Reducing graph coloring to stable set without symmetry

Transcription

1 Reducing graph coloring to stable set without symmetry Denis Cornaz (with V. Jost, with P. Meurdesoif) LAMSADE, Paris-Dauphine SPOC 11 Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 1 / 14

2 Coloring and perfect 0-1 matrices (P) = {Ax b} is TDI: min{y b : y A w } = min{y b : y A w, y integer} ( w integer) P = {x : Ax b} is integer: max{w x : x P} = max{w x : x P, x integer} ( w) A is 0-1: the clique-matrix {maximal cliques} {vertices} α(g) := max{1 x : Ax 1, x 0, x integer} =: ω(g) χ(g) := min{y 1 : y A 1, y 0, y integer} =: χ(g) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 2 / 14

3 Compact LP formulation χ(g) = min i x i s.t. x i x v i ( i, v) i x v i = 1 ( v) x u i + x v i 1 ( i, uv) x 0 x integer where G = (V, E) is a graph, i {1,..., V }, v V, uv E. Ĝ = G 1... G V (G) add u 1,..., u V (G) universal χ(g) + α(ĝ) = 2 V (G) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 3 / 14

4 Gallai identities For all G (without isolated vertex) ρ(g) + ν(g) = V (G) = α(g) + τ(g) If G has no 3-cycle χ(g) + α(l(g)) = V (G) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 4 / 14

5 Gallai identities For all G (without isolated vertex) ρ(g) + ν(g) = V (G) = α(g) + τ(g) If G has no 3-cycle χ(g) + α(l(g)) = V (G) = α(g) + χ(l(g)) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 5 / 14

6 The sandwich line-graphs S(G) of G Choose a orientation G of G without 3-dicycle (cliques are acyclic) S( G) is the line-graph L(G) of G minus the edges corresponding to simplicial pairs of arcs Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 6 / 14

7 Chromatic Gallai identities (a) (b) (c) (d) (e) (f) G is an orientation of G, without 3-dicycle χ(g) + α(s( G)) = V (G) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 7 / 14

8 Chromatic Gallai identities (a) (b) (c) (d) (e) (f) G is an orientation of G, without 3-dicycle χ(g) + α(s( G)) = V (G) = α(g) + χ(s( G)) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 7 / 14

9 The operator Φ Let β( ) be a graph parameter such that α(g) β(g) χ(g) ( G) Choose an orientation G of G without 3-dicycle and define Φ β (G) := V (G) β(s( G)) Corollary. α(g) Φ β (G) χ(g) Proof. V (G) χ(s( G)) V (G) β(s( G)) V (G) α(s( G)) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 8 / 14

10 Fractional clique-cover (co-chromatic) number χ f (G) = = max 1 x s.t. x(k) 1 ( clique K) x v 0 ( v V ) min s.t. 1 y K v y K = 1 ( v) y K 0 ( clique K) Lemma. y optimal for G x feasible for S( G) x uv := simplicial stars S K uv y K Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC 11 9 / 14

11 Lovász Theta function ϑ(g) = max v x v 2 s.t. x o 2 = 1 xo x v = x v 2 ( v) xu x v = 0 ( uv E) = min y o 2 s.t. y v 2 = 1 ( v) yo y v = 1 ( v) yu y v = 0 ( uv / E) Lemma. x R d m+1 optimal for S( G) y R nd n+1 feasible for G y o v := x o y o v if u = v x uv and y v u := x uv if uv E( G) 0 otherwise u:uv E( G) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC / 14

12 Improving Lovász s ϑ bound for coloring The sandwich theorem (Lovász 1979) α(g) ϑ(g) χ f (G) χ(g) ( G) Theorem (C. and Meurdesoif 2014) α(g) Φ χf (G) χ f (G) and ϑ(g) Φ ϑ (G) χ(g) ( G) V E min ρ := Φ ϑ ϑ ϑ mean ρ max ρ M % M % 27.3% 28.7% M % 27.6% 29.5% M % 27.8% 29.5% Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC / 14

13 Impact of orientation 1/4 1/4 1/3 1/6 1/6 1/3 1/6 (x2) 1/3 1/6 (max) (a) =2.5 (c) =3.75 (e) =3.5 (min) 2/5 1/5 1 1/ (x2) 1/5 3/5 (x2) (b) =2.2 =3 =3.5 (d) (f) wheel α χ f χ f χ orientation (a) Φ χ Φ χf Φ χ f Φ α orientation (b) Φ χ Φ χf Φ χ f = Φ α Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC / 14

14 Complexity Theorem. (Grötschel, Lovász and Schrijver 1981) ϑ(g) can be approximated in polynomial time (for any ε > 0) Theorem. (Gvozdenović and Laurent 2008) Computing χ f (G) is NP-hard In fact, [GL08] proved that, given any parameter β(g) such that β [χ f, χ], then computing β(g) is NP-hard. Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC / 14

15 Finding α with Φ χf (another proof for GL08) G k = G orient and add s 1,..., s k universal sources Feasible solution x for χ f (S( G k )) (max) x uv := { 1 α(g) if u = s i 0 otherwise Let β [χ f, χ]. Start with k := 1 and grow it until Φ β ( G k ) = k. Φ β ( G k ) = k α(g) = k Proof. k α(g) = α(g k ) Φ β ( G k ) Φ χf ( G V (G) k ) k + V (G) k α(g) Cornaz (with Jost and Meurdesoif) Coloring without symmetry SPOC / 14