GEOMETRIC GRAPHSAND

QUASI-PLANAR GRAPHS

Dafna Tanenzapf

1

Definition – Geometric Graph

A geometric graph is a graph drawn in the plane by straight-line segments .

There are no three points in which are collinear.

The edges of can be possibly crossing.

),( EVG

)(GV

)(GE

2

Forbidden Geometric Graphs

Given a class H of forbidden geometric graphs determine (or estimate) the maximum number of edges that a geometric graph with n vertices can have without containing a subgraph belonging to H.

),( nHt

3

Definition

Let be the class of all geometric graphs with vertices, consisting of pairwise disjoint edges ( ).

Example (k=2):

kDkk2

1k

4

Theorem

Let be the maximum number of edges that a geometric graph with n vertices can have without containing two disjoint edges (straight-line thrackle). Then for every :

nDt ,2

3n

nnDt ,2

5

Theorem (Goddard and others) Let be the maximum number of

edges that a geometric graph with n vertices can have without containing three disjoint edges. Then:

nnDt 3),( 3

nDt ,3

6

Definitions

Edge xy is Leftmost at x if we can rotate it by (180 degrees) counter-clockwise around x without crossing any other edge of x.

A vertex x is called pointed if it has an angle between two consecutive edges that is bigger than .

Π

Π

y z

x

Π

7

x

Proof

Let G be a geometric graph with n vertices and at least 3n+1 edges.

We will show that there exist three edges which are disjoint.

8

Proof - Continue

For each pointed vertex at G, delete the leftmost edge.

We will denote the new subgraph as G1. For each vertex at G1 delete an edge if

there are no two edges from its right (begin from the leftmost edge of every vertex).

9

Proof - continue

Examples:

|E|=1 |E|=2 |E|=3 |E|≥4

Number of deleted edges: 1 2 3 3

10

y z

x

z

x x x

yy y u z vu

Proof - continue For every vertex we

deleted at most 3 edges. The remaining graph has

at least one edge x0y0 (3n+1-3n=1). In G1 there were two edges

to the right of x0y0: x0y1 and x0y2. In G1 there were two edges

to the right of x0y0: y0x1 and y0x2.

11

Proof - continue

In G there was the edge x2y to the left of x2y0. In G there was the edge y2x to the left of y2x0. WLOG we can assume that the intersection of

x0y2 and y0x2 is on the same side of y2 (if we split the plane into two parts by x0y0). The edges y0x2, x0y1, y2x don’t intersect.

12

Conclusions

The upper bound is:

The lower bound is:

nnDt 3),( 3

45.2),( 3 nnDt

13

What’s next?

Dilworth Theorem

Pach and Torocsik Theorem

14

Definition – Partially ordered sets A partially ordered set is a pair . X is a set. is a reflexive, antisymmetric and

transitive binary relation on X. are comparable if or

. If any two elements of a subset are

comparable, then C is a chain. If any two elements of a subset are

incomparable, then C is an antichain.

),( X

Xyx yx xy

XC

XC

15

Theorem (Dilworth)

Let be a finite partially ordered set.1. If the maximum length of a chain is k, then

X can be partitioned into k antichains.2. If the maximum length of an antichain is k,

then X can be partitioned into k chains.

),( X

16

Explanation (Dilworth Theorem)1. If the maximum length of a chain is k, then X

can be partitioned into k antichains. For any x X, define the rank of x as the size of the

longest chain whose maximal element is x. 1≤rank(x)≤k The set of all elements of the same rank is an

antichain.

2. If the maximum length of an antichain is k, then X can be partitioned into k chains.

It can be shown by induction on k.

17

Definitions

Let uv and u’v’ be two edges. For any vertex v let x(v) be the x-coordinate

and let y(v) be the y-coordinate. Suppose that x(u)<x(v) and x(u’)<x(v’). uv precedes u’v’ (uv<<u’v’) if

x(u) x(u’) and x(v) x(v’). The edge uv lies below u’v’ if there is no

vertical line l that intersects both uv and u’v’ such that: y(l∩uv) y(l∩u’v’).

18

u’

v

v’

u

u’

l

v’

u v

Theorem (Pach and Torocsik) Let denote the maximum

number of edges that a geometric graph with n vertices can have without containing k+1 pairwise disjoint edges.Then for every k,n 1:

),( 1 nDt k

nknDt k4

1 ),(

19

Proof

Let G be a geometric graph with n vertices, containing no k+1 pairwise disjoint edges.

WLOG, no two vertices of G have the same x-coordinate.

Let uv and u’v’ be two disjoint edges of G such that uv lies below u’v’.

20

Proof - continue

We define four binary relations on E(G):1. uv <1 u’v’ if uv<<u’v’.2. uv <2 u’v’ if u’v’<<uv.3. uv <3 u’v’ if [x(u),x(v)] [x(u’),x(v’)].4. uv <4 u’v’ if [x(u’),x(v’)] [x(u),x(v)].

21

Proof - continue22

Proof - continue

We can conclude from the definitions that:

1. is a partially ordered set .

2. Any pair of disjoint edges is comparable by at least one of the relations .

iGE , 41 i

i

23

Proof - continue cannot contain a chain of length k+1. Otherwise, G has k+1 pairwise disjoint edges. According to the previous theorem, for any i, E(G)

can be partitioned into at most k anti-chains (classes), so that no two edges belonging to the same class are comparable by .

The edges can be partitioned into classes Ej , such that no two elements of Ej are comparable by any relation.

i

iGE ,

41 ki

4k

24

Proof - continue

From the second conclusion, any Ej does not contain two disjoint edges.

We saw earlier that . Then:

To sum up:

nnDt ),( 2

42 1 ),( kjnnDtE j

4

1

4k

jj nkEGE

25

Definition – Quasi Planar Graphs A graph is called quasi planar if it can

be drawn in the plane so that no three of its edges are pairwise crossing.

26

Motivation

We want to find an upper bound to the number of edges of quasi-planar graph.

Pach had shown that a quasi-planar graph with n vertices has edges.

For the general case, k-quasi-planar graph (a graph with no k pairwise crossing edges), the upper bound is:

We will prove a Theorem. Its conclusion will be that the upper bound is: (can be shown in induction).

)log( 2 nnO

)log( 42 nnO k

)log( 62 nnO k

27

Theorem (Agarwal, Aronov, Pach, Pollack and Sharir)

If G(V,E) is a quasi-planar graph (undirected, without loops or parallel edges), then |E|=O(|V|).

We will prove the theorem for the case that G has a straight-line drawing in the plane with no three pairwise crossing edges.

28

Definition

The arrangement A(E) of E(G) is a complex set consisted of:

1. Nodes: N=V(G) X(G)|X(G)=crossing points.

2. Segments: S=E’(G)|E’(G)=the edges between the vertices of N.

3. Faces F=the faces of the graph G’=(N,S).

29

Definition

The complexity |f| of f F is the number of segments in S on the boundary f of f.

If a segment is in the interior of f, then it contributes 2 to |f|.

30

Lemma2

Let G=(V,E) be a quasi-planar graph drawn in the plane.

The complexity of all f of A(E) such that:1. f is a non-quadrilateral face.2. f is quadrilateral face incident to at least on

vertex of G.

is O(|V|+|E|).

31

Definition

A graph is called overlap graph if its vertices can be represented by intervals on a line so that two vertices are adjacent if and only if the corresponding intervals overlap but neither of them contains the other.

32

Proof (Theorem)

Let G be a quasi-planar graph drawn in the plane with n vertices.

WLOG G is a connected graph. Let G0=(V,E0) be a spanning tree of G, |

E0|=n-1 E*=E\E0. G: G0:

33

Proof - continue

Each face of A(E0) is simply connected. By Lemma 2, the complexity of non-

quadrilateral and quadrilateral faces of A(E0) incident to a point of V is O(n).

We call the remaining faces of A(E0) crossing quadrilateral.

34

Proof - continue

For each edge e E*, let Ω(e) denote the set of segments of A(E0 e) that are contained in e.

Every s Ω(e) is fully contained in some face f A(E0) and its two endpoints lie on f.

35

Proof - continue

For each f A(E0) let X(f) denote the set of all segments in that are contained in f.

Any two segments in X(f) cross each other if and only if their endpoints alternate along f.

We “cut” the face in some vertex and get an open interval.

Two elements of X(f) cross each other if and only if the corresponding

intervals overlap.

*

)(Ee

e

36

Proof - continue

This defines a triangle-free overlap graph on the vertices X(f).

By Gyarfas and Kostochka, every triangle-free overlap graph can be colored by 5 colors.

Therefore, the segments of X(f) can be colored by at most 5 colors, so that no two segments with the same color cross each other.

37

Proof - continue

For each f A(E0) let H(f) denote the quasi- planar graph whose edges are X(f).

A monochromatic graph is a graph which all its edges are colored in one color.

Let f A(E0) be a face that is not crossing quadrilateral.

Let H1(f),…,H5(f) be the monochromatic subgraphs of H.

38

Proof - continue

We fix one of Hi, assume WLOG H1, and we reinterpret it to a new graph:every edge on the boundary of the face and vertices of V on the boundary will be a vertex, and all the interior segments will be the edges in the new graph.

The resulting graph H1* is planar.

39

a

b

c

de

a c

p

b

d

e p

Proof - continue

Face of H1*(f) is a digon if it is bounded by two edges of H1*(f).

Edge of H1*(f) is shielded if both of the faces incident to it are digons.

The remaining edges of H1*(f) are called exposed.

40

a

b

c

de

p

a c

b

d

e p

Proof - continue

By Euler’s formula (|E|≤3|V|-6), there are at most O(nf) exposed edges in H1*(f).

nf is the number of vertices of H1*(f). |nf|≤2|f|.

41

Proof - continue

We repeat this analysis for every Hi(f) (2≤i≤5).

The number of edges e E* containing at least one exposed segment is .

By Lemma2, this sum is O(n). We need to bound the number of e E*

with no exposed segments (shielded edges).

)||(f

fO

42

Lemma 343

There are no shielded edges.

Proof - continue

The total number of edges of E* is O(n).

The total number of edges of E is O(n).

44