Computational Geometry Seminar Lecture 5

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Computational Geometry Seminar Lecture 5
Circle Packings and Lipton-Tarjan Theorem
Maor Hizkiev
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Overview
Koebe Representation Theorem
Lipton-Tarjan Theorem
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Definition
Convex Body - A set S in a vector space over R is
called convex if the line segment joining any
pair of points of S lies entirely in S.
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Definition
Contact Graph Given a packing C of convex
bodies, the vertices are associated with the
members of C. 2 vertices are connected ? the
corresponding members touch each other.
Note if the bodies are circles, then the
connections are straight edges
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Koebe's Theorem
A finite graph is a contact graph of a circle
packing in the plane ? it is planar
We will use this theorem, in order to prove
Lipton-Tarjan separator theorem for planar graphs.
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In formal
Given any planar graph G, with vertex set V(G)
v1, , vn and edge set E(G)
e1, , en , We can find a packing of n (not
necessarily congruent) circular disks C
C1,,Cn in the plane with the property that Ci
and Cj touch each other ? vi,vj ? E(G) for 1 ? i
? n
V2
V1
V4
V3
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proof
It is sufficient to prove the theorem for maximal
planar graphs, AKA triangulation.
lt Assume that the resulting graph can be
represented by circular discs, satisfying the
conditions in the theorem, then, erasing the
discs corresponding to the new vertices, we
obtain a good representation of the original
graph, because the discs are non-overlapping.
V2
V1
V4
Note in this direction we do not demand a
maximal graph.
V3
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proof gt

• Let G be a fixed triangulated graph with vertex
  set V(G) v1, , vn , edge set E
  and face set F.
• By Eulers formula
• V - E F 2
• but since 3F 2E (triangulated graph), we
  get
• F 2V - 4 2n - 4

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• Let r (r1,,rn) be any vector of n positive
  reals such that r1 rn 1.

For each face vivjvk of G, consider a triangle
cut-out of cardboard, whose vertices are the
centers of 3 mutually tangent discs of radii
ri,rj and rk.
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Definition
Let ?r(vi) denote the sum of the angles at vi of
all triangles that have vi as one of their
vertices.
In case we are (extremely) lucky, than ?r(vi)
2? for every vertex G not in the exterior
triangle.
In this case, the triangles will perfectly fit
together in the plane, and we obtain a good
representation of G by discs of radii r1,,rn
NOTE the above statement is not as trivial as it
seems.
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In the case we are not lucky
The triangles overlap each other, thus,
is less than 2?
?r(vi)
vi
In any case we get that
Since we have F faces
m
??r(vi) F? (2n 4)?
i1
Recall that F 2n - 4
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Definition
Let S ? Rn denote the (n 1)-dimensional simplex
defined by
S r (r1,,rn) ri gt 0 for all i, and ? ri
1
And let
H x (x1,,xn) ? xi (2n 4)?
Consider the continuous mapping fS ? H. f(r)
(?r(v1),, ?r(vn))
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Assume WLOG that v1,v2,v3 are the vertices of the
exterior face. It is sufficient to show that, for
example, x (2?/3, 2?/3,2?/3, 2?, 2?, ,
2?) Lies in the image of the map f.
x implies that the circles of the exterior face,
will have equal sizes.
v1,v2,v3 donates 2?/3 to the sum, since each
regular triangle donates ?/3 to the sum, plus ?/3
from the exterior triangle.
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Claim A fS ? H is a one-to-one function. Proof
Well pick 2 distinct points r,r? S. Let I
denote the set of indices i which satisfy the
inequality ri lt ri. Obviously I ?? ? and I ?
1,,n.
Consider a triangle vivjvk determined by the
centers of 3 mutually touching discs with radii
of ri,rj,rk.
We have now 4 possibilities
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Case 1 2
Case 1 All vertices of the triangle does not
belong to I. Case 2 All vertices of the triangle
belong to I.
??r(vi)
In both cases, the sum does not
change.
i?I
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Case 3
Case 3 two vertices belongs to I, and one
doesnt. If we increase ri and rj but decrease or
leave unchanged the other radii so that the discs
remain tangent, then the angles of the triangle
at vi will decrease.
Note such a triangle must exist, since I ?
1,,n and I ? ?.
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Case 4
Case 4 one vertex belongs to I, and the other 2,
doesnt. If we increase ri but decrease or leave
unchanged the other 2 radii so that the discs
remain tangent, then the angle of the triangle at
vi will decrease.
? f(r) ? f(r)
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• Let s (s1,,sn) be a boundary point of the
  simplex S, and now let I denote the set of all
  indices for which si 0.

(0,0,1)
(1,0,0)
(0,0.5,0.5)
(0,1,0)

• Observe that if r tends to s, then in each
  triangle which has at least one vertex belonging
  to vi i ? I, the sum of the angles at these
  vertices tends to ? ,hence,

Where F(I) denotes the set of faces of G with at
least one vertex in vi i?I.
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• For any fixed r?S and for any nonempty proper
  subset I ? 1,,n, there exists a point
  s?BdS such that si 0 for all i?I and si gt ri
  for all i?I.

• If we move r toward s along a line, then, ?i?I
  ?r(vi) will increase. (from claim A)
• Taking our former result, we get that

?i?I?r(vi) lt F(I)?
In other words, the image of the map fS ? H lies
in the (n-1)-dimensional convex polytope P
determined by the relations
?i1xi (2n 4)?
?i?Ixi lt F(I)?
For all proper I?P(1,,n)\?
2 conditions on a vector of H, just like linear
programming
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• From the fact that fS?P is a one-to-one
  mapping, and all accumulation points of f(r), as
  r tends to BdS, lie in P (because we proved that
  ) , we get the
  following.

lim ??r(vi) F(I)?
i ? I
r ? s
Claim B fS?P is a surjective mapping, that is,
f(S) P.

• To complete the proof, we still have to show
  that the point x(x1,,xn)
• (defined earlier as x (2?/3, 2?/3,2?/3, 2?,
  2?, , 2?))
• belongs to P.
• Clearly,

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• For all proper I?P(1,,n)\?, If I n 1
  or n 2, then all faces of G have at least one
  vertex belonging to vi i? I, i.e.

F(I) F 2n-4
Therefore, in this cases,
?xi lt (2n 4)? F(I)?
i ? I
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Claim C for any subset I ? 1,,n with 1?I?n
3, G has more than 2I faces that have at
least one vertex belonging to vi i? I. For
any such I,
?xi ? 2?I lt F(I)?
i ? I
In conclusion, Thus, x satisfies all the
relations defining P, that is, x ? P f(S),
as required.
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Lipton-Tarjan Separator Theorem
We all know the divide-and-conquer paradigm the
idea is to divide a problem into 2 smaller
subproblems of the same type, which can be solved
recursively, and to combine the results of the
subproblems to obtain a solution to the original
problem.
The Lipton-Tarjan theorem, shows that any planar
graph can be separated into two much smaller
components by the removal of relatively few
vertices.
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In formal
Let G be a planar graph with n vertices. Then the
vertex set of G can be partitioned into three
parts, A, B and C, such that A, B ? 3n/4, C
lt 2??n, and no vertex in A is adjacent to any
vertex in B.
Note the Lipton and Tarjan bounded A and B
by 2n/3. We will prove a weaker result.
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Half-Space
x y 1
x y ? 1
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Helly's theorem
Suppose that X1,,Xn is a finite collection
of convex subsets of Rd ,where n gt d. If the
intersection of every d 1 of these sets is
nonempty, then the whole collection has a
nonempty intersection that is, ? Xj ? ?
n j1
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Lemma
for any n-element set P? ?Rd, there exists a
point q ? Rd with the property that any
half-space that does not contain q covers at most
dn/(d1) elements of P. (such a point q is called
a centerpoint of P).
For example, lets examine the case where d 2.
n9
q
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Lemma's Proof
Let H be the family of all half-spaces that cover
more than ??dn/(d1)? elements of P. we have to
show that the intersection of all members of H is
non-empty. By Hellys theorem, it is sufficient
to prove that any d 1 half-spaces H1,,Hd1?H
have a point in common. This is truly the case,
otherwise
d1
P ? ? (Rd Hi) ? P ? (d 1)(n - ?dn/(d1)?
- 1) lt n
i1
Which is a contradiction.
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Lipton-Tarjan Proof
Let S denote the unit sphere in R3 centered at
(0,0,1), whose poles are N (0,0,2) and
(0,0,0). The stereographic projection ?R2 ? S
maps any point p of the xy-plane, into the
intersection of the segment Np with S. the image
of any circular disc under ? will be a
. Conversely, the
pre-image of any spherical cap S, which
does not contain N in its
interior, is a
circular disc or a half-plane in the xy-plane.
spherical cap
N (0,0,2)
(0,0,0)
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How to make a stereographic projection?
N (0,0,2)
For a point p,
XY-plane
we will cross a line from N through p, until it
hits the xy-plane.
p
If we want to map N to the xy-plane, then we will
need an infinite line.
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Koebes Theorem implies that the vertices of G
can be represented by non-overlapping spherical
caps C1,,Cn ? S, such that 2 of them touch each
other ? the corresponding vertices are adjacent.
Pick a point pi in the interior of Ci, and set P
p1,pn. Let q ? R3 be the centerpoint of P,
satisfying the condition of the previous lemma.
Recall that we are working in R3, thus q should
cover at most 3n/4 vertices. By symmetry, we can
assume that q belongs to the segment connecting
the center and the south poles of S (we can
rotate the sphere).
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Suppose q does not coincide with (0,0,1) - the
center of S. Then
d q (0,0,1) ? 0
Let ?S ? S be a mapping defined by

?
If p ? N
?(p)
N if p N
Since ? is the composition of an inverse
stereographic projection, a dilatation of the
xy-plane, and a stereographic projection, it
carries spherical caps into spherical caps, and
it also preserves the incidences between them.
?(pi) will lie in the interior of the cap ?(Ci),
and it can shown that (0,0,1) will be a
centerpoint of ?(P).
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Now, we can assume WLOG, that (0,0,1) is a
centerpoint of P. This means that any plane
passing through (0,0,1) has at most 3n/4 elements
of P strictly on its right side (and left side).
Now we need to prove that theres a plane that
covers lt 2?n
Let ri denote the (spherical) radius of the cap
Ci. Since C1,,Cn form a packing, and the area of
Ci is at most ?ri2 , we have
Unit sphere area size
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Jensen's Inequality
The average of a convex function is bigger than
the function in the average of the points.
x1 xn
f( )
f(x1) f(xn)
?
n
n
If you use the convex function exp and x
log(a), we will get the inequality of arithmetic
and geometric means
a1 an
n
?
?a1 an
n
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By Jensens Inequality, we have
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Consider now any plane passing through (0,0,1),
and let u(H) denote the unit normal vector
sitting at (0,0,1). The location of the endpoint
of u(H), as H varies over all planes through
(0,0,1) that intersect a fixed spherical cap Ci,
will be a ringlike region Ri symmetric about a
great circle of S.
Ri 2ri


2ri
Ci
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Obviously, the area of Ri satisfies
A(Ri) lt (2ri)(2?) 4?ri
Hence, there exists a point of S which is covered
by at most
n
n
? A(ri)
? 4?ri
n
i1
i1
lt
? ri
n
2
lt

A(S)
4?
i1
Regions Ri. Equivalently, there is a plane H0
intersecting fewer than Caps of Ci.
n
2
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Let A, B and C denote the set of those vertices
of G, for which the corresponding spherical caps
Ci lie entirely on one side of H0, on the other
side of H0, or meet H0, respectively.
It follows directly from the above properties of
H0 that A, B ? 3n/4 and C lt
.
2
n
Clearly, no vertex in A can be adjacent to any
vertex in B, because two closed caps lying in two
complementary half-spaces can never touch each
other.