Graph Paths and Circuits

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Chapter 5

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ITCP 213 - Discrete Structure 1

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

An Euler path in graph G:

◦ Is a path that includes exactly once all the edges of G.



An Euler circuit in graph G:

◦ Likewise, but with same starting and ending vertices

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Def. 11.15 Let G=(V,E) be an undirected graph or multigraph with no isolated vertices. Then G is said to have an Euler circuit

if 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 this trail traverses each edge in G exactly once, the trail is called an

Euler trail.

Euler Paths and Circuits

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Vertex Degree: Euler Trails and Circuits

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

two odd degree vertices

a b add an edge

Theorem 11.4 A directed Euler circuit exists in G if and only if G 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?

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

Examples:

◦ The path a, b, c, d in (a) is an Euler circuit since all edges are included exactly once.

◦ The graph (b) has neither an Euler path nor circuit.

◦ The graph (c) has an Euler path a, b, c, d, e, f but not an Euler circuit.

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Historical note

◦ In Europe: Konigsberg 7-bridge problem

 Konigsberg, originally in Prussia, now in Russia

 Four sections, two rivers, seven bridges

 Euler solved this problem in 1736; the origin of graph theory

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Problem: Draw a path (or circuit) with a pencil in such a way that you trace over each bridge once and only once and you complete the 'plan' with one continuous pencil stroke

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Problem 2

Suppose they had decided to build one fewer bridge in Konigsberg, so that the map looked like this:



Problem 3

Does it matter which bridge you take away? What if you add bridges? Come up with some maps on your own, and try to 'plan your journey' for each one

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Problem: Draw a path (or circuit) with a pencil in such a way that you trace over each bridge once and only once and you complete the 'plan' with one continuous pencil stroke

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 Theorem:

◦ A connected multigraph has an Euler circuit if and only if the degree of each vertex is even.



Why ?

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 Theorem:

◦ A connected multigraph has an Euler circuit if and only if the degree of each vertex is even.



Proof (Basic idea) :

◦ For each vertex, if there is one “in”, there must be one “out”, because this is a circuit.

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All degrees are odd. Hence no Euler circuit

for the Konigsberg bridges problem.

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

Examples:

◦ Construct an Euler circuit for the following graph.

◦ Solution:

 The graph is connected and the degree of each vertex is even. So, it has an Euler circuit.

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◦ Procedure for constructing an Euler circuit:

 Select any vertex u, and construct a path P

₁

from u to u by randomly selecting unused edges for as long as

possible.

 e.g. if we start at G, we may construct the path:

P₁: G, h, E, d, C, e, F, g, E, j, H, k, G

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◦ Procedure (continued):

 Since the multigraph is connected, there must be a vertex in P

₁

that is incident with an edge not in P

₁

.

 In this case, the vertices E and H are such vertices.

 Arbitrarily choose one of these, say E, and construct a path P

₂

from E to E.

P₂: E, c, B, a, A, b, D, f, E

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◦ Procedure (continued):

 Enlarge P

₁

to include the path P

₂

by replacing any one occurrence of E in P

₁

by P

₂

.

 e.g. replace the first occurrence of E in P1:

P₁: G, h, E, c, B, a, A, b, D, f, E, d, C, e, F, g, E, j, H, k, G

 Repeat the above process.

 Construct a path P₃ from H to H and enlarge P₁ by P₃, we obtain the Euler circuit.

P₁: G, h, E, c, B, a, A, b, D, f, E, d, C, e, F, g, E, j, H, m, J, l, H, k, G

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 Theorem:

◦ A connected multigraph has an Euler path but not an Euler circuit if an only if it has exactly two

vertices of odd degree.



Why ?

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 Theorem:

◦ A connected multigraph has an Euler path but not an Euler circuit if an only if it has exactly two

vertices of odd degree.



Proof:

◦ If: add one edge connects the two vertices of odd degree

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K2

K3

K4

K5

K6

K8

K2: Euler path – ? Euler cycle – ? K3: Euler path – ? Euler cycle – ? K4: Euler path – ? Euler cycle – ? K5: Euler path – ? Euler cycle – ? K6: Euler path – ? Euler cycle – ? K8: Euler path – ? Euler cycle – ?

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K2

K3

K4

K5

K6

K8

K2: Euler path – Yes Euler cycle – No K3: Euler path – No Euler cycle – Yes K4: Euler path – No Euler cycle – No K5: Euler path – No Euler cycle – Yes K6: Euler path – No Euler cycle – No K8: Euler path – No Euler cycle – No

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Planar Graphs

Def. 11.17 A graph (or multigraph) G is called planar if G can be drawn 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.

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

Complete graphs - K

n

:

◦ How many edges of Kn? n(n-1)/2

K

1

K

2

K

3

K

4

K

5

Planar Graphs

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

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Planar Graphs

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

K5 is nonplanar

K₄

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Planar Graphs

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Planar Graphs

K_4,4

K_3,3 is not planar.

Therefore, any graph containing K₅ or K_3,3 is nonplanar.

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

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Planar Graphs

Def. 11.19 elementary subdivision (homeomorphic operation)

u w u v w

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

a b

c

d e

a b

c

d e

a b

c

d e

a b

c

d e

Two homeomorphic graphs are simultaneously planar or nonplanar.

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Planar Graphs

Theorem 11.5 (Kuratowski's Theorem) A graph is nonplanar if and only if it contains a subgraph that is homeomorphic to

either K₅ or K_3,3. Ex. 11.17 Petersen graph

a

b

d c

e f

g h

i j

a subgraph homeomorphic to K_3,3 j a

d

e f b

g h

i c

Petersen graph is nonplanar.

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Planar Graphs

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Planar Graphs

K₄

R1 R2

R3

R4

A planar graph divides the plane into 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

number

of vertices number of edges

number of regions

V = 4 ,e = 6 , r = 4 , v – e + r = 2

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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)=3 deg(R₃)=3,deg(R₄)=7

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

deg( R

_i

) deg( R ) | | E

i i

 i

      

1 4

5 8

18 2 9 2

abghgfda

a b

c

d f

g h

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Planar Graphs

. 6 - 3 or

, - 3 6 so 3, 3

+ 2 - +

-

= 2

theorem, Eulers

From .

3 2

and by

determined regions

the of

degrees the

of sum the

|=

| 2

= 2 ly, Consequent

3.

degree has

region each

Hence, edges.

e least thre at

contains region)

infinite the

(including region

each of

boundary

the , multigraph a

not is

and free

- loop a

is Since

: Proof

. 6 - 3

and 2

3 Then regions.

and 2,

>

|=

| ,

|=

| graph w ith

planar connected

free -

loop a

be ) , (

= Let 11.3

Corollary

v e

e e v

e v e

v r e v

r e

G

r E

e G v

e

e r

r e

E v V

E V G













Only a necessary condition, not sufficient.

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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 K_3,3, each region has at least 4 edges, hence 4r 2e. If K_3,3 is planar, r = e - v + 2 = 9 – 6 + 2 = 5.

So 20=4r 2e = 18, a contradiction.

 

. 6 - 3

and 2

3 Then regions.

and 2,

>

|=

| ,

|=

| graph with

planar connected

free - loop a

be ) , (

= Let

v e

e r

r e

E v V

E V G



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Dual graph

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Planar Graphs

A dual graph of a planar graph

a b

c d

e f

g 1

2

3 4 6 5

1

6 5

4 2

3

An edge in G corresponds with an edge in G^d.

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Planar Graphs

Def. 11.20 cut-set: a subset of edges whose removal increase the 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-sets in a dual graphs and vice versa.

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

A Hamilton path in graph G:

◦ Is a path that includes each vertex once and only once.

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

Examples:

◦ G

₁

has a Hamilton path: a, b, c, d, e, and cycle.

◦ G

₂

has only a Hamilton path: a, b, c, d.

◦ G

₃

has no.

◦ In general, no efficient method to find such a path

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Hamilton Paths and Cycles

a path or cycle that contain every vertex Unlike Euler circuit, there is no known necessary and sufficient condition for a

graph to be Hamiltonian.

Ex. 11.24

a b c

d e f

g h

i

There is a Hamilton path, but no Hamilton cycle.

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Hamilton Paths and Cycles

Ex. 11.25

x

y

y y y

x x

x y

y

start labeling from here

4x's and 6y's, since x and y must interleave in a Hamilton path (or cycle),

the graph is not Hamiltonian

The method works only for bipartite graphs.

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Hamilton Paths and Cycles

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

Consider K₁₇, a Hamilton cycle in K₁₇ corresponds to a seating arrangements. Each cycle has 17 edges, so we can have

(1/17)17(17-1)/2=8 different sittings.

1 2

3

4

5

6

17 16 15

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

1 2

3

4

5

6

17 16 15

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

1 2

3

4

5

6

17 16 15

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

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Hamilton Paths and Cycles

Theorem 11.7 Let be a complete directed graph, i. e. ,

has vertices and for any distinct pair , of vertices, exactly one 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. If

not, let be a vertex that doesn' t appear in

* *

*

K K

n x y

x y y x K

tournament

m p m

v v v v v v m n

v p

n n

n m

m m

m

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

.



1 2 2 3



1

case 1. v v₁ v₂ ...v_m

case 2. v₁ v₂ ...v_k v v_k+1 ...v_m case 3. v₁ v₂ ...v_m v

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Ex. 11.27 In a round-robin tournament each player plays every other player exactly once. We want to somehow rank the players

according to the result of the tournament.

not always possible to have a ranking where a player in a certain position 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 that each has beaten the next player on the list

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Hamilton Paths and Cycles

Theorem 11.8 Let = ( , ) be a loop - free graph with

| |= 2. If deg( ) + deg( ) -1 for all , , , then has a Hamilton path.

G V E

V n x y n x y V x y

G

   

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Corollary 11.4. If deg( ) -

for all vertices, then the graph has a Hamilton path.

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

, , then contains contains a Hamilton cycle.

v n

G V E

V n x y n

x y V G



 



1 2

deg

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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 connected either to or and then , . Let = ( ' , ' ) denote the resulting subgraph. Then | |=| ' |+ ( ) + ( ). Since | ' |= - ,

| ' | -

Consequently, -

| ' |+ ( ) + ( ) -

G V E

V n E n

G

a b V a b E

a b a b H V E

E E a b V n

E n n

E E a b

n

  





 

 

 













   



1

2 2

2

1

2 2

2

,

deg deg

. | | deg deg

2 1

2 2 2

2







 







   





 

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

- -

and has a Hamilton cycle.

deg a deg b deg a deg b

n n

n G

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A related problem: the traveling salesman problem a

b

c

d e 3

4 1

3

5 4

3

2

Find a Hamilton cycle of shortest total distance.

2

For example, a-b-e-c-d-a with total cost=

1+3+4+2+2=12.

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

Other Types of Graphs:

◦ Multigraphs : Section 11.1.

◦ Directed Graphs : Section 11.1.

◦ Directed Multigraphs : Section 11.1.



Paths and Circuits:

◦ Euler Paths and Circuits : Section 11.3.

◦ Planer graph : Section 11.4.

◦ Hamilton Paths and Circuits : Section 11.5.

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