Attempts to count Ksets

1
Attempts to count K-sets
2
K-set - Definition

• Consider an n-point set X ? Rd and integer k.
• A k-point subset S ? X is a k-set of X if there
  exists an open half-space ? such that S X ? ?
  .

S
An example of a 4-set.
3
K-facet - definition

• A k-facet is an oriented (d-1)-dimensional
  simplex with vertices x1, x2,, xd ? X, s.t. the
  hyperplane h determined by x1, x2,, xd has
  exactly k points of X (strictly) on its positive
  side.

An example for 2-facet in a planar set.
4
The halving facets

• The halving facets are the (n-d)/2-facets, where
  (n-d) is even. (have the same number of points on
  both directions).
• In R2 k-edges and halving edges.

5
Notation

• KFAC(X,k) denotes the number of k-facets of X.
• KFACd(n,k) is the maximum of KFAC(X,k) over all
  n-point sets X ? Rd in general position.

6
Levels, K-sets, K-facets

• Let X ? Rd be a finite point set in general
  position.
• Let H D(x) x ? X be the collection of
  hyperplanes dual to X.
• We assume that each k-set S ? X is cut off by a
  non-vertical hyperplane hs that does not pass
  through any point of x.

7
K-set - Duality
K-sets correspond to cells of Arr(H) of levels k
and n-k.
3-set
hs
S
ysD(hs)
Every other point ys lying in that cell, implies
that S S.
8
K-facet - Duality
K-facets correspond to vertices of Arr(H) of
levels k or n-k-d.
3-facet
hs
ysD(hs)
ys is the intersection of two lines and has
exactly 3 lines below it.
9
Estimating k-sets by k-facets

• Arr(H) has at most O(nd-1) unbounded cells.
• All but at most O(nd-1) cells of level k have a
  topmost vertex v (the level v is between k-d1
  and k).
• Every vertex is the topmost vertex of at most one
  cell of level k.
• The number of k-sets of X is at most O(nd-1)
  ?j0d-1 KFAC (X,k-j)

10
Example points in the plane
11
An immediate upper bound (using level at most k)

• The k-level has certainly no more verticesthan
  all the levels 0 through k togetherKFACd(n,k)
  O(nd/2(k1)d/2)
• (Clarksons theorem on levels)

12
An immediate lower bound

• Consider a subset S ? H of n/k hyperplanes, s.t.
  the lower unbounded cell in Arr(S) is a convex
  polyhedron with ?((n/k) d/2) vertices.
• We replace each of the hyperplanes by k very
  close and parallel hyperplanes.
• Each vertex of level 0 in Arr(S) gives rise to
  ?(kd-1) vertices of level k in the Arr(H).
• KFACd(n,k) ?(nd/2(k1)d/2-1)

13
Halving facets versus k-facets

• NotationHFACd(n) ½ KFACd(n, (n-d)/2), where
  (n-d) even.
• Claim KFACd(n,k) ? 2?HFACd(2nd).
• For providing asymptotic bounds on KFACd(n,k),
  (where 0 ? k ? n-d), it suffices to estimate the
  number of halving facets.

14
Theorem

• Suppose that for some d and for all n, HFACd(n)
  can be bounded by O(nd-cd) , for some constant
  cd gt 0 . Then we have, for all 0? k ? (n-d)/2,
• KFACd(n,k) O(nd/2(k1)d/2 - cd)

15
Proof

• We use the probabilistic method of the cutting
  lemma.
• We set r n/k, p r/n 1/k, and let S?H be a
  random sample obtained by independent Bernulli
  trials (p).
• T(S) denotes the bottom-vertex triangulation of
  the bottom unbounded cell of Arr(s).

16
Proof - Continue
lines of s T(S) Level k of H
17
Proof - Continue

• T(S) satisfies the following conditions
• ???T(S) , D(?)? ad (ad is a constant depending
  on d).
• ???T(S) D(?) ? S and S?I(?) ?
• ????S ? HT(S) and any subset S ? H, s.t. D(?) ? S
  and S?I(?) ? , we have ??T(S) .
• ? S ? H T(S) O(Sd/2)

18
Proof - Continue

• For every t?0, the expected number of simplices
  with excess at least t in T(S) is bounded as
  followsET(S) ? t O(2-t rd/2) O(2-t
  (n/k)d/2)
• Vk the set of vertices of level k in
  Arr(H).Vk(S) the vertices in Vk that have
  level 0 with respect to Arr(S).

19
Proof - Continue

• Claim
• EVk(S) ? ¼ Vk
• Proof
• ? v?Vk, the probability that v?Vk(S) (i.e. that
  non of the at most k hyperplanes below v goes
  into S), is at least (1-p)k (1- 1/k)k? ¼

20
Proof - Continue

• Now we bound EVk(S) from above.
• For ??T(S) , we let H? be the set of all
  hyperplanes of H intersecting ? in its interior.
• All the vertices in Vk(S)?? have the same level
  in Arr(H?).
• By the assumption in the theorem we have
  Vk(S)?? O(H?d-cd) O((t? n/r)d-cd)
  O((t?k)d-cd) , where t? is the excess of ?.

21
Proof - Continue

• Taking the sum over all ??T(S) implies
  EVk(S) ? O(kd-cd) ? ???T(S) t? d-cd
• It can be shown that ???T(S) t? d-cd O((n/k)
  d/2)
• Vk ? 4EVk(S) ? O(nd/2 kd/2 - cd))

22
Lower Bounds

• We are going to construct n-point planar sets
  with a superlinear number of halving edges.
• We present the construction on the dual setting,
  namely, present an arrangement of n lines with
  many vertices at level (n-2)/2.

23
Lower Bound I ?(nlogn)

• By induction on m, construct a set Lm of 2m lines
  with at least fm(m1)2m-2 vertices of the middle
  level.
• For m1 we take two non-vertical intersecting
  lines.

24
Lower Bound I Continue

• For m ? 1, we select a subset M ? Lm of 2m-1
  lines.
• To each line l ? M we assign a vertex v(l) of the
  middle level lying on l, in a way that v(l) ?
  v(l) for l ? l.
• We replace each line l ? Lm by a pair of lines
  almost parallel to l .
• For l ? M , we let the two lines replacing l
  intersect at v(l).
• Each of the remaining lines is replaced by two
  almost parallel lines (with an intersection far
  away from all vertices of Arr(Lm) ).

25
Example
L2 4 f2 (21)22-2 3
L1 2 f1 (11)21-2 1
l1
l1
l2
v
u
l2
w
l3
l4
L1 ? M
26
Example - continue
l1
l1
L3 8 f3 (31)23-2 8
l2
l2
l3
l3
l4
l4
27
Analysis

• A middle-level vertex of the form v(l) yields 3
  vertices of the new middle level (level 2m1 in
  Arr(Lm1)).
• Each of the other middle-level vertices yields 2
  vertices of the new middle level.
• The number of middle-level vertices in
  Arr(Lm1) is at least 2fm2m-1 2(m1)2m-2
  2m-1 (m2) 2m-1 fm1

28
Lower Bound II A Better Construction

• We show a lower bound of This bound is smaller
  than n1? for every ? gt 0, but much larger than
  n(logn)c , for any constant c.
• The construction is again inductive.
• We let L0 consist of 2 non-vertical intersecting
  lines.
• After m steps, we have a set of lines Lm, s.t
  Arr(Lm) has many vertices in its middle level.

29
Lower Bound II Continue
l

• We replace every line l ? Lm by am parallel
  lines. For a vertex v of the middle level of
  Arr(Lm) , we get am vertices of the new middle
  level.
• We add two new lines ?v and µv to obtain 2am
  vertices in the new middle level.

l
µv
?v
30
Problem!

• The new lines ?v and µv mess up the levels of the
  vertices.
• If we ignore the problem, and let nm Lm, and
  fm be the number of vertices in the middle level
  of Arr(Lm), the construction gives nm1 am?nm
  2fm fm1 ? 2am?fm
• With a suitable choice of am, it leads to the
  claimed bound.

31
Lower bound II
l2
l3
v
u
l4
l1
32
The new lines ?v and µv mess up the levels of the
vertices
We assume that am 2
l2
l3
?v
µv
v
u
?u
µu
The vertex has 6 vertices below, instead of 5.
l4
l1
33
The repaired construction
µv

• We want the auxiliary lines ?v and µv to be
  nearly parallel to l, by letting the spacing in
  the bundle of l be much smaller than the spacing
  in the bundle of l.
• If the lines of Lm are l1,,lnm, then the
  vertical spacing in the bundle of li is set to
  ?i, where ? gt 0 and very small.

l
?v
l
34
The repaired construction - continue

• Let li ? Lm, and let di denote the number of
  indices jlti, s.t lj intersect li in a vertex of
  the middle level.
• We obtain am lines of the bundle of li , and 2di
  lines of the form ?v and µv (almost parallel to
  li), di of them go above the bundle and di
  below.(li is replicated (am2di) times).
• All lines must have the same multiplicities. Let
  D max idi , for each i, we add 2(D-di) more
  lines parallel to li, half below li and half
  above it.

35
Replicating li D times
u1, u2 are on l2 d2 2
l3
w is contributed 4 lines above and only 2 below
w
l1
v1 is on l3 d3 1
?u1
?u2
l2
µu1
?v1
µu2
µv1
36
Summery

• Suppose we have already constructed Lm. We denote
  by Vm the set of the middle-level vertices in
  Arr(Lm).
• The number of vertices of Vm lying on li?Lm (i
  1,,m) is no more than Dm.
• We let ? ?m be sufficiently small , and replace
  each li by am parallel lines spaced by ?i.
• For each v?Vm we add the lines ?v and µv .
• Finally, we add for each i, 2(Dm-di) lines
  parallel to li.

37
d30, d41
l2
µv
?v
l3
?u
l4
d10, d21
µu
l1
38
Analysis

• nm1 (am 2Dm)nm fm1 Vm1 2amfm
• Each line in the bundle of li has exactly di
  vertices of Vm1. The lines ?v get 2am vertices
  of Vm1, and the remaining lines get non.
• Dm1 max(Dm, 2am)am 4Dm (am is a free
  parameter).D0 1, Dm 8m, am 4?8m

39
Analysis - Continue

• mn 2 ? 6m ? 8 12(m-1) fm 8m ? 8
  12(m-1)
• fm/nm since log(fm/nm) log(½ (8/6)m)
  ?(m).

40
An Upper Bound

• Definition
• Let X? Rd be an n-point set in general position
  with (n-d) even. Let T?X be a (d-1)-point subset
• VT x?X\T T?x is a halving facet of X

41
VT in R2 and in R3
T t

• In R2 , T has a single point and VT are the other
  endpoints of the halving edges emanating from it.
• In R3, conv(T) is a segment.

t
t1
T t1,t2
t2
42
Halving-Facet Interleaving Lemma

• Let h be a hyperplane containing T and no point
  of X\T. Since X\T is odd, h cannot have an
  equal number of points from both of its sides.
• LemmaEvery hyperplane h as above almost
  halves the halving facets containing T. More
  precisely, if r is the number of points of VT in
  the smaller half-space of h, then the larger
  half-space contains exactly r1 points of VT.
• (Namely, VT?h VT/2 , VT?h- VT/2 )

43
Proof

• For dgt2, we project T and VT to a plane ?
  orthogonal to T.
• The projection of T is a single point and is
  denoted by t. The projection of VT is VT .
• The halving facets containing T are projected to
  segments emanating from t.
• The hyperplane h is projected to a line h .

44
T, VT and h projected on ?
bt
b
t
z
at
a
h
45
Proof - Continue

• The idea is to show that for each two
  consecutive points a and b, the angle opposite to
  atb contains a point of VT (such as z). The
  line at has (n-d)/2 points of X from both its
  sides. By rotating it around t, the point a
  enters to at (or at-) . When reaching b, at (or
  at-) has again (n-d)/2 points. Hence, there was a
  moment when the number of points went from
  (n-d)/2 1 to (n-d)/2 , this must have been the
  moment when we reach z.

46
Proof - Continue

• Now we show that there is only one such point z
  If there were two, their opposite wedge would
  contain another point.
• (and then a and b are not consecutive).
• When we start rotating h around T, the first
  point of VT encountered must be in the larger
  half-space. Hence, the larger half-space has one
  more point of VT than the smaller half-space.

47
Corollary (Lovasz Lemma)

• Let X?Rd be an n-point set in general position,
  and let l be a line that is not parallel to any
  of the halving facets of X. Then l intersects the
  relative interior of at most O(nd-1) halving
  facets of X.

48
Sweeping by l the halving facets
Each vertex (boundary of a halving facet)
functions as T.
All vertices adjacent to the current vertex
functions as VT.
49
Proof

• First, we can move l s.t. it intersect only the
  interior of the same halving facets as before.
• We start translate l from infinity in a suitable
  chosen direction, s.t. we never cross any
  (d-3)-dimensional flat determined by the points
  of X.(There are infinite possible direction for
  I, hence it is always possible finding a plane
  passing through l, that avoiding all (d-3)-
  dimensional flats defined by X).

50
Proof - Continue

• As we translate l, the number of halving facets
  intersected by l may change only as l crosses the
  boundary of a halving facet F, (i.e. a
  (d-2)-dimentional face of F).
• By the interleaving lemma, when crossing such a
  face F, the number of intersected halving facets
  changes by 1.
• In the entire translation of l, it crosses no
  more than O(nd-1) boundaries (since there are
  O(nd-1) simplices of dimension (d-2) with
  vertices as X.)

51
Theorem (Upper Bound)

• For each d?2, the maximum number of halving
  facets satisfies
• HFACd(n) O(n(d - 1/Sd-1)), where Sd-1 is an
  exponent for which the statement of the second
  selection lemma holds in dimension d-1.
• In particular, in R2 we obtain HAFC2(n) O(n3/2)

52
Proof for the Planar Case

• We project the points of X vertically on the
  x-axis, obtaining a set Y.
• The projections of the halving edges of X define
  a set of intervals S with endpoints in Y.
• By Lovasz Lemma, any point p is contained in the
  interior of at most O(n) intervals from S.
  (Otherwise, a vertical line through p would
  intersect too many halving edges.)

53
Proof Continue

• We mark each q point in Y (q will be set later).
  We divide S into two classes
• S1 - Intervals containing some marked point in
  their interior.
• S2 - Intervals lying in a gap between two marked
  points of S.
• The number of intervals from S1 is at most O(n)
  per marked point. O(n2/q) in
  total.

54
Proof - Continue

• The number of intervals from S2 is no more than
  per gap, at most in
  total.
• Balancing both bounds by setting q we get that
  the total number of halving edges is O(n3/2).

55
First Selection Lemma

• Let X be an n-point set in Rd. Then there exists
  a point a?Rd (not necessarily belonging to X)
  contained in at least cdX-simplices, where cdgt0
  is a constant depending only on the dimension
  d.(cd (d1)-(d1) ).

56
Second Selection Lemma

• Let X be an n-point set in Rd, and let F be a
  family of ? X-simplices, where ??(0,1 is a
  parameter. Then there exists a point contained
  in at least c?sd X-simplices of F,
  where cc(d)gt0 and sd are constants.(sd
  (4d1)d1 )

57
A Point in the Interior of Many X-simplices

• Lemma
• Let X ?Rd be a set of n?d1 points in general
  position, meaning that no d1 points of X lie on
  a common hyperplane. Let H be the set of the
  hyperplanes determined by the points of X. Then
  no point a?Rd is contained in more than dnd-1
  hyperplanes of H.Consequently, at most O(nd)
  X-simplices have a on their boundary.

58
Proof for an Arbitrary Dimension

• We project the points of X vertically on the
  coordinate hyperplane xd0, obtaining a set Y
  (lying in Rd-1).
• Each facet of X projects to a (d-1)-dimensional
  Y-simplex in Rd-1.
• Let F be the family of those Y-simplices. F?
  , 0???1
• By the second selection lemma, there is a point a
  contained in at least c?sd-1 simplices of F.

59
Proof - Continue

• Only at most O(nd-1) of these contain a in their
  boundary (so the remaining ones have a in their
  interior).
• By the Lovasz lemma applied on the vertical line
  in Rd passing through a, we get c?sd-1
  O(nd-1) ? (1/n)1/sd-1
  F ? O(nd - 1/sd-1)