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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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!2" )
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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encil 'raing rolem-Euler aths
Graph heoretically/ 4hich o theolloing graphs has an Eulerpath5
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!2" 6
encil 'raing rolem-Euler aths
#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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!2" =
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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!2" ?
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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!2" 1)
%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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!2" 1"
%inding Euler Circuits
#/ @t contains a +erte o odddegree.
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%inding Euler Circuits
!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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!2" 1?
%inding Euler Circuits
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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%inding Euler Circuits*ohammedDs (cimitars
%ound a cycle ater starting rommiddle +erte.
'elete the cycle/
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%inding Euler Circuits*ohammedDs (cimitars
%ound a cycle ater starting rommiddle +erte.
'elete the cycle/
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!2" 2)
%inding Euler Circuits*ohammedDs (cimitars
%ound a cycle ater starting rommiddle +erte.
'elete the cycle/
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%inding Euler Circuits*ohammedDs (cimitars
%ound a cycle ater starting rommiddle +erte.
'elete the cycle/
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!2" 2"
%inding Euler Circuits*ohammedDs (cimitars
%ound a cycle ater starting rommiddle +erte.
'elete the cycle/
i di l i i
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%inding Euler Circuits*ohammedDs (cimitars
o try again 9say rom middle+erte:/
i di l Ci i
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%inding Euler Circuits*ohammedDs (cimitars
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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%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
1
%i di E l Ci i
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!2" 2?
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12
%i di E l Ci it
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!2" )0
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12)
%i di E l Ci it
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!2" )1
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3
%i di E l Ci it
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!2" )2
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3 "
%i di E l Ci it
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!2" ))
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3 "
6
%i di E l Ci it
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!2" )3
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3 "
6
8
%i di E l Ci it
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!2" )"
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3 "
6
8=
%i di E l Ci it
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!2" )6
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3 "
6
8=?
%i di E l Ci it
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!2" )8
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3 "
6
8=?
10
%i di E l Ci it
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!2" )=
%inding Euler Circuits*ohammedDs (cimitars
%ind another cycle rom middle+erte/
12) 3 "
6
8=?
10
11
%inding Euler Circuits
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!2" )?
%inding Euler Circuits*ohammedDs (cimitars
#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
%inding Euler Circuits
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%inding Euler Circuits*ohammedDs (cimitars
%inally, need to add the triangle.
Mse same naN+e approach loo&ing orcycle in remaining component/
12) 3 "
6
8=?
10
55
%inding Euler Circuits
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%inding Euler Circuits*ohammedDs (cimitars
%inally, need to add the triangle.
Mse same naN+e approach loo&ing orcycle in remaining component/
12) 3 "
6
8=?
10
55 11
%inding Euler Circuits
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%inding Euler Circuits*ohammedDs (cimitars
%inally, need to add the triangle.
Mse same naN+e approach loo&ing orcycle in remaining component/
12) 3 "
6
8=?
10
55 1112
%inding Euler Circuits
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!2" 3)
%inding Euler Circuits*ohammedDs (cimitars
%inally, need to add the triangle.
Mse same naN+e approach loo&ing orcycle in remaining component/
12) 3 "
6
8=?
10
55 1112
1)
%inding Euler Circuits
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!2" 33
%inding Euler Circuits*ohammedDs (cimitars
%inally, need to add the triangle.
Mse same naN+e approach loo&ing orcycle in remaining component/
12) 3 "
6
8=?
10
55 1112
1)13
%inding Euler Circuits
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!2" 3"
%inding Euler Circuits*ohammedDs (cimitars
%inally, need to add the triangle.
Mse same naN+e approach loo&ing orcycle in remaining component/
12) 3 "
6
8=?
10
1" 1112
1)13
%inding Euler Circuits
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%inding Euler Circuits*ohammedDs (cimitars
#malgamate the triangle cycleeteen edges ormerly laeled ?L 10/
12) 3 "
6
8=?
10
1" 1112
1)13
%inding Euler Circuits
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!2" 38
%inding Euler Circuits*ohammedDs (cimitars
#malgamate the triangle cycleeteen edges ormerly laeled ?L 10/
12) 3 "
6
8=?
55
55 5555
5555
%inding Euler Circuits
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%inding Euler Circuits*ohammedDs (cimitars
4e ound the Euler circuit
12) 3 "
6
8=?
1)
1= 131"
161810
11 12
Euler Circuit
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!2" 3?
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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!2" "0
Generaliing to Euler aths
F/ 'oes the olloing ha+e an Eulercircuit5
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!2" "1
Generaliing to Euler aths
#/ o, +ertices o odd degree/
F/ >ut hy does it ha+e an Euler path5
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!2" "2
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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!2" ")
Generaliing to Euler aths
#ll degrees no e+en so 7nd Eulercycle/
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!2" "3
Generaliing to Euler aths
o remo+e phantom edge otaining/
1 2
)
3
"
68
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!2" ""
Generaliing to Euler aths
Generalie/
1 2
)
3
"
6
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Generaliing to Euler aths
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*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
-Common
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!2" 62
Common*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
-Common
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!2" 6)
Common*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
-Common
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!2" 63
Common*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
-Common
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!2" 6"
Common*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
-Common
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!2" 66
Common*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
-Common
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!2" 68
Common*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
-Common
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!2" 6=
Common*isunderstanding
#/ >ecause it is isomorphic to agraph hich isplanar/
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!2" 6?
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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!2" 80
ro+ing lanarity)-Cue
ro+ing lanarity5
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!2" 81
ro+ing lanarity53-Cue
(eemingly not planar, ut hoould one pro+e this
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!2" 82
'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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!2" 8)
'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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!2" 8"
$egions
F/ Ho many regions does the )-cue ha+e5
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!2" 86
$egions
#/ 6 regions
1 2
)
36
"
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!2" 88
$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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!2" 8?
Euler Characteristic
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!2" =0
Euler Characteristic
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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!2" =1
Euler Characteristic
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
Euler Characteristic
SV S SE S r
r- SE S I SVS
1 0 1 2
#nimated @n+ariance o
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!2" =)
Euler Characteristic
SV S SE S r
r- SE S I SVS
2 1 1 2
#nimated @n+ariance o
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!2" =3
Euler Characteristic
SV S SE S r
r- SE S I SVS
) 2 1 2
#nimated @n+ariance o
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!2" ="
Euler Characteristic
SV S SE S r
r- SE S I SVS
3 ) 1 2
#nimated @n+ariance o
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!2" =6
Euler Characteristic
SV S SE S r
r- SE S I SVS
3 3 2 2
#nimated @n+ariance o
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!2" =8
Euler Characteristic
SV S SE S r
r- SE S I SVS
" " 2 2
#nimated @n+ariance o
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!2" ==
Euler Characteristic
SV S SE S r
r- SE S I SVS
6 6 2 2
#nimated @n+ariance o
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!2" =?
Euler Characteristic
SV S SE S r
r- SE S I SVS
8 8 2 2
#nimated @n+ariance o
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!2" ?0
Euler Characteristic
SV S SE S r
r- SE S I SVS
= = 2 2
#nimated @n+ariance o
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!2" ?1
Euler Characteristic
SV S SE S r
r- SE S I SVS
= ? ) 2
#nimated @n+ariance o
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!2" ?2
Euler Characteristic
SV S SE S r
r- SE S I SVS
= 10 3 2
#nimated @n+ariance o
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!2" ?)
Euler Characteristic
SV S SE S r
r- SE S I SVS
= 11 " 2
#nimated @n+ariance o
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!2" ?3
Euler Characteristic
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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!2" ?6
%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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!2" ?8
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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!2" 103
*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
(o add another color. ry to 7ll ine+ery country ith one o the tocolors.
* C l i
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!2" 108
*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" 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" 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/
%rom *ap Coloringt G h C l i
7/24/2019 Hamiltonian and Eulerian
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!2" 11"
to Graph Coloring
%or each region introduce a +erte/
%rom *ap Coloringt G h C l i
7/24/2019 Hamiltonian and Eulerian
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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/
%rom *aps to Graphst ' l G h
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!2" 11=
to 'ual Graphs
ual Graphs /1: ut +erte inside each region/
%rom *aps to Graphst ' l G h
7/24/2019 Hamiltonian and Eulerian
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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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!2" 120
'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
7/24/2019 Hamiltonian and Eulerian
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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.
%rom *ap Coloringto Graph Coloring
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!2" 123
to Graph Coloring
*ap not 2-colorale, so dual graphnot 2-colorale/
%rom *ap Coloringto Graph Coloring
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!2" 12"
to Graph Coloring
*ap not )-colorale, so graph not )-colorale/
%rom *ap Coloringto Graph Coloring
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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
7/24/2019 Hamiltonian and Eulerian
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!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
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
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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=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
130/141
!2" 1)0
(chedules
7/24/2019 Hamiltonian and Eulerian
131/141
!2" 1)1
(chedules
\and then compute thecomplementary graph/
1008
)1)8
)1"8
)20)
311"
)261
31"6
311=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
132/141
!2" 1)2
(chedules
\and then compute thecomplementary graph/
1008
)1)8
)1"8
)20)
311"
)261
31"6
311=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
133/141
!2" 1))
(chedules
$edra/
1008
)1)8
)1"8
)20)
311"
)261
31"6 311=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
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&
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
135/141
!2" 1)"
(chedules
ot 2-colorale ecause o triangle
1008
)1)8
)1"8
)20)
311"
)261
31"6 311=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
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=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
137/141
!2" 1)8
(chedules
4!
7/24/2019 Hamiltonian and Eulerian
138/141
!2" 1)=
(chedules
(o 31"6 must e >lue/
1008
)1)8
)1"8
)20)
311"
)261
31"6 311=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
139/141
!2" 1)?
(chedules
(o )261 and 311" must e $ed.
1008
)1)8
)1"8
)20)
311"
)261
31"6 311=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
140/141
!2" 130
(chedules
)1)8 and 1008 easy to color.
1008
)1)8
)1"8
)20)
311"
)261
31"6 311=
Graph Coloring and(chedules
7/24/2019 Hamiltonian and Eulerian
141/141
(chedules
(o need ) eam slots/
1008
)1)8
)1"8
)20)
311"
)261
31"6 311=
(lot 1
(lot 2
(lot )