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.
Subdivision of Graph
An H-subdivision (or subdivision of H) is a graph
obtained from a graph H by successive edge subdivisions.
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.
G
convex polygon
G
non-convex polygon
Kuratowski’s Theorem (if)
• A graph G is nonplanar if G has a Kuratowski
subgraph.
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. It’s a contradiction.
Subgraph of G
G
Kuratowski’s Theorem (only if)
Lemma 6.2.7: If G is a graph with fewest edges among
all nonplanar graphs without Kuratowski subgraphs,
then G is 3-connected.
Theorem 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.
This implies no nonplanar graph without Kuratowski
subgraph exists. That is,
A graph G is nonplanar only if G has a Kuratowski
subgraph.
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.
F = {a , b , c}
a
c
F = {a , b , c}

b
a
c
b
G
G
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.
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.
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. It’s
a contradiction.
one face of C1
C2
C1
C3
G
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
G1
smaller than 360/k degrees at v.
13. It implies G is a planar
G4
G2
v
graph. It’s a contradiction.
90。
G3
G
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 Hi=Gi∪xy.
2. Suppose that each Hi is planar.
3. Each Hi has an embedding with xy on the outside face.
Lemma 6.2.6
4.For each i >1, Hi can be attached to an embedding of
by embedding Hi in a face that has xy on its
boundary.
5. Deleting edge xy if it is not in G yields a planar
embedding of G.
6. It’s a contradiction.
x
x
H1=G1  edge xy H1
S-lobe
G1
y
G1
H2
G2
y
x
H1
H2
G1
G2
y
x
x
H1 H
H2 3
G 1 G2
G3
y
H1
H2
H3
G1 G2 G3
y
x
H1
H2
H3
G1 G2 G3
y
Lemma 6.2.7
If G is a graph with fewest edges among all nonplanar
graphs without Kuratowski subgraphs, then G is 3connected.
1. Deleting an edge of G cannot create a Kuratowski
subgraphs in G.
2. It implies that deleting one edge produces a planar
subgraph.
3. Therefore, G is a minimal nonplanar graph.
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 Slobe, 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.
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.
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.
6. {x, y, z} is a separating 3-set in G.
Lemma 6.2.9
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.
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.
Lemma 6.2.9
12. 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.
Branch Vertices
The branch vertices in a subdivision H’ of H are the
vertices of degree at least 3 in H’.
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.
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.
Lemma 6.2.10
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.
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.
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.
6. By the induction hypothesis, we obtain a convex
embedding of H = G．e with no three vertices on a line.
Theorem 6.2.11
7. In this embedding, the subgraph obtained by deleting
the edges incident to z has a face containing z. (perhaps
unbounded)
8. Since H – z is 2-connected, the boundary of this face is
a cycle C.
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.
Theorem 6.2.11
11. If all neighbors of y lie in the portion of C from xi to xi+1,
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 xxi+1, as show in the diagrams for Case 0.
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, xi+1 of x.
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, xixxi+1, and
xy form a subdivision of K3,3.
15. Since we are considering only graphs without
Kuratowski subgraph, in fact Case 0 must occur.
Theorem 6.2.11
Why do we need to put y at a point close to z in argument
11?
Xi+1
Xi+1
xi
b
a
a, b are neighbors of y
xi
b
a
Theorem 6.2.11
Why do we not show G has a convex embedding in the
plane but to show G has a convex embedding in the plane
with no three vertices on a line?
Xi+1
Xi+1
b
xi
xi
a, b are neighbors of y
a
a
b