Math236
Discrete Maths with Applications
P. Ittmann
UKZN, Pietermaritzburg
Semester 1, 2012
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2012
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Degree Sequences
Let
G
be a graph with vertex set
( ) = {v1 , v2 , v3 , . . . , vn }
V G
The
degree sequence
of
G
is the sequence of numbers
deg v , deg v , deg v , . . . , deg vn
1
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2
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3
2012
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Degree Sequences (cont.)
Example
Suppose we wish to nd the degree sequence of the following graph
There are two vertices of degree 1, two of degree 3 and one of degree 2
Hence the degree sequence is
3, 3, 2, 1, 1
Note we usually write degree sequences in non-increasing order
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Math236
2012
3 / 19
Degree Sequences (cont.)
Example
Can
3, 3, 3, 3, 3, 3, 3
be the degree sequence of some graph
G
?
No it cannot
Since the graph would have 7 vertices with odd degree and
X
v ∈V
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deg v 6= 2 · m
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2012
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Degree Sequences (cont.)
Denition
A sequence d1 , d2 , . . . , dn of non-negative integer is
degree sequence of some graph
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graphical
if it is a
2012
5 / 19
Degree Sequences (cont.)
Example
Consider the following sequences
3, 3, 3, 3, 1
2, 2, 2, 0
4, 4, 4
5, 5, 4, 2, 0
Are these graphical?
We now introduce a powerful tool to determine whether a particular
sequence is graphical due to Havel and Hakimi
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Math236
2012
6 / 19
Havel-Hakimi
Theorem
,
Let D be the sequence d1 d2
n
≥2
Let D
0
, . . . , dn
with d1
≥ d2 ≥ · · · ≥ dn
and
be the sequence obtained from D by
1
Discarding d1 , and,
2
Subtracting
That is, D
0
1 from each of the next d1
entries of D
is the sequence
d2
− 1, d3 − 1, . . . , dd1 +1 − 1, dd1 +2 , . . . , dn
Then, D is graphical if and only if D
0
is graphical
We give the proof later
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Math236
2012
7 / 19
Havel-Hakimi (cont.)
Example
Consider the sequence 5, 3, 3, 3, 2, 2, 1, 1
Is it graphical?
Applying the Havel-Hakimi theorem
5, 3, 3, 3, 2, 2, 1, 1 is graphical i
2, 2, 2, 1, 1, 1, 1 is graphical i
1, 1, 1, 1, 1, 1 is graphical
This last sequence is graphical!
Hence, 5, 3, 3, 3, 2, 2, 1, 1 is graphical
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Math236
2012
8 / 19
Havel-Hakimi (cont.)
Example
Is the sequence 6, 5, 5, 5, 4, 4, 2, 1 graphical?
Applying the Havel-Hakimi theorem
6, 5, 5, 5, 4, 4, 2, 1 is graphical i
4, 4, 4, 3, 3, 1, 1 is graphical i
3, 3, 2, 2, 1, 1 is graphical i
2, 1, 1, 1, 1 is graphical i
0, 0, 1, 1 is graphical
This last sequence is graphical!
Hence, 6, 5, 5, 5, 4, 4, 2, 1 is graphical
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Math236
2012
9 / 19
Havel-Hakimi (cont.)
Example
Is the the sequence 8, 7, 6, 6, 5, 3, 2, 2, 2, 1 graphical?
Applying the Havel-Hakimi theorem
8, 7, 6, 6, 5, 3, 2, 2, 2, 1 is graphical i
6, 5, 5, 4, 2, 1, 1, 1, 1 is graphical i
4, 4, 3, 1, 0, 0, 1, 1 is graphical i
4, 4, 3, 1, 1, 1, 0, 0 is graphical (re-arranging) i
3, 2, 0, 0, 1, 0, 0 is graphical i
3, 2, 1, 0, 0, 0, 0 is graphical (re-arranging) i
1, 0, −1, 0, 0, 0 is graphical
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Math236
2012
10 / 19
Havel-Hakimi (cont.)
Example
This last sequence is clearly not graphical
Hence, 8, 7, 6, 6, 5, 3, 2, 2, 2, 1 is not graphical
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Math236
2012
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Havel-Hakimi (cont.)
Given a sequence D , it is not necessarily easy to obtain a graph
with degree sequence D
G
We can use the Havel-Hakimi theorem in reverse
Suppose D 0 is the sequence formed by H-H and we know a graph
with degree sequence D 0
We can generate
necessary
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G
by adding a vertex to
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G
0
and adding
d1
G
0
edges as
2012
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Havel-Hakimi (cont.)
Example
Let D be the sequence 4, 4, 4, 3, 3, 2
We prove that
sequence D
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D
is graphical and then draw a graph
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G
with degree
2012
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Havel-Hakimi (cont.)
Example
Applying the Havel-Hakimi theorem
D : 4, 4, 4, 3, 3, 2 is graphical i
D1
: 3, 3, 2, 2, 2 is graphical i
D2
: 2, 1, 1, 2 is graphical i
D2
: 2, 2, 1, 1 is graphical i
D3
: 1, 0, 1 is graphical i
D3
: 1, 1, 0 is graphical i
D4
: 0, 0 is graphical
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Math236
2012
14 / 19
Havel-Hakimi (cont.)
Example
Clearly 0, 0 is graphical and hence 4, 4, 4, 3, 3, 2 is graphical
We now work in reverse to obtain a graph
D3
D2
D1
G
with degree sequence
D
D
Thus, we have found a graph with degree sequence 4, 4, 4, 3, 3, 2
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Math236
2012
15 / 19
Havel-Hakimi (cont.)
We now prove the Havel-Hakimi theorem
Proof.
Suppose that the sequence
Let
G1
be a graph of order
D
n
0
is graphical
− 1 with degree sequence
D
0
Then the vertices of G1 can be labelled as v2 , v3 , . . . , vn in such a way
that degG 0 (vi ) = di − 1 if 2 ≤ i ≤ d1 + 1 and d (vi ) = di if
d1 + 2 ≤ i ≤ n
We can now construct a new graph G from G1 by adding a new vertex
v1 and then joining v1 with an edge to each of v2 , v3 , . . . , vd +1
1
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Math236
2012
16 / 19
Havel-Hakimi (cont.)
Proof.
The degree of
v1
is
d1
The degree of the other vertices are the remaining values of
Thus, we have constructed a graph with degree sequence
is graphical
We show next that if
Assume that
D
D
is graphical, then
D
0
, and so
D
is graphical
is a graphical sequence
Therefore, there are one or more graphs of order
sequence D
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D
D
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n
with degree
2012
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Havel-Hakimi (cont.)
Proof.
Among all such graphs, let G be one such that
V (G ) = {v1 , v2 , . . . , vn }, where degG (vi ) = di for 1 ≤ i ≤ n , and the
sum of the degrees of the vertices adjacent with v1 is maximum
We show that in G , the vertex v1 must be adjacent with vertices
having degrees d2 , d3 , . . . , dd1 +1
Suppose this is not the case
There must exist two vertices vj and vk with dj > dk such that
adjacent to vk , but not to vj
v1
is
Since the degree of vj exceeds that of vk , there must be some vertex
v` such that v` is adjacent to vj , but not to vk
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Math236
2012
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Havel-Hakimi (cont.)
Proof.
Removing the edges v1 vk and vj v` and adding the edges v1 vj and vk v`
produces a new graph H that also has degree sequence D
However, in H the sum of the degrees of the vertices adjacent with
is larger than that in G , contradicting our choice of G
v1
This contradiction argument veries that our initial assumption that
v1 is not adjacent with vertices having degrees d2 , d3 , . . . , dd +1 ) was
1
false
Thus, as claimed,
, , . . . , dd1 +1
v1
must be adjacent with vertices having degrees
d2 d3
Hence, the graph obtained from G by removing v1 , together with all
the edges incident with v1 , produces a graph with degree sequence D 0 ,
so D 0 is graphical
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Math236
2012
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