MC0063(B)-unit9-fi

ModernApplicationsusingDiscreteMathematicalStructures Unit9

SikkimManipalUniversity PageNo:192

Unit9 TreesandAlgorithmsStructure

9.1 Introduction

Objectives

9.2 CharacterizationofTrees

9.3 RootedTreesandApplications

9.4 SpanningTrees

9.5 AlgorithmsforSpanningTrees

SelfAssessmentQuestions

9.6 Summary

9.7 TerminalQuestions

9.7 Answers

9.1Introduction

Trees are extensively used as models in area like computer science,

chemistry and in search procedures. These are also useful in design of

widerangeofalgorithms.Inthisunitweprovidesomecharacterizationsof

treeswithsuitable illustrations. Wealso introducedaspecial typeof trees

namely rooted treesand binary trees. Wegive fewapplicationsof these

trees.Thespanning treesare introducedand thealgorithmsforspanning

treesarealsoobtained.

Objectives

Attheendoftheunitthestudentmustbeableto

i) Study the connected graphs without circuits (called trees) and some

properties

ii) Characterizeatreeofaconnectedgraph.

iii) Knowfewapplicationsoftreesinrealm.

iv) Knowrootedtreesandsomeofitsapplications

v) Analyzeandwritealgorithmsforspanningtrees.



9.2CharacterizationofTrees

Theconcept ofa treeplaysa vital role in the theoryof graphs.Firstwe

introduce the concept of tree, study its properties and some of its

applications.Later,we introduce theconceptof spanning tree,andstudy

therelationshipsamongcircuitsandtrees.

9.2.1Definition

Aconnectedgraphwithoutcircuits iscalleda tree.Acollectionof trees is

calledaforest.

9.2.2Example

Treeswithone,two,threeandfourverticesaregivenintheFig.9.2.2.

9.2.3Observations

i) Atreecontainsatleastonevertex.

ii) Atreewithoutanyedgeisreferredtoasanulltree.

iii) Thetreesconsideredarefiniteandatreeisalwaysasimplegraph.

9.2.4Somenaturalexamples

i) Thelistoftheancestorsofafamilymayberepresentedbyatree.This

treereferredtoasafamilytree.

ii) Ariverwithitstributariesandsubtributariesmayberepresentedbya

tree.Thistreeisreferredtoasarivertree.

iii) Thesortingofmailaccordingtozipcodearedoneaccordingtoatree.

Thistreeiscalleddecisiontree(or)sortingtree.

Fig.9.2.2



9.2.5Theorem

Inatree T,thereisoneandonlyonepathbetweeneverypairofvertices.

Proof:Suppose T isatree.

Bydefinition, T isaconnectedgraphandcontainsnocircuits.

Since T isconnected,thereexistsatleastonepathbetweeneverypairof

verticesin T.Supposethatbetweentwovertices a and b of T,there

are two distinct paths. Now, the union of these two paths will contain a

circuitin T,acontradiction(since Tcontainsnocircuits).

Thisshowsthatthereexistsoneandonlyonepathbetweenagivenpairof

verticesin T.

9.2.6Theorem:

Ifthereisoneandonlyonepathbetweeneverypairofverticesin G,thenG

isatree.

Proof:Assume that there isoneandonlyonepathbetweeneverypairof

verticesinagraphG.ClearlyG isconnected.

Ifpossiblesupposethat G containsacircuit.Thenthereisatleastonepair

ofvertices a,b suchthattherearetwodistinctpathsbetween a and b,

whichisacontradictiontoourassumption.So G containsnocircuits.Thus

Gisatree.

9.2.7Theorem:

Atree G withnverticeshas(n1)edges.

Proof:(Useinductiononn):

Step(i): If n =1,then G containsonlyonevertexandnoedge.

Sothenumberofedgesin G is n1=11=0.

Step(ii):Inductionhypothesis:Thestatementistrueforalltreeswithless

thannvertices.



Step(iii):Letusconsideratreewithnvertices.

Letekbeanyedgein T whoseendverticesare vi and vj.

Since T isatree,byTheorem9.2.6, there isnootherpathbetween vi

and vj.

Sobyremoving ek from T,wegetadisconnectedgraph.

Also, Tek consistsofexactlytwocomponents(say T1 and T2).

Since T isatree,therewerenocircuitsin T andsotherewerenocircuits

inT1andT2.ThereforeT1 and T2 arealsotrees.

It isclearthat |V(T1)|+ |V(T2)| = |V(T)| where V(T) denotesthesetof

verticesin T.

Also|V(T1)|and|V(T2)|arelessthan n.

Thereforebytheinductionhypothesis,wehave

|E(T1)|=|V(T1)|1and|E(T2)|=|V(T2)|1.

Now|E(T)|1=|E(T1)|+|E(T2)|=|V(T1)|1+|V(T2)|1

Thus|E(T)|=|V(T1)|+|V(T2)|1=|V(T)|1= n1.

9.2.8Theorem:

Anyconnectedgraphwithnverticesand n1edgesisatree.

Proof:LetGbeaconnectedgraphwith n verticesand n1edges.

Itsufficestoshowthat Gcontainsnocircuits.

If possible, suppose that G contains a circuit. Let e be an edge in that

circuit.

Since einacircuit,wehavethat Ge isstillconnected.

Now Ge isconnectedwithnvertices,andso|Ge| n1edges,a

contradiction(tothefactthat Ge containonly(n2)edges,bytheorem

9.2.7).SoGcontainsnocircuits.Therefore G isatree.



9.2.9Definition:

Aconnectedgraphissaidtobeminimallyconnected if theremovalofany

oneedgefromthegraphprovidesadisconnectedgraph.

9.2.10Example

i) GraphgiveninFig.9.2.10Aisnotminimallyconnected.

ii) GraphgiveninFig.9.2.10B isminimallyconnected.

iii) Everycircuitisnotminimallyconnected.

iv) Everytreeisminimallyconnected.

9.2.11Problem

Agraph G isatreeifandonlyifitisminimallyconnected.

Solution: Part (i): Assume that G is a tree. If G is not minimally

connected,thenthereexistsanedgeesuchthat Ge isconnected.

Thatis,e isinsomecircuit,whichimplies G isnotatree,acontradiction.

HenceG minimallyconnected.

Part(ii):Supposethat G isminimallyconnected.

IfG containsacircuit,thenbyremovingoneoftheedgesinthecircuit,we

getaconnectedgraph,acontradiction(tothesupposition).Thisshowsthat

G containsnocircuits.ThusG isatree.

Observation:To join ngivendistinctpoints, theminimumnumberof line

segmentsneededis n1.

Fig.9.2.10A Fig.9.2.10B



9.2.12Problem::

Ifagraph G contains n vertices, n1edgesandnocircuits,then G isa

connectedgraph.

Proof: Let G beagraphwithnvertices, n 1 edgesandcontainsnocircuits.

Supposethat(inacontraryway)Gisdisconnected.

G consistsoftwoormorecircuitlesscomponents(say, g1, g2,, gk)(herek 2).

Chooseavertexvi ingi,for1 i k.

Addnewedges e1,e2,,ek1 where ei= 1 +iivv togetanewgraph G*.

Itisclearthat G* containsnocircuitsandconnected,andso G* isatree.

Now G*containsnverticesand

(n 1)+ (k 1)= (n+k 2) n edges,acontradiction (sincea tree

contains(n1)edges). Thisshowsthat G isconnected.

9.2.13CharacterizationforTree

Foragivengraph G,thefollowingconditionsareequivalent:

i) G isconnectedandiscircuitless,

ii) G isconnectedandhas n1edges,

iii) G iscircuitlessandhas n1edges,

iv) Thereisexactlyonepathbetweeneverypairofverticesin G,

v) G isaminimallyconnectedgraph,

vi) G isatree.

Proof: (i) (vi)isclear.

(vi) (ii)and(iii):Theorem9.2.7

(ii) (vi):Theorem9.2.8

(iii) (vi):Theorem9.2.12

(iv) (vi):Theorems9.2.5and9.2.6

(v) (vi):Problem9.2.11.



9.2.14Problem

If G is a tree (withminimum two vertices), then there exists at least two

pendantvertices.

Proof:LetGbeatreewith|V| 2.

Let v0e1v1e2v2e3 . . . vn1envn bea longestpath in G (since G isfinite

graph,itispossibletofindalongestpath).

Nowweshowthat d(v0)=1=d(vn).

If d(v0)>1,thenthereexistsatleastoneedge e withendpoint v0 such

that e e1.If e {e1,e2,en},then e = ei forsome i 1.

Soeither vi1= v0 or vi= v0 v0 repeatedinthepath,acontradiction.

Therefore e {e1, e2, e3 ,en}andsoe,e1, e2, e3 ,en isapathof

length n+1,acontradiction.

Thus d(v0)=1.Inasimilarway,wecanshowthat d(vn)=1.Hence

v0,vn aretwopendantvertices.

9.2.15Definition

LetGbeaconnectedgraph.Thedistancebetweentwoverticesvandu is

denotedby d(v,u)andisdefinedasthelengthoftheshortestpath[that

is, the number of edges in the shortest path] between v and u. In a

connectedgraph,wecanfindthedistancebetweenanytwogivenvertices.

9.2.16Example

i) ConsidertheconnectedgraphgiveninFig.9.2.16A.Heresomeofthe

pathsbetween v1 and v2 are(a,e),(a,b,f), (b,c,e),(b,f),(b,

g,h),(b,g,i,j),(b,g,i,k).Heretherearetwoshortestpaths(a,e)

and(b,f)eachoflength2.Hence d(v1,v2 )=2.



ii) ConsiderthetreegivenintheFig.9.2.16B.Here d(a,b)=1, d(a,c)=

2, d(a,d)=2, d(b,d)=1.

9.3RootedTreesandApplications

Rootedtreesareextensivelyusedinthecomputersearchmethods,binary

identificationproblems,andvariablelengthbinarycodes.

9.3.1Definition

Atreeinwhichonevertex(calledtheroot)isdistinguishedfromalltheother

vertices,iscalledarootedtree.Inarootedtree,therootisgenerallymarked

inasmalltriangle(orsmallcircle).

9.3.2Example

Distinctrootedtreeswithfourvertices,weregiveninFig.9.3.2

Generally, the term treemeans treeswithoutany root.However theyare

sometimescalledfreetrees(or) nonrootedtrees.Averityofrootedtrees

(calledtheBinaryrootedtrees)isofparticularinterest.

j

b

d

c

v2

v1i

h

k

g

e

a

fFig.9.2.16A

a

b cdFig9.2.16B

Fig.9.3.2



9.3.3Definition:

A tree in which there is exactly one vertex of degree 2, and all other

remainingverticesareofdegreeoneorthree,iscalledabinarytree.

i) TheFig.9.3.3representsabinarytree(sincetheonlyvertexv1isof

degree2,andallotherverticesareofdegreeeither1or3).

ii) Thevertexofdegree2 (that is, v1) is distinct fromallothervertices,

thisvertex v1 istheroot.

iii) Inabinary tree, the vertexwithdegree2 servesasa root.Soevery

binarytreeisarootedtree.

9.3.4 PropertiesofBinarytrees

Property(i):Thenumberofvertices n,inabinarytreeisalwaysodd.

Property(ii):Thenumberofpendentverticesis21n+ .

Property(iii):Numberofverticesofdegree3is =np1=n(2

1n+ )

1=2

3n- .

v3

v2

v4

v6v8

v5

v11

v7

v1

v10

v13

v9

v12

Fig9.3.3



9.3.5Example

InthegraphgiveninFig.9.3.3,wehavethat n=13, p=2

1n+ =

2

113+

=214

=7.Thereforenumberofverticesofdegree3is2

3n- =

2

313- =5.

9.3.6Definition:

Anonpendentvertexinatreeiscalledaninternalvertex.

Observation:

i) ThenumberofinternalverticesinaBinarytreeis

21n- =(p1)where p =thenumberofpendentvertices.

ii) InthebinarytreegiveninFig.9.3.3,theinternalverticesarev1,v3,v4,

v5,v6,v9.Theseare6(=71= p1)innumber.

9.3.7Definition:

Letv beavertexinabinarytree.Thenv issaidtobeatlevel I ifv isata

distanceofI fromtheroot.

9.3.8 Example:

i) A13vertex,4levelbinarytreewasgiveninFig.9.3.8.

level1

level4

level0

level2

level3

Fig.9.3.8



Here the number of vertices at levels 0, 1, 2, 3, 4 are 1, 2, 2, 4 and

4respectively.

9.3.9Definition:

Thesumofpath lengthsfromthe root toallpendentvertices iscalled the

pathlength(or) externalpathlengthofatree.

9.3.10Example:

i) ThepathlengthofthebinarytreegiveninFig.9.3.8is:

1+3+3+4+4+4+4=23.

ii) IntheFigures9.3.10A and B,therearetwo11vertexbinarytrees.

Thepathlengthofgraph(fig.9.3.10A):2+2+3+3+3+3=16.

Thepathlengthofgraph(fig.9.3.10B):1+2+3+4+5+5=20.

level5

level0

level1

level2

level4

level3

Fig9.3.10B

level 1

level 0

level 2

level 3

Fig9.3.10A



9.3.11Searchprocedures

Eachvertexofabinarytreerepresentsatestwithtwopossibleoutcomes.

Westartattheroot.Theoutcomeofthetestattherootsendsustooneof

thetwoverticesat thenext level,wherefurthertestsaremadeandsoon.

Reachingaspecifiedpendentvertex(thatvertexwhichrepresentsthegoal

ofthesearch),terminatesthesearch.

Forsuchsearchprocedures,itisoftenimportanttoconstructabinarytreein

which,foragivennumberofvertices n,thevertexfrothiestfromtherootin

asclosetotherootaspossible.

i) Therecanbeonlyonevertex(theroot)atlevel0.Numberofvertices

atleveloneisatmost2.Numberofverticesatleveltwoisatmost22

andsoon.So themaximumnumberofverticespossible inak level

binarytreeis20 +21 +22 +...+2k.

So n 20 +21 +22 +...+2k

ii) Themaximumnumberamongthelevelsoftheverticesinabinarytree

iscalledheightofthetree.Soheight=max{levelofavertex v /

v V}.

Thisheightisdenotedby lmax.

iii) Toconstructabinarytreeforagiven n suchthatthefarthestvertex

isasfaraspossiblefromtheroot,wemusthaveexactlytwoverticesat

eachlevel,exceptatthe0level.Somaxlmax =2

1n- .

9.3.12AnApplication:(CokeMachineProblem)

Supposethat there isa.Themachine is tohaveasequenceof tests (for

example, it should be capable of identifying the coin that is put into the

machine).We suppose that five rupees coin, two rupees coin, one rupee

coinandfiftypaisecoincangothroughtheslot.Sothemachinecanidentify

only these four coins.Every coin put in, is tobe testedby themachine.

Each test got the effect of partitioning the coins into two complementary



sets.[Supposeacoinisputintothemachine.Itshouldtestwhetherthecoin

isfiverupeecoin.Ifitisnotafiverupeescoin,thenitshouldtestwhetheritis

atworupeescoinandsoon].Wesupposethetimetakenforeachtestis.

TestPattern1:Onetypeof testingpatternwasshowninGraph(i), given

inFig.9.3.12A.

Supposethestatisticaldatatellsthat

w1 =probabilityofputtingaRs 5coin=0.5



w4 =probabilityofputtingaRs 0.5coin=0.1

Now

)v(lw ii = )v(lw 11 + )v(lw 22 + )v(lw 33 + )v(lw 44 =(0.5)(1) +(0.2)(2)+(0.2)(3)+(0.1)(4)=1.9

Soexpectedtimetobetakenbythemachinefortestingonecoinis1.9t.

Thus if themachinefollows (for its testingpattern) thebinary treegiven in

Graph(a),thentheexpectedtimefortestingonecoinisequalto1.9t.

Test Pattern2: Another type of testing pattern was given in the

Fig.9.3.12B.

Rs0.5coin

Rs1coin

Rs5coin

v30.2

v1

v2

v40.1

0.2

0.5

notRs1coin

notRs0.5coin

notRs2coin

notRs5coin

Fig9.3.12A

Rs2coin



)v(lw ii = )v(lw 11 + )v(lw 22 + )v(lw 33 + )v(lw 44 =(0.5)(2)+(0.2)(2)+(0.2)(2)+(0.1)(2)=2.0

Sohere,theexpectedtimetobetakenbythemachinefortestingonecoin

is2t.

Thus if themachinefollows (for its testingpattern) thebinary treegiven in

graph9.3.12Bthentheexpectedtimefortestingonecoinis2t.

9.3.13Definition:

Agraph G inwhicheveryvertexisassignedauniquenameorlabel(that

is,notwoverticeshavethesamelabel)iscalledalabeledgraph.Otherwise

issaidtobeanunlabeledgraph.

9.3.14Example:

Ifn =4,thenthereare16trees,shownbelow.

TakethevertexsetV ={A,B,C,D}.

v1 v2 v3 v40.10.2

0.20.5

Rs5or Rs2

Rs5 Rs2Rs1

Rs0.50

Rs1or Rs0.50

Fig9.3.12B



Thefollowingarethe16treesoffourlabeledvertices

B

C

A

D

A

DC

BA

C D

B

C

BA

D

A B

DC DC

BA

A B

C D

A B

C D

A

C

B

D

B

D

A

C

A B

C D

A B

C DA B

C D

A B

CD

A B

C D

A B

CD



Observations:

i) Whenwecountthenumberofdifferentgraphs,thedistinctionbetween

labeledandunlabeledgraphsisimportant.

ii) ConsiderthegraphsinExample9.3.14,the5th,6th,7th and8th,are

counted as four different trees (even through they are isomorphic),

becausetheyarelabeled.Ifthereisnodistinctionbetween A,B,C,

D,thenthesefourtreescountedasone.

iii) A careful inspectionof the 16 graphswill reveal that thenumberof

unlabeledtreeswithfourvertices(nodistinctionmadebetween A, B,

C, D),istwo.Thesetwographswere.

9.3.15CayleysTheorem:

Thenumberoflabeledtreeswith n vertices(n 2)is n(n2).

9.3.16 Example:Part(i):Suppose thegiven tree is T1 (Fig. 9.3.16A). a1 = thependentvertexwithsmallestlabel.So a1 isthevertex2.Now b1 =1.Afterremoving a1 andtheedge(a1,b1), intheremaininggraph,a2 = thependentvertexwithsmallestindex=4.

a1

1

2

6

5

3

79

4

8

Fig.9.3.16A



Now b2 =1. a3=1, b3=3, a4 =3, b4 =5, a5 =6, b5 =5, a6 =

7, b6=5, a7=5, b7=9.

Thereforewehavethesequence(b1, b2,b3, b4,b5, b6,b7)=(1,1,3,5,5,5,9).

Part(ii): (converseofpart(i)):wehavetoconstructatreewith n =9vertices.

Consider1,2,3,4,5,6,7,8,9.........(i)

Given(n2)tupleis(1,1,3,5,5,5,9).......(ii)

Observethesequencein (ii). First join1and3, 3and5, 5and9.

Thenweget thegraphgiven inFig.9.3.16B. Nowthe leastnumber in (i)

whichisnotin(ii),is2.Sowejoin2and1.Nextleastin(i)whichisnotin(ii)

is4.Sowejoin4and1.Cancel3in(i)and3in(ii).Alsocancel5in(i)

and5 in (ii) (in (ii), cancelingofonlyone 5 isallowed).Thenext least

whichisnotin(ii)is6.Sowejoin5and6.Thenextleastwhichisnotin(ii)

is 7. So we join 7 and 5. The next least which is not in (ii) is 8. The

remainingnumberavailablein(ii)is9.Sowejoin8and9.

13

5

9

Fig9.3.16B

8

1

2

6

5

3

7

9

4

Fig9.3.16C



Nowwewillstudythetreeasasubgraphofanothergraph.Agivengraph

mayhavenumeroussubgraphs.If e isthenumberofedgesin G,then

there are 2e distinct subgraphs are possible. Obviously some of these

subgraphs will be trees. Out of these trees we particularly interested in

certaintypeoftrees,calledspanningtrees.

9.4SpanningTrees

9.4.1Definition:

Atree T issaidtobeaspanningtreeofaconnectedgraph G if T isa

subgraphof G and T containsalltheverticesof G.

9.4.2 Example: Consider the graph G given in Fig. 9.4.2 A. Graph T

(giveninFig.9.4.2B)isaspanningtreeof G.

v3

v1

v4v2

v5

v6

v7b2

b3

b1

b6

b4

c5

c6

c8

c7

c1

b5

c2

c3

c4

Fig.9.4.2AGraphG

v7

v1

v2

v3v4

v5

v6

Fig.9.4.3BGraphT



Observations:

i) Spanning treesare the largest (with themaximumnumberofedges)

treesamongalltreesinG.Spanningtreeisalsocalledamaximaltree

subgraphormaximaltreeof G.

ii) Spanningisdefinedonlyforaconnectedgraph(sinceatreeisalways

connected).

iii) Eachcomponentofadisconnectedgraph,doeshaveaspanningtree.

Thus a disconnected graph with k components contains a spanning

forestconsistingof k spanningtrees.

9.4.3Theorem:

Everyconnectedgraphhasatleastonespanningtree.

9.4.4Definition:

i) Anedgeinaspanningtree T iscalledabranchof T.

ii) Anedgeof G that isnot inagivenspanningtree T iscalleda

chord.Inelectricalengineering,chordsometimesreferredtoastieora

link.

iii) Let T beanyspanningtreeofaconnectedgraph G,and T1 isthe

complementof T in G.Theneachedgein T iscalledabranch(with

respectto T),andthesetofalledgesin T iscalledthebranchset.

Eachedgein T1 iscalledachord(withrespectto T),andthesetof

edges in T1 is called thechord set (or)Tieset. T1 is calledas the

cotree.WemaywriteT insteadof T1.

Observation:

i) Branchesandchordsaredefinedonlywithrespecttoagivenspanning

tree.

ii) Anedgethatisabranchwithrespecttoonespanningtree T1 (of G)

maybeachordwithrespecttoanotherspanningtree T2.



9.4.5Theorem:

Withrespecttoanyofitsspanningtrees,aconnectedgraphofnvertices

andeedgeshasn1treebranchesand en+1chords.

Proof:LetG beanyconnectedgraphonnverticesand e edges.Let

T beanyspanningtreein G. Sinceeveryspanningtreeof G contains

allverticesof G,wehavethat|V(T)|= n andso|E(T)|=|V(T)|1= n

1.Sinceeveryedgeofaspanningtree T iscalledabranchofT,wehave

that Gcontains n1branches.SincethenumberofedgesinG is e,we

havethatthenumberofchordsof Tis e(n1)=en+1.

9.4.6Example:

Thereisaformconsistingofsixwalledplotsoflandasshowninthefigand

theseplotsarefullofwater.Wefindtheminimumnumberofwallsaretobe

brokensothatallthewatercanbedrainedout.

Consider the wall joints as vertices, and walls as edges. Then we can

consider itasagraph. Inthisgraphthenumberofvertices is n=10,and

thenumberofedgesis e=15.

Fig.9.4.6

If thereexistsacircuit, thenthewater inside thecircuitcannotbedrained

out.Sowehavetoremoveminimumnumberofedgessothatthegraphdo

not contain circuits. Tohave this, weshouldhave a spanning treewith

(n1)edges.Hencewehavetobreak e(n1)= en+1=1510

+1=6edges(walls)sothatallthewatercanbedrainedout.

If we add an edge between any two vertices of a tree, then a circuit is

created. This is because, therealreadyexistsonepathbetweenany two



verticesofa tree, addinganedge inbetween, createsanadditionalpath,

andhenceacircuit.Theconceptofafundamentalcircuithasanenormous

significanceinelectricalnetworkanalysis.

9.4.7Definition:

Let T beany spanning treeof a connectedgraph G.Addinganyone

chordto T willcreateexactlyonecircuit.Suchacircuitformedbyaddinga

chordtoaspanningtree,iscalledafundamentalcircuit.

9.4.8 Example: Consider the graph G (given in Fig. 9.4.8A), and its

spanningtree T(giveninFig.9.4.8B)of G.Now,ifweaddthechord c1

to T,wegetacircuitb1b2b3b5c1whichiscalledasfundamentalcircuit

(giveninFig.9.4.8C).

9.4.9Problem:AconnectedgraphG isatree ifandonly if addingan

edgebetweenanytwoverticesin G createsexactlyonecircuit.

Prof:SupposeGisatrue.Supposeweaddanewedge e= uv between

twovertices u and v in G.Since G isatree, G containsapathfrom u

and v.Byjoiningthisnewedge,therecreatesexactlyonecircuit.

c7c5

c6

c8

c4c3

c1

b2

b1

b4

b5b3b6

c2

Fig9.4.8AGraphG

b1

b2

b3

b4

b6b5

Fig.9.4.8BSpanningtreeT

c1b1

b2

b3

b4

b6b5

Fig.9.4.8CFundamentalcircuitF



Converse: Supposetheconversehypothesis.SupposeGcontainsacircuit:

v1e1v2e2env1.Byaddingnewedge e1= v1v2 ,wegettwocircuits v1e1

v2e1v1 and v1e1v2e2v3env1, acontradiction.So G containsnocircuits.

NowweshowthatGisconnected.Let u,v betwoverticesinG.Adda

newedge e*= uv toG.Thenbyconversehypothesis,therecreatesa

circuit.Supposethecircuitis ue*e1v1e2v2.eku.Now ve1v1e2v2ekuis

apathinbetweenv and u.Thisshowsthatthegraph G isconnected.

Hence G isatree.

9.4.10Problem:

ForatreeG isatree,byaddinganewedgebetweenanytwoverticesin G

creates,exactlyonefundamentalcircuit.

9.5AlgorithmsforSpanningTrees

9.5.1Definition:

i) AgraphG issaidtobeaweightedgraphifalltheedgese ofG were

assignedbyacorrespondingrealnumber w(e)(readastheweightofe).

ii) Let T beanyspanningtreeofaconnectedgraph G.thentheweight

of T,thatis, w(T)isdefinedasthesumofweightsofallbranches

inT.

iii) Aspanningtreewiththesmallestweightinaweightedgraphiscalleda

shortestspanningtree(or)minimalspanningtree(or)shortestdistance

spanningtree.

Observation:

i) Different spanning treesofG mayhavedifferentweights.Amongall

the spanning trees of G, one with the smallestweight is of practical

significance.

ii) Let G beagraphon n verticesinwhicheveryedgehasaunitweight.

Thenallthespanningtreeshavethesameweightof(n1)units.



9.5.2AnApplication:

i) Suppose that we are to connect n cities v1, v2,, vn through a

networkofroads.

ii) ThecostCij ofbuildingadirectroadbetween vi and vj isgivenfor

pairsofcities vi and vj whereroadscanbebuilt.(Notethatthere

maybepairsofcitiesbetweenwhichnodirectroadcanbebuilt).

iii) Nowtheproblemis:Tofindtheleastexpensivenetworkthatconnects

all n cities together. That is, to find a shortest spanning tree in a

connectedweightedgraphofnvertices.

iv) Heretheconnectednetworkmustbeatree.Otherwise,wecanalways

removesomeedgesandget a connectedgraphwith smallerweight.

Thus the problem of connecting n cities with a least expensive

network is the problem of finding a shortest spanning tree in a

connectedweightedgraphofnvertices.

9.5.3Problem: Let T bespanning tree (ofaweightedconnectedgraph

G).Thenthefollowingconditionsareequivalent:

i) T isashortestspanningtree(of G)and

ii) Thereexistsnospanningtree(of G)atadistanceonefrom T whose

weightissmallerthanthatof T.

9.5.4KruskalAlgorithm:(findingshortestspanningtree)

Step(i):ListalltheedgesofGinorderofnondecreasingweight.Nowwe

selectanedgee1ofGsuchthat w(e1)isassmallaspossibleande1isnot

aloop.

Step(ii):Selectnextsmallestedgefrom thesetofall remainingedgesof

G such that the selectededgedonot formacircuitwith theedges that

havealreadybeenchosen.

Step(iii):Wecontinuethisprocessof takingsmallestedgesamongthose

notalreadychosen,providednocircuitisformedwiththose,thathavebeen

chosenalready.



[If edges e1, e2, , ei have been chosen, then chose ei+1 from E

{e1,e2,,ei}insuchwaythatgraphwith{e1,e2,,ei+1}isacyclicand

w(ei+1)isassmallaspossible].

Step(iv):Ifagraph G hasnvertices,thenwewillstopthisprocess

afterchoosing n1edges.Theseedgesformasubgraph T,whichisnot

cyclic.(Thus T isashortestspanningtreeof G).

9.5.5 Example:

ConsidertheconnectedweightedgraphG.Listing:

)1(AB , )1(AD , )2(BC , )2(CD , )3(AE , )3(FE , )3(AF , )4(EB ,

)4(AC , )4(ED , )5(FD .

Step(i):Consideraspanningsubgraphof G withoutedges.

Select theedgewithminimumweight. For thisgraph theedgeAB isof

minimumweight.

Add this edge e1 = AB to the spanning null graph. Then we get the

subgraphG1.

Step(ii):Selecttheedge e2 suchthat G1+e2 containsnocircuitsand e2

gotminimumweightamongsuchedges.Forthisgraph e2=AD.Nowadd

e2 = ADto G1.ThenwegetthesubgraphG2.

Step(iii):Wecontinuethisprocess.Forthisexample,weadd e3 = BCto

G2 togetthesubgraphG3.



Nextweadd e4 =AE to G3 toget the subgraphG4. Nextweadd

e5 = EF toget the subgraphG5. Ifweaddanyedge to G5, weget a

circuit.Sotheprocessistobestoppedhere.

A

B

E

1F

12

Graph G5 or T

D C

33

F

A

B

E

1

12

SubgraphG4

DC

3

F

E

A

B

1

12

SubgraphG3

D CC

A

B

1

1 SubgraphG2

D

F

E

C

2

5 1

D

4

A

B

E

F

1

23 4

3

3

4

GraphG

A

B1

SubgraphG1

F

E

D

C



Thefinalgraph G5 isasubgraphof G andcontainsnocycles.Thus G5

isashortestspanningtreeofG.Write T = G5.Now w(T)=sumofthe

weightsofthebranchesin T =1+2+1+3+3= 10

9.5.6Prim'sAlgorithm:

Step(i):Chooseanyvertex v1 in G.

Step(ii):Chooseanedgee1=v1v2 of G suchthat v1 v2ande1has

the smallest weight among the edges of G incident with v1 . [For

convenience,wecanformatableindicatingtheweights].

Step(iii): Iftheedges e1,e2,,ei havebeenalreadychoseninvolving

endvertices v1,v2,,vi+1 ,chooseanedgeei+1,where ei+1=vjvk with

vj {v1,v2,,vi+1},and vk {v1,,vi+1}suchthat ei+1 hasthesmallest

weightamongtheedgesof G withpreciselyoneendin{v1,v2,,vi+1}.

Notethatafteraddingei+1 thegraphshouldbeacyclic.

Step(iv): The process will stop after choosing the n1 edges (otherwise,

repeatthestep(iii)).

9.5.7Illustration:Considertheconnectedweightedgraph G.Letusstart

with C.

5 1

D

4 2

A

C

B

E

F

1

23 4

33

4

GivengraphG GraphG1D

A

F

E

B

C



Step(i):Takethespanningnullsubgraph G0 of G.

Letuschoosev1=A.TheedgeAD isincidenton v1 andhasthesmallest

weightamongtheedgesincidentonv1=A.WriteG1 =(G0+AD)

Step(ii): InthisstepweselectDCandwriteG2=(G1+DC).

Step(iii): Weselect AB andwrite G3=(G2+ AB).

Step(iv):Weselect AF andwrite G4=(G3+AF).

Step(v):WeselectFEandwriteG5=(G4+FE).

Step(vi):Since G5 isaspanningtree,theprocessstopshere.

Write T = G5.

Hence,theshortestspanningtreeof G isG5=T.

Now w(T)=1+1+2+3+3=10.

A

B

E

1

F

1

2D

C

3

3

GraphG5=T

A

B1

F

1

2D

C

3

GraphG4

E

A

B1

1

2D

C

GraphG3

F

E

A

1

1

D

C

GraphG2

F

E

B



SelfAssessmentQuestions:

1. Drawtreeswith5vertices.

2. Findallspanningtreesofthegraph.

3. Findallspanningtreesofthefollowinggrpah.

4. The complete graph on n vertices Kn: has2 -nn different spanning

trees.GiveallthespanningtreesofK4.

5. Findaspanningtreeforeachofthefollowinggraphs

6. Whichconnectedsimplegraphshaveexactlyonespanningtree?

o oo

o

o

oo o

o o

(i)o

o

o

o o

o o

o o

(ii)

o

o

o

o

(iii)

o

o

o o

o

(iv)

o

o

ooo

o

o o

o



7. UsePrimsAlgorithmtofindaminimalspanningtreeforthefollowing

weightedgraph.

9.6Summary:

Thisunitdealtwithaspecialtypeofgraphsnamelytrees,whichareextensively

usedsearchproceduresanddesignofcomputeralgorithms.Thereaderwillbe

abletogetandideatofindthedifferentspanningtreesfromagivenconnected

graph. Different characterizations and properties of trees were given. Also

someapplicationsofbinarytreeswerediscussed.Spanningtreeplayavital

roleinmulticastingoverinternetprotocolnetworks.

9.7TerminalQuestions

1. Draw two different binary trees with five vertices having maximum

numberofleaves.

2. Drawagraphwiththegivenspecification.Incase,ifnotpossible,then

explainwhynosuchgraphsexists.

i) fullbinarytree,fiveinternalvertices

ii) fullbinarytree,fiveinternalvertices,seventerminalvertices

iii) fullbinarytree,twelvevertices

iv) fullbinarytree,ninevertices.

3. If T1 andT2 aretwospanningtreesinagraph,thenshowthat

i) Thenumberofedgesin T1 not in T2 isequaltothenumberof

edgesinT2 notin T1.(ii) d(T1,T2)= d(T2,T1).

4. LetN(g)=thenumberofedgesinagraph g.Then N(Ti Tj)=the

8

A

3

6

47

2

46

8 6

C

BE

D

o

o

o

o

o



numberofedgesinTi Tj ,and d(Ti,Tj)=Thedistancebetween Ti

and Tj.Showthat d(T1,T2)=21

N(T1 T2).

5. Considerthetableofairlinedistancesinmilesbetweensixoflargestcities

intheworld:London,Mexico,NewYork,Paris,PekingandTokyo.

L MC NY Pa Pe T

L 5558 3469 214 5074 5959

MC 2090 5725 7753 7035

NY 3636 6844 6757

Pa 5120 6053

Pe 1307

T

9.8Answers

SelfAssessmentQuestions

1.

2. The given graph has four vertices and so each spanning tree must

have41=3edges.Theyare

o

o

o

o

o

oo

o

o

o

o

o

o

o

o

o o

o

o

o

o o o o o o

o

o o

o

o

o

o

oo o oo

o o

ooo oo o

o

L



4. K4has244 - =16differentspanningtrees.Eachhavingthreeedges.

5.

6. Tree.

7. ChooseedgesAE,AC,DC,AB.

3.o o

o

o

oo o o o

o o o oo

oo o

o

(i) o

oo

o

o

o

o

o

o

(ii) o

oo o

(iii) o

o

o

o

o

(iv)

o

o o

o

o

oo

o

o



TerminalQuestions

1.

2.

ii) Not possible, since any full binary tree with five internal vertices

havesixterminalvertices,butnotseven.

iii) No,afullbinarytreehas2k+1verticeswherekisthenumberof

internalverticesbut2k+1isodd.

iv)

3. i) Let T1,T2 betwospanningtrees.Now|E(T1)|= n1=|E(T2)|.

Suppose T1 contains m edgeswhicharenotin T2.Nowthe

numberofedgesin T2 notin T1 is

(n1)[(n1)m]= m.followsfrom(i).

(i)

b

c

a

e

f g

j k

o

o

oo

o

o

o

d

h io o

o o

o

oo

o

oo

o

oo

o

o o

o o o

o

oo

o



4. N(T1 T2)=numberofedgesinT1 T2 =(numberofedgesinT1but

notinT2)+(numberofedgesinT1butnotinT2)= d(T1,T2)+d(T2,

T1)=2d(T1,T2) d(T1,T2)=21

N(T1 T2)

5.

214

2090

3469

5074

1307

MC

NY

T

L

Pa

Pe

FigB Shortest spanning tree

5725

6053

7753

6757

50743469

5120 3636

5558

2090

5959

1307

T

Pe

MC

NY

Pa

7035

6844

214

Fig.A

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MC0063(B)-unit9-fi