Applied Combinatorics, 4th Ed.
Alan Tucker
Section 1.4
Planar Graphs
Prepared by Michele Fretta
and Jessica Scheld
1
Planar Graphs
A graph is planar if it can be drawn on a plane without edges crossing.
A plane graph is a planar depiction of a planar graph.
Figure 1
a
Plane graph of Figure 1
a
b
f
e
e
c
d
d
b
c
f
2
Example 1: Map Coloring
• Question: How many different colors are needed to
color the countries on a map so that any pair of
countries with a common border have different
colors?
• A map is a planar graph, with edges as borders and
vertices where borders meet.
• Dual graph: obtained by making a vertex for each
country and an edge between vertices corresponding
to two countries with a common border
3
Dual Graph Example
a
b
c
d
4
Map coloring
E
A
C
B
D
F
We need 4 colors to color the map
without any bordering countries
having the same color.
G
It’s easier to see it in the second
representation of the graph, since it shows
that A,B,C,D are a separate subgraph from
D,E,F,G. Thus, we would need no more than
4 colors.
Note that this dual graph does not include an
outer region.
5
Circle-Chord Method
(one method of determining if a graph is planar)
• What we want is to prove that some graph is planar
(no edges cross).
• The circle-chord method:
– Step 1: Find a circuit that contains all the vertices of our
graph (draw it as a large circle)
• Note: Finding a circuit which includes all vertices is difficult.
• Recall: circuit is a path that ends where it began
– Step 2: The remaining non-circuit edges, called chords,
must be drawn either inside or outside the circle in a planar
drawing.
6
Step 1
a
b
c
d
Find a circuit. The orange
highlights the circuit on this graph.
e
f
g
a
h
f
Draw this circuit as a
circle.
e
c
b
h
g
d
7
Step 2
Step 2: Choose a chord to draw, either
inside or outside the circle. For example,
we will start with the chord (b,f),
drawing it outside the circle.
a
b
c
d
e
f
g
h
a
f
e
c
b
h
g
d
Since no lines cross
another, this graph is
planar!
8
Showing K 3,3 is Nonplanar
• K 3,3 indicates that the graph is a complete
bipartite graph consisting of two sets of 3
vertices with each vertex in one set adjacent to
all vertices in the other set.
1
3
5
1
3
5
1
2
6
5
2
4
6
2
4
3
6
4
9
K 3,3 Configuation
• Note: A subgraph is a K 3,3 configuration if it can
be obtained from a K 3,3 by adding vertices in the
middle of some edges. A K5 configuration is
defined similarly.
• Finding the configuration can be tricky.
1
2
6
5
3
4
10
Example of a K 3,3 configuration
1
2
8
K 3,3
3
7
A K 3,3configuration proves a
graph is nonplanar. It has 6
vertices of degree 3, plus some
number of vertices of degree 2
that subdivide the edges of it.
Here those vertices are 3 and 6.
4
6
5
In order to find a K 3,3 configuration, eliminate vertices of degree 2
that subdivide edges of it. The remaining graph looks like the
depiction of K 3,3 earlier.
11
Theorem 1 (Kuratowski, 1930)
• A graph is planar if and only if it does not contain a
subgraph that is K 3,3 or K5 configuration.
– Recall: K n is a graph on n vertices with an edge joining
every pair of vertices.( K3 is a triangle.) K m , n is a bipartite
graph with m and n vertices in its two vertex sets and all
possible edges between vertices in the two sets.
1
2
1
3
2
5
4
5
6
4
3
12
Theorem 2: Euler’s Formula
• If G is a connected planar graph, then any plane graph
depiction of G has r = e - v + 2 regions.
– Recall: Connected planar graphs have paths between each
pair of vertices.
– v = number of vertices
– e = number of edges
– r = number of regions
• This is important because there are many different
plane graph depictions that can be drawn for a planar
graph, however, the number of regions will not change.
13
Proof of Euler’s Formula
• Let’s draw a plane graph depiction of G, edge by
edge.
• Let Gn denote the connected plane graph after n
edges have been added.
• Let vn denote the number of vertices in Gn
• Let en denote the number of edges in Gn
• Let rn denote the number of regions in Gn
14
Proof of Euler’s Formula (cont’d)
• Let’s start by drawing
G1
v1  2
e1  1
r1  1
Euler’s formula is valid for
r=e-v+2
1=1-2+2
We obtain G2 from
G1 , since
G1 by adding an edge at one of the
vertices of G1 .
15
Proof of Euler’s Formula (cont’d)
th
n edge to
In general, we can obtain Gn from Gn by
1 adding an
one of the vertices of Gn 1.
• The new edge might link two vertices already in Gn 1 .
• Or, the new edge might add another vertex to Gn 1 .
We will use the method of induction to complete the proof:
We have shown that the theorem is true for G .
1
Next, let’s assume that it is true for Gn 1 for any n>1, and prove it
is true for Gn.
th
Let (x,y) be the n edge that is added to
There are two cases to consider.
Gn1 to get Gn .
16
Proof of Euler’s Formula (cont’d)
In the first case, x and y are both in Gn1 .
Then they are on the boundary of a common region K , possibly
an unbounded region.
y
Edge (x, y) splits K into two regions.
Then, rn  rn 1  1
K
en  en 1  1
vn  vn 1
x
Each side of Euler’s formula grows by one.
So, if the formula was true for Gn1, it will also be true for Gn .
17
Proof of Euler’s Formula (cont’d)
In the second case, one of the vertices x,y is not in Gn1 . Let’s say that it is x.
Then, adding (x,y) implies that x is also added, but that no new regions are
formed (no existing regions are split).
y
x
rn  rn 1
en  en 1  1
K
vn  vn 1  1
So, the value on each side of Euler’s equation is unchanged.
The validity of Euler’s formula for Gn1 implies its validity for Gn .
By induction, Euler’s formula is true for all Gn’s and the full graph G.
18
Example of Euler’s Formula
• How many regions would there be in a plane
graph with 10 vertices each of degree 3?
By Theorem 1, Sec. 1.3:
 (degrees of vertices)  2e
 (10*3)  30  2e
15= e
By Euler's Formula, r = e - v +2
r = 15-10+2 = 7
Answer: There are 7 regions.
19
Corollary
• If G is a connected planar graph with e > 1, then
e ≤ 3v – 6
• Proof:
– Define the degree of a region as the number of edges on its
boundary. If an edge occurs twice along the boundary, then
count it twice. The region K has degree 12.
K
20
Proof of Corollary continued
• Note that no region can be less than degree 3.
– A region of degree 2 would be bounded by two
edges joining the same pair of vertices (parallel
edges)
– A region of degree 1 would be bounded by a loop
edge.
• Neither of these is allowed, and so a region
must have at least degree 3.
21
Proof of Corollary continued
2
Since 2e =  (degrees of r), we know that 2e  3r  e  r
3
Also, we know r =e - v  2 from Euler's Formula.
Substitute Euler's Formula in to get:
2
eev2
3
1
=> 0  ( e  v  2) *3
3
=> 0  e - 3v  6
=> e  3v - 6
22
Example: Show K5 is nonplanar.
• Corollary: If G is a connected planar
graph with e > 1, then e ≤ 3v – 6
• Use the Corollary. K5 has 5 vertices
and 10 edges. Thus, we have 3v – 6 =
3 x 5 – 6 = 9. But, e ≤ 3v – 6 must be
true in a connected planar graph, thus,
K5 cannot be planar.
23
Exercise:
Is this graph planar?
a
Step 1:
b
f
Draw a
circuit
e
c
d
a–b–e–d–c–f–a
24
Exercise (cont’d)
Step 2: Draw the circuit as a circle.
Step 3: Add the remaining edges.
a
a
f
b
f
b
c
e
e
c
d
d
25