MC0063(B)-unit11-fi

ModernApplicationsusingDiscreteMathematicalStructures Unit11

SikkimManipalUniversity PageNo:234

Unit11 Planarity,ColoringandPartitioningStructure

11.1 Introduction

Objectives

11.2 PlanarandDualGraphs

11.3 Coloring

11.4 Partitioning

11.5 AlgorithmsforChromaticNumber

SelfAssessmentQuestions

11.6 Summary

11.7 TerminalQuestions

11.8 Answers

11.1Introduction

In this unit we consider the embedding (drawing without crossings) of

graphsonsurfaces,especiallytheplane.Westudythepropertiesofplanar

graphsandobservesomeKuratowskisgraphs,andthecharacterizationof

planar graphs. We also study the coloring of a graph, some of its

characterizations and properties. We provide algorithm for finding a

chromaticnumberofagivenconnectedgraph.

Objectives:

Attheendoftheunitthestudentmustbeableto:

i) Studytheplanerepresentationofgraphs.

ii) CharacterizetheKuratowskisgraphs.

iii) Findachromaticnumberofagivengraph.

iv) Apply the concepts to some real world problems like four color

problem.



11.2PlanarandDualGraphs

Consideragraph G1=(V,E, j).Supposevertexset V ={a,b,c,d,e}

andanedgesetE={1,2,3,4,5,6,7}.Therelationshipbetweenthetwo

sets V and E isdefinedbythemapping j.

Here 1 {a, c} means that object 1 from E is mapped onto the pair

(unordered){a,c}ofobjectsfrom V.G1=(V,E, j)canberepresentedby

meansofageometricrepresentationgiveninfigure11.1.

11.2.1Definition

i) AgraphG issaidtobeaplanargraph ifthereexistssomegeometric

representationofG whichcanbedrawnonaplanesuchthatnotwoof

itsedgesintersect.

ii) Agraphthatcannotbedrawnonaplanewithoutacrossoverbetween

itsedgesiscalledanonplanargraph.

iii) A drawing of a geometric representation of a graph on any surface

suchthatnoedgesintersectiscalledanembedding.AgraphGisnon

planar if all the possible geometric representations of G can not be

embeddedinaplane.

1 {a,c}5 {b,d} j =2 {c,d}6 {d,e}

3 {a,d}7 {b,e}4 {a,b}

b

3

a

c d

e

1

2

4

5

6

7Fig11.1



11.2.2Definition

AgeometricgraphGis planarifthereexistsagraphG*whichisisomorphic

toG thatcanbeembeddedinaplane.Otherwise, G isnonplanar.An

embeddingofaplanargraphGonaplaneiscalledaplanerepresentation

of G.Agraphisdrawnontheplanewithoutintersectingedgesiscalleda

planegraph.

11.2.3 Example

ConsiderthegraphgiveninFig.11.2.3A.

Ifwedraw theedge fout side thequadrilateralwhile theotheredgesare

unchanged,thenwegettheFig2.3B.ThegraphgiveninFig2.3Bcanbe

embeddedinaplane.SothegraphgiveninFig2.3Bisplanar.Therefore

thegraphgiven inFig2.3A isaplanargraph. Thegraph2.3A isnota

planegraphbutitisplanar.

11.2.4 Example

Observe the graphs (i) and (ii) given here. These two graphs are

isomorphictoeachother.Buttheyaredifferentgeometricrepresentationsof

thesamegraph.HereGraph(i)isaplanerepresentation,andtheGraph

(ii)isnotaplanerepresentation.

d

a

b

c

e

f

Fig.11.2.3A

fe

Fig11.2.3B



11.2.5 Example

Considerthefollowinggraphs G1,G2 andG3.ClearlyG1andG3areplanar

G3 is a geometric representation of G2 and G3 can be embedded in a

plane.Therefore G2 isalsoaplanargraph.

11.2.6Problem:Thecompletegraphof5vertices(denotedby K5)isa

nonplanargraph.

Proof:SupposethevertexsetofK5 is{v1, v2, v3, v4, v5}.

Sincethegraphiscomplete,wegetacircuitgoingfromv1 to v2 to v3 to

v4 to v5 tov1.Thatis,wehaveapentagon(giveninGraph(a)).

Nowthispentagonmustdividetheplaneofthepaperintotworegions,one

insideandtheotheroutside.

Graph(i) Graph(ii)

G1G2

G3



Wehavetoconnect v1 and v3 byanedge.Thisedgemaybedrawn

inside(or)outsidethepentagon(withoutintersectingthe5edgesofGraph

A).Letusselectalinefrom v1 to v3 insidethepentagon.(ifwechoose

outside, thenweendupwithasimilarargument).NowwehaveGraphB.

Nowwehavetodrawanedgefromv2 to v4,andanotheronefrom v2

to v5. Since neither of these edges can be drawn inside the pentagon

withoutcrossingovertheedgesthathavealreadydrawn,wehavetodraw

boththeseedgesoutsidethepentagon.

NowwehaveGraphC.Theedgeconnectingv3 andv5 cannotbedrawn

outsidethepentagonwithoutcrossingtheedgebetween v2and v4.So v3

and v5 havetobeconnectedwithanedgeinsidethepentagon.Nowwe

havetheGraphD.

v1

v2

v3

v4v5

GraphA

v2

v1v3

v4v5

GraphB

v2

v1 v3

v4v5

Graph C

v2

v1 v3

v4v5

Graph D

v2

v1 v3

v4v5

Graph E



Nowwehavetodrawanedgebetween v1 and v4. It isclearthatthis

edge can not be drawn either inside (or) outside the pentagon without a

crossover(ObserveGraphE).

Thusthisgraphcannotbeembeddedinaplane.Hencethecompletegraph

K5 on5verticesisnonplanar.

11.2.7Definition

A Jordancurve is the continuousnonselfintersecting curvewhoseorigin

andterminuscoincide.

11.2.8Example

i) C1isnotaJordancurve(sinceitintersectsitself).

ii) C2 is not a Jordan curve (because origin and terminus are not

coincide).

iii) C3 isaJordancurve.

11.2.9Definition

Considerthegraphs(i)and(ii)givenhere.Theseareregular(eachvertex

isofsamedegree)connectedgraphswith6verticesand9edges.These

two graphs are isomorphic and so they represent the same graph. This

graphiscalledastheKuratowskis2nd graph. Thesetwographsshowthe

twocommongeometricrepresentationsoftheKuratowski'ssecondgraph.

intersection

C1

origin terminus

C2

C3



TheKuratowskis2ndgraphisdenotedby K3,3.

11.2.10Problem:Provethat theKuratowskis 2nd graph (that isK3,3) isa

nonplanargraph.

Solution:ObservetheKuratowskis2ndgraphK3,3.

Itisclearthatthegraphcontainssixvertices vi,1 i 6andthereare

edges 21vv , 32vv , 43vv , 54vv , 65vv , 16vv .Nowwehave

aJordancurve.Soplaneofthepaperisdividedintotworegions,oneinside

andtheotheroutside.Since v1 isconnectedto v4,wecanaddtheedge

(i) (ii)

v3

v4

v1 v2

v5

v6

K3,3

v3

v4

v1 v2

v5

v6

Fig.11.2.10A

inside

outside

v3

v4

v1 v2

v5

v6

Fig.11.2.10B

v5

v3

v4

v1 v2

v6

Fig 11.2.10C



41vv in either inside or outside (without intersecting theedges already

drawn).Letusdraw 41vv inside.(Ifwechooseoutside,thenweend

upwiththesameargumentNowwehavetheFig.11.2.10(B))Nextwehave

todrawanedge 52vv andalsoanotheredge 63vv .Firstwedraw

52vv .Ifwedrawitinside,wegetacrossovertheedge 41vv .Sowe

drawitoutside.ThenwegettheFig.11.2.10(C).Stillwehavetodrawan

edgefrom v3 to v6.If 63vv drawninside,itcrosstheedge 41vv

(seetheFig.11.2.10(D)).

Sowecannot draw it inside. Soweselect the caseofdrawing 63vv

crosstheedge 52vv (seetheFig.11.2.10(E)).Thus 63vv cannotbe

drawneitherinsideoroutsidewithoutacrossover.Hencethegivengraph

isnotaplanargraph.

11.2.11PropertiesofKuratowskis1stand2ndgraphs:

i) Bothareregulargraphs.

ii) Botharenonplanner.

iii) Removalofoneedge(or)vertexmakeseachaplanargraph.

iv) The two graphs are nonplanar with the smallest number of edges.

Thusbotharethesimplestnonplanargraphs.

v3

v1 v2

v4v5

v6

Fig11.2.10(D)

v1 v2

v3

v4v5

v6

Fig.11.2.10(E)



Observation:Aplanegraph G dividestheplaneintonumberofregions

[alsocalledwindows,faces,ormeshes]asshowninfollowingfigure.

Aregioncharacterizedbythesetofedgesorthesetofverticesformingits

boundary.

Intheabovefigure,thenumbers1,2,3,4,5,6standfortheregions.

Aregionisnotdefinedinanonplanargraph.

Forexample,wecannotdefineregioninthegraphgivenbelow.

Observation:Theportionof theplane lyingoutsideagraphembedded in

plane (such as the region 4 in the first graph) is called an infinite [or

unbounded or outer or exterior] region for that particular plane

representation.

11.2.12Example

Observe thegraphgiven in theFig.11.2.12.Clearly ithas4 regions.The

region4isaninfiniteregion.

6

4

5

321

v6

v1

v2 v3

v4

v5

1 2

3

4

Fig.11.2.12



Ifweconsider twodifferentembeddingsofagivenplanargraph, then the

infiniteregionsofthesetworepresentationsmaybedifferent.

11.2.13Theorem:Agraphcanbeembeddedinthesurfaceofasphere if

andonlyifitcanbeembeddedonaplane.

11.2.14Theorem:Aplanargraphmaybeembedded inaplanesuch that

any specified region (that is, specified by the edges forming it) can be

madetheinfiniteregion.

11.2.15EulersFormula: Aconnectedplanargraphwithn verticesande

edgeshas en+2regions.

[LetG beaconnectedplanegraphandlet n, e,andf denotethenumber

ofvertices,edgesandfaces(orregions)of G respectively.Then ne

+f =2(or) f = en+2].

Proof: (UsingMathematicalInductiononthenumberoffacesf).

Part(i): Supposethat f=1.

ThenG hasonlyoneregion.IfG containsacycle,thenitwillhaveatleast

twofaces,acontradiction.SoG containsnocycles.

SinceG isconnected,wehavethatG isatree.

Weknowthatinatreen = e+1.

So en+2= e(e+1)+2=1=f.Sothestatementistruefor f =1.

Part(ii): InductionHypothesis:Supposethat f>1,andthetheoremistrue

forallconnectedplanegraphswiththenumberoffaceslessthan f.

Sincef>1,wehavethatG isnotatree(asatreecontainsonlyoneinfinite

region).

ThenG hasanedgek,whichisnotabridge[sinceG isatreeifandonlyif

everyedgeofG isabridge].



Sothesubgraph Gk isstillconnected.

Sinceanysubgraphofaplanegraph isalsoaplanegraph,wehave that

Gk isalsoaplanegraph.Since k isnotabridge,we

havethatk isapartofacycle[sinceanedgeeofG isabridgeifandonly

if e isnotpartofanycycleinG].

Soitseparatestwofaces F1and F2 of G fromeachother.

Thereforein Gk,thesetwofaces F1and F2 combinedtoformoneface

of Gk.Wecanobservethisfactinabovegraphs.

Let n(Gk), e(Gk), f(Gk)denotethenumberofvertices,edgesand

facesof Gk respectively.

Nowwehave n(Gk)= n, e(Gk)= e1and f(Gk)= f1.

Byourinductionhypothesis,wehavethatn(Gk)e(Gk)+f(Gk)=2

n(e1)+(f1)=2 n e+f =2 f = en+2.

HencebyMathematicalInduction,weconcludethatthestatementistruefor

allconnectedplanargraphs.

11.2.16Definition:Let f beafaceofaplanegraphG.Wedefinethedegree

of f (denotedby d(f))tobethenumberofedgesontheboundaryof f.For

anyinteriorface f ofasimpleplanegraph,wehavethat d(f) 3.

11.2.17Theorem:LetG beasimpleplanargraphwith nverticesand e

edges,wheren 3.Then(i)2e 3f(or e f23 )(ii) e (3n6)

F2

F1 Fig. 11. 2.15 B

Graph G k 3 faces

Fig 11.2.15 A Graph G 4 faces

k



11.2.18Example: Using theEulers theorem,weverify that Kuratowskis

1stgraph K5 (Thecompletegraphonfivevertices)isnonplanar.

Ifpossible,supposethatK5 isplanar.Observethegraph(giveninFig2.18)

for K5.

Here,n =5 3and e =10.ByEulertheorem,wehavethat3n6

e 9 e =10,acontradiction.Soweconcludethat K5 isanon

planargraph.

Fromthefollowingexamplewecanunderstandthat K3,3 isanonplanar

graphandsatisfiestheconditionthat e 3n6.

11.2.19Example: Showthat K3,3 isnonplanar.

UsingEulersFormula:Ifpossiblesupposethat K3,3 isplanar.Notethat n

=6and e =9.

In K3,3 everyfacehasatleastfouredgesonitsboundary.

So d(f) 4foreachface f.Write b = x f f)(d (i)where x denotesthesetofallfacesof G.

Fig. 11. 2.18

K3,3



Thenotationweuseis|x |=f.Now b= x f f )(d x f 4 =|x |

4=4f.

Sinceeachedgewascountedeitheronceor twice in (i) (if it iscounted)

(twiceifitoccurredasaboundaryedgefortwofaces),wehavethat b

2e.

Therefore2e b 4f e 2f e 2(e n+2)(sincef =en+2)

9 2(96+2) 9 2(5) 9 10,acontradiction.

Hence K3,3 isanonplanargraph.

11.2.20Definition:

Let G beaplanegraph. Wedefinethedualof G tobethegraph G*

constructedasfollows:

i) Toeachface(orregion) f of G thereisacorrespondingvertex f*

of G*.

ii) Toeachedge e of G,thereisacorrespondingedge e* of G* [if

theedge e occursontheboundaryofthetwofaces f and g,then

createanedge e* thatjoinsthecorrespondingvertices f* and g*

in G*].

[Iftheedge e isabridge,thenwetreatitasanedge,itoccurstwice

on the boundary of the face f in which face it lies. So the

correspondingedge e* isaloopincidentwiththevertex f* in G*].

11.2.21Constructionofadualfromagivengraph:

Step(i):Considerthegivenplanargraph andidentifytheregionsFi,1 i n.

[Considertheplanerepresentationof thegraphgiveninFig.11.2.21A. In

thisgraphthereare6faces(orregions)F1, F2, F3, F4, F5,and F6].

Step(ii): Place thepointsp1, p2,, pn on theplaneone for eachof the

regions.



[Intheexample,place6points p1, p2, p3, p4, p5,p6. Notethat pi isa

pointintheinterioroftheregion Fi].ObservetheFig.11.2.21B.

Step(iii): If two regions Fi and Fj are adjacent (that is, having a

commonedge),thendrawalinejoiningthecorrespondingpointspi and pj

thatintersectsthecommonedgebetween Fi and Fj exactlyonce.

[Ifthereismorethanoneedgecommonbetween Fi and Fj,thenwedraw

onelinebetweenthepoints pi and pj foreachofthecommonedges].

Weperformthisprocedureforall i, j {1,2,.,n}suchthat i j.

Step(iv):Supposea region Fk containsanedge e lyingentirely in it,

thenwedrawaselfloopatthepointpk intersecting e exactlyonce.

We perform this procedure for all such Fk. By using this procedure we

obtainanewgraph G*.Suchagraph G* obtainedhereiscalledadual

(orgeometricdual) of G.

[In this example, the dual G* was shown with broken lines (observe

Fig.11.2.21C).Thisdual G* consistof6vertices p1, p2, p3, p4, p5,

and p6 andtheedges(broken)joiningthesevertices].

F6F2

F3

F4

F1 F5

Fig. 11.2.21A Fig. 11.2.21C Fig 11.2.21 B

2p

6p

1p 5p3p

4p



11.2.22Observations

i) Fromtheconstructionof thedual G* of G, it isclearthat there isa

onetoonecorrespondencebetweenthesetofalledgesof G andthe

setofalledgesof G*.Alsooneedgeof G* intersectsoneedgeof G.

ii) AnedgeformingaselfloopinGyieldsapendentedgein G* anda

pendentedgein G yieldsaselfloopin G*.

iii) Edgesthatareinseriesin G produceparalleledgesin G*.

[If e1, e2 areinseriesin G,then 1*e , 2*e areparallelin G*].

iv) Paralleledgesin G produceedgesinseriesin G*.

v) It is a general observation that the number of edges constituting the

boundary of a region Fi in G is equal to the degree of the

correspondingvertexpi in G*,andviceversa.

vi) Thegraph G* isembeddedintheplane,andso G* isalsoaplanar

graph.

vii) Consider the process of drawing a dual G* from G. It can be

observedthat G isadual G*.Thereforeinsteadofcalling G*,a

dualof G,wecanusuallysaythat G and G* aredualgraphs.

11.3.Coloring

OneofthemostfamousproblemsinGraphtheoryisthefourcolorproblem.

Theproblemstatesthatanymaponaplaneoronthesurfaceofasphere

canbecoloredwithfourcolorsinsuchawaythatnotwoadjacentcountries

have the same color. This problem can be translated as pa problem in

GraphTheory.Werepresenteachcountrybyapointandjointwopointsby

a line if thecountriesareadjacent. Theproblem is tocolor thepoints in

suchwaythatadjacentpointshavedifferentcolors.Thisproblemwasfirst

posedin1852byFrancisGuthrie(astudentattheuniversitycollege,London).

Fourcolorconjectures(4CC)wasthemostfamousunsolvedproblemingraph

theory.Inthissectionwegivebriefideaofcoloringofagraph.



11.3.1Definition

i) LetG beagraph.Paintingall theverticesof thegraph G with the

given n colorssuchthatnotwoadjacentverticeshavethesamecolor,

iscalledapropercoloring(or)simplycoloringofthegraph G.

ii) Agraphinwhicheveryvertexhasbeenassignedacoloraccordingto

"propercoloring"iscalledaproperlycoloredgraph.

11.3.2Note:Agivengraph G canbeproperlycoloredindifferentways.

ThefollowingFigures2.2(a),(b),(c)showthreedifferentpropercoloringsof

thesamegraph G.

11.3.3Definition

A graphG that requires k different colors for its proper coloring andno

less,iscalleda kchromaticgraphandthenumberkiscalledthechromatic

number(orchromaticindex)of G.

11.3.4Observations:

i) ThegraphgivenintheFig.11.3.2(c)isa3chromaticgraph.

ii) IfthegraphG hasloopatthevertex v,thenv isadjacenttoitselfand

so no proper coloring for G is possible. So we assume that in any

coloring(thevertices)context,thegraphsconsideredhavenoloops.

v4 Pink

v5Yellow

v3Green

v2 Blue

v1 Red

Fig. 11.3.2(a) 5 colors used

v4 Red

v5Yellow

v3Green

v2 Blue

v1 Red

Fig. 11.3.2(b) 4 colors used

v4 Red

v5Yellow

v3Yellow

v2 Blue

v1 Red

Fig. 11.3.2(c) 3 colors used



iii) In graphs considering for coloring, there is no point in considering a

disconnected graph. Coloring vertices in one component of a

disconnected graph has no effect on the coloring of the other

components.Soweusuallystudythecoloringoftheconnectedgraphs

only.

iv) Paralleledgesbetweenanypairofverticescanbereplacedbyasingle

edgewithouteffectingtheadjacencyofthevertices.

v) For coloring problems, we need to consider only simple connected

graphs.

11.3.5Observations:

i) Agraphconsistingofonlyisolatedverticesis1chromatic.

ii) Agraphwithoneormoreedges isat least2chromatic. 2chromatic

graphsarealsocalledasbichromaticgraphs.

iii) A complete graph on n vertices is nchromatic, as every pair of

verticesareadjacent.Soagraphcontainingacompletegraphon r

verticesisatleast rchromatic.

[Soeverygraphcontainingatriangleisatleast3chromatic].

iv) A graph consisting of simply one circuit with n 3 vertices is 2

chromatic if n iseven,and3chromatic if n isodd. [Numberthe

verticesas1,2,,n (sequence)andassignonecolortooddvertices,

andanothertoeven.If n iseven,thennoadjacentverticeswillhave

thesamecolor.Thusif n iseven,thenthegraphis2chromatic.If n

isodd, then the nth andfirstvertexwillbeadjacentandhaving the

samecolor.Soitrequiresathirdcolorforitspropercoloring.Thusif

n isodd,thenthegraphis3chromatic].

Graph Kp Kpx pK Km,n C2n C2n+1

ChromaticNumber

p p1 1 2 2 3



11.3.6Theorem:Everytreewithtwoormoreverticesis2chromatic.

Proof:Selectanyvertexvinthegiventree.

Consider T as a rooted tree at vertex v. Paint v with color 1. Paint all

verticesadjacenttovwithcolor2.

Next,painttheverticesadjacenttothese(thosethatjusthavebeencolored

with2)usingcolor1.

ContinuethisprocesstilleveryvertexinThasbeenpainted.

NowinTwefindthatallverticesatodddistancesfromvhavecolor2while

v and vertices at even distances from v have color 1. (Observing the

followingtree).

NowalonganypathinTtheverticesareofalternatingcolors.

Sincethereisoneandonlyonepathbetweenanytwoverticesinatree,no

two adjacent vertices have the same color. Thus T has been properly

coloredwithtwocolors.Onecolorwouldnothavebeenenough.

11.3.7 Remark:

i) Every2chromaticgraphmaynotbeatree.

ii) Theutilitiesgraphisnotatreebutitis2chromatic

1

2

2

1

2

22

2

1

Fig. 11.3.6



Considerthefollowinggraph.

.

11.3.8 Definition

Let G beagraph.Write

dmax =max{d(v)/v isavertexof G}.

Thendmax iscalledthemaximumdegreeofverticesinthegraph G.

11.3.9Notation

i) dmax(G)=maximumofthedegreesoftheverticesinthegivengraph G.

ii) dmax(Gv)=maximumofthedegreesoftheverticesinthegraphGv.

iii) k(G)=thechromaticnumberof G.

11.3.10Theorem

ForanygraphG,chromaticnumber 1+dmax.(Thatis,k(G) 1+dmax).

11.4Partitioning:

11.4.1Definition:

Let G =(V,E)beabipartitegraphwiththepartition{V1,V2}of V such

thateveryedgein G joinsavertexin V1 andavertexin V2.Ifforanypair

ofpointsv1 V1and v2 V2 therecorrespondsanedge 21vv in G,thenwe

callthegraphascompletebipartitegraph.If V1 containsmverticesand V2

contains n vertices,thenthatcompletebipartitegraphisdenotedby Km,n.

11 1

22 2

Fig11.3.7BUtilitiesgraph

B

RR

B

Fig.11.3.7A2chromaticbutnotatree

BforBlueRforRed



11.4.2Example

i) Graph11.4.2A, is K3,3 which isabipartitegraphwith thebipartition

V1={v1,v2,v3}, V2={v4,v5,v6}.ThisisalsoknownasKurotoski's2nd

graph.

ii) Graph 11.4.2 B is K4,3 is a bipartite graph with the bipartition

V1={v1,v2,v3,v7},

V2={v4,v5,v6}.

11.4.3Observations:

i) Everytreeisabipartitegraph.

ii) Abipartitegraphcontainsnoselfloops.

iii) Asetofparalleledgesbetweenapairofverticescanbereplacedby

oneedgewithoutaffectingbipartitenessofagraph.

iv) Every2chromaticgraphisbipartite.

[Verification:LetG bea2chromaticgraphcoloredwithcolor1and

color2.

Write Vi =thesetofverticeswithcolori for i =1,2.

Now V1 V2 = f .Notwoverticesof V1 (or V2)areadjacent.So

everyedgeconnectsavertexof V1 andavertexof V2].

v) Every bipartite graph is 2chromatic with one trivial exception (the

exception isthatagraphof twoormoreisolatedverticesandwithno

edgesisabipartitegraph,butitis1chromatic).Sowecanconclude

thatabipartitegraphiseither1chromaticor2chromatic.

v v v

vvv

Graph 11.4.2 A

vvv

vvvv

Graph 11.4.2B



11.4.4Definition

A graph G is called ppartite if its vertex set can be partitioned into p

disjointsubsets V1, V2,,Vp suchthatnoedgein G joinstwoverticesof

thesamesubset Vi for1 i p.

11.4.5 Example

Thefollowinggraphisa3bipartitegraph.HereV1={v1,v2},V2={v3,v4}and

V3={v5}.ObservethatthereisnoedgeinGwhichjoinstwoverticesinthe

sameset Vi,1 i 3.

11.4.6Definition

A set of vertices in agraph is said tobe an independent set of vertices (or)

independentset(or)internallystableset ifnotwo verticesinthesetareadjacent.

11.4.7Problem:Thefollowingstatementsareequivalent:

i) Gis2colorable

ii) Gisbipartite

iii) EverycycleofGhasevenlength.

11.4.8Example:

i) Asetconsistingofasinglevertexformsanindependentset.

ii) ConsiderthegraphgiveninFig.11.4.8.

Heretheset{a,c,d}isanindependentset.

a

f

gd

e

b

c

Fig. 11.4.8

v2

v4

v1

v3

v5



11.4.9Definition

i) A maximal independent set (or) maximal internally stable set is an

independent set to which no other vertex can be added without

destroyingitsindependenceproperty.

[Inotherwords,amaximalindependentsetisamaximalelementinthe

collectionofallindependentsets].

ii) A graphmay havemanymaximal independent sets. Among these,

thenumberofverticesinthelargestindependentsetofagraph G is

calledtheindependentnumberofG(or)coefficientofinternalstability,

and thisnumber is denoteby b(G). Here largest setmeans the set

containingmorenumberofelements.

11.4.10Example

i) Thesets{a,c,d,f},{b,f},{b,g},{b,e}aremaximalindependentsets

ofthegraphgiveninFig4.8.Theindependencenumberofthisgraph

is4.

11.4.11Note:

Consider a kchromatic graph G of n vertices properly colored with k

differentcolors. Takeaset Aofverticeswithsamecolor. Thenno two

verticesinAcannotbeadjacent.SoAisalinearlyindependentset.So

|A| b(G).

Thelargestnumberofverticesin G withthesamecolor b(G)

k

n b(G).

11.5AlgorithmsforChromaticNumbers

11.5.1 Algorithm to find the independence number b (G) of a given

graph G:



Step(i):Findallthemaximalindependentsetsofthegivengraph.

Step(ii): Find out the number of elements in eachmaximal independent

set.

Step(iii):FindthelargestnumberamongthenumbersobtainedinStep(ii).

Thisnumberis b (G)whichdenotestheindependencenumberof G.

11.5.2Example

ConsiderthegraphinFigure11.5.2.

i) Themaximalindependencesetare{a,c,d,f},{b,g},{a,e},{a,c,d,g},

{b,f}.

ii) The largest number among the number of elements in the maximal

independentsetsis4.

Thustheindependencenumberof G is b (G)=4.

11.5.3 Algorithmtofindthechromaticnumber

k(G)ofagivengraph G.

Step(i):Findallthemaximalindependentsetsofthegivengraph G.

Step(ii):Express V astheunionofsomeofmaximalindependentsets.

Step(iii): In different expressions (obtained in Step(ii)), the number of

maximalindependentsetsusedaredifferent.Wefindtheminimumnumber

k(G)ofmaximalindependentsetswhoseunionisequalto V.Thisnumber

isthechromaticnumberof G.

a

f

gd

e

b

c

Fig. 11.5.2



11.5.4Example

ConsiderthegraphgiveninFig.11.5.2.

Step(i):Themaximalindependentsetare{a,c,d,f},{a,c,d,g},{b,g},{b,

f},{a,e}.

Step(ii):V={a,c,d,f} {b,g} {a,e},V={a,c,d,g} {b,f} {a,e}

V={a,c,d,f} {b,g} {a,e} {a,c,d,g}.

We can express V in different ways. Finally, we can observe that the

minimumnumberof independentsets thatcanbeusedis3. Thusforthis

example, k(G)=3.

11.5.5Definition:

i) Let G beasimpleconnectedgraph.Findapartitionofallthevertices

of G intothesmallestpossiblenumberofdisjoint,independentsets.

Finding suchapartition is knownas the chromaticpartitioning of the

graph.

ii) A graph that has only one chromatic partition is called a uniquely

colorablegraph.

11.5.6Example

ConsiderthegraphintheFig.11.5.2.

Thenthefourchromaticpartitioningof G aregivenbelow:

i) {{a,c,d,f},{b,g},{e}},(ii){{a,c,d,g},{b,f},{e}}

ii) {{c,d,f},{b,g},{a,e}},(iv){{c,d,g},{b,f},{a,e}}.

11.5.7Note

{{a},{b},{c},{d},{e},{f},{g}}isapartitionofthesetofverticesintodisjoint

independentsets.Itdoesnotcontaintheminimumnumberofsets.Hence

thisisnotachromaticpartition.



11.5.8Example

i) ThegraphinFigure11.5.2isnotuniquelycolorable.

(Since,thereareseveralchromaticpartitions(ObserveExample5.6)).

ii) ButthegraphinFigure11.5.8isuniquelycolorable.

(Since,{v1},{v5,v3},{v2,v4}isonlythechromaticpartition).

iii) isuniquelycolorable.

iv) Inthisgraph{v1,v4},{v2},{v3}and{v1},{v2,v4},{v3}aretwochromatic

partitioningexist.Sothisgraphisnotuniquelycolorable.

11.5.9Definition:

Let G beagraph, V thesetofverticesof G,and W V.

i) Theset W issaidtobeadominatingset(orexternallystableset)ifit

satisfies the following condition: v V either v W or there

exists u W suchthat u and v areadjacent.

ii) Wesaythat W dominatesavertex v V ifeitherv W orthere

exists u W suchthat u and v areadjacent.

iii) Aminimaldominatingsetisadominatingsetfromwhichnovertexcan

be removed without destroying its dominance property. [In other

words,aminimalelement in thesetofalldominatingsets iscalleda

minimaldominatingset].

11.5.10Note:

i) It is clear thatW is a dominating set for a graphG if and only ifW

dominateseveryvertex v of G.

ii) Dominatingsetneednotbeindependent.

iii) Thesetofallverticesofanygraph G isadominatingsetfor G.

v1v2

v3v4

v5

Fig. 11.5.8 v4

v2 v3

v1

v2 v3

v1



11.5.11 Example:ConsiderthegraphinFig.11.5.11.

i) The vertex sets {b, g}, {a, b, c, d, f} and {a, b, c, d, e, f, g} are

dominatingsets.

ii) Thevertexset{a,f,g}isnotadominatingset.

iii) Theset{a,b,c,d,e,f,g}isadominatingset.

iv) {a,f,g}isnotadominatingset.

11.5.12Example

ConsiderthegraphgiveninFig.11.5.11.Thevertexsets{a,c,d,f},

{a,d,e},{a,c,d,g},{b,e},and{b,f}aresomeminimaldominatingsets.

11.5.13Observations

i) Foranyvertex v inacompletegraph, theset{v} formsaminimal

dominatingset.

ii) Everydominatingsetcontainsatleastoneminimaldominatingset.

iii) A graph may have many minimal dominating sets (observe the

Example11.5.11).Thenumberofelements intheminimaldominating

setsmaydiffer. Thenumbermin{numberofelements in W /W isa

minimaldominatingsetofagraphG}iscalledthedominatingnumber

of G.Thisdominationnumberisdenotedby a (G).

So a (G) = min{number of elements in W / W is a minimal

dominatingset}.

iv) Aminimaldominatingsetmaynotbeindependent.

a

f

gd

e

b

c

Fig. 11.5.11



Forexample,wemayconsiderthegraphgivenintheFig.11.5.13A.

W={a,b,f}isadominatingsetfor G. W isaminimaldominatingset.

But W isnotanindependentset.

v) Asmallestminimaldominatingsetneednotbeindependent.

Forexample,considerthegraphgiveninFig.11.5.13B.

The set {b, c} is smallestminimal dominating set, but it is not an

independentset.

vi) Let M be an independent set. IfM is amaximal independent set,

thenMisadominatingset.

11.5.14Definition

i) Let G beagraphon n vertices.Supposethenumberofcolorsgiven

is l .ThenG may G beproperlycoloredinmanydifferentwaysby

using the given colors. We can express this number in terms of a

polynomial.Suchapolynomialissaidtobethechromaticpolynomial

ofG.ThispolynomialisdenotedbyPn(l ).

ii) Pn(l )denotesthenumberofwaysofpropercoloringthegraph G by

using l orfewercolors.

h a

f

gd

e

b

c

Fig. 11.5.13A

a b c d

Fig. 11.5.13B




1. Verifywhetherthefollowinggraphsareplanar.Ifso,redrawthegraphs

sothatnoedgescross.

2. Verifywhetherornot,thefollowinggraphsareplanar?

(i)

f

c

e

b

d

a o

o

o

oo

o

(ii) o

o oo

o

(iii)

a c

de

bo

o

o

o

o

(iv)

c

a

d

b

o

o

o

o

(i)o

o

o

o

o

of

d

c

e

ba(ii)o

o

o

o

o

o



3. Drawthedualofthefollowinggraphs

4. Statethefollowingtrueorfalse(oryesorno)

i) K5{e}isplanarforanyedgeeofK5

ii) K3,3{e}isplanarforanyedgeeofK3,3

iii) Anygraphon4orfewerverticesisplanar

iv) AcompletebipartitegraphKm,n isplanar.

5. Findtheminimumnumberofcolorsrequiredtopointeachmapgivenbelow.

(i)

o

o

o o

o

o

(ii)

o

o

o

o

o

o o

o

o

oo

o

o(i) (ii) ooo

o

oo

o

o

o o

o

o(iii)

o

o

o

o

(iv)

oo oo o o



11.6Summary

This unit provides some geometrical representations of graphs. Planar

graphs play vital role in real world problem. Interrelations between the

kpartitegraphsandcolorabilityofsuchgraphswereobserved.Thereader

abletoknowaboutthefamousfourcolorconjectureanditsimportance.

11.7TerminalQuestions

1. Findthechromaticnumberofeachgraph.

2. Theminimumnumberofcolorsrequiredinanedgecoloringofagraph

GiscalledtheChromaticindexofG.Findthechromaticindexofthe

followinggraphanddenoteit.

oo

o

oo

o

o(i) (iii) o

o

o

o

(iv)

o

o

o

o

(ii)

o

o o

o

o

(v)

o

o

o

o

o

(ii)

o

o

o

o

oo o

o

o

o

(i)



3. Determine which of the following statements are true and which are

false.

i) Everyplanargraphisaplanegraph

ii) Everyplanargraphisisomorphictoaplanegraph

iii) Anyplanegraphcanbeembeddedonthesurfaceofasphere

iv) Everysubgraphofaplanargraphisplanar

v) Thechromaticnumberofacompletegraphofpverticesisp.

vi) Thechromaticnumberofanytotallydisconnectedgraphis1

vii) Thechromaticnumberofanycycleis2

viii) Everyplanargraphis5colorable.

11.8Answers


1.

s

fb

d

e(i)

a o

oo o

o o

(ii)

o

o

o o o

c

b

a

de

(iii)

o o

oo

o

c

o

a b

d

(iv)

o

oo



2. i) Nonplanar,sinceK3,3 isasubgraph

ii) Nonplanar,sinceK3,3 isasubgraph.

3. i) Selfdual

ii) Selfdual

4. i) Yes

ii) Yes

iii) Yes

iv) No

5. i) Numberofverticesn=7,edgese=12,regionsr=7.

Theminimumnumberofcolorsrequiredis3.

ii) Minimumnumberofcolorsrequiredis3.

iii) Minimumnumberofcolorsrequiredis4.

iv) Minimumnumberofcolorsrequiredis3.

TerminalQuestions

1. Thechromaticnumberofeachgraphisasfollows.

i) 2

ii) 3

iii) 3

iv) 2

v) 5

2. Chromaticindexis3:

4

4

3

32 2

1

1o o

o

o

o

(i)



3. i) False

ii) True

iii) True

iv) True

v) True

vi) True

vii) False

viii) True

(ii)

Chromaticindexis5.

5

54

4

3

32

2

1

1

o

o

o

o

o

Documents

MC0063(B)-unit11-fi