MTH 231 - Lecture 24
Transcript of MTH 231 - Lecture 24
MTH 231 0
MTH 231 - Lecture 24
Yevgeniy Kovchegov
Oregon State University
MTH 231 1
Topics:
• Homeomorphic graphs.
• Planar graphs.
• Examples.
• Euler’s formula for planar graphs.
• Is K5 “planar” in the toral world?
MTH 231 2
Planar Graphs
K4
The above complete graph K4 is planar as illustrated on thefollowing picture:
a b
c d
MTH 231 3
Planar Graphs
However the complete graph K5 of five vertices is known to be
not planar.
MTH 231 4
Planar Graphs
There is another famous graph known to be not planar. It is
the complete bipartite graph K3,3 on 3 and 3 vertices. See
the figure above.
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Homeomorphic graphs
Graphs G1 and G2 are homeomorphic if G1 and G2 can bereduced to isomorphic graphs by performing a sequence of seriesreductions:
v
v
v
1
2
⇒
v
v
1
2
A series reduction consists of deleting a vertex v of degree 2
(if it has no incident loops or parallel edges), and replacing its
incident edges (v, v1) and (v, v2) by one edge (v1, v2).
MTH 231 6
Homeomorphic graphs
Graphs G1 and G2 are homeomorphic if G1 and G2 can bereduced to isomorphic graphs by performing a sequence of seriesreductions.
Example.
Graphs and
are homoemorphic.
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Planar Graphs
The Kuratowski’s Theorem (on p.724) states that a graph isnot planar if and only if it contains a subgraph homeomorphicto K5 or K3,3.
We will illustrate it with the following example:
g
h
a b
c d
e f
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Planar Graphs
g
h
a b
c d
e f
Example. Consider the above graph. In order for us to figureout if it is planar or not, we try to draw it so that no two edgesintersect. In this case, we try and fail. Now, we try to proveit is not planar. By Kuratowski’s Theorem, we need to find asubgraph homeomorphic to K5 or K3,3.
Here is how we find such subgraph.
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Planar Graphs
g
h
a b
c d
e f
⇒
g
h
a b
c d
e f
Example (continued). We notice that if we remove the edge
(g, h), the above subgraph will be homeomorphic to K3,3 as can
be seen after redrawing it as follows.
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Planar Graphs
g
h
a b
c d
e f
⇔
g
h
a b
c d
e f
Example (continued). The above subgraph is indeed homeo-
morphic to K3,3 as both vertices, g and h are of degree two, and
can be removed.
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Planar Graphs
g
h
a b
c d
e f
⇔
a b
c d
e f
Example (continued). Once g and h are removed, we get
K3,3.
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Planar Graphs
g
h
a b
c d
e f
Example (continued). Thus the above graph is not planar.
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Planar Graphs
Example.
g
a b
c d
e f
While another, slightly different graph above is planar as can be
seen by redrawing it in two steps.
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Planar Graphs
g
a b
c d
e f
⇔
a b
cd e
f
g
Example (continued). First, reposition vertex e.
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Planar Graphs
a b
cd e
f
g⇔
a
b
cd
e
f
g
Example (continued). Next, reposition vertex a, and no two
edges will intersect.
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Euler’s Formula
a b
c d
As we know K4 is planar, and can therefore be drawn as above.
Now, let us count the number of vertices, edges and faces we
get.
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Euler’s Formula
a b
c d
Here, the number of vertices is
v = 4,
the number of edges is
e = 6,and the number of faces is
f = 4
Thus the Euler’s formula for planar graphs holds for K4:
v − e + f = 2
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Euler’s Formula
As it is shown in Theorem 1 on p.720, the Euler’s formula
v − e + f = 2
holds for all planar graphs.
For example, consider another planar graph:
a
b
cd
e
f
g
Here v = 7, e = 11 and f = 6. So Euler’s Formula holds hereas well:
v − e + f = 7− 11 + 6 = 2
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Euler’s Formula
Theorem (Euler’s formula for graphs). If G is a connectedplanar graph with v vertices, e edges, and f faces, then
v − e + f = 2
holds for all planar graphs.
Proof: We will use induction on the number of vertices v.
Basis step: v = 1. If there are k loops, then e = k and f = k+1.Therefore
v − e + f = 1− k + (k + 1) = 2confirming the Euler’s formula for the basis step v = 1.
In general, by adding a loop, we always increase e and f by 1,thus keeping v − e + f constant.
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Euler’s Formula
Theorem (Euler’s formula for graphs). If G is a connectedplanar graph with v vertices, e edges, and f faces, then
v − e + f = 2
holds for all planar graphs.
Proof (continued): We will use induction on the number ofvertices v.
Induction step: Suppose the formula is known for v = n. Weconstruct a connected planar graph G of n + 1 vertices froma connected planar subgraph Gn of n vertices, by adding the(n + 1)-st vertex vn+1, and connecting it to Gn by edges, andadding loops.
n
n+1
G
v
First we add vertex vn+1 andan edge connecting it to Gn.
Note that adding one vertexand one edge and no faces, willnot change v − e + f .
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Euler’s Formula
Theorem (Euler’s formula for graphs). If G is a connectedplanar graph with v vertices, e edges, and f faces, then
v − e + f = 2
holds for all planar graphs.
Proof (continued): We will use induction on the number ofvertices v.
Induction step: Suppose the formula is known for v = n. Weconstruct a connected planar graph G of n + 1 vertices froma connected planar subgraph Gn of n vertices, by adding the(n + 1)-st vertex vn+1, and connecting it to Gn by edges, andadding loops.
n
n+1
G
v
Next we add more edges con-necting vn+1 to Gn, and loops.
Note that adding one edgeand one face will not changev − e + f .
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Is K5 “planar” in the toral world?
MTH 231 23
Is K5 “planar” in the toral world?
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Is K5 “planar” in the toral world?
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Is K5 “planar” in the toral world?
Yes, K5 is planar in the toral world. That is, it can be drawn ona torus, with no two edges intersecting each other.
However, if we count the vertices, edges, and faces, we see thatthere v = 5, e = 10, and f = 5. Obtaining for K5 on a torus,
v − e + f = 0