[PPT] - The Search for Moore Graphs: Beauty is Rare Daniel W. Cranston PowerPoint Presentation

SLIDE 1

The Search for Moore Graphs: Beauty is Rare

Daniel W. Cranston

Virginia Commonwealth University dcranston@vcu.edu LSU Student Colloquium 5 October 2011

SLIDE 2

Intro

Q: How many vertices can we have in a graph with low diameter?

SLIDE 3

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k?

SLIDE 4

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k

SLIDE 5

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2:

SLIDE 6

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2:

SLIDE 7

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2: 1 + k + k(k − 1) = 1 + k2

SLIDE 8

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2: 1 + k + k(k − 1) = 1 + k2 k = 2

SLIDE 9

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2: 1 + k + k(k − 1) = 1 + k2 k = 2 k = 3

SLIDE 10

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2: 1 + k + k(k − 1) = 1 + k2 k = 2 k = 3 k = 4 . . . ???

SLIDE 11

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2: 1 + k + k(k − 1) = 1 + k2 k = 2 k = 3 k = 4 . . . ??? Def: A Moore Graph is k-regular with k2 + 1 vertices and diam 2.

SLIDE 12

Intro

Q: How many vertices can we have in a graph with low diameter and maximum degree k? diam 1: 1 + k diam 2: 1 + k + k(k − 1) = 1 + k2 k = 2 k = 3 k = 4 . . . ??? Def: A Moore Graph is k-regular with k2 + 1 vertices and diam 2. Main Theorem: [Hoffman-Singleton 1960] Moore graphs exist only when k = 2, 3, 7, and (possibly) 57. When k ∈ {2, 3, 7}, the Moore graph is unique.

SLIDE 13

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph.

SLIDE 14

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A.

SLIDE 15

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A. ◮ Use linear algebra to simplify the equation to a polynomial.

SLIDE 16

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A. ◮ Use linear algebra to simplify the equation to a polynomial. ◮ Use rational root theorem to show k ∈ {2, 3, 7, 57}.

SLIDE 17

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A. ◮ Use linear algebra to simplify the equation to a polynomial. ◮ Use rational root theorem to show k ∈ {2, 3, 7, 57}.

Ex: A5 =       1 1 1 1 1 1 1 1 1 1      

SLIDE 18

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A. ◮ Use linear algebra to simplify the equation to a polynomial. ◮ Use rational root theorem to show k ∈ {2, 3, 7, 57}.

Ex: A5 =       1 1 1 1 1 1 1 1 1 1       A2

5 =

      2 1 1 2 1 1 1 2 1 1 1 2 1 1 2      

SLIDE 19

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A. ◮ Use linear algebra to simplify the equation to a polynomial. ◮ Use rational root theorem to show k ∈ {2, 3, 7, 57}.

Ex: A5 =       1 1 1 1 1 1 1 1 1 1       A2

5 =

      2 1 1 2 1 1 1 2 1 1 1 2 1 1 2       In fact, A2

5 + A5 − I5 = J5,

SLIDE 20

Outline

Our Plan

◮ Assume there exists a k-regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A. ◮ Use linear algebra to simplify the equation to a polynomial. ◮ Use rational root theorem to show k ∈ {2, 3, 7, 57}.

Ex: A5 =       1 1 1 1 1 1 1 1 1 1       A2

5 =

      2 1 1 2 1 1 1 2 1 1 1 2 1 1 2       In fact, A2

5 + A5 − I5 = J5, and more generally:

A2 + A − (k − 1)I = J

SLIDE 21

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J.

SLIDE 22

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J. (A2 + A − (k − 1)I) v = J v

SLIDE 23

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J. (A2 + A − (k − 1)I) v = J v k2 v + k v − (k − 1)) v = n v

SLIDE 24

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J. (A2 + A − (k − 1)I) v = J v k2 v + k v − (k − 1)) v = n v k2 + 1 = n

SLIDE 25

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J. (A2 + A − (k − 1)I) v = J v k2 v + k v − (k − 1)) v = n v k2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n

rthogonal eigenvectors.

SLIDE 26

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J. (A2 + A − (k − 1)I) v = J v k2 v + k v − (k − 1)) v = n v k2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n

rthogonal eigenvectors.

Let u be another eigenvector of A with eigenvalue r. (A2 + A − (k − 1)I) u = J u

SLIDE 27

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J. (A2 + A − (k − 1)I) v = J v k2 v + k v − (k − 1)) v = n v k2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n

rthogonal eigenvectors.

Let u be another eigenvector of A with eigenvalue r. (A2 + A − (k − 1)I) u = J u (r2 + r − (k − 1)) u =

SLIDE 28

Linear Algebra

Recall that A2 + A − (k − 1)I = J. Note that v = [1 . . . 1]T is an eigenvector of A and J. (A2 + A − (k − 1)I) v = J v k2 v + k v − (k − 1)) v = n v k2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n

rthogonal eigenvectors.

Let u be another eigenvector of A with eigenvalue r. (A2 + A − (k − 1)I) u = J u (r2 + r − (k − 1)) u = r2 + r − (k − 1) = 0

SLIDE 29

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

SLIDE 30

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries.

SLIDE 31

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0.

SLIDE 32

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational

SLIDE 33

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

SLIDE 34

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

SLIDE 35

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2

SLIDE 36

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

SLIDE 37

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational

SLIDE 38

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3.

SLIDE 39

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

,

SLIDE 40

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0

SLIDE 41

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0

SLIDE 42

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0 Rational Root Theorem ⇒ s ∈ {±1, ±3, ±5, ±15}

SLIDE 43

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0 Rational Root Theorem ⇒ s ∈ {±1, ±3, ±5, ±15}

SLIDE 44

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0 Rational Root Theorem ⇒ s ∈ {±1, ±3, ±5, ±15} s k n 2 5 3 3 10 5 7 50 15 57 3250

SLIDE 45

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0 Rational Root Theorem ⇒ s ∈ {±1, ±3, ±5, ±15} s k n 2 5 ← − 5-cycle 3 3 10 5 7 50 15 57 3250

SLIDE 46

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0 Rational Root Theorem ⇒ s ∈ {±1, ±3, ±5, ±15} s k n 2 5 ← − 5-cycle 3 3 10 ← − Petersen 5 7 50 15 57 3250

SLIDE 47

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0 Rational Root Theorem ⇒ s ∈ {±1, ±3, ±5, ±15} s k n 2 5 ← − 5-cycle 3 3 10 ← − Petersen 5 7 50 ← − ? 15 57 3250

SLIDE 48

Solving for k

r1 = −1+

√ 4k−3 2

and r2 = −1−

√ 4k−3 2

Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r1m1 + r2m2 = 0. Case 1: r1 and r2 are irrational 0 = k + (r1 + r2)n−1

2

= k + (−1)n−1

2

= k + (−1)k2

2 ⇒ k ∈ {0, 2}

Case 2: r1 and r2 are rational, so let s2 = 4k − 3. r1 = s−1

2

and r2 = −s−1

2

, so k + s−1

2 m + −s−1 2

(n − m − 1) = 0 s5 + s4 + 6s3 − 2s2 + (9 − 32m)s − 15 = 0 Rational Root Theorem ⇒ s ∈ {±1, ±3, ±5, ±15} s k n 2 5 ← − 5-cycle 3 3 10 ← − Petersen 5 7 50 ← − ? 15 57 3250 ← − ???

SLIDE 49

The Hoffman-Singleton Graph

Main Theorem: There exists a Moore graph when k = 7.

SLIDE 50

The Hoffman-Singleton Graph

Main Theorem: There exists a Moore graph when k = 7. Big idea: Partition the 50 vertices into ten 5-cycles:

A0 A1 A2 A3 A4 B0 B1 B2 B3 B4

SLIDE 51

The Hoffman-Singleton Graph

Main Theorem: There exists a Moore graph when k = 7. Big idea: Partition the 50 vertices into ten 5-cycles:

A0 A1 A2 A3 A4 B0 B1 B2 B3 B4

such that all remaining edges are between an Ai and a Bj, and such that ∀i, j the subgraph G[Ai ∪ Bj] ∼ = Petersen.

SLIDE 52

The Hoffman-Singleton Graph

Main Theorem: There exists a Moore graph when k = 7. Big idea: Partition the 50 vertices into ten 5-cycles:

A0 A1 A2 A3 A4 B0 B1 B2 B3 B4

such that all remaining edges are between an Ai and a Bj, and such that ∀i, j the subgraph G[Ai ∪ Bj] ∼ = Petersen. To avoid 3-cycles and 4-cycles, we can complete these 25 copies of the Petersen graph in exactly one way (up to isomorphism).

SLIDE 53

The Hoffman-Singleton Graph

Main Theorem: There exists a Moore graph when k = 7. Big idea: Partition the 50 vertices into ten 5-cycles:

A0 A1 A2 A3 A4 B0 B1 B2 B3 B4

such that all remaining edges are between an Ai and a Bj, and such that ∀i, j the subgraph G[Ai ∪ Bj] ∼ = Petersen. To avoid 3-cycles and 4-cycles, we can complete these 25 copies of the Petersen graph in exactly one way (up to isomorphism). So our desired Moore graph exists and is unique.

SLIDE 54

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A.

SLIDE 55

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A.

SLIDE 56

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5.

SLIDE 57

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25.

SLIDE 58

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. Lemma 2: G has 1260 5-cycles:

SLIDE 59

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50

SLIDE 60

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)

SLIDE 61

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)

SLIDE 62

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)

SLIDE 63

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)

SLIDE 64

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10

SLIDE 65

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10 1000 type ABABx

SLIDE 66

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10 1000 type ABABx 25(5)(4)(4)(1)/2

SLIDE 67

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10 1000 type ABABx 25(5)(4)(4)(1)/2 250 type AABBx

SLIDE 68

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10 1000 type ABABx 25(5)(4)(4)(1)/2 250 type AABBx 25(2)(5)(2)(1)/2

SLIDE 69

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10 1000 type ABABx 25(5)(4)(4)(1)/2 250 type AABBx 25(2)(5)(2)(1)/2 10 types AAAAx and BBBBx

SLIDE 70

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10 1000 type ABABx 25(5)(4)(4)(1)/2 250 type AABBx 25(2)(5)(2)(1)/2 10 types AAAAx 1260 − 1000 − 250 and BBBBx

SLIDE 71

Deducing Structure

Let A be the neighbors of some 5-cycle, and let B = V (G) \ A. Lemma 1: For all a ∈ A and b ∈ B we have dA(a) = dB(b) = 2 and dB(a) = dA(b) = 5. Pf: Note that dA(a) ≤ 2 and dA(b) ≤ 5 and |A| = |B| = 25. . . . . . . . . . Lemma 2: G has 1260 5-cycles: 50(7)(6)(6)(1)/10 1000 type ABABx 25(5)(4)(4)(1)/2 250 type AABBx 25(2)(5)(2)(1)/2 10 types AAAAx 1260 − 1000 − 250 and BBBBx Cor: G[A] and G[B] each consist of 5 disjoint 5-cycles.

SLIDE 72

Deducing Structure