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Lesson 41, Topic 2
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# Travelling Sales Person Problem

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We will be able to apply the branch-and-bound technique to instances of the traveling salesman problem if we come up with a reasonable lower bound on tour lengths. One very simple lower bound can be obtained by finding the smallest element in the intercity distance matrix D and multiplying it by the number of cities n. But there is a less obvious and more informative lower bound for instances with symmetric matrix D, which does not require a lot of work to compute. It is not difficult to show that we can compute a lower bound on the length l of any tour as follows. For each city i, 1 ≤ i ≤ n, find the sum si of the distances from city i to the two nearest cities; compute the sum s of these n numbers, divide the result by 2, and, if all the distances are integers, round up the result to the nearest integer:

lb = [s/2] —formula (a)

For example, for the instance in Figure a, formula (a) yields

lb = [[(1 + 3) + (3 + 6) + (1 + 2) + (3 + 4) + (2 + 3)]/2] = 14.

Moreover, for any subset of tours that must include particular edges of a given graph, we can modify lower bound (a) accordingly. For example, for all the Hamiltonian circuits of the graph in Figure a that must include edge (a, d), we get the following lower bound by summing up the lengths of the two shortest edges incident with each of the vertices, with the required inclusion of edges(a, d) and (d, a):

[[(1 + 5) + (3 + 6) + (1 + 2) + (3 + 5) + (2 + 3)]/2] = 16.

We now apply the branch-and-bound algorithm, with the bounding function given by formula (a), to find the shortest Hamiltonian circuit for the graph in Figure a. To reduce the amount of potential work, we take advantage of two observations. First, without loss of generality, we can consider only tours that start at a. Second, because our graph is undirected, we can generate only tours in which b is visited before c. In addition, after visiting n − 1 = 4 cities, a tour has no choice but to visit the remaining unvisited city and return to the starting one. The state-space tree tracing the algorithm’s application is given in Figure b. Figure (a) Weighted graph. (b) State-space tree of the branch-and-bound algorithmto find a shortest Hamiltonian circuit in this graph. The list of vertices ina node specifies a beginning part of the Hamiltonian circuits representedby the node.

To reiterate the main point: these state-space tree techniques enable us to solve many large instances of difficult combinatorial problems. As a rule, however, it is virtually impossible to predict which instances will be solvable in a realistic amount of time and which will not.