Question

P (b) H1 H2 H3 H4 (a) Figure 3: Nested hulls (c) P l

          P
(b)
H1
H2
H3
H4
(a)
Figure 3: Nested hulls
(c)
P
l

P
(b)
H1
H2
H3
H4
(a)
Figure 3: Nested hulls
(c)
P
l

Added by Ashley F.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Joel Mueller

06/01/2022

Step 1

A more robust method to describe the shape of a collection of points is by constructing nested layers of convex hulls, as described here: Given a planar point set P in general position (see Fig. 1(a)), let H1 be the convex hull of P. Remove the vertices of H1 Show more…

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The convex hull is a somewhat non-robust shape descriptor, since a few distant outlying points can dominate the shape of the hull. A more robust method to describe the shape of a collection of points is by constructing nested layers of convex hulls, as described here: Given a planar point set P in general position (see Fig. 1(a)), let H1 be the convex hull of P. Remove the vertices of H1 from P and compute the convex hull of the remaining points, call it H2. Repeat this until no more points remain, letting H1, ..., Hk denote the resulting hulls (see Fig. 1(b)). More formally, Hi = conv(P \ (S1 U S2 U ... U Sj)), where Sj represents the vertices of Hj. The final result is a collection of nested convex polygons, where the last one may degenerate to a single line segment or a single point. 1. Assuming that the points are in general position in R^2, as a function of n, what is the maximum number of hulls that can be generated by this process? (I am looking for an exact formula, not an asymptotic one. For every n, there should exist a point set that exactly achieves your bound.) Briefly justify your answer. 2. Given a set P of n points in the plane, devise an O(n^2) time algorithm to compute this iterated sequence of hulls. (FYI: O(n log n) is possible, but complicated.) 3. Prove the following lemma: Given a planar point set P in general position, let k denote the number of hulls generated by the repeated hull process. There exists a point q in the plane (it need not be in P) such that every closed halfplane whose bounding line l passing through q contains at least k points of P. (A closed halfplane is the region of the plane lying on or to one side of a line. An example of such a point for k = 4 is shown in Fig. 1(c).)

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Transcript

00:01 Okay, so these are the two things we want to show.

00:05 I did, b is the convex haul here.

00:09 I expressed it as it said.

00:17 So here, b, the convex hall is all y and a, such that y is equal to this linear combination of the xi, the xi are in a, and the sum of all the coefficients adds to one.

00:33 And so to prove that b is convex, we're going to let, we need to show about, we need to show if wz are in b then for all mu in the set, 0 to 1, mu, plus 1 minus mu, z, z is in there.

01:14 So let's start with that.

01:20 Let w z and b.

01:26 Then we can say that for i equals 1 to n lambda i equaling 1 and i equals 1 to n.

01:42 And i equals 1 to n.

01:48 That's definitely what's another let's use delta i is equal to 1.

01:55 We have lambda 1.

02:03 W1 excuse me let me say something else for x -in -a we have that so we know for a fact since b is the convex hull that w is equal to lambda 1 x1 plus dot dot dot plus lambda n x n and z is equal to delta 1 x1 plus dot dot dot delta n x -n so for um you have another guy how about mu again so for mu in 0 to 1 take mu w plus 1 minus mu z and that's equal to mu times lambda 1 x1 plus landa 1 x1 plus land n x n plus 1 minus e mu delta 1 x1 plus delta n x n okay we can rearrange this to be um new lambda 1 plus 1 minus mu delta 1 x1 plus dot dot dot plus mu lambda n plus one minus me delta n and we want to show that the sum from i equals one to n of mu lambda one plus excuse me mu lambda i plus one minus mu delta i is equal to one and that will show that the set is convex because all those coefficients sum to equal one but it's to show this because this is equal to mu i equals 1 and of lambda 1 plus 1 minus mu i equals 1 to n delta i i wrote lambda 1 here i meant to write lambda i sorry about that but this gives us exactly what we want because since the sum of the lambda i and the sum of the delta i equal 1 we get mu times 1 plus 1 minus mu times 1 which is just easily seen to be equal 1.

05:27 So there's our proof that maybe i'll write our conclusion.

05:36 We conclude that b is convex.

05:48 All right...

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