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Chapter 9: Geometric Selection Theorems 11/01/2013. 9.1 A Point in Many Simplices: The 1 st Selection Lemma. definition.
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Chapter 9: Geometric Selection Theorems 11/01/2013
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.
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.
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 nvery large, we may take 4
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
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 .
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
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
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
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.
9.1.2 Proof: • For each d-tupleSwhose hyperplane contains , we choose an inclusion-minimal set whose affine hull contains . • We claim that if then either or and share at most points.
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.
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.
9.2 The 2nd Selection Lemma • Same as the previous section but instead of considering allX-simplices, we consider some of all. • it turns out that still many of them must have a point in common.
9.2.1 Theorem (2nd Selection Lemma) • Let Xbe an n-point set in and let Fbe 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).
Definitions: • Hypergraphsare 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 hypergraphhas all edges of size k. • A k-partite hypergraphis one where the vertex set can be partitioned into k subsets, such that each edge contains at most one point from each subset.
9.2.1 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.
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+1vertex classes. • Example for [only 3 edged are drawn as a sample]
If tis 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.
Colored Tverberg’s Theorem • In such situation, if t is sufficiently large constant, the with r=d+1claims 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 : Factional Helly Theorem
9.2.2 Theorem (The Erdös-Simonovits Theorem) • Let d and t be positive numbers. Let Hbe a (d+1)-uniform hypergraph on n vertices and with edges, where for a certain sufficiently large constant C. Then Hcontains at least copies of , where is a constant.
9.2.1 …Proof • The given family Fcontains 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.
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 Fshare a common point, with . • The best explicit value is Factional Helly Theorem
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.
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.
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 .
i.e. In the planar: let , and be 3 points in . The orientation of the 3 points is:
Back to the order type: let be a point sequence in . The order type of pis 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.
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
To see this color each transversal ofby 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.
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.
9.3.1 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:
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
9.3.1 …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 9.3.2. 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
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 .
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 .
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 .
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.
9.3.3 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
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 .
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 , .
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 .
We haveand 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 .
We have • We want to show that the negative terms are not too large, using the assumption that the magical density of is maximum. (#)
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,
Now for any choice of , we have, • Therefore,
To estimate the term , we use random subsets and this time all of size . A similar calculation as before yields
(#) • From we obtain that is at least multiplied by the factor
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 nred points, n green points, and nbluepoints, 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