Route Inspection (Edexcel International A Level Maths): Revision Note
The Chinese Postman Problem
What is the Chinese postman problem?
The route inspection problem is also known as the Chinese postman problem
You are required to find the route of least weight that traverses every edge in a graph
The route must start and finish at the same vertex
Some edges may need to be traversed more than once
The challenge is to minimise the total weight of these repeated edges
Variations to the route inspection problem could involve
The start and finish vertices being different
Certain edges being disregarded
E.g. a road closure
Repetition being required on the route
E.g. a road sweeper covering both sides of the road
Networks with 0 or 2 Odd Vertices
How do I solve the Chinese postman problem in networks with 0 or 2 Odd Vertices?
If the graph is Eulerian
All of the vertices in the graph are even
I.e. there are 0 odd nodes
It will be possible to find an Eulerian circuit
I.e. a route that traverses each edge exactly once, starting and finishing at the same vertex
No route or edges will need repetition
The shortest route will be the sum of the weights of the edges in the network
If the graph is semi-Eulerian
There will be one pair of odd vertices
I.e. there will be 2 odd nodes
If the route has to start and end at the same vertex
The shortest path between the two odd nodes will need to be found
The edges making that shortest path will need to be repeated
The shortest route will be the sum of the weights of the edges in the network plus the weight of the repeated edges
If the route starts at one of the odd vertices and ends at the other
No path/edges will need repetition
The shortest route will be the sum of the weights of the edges in the network
What are the steps of the Chinese postman algorithm?
STEP 1
Inspect the degree of all of the vertices and identify any odd nodes
STEP 2
If all of the vertices are even go to STEP 3
If there is one pair of odd nodes, find the shortest path between them
The edges making this path will need to be repeated so add them to the graph
(Once they are added to the graph every node will be even, so an Eulerian circuit will exist)
STEP 3
Write down an Eulerian circuit of the adjusted graph to find a possible routeFind the sum of the edges traversed to find the total weight
Examiner Tips and Tricks
Look carefully for the shortest path when connecting two odd vertices
A path made up of several edges will sometimes be a shorter distance than a more direct route
Worked Example
The network shown below displays the distance, in metres, of the cables in a network between different connection points A, B, C, D, E and F.
Each length of cable must be inspected.

a) Find the shortest route that allows all the cables to be inspected, and that starts and finishes at connection point E.
STEP 1
Inspect the order of the nodes
A: 5 (odd)
B: 2 (even)
C: 3 (odd)
D: 2 (even)
E: 4 (even)
F: 2 (even)
STEP 2
There is exactly one pair of odd vertices, A and C
The shortest path between A and C is ABC so the edges AB and BC need to be repeated
An Eulerian circuit, starting and ending at connection point E, is now possible
STEP 3
Find an Eulerian circuit, starting and ending at connection point E
Shortest route: EFAEDABACBCE
b) State the total length of the shortest route.
Add together the lengths of the edges in the original graph
3 + 4 + 4 + 5 + 8 + 9 + 11 + 13 + 16 = 73
Add the result to the lengths of the repeated edges
73 + 5 + 9 = 87
Shortest route = 87 m
You've read 0 of your 5 free revision notes this week
Sign up now. It’s free!
Did this page help you?