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How do you show that the convex hull of a given set of points S, always has the minimum perimeter ? By perimeter i mean the length of the boundary of the hull

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Minimum with respect to what? To all convex sets containing S? Cause if its w.r.t. all sets containing S, then it is not true. Take for instance the 4 corners of a square. obviously, the square is the convex hull, its boundary has length $4Z$, $Z$ being the length of one side. But the set of corner points itself has zero length. – Raskolnikov Dec 5 '10 at 21:34
I mean that the hull has the shortest perimeter of all simple polygons that include every point of S. – user4444 Dec 5 '10 at 21:50

2 Answers 2

The steps for such demostration are indicated in the book of Berg "Computational Geometry. Algorithms and aplications". Exercise 1.1. The convex hull of a set S is defined to be the intersection of all convex sets that contain S. For the convex hull of a set of points it was indicated that the convex hull is the convex set with smallest perimeter. We want to show that these are equivalent definitions. a. Prove that the intersection of two convex sets is again convex. This implies that the intersection of a finite family of convex sets is convex as well. b. Prove that the smallest perimeter polygon P containing a set of points P is convex. c. Prove that any convex set containing the set of points P contains the smallest perimeter polygon P.

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Take an arbitrary simple polygon $\pi$ including every point of $S$, and take some segment $PQ$ of the convex hull. Extend it into a line $\ell$. Form a new polygon $\pi'$ by locating the "left"most and "right"most intersection points $L,R$ of $\ell$ with $\pi$, adding new vertices to $\pi$ at $L,R$, and replacing the part of $\pi$ between $L$ and $R$ by the single line segment $LR$. The resulting polygon $\pi'$ is a simple polygon including all points of $S$, and the triangle inequality implies that its perimeter is smaller or equal to the perimeter of $\pi$.

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Wow I didn't think it was that easy... thank you Yuval ! – user4444 Dec 6 '10 at 9:03

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