Eulerian Graphs

8/4/2019 Eulerian Graphs

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EULER GRAPHS,HAMILTONIONGRAPHS & TRAVELLING SALESMAN

PROBLEM

Lekshmi Krishna M.R

100609|Mtech-Technology Management

Department of Futures Studies

University of Kerala

Department of Futures Studies 1

1

2 3



Contents


2

I ntroduction

Euler Graphs

Theorems, Proof & Algorithms

Hamiltonian Graphs

Traveling Sales Man problems

Conclusion



I ntroduction


3

Configurations of nodes & connections occur in greatdiversity of applications

Such configurations are modeled by combinatorialstructures called Graphs

Consist of edges ,vertices & incidence relationbetween them

E.g. : Electrical circuits, road ways, organic moleculesetc



Euler Graphs


4

An Eulerian trail in a graph is a trail that contains

every edge of that graph

An Eulerian tour is a closed Eulerian trail

A graph that has an Euler tour (Circuit) is called an

Eulerian graph



Euler's Theorem 1


5

A

graph G contains an Eulerian circuit if and only ifthe degree of each vertex is even.



Proof


6

Suppose G contains an Eulerian circuit C. Then, for

any choice of vertex v, C contains all the edges that

are adjacent to v. Furthermore, as we traverse

along C, we must enter and leave v the samenumber of times, and it follows that deg(v) must be

even.



Example


7

1

2 3

Here all nodes are of degree 2(even degree) 1-2-3-1 Forms Euler graph

4

1

2 5

3

Node 3 is of odddegree(3)

No Eulerian path



Proof of Sufficiency


8

We prove by induction on the number of edges. Forgraphs with all vertices of even degree, the smallestpossible number of edges is 3 (i.e. a triangle) in thecase of simple graphs. In both cases, the graph

trivially contains an Eulerian circuit.

The Induction hypothesis then says:

Let H be a connected graph with k edges. If everyvertex of H has even degree, H contains an Euleriancircuit



V ariations of Eulerian Paths


9

1) Handshaking Lemma - Every graph has even

number of odd degree vertices.



Proof

Consider the sum of the degree of all the vertices,

S=�uv deg (u)

In this sum, every edge (a,b) in the graph gets counted twice: Once for a &

once for b.

Therefore S = 2m is an even number.

Now let Vodd (Veven) denote the subset of V that consists only of odd(even

respectively) degree vertices.

Since S =� uv deg (u) =� uv even deg (v) + � uv odd deg (w);

the number � wv odd deg (w) must be evenThus we can say that |Vodd| must be even


10




11

2

) Theorem2

.A

graph contains an Eulerian path ifand only if there are 0 or 2 odd degree vertices



Proof


12

Suppose a graph G contains an Eulerian path P.

Then, for every vertex v, P must enter and leave v

the same number of times, except when it is either

the starting vertex or the final vertex of P. Whenthe starting and final vertices are distinct, there are

precisely 2 odd degree vertices. When these two

vertices coincide, there is no odd degree vertex.



V ariation : 2 (Directed graphs)


13

Let D = (V;A) be a directed graph. Then D containsan Eulerian circuit if and only if, for every vertex u 2 V , indeg(u) = outdeg(v).

Furthermore, D contains an Eulerian path if and onlyif, there exists two vertices s and t such

that:

outdeg(s) = indeg(s) + 1

indeg(t) = outdeg(t) + 1

indeg(v) = outdeg(v)



Example:


14

B

C

T

S

F

E

According to the theoremoutdeg(s) = indeg(s) + 1indeg(t) = outdeg(t) + 1

indeg(v) = outdeg(v)

S-B-F-E-T-C-B-T-SEuler ian circuit

B

C

T

S

F

E

Indeg(s) = 0

Outdeg(s)= 1Indeg(t)=2Outdeg (t)=1Indeg(v) = 7

Outdeg (v)= 7 S-B-F-E-T-C-B-TEuler ian Path



Some facts«.


15

Euler's Theorems are examples of Existence theorems

Existence theorems tell whether or not something exists (e.g. Eulercircuit)

But doesn't tell us how to create it!

We want a constructive method for finding Euler paths and circuits

Methods (well-defined procedures, recipes) for construction arecalled algorithms

An algorithm for constructing an Euler circuit: Fleury's algorithm



F leury's algorithm


16

1. Check if the graph is connected, and every

vertex is of even degree. Reject otherwise.

2. Pick any vertex v(start) to start.

3. While the graph contains at least one edge:

(a) Pick an edge that is not a bridge.

(b) Traverse that edge, and remove it

from G.



Example:


17



F ormulation of Euler circuit using F leury·s

Algorithm


18

Marked Graph Reduced graph



Cont«.


19

The rest of the trip is obvious, and the complete Euler circuit is:(F,C, D,A,C, E,A, B, D, F)



Hamiltonian Graph


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A Hamiltonian path is a path in an undirected graph

that visits each vertex exactly once.

A Hamiltonian cycle (or Hamiltonian circuit) is a

cycle in an undirected graph which visits each vertex

exactly once and also returns to the starting vertex.

A graph that contains a Hamiltonian circuit is called a

Hamiltonian graph.



Example:


21

1

32

4

1

32

1-2-3

Hamiltonian graph

1-2-3-4-3-1

Not Hamiltonian

4

1

32

1-2-4-3

1-4-2-3 Many Hamiltonian

paths




22

1

32

54

Every vertex is connected to all other

vertex-Complete graph - Hamiltonian



O ptimal Graph Traversals


23

Eulerian Trails & Tours

Each edge be traversed at least once

Postman problems

Hamiltonian paths and Cycles

Each vertex be traversed at least once

Traveling Salesman problems



Hamiltonian Type Problems


24

Involve vertex based conditions

No simple characterization is known

Problems are notoriously time consuming

(NP hard)



Traveling Salesman Problem (TSP)


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Defined by W. R. Hamilton and Thomas Kirkman -

1800·s



As Graph Problem

Traveling salesman must travel to every city along the cheapest route

But he cannot visit a city more than once and he must come back where hestarted

� Modeled as undirected weighted graph

� Cities Vertices� Path- Edges� Path distance/cost = Edge length

26




F ive City Travel Problem


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Salesman wanted to travel 5 cities

Want to reach back to the starting city

Assumption : Possible to travel from one city to allthe other cities

Forms complete graph

Cost of travel from one city to another is denoted by¶C·

Problem :To minimize the cost of travel ????




28

51

4

3

2

Find the feasible solutions ?




29

1-3-5-4-2-1 5-1-4-3-2-5

5-4-2-1-3-5 3-2-5-1-4-3

51

4

3

2

51

4

3

2

� n nodes

� (n -1)! feasible

solutions



Graphical Representation


31

3

51

42

10

10 8

6

7

85

96

9

� n nodes

� (n ± 1)! feasible solution

� Complete graph

� Hamiltonian

� Need to find the optimal solution



Sub Tours


32

3

2

1 5

4

TSP

No Sub Tours

NeedCompleteTours



F ormulation of TSP


33

X ij = 1 | If the person goes from i to j|

Objective ± To minimize total cost/distance of

travel

Objective function : Min �� Cij Xij



F ormulation of TSP (Constrains)


34

= 1 ¥ i

= 1 ¥ j

From every city i person need to go j,and is going only one out of theremaining nodes

If salesman in the city j then hecomes to j from a unique city i

Xij = 0,1 Either go from i to j or don't go

from i to j



Sub Tour Elimination Constrains


35

Sub tours of length 1

X jj = 1

X15 = 1X51 = 1

n city TSP ± can have n ± 1 length sub toursNeed to eliminate these sub tours

Xij +X ji � 1

If X15 is in solution then X51 is not in solution

2

1 5

4

3

Sub tour of Length 1

Sub tour of Length 2




36

Eliminate sub tour of length 1

Xjj = 0 - Diagonal assignments Sub tours of length 1 ;hencewe can say that Cjj = infinity

Eliminate sub tour of length 2 Xij+Xji 1 nC2

Eliminate sub tour of length 3

Xij + Xjk + Xki 2 nC3

To eliminate sub tours of length k we need to do up to kterms k-1

nC2




37

Have large number of constrains

Exponentially increasing constrains

How to eliminate this????




38

Need to eliminate sub tours of length 1,2,3 & 4

If there is a sub tour of length 3,then automatically itshould be included in sub tours of 2 & 1 hence by

eliminating 1 & 2 Sub tour of length 3 also geteliminated Likewise sub tour of length 4 too

Hence for 5 city problem we need to eliminate subtours of length 1 & 2



5 CITY = Eliminate length of 1 & 2

7 CITY =Eliminate length of 1,2 & 3

Length of 1 is always eliminated by using theconstrain Cjj = infinity




42

Consider the sub tours 1-2-3-1 & 4-5-4

3

2

1 5

4




43

Ui - U j + n Xij � n-1 i = 1,2««..n-1 j = 2,3««...n-1

U1- U2 -5X12 � 4

U2- U3 +5X23 � 4 1-2-3-1U1-U3 + 10 � 8

U4 ± U5 + 5 � 4 4-5-4U5 ± U4 + 5 � 4

10 � 8

Need to find solution which satisfiesthis conditions

3

2

1 5

4



Solution


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Branch & Bound Algorithm

Heuristic Algorithm



Applications of TSP


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Logistics

Planning

DNA sequencing

Manufacture of Microchips




46

Thank you!!!!



Reference


47

Graph Theory & Its Applications : Jonathan Gross & Jay Yellan

Basic Graph Theory : K.R Parthasarathy

Discrete structures & Graph theory :G.S.S Bhishma Rao

http://www.austincc.edu/powens/+Topics/HTML/05-6/05-6.html

http://train-srv.manipalu.com/wpress/?p=138948

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