MC0063(B)-Unit10-fi

ModernApplicationsusingDiscreteMathematicalStructures Unit10

SikkimManipalUniversity PageNo:225

Unit10 TraversabilityStructure

10.1 Introduction

Objectives

10.2 EulerianGraphs

10.3 HamiltonianPathsandCircuits

10.4 TravelingSalesmanproblem

SelfAssessmentQuestions

10.5 Summary

10.6 TerminalQuestions

10.7 Answers

10.1IntroductionIn this unit, we discuss Euler graphs, Hamiltonian paths and Hamiltonian

circuits in a graphs G with special properties. A given graph G can be

characterizedandstudiedintermsofthepresenceorabsenceofthesesub

graphs.Manyphysicalproblemscanberepresentedbygraphsandsolved

byobservingtherelevantpropertiesofthecorrespondinggraphs.

ObjectivesInthisunit,weshallbedealingwith

Eulergraph

CharacteristicsofEulergraphs

Hamiltonianpaths,circuitsandGraphs

TravellingSalespersonsproblem.

10.2EulerianGraphsGraph theory was born in 1736 with Eulers famous paper. In the same

paper,Eulerposed(andthensolved)amoregeneralproblem.



WestatethefollowingresultsbeforegivingtheformaldefinitionsofEulers

graphs.

10.2.1 Problem: A graph G is disconnected its vertex set V can be

partitioned into two nonempty disjoint subsets V1 and V2 such that there

exists no edge in G, whose one endvertex is in V1 and the other end

vertexisinV2.

10.2.2Problem: Ifagraph(eitherconnectedordisconnected)hasexactly

two vertices of odd degree, then there exists a path joining these two

vertices.

10.2.3Theorem:A simplegraphwith n verticesand k components can

haveatmost(nk)(nk+1)/2edges.[orthemaximumnumberofedges

inasimplegraphwithnverticesandkcomponentsis(nk)(nk+

1)/2].

10.2.4Definition:LetGbeagraph.Aclosedwalkrunningthroughevery

edgeofthegraph GexactlyonceiscalledanEulerline.AgraphGthat

containsEulerlineiscalledanEulergraph.

10.2.5 Example:Thegraphsgiven intheFig.10.2.5.Aand fig10.2.5B

areEulergraphs.

4

6

8

1

2

35

7910

11

12

Fig10.2.5A

10

4

8 69

3

5

1 2

11

7

Fig10.2.5B



i. ConsiderFigA,123456789(10)(11)(12)13579(11)1isanEularline.

ii. ConsiderFigB,23456789(10)12(11)357(11)8(10)2isanEulerline.

AnEulergraphmaycontains isolatedvertices. IfG isEulergraphand it

containsno isolated vertices, then it is connected.Hereafter,we consider

onlythoseEulergraphs thatdonotcontain isolatedvertices.So theEuler

graphsthatweconsiderareconnected.

10.2.6 Characterization:

AgivenconnectedgraphGisanEulergraph alltheverticesofGare

ofevendegree.

Proof:SupposeGisanEulergraph.ThenGcontainsanEulerline.So

thereexistsaclosedwalkrunningthroughalltheedgesofGexactlyonce.

Letv VbeavertexofG.Nowintracingthewalkitgoesthroughtwo

incidentedgesonvwithoneenteredvandtheotherexited.

Thisistruenotonlyforalltheintermediateverticesofthewalk,butalsotrue

fortheterminalvertexbecauseweexitedandenteredatthesamevertexat

thebeginningandendingofthewalk.Thereforeifvoccursktimesinthe

Eulerline,thend(v)=2k.ThusifGisanEulergraph,thenthedegreeof

eachvertexiseven.

Converse:SupposealltheverticesofGareofevendegree.Nowtoshow

thatGisaEulergraph,wehavetoconstructaclosedwalkstartingatan

arbitraryvertexvandrunningthroughalltheedgesofGexactlyonce.To

finda closedwalk, letus start from the vertex v.Sinceevery vertex isof

evendegree,wecanexistfromeveryvertexweentered,thetracingcannot

stopat any vertexbut at v.Since v isalsoofevendegree,we shall

eventuallyreachvwhenthetracingcomestoanend.Ifthisclosedwalk

vFig10.2.6



(h,say)includesalltheedgesofG,thenGisanEulergraph.Suppose

the closed walk h does not include all the edges. Then the remaining

edges forma subgraph h1 of G. Sinceboth G and h haveall their

verticesofevendegree,thedegreesof theverticesof h1 arealsoeven.

Moreover, h1 must touch h at least one vertex a becauseG is

connected.Startingfromawecanagainconstructanewwalkingraphh1.

Sincealltheverticesofh1 areofevendegree,andthiswalkinh1 must

terminatesatthevertexa.Thiswalkinh1 combinedwithhformsanew

walkwhichstartsandendsatvertexvandhasmoreedgesthanthose

areinh.

Werepeatthisprocessuntilweobtainaclosedwalkthattravelsthroughall

theedgesof G. Inthisway,onecangetanEuler line.ThusG isan

EulerGraph.

10.2.7 Konigsberg Bridges Problem: In the graph of the Konigsberg

bridgesproblem,thereexistverticesofodddegree.Soalltheverticesare

notofevendegree.Henceby theTheorem10.2.6,we conclude that "the

graphrepresentingtheKonigsbergbridgesproblem"isnotanEulergraph.

Soweconcludethatitisnotpossibletowalkovereachofthesevenbridges

exactlyonceandreturntothestartingpoint.

A

B

DC Fig10.2.7



10.2.8Theorem:AconnectedgraphGisanEulergraphifandonlyifitcan

bedecomposedintocircuits.

Proof: SupposegraphG canbedecomposed into circuits that is,G isa

unionofedgedisjointcircuits.Sincethedegreeofeveryvertexinacircuit

istwothedegreeofeveryvertexinGiseven.HenceGisanEulergraph.

Conversely,letGbeanEulergraph.Consideravertexv1.Thereareatleast

two edges incident at v1. Let one of these edges be between v1 and v2.

Sincevertexv2 isalsoofevendegree, itmusthaveat leastanotheredge,

saybetweenv2andv3.Proceedinginthisfashion,weeventuallyarriveata

vertexthathaspreviouslybeentraversed, thusformingacircuit . Letus

remove from G. All vertices in the remaining graph (not necessarily

connected)mustalsobeofevendegree.Fromtheremaininggraphremove

anothercircuitinexactlythesamewayasweremoved fromG.Continue

thisprocessuntilnoedgesareleft.Hencethetheorem.

10.2.9ArbitrarilyTraceableGraphs:

ConsiderthegraphinFig.10.2.9,whichisanEulergraph.Supposethatwe

startfromvertexaandtracethepathabc.

Fig.10.2.9:Arbitrarilytraceablegraphfromc

Nowatcwehavethechoiceofgoingtoa,dore.Ifwetookthefirstchoice,

wewouldonly trace the circuitab ca,which is not anEuler line. Thus,

a b

c

d e



startingfroma,wecannottracetheentireEulerlinesimplybymovingalong

anyedgethathasnotalreadybeentraversed.

Such a graph is called an arbitrarily traceable graph from vertex v. For

instance,theEulergraphinFig.10.2.9isanarbitrarilytraceablegraphfrom

vertexC,butnotfromanyothervertex.

AnEulergraphGisarbitrarilytraceablefromvertexvinGifandonlyif

everycircuitinGcontainsv.

10.3HamiltonianPathsandCircuitsAnEuler line of a connected graph was characterized by the property of

beingaclosedwalkthattraverseseveryedgeofthegraph(exactlyonce).

10.3.1DefinitionAHamiltoniancircuitinaconnectedgraphisdefinedasaclosedwalkthattraverses every vertex of G exactly once, except of course the starting

vertex,atwhichthewalkalsoterminates.

AcircuitinaconnectedgraphGissaidtobeHamiltonianitincludesevery

vertexofG.HenceaHamiltoniancircuitinagraphofnverticesconsistsof

exactlynedges.Obviously,noteveryconnectedgraphhasaHamiltonian

circuit.

The resemblance between the problem of as Euler line and that of a

Hamiltonian circuit is deceptive. The latter is infinitely more complex.

AlthoughonecanfindHamiltoniancircuitsinmanyspecificgraphs.

10.3.2DefinitionHamiltonianPath:IfweremoveanyoneedgefromaHamiltoniancircuit,

we are left with apath. This path is calledaHamiltonianpath.Clearly, a

Hamiltonian path in a graph G traverses every vertex of G. Since a

Hamiltonianpath isasubgraphofaHamiltoniancircuit(which in turn isa

subgraphofanothergraph)everygraphthathasaHamiltoniancircuitalso



hasaHamiltonianpath.TherearemanygraphwithHamiltonianpathsthat

havenoHamiltoniancircuitsthelengthofaHamiltonianpath(ifitexists)in

aconnectedgraphofnverticesisn1.

InconsideringtheexistenceofaHamiltoniancircuit(orpath),weneedonly

consider simple graphs. This is because a Hamiltonian circuit (or path)

traverseseveryvertexexactlyonce.Henceitcannotincludeaselfloopora

setofparalleledges.Thusageneralgraphmaybemadesimplebyremoving

paralleledgesandselfloopsbeforelookingforaHamiltoniancircuitinit.

AgivengraphmaycontainmorethanoneHamiltoniancircuit. Of interest

areall theedgedisjointHamiltoniancircuits inagraph.Thedetermination

of the exact number of edgedisjoint Hamiltonian circuits (or paths) in a

graph in general is also an unsolved problem. However, the number of

edgedisjointHamiltonian circuits ina completegraphwithoddnumberof

vertices.

10.3.3Theorem:

Inacompletegraphwitha vertices thereareedgedisjointHamiltonian

circuits,ifnisanoddnumber 3.

10.4TravellingSalesmanProblemA problem closely related to the question of Hamiltonian circuits is the

travellingsalesman problem, stated as follows: A salesman is required to

visitanumberofcitiesduringatrip.Giventhedistancesbetweenthecities,

inwhatorder shouldhe travel soas to visitevery city preciselyonceand

returnhome,withtheminimummileagetraveled?

Representing the citiesby verticesand the roadbetween thembyedges,

wegetagraph.Inthisgraph,witheveryedgeei thereisassociatedareal

number(thedistancesinmiles,say), w(ei).Suchagraphiscalledweighted

graphw(ei)beingtheweightofedgeeI .



Inourproblem, ifeachofthecitieshasaroadtoeveryothercity,wehaveacompleteweightedgraph.ThisgraphhasnumerousHamiltoniancircuits,andwearetopicktheonethathasthesmallestsumofdistances(orweights).

ThetotalnumberofdifferentHamiltoniancircuits inacompletegraphofn

verticescanbeshowntobe2

)!1n( .Thisfollowsfromthefactthatstarting

fromanyvertexwehaven1edgestochoosefromthefirstvertex,n2fromthesecond,n3fromthethirdandsoon.Thesebeingindependentchoices,weget(n1)!possiblenumberofchoices.Thisnumberis,however,dividedby2,becauseeachHamiltoniancircuithasbeencountedtwice.

Theoretically, theproblemofthe travelingsalesmancanalwaysbesolvedby enumerating all (n1)!/2 Hamiltonian circuits, calculating the distancetraveled in each, and thenpicking the shortest one. However, for a largevalueofn,thelaborinvolvedistoogreatevenforadigitalcomputer.

SelfAssessmentQuestionsStateTrueorFalse.1) AgivenconnectedgraphGisanEulergraphifandonlyifallverticesof

Gareofodddegree.2) A connected graph G is an Euler graph if and only if it can be

decomposedintocircuits.3) IfweremoveanyonevertexfromHamiltoniancircuit,weareleftwitha

path.ThispathiscalledHamiltonianpath.

10.5Summary Ifsomeclosedwalkinagraphcontainsalltheedgesofthegraph,then

thewalkiscalledanEulerlineandthegraphanEulergraph. A connected graph G is an Euler graph if and only if it can be

decomposedintocircuits.



AHamiltoniancircuit inaconnectedgraphsisdefinedinaclosedwalkthattraverseseveryvertexofGexactlyonce.

10.6TerminalQuestions1. ProvethataconnectedgraphGisanEulergraphifandonlyifitcan

bedecomposedintocircuits.

2. WriteshortnotesonTravellingSalesmanProblem.

10.7AnswersSelfAssessmentQuestionsi)False ii)True iii)True

TerminalQuestions1. RefertoSection10.2.82. RefertoSection10.4

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MC0063(B)-Unit10-fi