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Year 12Mathematics
Contents
uLake Ltdu a e tduLake LtdInnovative Publisher of Mathematics Texts
• AchievementStandard .................................................. 2
• GraphTheoryBasics .................................................... 3
• EulerPathsandCircuits................................................. 6
• EvenandOddVertices................................................... 10
• Traversability.......................................................... 13
• HamiltonianCircuitsandPaths........................................... 18
• TravellingSalesmanProblem............................................. 22
• ShortestPathsandDijkstra’sAlgorithm.................................... 27
• MinimumSpanningTrees ............................................... 37
• Kruskal’sAlgorithm .................................................... 38
• TheChinesePostmanProblem............................................ 47
• TheKönigsbergBridgeProblem .......................................... 49
• PracticeInternalAssessment ............................................. 52
• Answers............................................................... 56
Robert Lakeland & Carl NugentUse Networks in Solving ProblemsIAS 2.5
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
2 IAS 2.5 – Networks
NCEA 2 Internal Achievement Standard 2.5 – NetworksThisachievementstandardinvolvesapplyingnetworkmethodsinsolvingproblems.
◆ ThisachievementstandardisderivedfromLevel7ofTheNewZealandCurriculum, andisrelatedtotheachievementobjective: ❖ chooseappropriatenetworkstofindoptimalsolutions intheMathematicsstrandoftheMathematicsandStatisticsLearningArea.
◆ Applynetworkmethodsinsolvingproblemsinvolves: ❖ selectingandusingmethods ❖ demonstratingknowledgeofconceptsandterms ❖ communicatingusingappropriaterepresentations.
Relationalthinkinginvolvesoneormoreof: ❖ selectingandcarryingoutalogicalsequenceofsteps ❖ connectingdifferentconceptsorrepresentations ❖ demonstratingunderstandingofconcepts ❖ formingandusingamodel; andalsorelatingfindingstoacontext,orcommunicatingthinkingusingappropriatemathematical statements.
Extendedabstractthinkinginvolvesoneormoreof: ❖ devisingastrategytoinvestigateasituation ❖ identifyingrelevantconceptsincontext ❖ developingachainoflogicalreasoning,orproof ❖ formingageneralisation; andalsousingcorrectmathematicalstatements,orcommunicatingmathematicalinsight.
◆ Problemsaresituationsthatprovideopportunitiestoapplyknowledgeorunderstandingof mathematicalconceptsandmethods.Situationswillbesetinreal-lifeormathematicalcontexts.
◆ Methodsincludeaselectionfromthoserelatedto: ❖ shortestpath ❖ traversability ❖ minimumspanningtree.
Achievement Achievement with Merit Achievement with Excellence• Applynetworkmethodsin
solvingproblems.• Applynetworkmethods,
usingrelationalthinking,insolvingproblems.
• Applynetworkmethods,usingextendedabstractthinking,insolvingproblems.
A B C
F E D
G H I
Start
End
3IAS 2.5 – Networks
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
Introduction
Graph Theory BasicsThegraphswewillstudyinthisAchievementStandardaredifferentfromthoseyouhavepreviouslystudiedanddrawnusinggridpaper.InGraphTheoryagraphconsistsofasetofvertices(nodes)andedges(straightorcurvedlines).Eachedgejoinsonevertex(node)toanother.
Thedegreeofavertex(node)isthenumberofedges(lines)whichstartorendatthevertex.
VertexAhasdegree3becauseithasthreeedges(lines)‘comingorgoing’fromitwhereasvertexChasdegree5becauseithasfiveedges(lines)‘comingorgoing’fromit.Insomegraphsitispossiblethattwolineswillcrossbutunlesstheyaremarkedasavertextheyshouldnotbetreatedassuch.Seethediagrambelow.TheintersectionoflinesACandDBisnotavertex.
Vertices(nodes)
Edges(lines)
Graph
A
B
DC
A
BD
C
Degree3
Degree5
Think of two lines that cross, but not at a vertex, as similar to an underpass on a motorway.
In this Achievement standard reference is made to graph theory and graphs. Networks are an application of graph theory. The authors use the term, graph, as this is the accepted practice, but it is possible to call these graphs, networks, in order to differentiate them from Cartesian graphs.©
uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
18 IAS 2.5 – Networks
Hamiltonian Paths and Circuits
AHamiltonianpath,namedaftertheIrishmathematicianSirWilliamRowanHamilton,isonethatvisitseachvertex(node)onlyonce.Eachedgedoesnothavetobetraversed.AHamiltoniancircuitorcycleisonethatvisitseachvertex(node)onlyoncebutalsoreturnstoitsstartingvertex(node).EachedgedoesnothavetobetraversedinaHamiltoniancircuit.AnexampleofaHamiltonianpathisdrawnbelow.
Apossiblepaththatvisitseachvertex(node)onlyoncebutdoesnotreturntothestartisdrawnbelow.NotealsothatnotalledgeshavebeentraversedwhichisnotarequirementofaHamiltonianpath.
AnexampleofaHamiltoniancircuitisdrawnbelow.
Apossiblepaththatstartsandendsatthesamepointandvisitseachvertex(node)isdrawnontheright.NotealsothatnotalledgeshavebeentraversedwhichisnotarequirementforaHamiltoniancircuit.
Hamiltonian circuits are useful in applications where it is necessary to visit each ‘vertex’ for example, a salesman’s clients in a town.
A circuit (cycle) is a closed path, i.e. it finishes where it starts.
A
BD
C
A B C
F E D
G H I
A
BD
C
Start/end
1 2
34
Unfortunately there is not an equivalent formula to Euler’s to identify whether or not a graph has a Hamiltonian path or circuit.
In an Euler path you can visit each vertex (node) more than once but each edge only once. In a Hamiltonian path you can only visit each node once.
A B C
F E D
G H I
Start
End
In problems involving Hamiltonian circuits there are many equivalent reversal routes. When we don’t want to count reversal routes we use the term ‘distinct’ routes.
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21IAS 2.5 – Networks
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
37.
DrawtheHamiltonianpathorcircuitbelow.Labelyourstartandendpoints. Alinethatyoumarkassolidwillbeonethatyouneednottraverse.
A
B
C
D
E
F G
A
B
C
D
E
F G
38.A
B
C
D
E
F
G
H
DrawtheHamiltonianpathorcircuitbelow.Labelyourstartandendpoints. Alinethatyoumarkassolidwillbeonethatyouneednottraverse.
A
B
C
D
E
F
G
H
39.
A B C D
EFGH DrawtheHamiltonianpathorcircuitbelow.Labelyourstartandendpoints. Alinethatyoumarkassolidwillbeonethatyouneednottraverse.
A B C D
EFGH
Circuitorpath?
Justification:
Circuitorpath?
Justification:
Circuitorpath?
Justification:
Path/Circuit:
Path/Circuit:
Path/Circuit:
I
I
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39IAS 2.5 – Networks
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
Everysuburb(node)ofourgraphisnowtouchedbyalink(fibreoptics).Thisisnowourminimumspanningtreewithaminimumcostof: 50000+50000+30000+70000+30000+110000=$340000
Kruskal’s Algorithm cont...
When using Kruskal’s Algorithm, if there are two equal cheapest links to choose from, it will not matter which one you select. The same minimum spanning tree will be generated but in a different order.
Kruskal’s Algorithm is often called a ‘Greedy Algorithm’. It repeatedly adds the next cheapest link that doesn’t produce a cycle. It was named after Joseph Kruskal an American mathematician who lived from 1928 – 2010.
E
7A
B
5
3
5
G
C
3 D
11
©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
41IAS 2.5 – Networks
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
WebeginbyfindingtheshortestpieceoftrackwhichisBtoD,withaweightof5(500metres).
D
13
F3
E
9
A
B C
G
3
5
5
cont...
Thecompletedminimumspanningtreehasatotalminimumweightof5+5+3+9+3+13=38.
ExampleAfarmerwishestodevelopasystemoftracksaroundhisfarmsothathe
canserviceallpartsofthefarm.Thedistanceinhundredsofmetresbetweenkeypointsonthefarmisgiveninthetablebelow.DrawupagraphfirstrepresentingthissituationandthenuseKruskal’sAlgorithmtofindaminimumspanningtreeforthefarmersohecandevelopasystemoftrackscoveringtheminimumdistance.
A B C D E F G HA - - - 9 8 - - -B - - 7 5 - - - -C - 7 - 8 - - 6 -D 9 5 8 - 10 12 11 13E 8 - - 10 - 15F - - 12 - - 8 10G - - 6 11 - 8 - -H - - - 13 15 10 - -
Drawingupagraphusingthedistancesfromthetablewegetthefollowing.
D
F
H
A
B
G
E
C
8
910 15
8
12116
8
57
10
13
D
B
5
ThenextshortestpieceoftrackisCtoGwithadistanceof600metressoweincludethislink.
D
B
G
C
6
5
ThenextshortestpieceoftrackisCtoBwithadistanceof700metressoweaddthislink.
D
B
G
C
6
57
ThenextshortestpiecesoftrackareCtoD,AtoEandGtoFallwithadistanceof800metres.WecannotaddCtoDaswewouldformacyclewithinthetreesowechoosetoaddGtoFandthenAtoE.
D
F
A
B
G
E
C
8
8
6
57
ThenextshortestpiecesoftrackisDtoAwithadistanceof900metressoweaddthislink.
©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
46 IAS 2.5 – Networks
64. Aplumberhastorunpipesfroma seriesofninepointsinabuilding. Atableoftheestimatedtime,inhours,tolay thepipesbetweeneachpointisgivenbelow. Firstdrawupagraphrepresentingthis situationandthenuseKruskal’sAlgorithmto findaminimumspanningtreeforthe pipingandthencalculateitsminimum layingtime.
A B C D E F G H IA - 5 - - - - - 9 -B 5 - 9 - - - - 12 -C - 9 - 8 - 5 - - 3D - - 8 - 10 15 - - -E - - - 10 - 11 - - -F - - 5 15 11 - 3 - -G - - - - - 3 - 2 7H 9 12 - - - - 2 - 8I - - 3 - - - 7 8 -
65. AinstallerhastolayTVcablingtoa neighbourhoodofninehouses. Atableoftheestimatedtimes,inhours, tolaythecablingbetweeneachhouseis givenbelow. Firstdrawupagraphrepresentingthis situationandthenuseKruskal’sAlgorithmto findaminimumspanningtreeforthe cablingandthencalculateitsminimum installingtime.
A B C D E F G H IA - 10 - 11 15 - - - -B 10 - 17 - 3 9 - - -C - 17 - - - 12 - - -D 11 - - - 14 - 21 - -E 15 3 - 14 - 16 19 4 -F - 9 12 - 16 - - 8 2G - - - 21 19 - - 5 -H - - - - 4 8 5 - 7I - - - - - 2 - 7 -©
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51IAS 2.5 – Networks
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
b) AlargewaterreservoiratvillageAisto supplywatertoallthevillages.Use Kruskal’salgorithmtofindaminimum spanningtreeforthenetworkofpipesand calculatethetotalamountofpiping required.
B F
A
C
E
G
I
D H
7.5km
10km
12.5km
4.2km 5km
4.2km3.3km 5km
7.5km 5.8km5km5.8km
2.5km 2.5km
4.2km 6.7km 5.8km 4.2km
c) AtrafficofficerleavesfromvillageA inthemorningandmustdriveallthe roadsbetweenthevillagesaspartofhis ‘run’.Givearouteforthetrafficofficerso thathevisitseachroadonlyonceandends upatvillageI.
d) Whatisthetotaldistancethetrafficofficer travelsinadayifhereturnstovillageA usingthequickestpossibleroute?
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57IAS 2.5 – Networks
IAS 2.5 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
Page 1222. Evenvertices:A,B,C,D,E,F,G Oddvertices:None Classification:Eulercircuit Reason:Allverticesareeven.23. Evenvertices:A,C,E,F,G Oddvertices:B,D Classification:Eulerpath Reason:Hastwooddvertices andsomeevenvertices.24. Evenvertices:A,B Oddvertices:C,D Classification:Eulerpath Reason:Hastwooddvertices andsomeevenvertices.25. Evenvertices:C,D Oddvertices:A,B,E,F Classification:Neither Reason:Morethantwoodd verticessoneither.Page 1426. Evenvertices:A,B,C,D,E,F,G H,I Oddvertices:None Traversable:Yes StartandEndpoints:StartI EndI
A B
D
C
EF
G
H
I1
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8
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54
3
2
Start/end
27. Evenvertices:C,D,E,H Oddvertices:A,B,F,G Traversable:No StartandEndpoints:N/A
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C
E
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H
I
A B1
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9 87
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3
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Start/end
Page 1528. Evenvertices:A,B,C,D,E,F,G H,I Oddvertices:None Traversable:Yes StartandEndpoints:StartA EndA
29. Evenvertices:A,D,E,H Oddvertices:B,C,F,G,I,J Traversable:No StartandEndpoints:N/A
Page 1630. Evenvertices:A,B,C,D,E,G, Oddvertices:F,H Traversable:Yes StartandEndpoints:StartF EndH
31. Evenvertices:A,B,C,D,F,H,K Oddvertices:E,G,I,J Traversable:No StartandEndpoints:N/A
DC
EFG1
H
A B
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Start
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Page 1732. Evenvertices:A,B,C,D,E,F,G H,I Oddvertices:None Traversable:Yes StartandEndpoints:StartB EndB
D CE
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A
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9
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Start/end
33. Evenvertices:A,B,D,E,H,J Oddvertices:C,F,G,I Traversable:No StartandEndpoints:N/A
Page 2034. Circuit/path:Hamiltoniancircuit Justification:Possibletovisit eachvertexand returntothestart.
Path:C,E,F,D,A,B,C.35. Circuit/path:Hamiltoniancircuit Justification:Possibletovisit eachvertexand returntothestart.
Path:A,B,D,E,F,C,A.
A
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2Start/end
A B
CD
EF
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Start/end
©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts
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