Investigating properties
of
Kneser Graphs
Modesty Briggs
California State University, Northridge
Sponsored by JPL/NASA Pair program;
Funded by NSA and NSF
What is a Kneser Graph?
 For n ≥ 2t + 1, the Kneser graph,
K( n, t), is the graph whose vertices
are the t subsets of an n-set.
Example: K (5, 2)
[n]=[5] = { 1, 2, 3, 4, 5}
t=2
{ 1, 2, 3, 4, 5}
{1,2}
{2,5}
{2,4}
{3,4}
{1,5}
{4,5}
{2,3}
{1,3}
{1,4}
{3,5}
What is a Kneser?
 For n ≥ 2t + 1, the Kneser graph,
K( n, t), is the graph whose vertices
are the t subsets of an n-set.
 Vertices are adjacent when
corresponding subsets are disjoint.
1,2
2,5
3,4
2,4
1,5
2,3
4,5
1,3
1,4
3,5
PETERSON GRAPH
is
K (5,2)
1,2
3,5
3,4
4,5
1,4
2,5
2,3
1,5
1,3
2,4
K (7,3)
35 vertices
Definitions
 Distance – the length of the shortest path
from vertex u to vertex v of a graph.
u
v
Definition
 Diameter – The longest distance in a graph
G.
u
v
Diameter
Fact
For n ≥ 3t -1,
diam(K ( n, t)= 2.
(2t +1 ≤ n < 3t -1)
Assumption
2 < diam(K ( n, t) ≤ t
Girth
(2t +1 ≤ n < 3t -1)
 The length of the shortest cycle
 Theorem: Let K(n,t) be a Kneser Graph
with n<3t-1.
4
n > 2t + 1
6
n = 2t + 1
girth K(n,t) =
WHY NOT 3?
Let A be subset of {1,2,…,n} containing
t elements. There exist a B subset
of {1,2,…,n}, such that A ∩ B =Ø.
Now consider subset C of {1,2,…,n}
containing t elements, such that
A ∩ C=Ø .
Then either B ∩ C=Ø or B ∩ C≠Ø .
 Assume B ∩ C = Ø. Then subsets A, B, C are
mutually disjoint.
A
B
C
|AυBυC|= |A| + |B| + |C|
=t+t+t
= 3t ≤ n
Then, 3t ≤ n < 3t -1
But 3t < 3t – 1 is a contradiction.
Therefore, B ∩ C ≠ Ø and there will
not exist a cycle of length three.
A
B
C
Hence, the girth K( n, t) > 3 when n<3t -1.
Assume n > 2t+1. WLOG, let A be the
subset {1,…,t} of n-set.
Since subsets B and C are both disjoint to A,
then B and C may be chosen such that
|B U C|= t +1(maintaining B ∩ A= C ∩ A=Ø).
So,
n> 2t +1= t + (t +1)
= |A| + | B U C |
=|AUBUC|
Then n > | A U B U C |.
Therefore, there are elements in n that are
not in A, B, or C.
Hence, another subset D can be composed
of t elements not in B or C.
Therefore, cycle length is four.
Hence the girth K(n,t)=4 when
2t+1< n< 3t-1
Assume n= 2t+1
As |A|=t and |B U C|= t+1, we have
n= 2t + 1= t + (t+1)
= |A| + |B U C|
= |A U B U C| .
So, n = |A U B U C|
{2,…,t, 2t+1}
{2,…, t, 2t}
 Assume D ∩ E = Ø.
So,
|D U E|= |D| + |E|
=t+t
= 2t
However, |D U E|= t + 2 when n = 2 t + 1.
Therefore, since t+2 < 2t, D ∩ E≠Ø and
there will not be a cycle of length five.
{2,…,t, 2t+1}
{2,…, t, 2t}
Hence, the girth K( n, t) > 5
when 2t + 1<3t -1.
{2,…,t, 2t+1}
{2,…, t, 2t}
{1, t+1,…, 2t-1}
What Next ?
Will the diameter equal t
as n gets closer to
2t + 1.
Special Thanks
• JPL/NASA PAIR PROGRAM
• NSF and NSA for funding
• Dr. Carol Shubin
• Dr. Cynthia Wyels (CAL Lutheran)
• Dr. Michael Neubauer
• Various Professors in the Math Department