Hamiltonian and Eulerian

7/24/2019 Hamiltonian and Eulerian

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Copyright ZephGrunschlag, 2001-2002.

Euler and Hamilton Cycles;lanar Graphs; Coloring.

Zeph Grunschlag


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#gendaEuler paths and cycles

Hamilton paths and cycles

lanar graphs $egions Euler characteristic Edge-%ace Handsha&ing Girth

Graph Coloring 'ual Graph (cheduling


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Euler and Hamilton aths-*oti+ation

#n pictorial ay to moti+ate thegraph theoretic concepts o

Eulerian and Hamiltonian pathsand circuits is ith to pules/

he pencil draing prolem

he taica prolem


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!2" 3

encil 'raing rolem-Euler aths

4hich o the olloing pictures cane dran on paper ithout e+erliting the pencil and ithout

retracing o+er any segment5


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Graph heoretically/ 4hich o theolloing graphs has an Eulerpath5


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#nser/ the let ut not the right.

start 7nish

1 2

)

3

"

6


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Euler aths and Circuits'e7nition

'E%/ #n Euler pathin a graph G isa simple path containing e+ery

edge in G. #n Euler circuit 9orEuler cycle: is a cycle hich is anEuler path.


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aica rolem-Hamilton aths

Can a taica dri+er mil& his haplesscustomer y +isiting e+eryintersection eactly once, hen

dri+ing rom point # to point > 5#

>


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aica rolem-Hamilton aths

Graph heoretically/ @s there aHamilton path rom # to > in theolloing graph5

9< in this case:#

>


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Circuits'e7nition

'E%/ # Hamilton pathin a graph G isa path hich +isits e+er +erte in Geactly once. # Hamilton circuit 9or

Hamilton cycle: is a cycle hich+isits e+ery +erte eactly once,except for the rst vertex, hich is

also +isited at the end o the cycle.


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@mplications to C(

%inding Hamilton paths is a +eryimportant prolem in C(.

EG/ Aisit e+ery city 9+erte: in aregion using the least trips 9edges:as possile.

EG/ Encode all it strings o a certain

length as economically as possileso that only change one it at atime. 9Gray codes:.


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@mplications to C(

#nalying diBculty o Euler +s.Hamilton paths is a great C( case

study.%inding Euler paths can e done inO 9n: time

%inding Hamilton paths is NP-

complete(light change in de7nition can result in

dramatic algorithmic iurcation


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%inding Euler aths

o 7nd Euler paths, eDll 7rst gi+e analgorithm or 7nding Euler cycles andthen modiy it to gi+e Euler paths.

H*/ #n undirected graph Ghas anEuler circuit i it is connected ande+ery +erte has e+en degree.


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%inding Euler Circuits

F/ 4hy does the olloing graphha+e no Euler circuit5


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#/ @t contains a +erte o odddegree.


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!etDs pro+e the theorem constructively.Constructi+e means that the proo illactually contain an algorithm or constructing

the Euler path, hen it eists.art 1: (uppose Gis connected and each

+erte has e+en degree. Construct an Eulercycle. 4e pro+e this y strong induction onm the numer o edges in G.

>ase case m 0/ (ince G is connected andcontains no edges, it must consist o a single+erte. he empty path is an Euler cycle.1


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%inding Euler Circuits@nduction step or mI1 edges,

assuming pro+ed this up to m 0.

C!#@*/ Gcontains a simple cycle.

Consider an edge e9since mI1 J 0:.@ eis a sel-loop, then is a simplecycle. (o can assume that e is a

loopless edge/ v v(ince deg9v : J 1 9all degrees are

e+en:, another edge e must e

incident ith v /v v

e


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%inding Euler Circuits(o can continue adding edges until e

7nd an edge hose ne endpointhas already een encountered during

the process. his endpoint is a+erte hich is seen tice so athich a simple cycle is ased

9his pro+es claim:Simplecycle


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Gi+es rise to a recursi+e prooKalgorithm/

1: %ind a simple cycle in connected graph ithmI1 edges.

2: 'elete all the edges rom the cycle and 7ndEuler cycles in each resulting component

): #malgamate Euler cycles together using thesimple cycle otaining anted Euler cycle.

!etDs see ho the amalgamation process or&s/


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%inding Euler Circuits*ohammedDs (cimitars


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%ound a cycle ater starting rommiddle +erte.

'elete the cycle/


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'elete the cycle/


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'elete the cycle/


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'elete the cycle/


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!2" 2"



'elete the cycle/

i di l i i


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o try again 9say rom middle+erte:/

i di l Ci i


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his time, ound a cycle starting andending at middle +erte/

#malgamate these cycles togetherrom a point o intersection, anddelete rom graph/

%i di E l Ci i


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!2" 2=


%ind another cycle rom middle+erte/

1

%i di E l Ci i


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12

%i di E l Ci it


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!2" )0



12)

%i di E l Ci it


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!2" )1



12) 3

%i di E l Ci it


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!2" )2



12) 3 "

%i di E l Ci it


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!2" ))



12) 3 "

6

%i di E l Ci it


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!2" )3



12) 3 "

6

8

%i di E l Ci it


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!2" )"



12) 3 "

6

8=

%i di E l Ci it


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!2" )6



12) 3 "

6

8=?

%i di E l Ci it


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!2" )8



12) 3 "

6

8=?

10

%i di E l Ci it


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!2" )=



12) 3 "

6

8=?

10

11



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#malgamate it to Euler cycle o deletedgraph, and delete it. eed to insertcycle eteen ormer edges 10 L 11/

12) 3 "

6

8=?

10

11



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%inally, need to add the triangle.

Mse same naN+e approach loo&ing orcycle in remaining component/

12) 3 "

6

8=?

10

55



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12) 3 "

6

8=?

10

55 11



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12) 3 "

6

8=?

10

55 1112



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!2" 3)




12) 3 "

6

8=?

10

55 1112

1)



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12) 3 "

6

8=?

10

55 1112

1)13



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!2" 3"




12) 3 "

6

8=?

10

1" 1112

1)13



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!2" 36


#malgamate the triangle cycleeteen edges ormerly laeled ?L 10/

12) 3 "

6

8=?

10

1" 1112

1)13



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!2" 38


#malgamate the triangle cycleeteen edges ormerly laeled ?L 10/

12) 3 "

6

8=?

55

55 5555

5555



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!2" 3=


4e ound the Euler circuit

12) 3 "

6

8=?

1)

1= 131"

161810

11 12

Euler Circuit


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Euler Circuit #ll 'egrees E+en

2ndhal o theorem says that an Euler circuitin a graph implies that all degrees are e+en/

@n a simple cycle, hene+er path enters+erte, must come out on dierent edge.

hus e+ery +isit o vcontriutes 2 to deg9v:.hus, i &eep only edges hich ere on thesimple cycle, degrees o resulting graph areall e+en. >ut in an Eulerian graph G, can

7nd a simple cycle containing all +erticesand conseOuently graph resulting rom cycleis G itsel


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Generaliing to Euler aths

F/ 'oes the olloing ha+e an Eulercircuit5


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#/ o, +ertices o odd degree/

F/ >ut hy does it ha+e an Euler path5


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Generaliing to Euler aths#/ PE( >ecause eactly 2 +ertices o odd

degree.

(o can add a phantom edge eteen odddegree +ertices/


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#ll degrees no e+en so 7nd Eulercycle/


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o remo+e phantom edge otaining/

1 2

)

3

"

68


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Generalie/

1 2

)

3

"

6


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H*/ #n undirected connectedgraph has an Euler path i thereare eactly to +ertices o odd

degree.


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>lac&oard Eercises or8."

1: ro+e y induction that Qnalayshas a Hamilton cycle or n J 1.

9his gi+es a Gray code:.2: ro+e that the olloing graph

has no Hamilton cycle/


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lanar Graphs

Planar graphsare graphs that can e dranin the plane ithout edges ha+ing to cross.

Mnderstanding planar graph is important/

#ny graph representation o mapsKtopographical inormation is planar. graph algorithms oten specialied to planar

graphs 9e.g. tra+eling salesperson:

Circuits usually represented y planargraphs

-Common


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-Common*isunderstanding

Qust ecause a graph is dran ithedges crossing doesnDt mean its

not planar.F/ 4hy canDt e conclude that the

olloing is non-planar5

-Common


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Common*isunderstanding

#/ >ecause it is isomorphic to agraph hich isplanar/

-Common


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-Common


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-Common


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-Common


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-Common


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-Common


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-Common


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-Common


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ro+ing lanarity

o pro+e that a graph is planaramounts to redraing the edges in

a ay that no edges ill cross.*ay need to mo+e +ertices aroundand the edges may ha+e to edran in a +ery indirect ashion.

E.G. sho that the )-cue is planar/

ro+ing lanarity


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ro+ing lanarity)-Cue

ro+ing lanarity5


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ro+ing lanarity53-Cue

(eemingly not planar, ut hoould one pro+e this


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'ispro+ing lanarity

he oo& gi+es se+eral methods. @Dlldescrie one method that ill alaysor& in eamples that youDll get onthe 7nal. Pou may also use any othe methods that the oo& mentions.

9


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'ispro+ing lanarity

he idea is to try to 7nd someinvariantOuantities possessed y

graphs hich are constrained tocertain +alues, or planar graphs.hen to sho that a graph is non-planar, compute the Ouantities andsho that they do not satisy theconstraints on planar graphs.


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$egions

he 7rst in+ariant o a planar graph ill e thenumer o regionsthat the graph de7nes inthe plane. # region is a part o the planecompletely disconnected o rom other partso the plane y the edges o the graph.

EG/ the car graph has 3 regions/

21

)3


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$egions

F/ Ho many regions does the )-cue ha+e5


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$egions

#/ 6 regions

1 2

)

36

"


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$egions

H*/ he numer o regions de7ned y aconnected planar graph is in+ariant o hoit is dran in the plane and satis7es theormula in+ol+ing edges and +ertices/

r SE S - SV S I 2EG/ Aeriy ormula or car and )-cue/

1 2

)

3612-=I2"

21

)36-3I2


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Euler Characteristic

he ormula is pro+ed y shoing thatthe Ouantity 9chi: r- SE S I SV Smust eOual 2 or planar graphs. iscalled the Euler characteristic. heidea is that any connected planargraph can e uilt up rom a +erte

through a seOuence o +erte andedge additions. %or eample, uild )-cue as ollos/


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hus to pro+e that is alays 2 orplanar graphs, one calculate or

the tri+ial +erte graph/ 1-0I1 2

and then chec&s that each possile

mo+e does not change .


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Chec& that mo+es donDt change /1: #dding a degree 1 +erte/

r is unchanged. SE S increases y 1. S

V S increases y 1. I 90-1I1:2: #dding an edge eteen pre-

eisting +ertices/

r increases y 1. SE S increases y 1. SV S unchanged. I 91-1I0:

EG/

EG/

#nimated @n+ariance o


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!2" =2


SV S SE S r

r- SE S I SVS

1 0 1 2



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!2" =)


SV S SE S r

r- SE S I SVS

2 1 1 2



84/141

!2" =3


SV S SE S r

r- SE S I SVS

) 2 1 2



85/141

!2" ="


SV S SE S r

r- SE S I SVS

3 ) 1 2



86/141

!2" =6


SV S SE S r

r- SE S I SVS

3 3 2 2



87/141

!2" =8


SV S SE S r

r- SE S I SVS

" " 2 2



88/141

!2" ==


SV S SE S r

r- SE S I SVS

6 6 2 2



89/141

!2" =?


SV S SE S r

r- SE S I SVS

8 8 2 2



90/141

!2" ?0


SV S SE S r

r- SE S I SVS

= = 2 2



91/141

!2" ?1


SV S SE S r

r- SE S I SVS

= ? ) 2



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!2" ?2


SV S SE S r

r- SE S I SVS

= 10 3 2



93/141

!2" ?)


SV S SE S r

r- SE S I SVS

= 11 " 2



94/141

!2" ?3


SV S SE S r

r- SE S I SVS

= 12 6 2


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%ace-Edge Handsha&ing

%or allgraphs handsha&ing theorem relatesdegrees o verticesto numer o edges.

%or planargraphs, can relate regions toedgesin similar ashion/

EG/ here are to ays to count thenumer o edges in )-cue/

1: Count directly/ 12

2: Count no. o edges aroundeach region; di+ide y 2/

93I3I3I3I3I3:K2 12 92 triangles peredge:


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%ace-Edge Handsha&ing

'E%/ he degreeof a regionF isthe numer o edges at itsoundary, and is denoted y

deg9F :.H*/ !et Ge a planar graph ith

region set R. hen/

=

RF

FE )deg(2

1||


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Girth

he girtho a graph is the length othe smallest simple cycle in the

graph.F/ 4hat the girth o each o the

olloing5


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!2" ?=

Girth

#/g 2 g 3 g 3

F/ 4hat the smallest possile girth or

simple ipartite graphs5


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!2" ??

Girth

#/ g 3 is the smallest possile girth/#ny cycle must start and end in thesame color, so must ha+e e+en

length. (ince simple, cannot ha+e a2-cycle, so 3-cycle is shortestpossile.

ro+ing that Q3is on-


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!2" 100

lanar

o e ha+e enough in+ariants to pro+ethat the 3-cue is non-planar.

1: Count the numer o +ertices and edges/

SVS 16 9tice the numer or )-cue:SES )2 9tice the numer or )-cue plus

numer o +ertices in )-cue:

2: (uppose 3-cue ere planar so y EulerDs

ormula numer o regions ould e/r )2-16I21=

ro+ing that Q3is on-


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!2" 101

lanar

): Calculate the girth/ g 33: #pply handsha&ing theorem to get a

loer ound on the numer o edges,

since the degree o each ace muste at least as large as the girth/

@n our case, this gi+e SE S TU1=U3)6contradicting SE S )2

hus 3-cue cannot e planar.

rggFE

RFRF 2

1

2

1)deg(

2

1|| ==

>lac&oard eercises or


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!2" 102

>lac&oard eercises or8.8

(ho that the olloing graphs arenon-planar/

1: K"2: K),)): Qnor n 3

h l i


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!2" 10)

Graph Coloring

Consider a 7ctional continent.

C l i


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*ap Coloring

(uppose remo+ed all orders utstill anted to see all thecountries. 1 color insuBcient.

* C l i


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!2" 10"

*ap Coloring

(o add another color. ry to 7ll ine+ery country ith one o the tocolors.

* C l i


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!2" 106

*ap Coloring


* C l i


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!2" 108

*ap Coloring


* C l i


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!2" 10=

*ap Coloring


* C l i


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!2" 10?

*ap Coloring

$!E*/ o adVacent countriesorced to ha+e same color. >orderunseen.

* C l i


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!2" 110

*ap Coloring

(o add another color/

* C l i


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!2" 111

*ap Coloring

@nsuBcient. eed 3 colors ecauseo this country.

* C l i


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!2" 112

*ap Coloring

4ith 3 colors, could do it.

3 C l h


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!2" 11)

3-Color heorem

H*/ #ny planar map o regions cane depicted using 3 colors so that

no to regions that share apositi+e-length order ha+e thesame color.

roo y Haa&en and #ppel usedehausti+e computer search.

%rom *ap Coloringt G h C l i


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!2" 113

to Graph Coloring

he prolem o coloring a map, cane reduced to a graph-theoreticprolem/



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!2" 11"

to Graph Coloring

%or each region introduce a +erte/



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!2" 116

to Graph Coloring

%or each pair o regions ith apositi+e-length common orderintroduce an edge/

%rom *aps to Graphst ' l G h


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!2" 118

to 'ual Graphs

$eally, could thin& o original map asa graph, and e are loo&ing at dualgraph/



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!2" 11=

to 'ual Graphs

ual Graphs /1: ut +erte inside each region/



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!2" 11?

to 'ual Graphs

ual Graphs /2: Connect +ertices across common

edges/

'e7nition o 'ual Graph


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'e7nition o 'ual Graph

'E%/ he dual graphG Wo aplanar graph G 9V! E! R:XAertices, Edges, $egionsY is the

graph otained y setting Aertices o G W/ V 9G W: R Edges o G W/ E 9G W: set o edges

o the orm F1,F2[ here F1andF2share a common edge.

%rom *aps to Graphsto 'ual Graphs


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!2" 121

to 'ual Graphs

(o ta&e dual graph/

%rom *ap Coloringto Graph Coloring


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!2" 122

to Graph Coloring

Coloring regions is eOui+alent tocoloring +ertices o dual graph.

'e7nition o Colorale


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!2" 12)

'e7nition o Colorale

'E%/ !et n e a positi+e numer. #simple graph is n -colorable i the+ertices can e colored using ncolorsso that no to adVacent +ertices ha+ethe same color.

he chromatic number o a graph issmallest numer nor hich it is n

-colorale.EG/ # graph is ipartite i it is 2-

colorale.



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!2" 123

to Graph Coloring

*ap not 2-colorale, so dual graphnot 2-colorale/



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!2" 12"

to Graph Coloring

*ap not )-colorale, so graph not )-colorale/



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!2" 126

to Graph Coloring

Graph is 3-colorale, so map is asell/

3-Color heoremGraph heory Aersion


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!2" 128

Graph heory Aersion

H*/ #ny planar graph is 3-colorale.

Graph Coloring and(chedules


128/141

!2" 12=

(chedules

EG/ (uppose ant to schedule some 7nal eams or C(courses ith olloing call numers/1008, )1)8, )1"8, )20), )261, 311", 311=, 31"6(uppose also that there are no common students in the

olloing pairs o courses ecause o prereOuisites/

1008-)1)81008-)1"8, )1)8-)1"81008-)20)1008-)261, )1)8-)261, )20)-)2611008-311", )1)8-311", )20)-311", )261-311"

1008-311=, )1)8-311=1008-31"6, )1)8-31"6, )1"8-31"6Ho many eam slots are necessary to schedule

eams5



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!2" 12?

(chedules

urn this into a graph coloring prolem.Aertices are courses, and edges arecourses hich cannote scheduledsimultaneously ecause o possile

students in common/

1008

)1)8

)1"8

)20)

311"

)261

31"6

311=



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!2" 1)0

(chedules


131/141

!2" 1)1

(chedules

\and then compute thecomplementary graph/

1008

)1)8

)1"8

)20)

311"

)261

31"6

311=



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!2" 1)2

(chedules

\and then compute thecomplementary graph/

1008

)1)8

)1"8

)20)

311"

)261

31"6

311=



133/141

!2" 1))

(chedules

$edra/

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=



134/141

!2" 1)3

(chedules

ot 1-colorale ecause o edge

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=

Suppose wanschedule some

exams or CS co

with ollowingnumbers:1007 !1!7 !1"7!#$1 %11" %11&



135/141

!2" 1)"

(chedules

ot 2-colorale ecause o triangle

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=



136/141

!2" 1)6

(chedules

@s )-colorale. ry to color y $ed,Green, >lue.

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=



137/141

!2" 1)8

(chedules

4!


138/141

!2" 1)=

(chedules

(o 31"6 must e >lue/

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=



139/141

!2" 1)?

(chedules

(o )261 and 311" must e $ed.

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=



140/141

!2" 130

(chedules

)1)8 and 1008 easy to color.

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=



141/141

(chedules

(o need ) eam slots/

1008

)1)8

)1"8

)20)

311"

)261

31"6 311=

(lot 1

(lot 2

(lot )

Documents

Hamiltonian and Eulerian