Chinese postman problem BY LAKSHMA REDDY KUSAM

Outline : Real world problem Graph construction Graph problem Graph traversability Solution

Real world problem The Chinese postman problem is a general real world problem for the postman’s delivering the letter’s as they jobs major role. The Chinese Postman algorithm was invented in 1962 and takes its name from the Chinese mathematician Kuan Mei-Ko . In this postman have to travel along all the streets in a town or city without missing even a single street. Now, the problem is that to find the minimum route possible and walking in a street only once. There is no graph problem if there are even vertices in the map network. The problem rasies when there are odd edges

Graph construction 5 6 6 9 B E 4 2 2 6 C 4 F 4 G 4 2 2 A D H I 8 14 7

Graph construction (cont.. B 6 E 10 2 2 13 6 F G C 4 10 2 2 A 8 D 14 H I 7

Graph traversability A network is said to be Traversable if you can draw it without removing your pen from the paper and without retracing the same arc twice. A graph is said to be closed if it is started and end at the same end. Are these Graphs Traversable and closed? Traversable? Yes Yes Yes Closed ? No Yes Yes

Euler’s path Euler found that the order of the vertices determines whether or not a graph is traversable. If it is possible to traverse a graph starting and finishing at the same point then the graph has an Eulerian trail. If it is possible to traverse a graph starting at one point and finishing at a different point then the graph has a semi-Eulerian trail.

Cont.. order 1 order 3 order 2 order 4 order 5 order 6 Even vertices Odd vertices order 3 order 1 order 2 order 4 order 5 order 6

In each case the graph is now closed. We need to pair the vertices together by adding an extra edge to make the order of each vertex four. We can join AB and CD, or AC and BD, or AD and BC. In each case the graph is now closed. [You only need to add one arc to be traversable]

Algorithm for Chinese postman problem Step 1: Sum all the vertices of the arcs in the graph. Step 2: Write down the degree of all the vertices to identify the odd vertices among them. Step 3: Pair all the nodes among themselves to make them even degree in such a way that the weight among them is minimum(matching up minimum possible distance). Step 4: Identify the odd vertices to be paired Step 5: Duplicate the vertices that are obtained from step4 .weight of the duplicated arc is same as the original arc. Add the weight of the duplicated arc to the original weight. Step 6: Now find the euler’s path of the obtained network.

Solution E B 6 Odd vertices 10 2 2 13 6 F G C 4 10 2 2 A 8 D 14 H I 7

Solution (cont..) Step 1: Total distance of the graph =7+10+10+2+4+2+14+8+13+2+2+6+2+6=88 Step 2: Degree of all the vertices A:2 ,B:3,C:3,D:3,E:3,F:3,G:4,H:3,I:2. From this the odd edges are B,C,D,E,F,H. Step 3: Possible combinations are BC, EF, DH BE,CF,DH. Those are the possible combinations now we calculate the distance between them. BC+EF+DH = 2+2+14=18 BE+CF+DH=6+6+14=26

Step 4: Now, from the above distance calculation we can see that 18<26 So we choose an extra duplicate path’s between BC,EF,DH. Step 5: Duplicate the edges between the BC,EF,DH. Now the total weight of the graph will be 88+18=106.

Now, we find out the euler’s route in the graph.

HDHIGFECDABCBEGH Now we have the shortest path between those vertices and thus gives the solution to the Chinese postman problem with minimum distance of 106 for the above network. .

ANY QUEIRES?

Thank you