I want to change the original TSP problem as follows: the cost to visit a city is not related to the previous city that it visited just now, but only on its position in the sequence. Is the problem of finding the minimal cost tour still NP-complete?
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Your problem is the same as minimum-weight perfect matching in a bipartite graph. If there are $n$ cities, consider the bipartite graph in which one side consists of vertices $p_1,\ldots,p_n$, the other side consists of vertices $c_1,\ldots,c_n$, and the weight of the edge $(p_i,x_j)$ is the cost of city $j$ at position $i$.
A minimum-weight perfect matching in a bipartite graph can be found efficiently (this is known as the assignment problem). In particular, the problem is not NP-hard unless P=NP.
Yuval Filmus
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