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Discrete Math by R.S. Chang, Dept. CSIE, NDHU 2

An Introduction to Graph Theory

Chapter 11


Chapter 11 An Introduction to Graph Theory11.1 Definitions and Examples

Undirected graph Directed graph

isolated vertex

adjacent

loop

multipleedges

simple graph: an undirected graph without loop or multiple edges

degree of a vertex: number of edges connected(indegree, outdegree)

G=(V,E)

For simple graphs, deg(v Eiv Vi

) | | 2



x ypath: no vertex can be repeated a-b-c-d-etrail: no edge can be repeat a-b-c-d-e-b-dwalk: no restriction a-b-d-a-b-c

closed if x=yclosed trail: circuit (a-b-c-d-b-e-d-a, one draw without lifting pen)closed path: cycle (a-b-c-d-a)

a

b

c

d

e

length: number of edges inthis (path,trail,walk)



Theorem 1.1 Let = ( , ) be an undirected graph, with , , . If there exists a trail from to , then there is

a path from to .

G V Ea b V a b a b

a b

a x b

remove any cycle on the repeatedvertices

Def 11.4 Let G=(V,E) be an undirected graph. We call G connected if there is a path between any two distinct vertices of G.

ab

cd

e ab

cd

e

disconnected withtwo components



Def. 11.5 For any graph = ( , ), the number of componentsof is denoted by ( ). 1 ( ) | |

Can you think of an algorithm to determine ( )?

G V EG G

G V

G

Def. 11.6multigraph of multiplicity 3

multigraphs


Chapter 11 An Introduction to Graph Theory11.2 Subgraphs, Complements, and Graph Isomorphism

Def. 11.7 If = ( , ) is a graph, then is calleda subgraph of if and where each edge ofin is incident with vertices in

11 1

G V E G V EG V V E E

E V

1 1 11

( , ),

.

a

b

c

d

e

a

b

c

d

e

b

c

d

ea

cd

spanning subgraph V1=V

induced subgraphinclude all edges of E in V1



Def. 11.11 complete graph: Kn

a

b

c

d

e

K5

Def. 11.12 complement of a graph

G Ga

b

c

d

e

a

b

c

d

e



Theorem: Any graph of six vertices contains a K3 or K3.(In a party of six, There exists 3 people who are eithermutually acquainted or mutually inacquainted.)

5 is not enough.

a

b

c

d

e

For 6 people, let's look from the point ofview of a:

From the pigeonhole principle, there are3 who know a or 3 who does not know a.

a

b c d

a

b c dK3 or K3. K3 or K3.



Ex. 11.7 Instant Insanity, 4 cubes, each of the six faces on acube is painted with one of the colors, red (R), white (W),blue (B), or Yellow (Y). The object is to place the cubes in acolumn of four such that all four colors appear on each of thefour sides of the column.

W R Y W

Y

B (1)

B B W Y

R

Y (2)

R B Y B

R

W (3)

W R B Y

W

W (4)

There are (3)(24)(24)(24)=41472 possibilities to consider.

the bottom cube 6 faces with 4 rotations



W R Y W

Y

B (1)

B B W Y

R

Y (2)

R B Y B

R

W (3)

W R B Y

W

W (4)

R W

Y B

13

11

2

4 2

3

4

4 3 2

Each edge correspondsto a pair of opposite faces.

R W

Y B1

2

3

4

R W

Y B2

4

1

3

Y

BRW

B

WBY

W

RYR

R

YWB

(1) (2) (3) (4)

Consider the subgraph of opposite column.



Graph Isomorphism

1 2

3 4

a b

c

d

w x y z

Def. 11.13 Let and be two undirected graphs. A function : is called a graphisomorphism if (a) is one - to - one and onto and (b) forall , ( , ) if and only if ( ( ), ( ))When such a function exists, and are called isomorphic graohs.

12

G V E G V Ef V V

fa b V a b E f a f b E

G G

1 1 1 2 2 21 2

1 21

( , ) ( , )

, .



Ex. 11.8 q r

wz xy

u t

v

a

b

cd

e f

gh

i

j

a-q c-u e-r g-x i-z b-v d-y f-w h-t j-s, isomorphic

Ex. 11.9

degree 2 vertices=2

degree 2vertices=3

Can you think of an algorithm for testing isomorphism?


Chapter 11 An Introduction to Graph Theory11.3 Vertex Degree: Euler Trails and Circuits

degree 1 vertex: pendant vertex

For simple graphs, deg(v Eiv Vi

) | | 2Theorem 11.2

Corollary 11.1 The number of vertices of odd degree must be even.

Ex. 11.11 a regular graph: each vertex has the same degreeIs it possible to have a 4-regular graph with 10 edges?

2|E|=4|V|=20, |V|=5 possible (K5)

with 15 edges?

2|E|=4|V|=30 not possible


11.3 Vertex Degree: Euler Trails and Circuits

Chapter 11 An Introduction to Graph Theory

Ex. 11.12 The Seven Bridge of Konigsberg

area a

area b area d

area ca

b

c

d

Find a way to walk about the city so as to crosseach bridge exactly once and then return to thestarting point.




Def. 11.15 Let G=(V,E) be an undirected graph or multigraphwith no isolated vertices. Then G is said to have an Euler circuitif there is a circuit in G that traverses every edge of the graph exactly once. If there is an open trail from a to b in G and thistrail traverses each edge in G exactly once, the trail is called anEuler trail.Theorem 11.3 Let G=(V,E) be an undirected graph or multigraphwith no isolated vertices. Then G has an Euler circuit if and onlyif G is connected and every vertex in G has even degree.

a

b

c

dAll degrees are odd. Hence no Euler circuitfor the Konigsberg bridges problem.




proof of Euler circuit theorem:

Euler circult connected and even degree

v for other vertices

s for starting vertexobvious

connected and even degree Euler circuit

by induction on the number of edges.

e=1 or 2 e=n find any circuit containing s

s




Corollary 11.2 An Euler trail exists in G if and only if G isconnected and has exactly two vertices of odd degree.

two odd degree verticesa b

add an edge

Theorem 11.4 A directed Euler circuit exists in G if and only ifG is connected and in-degree(v)=out-degree(v) for all vertices v.

one in, one out

Can you think of an algorithm to construct an Euler circuit?




Ex. 11.13 Complete Cycles (DeBruijn Sequences)If n is a positive integer and N=2n, a cycle of length N of 0's and 1's is called a complete cycle if all possible subsequences of 0's and 1's oflength n appear in this cycle. n=1 01,

n=2 0011,n=3 00010111,00011101n=4 16 complete cyclesIn general

For n=3:

00

01 10

11

vertex set={00,01,10,11}a directed edge from x1x2 to x2 x3

a

b

c

d

e

f

g

h

Find an Euler circuit:

00111010abgfcdeh 00101110abcdefgh

2 2 1n n


Chapter 11 An Introduction to Graph Theory11.4 Planar Graphs

Def. 11.17 A graph (or multigraph) G is called planar if G can bedrawn in the plane with its edges intersecting only at vertices of G.Such a drawing of G is called an embedding of G in the plane.

Ex. 11.14,11.15 K1,K2,K3,K4 are planar, Kn for n>4 are nonplanar.

K4 K5

applications: VLSI routing, plumbing,...


11.4 Planar Graphs


Def. 11.18 bipartite graph and complete bipartite graphs (Km,n)

K4,4

K3,3 is not planar.Therefore, any graph containing K5

or K4,4 is nonplanar.


11.4 Planar Graphs


Def. 11.19 elementary subdivision (homeomorphic operation)

u w u v w

G1 and G2 are called homeomorphic if they are isomorphicor if they can both be obtained from the same loop-free undirected graph H by a sequence of elementary subdivisions.

a b

cde

a b

cde

a b

cde

a b

cde

Two homeomorphic graphs are simultaneously planar or nonplanar.


11.4 Planar Graphs


Theorem 11.5 (Kuratowski's Theorem) A graph is planar ifand only if it contains a subgraph that is homeomorphic toeither K5 or K3,3.

Ex. 11.17 Petersen graph

a

b

cd

e f

gh

i

j

a subgraph homeomorphic to K3,3

j ad

e f b

g

h

ci

Petersen graph is nonplanar.


11.4 Planar Graphs


K4

R1R2

R3

R4

A planar graph divides the planeinto several regions (faces), one of them is the infinite region.

Theorem 11.6 (Euler's planar graph theorem)

For a connected planar graph or multigraph: v-e+r=2

numberof vertices

numberof edges

numberof regions

v=4,e=6,r=4, v-e+r=2


11.4 Planar Graphs


proof: The proof is by induction on e.

e=0 or 1 v=1r=1e=0

v=1r=2e=1

v=2r=1e=1

v-e+r=2

Assume that the result is true for any connected planar graph ormultigraph with e edges, where 0 e k

Now for G=(V,E) with |E|=k+1 edges, let H=G-(a,b) for a,b in V.

Since H has k edges, v e rH H H 2

And, v v e eG H G H , .1

Now consider the situation about regions.


11.4 Planar Graphs


case 1: H is connected

a(=b) a(=b) a

ba

b

v e r v e r v e rG G G H H H H H H( ) ( )1 1 2


11.4 Planar Graphs


case 2: H is disconnected

a

b

a b

a

b

H1

H2

a bH1H2

v v v e e e r r rv e r

v e r v e r v ve e r r v e r

v e r

H H G H H G H H G

H H H

H H H G G G H H

H H H H H H H

H H H

1 2 1 2 1 2

1 1 1

2 2 2 1 2

1 2 1 2 1 1 1

2 2 2

1 12

21 1

2 2 2 2 2

, , .,

. ( )( ) ( ) ( )

( )

And by the induction hypothesis, Therefore,


11.4 Planar Graphs


degree of a region (deg(R)): the number of edges traversed in a shortest closed walk about the boundary of R.

R1

R2

R3

R4 R5

R6

R7

R8

two different embeddings

deg(R1)=5,deg(R2)=3deg(R3)=3,deg(R4)=7

deg(R5)=4,deg(R6)=3deg(R7)=5,deg(R8)=6

deg( ) deg( ) | |R R Eii

ii

1

4

5

818 2 9 2

abghgfda

a b

c

d fg h


11.4 Planar Graphs


Corollary 11.3 Let = ( , ) be a loop - free connected planargraph with | |= , | |= > 2, and regions. Then and

- .Proof: Since is a loop - free and is not a multigraph, the boundary of each region (including the infinite region) containsat least three edges. Hence, each region has degree 3.Consequently, = | |= the sum of the degrees of the regions determined by and . From Euler' s theorem,

2 = - + - + so - , or - .

G V EV v E e r r e

e vG

e E rG e r

v e r v ee

ve

v e e v

3 23 6

2 22 3

2

3 36 3 3 6

,

Only a necessary condition, not sufficient.


11.4 Planar Graphs


Ex. 11.18 For K5, e=10,v=5, 3v-6=9<10=e. Therefore,by Corollary 11.3, K5 is nonplanar.

Ex. 11.19 For K3,3, each region has at least 4 edges, hence4r 2e. If K3,3 is planar, r=e-v+2=9-6+2=5. So 20=4r 2e=18,a contradiction.


11.4 Planar Graphs


A dual graph of a planar graph

a b

c

d

e f

g

1

23

456

1

6 5

4

2

3

An edge in G corresponds with an edge in Gd.

It is possible to have isomorphic graphs with respective duals thatare not isomorphic.


11.4 Planar Graphs


Def. 11.20 cut-set: a subset of edges whose removal increasethe number of components

Ex. 11.21

a

b

c

d

e

f

g

h

cut-sets: {(a,b),(a,c)},{(b,d),(c,d)},{(d,f)},...

a bridge

For planar graphs, cycles in one graph correspond to cut-setsin a dual graphs and vice versa.


Chapter 11 An Introduction to Graph Theory11.5 Hamilton Paths and Cycles

a path or cycle that contain every vertex

Unlike Euler circuit, there is no knownnecessary and sufficient condition for a graph to be Hamiltonian.

Ex. 11.24 a b c

d e f

g hi

There is a Hamilton path, but no Hamilton cycle.

an NP-complete problem



Ex. 11.25x

yy y y

x x

x

y

y

start labeling from here

4x's and 6y's, since x and y mustinterleave in a Hamilton path (or cycle),the graph is not Hamiltonian

The method works only for bipartite graphs.

The Hamilton path problem is still NP-complete when restrictedto bipartite graphs.



Ex. 11.26 17 students sit at a circular table, how many sittings arethere such that one has two different neighbors each time?

Consider K17, a Hamilton cycle in K17 corresponds to a seatingarrangements. Each cycle has 17 edges, so we can have (1/17)17(17-1)/2=8 different sittings.

12

3

4

5

6

17

16

15

1,2,3,4,5,6,...,17,1

12

3

4

5

6

17

16

15

1,3,5,2,7,4,...,17,14,16,1

12

3

4

5

6

17

16

15

1,5,7,3,9,2,...,16,12,14,1

14



Theorem 11.7 Let be a complete directed graph, i.e. , has vertices and for any distinct pair , of vertices, exactlyone of the edges ( , ) or ( , ) is in Such a graph (called a

) always contains a directed Hamilton path.Proof: Let 2 with a path containing -1 edges ( If = , we' re finished. Ifnot, let be a vertex that doesn' t appear in

* *

*

K Kn x y

x y y x Ktournament

m p mv v v v v v m n

v p

n n

n

mm m

m

.

, ), ( , ), , ( , )..

1 2 2 3 1

case 1. v v1 v2 ...vm

case 2. v1 v2 ...vk v vk+1 ...vm

case 3. v1 v2 ...vm v



Ex. 11.27 In a round-robin tournament each player plays everyother player exactly once. We want to somehow rank the playersaccording to the result of the tournament.

not always possible to have a ranking where a player in a certainposition has beaten all of the opponents in later positions

a b c

but by Theorem 11.7, it is possible to list the players such thateach has beaten the next player on the list



Theorem 11.8 Let = ( , ) be a loop - free graph with | |= 2. If deg( ) + deg( ) -1 for all , , , then

has a Hamilton path.

G V EV n x y n x y V x yG

Proof: First prove that G is connected. If not,

x yn1 vertices n2 vertices

deg( ) deg( ) ( ) ( )x y n n n n n n 1 2 1 2 1 21 1 2 1

a contradiction



Theorem 11.8 Let = ( , ) be a loop - free graph with | |= 2. If deg( ) + deg( ) -1 for all , , , then


G V EV n x y n x y V x yG

Assume a path pm with m vertices v1 v2 v3 ... vm

case 1. either v v1 or vm v

case 2. v1,v2,...,vm construct a cycle either v1 v2 v3 ... vm

or v1 v2 v3 ...vt-1 vt ... vm

otherwise assume deg(v1)=k, then deg(vm)<m-k.deg(v1)+deg(vm)<m<n-1, a contradiction

Therefore, v can be added to the cycle. v



Corollary 11.4. If deg( )-

for all vertices, then the graph


Theorem 11.9 Let = ( , ) be a loop - free undirected graph with | |= 3. If deg( ) + ( ) for all nonadjacent

, , then contains contains a Hamilton cycle.

vn

G V EV n x y n

x y V G

1

2

deg

Proof: Assume G does not contain a Hamilton cycle. We add edges to G until we arrive a subgraph H of Kn where H has no Hamiltoncycle, but for any edge e not in H, H+e has a Hamilton cycle.

For vertices a,b wher (a,b) is not an edge of H. H+(a,b) has aHamilton cycle and (a,b) is part of it.



a(=v1) b(=v2) v3 ... vn

If (b,vi) is in H, then (a,vi-1) cannot be in H. Otherwise,b vi vn a vi-1 vi-2 v3 is a Hamilton cycle in H.

Consequently, deg which meansdeg a contradiction.

Corollary 11.5 If deg( ) for all vertices, then the graph has a

Hamilton cycle.

H HG G

a b na b n

vn

( ) deg ( ) ,( ) deg ( ) ,

2



Corollary 11.6 If = ( , ) is a loop - free unirected graph with

| |= 3, and if | |-

then has a Hamilton cycle.

Proof: Let , where ( , ) . Remove all edges connectedeither to or and then , . Let = ( ' , ' ) denote the resultingsubgraph. Then | |=| ' |+ ( ) + ( ). Since | ' |= - ,

| ' |-

Consequently, -

| ' |+ ( ) + ( )

-

G V E

V n En

G

a b V a b Ea b a b H V E

E E a b V n

En n

E E a b

n

1

22

22

2

1

22

2

,

deg deg

. | | deg deg

21

22

2

2

+ ( ) + ( ). Therefore, ( ) + ( )

- -

and has a Hamilton cycle.

deg deg deg dega b a b

n nn G



A related problem: the traveling salesman problem

a

b

c

d

e3

41

3

5 4

32

Find a Hamilton cycle of shortest total distance.

2 graph problem vs. Euclidean plane problem (computational geometry)

Certain geometry properties (for example, the triangle inequality) sometimes (but not always) make it simpler.

For example, a-b-e-c-d-a with total cost=1+3+4+2+2=12.



Two famous computational geometry problems.

1. closest pair problem: which two points are nearest2. convex hull problem

the convex hull


Chapter 11 An Introduction to Graph Theory11.6 Graph Coloring and Chromatic Polynomials

Def. 11.22 If G=(V,E) is an undirected graph, a proper coloringof G occurs when we color the vertices of G so that if (a,b) is anedge in G, then a and b are colored with different colors. Theminimum number of colors needed to properly color G is calledthe chromatic number of G and is written (G).

b

c

d

e

a3 colors are needed.a: Redb: Greenc: Redd: Bluee: Red

In general, it's a very difficult problem (NP-complete).

(Kn)=n

(bipartite graph)=2



A related problem: color the map where two regions arecolored with different colors if they have same boundaries.

G

R e

B

BR

Y

Four colors are enough for any map. Remaina mystery for a century. Proved with the aid of computer analysis in 1976.

a b

c

df

a

b

c

d

e

f



P(G,): the chromatic polynomial of G=the number of waysto color G with colors.

Ex. 11.31 (a) G=n isolated points, P(G,)=n.(b) G=Kn, P(G,)=(-1)(-2)...(-n+1)=(n)

(c) G=a path of n vertices, P(G,)=(-1)n-1.(d) If G is made up of components G1, G2, ..., Gk, then P(G,)=P(G1,)P(G2,)...P(Gk,).

Ex. 11.32e

G eG eG'

coalescing the vertices



Theorem 11.10 Decomposition Theorem for Chromatic Polynomials.If G=(V,E) is a connected graph and e is an edge, then P(Ge,)=P(G,)+P(G'e,).

e

G eG eG'

coalescing the vertices

a

b

In a proper coloring of Ge:case 1. a and b have the same color: a proper coloring of G'e case 2. a and b have different colors: a proper coloring of G.Hence, P(Ge,)=P(G,)+P(G'e,).



Ex. 11.33

e = -

P(Ge,)P(G,) P(G'e,)

P(G,)=(-1)3-(-1)(-2)=4-43+62-3Since P(G,1)=0 while P(G,2)=2>0, we know that (G)=2.

Ex. 11.34

= - = -2

e e

P(G,)=(4)-2(4)= (-1)(-2)2(-3) (G)=4

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