Incidences and Many Faces via cuttings

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Incidences and Many Faces via cuttings

• Sivanne Goldfarb
• 5.6.07

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Outline

• The Many Faces Problem
• Point Line incidences
• Point Circle incidences

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The Many Faces Problem

• Input
• A set of m points
• A set L of n lines
• Notations
• A(L) the arrangement of the lines of L.
  Decomposition of the plane into cells.
• Combinatorial complexity of a single face The
  number of edges belonging to its boundary.
• - The combined combinatorial
  complexity of faces which contain points from P.

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The Many Faces Problem

• Assumption
• Each lies in unique cell.
• Problem
• Establish an upper bound for

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The Many Faces Problem

• We will show that
• The proof consists of 4 steps.
• The idea of the proof
• Partition the plane into sub regions, apply a
  weaker bound in each sub region and sum up the
  bounds.
• We will use the weaker upper bound derived from
  the forbidden complete graphs results

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Step 1 Construct a 1/r cutting

• Phase 1
• Choose a subset of size r.
• Each cell is decomposed into trapezoids by
  drawing a maximal vertical line segment through
  each vertex of A(R).
• We stop drawing a vertical line when it reaches a
  new line from R.

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Step 1 Construct a 1/r cutting

• Phase 2
• A trapezoid is called heavy if more than
  lines intersect it.
• Handle heavy trapezoids by further cutting them
  as was shown in the last lecture.
• As proven before, we have constructed a 1/r -
  cutting which consists of
• K O(r2) trapezoids. Each trapezoid is crossed
  by at most lines.

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Step 1 Construct a 1/r cutting

• Notations
• R for simplicity, the lines chosen for the
  cutting in both phases. Notice, R isnt of size r
  any more.
• For each trapezoid Di , ,
  . Pimi
• Li - a subset of lines that have a non empty
  intersection with Di .Lini
• Note that

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Step 2 Divide and conquer

• Using the 1\r - cutting of the arrangement A(L),
  we solve a sub problem in each trapezoid.
• In each Di, count the edges of A(Li) bounding the
  mi cells containing points of Pi.
• Summing the weaker bound in all trapezoids gives
  .

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Step 2 Divide and conquer

• What happens if the cell which contains pj
  intersects the boundary
• of Di?
• Solution count all edges of this cell.

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Step 3 Complexity of zones

• Inner zone of Di in A(Li ) the collection of
  for all cells c in A(Li ) which intersect
  the boundary of Di.

• Complexity of inner zone the total number of
  edges bounding its cells.
• The complexity of the inner zone of Di is O(ni)
  we will not prove this.
• Complexity of all inner zones

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Step 4 wrapping it up

• Summing the weaker bound in each trapezoid and
  taking inner zones into consideration gives
• Choose r to be , so it
  balances the two terms above, and we have
  - this is meaningful if
• According to our weaker upper bound, if
  then K(P, L) O(n).
• We recall that in the past we proved that
• Finally we get

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Outline

• The Many faces Problem
• Point Line incidences
• Point Circle incidences

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Point Line Incidences

• Let P be a set of m points
• Let L be a set of n lines
• We will show that
• As in the Many Faces problem, we will use the
  weaker bound derived from the forbidden complete
  graphs results
• The idea of the proof (again)
• Partition the plane into sub regions, apply the
  weaker bound in each sub region and sum up the
  bounds.

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Point Line Incidences

• We construct a 1/r - cutting as in the Many Faces
  proof.
• Decompose the plane into vertical
  trapezoids with pairwise disjoint interiors.
• Each trapezoid is crossed by at most
  lines of L.
• Notations
• For each trapezoid ,
• Li - subset of lines that have a non empty
  intersection with .
• R as in the Many Faces proof.
• For each trapezoid we get

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Point Line Incidences

• We recall that (ignoring
  points on the boundary of trapezoids)
• and we get
• We still need to consider points that lie on the
  boundary of the trapezoids!

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Point Line Incidences

• Let p be a point which lies on an edge of the
  cutting (but not on a vertex of
• the cutting). Let be a line
  which passes through p.

p lies on the boundary of at most two
trapezoids.
l crosses both trapezoids.
We assign p to one of the trapezoids and its
incidences with lines from L\R will be
counted within the sub-problem associated with
that trapezoid.
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Point Line Incidences

• Let p be a point as in the previous case.

p is incident to at most one line in R
There at most O(m) incidences of this kind
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Point Line Incidences

• 3. Let p be a point which is a vertex of the
  cutting and l a line incident to p.

l either crosses or bounds some adjacent
trapezoid Di
l can cross the boundary of a trapezoid in at
most two points.
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Point Line Incidences
We charge the incident (p, l) to the pair (l, Di
)
Recall that each trapezoid is crossed by at most
lines and that
The number of incidences involving vertices of
the cutting is at most
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Point Line Incidences

• We have shown that
• Choose for
• ? ? from weaker bound we
  get
• ? ? from weaker bound we get
• Putting all bounds together we get

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Outline

• The Many faces Problem
• Point Line incidences
• Point Circle incidences

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Point Circle Incidences

• Let P be a set of m points.
• Let C be a set of n circles.
• We will show that
• The idea is the same as earlier, we will use a
  weaker bound (again, can be derived from the
  forbidden complete graphs results)

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Point Circle Incidences

• Construct a 1/r cutting of the arrangement
  A(C). The construction is quite similar to the
  previous one, we will describe the changes in
  phase 1.
• Choose a subset of size r.
• Triangulate A(R)
• Draw vertical line segments through the left and
  rightmost points of all circles in R.
• Draw a maximal vertical line segment through
  every vertex of A(R) and stop when it reaches
  another circle from R.

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Point Circle Incidences

• A trapezoid is bounded by 4 edges
• 2 vertical edges to the left and the right
• 2 circle edges at the top and the bottom.
• The total number of trapezoids is K O(r2)
• As earlier, at most circles cross each
  trapezoid.

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Point Circle Incidences

• Notations
• For each trapezoid Di, , ,
• Ci- subset of circles that have a non empty
  intersection with Di. Cini
• R As earlier
• The number of incidences within the trapezoids

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Point Circle Incidences

• We also have to consider incidences between
  points which lie on the boundaries of trapezoids
  and all n given circles.

• Each circle c intersects the boundaries of a
  trapezoid at most eight
• times twice for each edge of the
  trapezoid.
• The number of incidences of this type is at
  most

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Point Circle Incidences

• There are at most m incidences between circles of
  R and points which are not vertices of A(R)

• Putting everything together we get

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Point Circle Incidences

• Choose for
  and we get
• for we get
• for we get
• Putting it all together we get

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Questions?