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.

Theorem 6.3.16. 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 5v

v3

v1

v5

v2

v4

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

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.

Color v with color 3. v5

Switch colors on this component.1 2 3 4 5

v

v3

v1

v5

v2

v4

This component of G1,3

doesn’t contain v1.

v

v

v3

v1

v2

v4

vv

Theorem 6.3.112. 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.

1 2 3 4 5v

v3

v1

v5

v2

v4

P2,5

P1,3

The Idea of Unavoidable Set

1. 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.

2. This suggests an approach to the Four Color Problem; we seek an unavoidable set of graphs that can’t be present!

3. We need only consider triangulations, since every simple planar graph is contained in a triangulation.

Unavoidable Set1. A configuration in a planar triangulation is a

separating cycle C (the ring) together with the portion of the graph inside C.

2. For the Four Color Problem, a set of configurations is unavoidable if a minimal counterexample must contain a member of it.

3. 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

Reducible Configuration1. A configuration is reducible if a planar graph

containing it cannot be a minimal counterexample.2. 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.

v

●3

Only 3 colors appears around v.

v

●4

Kempe-chain argument works as in Theorem 6.3.1.

Remark 6.3.41. 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.

2. 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.

3. We can eliminate color 1 from v1 unless P1,3 and P1,4 exist.

Remark 6.3.4

1 2 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 by P1,4.

v

v1

v5

v4 v3

v2

P1,4 P1,3

HH’

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.

HH’

Remark 6.3.41. The above argument is wrong because P1,3 and P1,4

can interwine, intersecting at a vertex with color 1 as shown below.

2. We can make the switch in H or in H’, but making them both creates a pair of adjacent vertices with color 2.

1 2 3 4

v

Switching colors 2 and 4 in H and colors 2 and 3 in H’ cannot get a proper coloring

v

HH’

Interwine

Theorem 6.3.6

Every planar graph is 4-colorable.

Year Authors #unavoidable sets

1976 Appel, Haken, Koch 1936

1983 1258

1996 Robertson, Sanders, Seymour, Thomas

633

Crossing Number

The crossing number ν(G) is the minimum number of crossings in a drawing of G in the plane.

ν(K5)=1

Example 6.3.21. Let H be the maximal planar subgraph.

2. Every edge not in H crosses some edge in H

3. So, the drawing has at least e(G)-e(H) crossings.

4. Find ν(K6).

5.

For K6, e(H) 12.≦

So, ν(K6)=3.Since e(K6)=15, ν(K6) 3.≧

Thus, ν(K3,2,2)=2.

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.≧

Proposition 6.3.13

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).4. Therefore, ν(G) m-k.≧

2

( ) .2 2

m mG

k

Proposition 6.3.135. 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,

1

2

2

( ) ( * ) ( 1) / 2

( )

2 2 2

.2 2

t

i

G m i k mt kt t

m m r k r

k k

m m

k

(Write m=tk+r and substitute t=(m-r)/k.)

Theorem 6.3.14

)(64

1)()(

80

1 3434 nOnKnOn n

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(K≧ n-1).

Theorem 6.3.144. We prove by induction on n that

45

1)(

nKn

5. Basis step: n=5, ν(K5)=1.

6. Induction Step: n>5:

45

1

4

1

5

1

4)(

4)( 1

nn

n

nK

n

nK nn

Theorem 6.3.14

12 3 4

5678

12 3 4 5

678

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.

Theorem 6.3.1410. 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.

x y

wz

Edges xw and yz cross

x y

z w

Edge yz winds around the can, and thus edges xw and yz do not cross.

Theorem 6.3.14

4

k

4

k

3

kk

x y

aw

z

)(64

1

)32

)(22

)(12

(24

1

)3)(2)(1(4

1

342)(

34 nOn

nnnn

kkkk

kk

kKn

Crossings

Crossings

Crossings

Example 6.3.15

• ν(Km,n).

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.

21

221

2)( ,

nnmmnmK

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