Lenses and How to Get Rid of Them

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Lenses and How to Get Rid of Them

• By Guy Harari

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Szekelys method reminder

• For m points and n lines we create a graph with
  the points as its vertices and edges connect
  vertices whose points are consecutive along a
  line.
• If each line contains at least one point then
  EI(P,L)-n (1)
• If Elt4V gt I(P,L)lt4mn (from (1))
• Else (crossing lemma)
• The graph should be (almost) simple (crossing
  lemma).

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plan

• We will improve the bound of number of subarcs
  into which we cut our arrangement from
  to .
• We will bound v(C) the size of maximum family
  of non-overlapping lenses and use which
  holds for pseudo-parabolas and pseudo-circles.

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Pairwise intersecting p-circles

• First step Each pair of n circles intersects
  (cross twice or tangent once)
• Then the number of empty lenses is O(n).
• Proof
• 1. Reduce to O(1) instances of a problem of
  counting number of empty lenses in a simpler
  problem (will not be shown).
• 2. Reduce each such problem to the problem of
  counting tangencies of pairwise intersecting
  pseudo parabolas.
• 3. Verify number of tangencies to be O(n) (will
  be shown later for similar problem).

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• The simpler problem
• Theorem The number of empty lenses in an
  arrangement of pairwise intersecting
  p-circles, no pair of which are tangent and no
  three concurrent, so that all their interiors are
  star shaped with respect to a point o, is at most
  2n-3.

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Maximum number of non-overlapping lenses

• For n pairwise intersecing pseudo-parabolas or
  pseudo-circles the size of 1-matching or maximal
  pairwise non-overlapping lenses is .
  
• Proof for pseudo-parabolas
• Assumption of general position No more than 2
  curves intersect at the same point.
• Lemma 1.1 If all lenses have pairwise disjoint
  interiors then there are O(n) pairwise
  non-overlapping lenses.

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• Proof of lemma 1.1
• s(l) number of edges of A(C) that lie in the
  interior of lens l.
• By induction on If S0 we have a theorem
  (empty lenses).
• Else for a lens L, we look at a curve C that
  intersects it. Up to symmetry There are two
  possibilities

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pairwise disjoint interiors

• 1. C intersects only one of the curves of L
  twice. We replace the lenses and thus reduce S
  and use induction hypothesis. We can remove all
  lenses of that type.
• 2. C intersects both lens arcs once or twice. We
  create a small lens inside the original one and
  again use induction hypothesis.

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• 1 2

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• Lemma 1.2 (non-overlapping lenses)
• X the number of crossing pairs of lenses.
• Y the number of nested pairs of lenses. If L is
  the set of all lenses then LO(nXY).
• Proof
• Removing a lens from crossing/nested pair
  decreases L by 1 and XY by at least 1 so the
  equality holds after removing.
• After remove them all we use Lemma 1.1.

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• Lemma 1.3
• Proof For each pair of lenses we count one point
  of intersection of their arcs.
• Non-overlapping gt O(1) counting
• We have intersections
• Hence, we have .

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• Lemma 1.4 If each lens from L (set of lenses) is
  crossed by at most k curves (kltn) then YO(kL)
• Proof for a fixed lens l, let l be a lens that
  fully contains l. We draw a vertical ray from
  the left vertex of l up to ls arc. This ray
  will cross up to k other curves before hitting
  that arc. Thus, only O(k) lenses can contain l
  so we have up to O(kL) lenses.
• (here we used general position assumption)

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• Theorem 1 For a pairwise intersecting collection
  of pseudo-parabolas C, the maximum size of a
  family of pairwise non-overlapping lenses L in
  A(C) is .
• Proof For some integer k, We first remove all
  lenses intersecting more than k curves (heavy).
  Each such lens contains at least k pairs of
  curves. From non-overlapping and O(n(n-1))
  intersecting points we get a bound for heavy
  lenses -

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• We now take randomly curves from C, each one with
  probability p.
• E(R)np E( lenses included)
  where L is the set of all lenses after
  removal.
• From lemmas 1.1 to 1.3
• gt
• We get

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Generalization

• Second step we now remove our assumption of
  pairwise intersection.
• Theorem 2 Let be a family of n
  pseudo-parabolas.
• Assumption of general position No more than 2
  curves intersect at the same point.

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• Proof We generate a graph by
  connecting intersections of curves and vertical
  line l iff their curves form a non-overlapping
  lens.

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• Partition
• Any vertex of is above any vertex of
  .
• G bipartite subgraph
• We have
• Any pair of edges from a circle of size 4 in G
  intersects even number of times (to be shown).
• By a result of Pinchasi and Radoicic 2003 for
  such graphs we conclude .

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• Define

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Even number of intersections

• Claim 2.1 Each pair of adjacent edges in G
  cross an even number of times.
• Claim 2.2 If (f,p,g,q) is a cycle of length four
  in G then curves f,p,g,q are pairwise
  intersecting.
• Claim 2.3 If (f,p,g,q) is a cycle of length four
  in G then the four lenses of this cycle are
  empty of other curves corresponding to the cycle.
• (some) of these claims will be proved.

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proof of claim 2.1
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• Proof of theorem
• Let (f,p,g,q) be a cycle in G.
• Claim 2.1 gt all adjacent pairs from the
  cycle intersect an even number of times.
• Claims 2.2, 2.3 gt f,p,g,q are pairwise
  intersecting but dont intersect with lenses
  (f,p) , (g,q) .
• We can think of tangency of (f,p) and (g,q).
• gt (f,p) , (g,q) intersect an even number of
  times. (to be proved)
• Similarly for (f,q) , (g,p).
• Hence we can use the result

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• Lemma 2.4 Let be four pairwise
  intersecting curves with tangencies
  and then exit the wedge of from
  the same arc.
• Proof

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• Lemma 2.5Let be four pairwise
  intersecting curves with tangencies
  and then it is impossible that the
  first exit point of from the wedge of
  and the first exit point of from the
  wedge of both lie on the bottom sides
  of respective wedges, or both lie on the top
  sides.
• Proof will not be shown

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• Then we get
• allowed forbidden

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The case of pseudo-circles

• The pseudo-circles are x-monotone.
• We cut n pseudo-circles to 2n pseudo-parabolas.
• No more than 4n non-overlapping lenses contain
  x-extreme points of the pseudo-circles in their
  boundary (non-overlapping and general position).
• Other lenses are fully contained in trimmed arcs
  and thus remain lenses.
• We thus have the bound for pseudo-parabolas.

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Bichromatic lenses

• Theorem 3.1 for n p-parabolas where and
  any p-parabolas of A intersects every p-parabola
  of B twice then the number of empty bichromatic
  lenses in is O(n).
• Proof we assume the arc from A lies above that
  of B.

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• For two disjoint curves a and a from A where a
  lies below a.
• gt no empty bichromatic lens between a and any b.
• gt so we can remove a.
• Similarly for b and b.
• gt so we can remove b.
• Hence, we remove them until all of them
  intersect.
• For that case we have a linear bound.

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• Claim 3.2 in the same configuration, if L is a
  family of pairwise non-overlapping bichromatic
  lenses with pairwise disjoint interiors then
  LO(n).
• Proof again we eliminate arcs as before.

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• Claim 3.3 If each pair of p-parabolas from
  disjoint A,B intersects twice then we can cut
  to arcs s.t any
  bichromatic pair of arcs intersect at most once.
• Proof replace lemma 1.1 with claim 3.2.
• Note in order that we should verify that
  needed properties of A,B hold for any random
  sample from it.

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• Theorem 3.4 For two families of circles A and B
  in the plane such that each circle in A
  intersects each circle in B and there is an
  interior point for all circles of A then the
  number of bichromatic empty lenses is .
• Proof will not be shown

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• Theorem for set of n circles with arbitrary
  radii C, for any .
• Proof

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Mapping circles

• Two circles intersect if
• We map

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• For circle c we call its mapped point and
  its mapped plane .
• Circles and intersect if lies
  between the following surfaces of
• Intersection condition
• Or iff the point of one of them is inside the
  wedge created by the surfaces of the other one.

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• L non-overlapping lenses.
• We partition C to A and B of equal size.
• We have bichromatic lenses (average).
• We map A to points ( ).
• We map B to surfaces ( ).

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• Theorem A set of n d-dimensional points can be
  partitioned into subsets (for
  some ) so that
• 1.
• 2. is contained in a simplex of dimension
  1,...,d.
• 3. No hyperplane crosses (intersects but doesnt
  not contain) more than simpleces
  (For the 4D case we use a bound of .For the
  case in the next section we will use ).

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• We partition A to so that
• 1.
• 2. Each hyperplane crosses no more than
  simpleces (d4).
• For some , let be the set of wedge pairs
  fully contain and is all
  corresponding circles.
• Each circle of intersect each circle of
  by construction.

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• c the circle of with minimal radius r
  (minimal radius of all circles in ).
• We place O(1) points in a concentric circle to c
  of radius 3r so that any circle from
  contains at least one point (we place p points).

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• We partition to .
• (superscript indicates common contained point).
• We take and .
• We count bichromatic lenses from A and B.

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• Light lenses
• - number of bichromatic light lenses (at
  most k intersecting curves).
• For a sample of n/k circles from we
  get and from theorem 3.4

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• Heavy lenses
• lens with more than k intersecting curves.
• heavy lenses

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• Number of bichromatic lenses
• gt
• gt
• gt

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• Lenses formed by a circle from and a circle
  c from B such that at least one wedge of c
  intersects .
• - the set of these circles from B.

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• Partition so that
• Each new subset is of size at most
• The number of sets is doubled.
• the number of new pairs is still .
• Apply next step to .
• -gt points in
• -gt wedges

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• We have

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• Theorem- for n circles and m points
• Where X is maximal number of intersections
  between two circles.
• For
• gt
• gt second term dominates (in I expression).

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• Proof
• We use the following lemma
• Lemma
• A set of n circles with at most X intersecting
  pairs we can cut it into
• arcs for any so that any two arcs
  intersect at most once.
• Will not be proved.

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• We cut all circles to arcs C.
• We draw a graph G as in Szekelys method.
• Arcs with one point contribute O(C).
• Circles with less than 3 points contribute at
  most 2n.
• From crossing lemma

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Small values of m

• We map
• Circle to
• Point to
• point p on circle c iff plane p contains c.
• P - new planes. C - new circles.
• Claim no three planes in P cross a common curve.

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• Partition C to where .
• Fix
• - corresponding circles
• - points which their dual plane crosses
  
• Choose incident of and
• p crosses or contains it.

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• For dimension d of
• d2 at most one plane contains .
• d1 at most two planes contain .
• At most 2n degenerate case incidences.
• Hence,

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• Hence, for
• Take
• for

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• for
• For we have .
• We get