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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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definition geometric graph
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
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
Definition
  • Let be the class of all geometric graphs with vertices, consisting of pairwise disjoint edges ( ).
  • Example (k=2):
theorem
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
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
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
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
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 continue1
Proof - continue
  • Examples:

|E|=1 |E|=2 |E|=3 |E|≥4

Number of deleted edges:

1 2 3 3

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proof continue2
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 continue3
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
Conclusions
  • The upper bound is:
  • The lower bound is:
what s next
What’s next?
  • Dilworth Theorem
  • Pach and Torocsik Theorem
definition partially ordered sets
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
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
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.
definitions1
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
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:

proof1
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 continue4
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 continue6
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 continue7
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 continue8
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
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
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
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.
definition1
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)}.
definition2
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
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|).

definition3
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
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 continue9
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 continue10
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 continue11
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 continue12
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 continue13
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 continue14
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 continue15
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 continue16
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 continue17
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
Lemma 3
  • There are no shielded edges.
proof continue18
Proof - continue
  • The total number of edges of E* is O(n).
  • The total number of edges of E is O(n).