The Search for Moore Graphs: Beauty is Rare Daniel W. Cranston Virginia Commonwealth University dcranston@vcu.edu LSU Student Colloquium 5 October 2011
Intro Q: How many vertices can we have in a graph with low diameter?
Intro Q: How many vertices can we have in a graph with low diameter and maximum degree k ?
Intro Q: How many vertices can we have in a graph with low diameter and maximum degree k ? diam 1: 1 + k
Intro Q: How many vertices can we have in a graph with low diameter and maximum degree k ? diam 1: 1 + k diam 2:
Intro Q: How many vertices can we have in a graph with low diameter and maximum degree k ? diam 1: 1 + k diam 2:
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 + k 2
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 + k 2 k = 2
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 + k 2 k = 2 k = 3
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 + k 2 ??? k = 2 k = 3 k = 4 . . .
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 + k 2 ??? k = 2 k = 3 k = 4 . . . Def: A Moore Graph is k -regular with k 2 + 1 vertices and diam 2.
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 + k 2 ??? k = 2 k = 3 k = 4 . . . Def: A Moore Graph is k -regular with k 2 + 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.
Outline Our Plan ◮ Assume there exists a k -regular Moore graph.
Outline Our Plan ◮ Assume there exists a k -regular Moore graph. ◮ Write an equation in terms of the adjacency matrix A .
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.
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 } .
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: 0 1 0 0 1 1 0 1 0 0 A 5 = 0 1 0 1 0 0 0 1 0 1 1 0 0 1 0
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: 0 1 0 0 1 2 0 1 1 0 1 0 1 0 0 0 2 0 1 1 A 2 A 5 = 0 1 0 1 0 5 = 1 0 2 0 1 0 0 1 0 1 1 1 0 2 0 1 0 0 1 0 0 1 1 0 2
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: 0 1 0 0 1 2 0 1 1 0 1 0 1 0 0 0 2 0 1 1 A 2 A 5 = 0 1 0 1 0 5 = 1 0 2 0 1 0 0 1 0 1 1 1 0 2 0 1 0 0 1 0 0 1 1 0 2 In fact, A 2 5 + A 5 − I 5 = J 5 ,
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: 0 1 0 0 1 2 0 1 1 0 1 0 1 0 0 0 2 0 1 1 A 2 A 5 = 0 1 0 1 0 5 = 1 0 2 0 1 0 0 1 0 1 1 1 0 2 0 1 0 0 1 0 0 1 1 0 2 In fact, A 2 5 + A 5 − I 5 = J 5 , and more generally: A 2 + A − ( k − 1) I = J
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that �
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that � ( A 2 + A − ( k − 1) I ) � v = J � v
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that � ( A 2 + A − ( k − 1) I ) � v = J � v k 2 � v + k � v − ( k − 1)) � v = n � v
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that � ( A 2 + A − ( k − 1) I ) � v = J � v k 2 � v + k � v − ( k − 1)) � v = n � v k 2 + 1 = n
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that � ( A 2 + A − ( k − 1) I ) � v = J � v k 2 � v + k � v − ( k − 1)) � v = n � v k 2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n orthogonal eigenvectors.
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that � ( A 2 + A − ( k − 1) I ) � v = J � v k 2 � v + k � v − ( k − 1)) � v = n � v k 2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n orthogonal eigenvectors. Let � u be another eigenvector of A with eigenvalue r . ( A 2 + A − ( k − 1) I ) � u = J � u
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that � ( A 2 + A − ( k − 1) I ) � v = J � v k 2 � v + k � v − ( k − 1)) � v = n � v k 2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n orthogonal eigenvectors. Let � u be another eigenvector of A with eigenvalue r . ( A 2 + A − ( k − 1) I ) � u = J � u ( r 2 + r − ( k − 1)) � u = � 0
Linear Algebra Recall that A 2 + A − ( k − 1) I = J . v = [1 . . . 1] T is an eigenvector of A and J . Note that � ( A 2 + A − ( k − 1) I ) � v = J � v k 2 � v + k � v − ( k − 1)) � v = n � v k 2 + 1 = n Spectral Theorem Every real symmetric n × n matrix has real eigenvalues and n orthogonal eigenvectors. Let � u be another eigenvector of A with eigenvalue r . ( A 2 + A − ( k − 1) I ) � u = J � u ( r 2 + r − ( k − 1)) � u = � 0 r 2 + r − ( k − 1) = 0
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries.
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r 1 m 1 + r 2 m 2 = 0.
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r 1 m 1 + r 2 m 2 = 0. Case 1: r 1 and r 2 are irrational
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r 1 m 1 + r 2 m 2 = 0. Case 1: r 1 and r 2 are irrational 0 = k + ( r 1 + r 2 ) n − 1 2
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r 1 m 1 + r 2 m 2 = 0. Case 1: r 1 and r 2 are irrational 0 = k + ( r 1 + r 2 ) n − 1 = k + ( − 1) n − 1 2 2
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r 1 m 1 + r 2 m 2 = 0. Case 1: r 1 and r 2 are irrational = k + ( − 1) k 2 0 = k + ( r 1 + r 2 ) n − 1 = k + ( − 1) n − 1 2 2 2
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r 1 m 1 + r 2 m 2 = 0. Case 1: r 1 and r 2 are irrational = k + ( − 1) k 2 0 = k + ( r 1 + r 2 ) n − 1 = k + ( − 1) n − 1 2 ⇒ k ∈ { 0 , 2 } 2 2
Solving for k √ √ r 1 = − 1+ 4 k − 3 and r 2 = − 1 − 4 k − 3 2 2 Fact: Sum of the eigenvalues equals sum of the diagonal entries. k + r 1 m 1 + r 2 m 2 = 0. Case 1: r 1 and r 2 are irrational = k + ( − 1) k 2 0 = k + ( r 1 + r 2 ) n − 1 = k + ( − 1) n − 1 2 ⇒ k ∈ { 0 , 2 } 2 2 Case 2: r 1 and r 2 are rational
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