Definition 6'1'7

1
Definition 6.1.7

• Dual Graph G of a Plane Graph
• A plane graph whose vertices corresponding to the
  faces of G.
• The edges of G corresponds to the edges of G as
  follows 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 in E(G) are the
  vertices x and y of G that represents the faces
  X and Y of G.

?
K4
2
Definitions
Proper Face-Coloring of a 2-Edge-Connected Plane
Graph An assignment of colors to its faces so
that faces having a common edge in their
boundaries have distinct colors. Tait Coloring A
proper 3-edge-coloring of a 3-regular graph. Four
Color Theorem Every planar graph is 4-colorable.
3
Face Coloring ? (Vertex) Coloring
4
Proof of Four Color Theorem

• Since adding edges does not make ordinary
  coloring easier, to prove the Four Color Theorem
  it suffices to prove that all triangulations are
  4-colorable. (A triangulation is a simple plane
  graph where every face boundary is a 3-cycle.)
• The Four Color Theorem reduces to showing all
  duals of triangulations are 4-face colorable.

5
Proof of Four Color Theorem

• The dual G of a triangulation G is a 3-regular,
  2-edge-connected plane graph. (Exercise 6.1.11)
• The Four Color Theorem reduces to showing all
  3-regular, 2-edge-connected plane graphs are
  4-face colorable.

6
Proof of Four Color Theorem
5. A simple 2-edge-connected 3-regular plane
graph is 3-edge-colorable if and only if it is
4-face-colorable. (Theorem 7.3.2). 6. The Four
Color Theorem reduces to showing all 3-regular,
2-edge-connected plane graphs are
3-edge-colorable (finding Tait colorings of all
2-edge-connected 3-regular planar graphs).
7
Proof of Four Color Theorem
7. All 2-edge-connected 3-regular simple planar
graphs are 3-edge-colorable if and only if all
3-connected 3-regular simple planar graphs are
3-edge-colorable (Theorem 7.3.4). 8. The Four
Color Theorem reduces to showing all 3-regular,
3-connected plane graphs are 3-edge-colorable
(finding Tait colorings of all 3-connected
3-regular planar graphs).
8
Proof of Four Color Theorem
9. Every Hamiltonian 3-regular has a Tait
coloring (Exercise 1) 10. The Four Color Theorem
reduces to showing that every 3-connected
3-regular planar graph is Hamiltonian.
9
Proof of Four Color Theorem
11. Grinberg proposed a necessary condition for a
Hamiltonian graph. 12. Tutte finds a 3-connected
3-regular planar graph, Tutte graph, which
violates Grinbergs condition. 13. The proof of
Four Color Theorem is not completed.
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Theorem 7.3.2
A simple 2-edge-connected 3-regular plane graph
is 3-edge-colorable if and only if it is
4-face-colorable. Proof (?) 1. Let G be a
4-face-colorable graph. 2. Let four colors be
denoted by c000, c101, c210, and c311. 3.
Color each edge between faces with colors ci and
cj the color obtained by ci cj (mod 2).
11
Theorem 7.3.2
4. Since G is 2-edge-connected, each edge bounds
two distinct faces, and hence the color 00 is
never used to color edge. 5. We have to check the
edges at a vertex receive distinct colors. 6. At
vertex v the faces bordering the three incident
edges must have distinct colors ci, cj, ck.
12
Theorem 7.3.2
7. If color 00 is not used in this set, the sum
of any two of these is the third.
13
Theorem 7.3.2
8. If ck00, ci and cj appear on two of the
edges, and the third receives color ci cj (mod
2), which is the color not in ci, cj, ck.
14
Theorem 7.3.2
(?) 9. Suppose that G has a proper
3-edge-coloring using colors a, b, c (shown bold,
solid, and dashed). 10. Let Ea, Eb, Ec be the
edge sets having colors a, b, c, respectively.
15
Theorem 7.3.2
11. Since G is 3-regular, each color appears at
every vertex, and the union of any two of Ea, Eb,
Ec is 2-regular, which makes it a union of
disjoint cycles.
16
Theorem 7.3.2
12. Let H1 Ea?Eb and H2 Ea?Ec. 13. Each face of
G is assigned the color whose ith coordinate
(i1,2) is the parity of the number of cycles in
H1 that contain it (0 for even, 1 for odd).
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Theorem 7.3.2
14. Faces F and F sharing an edge e are distinct
faces, since G is 2-edge-connected. 15. Edge e
belongs to a cycle C in at least one of H1 and H2
(in both if the edge has color a). 16. One of F
and F is inside C and the other is outside.
01
01
11
00
10
11
00
01
01
00
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Theorem 7.3.2
17. All other cycles in H1 and H2 fail to
separate F and F, leaving them on the same
side. 18. If e has color a, c, or b, then the
parity of the number of cycles containing F and
F is different in H1, in H2, or in both,
respectively.
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Lemma 7.3.3
If G is a 3-regular graph with edge-connectivity
2, then G has subgraphs G1, G2 and vertices
u1,v1?V(G1) and u2,v2 ?V(G2) such that
(u1,v1)?E(G), (u2,v2)?E(G), and G consists G1, G2
and a ladder of some length joining G1, G2 at u1,
v1, u2, v2 as shown below.
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Lemma 7.3.3
Proof. 1. If G has an edge cut of size 2 in which
the two edges are incident, then the third edge
incident to their common vertex is a cut-edge,
contracting k2.
1.1 Suppose that there is a path between c and d
after cd is deleted. 1.2 It implies there is a
path between b (or a) and d. 1.3 It implies
there is path between b (or a) and c after ac and
bc are deleted, a contradiction.
Since G is 3-regular, c has the third neighbor.
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Lemma 7.3.3
2. We assume that the four endpoints in our
minimum edge cut xu, yv are distinct. 3. If (x,y)
?E(G) and (u,v) ?E(G), then these are the four
desired vertices and the ladder has only these
two edges.
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Lemma 7.3.3
4. When (x,y) ?E(G), we extend the ladder (a
similar argument applies to (u,v) ?E(G)). 5. Let
w be the third neighbor of x and z the third
neighbor of y. 6. If wz, then the third edge
incident to this vertex is a cut-edge. 7. Hence
w?z and the ladder extends. 8. If (w,z) ? E(G),
this direction is finished otherwise, we repeat
the extension of the ladder.
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Theorem 7.3.4
All 2-edge-connected 3-regular simple planar
graphs are 3-edge-colorable if and only if all
3-connected 3-regular simple planar graphs are
3-edge-colorable. Proof. 1. Any 3-connected
3-regular simple plannar graph is a
2-edge-connected 3-regular simple planar
graph. 2. If all 2-edge-connected 3-regular
simple planar graphs are 3-edge-colorable, then
all 3-connected 3-regular simple planar graphs
are 3-edge-colorable. 3. It suffice to show the
if part if all 3-connected 3-regular simple
planar graphs are 3-edge-colorable, then all
2-edge-connected 3-regular simple planar graphs
are 3-edge-colorable .
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Theorem 7.3.4
4. Suppose that all 3-connected 3-regular simple
planar graphs are 3-edge-colorable. 5. We need to
show all 2-edge-connected 3-regular simple planar
graphs are 3-edge-colorable. 6. We use induction
on n(G).
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Theorem 7.3.4
7. Basis step (n(G)4) The only 2-edge-connected
3-regular simple planar graph with 4 vertices is
K4, which is 3-edge-colorable. 8. Induction step
Since ?(G)?(G) when G is 3-regular (Theorem
4.1.11), we may restrict our attention to
3-regular graphs with edge-connectivity 2. 9.
Lemma 7.3.3 gives us a decomposition of G into
G1 and G2 and a ladder joining them. The length
of the ladder is the distance from G1 to G2.
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Theorem 7.3.4
10. Both G1 u1v1 and G2 u2v2 are
2-edge-connected and 3-regular. 11. By induction
hypothesis, G1 u1v1 and G2 u2v2 are
3-edge-colorable. 12. Let f1 be a proper
3-edge-colorable of G1 u1v1, and f2 be a proper
3-edge-colorable of G2 u2v2.
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Theorem 7.3.4
13. Permute names of colors so that f1(u1v1)1
and so that f2(u2v2) is chosen from 1, 2 to
have the same parity as the length of the ladder.
G1
G2
u2
u1
v2
v1
color 1
color 2
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Theorem 7.3.4
14. Color each edge in G1 as in f1, and each edge
in G2 as in f2. 15. Beginning from the end of the
ladder of G1, color the paths forms the sides of
the ladder alternatively with 1 and 2. 16. Color
the rungs of the ladder with 3.
G1
G2
u2
u1
v2
v1
color 1
color 2
color 3
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Grinbergs Theorem
If G is a loopless plane graph having a
Hamiltonian cycle C, and G has fi faces of
length i inside C and fi faces of length i
outside C, then Si(i-2)(fi-fi)0. Proof. 1.
We can switch inside and outside by projecting
the embedding onto a sphere and puncturing a face
inside C.
2. We only need to show Si(i-2)fi is a constant.
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Grinbergs Theorem
3. We prove by induction on the number of inside
edges. 4. Basis When there are no inside edges,
Si(i-2)fi n-2. 5. Induction Hypothesis
Suppose that Si(i-2)fi n-2 when there are k
edges insice C.
6. Induction Step We can obtain any graph with
k1 edges inside C by adding an edge to such a
graph.
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Grinbergs Theorem
7. The added edge cuts a face of some length r
into two faces of lengths s and t. 8. str2,
because the new edge contributes to both new
faces. 9. Since (s-2)(t-2)(r-2), Si(i-2)fi
n-2.
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Tutte Graph

• Tutte graph is 3-connected 3-regular.
• Tutte graph is not Hamiltonian, as proved in the
  following.

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Non-Hamiltonian of Tutte Graph

• A Hamiltonian cycle must traverse one copy of H
  along a Hamiltonian path joining the other
  entrances to H.
• It suffices to show no Hamiltonian x, y-path
  exists in H.

y
H
x
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Non-Hamiltonian of Tutte Graph
3. Let H be the graph obtained by adding an x,
y-path of length two through a new vertex. 4. We
only have to show no Hamiltonian cycle exists in
H. 5. We show H violates Grinbergs condition.
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Non-Hamiltonian of Tutte Graph
6. H has five 5-faces, three 4-faces, and one
9-face. 7. Grinbergs condition becomes
2a43a57a90, where aifi-fi. 8. Since the
unbounded face is always outside (a9-1), the
equation reduces to 2a4 ? 7 (mod 3).
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Non-Hamiltonian of Tutte Graph
9. Since f4f43, the possibilities for a4 are
3, 1, -1, -3. 10. It implies a4-1. 11.
However, the 4-faces having a vertex of degree 2
cannot lie outside the cycle, since the edges
incident to the vertex of degree 2 separate the
face from the outside.
37
Non-Hamiltonian of Tutte Graph
12. We can reach a contradiction faster by
subdividing one edge incident to each vertex of
degree 2. 13. The resulting graph has seven
5-faces, one 4-faces, and one 11-face. 14. The
equation becomes 2?(?1) 9-3a5, which has no
solution since the left side is not a multiple of
3.
5
11

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