Subdivision of Edge

1
Subdivision of Edge
In a graph G, subdivision of an edge uv is the
operation of replacing uv with a path u,w,v
through a new vertex w.
2
Subdivision of Graph
An H-subdivision (or subdivision of H) is a graph
obtained from a graph H by successive edge
subdivisions.
3
Kuratowski Subgraph

• Kuratowski subgraph of G a subgraph of G that is
  a subdivision of K5 or K3,3.
• Minimal nonplanar graph a nonplanar graph such
  that every proper subgraph is planar.
• Convex embedding of a graph a planar embedding
  in which each face boundary is convex polygon.

4
Proposition 6.2.1

• If a graph G has a Kuratowski subgraph, G is
  nonplanar.

Proof. Suppose G is planar. ? Each subgraph of G
is planar. ? Subdivision of K5 or K3,3 is
planar. ? K5 or K3,3 is planar. Its a
contradiction.
Subgraph of G
5
Lemma 6.2.7 and Lemma 6.2.11
Lemma 6.2.7 If G is a graph with fewest edges
among all nonplanar graphs without Kuratowski
subgraphs, then G is 3-connected. Lemma 6.2.11
If G is a 3-connected graph without Kuratowski
subgraphs, then G has a convex embedding in the
plane with no three vertices on a line.
6
Theorem 6.2.2 (Kuratowski 1930)
A graph is planar if and only if it does not
contain a subdivision of K5 or K3,3.
1. We need to show that a graph is nonplanar if
and only if it contains a subdivision of K5 or
K3,3. 2. It suffices to show a nonplanar graph
contains a subdivision of K5 or K3,3 by
Proposition 6.2.1. 3. Suppose that there exists a
graph G which has fewest edges among all
nonplanar graphs without Kuratowski subgraph.
7
Theorem 6.2.2
4. G is 3-connected by Lemma 6.2.7. 5. G is
planar by Theorem 6.2.11. 6. It implies no such G
exists.
8
Lemma 6.2.4
If F is the edge set of a face in a planar
embedding of G, then G has an embedding with F
being the edge set of the unbounded face.
?
9
S-lobe
Let S be a set of vertices in a graph G. An
S-lobe of G is an induced subgraph of G whose
vertex set consists of S and the vertices of a
component of G S.
10
Lemma 6.2.5
Every minimal non-planar graph is 2-connected.
Proof. 1. Let G be a minimal non-planar graph. 2.
We need to show the following two cases are
impossible (1) G is disconnected. (2) G
is 1-connected.
11
Lemma 6.2.5
3. Case 1 G is disconnected. 4. Let G1, G2, ,
Gk be the components of G. 5. G1, G2, , Gk are
planar since G is minimal non-planar graph.
6. We can embed one component of G inside one
face of an embedding of the rest. 7. It implies
G is a planar graph. Its a contradiction.
12
Lemma 6.2.5
8. Case 2 G has a cut-vertex v. 9. Let G1, G2,
, Gk be the v-lobes of G. 10. G1, G2, , Gk
are planar since G is minimal non-planar
graph. 11. We can embed each Gi with v on the
outside face.
12. We squeeze each embedding to fit in an angle
smaller than 360/k degrees at v. 13. It implies
G is a planar graph. Its a contradiction.
13
Lemma 6.2.6
Let S x, y be a separating 2-set of G. If G
is nonplanar, then adding the edge xy to some
S-lobe of G yields a nonplanar graph. 1. Let
G1,, Gk be the S-lobes of G, and let HiGi U xy.
2. If Hi is planar, it has an embedding with xy
on the outside face. 3.For each i gt1, Hi can be
attached to an embedding of Uj1i-1 Hj by
embedding Hi in a face that has xy on its
boundary.
14
Lemma 6.2.6
Let S x, y be a separating 2-set of G. If G
is nonplanar, then adding the edge xy to some
S-lobe of G yields a nonplanar graph. 4.
Deleting edge xy if it is not in G yields a
planar embedding of G.
15
Lemma 6.2.7
If G is a graph with fewest edges among all
nonplanar graphs without Kuratowski subgraphs,
then G is 3-connected. 1. Deleting an edge of G
cannot create a Kuratowski subgraphs in G. 2.
Deleting one edge produces a planar subgraph. 3.
G is a minimal nonplanar graph.
16
Lemma 6.2.7

• 4. G is 2-connected by Lemma 6.2.5.
• 5. Suppose that G has a separating 2-set S x,
  y.
• 6. Since G is nonplanar, the union of xy with
  some S-lobe, H, is nonplanar by Lemma 6.2.6.
• 7. Since H has fewer edges than G, the minimality
  of G forces H to have a Kuratowski subgraph F.

17
Lemma 6.2.7

• 8. All of F appears in G except possibly the edge
  xy.
• 9. Since S is a minimal vertex cut, both x and y
  have neighbors in every S-lobe.
• 10. we can replace xy in F with an x, y-path
  through another S-lobe to obtain a Kuratowski
  subgraph of G. It is a contradiction.

18
Lemma 6.2.9
Every 3-connected graph G with at least five
vertices has an edge e such that G.e is
3-connected. Proof 1. We use contradiction and
extremality. 2. Consider an edge e with
endpoints x, y. 3. If G.e is not 3-connected, it
has a separating 2-set S. 4. Since G is
3-connected, S must include the vertex obtained
by shrinking e. 5. Let z denote the other vertex
of S and call it the mate of the adjacent pair x,
y.
19
Lemma 6.2.9
Every 3-connected graph G with at least five
vertices has an edge e such that G.e is
3-connected. 6. x, y, z is a separating 3-set
in G. 7. Suppose that G has no edges whose
contraction yields a 3-connected graph, so every
adjacent pair has a mate. 8. Among all the edges
of G, choose e xy and their mate z so that the
resulting disconnected graph G x, y, z has a
component H with the largest order.
20
Lemma 6.2.9
Every 3-connected graph G with at least five
vertices has an edge e such that G.e is
3-connected. 9. Let H be another component of G
x, y, z. 10. Since x, y, z is a minimal
separating set, each of x, y, z has a neighbor in
each of H, H. 11. Let u be a neighbor of z in
H, and let v be the mate of u, z.
21
Lemma 6.2.9

• G z, u, v is disconnected.
• 13. The subgraph of G induced by V(H) U x ,y is
  connected. Deleting v from this subgraph, if it
  occurs there, cannot disconnect it, since then G
  z, v would be disconnected.
• 14. GV(H) ? x ,y v is contained in a
  component of G z, u, v that has more vertices
  than H, which contradicts the choice of x, y, z.

22
Branch Vertices
The branch vertices in a subdivision H of H are
the vertices of degree at least 3 in H.
23
Lemma 6.2.10
If G has no Kuratowski subgraph, then also G.e
has no Kuratowski subgraph. Proof 1. We prove
the contrapositive. 2. If G . e contains a
Kuratowski subgraph H, so does G. 3. Let z be
the vertex of G.e obtained by contracting e
xy. 4. If z is not in H, H itself is a
Kuratowski subgraph of G.
24
Lemma 6.2.10
If G has no Kuratowski subgraph, then also G.e
has no Kuratowski subgraph. 5. If z ? V(H) but z
is not a branch vertex of H, we obtain a
Kuratowski subgraph of G from H by replacing z
with x or y or with the edge xy. 6. If z is a
branch vertex in H and at most one edge incident
to z in H is incident to x in G, then expanding z
into xy lengthens that path, and y is the
corresponding branch vertex for a Kuratowski
subgraph in G.
25
Lemma 6.2.10

• If G has no Kuratowski subgraph, then also G.e
  has no Kuratowski subgraph.
• 7. In the remaining case, H is a subdivision of
  K5 and z is a branch vertex, and the four edges
  incident to z in H consist of two incident to x
  and two incident to y in G.

26
Theorem 6.2.11
If G is a 3-connected graph without Kuratowski
subgraphs, G has a convex embedding in the plane
with no three vertices on a line. Proof 1. We
use induction on n(G). 2. Basis step n(G) 4.
The only 3-connected graph with at most four
vertices is K4, which has such an embedding. 3.
Induction step n(G) 5. There exists an edge e
such that G.e is 3-connected by Lemma 6.2.9. 4.
Let z be the vertex obtained by contracting e.
5. By Lemma 6.2.10, G.e has no Kuratowski
subgraph.
27
Theorem 6.2.11

• If G is a 3-connected graph without Kuratowski
  subgraphs, G has a convex embedding in the plane
  with no three vertices on a line.
• 6. By the induction hypothesis, we obtain a
  convex embedding of H G.e with no three
  vertices on a line.
• 7. In this embedding, the subgraph obtained by
  deleting the edges incident to z has a face
  containing z.
• 8. Since H z is 2-connected, the boundary of
  this face is a cycle C.

28
Theorem 6.2.11

• If G is a 3-connected graph without Kuratowski
  subgraphs, G has a convex embedding in the plane
  with no three vertices on a line.
• 9. All neighbors of z lie on C they may be
  neighbors in G of x or y or both, where x and y
  are the original endpoints of e.
• 10. The convex embedding of H includes straight
  segments from z to all its neighbors. Let x1, ,
  xk be the neighbors of x in cycle order on C.

29
Theorem 6.2.11
11. If all neighbors of y lie in the portion of C
from xi to xi1, then we obtain a convex
embedding of G by putting x at z in H and putting
y at a point close to z in the wedge formed by
xxi and xxi1, as show in the diagrams for Case 0.
30
Theorem 6.2.11
12. If this does not occur, then either 1) y
shares three neighbors u, v, w with x, or 2) y
has neighbors u, v that alternate on C with
neighbors xi, xi1 of x.
31
Theorem 6.2.11

• 13. In Case 1, C together with xy and the edges
  from x, y to u, v, x form a subdivision of
  K5.
• 14. In case 2, C together with the paths uyv,
  xixxi1, and xy form a subdivision of K3,3.
• 15. Since we are considering only graphs without
  Kuratowski subgraph, in fact Case 0 must occur.