9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle 3. can’t have “too many 6 − -vertices near each other”
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle 3. can’t have “too many 6 − -vertices near each other” if so, then contract some non-adjacent pairs of nbrs; color smaller graph by induction, then extend to G
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle 3. can’t have “too many 6 − -vertices near each other” if so, then contract some non-adjacent pairs of nbrs; color smaller graph by induction, then extend to G Use discharging method to contradict (1), (2), or (3).
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle 3. can’t have “too many 6 − -vertices near each other” if so, then contract some non-adjacent pairs of nbrs; color smaller graph by induction, then extend to G Use discharging method to contradict (1), (2), or (3). ◮ each v gets ch ( v ) = d ( v ) − 6, so � v ∈ V ch ( v ) = − 12
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle 3. can’t have “too many 6 − -vertices near each other” if so, then contract some non-adjacent pairs of nbrs; color smaller graph by induction, then extend to G Use discharging method to contradict (1), (2), or (3). ◮ each v gets ch ( v ) = d ( v ) − 6, so � v ∈ V ch ( v ) = − 12 ◮ redistribute charge, so every vertex finishes nonnegative
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle 3. can’t have “too many 6 − -vertices near each other” if so, then contract some non-adjacent pairs of nbrs; color smaller graph by induction, then extend to G Use discharging method to contradict (1), (2), or (3). ◮ each v gets ch ( v ) = d ( v ) − 6, so � v ∈ V ch ( v ) = − 12 ◮ redistribute charge, so every vertex finishes nonnegative ◮ Now − 12 = � v ∈ V ch ∗ ( v ) ≥ 0, v ∈ V ch ( v ) = �
9 2 -Coloring Planar Graphs Thm: Every planar graph has a homomorphism to K 9:2 . Pf: Assume not. A minimal counterexample G : 1. has minimum degree 5 2. has no separating triangle 3. can’t have “too many 6 − -vertices near each other” if so, then contract some non-adjacent pairs of nbrs; color smaller graph by induction, then extend to G Use discharging method to contradict (1), (2), or (3). ◮ each v gets ch ( v ) = d ( v ) − 6, so � v ∈ V ch ( v ) = − 12 ◮ redistribute charge, so every vertex finishes nonnegative ◮ Now − 12 = � v ∈ V ch ∗ ( v ) ≥ 0, Contradiction! v ∈ V ch ( v ) = �
Too many 6 − -vertices near each other
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 .
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 . If v has 5 allowable colors and each u i has 3 allowable colors, then we can color each vertex with 2 colors, such that no color appears on both ends of an edge.
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 . If v has 5 allowable colors and each u i has 3 allowable colors, then we can color each vertex with 2 colors, such that no color appears on both ends of an edge. Pf: Give v a color available for at most one u i , say u 1 .
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 . If v has 5 allowable colors and each u i has 3 allowable colors, then we can color each vertex with 2 colors, such that no color appears on both ends of an edge. Pf: Give v a color available for at most one u i , say u 1 . 2(5) > 3(3)
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 . If v has 5 allowable colors and each u i has 3 allowable colors, then we can color each vertex with 2 colors, such that no color appears on both ends of an edge. Pf: Give v a color available for at most one u i , say u 1 . 2(5) > 3(3) Now give v another color not available for u 1 .
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 . If v has 5 allowable colors and each u i has 3 allowable colors, then we can color each vertex with 2 colors, such that no color appears on both ends of an edge. Pf: Give v a color available for at most one u i , say u 1 . 2(5) > 3(3) Now give v another color not available for u 1 . Now color each u i .
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 . If v has 5 allowable colors and each u i has 3 allowable colors, then we can color each vertex with 2 colors, such that no color appears on both ends of an edge. Pf: Give v a color available for at most one u i , say u 1 . 2(5) > 3(3) Now give v another color not available for u 1 . Now color each u i . B A u 1 u 2 v A B
Too many 6 − -vertices near each other Key Fact: Denote the center vertex of by v and the other vertices by u 1 , u 2 , u 3 . If v has 5 allowable colors and each u i has 3 allowable colors, then we can color each vertex with 2 colors, such that no color appears on both ends of an edge. Pf: Give v a color available for at most one u i , say u 1 . 2(5) > 3(3) Now give v another color not available for u 1 . Now color each u i . B u 2 B B C A u 1 u 2 A D v v u 1 u 3 A B C D A B
Coloring the Plane
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors.
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph.
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph. Min number of colors needed is χ ( R 2 ). [Nelson ’50]
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph. Min number of colors needed is χ ( R 2 ). [Nelson ’50] What’s known?
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph. Min number of colors needed is χ ( R 2 ). [Nelson ’50] What’s known? ? 1 3 2 3 2 1 (a) The Moser spindle
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph. Min number of colors needed is χ ( R 2 ). [Nelson ’50] What’s known? ? 1 3 2 3 2 1 (a) The Moser spindle
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph. Min number of colors needed is χ ( R 2 ). [Nelson ’50] What’s known? ? 2 3 1 ? 3 ? 2 3 3 1 2 2 ? 1 2 3 (a) The Moser spindle (b) The Golomb graph
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph. Min number of colors needed is χ ( R 2 ). [Nelson ’50] What’s known? ? 2 3 1 ? 3 ? 2 3 3 1 2 2 ? 1 2 3 (a) The Moser spindle (b) The Golomb graph
Coloring the Plane Goal: Color the plane so points at distance 1 get distinct colors. ◮ vertices are points of R 2 ◮ two vertices adjacent if points are at distance 1 Unit distance graph is any subgraph of this graph. Min number of colors needed is χ ( R 2 ). [Nelson ’50] What’s known? ? 2 3 1 ? 3 ? 2 3 3 1 2 2 ? 1 2 3 (a) The Moser spindle (b) The Golomb graph So χ ( R 2 ) ≥ 4
Coloring the Plane: an Upper Bound
Coloring the Plane: an Upper Bound Also, χ ( R 2 ) ≤ 7 [Isbell early ’50s] 1 6 4 2 7 5 4 2 7 5 3 2 7 5 3 1 6 5 3 1 6 4 3 1 6 4 2 7 6 4 2 7 5 4 2 7 5 3 1 7 5 3 1 6 5 3 1 6 4 2 1 6 4 2 7 6 4 2 7 5 3 2 7 5 3 1 7 5 3 1 6 4 3 1 6 4 2 1 6 4 2 7 5 4 2 7 5 3
Fractional Coloring, Revisited
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ).
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). | V ( G ) |
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � | V ( G ) | = w I v ∈ V I ∋ v
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � | V ( G ) | = w I = w I | I | v ∈ V I ∋ v I ∈I
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � � | V ( G ) | = w I = w I | I | ≤ α ( G ) w I v ∈ V I ∋ v I ∈I I ∈I
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � � | V ( G ) | = w I = w I | I | ≤ α ( G ) w I = α ( G ) χ f ( G ) . v ∈ V I ∋ v I ∈I I ∈I
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � � | V ( G ) | = w I = w I | I | ≤ α ( G ) w I = α ( G ) χ f ( G ) . v ∈ V I ∋ v I ∈I I ∈I 1,4 3,2 5,7 5,6 3,7 4,6 1,2
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � � | V ( G ) | = w I = w I | I | ≤ α ( G ) w I = α ( G ) χ f ( G ) . v ∈ V I ∋ v I ∈I I ∈I 1,4 3,2 5,7 5,6 3,7 4,6 1,2
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � � | V ( G ) | = w I = w I | I | ≤ α ( G ) w I = α ( G ) χ f ( G ) . v ∈ V I ∋ v I ∈I I ∈I 1,4 3,5 3,2 1,2 5,7 1,4 2,4 2,3 1,5 5,6 3,7 4,6 3,4 4,5 1,2 2,5 1,3
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � � | V ( G ) | = w I = w I | I | ≤ α ( G ) w I = α ( G ) χ f ( G ) . v ∈ V I ∋ v I ∈I I ∈I 1,4 3,5 3,2 1,2 5,7 1,4 2,4 2,3 1,5 5,6 3,7 4,6 3,4 4,5 1,2 2,5 1,3
Fractional Coloring, Revisited Prop. χ f ( G ) ≥ | V ( G ) | /α ( G ). � � � � | V ( G ) | = w I = w I | I | ≤ α ( G ) w I = α ( G ) χ f ( G ) . v ∈ V I ∋ v I ∈I I ∈I 1,4 3,5 3,2 1,2 5,7 1,4 2,4 2,3 1,5 5,6 3,7 4,6 3,4 4,5 1,2 2,5 1,3 More generally, for every weight function µ , χ f ( G ) ≥ | V µ ( G ) | /α µ ( G ) .
A Computational Approach
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5.
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5.
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact;
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally.
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; 3 3 4 7 4 3 7 7 3 3 4 3
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; 3 3 4 7 4 3 7 7 3 3 4 3
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; 3 3 4 7 4 3 7 7 3 3 4 3
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; 3 3 4 7 4 3 7 7 3 3 4 3
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; 3 3 4 7 4 3 7 7 3 3 4 3
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; solve for best weights. 3 3 4 7 4 3 7 7 3 3 4 3
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; solve for best weights. 3 3 4 7 4 3 7 7 3 3 4 3 Core weights above, spindle weights 1, total weight: 51 + 45 = 96.
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; solve for best weights. 3 3 4 7 4 3 7 7 3 3 4 3 Core weights above, spindle weights 1, total weight: 51 + 45 = 96. Max independent set weight: 27.
A Computational Approach Goal: Find unit distance H with χ f ( H ) > 3 . 5. Idea: Recall χ f (spindle) = 3 . 5. Find graph with many spindles that interact; at least one colored suboptimally. Core vertices from triangular lattice; attach many spindles; solve for best weights. 3 3 4 7 4 3 7 7 3 3 4 3 Core weights above, spindle weights 1, total weight: 51 + 45 = 96. Max independent set weight: 27. So [Fisher–Ullman ’92] χ f ( H ) ≥ 96 / 27 = 32 / 9 = 3 . 5555 . . .
Bigger Cores
Bigger Cores 3 3 4 7 4 4 8 8 4 3 7 8 7 3 3 4 4 3 Spindle weight 1 gives χ f ≥ 168 47 ≈ 3 . 5744
Bigger Cores 5 5 3 3 6 12 6 4 7 4 7 16 16 7 4 8 8 4 6 16 20 16 6 3 7 8 7 3 5 12 16 16 12 5 3 4 4 3 5 6 7 6 5 Spindle weight 1 gives Spindle weight 2 gives χ f ≥ 168 χ f ≥ 491 47 ≈ 3 . 5744 137 ≈ 3 . 5839
Our Biggest Core
Our Biggest Core 6 6 11 21 11 9 26 26 9 9 19 21 19 9 9 18 18 18 18 9 9 19 18 19 18 19 9 11 26 21 18 18 21 26 11 6 21 26 19 18 19 26 21 6 6 11 9 9 9 9 11 6 Spindle weight 3 gives χ f ≥ 1732 481 ≈ 3 . 6008
A “By Hand” Approach
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