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)

• f 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

• f 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 = λ1 ≥ λ2 ≥ · · · ≥ λn

• 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 Kn is class 1 iff n even
• (K¨
• nig) 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

2n, then G is class 1

Theorem (Csaba, K¨ uhn, Lo, Osthus, Treglown) True if n is large Theorem (Cariolaro, Hilton) True if k ≥ 0.823n Theorem True if the complement of G is bipartite

Theorem The complement of a regular bipartite graph with k ≥ 1

2n is class 1

Proof If m = n/2 is even then J−Im N N⊤ J−Im

• =

O N N⊤ O

• +

J−Im O O J−Im

• class 1

class 1 If m is odd then w.l.o.g. Ni,i = 1 for i = 1, . . . , m, and J−Im N N⊤ J−Im

• =
• O

N−Im N⊤−Im O

• +

J−Im Im Im J−Im

• 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) k

✖✕ ✗✔ ✉

• ❅

❅ ❅

λ

✖✕ ✗✔ ✉ ✉

• ❅

❅ ❅

µ

✖✕ ✗✔ ✉ ✉

Complement is an SRG(n, n − k − 1, n − 2k + µ − 2, n − 2k + λ) The eigenvalues of an SRG(n, k, λ, µ) are k = λ1 ≥ λ2 = · · · = λm > λm+1 = · · · = λn k − µ = −λ2λn, λ − µ = λ2 + λn

An SRG G is imprimitive if G = mKℓ (class 1 iff ℓ is even), or G = Kℓ,...,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 − λ2)/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/λn Equality implies that all color classes have equal cardinality, and that each vertex has exactly −λn neighbors in each other color class Theorem If χ(G) = 1 − k/λn and even, then G is class 1, and so is its complement (provided G = Kℓ,...,ℓ)

Proof     O N1,2 N1,3 N1,4 N2,1 O N2,3 N2,4 N3,1 N3,2 O N3,4 N4,1 N4,2 N4,3 O     =     O N1,2 O O N2,1 O O O O O O N3,4 O O N4,3 O     +

Proof     O N1,2 N1,3 N1,4 N2,1 O N2,3 N2,4 N3,1 N3,2 O N3,4 N4,1 N4,2 N4,3 O     =     O N1,2 O O N2,1 O O O O O O N3,4 O O N4,3 O     + class 1     O O N1,3 O O O O N2,4 N3,1 O O O O N4,2 O O     +     O O O N1,4 O O N2,3 O O N3,2 O O N4,1 O O O     class 1 class 1

Complement     J-I N1,2 N1,3 N1,4 N2,1 J-I N2,3 N2,4 N3,1 N3,2 J-I N3,4 N4,1 N4,2 N4,3 J-I     =     J-I N1,2 O O N2,1 J-I O O O O J-I N3,4 O O N4,3 J-I     + class 1     O O N1,3 O O O O N2,4 N3,1 O O O O N4,2 O O     +     O O O N1,4 O O N2,3 O O N3,2 O O N4,1 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 m2 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 (m2, (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 k0 and n0, such that if k > k0, n > n0, n even, and max{λ2, −λn} < k1−ε then χ′(G) = k Corollary There exists an integer n0 such that every primitive SRG of even order n > n0, which is not the block graph of a Steiner 2-design or its complement, is class 1

Proof Suppose G is an primitive SRG

• n ≤ k2 + 1, so we can delete the condition k > k0
• (Neumaier) If G is not a Latin square graph or the

block graph of a Steiner 2-design, then λ2 ≤ λ6

n

• Applied to G and its complement G this gives

max{λ2, −λn} ≤ k6/7, when G nor G is a Latin square graph or the block graph of a Steiner 2-design

• Apply Ferber and Jain with ε = 1/7
• Latin square graphs of even order and their

complements are class 1

All SRGs with n even and k ≤ 18 are known (Brouwer) (10,3,0,1 ) 1 (16,5,0,2) 1 (16,6,2,2) 2 (26,10,3,4) 10 (28,12,6,4) 4 (36,10,4,2) 1 (36,14,4,6) 180 (36,14,7,4) 1 (36,15,6,6) 32548 (40,12,2,4) 28 (50,7,0,1) 1 (56,10,0,2) 1 (64,14,6,2) 1 (64,18,2,6) 167 (100,18,8,2) 1

The primitive SRGs with n even and k ≤ 18, k < n/2

Also the complements of the first column have k ≤ 18; the first parameter set belongs to the Petersen graph

Using the description of the graphs in the table given by Spence we searched (using SageMath) for a 1-factorization in each of these graphs and their complements, and found one in all cases, except for the Petersen graph Theorem Except for the Petersen graph, a primitive SRG of even order and degree at most 18 is class 1, and so is its complement

The block graph of a Steiner 2-(v, m, 1) design is an SRG( v(v−1)

m(m−1), m(v−m) m−1 , (m − 1)2 + v−2m+1 m−1 , m2)

defined on the blocks of the design, where two blocks are adjacent whenever they intersect. The graph is primitive if v > m2 If m = 2, then G is the line graph of Kv, better known as the triangular graph T(v), which has even order if v ≡ 0 or 1 mod 4

Theorem T(v) is class 1 if v ≡ 1 mod 4 Theorem The complement of T(v) is class 1 if the

• rder is even and v = 5

Proof Computer search for v ≤ 21; Cariolaro and Hilton if v ≥ 24

Conjecture Except for the Petersen graph, every connected SRG of even order is class 1 Problems

• Prove the conjecture
• Prove the conjecture for n > n0
• Prove the conjecture for the complement of the block

graph of a Steiner 2-design

• Prove that T(v) is class 1 if v ≡ 0 mod 4
• Find a second connected SRG with n even which is

class 2

• Find two cospectral SRGs, where one is class 1 and

the other class 2