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Computational GeometryComputational GeometrySeminarSeminar

Lecture 9Lecture 9

Eyal ZurEyal Zur

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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.

These edges are in convex position

Also…

These edges aren’t in convex position

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Our motivationKupitz ConjectureAn improper geometric graph with n vertices has at most 2n-2edges.

An improper geometric graph on 7 vertices and 12 edges.

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What we are going to see?

Theorem (Katchalski and Last)

An improper geometric graph with n vertices has at most 2n-1 edges.

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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 Davenport–Schinzel 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) – aren’t

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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) - aren’t

v8

v7

v0

v6

v2v5

v1

v4

v3

v9

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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 1The length of a circular Davenport–Schinzel 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 2Let Ai, Aj, Ak, Al be four points appearing in this order on a convex curve δ. Let P, Q be two points inside δ.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, Qand 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)

Aj

Ai

Ak

Al

P

Q

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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)

A

D

B

C

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Proof of lemma 2Let l=l(P,Q) and let l1 and l2 be two half-planes such that l1 l2 = l.Let δ1(δ2) be the boundary of the convex hull of l1 δ(of l2 δ).

Each δi contains l also.

Aj

Ai

Ak

Al

P

Q

l

l1l2

δ2

δ1

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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

Ai

Ak

Al

P

Q

l l1

l2

δ2

δ1

Aj

Ai

Ak

Al

P Q

l l1

l2

δ2

δ1

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

AndAi lie in the interior of l2

Or

Ai and Aj lie in the interior of l1

AndAk 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 δ1, either PAi and QAj are in convex position or PAk andQAj are in convex position.

Ai

Al

Aj

Ak

Q

P

l

l1

l2

δ2

δ1

Ai

Al

Aj

Ak

Q

P

l

l1

l2

δ2

δ1

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 δ is Ai, Aj, Al, Ak , a contradiction.

Ai

Al

Aj

Ak

Q

P

l

l1

l2δ2

δ1

Ai

Al

Aj

Ak

P

Q

l

l1

l2

δ2

δ1

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Proof of the TheoremLet 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:

andcij jivva

cji ijvva

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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) = (α41, α42, α43, α23, α12, α14, α24, α34, α32, α21)

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Proof of the Theorem (cont.)Color the points of D(G) with n colors such that αij receives the color i :Point αij has a dark color i if vivj is an interior edge of vj. Otherwise, αij has a light color i .

D(G) = (α41, α42, α43, α23, α12,

α14, α24, α34, α32, α21)

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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) = (α41,α42,α43,α23,α12,α14,α24,α34,α32,α21) The arcs of G are (α41, α42, α43),(α23),(α12, α14),(α24),(α34, α32),(α21)

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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 3PS(G) is a circular Davenport-Schinzel sequece of order 2.

Lemma 4An arc of D(G) contains at most one point with dark color.

We’ll 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 obtaine ≤ 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 3If PS(G) is not a circular Davenport–Schinzel sequence of order 2, then there are four points, αa1, αb2, αa3, αb4 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 containsan endpoint of the other.(If exists then there are 3 edges on the same line, and the vertices aren’t in general position)

αb2

αa3

αa1

αb4

u2

u1

u3u4 vb

va

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Proof of lemma 3 (cont.)Therefore by Lemma 2, two of the segmentsvaαa1, vbαb2, vaαa3, vbαb4 , are in convex position. Since every edge vxvy is contained in the segment vxαxy, two of the segments vau1, vbu2, vau3, vbu4 Are in convex position, a contradiction.

In the example: vau3 and vbu4 are in convex position

αb2

αa3

αa1

αb4

u2

u1

u3u4 vb

va

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Proof of lemma 4Suppose that the points αab, αac 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 α1, α2, α3, α4 be the following angles:α1 = conv(vbαba υυ r)

α2 = conv(r υυ vbs)

α3 = conv(vbs υυ vbαab)

α4 = conv(vct υυ vcαac)

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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 α1 or α2 or α3. If vx α1, then the points αac, αxb, αab, αbx are in that order on C. Therefore αab, αac are not on the same arc, a contradiction. If vx α2, then vbvx and vcva are inconvex position, a contradiction.Therefore vbvx is in α3 and by symmetry vcvy is in α4, implying vbvx and vcvy are inconvex position, a contradiction.

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The first end…The first end…

Get out for a BREAKGet 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 convexposition.

These edges are parallel

Also…

These edges aren’t 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 nok pairwise parallel edges has at most O(n) edges.

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Generalized Davenport–Schinzel 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) – isn’t 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 i=1,2,…,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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ConclusionA 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).

ClaimIt 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 4Let 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 don’t 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), where0 < dir(e1) < dir(e2) < … < dir(em)<

Π

v6 v8

v1

e1 e3e4

e2e5

e6 v3

v4

v2

v5

v7

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Proof of theorem 1 (cont.)DefinitionLet 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)

v6 v8

v1

e1 e3e4

e2e5

e6 v3

v4

v2

v5

v7

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Lemma 5For 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.

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3 3

3

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Proof of lemma 5We 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 l=3, 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-j1+1 terms.

Let |T| denote the length of a sequence T , and I[T] 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| = 12I[A] = {1,3,5,2}

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Claim 7For each j = 1, 2, …, m :|B(P1,[1, j], l)| + |B(P2,[1, j], l)| ≥ j / (2l)

ProofFirst, consider two integers j1,j2 such that 1≤j1≤j2≤m. Obviously,{ej1,ej1+1 …,ej2} {(va,vb)|a I(P1,[j1,j2]),b I(P2,[j1,j2])}

|{ej1,ej1+1 …,ej2}| ≤ |{(va,vb)|a I(P1,[j1,j2]),b I(P2,[j1,j2])}|

j2-j1+1 ≤ |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])|

2≥ j2-j1+1

(1)

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 < j0 ≤ m and that Claim 7 holds for j = 1, 2, …, j0-1.Since for i=1,2 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, it’s 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 equation’s 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 5Lemma 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 6First, we apply the following consequence of Dilworth’sTheorem:

Theorem 8If 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 8Let , , , be the three partial orderings on I. If (I, ) does not contain a chain of length k then, by Dilworth’s 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 Dilworth’s theorem gives k elements in A which form a chain with respect to or to .

(but it’s not our main issue…)

1 2 31

12

2 3

2 3

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Proof of Lemma 6Because 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 subsequenceS = s1, s2, …, s3k -2

of P1 such that the integers s1,s2,…sk are pairwise different and that, for i=1,2,…,k , si=s2k -i=s(2k -2)+i . For simplicity of notation, suppose that si=i (i=1,…,k ) and that S=P1,[1,3k -2].We obtain a contradiction by showing that kof the edges e1,e2,…,e3k -2 are pairwise parallel.

3

3

3

3

3

3

3

3

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Definition 9Let i,j I, and let dir(vi,vj) denote the direction of the vector vi,vj.Then:(i) i j, if i<j and dir(vi,vj) [dir(ek ), ](ii) i j, if i<j and dir(vi,vj) [ , + dir(e2k -1)](iii) i j, if i<j and dir(vi,vj) [ + dir(e2k -1), 2 ] U

(0, dir(ek ))

123

3

Π

ΠΠ

3

ΠΠ

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.

1 2 3

1 2 3

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Definition 9 (cont.)If i1 i2 …. ik, then the edges ei1, ei2, …, eik are pairwise parallel.Indeed, if 1 ≤ j < j’ ≤ k then the inequalities0 ≤ dir(eij) < dir(eij’) ≤ dir(ek ) ≤ dir(vij,vij’) <show that the edges eij, eij’ are parallel.

1

2

3

1 1

Π3

vij

vij’

eij

eij’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.

2 23 3 3

3 33 3 3

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The End…The End…