1. Theorem 6.3.1 Every planar graph is 5-colorable.
Proof. 1. We use induction on n(G), the number of nodes in G.
2. Basis Step: All graphs with n(G) ≤ 5 are 5-colorable.
3. Induction Step: n(G) > 5.
4. G has a vertex, v, of degree at most 5 because e(G) ≤ 3n(G)-6 (Theorem ).
5. G-v is 5-colorable by Induction Hypothesis.

2. Theorem 6.3.1 6. Let f be a proper 5-coloring of G-v.
7. If G is not 5-colorable, f assigns each color to some neighbor of v, and hence d(v)=5.
8. Let v1, v2, v3, v4, and v5 be the neighbors of v in clockwise order around v, and name the colors so that f(vi)=i. 1 2 3 4 5 v v3 v1 v5 v2 v4

3. Theorem 6.3.1
9. Let Gi,j denote the subgraph of G-v induced by the vertices of colors i and j.
10. Switching the two colors on any component of Gi,j yields another proper coloring of G-v. 1 2 3 4 5 v v3 v1 v5 v2 v4
Switching colors on
G2,5 yields another
proper coloring v v3 v1 v5 v2 v4

4. Theorem 6.3.1
11. If the component of Gi,j containing vi does not contain vj, then
we can switch the colors on it to remove color i from N(v). 1 2 3 4 5 v
This component of G3,1 doesn’t contain v1. v3 v1 v5 v2 v4
Step 1: switches colors on this component.
Step 2: v uses color 3. v v3 v1 v5 v2 v4

5. Theorem 6.3.1
12. G is 5-colorable unless, for each i and j, the component of Gi,j containing vi also contains vj.
13. Let Pi,j be a path in Gi,j from vi to vj.
14. By the Jordan Curve Theorem, the path P1,3 must cross P2,5.
15. Since G is planar, paths can cross only at shared vertices, which is impossible because the vertices of P1,3 all have color 1 or 3, and the vertices of P2,5 all have color 2 or 5. 1 2 3 4 5 v v3 v1 v5 v2 v4
P2,5
P1,3

6. The Idea of Unavoidable Set
In proving Five Color Theorem inductively, we argue that a minimal counterexample contains a vertex of degree at most 5 and that a planar graph with such a vertex cannot be a minimal counterexample.
This suggests an approach to the Four Color Problem; we seek an unavoidable set of graphs that can’t be present!
We need only consider triangulations, since every simple planar graph is contained in a triangulation.

7. Unavoidable Set
A configuration in a planar triangulation is a separating cycle C (the ring) together with the portion of the graph inside C.
For the Four Color Problem, a set of configurations is unavoidable if a minimal counterexample must contain a member of it.
Because (G)<=5 for every simple planar graph G and every vertex has degree at least 3 in triangulation, the set of three configurations below is unavoidable. ●3 ●4 ●5

8. Reducible Configuration
A configuration is reducible if a planar graph containing it cannot be a minimal counterexample.
Kempe proves Four Color Theorem by showing configurations ●3, ●4, and ●5 each are reducible by extending a 4-coloring of G-v to complete a 4-coloring of G as in Theorem v ●4
Kempe-chain argument works as in Theorem v ●3
Only 3 colors appears around v.

9. Remark 6.3.4
Consider ●5. When d(v)=5, the repeated color on N(v) in the propoer 4-coloring of G-v appears on nonconsecutive neighbors of v in triangulations.
Let v1, v2, v3, v4, and v5 be the neighbors of v in clockwise order. In the 4-coloring f of G-v, we assume by symmetry that f(v5)=2 and that f(vi)=i for 1<=i<=4.
We can eliminate color 1 from v1 unless P1,3 and P1,4 exist.

10. Remark 6.3.4 v v1 v5 v4 v3 v2 P1,4 P1,3 H H’ v v1 v5 v4 v3 v2 P1,4
The component H of G2,4 containing v2 is separated from v4 and v5 by P1,3 v v1 v5 v4 v3 v2
P1,4
P1,3 H H’ v v1 v5 v4 v3 v2
P1,4
P1,3
Switching colors 2 and 4 in H and colors 2 and 3 in H’. Then assign color 2 to v. H H’
The component H’ of G2,3 containing v5 is separated from v2 and v3 by P1,4. 1 2 3 4

11. Remark 6.3.4
The above argument is wrong because P1,3 and P1,4 can interwine, intersecting at a vertex with color 1 as shown below.
We can make the switch in H or in H’, but making them both creates a pair of adjacent vertices with color 2. v H H’
Interwine v
Switching colors 2 and 4 in H and colors 2 and 3 in H’ cannot get a proper coloring 1 2 3 4

12. Robertson, Sanders, Seymour, Thomas
Theorem 6.3.6
Every planar graph is 4-colorable.
Year
Authors
#unavoidable sets
1977
Appel, Haken, Koch
1936
1983
1258
1996
Robertson, Sanders, Seymour, Thomas
633

13. Crossing Number
The crossing number ν(G) is the minimum number of crossings in a drawing of G in the plane.
ν(K5)=1

14. Example 6.3.2
Let H be the maximal planar subgraph. Every edge not in H crosses some edge in H so the drawing has at least e(G)-e(H) crossings.
ν(K6)=3.
ν(K3,2,2)=2.
e(K6)=15; e(H) ≦ 12.
Hence ν(K6) ≧ 3.
e(K3,4)=12; e(H) ≦10 because K3,4 is triangle-free. Hence ν(K3,4) ≧ 2. It implies ν(K3,2,2) ≧ 2.

15. Proposition
Let G be an n-vertex graph with m edges, If k is the maximum number of edges in a planar subgraph of G, then ν(G)≧m-k. Furthermore,
Proof: 1. Let H be maximal planar subgraph.
2. Every edge not in H crosses at least one edge in H; otherwise it can be added to H.
3. At least m-k crossings between edges of H and edges of G-E(H) because e(H) ≦k.
4. Therefore, ν(G)≧m-k.

16. Proposition
5. After discarding E(H), we have at least m-k edges remaining. The same argument yields at least (m-k)-k crossings in the drawing of the remaining graph.
6. Iterating the argument yields at least Σti=1(m-i*k) crossings, where t=m/k. Write m=tk+r.
Substitute t=(m-r)/k

17. Theorem
Proof: 1. A drawing of Kn with fewest crossings contains n drawings of Kn-1, each obtained by deleting one vertex.
2. Each subdrawing has at least ν(Kn-1) crossings. The total count is at least n*ν(Kn-1).
3. Each crossing in the full drawing has been counted n-4 times. We conclude that (n-4)*ν(Kn) ≧ n*ν(Kn-1).

18. Theorem 6.3.14 4. We prove by induction on n that
5. Basis step: n=5, ν(K5)=1.
6. Induction Step: n>5:
7. The denominator of the quartic term in the lower bound can be improved from 120 to 80 by considering copies of K6,n-6, which has crossing number 6(n-6)/2 (n-7)/2.

19. We draw edges to wind around the can as little as possible.
Theorem
8. A better drawing lowers the upper bound. Consider n=2k. Drawing kn in the plane is equivalent to drawing it on a sphere or on the surface of a can. 1 2 3 4 5 6 7 8
We draw edges to wind around the can as little as possible.
Put k vertices on the top rim of the can.
The others are placed on the bottom rim.

20. Theorem
9. For top vertices x,y and bottom vertices z,w, where xz has smaller positive displacement than xw, we have a crossing for x,y,z,w if and only if the displacements to y,z,w are distince positive values in increasing order. y y x x w w a z a z
Edge ya winds around the can, and thus edges xz and, yw do not cross.
Edges xz and, yw cross

21. Theorem
Crossings y x
Crossings w a z
Crossings

22. Example
ν(Km,n).
Adding up the four types of crossings generated when we join each vertex on the x-axis to every vertex on the y-axis yields
Put m/2 vertices along the positive y axis and m/2 along the negative y axis. Similarly, split n vertices and put them along the x axis.