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Show that the following problem is NP-complete: LP={⟨G,k⟩∣G is an undirected graph with a simple path of length at least k}.

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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:

  1. LP is in NP.
  2. 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.

User Elar
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