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IntroductionThe Chromatic Number of the Plane
The Problem
Coloring Problems
Thomas Chartier
October 13, 2011
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Acknowledgments
I would like to thank the Mathematics department of BoiseState University for providing support in the form of a TeachingAssistantship. I would also like to thank Andres Caicedo for allhis help and encouragement.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Table of contents
1 Introduction
2 The Chromatic Number of the Plane
3 The Problem
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Coloring Problems
Graph Coloring is the assignment of labels or colors to the edgesor vertices of a graph. Problems in this area have given impetusto Ramsey theory and the partition calculus; the results oftenhave application in number theory and analysis, among others.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Four Color Theorem
Consider the following question, originally posed by FrancisGuthrie in 1852.
Question
Is it possible to color any planar map using four colors in sucha way that regions sharing a common boundary, excludingboundaries which are comprised of a single point, do not sharethe same color?
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Four Color Theorem
Consider the following question, originally posed by FrancisGuthrie in 1852.
Question
Is it possible to color any planar map using four colors in sucha way that regions sharing a common boundary, excludingboundaries which are comprised of a single point, do not sharethe same color?
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Four Color Theorem
An affirmative answer was obtained by Kenneth Appel andWolfgang Haken in 1976.
This was the first major result which made use of computers inan essential way.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Four Color Theorem
An affirmative answer was obtained by Kenneth Appel andWolfgang Haken in 1976.This was the first major result which made use of computers inan essential way.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Question
(Edward Nelson, 1950) What is the smallest number of colorssufficient for coloring the plane in such a way that no twopoints of the same color are unit distance apart?
Denote the chromatic number of the plane by χ.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Question
(Edward Nelson, 1950) What is the smallest number of colorssufficient for coloring the plane in such a way that no twopoints of the same color are unit distance apart?
Denote the chromatic number of the plane by χ.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proposition
χ ≥ 4.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B
C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′
D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proof.
A
B C
A′
√3
A′D
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The Chromatic Number of the Plane
Proposition
χ ≤ 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof.
2
√7
22.1≈ 0.95
√7
2.1≈ 1.26
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Introduction of the Problem
Question
Is it possible to color Z+ using n colors, in such a way that forany a, the numbers a, 2a, 3a, ..., na are colored differently?
This question was posted to MathOverflow.net by PalvolgyiDomotor on May 29, 2010. If such a coloring exists, call it asatisfactory coloring.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Introduction of the Problem
Question
Is it possible to color Z+ using n colors, in such a way that forany a, the numbers a, 2a, 3a, ..., na are colored differently?
This question was posted to MathOverflow.net by PalvolgyiDomotor on May 29, 2010. If such a coloring exists, call it asatisfactory coloring.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
To begin, we shall consider the case when n = 5. That is, wewill first attempt to construct a satisfactory coloring using 5colors. To do so let’s introduce some notation.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Notation continued
The table is read as, 1 is being colored magenta, 2 is beingcolored red, 3 is being colored blue, 4 is being colored green and5 is being colored orange. In functional notation we havec(1) =magenta, c(2) =red, c(3) =blue, c(4) =green andc(5) =orange.
1 magenta 2
red
3
blue
4
green
5
orange
6 7 8 9 102 red 4
green
6 8 10 12 14 16 18 203 blue 6 9 12 15 18 21 24 27 304 green 8 12 16 20 24 28 32 36 405 orange 10 15 20 25 30 35 40 45 50
Note that the ith column is the column with entriesi, 2i, 3i, 4i, 5i.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Notation continued
The table is read as, 1 is being colored magenta, 2 is beingcolored red, 3 is being colored blue, 4 is being colored green and5 is being colored orange. In functional notation we havec(1) =magenta, c(2) =red, c(3) =blue, c(4) =green andc(5) =orange.
1 magenta 2
red
3
blue
4
green
5
orange
6 7 8 9 102 red 4
green
6 8 10 12 14 16 18 203 blue 6 9 12 15 18 21 24 27 304 green 8 12 16 20 24 28 32 36 405 orange 10 15 20 25 30 35 40 45 50
Note that the ith column is the column with entriesi, 2i, 3i, 4i, 5i.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Notation continued
The table is read as, 1 is being colored magenta, 2 is beingcolored red, 3 is being colored blue, 4 is being colored green and5 is being colored orange. In functional notation we havec(1) =magenta, c(2) =red, c(3) =blue, c(4) =green andc(5) =orange.
1 magenta 2 red 3 blue 4 green 5 orange 6 7 8 9 102 red 4 green 6 8 10 12 14 16 18 203 blue 6 9 12 15 18 21 24 27 304 green 8 12 16 20 24 28 32 36 405 orange 10 15 20 25 30 35 40 45 50
Note that the ith column is the column with entriesi, 2i, 3i, 4i, 5i.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Notation continued
The table is read as, 1 is being colored magenta, 2 is beingcolored red, 3 is being colored blue, 4 is being colored green and5 is being colored orange. In functional notation we havec(1) =magenta, c(2) =red, c(3) =blue, c(4) =green andc(5) =orange.
1 magenta 2 red 3 blue 4 green 5 orange 6 7 8 9 102 red 4 green 6 8 10 12 14 16 18 203 blue 6 9 12 15 18 21 24 27 304 green 8 12 16 20 24 28 32 36 405 orange 10 15 20 25 30 35 40 45 50
Note that the ith column is the column with entriesi, 2i, 3i, 4i, 5i.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
Letting c(i) = i for i = 1, 2, 3, 4, 5 the first column becomes.
Now, we can fill in the values which are determined by theassignment of column 1. Which gives us:
1
1
2
2
3
3
4
4
5
5
6 7 8 9 102
2
4
4
6 8 10 12 14 16 18 203
3
6 9 12 15 18 21 24 27 304
4
8 12 16 20 24 28 32 36 405
5
10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
Letting c(i) = i for i = 1, 2, 3, 4, 5 the first column becomes.
Now, we can fill in the values which are determined by theassignment of column 1. Which gives us:
1 1 2
2
3
3
4
4
5
5
6 7 8 9 102 2 4
4
6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
Letting c(i) = i for i = 1, 2, 3, 4, 5 the first column becomes.Now, we can fill in the values which are determined by theassignment of column 1. Which gives us:
1 1 2
2
3
3
4
4
5
5
6 7 8 9 102 2 4
4
6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
Letting c(i) = i for i = 1, 2, 3, 4, 5 the first column becomes.Now, we can fill in the values which are determined by theassignment of column 1. Which gives us:
1 1 2 2 3 3 4 4 5 5 6 7 8 9 102 2 4 4 6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 7 8 9 102 2 4 4 6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 7 8 9 102 2 4 4 6 8 10 12 14 16 18 203 3 6 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 9 102 2 4 4 6 1 8 10 12 14 16 18 203 3 6 1 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 9 102 2 4 4 6 1 8 10 12 14 16 18 203 3 6 1 9 12 15 18 21 24 27 304 4 8 12 16 20 24 28 32 36 405 5 10 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8
5
9 10 32 2 4 4 6 1 8
5
10 3 12 14 16 18 203 3 6 1 9 12 15 18 21 24 27 304 4 8
5
12 16 20 24 28 32 36 405 5 10 3 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 10 32 2 4 4 6 1 8 5 10 3 12 14 16 18 203 3 6 1 9 12 15 18 21 24 27 304 4 8 5 12 16 20 24 28 32 36 405 5 10 3 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 10 32 2 4 4 6 1 8 5 10 3 12 14 16 18 203 3 6 1 9 12 15 18 21 24 27 304 4 8 5 12 16 20 24 28 32 36 405 5 10 3 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 18 203 3 6 1 9 12 2 15 18 21 24 27 304 4 8 5 12 2 16 20 24 28 32 36 405 5 10 3 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 18 203 3 6 1 9 12 2 15 18 21 24 27 304 4 8 5 12 2 16 20 24 28 32 36 405 5 10 3 15 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9
5
10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 18 203 3 6 1 9
5
12 2 15 4 18 21 24 27 304 4 8 5 12 2 16 20 24 28 32 36 405 5 10 3 15 4 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 18 203 3 6 1 9 5 12 2 15 4 18 21 24 27 304 4 8 5 12 2 16 20 24 28 32 36 405 5 10 3 15 4 20 25 30 35 40 45 50
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 40 2 45 2 50 4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
Now we make the following observation:
1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 40 2 45 2 50 4
This follows from the fact that 7 is relatively prime with allnumbers less than or equal to 5 or, equivalently, with2 · 3 · 5 = 30. In fact, for every column x for which (x, 30) = 1we have that c(x), c(2x), c(3x), c(4x) and c(5x) are notdetermined by c(y) for any y < x.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
Now we make the following observation:1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 40 2 45 2 50 4
This follows from the fact that 7 is relatively prime with allnumbers less than or equal to 5 or, equivalently, with2 · 3 · 5 = 30. In fact, for every column x for which (x, 30) = 1we have that c(x), c(2x), c(3x), c(4x) and c(5x) are notdetermined by c(y) for any y < x.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
n = 5
Now we make the following observation:1 1 2 2 3 3 4 4 5 5 6 1 7 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 40 2 45 2 50 4
This follows from the fact that 7 is relatively prime with allnumbers less than or equal to 5 or, equivalently, with2 · 3 · 5 = 30. In fact, for every column x for which (x, 30) = 1we have that c(x), c(2x), c(3x), c(4x) and c(5x) are notdetermined by c(y) for any y < x.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The n-core
Definition
The n-core, or simply the core if n is understood, is the set Kn
of all positive integers whose prime decomposition only involvesprimes less than or equal to n.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7
1
8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14
2
16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21
3
24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28
4
32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35
5
40 2 45 2 50 4
7 1 14
2
21
3
28
4
35
5
42
1
49
1
56
5
63
5
70
3
14 2 28
4
42
1
56
5
70
3
84
2
98
2
112
3
126
3
140
1
21 3 42
1
63
5
84
2
105
4
126
3
147
3
168
4
189
4
210
5
28 4 56
5
84
2
112
3
140
1
168
4
196
4
224
1
252
1
280
2
35 5 70
3
105
4
140
1
175
2
210
5
245
5
280
2
315
2
350
4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14
2
21
3
28
4
35
5
42
1
49
1
56
5
63
5
70
3
14 2 28
4
42
1
56
5
70
3
84
2
98
2
112
3
126
3
140
1
21 3 42
1
63
5
84
2
105
4
126
3
147
3
168
4
189
4
210
5
28 4 56
5
84
2
112
3
140
1
168
4
196
4
224
1
252
1
280
2
35 5 70
3
105
4
140
1
175
2
210
5
245
5
280
2
315
2
350
4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14 2 21
3
28
4
35
5
42
1
49
1
56
5
63
5
70
3
14 2 28 4 42
1
56
5
70
3
84
2
98
2
112
3
126
3
140
1
21 3 42 1 63
5
84
2
105
4
126
3
147
3
168
4
189
4
210
5
28 4 56 5 84
2
112
3
140
1
168
4
196
4
224
1
252
1
280
2
35 5 70 3 105
4
140
1
175
2
210
5
245
5
280
2
315
2
350
4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14 2 21 3 28
4
35
5
42
1
49
1
56
5
63
5
70
3
14 2 28 4 42 1 56
5
70
3
84
2
98
2
112
3
126
3
140
1
21 3 42 1 63 5 84
2
105
4
126
3
147
3
168
4
189
4
210
5
28 4 56 5 84 2 112
3
140
1
168
4
196
4
224
1
252
1
280
2
35 5 70 3 105 4 140
1
175
2
210
5
245
5
280
2
315
2
350
4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14 2 21 3 28 4 35
5
42
1
49
1
56
5
63
5
70
3
14 2 28 4 42 1 56 5 70
3
84
2
98
2
112
3
126
3
140
1
21 3 42 1 63 5 84 2 105
4
126
3
147
3
168
4
189
4
210
5
28 4 56 5 84 2 112 3 140
1
168
4
196
4
224
1
252
1
280
2
35 5 70 3 105 4 140 1 175
2
210
5
245
5
280
2
315
2
350
4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14 2 21 3 28 4 35 5 42 1 49
1
56 5 63 5 70 314 2 28 4 42 1 56 5 70 3 84 2 98
2
112 3 126 3 140 121 3 42 1 63 5 84 2 105 4 126 3 147
3
168 4 189 4 210 528 4 56 5 84 2 112 3 140 1 168 4 196
4
224 1 252 1 280 235 5 70 3 105 4 140 1 175 2 210 5 245
5
280 2 315 2 350 4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14 2 21 3 28 4 35 5 42 1 49
1
56 5 63 5 70 314 2 28 4 42 1 56 5 70 3 84 2 98
2
112 3 126 3 140 121 3 42 1 63 5 84 2 105 4 126 3 147
3
168 4 189 4 210 528 4 56 5 84 2 112 3 140 1 168 4 196
4
224 1 252 1 280 235 5 70 3 105 4 140 1 175 2 210 5 245
5
280 2 315 2 350 4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14 2 21 3 28 4 35 5 42 1 49 1 56 5 63 5 70 314 2 28 4 42 1 56 5 70 3 84 2 98 2 112 3 126 3 140 121 3 42 1 63 5 84 2 105 4 126 3 147 3 168 4 189 4 210 528 4 56 5 84 2 112 3 140 1 168 4 196 4 224 1 252 1 280 235 5 70 3 105 4 140 1 175 2 210 5 245 5 280 2 315 2 350 4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 1 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 5 40 2 45 2 50 4
7 1 14 2 21 3 28 4 35 5 42 1 49 1 56 5 63 5 70 314 2 28 4 42 1 56 5 70 3 84 2 98 2 112 3 126 3 140 121 3 42 1 63 5 84 2 105 4 126 3 147 3 168 4 189 4 210 528 4 56 5 84 2 112 3 140 1 168 4 196 4 224 1 252 1 280 235 5 70 3 105 4 140 1 175 2 210 5 245 5 280 2 315 2 350 4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Back to n = 5
1 1 2 2 3 3 4 4 5 5 6 1 7 5 8 5 9 5 10 32 2 4 4 6 1 8 5 10 3 12 2 14 2 16 3 18 3 20 13 3 6 1 9 5 12 2 15 4 18 3 21 3 24 4 27 4 30 54 4 8 5 12 2 16 3 20 1 24 4 28 4 32 1 36 1 40 25 5 10 3 15 4 20 1 25 2 30 5 35 1 40 2 45 2 50 4
7 5 14 2 21 3 28 4 35 1 42 5 49 5 56 1 63 1 70 314 2 28 4 42 5 56 1 70 3 84 2 98 2 112 3 126 3 140 521 3 42 5 63 1 84 2 105 4 126 3 147 3 168 4 189 4 210 128 4 56 1 84 2 112 3 140 5 168 4 196 4 224 5 252 5 280 235 1 70 3 105 4 140 5 175 2 210 1 245 1 280 2 315 2 350 4
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Cardinality of the set of of satisfactory colorings
Theorem
For n > 1, if there is one satisfactory n-coloring of the core,then there are as many satisfactory n-colorings of the positiveintegers as there are real numbers.
Proof Outline
Note that there are infinitely many numbers relatively prime tothe elements of the core. So there are infinitely many columnswhich are not determined by the coloring of the core. For eachx relatively prime with the core there are at least n! ways inwhich we can assign a satisfactory coloring to x ·Kn. It followsthat the set of colorings has cardinality |n!||N| which is |R|.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Cardinality of the set of of satisfactory colorings
Theorem
For n > 1, if there is one satisfactory n-coloring of the core,then there are as many satisfactory n-colorings of the positiveintegers as there are real numbers.
Proof Outline
Note that there are infinitely many numbers relatively prime tothe elements of the core. So there are infinitely many columnswhich are not determined by the coloring of the core. For eachx relatively prime with the core there are at least n! ways inwhich we can assign a satisfactory coloring to x ·Kn. It followsthat the set of colorings has cardinality |n!||N| which is |R|.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong representations
Theorem
Let n, k ∈ N. If p = nk + 1 is prime, and 1k, 2k, 3k, . . . , nk aredistinct modulo p, then the following produces a satisfactorycoloring: For m ∈ Z+, write it as m = apr where (a, p) = 1 andr ∈ N. Now define
c(m) = ak (mod p).
Proof.
¨
Let’s do an example first.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong representations
Theorem
Let n, k ∈ N. If p = nk + 1 is prime, and 1k, 2k, 3k, . . . , nk aredistinct modulo p, then the following produces a satisfactorycoloring: For m ∈ Z+, write it as m = apr where (a, p) = 1 andr ∈ N. Now define
c(m) = ak (mod p).
Proof.
¨
Let’s do an example first.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong representations
Theorem
Let n, k ∈ N. If p = nk + 1 is prime, and 1k, 2k, 3k, . . . , nk aredistinct modulo p, then the following produces a satisfactorycoloring: For m ∈ Z+, write it as m = apr where (a, p) = 1 andr ∈ N. Now define
c(m) = ak (mod p).
Proof.
¨
Let’s do an example first.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong representations
Theorem
Let n, k ∈ N. If p = nk + 1 is prime, and 1k, 2k, 3k, . . . , nk aredistinct modulo p, then the following produces a satisfactorycoloring: For m ∈ Z+, write it as m = apr where (a, p) = 1 andr ∈ N. Now define
c(m) = ak (mod p).
Proof.
_
Let’s do an example first.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
Consider n = 5 and k = 2. Then p = 5 · 2 + 1 = 11. We need tocheck if 1k, 2k, 3k, 4k and 5k are distinct modulo p. A quickcalculation gives us:
12 ≡ 1 (mod 11),
22 ≡ 4 (mod 11),
32 ≡ 9 (mod 11),
42 ≡ 5 (mod 11),
52 ≡ 3 (mod 11).
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
Consider n = 5 and k = 2. Then p = 5 · 2 + 1 = 11. We need tocheck if 1k, 2k, 3k, 4k and 5k are distinct modulo p. A quickcalculation gives us:
12 ≡ 1 (mod 11),
22 ≡ 4 (mod 11),
32 ≡ 9 (mod 11),
42 ≡ 5 (mod 11),
52 ≡ 3 (mod 11).
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
To see how we construct the colorings consider the followingexamples. Since
11 = 11 · 1,
c(11) = c(1) = 12 (mod 11) = 1
22 = 11 · 2,
c(22) = c(2) = 22 (mod 11) = 4
24 = 110 · 24,
c(24) = c(24) = 242 (mod 11) = 2
100 = 110 · 100,
c(100) = c(100) = 1002 (mod 11) = 3
3630 = 112 · 30,
c(3630) = c(112 · 30) = 302 (mod 11) = 9
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
To see how we construct the colorings consider the followingexamples. Since
11 = 11 · 1, c(11) = c(1) = 12 (mod 11) = 122 = 11 · 2,
c(22) = c(2) = 22 (mod 11) = 4
24 = 110 · 24,
c(24) = c(24) = 242 (mod 11) = 2
100 = 110 · 100,
c(100) = c(100) = 1002 (mod 11) = 3
3630 = 112 · 30,
c(3630) = c(112 · 30) = 302 (mod 11) = 9
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
To see how we construct the colorings consider the followingexamples. Since
11 = 11 · 1, c(11) = c(1) = 12 (mod 11) = 122 = 11 · 2, c(22) = c(2) = 22 (mod 11) = 424 = 110 · 24,
c(24) = c(24) = 242 (mod 11) = 2
100 = 110 · 100,
c(100) = c(100) = 1002 (mod 11) = 3
3630 = 112 · 30,
c(3630) = c(112 · 30) = 302 (mod 11) = 9
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
To see how we construct the colorings consider the followingexamples. Since
11 = 11 · 1, c(11) = c(1) = 12 (mod 11) = 122 = 11 · 2, c(22) = c(2) = 22 (mod 11) = 424 = 110 · 24, c(24) = c(24) = 242 (mod 11) = 2100 = 110 · 100,
c(100) = c(100) = 1002 (mod 11) = 3
3630 = 112 · 30,
c(3630) = c(112 · 30) = 302 (mod 11) = 9
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
To see how we construct the colorings consider the followingexamples. Since
11 = 11 · 1, c(11) = c(1) = 12 (mod 11) = 122 = 11 · 2, c(22) = c(2) = 22 (mod 11) = 424 = 110 · 24, c(24) = c(24) = 242 (mod 11) = 2100 = 110 · 100, c(100) = c(100) = 1002 (mod 11) = 33630 = 112 · 30,
c(3630) = c(112 · 30) = 302 (mod 11) = 9
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representations
To see how we construct the colorings consider the followingexamples. Since
11 = 11 · 1, c(11) = c(1) = 12 (mod 11) = 122 = 11 · 2, c(22) = c(2) = 22 (mod 11) = 424 = 110 · 24, c(24) = c(24) = 242 (mod 11) = 2100 = 110 · 100, c(100) = c(100) = 1002 (mod 11) = 33630 = 112 · 30, c(3630) = c(112 · 30) = 302 (mod 11) = 9
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
This coloring gives us the following table:
1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
This coloring gives us the following table:1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
Note the following:
1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
Note the following:1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
Note the following:1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
Note the following:1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
Note the following:1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
p = 11
Note the following:1 1 2 4 3 9 4 5 5 3 6 3 7 5 8 9 9 4 10 12 4 4 5 6 3 8 9 10 1 12 1 14 9 16 3 18 5 20 43 9 6 3 9 4 12 1 15 5 18 5 21 1 24 4 27 3 30 94 5 8 9 12 1 16 3 20 4 24 4 28 3 32 1 36 9 40 55 3 10 1 15 5 20 4 25 9 30 9 35 4 40 5 45 1 50 3
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Since p = nk + 1 is prime, Z∗p is cyclic.
Let g be a generator of
Z∗p. Then gk, (gk)2, . . . , (gk)n are distinct modulo p. In fact, any(non-zero) kth power is one of them. In other words, there aren pairwise incongruent kth power residues modulo p. It followsthat, for any b with (b, p) = 1, bk = jk (mod p) for somej ∈ {1, 2, . . . , n}.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Since p = nk + 1 is prime, Z∗p is cyclic. Let g be a generator of
Z∗p.
Then gk, (gk)2, . . . , (gk)n are distinct modulo p. In fact, any(non-zero) kth power is one of them. In other words, there aren pairwise incongruent kth power residues modulo p. It followsthat, for any b with (b, p) = 1, bk = jk (mod p) for somej ∈ {1, 2, . . . , n}.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Since p = nk + 1 is prime, Z∗p is cyclic. Let g be a generator of
Z∗p. Then gk, (gk)2, . . . , (gk)n are distinct modulo p.
In fact, any(non-zero) kth power is one of them. In other words, there aren pairwise incongruent kth power residues modulo p. It followsthat, for any b with (b, p) = 1, bk = jk (mod p) for somej ∈ {1, 2, . . . , n}.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Since p = nk + 1 is prime, Z∗p is cyclic. Let g be a generator of
Z∗p. Then gk, (gk)2, . . . , (gk)n are distinct modulo p. In fact, any(non-zero) kth power is one of them. In other words, there aren pairwise incongruent kth power residues modulo p.
It followsthat, for any b with (b, p) = 1, bk = jk (mod p) for somej ∈ {1, 2, . . . , n}.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Since p = nk + 1 is prime, Z∗p is cyclic. Let g be a generator of
Z∗p. Then gk, (gk)2, . . . , (gk)n are distinct modulo p. In fact, any(non-zero) kth power is one of them. In other words, there aren pairwise incongruent kth power residues modulo p. It followsthat, for any b with (b, p) = 1, bk = jk (mod p) for somej ∈ {1, 2, . . . , n}.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Now let d, e ∈ {1, 2, . . . , n}.
We need to argue thatc(dm) = c(em) if and only if d = e. To see this considerdm = adpr. Thus c(dm) = (ad)k (mod p). Similarly, we havethat c(em) = (ae)k (mod p). But c(dm) = c(em) implies that(ad)k ≡ (ae)k (mod p) and therefore dk ≡ ek (mod p), so d = eby assumption.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Now let d, e ∈ {1, 2, . . . , n}. We need to argue thatc(dm) = c(em) if and only if d = e.
To see this considerdm = adpr. Thus c(dm) = (ad)k (mod p). Similarly, we havethat c(em) = (ae)k (mod p). But c(dm) = c(em) implies that(ad)k ≡ (ae)k (mod p) and therefore dk ≡ ek (mod p), so d = eby assumption.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Now let d, e ∈ {1, 2, . . . , n}. We need to argue thatc(dm) = c(em) if and only if d = e. To see this considerdm = adpr.
Thus c(dm) = (ad)k (mod p). Similarly, we havethat c(em) = (ae)k (mod p). But c(dm) = c(em) implies that(ad)k ≡ (ae)k (mod p) and therefore dk ≡ ek (mod p), so d = eby assumption.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Now let d, e ∈ {1, 2, . . . , n}. We need to argue thatc(dm) = c(em) if and only if d = e. To see this considerdm = adpr. Thus c(dm) = (ad)k (mod p).
Similarly, we havethat c(em) = (ae)k (mod p). But c(dm) = c(em) implies that(ad)k ≡ (ae)k (mod p) and therefore dk ≡ ek (mod p), so d = eby assumption.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Now let d, e ∈ {1, 2, . . . , n}. We need to argue thatc(dm) = c(em) if and only if d = e. To see this considerdm = adpr. Thus c(dm) = (ad)k (mod p). Similarly, we havethat c(em) = (ae)k (mod p).
But c(dm) = c(em) implies that(ad)k ≡ (ae)k (mod p) and therefore dk ≡ ek (mod p), so d = eby assumption.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Proof of Theorem
Proof.
Now let d, e ∈ {1, 2, . . . , n}. We need to argue thatc(dm) = c(em) if and only if d = e. To see this considerdm = adpr. Thus c(dm) = (ad)k (mod p). Similarly, we havethat c(em) = (ae)k (mod p). But c(dm) = c(em) implies that(ad)k ≡ (ae)k (mod p) and therefore dk ≡ ek (mod p), so d = eby assumption.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Strong Representatives
When p = nk + 1 satisfies the hypothesis of the theorem, thenwe say p is a strong representative.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Where do we stand?
There are two cases when this assumption holds automatically.
1 When p = n+ 1 is prime.
2 When p = 2n+ 1 is prime.
We call these p trivial strong representatives.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Where do we stand?
There are two cases when this assumption holds automatically.
1 When p = n+ 1 is prime.
2 When p = 2n+ 1 is prime.
We call these p trivial strong representatives.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Where do we stand?
There are two cases when this assumption holds automatically.
1 When p = n+ 1 is prime.
2 When p = 2n+ 1 is prime.
We call these p trivial strong representatives.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Where do we stand?
There are two cases when this assumption holds automatically.
1 When p = n+ 1 is prime.
2 When p = 2n+ 1 is prime.
We call these p trivial strong representatives.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1
1 2
2
1 3
3
2 7
4
1 5
5
2 11
6
1 7
8
2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2
1 3
3
2 7
4
1 5
5
2 11
6
1 7
8
2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3
2 7
4
1 5
5
2 11
6
1 7
8
2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4
1 5
5
2 11
6
1 7
8
2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5
2 11
6
1 7
8
2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6
1 7
8
2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8
2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9
2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10
1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11
2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12
1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14
2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15
2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16
1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18
1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20
2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21
2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22
1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23
2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26
2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28
1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28 1 29
29
2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28 1 29
29 2 59
30
1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28 1 29
29 2 59
30 1 31
33
2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28 1 29
29 2 59
30 1 31
33 2 67
35
2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28 1 29
29 2 59
30 1 31
33 2 67
35 2 71
36
1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28 1 29
29 2 59
30 1 31
33 2 67
35 2 71
36 1 37
39
2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 or 2n + 1 is prime
The following lists the values of n ≤ 39 for which a trivialstrong representative exists.
n k p
1 1 2
2 1 3
3 2 7
4 1 5
5 2 11
6 1 7
8 2 17
9 2 19
10 1 11
n k p
11 2 23
12 1 13
14 2 29
15 2 31
16 1 17
18 1 19
20 2 41
21 2 43
22 1 23
n k p
23 2 47
26 2 53
28 1 29
29 2 59
30 1 31
33 2 67
35 2 71
36 1 37
39 2 79
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime.
However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94,
thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime,
and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
What about 7?
n = 7 is the first case in which neither n+ 1 = 8 nor2n+ 1 = 15 is prime. However, if we take k = 94, thenp = 7 · 94 + 1 = 659 is prime, and
194 ≡ 1 (mod 659),
294 ≡ 307 (mod 659),
394 ≡ 144 (mod 659),
494 ≡ 12 (mod 659),
594 ≡ 389 (mod 659),
694 ≡ 55 (mod 659),
794 ≡ 410 (mod 659),
are all different, so 659 is a strong representative of order 7.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7
94 659
13
198364 2578733
17
2859480 48611161
19
533410 10134791
24
56610508 1358652193
25
1170546910 29263672751
27
6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13
198364 2578733
17
2859480 48611161
19
533410 10134791
24
56610508 1358652193
25
1170546910 29263672751
27
6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17
2859480 48611161
19
533410 10134791
24
56610508 1358652193
25
1170546910 29263672751
27
6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17 2859480 48611161
19
533410 10134791
24
56610508 1358652193
25
1170546910 29263672751
27
6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17 2859480 48611161
19 533410 10134791
24
56610508 1358652193
25
1170546910 29263672751
27
6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17 2859480 48611161
19 533410 10134791
24 56610508 1358652193
25
1170546910 29263672751
27
6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17 2859480 48611161
19 533410 10134791
24 56610508 1358652193
25 1170546910 29263672751
27
6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17 2859480 48611161
19 533410 10134791
24 56610508 1358652193
25 1170546910 29263672751
27 6700156678 180904230307
31
27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17 2859480 48611161
19 533410 10134791
24 56610508 1358652193
25 1170546910 29263672751
27 6700156678 180904230307
31 27184496610 842719394911
32
162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
When n + 1 and 2n + 1 are not prime
The following table lists the values of n ≤ 32 that do not admita trivial strong representative and the smallest strongrepresentative of order n.
n k p
7 94 659
13 198364 2578733
17 2859480 48611161
19 533410 10134791
24 56610508 1358652193
25 1170546910 29263672751
27 6700156678 180904230307
31 27184496610 842719394911
32 162802814486 5209690063553
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The search for strong representatives
To find the strong representative for n = 32 required thefollowing:
1 Buying a new computer,
2 writing a Maple procedure,
3 letting the procedure run for 30 days and nights.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The search for strong representatives
To find the strong representative for n = 32 required thefollowing:
1 Buying a new computer,
2 writing a Maple procedure,
3 letting the procedure run for 30 days and nights.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The search for strong representatives
To find the strong representative for n = 32 required thefollowing:
1 Buying a new computer,
2 writing a Maple procedure,
3 letting the procedure run for 30 days and nights.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The search for strong representatives
To find the strong representative for n = 32 required thefollowing:
1 Buying a new computer,
2 writing a Maple procedure,
3 letting the procedure run for 30 days and nights.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
A new approach
Since finding (let alone guaranteeing the existence of) nontrivialstrong representatives is difficult, we need a new way to color.This is where what we call partial homomorphisms come intoplay.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Partial homomorphisms
Definition
A partial homomorphism of order n is a bijective map,
h : {1, 2, . . . , n} → Zn,
such thath(a · b) = h(a) + h(b) (mod n),
whenever a, b ∈ {1, 2, . . . , n} and a · b ≤ n.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
Example
Consider n = 5. Let k = 2, so p = 11. Take g = 2 (since 2 is agenerator of Z∗11). Then
g, g2, g3, g4, g5, g6, g7, g8, g9, g10
are all distinct (mod 11). In particular,
g, g2, g3, g4, g5, g6, g7, g8, g9, g10
are all distinct. But these are precisely the kth powers(mod 11).
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
Example
Consider n = 5. Let k = 2, so p = 11. Take g = 2 (since 2 is agenerator of Z∗11). Then
g, g2, g3, g4, g5, g6, g7, g8, g9, g10
are all distinct (mod 11). In particular,
g, g2, g3, g4, g5, g6, g7, g8, g9, g10
are all distinct. But these are precisely the kth powers(mod 11).
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1
0
2
1
3
3
4
2
5
4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02
1
3
3
4
2
5
4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13
3
4
2
5
4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13
3
4 25
4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25
4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2
= (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1
= (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4
= (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2
= (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6
= (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3
= (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8
= (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4
= (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10
= (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5
= (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An Example
x h(x)
1 02 13 34 25 4
g2 = (22)1 = (22)h(2)
g4 = (22)2 = (22)h(4)
g6 = (22)3 = (22)h(3)
g8 = (22)4 = (22)h(5)
g10 = (22)5 = (22)h(1)
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Partial homomorphisms
Note that if we have a coloring from a strong representative, wecan get a partial homomorphism, h, as a kind of logarithm. Tosee this note that every element of Z∗p is gt for some g. So each
ak = gkt. In this case, we are taking h(a) = t.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Partial homomorphisms
The point is we do not have to have a strong representative todo this. Instead, we can try to construct a partialhomomorphism and use it to get a coloring.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The process
Now to construct a satisfactory coloring we do the following:
1 Construct a partial homomorphism.
2 Color the first column by the partial homomorphism fromstep 1.
3 Define the coloring of the core byc(ai) = h(ai) = h(a) + h(i) (mod n) for a ∈ {1, 2, . . . , n}and i > 1.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The process
Now to construct a satisfactory coloring we do the following:
1 Construct a partial homomorphism.
2 Color the first column by the partial homomorphism fromstep 1.
3 Define the coloring of the core byc(ai) = h(ai) = h(a) + h(i) (mod n) for a ∈ {1, 2, . . . , n}and i > 1.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
The process
Now to construct a satisfactory coloring we do the following:
1 Construct a partial homomorphism.
2 Color the first column by the partial homomorphism fromstep 1.
3 Define the coloring of the core byc(ai) = h(ai) = h(a) + h(i) (mod n) for a ∈ {1, 2, . . . , n}and i > 1.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 1: Construct a partial homomorphism
1 02 13 34 25 56 47 6
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 2: Color the first column:1 0 2 3 4 5 6 72 1 4 6 8 10 12 143 3 6 9 12 15 18 214 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 10 12 143 3 6 4 9 12 15 18 214 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 12 143 3 6 4 9 12 15 18 214 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 6 12 143 3 6 4 9 12 15 18 214 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 6 12 5 143 3 6 4 9 12 15 18 214 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 6 12 5 14 03 3 6 4 9 12 15 18 214 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 6 12 5 14 03 3 6 4 9 6 12 15 18 214 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 6 12 5 14 03 3 6 4 9 6 12 5 15 1 18 0 21 24 2 8 12 16 20 24 285 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 6 12 5 14 03 3 6 4 9 6 12 5 15 1 18 0 21 24 2 8 3 12 5 16 4 20 0 24 6 28 15 5 10 15 20 25 30 356 4 12 18 24 30 36 427 6 14 21 28 35 42 49
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Step 3: Color the remaining columns:1 0 2 1 3 3 4 2 5 5 6 4 7 62 1 4 2 6 4 8 3 10 6 12 5 14 03 3 6 4 9 6 12 5 15 1 18 0 21 24 2 8 3 12 5 16 4 20 0 24 6 28 15 5 10 6 15 1 20 0 25 3 30 2 35 46 4 12 5 18 0 24 6 30 2 36 1 42 37 6 14 0 21 2 28 1 35 4 42 3 49 5
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
A second partial homomorphism.
12345678910
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02345678910
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 1345678910
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 134 25678910
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 134 25678 3910
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 13 44 25678 3910
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 13 44 256 578 3910
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 13 44 256 578 39 810
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 13 44 25 66 578 39 810
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 13 44 25 66 578 39 810 7
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
1 02 13 44 25 66 57 98 39 810 7
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1
0 0
2
1 1
3
5 5
4
22
5
8 8
6
6 6
7
11 11
8
33
9
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16
44
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1
0
02
1 1
3
5 5
4
22
5
8 8
6
6 6
7
11 11
8
33
9
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16
44
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2
1
13
5 5
4
22
5
8 8
6
6 6
7
11 11
8
33
9
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16
44
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3
5 5
4
2
25
8 8
6
6 6
7
11 11
8
33
9
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16
44
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3
5 5
4 2
2
5
8 8
6
6 6
7
11 11
8
3
39
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16
44
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3
5 5
4 2
2
5
8 8
6
6 6
7
11 11
8 3
3
9
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16
4
417
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3
5
54 2
2
5
8 8
6
6 6
7
11 11
8 3
3
9
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5
8 8
6
6
67
11 11
8 3
3
9
10 10
10
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5
8 8
6 6
6
7
11 11
8 3
3
9
10
1010
9 9
11
14 14
12
7 7
13
15 15
14
12 12
15
13 13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5
8 8
6 6
6
7
11 11
8 3
3
9 10
10
10
9 9
11
14 14
12
7
713
15 15
14
12 12
15
13 13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5
8
86 6
6
7
11 11
8 3
3
9 10
10
10
9 9
11
14 14
12 7
7
13
15 15
14
12 12
15
13 13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5 8
8
6 6
6
7
11 11
8 3
3
9 10
10
10
9
911
14 14
12 7
7
13
15 15
14
12 12
15
13 13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5 8
8
6 6
6
7
11 11
8 3
3
9 10
10
10 9
9
11
14 14
12 7
7
13
15 15
14
12 12
15
13
1316 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5 8
8
6 6
6
7
11
118 3
3
9 10
10
10 9
9
11
14 14
12 7
7
13
15 15
14
12 12
15 13
13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5 8
8
6 6
6
7 11
11
8 3
3
9 10
10
10 9
9
11
14 14
12 7
7
13
15 15
14
12
1215 13
13
16 4
4
17
16 16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5 8
8
6 6
6
7 11
11
8 3
3
9 10
10
10 9
9
11
14
1412 7
7
13
15
1514 12
12
15 13
13
16 4
4
17
16
16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
An example
Let’s construct one more partial homomorphism.
1 0
0
2 1
1
3 5
5
4 2
2
5 8
8
6 6
6
7 11
11
8 3
3
9 10
10
10 9
9
11 14
14
12 7
7
13 15
15
14 12
12
15 13
13
16 4
4
17 16
16
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Where do we stand?
As you have seen, constructing partial homomorphisms is aboutas hard as an easy Sudoku puzzle. However, guaranteeing theirexistence is not trivial. Currently we have partialhomomorphisms for n ≤ 54.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Although the problem is easy to state, it seems to be inherentlyhard. As evidence of this consider the following conjecture ofRonald Graham from 1970.
Conjecture
(Graham, 1970)Let 0 < a1 < a2 < . . . < an be integers. Then
maxi,j
{ai
(ai, aj)
}≥ n.
An affirmative answer was given by Balasubramanian andSoundararajan in 1996.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Although the problem is easy to state, it seems to be inherentlyhard. As evidence of this consider the following conjecture ofRonald Graham from 1970.
Conjecture
(Graham, 1970)Let 0 < a1 < a2 < . . . < an be integers. Then
maxi,j
{ai
(ai, aj)
}≥ n.
An affirmative answer was given by Balasubramanian andSoundararajan in 1996.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Theorem
If satisfactory colorings exist for all n, then theBalasubramanian-Soundararajan Theorem holds for all n. Infact, if there is a satisfactory coloring with m− 1 colors, thenthe Balasubramanian-Soundararajan Theorem holds for n = m.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Proof.
Suppose that there exists some m and a set B = {b1, . . . , bm}
with 0 < b1 < . . . < bm such that maxi,j
{bi
(bi, bj)
}< m.
Suppose
i 6= j and let M = (bi, bj). Let biM = ai and
bjM = aj . By our
hypothesis we have that ai < m, aj < m, and ai 6= aj . Sincebi = aiM, bj = ajM, in any satisfactory coloring with at leastm− 1 colors, we must have that bi is colored differently than bj .But |B| = m so we need, at least, m colors.
Note that the Balasubramanian-Soundararajan Theorem is avery particular case of our coloring problem.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Proof.
Suppose that there exists some m and a set B = {b1, . . . , bm}
with 0 < b1 < . . . < bm such that maxi,j
{bi
(bi, bj)
}< m. Suppose
i 6= j and let M = (bi, bj).
Let biM = ai and
bjM = aj . By our
hypothesis we have that ai < m, aj < m, and ai 6= aj . Sincebi = aiM, bj = ajM, in any satisfactory coloring with at leastm− 1 colors, we must have that bi is colored differently than bj .But |B| = m so we need, at least, m colors.
Note that the Balasubramanian-Soundararajan Theorem is avery particular case of our coloring problem.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Proof.
Suppose that there exists some m and a set B = {b1, . . . , bm}
with 0 < b1 < . . . < bm such that maxi,j
{bi
(bi, bj)
}< m. Suppose
i 6= j and let M = (bi, bj). Let biM = ai and
bjM = aj .
By ourhypothesis we have that ai < m, aj < m, and ai 6= aj . Sincebi = aiM, bj = ajM, in any satisfactory coloring with at leastm− 1 colors, we must have that bi is colored differently than bj .But |B| = m so we need, at least, m colors.
Note that the Balasubramanian-Soundararajan Theorem is avery particular case of our coloring problem.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Proof.
Suppose that there exists some m and a set B = {b1, . . . , bm}
with 0 < b1 < . . . < bm such that maxi,j
{bi
(bi, bj)
}< m. Suppose
i 6= j and let M = (bi, bj). Let biM = ai and
bjM = aj . By our
hypothesis we have that ai < m, aj < m, and ai 6= aj .
Sincebi = aiM, bj = ajM, in any satisfactory coloring with at leastm− 1 colors, we must have that bi is colored differently than bj .But |B| = m so we need, at least, m colors.
Note that the Balasubramanian-Soundararajan Theorem is avery particular case of our coloring problem.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Proof.
Suppose that there exists some m and a set B = {b1, . . . , bm}
with 0 < b1 < . . . < bm such that maxi,j
{bi
(bi, bj)
}< m. Suppose
i 6= j and let M = (bi, bj). Let biM = ai and
bjM = aj . By our
hypothesis we have that ai < m, aj < m, and ai 6= aj . Sincebi = aiM, bj = ajM, in any satisfactory coloring with at leastm− 1 colors, we must have that bi is colored differently than bj .
But |B| = m so we need, at least, m colors.
Note that the Balasubramanian-Soundararajan Theorem is avery particular case of our coloring problem.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Evidence of the underlying difficulty
Proof.
Suppose that there exists some m and a set B = {b1, . . . , bm}
with 0 < b1 < . . . < bm such that maxi,j
{bi
(bi, bj)
}< m. Suppose
i 6= j and let M = (bi, bj). Let biM = ai and
bjM = aj . By our
hypothesis we have that ai < m, aj < m, and ai 6= aj . Sincebi = aiM, bj = ajM, in any satisfactory coloring with at leastm− 1 colors, we must have that bi is colored differently than bj .But |B| = m so we need, at least, m colors.
Note that the Balasubramanian-Soundararajan Theorem is avery particular case of our coloring problem.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Final remarks
In closing, we can guarantee a satisfactory n-coloring whenever:
1 n+ 1 is prime,
2 2n+ 1 is prime,
3 nk + 1 is prime and 1k, 2k, . . . , nk are distinct(mod nk + 1),
4 a partial homomorphism of order n exists.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Final remarks
In closing, we can guarantee a satisfactory n-coloring whenever:
1 n+ 1 is prime,
2 2n+ 1 is prime,
3 nk + 1 is prime and 1k, 2k, . . . , nk are distinct(mod nk + 1),
4 a partial homomorphism of order n exists.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Final remarks
In closing, we can guarantee a satisfactory n-coloring whenever:
1 n+ 1 is prime,
2 2n+ 1 is prime,
3 nk + 1 is prime and 1k, 2k, . . . , nk are distinct(mod nk + 1),
4 a partial homomorphism of order n exists.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Final remarks
In closing, we can guarantee a satisfactory n-coloring whenever:
1 n+ 1 is prime,
2 2n+ 1 is prime,
3 nk + 1 is prime and 1k, 2k, . . . , nk are distinct(mod nk + 1),
4 a partial homomorphism of order n exists.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
Final remarks
In closing, we can guarantee a satisfactory n-coloring whenever:
1 n+ 1 is prime,
2 2n+ 1 is prime,
3 nk + 1 is prime and 1k, 2k, . . . , nk are distinct(mod nk + 1),
4 a partial homomorphism of order n exists.
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
I would like to thank you all for coming. Before takingquestions I would like to say the following:
May the math be with you!Do you have any questions?
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
I would like to thank you all for coming. Before takingquestions I would like to say the following:
May the math be with you!
Do you have any questions?
Thomas Chartier Coloring Problems
IntroductionThe Chromatic Number of the Plane
The Problem
I would like to thank you all for coming. Before takingquestions I would like to say the following:
May the math be with you!Do you have any questions?
Thomas Chartier Coloring Problems