The chromatic number of random regular graphs

Transcription

1 The chromatic number of random regular graphs Xavier Pérez Giménez Stochastics Numerics Seminar

2 Models of random graphs Erdős & Rényi classic models (G n,p, G n,m ) Fixed degree sequence Regular graphs (G n,d ) Drawn u.a.r. from the set of d-regular graphs on n vertices. Asymptotic statements as n. (Assume dn is even.)

3 The problem Chromatic number: χ(g) = 3 What is χ(g n,d )?

4 Some results Thm (Molloy & Reed 1992) Let d be constant. If q(1 1/q) d/2 < 1, then a.a.s. χ(g n,d ) > q. Thm (Frieze & Łuczak 1992) Let d < n 1/3 ɛ. Then a.a.s. χ(g n,d ) = Thm (Cooper, Frieze, Reed & Riordan 2002) Same for d n 1 ɛ. ( ) d 2 log d + O d log log d. log 2 d Thm (Krivelevich, Sudakov, Vu & Wormald 2001) Let n 6/7+ɛ d 0.9n. Then a.a.s. χ(g n,d ) n 2 log n/(n d) d.

5 Some results Thm (Ben-Shimon & Krivelevich 2009) For d = o(n 1/5 ), χ(g n,d ) is a.a.s. two-point concentrated. Thm (Achlioptas & Moore 2004) (For fixed d 3.) Let k be the smallest integer s.t. d < 2(k 1) log(k 1). Then a.a.s. χ(g n,d ) {k 1, k, k + 1}. If moreover d > (2k 3) log(k 1), then a.a.s. χ(g n,d ) {k, k + 1}. Thm (Shi & Wormald 2007) A.a.s. d , 8, 9 10 χ(g n,d ) 3 {3, 4} 4 {4, 5} {5, 6}

6 Some results Thm (Díaz, Kaporis, Kemkes, Kirousis, P.G. & Wormald 2009) A.a.s. χ(g n,5 ) = 3... unless there are kangaroos on the moon. Thm (Kemkes, P.G. & Wormald 2010) (For fixed d 3.) Let k be the smallest integer s.t. d < 2(k 1) log(k 1). Then a.a.s. χ(g n,d ) {k 1, k}. If moreover d > 2(k 3) log(k 1), then a.a.s. χ(g n,d ) = k. So e.g. χ(g n,10 ) = 5 and χ(g n,10 6) =

7 The configuration model P n,d G n,d

8 The second moment method Lemma Let X 0 be a discrete random variable (e.g. # of proper colourings). Then P(X > 0) (EX)2 EX 2 Thus, if EX 2 (EX) 2 then a.a.s. X > 0. Problem! EX 2 = Θ ( (EX) 2). Variance is too large! This only shows that X > 0 with probability bounded away from 0.

9 The small subgraph conditioning method Why?? Let Z = (Z 1, Z 2, Z 3,...) be the cycle counts. Law of total variance: VarX = E(Var(X Z )) + Var(E(X Z )) Unexplained / explained variance Thm (Janson 1995; Robinson & Wormald 1992) (i) Let λ 1, λ 2,... 0 and δ 1, δ 2,... 1 with i λ iδ 2 i < ; (ii) Z 1,..., Z k asymptotically indep. Poisson with EZ i λ i ; (iii) E(X[Z 1 ] j1 [Z k ] jk k i=1 (λ i(1 + δ i )) j i ( ; and ) (iv) EX 2 /(EX) 2 exp i λ 2 iδ i. ( Then P(X > 0) exp ) δ i = 1 λ i and [... ] (contiguity)

10 The moments First moment: EX = { (G, C) G P n,d, C = G } P n,d Second moment: { EX 2 (G, C 1, C 2 ) G P n,d, C 1, C 2 = G } = P n,d Joint moments: E(X[Z 1 ] j1 [Z k ] jk ) = { (c 1,..., c k, G, C) } P n,d

11 Asymptotics moment M Polytope D dim(d) = r ( ) r I = D 1 n Z Exact expression: M = T i i I Asymptotic expression: M poly i (n)( F i ) n i I Exponential behaviour: If F i0 F i ( i I) then M ( F i0 ) n. (Global non-linear optimisation) Polynomial factors: M = Θ (n r/2) poly i0 (n)( F i0 ) n (Laplace integration technique/ Saddlepoint method)

12 Asymptotics moment M Polytope D dim(d) = r ( ) r I = D 1 n Z Exact expression: M = T i i I Asymptotic expression: M poly i (n)( F i ) n i I Exponential behaviour: If F i0 F i ( i I) then M ( F i0 ) n. (Global non-linear optimisation) Polynomial factors: M = Θ (n r/2) poly i0 (n)( F i0 ) n (Laplace integration technique/ Saddlepoint method)

13 Asymptotics moment M Polytope D dim(d) = r ( ) r I = D 1 n Z Exact expression: M = T i i I Asymptotic expression: M poly i (n)( F i ) n i I Exponential behaviour: If F i0 F i ( i I) then M ( F i0 ) n. (Global non-linear optimisation) Polynomial factors: M = Θ (n r/2) poly i0 (n)( F i0 ) n (Laplace integration technique/ Saddlepoint method)

14 Asymptotics moment M Polytope D dim(d) = r ( ) r I = D 1 n Z Exact expression: M = T i i I Asymptotic expression: M poly i (n)( F i ) n i I Exponential behaviour: If F i0 F i ( i I) then M ( F i0 ) n. (Global non-linear optimisation) Polynomial factors: M = Θ (n r/2) poly i0 (n)( F i0 ) n (Laplace integration technique/ Saddlepoint method)

15 What about d = 5? We know χ(g n,5 ) {3, 4}. Is it 3? Problem! If X = # of proper 3-colourings, then EX 2 = ω((ex) 2 ). So the second moment method is useless. Idea: Balanced Rainbow 3-Colourings Balanced: n/3 vertices of each colour. Rainbow: No single vertex can legally change colour. (No vertex has mono-chromatic neighbors).

16 It works Thm EX (2πn) 2 ( ) 25 n 24 If no kangaroos... EX (2πn) 2 All other required conditions for the small subgraph conditioning method are satisfied. ( ) 25 n 24

17 Kangaroos on the moon? Maximisation of F in a domain D, dim(d) = 301. We have a local maximum candidate. Is it the global too? We transform the problem (Lagrange multipliers) max D ( F ) max N (F) Good: dim N = 4. Bad: F is implicit. We find max N (F) numerically Fine sweep of the domain (step = 10 4 ). IBM-Mare Nostrum (Barcelona Supercomputing Center): dual 64-bit processor blade nodes, GHz PPC970FX processors.

18 Maximisation of F Ê Òµ µ ¾ ѵ Òµ D R 324, dim D = 301 M R 36, dim M = 13 N R 9, dim N = 4 M(n) R 36, dim M(n) = 9 log F concave in M(n) Æ Å Å Òµ

19 Open Issues Case d = 5. Are there kangaroos on the moon? Close the gap for χ(g n,d ) for general constant d. Can we extend the idea of rainbow balanced colourings? Extend results to some d.

20 Thank you!