Tracking down graphs that revolt against upper bounds on the - - PowerPoint PPT Presentation

The Combinatorial Optimization Workshop 04.-08. January 2016 Aussois

Tracking down graphs that revolt against upper bounds

• n the maximum degree

Vera Weil

Chair of Management Science Group Schrader School of Business and Economics Department of Computer Science RWTH Aachen University University of Cologne

Introduction

G

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Introduction

G

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Introduction

G

maximum clique size: ω(G)

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Introduction

G

maximum clique size: ω(G)

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Introduction

G

maximum clique size: ω(G) degree: at least ω(G) − 1

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Introduction

G

maximum clique size: ω(G) degree: at least ω(G) − 1

⇒ ω(G) − 1 ≤ ∆(G)

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Introduction

G

maximum clique size: ω(G) degree: at least ω(G) − 1

⇒ ω(G) − 1 ≤ ∆(G) ω ≤ ∆ + 1

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙

• Vera Weil

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙

• ∆ + 1 ≤ ω

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙

• ∆ + 1 ≤ ω

⊙ ⊙

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙

• ∆ + 1 ≤ ω

⊙ ⊙

• ∆ = 2, ω = 2

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙

• ∆ + 1 ≤ ω

⊙ ⊙

• ∆ = 2, ω = 2

What about ∆ + 1 ≤ ω + 1 ?

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙

• ∆ + 1 ≤ ω

⊙ ⊙

• ∆ = 2, ω = 2

What about ∆ + 1 ≤ ω + 1 ?

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙

• ∆ + 1 ≤ ω

⊙ ⊙

• ∆ = 2, ω = 2

What about ∆ + 1 ≤ ω + 1 ? ∆ = 3, ω = 2

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Introduction

ω ≤ ∆ + 1 ⊙ ⊙ ∆ + 1 ≤ ω ⊙ ⊙

• ∆ = 2, ω = 2

What about ∆ + 1 ≤ ω + 1 ? ∆ = 3, ω = 2 . . .

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Introduction

Can we find a function f such that for all graphs G, ∆(G) ≤ f (ω(G)) holds?

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Introduction

Can we find a function f such that for all graphs G, ∆(G) ≤ f (ω(G)) holds?

No!

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Introduction

Can we find a function f such that for all graphs G, ∆(G) ≤ f (ω(G)) holds?

No!

For every function f with ∆(G) ≤ f (ω(G)) we find this opponent:

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Introduction

Can we find a function f such that for all graphs G, ∆(G) ≤ f (ω(G)) holds?

No!

For every function f with ∆(G) ≤ f (ω(G)) we find this opponent: K1,f (2)+1 . . . ∆(G) > f (ω(G)) f (2) + 1 f (2)

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Introduction

For all functions f we find an infinite number of rebels G that deny ∆(G) ≤ f (ω(G)).

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Introduction

For all functions f we find an infinite number of rebels G that deny ∆(G) ≤ f (ω(G)). For example: K1,f (2)+1 K1,f (2)+2 K1,f (2)+3 . . .

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Introduction

For all functions f we find an infinite number of rebels G that deny ∆(G) ≤ f (ω(G)). For example: K1,f (2)+1 ← is an inclusionwise minimal rebel! K1,f (2)+2 K1,f (2)+3 . . .

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Introduction

For all functions f we find an infinite number of rebels G that deny ∆(G) ≤ f (ω(G)). For example: K1,f (2)+1 ← is an inclusionwise minimal rebel! K1,f (2)+2 K1,f (2)+3 . . . Is the number of inclusionwise minimal rebels finite?

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Is the number of inclusionwise minimal rebels finite?

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Is the number of inclusionwise minimal rebels finite?

Example:

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Is the number of inclusionwise minimal rebels finite?

Example: Answer is YES for f (ω) = ω + k − 1, for all k ∈ N0. (Schaudt & Weil 2015)

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Is the number of inclusionwise minimal rebels finite?

Example: Answer is YES for f (ω) = ω + k − 1, for all k ∈ N0. (Schaudt & Weil 2015) ⇓ Polynomial time algorithm to recognize graphs G where ∆(H) ≤ ω(H) + k − 1 for every H ⊆ G

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Is the number of inclusionwise minimal rebels finite?

Example: Answer is YES for f (ω) = ω + k − 1, for all k ∈ N0. (Schaudt & Weil 2015) ⇓ Polynomial time algorithm to recognize graphs G where ∆(H) ≤ ω(H) + k − 1 for every H ⊆ G ⇓ for some special multigraphs, the chromatic index can be computed in polynomial time

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G we find p ∈ N such that K1,p is a rebel of G.

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G we find p ∈ N such that K1,p is a rebel of G. But also the reverse is true: K1,p-rebel generates a rule for the whole graph class!

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G we find p ∈ N such that K1,p is a rebel of G. But also the reverse is true: K1,p-rebel generates a rule for the whole graph class! Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G.

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G.

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G.

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k}

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1)

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p G ∈ G with ω(G) ≤ k

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p G ∈ G with ω(G) ≤ k and ∆(G) ≥ R(k − 1, p − 1)

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p G ∈ G with ω(G) ≤ k and ∆(G) ≥ R(k − 1, p − 1) for some k ∈ N

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p G ∈ G with ω(G) ≤ k and ∆(G) ≥ R(k − 1, p − 1) for some k ∈ N ∆

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p G ∈ G with ω(G) ≤ k and ∆(G) ≥ R(k − 1, p − 1) for some k ∈ N ∆ ω ≤ k − 1

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p G ∈ G with ω(G) ≤ k and ∆(G) ≥ R(k − 1, p − 1) for some k ∈ N ∆ ω ≤ k − 1 α ≥ p

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Is the number of inclusionwise minimal rebels finite?

Let G be a hereditary graph class. For every p ∈ N such that K1,p is a rebel of G we find a monotone function f s.t. ∆(G) ≤ f (ω(G)) is a rule in G. f (k) = sup{∆(G) : G ∈ G, ω(G) ≤ k} < R(k − 1, p − 1) ω ≥ k or α ≥ p G ∈ G with ω(G) ≤ k and ∆(G) ≥ R(k − 1, p − 1) for some k ∈ N ∆ ω ≤ k − 1 α ≥ p

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: K1,f (2)+1

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: Rest of the pack: K1,f (2)+1 K1,f (2)+1-free

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: Rest of the pack: For every G in the pack: K1,f (2)+1 K1,f (2)+1-free ∆(G) = f (ω(G)) + 1

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: Rest of the pack: For every G in the pack: K1,f (2)+1 K1,f (2)+1-free ∆(G) = f (ω(G)) + 1 ∆(G) ≥ R(ω(G) − 1, f (2))

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: Rest of the pack: For every G in the pack: K1,f (2)+1 K1,f (2)+1-free ∆(G) = f (ω(G)) + 1 ∆(G) ≥ R(ω(G) − 1, f (2)) ∆

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: Rest of the pack: For every G in the pack: K1,f (2)+1 K1,f (2)+1-free ∆(G) = f (ω(G)) + 1 ∆(G) ≥ R(ω(G) − 1, f (2)) ∆ ω ≤ ω(G) − 1

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: Rest of the pack: For every G in the pack: K1,f (2)+1 K1,f (2)+1-free ∆(G) = f (ω(G)) + 1 ∆(G) ≥ R(ω(G) − 1, f (2)) ∆ ω ≤ ω(G) − 1 α ≥ f (2) + 1

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Is the number of inclusionwise minimal rebels finite?

Leader of the pack: Rest of the pack: For every G in the pack: K1,f (2)+1 K1,f (2)+1-free ∆(G) = f (ω(G)) + 1 ∆(G) ≥ R(ω(G) − 1, f (2)) ∆ ω ≤ ω(G) − 1 α ≥ f (2) + 1 If there exists ω0 ∈ N such that for all ω ≥ ω0 f (ω) > R(ω − 1, f (2)) − 1 then the number of minimal rebels is finite!

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Take aways

Is the number of inclusionwise minimal rebels against ∆ ≤ f (ω) finite?

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Take aways

Is the number of inclusionwise minimal rebels against ∆ ≤ f (ω) finite? We know: If f grows quickly or does not grow at all: Answer is YES.

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Take aways

Is the number of inclusionwise minimal rebels against ∆ ≤ f (ω) finite? We know: If f grows quickly or does not grow at all: Answer is YES. We think: If f is linear: Answer is YES.

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Take aways

Is the number of inclusionwise minimal rebels against ∆ ≤ f (ω) finite? We know: If f grows quickly or does not grow at all: Answer is YES. We think: If f is linear: Answer is YES. We ask: Can K1,p help us to track down the rest? Can we close the gap?

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Take aways

Is the number of inclusionwise minimal rebels against ∆ ≤ f (ω) finite? We know: If f grows quickly or does not grow at all: Answer is YES. We think: If f is linear: Answer is YES. We ask: Can K1,p help us to track down the rest? Can we close the gap? Thank you!

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