Injective coloring of sparse graphs Daniel W. Cranston DIMACS, - PowerPoint PPT Presentation

Injective coloring of sparse graphs Daniel W. Cranston DIMACS, Rutgers and Bell Labs joint with Seog-Jin Kim and Gexin Yu dcransto@dimacs.rutgers.edu AMS Meeting, University of Illinois March 28, 2009

Definitions and Examples Def. injective coloring: vertex coloring such that if u and v have a common neighbor, then c ( u ) � = c ( v ).

Definitions and Examples Def. injective coloring: vertex coloring such that if u and v have a common neighbor, then c ( u ) � = c ( v ). injective chromatic number: χ i ( G ) is minimum k such that G has an injective coloring with k colors.

Definitions and Examples Def. injective coloring: vertex coloring such that if u and v have a common neighbor, then c ( u ) � = c ( v ). injective chromatic number: χ i ( G ) is minimum k such that G has an injective coloring with k colors.

Definitions and Examples Def. injective coloring: vertex coloring such that if u and v have a common neighbor, then c ( u ) � = c ( v ). injective chromatic number: χ i ( G ) is minimum k such that G has an injective coloring with k colors. Easy bounds: ∆ ≤ χ i ( G ) ≤ ∆ 2 − ∆ + 1

Definitions and Examples Def. injective coloring: vertex coloring such that if u and v have a common neighbor, then c ( u ) � = c ( v ). injective chromatic number: χ i ( G ) is minimum k such that G has an injective coloring with k colors. Easy bounds: ∆ ≤ χ i ( G ) ≤ ∆ 2 − ∆ + 1 (greedy)

Definitions and Examples Def. injective coloring: vertex coloring such that if u and v have a common neighbor, then c ( u ) � = c ( v ). injective chromatic number: χ i ( G ) is minimum k such that G has an injective coloring with k colors. Easy bounds: ∆ ≤ χ i ( G ) ≤ ∆ 2 − ∆ + 1 (greedy)

Definitions and Examples Def. injective coloring: vertex coloring such that if u and v have a common neighbor, then c ( u ) � = c ( v ). injective chromatic number: χ i ( G ) is minimum k such that G has an injective coloring with k colors. Easy bounds: ∆ ≤ χ i ( G ) ≤ ∆ 2 − ∆ + 1 (greedy)

Sparse graphs and Mad(G) When can we prove χ i ( G ) ≤ ∆ + c for c ∈ { 0 , 1 , 2 } ? Ques.

Sparse graphs and Mad(G) When can we prove χ i ( G ) ≤ ∆ + c for c ∈ { 0 , 1 , 2 } ? Ques. We will study sparse graphs.

Sparse graphs and Mad(G) When can we prove χ i ( G ) ≤ ∆ + c for c ∈ { 0 , 1 , 2 } ? Ques. We will study sparse graphs. 2 E ( H ) Def. maximum average degree of G Mad( G ) = max H ⊆ G V ( H )

Sparse graphs and Mad(G) When can we prove χ i ( G ) ≤ ∆ + c for c ∈ { 0 , 1 , 2 } ? Ques. We will study sparse graphs. 2 E ( H ) Def. maximum average degree of G Mad( G ) = max H ⊆ G V ( H ) Ex. planar graphs Mad( G ) < 6

Sparse graphs and Mad(G) When can we prove χ i ( G ) ≤ ∆ + c for c ∈ { 0 , 1 , 2 } ? Ques. We will study sparse graphs. 2 E ( H ) Def. maximum average degree of G Mad( G ) = max H ⊆ G V ( H ) Ex. planar graphs Mad( G ) < 6 forests Mad( G ) < 2

Sparse graphs and Mad(G) When can we prove χ i ( G ) ≤ ∆ + c for c ∈ { 0 , 1 , 2 } ? Ques. We will study sparse graphs. 2 E ( H ) Def. maximum average degree of G Mad( G ) = max H ⊆ G V ( H ) Ex. planar graphs Mad( G ) < 6 forests Mad( G ) < 2 2 g planar, girth ≥ g Mad( G ) < g − 2

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. 2 1 1 2

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v .

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v . 3-vertex not adjacent to 2-vertex: 3-vertex adjacent to 2-vertex: 2-vertex:

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v . 3-vertex not adjacent to 2-vertex: µ ∗ ( v ) ≥ 3 − 3( 1 13 ) = 36 13 3-vertex adjacent to 2-vertex: 2-vertex:

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v . 3-vertex not adjacent to 2-vertex: µ ∗ ( v ) ≥ 3 − 3( 1 13 ) = 36 13 3-vertex adjacent to 2-vertex: µ ∗ ( v ) = 3 − 3 13 = 36 13 2-vertex:

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v . 3-vertex not adjacent to 2-vertex: µ ∗ ( v ) ≥ 3 − 3( 1 13 ) = 36 13 3-vertex adjacent to 2-vertex: µ ∗ ( v ) = 3 − 3 13 = 36 13 2-vertex: 2 + 2( 3 13 ) + 4( 1 13 ) = 36 13

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v . 3-vertex not adjacent to 2-vertex: µ ∗ ( v ) ≥ 3 − 3( 1 13 ) = 36 13 3-vertex adjacent to 2-vertex: µ ∗ ( v ) = 3 − 3 13 = 36 13 2-vertex: 2 + 2( 3 13 ) + 4( 1 13 ) = 36 13 And the theorem is best possible.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v . 3-vertex not adjacent to 2-vertex: µ ∗ ( v ) ≥ 3 − 3( 1 13 ) = 36 13 3-vertex adjacent to 2-vertex: µ ∗ ( v ) = 3 − 3 13 = 36 13 2-vertex: 2 + 2( 3 13 ) + 4( 1 13 ) = 36 13 And the theorem is best possible.

∆ = 3 and Mad( G ) < 36 13 If ∆ = 3 and Mad( G ) < 36 Thm. 13 , then χ i ( G ) ≤ 5. Pf. Assume G is min counterexample. Forbidden configs. Discharging rules 3 R1) Each 3-vertex gives 13 to each 2-vertex at distance 1. 1 R2) Each 3-vertex gives 13 to each 2-vertex at distance 2. µ ( v ) = d ( v ) and we check that µ ∗ ( v ) ≥ 36 13 for all v . 3-vertex not adjacent to 2-vertex: µ ∗ ( v ) ≥ 3 − 3( 1 13 ) = 36 13 3-vertex adjacent to 2-vertex: µ ∗ ( v ) = 3 − 3 13 = 36 13 2-vertex: 2 + 2( 3 13 ) + 4( 1 13 ) = 36 13 And the theorem is best possible.

∆ = 3 and Mad( G ) < 5 2 If ∆ = 3 and Mad( G ) < 5 2 , then χ i ( G ) ≤ 4. Thm.

∆ = 3 and Mad( G ) < 5 2 If ∆ = 3 and Mad( G ) < 5 2 , then χ i ( G ) ≤ 4. Thm. Assume G is min counterexample. Forbidden configs. Pf.

∆ = 3 and Mad( G ) < 5 2 If ∆ = 3 and Mad( G ) < 5 2 , then χ i ( G ) ≤ 4. Thm. Assume G is min counterexample. Forbidden configs. Pf.