Investigating properties of Kneser Graphs

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Investigating properties ofKneser Graphs

• Modesty Briggs
• California State University, Northridge
• Sponsored by JPL/NASA Pair program
• Funded by NSA and NSF

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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.

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Example K (5, 2)

• n5 1, 2, 3, 4, 5
• t2

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

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1,2
2,5
3,4
2,4
2,3
1,5
4,5
1,3
1,4
3,5
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PETERSON GRAPH

• is
• K (5,2)

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1,2
3,5
3,4
4,5
1,4
2,5
2,3
1,3
2,4
1,5
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K (7,3) 35 vertices
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FACTS
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Definitions

• Distance the length of the shortest path from
  vertex u to vertex v of a graph.

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Definition

• Diameter The longest distance in a graph G.

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Diameter (2t 1 n lt 3t -1)

• Fact
• For n 3t -1, diam(K ( n, t) 2.

• Assumption
• 2 lt diam(K ( n, t) t

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Girth (2t 1 n lt 3t -1)

• The length of the shortest cycle
• Theorem Let K(n,t) be a Kneser Graph
  with nlt3t-1.
• 4 n gt 2t
  1
  girth K(n,t)
  6 n 2t 1

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WHY NOT 3?
PROOF
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• Let A be subset of 1,2,,n containing
  t elements. There exist a B subset
  of 1,2,,n, such that A n B Ø.
• Now consider subset C of 1,2,,n
  containing t elements, such that
  A n CØ .
• Then either B n CØ or B n C?Ø .

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• Assume B n C Ø. Then subsets A, B, C are
  mutually disjoint.

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• A?B?C A B C
• t t t
• 3t n
• Then, 3t n lt 3t -1
• But 3t lt 3t 1 is a contradiction.
• Therefore, B n C ? Ø and there will not exist a
  cycle of length three.

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Hence, the girth K( n, t) gt 3 when nlt3t -1.
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CASE 1
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• Assume n gt 2t1. 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 n A C n AØ).
• So,
• ngt 2t 1 t (t 1)
• A B U C
• A U B U C

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• Then n gt 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.

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1,,t
t1,,2t
t1,, 2t-1, 2t1
1,, t-1, 2t2
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Therefore, cycle length is four. Hence the girth
K(n,t)4 when 2t1lt nlt 3t-1
CASE 2
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• Assume n 2t1
• As At and B U C t1, we have
• n 2t 1 t (t1)
• A B U C
• A U B U C .
• So, n A U B U C

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2,,t, 2t1
2,, t, 2t

• Assume D n E Ø.

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• So,
• D U E D E
• t t
• 2t
• However, D U E t 2 when n 2 t 1.
• Therefore, since t2 lt 2t, D n E?Ø and
  there will not be a cycle of length five.

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2,,t, 2t1
2,, t, 2t
Hence, the girth K( n, t) gt 5 when 2t 1lt3t -1.
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2,,t, 2t1
2,, t, 2t
1, t1,, 2t-1
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What Next ?

• Will the diameter equal t as n gets closer to
  2t 1.

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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