Edge-coloring Multigraphs Daniel W. Cranston Virginia Commonwealth - - PowerPoint PPT Presentation

Edge-coloring Multigraphs

Daniel W. Cranston

Virginia Commonwealth University dcranston@vcu.edu

Cumberland Conference 20 May 2017

Edge-coloring Examples

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors;

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible.

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G).

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

1 3 4 2 1 3 4 3 4 2

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

4 3 3 2 1 4 3 1 2 4 1 3 4 2 1 3 4 3 4 2

Equivalent to coloring vertices of line graph L(G) of G.

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

4 3 3 2 1 4 3 1 2 4 1 3 4 2 1 3 4 3 4 2

Equivalent to coloring vertices of line graph L(G) of G. Ex 2: Simple graphs with 0(G) ∆(G) + 1

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

4 3 3 2 1 4 3 1 2 4 1 3 4 2 1 3 4 3 4 2

Equivalent to coloring vertices of line graph L(G) of G. Ex 2: Simple graphs with 0(G) ∆(G) + 1 Let G be k-regular on 2t vertices.

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

4 3 3 2 1 4 3 1 2 4 1 3 4 2 1 3 4 3 4 2

Equivalent to coloring vertices of line graph L(G) of G. Ex 2: Simple graphs with 0(G) ∆(G) + 1 Let G be k-regular on 2t vertices. Form b G from G by subdividing one edge.

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

4 3 3 2 1 4 3 1 2 4 1 3 4 2 1 3 4 3 4 2

Equivalent to coloring vertices of line graph L(G) of G. Ex 2: Simple graphs with 0(G) ∆(G) + 1 Let G be k-regular on 2t vertices. Form b G from G by subdividing one edge. b G has kt + 1 edges, but each color class has size at most t.

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

4 3 3 2 1 4 3 1 2 4 1 3 4 2 1 3 4 3 4 2

Equivalent to coloring vertices of line graph L(G) of G. Ex 2: Simple graphs with 0(G) ∆(G) + 1 Let G be k-regular on 2t vertices. Form b G from G by subdividing one edge. b G has kt + 1 edges, but each color class has size at most t. Thus, 0(b G) ⌃ kt+1

t

⌥ = k + 1.

Edge-coloring Examples

Goal: Assign colors to the edges of a graph so that edges with a common endpoint get distinct colors; use as few colors as possible. For a graph G, minimum number of colors is 0(G). Ex 1:

4 3 3 2 1 4 3 1 2 4 1 3 4 2 1 3 4 3 4 2

Equivalent to coloring vertices of line graph L(G) of G. Ex 2: Simple graphs with 0(G) ∆(G) + 1 Let G be k-regular on 2t vertices. Form b G from G by subdividing one edge. b G has kt + 1 edges, but each color class has size at most t. Thus, 0(b G) ⌃ kt+1

t

⌥ = k + 1. b G is an overfull graph.

Easy Theorems for Simple Graphs

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1.

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G).

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Easy Theorems for Simple Graphs

I K¨

• nig: If G is bipartite, then 0(G) = ∆(G).

I Vizing: Always ∆(G)  0(G)  ∆(G) + 1. I Holyer: NP-hard to decide if 0(G) = ∆(G). I Erd˝

• s–Wilson: Almost always 0(G) = ∆(G).

Proof of K¨

• nig’s Theorem:

Rem: Kempe swaps are fundamental tool for edge-coloring.

Harder Theorems for Simple Graphs

Vizing’s Planar Graph Conjecture: If G is planar and ∆(G) 6, then 0(G) = ∆(G).

Harder Theorems for Simple Graphs

Vizing’s Planar Graph Conjecture: If G is planar and ∆(G) 6, then 0(G) = ∆(G). True for ∆(G) 7 (Sanders–Zhao; Zhang).

Harder Theorems for Simple Graphs

Vizing’s Planar Graph Conjecture: If G is planar and ∆(G) 6, then 0(G) = ∆(G). True for ∆(G) 7 (Sanders–Zhao; Zhang). False for ∆(G)  5.

Harder Theorems for Simple Graphs

Vizing’s Planar Graph Conjecture: If G is planar and ∆(G) 6, then 0(G) = ∆(G). True for ∆(G) 7 (Sanders–Zhao; Zhang). False for ∆(G)  5. Ex 2, starting from 4-cycle, cube, octahedron, and icosahedron.

Harder Theorems for Simple Graphs

Vizing’s Planar Graph Conjecture: If G is planar and ∆(G) 6, then 0(G) = ∆(G). True for ∆(G) 7 (Sanders–Zhao; Zhang). False for ∆(G)  5. Ex 2, starting from 4-cycle, cube, octahedron, and icosahedron. 4 Color Theorem: If G is 3-regular, has no overfull subgraph, and is planar, then 0(G) = 3.

Harder Theorems for Simple Graphs

Vizing’s Planar Graph Conjecture: If G is planar and ∆(G) 6, then 0(G) = ∆(G). True for ∆(G) 7 (Sanders–Zhao; Zhang). False for ∆(G)  5. Ex 2, starting from 4-cycle, cube, octahedron, and icosahedron. 4 Color Theorem: If G is 3-regular, has no overfull subgraph, and is planar, then 0(G) = 3. Tutte’s Edge-coloring Conj (proved!): If G is 3-regular, has no overfull subgraph, and has no subdivision of the Petersen graph, then 0(G) = 3.

Simple Graphs with χ0(G) = ∆

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆.

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆.

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆. I Does ∆(G∆)  2 imply 0(G) = ∆?

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆. I Does ∆(G∆)  2 imply 0(G) = ∆? I No. G could be overfull.

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆. I Does ∆(G∆)  2 imply 0(G) = ∆? I No. G could be overfull. I Does ∆(G∆)  2 imply 0(G) = ∆ if G is not overfull?

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆. I Does ∆(G∆)  2 imply 0(G) = ∆? I No. G could be overfull. I Does ∆(G∆)  2 imply 0(G) = ∆ if G is not overfull? No.

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆. I Does ∆(G∆)  2 imply 0(G) = ∆? I No. G could be overfull. I Does ∆(G∆)  2 imply 0(G) = ∆ if G is not overfull? No.

Hilton–Zhao Conjecture: If ∆(G∆)  2 and G 6= P⇤, then 0(G) > ∆ iff G is overfull.

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆. I Does ∆(G∆)  2 imply 0(G) = ∆? I No. G could be overfull. I Does ∆(G∆)  2 imply 0(G) = ∆ if G is not overfull? No.

Hilton–Zhao Conjecture: If ∆(G∆)  2 and G 6= P⇤, then 0(G) > ∆ iff G is overfull. Cariolaro–Cariolaro: True for ∆ = 3.

Simple Graphs with χ0(G) = ∆

Def: Let G∆ be subgraph induced by ∆-vertices.

I If G∆ has no cycles, then 0(G) = ∆. I Does ∆(G∆)  2 imply 0(G) = ∆? I No. G could be overfull. I Does ∆(G∆)  2 imply 0(G) = ∆ if G is not overfull? No.

Hilton–Zhao Conjecture: If ∆(G∆)  2 and G 6= P⇤, then 0(G) > ∆ iff G is overfull. Cariolaro–Cariolaro: True for ∆ = 3. C.–Rabern: True for ∆ = 4.

Multigraphs

Obs: Now 0(G)  ∆(G) + 1 may not hold!

Multigraphs

Obs: Now 0(G)  ∆(G) + 1 may not hold! Ex 4:

Multigraphs

Obs: Now 0(G)  ∆(G) + 1 may not hold! Ex 4: Let W(G) = max

H ⊆ G |H| ≥ 3

|E(H)| b|V (H)|/2c.

Multigraphs

Obs: Now 0(G)  ∆(G) + 1 may not hold! Ex 4: Let W(G) = max

H ⊆ G |H| ≥ 3

|E(H)| b|V (H)|/2c. Since 0(G) 0(H) for every subgraph H, 0(G) dW(G)e.

Multigraphs

Obs: Now 0(G)  ∆(G) + 1 may not hold! Ex 4: Let W(G) = max

H ⊆ G |H| ≥ 3

|E(H)| b|V (H)|/2c. Since 0(G) 0(H) for every subgraph H, 0(G) dW(G)e. Goldberg–Seymour Conj: Every multigraph G satisfies 0(G)  max{∆(G) + 1, dW(G)e}.

Multigraphs

Obs: Now 0(G)  ∆(G) + 1 may not hold! Ex 4: Let W(G) = max

H ⊆ G |H| ≥ 3

|E(H)| b|V (H)|/2c. Since 0(G) 0(H) for every subgraph H, 0(G) dW(G)e. Goldberg–Seymour Conj: Every multigraph G satisfies 0(G)  max{∆(G) + 1, dW(G)e}. Thm: G–S Conj is true asymptotically, and for ∆(G)  23.

Multigraphs

Obs: Now 0(G)  ∆(G) + 1 may not hold! Ex 4: Let W(G) = max

H ⊆ G |H| ≥ 3

|E(H)| b|V (H)|/2c. Since 0(G) 0(H) for every subgraph H, 0(G) dW(G)e. Goldberg–Seymour Conj: Every multigraph G satisfies 0(G)  max{∆(G) + 1, dW(G)e}. Thm: G–S Conj is true asymptotically, and for ∆(G)  23. Always 0(G)  max{∆ +

3

p ∆/2, dW(G)e}.

Strengthening Brooks’ Theorem for Line Graphs

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3}

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph; this is best possible Ex 5:

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph; this is best possible Ex 5: ∆(G) = 3k 1

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph; this is best possible Ex 5: ∆(G) = 3k 1, (G) = ⌃ 5k

2

⌥

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph; this is best possible Ex 5: ∆(G) = 3k 1, (G) = ⌃ 5k

2

⌥ , 5(3k1)+8

6

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph; this is best possible Ex 5: ∆(G) = 3k 1, (G) = ⌃ 5k

2

⌥ , 5(3k1)+8

6

= 5k+1

2

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph; this is best possible Ex 5: ∆(G) = 3k 1, (G) = ⌃ 5k

2

⌥ , 5(3k1)+8

6

= 5k+1

2

= ⌃ 5k

2

⌥

Strengthening Brooks’ Theorem for Line Graphs

I Brooks: (G)  max{!(G), ∆(G), 3} I Vizing: (G)  !(G) + 1 for line graph of simple graph I Kierstead: (G)  !(G) + 1 for {K1,3, K5 e}-free I C.–Rabern: (G)  max{!(G), 5∆(G)+8 6

} for line graph of a multigraph; this is best possible Ex 5: ∆(G) = 3k 1, (G) = ⌃ 5k

2

⌥ , 5(3k1)+8

6

= 5k+1

2

= ⌃ 5k

2

⌥

Kierstead Paths

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma):

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma): Induction on |E(G)|.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma): Induction on |E(G)|. Let k = ∆(G) + 1.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma): Induction on |E(G)|. Let k = ∆(G) + 1. Base case: at most ∆(G) + 1 edges.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma): Induction on |E(G)|. Let k = ∆(G) + 1. Base case: at most ∆(G) + 1 edges. Induction: Given k-edge-coloring of G e, get long Kierstead path.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma): Induction on |E(G)|. Let k = ∆(G) + 1. Base case: at most ∆(G) + 1 edges. Induction: Given k-edge-coloring of G e, get long Kierstead path.

u0 u1 u2 u3 vk 2 2 1 1 1 1

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma): Induction on |E(G)|. Let k = ∆(G) + 1. Base case: at most ∆(G) + 1 edges. Induction: Given k-edge-coloring of G e, get long Kierstead path.

u0 u1 u2 u3 vk 2 2 1 1 1 1

By Pigeonhole, two vertices miss the same color.

Kierstead Paths

Def: Fix G, u0u1 2 E(G), k ∆(G) + 1, and ' a k-edge-coloring

• f G u0u1. A Kierstead Path is a path u0, u1, . . . , u` where for

each i, '(uiui1) is missing at uj for some j < i.

u0 u1 u2 u3 u4 1,2 3,4 5 6 2 2 3 5

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Vizing’s Theorem: If G is simple, then 0(G)  ∆(G) + 1. Pf (using Key Lemma): Induction on |E(G)|. Let k = ∆(G) + 1. Base case: at most ∆(G) + 1 edges. Induction: Given k-edge-coloring of G e, get long Kierstead path.

u0 u1 u2 u3 vk 2 2 1 1 1 1

By Pigeonhole, two vertices miss the same color.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 I Case 2: i = j 1 I Case 3: i < j 1

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 I Case 2: i = j 1 I Case 3: i < j 1

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 I Case 2: i = j 1 I Case 3: i < j 1

ui uj ↵ ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 I Case 3: i < j 1

ui uj ↵ ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 I Case 3: i < j 1

ui uj ↵ ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 I Case 3: i < j 1

ui uj ↵ ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 I Case 3: i < j 1

ui uj ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 I Case 3: i < j 1

ui uj ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui uj ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui ui+1 uj

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui ui+1 uj ↵

• ↵

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui ui+1 uj ↵

• ↵

Do ↵, swap at ui+1. Three places path could end.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui ui+1 uj ↵

• ↵

Do ↵, swap at ui+1. Three places path could end.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui ui+1 uj ↵ ↵ ↵

Do ↵, swap at ui+1. Three places path could end.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui ui+1 uj ↵ ↵

• Do ↵, swap at ui+1. Three places path could end.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1

ui ui+1 uj

• ↵

↵

Do ↵, swap at ui+1. Three places path could end.

Proof of Key Lemma

Key Lemma: If a Kierstead Path has distinct ui and uj with color ↵ missing at both, then G has a k-coloring. Pf: Double induction, first on path length `; next on distance between ui and uj. Assume i < j. Three cases:

I Case 1: i = 0, j = 1 X I Case 2: i = j 1 X I Case 3: i < j 1 X

ui ui+1 uj

• ↵

↵

Do ↵, swap at ui+1. Three places path could end. In each case, win by induction hypothesis.

Tashkinov Trees

Tashkinov Trees

1, 2 3, 4

Tashkinov Trees

1, 2 3, 4 1 5

Tashkinov Trees

1, 2 3, 4 1 5 3 6

Tashkinov Trees

1, 2 3, 4 1 5 3 6 6 7

Tashkinov Trees

1, 2 3, 4 1 5 3 6 6 7 1 8

Tashkinov Trees

1, 2 3, 4 1 5 3 6 6 7 1 8 4 9

Tashkinov Trees

1, 2 3, 4 1 5 3 6 6 7 1 8 4 9 3 10

Tashkinov Trees

1, 2 3, 4 1 5 3 6 6 7 1 8 4 9 3 10 8 6

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7.

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6.

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Multigraphs: Now 0(G) can be much bigger than ∆

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Multigraphs: Now 0(G) can be much bigger than ∆

I Goldberg–Seymour: If 0(G) > ∆ + 1, then 0 determined by

most overfull subgraph;

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Multigraphs: Now 0(G) can be much bigger than ∆

I Goldberg–Seymour: If 0(G) > ∆ + 1, then 0 determined by

most overfull subgraph; true for ∆  23 and asymptotically

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Multigraphs: Now 0(G) can be much bigger than ∆

I Goldberg–Seymour: If 0(G) > ∆ + 1, then 0 determined by

most overfull subgraph; true for ∆  23 and asymptotically

I For line graph of multigraph, (G)  max{!(G), 5 6∆(G) + 4 3}

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Multigraphs: Now 0(G) can be much bigger than ∆

I Goldberg–Seymour: If 0(G) > ∆ + 1, then 0 determined by

most overfull subgraph; true for ∆  23 and asymptotically

I For line graph of multigraph, (G)  max{!(G), 5 6∆(G) + 4 3}

Tools:

I Kempe swaps

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Multigraphs: Now 0(G) can be much bigger than ∆

I Goldberg–Seymour: If 0(G) > ∆ + 1, then 0 determined by

most overfull subgraph; true for ∆  23 and asymptotically

I For line graph of multigraph, (G)  max{!(G), 5 6∆(G) + 4 3}

Tools:

I Kempe swaps, Kierstead paths

Summary

Simple Graphs: 0(G) = ∆ or 0(G) = ∆ + 1

I To get 0 = ∆ must avoid overfull subgraphs I Often this is enough; also watch out for Petersen

I 4 Color Theorem: 3-regular planar I Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision I Vizing’s Planar Graph Conj: Planar with ∆ 7. Open for 6. I Hilton–Zhao Conj: ∆(G∆)  2, proved for ∆  4

Multigraphs: Now 0(G) can be much bigger than ∆

I Goldberg–Seymour: If 0(G) > ∆ + 1, then 0 determined by

most overfull subgraph; true for ∆  23 and asymptotically

I For line graph of multigraph, (G)  max{!(G), 5 6∆(G) + 4 3}

Tools:

I Kempe swaps, Kierstead paths, Tashkinov trees