Geometric Graphs and Quasi-Planar Graphs

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Geometric Graphs and Quasi-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.

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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 .

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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
• Examples:

|E|=1 |E|=2 |E|=3 |E|≥4

Number of deleted edges:

1 2 3 3

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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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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).