DCG19 Convex Hulls - Graz University of Technology · Irene Parada Discrete and Computational...

Irene Parada Discrete and Computational Geometry 2019. Convex Hulls1

Convex Hulls

Discrete & Computational Geometry WS 2019/20

Slides by Irene Parada and Vera Sacristan

Outline

• Convexity.

• Caratheodory’s Theorem.

• Convex hulls in 2D.

• Doubly connected edge list (DCEL).

• Convex hulls in 3D.

• Convex hulls in higher dimensions.

• Convex hulls of simple polygons.

In this lecture: general position (no three collinear points)and worst-case complexity.

Convexity I

Let S be a set of points in Ed (real Euclidean space of dim. d).

Linear combination of points in S:k∑i=1

λipi, with k ∈ N, pi ∈ S, λi ∈ R.

Affine combination of points in S:

linear combination s.t.k∑i=1

λi = 1.

The set of all affinecombinations of S is calledaffine hull of S.

Convex combination of points in S:affine combination s.t. λi ≥ 0 ∀ i.

The set of all convexcombinations of S is calledconvex hull of S, conv(S).

A set is called convex if it is stable under convex combinations.

Convexity I

λi = 1.

Exercise: Find the set of allcombinations of each type of thesetwo points in the Euclidean plane.

Let S be a set of points in Ed.

Convexity I

λi = 1.

Let S be a set of points in Ed.Exercise: Find the set of all

combinations of each type of thesethree points in the plane.

Convexity II

Prop. A set S is convex iff it contains segment pq for all p, q ∈ S.

convexnot convex

Convexity II

Proof :(⇒) If S is convex, it is stable under convex combinations and, inparticular, it contains all segments with endpoints in S.

(⇐) By induction on the number of points k of the convexcombination. For k = 2, we have the statement.Consider now a convex combination with k + 1 points. We have

k∑i=0

λipi = λ0p0 +k∑i=1

λipi = λ0p0 + (1− λ0)k∑i=1

λi1− λ0

pi ∈ S.

k∑i=1

λipi ∈ S, for all pi ∈ S, λi ≥ 0,∑ki=1 λi = 1.

Want to show ∈ S Since∑ki=1

1−λ0= 1−λ0

1−λ0= 1, by ind. hip. ∈ S

Convexity II

Corollaries:• The intersection of convex sets is convex.

• The convex hull conv(S) of a set S is the smallest convex setcontaining S.

• The convex hull conv(S) of a set S is the intersection of all theconvex sets containing S.

Caratheodory’s Theorem

Caratheodory’s Thm. Let S be a set of points in Ed. Every pointp ∈ conv(S) can be written as the convex combination of at mostd+ 1 points of S.

Proof. Suppose p is a convex combination of k points of S and k is

the smallest possible: p =k∑i=1

λipi with pi ∈ S, λi > 0,k∑i=1

λi = 1.

Assume k > d+ 1. Then the k − 1 > d vectors p2 − p1, . . . , pk − p1are linearly dependent:

∑ki=2 µi(pi − p1) = 0 with not all µi = 0.

We define µ1 = −∑ki=2 µi. Then

k∑i=1

µipi = 0 andk∑i=1

µi = 0.

Thus, for any α ∈ R, p =k∑i=1

λipi − αk∑i=1

µipi =k∑i=1

(λi − αµi)pi.

Set α = min1≤j≤k

µj: µj > 0

⇒ p is a convex comb. of < k points.

Computing the Convex Hull in 2D

Input: S = p1, . . . , pn.Output: set of the extreme points.

Procedure:For each i,

For each j, k, l 6= i,If pi lies in the triangle pj , pk, pl,eliminate pi.

Return the set of surviving pi’s.

Algorithm 1: Computing the extreme points by brute force.

Given S = p1, . . . , pn, the point pi belongs to the boundary ofconv(S) iff pi does not lie in any of the triangles pjpkpl.

Running time: O(n4)

Exercise: Does this approachwork in d dimensions? If yes,

what is the running time?

Yes, using Caratheodory’stheorem and the same approachwe get an O(nd+2) algorithm.

(Can be tightened more, thoughnot in the exponent.)

Still, this is not an efficientalgorithm!

Algorithm 2: Computing the extreme segments by brute force.

Given S = p1, . . . , pn, the segment pipj belongs to the boundaryof conv(S) iff all other pi lie in the same halfplane defined by pipj .

Input: S = p1, . . . , pn.Output: set of extreme segments.

Procedure:For each i, j,

Check whether all pk, k 6= i, jlie on the same halplanedefined by pipj .

If true, return the segment pk.

Running time: Θ(n3)

Algorithm 3: Jarvis’ march (gift wrapping).

Given S = p1, . . . , pn, the segment pipj belongs to the boundaryof conv(S) iff all other pk lie to the left of pipj (oriented).

Input: S = p1, . . . , pn.Output: l: ccw points on ∂ conv(S).

Procedure:

Add extreme point to l.Set v = Last(l).While |l| = 1 < n or v 6= First(l):

Detect angularly rightmost pointpj ∈ P wrt v and add it to l.

Procedure:

Running time: Θ(hn) = O(n2)

Initialization:- Push v extremal in l, del. from S.- Angularly sort pts. in S around v.- Push first pt. in l and del. from S.

Advance:While ∃pi ∈ S to be explored:p = last(l); p− = prev(last(l))- If p−ppi is a left turn:

Push pi in l, del. from S.- Else: Pop p from l.

Algorithm 4: Graham’s scan (succesive local repair).

Running time: O(n log n)

To play:http://animalgo.ist.tugraz.at

O(n log n)

Algorithm 5: Incremental algorithm.

Initialization: l = p1, p2, p3

Advance: From i = 4 to n, do:If pi lies out of the polygon def. by l:

- Compute the points pl and prof tangency.

- Replace pl, . . . , pr in lwith pl, pi, pr.

By storing l in a structure allowingbinary search and updates (insertionsand deletions) in O(log n) time.

Algorithm 6: Divide and conquer.

Initialization: Sort points by x-coord.

Division: Divide the points (xi, yi) intotwo subsets wrt the median x-coord.

Recursion: Recursively compute theconvex hull of the two subsets.

Merge: Compute the external commontangents of the two convex polygons.Discard int. points between tangents.

Question: How to compute the tangentsefficiently?

Question: Are the topmostand bottommost points onthe tangents?

O(n log n)

T (n) = 2T(n2

)+O(n)

= O(n log n)

Convex Hulls in 2D: Lower Bound

Input: n real numbersx1, . . . , xn real numbers

Input: n pointsp1, . . . , pn, with pi = (xi, x

Output: convex hull of the pointsSorted list of the vertices p1, . . . , pn

Output: sorting the numbersSorted list of the numbers x1, . . . , xn

Ω(n log n)

So, this is it? AreO(n log n) algorithms thebest possible?

Algorithm 3: Jarvis’ march (gift wrapping). Θ(hn) = O(n2)

O(n log n)Algorithm 4: Graham’s scan (succesive local repair).

O(n log n)

Algorithm 2: Computing the extreme segments by brute force. Θ(n3)

Algorithm 1: Computing the extreme points by brute force. O(n4)

Algorithm 7: Chan’s algorithm. O(n log h)

Algorithm 7: Chan’s algorithm.

Input: S = p1, . . . , pn, H ≤ n.Output: If there are at most H extremepoints, l: ccw points on ∂ conv(S).

Division:• Divide S into k = dn/He setsS1, . . . , Sk with |Si| ≤ H.

• Construct conv(Si) for all i.

Conquest: Use gift wrapping (Jarvi’smarch) on the small convex hulls tofind the first H points on ∂ conv(S)ccw from the lex. min. point in S.

To see it working:http://chrisgregory.me/

projects/web/chans/

Running time: H ·O(k logH) = O(n logH)

To see it working:http://chrisgregory.me/

projects/web/chans/

O(n logH)For a extremal point we canfind the tangent point qi withone of the k = dn/He sets inO(logH) time.

We can find the next extremalpoint among the qi’s in O(k)time.⇒ Each wrapO(k logH + k) = O(k logH).

Running time:O(n logH)

If we knew h, then we could make H = h, but we don’t know h...

We call the algorithm with H = min22t , n for t = 0, 1, . . . until weget the extremal points (wrap around) when t = s.

Total running time:

s∑t=0

cn log 22t

= cns∑t=0

2t = cn(2s+1 − 1) < 4cn log h = O(n log h)

for some constant c.

Algorithm 3: Jarvis’ march (gift wrapping). Θ(hn) = O(n2)

O(n log n)Algorithm 4: Graham’s scan (succesive local repair).

O(n log n)

Algorithm 2: Computing the extreme segments by brute force. Θ(n3)

Algorithm 1: Computing the extreme points by brute force. O(n4)

Algorithm 7: Chan’s algorithm. O(n log h)

Other algorithms: Quick hull.Kirkpatrick–Seidel algorithm (marriage-before-conquest).

O(n2)O(n log h)

• For each vertex v: Its coordinates (x, y, z) An edge e(v) incident to it

• For each face f : An edge e(f) incident to it

• For each (oriented) edge e: Its starting and ending vertices vS(e), vE(e) Its left and right incident faces fL(e), fR(e) Its counterclockwise previous and next edges eP (e), eN (e)

The doubly connected edge list (DCEL), is a data structure torepresent an embedding of a planar graph (and polyhedra).

Why do we use it?

• Time: Sorted lists of vertices and edges of

any face in time ∝ size of the face. Sorted lists of edges and faces indicent

to any vertex in time ∝ its degree.

• Space: linear in the number of vertices V .

Proof: Since the graph is planar, V + F = E + 2.Since the faces are k-gons (with k ≥ 3), 2E ≥ 3F . Therefore:

2V + 2F = 2E + 4 ≥ 3F + 4 =⇒ F ≤ 2V − 4.

3E + 6 = 3V + 3F ≤ 3V + 2E =⇒ E ≤ 3V − 6.

Convex Hull in 3D

Algorithm: Divide and conquer.

Input: S = p1, . . . , pn.Output: conv(p1, . . . , pn).

Division: Divide the points into twoequal subsets by a vertical plane.

Merge: Compute the convex hull of theunion of the convex hulls of the subsets.

O(n log n)

T (n) = 2T(n2

)+O(n)

= O(n log n)

We must merge in O(n) time!

Convex Hull in 3D

Algorithm: Divide and conquer. Running time: O(n log n)

Merging

Notation:

• Faces of C1 and C2 that appearin C are called blue.

• Faces of C1 and C2 that do notappear in C are called red.

• Edges and vertices are coloredwith the intersection color (blue,red and purple) of their incidentfaces.

Convex Hull in 3D

Merging

Careful!

• Purple might not be connected.

Convex Hull in 3D

Merging

Careful!

• Purple might not be connected.a

Convex Hull in 3D

Merging

Careful!

• Purple might not be connected.• There might be no blue faces.

Convex Hull in 3D

Merging

Careful!

• Purple might not be connected.• There might be no blue faces.

Edge pq is incident to two red faces.

Vertex p is incident to more thantwo purple edges.

Convex Hull in 3D

Merging

Convex polygon C0.

Merging algorithm:

• Find an edge of C \ C1 ∪ C2.• Find the remaining new faces

and edges in the order inducedby C0.

• Identify the red faces, edges andvertices and update the DCEL.

Convex Hull of a Simple Polygon

Given a simple polygon with n vertices, how fast can we compute itsconvex hull?

Exercise: Does Graham scan work for a simple polygon using theorder of its vertices?

Given a simple polygon with n vertices, we can compute its convexhull in O(n) time.

• 1972 Sklansky: Incorrect.• 1975 Shamos: Incorrect.• 1979 McCallum, Avis: Correct.• 1980 Kim: Incorrect.• 1982 Sklansky: Incorrect.• 1983 Lee: Correct.• 1983 Graham, Yao: Correct.• 1983 ElGindy, Avis, Toussaint: Correct.• 1983 Ghosh, Shyamasundar : Incorrect.

• 1984 Bhattacharya,ElGindy: Correct.

• 1985 Preparata, Shamos:Correct.

• 1985 Orlowski: Correct.• 1986 Shin, Woo: Correct.• 1986 Werner: Incorrect.• 1987 Melkman: Correct.• 1989 Chen: Incorrect.

(From http://http://cgm.cs.mcgill.ca/ athens/cs601/)

It is an on-line algorithm!Let’s see how it works:https://maxgoldste.in/melkman/

Exercise: Can we compute the convex hull of a monotone polygon ino(n) time? Design an optimal-time algorithm.

DCG19 Convex Hulls - Graz University of Technology · Irene Parada Discrete and Computational...

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