Contents
Why is it called the Chinese postman problem?
A similar problem is called Chinese Postman Problem (after the Chinese mathematician, Kwan Mei-Ko, who discovered it in early 1960’s). 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.
Is Chinese postman problem NP hard?
The Chinese postman problem (CPP) is a special case of the routing problem, which has many potential applications. We propose to solve the MCPP (special NP-hard case of CPP, which defined on mixed multigraph) using the reduction of the original problem into General Travelling Salesman Problem (GTSP).
How can we solve the postman problem?
Solving the chinese postman problem
- List all odd vertices.
- List all possible pairings of odd vertices.
- For each pairing find the edges that connect the vertices with the minimum weight.
- Find the pairings such that the sum of the weights is minimised.
- On the original graph add the edges that have been found in Step 4.
What is rural postman problem?
The rural postman problem (RPP) is a practical extension of the well-known Chinese postman problem (CPP), in which a subset of the edges (streets) from the road network are required to be traversed at a minimal cost. The RPP is NP-complete if this subset does not form a weakly connected network.
What is Fleury’s algorithm?
Fleury’s algorithm is an elegant but inefficient algorithm that dates to 1883. Consider a graph known to have all edges in the same component and at most two vertices of odd degree. The algorithm starts at a vertex of odd degree, or, if the graph has none, it starts with an arbitrarily chosen vertex.
How do I get Euler path?
To find the Euler path (not a cycle), let’s do this: if V1 and V2 are two vertices of odd degree,then just add an edge (V1,V2), in the resulting graph we find the Euler cycle (it will obviously exist), and then remove the “fictitious” edge (V1,V2) from the answer.