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By making use of the fact that sorting $n$ numbers requires $(n \log n)$ steps for any optimal algorithm (which uses 'comparison' for sorting), how can I prove that finding the convex-hull of $n$ points is bounded by $\Omega (n \log n)$ steps?

By making use of the fact that sorting $n$ numbers requires $\Omega(n \log n)$ steps for any optimal algorithm (which uses 'comparison' for sorting), how can I prove that finding the convex-hull of $n$ points is bounded by $\Omega (n \log n)$ steps?

By making use of the fact that sorting $n$ numbers requires $(n \log n)$ steps for any optimal algorithm (which uses 'comparison' for sorting), how can I prove that finding the convex-hull of $n$ points has the lower bound of $\Omega (n \log n)$ steps?

By making use of the fact that sorting $n$ numbers requires $(n \log n)$ steps for any optimal algorithm (which uses 'comparison' for sorting), how can I prove that finding the convex-hull of $n$ points is bounded by $\Omega (n \log n)$ steps?

By making use of the fact that sorting $n$ numbers requires $(n log n)$ steps for any optimal algorithm (which uses 'comparison' for sorting), how can I prove that finding the convex-hull of $n$ points has the lower bound of $ (n log n)$ steps?

By making use of the fact that sorting $n$ numbers requires $(n \log n)$ steps for any optimal algorithm (which uses 'comparison' for sorting), how can I prove that finding the convex-hull of $n$ points has the lower bound of $\Omega (n \log n)$ steps?