Edge-coloring Multigraphs
Edge-coloring Multigraphs Daniel W. Cranston Virginia Commonwealth University [email protected] 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 4
3 2
3 2 4
4 3 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
1 2
4
3
4 2
3
3 3 4
2 3 1
2 4
4
4 3 1
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:
1
1 2
4
3
4 2
3
3 3 4
2
2 3
4 4
4 3 1
1
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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
4 3 1
1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) Let G be k-regular on 2t vertices.
(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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G 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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G b G by subdividing one edge. 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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G b G by subdividing one edge. G has kt + 1 edges, but each ⌃color ⌥class has size at most t. Thus, kt+1 0 (G b) = k + 1. 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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G b G by subdividing one edge. G has kt + 1 edges, but each ⌃color ⌥class has size at most t. Thus, kt+1 0 (G b) b is an overfull graph. = k + 1. G t
Easy Theorems for Simple Graphs
Easy Theorems for Simple Graphs I
K¨ onig: If G is bipartite, then
0 (G )
=
(G ).
Easy Theorems for Simple Graphs I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
0 (G )
=
(G ).
(G ) + 1.
Easy Theorems for Simple Graphs I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
(G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Proof of K¨ onig’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
0
(G ) =
Simple Graphs with Def: Let G
0
(G ) =
be subgraph induced by
-vertices.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
-vertices. .
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
-vertices. .
Simple Graphs with Def: Let G I I
If G Does
0
(G ) =
be subgraph induced by
has no cycles, then (G ) 2 imply
0 (G ) 0 (G )
-vertices.
=
=
. ?
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
I
Does
I
No. G could be overfull.
(G ) 2 imply
0 (G ) 0 (G )
-vertices.
=
=
. ?
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull?
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull? No.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
Hilton–Zhao Conjecture: If (G ) 2 and G 6= P ⇤ , then
-vertices.
0 (G )
. ?
if G is not overfull? No.
>
i↵ G is overfull.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull? No.
Hilton–Zhao Conjecture: If (G ) 2 and G 6= P ⇤ , then 0 (G ) > Cariolaro–Cariolaro: True for = 3.
i↵ G is overfull.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull? No.
Hilton–Zhao Conjecture: If (G ) 2 and G 6= P ⇤ , then 0 (G ) > Cariolaro–Cariolaro: True for = 3. C.–Rabern: True for = 4.
i↵ G is overfull.
Multigraphs Obs: Now
0 (G )
(G ) + 1 may not hold!
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
|E (H)| . b|V (H)|/2c
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
for every subgraph H,
0 (G )
dW(G )e.
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
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 Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
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 Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
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 p Always 0 (G ) max{ + 3 /2, dW(G )e}.
(G ) 23.
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:
I
Vizing:
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5
e}-free
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: multigraph
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G ) max{!(G ), 5
(G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
Ex 5:
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ) !(G ) + 1 for line graph of simple graph
Ex 5:
(G ) = 3k
(G ), 3}
1
(G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5
Ex 5:
(G ) = 3k
(G ), 3}
1, (G ) =
⌃ 5k ⌥ 2
(G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 = 5k+1 6 2
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 = 5k+1 6 2 =
⌃ 5k ⌥ 2
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 = 5k+1 6 2 =
⌃ 5k ⌥ 2
Kierstead Paths
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k of G u0 u1 .
(G ) + 1, and ' a k-edge-coloring
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i.
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
5
u3
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
5
u3
2 u4
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 Pf (using Key Lemma):
0 (G )
(G ) + 1.
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6
5
u3
2 u4
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 ) Pf (using Key Lemma): Induction on |E (G )|.
(G ) + 1.
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
2
5
u3
u4
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. 2
2
1
1
u0
u1
u2
u3
1
1 vk
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
2
5
u3
u4
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. 2
2
1
1
u0
u1
u2
u3
1
By Pigeonhole, two vertices miss the same color.
1 vk
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
2
5
u3
u4
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. 2
2
1
1
u0
u1
u2
u3
1
By Pigeonhole, two vertices miss the same color.
1 vk
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
1X
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
1X
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
1X
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
1X
I
Case 3: i < j
1 ↵ ui
Do ↵,
↵ ui+1
swap at ui+1 . Three places path could end.
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
1X
I
Case 3: i < j
1 ↵ ui
Do ↵,
↵ ui+1
swap at ui+1 . Three places path could end.
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
1X
I
Case 3: i < j
1
Do ↵,
↵
↵
↵
ui
ui+1
uj
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
1X
I
Case 3: i < j
1
Do ↵,
↵
↵
ui
ui+1
swap at ui+1 . Three places path could end.
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
1X
I
Case 3: i < j
1
ui
Do ↵,
↵
↵
ui+1
uj
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
1X
I
Case 3: i < j
1X
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
3, 4
1, 2
Tashkinov Trees
5 1 3, 4
1, 2
Tashkinov Trees
3 5 1 3, 4
1, 2
6
Tashkinov Trees
7
6
3 5 1 3, 4
1, 2
6
Tashkinov Trees 8 1
7
6
3 5 1 3, 4
1, 2
6
Tashkinov Trees 8 1
7
6
3 5 1 3, 4
1, 2
6
4
9
Tashkinov Trees 8 1
10
3
7
6
3 5 1 3, 4
1, 2
6
4
9
Tashkinov Trees 6
8
8
10
3
7
1
6
3 5 1 3, 4
1, 2
6
4
9
Summary Simple Graphs:
0 (G )
=
or
0 (G )
=
+1
Summary Simple Graphs: I
To get
0
=
0 (G )
=
or
0 (G )
=
+1
must avoid overfull subgraphs
Summary Simple Graphs: I I
0
0 (G )
=
or
0 (G )
=
+1
To get = must avoid overfull subgraphs Often this is enough
Summary Simple Graphs: I I
0
0 (G )
=
or
0 (G )
=
+1
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
0
+1
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I
4 Color Theorem: 3-regular planar
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7.
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6.
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
Summary Simple Graphs: I I
0 (G )
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now I
=
0 (G )
can be much bigger than
Goldberg–Seymour: If 0 (G ) > most overfull subgraph;
+ 1, then
0
determined by
Summary Simple Graphs: I I
0 (G )
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now I
=
0 (G )
can be much bigger than
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
+ 1, then 0 determined by 23 and asymptotically
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
+ 1, then 0 determined by 23 and asymptotically
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Tools: I
Kempe swaps
+ 1, then 0 determined by 23 and asymptotically
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Tools: I
Kempe swaps, Kierstead paths
+ 1, then 0 determined by 23 and asymptotically
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
+ 1, then 0 determined by 23 and asymptotically
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Tools: I
Kempe swaps, Kierstead paths, Tashkinov trees
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 4
3 2
3 2 4
4 3 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
1 2
4
3
4 2
3
3 3 4
2 3 1
2 4
4
4 3 1
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:
1
1 2
4
3
4 2
3
3 3 4
2
2 3
4 4
4 3 1
1
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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
4 3 1
1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) Let G be k-regular on 2t vertices.
(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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G 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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G b G by subdividing one edge. 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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G b G by subdividing one edge. G has kt + 1 edges, but each ⌃color ⌥class has size at most t. Thus, kt+1 0 (G b) = k + 1. 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:
1
1 2
4
3
4 2
3
3 3 4
2 3
2 4
4
1
4 3 1
Equivalent to coloring vertices of line graph L(G ) of G . Ex 2: Simple graphs with 0 (G ) (G ) + 1 b from Let G be k-regular on 2t vertices. Form G b G by subdividing one edge. G has kt + 1 edges, but each ⌃color ⌥class has size at most t. Thus, kt+1 0 (G b) b is an overfull graph. = k + 1. G t
Easy Theorems for Simple Graphs
Easy Theorems for Simple Graphs I
K¨ onig: If G is bipartite, then
0 (G )
=
(G ).
Easy Theorems for Simple Graphs I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
0 (G )
=
(G ).
(G ) + 1.
Easy Theorems for Simple Graphs I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
(G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
Proof of K¨ onig’s Theorem:
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Easy Theorems for Simple Graphs I I I I
K¨ onig: If G is bipartite, then Vizing: Always
(G )
0 (G )
Holyer: NP-hard to decide if Erd˝ os–Wilson: Almost always
0 (G )
0 (G )
=
(G ).
(G ) + 1. =
0 (G )
=
(G ). (G ).
Proof of K¨ onig’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
0
(G ) =
Simple Graphs with Def: Let G
0
(G ) =
be subgraph induced by
-vertices.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
-vertices. .
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
-vertices. .
Simple Graphs with Def: Let G I I
If G Does
0
(G ) =
be subgraph induced by
has no cycles, then (G ) 2 imply
0 (G ) 0 (G )
-vertices.
=
=
. ?
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
I
Does
I
No. G could be overfull.
(G ) 2 imply
0 (G ) 0 (G )
-vertices.
=
=
. ?
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull?
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull? No.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
Hilton–Zhao Conjecture: If (G ) 2 and G 6= P ⇤ , then
-vertices.
0 (G )
. ?
if G is not overfull? No.
>
i↵ G is overfull.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull? No.
Hilton–Zhao Conjecture: If (G ) 2 and G 6= P ⇤ , then 0 (G ) > Cariolaro–Cariolaro: True for = 3.
i↵ G is overfull.
Simple Graphs with Def: Let G I
If G
0
(G ) =
be subgraph induced by
has no cycles, then
0 (G )
=
(G ) 2 imply
0 (G )
=
(G ) 2 imply
0 (G )
=
I
Does
I
No. G could be overfull.
I
Does
-vertices. . ?
if G is not overfull? No.
Hilton–Zhao Conjecture: If (G ) 2 and G 6= P ⇤ , then 0 (G ) > Cariolaro–Cariolaro: True for = 3. C.–Rabern: True for = 4.
i↵ G is overfull.
Multigraphs Obs: Now
0 (G )
(G ) + 1 may not hold!
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
|E (H)| . b|V (H)|/2c
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
for every subgraph H,
0 (G )
dW(G )e.
Multigraphs Obs: Now Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
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 Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
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 Ex 4:
0 (G )
(G ) + 1 may not hold!
Let W(G ) = max
H ✓G |H| 3
Since
0 (G )
0 (H)
|E (H)| . b|V (H)|/2c
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 p Always 0 (G ) max{ + 3 /2, dW(G )e}.
(G ) 23.
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:
I
Vizing:
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5
e}-free
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: multigraph
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G ) max{!(G ), 5
(G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
Ex 5:
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ) !(G ) + 1 for line graph of simple graph
Ex 5:
(G ) = 3k
(G ), 3}
1
(G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5
Ex 5:
(G ) = 3k
(G ), 3}
1, (G ) =
⌃ 5k ⌥ 2
(G )+8 } 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 6
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 = 5k+1 6 2
e}-free
for line graph of a
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 = 5k+1 6 2 =
⌃ 5k ⌥ 2
Strengthening Brooks’ Theorem for Line Graphs I
Brooks:
I
Vizing:
I
Kierstead:
I
C.–Rabern: (G ) max{!(G ), 5 multigraph; this is best possible
(G ) max{!(G ),
(G ), 3}
(G ) !(G ) + 1 for line graph of simple graph (G ) !(G ) + 1 for {K1,3 , K5 (G )+8 } 6
e}-free
for line graph of a
Ex 5:
(G ) = 3k
1, (G ) =
⌃ 5k ⌥ 2
,
5(3k 1)+8 = 5k+1 6 2 =
⌃ 5k ⌥ 2
Kierstead Paths
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k of G u0 u1 .
(G ) + 1, and ' a k-edge-coloring
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i.
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
5
u3
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
5
u3
2 u4
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 Pf (using Key Lemma):
0 (G )
(G ) + 1.
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6
5
u3
2 u4
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 ) Pf (using Key Lemma): Induction on |E (G )|.
(G ) + 1.
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5 u2
3
6 u3
5
2 u4
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 , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
2
5
u3
u4
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. 2
2
1
1
u0
u1
u2
u3
1
1 vk
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
2
5
u3
u4
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. 2
2
1
1
u0
u1
u2
u3
1
By Pigeonhole, two vertices miss the same color.
1 vk
Kierstead Paths Def: Fix G , u0 u1 2 E (G ), k (G ) + 1, and ' a k-edge-coloring of G u0 u1 . A Kierstead Path is a path u0 , u1 , . . . , u` where for each i, '(ui ui 1 ) is missing at uj for some j < i. 1,2
3,4
u0
u1
2
5
3
u2
6
2
5
u3
u4
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. 2
2
1
1
u0
u1
u2
u3
1
By Pigeonhole, two vertices miss the same color.
1 vk
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
1X
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
1X
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
1X
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
1X
I
Case 3: i < j
1 ↵ ui
Do ↵,
↵ ui+1
swap at ui+1 . Three places path could end.
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
1X
I
Case 3: i < j
1 ↵ ui
Do ↵,
↵ ui+1
swap at ui+1 . Three places path could end.
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
1X
I
Case 3: i < j
1
Do ↵,
↵
↵
↵
ui
ui+1
uj
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
1X
I
Case 3: i < j
1
Do ↵,
↵
↵
ui
ui+1
swap at ui+1 . Three places path could end.
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
1X
I
Case 3: i < j
1
ui
Do ↵,
↵
↵
ui+1
uj
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
1X
I
Case 3: i < j
1X
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
3, 4
1, 2
Tashkinov Trees
5 1 3, 4
1, 2
Tashkinov Trees
3 5 1 3, 4
1, 2
6
Tashkinov Trees
7
6
3 5 1 3, 4
1, 2
6
Tashkinov Trees 8 1
7
6
3 5 1 3, 4
1, 2
6
Tashkinov Trees 8 1
7
6
3 5 1 3, 4
1, 2
6
4
9
Tashkinov Trees 8 1
10
3
7
6
3 5 1 3, 4
1, 2
6
4
9
Tashkinov Trees 6
8
8
10
3
7
1
6
3 5 1 3, 4
1, 2
6
4
9
Summary Simple Graphs:
0 (G )
=
or
0 (G )
=
+1
Summary Simple Graphs: I
To get
0
=
0 (G )
=
or
0 (G )
=
+1
must avoid overfull subgraphs
Summary Simple Graphs: I I
0
0 (G )
=
or
0 (G )
=
+1
To get = must avoid overfull subgraphs Often this is enough
Summary Simple Graphs: I I
0
0 (G )
=
or
0 (G )
=
+1
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
0
+1
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I
4 Color Theorem: 3-regular planar
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7.
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6.
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
Summary Simple Graphs: I I
0 (G )
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now I
=
0 (G )
can be much bigger than
Goldberg–Seymour: If 0 (G ) > most overfull subgraph;
+ 1, then
0
determined by
Summary Simple Graphs: I I
0 (G )
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now I
=
0 (G )
can be much bigger than
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
+ 1, then 0 determined by 23 and asymptotically
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
+ 1, then 0 determined by 23 and asymptotically
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Tools: I
Kempe swaps
+ 1, then 0 determined by 23 and asymptotically
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Tools: I
Kempe swaps, Kierstead paths
+ 1, then 0 determined by 23 and asymptotically
Summary Simple Graphs: I I
0 (G )
=
or
0 (G )
=
+1
0
To get = must avoid overfull subgraphs Often this is enough; also watch out for Petersen I I I I
4 Color Theorem: 3-regular planar Tutte’s Edge-Coloring: 3-regular with no Petersen subdivision Vizing’s Planar Graph Conj: Planar with 7. Open for 6. Hilton–Zhao Conj: (G ) 2, proved for 4
Multigraphs: Now
0 (G )
can be much bigger than
I
Goldberg–Seymour: If 0 (G ) > most overfull subgraph; true for
+ 1, then 0 determined by 23 and asymptotically
I
For line graph of multigraph, (G ) max{!(G ), 56 (G ) + 43 }
Tools: I
Kempe swaps, Kierstead paths, Tashkinov trees
