Crossing Lemma - Part I1 Computational Geometry Seminar Lecture 7 The “Crossing Lemma” and...
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Transcript of 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
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:
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:
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
Crossing Lemma - Part I5
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.
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
Crossing Lemma - Part I7
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
Crossing Lemma - Part I8
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 (*)
Crossing Lemma - Part I9
Example
Consider K6. we have that:
Cr(K6)≥15-18+6=3
Indeed, we have a drawing with just 3 crossings:
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?
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
m
n
Known as the “Crossing lemma”
K(n,m) is the same as cr(G)
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)
64n ncr G m m n
Then, if (2) is proven, then (1) will follow immediately
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 /( )) 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.
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( ) ( ) ( /( ))
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)
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 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
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)
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)
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 G
c n
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)
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 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
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 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
Crossing Lemma - Part I22
Crossing Lemma (using probability)
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.
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 G
n
Crossing Lemma - Part I24
Improving the constant
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
Crossing Lemma - Part I25
Improving the constant
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
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
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
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
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.
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
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
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
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)
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 c
n
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
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 c
n
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
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
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
e
A n
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
e
A n
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 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
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
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 en
n
e n ec A k en
n e
)m=cnlog2n(
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 n
A n An
j M
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( ) ( ) ( )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