Curve

1
Curve

• Curve The image of a continous map from 0,1 to
  R2.
• Polygonal curve A curve composed of finitely
  many line segments.
• Polygonal u,v-curve A polygonal curve that
  starts at u and ends at v.

2
Drawing of G

• Drawing of a graph G A function f defined on
  V(G)?E(G) that assigns each vertex v a point f(v)
  in the plane and assigns each edge with endpoints
  u,v a polygonal f(u),f(v)-curve.
• Crossing A point in f(e)?f(e) that is not a
  common endpoint.

3
Planar Embedding of G

• Planar graph A graph has a drawing without
  crossings.
• Planar embedding of planar graph G A drawing of
  G that has no crossings.
• Plane graph A particular planar embedding of a
  planar graph.
• Closed curve A curve whose first and last points
  are the same.
• Simple closed curve A closed curve that has no
  repeated points except possibly firstlast.

4
Chord

• Chord of cycle C An edge not in C whose
  endpoints lie in C

5
Proposition 6.1.2

• K5 and K3,3 cannot be drawn without crossings.
• Proof. 1. Let C be a spanning cycle in a drawing
  of K5 or K3,3.
• 2. If the drawing does not have crossing edges, C
  is drawn as a closed curve.

6
Proposition 6.1.2
3. Two chords conflict if their endpoints on C
occur in alternating order. 4. When two chords
conflict, we can draw only one inside C and one
outside C.
Two Chords Conflict
7
Proposition 6.1.2
5. K3,3 has three pairwise conflict chords.
We can put at most one inside and one outside, so
it is not possible to complete the embedding.
8
Proposition 6.1.2
5. In K5, at most two chords can go outside or
inside. Since there are five chords, it is
not possible to complete the embedding.
Red Line chord
9
Face

• Open set A set U that belongs to R2.
• Region An open set U that contains a polygonal
  u, v-curve for every pair u, v belongs to U.
• Face A maximal region of the plane that contains
  no point used in the embedding.

10
Dual Graph

• Dual graph G of a plane graph G
• 1. The vertices of G correspond to the faces of
  G.
• 2. The edges of G correspond to the edges of G
• If e is an edge of G with face X on one side and
  face Y on the other side, then the endpoints of
  the dual edge e?E(G) are the vertices x, y of
  G that represent the faces X, Y of G.
• The order in the plane of the edges incident to
  x?V(G) is the order of the edges bounding the
  face X of G in a walk around its boundary.

G
11
Dual Graph

• Two embeddings of a planar graph may have
  non-isomorphic duals
• 1. Each embedding shown below has 3 faces, so in
  each case the dual has 3 vertices.
• 2. In the embedding on the right, the dual vertex
  corresponding to the outside face has degree 4.
• 3. In the embedding on the left, no dual vertex
  has degree 4, so the duals are no isomorphic.

12
Face Length
Length of a face in a plane graph G The total
length of the closed walk in G bounding the face.
1
2
3
7
4
5
6
13
Example 6.1.12
1. A cut-edge belongs to the boundary of only one
face, and it contributes twice to its length. 2.
In the embedding on the left the lengths of three
faces are 3, 6, 7 on the right they are 3, 4,
9. 3. The sum of the lengths is 16 in each case,
which is twice the number of edges.
F1
F2
F1
F0
F2
F0
14
Proposition 6.1.13

• If L(Fi) denotes the length of face Fi in a plane
  graph G, then 2e(G) ??L(Fi).

15
Eulers Formula

• If a connected plane graph G has exactly n
  vertices, e edges, and f faces, then n e f
  2.

n 7 e 8 f 3 n e f 2
16
Eulers Formula

• Proof. 1. Use induction on n (number of
  vertices).
• 2. Basis (n 1)
• G is a bouquet of loops, each a closed curve in
  the embedding. If e 0, then f 1, and the
  formula holds.
• Each added loop passes through a face and cuts it
  into 2 faces (by the Jordan Curve Theorem). This
  augments the edge count and the face count each
  by 1. Thus the formula holds when n 1 for any
  number of edges.

17
Eulers Formula

• 3. Induction step (ngt1)
• There exists an edge e that is not a loop because
  G is connected.
• Obtain a plane graph G with n vertices, e
  edges, and f faces by contracting e.
• Clearly, nn1, ee1, and ff.
• n-ef2 by the induction hypothesis, implying n
  e f 2.

18
Theorem 6.1.23

• If G is a simple planar graph with at least three
  vertices, then e(G)3n(G)6.
• If also G is triangle-free, then e(G) 2n(G)4.
• Proof. 1. It suffices to consider connected
  graphs otherwise, we could add edges.
• 2. If n(G) ? 3, every face boundary in a simple
  graph contains at least three edges (?L(Fi)? 3f).
• 3. By Proposition 6.1.13, 2e?L(Fi), implying
  2e?3f.
• (2e?4f)
• 4. By Eulers Formula, nef2, implying e 3n
  6.
• (e 2n4)

(?L(Fi)? 4f)
19
Nonplanarity of K5 and K3,3

• These graphs have too many edges to be planar.
• For K5, we have e 10gt9 3n-6.
• Since K3,3 is triangle-free, we have e 9gt8
  2n-4.