Eulerian Graph,Chinese postman problem

1. PRESENTATION
On
Topic: Graph Theory :Eulerian Graph

2. Seven Bridges of Konigsberg
 The Konigsberg bridge problem:
 Starting and ending at the same point, is it possible to cross all seven bridges
just once and return to the starting point.
 This problem can be represented by a graph.
 Edges represent bridges and each vertex represents a region.
 Begun in 1735.
 Mentioned in Leonhard Euler's paper on “Seven Bridges of Konigsberg ” .
Problem : Walk all 7 bridges without crossing a bridge twice.

3. Eulerian Graph
If there is a path joining any two vertices in a graph, that graph is
said to be connected.
 A path that begins and ends at the same vertex
without traversing any edge more than once is called a circuit, or a
closed path.
A circuit that follows each edge exactly once while visiting every
vertex is known as an Eulerian circuit, and the graph is called an
Eulerian graph.
Note:
An Eulerian graph is connected and, in addition, all its vertices have
even degree.

4. Chinese Postman Problem
discovered by Chinese Mathematician Kwan Mei-Ko.
Algorithm:
 step 1 : If graph is Eulerian, return sum of all edge weights. Else do following steps.
 step 2 : We find all the vertices with odd degree .
 step 3 : List all possible pairings of odd vertices For n odd vertices total number of
pairings possible are, (n-1) * (n-3) * (n -5)... * 1 .
 step 4 : For each set of pairings, find the shortest path connecting them.
 step 5 : Find the pairing with minimum shortest path connecting pairs.
 step 6 : Modify the graph by adding all the edges that have been found in step 5.
 step 7 : Weight of Chinese Postman Tour is sum of all edges.
 step 8 : Draw Euler Circuit of the modified graph. This Euler Circuit is Chinese
Postman Tour.

5. Chinese Postman problem:
a b
e f
d
c
3
1
1
4
1
2
5 6
Illustration:

6.  As we see that graph does not contain Eulerian circuit because is has odd degree
vertices [a, b, e, f].
 First we make all possible pairs of odd degree vertices:[ae, bf], [ab, ef], [af, eb]
 so pairs with min sum of weight are [ae, bf] :
 ae = (ac + ce = 3 ), bf = ( bd + df = 2 ) Total : 5
 We add edges ac, ce , bd & df to the original graph and create a modified graph.
a b
e f
d
c
3
1
1
4
1
2
5 6

7. Solution
Graph with adding new edges:ac, ce , bd & df
Optimal Chinese postman route is of length : 5 + 23 = 28
 Chinese Postman Route : a - b - d - f - d - b - f - e - c - a - c - e - a
This route is Euler Circuit of the modified graph.
a b
e f
d
c
3
1
1
1
2
1
2
1
1
5 6
4

8. THANK YOU