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Seminar of computational geometry. Lecture #1 Convexity. Example of coordinate-dependence. Point p. What is the “sum” of these two positions ?. Point q. If you assume coordinates, …. p = (x 1 , y 1 ). The sum is (x 1 +x 2 , y 1 +y 2 ) Is it correct ? Is it geometrically meaningful ?.
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Seminar of computational geometry Lecture #1 Convexity
Example of coordinate-dependence Point p • What is the “sum” of these two positions ? Point q
If you assume coordinates, … p = (x1, y1) • The sum is (x1+x2, y1+y2) • Is it correct ? • Is it geometrically meaningful ? q = (x2, y2)
If you assume coordinates, … p = (x1, y1) • Vector sum • (x1, y1) and (x2, y2) are considered as vectors from the origin to p and q, respectively. (x1+x2, y1+y2) q = (x2, y2) Origin
If you select a different origin, … p = (x1, y1) • If you choose a different coordinate frame, you will get a different result (x1+x2, y1+y2) q = (x2, y2) Origin
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
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 q vector (q - p) Point p
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
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
Addition p + w u + v w v u p u + v is a vector p + w is a point u,v, w : vectors p, q : points
Subtraction p p - w p - q -w u - v u v q p u - v is a vector p - q is a vector p - w is a point u,v, w : vectors p, q : points
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
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.
Convexity p q p convex non-convex q A set S is convex if for any pair of points p,q S we have pq S.
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
Convex hulls Extreme point Int angle < pi p6 p9 p5 p12 p7 p4 p11 p1 p8 p2 p0 Extreme edge Supports the point set
(0, 1) (1, 1) (¼, ¼) (0, 0) (1, 0) Caratheodory's theorem
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=∅.
h C p q C Example for separation
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.
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.
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.
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.
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.
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
a1 C2 a3 a4 C1 a2 C4 C3 Example to Helly’s theorem
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
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.
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).
conv(X⋂η) η 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
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.
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.
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.