Definition 6.1.7
Dual Graph G* of a Plane Graph:
(1) A plane graph whose vertices corresponding to the
faces of G.
(2) 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
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 3regular graph.
Four Color Theorem: Every planar graph is 4colorable.
Face Coloring  (Vertex) Coloring
Proof of Four Color Theorem
1. 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.)
2. The Four Color Theorem reduces to showing all
duals of triangulations are 4-face colorable.
Proof of Four Color Theorem
3. The dual G* of a triangulation G is a 3-regular, 2edge-connected plane graph. (Exercise 6.1.11)
4. The Four Color Theorem reduces to showing all 3regular, 2-edge-connected plane graphs are 4-face
colorable.
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-facecolorable. (Theorem 7.3.2).
6. The Four Color Theorem reduces to showing all 3regular, 2-edge-connected plane graphs are 3-edgecolorable (finding Tait colorings of all 2-edgeconnected 3-regular planar graphs).
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 3regular simple planar graphs are 3-edge-colorable
(Theorem 7.3.4).
8. The Four Color Theorem reduces to showing all 3regular, 3-connected plane graphs are 3-edgecolorable (finding Tait colorings of all 3-connected
3-regular planar graphs).
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.
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 Grinberg’s condition.
13. The proof of Four Color Theorem is not completed.
Theorem 7.3.2
A simple 2-edge-connected 3-regular plane graph is 3edge-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 c0=00, c1=01, c2=10,
and c3=11.
3. Color each edge between faces with colors ci and cj
the color obtained by ci + cj (mod 2).
0110=11
01
10
1110=01
0111=10
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}.
01
10
11
Theorem 7.3.2
7. If color 00 is not used in this set, the sum of any two
of these is the third.
0110=11
01
10
1110=01
0111=10
11
Theorem 7.3.2
8. If ck=00, 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}.
0111=10
01
11
1100=11
0100=01
00
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.
Ea
Eb
Ec
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 2regular, which makes it a union of disjoint cycles.
Ea
Eb
Ec
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 (i=1,2) is the parity of the number of
cycles in H1 that contain it (0 for even, 1 for odd).
This face is contained
in 1 cycle in H1, so the
first bit is 1.
01
11
Ea
Eb
Ec
01
00
01
10
00
01
11
This face is contained
in 2 cycles in H1, so the
first bit is 1.
00
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
11
01
00
01
Ea
Eb
Ec
10
00
01
11
00
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.
01
11
01
00
01
10
00
01
11
00
Ea
Eb
Ec
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.
G1
u1
v1
u2
G2
v2
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 k’=2.
d
Since G is 3-regular, c
has the third neighbor.
c
a
b
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.
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.
x
G1
y
u
G2
v
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 w=z, 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.
x
w=z
w
u
G2
G1
y
v
x
u
G1
z
G2
y
v
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 3regular simple planar graphs are 3-edge-colorable, then all
2-edge-connected 3-regular simple planar graphs are 3edge-colorable .
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).
Theorem 7.3.4
7. Basis step (n(G)=4): The only 2-edge-connected 3regular simple planar graph with 4 vertices is K4, which is
3-edge-colorable.
8. Induction step: Since κ(G)=κ’(G) when G is 3-regular
(Theorem 4.1.11), we may restrict our attention to 3regular 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.
Theorem 7.3.4
10. Both G1+ u1v1 and G2+ u2v2 are 2-edgeconnected 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.
G1
u1
v1
u2
G2
v2
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
u1
v1
u2
G2
v2
G1
u1
v1
: color 1
: color 2
u2
G2
v2
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
u1
v1
u2
G2
v2
G1
u1
v1
: color 1
: color 2
: color 3
u2
G2
v2
Grinberg’s Theorem
If G is a loopless plane graph having a Hamiltonian
cycle C, and G has f’i faces of length i inside C
and f”i faces of length i outside C, then
Σi(i-2)(f’i-f”i)=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
Σi(i-2)f’i is a constant.
Grinberg’s Theorem
3. We prove by induction on the number of inside
edges.
4. Basis: When there are no inside edges, Σi(i-2)f’i =
n-2.
5. Induction Hypothesis: Suppose that Σi(i-2)f’i = n-2
when there are k edges insice C.
6. Induction Step: We can obtain
any graph with k+1 edges inside
C by adding an edge to such a
graph.
Grinberg’s Theorem
7. The added edge cuts a face of some length r into two
faces of lengths s and t.
8. s+t=r+2, because the new edge contributes to both
new faces.
9. Since (s-2)+(t-2)=(r-2), Σi(i-2)f’i = n-2.
Tutte Graph
1. Tutte graph is 3-connected 3-regular.
2. Tutte graph is not Hamiltonian, as proved in the
following.
Non-Hamiltonian of Tutte Graph
1. A Hamiltonian cycle must traverse one copy of H along
a Hamiltonian path joining the other entrances to H.
2. It suffices to show no Hamiltonian x, y-path exists in H.
y
H
x
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 Grinberg’s condition.
x
y
Non-Hamiltonian of Tutte Graph
6. H’ has five 5-faces, three 4-faces, and one 9-face.
7. Grinberg’s condition becomes 2a4+3a5+7a9=0, where
ai=f’i-f”i.
8. Since the unbounded face is always outside (a9=-1), the
equation reduces to 2a4  7 (mod 3).
Non-Hamiltonian of Tutte Graph
9. Since f’4+f”4=3, 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.
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
5