Computational Geometry Seminar Lecture 1

1
Computational Geometry Seminar Lecture 1

• Drawing Planar Graphs

2
Graph

• Definition A Simple Undirected Graph G consists
  of
• A finite set of vertices V(G).
• A set E(G) of sets of 2 vertices, called edges.
• Note From now on we will call this type of
  graphs by the general name Graph.

3
Degree

• Definition In a graph G, the vertices u, v are
  adjacent iff the edge uv belongs to E(G).
• The number of adjacent vertices to a vertex u is
  called the degree of u, denoted d(u).

4
Subgraph

• Definition A graph H is a subgraph G (written H
  G) iff V(H) V(G) and E(H) E(G).
• We say that a graph H is the subgraph of G
  induced by a set of vertices U V(G) if
  V(H) U and E(H) is the set of all the edges
  of G connecting vertices of U.

5
Paths and Cycles

• Definition A sequence of k distinct vertices, in
  which every consecutive vertices are adjacent, is
  called a path of length k-1.
• Definition A path of length k-1 with the
  addition of an edge from the last vertex to the
  first vertex of the path is called a cycle of
  length k.

6
Connectivity

• Definition A maximal set of vertices, for which
  there exists a path from every vertex in the set
  to another, is called a connected component.
• A graph composed of only one connected component
  is called connected.

7
Graphs in the plane

• Can represent graphs in the plane by assigning
  distinct points to vertices and drawing
  continuous non-self-intersecting curves (Jordan
  arcs) between adjacent vertices.

8
Graphs in the plane

• But sometimes we want the drawing to be simple or
  satisfy other requirements, such as straight
  line segments as arcs or avoiding crossing arcs.

• Can we always achieve that?

9
Planar Graphs

• Definition A graph that can be represented in
  the plane so that no two arcs meet at a point
  other than their endpoints is called a Planar
  graph.
• Such representation of a planar graph is called a
  Plane graph or Planar embedding of the graph.

10
Nonplanar Graphs
K5 and K3,3 are not planar
K5
K3,3
11
K3,3 is not planar
u1
v3
v2
u2
u3
v1
12
Subdivision

• A subdivision of a graph is obtained by repeating
  the operation of removing an edge and introducing
  a new vertex connected to the endpoints of the
  edge removed.

13
Kuratowskis Theorem

• The theorem states that a graph is not planar iff
  it has a subgraph which is a subdivision of K5 or
  K3,3.

14
Straight Line Embedding

• Deleting any edge from K5 will result in a planar
  graph. Moreover this graph can be embedded in the
  plane by using straight line segments.

• Does every planar graph have a straight line
  embedding???

15
Faces

• Definition A plane graph divides (with its arcs)
  the plane into connected regions called faces.
• Exactly one of these faces is unbounded and is
  called the exterior face.
• We denote the number of faces of a plane graph G
  by f(G).

Exterior Face
16
Dual Graph

• Definition For a plane graph G we construct G,
  the dual of G as follows. A vertex is placed in
  each face of G. These are the vertices of G. For
  each edge e of G we draw an edge e, called the
  dual edge of e, which crosses e (and no other
  edge of G) and joins the vertices corresponding
  to the faces, whose boundary consists of e.

17
Eulers Formula

• For a connected plane graph G

v 5 e 7 f 4
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Eulers Formula (Proof)

• Proof By induction on f.
• If f(G) 1 then G has no cycles (a tree), thus
  e(G) v(G) 1.

• Assume for f(G) 2 that the theorem is correct
  for connected plane graphs with fewer than f(G)
  faces.

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Eulers Formula (Proof)

• Delete an edge e that belongs to a cycle in G.
  For the resulting connected plane graph G-e we
  get f(G-e) f(G) 1. Therefore, using
  the induction hypothesis on G-e we obtain
• v(G) (e(G) 1) (f(G) 1) 2. ?

20
Eulers Formula

• Note When dealing with disconnected graphs we
  can add edges between connected components
  without adding more faces, eventually creating a
  connected graph. Thus, the formula becomes (C is
  the number of connected components)

21
Bridges

• Definition An edge which is a boundary to only
  one face is called a bridge.
• Same bridges as in graph theory (edges not
  contained in cycles).

22
Sides

• Definition For a face f of G, the number of
  sides of f is the number of edges belonging to
  the boundary of f, where bridges are counted
  twice. Denoted s(f).

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Sides

• Every non-bridge is in a boundary of exactly two
  faces. Therefore

24
Eulers Formula

• Note According to Eulers formula the number of
  faces is independent of the embedding we choose
  for the graph. However, the number of sides of
  the faces is not.

25
Triangulation

• Definition A face f for which s(f) 3 is
  called a triangle. If all faces of G are
  triangles, G is called a Triangulation.

Triangle
Triangulation
26
Triangulation

• Every graph can be extended to a triangulation by
  the addition of new edges between existing
  vertices.

• A triangulation is maximal in the sense that no
  more edges can be added without violating its
  planarity.

27
Eulers Formula

• Corollary 1 For a plane graph G with at least 3
  vertices
• e(G) 3v(G) 6
• f(G) 2v(G) 4
• The equalities hold iff G is a triangulation.

28
Corollary 1

• Proof Sufficient to prove for connected plane
  graphs since otherwise number of edges and faces
  only decreases.
• For every face f of G, s(f) 3.
• Thus
• By Eulers formula we obtain
• v(G) e(G) e(G) 2
• v(G) f(G) f(G) 2 ?

29
Chromatic Number

• Definition The chromatic number ?(G) of a graph
  G is the minimum number of colors required to
  color the vertices of G so that no adjacent
  vertices are of the same color.

30
The Four-Color Theorem

• According to the four-color-theorem of Appel and
  Haken the chromatic number of a planar graph is
  at most 4.
• This bound cannot be
• improved.
• Proof is quite complicated. Well prove a weaker
  statement deduced from Corollary 1.

31
Corollary 2

• Corollary 2 If G is a planar graph, then ?(G)
  5.
• Proof By induction on v(G).
• If v(G) 5, we can assign every vertex a
  different color.
• Assume that for v(G) 6 we proved the statement
  for graphs of size smaller than v(G).

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Corollary 2 (Proof)

• From Corollary 1 G must have a vertex u with
  d(u) 5.
• Otherwise for every vertex u, d(u) 6. Thus

In contradiction to corollary 1.
33
Corollary 2 (Proof)

• If d(u) 4 then color the rest of the graph G-u
  with 5 colors using the induction, and then color
  u with a color different from its neighbors.

34
Corollary 2 (Proof)

• If u has 5 neighbors wi (1 i 5)
• Since G is planar it does not contain K5 as a
  subgraph. Thus, assume WLOG that w1 and w2 are
  not adjacent.

Merge
Let G be the graph obtained from G-u by merging
w1 and w2 to w, which is adjacent to neighbors
of w1 or w2.
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Corollary 2 (Proof)

• G is a planar graph, hence we apply the
  induction to obtain a 5-coloring of G. If we
  use the same coloring on G-u, where w1 and w2 are
  assigned the color of w, the neighbors of u are
  colored in at most 4 colors. Therefore, we can
  assign a color for u different from its
  neighbors. ?

36
Straight Line Drawing

• We will now prove that every planar graph has and
  embedding with straight line segments, called the
  straight-line embedding.

37
Straight Line Drawing

• Let us start off with the following lemma
• Lemma 1 Let G be a plane graph whose exterior
  face is bounded by a cycle u1, , uk. Then
  exists up (p ? 1, k) not adjacent to any uj other
  than up-1 and up1.

u2
u3
u4
u5
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Lemma 1 (Proof)

• Proof If no two non-consecutive vertices in the
  exterior boundary are adjacent, then the lemma is
  trivial. Otherwise
• Pick two non-consecutive adjacent vertices ui, uj
  (j gt i1) for which j-i is minimal.

u2
i 1 j 3
u3
u4
u5
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Lemma 1 (Proof)

• ui1 cannot be adjacent to u1, , ui-1, , uj1 ,
  , uk by planarity.

u2
i 1 j 3
u3
crossing
u4
u5

• ui1 cannot be adjacent to any ui, , uj other
  than ui and ui2 because of minimality of j-i. ?

40
Canonical Construction of Triangulations

• Let G be a triangulation with an exterior face
  uvw and a labelling u1 u, u2 v, u3, , un w
  of the vertices of G. Denote by Gk the subgraph
  of G induced by u1, , uk and by Ck the exterior
  boundary of Gk.

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Canonical Construction of Triangulations

• There exists a labelling such that for every 4
  k n

1. Gk-1 is internally triangulated.
2. The edge uv is in Ck-1.
3. uk is in the exterior face of Gk-1 the neighbors
   of uk in V(Gk-1) are consecutive on
   Ck-1.

• This kind of labelling is called a canonical
  labelling.

42
Canonical Construction of Triangulations (Proof)

• Proof We define un, un-1, , u3 by reverse
  induction.
• For un w

w
u
v
43
Canonical Construction of Triangulations (Proof)

• For 4 k n Assume we defined un, , uk
  correctly. Applying Lemma 1 to Gk-1, the subgraph
  induced by the remaining vertices, we know that
  there is a vertex on Ck-1, other than u and v,
  that is only adjacent to its preceding and
  subsequent vertices on Ck-1. Let uk-1 be that
  vertex. Proof of I.-III. is the same as with
  un. ?

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Canonical Labelling (Example)
w
u
v
45
Canonical Labelling (Example)
w u7
u2v
u u1
46
Canonical Labelling (Example)
u6
u2v
u u1
47
Canonical Labelling (Example)
u5
u2v
u u1
48
Canonical Labelling (Example)
u4
u2v
u u1
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Canonical Labelling (Example)
u3
u2v
u u1
50
Straight Line Drawing

• Corollary 3 Every planar graph has a straight
  line embedding in the plane.
• Proof It is sufficient to show that the
  statement is true for any maximal planar graph,
  i.e. a graph that can be represented as a
  triangulation.

51
Corollary 3 (Proof)

• Let G be a triangulation with the canonical
  labelling u1 u, u2 v, u3, , un w as
  described earlier. We will determine the
  positions p(uk) (x(uk), y(uk)) of the vertices
  by induction on k.

52
Corollary 3 (Proof)

• Set p(u1) (0,0) , p(u2) (2,0) , p(u3) (1,1).

• For k 4 assume that p(u1), , p(uk-1) have
  already been defined s.t., by connecting the
  images of adjacent vertices with segments, we
  obtain a straight line embedding of Gk-1, whose
  exterior face is bounded by the segments
  corresponding to the edges of Ck-1.

53
Corollary 3 (Proof)

• Let v1 u, v2, , vm v denote the vertices of
  Ck-1 listed in the order they appear on the
  cycle.
• Suppose further that
• (1) x(v1) lt x(v2) lt lt x(vm)
• y(vi) gt 0 for 1 lt i lt m

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Corollary 3 (Proof)

• By condition III. of the canonical labelling, uk
  is connected to a series of consecutive vertices
  of Ck-1, vq, , vr (1 q lt r m). Choose x(uk)
  s.t. x(vq) lt x(uk) lt x(vr). Now choose
  y(uk) gt 0 large enough. The new position p(uk)
  meets all the requirements (including the
  assumption (1)). ?

55
Straight Line Drawing (Example)
The starting graph
56
Straight Line Drawing (Example)
Making the triangulation
57
Straight Line Drawing (Example)
w u6
u3
u5
u4
v u2
u u1
Canonical labelling
58
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
v u2
u u1
x
2
0
1
Setting p(u1) p(u2) p(u3)
59
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
v2
v u2
v3
u u1
v1
x
2
0
1
60
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
v2
v3
v u2
v4
u u1
v1
x
2
0
1
61
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
v2
v3
v4
v u2
v5
u u1
v1
x
2
0
1
62
Straight Line Drawing (Example)
Delete added edges
63
Straight Line Drawing (Example)
64
Straight Line Drawing

• Note By using this method we can also establish
  the existence of straight line embedding of other
  representations. For example, straight line
  embeddings of the following representation, found
  by Rosenstiehl and Tarjan.

65
Straight Line Drawing

• Corollary 4 The vertices and edges of any planar
  graph can be represented by horizontal and
  vertical segments, respectively, s.t.
• I. Segments dont have interior points in
  common.
• II. Horizontal segments connected by a vertical
  segment represent adjacent vertices.

66
Corollary 4 (Proof)

• Proof As earlier it is sufficient to establish
  the statement for triangulations.
• Let G be a triangulation with the canonical
  labelling u1 u, u2 v, u3, , un w. To every
  uk we will assign a segment s(uk) whose endpoints
  are (xk, k) and (x'k, k).

67
Corollary 4 (Proof)

• Set x10, x'12, x22, x'24, x31, x'33.

• For k 4 assume that s(u1), , s(uk-1) have
  already been defined s.t., the subgraph Gk-1 has
  a representation satisfying conditions I. and II.

68
Corollary 4 (Proof)

• We define the upper envelope of s(u1), , s(uk-1)
  as the set of points for which no other point is
  directly above them. Denote this by UEk-1.

69
Corollary 4 (Proof)

• Let v1 u, v2, , vm v denote the vertices of
  Ck-1 listed in the order they appear on the
  cycle.
• Suppose further that UEk-1 consists of portions
  of s(v1), , s(vm), in this order
  (from left to right).

s(v2)
s(v3)
s(v4)
s(v1)
70
Corollary 4 (Proof)

• By condition III. of the canonical labelling, uk
  is connected to a series of consecutive vertices
  of Ck-1, vq, , vr (1 q lt r m). Choose xk to
  be some point of s(vq) which belongs to UEk-1 and
  choose x'k to be some point which belongs to
  UEk-1. The resulting representation of Gk meets
  the requirements. ?

71
Straight Line Drawing (Example)
w u6
u3
u5
u4
v u2
u u1
Canonical labelling
72
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
v u2
u u1
x
4
0
2
Setting s(u1) s(u2) s(u3)
73
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
s(v2)
s(v3)
v u2
u u1
s(v1)
x
4
0
2
74
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
s(v2)
s(v3)
s(v4)
v u2
u u1
s(v1)
x
4
0
2
75
Straight Line Drawing (Example)
Left
u3
w u6
u4
u5
s(v2)
s(v3)
s(v4)
s(v5)
v u2
u u1
s(v1)
x
4
0
2
76
Straight Line Drawing (Example)
Delete added edges
77
Straight Line Drawing (Example)
78
3D Implementations

• So far weve seen embeddings of graphs in 2D (on
  the plane).
• We will now see how to embed planar graphs in 3D.
• We will also get to know new shapes which allow
  use to draw graphs which are not planar.

79
The Sphere

• Every planar graph can be embedded on the sphere.
  
• Likewise, every embedding on the sphere can be
  transformed to an embedding on the plane.

80
The Sphere

• To transform an embedding on the sphere to an
  embedding on the plane
• We choose a face and puncture a hole in it. Then
  stretch the face using the hole until we get a
  flat surface. That face will be the exterior face.

• This procedure is reversible.

81
Handle

• Definition A handle is a tube joining two holes
  cut in a surface.

82
Genus

• Definition The genus of a surface obtained by
  adding handles to a sphere is the number of
  spheres added.

83
Genus
Genus 1 Torus
Genus 0 Sphere
Genus 2 Double-Torus
Genus 3 Pretzel
84
Graphs

• The genus of a graph is the minimum ? such that
  the graph can be embedded on a surface of genus
  ?.
• As stated planar graphs can be embedded on the
  sphere. Therefore, they are of genus 0.

85
Graphs

• K5 and K3,3 are of genus 1.