Crossing Lemma - Part I1 Computational Geometry Seminar Lecture 7 The “Crossing Lemma” and...

Crossing Lemma - Part I1

Computational GeometryComputational GeometrySeminarSeminar

Lecture 7Lecture 7

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

Ori OrenbachOri 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

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

For Example K5:


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:


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


Definitions

Consider 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.


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


Immediate lower bound

Suppose 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


Immediate lower bound

Obviously 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 (*)


Example

Consider K6. we have that:

Cr(K6)≥15-18+6=3

Indeed, we have a drawing with just 3 crossings:


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?


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

m

n

Known as the “Crossing lemma”

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


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)

64n ncr G m m n

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



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 /( )) 6

64

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.



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( ) ( ) ( /( ))

64n n

xcr 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)



And now:

( )

1( ) ( )

4 x V G

cr G cr G xn

134

1 3( )2

( )1 3

4 64 ( )

n

xnx V G

mn

134

1 32

( )1 3 ( 2)( )

4 64 ( )

n

n

m nn

n n

34 2

3( )( /( ))

64n nm 3 21

( / ) n 9100

m n

)By*)

)By**)

)And Jensen (inequality


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)


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)


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 G

c n


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)


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 G

p 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



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 G

p p c n m c n m

m m m c

c n c n n c

2 3

3( )

m ncr G

p p



For example if m≥4n then p=m/4n≤1

3

2

1( )

64

mcr G

n

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


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 G

n



Corollary: 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



Proof (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 G

n


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


Tightness

Cr(Kt) = O(n4)

Kt Kt Kt

…

e(Kt) = O(n2)

m = t2*n/t=nt

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


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


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.


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


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


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


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)



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 c

n



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



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 c

n



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


The “Decomposition algorithm”

k steps



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



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

e

A n



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

e

A n



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 imm

i i ij i j

j j

ai

a

cr G m cr G cr G

c e

n

1 1

m m

j jj j

a m a



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



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 en

n

e n ec A k en

n e

)m=cnlog2n(



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 n

A n An

j M



Proof continued:

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

follows that:

1( ) ( ) ( )2

k a k kj j j

ee G An G An G

An

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 G

An

A contradiction !

And the theorem is proved

Crossing Lemma - Part I1 Computational Geometry Seminar Lecture 7 The “Crossing Lemma” and...

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