What he wishes to travel along every road in

What is the Chinese Postman Problem?

The Chinese
Postman Problem is named after the Chinese mathematician, Kwan Mei-Ko, who
discovered it in early 1960’s. The background of Chinese postman problem is
interesting. It is the problem that the Chinese Postman faces: he wishes to
travel along every road in a city in order to deliver letters, with the least
possible distance.

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The Chinese
Postman Problem or Route Inspection Problem is about visiting
each route between cities at least once while returning to the original city
and taking the shortest route among all possible routes that fulfill this
criterium (if such a route exists). A solution that takes each route exactly
once is automatically optimal and called an Eulerian Cycle. Finding such a
cycle is easily possible. The true problem, however is how to find a shortest
closed walk of the graph in which each edge is traversed at least once,
rather than exactly once. In graph theory, an Euler cycle in a connected,
weighted graph is called the Chinese Postman problem. However, the Chinese
Postman Problem differs from Euler circuit problems.

 

How do they differ?

The difference between Euler’s
circuit, which involved crossing each edge once and once only in order to be
solved, and the Chinese Postman problem, which allowed an edge to be crossed
more than once, however, the weight of the circuit is considered instead. This
problem differs as well from the problem of the travelling salesman.

The travelling salesman needs to
visit every node (or vertex) whereas the Postman needs to travel along each arc
(or edge). Therefore, it is required that the network is traversable
(Eulerian). The problem is therefore to obtain “a closed trail of minimum weight
containing every arc. Another way to put it is to “find a route of minimum
weight which traverses every edge at least once, returning to the start
vertex”.

 

 

 

 

 

 

 

Examples (A):

Here are two graphs, one
which can be solved with a Euler circuit and one that is considered as a
Chinese Postman Problem. The Euler circuit does not have a numerical weight on
it, and therefore its required to be solved by going over every edge once and
once only. The Chinese Postman problem, however, includes a numerical weight on
each individual edge, and must be solved by having the lowest possible weight,
all while being able to visit each vertex and then return to the starting
vertex.