Definition 6.1.7Dual 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 3-regular graph.

Four Color Theorem: Every planar graph is 4-colorable.

Face Coloring (Vertex) Coloring

Proof of Four Color Theorem1. 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 Theorem3. The dual G* of a triangulation G is a 3-regular, 2-

edge-connected plane graph. (Exercise 6.1.11)

4. The Four Color Theorem reduces to showing all 3-regular, 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-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).

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).

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 7.3.2A simple 2-edge-connected 3-regular plane graph is 3-e

dge-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).

11

01 10

0110=11

1110=010111=10

Theorem 7.3.24. Since G is 2-edge-connected, each edge bounds two d

istinct 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}.

11

01 10

Theorem 7.3.27. If color 00 is not used in this set, the sum of any two

of these is the third.

11

01 10

0110=11

1110=010111=10

Theorem 7.3.28. 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}.

00

01 11

0111=10

1100=110100=01

Theorem 7.3.2() 9. Suppose that G has a proper 3-edge-coloring usin

g colors a, b, c (shown bold, solid, and dashed).

10. Let Ea, Eb, Ec be the edge sets having colors a, b, c, respectively.

EaEbEc

Theorem 7.3.211. Since G is 3-regular, each color appears at every vert

ex, and the union of any two of Ea, Eb, Ec is 2-regular, which makes it a union of disjoint cycles.

EaEbEc

Theorem 7.3.212. Let H1= EaEb and H2= EaEc.

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).

EaEbEc

This face is contained in 1 cycle in H1, so the first bit

is 1.

10

0100 11

0100

01

11

01

00

This face is contained in 2 cycles in H1, so the first bit

is 1.

Theorem 7.3.214. 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.

EaEbEc

10

0100 11

0100

01

11

01

00

Theorem 7.3.217. 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.

EaEbEc

10

0100 11

0100

01

11

01

00

Lemma 7.3.3

If G is a 3-regular graph with edge-connectivity 2, then G has subgraphs G1, G2 and vertices u1,v1V(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 G2

u1 u2

v1 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

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.

Since G is 3-regular, c has the third

neighbor.

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.

G1 G2

x u

y v

Lemma 7.3.34. When (x,y) E(G), we extend the ladder (a simila

r argument applies to (u,v) E(G)).5. Let w be the third neighbor of x and z the third nei

ghbor of y.6. If w=z, then the third edge incident to this vertex i

s a cut-edge. 7. Hence w≠z and the ladder extends.8. If (w,z) E(G), this direction is finished; otherwis

e, we repeat the extension of the ladder.

G2

u

v

x

y

G1

w

z

G2

u

v

x

y

w=z

G1

Theorem 7.3.4All 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 .

Theorem 7.3.44. 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 7.3.47. 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 κ(G)=κ’(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.

Theorem 7.3.410. 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.

G1 G2

u1 u2

v1 v2

Theorem 7.3.413. 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

u1 u2

v1 v2

G1 G2

u1u2

v1v2

: color 1

: color 2

Theorem 7.3.414. 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

u1 u2

v1 v2

G1 G2

u1u2

v1v2

: color 1

: color 2

: color 3

Grinberg’s TheoremIf G is a loopless plane graph having a Hamiltonian cy

cle 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 Theorem3. We prove by induction on the number of inside ed

ges.

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 Theorem7. 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 Graph1. 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.

Hx

y

Non-Hamiltonian of Tutte Graph3. Let H’ be the graph obtained by adding an x, y-path of le

ngth 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 Graph6. 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 Graph9. 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 Graph12. 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

5

11