Question

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

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

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.

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.

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.