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
caSorbonneUniversités,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}
thesetofvariablesandby 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.
xi or its negation
¬
xi.Aρ
-approximation algorithm for maxsat (withρ
<
1) is an algorithm that finds an assignment satisfyingatleastafractionρ
ofthemaximalnumberofsimultaneouslysatisfiedclauses.Thebestknownratioguaranteed by apolynomialtime approximationalgorithmisα
=
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 Thislatterresultmotivatesthestudyofexponentialtimeapproximationalgorithms.
Dealingwithexactsolution,
[10]
givesanexactalgorithmworkingintime O∗(
1.
3247m)
whichisthebestknownboundsofarwrt.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 clausesand negatively in exactly j clauses. Forany instance C of maxsat, we will denote by OPT
(
C)
(or OPT if no ambiguity occurs)anoptimalsetofsatisfiedclauses.Finally,wedenotebyα
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, h1, can be removed from the instance. This is correct becausewecansetthisliteralto
TRUE
orFALSE
andsatisfytheclausesthatcontainit.Rule2. Any
(
1,
1)
-literalcanbe removed too.LetC1=
x1∨
x2∨ · · · ∨
xp andC2= ¬
x1∨
x2∨ · · · ∨
xq betheonly twoclausescontainingthevariablex1.Ifthereexisttwooppositeliterals
and
¬
inresp.C1andC2,thenwecansatisfybothclausesbysettingx1 to
TRUE
andto
FALSE
andthereforewecanremovetheseclauses.Otherwise,wecanreplacetheseclauses by C
=
x2∨ · · · ∨
xp∨
x2∨ · · · ∨
xq. The optimum inthe initial instanceis the optimumin the reducedinstanceplus 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 (withratioα
) for max satintoan approximationalgorithmwithratioρ
(foranyα
≤
ρ
≤
1) andrunningtime O∗(
1.
618(ρ−α)(1−α)−1m)
.Thebasic idea ofthisalgorithmistobuild thesametreeasinAlgorithm 1
uptothefactthatwestop thebranching whenenough clauses aresatisfied.Then theα
-approximationpolynomialalgorithmisappliedontheresultingsub-instances. Asalready mentioned, the bestvalue ofα
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 timeO∗
((
2−
c,k)
n)
wherec,k=
2/(
k(
1+
))
.For max-2-sat,thecomplexityisimproveddowntoO∗((
2−
3/(
1+
3))
n)
.The paperisorganized asfollows.InSections
2
and3
weproposesome improvements(foranyratioρ
)oftheresults of [14].Fig.
1illustrates therelationship approximation ratio-running time ofthedifferentmethods we develop inthese sectionsandcomparethemwiththeresultin[14]
.Moreprecisely,inSection2
twofirstresultsarepresented:thefirstone1 Thishypothesis[21]saysthat maxsatwhereeachclausehasthreeliteralsisnotsolvableinsubexponentialtimewrt.thenumberofvariables. 2 WeusethestandardnotationO∗(f)todenote f×p(m+n)forsomepolynomialp.
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]
(leadingtoAlgorithm 2
inFig. 1
)whilethesecondoneusesadifferentapproach consistingofsplittingtheinstancein“small”sub-instances(Algorithm 3inFig. 1
).InSection3
,wefurtherimprovethese resultsforsome ratiosusinganothertechnique consistingofapproximatelypruningasearch tree(Algorithm 5inFig. 1
). NotethatFig. 1
isdrawn usingα
=
0.
796 asthe bestpolynomial timeapproximation ratio,buta figureofsimilar shape wouldfollowwithotherpossiblevaluesofα
.Inthesesections,we alsoshow thatsimilar resultscanbe derived forFPT approximationalgorithmswhere,givenaratioρ
andanintegerk,onehaseithertooutputasolutionthatsatisfiesatleastρ
k clausesortoassertthatnosolutionsatisfies(atleast)k clauses.Alltheseresultsdealwithcomplexitydependingonthenumberofclauses.InSection
4
,weconsidercomplexity depend-ingonthenumberofvariablesandimprovetheresultsof[20]
(seeFig. 2
) bygivingforanyratioρ
≤
1 aρ
-approximate algorithmthatis(1)deterministic,(2)validfor maxsat(norestrictionontheclauses length)and(3)withamuchsmaller runningtime(foranyratioρ
).WeconcludethearticleinSection5
wherewealsobrieflydiscussthe minsatproblem.2. Firstresults
Weprovide inthissection twofirst improvementsoftheresultgivenin
[14]
.The firstone,givenin Section2.1
,uses thesameideaas[14]
whilethesecondoneusesacompletelydifferenttechniqueandachieveimprovedrunningtimes(for someapproximationratios)bysplittingtheinitialinstanceinsub-instancesofsmallersize.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]
insteadofthesearchtreepresentedinSection1
.Wesoderivethefollowing algorithm,thatisstrictlybetterthantheoneof[14]
(seeFig. 1
inSection1
).Algorithm2. Buildasearch-treeastheparameterized algorithmof
[10]
does.3 Stop thedevelopment ofthistreeateachnode where atleast
(
m(
ρ
−
α
)/(
1−
α
))
clausesare satisfied (recallthatα
isthe bestknown polynomial approximation ratiofor maxsat),orwhentheinstanceisempty.Foreachleafoftheso-prunedtree,applyapolynomialα
-approximation algorithmtocompletetheassignmentoftheremainingvariables;thus,eachleafofthetreecorrespondstoacompletetruth assignment.Returntheassignmentsatisfyingthelargestnumberofclauses.Proposition1.Forany
ρ
suchthatα
≤
ρ
≤
1,Algorithm2achievesapproximationratioρ
intimeO∗(
1.
3695m(ρ−α)/(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(ρ−α)/(1−α))
.Wenowhandletheapproximationratio.First,ifthenumber
|
OPT|
ofclausessatisfiedbyanoptimalsolutionOPT isless thanm(
ρ
−
α
)/(
1−
α
)
,thenAlgorithm 2
obviouslyfindsanoptimumsolution.Otherwise,letusconsiderthebranchofthe branchingtreewheretheleafcorrespondstoapartialoptimaltruthassignmentsatisfyingclausesinOPT.Denotebyk0 thenumberof clausessatisfiedin thisleaf (k0
m(
ρ
−
α
)/(
1−
α
)
),i.e.by thepartialassignment corresponding tothisleaf.Usingthisassignment,wegetaresultinginstanceinwhichitispossibletosatisfy
|
OPT| −
k0clauses (becausetheoptimalassignmentsatisfies
|
OPT|
clauses).Consequently,theα
-approximationalgorithmcalledbyAlgorithm 2
willsatisfyatleastα
(
|
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 intoaρ
-approximation algorithm,foranyrationalρ
,andwithrunningtimeexponentiallyincreasingwithρ
.Algorithm3.Let p
,
q betwointegerssuchthatρ
=
p/
q.Splitthesetofclausesintoq pairwisedisjointsubsets A1,
· · · ,
Aqofsizem
/
q (atmostm
/
qifm
/
q isnotaninteger).Then,considertheq subsetsCi=
Ai∪
Ai+1∪ · · · ∪
Ai+p−1 (iftheindexis 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
Proposition2.Givenanexactalgorithmfor maxsatrunningintimeO∗
(
γ
m)
,Algorithm3achievesapproximationratioρ
intimeO∗
(
γ
ρm)
.Proof. Algorithm 3callsq timesanexactalgorithm(whoserunningtimeis O∗
(
γ
m)
).ThentheboundoftherunningtimeeasilyfollowsfromthefactthateachsubsetCi 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)
|
.Bythediscussionabove,
qi=1|
OPT(
Ci)
|
p× |
OPT(
C)
|
.Sinceqi=1|
OPT(
Ci)
|
q×
maxqi=1|
OPT(
Ci)
|
,thenmaxqi=1|
OPT(
Ci)
|
qp|
OPT(
C)
|
.2
NotethatifweconsideranFPTalgorithmworkingintimeO∗
(
γ
k)
,usingitinAlgorithm 3
withparameterρ
k (insteadofanexactone)leadstoa
ρ
approximateFPTalgorithmworkingintime O∗(
γ
ρk)
.Also,itisworthnoticingthatAlgorithm 3
isfasterthan
Algorithm 2
forratioscloseto1(seeFig. 1
inSection1
).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 datumdforinclusioninthesolution,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 otherclause,wepicktwoclauses C2 andC3suchthatC2 containsaliteral
andC3containstheliteral
¬
,andweremovethesetwoclauses. 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
=
1ρ−−21ρ . 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))
andsatisfiesatleastρ
· |
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
.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,denotebymithenumberofclausesleftinthesurvivingsub-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−
2pandti
=
mi−
2p/
q(
q−
1)
=
mi−
2p+
2p/
q.Finally,tj=
ti−
1−
2p/
q.Therefore,inbothcases i) and ii),tjti−
1−
2p/
q.Of course, by a similar argument, ifwe satisfy g clauses, then the quantity ti is reducedby g
(
1+
2p/
q)
. Thisleads toa 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 getT
(
m)
=
1.
618m/(1+2p/q).Takingintoaccountthat p/
q= (
ρ
−
1)/(
1−
2ρ
)
,wegetimmediately1/(
1+
2p/
q)
=
2ρ
−
1.2
Algorithm 4canbeimprovedifinsteadofusingthesimplebranchingruleinthetree,themoreinvolvedcaseanalysis of
[10]
isused.AsalreadynotedinSection2.1
,weuseonlythesearch-treeofthealgorithmof[10]
,thatensures thatat leastonebranchofthetreecorresondstoapartialoptimalassignment.Thisderivesthefollowingalgorithm.Algorithm5.Letp andq betwointegerssuchas pq
=
1ρ−−21ρ .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 by1
+
2p/
q inAlgorithm 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 becomesT
(
m)
=
T(
m−
2−
2χ
)
+
T(
m−
3−
2χ
)
withχ
=
2p/
q intheanalysisofAlgorithm 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)
.ForAlgorithm 5
,this gives T(
m)
T
(
m−
2−
4p/
q)
+
T(
m−
3−
4p/
q)
.Tosimplifytheseresults,setχ
=
2p/
q.Then T(
m)
T
(
m−
2−
2χ
)
+
T
(
m−
3−
2χ
)
,whichleadstoT(
m)
=
O∗(
γ
m)
withγ
thelargestrealsolutionoftheequationγ
2χ+3−
γ
−
1=
0.Fortheother cases,acomparativestudybetweenthealgorithmof
[10]
andAlgorithm 5
issummarizedinTable 1
.Its thirdcolumngivesequationswhoselargestrealsolutionsaretheworstcaserunningtimesforAlgorithm 5
.Dependingontheratio
ρ
weseek,therunningtimeofthealgorithmisgivenbytheworstcaseofallthecasesgiveninTable 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)
= (
4χ
+
5)
X4χ+4− (
3χ
+
4)
X3χ+3>
0;hence,function fFig. 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=
Xχ+2+
Xχ+1⇔
X4χ+5=
X2χ+3+
X2χ+2⇔
X4χ+5=
X2χ+2+
X+
1⇒
X4χ+5−
X3χ+4−
1=
X2χ+2+
X−
Xχ+2−
Xχ+1⇒
f4.0(
X)
=
X2χ+2+
X−
Xχ+2−
Xχ+1⇒
f4.0(
X)
=
Xχ+1−
1Xχ+1−
X0andtheresultfollows.
Now,theclaimof
Theorem 1
dealingwiththeapproximationratioofAlgorithm 4
identicallyappliesalsoforAlgorithm 5
. Puttingalltheabovetogether,thefollowingtheoremholds.Theorem2.Forany
ρ
<
1,Algorithm5achievesapproximationratioρ
on maxsatwithrunningtimeT(
m)
=
O∗(
γ
m)
,whereγ
isthelargestrealsolutionoftheequationX2α+3
−
X−
1=
0 andα
=
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, maxsatisapproximablewithinratioρ
intimeO∗(
1.
3695(2ρ−1)k)
,wherek isthemaximumnumberofsatisfiedclausesintheinstance.
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 cnwithc
<
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,asforAlgorithm 2
inSection2.1
,itcutsthetreeatsomepoint andcompletes variables’assignmentusingapolynomialapproximationalgorithm.Algorithm6.Letp andq betwointegerssuchthatp
/
q= (
ρ
−
α
)/(
1−
α
)
.Buildq subsetsX1,
· · · ,
Xqofvariables,eachonecontaining roughly p
/
q×
n variables,whereeach variableappearsinexactly p subsets(asinAlgorithm 3
inSection 2.2). Foreachsubset Xi,constructacompletesearchtree,consideringonlythevariablesinthesubset(i.e.,thedepthofeachofthesetreesisexactly
|
Xi|
p/
q×
n).Foreachoftheleavesofthesetrees,runapolynomialtimealgorithmguaranteeingaratio
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 forwhichx or
¬
x isthepickedliteral,i.e.,∀
x∈
X ,C(
x)
= {
Ci∈
OPT/
Var(
Ci)
=
x}
.Baseduponthis,OPT=
x∈XC(
x)
.Inthetreeobtainedontheset Xi,denoteby
Λ
ithesetofsatisfiedclausesonsomeoptimalleafandsetλ
i= |Λ
i|
.Then,x∈XiC
(
x)
⊆ Λ
i and,byconstruction,∀
i,
j,C(
xi)
∩
C(
xj)
= ∅
.Wesohave:λi
x∈Xi C(
x)
q i=1λi
q i=1 x∈Xi C(
x)
Aseveryx belongstoexactlyp subsetsamongtheq setsXi,itholdsthat: q
i=1|λ
i|
p×
x∈X C(
x)
=
p× |
OPT|
(1)From
(1)
,itisimmediatelyderivedthat:q max i=1
|λ
i|
1 q q i=1|λ
i|
p q×
x∈X C(
x)
=
p q× |
OPT|
thatconcludestheproof.2
Proposition4.Algorithm6achievesapproximationratio
ρ
.Proof. By
Lemma 1
,amongalltheoptimalleaves,atleastonesatisfiesλ
pq
×|
OPT|
clauses.Asanoptimalleafcorresponds toanoptimaltruthassignment,itispossibletocompletethisassignmentintoanoptimal(global)solution.Inotherwords, thereexist|
OPT|
− λ
remainingclausesthatbecometrueonthesurvivingsub-instance.Ifthepolynomialalgorithmcalled byAlgorithm 6
achievesapproximationratioα
,itwill computea solutionthat satisfiesatleastα
× (|
OPT− |λ|)
clauses. Hence,thenumberofsatisfiedclauseswillbeatleast:|λ| +
α
×
|
OPT− |λ|
=
α
|
OPT| + (
1−
α
)λ
α
|
OPT| + (
1−
α
)
p q|
OPT|
thatleadstoanapproximationratioofα
+ (
1−
α
)
pq=
ρ
.2
Puttingalltheabovetogether,thefollowingtheoremholds.
Theorem3.Algorithm6achievesratio
ρ
intimeO∗(
2n(ρ−α)/(1−α))
,foranyρ
≤
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−
3ρ
))
n)
. Interestingly enough,(
2−
3(
1−
ρ
)/(
4−
3ρ
))
is greater than 2(ρ−α)/(1−α) (withα
=
0.
796) for anyρ
<
1. Finally,note thatfor max-2-sat onecanuseTheorem 3
withthebestpolynomialtimeapproximationratioknownforthisproblem,i.e.,α
=
0.
931[17] (notethat max-2-sat is notapproximable inpolynomial timewithin ratio0.955 unless P=
NP [19]). SeeFig. 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)
(forsomeγ
<
2), is itpossible tofind aρ
approximation algorithm intimeO∗
(
γ
nexactone wouldallowtotake advantageofanypossibleimprovementofthe exactsolutionof maxsat,whichisnotthe casein
Algorithm 6
.Notethatfindinganexactalgorithmintime O∗(
γ
n)
forsomeγ
<
2 isafamousopenquestionfor maxsat(cf.thestrongexponentialtimehypothesis
[21]
)aswellasforsomeothercombinatorialproblems.Ithasveryrecently receivedapositiveanswerfortheHamiltoniancycleproblemin[4]
.Indeed, we propose in what follows a
ρ
-approximation algorithms working in time O∗(
γ
nρ
)
withγ
ρ<
γ
for anyρ
∈ ]
α
,
1[
. Wefirst give a simple solution (Algorithm 7) that we improvelater (Algorithm 8). Algorithm 7moves inthe samespiritasAlgorithm 6
but,insteadofbuildingafullbranchingtreeon Xi,callsanexactalgorithmonthesub-instanceinducedbytheset Xi.
Algorithm7.Let
ρ
∈ Q
and p and q two integers such that p/
q=
ρ
. Buildq subsets ofvariables, each one containingp
/
q×
n variables(asinAlgorithm 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 (andinparticulartheclausesin C
(
x)
) are not removed fromtheinstance.The optimalbranching wouldthen satisfy atleast x∈Xi|
C(
x)
|
clauses and, obviously,theexactalgorithmwouldsatisfyevenmore.Hence,thefollowingresultholds.Proposition5.Algorithm7achievesapproximationratio
ρ
intimeO∗(
γ
ρn)
,whereO∗(
γ
n)
istherunningtimeofanexactalgorithmfor maxsat.
As onecan see,in stepc) ofAlgorithm 7,variables outside Xi, i
=
1,
. . . ,
q, areassignedarbitrarily, so,atworst theirtruthvaluemaysatisfynoadditionalclause.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
,thatisanimprovementofAlgorithm 7
.Foranyρ
∈]
α
,
1]
,itachievesapproximationratioρ
andrunsintime O∗(
γ
nρ
)
.Algorithm8.Letp
,
q∈ Q
besuchthat p/
q=
2ρ
−
1.Buildq subsetsofvariables X1,
. . . ,
Xq,eachonecontaining p/
q×
nvariables(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 issettoTRUE
,elseit is setto
FALSE
(and the instanceis modified accordingly)and returnthe best among thetruth-assignmentsso-produced.
Lemma2.Ifthereisa maxsat-algorithmworkingintimeO∗
(
γ
n)
,thentheinstancesof maxweightedsatinAlgorithm8canbesolvedwiththesameboundontherunningtime.
Proof. Notethat theonlyweights assignedby Algorithm 8 are1and2.In sucha weighted instance,we canadd anew variablex0andreplaceeachclausec ofweight2bythreenewclauses:c,c
∨
x0 andc∨ ¬
x0.Thus,ifc issatisfied,thenitwillcountinthenewinstanceasthreesatisfiedclauses.Otherwise,exactlyoneofthethreenewclauseswillbesatisfied. Thus,theso-builtinstanceof maxsatisequivalenttotheinitial maxweightedsat-instancebuiltby
Algorithm 8
.2
Theorem4.Algorithm8achievesapproximationratioρ
intimeO∗(
γ
(2ρ−1)n)
,whereO∗(
γ
n)
istherunningtimeofanexactalgo-rithmfor maxsat.
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 Xisuchthatx∈Xi|
C(
x)
|
p
q
|
OPT|
.Considernowsucha Xi,anddenoteby(seeFig. 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 variablein 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 areexactlythoseinB
∪
C .Withthesenotations,OPT=
A+∪
B1+∪
B2+∪
C+andx∈Xi|
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 thesubsetofsatisfiedclausesfromC .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+|
2So,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 usedcomingeitherfromthepolynomialapproximationorfromtheexactcomputation.Furthermore,
Algorithm 8
inSection4
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
numberofsatisfiedclauses.A
ρ
-approximationalgorithmfor minsat(withρ
>
1)isanalgorithmthatfindsanassignment satisfyingatmostρ
timestheminimalnumberofsimultaneouslysatisfiedclauses.In
[11]
an approximability-preservingreduction between minvertexcoverand minsatispresented transforminganyρ
-approximation for the former problem into aρ
-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
,ifthereexistsasolutionfor minsatsatisfyingatmostk clauses,itispossibletodeterminewithcomplexityO∗
(
1.
28rk)
a2−
r-approximationofit.
Wealsonotethatthemethodusedin
Algorithm 6
canbeappliedaswellto minsatwiththefollowingmodificationof thealgorithm.Letp,
q∈ Q
besuchthat p/
q=
2ρ
−
1.Buildq subsetsofvariables,eachonecontaining p/
q×
n variables.Foreachsubset,constructasearchtree,consideringonlythevariablesinthesubset(thedepthofthetreesisp
/
q×
n).For eachleafofanyoftheso-builttrees,usesomepolynomialalgorithmwithratioα
onthesurvivingsub-instance.Returnthe bestofthetruthassignmentscomputed.The complexityof themodificationjustdescribed isthe sameasthatof
Algorithm
6,i.e., O∗(
2n(α−ρ)/(α−1))
(thebest knownratioisα
=
2),andasimilaranalysisderivesanapproximationratioα
− (
α
−
1)
qp=
ρ
.References
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