Precoloring Extension Involving
Pairs of Vertices of Small Distance
Akira Saito
(Nihon University, Japan)
This is a joint work with Chihoko Ojima and Kazuki Sano
(Nihon University)
1. What is precoloring?
2. Background and Motivation
3. Pairs of Vertices of Distance Three
4. Pairs of Vertices of Distance Two
1. What is precoloring?
2. Background and Motivation
3. Pairs of Vertices of Distance Three
4. Pairs of Vertices of Distance Two
Job Scheduling
{ job1, job2 , job3, job4, job5, job6 }
•
Multi-processors
•
concurrent environment
•
It takes a unit time to finish each job.
1
6
2
5
3
4
Job Scheduling
{ job1, job2 , job3, job4, job5, job6 }
•
multi processors
•
concurrent environment
•
It takes a unit time to finish each job.
•
If there is no constraint, we can finish
all the jobs in 1 unit time (with 6 processors).
1
6
2
5
3
4
Job Scheduling
{ job1, job2 , job3, job4, job5, job6 }
•
But some jobs may not be processed
at the same time
1
6
• They use the same resources
•
restrictions of the processors
2
5
We join pairs of jobs which cannot be
processed concurrently
We obtain a graph.
3
4
Job Scheduling
{ job1, job2 , job3, job4, job5, job6 }
We can color this graph in 3 colors.
1
All the jobs can be
finished in 3 unit times.
6
2
5
3
4
Precoloring
We can color this graph in 3 colors.
1
All the jobs can be
finished in 3 unit times.
In some applications, the schedule of several
jobs may already be fixed.
6
2
5
3
4
Precoloring
We can color this graph in 3 colors.
1
All the jobs can be
finished in 3 unit times.
In some applications, the schedule of several
jobs may already be fixed.
6
2
5
Some vertices are colored (precolored).
We must find a coloring which extends the
given precoloring.
We may need more colors than the
chromatic number.
3
4
Precoloring
We can color this graph in 3 colors.
1
All the jobs can be
finished in 3 unit times.
In some applications, the schedule of several
jobs may already be fixed.
6
2
5
Some vertices are colored (precolored).
We must find a coloring which extends the
given precoloring.
We may need more colors than the
chromatic number.
3
4
Formulation
Given :
The subgraph of 𝐺 induced by 𝑃
• 𝐺 : 𝑟-colorable graph
• 𝑃 ⊂ 𝑉(𝐺)
• 𝑓0 ∶ 𝑃 → {1, 2, … , 𝑡0} : a coloring of 𝐺 𝑃
(We call 𝑓0 a 𝑡0-coloring of 𝑃.)
Want :
𝑓 ∶ 𝑉(𝐺) → {1, 2, … , 𝑡} : a coloring of 𝐺 with 𝑓|𝑃 = 𝑓0 and 𝑡 small
The coloring 𝑓 is called an extension of 𝑓0.
1. What is precoloring?
2. Background and Motivation
3. Pairs of Vertices of Distance Three
4. Pairs of Vertices of Distance Two
Extension in 𝜒(𝐺) colors – No hope
𝐺:
a path of even order
χ 𝐺 = 2 (𝐺 is a bipartite graph.)
Extension in 𝜒(𝐺) colors – No hope
precolor
𝐺:
a path of even order
Extension in 𝜒(𝐺) colors – No hope
precolor
𝐺:
a path of even order
We need the third color.
• We need three colors to extend the given precoloring.
• Only two vertices are precolored in only one color.
• The distance between the precolored vertices can be very large.
Extension in 𝜒 𝐺 + 1 colors
Extension in 𝜒(𝐺) colors : No hope
Extension in 𝜒(𝐺) + 1 colors?
If the vertices in 𝑃 are “sparsely” distributed, we have a better chance.
How do we define?
Minimum Distance
𝑃 ⊂ 𝑉(𝐺), |𝑃| ≥ 2
𝑑 𝑃 = min{ 𝑑𝐺 𝑥, 𝑦 ∶ 𝑥, 𝑦 ⊂ 𝑃, 𝑥 ≠ 𝑦}
minimum distance
Problem (Thomassen 1997)
Let 𝐺 be a planar graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) is sufficiently large
(say 𝑑(𝑃) ≥ 100), then a 5-coloring of 𝑃 extends to a 5-coloring of 𝐺.
Answer by Albertson (1998)
Theorem A (Albertson, 1998)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺) with |𝑃| ≥ 2. If 𝑑(𝑃) ≥ 4,
then every (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 1)-coloring of 𝐺.
He does not assume planarity of 𝐺.
Albertson’s Idea
Theorem A (Albertson, 1998)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺) with |𝑃| ≥ 2. If 𝑑(𝑃) ≥ 4,
then every (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 1)-coloring of 𝐺.
Color 𝐺 in 𝑟 colors, ignoring the given (𝑟 + 1)-coloring of 𝑃 (say 𝑓0).
A vertex 𝑣 in 𝑃 may receive a color which is different from the one
specified by 𝑓0.
Adjust the coloring of G so that 𝑣 receives 𝑓0(𝑣), using the extra color 𝑟 + 1.
Albertson’s Idea
𝑃
𝑃
3
3
3
2
𝐺
2
𝐺
Color G, ignoring
the precolor.
1
2
1
1
𝑃 is precolored.
1
1
We want to re-color
this vertex in color 3.
Albertson’s Idea
3
3
3
3
𝑟+1
We want to re-color
this vertex in color 3.
Assign the color 𝑟 + 1.
𝑟+1
𝑟+1
3
Swap colors.
Albertson’s Idea
𝑃
𝑃
3
3
3
2
𝐺
2
𝐺
Color G, ignoring
the precolor.
1
2
1
1
𝑃 is precolored.
1
1
We want to re-color
this vertex in color 3.
Albertson’s Idea
We repeat this process, spreading the new
color 𝑟 + 1 around the vertices in 𝑃.
𝑃
𝑃
3
3
3
2
𝐺
2
𝐺
Color G, ignoring
the precolor.
3
2
1
1
𝑃 is precolored.
1
1
Albertson’s Idea
We repeat this process, spreading the new
color 𝑟 + 1 around the vertices in 𝑃.
𝑃
𝑃
3
3
3
2
𝐺
2
𝐺
Color G, ignoring
the precolor.
3
3
1
1
𝑃 is precolored.
1
1
Where does 𝑑(𝑃) ≥ 4 come in?
If the vertices in P are too close,…
1
2
1
3
2
3
1
𝑃
It should be 3.
3
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If the vertices in P are too close,…
𝑟+1
2
1
3
2
3
1
𝑃
It should be 3.
3
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If the vertices in P are too close,…
3
2
1
𝑟+1
2
𝑟+1
1
𝑃
It should be 3.
3
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If the vertices in P are too close,…
3
2
1
𝑟+1
2
𝑟+1
1
𝑃
It should be 3.
𝑟+1
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If the vertices in P are too close,…
2
3
1
𝑟+1 𝑟+1
𝑃
𝑟+1
It should be 3.
2
𝑃
1
It should be 2.
Adjacent vertices receive the color 𝑟 + 1!
Where does 𝑑(𝑃) ≥ 4 come in?
If 𝑑(𝑃) ≥ 4, this does not happen.
1
2
1
3
2
3
1
𝑃
It should be 3.
3
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If 𝑑(𝑃) ≥ 4, this does not happen.
𝑟+1
2
1
3
2
3
1
𝑃
It should be 3.
3
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If 𝑑(𝑃) ≥ 4, this does not happen.
3
2
1
𝑟+1
2
𝑟+1
1
𝑃
It should be 3.
3
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If 𝑑(𝑃) ≥ 4, this does not happen.
3
2
1
𝑟+1
2
𝑟+1
1
𝑃
It should be 3.
𝑟+1
𝑃
It should be 2.
Where does 𝑑(𝑃) ≥ 4 come in?
If 𝑑(𝑃) ≥ 4, this does not happen.
2
3
𝑟+1
1
𝑟+1
𝑃
𝑟+1
It should be 3.
2
𝑃
1
It should be 2.
Answer by Albertson (1998)
Theorem A (Albertson, 1998)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺) with |𝑃| ≥ 2. If 𝑑(𝑃) ≥ 4,
then every precoloring of 𝑃 in 𝑟 + 1 colors extends to an (𝑟 + 1)-coloring
of 𝐺.
The condition on 𝑑(𝑃) is sharp.
Theorem B (Albertson, 1998)
For every 𝑟 ≥ 2, there exist infinitely many triples (𝐺, 𝑓, 𝑃) such that
• 𝐺 is an 𝑟-colorable graph,
• 𝑃 ⊂ 𝑉(𝐺), |𝑃| ≥ 2, 𝑑(𝑃) ≥ 3,
• 𝑓 ∶ 𝑃 → {1, … , 𝑟 + 1} is a precoloring of 𝑃, and
• 𝑓 does not extend to an (𝑟 + 1)-coloring of 𝐺.
What happens if 𝑃 contains a pair
of vertices of distance less than four?
In the real-world application, the assumption 𝑑(𝑃) ≥ 4 may not hold.
If 𝑃 contains pairs of vertices of distance less than four, then Albertson’s
Theorem says nothing.
We may need more than 𝑟 + 1 colors to extend an (𝑟 + 1)-coloing of 𝑃.
But if the number of such pairs is small, we probably do not need
many additional colors
motivation
Motivation
𝐷𝐺 (𝑃, 𝑘) = { {𝑥, 𝑦} ⊂ 𝑃 ∶ 𝑥 ≠ 𝑦, 𝑑𝐺 (𝑥, 𝑦) ≤ 𝑘 }
𝑡 = |𝐷𝐺 (𝑃, 3)|
χ G ≤ 𝑟, 𝑃 ⊂ V G
How many colors are sufficient to extend an (𝑟 + 1)-coloring of
𝑃 to a coloring of 𝐺?
Albertson’s Theorem : 𝑟 + 1 colors if 𝑡 = 0
1. What is precoloring?
2. Background and Motivation
3. Pairs of Vertices of Distance Three
4. Pairs of Vertices of Distance Two
Result 1
Theorem 1.
Let 𝑟 ≥ 2, 𝑘 ≥ 1, and let 𝐺 be an 𝑟-colorable graph. Let 𝑃 ⊂ 𝑉(𝐺). If
1
|𝐷𝐺 (𝑃, 3)| ≤ 2 𝑘(𝑘 + 1), then an (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 𝑘)coloring of 𝐺.
According to this theorem, if |𝐷𝐺 (𝑃, 3)| = 𝑡, then 𝑟 + 𝑂 𝑡
colors suffice to extend an (𝑟 + 1)-coloring of 𝑃 to a coloring of 𝐺.
Idea of the Proof : Use coloring
Theorem 1.
Let 𝑟 ≥ 2, 𝑘 ≥ 1, and let 𝐺 be an 𝑟-colorable graph. Let 𝑃 ⊂ 𝑉(𝐺). If |𝐷𝐺 (𝑃, 3)| ≤
1
𝑘(𝑘
2
+ 1), then an (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 𝑘)-coloring of 𝐺.
Construct an auxiliary graph 𝐻.
𝑉(𝐻) = 𝑃
𝐸 𝐻 = 𝐷𝐺 𝑃, 3 = {𝑥𝑦 ∶ 𝑑 𝐺 𝑥, 𝑦 ≤ 3}
1
|𝐸(𝐻)| ≤ 𝑘(𝑘 + 1)
2
𝑃1, … , 𝑃𝑘 : color classes
𝐻 is (almost) 𝑘-colorable.
Idea of the Proof : Use coloring
Construct an auxiliary graph 𝐻.
𝑉(𝐻) = 𝑃
𝐸 𝐻 = 𝐷𝐺 𝑃, 3 = {𝑥𝑦 ∶ 𝑑 𝐺 𝑥, 𝑦 ≤ 3}
|𝐸(𝐻)| ≤
1
𝑘(𝑘 + 1)
2
𝐻 is (almost) 𝑘-colorable.
𝑃1, … , 𝑃𝑘 : color classes
𝑃𝑖 is independent in 𝐻.
𝑑(𝑃𝑖) ≥ 4
Apply Albertson’s proof technique to each 𝑃𝑖, using a new
color 𝑟 + 𝑖 (1 ≤ 𝑖 ≤ 𝑘).
𝐺
: uncolored vertices
: precolored vertices
Apply Albertson’s technique.
𝐺
𝑃1
𝑟+1
: uncolored vertices
: precolored vertices
𝑟+2
𝑃2
𝑟+3
𝑃3
𝑑(𝑃𝑖) ≥ 4
𝐺
𝑃1
𝑃2
𝑃3
𝑑(𝑃𝑖) ≥ 4
: precolored vertices
Result 1
Theorem 1.
Let 𝑟 ≥ 2, 𝑘 ≥ 1, and let 𝐺 be an 𝑟-colorable graph. Let 𝑃 ⊂ 𝑉(𝐺). If
1
|𝐷𝐺 (𝑃, 3)| ≤ 2 𝑘(𝑘 + 1), then an (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 𝑘)coloring of 𝐺.
1. What is precoloring?
2. Background and Motivation
3. Pairs of Vertices of Distance Three
4. Pairs of Vertices of Distance Two
Albertson & Moore 1999
Theorem A (Albertson, 1998)
𝑑(𝑃) ≥ 4, 𝜒(𝐺) = 𝑟
Every (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 1)-coloring 𝐺.
What can we say under the assumption of 𝑑(𝑃) ≥ 3 instead of 𝑑(𝑃) ≥ 4 ?
Albertson & Moore 1999
Theorem B (Albertson and Moore , 1999)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) ≥ 3, then every
(𝑟 + 1)-coloring of 𝑃 extends to a
3𝑟+1
2
-coloring of 𝐺.
Motivation 2
Theorem B (Albertson and Moore , 1999)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) ≥ 3, then every
(𝑟 + 1)-coloring of 𝑃 extends to a
𝑑(𝑃) ≥ 3
3𝑟+1
2
-coloring of 𝐺.
𝐷𝐺 𝑃, 2 = ∅
What happens if 𝐷𝐺 (𝑃, 2) ≠ ∅ ?
Motivation 2
Theorem B (Albertson and Moore , 1999)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) ≥ 3, then every
(𝑟 + 1)-coloring of 𝑃 extends to a
3𝑟+1
2
-coloring of 𝐺.
Even if 𝑃 is colored in 𝑟 + 1 colors, we need more than 𝑟 + 1 colors to extend it.
This assumption is not important in this case.
We start from the assumption that 𝑃 is colored in 𝑟 + 𝑘 colors.
Result 2
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
Comparison
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
can be extended to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 can be extended to a
3𝑟+𝑘+1
-coloring
2
If we put 𝑘 = 1,…
of 𝐺.
Comparison
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 1 is even and |𝐷𝐺 (𝑃, 2)| < 2𝑟, then every (𝑟 + 1)-coloring of 𝑃 extends to a
3𝑟+1
-coloring
2
of 𝐺.
(2) If 𝑟 + 1 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3𝑟, then every (𝑟 + 1)-coloring of 𝑃
extends to a
3𝑟+2
-coloring
2
of 𝐺.
Theorem B (Albertson and Moore , 1998)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝐷𝐺 𝑃, 2 = ∅, then every
(𝑟 + 1)-coloring of 𝑃 extends to a
3𝑟+1
2
-coloring of 𝐺.
Theorem 2 (almost) contains Theorem B.
Idea of Albertson - Moore
Theorem B (Albertson and Moore , 1999)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) ≥ 3, then every (𝑟 + 1)-coloring of 𝑃
extends to a
(𝑟:odd)
3𝑟+1
2
-coloring of 𝐺.
Color 𝐺 − 𝑃 in 𝑟 colors.
𝑉(𝐺) − 𝑃 is partitioned into 𝑟
independent sets (color classes)
𝑈1 , … , 𝑈 𝑟 : color classes of 𝐺 − 𝑃
We assign 2 old colors or 1 new color to each 𝑈𝑖 to accommodate the
request from the precoloring of 𝑃.
Idea of Albertson - Moore
We partition uncolored vertices into 𝑟 independent sets
𝑈1
𝑈𝑟+1
𝑈2
1
𝑈𝑟+3
2
1
3
6
𝑃
2
𝑟
𝑟+1
precolored in 𝑟 + 1 colors
𝑈r
Idea of Albertson - Moore
new colors
Existing colors
1, 2
3, 4
r, 𝑟 + 1
𝑟+2
3𝑟 + 1
2
𝑈1
𝑈2
𝑈𝑟+1
𝑈𝑟+3
𝑈r
1
2
1
3
6
𝑃
2
𝑟
𝑟+1
precolored in 𝑟 + 1 colors
Idea of Albertson - Moore
available colors : 2𝑖 − 1, 2𝑖
If two existing colors are
assigned to 𝑈𝑖,…
𝑈𝑖
𝑃
1
6
2𝑖 − 1 2𝑖 − 1 2𝑖
Idea of Albertson - Moore
available colors : 2𝑖 − 1, 2𝑖
If two existing colors are
assigned to 𝑈𝑖,…
𝑈𝑖
𝑃
1
6
2𝑖 − 1 2𝑖 − 1 2𝑖
If a vertex in 𝑈𝑖 is not adjacent to a vertex in 𝑃 in color 2𝑖 − 1, then color it
in 2𝑖 − 1.
Idea of Albertson - Moore
available colors : 2𝑖 − 1, 2𝑖
If two existing colors are
assigned to 𝑈𝑖,…
𝑈𝑖
𝑃
1
6
2𝑖 − 1 2𝑖 − 1 2𝑖
If a vertex in 𝑈𝑖 is not adjacent to a vertex in 𝑃 in color 2𝑖 − 1, then color it
in 2𝑖 − 1.
Idea of Albertson - Moore
available colors : 2𝑖 − 1, 2𝑖
If two existing colors are
assigned to 𝑈𝑖,…
𝑈𝑖
𝑃
1
6
2𝑖 − 1 2𝑖 − 1 2𝑖
If a vertex in 𝑈𝑖 is adjacent to a vertex in 𝑃 in color 2𝑖 − 1, then color it in 2𝑖.
Idea of Albertson - Moore
available colors : 2𝑖 − 1, 2𝑖
If two existing colors are
assigned to 𝑈𝑖,…
𝑈𝑖
𝑃
1
6
2𝑖 − 1 2𝑖 − 1 2𝑖
If a vertex in 𝑈𝑖 is adjacent to a vertex in 𝑃 in color 2𝑖 − 1, then color it in 2𝑖.
Idea of Albertson - Moore
available colors : 2𝑖 − 1, 2𝑖
If two existing colors are
assigned to 𝑈𝑖,…
𝑈𝑖
𝑑(𝑥, 𝑦) ≤ 2
𝑃
𝑥
𝑦
2𝑖 − 1 2𝑖
What happens if a vertex in 𝑈𝑖 is adjacent to vertices of colors 2𝑖 − 1
and 2𝑖 ?
It does not occur since 𝑑(𝑃) ≥ 3.
Idea of Albertson - Moore
available color : 𝑟 + 2
If one new color is
assigned to 𝑈𝑖,…
𝑈𝑟+3
2
𝑃
1
6
𝑟
𝑟
𝑟+1
Idea of Albertson - Moore
available color : 𝑟 + 2
If one new color is
assigned to 𝑈𝑖,…
𝑈𝑟+3
2
𝑃
1
6
Color each vertex in color 𝑟 + 2
𝑟
𝑟
𝑟+1
Idea of Albertson - Moore
available color : 𝑟 + 2
If one new color is
assigned to 𝑈𝑖,…
𝑈𝑟+3
2
𝑃
1
6
Color each vertex in color 𝑟 + 2
𝑟
𝑟
𝑟+1
Idea of Albertson - Moore
new colors
Existing colors
1, 2
3, 4
r, 𝑟 + 1
𝑟+2
3𝑟 + 1
2
𝑈1
𝑈2
𝑈𝑟+1
𝑈𝑟+3
𝑈r
1
2
1
3
6
𝑃
2
𝑟
𝑟+1
We use
3𝑟+1
2
colors.
Albertson & Moore 1999
Theorem B (Albertson and Moore , 1999)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) ≥ 3, then every
(𝑟 + 1)-coloring of 𝑃 extends to a
3𝑟+1
2
-coloring of 𝐺.
Idea of the Proof : input from matching
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
can be extended to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 can be extended to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
Idea of the Proof : input from matching
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
can be extended to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 can be extended to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
Idea of the Proof : input from matching
available colors : 2𝑖 − 1, 2𝑖
If two existing colors are
assigned to 𝑈𝑖,…
𝑈𝑖
𝑑(𝑥, 𝑦) ≤ 2
𝑃
𝑥
𝑦
2𝑖 − 1 2𝑖
This does not occur if 𝑑 𝑃 ≥ 3 (Albertson-Moore).
In our case, it can happen since possibly 𝐷𝐺 (𝑃, 2) ≠ ∅
But not many times since |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1).
Idea of the Proof : input from matching
new colors
Existing colors
1, 2
3, 4
r + 𝑘 − 1, 𝑟 + 𝑘
𝑈1
𝑈2
𝑈𝑟+𝑘
1
𝑟+𝑘+1
𝑈𝑟+𝑘+1
2
1
3
6
𝑃
𝑟+k-1 𝑟 + 𝑘
There are many ways to assign two colors to 𝑈𝑖.
2
3𝑟 + 𝑘
2
𝑈r
Idea of the Proof : input from matching
Suppose :
𝐷𝐺 𝑃, 2 = { 𝑢1, 𝑢2 , 𝑢2, 𝑢3 , 𝑢4, 𝑢5 , 𝑢6, 𝑢7 }
𝑓0 ∶ 𝑃 → {1,2,3, … } : precoloring of 𝑃
𝑓0(𝑢1) = 1, 𝑓0(𝑢2) = 2, 𝑓0(𝑢3) = 3, 𝑓0(𝑢4) = 3, 𝑓0(𝑢5) = 4,
𝑓0(𝑢6) = 3, 𝑓0(𝑢7) = 4
Idea of the Proof : input from matching
Re-assign colors.
1, 2
3, 4
𝑈1
𝑈2
No colors are available for
them.
1
𝑢1
2
𝑢2
3
𝑢3
3
𝑢4
𝑃
4
𝑢5
4
3
𝑢6
𝑢7
𝑓0(𝑢1) = 1, 𝑓0(𝑢2) = 2, 𝑓0(𝑢3) = 3, 𝑓0(𝑢4) = 3, 𝑓0(𝑢5) = 4,
𝑓0(𝑢6) = 3, 𝑓0(𝑢7) = 4
Idea of the Proof : input from matching
2, 3
1, 4
𝑈1
𝑈2
We can give them color 1.
1
𝑢1
2
𝑢2
3
𝑢3
3
𝑢4
𝑃
4
𝑢5
4
3
𝑢6
𝑢7
𝑓0(𝑢1) = 1, 𝑓0(𝑢2) = 2, 𝑓0(𝑢3) = 3, 𝑓0(𝑢4) = 3, 𝑓0(𝑢5) = 4,
𝑓0(𝑢6) = 3, 𝑓0(𝑢7) = 4
Idea of the Proof : input from matching
Re-assign colors.
2, 3
Still neither color
is available.
1, 4
𝑈1
𝑈2
We can give them color 1.
1
𝑢1
2
𝑢2
3
𝑢3
3
𝑢4
𝑃
4
𝑢5
4
3
𝑢6
𝑢7
𝑓0(𝑢1) = 1, 𝑓0(𝑢2) = 2, 𝑓0(𝑢3) = 3, 𝑓0(𝑢4) = 3, 𝑓0(𝑢5) = 4,
𝑓0(𝑢6) = 3, 𝑓0(𝑢7) = 4
Idea of the Proof : input from matching
1, 4
2, 3
𝑈1
𝑈2
We can give them color 2.
1
𝑢1
2
𝑢2
3
𝑢3
3
𝑢4
𝑃
4
𝑢5
4
3
𝑢6
𝑢7
𝑓0(𝑢1) = 1, 𝑓0(𝑢2) = 2, 𝑓0(𝑢3) = 3, 𝑓0(𝑢4) = 3, 𝑓0(𝑢5) = 4,
𝑓0(𝑢6) = 3, 𝑓0(𝑢7) = 4
Idea of the Proof : input from matching
1, 4
We can give it
color 4.
2, 3
𝑈1
𝑈2
We can give them color 2.
1
𝑢1
2
𝑢2
3
𝑢3
3
𝑢4
𝑃
4
𝑢5
4
3
𝑢6
𝑢7
𝑓0(𝑢1) = 1, 𝑓0(𝑢2) = 2, 𝑓0(𝑢3) = 3, 𝑓0(𝑢4) = 3, 𝑓0(𝑢5) = 4,
𝑓0(𝑢6) = 3, 𝑓0(𝑢7) = 4
Idea of the Proof : input from matching
This assignment works.
1, 4
We can give it
color 4.
2, 3
𝑈1
𝑈2
We can give them color 2.
1
𝑢1
2
𝑢2
3
𝑢3
3
𝑢4
𝑃
4
𝑢5
4
3
𝑢6
𝑢7
𝑓0(𝑢1) = 1, 𝑓0(𝑢2) = 2, 𝑓0(𝑢3) = 3, 𝑓0(𝑢4) = 3, 𝑓0(𝑢5) = 4,
𝑓0(𝑢6) = 3, 𝑓0(𝑢7) = 4
Idea of the Proof : input from matching
new colors
Existing colors
1, 2
3, 4
r + 𝑘 − 1, 𝑟 + 𝑘
𝑈1
𝑈2
𝑈𝑟+𝑘
1
2
1
3
𝑟+𝑘+1
𝑈𝑟+𝑘+1
2
6 𝑟+𝑘−1 𝑟+𝑘
𝑃
There are many ways to assign two colors to 𝑈𝑖.
3𝑟 + 𝑘
2
𝑈r
Idea of the Proof : input from matching
new colors
Existing colors
𝑐1, 𝑐2
𝑐3, 𝑐4
𝑐r + 𝑘 − 1, 𝑐𝑟 + 𝑘
𝑈1
𝑈2
𝑈𝑟+𝑘
1
2
1
3
𝑟+𝑘+1
𝑈𝑟+𝑘+1
2
6 𝑟+𝑘−1 𝑟+𝑘
𝑃
{𝑐1, 𝑐2, … , 𝑐𝑟 + 𝑘 − 1, 𝑐𝑟 + 𝑘} = {1, 2, … , 𝑟 + 𝑘 − 1, 𝑟 + 𝑘}
3𝑟 + 𝑘
2
𝑈r
Idea of the Proof : input from matching
new colors
Existing colors
𝑐1, 𝑐2
𝑐3, 𝑐4
𝑐r + 𝑘 − 1, 𝑐𝑟 + 𝑘
𝑈1
𝑈2
𝑈𝑟+𝑘
2
𝑟+𝑘+1
𝑈𝑟+𝑘+1
2
{𝑐1, 𝑐2, … , 𝑐𝑟 + 𝑘 − 1, 𝑐𝑟 + 1 } = {1, 2, … , 𝑟 + 𝑘 − 1, 𝑟 + 𝑘}
Find suitable pairs {𝑐1, 𝑐2}, {𝑐3, 𝑐4}, … , {𝑐𝑟 + 𝑘 − 1, 𝑐𝑟 + 𝑘 }.
matching problem!
3𝑟 + 𝑘
2
𝑈r
Idea of the Proof : input from matching
𝐶 = {1, 2, … , 𝑟 + 𝑘}
𝐾 : the complete graph on 𝐶 (complete graph on colors)
𝑒 = 𝑖𝑗 : an edge in 𝐾 (a pair of colors)
𝑓0 : a precolor of 𝑃
𝑒 = 𝑖𝑗 is a green edge if there is no pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with {𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a yellow edge if there is exactly one pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with
{𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a red edge if there are two or more pairs 𝑢, 𝑣 , {𝑢′ , 𝑣 ′ } ∈ 𝐷𝐺 𝑃, 2 with
𝑓0 𝑢 , 𝑓0 𝑣 = {𝑓0 𝑢′ , 𝑓0 𝑣 ′ } = {𝑖, 𝑗}
Idea of the Proof : input from matching
𝑒 = 𝑖𝑗 is a green edge if there is no pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with {𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a yellow edge if there is exactly one pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with
{𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a red edge if there are two or more pairs 𝑢, 𝑣 , {𝑢′ , 𝑣 ′ } ∈ 𝐷𝐺 𝑃, 2 with
𝑓0 𝑢 , 𝑓0 𝑣 = {𝑓0 𝑢′ , 𝑓0 𝑣 ′ } = {𝑖, 𝑗}
1
2
1
3
4
2
3
1
4
Idea of the Proof : input from matching
𝑒 = 𝑖𝑗 is a green edge if there is no pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with {𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a yellow edge if there is exactly one pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with
{𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a red edge if there are two or more pairs 𝑢, 𝑣 , {𝑢′ , 𝑣 ′ } ∈ 𝐷𝐺 𝑃, 2 with
𝑓0 𝑢 , 𝑓0 𝑣 = {𝑓0 𝑢′ , 𝑓0 𝑣 ′ } = {𝑖, 𝑗}
1
2
1
3
4
2
3
1
4
Idea of the Proof : input from matching
𝑒 = 𝑖𝑗 is a green edge if there is no pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with {𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a yellow edge if there is exactly one pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with
{𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a red edge if there are two or more pairs 𝑢, 𝑣 , {𝑢′ , 𝑣 ′ } ∈ 𝐷𝐺 𝑃, 2 with
𝑓0 𝑢 , 𝑓0 𝑣 = {𝑓0 𝑢′ , 𝑓0 𝑣 ′ } = {𝑖, 𝑗}
1
2
1
3
4
2
3
1
4
Idea of the Proof : input from matching
𝑒 = 𝑖𝑗 is a green edge if there is no pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with {𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a yellow edge if there is exactly one pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with
{𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a red edge if there are two or more pairs 𝑢, 𝑣 , {𝑢′ , 𝑣 ′ } ∈ 𝐷𝐺 𝑃, 2 with
𝑓0 𝑢 , 𝑓0 𝑣 = {𝑓0 𝑢′ , 𝑓0 𝑣 ′ } = {𝑖, 𝑗}
1
2
1
3
4
2
3
1
4
Idea of the Proof : input from matching
𝑒 = 𝑖𝑗 is a green edge if there is no pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with {𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a yellow edge if there is exactly one pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with
{𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a red edge if there are two or more pairs 𝑢, 𝑣 , {𝑢′ , 𝑣 ′ } ∈ 𝐷𝐺 𝑃, 2 with
𝑓0 𝑢 , 𝑓0 𝑣 = {𝑓0 𝑢′ , 𝑓0 𝑣 ′ } = {𝑖, 𝑗}
1
2
1
3
4
2
3
1
4
Idea of the Proof : input from matching
𝑒 = 𝑖𝑗 is a green edge if there is no pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with {𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a yellow edge if there is exactly one pair {𝑢, 𝑣} ∈ 𝐷𝐺 𝑃, 2 with
{𝑓0(𝑢), 𝑓0(𝑣)} = {𝑖, 𝑗}
𝑒 = 𝑖𝑗 is a red edge if there are two or more pairs 𝑢, 𝑣 , {𝑢′ , 𝑣 ′ } ∈ 𝐷𝐺 𝑃, 2 with
𝑓0 𝑢 , 𝑓0 𝑣 = {𝑓0 𝑢′ , 𝑓0 𝑣 ′ } = {𝑖, 𝑗}
1
2
1
3
4
2
3
1
4
Idea of the Proof : input from matching
𝐶 = {1, 2, … , 𝑟 + 𝑘}
𝐾 : the complete graph on 𝐶 (complete graph on colors)
𝑒 = 𝑖𝑗 : an edge in 𝐾 (a pair of colors)
𝑓0 : a precolor of 𝑃
Lemma 1
If 𝐾 has a perfect matching 𝑀 which contains no red edge and at most one yellow
edge, then we can assign edges of 𝑀 to color classes of 𝐺 − 𝑃 so that it extends to a
3𝑟+𝑘
–coloring
2
of G.
Idea of the Proof : input from matching
𝐶 = {1, 2, … , 𝑟 + 𝑘}
𝐾 : the complete graph on 𝐶 (complete graph on colors)
𝑒 = 𝑖𝑗 : an edge in 𝐾 (a pair of colors)
𝑓0 : a precolor of 𝑃
The edges of 𝐾 are factorized into 𝑟 + 𝑘 − 1 perfect matchings.
Lemma 2
If |𝐷𝐺 𝑃, 2 | < 2(𝑟 + 𝑘 − 1), then at least one of the factorized matchings has no
red edge and at most one yellow edge.
Idea of the Proof : input from matching
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
Idea of the Proof : input from matching
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
can be extended to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
More complicated, but similar strategy
of 𝐺.
Comparison
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and 𝐷𝐺 𝑃, 2 < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and 𝐷𝐺 𝑃, 2
coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
< 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-
of 𝐺.
We have different bounds for |𝐷𝐺 (𝑃, 2)| depending on the parity of 𝑟 + 𝑘.
But both bounds are best-possible.
Sharpness
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and 𝐷𝐺 𝑃, 2 < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to a
3𝑟+𝑘
-coloring
2
of 𝐺.
Theorem 3
For every integer 𝑟 and 𝑘 with 𝑟 ≥ 2, 1 ≤ 𝑘 ≤ 𝑟 and 𝑟 + 𝑘 ≡ 0 (mod 2), there exist
infinitely many triples (𝐺, 𝑃, 𝑓) such that
(1) 𝐺 is an 𝑟-colorable graph,
(2) 𝑃 ⊂ 𝑉(𝐺), |𝐷𝐺 (𝑃, 2)| = 2(𝑟 + 𝑘 − 1),
(3) 𝑓 ∶ 𝑃 → {1, … , 𝑟 + 𝑘} is an (𝑟 + 𝑘)-coloring of 𝑃, and
(4) 𝑓 does not extend to a
3𝑟+𝑘
-coloring
2
of 𝐺.
Comparison
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
Theorem 4
For every integer 𝑟 and 𝑘 with 𝑟 ≥ 2, 1 ≤ 𝑘 ≤ 𝑟 and 𝑟 + 𝑘 ≡ 1 (mod 2), there exist
infinitely many triples (𝐺, 𝑃, 𝑓) such that
(1) 𝐺 is an 𝑟-colorable graph,
(2) 𝑃 ⊂ 𝑉(𝐺), |𝐷𝐺 (𝑃, 2)| = 3(𝑟 + 𝑘 − 1),
(3) 𝑓 ∶ 𝑃 → {1, … , 𝑟 + 𝑘} is an (𝑟 + 𝑘)-coloring of 𝑃, and
(4) 𝑓 does not extend to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
For large k…
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝒌 ≤ 𝒓.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
What happens if 𝑘 > 𝑟 ?
The situation is quite different.
For large k…
1
𝑔1(𝑟, 𝑘) = min{2 𝑘 + 3𝑟 − 4 𝑘 − 𝑟 + 3 , (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
1
𝑔2 𝑟, 𝑘 = min{2 (𝑘 + 1)(𝑘 + 2), (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
Theorem 5
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺).
(1) If 𝑟 < 𝑘 ≤
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔1(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to an (𝑟 + 𝑘)-coloring of 𝐺.
(2) If 𝑘 >
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔2(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃 extends to
an (𝑟 + 𝑘)-coloring of 𝐺.
For large k…
1
𝑔1(𝑟, 𝑘) = min{2 𝑘 + 3𝑟 − 4 𝑘 − 𝑟 + 3 , (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
1
𝑔2 𝑟, 𝑘 = min{2 (𝑘 + 1)(𝑘 + 2), (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
Theorem 5
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺).
(1) If 𝑟 < 𝑘 ≤
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔1(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to an (𝑟 + 𝑘)-coloring of 𝐺.
(2) If 𝑘 >
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔2(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃 extends to
an (𝑟 + 𝑘)-coloring of 𝐺.
• No additional colors
For large k…
1
𝑔1(𝑟, 𝑘) = min{2 𝑘 + 3𝑟 − 4 𝑘 − 𝑟 + 3 , (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
1
𝑔2 𝑟, 𝑘 = min{2 (𝑘 + 1)(𝑘 + 2), (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
Theorem 5
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺).
(1) If 𝑟 < 𝑘 ≤
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔1(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to an (𝑟 + 𝑘)-coloring of 𝐺.
(2) If 𝑘 >
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔2(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃 extends to
an (𝑟 + 𝑘)-coloring of 𝐺.
• No additional colors
• The bound of |𝐷𝐺 (𝑃, 2)| jumps from a linear function to a quadratic function of 𝑘.
Conclusion
Theorem A (Albertson, 1998)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺) with |𝑃| ≥ 2. If 𝑑(𝑃) ≥ 4,
then every (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 1)-coloring of 𝐺.
𝐷𝐺 (𝑃, 𝑘) = { {𝑥, 𝑦} ⊂ 𝑃 ∶ 𝑥 ≠ 𝑦, 𝑑𝐺 (𝑥, 𝑦) ≤ 𝑘 }
𝑡 = |𝐷𝐺 (𝑃, 3)|
Theorem 1.
Let 𝑟 ≥ 2, 𝑘 ≥ 1, and let 𝐺 be an 𝑟-colorable graph. Let 𝑃 ⊂ 𝑉(𝐺). If
1
|𝐷𝐺 (𝑃, 3)| ≤ 2 𝑘(𝑘 + 1), then an (𝑟 + 1)-coloring of 𝑃 extends to an (𝑟 + 𝑘)coloring of 𝐺.
If 𝑡 ≠ 0, 𝑟 + 𝑂
coloring of 𝐺.
𝑡 new colors suffice to extend an (𝑟 + 1)-coloring of 𝑃 to a
Conclusion
Theorem B (Albertson and Moore , 1999)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) ≥ 3, then every
(𝑟 + 1)-coloring of 𝑃 extends to a
3𝑟+1
2
-coloring of 𝐺.
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
Conclusion
Theorem B (Albertson and Moore , 1999)
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). If 𝑑(𝑃) ≥ 3, then every
(𝑟 + 1)-coloring of 𝑃 extends to a
3𝑟+1
2
-coloring of 𝐺.
Theorem 2
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺). Suppose 𝑘 ≤ 𝑟.
(1) If 𝑟 + 𝑘 is even and |𝐷𝐺 (𝑃, 2)| < 2(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to a
3𝑟+𝑘
-coloring
2
of 𝐺.
(2) If 𝑟 + 𝑘 is odd, 𝑟 + 𝑘 ≥ 13 and |𝐷𝐺 (𝑃, 2)| < 3(𝑟 + 𝑘 − 1), then every (𝑟 + 𝑘)coloring of 𝑃 extends to a
3𝑟+𝑘+1
-coloring
2
of 𝐺.
Conclusion
1
𝑔1(𝑟, 𝑘) = min{2 𝑘 + 3𝑟 − 4 𝑘 − 𝑟 + 3 , (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
1
𝑔2 𝑟, 𝑘 = min{2 (𝑘 + 1)(𝑘 + 2), (𝑘 − 𝑟 + 2)(𝑘 + 𝑟 − 1)}
Theorem 5
Let 𝐺 be an 𝑟-colorable graph and let 𝑃 ⊂ 𝑉(𝐺).
(1) If 𝑟 < 𝑘 ≤
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔1(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃
extends to an (𝑟 + 𝑘)-coloring of 𝐺.
(2) If 𝑘 >
3𝑟−7
2
and |𝐷𝐺 (𝑃, 2)| ≤ 𝑔2(𝑟, 𝑘), then every (𝑟 + 𝑘)-coloring of 𝑃 extends to
an (𝑟 + 𝑘)-coloring of 𝐺.
• No additional colors
• The bound of |𝐷𝐺 (𝑃, 2)| jumps from a linear function to a quadratic function of 𝑘.
Thank you!
ありがとうございました!