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Seminar of computational geometry

Lecture #1 Convexity

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Example of coordinate-dependence

What is the “sum” of these two positions ?

Point p

Point q

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If you assume coordinates… ,

The sum is (x1+x2, y1+y2) Is it correct ? Is it geometrically meaningful ?

p = (x1, y1)

q = (x2, y2)

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If you assume coordinates… ,

p = (x1, y1)

q = (x2, y2)

Origin

(x1+x2, y1+y2)

Vector sum (x1, y1) and (x2, y2) are considered as vectors from the

origin to p and q, respectively.

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If you select a different origin… ,

p = (x1, y1)

q = (x2, y2)

Origin

(x1+x2, y1+y2)

If you choose a different coordinate frame, you will get a different result

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Vector and Affine Spaces

Vector space Includes vectors and related operations No points

Affine space Superset of vector space Includes vectors, points, and related operations

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Points and Vectors

A point is a position specified with coordinate values. A vector is specified as the difference between two points. If an origin is specified, then a point can be represented by a vector

from the origin. But, a point is still not a vector in coordinate-free concepts.

Point p

Point qvector (q - p)

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Vector spaces

A vector space consists of Set of vectors, together with Two operations: addition of vectors and

multiplication of vectors by scalar numbers

A linear combination of vectors is also a vectorVV wvuwvu ,,

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Affine Spaces

An affine space consists of Set of points, an associated vector space, and Two operations: the difference between two

points and the addition of a vector to a point

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Addition

u

vu + v

p

p + w

u + v is a vector p + w is a point

w

u, v, w : vectorsp, q : points

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Subtraction

v

u - vu

q

p

u - v is a vector p - q is a vector

p - q

p

p - w

p - w is a point

-w

u, v, w : vectorsp, q : points

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Linear Combination

vvvvv

NN

N

iii cccc 1100

0

A linear space is spanned by a set of bases Any point in the space can be represented as

a linear combination of bases

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Affine Combination

)()( 010

0

11000

ppp

pppp

N

iii

N

ii

NN

N

iii

cc

cccc

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General position

"We assume that the points (lines, hyperplanes,. . . ) are in general position."

No "unlikely coincidences" happen in the considered configuration. No three randomly chosen points are collinear. Points in lR^d in general position, we assume similarly that no

unnecessary affine dependencies exist: No k<=d+1 points lie in a common (k-2)-flat.

For lines in the plane in general position, we postulate that no 3 lines have a common point and no 2 are parallel.

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Convexity

A set S is convex if for any pair of points p,q S we have pq S.

p

q

non-convex

q

p

convex

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Convex Hulls : Equivalent definitions

The intersection of all covex sets that contains P The intersection of all halfspaces that contains P. The union of all triangles determined by points in P. All convex combinations of points in P.

P here is a set of input points

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Convex hulls

p0

p1p2

p4

p5

p6

p7

p8

p9

p11

p12

Extreme pointInt angle < pi

Extreme edgeSupports the point set

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Caratheodory's theorem

Let X . Then each point of ( ) is a convex combination

of at most 1 points of .

d conv X

d X

)0 ,1( (1, 1)

(1, 0))0 ,0(

(¼ ,¼)

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Separation theorem

Let C, D⊆ℝd be convex sets with C∩D=∅. Then there exist a hyperplane h such that C lies in one of the closed half-spaces determined by h, and D lies in the opposite closed half-space.In other words, there exists a unit vector a∈ℝd and a number b∈ℝ such that for all x∈C we have <a, x>≥b, and for all x∈D we have <a, x>≤b.

If C and D are closed and at least one of them is bounded, they can be separated strictly; in such a way that C∩h=D∩h=∅.

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Example for separation

C

C

h

q

p

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Sketch of proof

We will assume that C and D are compact (i.e., closed and bounded). The cartesian product C x D∈ℝ2d is a compact set too.

Let us consider the function f : (x, y)→||x-y|| , when (x, y) ∈ C x D. f attains its minimum, so there exist two points a∈C and b∈D such that ||a-b|| is the possible minimum.

The hyperplane h perpendicular to the segment ab and passing through its midpoint will be the one that we are searching for.

From elementary geometric reasoning, it is easily seen that h indeed separates the sets C and D.

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Farkas lemma

For every d x n real matrix A, exactly one of the following cases occurs: There exists an x∈ℝn such that Ax=0 and x>0 There exists a y∈ℝd such that yT A<0. Thus, if we multiply

j-th equation in the system Ax=0 by yi and add these equations together, we obtain an equation that obviously has no nontrivial nonnegative solution, since all the coefficients on the left-hand sides are strictly negative, while the right-hand side is 0.

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Proof of Farkas lemma

Another version of the separation theorem. V⊂ℝd be the set of n points given bye the column

vectors of the matrix A. Two cases:

0∈conv(V)0 is a convex combination of the points of V.The coefficients of this convex combination determine a nontrivial nonnegative solution to Ax=0

0∉conv(V)Exist hyperplane strictly separation V from 0, i.e., a unit vector y∈ℝn such that <y, v> < <y, 0> = 0 for each v∈V.

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Radon’s lemma

Let A be a set of d+2 points in ℝd. Then there exist two disjoint subsets A1, A2⊂A such that conv(A1) ∩ conv(A2)≠∅

A point x ∈ conv(A1) ∩ conv(A2) is called a Radon point of A.

(A1, A2) is called Radon partition of A.

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Helly’s theorem

Let C1, C2, …, Cn be convex sets in ℝd, n≥d+1. Suppose that the intersection of every d+1 of these sets is nonempty. Then the intersection of all the Ci is nonempty.

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Proof of Helly’s theorem Using Radon’s lemma. For a fixed d, we proceed by induction on n. The case n=d+1 is clear. So we suppose that n ≥d+2 and the statement of Helly’s theorem holds for smaller n. n=d+2 is crucial case; the result for larger n follows by a simple reduction. Suppose C1, C2, …, Cn satisfying the assumption. If we leave out any one of these sets, the remaining sets have a nonempty

intersection by the inductive assumption. Fix a point ai ∈ ⋂i≠jCj and consider the points a1,a2, …, ad+2

By Radon’s lemma, there exist disjoint index sets I1, I2 ⊂{1, 2, …, d+2} such that

1 2({ : }) ({ : })i iconv a i I conv a i I

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Example to Helly’s theorem

a2

a1

a4a3

C2

C1

C3 C4

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Continue proof of Helly’s theorem

Consider i∈{1, 2, …, n}, then i∉I1 or i∉I2 If i∉I1 then each aj with j ∈ I1 lies in Ci

and so x∈conv({aj : j ∈ I1})⊆Ci

If i∉I2 then each aj with j ∈ I2 lies in Ci

and so x∈conv({aj : j ∈ I2})⊆Ci

Therefore x ∈ ∩i=1nCi

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Infinite version of Helly’s theorem

Let C be an arbitrary infinite family of compact convex sets in ℝd such that any d+1 of the sets have a nonempty intersection. Then all the sets of C have a nonempty intersection.

Proof

Any finite subfamily of C has a nonempty intersection. By a basic property of compactness, if we have an arbitrary family of compact sets such that each of it’s finite subfamilies has a nonempty intersection, then the entire family has a nonempty intersection.

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Centerpoint

Definition 1: Let X be an n-point set in ℝd. A point x ∈ ℝd is called a centerpoint of X if each closed half-space containing x contains at least n/(d+1) points of X.

Definition 2: x is a centerpoint of X if and only if it lies in each open half space η such that |X⋂η|>dn/(d+1).

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Centerpoint theorem

Each finite point set in ℝd has at least one centerpoint. Proof

Use Helly’s theorem to conclude that all these open half-spaces intersect.

But we have infinitely many half-spaces η which are unbound and open.

Consider the compact convex set conv(X⋂η)⊂η

ηconv(X⋂η)

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Centerpoint theorem(2)

Run η through all open-spaces with |X⋂η|>dn/(d+1) We obtain a family C of compact convex sets. Each Ci contains more than dn/(d+1) points of

X. Intersection of any d+1 Ci contains at least one

point of X. The family C consists of finitely many distinct

sets.(since X has finitely many distinct subsets).

By Helly’s theorem C⋂ ≠∅, then each point in this intersection is a centerpoint.

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Ham-sandwich theorem

Every d finite sets in ℝd can be simultaneously bisected by a hyperplane. A hyperplane h bisects a finite set A if each of the open half-spaces defined by h contains at most |⌊A|/2 points of A.⌋

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Center transversal theorem

Let 1≤k≤d and let A1, A2, …, Ak be finite point sets in ℝd. Then there exists a (k-1)-flat f such that for every hyperplane h containing f, both the closed half-spaces defined by h contain at least |Ai|/(d-k+2) points

of Ai i=1, 2, …, k. For k=d it’s ham-sandwich theorem. For k=1 it’s the centerpoint theorem.