Theorem 6.3.1

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 (Theorem 6.1.23). 5. G-v is a planar graph. G-v is 5-colorable by Induction Hypothesis.

2. v5 v4 v1 1 2 3 4 5 v v3 v2 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.

3. v5 v5 Switching colors on G2,5 yields another proper coloring v4 v1 v4 v1 v v 1 2 3 4 5 v3 v3 v2 v2 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.

4. 1 2 3 4 5 Theorem 6.3.1 11. If the component of Gi,j containing vi does not contain vj, we can switch the colors on it to remove color i from N(v), and then color v with color i. v5 Color v with color 3. v5 v4 v1 v4 v1 v v3 v2 v v3 v2 This component of G1,3 doesn’t contain v1. Switch colors on this component. v v v

5. P2,5 v5 v4 v1 v 1 2 3 4 5 v3 v2 P1,3 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. 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.

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. ●3 ●4 ●5 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.

8. v ●3 v ●4 Only 3 colors appears around v. 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 6.3.1. Kempe-chain argument works as in Theorem 6.3.1.

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. P1,4 P1,3 P1,4 P1,3 v1 v1 H’ H H’ H v2 v5 v2 v5 v v v4 v3 v4 v3 Switching colors 2 and 4 in H and colors 2 and 3 in H’. Then assign color 2 to v. 1 2 3 4 Remark 6.3.4 The component H of G2,4 containing v2 is separated from v4 and v5 by P1,3 The component H’ of G2,3 containing v5 is separated from v2 and v3 byP1,4.

11. H’ H Switching colors 2 and 4 in H and colors 2 and 3 in H’ cannot get a proper coloring v v Interwine 1 2 3 4 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.

12. Every planar graph is 4-colorable. Theorem 6.3.6

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. • Find ν(K6). 5. For K6, e(H) ≦ 12. Since e(K6)=15, ν(K6) ≧ 3. So, ν(K6)=3. Find ν(K3,2,2). For K3,4, e(H) ≦ 10. Since e(K3,4)=12, ν(K3,4) ≧ 2. It implies ν(K3,2,2) ≧ 2. Thus, ν(K3,2,2)=2.

15. 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, Proposition 6.3.13 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). 4. Therefore, ν(G)≧m-k.

16. 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. 7. Hence, Proposition 6.3.13 (Write m=tk+r and substitute t=(m-r)/k.)

17. Theorem 6.3.14 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. 4. 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:

19. Theorem 6.3.14 8 7 6 1 5 3 2 4 8 7 6 1 5 2 3 4 7. A better drawing lowers the upper bound. 8. Drawing kn in the plane is equivalent to drawing it on a sphere or on the surface of a can. 9. Consider n=2k. Put k vertices on the top rim of the can. The others are placed on the bottom rim.

20. Theorem 6.3.14 10. For top vertices x, y and bottom vertices z, w, where xz has smaller positive displacement than xw, we have a crossing for edges xw and yz if and only if the displacements to y, z, w are distinct positive values in increasing order. y y x x w z z w Edge yz winds around the can, and thus edges xw and yz do not cross. Edges xw and yz cross

21. Theorem 6.3.14 Crossings y x Crossings w a z Crossings

22. Example 6.3.15 • 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.