# 1 Seminar of computational geometry Lecture #1 Convexity.

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1 Seminar of computational geometry Lecture #1 Convexity

2 Example of coordinate-dependence What is the “sum” of these two positions ? Point p Point q

3 If you assume coordinates, … The sum is (x 1 +x 2, y 1 +y 2 ) Is it correct ? Is it geometrically meaningful ? p = (x 1, y 1 ) q = (x 2, y 2 )

4 If you assume coordinates, … p = (x 1, y 1 ) q = (x 2, y 2 ) Origin (x 1 +x 2, y 1 +y 2 ) Vector sum (x 1, y 1 ) and (x 2, y 2 ) are considered as vectors from the origin to p and q, respectively.

5 If you select a different origin, … p = (x 1, y 1 ) q = (x 2, y 2 ) Origin (x 1 +x 2, y 1 +y 2 ) If you choose a different coordinate frame, you will get a different result

6 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

7 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 q vector ( q - p )

8 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 vector

9 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

10 Addition u v u + v p p + w u + v is a vectorp + w is a point w u, v, w : vectors p, q : points

11 Subtraction v u - v u q p u - v is a vectorp - q is a vector p - q p p - w p - w is a point -w u, v, w : vectors p, q : points

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

13 Affine Combination

14 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.

15 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

16 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

17 Convex hulls p0p0 p1p1 p2p2 p4p4 p5p5 p6p6 p7p7 p8p8 p9p9 p 11 p 12 Extreme point Int angle < pi Extreme edge Supports the point set

18 Caratheodory's theorem (0, 1)(1, 1) (1, 0)(0, 0) (¼, ¼)

19 Separation theorem Let C, D⊆ℝ d 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∈ℝ d and a number b∈ℝ such that for all x∈C we have ≥b, and for all x∈D we have ≤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=∅.

20 Example for separation C C h q p

21 Sketch of proof We will assume that C and D are compact (i.e., closed and bounded). The cartesian product C x D ∈ℝ 2d 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.

22 Farkas lemma For every d x n real matrix A, exactly one of the following cases occurs: There exists an x ∈ℝ n such that Ax=0 and x >0 There exists a y ∈ℝ d such that y T A<0. Thus, if we multiply j-th equation in the system Ax=0 by y i 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.

23 Proof of Farkas lemma Another version of the separation theorem. V ⊂ℝ d 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∈ℝ n such that = 0 for each v∈V.

24 Radon ’ s lemma Let A be a set of d+2 points in ℝ d. Then there exist two disjoint subsets A 1, A 2 ⊂ A such that conv(A 1 ) ∩ conv(A 2 )≠∅ A point x ∈ conv(A 1 ) ∩ conv(A 2 ) is called a Radon point of A. (A 1, A 2 ) is called Radon partition of A.

25 Helly ’ s theorem Let C 1, C 2, …, C n be convex sets in ℝ d, n≥d+1. Suppose that the intersection of every d+1 of these sets is nonempty. Then the intersection of all the C i is nonempty.

26 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 C 1, C 2, …, C n 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 a i ∈ ⋂ i≠j C j and consider the points a 1,a 2, …, a d+2 By Radon’s lemma, there exist disjoint index sets I 1, I 2 ⊂{1, 2, …, d+2} such that

27 Example to Helly ’ s theorem a2a2 a1a1 a4a4 a3a3 C2C2 C1C1 C3C3 C4C4

28 Continue proof of Helly ’ s theorem Consider i ∈{1, 2, …, n}, then i∉I 1 or i∉I 2 If i∉I 1 then each a j with j ∈ I 1 lies in C i and so x ∈conv({a j : j ∈ I 1 })⊆C i If i∉I 2 then each a j with j ∈ I 2 lies in C i and so x ∈conv({a j : j ∈ I 2 })⊆C i Therefore x ∈ ∩ i=1 n C i

29 Infinite version of Helly ’ s theorem Let C be an arbitrary infinite family of compact convex sets in ℝ d 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.

30 Centerpoint Definition 1: Let X be an n-point set in ℝ d. A point x ∈ ℝ d 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).

31 Centerpoint theorem Each finite point set in ℝ d 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 η conv(X ⋂ η)

32 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 C i contains more than dn/(d+1) points of X. Intersection of any d+1 C i 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.

33 Ham-sandwich theorem Every d finite sets in ℝ d 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.

34 Center transversal theorem Let 1 ≤k≤d and let A 1, A 2, …, A k be finite point sets in ℝ d. 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 | A i |/(d-k+2) points of A i i=1, 2, …, k. For k=d it’s ham-sandwich theorem. For k=1 it’s the centerpoint theorem.