Final answer:
The problem LP, where LP={⟨G,k⟩∣G is an undirected graph with a simple path of length at least k}, can be shown to be NP-complete by proving it is in NP and NP-hard.
Step-by-step explanation:
The problem LP, where LP={⟨G,k⟩∣G is an undirected graph with a simple path of length at least k}, can be shown to be NP-complete.
To prove this, we need to show two things:
- LP is in NP.
- LP is NP-hard.
To show that LP is in NP, we need to show that given a certificate (representing a simple path of length at least k), we can verify it in polynomial time.
We can do this by checking that the certificate is a valid simple path in the given graph G and that its length is at least k. Both of these checks can be done in polynomial time.
To show that LP is NP-hard, we can reduce a known NP-complete problem, such as the Hamiltonian Path problem, to LP. This reduction shows that if we had a polynomial-time algorithm for solving LP, we could use it to solve the Hamiltonian Path problem in polynomial time, which would imply that P=NP.
Therefore, since LP is in NP and is also NP-hard, it is NP-complete.