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Crossing Lemma - Part I 1 Computational Geometry Computational Geometry Seminar Seminar Lecture 7 Lecture 7 The “Crossing Lemma” and The “Crossing Lemma” and applications applications Ori Orenbach Ori Orenbach

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Computational Geometry Seminar. Lecture 7 The “Crossing Lemma” and applications Ori Orenbach. What did we see till now ?. Every planar graph with |V|=n vertices has at most 3n-6 edges (Euler’s formula). So, a graph with more than 3n-6 edges must have at least one crossing. For Example K 5 :. - PowerPoint PPT Presentation

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Page 1: Computational Geometry Seminar

Crossing Lemma - Part I1

Computational GeometryComputational GeometrySeminarSeminar

Lecture 7Lecture 7

The “Crossing Lemma” and The “Crossing Lemma” and applicationsapplications

Ori OrenbachOri Orenbach

Page 2: Computational Geometry Seminar

Crossing Lemma - Part I2

What did we see till now? Every planar graph with |V|=n vertices has at most 3n-6 edges (Euler’s formula)

So, a graph with more than 3n-6 edges must have at least one crossing

Ok, but how many crossings are there ? Can we minimize the crossings efficiently ?

For Example K5:

Page 3: Computational Geometry Seminar

Crossing Lemma - Part I3

Why is that important? In real life crossings are very common.

“The brick factory problem” (Turan)

Minimizing crossings is important in many fields, for example VLSI chip area:

Page 4: Computational Geometry Seminar

Crossing Lemma - Part I4

What will we see today?Crossing number definitions

Improving the lower bound using pre-assumptions on the graph (bisection width)

First lower bound on the crossing number

Improved constant on the lower bound using the probability method (The crossing lemma)

Tightness of the lower bound

Page 5: Computational Geometry Seminar

Crossing Lemma - Part I5

DefinitionsConsider a simple graph G=(V,E) with n vertices and m edges (m>3n-6)

Crossing of more than two edges in one point is not allowed

We want to embed G into the plane (just as we did for planar graphs)

Only now we know that we would have at least one crossing.

The crossing number of G: cr(G) is the smallest number of crossings among all drawings of G.

Page 6: Computational Geometry Seminar

Crossing Lemma - Part I6

DefinitionsIn such a minimal drawing the following three situations are ruled out:

No edge can cross itself

Edges with a common end vertex cannot cross

No two edges cross twice

Page 7: Computational Geometry Seminar

Crossing Lemma - Part I7

Immediate lower boundSuppose that G is drawn in the plane with cr(G) crossings.

Consider the following graph H:

The vertices of H are those of G together with all crossing points

The edges are all pieces of the original edges as we go along from crossing point to crossing point

G H

Page 8: Computational Geometry Seminar

Crossing Lemma - Part I8

Immediate lower boundObviously H is a planar graph (and simple)

|EH|= m+2cr(G) because every new vertex has a degree of 4.

|VH|= n+ cr(G)

And from Euler we get: m+2cr(G) ≤ 3(n+cr(G))-6

Cr(G) ≥ m-3n+6 (*)

Page 9: Computational Geometry Seminar

Crossing Lemma - Part I9

ExampleConsider K6. we have that:

Cr(K6)≥15-18+6=3

Indeed, we have a drawing with just 3 crossings:

Page 10: Computational Geometry Seminar

Crossing Lemma - Part I10

Can we know cr(G) for every graph?

The problem is believed to be NP complete

So, we cannot calculate the crossing number of a graph G efficiently, but can we bound it efficiently?

The bound proven before is good enough when m is linear in n, but not when m is larger compared to n

For example if m≥4n then m-3n+6≥n+6

How can we improve the bound?

Page 11: Computational Geometry Seminar

Crossing Lemma - Part I11

Theorem (Ajtai et al, Leighton, Chavatal, Newborn and others)

Let K(n,m) denote the minimum number of crossing pairs of edges in a graph with n vertices and m edges,

If m≥4n then

K(n,m)3

2

1 (1)100

mn

Known as the “Crossing lemma”

K(n,m) is the same as cr(G)

Page 12: Computational Geometry Seminar

Crossing Lemma - Part I12

The Crossing Lemma - Proof

We will prove by induction on n that:

If n≤8 then G cannot have 4n edges, so it must be the case that n≥9 (notice that if n=9 then G=K9)

34 2

3( ) ( ) ( /( )) 4 (2)64

n ncr G m m n

Then, if (2) is proven, then (1) will follow immediately

Page 13: Computational Geometry Seminar

Crossing Lemma - Part I13

The Crossing Lemma - Proof

The induction base case:

For n=9 we get

and (2) follows from (1)

We can see that (2) follows from (*) for every n≥10 and 4n≤m≤5n

(Notice that the right side of (2) cannot exceed m-3n)

9 9 34 2

3 *( )*(36 /( )) 664

Hence, We can assume that G is a graph with n≥10 vertices and m>5n edges and that (2) is valid for all graphs with fewer than n vertices.

Page 14: Computational Geometry Seminar

Crossing Lemma - Part I14

The Crossing Lemma - Proof

We notice that

Since G-x has at least m-(n-1) ≥5n-(n-1)>4(n-1) edges, we can use the induction on G-x, so we get:

( )

( ) ( 4) ( )x V G

cr G x n cr G

We can show that (Why?)

1 1 34 2

3( ) ( ) ( /( ))64

n nxcr G x m

( )

( 2)xx V G

m m n

(mx is the number of edges in G-x)

)*)

(**)

Denote by (G-x) the graph obtained from G by removing a vertex x and all connected edges)

Page 15: Computational Geometry Seminar

Crossing Lemma - Part I15

The Crossing Lemma - Proof

And now:

( )

1( ) ( )4 x V G

cr G cr G xn

134

1 3( )2

( )1 34 64 ( )

n

xnx V G

mn

134

1 32

( )1 3 ( 2)( )4 64 ( )

n

n

m nnn n

34 2

3 ( )( /( ))64

n nm 3 21 ( / ) n 9100

m n

)By*)

)By**)

)And Jensen (inequality

Page 16: Computational Geometry Seminar

Crossing Lemma - Part I16

How can we improve the bound?

Improve the constant

Improve the order of magnitude

Can be done only by using some assumptions on the graph (will be shown later)

Page 17: Computational Geometry Seminar

Crossing Lemma - Part I17

Improving the constant on the lower bound

Many proofs given

We will see a general proof using the “Probabilistic method *”, and an example in case that m≥4n

(The probability method used to prove an existence of an object in a collection by showing that the probability it exists is positive)

Page 18: Computational Geometry Seminar

Crossing Lemma - Part I18

The “Crossing Lemma” – An improved constantG=(V,E), |V|=n, |E|=m≥cn (for any c>3)

Then we can show a lower bound on the number of crossings in G, such that:

3

3 2

3( ) c mcr Gc n

Page 19: Computational Geometry Seminar

Crossing Lemma - Part I19

Crossing Lemma (using probabilistic method)Consider a minimal drawing of G (containing minimal number of crossings)

Let ‘p’ be a number between 0 and 1 (p will be chosen later)

Generate a subgraph of G: H

VH = Vertices of G chosen with probability p each

EH = All the edges uv in G, that both u and v were chosen in VH (we get that any edge remains with probability p2)

crH = Crossings in G that all four (distinct) vertices involved were chosen in VH (probability p4)

Page 20: Computational Geometry Seminar

Crossing Lemma - Part I20

Crossing Lemma (using probability)

Let np, mp crp be the random variables counting the number of vertices, edges and crossings in H

Since cr(G) ≥m-3n+6>m-3n for any graph we have that:

E(crp-mp+3np)≥0 E(crp)-E(mp)+E(3np)≥0

p4cr(G) –p2m +3pn≥0 *

And we get that:2

4 2 3

3 3( ) p m pn m ncr Gp p p

*Note that p4cr(G) is the expected number of crossings inherited from G, but the expected number of crossings in H is even smaller

Page 21: Computational Geometry Seminar

Crossing Lemma - Part I21

Crossing Lemma (using probability)

Now, set p= cn/m (which is at most 1 by our assumption)

And we get:

2 3 2 2 2 3 3 3

3 3 3

2 2 3 2 2 3

3 3( )/ /

3 3

m n m ncr Gp p c n m c n m

m m m cc n c n n c

2 3

3( ) m ncr Gp p

Page 22: Computational Geometry Seminar

Crossing Lemma - Part I22

Crossing Lemma (using probability)

For example if m≥4n then p=m/4n≤1

3

2

1( )64

mcr Gn

We can improve the constant by using different assumptions on the edges.

Page 23: Computational Geometry Seminar

Crossing Lemma - Part I23

Improving the constant

Theorem (Pach and Toth): let G be a simple graph drawn in the plane with cr(G) crossings, then

First we will mention the following corollary:

Corollary: The crossing number of any simple graph with at least 3 vertices satisfies:

cr(G)≥5e(G)-25V(G)+50

3

2

1( )33.75

mcr Gn

Page 24: Computational Geometry Seminar

Crossing Lemma - Part I24

Improving the constantCorollary: The crossing number of any simple graph G with at least 3 vertices satisfies:

cr(G)≥5e(G)-25V(G)+50

Proof Idea:

A graph with e(G)≥ (k+3)(v(G)-2) must have one edge crossing at least (k+1) other edges, for 0≤ k≤4.

By induction on e, we delete such an edge and the minimum number of crossings is:

4

0

( ) [ ( ) ( 3)( ( ) 2)]

5 ( ) 25 ( ) 50k

cr G e G k v G

e G v G

Page 25: Computational Geometry Seminar

Crossing Lemma - Part I25

Improving the constantProof (Theorem) : Again, we will use the probabilistic method. We have that e≥7.5n>(4+3)(n-2)

Using the corollary: cr(G)≥5e(G)-25V(G)+50

And we have that:

p4cr(G)≥5p2m-25pn

*p=7.5n/m<=1

3

2

1( )33.75

mcr Gn

Page 26: Computational Geometry Seminar

Crossing Lemma - Part I26

Tightness

We want To show that the lower bound is tight (we cant do better than m3/n2)

Consider n/t copies of Kt drawn on the plane:

Kt Kt Kt

n/t

Page 27: Computational Geometry Seminar

Crossing Lemma - Part I27

Tightness

Cr(Kt) = O(n4)

Kt Kt Kt

e(Kt) = O(n2)

m = t2*n/t=nt

cr = t4*n/t = t3n = m3/n2

Page 28: Computational Geometry Seminar

Crossing Lemma - Part I28

Improving the order of magnitude

The key is to have some pre assumptions on the graph, and use them to find a better bound

Bisection width and crossing number

Crossing number in graphs with monotone property

Page 29: Computational Geometry Seminar

Crossing Lemma - Part I29

Definitions

A graph property P is said to be monotone if:

1) Every subgraph of a graph G with the property also satisfies P

2) Whenever G1 and G2 satisfies P, their disjoint union also satisfies P.

Page 30: Computational Geometry Seminar

Crossing Lemma - Part I30

Definitions

The bisection width of a graph b(G) is defined to be:

The minimum is taken over all partitions of V(G) to 2 disjoint groups each n/3

1 2| |,| | / 3 1 2( ) min | ( , ) |v v nb G E V V

Page 31: Computational Geometry Seminar

Crossing Lemma - Part I31

Crossing number and bisection width

Theorem: let G be a graph with bounded degree, then:

Proof:

Consider a drawing of G with cr(G) crossings. Like before, we introduce a new vertex at each crossing, so we obtain a graph with n+cr(G) crossings

2( ) ( ( ) )cr G n b G

Page 32: Computational Geometry Seminar

Crossing Lemma - Part I32

Crossing number and bisection width

Proof continued:

We obtain a graph H with n+cr(G) vertices

We weight each new vertex with 0 weight and assign a weight of 1/n for old vertices, and we get by the planar separation theorem, that by deletion of at most:

O((n+cr(G))1/2)

vertices, H can be separated into H1 and H2 such that each one has at least n/3 elements and we get:

b(G) ≤ O((n+cr(G))1/2) •Note that this is a revised version on the planar separation theorem using weights

Page 33: Computational Geometry Seminar

Crossing Lemma - Part I33

Crossing number and monotone property

For any monotone property P, let ex(n, P) denote the max number of edges that a graph on n vertices can have if it satisfies P

If the property P is “G does not contain a subgraph isomorphic to a fixed forbidden subgraph H”, we write:

ex(n, H) for ex(n, P)

Page 34: Computational Geometry Seminar

Crossing Lemma - Part I34

Crossing number and monotone property

Theorem:

Let P be a monotone property with ex(n, P)=O(n1+a) for some a>0 then there exists two constants c, c’>0 such that the crossing number of any graph G with property P, which has n vertices and e≥cnlog2n edges satisfies:

2 1/

1 1/( ) 'a

a

ecr G cn

Page 35: Computational Geometry Seminar

Crossing Lemma - Part I35

Crossing number and monotone property

proof:

We will use a slightly different version of the bisection width lower bound:

Let G be a graph of n vertices, whose degrees are d1,…,dn

Then:

2

1

( ) 10 ( ) 2n

ii

b G cr G d

Will be used without a proof

Page 36: Computational Geometry Seminar

Crossing Lemma - Part I36

Crossing number and monotone property

Proof continued:

Let P be a monotone graph property with ex(n, P)

For some A,a>0

Let G be a graph with |V(G)|=n, |E(G)|=e

Suppose G satisfies P, and e≥cnlog2n

We assume by contradiction that:

1 aAn

2 1/

1 1/( ) 'a

a

ecr G cn

Page 37: Computational Geometry Seminar

Crossing Lemma - Part I37

Crossing number and monotone property

Proof idea:

We will use a decomposition algorithm:

On every step i of the algorithm we will look at the Mi components of the graph: Gi

1, …, GiMi

On every step of the algorithm every Gij falls into 2

components, each at most (2/3)n(Gij) vertices

We will stop when the number of vertices in each component is small enough

Page 38: Computational Geometry Seminar

Crossing Lemma - Part I38

The “Decomposition algorithm”

k steps

Page 39: Computational Geometry Seminar

Crossing Lemma - Part I39

Crossing number and monotone property

The algorithm:

Step 0: let G0=G, G10=G, M0=1, m0=1

On every step of the algorithm we will look at the Mi components of the graph: Gi

1, …, GiMi

Each component has at most (2/3)in vertices

We can assume (without loss of generality) that:

The first mi components of Gi have at least (2/3)i+1n vertices and the remaining Mi-mi has fewer

Page 40: Computational Geometry Seminar

Crossing Lemma - Part I40

Crossing number and monotone property

Proof continued:

We have that

The stopping rule:

Else: delete for j=1,…,mi b(Gji) edges from Gi

j such that Gij

falls into 2 components, each at most (2/3)n(Gij) vertices

1i

1

(2 / 3) ( ) ( ) (2 / 3) ( ) (j=1,2,...,m )

we have that: m (3/ 2)

i i ij

ii

n G n G n G

and

1/

1/ 1 1/

1(2 / 3)2

ai

a a

eA n

Page 41: Computational Geometry Seminar

Crossing Lemma - Part I41

Crossing number and monotone property

Let Gi+1 denote the resulting graph on the original set.

Each component of Gi+1 has at most (2/3)i+1n vertices

Suppose the decomposition algorithm terminates in step k+1, then if k>0:

1/1

1/ 1 1/

1(2 / 3) (2 / 3)(2 )

ak k

a a

eA n

Page 42: Computational Geometry Seminar

Crossing Lemma - Part I42

Crossing number and monotone property

Proof continued:

Using that for any non-negative real number:

We get:

1

1 1

2 1/1

1 1/

( ) ( ) (3 / 2) ( )

'(3 / 2)

i immi i ij i j

j j

ai

a

cr G m cr G cr G

c en

1 1

m m

j jj j

a m a

Page 43: Computational Geometry Seminar

Crossing Lemma - Part I43

Crossing number and monotone property

Proof continued:

Denoting the degree of a vertex v in Gij by d(v, Gi

j)

We get:

1

( )

1

(3 / 2) max ( , ) ( , )

(3 / 2) (2 / 3) (2 ) 3

i i

i

i

v V G

i i

d v G d v G

n e en

2 1 2

1 ( ) ( )

( , ) (3 / 2) ( , )i

i i

mi i ij

j v C G v V G

d v G d v G

Page 44: Computational Geometry Seminar

Crossing Lemma - Part I44

Crossing number and monotone property

Proof continued:

Now we use previous theorem about b(G), and we get that the total number of edges deleted during the procedure is:

1 1 12

0 1 0 1 0 1 ( )

2 1/ 1

1 1/0

2 1/ 1 1/1/

1 1/ 1/

( ) 10 ( ) 2 ( , )

10 ' (3 / 2) 2 3

250 ' (2 ) 2 32

i i

ij

m mmk k ki i ij j j

i j i j i j v V G

a ki

ai

a aa

a a

b G cr G d v G

ec k enn

e n ec A k enn e

)m=cnlog2n(

Page 45: Computational Geometry Seminar

Crossing Lemma - Part I45

Crossing number and monotone property

Proof continued:

And finally we get that e(Gk)≥e/2

Next we will find an upper bound on e(Gk):

The number of vertices of each connected component is:1/

1/1/ 1 1/

1( ) (2 / 3) ( )(2 ) 2

1,2,...,

ak k aj a a

k

e en G n nA n An

j M

Page 46: Computational Geometry Seminar

Crossing Lemma - Part I46

Crossing number and monotone property

Proof continued:

But each Gkj has the property P, since it is monotone, so it

follows that:1( ) ( ) ( )

2k a k kj j j

ee G An G An GAn

And from this, the total number of edges in Gk is:

1 1

( ) ( ) ( )2 2

k kM Mk k k

j jj j

e ee G e G A n GAn

A contradiction !

And the theorem is proved