Computational Geometry Seminar

1
Computational GeometrySeminar

• Lecture 7
• The Crossing Lemma and applications
• Ori Orenbach

2
What did we see till now ?
Every planar graph with Vn vertices has at
most 3n-6 edges (Eulers formula)
So, a graph with more than 3n-6 edges must have
at least one crossing
For Example K5
Ok, but how many crossings are there ? Can we
minimize the crossings efficiently ?
3
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
4
What will we see today?
Crossing number definitions
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
Improving the lower bound using pre-assumptions
on the graph (bisection width)
5
Definitions
Consider a simple graph G(V,E) with n vertices
and m edges (mgt3n-6)
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 of more than two edges in one point is
not allowed
6
Definitions
In 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
7
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
8
Immediate lower bound
Obviously H is a planar graph (and simple)
EH m2cr(G) because every new vertex has a
degree of 4.
VH n cr(G)
And from Euler we get m2cr(G) 3(ncr(G))-6
Cr(G) m-3n6 ()
9
Example
Consider K6. we have that
Cr(K6)15-1863
Indeed, we have a drawing with just 3 crossings
10
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 m4n then m-3n6n6
How can we improve the bound?
11
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 m4n then K(n,m)
Known as the Crossing lemma
K(n,m) is the same as cr(G)
12
The Crossing Lemma - Proof
We will prove by induction on n that
If n8 then G cannot have 4n edges, so it must be
the case that n9 (notice that if n9 then GK9)
Then, if (2) is proven, then (1) will follow
immediately
13
The Crossing Lemma - Proof
The induction base case For n9 we get and (2)
follows from (1)
We can see that (2) follows from () for every
n10 and 4nm5n (Notice that the right side of
(2) cannot exceed m-3n)
Hence, We can assume that G is a graph with n10
vertices and mgt5n edges and that (2) is valid for
all graphs with fewer than n vertices.
14
The Crossing Lemma - Proof
Denote by (G-x) the graph obtained from G by
removing a vertex x and all connected edges)
We notice that
((
Since G-x has at least m-(n-1) 5n-(n-1)gt4(n-1)
edges, we can use the induction on G-x, so we get
(mx is the number of edges in G-x)
()
We can show that (Why?)
15
The Crossing Lemma - Proof
And now
(By (
(By ( (And Jensen (inequality
16
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)
17
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 m4n
(The probability method used to prove an
existence of an object in a collection by showing
that the probability it exists is positive)
18
The Crossing Lemma An improved constant
G(V,E), Vn, Emcn (for any cgt3)
Then we can show a lower bound on the number of
crossings in G, such that
19
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)
20
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-3n6gtm-3n for any graph we have
that
E(crp-mp3np)0 ? E(crp)-E(mp)E(3np)0 ? p4cr(G)
p2m 3pn0 And we get that
Note that p4cr(G) is the expected number of
crossings inherited from G, but the expected
number of crossings in H is even smaller
21
Crossing Lemma (using probability)
Now, set p cn/m (which is at most 1 by our
assumption)
And we get
22
Crossing Lemma (using probability)
For example if m4n then pm/4n1
We can improve the constant by using different
assumptions on the edges.
23
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
24
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) (k3)(v(G)-2)
must have one edge crossing at least (k1) other
edges, for 0 k4. By induction on e, we delete
such an edge and the minimum number of crossings
is
25
Improving the constant
Proof (Theorem) Again, we will use the
probabilistic method. We have that
e7.5ngt(43)(n-2) Using the corollary
cr(G)5e(G)-25V(G)50
And we have that p4cr(G)5p2m-25pn
? p7.5n/mlt1
26
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

n/t
27
Tightness

Cr(Kt) O(n4)
m t2n/tnt
cr t4n/t t3n m3/n2
e(Kt) O(n2)
28
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
29
Definitions

• A graph property P is said to be monotone if
• Every subgraph of a graph G with the property
  also satisfies P
• Whenever G1 and G2 satisfies P, their disjoint
  union also satisfies P.

30
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
31
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
ncr(G) crossings
32
Crossing number and bisection width
Proof continued We obtain a graph H with ncr(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((ncr(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((ncr(G))1/2)

• Note that this is a revised version on the planar
  separation theorem using weights

33
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)
34
Crossing number and monotone property
Theorem Let P be a monotone property with ex(n,
P)O(n1a) for some agt0 then there exists two
constants c, cgt0 such that the crossing number
of any graph G with property P, which has n
vertices and ecnlog2n edges satisfies
35
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
Will be used without a proof
36
Crossing number and monotone property
Proof continued Let P be a monotone graph
property with ex(n, P) For some A,agt0 Let G be a
graph with V(G)n, E(G)e Suppose G satisfies
P, and ecnlog2n We assume by contradiction
that
37
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 Gi1,
, 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
38
The Decomposition algorithm
k steps
39
Crossing number and monotone property
The algorithm Step 0 let G0G, G10G, M01,
m01 On every step of the algorithm we will look
at the Mi components of the graph Gi1, ,
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)i1n vertices and the
remaining Mi-mi has fewer
40
Crossing number and monotone property
Proof continued We have that The stopping
rule Else delete for j1,,mi b(Gji) edges
from Gij such that Gij falls into 2 components,
each at most (2/3)n(Gij) vertices
41
Crossing number and monotone property
Let Gi1 denote the resulting graph on the
original set. Each component of Gi1 has at most
(2/3)i1n vertices Suppose the decomposition
algorithm terminates in step k1, then if kgt0
42
Crossing number and monotone property
Proof continued Using that for any non-negative
real number We get
43
Crossing number and monotone property
Proof continued Denoting the degree of a vertex
v in Gij by d(v, Gij) We get
44
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
(mcnlog2n)
45
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
46
Crossing number and monotone property
Proof continued But each Gkj has the property
P, since it is monotone, so it follows that
And from this, the total number of edges in Gk is
A contradiction ! And the theorem is proved