THE CHROMATIC INDEX OF STRONGLY REGULAR GRAPHS Willem H. Haemers - PowerPoint PPT Presentation

THE CHROMATIC INDEX OF STRONGLY REGULAR GRAPHS Willem H. Haemers joint work with Sebastian M. Cioab˘ a and Krystal Guo

The chromatic index (edge-chromatic number) χ ′ ( G ) of a graph G is the minimum number of colors needed to color the edges of G such that intersecting edges have different colors ✈ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ✈ ✈ ☎ ❉ ❚ ✔ ★ ❝ ❚ ❚ ✔ ★ ❝ ❚ ✔ ✔ ✈ ❚ ★★★★ ❝ ❚ ✔ ✏✏✏✏✏✏✏ ✔ P P ❝ ❚ P ✔ P ❝ P ❚ P ✈ ✈ ✔ ❝ P ❚ χ ′ ( G ) = 4

The chromatic number χ ( G ) of a graph G is the minimum number of colors needed to color the vertices of G such that adjacent vertices have different colors ✈ ✔ ☎ ❚ ❉ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ☎ ❉ ❚ ✔ ✈ ✈ ☎ ❉ ❚ ✔ ★ ❝ ❚ ❚ ✔ ★★ ❝ ❚ ✔ ✔ ✈ ❚ ❝ ❚ ✔ ✏✏✏✏✏✏✏ ✔ P ★★★ P ❝ ❚ P ✔ P ❝ P ❚ P ✈ ✈ ✔ ❝ P ❚ χ ( G ) = 3 χ ′ ( G ) = χ ( L ( G )), where L ( G ) is the line graph of G

Theorem (Vizing) If ∆( G ) is the maximum degree in G then χ ′ ( G ) = ∆( G ) ( G is class 1), or χ ′ ( G ) = ∆( G ) + 1 ( G is class 2) Theorem (Holyer) To decide that a graph is class 1 is NP-complete

From now on G is a k -regular graph of order n with eigenvalues k = λ 1 ≥ λ 2 ≥ · · · ≥ λ n

• If G is class 1, and edge-colored with k colors, then each color class is a perfect matching (1-factor), and the partition into color classes is a 1-factorization • G is class 1 iff G is 1-factorable • If n is odd, then G is class 2 • The complete graph K n is class 1 iff n even • (K¨ onig) A regular bipartite graph is class 1 • The Petersen graph is class 2

  0 1 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0     0 1 0 1 0 0 0 1 0 0     0 0 1 0 1 0 0 0 1 0     1 0 0 1 0 0 0 0 0 1     1 0 0 0 0 0 0 1 1 0     0 1 0 0 0 0 0 0 1 1      0 0 1 0 0 1 0 0 0 1      0 0 0 1 0 1 1 0 0 0   0 0 0 0 1 0 1 1 0 0 Petersen graph

  0 1 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0     0 1 0 1 0 0 0 1 0 0     0 0 1 0 1 0 0 0 1 0     1 0 0 1 0 0 0 0 0 1     1 0 0 0 0 0 0 1 1 0     0 1 0 0 0 0 0 0 1 1      0 0 1 0 0 1 0 0 0 1      0 0 0 1 0 1 1 0 0 0   0 0 0 0 1 0 1 1 0 0 Petersen graph

  0 1 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0     0 1 0 1 0 0 0 1 0 0     0 0 1 0 1 0 0 0 1 0     1 0 0 1 0 0 0 0 0 1     1 0 0 0 0 0 0 1 1 0     0 1 0 0 0 0 0 0 1 1      0 0 1 0 0 1 0 0 0 1      0 0 0 1 0 1 1 0 0 0   0 0 0 0 1 0 1 1 0 0 Petersen graph

  0 1 0 0 1 1 0 0 0 0 1 0 1 0 0 0 1 0 0 0     0 1 0 1 0 0 0 1 0 0     0 0 1 0 1 0 0 0 1 0     1 0 0 1 0 0 0 0 0 1     1 0 0 0 0 0 0 1 1 0     0 1 0 0 0 0 0 0 1 1      0 0 1 0 0 1 0 0 0 1      0 0 0 1 0 1 1 0 0 0   0 0 0 0 1 0 1 1 0 0 χ ′ (Petersen graph) = 4

Conjecture (Dirac) If n is even and k ≥ 1 2 n , then G is class 1 Theorem (Csaba, K¨ uhn, Lo, Osthus, Treglown) True if n is large Theorem (Cariolaro, Hilton) True if k ≥ 0 . 823 n Theorem True if the complement of G is bipartite

Theorem The complement of a regular bipartite graph with k ≥ 1 2 n is class 1 Proof If m = n / 2 is even then � J − I m � O � J − I m � � � N N O = + N ⊤ O N ⊤ J − I m J − I m O class 1 class 1 If m is odd then w.l.o.g. N i , i = 1 for i = 1 , . . . , m , and � J − I m � J − I m � � � � N O N − I m I m = + N ⊤ N ⊤ − I m J − I m O I m J − I m class 1

  0 1 1 1 1 1 0 0 0 0 1 0 1 1 1 0 1 0 0 0     1 1 0 1 1 0 0 1 0 0     1 1 1 0 1 0 0 0 1 0     1 1 1 1 0 0 0 0 0 1     1 0 0 0 0 0 1 1 1 1     0 1 0 0 0 1 0 1 1 1      0 0 1 0 0 1 1 0 1 1      0 0 0 1 0 1 1 1 0 1   0 0 0 0 1 1 1 1 1 0

  1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0     1 1 1 1 1 0 0 0 0 0     1 1 1 1 1 0 0 0 0 0     1 1 1 1 1 0 0 0 0 0     0 0 0 0 0 1 1 1 1 1     0 0 0 0 0 1 1 1 1 1      0 0 0 0 0 1 1 1 1 1      0 0 0 0 0 1 1 1 1 1   0 0 0 0 0 1 1 1 1 1

  1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0     1 1 1 1 1 0 0 0 0 0     1 1 1 1 1 0 0 0 0 0     1 1 1 1 1 0 0 0 0 0     0 0 0 0 0 1 1 1 1 1     0 0 0 0 0 1 1 1 1 1      0 0 0 0 0 1 1 1 1 1      0 0 0 0 0 1 1 1 1 1   0 0 0 0 0 1 1 1 1 1

  0 1 1 1 1 1 0 0 0 0 1 0 1 1 1 0 1 0 0 0     1 1 0 1 1 0 0 1 0 0     1 1 1 0 1 0 0 0 1 0     1 1 1 1 0 0 0 0 0 1     1 0 0 0 0 0 1 1 1 1     0 1 0 0 0 1 0 1 1 1      0 0 1 0 0 1 1 0 1 1      0 0 0 1 0 1 1 1 0 1   0 0 0 0 1 1 1 1 1 0

Strongly regular graph SRG( n , k , λ, µ ) ✗✔ ✗✔ ✗✔ λ µ k ✖✕ ✖✕ ✖✕ � ❅ � ❅ � ❅ � ❅ ✉ ✉ ✉ ✉ ✉ � ❅ � ❅ Complement is an SRG( n , n − k − 1 , n − 2 k + µ − 2 , n − 2 k + λ ) The eigenvalues of an SRG( n , k , λ, µ ) are k = λ 1 ≥ λ 2 = · · · = λ m > λ m +1 = · · · = λ n k − µ = − λ 2 λ n , λ − µ = λ 2 + λ n An SRG G is imprimitive if G = mK ℓ (class 1 iff ℓ is even), or G = K ℓ,...,ℓ (class 1 iff n is even); otherwise G is primitive

Why SRGs? Suitable for eigenvalue techniques Worked well for the chromatic number Theorem (WHH) • For a given number χ , there exist finitely many primitive SRGs with chromatic number χ • The SRGs with chromatic number ≤ 4 are known Question (Rosa) Whan can be said about the chromatic index of SRGs Can eigenvalues be helpful for the chromatic index?

Theorem (Brouwer and WHH) If n is even, G has at least ⌊ ( k + 1 − λ 2 ) / 2 ⌋ edge-disjoint perfect matchings Corollary Every connected SRG of even order has a perfect matching An easy to handle sufficient condition for being class 2, is the absence of a perfect matching. This condition will never work for a connected SRG of even order

Theorem (Hoffman) χ ( G ) ≥ 1 − k /λ n Equality implies that all color classes have equal cardinality, and that each vertex has exactly − λ n neighbors in each other color class Theorem If χ ( G ) = 1 − k /λ n and even, then G is class 1, and so is its complement (provided G � = K ℓ,...,ℓ )

Proof     O N 1 , 2 N 1 , 3 N 1 , 4 O N 1 , 2 O O N 2 , 1 O N 2 , 3 N 2 , 4 N 2 , 1 O O O      =  +     N 3 , 1 N 3 , 2 O N 3 , 4 O O O N 3 , 4   N 4 , 1 N 4 , 2 N 4 , 3 O O O N 4 , 3 O

Proof     O N 1 , 2 N 1 , 3 N 1 , 4 O N 1 , 2 O O N 2 , 1 O N 2 , 3 N 2 , 4 N 2 , 1 O O O      =  +     N 3 , 1 N 3 , 2 O N 3 , 4 O O O N 3 , 4   N 4 , 1 N 4 , 2 N 4 , 3 O O O N 4 , 3 O class 1     O O N 1 , 3 O O O O N 1 , 4 O O O N 2 , 4 O O N 2 , 3 O      +     N 3 , 1 O O O O N 3 , 2 O O    O N 4 , 2 O O N 4 , 1 O O O class 1 class 1

Complement  J - I N 1 , 2 N 1 , 3 N 1 , 4   J - I N 1 , 2 O O  N 2 , 1 J - I N 2 , 3 N 2 , 4 N 2 , 1 J - I O O      =  +     N 3 , 1 N 3 , 2 J - I N 3 , 4 O O J - I N 3 , 4   N 4 , 1 N 4 , 2 N 4 , 3 J - I O O N 4 , 3 J - I class 1     O O N 1 , 3 O O O O N 1 , 4 O O O N 2 , 4 O O N 2 , 3 O      +     N 3 , 1 O O O O N 3 , 2 O O    O N 4 , 2 O O N 4 , 1 O O O class 1 class 1

Corollary The Latin square graphs of even order, and their complements are class 1 The Latin square graph LS ( m , t ) is defined on the m 2 entries of t ≥ 0 MOLS of order m , where two vertices are adjacent if they have the same row, column, or symbol in one of the squares LS ( m , t ) has parameters ( m 2 , ( t + 2)( m − 1) , m − 2 + t ( t + 1) , ( t + 1)( t + 2)) , and is primitive if 0 ≤ t < m − 2

Theorem (Ferber and Jain) For every ε > 0 there exist integers k 0 and n 0 , such that if k > k 0 , n > n 0 , n even, and max { λ 2 , − λ n } < k 1 − ε then χ ′ ( G ) = k Corollary There exists an integer n 0 such that every primitive SRG of even order n > n 0 , which is not the block graph of a Steiner 2-design or its complement, is class 1