In the bottleneck traveling-salesman problem , we wish to find the hamiltonian c

Question

In the bottleneck traveling-salesman problem, we wish to find the hamiltonian cy- cle that minimizes the cost of the most costly edge in the cycle. Assuming that the cost function satisfies the triangle inequality, show that there exists a polynomial- time approximation algorithm with approximation ratio 3 for this problem. (Hint: Show recursively that we can visit all the nodes in a bottleneck spanning tree, as discussed in Problem 23-3, exactly once by taking a full walk of the tree and skip- ping nodes, but without skipping more than two consecutive intermediate nodes. Show that the costliest edge in a bottleneck spanning tree has a cost that is at most the cost of the costliest edge in a bottleneck hamiltonian cycle.)

Explanation / Answer

You require proving by induction the next claim:

Consent to T is a tree rooted at r with as a minimum two vertices. You are able to file the vertices of T in a series such that the distance as in T between two adjacent vertices is at most 33, and also the vertex following r is adjacent to r in T.

The induction is lying on the number of vertices.

Known this, you argue as follows. Assume that there is a Hamiltonian cycle by heaviest edge contains weight w. next there is a spanning tree whose heaviest edge have weight w. thus the heaviest edge in the block spanning tree have weight at most w. relate the claim to that tree, and believes the corresponding Hamiltonian cycle. As the tree space between adjacent vertices is at most 3 and be grateful to the triangle inequality, the heaviest edge in the resultant cycles weighs at most 3w.

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