Lower Envelopes of Algebraic Surfaces in 3D

1
Lower Envelopes of Algebraic Surfaces in 3D
2
Lower envelope Definition

• ? ?1, , ?n surface patches in R3 .
• The lower envelope of ?, projected onto the
  xy-plane (minimization diagram) is a planar map.
• Above each face there is one (or none) patch.
• Above each edge there are two intersecting
  patched (or a bounding curve of one patch).
• Above each vertex there are three intersecting
  patched (or two boundaries of two respective
  patches).

3
Motivation

• Central problem in computation geometry
• Voronoi diagrams.
• Moving points
• Convex hull
• Nearest neighbors
• Voronoi diagram
• Union of fat objects in R3.

4
Known results R2

• Input Lower envelope
• n continuously-defined
  O(?s(n)) functions, each pair of whichintersect
  in at most s points
• n lines, parabolas
  O(n)
• n partially-defined
  O(?s2(n)) functions, each pair of
  whichintersect in at most s points
• n line-segments
  T(n?(n))
• n circular-arcs
  O(n2?(n))

Maximum length of Davenport-Schinzel sequence of
order s with n symbols
5
The input surfaces Assumptions

• Each ??? is of constant description complexity
• ? is a portion of an algebraic surface of
  constant degree.
• ? is xy-monotone.
• The vertical projection of ? onto the xy-plane is
  a simple planar region.
• The relative interior of any triple of the
  surface patches intersect in at most s (constant)
  points.
• The surface patches in ? are in general position.

6
Known results R3 (and Rd1)

• Input lower envelope
• n triangles
  T(n2?(n))
• n (partially-defined) bivariate
  O(n2e)
• functions
• n (partially-defined) bivariate
• functions with s2
• Rd1
• n d-simplices
  T(nd?(n))
• n (partially-defined) d-variate
  O(nde)
• functions

7
Our problem

• Additional assumptions
• Each ??? is the graph of a continuously totally
  defined bivariate function in R3.
• s3 .

8
Our result

• The combinatorial complexity of the lower
  envelope of the surfaces in ? is O(n2?(n)) .
• Upper bound O(n2) ?

9
Proof

• Intersect ?i?? with ?j??, j?i , obtaining a set
  of planar regions ?i,j ?i ? ?j on ?i .
• Let ? ?i,jj?i .

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The portion of H on the lower envelope
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Proof

• The portions of ?i appearing on the lower
  envelope belong to the complementary of ???? ? .
• It is sufficient to show that the combinatorial
  complexity of the union ???? ? is O(n?(n)) .

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Proof The regions in ?

• Each ???
• is either closed or unbounded
• has constant description complexity.
• The boundaries of each pair ?, ??? intersect in
  at most three points.

13
Proof Classification of ?

• Partition ? into two subcollections R, B, s.t.,
• R ??? the boundary of ? is an unbounded
  curve
• B ??? the boundary of ? is a closed curve
• Each vertex v on the boundary of ???? ? is
  classified as
• red-red, if the two regions form the
  intersection are red.
• blue-blue, if these two regions are blue.
• red-blue, if one of these regions is red and the
  other is blue.

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Proof The number of red-red vertices

• Edelsbrunner et al. 1989
• The combinatorial complexity of the union of n
  planar regions delimited by unbounded Jordan
  curves, with at most three intersections per
  pair, is O(n?(n)) .
• The number of red-red vertices of the union is
  O(n?(n)) .

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Proof The number of blue-blue vertices

• For each ??B, ?? is a closed Jordan curve.
• Each pair ?, ?? B must intersect in an even
  number of points.
• Since s3, ?, ? intersect in at most two points.
  
• B is a collection of pseudo-discs.
• Kedem et al. 1986 The combinatorial complexity
  of ???B ? is O(n).
• The number of blue-blue vertices of the union is
  O(n) .

pseudo-discs
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Proof The number of red-blue vertices

• ER all red edges appearing on the boundary of
  ???R ? ,
• EB all blue edges appearing on the boundary of
  ???B ? .
• Define the graph G(V,E)
• V ER ? EB ,
• E e(cr, cb) cr?ER , cb?EB intersect along
  the boundaryof ???? ? .

A region that does not appear on the boundary of
???? ? is ignored.
If cr, cb intersect in two points, e would visits
only one of these points.
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Proof Geometric realization of G
p, p are planted points on the boundary of the
union
?1
?2
w
?2
?1
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Proof The number of red-blue vertices

• V is proportional to the number of red-red
  vertices the number of blue-blue vertices of
  ???? ? . V O(n?(n))
• E is proportional to the number of red-blue
  vertices of ???? ?
• Our goal Show that G is planar.

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ProofThe number of red-blue vertices

• Hanani-Tutte's theorem 1970
• If a graph G can be drawn in the plane so that
  any pair of its edges cross an even number of
  times, then it can also be drawn without any
  crossing.That is, G is planar.
• It is sufficient to consider only the pairs of
  edges that do not share a common vertex.

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ProofThe number of red-blue vertices

• Split each edge e of G at the red-blue vertex
  that it visits. obtain a red
  half-edge and a blue half-edge.
• For every pair er , er (eb , eb) of half-edges
  are disjoint in their interiors.
• We show that
• For every pair er , eb of half-edges, er, eb are
  either disjoint or cross twice.

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ProofThe number of red-blue vertices

• er ? ? ?r and eb ? ? ?b intersect in at most two
  points (since ?r and ?b intersect at most twice).
• Claim er and eb cannot have a single crossing.

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Proof Proving the claim
Assume, for the contrary, that er , eb intersect
in a single point p.
v
p
w
er
?b
?r
er
p
p
w
v
?r and ?b intersect in three points. A
contradiction!
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Proof Summarize

• Every pair of original edges of G intersect in an
  even number of points.
• G is planar.
• E O(n?(n)) .
• The number of red-blue vertices of ???? ? is
  O(n?(n)) .
• The number of red-red vertices the number of
  blue-blue vertices of ???? ? is O(n?(n)) .
• The combinatorial complexity of ???? ? is
  O(n?(n)) .
• The combinatorial complexity of the lower
  envelope is O(n2?(n)) .

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The sandwich region
The sandwich region
The sandwich region
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An Upper bound on the sandwich region

• Our proof concerns the union of all 3D-regions
  bounded by the surfaces in ? .
• In particular, it is applied for the sandwich
  region.
• The combinatorial complexity of the sandwich
  region is O(n2?(n)) .

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Tight bound?

• Lower bound of ?(n2?(n)) ?
• The standard construction of ?(n2?(n)) on the
  lower envelope of triangles is not applied here.
• Lower bound of ?(n2)
• Balls in R3.
• Extend each ball into a semi-unbounded prism.

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Improve the upper bound (?) The technique of
Tagansky

• Example Lower envelope of curves in 2D with s3
• Classification of the vertices of the envelope
• External vertex.Rightmost or leftmost
  intersection vertex of two intersecting curves.
  Their overall number is O(n) .
• Central vertex.The middle intersection vertex
  of two intersecting curves.

external
central
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The technique of TaganskyThe charging scheme

• Enter into the envelope from a 0-level vertex v.
• Advance left along a 1-level edge emanating from
  v.
• Stop as soon as
• We reach a 1-level vertex
• We reach another 0-level vertex.This vertex is
  external.

1-level vertex
0-level vertex
v1
v
v
v
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The technique of Tagansky Lower envelope in 2D
with s3

• C - a set of of n curves.
• Vk(C) the number of k-level central vertices in
  the arrangement of C.
• V0(C) ? V1(C) O(n)
• V0(n) O(n log n) .
• (The bound is not tight.)

The number of 1-level external vertices is O(n)
The maximum of V0(C) over all sets C.
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The technique of TaganskyLower envelope of
triangles in R3

• 4/3 V0(C) ? V1(C) O(n2?(n))
• V0(n) O(n2?(n)) .

The ratio between the 1-level vertices and the
0-level vertices gt 1 .
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The technique of TaganskyOur Problem Trivial
bound.

• Enter into the envelope from a 0-level vertex v.
• Advance left along a 1-level edge emanating from
  v.
• Stop as soon as
• We reach a 1-level vertex
• We reach another 0-level vertex.This vertex is
  external. Their overall number is O(n2).
• V0(?) ? V1(?) O(n2) V0(n)
  O(n2 log n) .

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The technique of TaganskyOur Problem ?

• 4/3 V0(?) ? V1(?) O(n2) (?)

23y
13x
12y
v112x
23x
v113y
12gt3
Is it possible that all the black vertices are at
0-level ?
13gt2
v0123
23gt1
v123z
12z
13z
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Application Dynamic Voronoi diagram
The circle is empty.
pi(t) (xi(t), yi(t))
Polynoms in t of degree s.
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Application Dynamic Voronoi diagram

• For each cell Vor(pi(t)) we count the number of
  combinatorial changes where -? lt t lt ? .
• Topological change Four points lie on one empty
  circle.

An edge is shrink to a vertex
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Topological change

• Fix a point p0.
• Fi(?,t) the radius r of the smallest circle
  emanating from p0(t) in direction ? ,
• and touches pi(t).
• Fi(?,t) Fj(?,t) Fk(?,t)
• p0(t), pi(t), pj(t), pk(t) lie on a common
  circle.
• An intersection point of triple of surfaces
  matches the event of a topological change.

r
pi(t)
?
p0(t)
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Overall number of topological changes.

• Why lower envelope?
• We consider only empty circles.
• An intersection point of Fi(?,t), Fj(?,t),
  Fk(?,t) does not lie on the lower envelope
• The circle is not empty.
• The number of combinatorial changes in
  Vor(po(t)) is proportional to the overall
  complexity of the lower envelope of
  Fi(?,t)i1,,n

r
pl(t)
?
p0(t)
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Number of intersections of triple of surfaces.

• Fi(?,t), Fj(?,t), Fk(?,t) intersect if and only
  if
• 1 x0(t) y0(t) x02(t) y02(t)
• 1 xi(t) yi(t) xi2(t) yi2(t)
  0
• 1 xj(t) yj(t) xj2(t) yj2(t)
• 1 xk(t) yk(t) xk2(t) yk2(t)
• The number of intersections is the number of
  solutions for the above equality.

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Special case s3

• xi(t) ai t bi
• yi(t) c xi(t) di
• The degree of the determinant is three.
• The number of combinatorial changes of
• Vor(pi(t)) is O(n2?(n)) .
• The overall number of changes is O(n3?(n)) .