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## PowerPoint Slideshow about 'Geometric Graphs and Quasi-Planar Graphs' - nardo

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### Geometric GraphsandQuasi-Planar Graphs

Dafna Tanenzapf

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.

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.

Definition

- Let be the class of all geometric graphs with vertices, consisting of pairwise disjoint edges ( ).
- Example (k=2):

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 :

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:

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 .

z

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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.

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).

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.

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.

Conclusions

- The upper bound is:
- The lower bound is:

What’s next?

- Dilworth Theorem
- Pach and Torocsik Theorem

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.

Theorem (Dilworth)

- Let be a finite partially ordered set.
- If the maximum length of a chain is k, then X can be partitioned into k antichains.
- If the maximum length of an antichain is k, then X can be partitioned into k chains.

Explanation (Dilworth Theorem)

- 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.
- 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.

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’).

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v’

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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:

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’.

Proof - continue

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

Proof - continue

- We can conclude from the definitions that:
- is a partially ordered set .
- Any pair of disjoint edges is comparable by at least one of the relations .

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.

Proof - continue

- From the second conclusion, any Ej does not contain two disjoint edges.
- We saw earlier that .
- Then:
- To sum up:

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.

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).

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.

Definition

- The arrangement A(E) of E(G) is a complex set consisted of:
- Nodes: N={V(G) X(G)|X(G)=crossing points}.
- Segments: S={E’(G)|E’(G)=the edges between the vertices of N}.
- Faces F={the faces of the graph G’=(N,S)}.

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|.

Lemma2

- Let G=(V,E) be a quasi-planar graph drawn in the plane.
- The complexity of all f of A(E) such that:
- f is a non-quadrilateral face.
- f is quadrilateral face incident to at least on vertex of G.

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

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.

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:

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.

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.

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.

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.

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.

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.

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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.

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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|.

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).

Lemma 3

- 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).

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