAX-k-SAT,DiscreteAppl.Math.130 (2)(2003)173–184.

[21]R.Impagliazzo,R.Paturi,F.Zane,Whichproblemshavestronglyexponentialcomplexity?,J.Comput.SystemSci.63 (4)(2001)512–530. [22]D.Moshkovitz,R.Raz,Twoquerypcpwithsub-constanterror,in:Proc.FOCS’08,2008,pp. 314–323.

Figure

Fig. 2. Comparison between the algorithm of [20] for max-2-sat (upper curve), the algorithms for max sat (intermediate curve) and max-2-sat (lower curve) that will be given in Section 4 .
Fig. 3. Forming the q subsets of clauses.
Fig. 4. Illustration of Algorithm 6 .
Fig. 5. Division of clauses according to a subset X i of variables.

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