Geometric Graphs and QuasiPlanar Graphs

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Geometric GraphsandQuasi-Planar Graphs

• Dafna Tanenzapf

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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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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.

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Definition

• Let be the class of all geometric graphs
  with vertices, consisting of pairwise
  disjoint edges ( ).
• Example (k2)

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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
  

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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

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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 3n1 edges.
• We will show that there exist three edges which
  are disjoint.

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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).

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Proof - continue

• Examples
• E1 E2 E3
  E4
• 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
• (3n1-3n1).
• 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.

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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 dont intersect.

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Conclusions

• The upper bound is
• The lower bound is

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Whats next?

• Dilworth Theorem
• Pach and Torocsik Theorem

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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.

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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.

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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.
• 1rank(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.

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Definitions

• Let uv and uv 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)ltx(v) and x(u)ltx(v).
• uv precedes uv (uvltltuv) if
• x(u) x(u) and x(v) x(v).
• The edge uv lies below uv if there is no
  vertical line l that intersects both uv and uv
  such that y(lnuv) y(lnuv).

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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 k1 pairwise disjoint
  edges.
• Then for every k,n 1

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Proof

• Let G be a geometric graph with n vertices,
  containing no k1 pairwise disjoint edges.
• WLOG, no two vertices of G have the same
• x-coordinate.
• Let uv and uv be two disjoint edges of G such
  that uv lies below uv.

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Proof - continue

• We define four binary relations on E(G)
• uv lt1 uv if uvltltuv.
• uv lt2 uv if uvltltuv.
• uv lt3 uv if x(u),x(v) x(u),x(v).
• uv lt4 uv if x(u),x(v) x(u),x(v).

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Proof - continue
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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 .

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Proof - continue

• cannot contain a chain of length
  k1.
• Otherwise, G has k1 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.

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Proof - continue

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

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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.

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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).
  

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

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

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Definition

• The arrangement A(E) of E(G) is a complex set
  consisted of
• Nodes NV(G) X(G)X(G)crossing points.
• Segments SE(G)E(G)the edges between the
  vertices of N.
• Faces Fthe faces of the graph G(N,S).

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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.

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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(VE).

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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.

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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, E0n-1
• EE\E0.
• G G0

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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.

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Proof - continue

• For each edge e E, let ?(e) denote the set of
  segments of A(E0 e) that are contained in e.
• Every s ?(e) is fully contained in some face
  f A(E0) and its two endpoints lie on
  f.

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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.

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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.

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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.

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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 Eulers formula (E3V-6), there are at
  most O(nf) exposed edges in H1(f).
• nf is the number of vertices of H1(f).
• nf2f.

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Proof - continue

• We repeat this analysis for every Hi(f) (2i5).
• 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).

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Lemma 3

• There are no shielded edges.

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Proof - continue

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