CONDITIONAL COLORING by Mark R. Dillon B.S., Purdue ...

CONDITIONALCOLORING
MarkR.Dillon by
B.S.,PurdueUniversity,1982,1985 M.S.,PurdueUniversity,1984
UniversityofColoradoatDenver Athesissubmittedtothe
oftherequirementsforthedegreeof inpartialfulllment
AppliedMathematics DoctorofPhilosophy
1998

ThisthesisfortheDoctorofPhilosophy
MarkR.Dillon degreeby
hasbeenapproved by
KathrynL.Fraughnaugh
DavidC.Fisher
J.RichardLundgren
TomRussell
TomAltman
Date

Dillon,MarkR.(Ph.D.,AppliedMathematics) ConditionalColoring ThesisdirectedbyAssociateProfessorKathrynL.Fraughnaugh
Theconditionalchromaticnumber(G;P)ofagraphGwithrespecttoa ABSTRACT
graphicalpropertyPistheminimumnumberofcolorsneededtocolorthevertices
write(G;:F).Theconditionalchromaticnumberofagraphhasbeenstudied ofGsuchthateachcolorclassinducesasubgraphofGwithpropertyP.When
byvariousauthorssince1968.Wefocusontwoconditionalchromaticnumbers, PisthepropertythatagraphcontainsnosubgraphisomorphictoagraphF,we
specically(G;:Cj)and(G;:Pj),whereCjisacycleoflengthjforsomexed
2j?5edges.Toaccomplishthis,wecharacterizeallHamiltoniangraphsoforder jforgraphsmissingatmostj?1edgesand(G;:Pj)forgraphsmissingatmost 3andPjisapathoflengthj?1forsomexedj 2.Wend(G;:Cj)
atleastn2?(2n?5)edges.Wedeterminebothconditionalchromaticnumbers nwithatleastn2?(n?1)edgesandallgraphswithnoHamiltonianpathswith forallgraphswithacycliccomplements.Wealsodeterminealowerboundon if(G;:P3) (G;:Pj)intermsofthesizeofG.Finally,weshowtheproblemofdetermining
Thisabstractaccuratelyrepresentsthecontentofthecandidate'sthesis.Irec- k,forsomek 0,isNP-complete.
ommenditspublication. Signed KathrynL.Fraughnaugh
iii

DEDICATION
Tomywife,WendyTwetenDillon,whoselove,understanding,patience,and
encouragementmadeitpossibletocompletethisdissertation.

Fraughnaughforhertirelessguidance,patience,inspirationandencouragement Iwouldliketoexpressmydeepgratitudetomyadvisor,ProfessorKathryn ACKNOWLEDGEMENTS
tation.MysincerethankstoProfessorRichardLundgrenandProfessorDavidduringthecourseofmyresearchandthroughoutthepreparationofthisdisser- Fisherwhowerealwaystheretoanswerquestions. alwaysgavemethesupportIneededandmybrothersandsister,Steve,Dave, Scott,LauraandPaulDillonfortheirconstantencouragement. Iwouldalsoliketothankmymotherandfather,PatandRussellDillonwho
persontointerestmeintheeldofmathematicsandmyclosefriendsDoug Mckissack,ArthurShulmanandKevinWilliamsfortheirconstantencouragement. IalsothankMr.JohnDrubert,mysixthgradeteacher,whowastherst
pleted.AircraftCompanywithoutwhosesupport,thisthesiswouldneverhavebeencom-
Finally,Ireceived,andgratefullyacknowledgenancialassistancefromHughes

Contents 1Introductiontocoloringanditsapplications 1.1OverviewofThesisResults...................... 1
1.2DenitionsandNotation....................... 54
2BasicResultsfor(G;:Pj)and(G;:Cj) 1.3ConditionalColoring......................... 169
3NP-complete 4Determiningtheconditionalchromaticnumberforgraphswith22
acycliccomplements 4.1Determining(G;:Cj)whenGisacyclic............. 29
4.2Determining(G;:Pj)whenGisacyclic............. 35 29
4.3Thedierencebetweenthe:Cj-and:Pj-chromaticnumbersofa
5Determining(G;:Cj) graph.................................. 39 37
5.1Preliminaries............................. 39
vi

6Determining(G;:Pj) 5.2Determining(G;:Cj)whene(G) n2?(j?1)........ 58 46
6.1DeterminingwhichgraphsoflargesizehaveaHamiltonianpath. 6.2Graphsmissingcompletesubgraphsorastar............ 73 58
6.3Determining(G;:Pj)whene(G)islarge............. 6.4Determiningboundson(G;:Pj)giventhenumberofedgesina 75
AAppendix graph.................................. 84 79
BReferences 95
vii

1LetGbeagraph.AvertexcoloringofGisanassignmentofcolorstoitsvertices Introductiontocoloringanditsapplications
sothatnotwoadjacentverticesreceivethesamecolor.Thechromaticnumber ofthechromaticnumberistheminimumintegerksuchthatthereisapartitionof theverticesintoksetssothatthesubgraphinducedbyeachsetisanindependent (G)istheminimumnumberofcolorsneededtocolorG.Anequivalentdenition
following.IntheUnitedStatesGovernment,therearecongressionalcommittees set.Vertexcoloringhasawidevarietyofapplications.Onesuchapplicationisthe
ingtimesforthesecommittees,onemustnotschedulesimultaneousmeetingsfortwocommitteesthathaveacommonmember.AschedulesolutionisfoundbydewithmembersofCongressservingonmultiplecommittees.Whenassigningmeet- committees.Wedrawanedgebetweentwoverticesifthecommitteesrepresented WecanmodelthisproblembyconstructingagraphGwhoseverticesrepresent terminingtheminimumnumberoftimeslotsrequiredforthecommitteestomeet.
bytheseverticeshaveacommonmember.Determiningtheminimumnumberof Forfurtherdetails,seeRoberts[47]orBodinandFriedman[8]. timeslotsrequiredforthecommitteestomeetisequivalenttodetermining(G).
geographicallyclose,say50miles,receivedierentfrequencyassignments.The regionaretobeassignedtransmittingfrequenciessothatradiostationswhichare Anotherapplicationisthechannelassignmentproblem.Radiostationsina
problemofassigningfrequenciesisagraphcoloringproblem.Leteachvertex 1

epresentaradiostationanddrawanedgebetweentwoverticesiftheradiostationsrepresentedbythoseverticesarewithin50miles.Thenumberoffrequenciessignmentproblem,seeCozzensandRoberts[20],Hale[28],OpsutandRobertschromaticnumberofthisgraph.Formoreinformationregardingthechannelas- requiredsothattheradiostationsdonotinterferewitheachotheristhevertex
[43],orPennotti[46].
textbooks.Forexample,seeBondyandMurty[10],Harary[29],Chartrandand problem.Thisisamuchstudiedproblemandisdiscussedinmostgraphtheory Thelastapplicationofvertexcoloringwewilldiscussistheclassicmapcoloring
Lesniak[17],orRoberts[47].Givenamapwithvariouscountries,wewouldlike tocolorthecountriesinsuchawaythatweusethefewestnumberofcolors,andif twocountriesshareacommonborder,theyreceiveadierentcolor.Thisproblem canbetranslatedintoagraphcoloringproblembybuildingagraphwitheach
minimumnumberofcolorsrequiredtocolorthemap,wend(G).Thisgraph countriesrepresentedbythoseverticeshaveacommonborder.Todeterminethe vertexrepresentingacountryanddrawinganedgebetweentwoverticesifthe
graphscanbecoloredusingatmostfourcolors. hasthepropertythatitisplanar.TheFourColorTheoremstatesthatallplanar
classandallowcertaincongurationsofedges.Forexample,inthecommittee schedulingapplication,wecouldalloweachcommitteetohaveatimeconictwith Now,supposewerelaxtheconditionthattherebenoedgesineachcolor
atmostoneothercommitteewithacommonmember.Inthiscase,eachcolor committeeschedulingproblem. classcancontainisolatedverticesandedges.Wewillcallthisproblemtherelaxed
2

graphcoloring.Wedenetheconditionalchromaticnumber(G;P)ofGwith respecttoagraphtheoreticalpropertyPtobetheminimumintegerksuchthat Thisisanexampleofageneralizationforgraphcoloringcalledconditional
eachsethasthepropertyP. thereisapartitionoftheverticesintoksetssothatthesubgraphinducedby
lem.Adesignerdrawsanelectricalcircuittobemanufactured.Severalcircuit boardssandwichedoneontopoftheothermayberequiredtobuildtheentire Anotherapplicationforconditionalcoloringisthecircuitmanufacturingprob-
circuitsince,forthiscircuittoworkproperly,eachcircuitboardmustbebuilt withoutintersectingedges.Now,themanufacturerwouldliketodeterminethe
tioninthecircuit.Drawanedgebetweentwoverticesinthegraphifthejunctions minimumnumberofrequiredcircuitboards.Theproblemcanbemodeledasa conditionalgraphcoloringproblem.Leteachvertexinourgraphrepresentajunc- representedbythoseverticesareconnectedinthecircuitdesign.Determiningthe
i.e.,isplanar.Thisparticularconditionalchromaticnumberisalsoknownasthe wherePisthepropertythateachcolorclassbedrawnwithnointersectingedges, minimumnumberofcircuitboardsrequiredisequivalenttodetermining(G;P),
BeinekeandWhite[7]orCimikowski[18].Forotherelectricalcircuitproblems, seeHutchinson[37]orGareyandJohnson[26]. vertexthicknessofagraph.Formoreinformationregardingvertexthickness,see
nopathsoforderjinacolorclass,andthesecondpermitsnocyclesoforder jinacolorclass.Therelaxedcommitteeschedulingproblemisanexampleof Wewillstudytwotypesofconditionalchromaticnumbers.Therstpermits
conditionalcoloringwiththepropertythatnocolorclasscontainsP3.Studying 3

thesetwonumbersmaygiveinsightintoand/orboundsonotherconditional understandingofageneralizationleadstoabetterunderstandingoftheoriginal chromaticnumbersortheoriginalchromaticnumber.Itisoftenthecasethatthe concept. 1.1 Thisthesiswillpresentsomenewresultsabouttheconditionalchromaticnumber OverviewofThesisResults
tothepropertyofhavingnopathsofaxedlengthj?1. somenewresultsabouttheconditionalchromaticnumber(G;:Pj)withrespect (G;:Cj)withrespecttothepropertyofhavingnocyclesofaxedlengthjand
reviewtheexistingliteratureregarding(G;:Cj)and(G;:Pj). Therstchapterwillprovidesomebasicdenitionsfromgraphtheoryand
and(G;:Pj).Someoftheseresultswillbeneededinlaterchapters. Thesecondchapterwillpresentandprovesomebasicresultsregarding(G;:Cj)
:P3-coloredusingkcolorsisNP-complete. Thethirdchapterwillshowthattheproblemofdeterminingifagraphcanbe
complement,whatis(G;:Cj)and(G;:Pj)?Aconstructionwhichillustrates that(G;:Pj)?(G;:Cj) Thefourthchapterwillanswerthequestion:givenanygraphwithacyclic
andFraughnaugh[21]characterized(G;:Cj)whenagraphismissingatmost Thefthchapterwilldetermine(G;:Cj)forallgraphsoflargesize.Dargen aforanypositiveintegeraisalsoprovided.
j?2edges.Inthischapter,weextendthisresulttographsmissingatmostj?1 edges.Toaccomplishthis,wecharacterizeallHamiltoniangraphsofordernwith atleastn2?(n?1)edges.
4

determine(G;:Pj)forallgraphsoflargesize,wewillcharacterizeallgraphsof determineanupperboundonthesizeofGgivenaboundfor(G;:Pj).To Thesixthchapterwilldetermine(G;:Pj)forallgraphsoflargesizeand
1.2 ordernwithnoHamiltonianpathshavingatleastn2?(2n?5)edges.
AgraphGconsistsofanitenonemptysetV=V(G)ofverticesandacollection DenitionsandNotation
orderornumberofverticesofGisdenotedbyn=n(G).Agraphissimpleif Gbeagraph.ThesizeornumberofedgesofGisdenotedbye=e(G),andthe E=E(G)ofdistinctpairsofvertices,callededges.Throughoutthispaper,let
thereisatmostoneedgebetweenanydistinctpairofverticesandthereareno
andvneighbors.TheopenneighborhoodN(u)ofavertexuisthesetofneighbors loops.Forthispaper,wewillassumethatallgraphsaresimple.
ofu,andtheclosedneighborhoodN[u]ofuisN(u)[fug.ThedegreedG(v)ofa IfuandvareverticesofG,wewritetheedgejoininguandvasuvandcallu
vertexisdenedtobethenumberofedgesinGincidentwiththatvertex.When itcleartowhichgraphwearereferring,wemaywrited(v).Theminimumdegree
allverticeshavethesamedegreeandGisregularofdegreerorr-regular.A (G)istheminimumdegreeamongallverticesofGwhilethemaximumdegree
3-regulargraphisacubicgraph.Ifavertexisnotincidenttoanyedge,thenthis (G)isthemaximumdegreeamongallverticesofG.If(G)=(G)=r,then
vertexisanisolatedvertex. wewriteH AsubgraphHofGisagraphhavingallofitsverticesandedgesinG,and G.ForanysetSofverticesinG,theinducedsubgraphhSiisthe
5

maximalsubgraphofGwithvertexsetS.IfSisanonemptysetofedgesinG, thenhSiisthesubgraphwhoseedgesetisSandwhosevertexsetisthesetof endsofedgesinS.Agraphiscompleteifeverypairofverticesisjoinedbyan edge.ThecompletegraphonnverticesisKn.Thegraphonnverticeswithno edgesisIn. apathPn.Wesaytheverticesv1andvnareconnectedbythepathPn.We sometimescallthispatha(v1;vn)-path.Thelengthofapathisthenumberof Thegraphwithdistinctverticesv1;:::;vnandedgesv1v2;v2v3;:::;vn?1vnis
edgesinit.Thedistanced(u;v)fromutovinGisthelengthofashortestpath fromutov.ThediameterofGisthemaximumdistancebetweenanytwovertices cycle.ThegraphwhichconsistsofacycleonjverticesisCj,andagraphwhich ofG.Ifu=vandthereareatleastthreeverticesonthepath,thenthispathisa containsnocyclesisacyclic.AgraphGisHamiltonianifithasacyclecontaining alltheverticesofG.AgraphGhasaHamiltonianpathifithasapathcontaining everyvertexofG.AgraphGisHamiltonianconnectedifforeverypairuandv ofdistinctverticesofG,thereexistsaHamiltonian(u,v)-path. tov.AmaximalconnectedsubgraphofGiscalledacomponentofG.IfGis notconnected,thenwesayGisdisconnected.Theconnectivity(G)ofGisthe Agraphisconnectedifforanytwoverticesuandv,thereisapathfromu
graphisatree,andaforestisagrapheachofwhosecomponentsisatree. graph.IftheconnectivityofGis,wesayGis-connected.Aconnectedacyclic minimumnumberofverticeswhoseremovalresultsinadisconnectedortrivial
vertex.Oncewechoosearootu,thelevelofavertexvisd(u;v).Therootisthe InatreeT,wecanmakeTarootedtreebydesignatinganyvertexastheroot
6

onlyvertexatlevel0.Further,alladjacentverticesdierbyexactlyonelevel, maximumlevelistheheighthu(T)ofthetree. andeachvertexatleveli+1isadjacenttoexactlyonevertexatleveli.The
adjacentinGifandonlyiftheyarenotadjacentinG. ThecomplementGofGalsohasV(G)asitsvertexset,buttwoverticesare
themaximumorderamongallcliquesofG.AnindependentsetofverticesofG isasetI AcliqueofGisacompletesubgraphofG.Thecliquenumber!(G)ofGis
hasnotwoofitsedgesincident,andsuchasetisamatching. Next,wewilldiscusssomeoperationsdenedongraphs.LetG1andG2
Vsuchthatxy=2Eforallx;y2I.AnindependentsetofedgesofG
betwographswithdisjointvertexsets.TheunionG1+G2ofG1andG2has mGisthepairwisevertexdisjointunionofmcopiesofG.ThejoinG1_G2 V(G1+G2)=V(G1)[V(G2)andE(G1+G2)=E(G1)[E(G2).Ingeneral, oftwographsG1andG2hasV(G1_G2)=V(G1)[V(G2)andE(G1_G2)= E(G1)[E(G2)[fuv:u2V(G1)andv2V(G2)g. f(u;v):u2V(G1)andv2V(G2)g,and(u1;v1)and(u2;v2)areadjacentwhenevereitheru1u22E(G1)andv1=v2,oru1=u2andv1v22E(G2).Thestrong ThecartesianproductG1 G2oftwographsG1andG2hasV(G1 G2)=
productG1G2oftwographsG1andG2hasV(G1G2)=f(u;v):u2V(G1)and v2V(G2)g,and(u1;v1)and(u2;v2)areadjacentwhenevereitheru1u22E(G1) andv1=v2,oru1=u2andv1v22E(G2),oru1u22E(G1)andv1v22E(G2). TheconjunctionG1^G2oftwographsG1andG2hasV(G1^G2)=f(u;v): u2V(G1)andv2V(G2)g,and(u1;v1)and(u2;v2)areadjacentwhenever u1u22E(G1)andv1v22E(G2). 7

twoadjacentverticesreceivedierentlabels.Wethinkofeachdistinctlabelasa colorandcalleachsetofverticesassignedaxedcoloracolorclass.Thevertex AvertexcoloringisanassignmentoflabelstotheverticesofGsothatany
chromaticnumber(G)ofGistheminimumkforwhichGhasak-coloring,i.e., oneusingkcolors.Agraphisk-colorableifwecancoloritusingkcolors.
ofagraphisdenedtobetheminimumnumberofcolorsneededtoedgecolor thatanytwoincidentedgesreceiveadierentlabel.Theedgechromaticnumber Anedgecoloringisanassignmentoflabels(orcolors)totheedgesofGso
G.Ifweusetheterm\chromaticnumber"or\coloring"inthispaper,wemean vertexchromaticnumberorvertexcoloring.
planegraphhasalreadybeenembeddedintheplane.Wewillrefertotheregions notwoedgesintersect.Agraphisplanarifitcanbeembeddedintheplane;a AgraphissaidtobeembeddedinasurfaceSwhenitisdrawnonSsothat
planargraphis4-colorable.Foraproofofthisresult,seeAppelandHaken[5]. denedbyaplanegraphasitsfaces.TheFourColorTheoremstatesthatevery
subsetsV1andV2suchthateveryedgeofGjoinsV1andV2.Thebipartition numberb(G)ofagraphGisgivenbyb(G)=maxfe(B):B AbipartitegraphGisagraphwhosevertexsetVcanbepartitionedintotwo
bipartiteg. Forotherbasicgraphtheorydenitionsandterminology,thereaderisreferred GandBis
toHarary[29].
8

Anequivalentdenitionofvertexcoloringisapartitionofthevertexsetsothat 1.3 ConditionalColoring
thesubgraphinducedbyeachsetofthispartitionisanindependentset.Stated dierently,eachcolorclasscontainsnopathontwovertices. colorings.LetPbeanygraphtheoreticproperty.Forexample,Pcouldbethe propertythatagraphcontainsacliqueofacertainorder,thatagraphdoesnot Thisledseveralauthors[15],[16],[35]toamoregeneralconceptofvertex
containacycleoforder6,thatagraphdoesnotcontainaninducedcycleoforder 6,orthemaximumdegreeofagraphis5.
conditionalchromaticnumber(G;P)ofG(orbrieyP-chromaticnumber)isthesothatthesubgraphinducedbyeachcolorclasssatisesthepropertyP.TheP- WedeneaP-coloringofagraphtobeanassignmentofcolorstoitsvertices
minimumkforwhichGhasaP-coloringwithkcolors.WhenPistheproperty usualchromaticnumber. thatagraphconsistsentirelyofisolatedvertices,theP-chromaticnumberisthe
induced)isomorphictoagraphF,wewrite(G;:F)fortheP-chromaticnumber andrefertoaP-coloringasa:F-coloringandtheP-chromaticnumberasthe WhenPisthepropertythatagraphcontainsnosubgraph(notnecessarily
Pisthepropertythatagraphcontainsnoinducedsubgraphisomorphictoagraph :F-chromaticnumber.If(G;:F) F,wewrite(G;:F!)fortheP-chromaticnumberandrefertoaP-coloringasa k,thenwesayGis:Fk-colorable.When
:F!-coloringandtheP-chromaticnumberasthe:F!-chromaticnumber. inducedbytheverticesineachsetinthepartitionhasthepropertyPseemsto Theconceptofpartitioningthevertexsetofagraphsothatthesubgraph
9

havebeenindependentlydiscoveredbyseveralauthorsaround1968(see[15],[16], [42]).Mathematicianshavestudiedvariousconditionalchromaticnumberssince [35])andagainbyseveralauthorsaround1985(see[3],[4],[12],[14],[23],[30],
1968.In1968,Chartrand,Kronk,andWall[16]studiedaconditionalcoloring numbercalledthearboricityofagraphwherecolorclassesareacyclic.Hedetniemi [35]provedthatthearboricityofaplanargraphisatmostthree. (G;:Pj).Forexample,ifthediameterofGisd,then(G;:Pj) In1968,Chartrand,GellerandHedetniemi[15]provedseveralresultsabout
suchthat(G;:Pj)=4.Thisprovedthatthereisnostrongerresultthanthe forj 2.Further,foreveryj 2,theauthorsconstructedaplanargraphG d?j+3
FourColorTheoremfor(G;:Pj)forj [48]provedthatgivenj 2andK k2.Also,in1968,SachsandSchauble isomorphictoKj?1. (G;:Kj)=kanda:KjK-coloringofGcontainingatleastkcolorclasses 1,thereexistsagraphGwith
thatthesubgraphinducedbyeachcolorclasshasminimumdegreejforsome xedj In1969,KramerandKramer[39]discussed(G;P),wherePistheproperty
jsameconditionalchromaticnumber.In1975,Cook[19]studied(G;:K1;j)for 1. 0.In1970,LickandWhite[41]publishedapaperwithresultsforthe
Lesniak-FosterandStraight[40]publishedresultsfor(G;P),wherePisthe propertythateachcolorclassinducesacompletegraphoragraphwithnoedges. In1977,HararyandKainen[34]discussed(G;:K3)forplanargraphs.Also,
Sampathkumar,Prabha,Neeralagi,andVenkatachalam[50]publishedresultsfor 10

ofthicknessisequivalentto(G;P),wherePisthepropertythateachcolor (G;:Pj)forj In1978,BeinekeandWhite[7]discussedthethicknessofagraph.Theconcept 2.
classisplanar.Theauthorsdeterminedthethicknessofcompleteandcomplete bipartitegraphs.
bersforseveraldierentpropertiesandhasbecomethestandardforconditionalcoloring.Thispaperdiscussedtheconditionalvertexandedgechromaticnum- In1985,Harary[30]publishedapaperprovidinganoverviewofconditional
vertexdegreeineachcolorclassisatmostt.Theauthorsrelated(G;t) coloringterminology.
to(G)byprovingthat(G;t) In1985,AndrewsandJacobson[3]studied(G;t),wherethemaximum
andFraughnaugh[31]alsopublishedresultsfor(G;t)in1985. (G;t) n2=(tn+n2?2e),wherenandearetheorderandsizeofG.Harary (G)=(t+1).Furthermore,theyshowed
respecttothepropertythateachcolorclassisadisjointunionofcompletesubgraphs.Theyalsostudiedtheproblemofndingagraphsubjecttocertainrestric Alsoin1985,MynhardtandBroere[42]discussedconditionalcoloringwith
tionsforwhichtheconditionalchromaticnumberisarbitrarilylarge.Mynhardt andBroere[12]alsostudiedconditionalcoloringwithrespecttothepropertythat eachcolorclasshasnoinducedsubgraphisomorphictoagraphF. numbertothatofaconditionalbipartitionnumber.Specically,givenaproperty P,agraphisconditionallybipartitewithrespecttoPifV(G)isthedisjointunion In1986,HararyandFraughnaugh[32]generalizedtheconceptofbipartition
ofsetsXandYwheretheinducedsubgraphshXiandhYibothhaveproperty 11

P.Theconditionalbipartitionnumberb(G;P)ismaxfe(B):B conditionallybipartitewithrespecttoPg.Theauthorsstudiedb(G;P)forseveral minimumandmaximumdegreeproperties. GandBis
wherePisthepropertythatthesubgraphinducedbyeachcolorclassisacomplete r-partitegraphforanyrand(G;Q),whereQisthepropertythateachcolor Alsoin1986,Domke,Laskar,Hedetniemi,andPeters[23]discussed(G;P),
classisadisjointunionofcompletesubgraphs.
thepropertythateachcolorclasshasnoinducedsubgraphisomorphictoasetof Alsoin1987,BrownandCorneil[14]studiedconditionalcoloringwithrespectto In1987,AndrewsandJacobson[4]publishedapaperwithresultsfor(G;t).
maticnumber.Thek-pathchromaticnumber(G;Pk)ofGisthesmallestnumber graphs.
cofdistinctcolorswithwhichV(G)canbecoloredsuchthateachconnectedcomIn1989,Akiyama,Era,Gervacio,andWatanabe[1]discussedthek-pathchro- equivalenttoour:P3-chromaticnumber,weimmediatelygetthat(G;:P3) ponentofhViiisapathoforderatmostk,1 (G;Pk) lr+1 2mforr-regulargraphs.Sincethe2-pathchromaticnumberis i c.Theauthorsprovedthat
forcubicgraphs.Alsoin1989,Albertson,Jamison,Hedetniemi,andLocke[2] discussedconditionalcoloringwithrespecttothepropertythateachcolorclass 2
isadisjointunionofcompletesubgraphs.In1990,Baldi[6]publishedresultsfor
maticnumberoftheCartesianproduct,join,strongproduct,andconjunctionof (G;:Pj)forj In1991,HararyandHsu[33]providedtheoremsrelatingtheconditionalchro- 2.
twographstotheconditionalchromaticnumberoftheoriginalpairofgraphs. 12

oftheverticesofaplanargraphisk-cyclicifwhenevertwoverticeslieinthe boundaryofthesamefaceofsizeatmostk,theircolorsaredierent. In1992,Borodin[11]discussedthek-cyclicchromaticnumber.Acoloring
classisnotpermittedtocontainsomeinducedsubgraph.Forexample,Brownand Corneil[13],discusseduniquely:H!k-colorablegraphswhereHisanygraph.A Therearealsoconditionalcoloringpaperspublishedin1992whereacolor
graphGisuniquelyk-colorableifGisk-colorableandthereisonlyonek-coloring (uptoapermutationofcolors).Infact,BrownandCorneilconjecturedthat forallgraphsoforderatleasttwoandforallnonnegativeintegersk,thereexist is2-connectedorGis2-connected. uniquely:H!k-colorablegraphs.Sofar,theyhaveshownthisresultwheneverG
integersl In1992,JohnsandSaba[38]considered(G;:Pj).Theyprovedthatgiven
Kj(j+1)l. K(j?1)l.Also,thereexistsagraphsuchthat(G;:Pj)?(G;:Pj+1)=l,namely 1andj 2,thereexistsagraphGsuchthat(G;:Pj)=l,namely
inguptoj?2edgesandSampathkumar[49]discussed(G;P),wherePisthe propertythatthesubgraphinducedbyeachcolorclasshasindependencejfor In1993,DargenandFraughnaugh[21]determined(G;:Cj)forgraphsmiss-
somexedj printedcircuitboardsforelectricalshorts. In1995,Cimikowski[18]presentedsomeheuristicsforthegraphthickness 0.Alsoin1993,Hutchinson[37]appliedthicknessresultstotesting
subgraphofanonplanargraph.HealsoprovedthatT(G) problem,i.e.,decomposingagraphintotheminimumnumberofplanarsubgraphs.Theheuristicsarebasedonsomealgorithmsforndingamaximalplanar 13
bq2e=3+3=2c,

whereT(G)isthethicknessofGandeisthesizeofG. werestudied,seeTable1.1. Forasurveyofdierentpropertiesstudied,whostudiedthem,andwhenthey
disconnectedortrivial ColorClassProperty Table1.1.ConditionalColoringPropertiesstudied
acyclic Year 1968Hedetniemi[35] 1968Chartrand,Kronk,Wall[16] 1970Hedetniemi[36] Reference
hasnoK3 hasnoKjforsomexedj hasnoPjforsomexedj 1968Sachs,Schauble[48] 1968Chartrand,Geller,Hedetniemi[15] 1968Hedetniemi[35] 1977Harary,Kainen[34] 1977Sampathkumar,Prabha,Neeralagi,
completeoragraphwithoutedges1977Lesniak-Foster,Straight[40] 1990Baldi[6] 1992Johns,Saba[38] Venkatachalam[50]
hasmaximumdegreejforsome xedj 1985Harary,Fraughnaugh[31]
completer-partitegraphforanyr 1985Andrews,Jacobson[3] 1986Harary,Fraughnaugh[32]
hasnoinducedsubgraph 1986Domke,Laskar,Hedetniemi,Peters 1987Andrews,Jacobson[4]
isomorphictographF 1985Broere,Mynhardt[12] 1985Mynhardt,Broere[42] 1987Brown,Corneil[14] [23]
graphsF phictoanygraphFinasetofhasnoinducedsubgraphisomor- disjointunionofcomplete 1985Mynhardt,Broere[42] 1987Brown,Corneil[14]
subgraphs 1986Domke,Laskar,Hedetniemi,Peters
containsnoK1;j 1989Albertson,Jamison,Hedetniemi, 1975Cook[19] Locke[2] [23]
minimumdegree jforxedj 1969Kramer,Kramer[39] 1970Lick,White[41] 14

disjointunionofpathsoforderat ColorClassProperty Table1.1.ConditionalColoringPropertiesstudied(cont.)
jindependentforxedj mostk 1993Sampathkumar[49] Year 1989Akiyama,Era,Gervacio,Watanabe [1] Reference
hasnoCjforxedj planar 1993Dargen,Fraughnaugh[21] 1978Beineke,White[7] 1993Hutchinson[37]
phictoH k-cyclic containsnoinducedsubgraphisomor- 1992Borodin[11] 1995Cimikowski[18]
surveyofproperties 1985Harary[30] 1992Brown,Corneil[13]
orderj (G;:Cj)whereCjisacycleoforderj Inthisthesis,weconcentrateprimarilyontwoconditionalchromaticnumbers, 2. 3and(G;:Pj)wherePjisapathof
15

2ThischapterprovidesthenecessarybackgroundforthetopicspresentedinChap BasicResultsfor (G;:Pj)and (G;:Cj)
ters3,4,5,and6.Westartbyprovingsomebasicrelationshipsfor(G;:Cj) [15]. and(G;:Pj).Thefollowingstraightforwardresulthasbeenknownsince1968
Proof.Letn=a(j?1)+r,where0 Theorem2.1LetGbeagraphofordern.Ifj r0.Hence(G;:Pj) ln j?1m.
Proof.SinceacompletesubgraphoforderatleastjcontainsPj,eachcolorclass 2,then(Kn;:Pj)=ln j?1m:
equalityfollowsfromTheorem2.1. ofKncancontainatmostj?1vertices.Therefore,(Kn;:Pj) ln j?1mand
Thenexttheoremprovidesthemostbasicrelationshipbetween(G;:Cj)and 2
Theorem2.2Ifj (G;:Pj).
Proof.LetCbeanyminimum:Pj-coloringofG.Sinceacolorclassthat 3andGisagraph,then(G;:Cj) (G;:Pj):
containsnoPjcontainsnoCj,thecoloringCisalsoa:Cj-coloringofG.So, (G;:Cj) (G;:Pj): 16
2

(G;:Cj)and(G;:Pj),wecanndanupperboundfor(G;:Cj). Nowthatwehaveanupperboundfor(G;:Pj)andarelationshipbetween
Corollary2.2Ifj Further,(Kn;:Cj)=ln 3andGisagraphofordern,then(G;:Cj) j?1m: ln j?1m:
ln Proof.ByTheorem2.2andTheorem2.1,weget(G;:Cj) andthereforecannotbeacolorclassofKn.Hence,(Kn;:Cj) j?1m.LetG=Kn.AsubgraphinducedbyjormoreverticesofKncontainsCj, ln(G;:Pj)
Thenexttheoremrelatestwodierentconditionalcoloringnumbers. j?1m. 2
Theorem2.3Ifj Proof.LetCbeanyminimum:Pk-coloringofG.Sinceacolorclassthat k 2,then(G;:Pj) (G;:Pk):
containsnoPkcontainsnoPjfork G.Therefore,(G;:Pj) (G;:Pk): j,thecoloringCisalsoa:Pj-coloringof
Onemightaskiftheabovetypeoftheoremistruefor(G;:Cj).Infact,there 2
onegraph.Further,(C3;:C4)=1and(C3;:C3)=2,whichimpliesthat isnorelationshipingeneralineitherdirection.Forexample,(C4;:C4)=2 and(C4;:C3)=1,whichimpliesthat(G;:C4)6 (G;:C4)6 (G;:C3)foratleastonegraph. (G;:C3)foratleast
numbersofagraphanditssubgraphs. Thenexttheoremprovidesarelationshipbetweentheconditionalchromatic
Theorem2.4LetGbeagraphwithH (G;:Pj).Ifj 3,then(H;:Cj) (G;:Cj): G.Ifj 2,then(H;:Pj)
17

Proof.LetCbeanyminimum:Cj(:Pj)-coloringofG.SinceH coloringCisalsoa:Cj(:Pj)-coloringofH.So,(H;:Cj) (H;:Pj) (G;:Pj): (G;:Cj)and G,the
Nowthatwehaveresultsforsubgraphsfor(G;:Cj)and(G;:Pj),wecan 2
Theorem2.5IfGisagraphandj deriveanelementarylowerboundfor(G;:Pj)and(G;:Cj).
Proof.Firstofall,K!(G) G.ByCorollary2.2,(K!(G);:Cj)=l!(G) 3,then(G;:Pj) (G;:Cj) l!(G) j?1m.By j?1m.
Theorem2.4,weget(G;:Cj) have(G;:Pj) (G;:Cj): (K!(G);:Cj)=l!(G) j?1m.ByTheorem2.2,we
ThisboundisattainedwhenG=Kn,andthereforetheboundistight.In 2
ordertoderivesomeupperboundsontheorderofacolorclassofsomegraph G,weneedsomewellknownresultsregardingHamiltoniangraphs,Hamiltonian connectedgraphs,andgraphswhichhaveaHamiltonianpath.First,westate Ore'sTheorem,Dirac'sTheorem,andanothertheoremwhichappearsasCorollary
Theorem2.6(Ore[44])IfGisagraphofordern 4.6inBondyandMurty[10].
nonadjacentverticesuandv,d(u)+d(v) n,thenGisHamiltonian. 3suchthatforalldistinct
Theorem2.7(Dirac[22])IfGisagraphofordern degreeatleastn2,thenGisHamiltonian. 3andeachvertexhas
Theorem2.8(Ore[45],Bondy[9])IfGisagraphofordern nverticesandn2?3n+4 n2?3n+6 2 ,thenGisHamiltonian.Moreover,theonlynon-Hamiltoniangraphswith 3ande(G)
2 edgesareK1_(K1+Kn?2)andK2_I3. 18

exerciseinWest[51]andpart(ii)issimplyarestatementoftherstsentencein Theorem2.8. Thefollowingtheoremisalsowellknown.Parts(i)and(iii)appearasan
Theorem2.9LetGbeagraphofordern. Ifn 2ande(G) 3ande(G) n2?(n?4),thenGisHamiltonianconnected: n2?(n?3),thenGisHamiltonian. (ii) (i)
Proof.(i).(byinductiononn).Ifn Ifn 2ande(G) n2?(n?2),thenGhasaHamiltonianpath: 3,then(i)isvacuouslytrue.If (iii)
n=4,thenG=K4andacompletegraphisHamiltonianconnected.Ifn=5, thenG2fK5;K5?egwhereeisanedge.ThesetwographsareHamiltonian connected.Ifn=6,thenG2fK6;K6?e;K6?E(2K2);K6?E(P3)g=Gwhere appropriategraphsonn?1vertices. eisanedge.EachG2GisHamiltonianconnected.Soassume(i)holdsfor
GisHamiltonianconnected.Letfu;vg LetGbeagraphonn 7verticesande(G) V(G)withu6=v. n2?(n?4).Wewillshow
n?1 2?(n?5)and,bytheinductionhypothesis,G?uisHamiltonianconnected. Ifd(u) n?2,thene(G?u) e(G)?(n?2) n2?(n?4)?(n?2)= Nowe(G) leastn?3edgesfromacompletegraphinorderfor(G)=2.Sinced(u) n2?(n?4)impliesthat(G) wecanchoosez2N(u)?fvgandadduztoaHamiltonian(z,v)-pathinG?u 3forwewouldneedtoremoveat
toformaHamiltonian(u,v)-pathinG. 3,
e(G)?(n?1) IfuisadjacenttoeveryothervertexinG(i.e.,d(u)=n?1),thene(G?u)= n2?(n?4)?(n?1)=n?1 19
2?(n?4)and,byTheorem2.8,

G?uisHamiltonian.Breakanedgeinvolvingv(sayvw)ontheHamiltoniancycle
(ii).ForaproofofOre'sTheorem,seeRoberts[47]. case,GhasaHamiltonian(u,v)-path.Therefore,GisHamiltonianconnected. inG?uandaddtheedgewutoobtainaHamiltonian(u,v)-pathinG.Ineither
Sincee(G) (iii).IfGiscomplete,thenGisHamiltonian.So,assumeGisnotcomplete. ConsiderG+uv.Now,e(G+uv)=e(G)+1 n2?(n?2),wegetn G+ehasaHamiltoniancycle.Therefore,GhasaHamiltonianpath. 3.LetuvbeanymissingedgeofG. n2?(n?3).By(ii),thegraph
Wecanseethatallthreestatementsintheprevioustheoremareexactsince 2
K2_(K1+Kn?3)hasn2?(n?3)edgesandisnotHamiltonianconnected,
hasn2?(n?1)edgesandcontainsnoHamiltonianpath.Lastly,westatea K1_(K1+Kn?2)andK2_I3haven2?(n?2)edgesandarenotHamiltonian(in
sucientconditiononthesizeofagraphtoobtaincyclesofallorders. fact,Theorem2.8pointsoutthatthesearetheonlysuchgraphs),andK1+Kn?1
Theorem2.10IfGisagraphwithordern Ck Gforallk=3;4:::;n. 3ande(G) n2?(n?3),then
Proof.(byinductiononn).Ifn=3,thenG=K3andtheresultfollows immediately.Assumethatn n?1 2?(n?4)edgescontainsCjforj=3;4;:::;n?1. LetGbeagraphofordernwithe(G) 4andthatanygraphofordern?1withatleast
wegetCn G.IfGiscomplete,theresultfollowsimmediately.IfGisnot n2?(n?3).ByTheorem2.9, complete,then(G) G.ConsiderG?v.Thene(G?v)=e(G)?(G) n?2.Chooseavertexv2V(G)ofminimumdegreein
20
n2?(n?3)?(n?2)=

subgraphs.Therefore,GcontainsthesameC3;:::;Cn?1assubgraphs. n?1 2?(n?4).Bytheinductionhypothesis,G?vcontainsC3;C4;:::;Cn?1as
Nowthatwehavesomebasicresults,wediscussthecomputationalcomplexity 2
ofthefollowingproblem:GivenagraphGandapositiveintegerk,canGbe NP-complete. :P3-coloredusingkcolors?Inthenextchapter,weshowthatthisproblemis
21

3Apolynomialtimealgorithmisanalgorithmwhoseworst-caserunningtimeis NP-complete
O(nk),wheretheinputtothealgorithmisofcardinalitynandkissomecon- instancesandasetSofproblemsolutions.Forexample,considertheproblem SHORTEST-PATHofndingashortestpathbetweentwogivenverticesinG. stant.Aproblem isdenedtobeabinaryrelationonasetIofproblem
tices.AsolutionisasequenceofverticesinthegraphGwithperhapstheemptyAninstanceforSHORTEST-PATHisatripleconsistingofagraphandtwover- sequencedenotingthatnopathexists.TheproblemSHORTEST-PATHitselfisa relationthatassociateseachinstanceofagraphandtwoverticeswithashortest pathinthegraphthatconnectsthetwovertices.Forsimplicity,thetheoryofNP- Inthiscase,wecanviewanabstractdecisionproblemasafunctionthatmapsthe instancesetItothesolutionsetf0;1g.Forexample,adecisionproblemrelated completenessrestrictsitselftodecisionproblems,thosehavingayes/nosolution.
toSHORTEST-PATHisasfollows:GivenagraphG,twoverticesfu;vg andapositiveintegerk,doesthereexistapathoflengthatmostk? Therearedecisionproblemswhichcanbesolvedinpolynomialtimeandthose V(G),
whichcanbesolvedusingadeterministicpolynomialtimealgorithm. whichrequiresuperpolynomialtime.Apolynomialtimesolvableproblemisone
completenessisthatofapolynomialtransformation.Let1and2denotetwowecanverifythissolutioninpolynomialtime.AbasicideainthetheoryofNP- ThecomplexityclassNPistheclassofproblemswheregivena\yes"solution,
22

decisionproblems.Wesaythatthereisapolynomialtransformationfrom1to 2,written1/2,ifthefollowingtwoconditionshold:
answertoF(I)is\yes"withrespectto2. F(I)of2suchthattheanswertoIwithrespectto1is\yes"ifandonlyifthe (a)ThereexistsafunctionFtransforminganyinstanceIof1toaninstance
Adecisionproblem 0/. (b)ThereexistsanpolynomialtimealgorithmtocomputeF(I).
Thek-COLORINGproblemisstatedasfollows:givenagraphG=(V;E)and isNP-completeif 2NPandforeveryproblem02NP,
integer3 istodeterminehowdicultitistosolvetheproblemofconditionallycoloringa thek-COLORINGproblemisNP-completefork k jVj,isGk-colorable?WeknowfromGareyandJohnson[27]that
graph.Observethatthe:P2k-COLORINGproblemistheusualk-COLORING 3.Thegoalofthischapter
toeachcolorclasscontainingnoPjforj ofthecolorabilityproblem?Wewillshowtheansweris\no"whenj=3.The problem.Doesrelaxingtheconditionforeachcolorclassbeinganindependentset
:P3k-COLORINGproblemisstatedasfollows:givenagraphG=(V;E)and 3changethecomputationalcomplexity
withatmostljVj integer3 k ljVj
eachsubgraphinducedbyasetinthepartitiondoesnotcontainP3asasubgraph 2mcolors),doesthereexistak-partitionofthevertexsetsothat 2m(Note:ByTheorem2.1,agraphcanalwaysbe:P3-colored
Theorem3.1The:P3k-COLORINGproblem (notnecessarilyinduced)?
Proof.LetG=(V;E)beagraphofordern,andCa:P3k-coloringofG.The isNP-completefork 3.
followingisapolynomialtimealgorithmwhichcanbeusedtocheckthatCisa 23

valid:P3-coloringofG.Examinealltriplesofdistinctverticesineachcolorclass anddeterminewhetherthegraphinducedbyeachtriplecontainsP3.Determining
that ifatriplecontainsP3isO(1).Sinceeachsetinthepartitioncancontainatmost nvertices,thereareO(n3)triples.Thisalgorithmisofordern3andthisshows
.ConstructagraphF(G)fromG=(V;E)asfollows:foreachv2V,letF(G) Next,wewillshowthataknownNP-completeproblemcanbetransformedto 2NP.
containtwoverticesv1andv2.Wesaythatv1andv2areassociatedwiththe
3.1forthetransformationofG=C4. vandjoinv1tov2.Observethatthistransformationispolynomial.SeeFigure vertexv.InF(G),joinv1andv2totheverticesassociatedwitheachneighborof
Wewillshowthat(G) kimpliesthat(F(G);:P3) k.Assume(G)
us sv sw xs x1s
sx2 sw1
C4 u1s su2w2 ??? H Hv2s
H@H
A sHAAH@A
AA@Asv1
? @@@ HHHHHHH AAAAAAA ? ?
Figure3.1.TheconstructionofF(C4). F(C4)
k.LetH=F(G)andCbeak-coloringofG.ColorHasfollows:Foreachvertex v2V,assignC(v)tothetwoverticesassociatedwithvinH.Thiscoloringuses atmostkcolorstocolorH.Toseethatthiscoloringisavalid:P3-coloringofH, 24

suppose,bywayofcontradiction,thatacolorclassofHcontainsP3=uvw.By construction,ab2E(H)ifandonlyifeitheraandbareassociatedwiththesame vertexinGorthereexistscd2E(G)suchthataisassociatedwithcandbis
k-coloring.Thus,uv2E(H)impliesthatthereexistsc2V(G)suchthatuand color,thentheymustbeassociatedwiththesamevertexinGsinceGhasavalid associatedwithd.Therefore,iftwoadjacentverticesinHareassignedthesame
construction,thereareexactlytwoverticesinHassociatedwithavertexinG. withcimplythatvandwareassociatedwithc.Thisisacontradictionsince,by vareassociatedwithc.Similarly,vw2E(H)andvbeingavertexassociated
f1;2;:::;kgbea:P3k-coloringofH.WecolortheverticesofGusingAlgorithm Therefore,(H;:P3) Next,wewillshowthat(H;:P3) k.
3.1presentedinFigure3.1.Letc:V(G)!f1;2;:::;kgbethefunctionthat kimpliesthat(G) k.Letd:V(H)!
assignscolorstotheverticesofGbasedonAlgorithm3.1. thecolorsassignedtothetwoverticesinHassociatedwithz. Note1:ObservethatAlgorithm3.1assignsacolortoz2V(G)fromoneof
onew2NH(v1;v2)coloredc(v).Supposebywayofcontradictionthatthereexist x;y2V(H)suchthatfx;yg Note2:Foreveryv2V(G)withassociatedverticesv1andv2,thereisatmost
knowthateitherv1orv2iscoloredc(v),sayv1.Butxv1yformsP3inH,which isacontradiction. NH(v1;v2)andd(x)=d(y)=c(v).ByNote1,we
eitherLine9orLine17.Sincethenumberofverticesinagraphisnite,allthe verticesofGareassignedacolor.Wewillshowthatcisavalidk-coloringofG. Eachtimethroughtheloop(Lines7-22),avertexinGisassignedacolorin
25

Input:GraphsGandH,anda:P3-coloringd:V(H)!f1;2;:::;kgofH. Output:Ak-coloringc:V(G)!f1;2;:::;kgofG. 1.Foreveryz2V(G)withassociatedverticesz1andz2inH. 2. 3. 4. Ifd(z1)=d(z2),thenletc(z)=d(z1). Ifd(z1)62d(N(z1))andzisuncolored,thenletc(z)=d(z1).
7. 6. 5.IfGhasuncoloredvertices,then Ifd(z2)62d(N(z2))andzisuncolored,thenletc(z)=d(z2).
8. Repeatsteps8-22untilallverticesofGarecolored. Letpreviouscolor=?1andzbeanuncoloredvertexinG.
10. 9. Ifpreviouscolor6=d(z1),then
11. letpreviouscolor=d(z1),andwbetheuniquevertexinV(G)with letc(z)=d(z1),
12. Ifwisuncolored,then whichthevertexinNH(z1)withcolord(z1)isassociated.
16. 13. 14. elseifthereareuncoloredverticesinG,then letz=w.
17. 18. else(previouscolor=d(z1)) letc(z)=d(z2), letpreviouscolor=d(z2),andwbetheuniquevertexinV(G)with letpreviouscolor=?1andzbeanuncoloredvertexinG.
19 20. Ifwisuncolored,then whichthevertexinNH(z2)withcolord(z2)isassociated.
23.Outputcandstop. 21. 22. elseifthereareuncoloredverticesinG,then letz=w. letpreviouscolor=?1andzbeanuncoloredvertexinG. Figure3.1.Algorithm3.1
26

Letv2V(G)withc(v)=a,u2N(v),v1andv2betheverticesinHassociated c(u)6=a. withv,andu1andu2betheverticesinHassociatedwithu.Wewillshowthat
assignedcoloraorelsethesubgraphinducedbythecolorclassinHcontainingv1, v2,andoneoftheverticesassociatedwithuwouldcontainP3.Thus,allvertices AssumevwasassignedcolorainLine2ofAlgorithm3.1.Now,ucannotbe
assignedacolorinLine2ofAlgorithm3.1receiveavalidcolor. andtheneighborsofv1(whichincludeu1andu2)arenotcoloreda.Sinceuis assigneditscolorfromoneofthecolorsassignedtou1oru2(Note1),wehave AssumevwasassignedcolorainLine3ofAlgorithm3.1,thatis,d(v1)=a
argumentprovesthateveryvertexcoloredinLine4receivesavalidcolor. c(u)6=a.Thus,allverticescoloredinLine3receiveavalidcolor.Asimilar
3.1.Sinceonlyonevertexiscoloredatattime,let'sassumethatuwasalready c(u)6=c(v).SoassumeuandvwereassignedcolorsinLines9or17ofAlgorithm WehavejustshownthatifuorvwereassignedacolorinLines2,3,or4,then
coloredwhenviscolored.
d(u1)=aandweletpreviouscolor=a(Line10).Sinceweassumedthat Algorithm3.1.Intheloop(Lines7-22)withz=u,inLine9wehavec(u)= Supposethatc(u)=c(v)=a.AssumeuwasassignedcolorainLine9of
c(v)=a,eitherv1orv2iscoloreda(Note1).Further,sincethereisatmostone neighborofu1coloreda(Note2),wemusthavez=v(Line12),i.e.,visthenext vertextobecoloredintheloop(Lines7-22).Whenwegothroughtheloopto colorv,eitheritgetscoloredinLine9or17.IfviscoloredinLine9,thensince previouscolor=a,weknowthatthecolorofc(v1)6=a(Line8)andc(v)6=a 27

(Line9).SovmustbecoloredinLine17andweknowthatd(v1)=a(Line16) andc(v)=d(v2)=a(Line17).Now,d(v2)6=aorelsevwouldhavebeencolored inLine2.Thusc(v)6=a,whichcontradictstheassumptionthatc(v)=a.By applyingthepreviousargumentreplacingx1withx2andstartingwithz=xin Line17,weagainreachacontradiction.
completeproblemtoourproblem.Therefore,the:P3k-COLORINGproblemisAlgorithm3.1producesak-coloringofG.WehavetransformedaknownNP- Thus,foreveryadjacentu;v2V(G),wehavec(u)6=c(v),whichimpliesthat
NP-complete. ThisconstructionforHdoesnothelptoprovethatthe:Pjk-colorability 2
researchincludedeterminingwhetherthe:Pjk-colorabilityproblem,the:Cj
problemisNP-completeforj forthe:Pjk-colorabilityproblemwithnosuccess.Suggestedareasforfuture 4.WehavetriedotherconstructionsforH
withacycliccomplements. jk-colorabilityproblem,andthe:Kjk-colorabilityproblemareNP-completefor 3.Now,weturnourattentiontonding(G;:Pj)and(G;:Cj)forgraphs
28

4 Determiningtheconditionalchromaticnum-
Inthischapterwewillshowthat(G;:Cj)and(G;:Pj)mayhavedierent berforgraphswithacycliccomplements
graphswhosecomplementsareacyclicandthenperformasimilaranalysisfor valuesformanygraphs.First,wewillexaminethevaluesof(G;:Cj)for
between(G;:Cj)and(G;:Pj)canbemadearbitrarilylargeinaparticular familyofgraphs. (G;:Pj).Finally,themaintheoremofthischaptershowsthatthedierence
Todetermine(G;:Cj)forj 4.1 Determining(G;:Cj)whenGisacyclic
colorclasscanbeinaminimum:Cj-coloringofG.Ifj=3,thenwewillseethe largestcolorclassmustbeofsize4orlessandifj 3,itisnecessarytodeterminehowlargeany
mustbeofsizejorless.Therefore,wedetermine(G;:C3)and(G;:Cj)for j 4intwoseparatetheorems.Werstaddressthesizeofthelargestpossible 4,thenthelargestcolorclass
colorclassfor(G;:C3)withthefollowinglemma. Lemma4.1IfGisagraphofordern asasubgraph. 5andGisacyclic,thenGcontainsC3
Proof.LetGbeagraphofordern asubgraphofGwithvevertices.WemayassumeHisatree(otherwisewe removeedgesfromHuntilHisatree).IfHhasavertexvsuchthatdH(v) 5withacycliccomplement.LetHbe
thensinceHcontainsnocycles,NH(v)isanindependentsetinHofsizeatleast 3,
29

3.IfeveryvertexinHhasdegreeatmost2,thenH=P5,whichclearlycontains anindependentsetofsize3.Ineithercase,theverticesfromtheindependentset inHformsC3inH.SinceH Thefollowingtheoremevaluates(G;:C3)whenGisacyclic. G,thegraphGcontainsC3. 2
Theorem4.1LetGbeagraphwithacycliccomplement.Letmbethenumber
Proof.LetAbeacolorclassina:C3-coloringofG.ByLemma4.1,jAj ofedgesinamaximummatchingofG.Ifm 0,then(G;:C3)=ln?m 2m.
size3.TheonlyC3-freegraphsoforder4whosecomplementsareacyclicareP4 Letabethenumberofcolorclassesofsize4andbthenumberofcolorclassesof 4.
edges.TheonlyC3-freegraphsoforder3whosecomplementsareacyclicareP3 subgraphinducedbyeachcolorclassofsize4mustbemissingtwoindependent andC4.Eachofthesegraphsismissingtwoindependentedges.Therefore,the
andK2+K1.Eachofthesegraphsismissinganedge.Therefore,thesubgraph inducedbyeachcolorclassofsize3mustbemissinganedge. ofedgesfromeachcolorclassofsize4andonefromeachcolorclassofsize3. ThusjM1j=2a+b,andsincemisthesizeofamaximummatching,weget Sincecolorclassesaredisjoint,wecanformamatchingM1inGwithapair
2a+b=jM1j m.Thus,(G;:C3) ln?4a?3b 2 m+a+b=ln?2a?b
manycolorclassesofsize4aspossiblebyusingtheendpointsoftwoedgesinM Toconstructacoloring,letMbeamaximummatchinginG.Weformas 2mln?m 2m.
ineach,thenbasedontheparityofm,zero(ifmiseven)orone(ifmisodd) colorclassofsize3usingtheendpointsofoneedgeinMandanadditionalvertex, andnallyasmanycolorclassesofsize2aspossible.Withthisconstruction,the 30

C3-freeandarevalidcolorclasses. inducedbyacolorclassofsize3iseitherP3orK2+K1.Eachofthesegraphsis graphinducedbyacolorclassofsize4iseitherC4orP4.Further,thegraph
2.Withthiscoloring,weget(G;:C3) Ifmiseven,therearem2colorclassesofsize4,andtherestareofsizeatmost ln?4(m2)
ofsize3.Therestareofsizeatmost2.Withthiscoloring,weget(G;:C3) Ifmisoddandn>2m,therearem?1 2colorclassesofsize4andonecolorclass 2m+m2=ln?m 2m.
n?4(m?1 Ifmisoddandn=2m,therearem?1 2 2)?3+m?1
2+1=ln?m 2m.
2Nextwewillcompletethestudyofgraphswhosecomplementsareacyclicby
classofsize2.Withthiscoloring,(G;:C3) 2colorclassesofsize4andonecolor m?1 2+1=m+1 2=lm2m=ln?m 2m.
determining(G;:Cj)forj 2.10guaranteescyclesofallordersifagraphismissingatmostj?3edges.The colorclasscanbeinaminimum:Cj-coloringofagraph.RecallthatTheorem 4.Again,weneedtodeterminehowlargeany
ton?1edges. followingtheoremshowsthatwestillhavelongcyclesforsomegraphsmissingup
Cn?1 Lemma4.2LetGbeagraphofordern G.Furthermore,ifK1;n?26G,thenGisHamiltonian. 5withacycliccomplement.Then
Proof.WemayassumeG=TwhereTisatreeandK1;n?2 G.(Otherwise,removeedgesfromGuntilwegetagraphwhosecomplementisa tree.Sincen 5,thiscanbedonewithoutcreatingK1;n?2inthecomplement.) TonlyifK1;n?2
Choosearootvertexv0ofTsothattheheightofthetreehv0(T)ismaximum. 31

Sincen Further,dT(v0)=1andthereisexactlyonevertexv1atlevel1.LetTEbethe subgraphofGinducedbytheverticesonevenlevelsofT,andTObethesubgraph 5,thediameterofthetreeTisatleast2andtherefore,hv0(T) 2.
ofGinducedbytheverticesonoddlevelsofT.
onlevelsi?2;i?3;:::;1;0inG.Ineachofthefollowingcases,wewillshow Hamiltonianconnected.Further,allverticesonleveliareadjacenttoallvertices SinceTEandTOarecompletegraphs,everyinducedsubgraphofTEorTOis
Cn?1 Case1.hv0(T) Now,aHamiltonian(v0,v4)-pathfromTE?v2togetherwithv4v1together GandeitherK1;n?2 4.Letvibeavertexonthetreeatlevelifori=2;3;4. TorCn G.
Hamiltonian(v1,v3)-pathfromTOtogetherwithv3v0formsCn Further,aHamiltonian(v0,v4)-pathfromTEtogetherwithv4v1togetherwitha withaHamiltonian(v1,v3)-pathfromTOtogetherwithv3v0formsCn?1 G. G.
Case2.hv0(T)=3.Letw1;w2;:::betheverticesonlevel2andx1;x2;:::the verticesonlevel3.
aHamiltonian(x1,x2)-pathinTOtogetherwithx1v0x2.Inthiscase,wehave atleasttwoverticesx1andx2onlevel3.ThenCn?1isformedinGbyusing Case2A.Thenumberofverticesinlevel2is1.Sincen 5,thereare
K1;n?2 Case2B.Thenumberofverticesinlevel2isatleast2.Sincehv0(T)=3, thereisatleastonevertexx1onlevel3.Assumewithoutlossofgeneralitythat Tinducedbytheverticesonlevels1,2,and3.
w1x12E(T). getherwithw2x1v0formsCn?1Iflevel3hasonlyonevertex,thenaHamiltonian(v0,w2)-pathfromTEto- G.Further,K1;n?2 32
Tisinducedbythe

verticesonlevels0,1,and2. E(T),thenformCn?1 withw2x2togetherwithaHamiltonian(x2,x1)-pathinTO?v1togetherwithx1v0. Assumelevel3hasatleasttwoverticesx1,x2(andw1x12E(T)).Ifx2w12
AHamiltoniancycleisformedinGbyusingaHamiltonian(v0,w2)-pathinTE
GbyusingaHamiltonian(v0,w2)-pathinTEtogether
x1v0.Otherwise,withoutlossofgeneralityassumex2w22E(T),andwecanform togetherwithw2x2togetherwithaHamiltonian(x2,x1)-pathinTOtogetherwith
Cn?1 withaHamiltonian(x2,x1)-pathinTO?v1togetherwithx1v0.AHamiltonian cycleisformedinGbyusingaHamiltonian(v0,w1)-pathinTEtogetherwith GbyusingaHamiltonian(v0,w1)-pathinTEtogetherwithw1x2together
w1x2togetherwithaHamiltonian(x2,x1)-pathinTOtogetherwithx1v0. Case3.hv0(T)=2.ThenG=K1;n?1andTE=Kn?1whichcontainsCn?1.2
inaminimum:Cj-coloringofG.Wecompleteouranalysisofgraphswhose complementsareacyclicwiththefollowingtheoremwhichdetermines(G;:Cj) Weusetheaboveresulttogetanupperboundonthesizeofacolorclass
Theorem4.2LetGbeagraphofordernwithacycliccomplement.Letm whenj 4.
then(G;:Cj)=maxln?m bethemaximumnumberofpairwisevertexdisjointcopiesofK1;j?2inG.Ifj 4, 0
Proof.LetAbeacolorclassina:Cj-coloringofG.NoticethathAiisacyclic j?1m;lnjm.
contradiction.HencejAj sinceGisacyclic.IfjAjj,anditfollowsthat(G;:Cj) j+1,thenbyLemma4.2,wegetCj lnjm. hAi,a
33

wemusthaveK1;2 order4withacycliccomplementcontainseitherC4orK1;2.SinceAisacolorclass, SupposejAj=j.WewillshowK1;j?2 hAi.Ifj 5,thenbyLemma4.2,wehaveK1;j?2
hAi.Supposej=4.Everygraphof
disjoint,wegeta Letabethenumberofcolorclassesofsizej.Sincecolorclassesarevertex m.Therefore,(G;:Cj) ln?aj hAi.
(G;:Cj) j?1m+a=ln?a j?1m ln?m
Toshowtheinequalityintheotherdirection,weproduceaminimum:Cj- maxln?m j?1m;lnjm. j?1mand
coloring.ConsiderfU1;U2;:::;Umg,whereeachUiisaK1;j?2inGandtheUi's Weconsidertwocases:m arepairwisevertexdisjoint.LetS=V(G)?Smi=1V(Ui).Now,n=(j?1)m+s.
letV(Ui)togetherwithavertexfromSformacolorclassofsizej(thesubgraph Ifm s,thenn mj,whichimpliesthatln?m sand0 s

Now, jBj=jTj?$nj% =s+r(j?1)?(m?r)

whereeisanedge.AllgraphscontainP4exceptK4?E(K1;3),whichcontains P3.So,assumen andthereforePn?1 onCn?1canbeaddedtoCn?1toformPn
5.ByLemma4.2,thegraphGcontainsCn?1asasubgraph G.IfG6=K1;n?1,then(G) G. 1andthevertexinGnot
Thefollowingtheoremdetermines(G;:Pj)forgraphswhosecomplements 2
areacyclic. Theorem4.3LetGbeagraphofordernwithacycliccomplement.Letm then(G;:Pj)=ln?m bethemaximumnumberofpairwisevertexdisjointcopiesofK1;j?1inG.Ifj 2, 0
Proof.LetAbeacolorclassina:Pj-coloringofG.NoticethathAiisacyclic j?1m.
sinceGisacyclic.IfjAj>j,thenbyLemma4.3,wegetPj contradictstheassumptionthatAisacolorclass.Therefore,jAj SupposejAj=j.Now,hAiisacyclicandisPj-free.Therefore,byLemma4.3, j. hAi,which
classesarevertexdisjoint,wegeta ln?a wegetK1;j?1=hAi.Letabethenumberofcolorclassesofsizej.Sincecolor
j?1m m.Therefore,(G;:Pj) ln?aj
Toshowtheinequalityintheotherdirection,weproduceaminimum:Pj- ln?m j?1m. j?1m+a=
coloring.ConsiderfU1;U2;:::;Umg,whereeachUiisaK1;j?1inGandtheUi's V(Ui)(thesubgraphinducedbyeachsuchcolorclassisPj-freesinceitcontainsa vertexofdegreezero).Withtheremainingvertices,formasmanycolorclassesof arepairwisevertexdisjoint.Foreachi=1;:::;m,formacolorclassofsizejusing
Therefore,(G;:Pj)=ln?m sizej?1aspossible.Withthiscoloring,weget(G;:Pj) j?1m. ln?mj j?1m+m=ln?m j?1m.
36
2

statedhereforreference. Thefollowingcorollarywillbeusedseveraltimesinupcomingchaptersandis
Corollary4.1IfGisagraphwithhE(G)i=K1;mandj (G;:Pj)=8>: 2,then ln ln?1 j?1mif0 j?1mifj?1m

s
sss BBBBBB
s
sss BBBBBB
s
sss BBBBBB
s
sss BBBBBB Figure4.1.TheconstructionofGforj=5,a=1andn=24 ss
questions.Whathappensifweremovetherestrictionthatthecomplementis acyclicandexamineallgraphsmissingn?1edgesorexaminegraphswhose Somepossibledirectionsforfurtherresearchincludeansweringthefollowing
Also,aretherefamiliesofgraphswhere(G;:Pj)?(G;:Cj)isO(nk)fork complementsarebipartite?Willthedierenceincreaseand,ifso,byhowmuch?
n,howmuchcanthesetwoconditionalchromaticnumbersdierifweremovehalf Thefollowingquestionsalsoremainopen.Forthesetofgraphsofaxedorder 1?
oftheedgesfromthecompletegraph?And,ingeneral,giventhesizeofG,what areupperandlowerboundsfor(G;:Pj)?(G;:Cj)?Whichgraphsattain thesebounds?
38

Nextweaddresstheproblem:givenagraphofordernwitheedges,whatarethe 5 Determining (G;:Cj)
boundsontheconditionalchromaticnumber?In1995,DargenandFraughnaugh
Theorem5.1(DargenandFraughnaugh,[21])Ife(G) publishedthefollowingtheorem.
(G;:Cj)=8>:
n2?(j?2),then ln?1 ln j?1m j?1mifG=K1;j?2 orG=K3andj=5
fromacompletegraphtoobtainagraphwhose:Cj-conditionalchromaticnumber Wecandeducefromthistheoremthatweneedtoremoveatleastj?2edges otherwise.
sothat(G;:Cj)=ln?2 removingonlytwoedgesfromacompletegraphwhenj=3. isln?1 j?1m.Anaturalquestiononemayaskishowmanyedgesdoweneedtoremove
Inthischapter,wewilldeterminewhichgraphsareHamiltoniangiventhat j?1m?Wewillshowthatthisnumbercanbeobtainedby
thesizeoftheircomplementisatmostn?1.Usingthisinformation,thenal graphsmissingexactlyj?1edges. theoremofthischapterwillexpandtheknowledgeof(G;:Cj)totheclassof
5.1 Toprovethenaltheorem,wewillrstaddresssomespecialcases.Wewillsoon Preliminaries
seethatthespecialcasesforthenaltheoremaregraphsmissingastarofa particularorderandgraphsmissingpairwisevertexdisjointcompletesubgraphs. First,wewilladdressgraphsmissingastarofaparticularorder. 39

Theorem5.2LetGbeagraphofordern.Ifj andK1;j?2 G,then(G;:Cj)=ln?1 j?1m. 3ande(G) n2?(2j?6)
Proof.LetXbeasetofverticeswhichinduceK1;j?2 ande(G) Therefore,v02X.LetAbeacolorclassina:Cj-coloringofG. 2j?6,thereisexactlyonevertexv0ofdegreeatleastj?2inG. G.SinceK1;j?2 G
andbyTheorem2.10,wegetCj musthavejAj Supposev062A.IfjAj j,thenhAiwouldbemissingatmostj?4edges
thenbythesameargument,wegetCj j?1.Supposev02A,andconsiderA?v0.IfjA?v0j hAi.Therefore,forAtobeacolorclass,we
whichimpliesjAj j.Thus,therecanbeatmostonecolorclassofsizej(one hA?v0i.Therefore,jA?v0j j?1, j,
containingv0),and(G;:Cj) ln?j
jbyusingtheverticesinXtogetherwithonevertexfromG?Xandformas Toshowtheinequalityintheotherdirection,formonecolorclassBofsize j?1m+1=ln?1 j?1m.
manyothercolorclassesaspossibleofsizej?1withtheremainingverticesin hBi.Therefore,(G;:Cj) G?X.NowhBiisCj-freesincev0isadjacenttoatmostoneothervertexin
Next,wewouldliketodetermine(G;:Cj)whenGismissingasetofmutu- ln?j j?1m+1=ln?1 j?1m. 2
allydisjointcompletegraphsofaspecicorder.Toaccomplishthis,weusethe
Lemma5.1LetGbeagraphofordern followingtwolemmas.
(ri 1;p 1)foreachi.TheneitherGisHamiltonianormaxifrig 3suchthatG=Wpi=1SiwhereSi=Iri
Proof.Letv2V(G).Thenv2Siforsomeiandd(v)=(n?1)?(ri?1)=n?ri. ln+1 2m.
Therefore,sinceeachvertexinGisinsomeSi,weget(G)=n?maxifrig.If 40

Dirac'sTheorem(Theorem2.7),GisHamiltonian. maxifrign?ln+1 2m=jn?1 2k.Thus,(G) n=2andby
Lemma5.2LetGbeagraphofordern 4suchthatG=Wpi=1SiwhereSi=Iri 2
existr1andr2suchthatr1=r2=ln2m. (1 ri ln2m;p 1)foreachi.TheneitherGcontainsan(n?1)-cycleorthere
ThenG0=Ir1?1_(Wpi=2Si)andmaxi2fri;r1?1gj+1.Let 2m=j+1 2. 2mm.
BthathBicancontainatmosttwocopiesofIj+1 1 q2 AsuchthatjBj=j+2.ThenhBi=Wki=1Iqiwhere1 (j+1)=2,and1 qi (j?1)=2for3 i ksincejBj=j+2implies q1 (j+1)=2,
ofIq1andIq2,formC BsuchthatjCj=j.Now,hCi=Iq1?1_Iq2?1_Wki=3Iqi, 2.Byremovingonevertexfromeach
41

andmaxifqig assumptionthathAiisCj-free.SojAj j?1
IfjAj=j+1,thenbyLemma5.2,either2Ij+1 2.ByLemma5.1,hCiisHamiltonian.Thiscontradictsthe j+1.
jAj=j,thenbyLemma5.1,wehaveIj+1 j-cycle.BecausehAicannotcontainCj,wemusthave2Ij+1 2 hAiorhAicontainsa
classesofsizej+1andbbethenumberofcolorclassesofsizejina:Cj- 2 hAi.Letabethenumberofcolor 2 hAi.Now,if
coloringofG.Sincecolorclassesarevertexdisjoint,weget2a+b (G;:Cj) Toshowtheinequalityintheotherdirection,wesplittheproofintothree ln?(j+1)a?jb j?1 m+a+b=ln?2a?b j?1m ln?m j?1m. m.Now,
cases.RecallthatG=Wm+n?(j+1 i=1 2)m
Case1.miseven.ColorGasfollows:For1 Si=I1fori=m+1;m+2;:::;n?j+1 SiwhereSi=Ij+1 2m. l m=2,formacolorclassof 2fori=1;2;:::;mand
Sincebipartitegraphsonlycontainevencycles,MisCj-free.Withtheremaining sizej+1byusingalltheverticesfromS2l?1andS2l.Thesubgraphinducedby eachofthesecolorclassesisisomorphictoM=Ij+1 vertices,formasmanyothercolorclassesofsizej?1aspossible.Withthis 2_Ij+1 2,whichisbipartite.
Case2.misoddandn coloring,weget(G;:Cj) mj+1 ln?(j+1)m2 j?1m+m2=ln?m 2+j?1 2.ColorGasfollows:For1 j?1m. l m?1
sincethesubgraphinducedbyeachofthesecolorclassesisbipartite,itisCj-free. formacolorclassofsizej+1byusingalltheverticesfromS2l?1andS2l.Again, 2,
assignedtocolorclasses.FormonecolorclassCofsizejusingalltheverticesfrom Smtogetherwithj?1 Observethatn mj+1
2verticesfromV(G)?Smi=1V(Si).Now,hCi=Ij+1 2+j?1 2guaranteesthatthereareatleastjverticesnotyet
thelargestcyclewecanformisCj?1obtainedbyalternatelytraversingbetween 2_Kj?1 2and
42

j?1aspossible.Then(G;:Cj) theverticesofKj?1 isCj-free.Withtheremainingvertices,formasmanyothercolorclassesofsize 2andIj+1 2untilweusealltheverticesinKj?1 n?m?1 2(j+1)?j j?1 +m?1 2+1=ln?m 2.Therefore,hCi
Case3.misoddandn

Band1 LetAbeacolorclassina:Cj-coloringofG.SupposethatjAj>j.Let
mostonecopyofIj+2 AsuchthatjBj=j+1.ThenhBi=Wki=1Iqiwhere1 qi j=2for2 i ksincejBj=j+1impliesthathBicancontainat q1 (j+2)=2,
isHamiltonian.ThiscontradictstheassumptionthathAiisCj-free.SojAj jCj=j.Now,hCi=Iq1?1_Wki=2Iqi,andmaxifqig 2.ByremovingonevertexfromIq1,formC j2.ByLemma5.1,hCi Bsuchthat
and(G;:Cj) IfjAj=j,thensincehAiisCj-free,byLemma5.1wehaveIj+2 lnjm. j
bethenumberofcolorclassesofsizejina:Cj-coloringofG.Sincecolorclasses arevertexdisjoint,wegeta m.Therefore,(G;:Cj) ln?jm2
hAi.Leta
Hence,(G;:Cj) Toshowtheinequalityintheotherdirection,weproduceaminimum:Cj- maxlnjm;ln?m j?1m. j?1m+m=ln?m j?1m.
coloring.LetU=V(G)?Smi=1V(Si).Now,n=(j+2 cases:j?2 Ifj?2 2m 2m s,thenn sand0 s

T=U[(Smi=m?r+1V(Si)).For1 If0 s

Todeterminethe:Cj-chromaticnumberwhenj 5.2 Determining(G;:Cj)whene(G) 8forgraphsgiventhesizeof n2?(j?1)
thecomplementisatmostj?1,werstdeterminewhichgraphsofordern areHamiltoniangiventhesizeoftheircomplementisatmostn?1. 8
Theorem5.4LetGbeagraphofordern K1;n?2 GorGisHamiltonian. 8withe(G) n?1.Theneither
Proof.AssumeK1;n?26G.IfG=Kn,thenGisobviouslyHamiltonian.So K1;n?26G,wemusthaved(u) assumeGisnotcompleteandletuandvbeapairofnonadjacentvertices.Since theproofintocasesbasedonthenumberofedgesinGincidenttouorv. Case1.SupposethatthenumberofedgesinGincidenttouorvis 2andd(v) 2.LetH=G?fu;vg.Webreak
n?1.Sincee(G)=n?1,Hiscomplete. Theseverticesexistsinced(u) Hamiltonian(a,x)-pathfromH?fy;bgformsCn.SoassumejN(u)\N(v)j IfjN(u)\N(v)j=0,thenletx;y2N(u)anda;b2N(v)bedistinctvertices. 2andd(v) 2.Nowxuybvatogetherwitha
thatw2N(u)\N(v),x2N(u)andy2N(v).Thenxuwvytogetherwithan IfjN(u)\N(v)j=1,thenletw,x,andybedistinctverticesinHsuch 1.
(y,x)-pathfromH?fwgformsCn. intoHoutofapossible2(n?2),theremustben?2edgesfromfu;vgintoH. Sincen SoassumejN(u)\N(v)j 8,thereareatleastsixedgesfromfu;vgintoH.Therefore,wecan 2.Sincetherearen?2edgesmissingfromfu;vg
assumed(u)ord(v)isatleast3.Withoutlossofgenerality,assumed(u) Letw;x;y2N(u),andw;x2N(v).ThenxvwuytogetherwithaHamiltonian 3.
46

(y,x)-pathfromH?fwgformsCn.Therefore,wecanassumethateverysuch Case2.SupposethatthenumberofedgesinGincidenttouorvis pairofverticescanbeincidenttonomorethann?2edgesinG. exactlyn?2.Sincetherearen?3missingedgesfromfu;vgintoH,His missingatmostoneedge,andthereare2(n?2)?(n?3)=n?1edgesfrom fu;vgintoH. HandjN(u)\N(v)j=0,thesubgraphHneedstohaveatleastn?1vertices. ButHonlycontainsn?2vertices,andwereachacontradiction. SupposejN(u)\N(v)j=0.Theninordertohaven?1edgesfromfu;vginto
J=G?fu;v;wg.ThenJisagraphofordern?3missingatmostoneedge. Sincen Therefore,wecanassumejN(u)\N(v)j 8,wehavee(J) n?3 1.Letw2N(u)\N(v).Consider
connectedbyTheorem2.9. 2?1 n?3
IfN(u)6=N(v),thensinced(u) 2,d(v) 2?(n?7)andJisHamiltonian
verticesx;y2V(J)suchthatx2N(u)andy2N(v).Nowxuwvytogetherwith aHamiltonian(x,y)-pathinJformsCn. 2,andn 8,thereexistdistinct
fu;vgintoH.Therefore,d(u)=d(v)=n?1 SowecanassumethatN(u)=N(v).Recallthattherearen?1edgesfrom
thereexistdistinctverticesx;y2V(J)suchthatx;y2N(u)\N(v),x6=w, forn?1 2tobeaninteger.Therefore,n 9andd(u)=d(v) 2.Theintegernmustbeoddinorder
andy6=w.ThenxuwvytogetherwithaHamiltonian(x,y)-pathinJformsCn. 4.Thisimpliesthat
Thus,wemayassumethatforeverynonadjacentpairofverticesuandv,the numberofedgesinGincidenttouorvisatmostn?3.
47

Case3.SupposethatthenumberofedgesinGincidenttoevery Thereareatmostn?4edgesmissingfromfu;vgintoH.Thereforethereare nonadjacentpairofofverticesu,visatmostn?3. atleast2(n?2)?(n?4)=nedgesfromfu;vgintoH.So,d(u)+d(v) andsinceuandvarearbitrarynonadjacentvertices,byOre'sTheorem(Theorem 2.6)GisHamiltonian. 2n
determining(G;:Cj)whene(G) (G;:Cj)whenj Now,armedwithTheorems4.1,4.2,5.2,5.3,and5.4,wecanproceedwith 8. n2?(j?1).Webeginbydetermining
Theorem5.5Ifj 8ande(G) n2?(j?1),then (G;:Cj)=8>: ln?1 ln j?1motherwise. j?1mifK1;j?2 G
Proof.SupposeK1;j?2 e(G)=n2?(j?1) ln?1 n2?(2j?6),andbyTheorem5.2,wehave(G;:Cj)= G.Wehavej?1 2j?6sincej 8.Therefore,
showthatjAj

Theorem5.6LetGbeagraphofordern Hamiltonianexceptinthefollowingcases: K1;n?2 G 3withe(G) n?1.ThenGis
K3 hE(G)i=C4andn=5 hE(G)i=K4?ewhereeisanedgeandn=6 Gandn=5 (iii) (iv) (ii) (i)
Proof.LetGbeagraphofordern hE(G)i=K4andn=7 3withe(G) n?1.Ife(G) (v)
thenGisHamiltonianbyTheorem2.10.Soassumen?2 e(G) n2?(n?3),
isexceptionalcase(i).Soassume(G) (G) 1(orequivalentlyK1;n?2 G),thenclearlyGisnotHamiltonian.This 2.Ifn=3,then(G) n?1.If
thatG=C3,whichisHamiltonian.Ifn=4,then(G) G2fK4;K4?e;C4g,allofwhichareHamiltonian.Ifn 8,thenbyTheorem 2impliesthat 2implies
5.4,GisHamiltonian.So,assume5 andvbeapairofnonadjacentverticesinG.LetH=G?fu;vg.Next,webreak Nown?2 e(G) n?1and5 n n7. theproofintocasesbasedonthenumberofedgesinGincidenttouorv. 7implyGisnotcomplete.Soletu
Case1.SupposethatthenumberofedgesinGincidenttouorvis n?1.SincethetotalnumberofedgesinGisatmostn?1,Hiscomplete. contradictstheassumptionthat(G) Ifn=6,thentherearefouredgesfromfu;vgtoHinG.Moreover,(G) Ifn=5,thenhavingfouredgesincidenttouorvinGforces(G) 2. 1,which
impliesthat(G) inFigure5.1.IfGisgraph(a)inFigure5.1,thenuwyvzxuformsaHamiltonian 3andThereforeGis,uptoisomorphism,oneofthegraphs 2
cycle.IfGisgraph(b)inFigure5.1,thenuwyzvxuformsaHamiltoniancycle. IfGisgraph(c)inFigure5.1,thenclearlyGisnotHamiltoniansinceN(u)= 49

z r??@@ uv
y r rxrr
(a) r wz
rrrr ??@@ uv y(b)
rxr
wz
rrrr ??uv y(c)
rxr
w
Figure5.1.ThenonisomorphicpossibilitiesforGwhenthenumberof
N(v)=fw;xgandthelargestcyclecontainingverticesuandvisC4.Further, edgesincidenttouorvis5andn=6.
hE(G)i=K4?e,whereeisanedge,andwegetexceptionalcase(iv).
za??@@
rrrr uv
yrAAAwx
rrrr (a) za??@@
uv
yr
wx rrrr (b) ? ??? ? r uv za y wx
Figure5.2.ThenonisomorphicpossibilitiesforGwhenthenumberof edgesincidenttouorvis6andn=7. (c)
inFigure5.2.IfGisgraph(a)or(b)inFigure5.2,thenuwavzyxuforms impliesthat(G) Ifn=7,thenthereareveedgesfromfu;vgtoHinG.Moreover,(G) 4.ThereforeGis,uptoisomorphism,oneofthegraphs 2
aHamiltoniancycle.IfGisgraph(c)inFigure5.2,thenuwvazyxuformsa edgesincidenttofu;vgisn?1. Hamiltoniancycle.Thisaddressesallthegraphswherethenumberofmissing Case2.SupposethatthenumberofedgesinGincidenttouorvis exactlyn?2.SincethetotalnumberofedgesinGisatmostn?1,H= 50

G?fu;vgismissingatmostoneedge. implies(G) 5.3.ThetwononisomorphicpossibilitiesforGwhenGhasthreeedgesincident Ifn=5,thentherearetwoedgesfromfu;vgtoHinG.Moreover,(G) 2.ThereforeGis,uptoisomorphism,oneofthegraphsinFigure 2
y ??@@ rrrr uv x
y ??uv r(a) r wy
rrrr ??@@ uv
xr
r rr
x
(d) wy
rrrr ??uv (b) r w @@ rrrr ??@@ r
x(e)
r wy
rrrr ??uv xr(f)
w
uv y x(c)
w
Figure5.3.ThenonisomorphicpossibilitiesforGwhenthenumberof
touorvande(H)=0aredepictedintherstcolumninFigure5.3.Asstated edgesincidenttouorvis3andn=5.
previously,thesubgraphHcanbemissingatmostoneedge.Thegraphsinthe secondandthirdcolumndepictallpossiblenonisomorphicplacementsofthat additionaledge.
vertexinG.Further,hE(G)i=C4andwegetexceptionalcase(iii).IfGisgraph IfGisgraph(b)inFigure5.3,thenclearlyGisnotHamiltoniansincexisacut IfGisgraph(a)or(c)inFigure5.3,thenuxvywuformsaHamiltoniancycle.
(d),(e),or(f)inFigure5.3,thenGisnotHamiltoniansinceN(u)[N(v)=fw;xg. Further,K3 implies(G) Ifn=6,thentherearethreeedgesfromfu;vgtoHinG.Moreover,(G) Gandwegetexceptionalcase(ii). 3.ThereforeGis,uptoisomorphism,oneofthegraphsinFigure 2
51

5.4.ThetwononisomorphicpossibilitiesforGwhenGhasfouredgesincident
z rrrr ??@@ uv yr(f)
xr
wz
rrrr ??@@ uv y(g)
rxr
w HHH rrrr ??@@ rr
uv z y(h)
x w rrrr ??@@ @@rr uv z rrrr ??uv y(a)
rxr
wz
rrrr ??uv y(b)
rxr
w rrrr ??@@r r
z y(i)
x wz
rrrr ??@@?? uv yr(j)
xr
w
uv z y(c)
x wz
rrrr ??uv y(d)
rxr
wz
rrrr ???? uv y(e)
rxr
w
Figure5.4.ThenonisomorphicpossibilitiesforGwhenthenumberof edgesincidenttouorvis4andn=6.
tofu;vgande(H)=0aredepictedintherstcolumninFigure5.4.Asstated previously,thesubgraphHcanbemissingatmostoneedge.Thegraphsinthe secondandsubsequentcolumnsdepictallpossiblenonisomorphicplacementsof thatadditionaledge. inFigure5.1.WehavealreadyshownthatthisgraphisnotHamiltonianand isexceptionalcase(iv).IfGisgraph(a),(b),(d),or(e)inFigure5.4,then IfGisgraph(c)inFigure5.4,thenthisgraphisisomorphictograph(c)
uwvzyxuformsaHamiltoniancycle.IfGisgraph(f),(g),or(h)inFigure5.4, thenuwzvyxuformsaHamiltoniancycle. thenuwyzvxuformsaHamiltoniancycle.IfGisgraph(i)or(j)inFigure5.4,
implies(G) 5.5. Ifn=7,thentherearefouredgesfromfu;vgtoHinG.Moreover,(G) 4.ThereforeGis,uptoisomorphism,oneofthegraphsinFigure 2
52

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za y qqqq ?@ uvwx (j) za y(k)
q
uvwx za y qqqq (c) za ?uvwx y(d)
q za qqqq ??? uvwx y(e)
q za qq ?uvwx y(f)
q q
Figure5.5.ThenonisomorphicpossibilitiesforGwhenthenumberof edgesincidenttouorvis5andn=7.
53

varedepictedintherstcolumninFigure5.5.Sincethetotalnumberofedges inGisatmost6andthenumberofedgesinHincidenttouorvisexactly5,H ThevenonisomorphicpossibilitiesforGwhenGhas5edgesincidenttouor
depictallpossiblenonisomorphicplacementsofthatadditionaledge. ismissingatmostoneedge.Thegraphsinthesecondandsubsequentcolumns
verticesofK3andI4untilweusealltheverticesinK3.Further,hE(G)i=K4 thelargestcyclewecanformisC6obtainedbyalternatelytraversingbetweenthe IfGisgraph(n)inFigure5.5,thenclearlyG=I4_K3isnotHamiltoniansince
andwegetexceptionalcase(v). Hamiltoniancycle.IfGisgraph(e)or(f)inFigure5.5,thenuwvazyxuforms aHamiltoniancycle.IfGisgraph(g),(h),(i),(k),(p),(q),(r),or(t)inFigure IfGisgraph(a),(b),(c),(d),or(j)inFigure5.5,thenuwyzvaxuformsa
5.5,thenuxvazywuformsaHamiltoniancycle.IfGisgraph(l),(m),or(o)in Figure5.5,thenuwvyzaxuformsaHamiltoniancycle.IfGisgraph(s)or(u) inFigure5.5,thenuwzyavxuformsaHamiltoniancycle.IfGisgraph(v),(w), (x),or(y)inFigure5.5,thenuwzxavyuformsaHamiltoniancycle.
andvinGthatthenumberofedgesinGincidenttouorvisatmostn?3. uorvisatleastn?2.Therefore,wecanassumeforallnonadjacentverticesu Thisaddressesallthegraphswherethenumberofmissingedgesincidentto
Case3.SupposethatthenumberofedgesinGincidenttoevery nonadjacentpairofofverticesu,visatmostn?3. atleast2(n?2)?(n?4)=nedgesfromfu;vgintoH.So,dG(u)+dG(v) andbyOre'sTheorem(Theorem2.6),GisHamiltonian. Thereareatmostn?4edgesmissingfromfu;vgintoH.Therefore,thereare 2n
54

forallj Nowwecanproceedwithdetermining(G;:Cj)whene(G) 3. n2?(j?1)
Theorem5.7Ife(G) n2?(j?1),then
(G;:Cj)= 8 >< ln?1 ln?2 j?1mifhE(G)i=2K2andj=3, j?1mifhE(G)i=P3andj=3, K1;j?2 K3 Gandj=5, Gandj 4,
>: ln hE(G)i=C4andj=5,
j?1motherwise. orhE(G)i=K4andj=7,and hE(G)i=K4?ewhereeisanedgeandj=6,
Proof.Ife(G)>n2?(j?1),thentheresultfollowsfromTheorem5.1.So assumee(G)=n2?(j?1).Further,theresultfollowswhenj 5.5.Therefore,weonlyneedtodetermine(G;:Cj)whenj=3,4,5,6or7. Ifj=3,thenhE(G)i2f2K2;P3g.IfhE(G)i=2K2,then(G;:C3)=ln?2 8byTheorem
byTheorem4.1.IfhE(G)i=P3,then(G;:C3)=ln?1 Ifj=4,thenhE(G)i2f3K2;P3+K2;K1;3;P4;K3g.IfhE(G)i=3K2,then j?1mbyTheorem4.1. j?1m
and(G;:C4)=lnjmifn

thenwegetCj e(hBi) e(hBi)=j+1 j+1 2?(j?1).ButsinceGismissingj?1edges,wemusthave hBibyTheorem2.10andreachacontradiction.Therefore,
whichimpliesj If(hBi) 2?(j?1).
(hBi) 2,whichimplies(hBi) 1,thene(hBi) 3.Butthiscontradictstheassumptionthatj (j+1)=2.Now,j?1=e(hBi) j?2.Choosev2V(hBi)sothatdhBi(v) 5.Therefore, (j+1)=2,
j?2.ConsiderhBi?v.Now,e(hBi?v)=e(hBi)?d(v) j+1
Therefore,jAj j2?(j?3)andbyTheorem2.10,wehaveCj j. hBiandreachacontradiction. 2?(j?1)?(j?2)=
acolorclassofsizej.ByTheorem2.10,hAimustbemissingatleastj?2edges. SinceGismissingatmostj?1edges,wemusthavea(j?2) Letabethenumberofcolorclassesofsizejina:Cj-coloringofGandAbe
asizej,whichimpliesthat(G;:Cj) 1.Therefore,all:Cj-coloringsofGcancontainatmostonecolorclassof ln?j j?1,orsimply
Hamiltonian.IfGcontainsoneofthesenon-Hamiltoniangraphsasaninduced ByTheorem5.6,weknowwhichgraphsmissingatmostj?1edgesarenon- j?1m+1=ln?1 j?1m.
subgraph,thenwecancolorGsothatthereisacolorclassofsizejandget (G;:Cj)=ln?1 Theprevioustheoremproducesanupperboundfordetermininghowmany j?1m,andifnot,then(G;:Cj)=ln j?1m. 2
believe,usingtheabovemethods.However,extrapolatingonthecomplexityofthe edgesneedtoremovedfromacompletegraphtoobtainagraphwith(G;:Cj)= ln?2
proofsforremovingj?1edgesversusthecomplexityoftheproofforremoving j?1m.Determining(G;:Cj)whenGismissingjedgescanbeaccomplished,we
onlyj?2edgesleadsustobelievethattheproofforremovingjedgeswill 56

eextremelylongandcomplicated.Wehavealreadydeterminedtheminimum inTheorem5.7.Removingjedgesfromacompletegraphwillprobablynotallow theconditionalchromaticnumbertodecreaseagain.Thus,thereislittletobe numberofedgestoberemovedtoobtainaconditionalchromaticnumberofln?2 j?1m
gainedbyprovingtheabovetypeofresultforgraphsmissingjedges.
toattainaparticularconditionalchromaticnumber. determininggeneralupperandlowerboundsonthenumberofedgesforagraph Infact,furtherresearchintheareaofdetermining(G;:Cj)shouldfocuson
57

WediscoveredinTheorem5.7thatitisnecessarytoremovej?2edgesfrom 6 Determining (G;:Pj)
:Cj-chromaticnumberwhenj=3.Wewouldliketodeterminetheminimum acompletegraphtodecreasethe:Cj-chromaticnumberfromln andthatitisnecessarytoremovej?1edgestoattainavalueofln?2 j?1mtoln?1 j?1mforthe j?1m
fromacompletegraphtoattainavalueofln?2 :Pj-chromaticnumberanddeterminetheminimumnumberofedgestoremove numberofedgestoremovefromacompletegraphtoattainavalueofln?1
j?1mforthe:Pj-chromaticnumber. j?1mforthe
:Pj-chromaticnumberforalargesetofgraphs.Webeginbydeterminingwhich Whileattemptingtondthesenumbers,wewillobtaininformationaboutthe graphsoflargesizehaveaHamiltonianpath,thendeterminetheminimumnum-
graphsmissing2j?5orfeweredges.Finally,wewilldeterminealowerboundon berofedgeswhichneedtoberemovedfromacompletegraphfor(G;:Pj)to decreasefromln
(G;:Pj)intermsofthesizeofG. j?1mtoln?1 j?1m,andthendeterminethechromaticnumberforall
6.1 DeterminingwhichgraphsoflargesizehaveaHamil-
Todeterminethe:Pj-chromaticnumberforgraphswhosecomplementshavesize tonianpath
atmost2j?5,weneedtodeterminewhichgraphsofordernmissingatmost 2n?5edgeshaveaHamiltonianpath.Thefollowingisaneasyandusefulresult.
58

Lemma6.1LetGbeagraphofordernandS path,thenthenumberofcomponentsinG?SisatmostjSj+1. V(G).IfGhasaHamiltonian
G.IfweremoveSfromG,thenHbreaksapartintoatmostjSj+1subpaths. Proof.LetS Therefore,G?ScontainsatmostjSj+1components. V(G)with0 jSj n,andletHbeaHamiltonianpathin
Thefollowingsetswillbeusedrepeatedlyinthissection.Let 2
F=fK2_(K2+I3),K1_(Kn?3+I2),Ir_Kr?2,wherer G=fGjGisaspanningsubgraphofagraphinFg,and H=fK2_(K2+I3),K1_(Kn?3+I2),I4_K2,(I4_K2)?e,I5_K3, 4andn 4g,
(I5_K3)?e,I6_K4,whereeisanedgeandn Lemma6.2IfagraphG2G,thenGdoesnothaveaHamiltonianpath. 4g.
Proof.LetG=F_(K2+I3)whereF=K2.LetS=V(F).Thenthenumber aHamiltonianpath. ofcomponentsinG?Sis4and4>3=jSj+1.ByLemma6.1,Gdoesnothave
componentsinG?Sis3and3>2=jSj+1.ByLemma6.1,Gdoesnothave aHamiltonianpath. LetG=F_(Kn?3+I2)whereF=K1.LetS=V(F).Thenthenumberof
nothaveaHamiltonianpath. ofcomponentsinG?Sisrandr>(r?2)+1=jSj+1.ByLemma6.1,Gdoes LetG=Ir_FwhereF=Kr?2andr 4.LetS=V(F).Thenthenumber
2inFdoesnotcontainaHamiltonianpath,FcannotcontainaHamiltonianpath. LetG=F,whereFisaspanningsubgraphofagraphinF.Sinceeachgraph
59

2n?5edges.Sinceweareinterestedindeterminingwhichgraphsmissingatmost 2n?5edgeshavenoHamiltonianpath,wewilldeterminethelargestsubsetof NowGcontainsmoregraphsthanjustthosegraphsofordernmissingatmost
graphsoforderninGthataremissingatmost2n?5edges.
ifandonlyifG2H. Lemma6.3LetGbeagraphofordern 1withe(G) 2n?5.ThenG2G
Proof.LetGbeagraphofordern IfGisaspanningsubgraphofK2_(K2+I3),thensinceK2_(K2+I3)ismissing 9=2(7)?5edges,wemusthavethatG=K2_(K2+I3)2H.IfGisaspanning 1withe(G) 2n?5.AssumethatG2G.
wemusthavethatG=K1_(Kn?3+I2)2H.SoassumethatGisaspanning subgraphofK1_(Kn?3+I2),thensinceK1_(Kn?3+I2)ismissing2n?5edges, subgraphofIr_Kr?2,wherer or(r?6)(r?3) n2?r2,orr2 2n?5=2(2r?2)?5.Simplifying,wegetr2?9r+18 4.Nown2?(2n?5) e(G) e(Ir_Kr?2)=
thatr 4.IfGisaspanningsubgraphofI4_K2,thensinceI4_K2ismissing 0.Thisimpliesthat3 r 6.Wehavealreadyassumed0
6=2(6)?6edges,Gcanbemissinguptoonemoreedge.Thus,G=I4_K2or G=(I4_K2)?ewhereeisanedge.BothofthesegraphsareinH.IfGisa spanningsubgraphofI5_K3,thensinceI5_K3ismissing10=2(8)?6edges, Gcanbemissinguptoonemoreedge.Thus,G=I5_K3orG=(I5_K3)?e whereeisanedge.BothofthesegraphsareinH.IfGisaspanningsubgraph ofI6_K4,thensinceI6_K4ismissing15=2(10)?5edges,wemusthavethat G=I6_K42H.Therefore,ifG ByLemma6.2,Hconsistsofgraphsofordernmissingatmost2n?5edges H.Clearly,H G. 2
withnoHamiltonianpath.Thenextstepistoprovethataconnectedgraph 60

this,weneedthefollowingtheoremwhichappearsinChartrandandLesniak[17]. missingatmost2n?5edgesiseitherinHorhasaHamiltonianpath.Toprove
tonian.Ifforalldistinctnonadjacentverticesuandv,d(u)+d(v)Theorem6.1LetGbeaconnectedgraphoforder3ormorethatisnotHamil- misapositiveinteger,thenPm+1 G. m,where
ingatmost2n?5edgeshaveaHamiltonianpath.Thefollowingtheoremcompletelycharacterizeswhichgraphsofordernmiss- Theorem6.2IfGisaconnectedgraphofordern eitherG2HorGhasaHamiltonianpath. 1ande(G) 2n?5,then
Proof.LetGbeaconnectedgraphofordernwithe(G) thenthetheoremisvacuouslytrue.Ifn=3,thenG2fP3;K3g,bothofwhich haveaHamiltonianpath.Ifn=4,thensincee(G) 3andGisconnected,G2 2n?5.Ifn 2,
fK1;3;P4;K1_(K1+K2);C4;K4?e;K4g.IfG=K1;3,thenG=K1_(K1+I2)2H. obviouslyGhasaHamiltonianpath.SoassumeGisnotcomplete.Letuandv Otherwise,GhasaHamiltonianpath.Soassumen beapairofnonadjacentverticesinGandH=G?fu;vg.SinceGisconnected, 5.IfG=Kn,then
eachofuandvhasaneighbor.IfN(u)=N(v)=fwg,thenthenumberofedges
edgesincidenttouorv.ThisimpliesthatG=K1_(Kn?3+I2)2H.Therefore, Hisn?3.Sinceweassumedthatuv62E(G),wehaveatotalof2n?5missing missingbetweenuandHisn?3andthenumberofedgesmissingbetweenvand
wecanassumethatthereexista;b2V(H)suchthata2N(u),b2N(v)and a6=b.WebreaktheproofintocasesbasedonthenumberofedgesinGincident touorv. 61

Case1.SupposethatthenumberofedgesinGincidenttouorvisn+1 ormore.ThenHisagraphonn?2verticesmissingatmost(2n?5)?(n+1)= (n?2)?4edges.ByTheorem2.9,HisHamiltonianconnected,andhencethere isaHamiltonian(a,b)-pathinH.So,uatogetherwiththeHamiltonian(a,b)-path inHtogetherwithbvisaHamiltonianpathinG.Therefore,wecanassumethat
n.ThenHismissingatmost(2n?5)?n=(n?2)?3edges.ByTheorem Case2.SupposethenumberofedgesinGincidenttouorvisexactly everypairofnonadjacentverticesisincidenttonomorethannedgesinG.
2.9,thesubgraphHhasaHamiltoniancycleC.LetC=u1u2:::un?2u1.Ifu (orv)isadjacenttoconsecutiveverticesonC,thenwecouldconstructa(n?1)cycleusingadetourthroughu(orv)andcouldthenattachv(oru)tothiscycle thepigeonholeprinciple,u(orv)wouldbeadjacenttoconsecutiveverticesonC. andproduceaHamiltonianpathinG.Soassumeneitherunorvisadjacentto consecutiveverticesonC.Thus,d(u) n?2 2andd(v) n?2
Next,wedetermineallpossiblevaluesforthedegreesofuandvinG.There 2;forotherwise,by
are2(n?2)possibleedgesfromfu;vgtoHandsincen?1oftheseedgesare
even,d(u)=n?4 missing,thereareexactlyn?3edgesfromfu;vgtoH.Nowd(u)+d(v)=n?3 andthefactthateachofuandvhasdegreeatmostn?2 2,andd(v)=n?2 2(Case2A);ornisoddandd(u)=d(v)=n?3 2impliesthateithernis
Case2A.Assumeniseven,neitherunorvisadjacenttoapairofconsecutive (Case2B). 2
verticesonCandd(u)=n?4
neighborsadjacentonC,thevertexvmustbeadjacenttoalternatingvertices ForvtobeadjacenttohalfoftheverticesinHandnothavetwoofits 2andd(v)=n?2 2.
62

onC,saytheverticesonCwithoddsubscripts.Sinced(u)=n?4
2,wecansay
withoutlossofgeneralitythatthevertexuisnotadjacenttou1.Nowwebreak
thiscaseintosubcasesbasedonwhetherornotN(u) N(v).
Case2A1.jN(u)\N(v)j=n?4
2.FirstwewillshowthatifthereisanedgeinH
betweenapairofverticeswithevenindices,thenwecanconstructaHamiltonian
pathinG.Ifu2u42E(G),thenwecanconstructaHamiltonianpathinGasfol-
lows:u4u2u1vu3uifn=6andu4u2u1vu3uu5:::un?2ifn>6.Ifu2un?22E(G),
thenwecanconstructaHamiltonianpathinGasfollows:uu3u4:::un?2u2u1v.
Ifu2u2k2E(G)forsome2

wecanconstructaHamiltonianpathinGasfollows:
Ifu2ku2k+22E(G)forsome1

Case2A2.jN(u)\N(v)j

totwoconsecutiveverticesonC,sayuu162E(G)anduu262E(G).Forutohave degreen?3 Sinced(u)=n?3
2andnothavetwoadjacentneighborsonC,thevertexumustbe 2andtherearen?2verticesonC,thevertexuisnotadjacent
adjacenttotheremainingverticesonCwithodd(oreven)subscripts,sayodd. Thus,wecanassumethatN(u)=fu3;u5;:::;un?2g.Nextwebreakthiscase intosubcasesbasedonwhetherornotN(u)=N(v). Case2B1.N(u)=N(v).FirstwewillshowthatifthereisanedgeinH betweenapairofverticeswithevenindices,thenwecanconstructaHamiltonian pathinG.Ifu2u2i2E(G)forsome2 HamiltonianpathinGasfollowsu1u2u4u3vu5uifn=7and i (n?3)=2,thenwecanconstructa
ifn>7(seeFigure6.3). u1u2u2iu2i?1u2i?2:::u3vu2i+1uun?2un?3:::u2i+2
Ifu4u2l2E(G)forsome2

su4
pps
u3
qsu2
qs
u1pp
p
p
pp
p
p
pp
p
pp
p
p
pp
p
p
pp
p
p
p
p
p
pp
p
p
pp
p
p
pp
p
p
p
p
p
pp
p
p
pp
p
p
pp
p
p
p
p
p
pp
p
p
ps
un?2q
q
q
q
q
q
q
q
q
q
s
u2l+1 qqs
u2lpps
u2l?1 qs
u2l?2qqqqqqqqqqs
u2k+1
pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppsu2k
qsu2k?1
qqqqqqqqqq
sv
qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq
ppppppppppp
p p qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq
ppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
s
u p
q
p q pp
p
q
Figure6.4.ThegraphGwhennisoddandu2ku2l2E(G)forsome
2

Case2B2.N(u)6=N(v).RecallthatN(u)=fu3;u5;:::;un?2g.Noweither adjacenttoconsecutiveverticesonC.Ifvu12E(G),thenvu1u2:::un?2uisa vu12E(G)orvu22E(G),otherwise,bythepigeonholeprinciple,vwouldbe HamiltonianpathinG.Ifvu22E(G),thenvu2u1un?2:::u4u3uisaHamiltonian pathinG.Therefore,wecanassumethateverypairofnonadjacentverticesis Case3.SupposethatthenumberofedgesinGincidenttouorv incidenttonomorethann?1edgesinG. isexactlyn?1.Thereare2(n?2)possibleedgesfromfu;vgtoH,and sincen?2oftheseedgesaremissing,wegetd(u)+d(v)=n?2.Thegraph
H=K1_(K1+Kn?4)forn Hismissingatmost(2n?5)?(n?1)=n?4edges.Therefore,e(H) n?2
Case3A.AssumeHisHamiltonian.Again,letC=u1u2:::un?2u1beaHamil- 2?(n?4)=(n?2)2?3(n?2)+4 2 5,orH=K2_I3. edges.ByTheorem2.8,eitherHisHamiltonian,
toniancycleinH.Ifu(orv)isadjacenttoconsecutiveverticesonC,thenwe couldconstructa(n?1)-cycleusingadetourthroughu(orv)andcouldthen attachv(oru)tothiscycleandproduceaHamiltonianpathinG.Soassume
bythepigeonholeprinciple,uwouldbeadjacenttoconsecutiveverticesonC. otherwiseeitheruorvwouldbeadjacenttoatleastn?1 neitherunorvisadjacenttoconsecutiveverticesonC.Nownmustbeeven,for
Soassumethatniseven,neitherunorvisadjacenttoapairofconsecutive 2verticesinH,sayu,and
verticesonCanduisnotadjacenttoapairofconsecutiveverticesonC,thevertex uisadjacenttoeitheralltheverticesonCwithoddsubscriptsorallthevertices verticesonC,andd(u)=d(v)=n?2 2.Sinceuisadjacenttoexactlyhalfofthe
onCwithevensubscripts.Thesameistrueforv.Bysymmetry,therearetwo 68

casestoconsider:N(u)=fu2;u4;:::;un?2gandN(v)=fu1;u3;u5;:::;un?3g (Case3A1);andN(u)=N(v)=fu1;u3;u5;:::;un?3g(Case3A2). Case3A1.N(u)=fu2;u4;:::;un?2gandN(v)=fu1;u3;u5;:::;un?3g.Inthis Case3A2.N(u)=N(v)=fu1;u3;u5;:::;un?3g.Firstwewillshowthatifthere case,vu1u2:::un?2uisaHamiltonianpathinG. isanedgeinHbetweenapairofverticeswithevenindices,thenwecanconstruct aHamiltonianpathinG.Ifu2kun?22E(G)forsome1 aHamiltonianpathinG.Ifun?4un?22E(G),thenun?4un?2un?3uu1:::un?5vis
u2kun?2un?3uu1:::u2k?1vu2k+1u2k+2:::un?4 k

qqun?2q
qqqqqq
u1qs qs qsu2
u2l u2l+1ppspsq
p qsu3
u2l?1 qq
pqs us pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp
qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq
sv pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppp qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq
p
q su4
u2l?2qqqqqqqqqqs ps u2k+1 pppp pppppppppppppppppppppppppppppppppppppppppppppppppppppppppppppsu2k psu2k?1 qqqqqqqqqq p
Figure6.5.ThegraphGwhennisevenandu2ku2l2E(G)forsome 1k

adjacenttoavertexinU,sayu1.Sinced(v) N(v) SupposethatN(u) U.
where1

casethereareveedgesbetweenfu;vgandtheverticesofH.Withoutlossof generality,eitherd(u)=1andd(v)=4,ord(u)=2andd(v)=3. Case3C1.Supposethatd(u)=1andd(v)=4.Thenbythesymmetryof H,wecanassumethateithervu562E(G)orvu162E(G).Ifvu5=2E(G),then u1u5u2u3vu4u1formsC6andwecanattachutothiscycletoformP7 vu1=2E(G),thenu1u4vu3u2u5u1formsC6andwecanattachutothiscycleto formP7 G. G.If
leasttwooftheverticesinVorvisadjacenttoatexactlyonevertexinV. Case3C2.Supposethatd(u)=2andd(v)=3.Theneithervisadjacenttoat
u1u5vu4u2u3u1formsC6andwecanaddutoformP7 IfvisadjacenttoatleasttwooftheverticesinV,sayu4andu5,then
Now,ifuisadjacenttoatleastoneofu3oru4,sayu3(asimilarconstruction IfvisadjacenttoexactlyonevertexinV,sayu5,thenN(v)=fu1;u2;u5g. G.
eitherN(u)=fu1;u2gorN(u)=fu1;u5g.IfN(u)=fu1;u2g,thenG= worksforu4)thenuu3u1u4u2u5visaHamiltonianpathinG.Ontheotherhand, ifuisnotadjacenttoatleastoneofu3oru4,thenwithoutlossofgenerality, hu1;u2i_(hv;u5i+u+u3+u4)=K2_(K2+I3)2H.IfN(u)=fu1;u5g,then Case4.SupposethatthenumberofedgesinGincidenttouorvisat mostn?2.IfGisHamiltonian,thenobviouslyGhasaHamiltonianpath.So uu5vu1u4u2u3isaHamiltonianpathinG.
assumeGisnotHamiltonian.Thereareatmostn?3edgesmissingfromfu;vg intoH.Therefore,thereareatleast2(n?2)?(n?3)=n?1edgesfromfu;vg intoH.So,d(u)+d(v) n?1.ByTheorem6.1,Pn G. 2
72

ordernmissingatmost2n?5edgeshaveaHamiltonianpath. Thenexttheorembuildsontheprevioustheorembystatingwhichgraphsof
aHamiltonianpathifandonlyif(G)>0andG62H. Theorem6.3LetGbeagraphofordern 1withe(G) 2n?5.ThenGhas
Proof.LetGbeagraphofordern thatGisnotconnected.LetAbeacomponentofGwithjAj=aandB=G?A. andG62H.ToapplyTheorem6.2,wemustshowthatGisconnected.Suppose 1withe(G) 2n?5.Assumethat(G)>0
minimumforthisquadraticfunctionoccurswhena=2ora=n?2andweget Since(G)>0,wehave2 e(G) an?a2 2n?4.Thus,thesizeofGmustbeatleast2n?4,which a n?2.Now,e(G) a(n?a)=an?a2.The
contradictstheassumptionthate(G) Theorem6.2,GhasaHamiltonianpath. AssumethatGhasaHamiltonianpath.Now(G)>0orelseGwouldnot 2n?5.Therefore,Gisconnected.By
haveaHamiltonianpath.ByLemma6.2andLemma6.3,G62H. NowthatweknowwhichgraphsoflargesizehaveaHamiltonianpath,wewill 2
6.2 addresstheproblemofdetermining(G;:Pj)whene(G) Graphsmissingcompletesubgraphsorastar 2j?5.
Aswehaveseeninthepreviouschapter,graphsoflargesizemissingastaror termining(G;:Cj).Thesamewillprovetobetruewhendetermining(G;:Pj)completesubgraphsarethegraphswhichneedtohandledasspecialcaseswhende- forgraphsoflargesize.First,let'sprovearesultforagraphwhosecomplement containsastarasasubgraph. 73

K1;j?1 Theorem6.4LetGbeagraphofordern.Ifj G,then(G;:Pj)=ln?1 j?1m. 2,e(G) n2?(2j?5),and
asetofverticeswhichinduceK1;j?1 Proof.Ifj=2,thenthetheoremisvacuouslytrue.Soassumej thereisexactlyonevertexv0ofdegreeatleastj?1inG.Therefore,v02X. G.SinceK1;j?1 Gande(G) 3.LetXbe
LetAbeacolorclassina:Pj-coloringofG. 2j?5,
j?4edges,andbyTheorem2.9,weknowthathAihasaHamiltonianpath. Therefore,forhAitobePj-free,wemusthavejAj Supposev062A.IfjAj j,thenhAiwouldbemissingatmost2j?5?(j?1)=
considerA?v0.IfjA?v0j Therefore,jA?v0j j,thenbythesameargument,wegetPj j?1.Supposev02A,and
onecolorclassofsizej(onecontainingv0),and(G;:Pj) j?1,whichimpliesjAj j.Thus,therecanbeatmost ln?j hA?v0i.
jbyusingtheverticesinXtogetherwithonevertexfromG?Xandformas Toshowtheinequalityintheotherdirection,formonecolorclassBofsize j?1m+1=ln?1 j?1m.
manyothercolorclassesaspossibleofsizej?1withtheremainingverticesin G?X.NowhBiisPj-freesincev0isnotadjacenttoanyothervertexinhBi. Therefore,(G;:Pj) Theprevioustheoremalongwithmostofthefollowingtheoremswillbeused ln?j j?1m+1=ln?1 j?1m: 2
inthenextsection.Thefollowingtheoremsaremoregeneralthanweneed,butwe wishtopresentthemintheirentiregenerality.Theirproofsarestraightforward andusethesameproofmethodology.Therefore,toquicklyreachthemainresults ofthischapter,theremainingtheoremsinthissectionwillbepresentedhere withoutproof.TheirproofscanbefoundinAppendixA.
74

Theorem6.5LetGbeagraphofordern hE(G)i=mKj+2 2,then(G;:Pj)=maxln?m j?1m;lnjm. 1.Ifj 2iseven,m 0,and
whereeisanedge,then(G;:Pj)=ln?1 Theorem6.6LetGbeagraphofordern.Ifj j?1m. 3isoddandhE(G)i=Kj+3 2?e,
Theorem6.7LetGbeagraphofordern.Ifj then(G;:Pj)=ln?1 j?1m. 4andhE(G)i=Ij?3_K2,
Theorem6.8LetGbeagraphofordern hE(G)i=mKj+4 2,then(G;:Pj)=maxln?2m j?1m;ln 1.Ifj j+1m. 4iseven,m 0,and
Theorem6.9LetGbeagraphofordern.Ifj Kj+4 2?e,whereeisanedge,then(G;:Pj)=ln?1 j?1m. 2isevenandhE(G)i=
Theorem6.10LetGbeagraphofordern.Ifj K1_(Kj2+K1),then(G;:Pj)=ln?1 j?1m. 6isevenandhE(G)i=
Theorem6.11LetGbeagraphofordern.Ifj Kj+2 2+K2,then(G;:Pj)=ln?1 j?1m. 6isevenandhE(G)i=
6.3Wewillusemostofthesetheoremstodetermine(G;:Pj)whene(G)islarge. Firstofall,wecanremovej?2edgesandshowthat(G;:Pj)doesnotdecrease Determining(G;:Pj)whene(G)islarge
byapplyingawellknownresultaboutHamiltonianpaths.Asanaside,ifj toproveTheorem6.12infullgenerality.However,itwaseasiertoproveTheorem thenTheorem6.12isanimmediateresultofTheorem6.13.Thuswedidnotneed 3,
6.12thisway. 75

Theorem6.12LetGbeagraphofordern.Ifj then(G;:Pj)=ln j?1m. 2ande(G) n2?(j?2),
LetB Proof.LetAbeacolorclassina:Pj-coloringofG.SupposethatjAj 2.9,wegetPj AwithjBj=j.NowhBiismissingatmostj?2edges.ByTheorem hBi,acontradiction.Therefore,jAj< j?1mifG=Kn?5_(K2+I3)andj=7, K1;j?1 G=Kn?(j?1)_(Kj?3+I2)andj G=(Kn?4_I4)?eandj=6, G=Kn?4_I4andj=6, G, 4,
>: ln G=(Kn?5_I5)?eandj=8, G=Kn?5_I5andj=8,
j?1m otherwise. orG=Kn?6_I6andj=10,and
76

Proof.Letj j=2,thenthetheoremisvacuouslytrue.Soassumej 2andGbeagraphofordernwithe(G) 3. n2?(2j?5).If
eisanedge,andbyTheorem6.6,weget(G;:P7)=ln?1 SupposethatK1;j?1
SupposethatG=Kn?5_(K2+I3)andj=7.ThenhE(G)i=K5?e,where G.ThenbyTheorem6.4,weget(G;:Pj)=ln?1 j?1m.
andbyTheorem6.7,weget(G;:Pj)=ln?1 SupposethatG=Kn?(j?1)_(Kj?3+I2)andj j?1m. 4.ThenhE(G)i=Ij?3_K2, j?1m.
G=Kn?6_I6andj=10.ThenhE(G)i=Kj+2 SupposethatG=Kn?4_I4andj=6,orG=Kn?5_I5andj=8,or
andonlyifn (G;:Pj)=ln?1 j?1mifn j.But,whenn

LetB aHamiltonianpath.SinceK1;j?16G,(hBi)>0. IfhBi=K2_(K2+I3),thenj=7.Now9=2(7)?5=e(hBi)and AsuchthatjBj=j.WewanttouseTheorem6.3toshowthathBihas
e(G) thatG=Kn?(j?1)_(Kj?3+I2),acontradiction. IfhBi=K1_(Kj?3+I2),thenj 2j?5implythatG=Kn?5_(K2+I3),acontradiction.
IfhBi=I4_K2,thenj=6and2j?5=7=e(hBi)+1.IfhBi= 4.Now2j?5=e(hBi),whichimplies
(I4_K2)?e,thenj=6and2j?5=7=e(hBi).NoweitherG=Kn?4_I4or
(I5_K3)?e,thenj=8and2j?5=11=e(hBi).NoweitherG=Kn?5_I5
G=(Kn?4_I4)?e,andwereachacontradiction.
orG=(Kn?5_I5)?e,andwereachacontradiction. IfhBi=I5_K3,thenj=8and2j?5=11=e(hBi)+1.IfhBi=
thatG=Kn?4_I6,acontradiction. ThereforehBi62HandbyTheorem6.3,hBihasaHamiltonianpath,a IfhBi=I6_K4,thenj=10.Now2(10)?5=15=e(hBi),whichimplies
contradiction.ThusjAj

coloring.Therefore,weabandonthisapproachtond(G;:Pj)forallgraphs
2j?2edges. oflargesizewiththeknowledgethatinordertoproduceagraphwhose:Pjchromaticnumberisln?2 j?1m,weneedtoremovesomewherebetween2j?4and
6.4 Determiningboundson(G;:Pj)giventhenumberof
Inthissection,wendanupperboundonthesizeofagraphgiventheconstraint edgesinagraph
Pj-freegraphofordern.TherstupperboundonthesizeofaPj-freegraph thatthe:Pj-chromaticnumberisatmostjn?k Tondtheupperbound,werstneedtodeterminethemaximumsizeofa j?1k.
wasdiscoveredin1959,byP.ErdosandT.Gallai[24].TheirtheoremstatesifG isaPj-freegraph,thene(G) improvedonthisboundforthemaximumsizeofaPj-freegraph.Infact,they foundtheleastupperboundanddeterminedwhichgraphsmetthatbound.The (j?2 2)n.In1975,R.FaudreeandR.Schelp[25]
theoremisrestatedhereforcompletenessandreference. Theorem6.14(FaudreeandSchelp,[25])IfGisagraphofordern=q(j?1)+r (0 equalityifandonlyifG=qKj?1+Kror(whenjiseven,q>0,andr=j=2or (j?2)=2),G=lKj?1+Kj?2 q,0 r

Corollary6.1IfGisaPj-freegraphofordern 1)(j?1). j 2,thene(G) (n?j+
Proof.LetmnbetheminimumnumberofedgesinthecomplementofaPj-free graphofordern=q(j?1)+r,whereq inductionthatforn statementofthecorollary.Nown j,wehavemn jimpliesthatq (n?j+1)(j?1),whichestablishesthe 0and0 r1.ByTheorem2.9(iii), j?2.Wewillshowby
mj mn= j?1.ByTheorem6.14,
= n2!?q q(j?1)+r j?1 2!? r2!
2 !?q =(q(j?1)+r)(q(j?1)+r?1)?q(j?1)(j?2)?r(r?1) j?1 2!? r2!
=q2(j?1)2+rq(j?1)+q(j?1)(r?1)?q(j?1)(j?2) 22
=q(j?1)[(q?1)(j?1)+2r] =q(j?1)[q(j?1)+2r?1?j+2] 2 2
Whenr

=q(j?1)2?q(j?1)(j?2) =q(j?1)
Nowbytheinductionhypothesis,(mn+1?mn)+mn j?1:
(n+1?j+1)(j?1). j?1+(n?j+1)(j?1)=
Wegetthefollowingimmediateresultforthe:Pj-chromaticnumberfrom2
Theorem6.14. Theorem6.15Letj 0 rkj?1 2.IfGisagraphofordern=k(j?1)+r(0 2+r2,then(G;:Pj) 2. k,
Therefore,(G;:Pj) Proof.Letj 0 rkj?1 2andletGbeagraphofordern=k(j?1)+r(0 2. 2+r2.ByTheorem6.14,wegetPj G. k,
NextweuseCorollary6.1toderiveanupperboundonthesizeofGgiven 2
Theorem6.16LetGbeagraphofordern (G;:Pj).
jn?k j?1k,thene(G) n2?k(j?1). 1,j 2andk 0.If(G;:Pj)
maximumnumberofedges.NowGmustcontainalledgesbetweenanytwocolor Proof.LetGbeagraphofordernsatisfying(G;:Pj) jn?k
classes.For,ifnot,wecouldaddanedgebetweentwocolorclassestoformanew j?1kwiththe
thiscontradictsthemaximalityofG.Therefore,wecanassumeallmissingedges graphHwith(H;:Pj) jn?k j?1kusingthesamecoloringweusedforG.But
81

esidewithinthecolorclasses.Furthermore,bymaximality,anycolorclassofsize atmostj?1isnotmissinganyedges. thenumberofcolorclassesofsizeifori=1;:::;j?1.Tominimizethenumber ofmissingedgesinourgraphG,weneedtominimizethenumberofmissingedges Letaj+ibethenumberofcolorclassesofsizej+ifori=0;:::n?jandbibe
mustbemissingatleast(i+1)(j?1)edgesforalli=0;:::;n?j. inallcolorclassesofsizeatleastj.ByCorollary6.1,acolorclassofsizej+i
1)aj+1+:::+(n?j+1)(j?1)aj+n?j=(j?1)Pn?j foramomentandderiveanotherinequality. Therefore,wewanttondalowerboundforthefunction(j?1)aj+2(j? i=0(i+1)aj+i.Let'sleavethis
n=j?1
=j?1 Xi=1ibi+n?j Xi=1(j?1)bi?j?1 Xi=0(j+i)aj+i
=(j?1)(j?1 Xi=1bi+n?j Xi=1[(j?1)?i]bi+n?j Xi=0aj+i)?j?1 Xi=1[(j?1)?i]bi+n?j Xi=0(i+1)aj+i+n?j Xi=0(i+1)aj+i
Xi=0(j?1)aj+i
=(j?1)$n?k n?k?j?1 Xi=1[(j?1)?i]bi+n?j j?1%?j?1 Xi=1[(j?1)?i]bi+n?j Xi=0(i+1)aj+i: Xi=0(i+1)aj+i
Now,rearrangingtheaboveequation,weget
sincePj?1 n?j
i=1[(j?1)?i]bi Xi=0(i+1)aj+i 0. k+j?1 Xi=1[(j?1)?i]bi k
thate(G) thenumberofmissingedgesinallcolorclassesisatleastk(j?1),whichimplies Usingtheaboveinequality,weget(j?1)Pn?j
n2?k(j?1). i=0(i+1)aj+i (j?1)k.Therefore,
82
2

LetGbeagraphofordernwithG=kK1;j?1,wherek Itisimmediatelyseenfromthefollowingexamplethatthisboundisattainable.
get(G;:Pj)=ln?k n=kj=k(j?1)+k,wehavethatj?1dividesn?kandbyTheorem4.3,we 0andj 2.Since
givenaboundfor(G;:Pj)forj thesizeofGgivenaparticular:Pj-chromaticnumber. j?1m=jn?k j?1k.WehavefoundanupperboundonthesizeofG 2.Thenexttheoremgivesalowerboundon
thene(G) Theorem6.17LetGbeagraphofordernandj n2?n2(j?2). 2.If(G;:Pj)>ln?1 j?1m,
1) Proof.Wewillprovethecontrapositive.LetGbeagraphwithe(G)

AAllthesetheoremsdeterminethe:Pj-chromaticnumberforagraphwhenits Appendix
complementcontainsaveryspecicsetofedges.Therstthreetheoremsand
Theorem6.5LetGbeagraphofordern thenaltwotheoremsareusedintheproofofTheorem6.13.
hE(G)i=mKj+2 2,then(G;:Pj)=maxln?m j?1m;lnjm. 1.Ifj 2iseven,m 0,and
Proof.Ifj=2,thenhE(G)i=mK2,whichimpliesn Wewillshowthereisnocolorclassofsizeatleast3.IfAwereacolorclassofsize 2m,orn?m ln2m.
mostoneedgeandP3 atleast3,thenthegraphinducedbyanythreeverticesinAwouldbemissingat consistoftheendpointsofanedgeinG.Therefore,(G;:P2) hAi,acontradiction.Also,acolorclassofsize2must
byTheorems5.3and2.2,weget(G;:Pj) n 2m,wegetn?m ln2m.Thus,(G;:P2)=maxn?m;ln2m.Ifj n?m.Since
Toshowtheinequalityintheotherdirection,weproduceaminimum:Pj- (G;:Cj)=maxln?m j?1m;lnjm. 4,then
coloring.ObservethatG=Wm+n?(j+2 i=1 2)m
Now,n=j+2 andSi=I1fori=m+1;m+2;:::;n?j+2 SiwhereSi=Ij+2
2m+s.Weconsidertwocases:j?2 2m.LetS=V(G)?Smi=1V(Si). 2m sand0 2fori=1;2;:::;m
colorclassofsizejbyusingalltheverticesfromSitogetherwithj?2 Ifj?2 2m s,thenn mjandln?m j?1m lnjm.Foreachi=1;:::;m,forma s

possible.Thus,wehave(G;:Pj) colorclassesareisomorphictoIj+2 partitiontheremainings?j?2 2mverticesintoasmanycolorclassesofsizej?1as 2_Kj?2 m+s?(j?2 2,whichbyLemma6.2,isPj-free.We
j?1 2)m=(j+2
2)m+s?m j?1 =ln?m
andT=S[Smi=m?r+1V(Si).Foreachi=1;:::;m?r,formacolorclassof If0 s

Theorem6.6LetGbeagraphofordern.Ifj whereeisanedge,then(G;:Pj)=ln?1 j?1m. 3isoddandhE(G)i=Kj+3 2?e,
Proof.ObservethathE(G)i=Kj+3 coloringofG.SupposethatjAj whereR=Kn?(j+3 2),S=Ij?1 2andT=K2.LetAbeacolorclassina:Pj- j+1.LetB 2?eifandonlyifG=R_(S+T),
1.Now,Bcontainsuptoj+3 2verticesfromS[Tandatleastj?1 AsuchthatjBj=j+
Kj?1 fromR.Forallcombinationsofverticesfromthesesets,wegethCi=Ij+3 2vertices
hCiandW=fw1;w2;:::;wj?1 2 hBi.LetU=fu1;u2;:::;uj+3 2gbethesetofverticesofdegreejinhCi.Now, 2gbethesetofverticesofdegreej?1 2in 2_
u1w1u2w2:::uj?1 fromS[T.SupposebywayofcontradictionthatacolorclassAofsizejcontains Now,eachcolorclassofsizejmustcontainstrictlymorethanj+3 2wj?1 2uj+1 2formsPj hCi,acontradiction.Therefore,jAj 4vertices j.
andatleast3j?3 by4.Ifj+3isdivisibleby4,thenAcontainsatmostj+3 atmostj+3 4verticesfromS[T.Sincejisodd,eitherj+3orj+1isdivisible
wegethDi=Ij+3 4verticesfromR.Forallcombinationsofverticesfromthesesets, 4verticesfromS[T
x1y1x2y2:::xj+3 3j?3 4inhDiandy1;y2;:::;y3j?3 4_K3j?3 4 4betheverticesofdegreej?1inhDi.Now, hAi.Letx1;x2;:::;xj+3
4yj+3 4yj+7 4yj+11 4:::y3j?3 4formsPj hDi,acontradiction.There- 4betheverticesofdegree
Amustcontainatleast3j?1 fore,j+1mustbedivisibleby4,Acontainsatmostj+1
thesesets,wegethDi=Ij+1 4verticesfromR.Forallcombinationsofverticesfrom 4verticesfromS[T,and
ofdegree3j?1 Now,x1y1x2y2:::xj+1 4inhDiandy1;y2;:::;y3j?1 4_K3j?1 4 4betheverticesofdegreej?1inhDi. hAi.Letx1;x2;:::;xj+1
4yj+1 4yj+5 4yj+9 4:::y3j?1 4formsPj hDi,acontradiction. 4bethevertices
Therefore,eachcolorclassofsizejmustcontainstrictlymorethanj+3 86
4vertices

therecanbeatmostonecolorclassofsizejand(G;:Pj) fromtheS[T,whichisstrictlymorethanhalftheverticesofS[T.Therefore, ln?j
formonecolorclassAofsizejbyusingalltheverticesinStogetherwithj?3 Next,wewillshowtheinequalityintheotherdirection.ColorGasfollows: j?1m+1=ln?1 j?1m.
sincehAicontainsj?1 verticesfromRandbothverticesfromT.Now,hAi=Kj?3 2verticesofdegreej?3 2withexactlyj?3 2commonneighbors, 2_Ij?1 2+K2and 2
hAiisPj-free.Withtheremainingvertices,formasmanycolorclassesofsize j?1aspossible.Withthiscoloring,weget(G;:Pj) (G;:Pj)=ln?1 j?1m. ln?j j?1m+1=ln?1 j?1m.So,
Theorem6.7LetGbeagraphofordern.Ifj 4andhE(G)i=Ij?3_K2, 2
Proof.Letj then(G;:Pj)=ln?1 4andGbeagraphofordernsuchthathE(G)i=Ij?3_ j?1m.
fr1;r2;:::;rn?j+1gbetheverticesofR,fs1;s2;:::;sj?3gbetheverticesofS, K2.ThenG=R_(S+T)whereR=Kn?(j?1),S=Kj?3,T=I2.Let andft1;t2gbetheverticesofT.LetAbeacolorclassina:Pj-coloringof andt262B,thenhBiiscomplete,whichimpliesthatPj G.SupposethatjAj tion.Ift12Bandt262B,thenB=ft1;r1;:::;rp;s1;s2;:::;sj?1?pgforsome j+1.LetB AsuchthatjBj=j+1.Ift162B
3 p j?1.Now,t1r1r2r3:::rps1s2:::sj?1?pformsPj
hBi,acontradiction.Therefore,wemusthaveft1;t2g
verticesfromS,thesubgraphhBimustcontainatleasttwoverticesr1andr2 B.SinceBcancontainatmostj?3 hBi,acontradic-
fromR.SoB=ft1;t2;r1;:::;rt;s1;s2;:::;sj?1?pgforsome2 Now,t1r1t2r2r3:::rps1s2:::sj?1?pformsPj hBi,whichagainisacontradic- p j?1.
87

tion.Therefore,wemusthavejAj Gismissingatmost2j?5edges,therecanbeatmostonecolorclassofsize hAiismissingfewerthanj?1edges,thenhAihasaHamiltonianpath.Since j.SupposejAj=j.ByTheorem2.9,if
j.Thus,(G;:Pj) coloring.FormonecolorclassAofsizejwithA=ft1;t2;r1;s1;s2;:::;sj?3g. ln?j
NowhAi=K1_(Kj?3+I2),whichbyLemma6.2,isPj-free.Wepartitionthe j?1m+1=ln?1 j?1m.Nextweproduceaminimum:Pj-
remainingn?jverticesintoasmanycolorclassesofsizej?1aspossible.Thus, wehave(G;:Pj) Inearliertheoremswehavederivedavalueofln?m ln?j j?1m+1=ln?1 j?1m. j?1mforthe:Pj-chromatic 2
numberforgraphsofordernwhosecomplementsconsistofmpairwisevertex
whosecomplementsconsistofmpairwisevertexdisjointcopiesofacomplete disjointcopiesofacompletegraphofaparticularorder.Thefollowingtheorem
graphofaneworder. derivesanewvalueforthe:Pj-chromaticnumber,namelyln?2m j?1m,forgraphs
Theorem6.8LetGbeagraphofordern hE(G)i=mKj+4 2,then(G;:Pj)=maxln?2m j?1m;ln 1.Ifj j+1m. 4iseven,m 0,and
andSi=I1fori=m+1;m+2;:::;n?j+4 Proof.ObservethatG=Wm+n?(j+4 i=1 2)m
coloringofG.SupposethatjAj>j+1.LetB SiwhereSi=Ij+4 2m.LetAbeacolorclassina:Pj- AsuchthatjBj=j+2.Then 2fori=1;2;:::;m
forall3 orhBi=Wki=1Iqiwhere1 eitherhBi=Wki=1Iqiwhereq1=(j+4)=2,and1 q1 (j+2)=2,1 q2 qi(j+2)=2,and1 j=2forall2 iqik(1), verticesfromIq1.If(2),thenformC2 i k(2).If(1),thenformC1BsuchthatjC2j=jbyremovingone BsuchthatjC1j=jbyremovingtwo j=2
88

Also,hC2i=Iq1?1_Iq2?1_Wki=3Iqi,andmaxifqig vertexfromeachofIq1andIq2.Now,hC1i=Iq1?2_Wki=2Iqi,andmaxifqig j2.ByLemma5.1,hC1i j2. andhC2iareHamiltonian.ThiscontradictstheassumptionthathAiisPj-free. SojAj LetAbeacolorclassofsizej+1.WewillshowthatIj+4 j+1and(G;:Pj) ln j+1m.
and1 wayofcontradictionthatIj+4 26hAi.ThenhAi=Wki=1Iqiwhereq1 2 hAi.Supposeby
onevertexfromIq1.Now,hBi=Iq1?1_Wki=2Iqi,andmaxifqig qi j=2forall2 i k.FormB AsuchthatjBj=jbyremoving j2.ByLemma (j+2)=2,
5.1,hBiisHamiltonian.ThiscontradictstheassumptionthathAiisPj-free.So Ij+4 Ij+2 2LetCbeacolorclassofsizej.ByLemma5.1,eitherhCiisHamiltonianor hAi.
classesofsizejina:Pj-coloringofG.Sinceeachcolorclassofsizej+1must 2Letabethenumberofcolorclassesofsizej+1andbbethenumberofcolor hCi.SincehCiisnotHamiltonian,wemusthavethatIj+2 2 hCi.
containalltheverticesfromSiforsome1 vertexdisjoint,wemusthavea sizejmustcontainj+2 m.Wehavejustshownthateachcolorclassof i m,andsincecolorclassesare
fromsomeSi(1 usetheverticesofSi(1 i2independentvertices,whichismorethanhalfthevertices Sinceacolorclassofsizeatleastjrequiresmorethanhalftheverticesfromsome m)sincej+2 i m)toformtwodierentcolorclassesofsizej. 2>12j+4 2 whenj 2.Thuswecannot
Si(1 inequalityanda Now, i m)andcolorclassesarevertexdisjoint,wegeta+b mimplythat2a+b 2m. m.This
(G;:Pj) &n?a(j+1)?bj j?1 '+a+b=&n?2a?b 89
j?1'&n?2m j?1'

coloring.RecallthatG=Wm+n?(j+4 Hence,(G;:Pj) Toshowtheinequalityintheotherdirection,weproduceaminimum:Pj- maxln j+1m;ln?2m
i=1 2)m j?1m.
Si=I1fori=m+1;m+2;:::;n?j+4 n=j+4 SiwhereSi=Ij+4
2m+s.Weconsidertwocases:j?2 2m.LetS=V(G)?Smi=1V(Si).Now, 2m sand0 2fori=1;2;:::;mand
formacolorclassofsizej+1byusingalltheverticesfromSitogetherwithj?2 Ifj?2 2m s,thenn m(j+1)andln?2m j?1m ln j+1m.Foreachi=1;:::;m, s

thosevertices.ToshowhBiisPj-free,wewillrstshowthatjBj jBj=jTj?$n j+1%j?2 2 j.Now,
=s+r(j+1)?mj?2 =s+rj+4 2 ?(m?r)j?2 2 2
j.LetB 2),S=Ij2,andT=K2.LetAbeacolorclassina:Pj-coloring 2?eifandonlyifG=R_(S+T),where
containsuptoj+4 2verticesfromS[Tandatleastj?2 AsuchthatjBj=j+1.Now,B
Letu1;u2;:::;uj2betheverticesofdegreej?2 combinationsofverticesfromthesesets,wegethCi=Ij2+K2_Kj?2 2inhCi,v1;v2betheverticesof 2verticesfromR.Forall 2 hBi.
91

u1w1u2w2:::uj?2 degreej2inhCi,andw1;w2;:::;wj?2 2wj?2 2v1v2formsPj2betheverticesofdegreejinhCi.Now, S[T.SupposebywayofcontradictionthatacolorclassAofsizejcontains Noweachcolorclassofsizejmustcontainstrictlymorethanj+4 hCi,acontradiction.Therefore,jAj 4verticesfrom j.
andtherefore,atleast3j?4 by4.Ifjisdivisibleby4,thenAcontainsatmostj+4 atmostj+4 4verticesfromS[T.Sincejiseven,eitherjorj+2isdivisible
thesesets,wegethDi=Ij+4 4verticesfromR.Forallcombinationsofverticesfrom 4verticesfromS[T
Now,x1y1x2y2:::xj+4 ofdegree3j?4 4inhDiandy1;y2;:::;y3j?4 4_K3j?4 4 4betheverticesofdegreej?1inhDi. hAi.Letx1;x2;:::;xj+4
4yj+4 4yj+8 4yj+12 4:::y3j?4 4formsPj hDi,acontradiction.So 4bethevertices
mustcontainatleast3j?2 thesesets,wegethDi=Ij+2 j+2mustbedivisibleby4,Acontainsatmostj+2 4verticesfromR.Forallcombinationsofverticesfrom 4verticesfromS[T,andA
Now,x1y1x2y2:::xj+2 ofdegree3j?2 4inhDiandy1;y2;:::;y3j?2 4_K3j?2 4 4betheverticesofdegreej?1inhDi. hAi.Letx1;x2;:::;xj+2
4yj+2 4yj+6 4yj+10 4:::y3j?2 4formsPj hDi,acontradiction. 4bethevertices
fromS[T,whichisstrictlymorethanhalftheverticesinS[T.Therefore,there canbeatmostonecolorclassofsizejand(G;:Pj) Therefore,eachcolorclassofsizejmustcontainstrictlymorethanj+4 ln?j j?1m+1=ln?1 4vertices
hE(G)i=P3.Formonecolorclassofsize2usingthetwononadjacentver- Next,wewillshowtheinequalityintheotherdirection.Ifj=2,then j?1m.
ticesinGandputtheremainingn?2verticesinton?2colorclasses.Therefore, sizejbyusingalltheverticesinStogetherwithj?4 verticesfromT.Now,hAi=Kj?4 (G;:Pj) n?1.Ifj 4,thencolorGasfollows:formonecolorclassAof
2_Ij2+K2andhAiisaspanningsubgraphof 2verticesfromRandboth
92

So,(G;:Pj)=ln?1 Ij+2 manycolorclassesofsizej?1aspossible.Weget(G;:Pj) 2_Kj?2 2.ByLemma6.2,hAiisPj-free.Withtheremainingvertices,formas
j?1m. ln?j j?1m+1=ln?1 j?1m.
Theorem6.10LetGbeagraphofordern.Ifj 6isevenandhE(G)i= 2
Proof.Assumej K1_(Kj2+K1),then(G;:Pj)=ln?1 6isevenandGisagraphofordernsuchthathE(G)i= j?1m.
T=I1.Letfr1;r2;:::;rn?(j2+2)gbetheverticesofR,fs1;s2;:::;sj2gbethe verticesofdegree1inS,sbethevertexofdegreej2inS,andtbethevertex K1_(Kj2+K1).ThenG=R_(S+T)whereR=Kn?(j2+2),S=K1;j2and
inT.NowH=Kn?(j2+2)_(K2+Ij2) (G;:Pj) Toshowtheinequalityintheotherdirection,colorGasfollows:formone (H;:Pj)=ln?1 j?1m. GandbyTheorems6.9and2.4,
colorclassAofsizejwithA=fs1;s2;:::;sj2;s;r1;r2:::;rj?4 Kj?4 6.2,isPj-free.Withtheremainingverticesformasmanycolorclassesofsizej?1 2_(K1;j2+I1),whichisaspanningsubgraphofIj+2 2_Kj?2 2;tg.NowhAi=
aspossible.Withthiscoloring,weget(G;:Pj) ln?1 j?1m. 2andbyLemma
Theorem6.11LetGbeagraphofordern.Ifj Kj+2 2+K2,then(G;:Pj)=ln?1 j?1m. 6isevenandhE(G)i=
LetAbeacolorclassina:Pj-coloringofG.SupposethatjAj Proof.Assumej Kj+2 2+K2.ThenG=R_(S_T)whereR=Kn?(j+2 6isevenandGisagraphofordernsuchthathE(G)i=
B AwithjBj=j+1.NowBcontainsuptoj+2 2+2verticesfromS[Tand 2+2),S=Ij+2 2andT=I2. j+1.Let
93

atleastj?4 getC=Kj?4 2verticesfromR.Forallcombinationsofverticesfromthesesets,we
degreej?1inC.Now,y1x1y2x2:::yj?4 inC,y1;y2;:::;yj+2 2_(Ij+2 2betheverticesofdegreej2inC,andz1;z2betheverticesof 2_I2) hBi.Letx1;x2;:::;xj?4
2xj?4 2yj?2 2z1yj2z2yj+2 2betheverticesofdegreej
jAjToformacolorclassofsizej,wemustusealltheverticesinS.ForifA j. 2formsPj C.Hence
containsatmostj2verticesfromS,thenforv2A,wehavedhAi(v) byDirac'sTheorem(Theorem2.7),hAiisHamiltonian,acontradiction.Thus, j2and,
ln?j therecanbeatmostonecolorclassofsizej.Therefore,weget(G;:Pj)
colorclassofsizejbyusingallj+2 j?1m+1=ln?1 Toshowtheinequalityintheotherdirection,colorGasfollows:formone j?1m.
verticesfromR.NowhAi=Ij+2 theremainingverticesformasmanycolorclassesofsizej?1aspossible.With 2_Kj?2 2verticesfromS,onevertexfromT,andj?4 2,whichbyLemma6.2,isPj-free.With 2
thiscoloring,weget(G;:Pj) ln?1 j?1m. 2
94

B1.J.Akiyama,H.Era,S.GrevacioandM.Watanabe,Pathchromaticnumbers References
2.M.O.Albertson,R.Jamison,S.T.HedetniemiandS.C.Locke,Thesub- ofgraphs,J.GraphTheory13(5):569{575(1989).
3.J.A.AndrewsandM.S.Jacobson,Onageneralizationofchromaticnumber, chromaticnumberofagraph,Discr.Math.74:33{49(1989).
4.J.A.AndrewsandM.S.Jacobson,Onageneralizationofchromaticnumber Congr.Numer.47:33{48(1985).
5.K.I.AppelandW.Haken,Thesolutionofthefour-color-mapproblem, andtwokindsofRamseynumbers,ArsCombin.23:97{102(1987).
6.P.Baldi,Onageneralizedfamilyofcolorings,GraphsandCombinatorics Freeman,California,1977.
7.L.W.BeinekeandA.T.White,SelectedTopicsinGraphTheory,Academic 6:95{110(1990).
8.L.D.BodinandA.J.Friedman,SchedulingofcommitteesfortheNewYork Press,London,1978.
9.J.A.Bondy,VariationsontheHamiltoniantheme,Can.Math.B.15:57{62 StateAssembly,Technicalreport,StateUniversityofNewYork,1971.
10.J.A.BondyandU.S.R.Murty,GraphTheorywithApplications,Elsevier (1972).
NorthHolland,Inc.,NewYork,1976. 95

11.O.Borodin,Cycliccoloringofplanegraphs,Discr.Math.100:281{289
12.I.BroereandC.M.Mynhardt,Generalizedcoloringsofouterplanarandpla- (1992).
nargraphs,inGraphTheorywithApplicationstoAlgorithmsandComputer13.J.I.BrownandD.G.Corneil,Onuniquely?Gk-colorablegraphs,Quaes-
Science,WileyIntersci.Publ.,NewYork,1985.
14.J.I.BrownandD.G.Corneil,Ongeneralizedgraphcolorings,J.Graph tionesMathematicae15:477{487(1992).
15.G.Chartrand,D.P.GellerandS.T.Hedetniemi,Ageneralizationofthe Theory11:87{99(1987).
16.G.Chartrand,H.V.KronkandC.E.Wall,Thepoint-arboricityofagraph, chromaticnumber,Proc.Camb.Philos.Soc.64:265{271(1968).
17.G.ChartrandandL.Lesniak,GraphsandDigraphs,Wadsworth,Inc.,Cal- Isr.J.Math.6:169{175(1968).
18.R.Cimikowski,Onheuristicsfordeterminingthethicknessofagraph,Inf. ifornia,1986.
19.R.J.Cook,Pointpartitionnumbersandgirth,Proc.Am.Math.Soc. Sciences85(1):87(1995).
20.M.B.CozzensandF.S.Roberts,T-coloringsofgraphsandthechannel 49:510{514(1975).
assignmentproblem,Congr.Numer.35:191{208(1982). 96

21.K.C.DargenandK.Fraughnaugh,Conditionalchromaticnumberswith
22.G.A.Dirac,Sometheoremsonabstractgraphs,Proc.LondonMath.Soc. forbiddencycles,LinearAlgebraitsAppl.217:53{66(1995).
23.G.S.Domke,S.T.Laskar,S.T.HedetniemiandK.Peters,Thepartite- 2:69{81(1952).
24.P.ErdosandT.Gallai,Onmaximalpathsandcircuitsofgraphs,Acta chromaticnumberofagraph,Congr.Numer.53:235{246(1986).
25.R.J.FaudreeandR.H.Schelp,PathRamseynumbersinmulticolorings,J. Math.Acad.Sci.Hungar10:337{356(1959).
26.M.R.GareyandD.S.Johnson,Anapplicationofgraphcoloringtoprinted Comb.Th.B19:150{160(1975).
27.M.R.GareyandD.S.Johnson,ComputersandIntractability,AGuideto circuittesting,IEEETrans.CAS23:591{599(1976).
28.W.K.Hale,Frequencyassignment:theoryandapplications,Proc.IEEE theTheoryofNP-Completeness,W.H.Freeman,NewYork,1979.
29.F.Harary,GraphTheory,Addison-Wesley,1969. 68:1497{1514(1980).
30.F.Harary,Conditionalcolorabilityingraphs,inGraphsandApplications, Proc.FirstColo.Symp.GraphTheory(F.HararyandJ.Maybee,eds.), WileyIntersci.Publ.,NewYork,1985.
97

31.F.HararyandK.Fraughnaugh(Jones),ConditionalcolorabilityII:bipartite
32.F.HararyandK.Fraughnaugh(Jones),Degreeconditionalbipartitionnum- variations,Congr.Numer.50:205{218(1985).
33.F.HararyandL.Hsu,Conditionalchromaticnumberandgraphoperations, bersingraphs,Congr.Numer.55:39{50(1986).
34.F.HararyandP.C.Kainen,Ontriangularcoloringsofaplanargraph,Bull. Bull.Inst.Math.Acad.Sinica19(2):125{134(1991).
35.S.T.Hedetniemi,Onpartitioningplanargraphs,Can.Math.B.11(2):203- CalcuttaMathSoc.69:393{395(1977).
36.S.T.Hedetniemi,Disconnectedcoloringsofgraphs,inCombinatorialStruc- 211(1968).
turesandTheirApplications,Proc.CalgaryInternat.Conf.(R.Guy,ed.),
37.J.P.Hutchinson,Coloringordinarymaps,mapsofempires,andmapsof GordonandBreach,NewYork,1970.
38.G.JohnsandF.Saba,Onthepath-chromaticnumberofagraph,inGraph themoon,Math.Mag.66(4):211{225(1993).
1986. TheoryanditsApplications:EastandWest(M.F.Capobianco,ed.),Jinan,
39.F.KramerandH.Kramer,Einfarbungsproblemderknotenpunkteeines graphenbezuglichderdistanzp,Rev.RoumaineMath.PuresAppl.14(2):1031{ 1038(1969). 98

40.L.Lesniak-FosterandJ.Straight,Thecochromaticnumberofagraph,Ars
41.D.R.LickandA.R.White,k-degenerategraphs,Can.J.Math.22(5):1082{ Combin.3:39{45(1977).
42.C.MynhardtandI.Broere,Generalizedcoloringsofgraphs,inGraphTheory 1096(1970).
withApplicationstoAlgorithmsandComputerScience(Y.Alavi,ed.),Wiley
43.R.J.OpsutandF.S.Roberts,Ontheeetmaintenance,mobileradio Intersci.Publ.,NewYork,1985.
frequency,taskassignment,andtracphasingproblems,inTheTheory York,1981. andApplicationsofGraphs(G.Chartrand,ed.),WileyIntersci.Publ.,New
44.O.Ore,NoteonHamiltoniangraphs,Am.Math.Mo.67:55(1960). 45.O.Ore,Arccoveringsofgraphs,Ann.Mat.Pura.Appl.55:315{321(1961). 46.R.J.Pennotti,ChannelAssignmentinMobileCommunicationSystems,
47.F.S.Roberts,AppliedCombinatorics,PrenticeHall,1984. PhDThesis,PolytechnicInstituteofNewYork,1976.
48.H.SachsandM.Schauble,KonstructionvonGraphenmitgewissenfarbungseigenschafte,inBeitragezurgraphentheorie,(Kolloquium,Maneback,1967)(H. 49.E.Sampathkumar,Generalizationsofindependenceandchromaticnumbers Sachs,H.VossandH.Walter,eds.)Teubner,Leipzig,1968.
ofagraph,Discr.Math.115:245{251(1993). 99

50.E.Sampathkumar,S.Prabha,S.NeeralagiandV.Venkatachalam,AgennatakUniv.Sci.22:44{49(1977).eralizationofthechromaticandlinechromaticnumbersofagraph,J.Kar 51.D.B.West,IntroductiontoGraphTheory,PrenticeHall,NewJersey,1996.
100