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Chapter 9: Problem 12

Show that the reduction of the Hamiltonian Circuits Decision problem to the Traveling Salesperson (Undirected) Decision problem can be done in polynomial time.

Short Answer

Expert verified
Yes, the Hamiltonian Circuits Decision problem can be reduced to the Traveling Salesperson Decision problem in polynomial time. This is done by setting the weight of each edge to 1 and setting the budget B equal to the number of nodes.

Step by step solution

01

Understanding the Problems Involved

Firstly, make sure to understand both the Hamiltonian circuit decision problem and the Traveling Salesperson problem. A Hamiltonian circuit is a closed loop on a graph where every node (vertex) is visited exactly once. A Traveling Salesperson problem is an optimization problem that aims to determine the shortest possible route that a traveling salesperson can take to visit each city once and return to the original city. The version we are dealing with here is a decision problem, where the question is whether such a tour exists within a specified budget B.
02

Defining The Reduction

Now describe a method for converting any instance of the Hamiltonian Circuit Decision problem to an instance of the Traveling Salesperson Decision problem, such that the former has a Hamiltonian circuit if and only if the latter has a tour of cost at most B. As the weights of the edges are not part of the Hamiltonian Circuit problem, assign a weight of 1 to every edge in the graph. Then, for the Traveling Salesman problem, set the budget B equal to the number of nodes in the graph, which is the number of edges in the Hamiltonian circuit.
03

Proving The Reduction

Finally, the Hamiltonian circuit decision problem is reducible to the Traveling Salesperson Decision problem if an algorithm exists that performs this reduction in polynomial time. The reduction described simply involves setting the weight of each edge to 1 and setting the budget B equal to the number of nodes, which can be done in linear time with respect to the number of edges. Therefore, this is a polynomial time reduction.

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Most popular questions from this chapter

Suppose that problem \(A\) and problem \(B\) are two different decision problems. Furthermore, assume that problem \(A\) is polynomial-time manyone reducible to problem \(B\). If problem \(A\) is \(N P\) complete, is problem \(B N P\) complete? Justify your answer.

Given a list of \(n\) positive integers \((n \text { even }),\) divide the list into two sublists such that the difference between the sums of the integers in the two sublists is minimized. Is this problem an \(N P\) complete problem? Is this problem an \(N P\) -hard problem?

Show that the reduction of the Traveling Salesperson (Undirected) Decision problem to the Traveling Salesperson Decision problem can be done in polynomial time.

Show that a problem is \(N P\) -easy if and only if it reduces to an \(N P\) -complete problem.

Is the Towers of Hanoi problem an \(N P\) -complete problem? Is it an \(N P\) easy problem? Is it an \(N P\) hard problem? Is it an \(N P\) -equivalent problem? Justify your answers. This problem is presented in Exercise 17 in Chapter
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