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1/22/03 Tucker, Applied Combinato rics, Section 1.3 1 EDGE COUNTING TUCKER, APPLIED COMBINATORICS, SECTION 1.3, GROUP B Michael Duquette & Amanda Dargie

1/22/03Tucker, Applied Combinatorics, Section 1.3 1 EDGE COUNTING TUCKER, APPLIED COMBINATORICS, SECTION 1.3, GROUP B Michael Duquette & Amanda Dargie

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Page 1: 1/22/03Tucker, Applied Combinatorics, Section 1.3 1 EDGE COUNTING TUCKER, APPLIED COMBINATORICS, SECTION 1.3, GROUP B Michael Duquette & Amanda Dargie

1/22/03 Tucker, Applied Combinatorics, Section 1.3

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EDGE COUNTING

TUCKER, APPLIED COMBINATORICS, SECTION 1.3, GROUP B

Michael Duquette & Amanda Dargie

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1/22/03 Tucker, Applied Combinatorics, Section 1.3

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General Formula for counting edges

Theorem 1: In any graph, the sum of the degrees of all vertices is equal to twice the number of edges.

Vertex Degree

A 3B 2C 3D 2

Total: 10

E.g.

Proof of Theorem 1:

Summing the degrees of all the vertices in any graph counts the number of times an edge is incident with some vertex. Since each edge is incident with two vertices the total number of edge-vertex incidences is twice the number of edges.

A

BD

C

10/2 = 5 Edges

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In order for the theorem to work the sum of the degrees must be even. Therefore there must be an even number of odd integers in the sum. This conclusion leads us to a corollary for Section 1.3.

Corollary: In any graph, the number of vertices of odd degree is even.

Reminder

Within a bipartite graph you can separate all the vertices into two subsets where all the edges go between a vertex in one subset to a vertex in the other. There are no edges between vertices in the same subset.

Bipartite graphs have the property that all circuits (paths that end where they start) found within them have even length, where the length is equal to the number of edges found within the cycle.

The number of edges is equal to 6

E.g.

Start/Finish

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Theorem 2: A graph G is bipartite if and only if any circuit in G has even length

Proof of Theorem 2:

If and Only If Proofs are separated into two separate proofs. First, assume the first statement and prove the second. Lastly, assume the second statement and prove the first.

• Assume that G is bipartite. Must show that a circuit in G also has even length.

• Since G is bipartite all of its edges must connect a left vertex with a right vertex. This means that any circuit found within G will alternate back and forth from left to right vertices.

• Therefore, any circuit will contain an even number of vertices.

• Since within a circuit the number of edges is equal to the number of vertices, then the number of edges must also be even.

• Therefore, since the number of edges is even by definition G has even length.

Recall:

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Assume that every circuit in G has even length. Must show that G is bipartite.

• Take any vertex, lets start with A from the example circuit. Put it on the left of your new graph.

• Put all vertices of odd length away from A on the right side of your graph.

• Next, put all the vertices of even length away from A on the left side of your new graph near A.

• Lastly use your circuit to connect the vertices in your new graph. As you can see from the example a bipartite graph G has been constructed where no two vertices on the right or left are adjacent. If two vertices on the same side were by chance connected then our circuit would have had an odd length.

E.g. Graph G

A

CD

B

A B

D

C

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WORKED EXAMPLE: See Book Example #4, pg 29

Scenario:

Two people start at locations A and Z at the same elevation on opposite sides of a mountain range whose summit is labeled M.

Question:

Is it possible for the two people to move along the range in order to meet at M in a fashion so that they are always at the same altitude every moment? Assume that there is no point lower then A or Z and no point higher than M.

Answer:

Yes. Use a range graph to model the answer, where the vertices are a pair of points located at the same altitude such that one of the points is either a local peak or valley.

Possible Degrees for Range Graph: Degree 1: Starting and summit points.

Degree 2: One point is a peak or valley while the other point is neither a peak or a valley.

Degree 4: Both points are peaks or valleys.

Degree 0: One point is a peak while the other is a valley.

BA D

EC

M

U

T VW

X

Y

Z

(A,Z)(D,Y)

(C,X)(E,W)

(B,U)

(C,V)(C,T)

(D,U)

(M,M)

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WORKED EXAMPLE:

See Book Example #1, pg 29

Question:

Suppose we want to construct a graph with 20 edges and have all vertices of degree 4. How many vertices must the graph have?

Answer:

Let v denote the number of vertices. Since each vertice has degree four the sum of the degrees will be equal to 4v. From the first theorem we conclude that the sum of the degrees of all vertices is equal to twice the number of edges so we are left with the following equation:

4v = 2(20 edges); Therefore v = 10

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FOR CLASS TO TRY:

See Book Example #3. Pg 29

Question:

Is it possible to have a group of seven people such that each person knows exactly three other people in the group?

 

Hint: Use the Corollary to Theorem 1

Corollary: In any graph, the number of vertices of odd degree is even.

Answer:

If we try to model this problem using a graph with a vertex for each person and an edge between each pair of people who know each other, then we would have a graph with 7 vertices all of degree 3. But this is impossible by the corollary that states that the number of vertices of odd degree must be even. Therefore, no such set of seven people can exist.