Computational%20Geometry%20Seminar

1
Computational GeometrySeminar
Lecture 9 Eyal Zur
2
These edges are in convex position

• Definitions
• Two edges of a geometric graph are in convex
  position if they are disjoint edges of a convex
  quadrilateral.
• A geometric graph is called proper if it has two
  edges in convex position.
• Otherwise, it is called improper.

Also
These edges arent in convex position
3
Our motivation
Kupitz Conjecture An improper geometric graph
with n vertices has at most 2n-2 edges.
An improper geometric graph on 7 vertices and 12
edges.
4
What we are going to see?
Theorem (Katchalski and Last) An improper
geometric graph with n vertices has at most 2n-1
edges.
5

• Definitions
• A circular sequence is a sequence whose first and
  last term are considered adjacent.
• (1,2,3,4,5) is a circular sequence ? 1 and 5 are
  adjacent.
• A circular sequence from a set of n symbols shall
  be called a circular DavenportSchinzel sequence
  of order 2 if
• No two adjacent terms are identical
• and
• If it does not contain a circular subsequence of
  type abab.
• (a,b,c,d) is a sequence like this,
• (1,2,3,4,2,3,1) , (1,2,3,2,3,4) , (a,b,c,a,b)
  arent

6

• Definition (cont.)
• A convex curve is the boundary of a compact
  convex planar set with nonempty interior.
• (v1,v2,v4,v5,v6,v8,v0) is convex curve
• (v1,v2,v3) , (v3,v4,v5,v6) - arent

v8
v0
v9
v1
v7
v3
v4
v2
v5
v6
7
We shall need a known upper bound on the length
of such circular sequences of order 2 which is
proved by induction on n, the number of
symbols Lemma 1 The length of a circular
DavenportSchinzel sequence of order 2 on n
symbols for n 2 is at most 2n-2. not our main
issue, so without proof.
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Lemma 2 Let Ai, Aj, Ak, Al be four points
appearing in this order on a convex curve d.
Let P, Q be two points inside d. Consider the
four (closed) segments PAi, QAj, PAk, QAl, and
assume that among them there is no segment s
such that s contains only one of the points P,
Q and the line supporting s contains both of
them. Then two of the four segments are in
convex position. (2) Implies that if one of the
points P,Q lies on one of the gray lines, then
the other one also lies on the same gray line.
(1)
(2)
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Observation If points A, B, C, D in general
position lie in this order on a convex curve,
then the segments AB and CD are in convex
position.
(3)
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Proof of lemma 2 Let ll(P,Q) and let l1 and l2
be two half-planes such that l1 l2 l. Let
d1(d2) be the boundary of the convex hull of l1
d (of l2 d). Each di contains l
also.
l
Aj
Ak
Q
d1
l2
l1
d2
Ai
P
Al
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Proof of lemma 2 (cont.) It is easy to check,
using (2) and (3) that if one of the four points
Ai, Aj, Ak, Al lies on l, then two of the
segments are in convex position.
Aj
Ak
Aj
Ak
l1
l
d1
l1
l
d1
l2
l2
Q
d2
d2
Ai
Q
P
Ai
P
Al
Al
Here QAj and PAk and are in convex position
Here QAl and PAi and are in convex position
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Proof of lemma 2 (cont.) Assume therefore without
loss of generality that either Ai, Aj and Ak
lie in the interior of l1 And Ai lie in the
interior of l2 Or Ai and Aj lie in the
interior of l1 And Ak and Al lie in the interior
of l2
(4)
(5)
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Proof of lemma 2 (cont.) In case (4), by (3)
applied to d1, either PAi and QAj are in convex
position or PAk and QAj are in convex position.
l
l
Aj
Aj
Ai
Ai
l1
d1
d1
l1
Q
l2
l2
P
d2
d2
P
Al
Al
Q
Ak
Ak
Here QAj and PAi and are in convex position
Here QAj and PAk and are in convex position
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Proof of lemma 2 (cont.) In case (5) if PAi , QAj
are not in convex position and PAk, QAl are not
in convex position, then by (3) the order of the
points on d is Ai, Aj, Al, Ak , a contradiction.
l
Aj
Ai
l
Aj
Ai
d1
d1
l1
l1
Q
l2
P
l2
d2
d2
Al
Al
Q
P
Ak
Ak
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Proof of the Theorem Let v1,,vn be the vertices
of G and e the number of its edges. Assume that
G is improper with e 1. Let C be a circle
containing v1,, vn in its interior. For any two
vertices vi and vj joined by an edge vivj define
two points on C
and
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Proof of the Theorem (cont.) Arrange the 2e
points on C in a circular sequence according to
the order of their appearance on C. Let D(G) be
the circular sequence thus obtained. D(G)
(a41, a42, a43, a23, a12, a14, a24, a34, a32,
a21)
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Proof of the Theorem (cont.) Color the points of
D(G) with n colors such that aij receives
the color i Point aij has a dark color i if
vivj is an interior edge of vj. Otherwise, aij
has a light color i . D(G) (a41, a42, a43,
a23, a12, a14, a24, a34, a32, a21)
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Proof of the Theorem (cont.) Divide the sequence
D(G) into arcs where an arc is a maximal
subsequence of consecutive points of D(G) having
the same color i . Note that a dark i and a
light i may belong to the same arc. For our
example D(G) (a41,a42,a43,a23,a12,a14,a24,a34,a
32,a21) ? The arcs of G are (a41, a42,
a43),(a23),(a12, a14),(a24),(a34, a32),(a21)
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Proof of the Theorem (cont.) The circular
sequence obtained from D(G) by contracting each
arc to one of its points and then replacing the
point by its color i is called the pattern
sequence of G, or PS(G). For our
example PS(G) (4,2,1,2,3,2)
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Lemma 3 PS(G) is a circular Davenport-Schinzel
sequece of order 2. Lemma 4 An arc of D(G)
contains at most one point with dark
color. Well see proofs for the lemmas
soon But for now we have, D(G) 2e of
light colored points of dark colored
points
(6)
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Proof of the Theorem (cont.) Each vertex of G has
at most one rightmost edge and at most one
leftmost edge incident to it, so that the number
of light colored points in D(G) is bounded by 2n.
By Lemmas 1, 3, and 4, of dark colored points
PS(G) 2n-2 Substitute all of the above in
(6) to obtain e 2n 1 D(G) 2e of
light colored points of dark colored points
2n 2n 2
4n 2 It remains to prove lemmas 3 and 4.
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Proof of lemma 3 If PS(G) is not a circular
DavenportSchinzel sequence of order 2, then
there are four points, aa1, ab2, aa3, ab4
appearing in that order on C. Since
v1,,vn are in general position there are no two
disjoint segments vxvy , vzvt such that a line
through one of them contains an endpoint of the
other. (If exists then there are 3 edges on the
same line, and the vertices arent in general
position)
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Proof of lemma 3 (cont.) Therefore by Lemma 2,
two of the segments vaaa1, vbab2, vaaa3, vbab4 ,
are in convex position. Since every edge vxvy is
contained in the segment vxaxy, two of the
segments vau1, vbu2, vau3, vbu4 Are in convex
position, a contradiction. In the example
vau3 and vbu4 are in convex position
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Proof of lemma 4 Suppose that the points aab, aac
belong to the same arc and are dark colored.
Assume without loss of generality that vavc is
to the right of vavb. The edges vavb, vavc are
interior edges of vb and vc, respectively. So
let vbvx be to the right of vbva and let vcvy be
to the left of vcva.
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Proof of lemma 4 (cont.) Let st be the chord of C
that contains vbvc, so that vb lies in vcs and vc
in vbt. Let r be the ray with apex vb and
parallel to vcva. Let a1, a2, a3, a4 be the
following angles a1 conv(vbaba ? r) a2
conv(r ? vbs) a3 conv(vbs ? vbaab) a4
conv(vct ? vcaac)
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Proof of lemma 4 (cont.) Since vbvx is to the
right of vbva, vx must lie in at least one of the
angles a1 or a2 or a3. If vx a1, then the
points aac, axb, aab, abx are in that order on C.
Therefore aab, aac are not on the same arc, a
contradiction. If vx a2, then vbvx and vcva
are in convex position, a contradiction. Therefore
vbvx is in a3 and by symmetry vcvy is in a4,
implying vbvx and vcvy are in convex position, a
contradiction.
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The first end Get out for a BREAK
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• Definition
• Two edges of a geometric graph are said to be
  parallel if they are opposite sides of a convex
  quadrilateral.
• Actually (as you see) there is no different
• between parallel edges and edges in convex
• position.

These edges are parallel
Also
These edges arent parallel
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What we are going to see?
Theorem (Valtr) Let k 2 be a constant, Then
any geometric graph on n vertices with no k
pairwise parallel edges has at most O(n) edges.
9
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• Generalized DavenportSchinzel Sequences,
  definitions
• For l 1, a sequence is called l-regular, if any
  l consecutive
• terms are pairwise different.
• Examples
• (1,2,3,4,5) is 4-regular sequence.
• (1,2,3,4,2,5) isnt 4-regular sequence, but is
  3-regular.
• For l 2, a sequence S s1, s2, , s3l-2
• of length 3l-2 is said to be of type
  up-down-up(l),
• if the first l terms are pairwise different and,
  
• for i1,2,,l si s2l-i s(2l-2)i
• Example (1,2,3,4,5,4,3,2,1,2,3,4,5) is
  up-down-up(5) sequence.

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Conclusion A sequence is of type up-down-up(2) ?
it is an alternating sequence of length
4. Meaning, in the form of (a,b,a,b). Claim It
is known (and without proof here) that any
2-regular sequence over an n-element alphabet
containing no alternating subsequence of length 4
has length at most 2n-1.
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In the proof of Theorem 1 we apply the following
related result Theorem 4 Let l 2 be a
constant. Then the length of any l-regular
sequence over an n-element alphabet containing no
subsequence of type up-down-up(l) is at most
O(n). (also not our main issue)
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Proof of Theorem 1

• Let G)V,E) be a geometric graph on n vertices,
  with no k pairwise parallel edges.
• Let V(v1,v2,,vn).
• Without loss of generality, we assume that no two
  points lie on a horizontal line.
• If necessary,
• we perturb the vertices of G
• to make the directions of edges
• of G pairwise different.
• We dont allow the
• red edges exist that way

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• Proof of theorem 1 (cont.)
• Let e E.
• An oriented edge e is defined as the edge e
  oriented upward.
• The direction of e, dir(e), is defined as the
  direction of the vector vivj, where e (vi,vj),
• Let E e1,e2,,em), where
• 0 lt dir(e1) lt dir(e2) lt lt dir(em) lt

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Proof of theorem 1 (cont.) Definition Let P1 and
P2 be the sequences of m integers obtained from
the sequence e1, e2, , em by replacing each
edge ek (vi,vj) by integer i and by integer j ,
respectively. We call the sequences P1, P2 the
pattern sequences of G. P1(5,3,7,4,2,3) P2(6,8,
6,8,1,2)
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Lemma 5 For each l 1, at least one of the
pattern sequences P1, P2 contains an l-regular
subsequence of length at least E/(4l)
m/(4l). Lemma 6 Neither of the pattern
sequences P1, P2 contains a subsequence of type
up-down-up(k ). Before proving Lemmas 5 and 6,
we complete the proof of Theorem 1.
3
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Proof of Theorem 1 (cont.) According to Lemma 5,
at least one of the sequences P1, P2 contains a k
-regular subsequence S of length at least E/(4k
). According to Lemma 6, the sequence S contains
no subsequence of type up-down-up(k ). Theorem 4
implies that the length of S is at most O(n).
Consequently, E 4k O(n) O(n). It
remains to prove Lemmas 5 and 6.
3
3
3
3
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Proof of lemma 5 We apply a simple greedy
algorithm which, for given integer l 1 and
finite sequence A, returns an l-regular
subsequence B(A,l) of A. In the first step, an
auxiliary sequence B is taken empty. Then the
terms of A are considered one by one from left to
right, and in each step the considered term is
placed at the end of B iff this does not violate
the l-regularity of B. Finally, the obtained
l-regular subsequence B of A is taken for
B(A,l). For example A(1,3,1,3,5,2,2,5,1,5,1,2)
and l3, then the algorithm returns the
sequence B(A,3)(1,3,5,2,1,5,2).
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Proof of lemma 5 (cont.) Let l 1 be given. For
i 1,2 and for 1 j1 j2 m, let Pi,j1,j2
denote the part of Pi starting with the j1th term
and ending with the j2th term. Thus, Pi, j1, j2
consists of j2-j11 terms. Let T denote the
length of a sequence T , and IT the set of
integers (different numbers) appearing in T
. For example A(1,3,1,3,5,2,2,5,1,5,1,2) A
12 IA 1,3,5,2
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Claim 7 For each j 1, 2, , m B(P1,1, j,
l) B(P2,1, j, l) j / (2l) Proof First,
consider two integers j1,j2 such that 1j1j2m.
Obviously, ej1,ej11 ,ej2 (va,vb)a
I(P1,j1,j2),b I(P2,j1,j2) ? ej1,ej11
,ej2 (va,vb)a I(P1,j1,j2),b
I(P2,j1,j2) ? j2-j11 I(P1,j1,j2) x
I(P2,j1,j2)
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Proof Claim 7 (cont.) By the inequality between
algebraic and geometric means,
I(P1,j1,j2) I(P2,j1,j2)
(1)

j2-j11
2
We can now prove the claim by induction on j. If
j min(16l ,m), then by (1) and by j
16l B(P1,1,j,l) B(P2,1,j,l)
I(P1,1,j) I(P2,1,j) 2 j
2j/ j 2j/4l j/(2l)
2
2
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Proof Claim 7 (cont.) Suppose now that 16l lt j0
m and that Claim 7 holds for j 1, 2, ,
j0-1. Since for i1,2 each integer of I(Pi,j0-4l
1, j0) not appearing among the last l-1 terms
in B(Pi,1,j0-4l ,l) appears more times in
B(Pi,1, j0,l) than in B(Pi,1, j0-4l
,l), we have B(Pi,1,j0,l) B(Pi,1,j0-4l
,l) I(Pi,j0-4l 1, j0) - (l-1) From our
the greedy algorithm, its obvious
that B(Pi,1,j0,l) B(Pi,1,j0-4l
,l) And if I(Pi,j0-4l 1, j0) contains
more than l-1 integers, then each integer which
not appear in B(Pi,1,j0-4l ,l) adds at
least one to the equations left side.
2
2
2
2
2
2
2
2
2
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Proof Claim 7 (cont.) Consequently, by the
inductive hypothesis and by (1), B(P1,1,j0,l)
B(P2,1,j0,l) (j0-4l )/(2l) 2 4l -
2(l-1) j0/(2l)
2
2
Proof of Lemma 5 Lemma 5 follows easily from
claim 7 (with j m) and from the pigeon-hole
principle B(P1,1,m, l) B(P2,1,m, l) m
/ (2l) , And therefore, at least one of the
l-regular sequences is at length of m/(4l).
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Proof of Lemma 6 First, we apply the following
consequence of Dilworths Theorem Theorem 8 If
the union of three partial orderings on a set I
of size at least (k-1) 1 is a linear ordering
on I , then at least one of the partial orderings
contains a chain of length k. Remainder By
linear ordering we mean If a b and b a then
a b ( antisymmetry) If a b and b c then a
c (transitivity) a b or b a (totally).
3
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Proof of Theorem 8 Let , , , be the
three partial orderings on I. If (I, ) does
not contain a chain of length k then, by
Dilworths theorem, it can be covered by at most
k-1 antichains. Consequently, There is an
antichain A of size (k-1) 1 in (I, ). If we
restrict our attention to A and to orderings
, another application of Dilworths theorem
gives k elements in A which form a chain with
respect to or to . (but its not our
main issue)
2
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Proof of Lemma 6 Because of symmetry, it suffices
to prove Lemma 6 for the pattern sequence P1.
Suppose to the contrary that P1 contains a
subsequence of type up-down-up(k ). Thus, there
is a subsequence S s1, s2, , s3k -2 of P1
such that the integers s1,s2,sk are pairwise
different and that, for i1,2,,k , sis2k
-is(2k -2)i . For simplicity of notation,
suppose that sii (i1,,k ) and that SP1,1,3k
-2. We obtain a contradiction by showing that
k of the edges e1,e2,,e3k -2 are pairwise
parallel.
3
3
3
3
3
3
3
3
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• Definition 9
• Let i,j I, and let dir(vi,vj) denote the
  direction of the vector vi,vj.
• Then
• (i) i j, if iltj and dir(vi,vj) dir(ek ),
  
• (ii) i j, if iltj and dir(vi,vj) ,
  dir(e2k -1)
• i j, if iltj and dir(vi,vj) dir(e2k
  -1), 2 U
• (0, dir(ek ))

3
3
3
3
Since the union of , , , is a linear
ordering on I, Theorem 8 implies that one of the
orderings , , contains a chain
i1,i2,..,ik of length k. We distinguish the
corresponding three possible cases.
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Definition 9 (cont.) If i1 i2 . ik, then
the edges ei1, ei2, , eik are pairwise
parallel. Indeed, if 1 j lt j k then the
inequalities 0 dir(eij) lt dir(eij) dir(ek )
dir(vij,vij) lt show that the edges eij, eij
are parallel.
3
Similarly, if i1 i2 ik, then the edges
e(2k -2)i1, e(2k -2)i2,,e(2k -2)ik are
pairwise parallel, and if i1 i2 ik then
the edges e2k i, e2k -i, , e2k -i are
pairwise parallel.
eij
3
3
3
vij
3
3
3
eij
vij
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The End