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Transcript of 9.1 A Point in Many Simplices: The 1 st Selection Lemma Consider n points in the plane in general...
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Chapter 9:
Geometric Selection Theorems
11/01/2013
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9.1 A Point in Many Simplices: The 1st Selection Lemma
• 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
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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.
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• 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
9.1.1 Theorem (1st Selection Lemma)
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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
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• 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
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• 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
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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
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• 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
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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.
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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.
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• 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.
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• 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.
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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.
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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).
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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.16
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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.
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• 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]
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• 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.
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• 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
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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.
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9.2.1… 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.
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• 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
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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.
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• 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.
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• 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 .
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• i.e. In the planar: let ,
and be 3 points in . The
orientation of the 3 points is:
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• 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.
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• 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
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• 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.
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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.
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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:
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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. 33
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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
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• 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
.
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• 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 .
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• 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 .
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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.
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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.
39
Erdös-Szekeres theorem
Erdös-Szekeres theorem
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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 .40
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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 , .
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The following scheme illustrates the situation.
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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 . 43
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• 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 .
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• We have
• We want to show that the negative terms are not too large,
using the assumption that the magical density of
is maximum.
)#(
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• 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,
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• Now for any choice of , we have,
• Therefore,
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• To estimate the term , we
use random subsets and
this time all of size . A similar calculation as before yields
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• From we obtain that is at least
multiplied by the factor
)#(
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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
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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
.
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9.5.1 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
.
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• 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 9.4.1 ) 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 .53
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• 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).
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♦THE END ♦
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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 9.1.1
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Tverberg’s Theorem:
Back to 9.1.1 (i)
• Let be points in , .
Then there is a partition of such
that
.
Back to 9.1.1 (ii)
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Colorful Carathẽodory’s Theorem :
• Let be d+1 sets in . Suppose that
.
Then there are such that
. Back to 9.1.1
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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 9.2.1 (i)
Back to 9.2.1 (ii)
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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
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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
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• 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
Separation Theorem :
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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).
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Back to 9.3.3