Answer:
A) To show that the punching set problem is in NP, we need to demonstrate that there is a polynomial-time verifier for a proof of yes instances. For every input, the verifier performs the following steps:
1.Take the input set E, S, and k.
2.Choose any subset P of E with |P|≤k.
3.Check if P punches every set in S.
To check if P punches every set in S, we can go through each set one by one and check if the intersection is empty. Since the number of sets in S is polynomial in the size of the input, and the size of E is fixed, this step can be performed in polynomial time. Therefore, the punching set problem is indeed in NP.
B) To show that the punching set problem is NP-hard, we will reduce the well-known NP-hard problem Vertex Cover to it. Given an undirected graph G=(V,E) and an integer k, we will construct a punching set problem instance with E and S of polynomial size.
The set of elements E will consist of the vertices of G and k additional elements, where each element corresponds to an edge of G. For each edge (v,w) in G, we will add the element corresponding to E to S and to the punching set. The set S will thus consist of the subsets of E corresponding to the vertex covers of G. The goal is to find a punching set P of E of size at most k such that P punches every subset of S.
To see that such a P exists if and only if G has a vertex cover of size at most k, observe that P punches a subset of S corresponding to a vertex cover of G if and only if P contains no element corresponding to an edge not in the vertex cover. Therefore, a punching set of size at most k exists if and only if G has a vertex cover of size at most k. Thus the punching set problem is NP-hard