02/01/05 Tucker, Sec. 1.3 1

Applied Combinatorics, 4th Ed.Alan Tucker

Section 1.3

Edge Counting

Prepared by Joshua Schoenly and Kathleen McNamara

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Theorem 1

Statement:

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

where E is the number of edges, d is the degree of any vertex, nd is the number of vertices of degree d, and v is any vertex in the graph.

2 deg( )dd v

E n d v

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Example 1: Use of Theorem 1

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

2 dd

E n d

4

4

20

all 4

2 20 4

10

E

d

n

n

It must have 10 vertices!

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Proof:

Summing the degrees of all vertices counts all instances of some edge being incident to some vertex. But, each edge is incident with two vertices, and so the total number of such edge-vertex incidences is simply twice the number of edges.

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Corollary:

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

Proof of Corollary:

Because twice the number of edges must be an even integer, the sum of degrees must be an even integer. For the sum of degrees to be an even integer, there must be an even number of odd integers in the sum.

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Definitions• Component- connected pieces of a graph

Component 1 Component 3Component 2

• Length- the number of edges in a path.

a

b

c

d

e

Path abde has length 3.

Path eabcd has length 4.

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Definitions• Bipartite- a graph G is bipartite if its vertices can be partitioned into two sets, v1 and v2, such that every edge joins a vertex in v1 with a vertex v2. v1 v2

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Definitions• Complete Bipartite- a bipartite graph in which every vertex on one side is connected to every vertex on the other side.

For Example:

K3,4

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Theorem 2

Statement:

A graph G is bipartite if and only if every circuit in G has an even length.

L=4

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Proof: Bipartite implies even length circuits

a b

d

fe

c

a

f

b

d

ceG

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Proof (part 2): Even length circuits imply bipartite

f

b

d

ceG

a b

d

fe

c

a

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Example 4

A

B

C

D

E

M

T

U

VW

X

Y

Z

Rules:1. Hikers start at points A and Z2. Hikers must stay at the same altitude at all times3. Stops must have at least one hiker on a peak or valley4. Objective: Hikers must reach M at the same time

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The hikers must be at the same altitude, so when hiker A climbs to point C, hiker Z must climb to point X. Then when hiker A descends to D, hiker Z must then descend to Y in order to be at the same altitude.

A

B

C

D

E

M

T

U

V

W

X

Y

Z

(A,Z)

(C,X)

(D,Y)

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Solution to Example 4

(A,Z)

(C,X)

(D,Y) (D,U)

(E,W)

(B,U)

(M,M)

(C,V)(C,T)

A

B

C

D

E

M

T

U

V

W

X

Y

Z

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Class Problem – 1.3 / 3

What is the largest possible number of vertices in a graph with 19 edges and all vertices of degree at least 3? (Hint – use Theorem 1)

2 dd

E n d

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Solution to Problem

• First try with only degree 3:

• But 38 isn’t divisible by 3, so that doesn’t work.

3

3

2 19 38 3 1 4

34 3

n

n

32 19 38 3n

3

3

3 4

2 19 38 3 2 4

30 3

10; 2

10 2 12

n

n

n n

n

• Next try with all vertices having degree 3, except one with degree 4:

• But 34 isn’t divisible by 3, so that doesn’t work either.

• Next try with all vertices having degree 3, except two with degree 4:

• That works! So the largest number of vertices is 12.