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Transcript
```MC302 GRAPH THEORY
Thursday, 9/5/13
Today:
Graph degrees and edges
The Havel-Hakimi Theorem
[CH] 1.4
[HR] Rest of 1.1, 1.2
Hand-in HW #1 will be on our
website by this evening
Exercises:
Due Friday 9/13/13 by 3 PM
[CH] 1.4.2, 1.4.9
[HR] 1.1.8, 1.1.9
Thursday, 9/5/13, Slide #1
Exercise: Counting edges and degrees
1. How many edges are there in the complete
graph Kn, as a function of n?
2. How many edges are there in the complete
bipartite graph Kp,q, as a function of p and q?
3. For each of Kn and Kp,q, what are all the
vertex degrees?
4. If G is any graph, and we add up the vertex
degrees, how are the answers related to the
numbers of edges in G?
Thursday, 9/5/13, Slide #2
The Degree-Sum Theorem
First Theorem of Graph Theory (Degree-Sum
Theorem): The sum of all the vertex degrees equals
twice the number of edges. Symbolically,
n
∑ d ( v ) = 2e
k
k =1
Is this also true for non-simple graphs, i.e., graphs
with parallel edges and loops?
Corollary: In any graph there is an even number of
vertices with odd degree.
Proof?
Thursday, 9/5/13, Slide #3
Regular Graphs
A regular graph is one in which all vertices have
the same degree.
Exercises:
1. Is Kn regular? Is Kp, q regular?
2. If a graph with n vertices is d-regular, how
many edges does it have?
3. If two graphs on n vertices are both d-regular
for some d, must they be isomorphic?
4. For each k = 0, ..., 5, draw a k-regular graph
on 5 vertices, or explain why one cannot exist.
Thursday, 9/5/13, Slide #4
Degree Sequences
The degree sequence of a graph is the sequence
of its vertex degrees, usually written in decreasing
order.
Exercises:
1. What’s the degree sequence of the graph below?
2. A graphic sequence is a sequence that is the degree
sequence of some graph. Which of the following are graphic
sequences?
5, 4, 3, 2, 2
5, 4, 3, 2, 2, 1
5, 4, 3, 3, 2, 2, 1
Thursday, 9/5/13, Slide #5
Havel-Hakimi Theorem
Theorem.
Let S =(d1, d2, …, dn) be a sequence of n integers written in
decreasing order: d1 ≥ d2 ≥ … ≥ dn-1 ≥ dn.
Then S is a graphic sequence if and only if the following
sequence S’ of n-1 integers is graphic, where k = d1:
S’ = (d2-1 , d3-1 , … , dk+1-1 , dk+2 ,…, dn-1 , dn)
Notes:
S’ is obtained from S by removing the first number d1, and then
subtracting 1 from each of the next d1 numbers of S.
To apply the theorem, S must be in decreasing order. However,
the set S’ may not be in decreasing order.
Exercise:
For the graph on the right,
compute S and S’.
Thursday, 9/5/13, Slide #6
Applying Havel-Hakimi
Given sequence S with n = 9:
S = (6, 5, 5, 4, 3, 3, 2, 2, 2)
Apply H-H to S to get S’ with n = 8:
S’ = (5-1, 5-1, 4-1, 3-1, 3-1, 2-1, 2, 2)
= (4, 4, 3, 2, 2, 1, 2, 2)
Put S’ in decreasing order :
S’ = (4, 4, 3, 2, 2, 2, 2, 1)
Apply H-H again, getting set with n= 7.
Keep repeating until answer is clearly yes or
clearly no!
Thursday, 9/5/13, Slide #7
Generating a graph with a given degree
sequence
If we reach a sequence S’ that is graphic, we can
reverse the process of H-H, adding vertices as we
go:
When we add back a vertex corresponding to a
number we’ve removed, we must remember to
attach it to the corresponding vertices whose
degrees were reduced.
4, 4, 3, 2, 2, 2, 2, 1- - >
3, 2, 1, 1, 2, 2, 1
= 3, 2, 2, 2, 1, 1, 1- - >
1, 1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1
Thursday, 9/5/13, Slide #8
“If and only if” theorems
If a theorem says “A if and only if B,”
where A and B are statements, this means:
“If A then B” and “If B then A”
“If A then B” is called “necessity:”
“If B then A” is called “sufficiency:”
If A is true, then B must (necessarily) be true.
If B is true that is sufficient evidence that A is also true.
Havel-Hakimi says “S is graphic if and
only if S’ is graphic”
The construction on previous slide give idea of one
proof direction – which one?
Thursday, 9/5/13, Slide #9
Proof of Necessity:
If S is graphic, then S’ is graphic
It would be nice if we could say: If S is
graphic, just delete the highest degree
vertex, leaving a graph with degree
sequence S’. Thus S’ is graphic
This might be true, but it doesn’t have to be.
For the graph below, compute S and S’, and
then delete the highest degree vertex and
compute the degree sequence of the new
graph.
Thursday, 9/5/13, Slide #10
Proof of Necessity (If S is graphic, then
S’ is graphic)
Assume S = (d1 ≥ d2 ≥ … ≥ dn) is graphic.
Prove that S’ = (d2-1 , d3-1 , … , dk+1-1 ,
…, dn-1 , dn) is graphic.
By assumption, there is some graph G with
V(G) = {v1, v2, …, vn} such that deg(vi)
= di for each i = 1, …, n.
Let k = d1 = deg(v1). If N(v1) = {v2, …,
vk+1}, we’re done.
Otherwise, repeatedly “adjust” G to get
new graph with N(v1) = {v2, …, vk+1}
Thursday, 9/5/13, Slide #11
1.
2.
3.
4.
5.
6.
Assume: v1 is not adjacent to at least one vi in
{v2, ..., vk}.
Then: There must be some vj in {vk+1, ..., vn} that
is adjacent to v1 – why?
If deg(vi) = deg(vj), we just switch their labels to