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Theorem 34.12: VC $\in$ NPC



Proof Sketch:

1.
VC $\in$ NP
Given V', check $\mid$V'$\mid$ = k, and for each edge (u,v) $\in$ E, check that either u $\in$ V' or v $\in$ V'.

2.
L' = CLIQUE

3.
CLIQUE \(\leq_P\) VC
If graph G = (V, E) has clique V', then graph \(\overline{G}\) has vertex cover V - V'.
\(\overline{G}\) = (V, \(\overline{E}\)) is the complement of G = (V,E), where \(\overline{E} = \{ (u,v) \vert (u,v) \not\in E \}\)
Reduction: G $\rightarrow$ \(\overline{G}\) (poly-time)

4.
x $\in$ CLIQUE(G) = V' $\longrightarrow$ f(x) $\in$VC( \(\overline{G}\)) = V - V' ( | V' | = k)
Every edge (u,v) $\in$ \(\overline{E}\) implies (u,v) \(\not\in\) E, thus at least one of u and v \(\not\in\) V'. Thus, at least one of u,v belongs to V - V', which means edge (u,v) is covered by V - V'. Similar argument for other direction.


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