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Thursday, 9/5/13
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
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,
∑ d ( v ) = 2e
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.
Thursday, 9/5/13, Slide #3
Regular Graphs
A regular graph is one in which all vertices have
the same degree.
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
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
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
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)
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.
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
If we reach a sequence S’ that is graphic, we can
reverse the process of H-H, adding vertices as we
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
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
Details of “adjusting” G
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
make v1 adjacent to vi.
Otherwise, deg(vi) > deg(vj) – why? So there is
another vertex w, adjacent to vi, but not to vj.
Now we remove and add edges to make v1 adjacent
to vi, without changing the degree of any vertex.
Keep repeating this process as long as v1 is not
adjacent to some vertex in S.
See file HavelHakimi.pdf for write-up of this proof.
Thursday, 9/5/13, Slide #12