1 Olaf Hall-Holt Matthew J.Katz Piyush KummarJoseph S.B. Mitchell Zhao Zhang Department of Computer Science and Engineering Texas A&M University April.

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Olaf Hall-Holt Matthew J.Katz Piyush Kummar Joseph S.B. Mitchell

Zhao Zhang

Department of Computer Science and EngineeringTexas A&M University

April 9, 2009/ CPSC-669

1. Introduction and Related work2. Approximately Peeling Potatoes

The Biggest Stick Problem(PTAS) A PTAS for Largest Triangles Peeling an ellipse for well sampled curves

3. Future work

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Definition: compute a largest convex body inside a given shape has been called the “potato peelers problem”

Goodman (81): How does one optimally “peel” a nonconvex potato P to obtain a convex potato, while wasting as little as possible of the original potato?

Typical Subproblems: Longest Stick(Line) Biggest Rectangles (French Fry) Biggest Potato(polygon) Biggest Ellipse (smooth closed curves)

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Mostly used in computer graphics application Application: 1. Natural Optimization Problems 2. Shape Approximation 3. Visibility Culling for Computer Graphics

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[Koltam, Chrysanthon and Gohen-Or]: Virtual Occluders for PVS

[Fausto, J. El-sana, J. T. Klosowski] Directional Discretized Occluders for Accelerated Occlusion Culling

[Bischoff and Kobbelt] Packing ellipsoids by inflation

[Brunet, Navazo, J.Bossignac,Carlos Saona] Hoops: 3D Curves as Conservative Occluders for Cell-Visibility

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Maximum area convex k-gon inside a convex n-gon [A. Aggarwal ,Algorithmica 87’] O(kn+nlogn), Matrix searching

Equilateral triangles or squares within a simple polygon [A.DePano, CCC 87’] O(n3) [L.P.Chew, Symbolic Comput. 90’]O(n2logn) ,motion planning method

Maximum area triangle inside a simple polygon [E.A. Melissaratos SIAM 92’] O(n3)

Maximum-area convex polygon inside a simple polygon [J.Chang C.K Yap, Geom. 86’] O(n7)

Longest stick Problem O(n1.99) [B.Chazelle Symbolic Comput 90’] O(n 8/5+ε ) [Agarwal, Sharir Algorithms 94’] parametric searching O(n 17/11+ε) [Agarwal, Sharir Discrete Comput 96’] O(n 3/2+ε ) [Agarwal, Aronov SIAM 97’]

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In simple n-gons: Longest stick: ½-approx,

O(n logn) time

Longest stick (1-ε)-approx O(n logn2) time

Biggest triangle: O(1)-approx O(n logn)

Biggest triangle:(1- ε)-approx O(n) time

Maximum-area Ellipse within well-sampled smooth closed curves:

O(n2) time

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S*

a

b c

db

c

f

d

e

i

f ie

At least length(S*)/2

Step 1 : Hierarchical decomposition(Chazelle Cuts 91’): O(n) time Assume S* is the longest line in the polygon.

There must at least one Chazelle Cut intersect S*. The first intersection cut S* into two part and one of them >length(S*)/2

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Theorem:

One can compute a ½-approximation for longest stick in a simple polygon in O(nlogn) time.

Algorithm:

At each level of the recursive decomposition of P, compute longest anchored sticks from each diagonal cut: O(n) per level.

Longest Anchored stick is at least ½ the length of the longest stick.

Step 2: Compute weak visibility region m from anchor edge (diagonal) e. time :O(n)87’ Step 3: Compute the longest length from the e to m. time:O(1)

There are logn level of the hierarchy, the total time consuming O(nlogn)

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A mega-square is an axis-parallel square.

Step 1: Set l0 equals to the longest edge of P. l* is the optimal solution.

Make three additional copies.

Claim: P can be covered by linear time of mega-squares.Step 2: Mark those grid cells that are intersected by P’s boundary.Step 3: Divide the mega-square into pixels whose length is εl0 /c0 , .

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Step 4: For each pair of pixel sides s1, s2, check whether there exist a line segment in σ∩P whose endpoint lie on s1 and s2.

If σ is a good mega-square, then we can have a ( 1- ε)-approximation of L*, since we just shorten it by only O(εl0) by clipping its ends at the boudaries of the pixels containing them.

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s1 an s2 are pixel sides and assume horizontal.

R(ai) – the range on s2 that is seen from ai.R(bj) – the range on s1 that is seen from bj.

If R(ai)>bj and R(bj)>ai => ai and bj can see each other.

We can compute R(ai) using shortest path queries from L.J. Guibas. 【 Optimal shortest path queries in a simple ploygon 】 in logarithmic time

Given a range R(ai) of an interval ai on s1, determining whether there is bj on s2 can be seen by ai takes O(log2 k2)(basing one a pecial tree stucture)

There are O(1/ε4 ) pairs of pixels.

The total time consuming is O(nlogn)

Step 1: Recursive decomposition of the polygon using Chazelle cuts.Step 2: Traversing the boundary of the weakly visible polygon of each

cut

There are O(logn) cuts and each traversing takes O(n)

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Theorem 3.1 In a simple polygon P with a n vertices one can compute an O(1)-approximation of the maximum-area triangle within P in time O(nlogn)

Actually ,one can compute (3/32)- approx for the maximum area trangle

A triangle is δ-fat if all three of its angles are at least some specified constantδ.

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CLAIM 1. Assume that δ <60 and that φ>=α,β. Then , for any ε>0, there exists ε 0 >0 such that one can place a δ –fat triangle inside P with angles α+ ε 0 , β+ ε 0 ,φ- 2ε

0 and whose area is at least (1- ε) area(Δ*).

Let a* be the length of the shortest side of Δ*. We can get a constant factor approximation a0 of a*, such that a0 <a*<c1* a0 for some small constant c1. [using the Voronoi diagram of P in O(n) from Chin 99’]

A mega-square is an axis-parallel square of side length O(a0 )

If one mega-square can place a δ-fat triangle in P of area(Δ*) ， we say it’s good.

Step 1: For each vertex pi in P, draw a mega-square of side length c2a0 around p.

Step 2:Divid each mega-square into pixels with length ε0a0/c3.

ε0 can be calculated from CLAIM 1C3 =(3+3/tan δ)

Step 3:Consider all the pixels intersect with the boundary. Select all the pixels inside the P.Then get the set of rectilinear polygons Q inside P.

We claim the Q contains a triangle of area (1-ε0)2 area (Δ*)

CLAIM 2. There exists a constant c2 , such that, for at least one of the vertices pi of P, the mega-square of side length c2a0 centered at pi is good.

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Step 4:Apply CK-algorithm to return a fat triangle within factor(1-ε0 )CK-algorithm is from Chew an Kedem find_largest_copy(Q,P)( 93’) O(1) time

CLAIM 3. The area of the returned triangle is at least (1-ε0)3 area (Δ*)

(1-ε0)3 => (1-ε)

THEOREM 2. One can compute a (1-ε)-approximation of the largest fat triangle inside a simple polygon P in time O(n).

A curve S is said to be α-sampled if the length of the curve between any tow consecutive sample points is no more than α.

Let Q* denote a maximum-area empty ellipse contained in the smooth closed curve S.

Let we get Q in this way: Scale Q* until it hits three sample point p1, p2, p3 and the triangle p1p2p3 contains

the center of Q*. Let Qin = (1- α /b)Q. Then area(Qin)<=area(Q*)<=area(b/ （ b- α ） Qin)

The goal is to determine an ellipse that is close in some sense to Q.

If Q and S both pass through the sample point p, then the angle between their normals at p is O(α).

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Step 1: Linearization LL = {(x2, y2, 1/2xy, x, y)|(x, y) ∈L*}

Step 2: Get the Convex Hull of L

Step 3: Pick any three points p, q, r belong to L, define one 2-dimensional space.

Step 4: Compute ellipse Qpqr passing through p, q and r, whose normal at p, q, r each make a small angle with the corresponding normals of S, whose small is contained in the triangle pqr.

Step 5: Return the area that is maximum over all area computed for feasible ellipses.

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• PTAS for largest triangle ?

• Find exact solutions/approximations for

biggest potato ?

Questions?

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I would like to thank Olaf and Piyush, the authors, for helping me understand the paper and providing their slides on SODA.