Introduction The Chromatic Number of the Plane The Problem€¦ · Introduction The Chromatic...

IntroductionThe Chromatic Number of the Plane

The Problem

Coloring Problems

Thomas Chartier

October 13, 2011

Thomas Chartier Coloring Problems

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.

The Problem

Table of contents

1 Introduction

2 The Chromatic Number of the Plane

3 The Problem

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.

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?

The Problem

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?

The Problem

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.

The Problem

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.

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.

The Problem

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.

The Problem

Proposition

χ ≥ 4.

The Problem

Proof.

The Problem

Proof.

The Problem

Proof.

The Problem

Proof.

The Problem

Proof.

The Problem

Proof.

The Problem

Proof.

The Problem

Proof.

The Problem

Proof.

The Problem

Proposition

χ ≤ 7.

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

The Problem

Proof.

22.1≈ 0.95

2.1≈ 1.26

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.

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.

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.

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

orange

6 7 8 9 102 red 4

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.

The Problem

Notation continued

1 magenta 2

orange

6 7 8 9 102 red 4

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

The Problem

Notation continued

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

The Problem

Notation continued

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

The Problem

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:

6 7 8 9 102

6 8 10 12 14 16 18 203

6 9 12 15 18 21 24 27 304

8 12 16 20 24 28 32 36 405

10 15 20 25 30 35 40 45 50

The Problem

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:

6 7 8 9 102 2 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

The Problem

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:

6 7 8 9 102 2 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

The Problem

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

The Problem

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

The Problem

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

The Problem

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

The Problem

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

The Problem

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

9 10 32 2 4 4 6 1 8

10 3 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 3 15 20 25 30 35 40 45 50

The Problem

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

The Problem

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

The Problem

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

The Problem

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

The Problem

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 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

The Problem

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

The Problem

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

The Problem

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.

The Problem

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

The Problem

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

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.

The Problem

Back to 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

7 1 14

14 2 28

21 3 42

28 4 56

35 5 70

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

14 2 28

21 3 42

28 4 56

35 5 70

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

14 2 28 4 42

21 3 42 1 63

28 4 56 5 84

35 5 70 3 105

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

14 2 28 4 42 1 56

21 3 42 1 63 5 84

28 4 56 5 84 2 112

35 5 70 3 105 4 140

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

14 2 28 4 42 1 56 5 70

21 3 42 1 63 5 84 2 105

28 4 56 5 84 2 112 3 140

35 5 70 3 105 4 140 1 175

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

56 5 63 5 70 314 2 28 4 42 1 56 5 70 3 84 2 98

112 3 126 3 140 121 3 42 1 63 5 84 2 105 4 126 3 147

168 4 189 4 210 528 4 56 5 84 2 112 3 140 1 168 4 196

224 1 252 1 280 235 5 70 3 105 4 140 1 175 2 210 5 245

280 2 315 2 350 4

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

56 5 63 5 70 314 2 28 4 42 1 56 5 70 3 84 2 98

112 3 126 3 140 121 3 42 1 63 5 84 2 105 4 126 3 147

168 4 189 4 210 528 4 56 5 84 2 112 3 140 1 168 4 196

224 1 252 1 280 235 5 70 3 105 4 140 1 175 2 210 5 245

280 2 315 2 350 4

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

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

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

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|.

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|.

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.

The Problem

Theorem

c(m) = ak (mod p).

Proof.

The Problem

Theorem

c(m) = ak (mod p).

Proof.

The Problem

Theorem

c(m) = ak (mod p).

Proof.

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).

The Problem

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).

The Problem

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

The Problem

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

The Problem

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

The Problem

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

The Problem

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

The Problem

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

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

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

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

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

The Problem

p = 11

The Problem

p = 11

The Problem

p = 11

The Problem

p = 11

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}.

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}.

The Problem

Proof of Theorem

Proof.

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}.

The Problem

Proof of Theorem

Proof.

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}.

The Problem

Proof of Theorem

Proof.

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}.

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.

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.

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.

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.

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.

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.

The Problem

Strong Representatives

When p = nk + 1 satisfies the hypothesis of the theorem, thenwe say p is a strong representative.

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.

The Problem

Where do we stand?

The Problem

Where do we stand?

The Problem

Where do we stand?

The Problem

When n + 1 or 2n + 1 is prime

The following lists the values of n ≤ 39 for which a trivialstrong representative exists.

The Problem

5 2 11

The Problem

5 2 11

The Problem

5 2 11

8 2 17

The Problem

5 2 11

8 2 17

9 2 19

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

12 1 13

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

12 1 13

14 2 29

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

12 1 13

14 2 29

15 2 31

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

12 1 13

14 2 29

15 2 31

16 1 17

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

12 1 13

14 2 29

15 2 31

16 1 17

18 1 19

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

12 1 13

14 2 29

15 2 31

16 1 17

18 1 19

20 2 41

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

11 2 23

12 1 13

14 2 29

15 2 31

16 1 17

18 1 19

20 2 41

21 2 43

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

26 2 53

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

26 2 53

28 1 29

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

26 2 53

28 1 29

29 2 59

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

26 2 53

28 1 29

29 2 59

30 1 31

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

26 2 53

28 1 29

29 2 59

30 1 31

33 2 67

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

26 2 53

28 1 29

29 2 59

30 1 31

33 2 67

35 2 71

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

23 2 47

26 2 53

28 1 29

29 2 59

30 1 31

33 2 67

35 2 71

36 1 37

The Problem

5 2 11

8 2 17

9 2 19

10 1 11

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

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

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.

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),

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,

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),

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),

The Problem

What about 7?

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),

The Problem

What about 7?

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),

The Problem

What about 7?

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),

The Problem

What about 7?

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),

The Problem

What about 7?

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),

The Problem

What about 7?

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),

The Problem

What about 7?

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),

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.

94 659

198364 2578733

2859480 48611161

533410 10134791

56610508 1358652193

1170546910 29263672751

6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

198364 2578733

2859480 48611161

533410 10134791

56610508 1358652193

1170546910 29263672751

6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

13 198364 2578733

2859480 48611161

533410 10134791

56610508 1358652193

1170546910 29263672751

6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

13 198364 2578733

17 2859480 48611161

533410 10134791

56610508 1358652193

1170546910 29263672751

6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

13 198364 2578733

17 2859480 48611161

19 533410 10134791

56610508 1358652193

1170546910 29263672751

6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

13 198364 2578733

17 2859480 48611161

19 533410 10134791

24 56610508 1358652193

1170546910 29263672751

6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

13 198364 2578733

17 2859480 48611161

19 533410 10134791

24 56610508 1358652193

25 1170546910 29263672751

6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

13 198364 2578733

17 2859480 48611161

19 533410 10134791

24 56610508 1358652193

25 1170546910 29263672751

27 6700156678 180904230307

27184496610 842719394911

162802814486 5209690063553

The Problem

7 94 659

13 198364 2578733

17 2859480 48611161

19 533410 10134791

24 56610508 1358652193

25 1170546910 29263672751

27 6700156678 180904230307

31 27184496610 842719394911

162802814486 5209690063553

The Problem

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

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.

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.

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.

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).

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).

The Problem

An Example

x h(x)

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)

The Problem

An Example

x h(x)

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)

The Problem

An Example

x h(x)

1 02 13

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)

The Problem

An Example

x h(x)

1 02 13

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)

The Problem

An Example

x h(x)

1 02 13 34 25

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)

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)

The Problem

An Example

x h(x)

1 02 13 34 25 4

= (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)

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)

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)

The Problem

An Example

x h(x)

1 02 13 34 25 4

g2 = (22)1 = (22)h(2)

= (22)2 = (22)h(4)

g6 = (22)3 = (22)h(3)

g8 = (22)4 = (22)h(5)

g10 = (22)5 = (22)h(1)

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)

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)

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)

= (22)3 = (22)h(3)

g8 = (22)4 = (22)h(5)

g10 = (22)5 = (22)h(1)

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)

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)

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)

= (22)4 = (22)h(5)

g10 = (22)5 = (22)h(1)

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)

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)

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)

= (22)5 = (22)h(1)

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)

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)

The Problem

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.

The Problem

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.

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.

The Problem

The process

The Problem

The process

The Problem

An example

Step 1: Construct a partial homomorphism

1 02 13 34 25 56 47 6

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

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

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

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

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

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

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

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

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

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

The Problem

An example

A second partial homomorphism.

12345678910

The Problem

An example

1 02345678910

The Problem

An example

1 02 1345678910

The Problem

An example

1 02 134 25678910

The Problem

An example

1 02 134 25678 3910

The Problem

An example

1 02 13 44 25678 3910

The Problem

An example

1 02 13 44 256 578 3910

The Problem

An example

1 02 13 44 256 578 39 810

The Problem

An example

1 02 13 44 25 66 578 39 810

The Problem

An example

1 02 13 44 25 66 578 39 810 7

The Problem

An example

1 02 13 44 25 66 57 98 39 810 7

The Problem

An example

Let’s construct one more partial homomorphism.

The Problem

An example

The Problem

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The Problem

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An example

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1316 4

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1215 13

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1412 7

1514 12

The Problem

An example

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.

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, aj)

}≥ n.

An affirmative answer was given by Balasubramanian andSoundararajan in 1996.

The Problem

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, aj)

}≥ n.

An affirmative answer was given by Balasubramanian andSoundararajan in 1996.

The Problem

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.

The Problem

Proof.

Suppose that there exists some m and a set B = {b1, . . . , bm}

with 0 < b1 < . . . < bm such that maxi,j

(bi, bj)

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.

The Problem

Proof.

(bi, bj)

}< m. Suppose

i 6= j and let M = (bi, bj).

Let biM = ai and

bjM = aj . By our

The Problem

Proof.

(bi, bj)

}< m. Suppose

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.

The Problem

Proof.

(bi, bj)

}< m. Suppose

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.

The Problem

Proof.

(bi, bj)

}< m. Suppose

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.

The Problem

Proof.

(bi, bj)

}< m. Suppose

bjM = aj . By our

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.

The Problem

Final remarks

1 n+ 1 is prime,

2 2n+ 1 is prime,

The Problem

Final remarks

1 n+ 1 is prime,

2 2n+ 1 is prime,

The Problem

Final remarks

1 n+ 1 is prime,

2 2n+ 1 is prime,

The Problem

Final remarks

1 n+ 1 is prime,

2 2n+ 1 is prime,

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?

The Problem

May the math be with you!

Do you have any questions?

The Problem

May the math be with you!Do you have any questions?

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