Topology and measurements:
Counting on a line. Counting
on a curve.
Keep count on the number of vertices and segments for a
graph. Compare:
Vertices |
Edges |
V-E |
|
Line
segment I |
2 |
1 |
1 |
circle
O |
1 |
1 |
0 |
8 | 1 |
2 |
-1 |
9 | 2 |
2 |
0 |
B | 2 |
3 |
-1 |
Curves will not be topologically
equivalent
if the number V-E is not the same.
However, as the table shows,
there are
curves that have the same number V-E which are also not equivalent.
These can be distinguished by other criterion, such as connectivity when a point is removed.
The number V-E is a
characteristic of the curve
that will be the same for any topologically equivalent curve.
However, this number does not completely classify the curves topologically, since curves can have the same number V-E without being equivalent.
Review from last class:
Counting in the plane or Counting
on a surface.
Vertices, edges, regions.
Vertices |
Regions | Edges | |
Line segment | 2 |
1 |
1 |
circle | 1 |
2 |
1 |
8 | 1 |
3 |
2 |
9 | 2 |
2 |
2 |
B | 2 |
3 |
3 |
6 |
8 |
12 |
Counting in the plane or Counting on a sphere.
Euler's
Formula for the plane or the sphere:
Theorem: For any connected graph in the plane,
The number 2 is called the "Euler
characteristic for the plane." ( and the sphere).
The idea of the Proof:
Deconstruct the graph.
The relation between V+R and E+ 2 stays the same. .... [First to
obtain
one region, then to obtain a single vertex... staying connected at all
times!]
Finally we have just one vertex, no edges, and one region.
So the relationship is that V+R = E + 2!
Proof:
Choose a point P on the sphere not on the graph.
Place a plane touching the sphere opposite to P.
Then project the graph onto the plane.
The projected graph will have the same number of vertices, edges and
regions.
The counting in the plane shows that the formula is true for the
projected graph, and thus for the original graph on the sphere!
What about graphs on the Torus?
We have a connected graph on the torus with one vertex, two edges and
only on region. But this information does not match for the euler
formula
for the plane or the sphere. SO.. the torus must be topologically
different
from the sphere or the plane!
In fact for the torus we see that it is possible for V+R= E or
V-E+R= 0
Summary: We have learned about
(and proved) Euler's
Formula for the plane or the sphere:
The number 2 is called the "Euler
characteristic for the plane." ( and the sphere).
This formula can help solve some problems
on the
plane and elsewhere.
We'll continue to examine more applications of this topological
property of graphs (networks) in the plane and eventually use it to
prove a result
about coloring!
The Utilities problem:
There are three houses being planned on a new development owned by
Aaron(A), Beth(B), and Carol (C).
The phone company(P), cable company(L), and
electric company(E) are planning to put in underground lines to
the three
houses and to minimize interference between the services, they would
like
to set up separate lines from there connection boxes to the
houses that do not cross on the map of the development.
Draw a design that will achieve this objective.
Discussion: Analysis. Suppose this were
possible and consider the consequences.
If the graph were
drawn it would have V = 6 vertices and E=?__ _
edges.
Based on Euler's formula, the graph would therefore have to have
R=?_
__ regions in the plane.
By design each region would have to have
at least 4 edges.
Thus if we count the edges by the regions we would have at least 4R
edges counting each edge twice- because an edge is a boundary for
exactly
two regions.
Thus the number of edges must be more than 2R.
Is this possible? What can we conclude?
Complete graphs. A complete graph is a graph in which each vertex of the graph is connected by an edge to every other vertex.
For example:
2 vertices: 1 edge - a line segment
3 vertices: 3 edges - a triangle
4 vertices: 6 edges - a tetrahedron
Notice these all have planar representations.
The technical way this is decribed is to say these graphs can be imbedded in the plane.
What about 5 vertices?
Draw a complete graph using 5 vertices.
Discussion: Analysis. Suppose this were
possible and consider the consequences.
If the graph were drawn it would have V = 5 vertices and E= ?_
__ edges.
Based on Euler's formula, the graph would therefore have to have R=?
regions in
the
plane.
By design each region would have to
have
at least 3 edges.
Thus if we count the edges by the regions we would have at least
3R=21
edges counting each edge twice- because an edge is a boundary for
exactly
two regions.
Thus the number of edges must be more than 3/2R=10 1/2.
Is this possible? What can we conclude?
Coloring problems: Suppose we have a map
that consists of a graph that determines R regions (countries) with E
edges
(borders). How many colors are needed
to color the regions so that countries that share a border (not merely
a
vertex) have different colors.
The
Junkyard on the color problem.