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Chapter 9: Problem 17

Show that if a problem is not in \(N P\), it is not \(N P\) -easy. Therefore, Presburger Arithmetic and the Halting Problem are not \(N P\) -easy.

Short Answer

Expert verified
The results show that if a problem is not in NP, it cannot be NP-Easy as it would contradict the definition of NP. Also, it has been established that Presburger arithmetic and the Halting problem are not NP-Easy because they are undecidable and their solutions cannot be verified in polynomial time, as required to be in NP.

Step by step solution

01

STEP 1: UNDERSTAND NP and NP-Easy Problems

An NP (nondeterministic polynomial time) problem is a class of problems whose solution can be verified in polynomial time by a deterministic Turing machine. On the other hand, NP-Easy problems are the subsets of NP problems which can be reduced to another NP problem in polynomial time.
02

STEP 2: Prove Non-NP Problems are not NP-Easy

Suppose to the contrary that a non-NP problem is NP-Easy. By definition of NP-Easy, this problem can be reduced to another problem in polynomial time. However, since the original problem is not in NP, it cannot be verified in polynomial time. This creates a contradiction, thus a non-NP problem cannot be NP-Easy.
03

STEP 3: Show that Presburger Arithmetic and the Halting Problem are not NP-easy

Presburger arithmetic and the Halting problem are both known to be undecidable problems. This means there is no algorithm that can determine the correct solution for all inputs in a finite amount of time. Since by definition NP problems and by extension, NP-Easy problems, must have a solution that can be verified in finite time (specifically polynomial time), it follows that neither Presburger arithmetic nor the Halting problem can be NP-Easy.

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Most popular questions from this chapter

Given a list of \(n\) positive integers \((n\) 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 Hamiltonian Circuits Decision Problem to the Traveling Salesperson (Undirected) Decision Problem can be done in poly. nomial time.

When all instances of the CNF-Satisfiability Problem have exactly three literals per clause, it is called the 3 -Satisfiability Problem. Knowing that the 3-Satisfiability Problem is \(N P\) -complete, show that the Graph 3 -Coloring Problem is also \(N P\) -complete.

Is the Towers of Hanoi Problem an NP-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 2.

Show that a graph problem using the number of vertices as the measure of the size of an instance is polynomially equivalent to one using the number of edges as the measure of the size of an instance.
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