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Chapter 9: Geometric Selection Theorems 11/01/2013

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1 Chapter 9: Geometric Selection Theorems 11/01/2013

2 9.1 A Point in Many Simplices: The 1st Selection Lemma
definition Consider n points in the plane in general position, and draw all the triangles with vertices at the given points. Then there exists a point of the plane common to at least of these triangles. Here is the optimal constant.

3 Definition: If is a finite set, an X-simplex is the convex hull of some (d+1)-tuple of points in X. Convention: X-simplices are in bijective correspondence with their vertex sets.

4 9.1.1 Theorem (1st Selection Lemma)
Let X be an n-point set in Then there exists a point contained in at least X-simplices, where is a constant depending only on the dimension d. For n very large, we may take 4

5 9.1.1 The 1st Proof: from Tverberg and colorful Carathẽodory
We may suppose that n is sufficiently large. ( by ) Put There exist r pairwise disjoints sets whose convex hull have a point in common: call this point Tverberg’s Theorem

6 Colorful Carathẽodory’s Theorem
Let be a set of (d+1) indices. We apply for the (d+1) “color” sets, which all contain in their convex hull This yields a rainbow X-simplex containing and having one vertex from each Colorful Carathẽodory’s Theorem

7 If are two (d+1)-tuples of indices, then
If are two (d+1)-tuples of indices, then Hence the number of X-simplices containing the point is: For n sufficiently large, say , this is at least

8 9.1.1 The 2nd Proof: from Fractional Helly
Let F denote the family of all X-simplices. Put We want to apply to F. A (d+1)-tuple of sets of F is good if its d+1 sets have a common point. It suffices to show that there are at least good (d+1)-tuple for some independent of n (since then the theorem provides a point common to at least members of F). Fractional Helly Theorem

9 Set and consider a t-point set
Set and consider a t-point set Using we find that Y can be partitioned into d+1 pairwise disjoint sets, of size d+1 each, whose convex hulls have a common point. Therefore, each t-point provides at least one good (d+1)-tuple of members of F. Moreover, the members of this good (d+1)-tuple are pairwise vertex-disjoint, and therefore the (d+1)-tuple uniquely determines Y. It follows that the number of good (d+1)-tuples is at least: Tverberg’s Theorem

10 9.1.2 A Point in the interior of many X-simplices. Lemma:
Let be a set of points in general position, and let H be the set of the hyperplanes determined by the points of X. Then no point is contained in more than hyperplanes of H. consequently, at most X-simplices have on their boundary. General places - add

11 9.1.2 Proof: For each d-tuple S whose hyperplane contains , we choose an inclusion-minimal set whose affine hull contains . We claim that if then either or and share at most points.

12 If and , , then the affine hulls of and are distinct, for otherwise, we would have k+1 points in a common (k-1)-flat, contradicting the general positions of X. But then the affine hulls intersect in the (k-2)-flat generated by and containing , and are not inclusion-minimal.

13 Therefore, the first k-1 points of determine the last one uniquely, and the number of distinct sets of the form of cardinality k is at most The number of hyperplanes determined by X and containing a given k-point set is at most and the lemma follows by summing over k.

14 9.2 The 2nd Selection Lemma Same as the previous section but instead of considering all X-simplices, we consider some of all. it turns out that still many of them must have a point in common.

15 9.2.1 Theorem (2nd Selection Lemma)
Let X be an n-point set in and let F be a family of X-simplices, where is a parameter. Then there exists a point contained in at least X-simplices of F, where and are constants (depending on d).

16 Definitions: Hypergraphs are a generalization of graphs where edges can have more than 2 points. A hypergraph is a pair , where is the vertex set and is a system of subsets of , the edge set. A k-uniform hypergraph has all edges of size k. A k-partite hypergraph is one where the vertex set can be partitioned into k subsets, such that each edge contains at most one point from each subset.

17 Proof: We can view F as a (d+1)-uniform hypergraph: we regard X as the vertex set and each X-simplex corresponds to an edge. First, let us concentrate on the simpler task of exhibiting at least one good (d+1)-tuple.

18 Hypergraphs with many edges need not contain complete hypergraphs, but they have to contain complete multipartite hypergraphs. Let denote the complete (d+1)-partite (d+1)-uniform hypergraph with t vertices in each of its d+1 vertex classes. Example for [only 3 edged are drawn as a sample]

19 If t is a constant and we have a (d+1)-uniform hypergraph on n vertices with sufficiently many edges, then it has to contain a copy of as a subhypergraph. In geometric language, given a family F of sufficiently many X-simplices, we can color different sets of t points in (d+1) colors in such a way that all the rainbow X-simplices on the (d+1)t colored points are present in F.

20 Colored Tverberg’s Theorem Factional Helly Theorem
In such situation, if t is sufficiently large constant, the with r=d+1 claims that we can find a (d+1)-tuple of vertex-disjoint rainbow X-simplices whose convex hull intersected. And so there is a good (d+1)-tuple. For the we need not only one but many good (d+1)-tuples. We use an appropriate stronger hypergraph result, saying that if a hypergraph has enough edges, then it contains many copies of : Colored Tverberg’s Theorem Factional Helly Theorem

21 9.2.2 Theorem (The Erdös-Simonovits Theorem)
Let d and t be positive numbers. Let H be a (d+1)-uniform hypergraph on n vertices and with edges, where for a certain sufficiently large constant C. Then H contains at least copies of , where is a constant.

22 …Proof The given family F contains copies of Each such copy contributes at least one good (d+1)-tuple of vertex-disjoint X-simplices of F. On the other hand, d+1 vertex disjoint X-simplices have together vertices and hence their vertex set can be extended to a vertex set of some (with t(d+1) vertices) in at most ways.

23 Factional Helly Theorem
is the maximum number of copies of that can give rise to the same good (d+1)-tuple. Hence there are at least good (d+1)-tuples of X-simplices of F. By at least X-simplices of F share a common point, with The best explicit value is Factional Helly Theorem

24 9.3 Order Types and the Same-Type Lemma - Definitions:
The are infinitely many 4-point sets in the plane in general positions, but there are only two “combinatorially distinct” types of such sets.

25 What is “combinatorially the same”
What is “combinatorially the same”? Let’s see an explanation for this notation for planar configuration in general position: let and be two sequences of points in , both in general positions. Then p and q have the same order type if for any indices we turn in the same direction when going from to via and then going from to via We say that the triples and have the same orientation.

26 The order type of a set. Let p and q be two sequences of points in
The order type of a set. Let p and q be two sequences of points in We require that every (d+1)-element subsequence of p have the same orientation as the corresponding subsequence of q. Then p and q have the same-order type. notion of orientation: if are vectors in Let matrix A be the matrix that has vectors as the columns. The orientation of is defined as the sign of det(A) (it can be +1, -1, or 0). For a (d+1)-tuple of points , we define the orientation of the d vectors

27 i. e. In the planar: let , and be 3 points in
i.e. In the planar: let , and be 3 points in The orientation of the 3 points is:

28 Back to the order type: let be a point sequence in
Back to the order type: let be a point sequence in The order type of p is defines as the mapping assigning to each (d+1)-tuple of indices, , the orientation of the (d+1)-tuple Thus, the order type of p can be described as a sequence of +1’s, -1’s and 0’s with terms.

29 Same-type transversals: let be an m-tuple of finite sets in
Same-type transversals: let be an m-tuple of finite sets in By a transversal of this m-tuple we mean any m-tuple such that for all i. we say that has same-type transversals if all of its transversals have the same order type. Example of 4 planar sets with same-type transversals

30 To see this  color each transversal of by its order type
To see this  color each transversal of by its order type. Since the number of possible order types of an m-point set in general position cannot be greater than , we have a coloring of the edges of the complete m-partite hypergaph on by r colors. By Erdös-Simonovits theorem (9.2.2), there are sets , not too small, such that all edges induces by have the same color, meaning – have same-type transversals.

31 9.3.1 Same-type Lemma (Theorem)
For any integers , there exists such that the following holds. Let be finite sets in such that is in general position*. Then there are such that the m-tuple has same-type transversals and for all This is shorthand for saying that for all and is in general position.

32 Proof: It sufficient to prove the same-type lemma for If is the current m-tuple of sets, we go through all (d+1)-tuple of indices, and we apply the same-type lemma to the (d+1)-tuple These sets are replaced by smaller sets such that this (d+1)-tuple has same-type transversals. After executing this for all (d+1)-tuples of indices, the resulting current m-tuple of sets has the same-type transversal. This method gives a smaller bound:

33 9.3.2 Lemma (handling the m=d+1 case):
Let be convex sets. The following two conditions are equivalent: i) there is no hyperplane simultaneously intersecting all of ii) for each nonempty index set , the sets and can be by a hyperplane. Moreover, if are finite sets such that the sets have property (i) (and (ii)), then has the same-type transversals. strictly separated

34 …Proof: To prove the same-type lemma for the case , it suffices to choose the sets in such a way that their convex hulls are separated in the sense of (ii) in Lemma this can be done by an iterative application of the Suppose that for some nonempty index set the sets and cannot be separated by a hyperplane. Lets assume that Let h be a hyperplane simultaneously bisecting whose existence is guaranteed by Let be a closed half-space bounded by h and containing at least half of the points of Ham-Sandwich Theorem Ham-Sandwich Theorem

35 For all we discard the points of not lying in , and for j we throw away the points of that lie in the interior of .

36 We claim that union of the resulting sets with indices in I is now strictly separated from the union of the remaining sets. If h contains no points of the sets, then it is a separating hyperplane. Otherwise, let the points contained in h be We have by the general position assumption. For each , choose a point very near to If lies in some with , then is chosen in the complement of . We let h’ be a hyparplane passing through and lying very close to h. Then h’ is the desired separating hyperplane, provided that the are sufficiently close to the corresponding

37 The size of a set is reduces from to at least
The size of a set is reduces from to at least We can continue with the other index sets in the same manner. After no more than halvings, we obtain sets satisfying the separation condition and thus having same-type transversals. The same-type lemma is proved. The lower bound for is doubly exponential, roughly

38 9.3.3 Theorem (Positive-Fraction Erdös-Szekeres Theorem):
For every integer there is a constant such that every sufficiently large finite set in general positions contains k disjoints subsets , of size at least each, such that each transversal of is in convex position.

39 Erdös-Szekeres theorem Erdös-Szekeres theorem
Proof: Let be the number as in the We partition X into n sets of almost equal size, and we apply the same-type lemma to them, obtaining sets , , with the same-type transversals. Let be a transversal of By the , there are such that are in convex positions. Then are as required in the theorem. Erdös-Szekeres theorem Erdös-Szekeres theorem

40 9.4 Hypergraph Regularity Lemma:
Let be a k-partite hypergraph whose vertex set is in the union of k pairwise disjoint n-element sets , and whose edges are k-tuples containing precisely one element from each For subsets , , let denote the number of edges of H contained in In this notation, the total number of edges of H is equal to Let denote the density of the graph induces by the .

41 9.4.1 Theorem (Weak regularity lemma for hypergraphs):
Let H be a k-partite hypergraph as defined before, and suppose that for some Let Suppose that n is sufficiently large in terms of k, and . Then there exist subsets of equal size , , such that: (i) (High density) , and (ii) (Edges on all large subsets) for any with ,

42 The following scheme illustrates the situation.

43 9.4.1 proof We look at a modified density parameter that slightly favors larger sets. Thus, we define the magical density : We choose , , as sets of equal size that have the maximum possible magical density We denote the common size by s. First we derive condition (i) in the theorem for this choice of the

44 We have and so , which verifies (i). Since obviously , we have
We have and so , which verifies (i). Since obviously , we have Combining with , we also obtain that Since is a large number by assumption, rounding it up to an integer doesn’t matter. We will assume is an integer, and let be a -element sets. We want to prove

45 We have We want to show that the negative terms are not too large, using the assumption that the magical density of is maximum. (#)

46 The problem is that maximize the magical density only among the sets of equal size, while we have sets of different sizes in the terms. To get back to equal size, we use the following observation. If, say, is a randomly chosen subset of of some given size r, we have For estimating the term , we use random subsets of size of respectively. Thus,

47 Now for any choice of , we have,
Therefore,

48 To estimate the term , we use random subsets and this time all of size
To estimate the term , we use random subsets and this time all of size A similar calculation as before yields

49 From we obtain that is at least multiplied by the factor
(#)

50 9.5 A Positive-Fraction Selection Lemma :
Here we discuss a stronger version of the first selection lemma. The theorem below shows that we can even get a large collection of simplices with a quite special structure. For example, in the plane, given n red points, n green points, and n blue points, we can select red, green, and blue points in such a way that all the red-green-blue triangles for the resulting sets have a point in common. Here is the d-dimensional generalization 

51 9.5.1 theorem (Positive-Fraction Selection Lemma):
For all natural numbers d, there exists with the following property. Let be finite sets of equal size, with in general positions. Then there is a point and subsets , with , such that the convex hull of every transversal of contains .

52 Proof: Let We may suppose that all the are large. Let be the set of all “rainbow” X-simplices, i.e., of all the transversals of , where the transversals are formally considered as sets for the moment. The size of is, for d fixed, at least a constant fraction of ( are of equal size). Therefore, by the second selection lemma there is a subset of at least X-simplices containing a common point , where

53 For the subsequent argument we need to apply Lemma 9. 1
For the subsequent argument we need to apply Lemma 9.1.2, which guarantees that lies on the boundary of at most of the X-simplices of So we let be the X-simplices containing in the interior, and for a sufficiently large n we still have We consider the (d+1)-partite hypergraph H with vertex set X and edge set We let , where is as in the same-type lemma, and we apply the weak regularity theorem (theorem ) to H. This yields sets , whose size is at least a fixed fraction of the size of , and such that any subsets of size at least induce an edge. Meaning: there is a rainbow X-simplices with vertices in the and containing .

54 The argument is finished by applying the same-type lemma with d+2 sets and We obtain sets and with same-type transversals, and with for Now either all transversals of contain the point in their convex hull or none does. But the latter possibility is excluded by the choice of the (by the weak regularity lemma).

55 ♦ THE END ♦

56 Notation– General Position:
For points in in general position, we assume that no unnecessary affine dependencies exist: no points lie in a common (k-2)-flat. i.e.: for lines in the plane in general position we postulate, that no 3 lines have a common point and no 2 are parallel. Back to

57 Tverberg’s Theorem: Let be points in , Then there is a partition of such that Back to (i) Back to (ii)

58 Colorful Carathẽodory’s Theorem:
Let be d+1 sets in Suppose that Then there are such that Back to 9.1.1

59 Fractional Helly Theorem:
For every there exists with the following property. Let be convex sets in , , and at least of the collection of sets of size d+1 have nonempty intersection, so there exists a point contained in at least sets. Back to 9.1.1 Back to (i) Back to (ii)

60 Colored Tverberg’s Theorem:
For every d and r there exists such that that for every set of cardinality (d+1)t , partitioned into t-point subsets , there exist r disjoint sets that are rainbow, meaning that for every i, j, and whose convex hulls all have a common point. Back to 9.2.1

61 Ham-Sandwich Theorem:
Every d finite sets in can be simultaneously bisected by hyperplane. A hyperplane h bisects a finite set A if each of the open half-spaces defined by h contains at most points of A. Back to 9.3.2

62 Separation Theorem: Let be convex sets with Then there exists 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. If C and D are closed and at least one of them is bounded, they be separated strictly; in such a way that Back to 9.3.2

63 Erdös-Szekeres Theorem:
For any natural number k there is a natural number such that any n-point set in the plane in general position contains a subset of k points in convex position (forming the vertices of a convex k-gon). Back to 9.3.3


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