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 t=2

{1,2} {1,3} {1,4} {1,5}{2,4} {2,3} {4,5} {3,4} {2,5} {3,5} { 1, 2, 3, 4, 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 2,42,3 4,5 1,4 1,3 3,5 1,5 3,4

PETERSON GRAPH is K (5,2)

1,2 2,5 2,4 2,3 4,5 1,4 1,3 3,5 1,5 3,4

K (7,3) 35 vertices

Definitions  Distance – the length of the shortest path from vertex u to vertex v of a graph. uv

Definition  Diameter – The longest distance in a graph G. u v

Diameter (2t +1 ≤ n < 3t -1) Fact  For n ≥ 3t -1, diam(K ( n, t)= 2. 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 girth K(n,t) = 6n = 2t n > 2t + 1 girth K(n,t) = 6n = 2t + 1

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 BC

|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.

Hence, the girth K( n, t) > 3 when n 3 when n<3t -1. A BC

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 | = | A U B U C |

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.

{1,…,t} {t+1,…,2t} {t+1,…, 2t-1, 2t+1} {1,…, t-1, 2t+2}

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|

{1,…,t} {t+1,…,2t} {t+1,…, 2t-1, 2t+1} {2,…, t, 2t} {2,…,t, 2t+1}  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.

{1,…,t} {t+1,…,2t} {t+1,…, 2t-1, 2t+1} {2,…, t, 2t} {2,…,t, 2t+1} Hence, the girth K( n, t) > 5 when 2t when 2t + 1<3t -1.

{1, t+1,…, 2t-1} {1,…,t} {t+1,…,2t} {t+1,…, 2t-1, 2t+1} {2,…, t, 2t} {2,…,t, 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