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We define B as:

$B = \{ <G> | \text{ G is an undirected graph in which} \\ \text{the number of vertices in the largest independent set} \\ \text{is greater than the number of vertices in the largest clique} \}$

Can you prove that B is NP-Hard? I have tried to polynomialy reduce from $CLIQUE$ but couldn't get it to work (I have tried to look at the complement graph, and add vertices)

Shlomiz
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1 Answers1

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To prove the NP-hardness of the problem B under the polynomial-time many-one reduction, we can reduce from the maximum independent set problem.

We are given a graph of size $n$ and an integer $k$, to determine if the graph contains an independent of size at least $k$. The reduction algorithm construct an instance of the problem B by adding $n$ isolated vertices and an $(n+k)$-clique to the original graph. Then,

  1. There exists an independent set of size $k$ in the the original graph if and only if there exists an independent set of size $n+k+1$ in the new graph.
  2. The maximum clique size of the new graph is $n+k$.
pcpthm
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