theorem 6.3.1 every planar graph is 5-colorable. proof. 1. we use induction on n(g), the number of...
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Theorem 6.3.1
Every planar graph is 5-colorable.Proof. 1. We use induction on n(G), the number of node
s 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 6.1.23).5. 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 n
eighbor of v, and hence d(v)=5.8. Let v1, v2, v3, v4, and v5 be the neighbors of v in clock
wise 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, 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
Theorem 6.3.112. G is 5-colorable unless, for each i and j, the component o
f 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 P
2,5.15. 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 contain
ing it cannot be a minimal counterexample.2. Kempe proves Four Color Theorem by showing conf
igurations ●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
●4Kempe-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
1977 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. Every edge
not in H crosses some edge in H so the drawing has at least e(G)-e(H) crossings.
2. ν(K6)=3.
3. ν(K3,2,2)=2.
e(K6)=15; e(H) 12.≦Hence ν(K6) 3.≧
e(K3,4)=12; e(H) 10 because K≦ 3,4 is triangle-free. Hence ν(K3,4) 2. It implies ν(K≧ 3,2,2) 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; ot
herwise 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.≧
2
( ) .2 2
m mG
k
Proposition 6.3.135. After discarding E(H), we have at least m-k edges re
maining. 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) cros
sings, where t=m/k. Write m=tk+r.
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
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. 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
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.
Theorem 6.3.14
12 3 4
5678
12 3 4 5
678
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.
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.
Theorem 6.3.149. 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.
x y
aw
z
x y
aw
z
Edges xz and, yw cross Edge ya winds around the can, and thus edges xz and, yw 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