The Beauty of Combinatorics November 1, 2012 () The Beauty of - - PowerPoint PPT Presentation

The Beauty of Combinatorics

November 1, 2012

() The Beauty of Combinatorics November 1, 2012 1 / 19

Moshe Rosenfeld Institute of Technology University of Washington and Vietnam National University Hanoi University of Science

() The Beauty of Combinatorics November 1, 2012 1 / 19

Moshe Rosenfeld Institute of Technology University of Washington and Vietnam National University Hanoi University of Science East China Normal University Shanghai, Nov. 1, 2012

() The Beauty of Combinatorics November 1, 2012 1 / 19

Introduction

1

What is mathematics?

() The Beauty of Combinatorics November 1, 2012 2 / 19

Introduction

1

What is mathematics?

2

"Mathematics is the study of numbers, shapes and patterns."

() The Beauty of Combinatorics November 1, 2012 2 / 19

Introduction

1

What is mathematics?

2

"Mathematics is the study of numbers, shapes and patterns."

3

I also like the following quote by Sun-Yung Alice Chang Professor of Mathematics, Princeton University:

() The Beauty of Combinatorics November 1, 2012 2 / 19

Introduction

1

What is mathematics?

2

"Mathematics is the study of numbers, shapes and patterns."

3

I also like the following quote by Sun-Yung Alice Chang Professor of Mathematics, Princeton University:

4

"Mathematics is a language like music. To learn it systematically, it is necessary to master small pieces and gradually add another piece and then another. Mathematics is like the classical Chinese language - very polished and very elegant. Sitting in a good mathematics lecture is like sitting in a good opera. Everything comes together."

() The Beauty of Combinatorics November 1, 2012 2 / 19

Introduction

1

What is mathematics?

2

"Mathematics is the study of numbers, shapes and patterns."

3

I also like the following quote by Sun-Yung Alice Chang Professor of Mathematics, Princeton University:

4

"Mathematics is a language like music. To learn it systematically, it is necessary to master small pieces and gradually add another piece and then another. Mathematics is like the classical Chinese language - very polished and very elegant. Sitting in a good mathematics lecture is like sitting in a good opera. Everything comes together."

5

I am not planning to sing in this lecture...

() The Beauty of Combinatorics November 1, 2012 2 / 19

“Elementary problems, a sample”

Not too long ago Graph Theory was dubbed as the The Slums of Topology combinatorics was treated as a curious side issue in mathematics.

() The Beauty of Combinatorics November 1, 2012 3 / 19

“Elementary problems, a sample”

Not too long ago Graph Theory was dubbed as the The Slums of Topology combinatorics was treated as a curious side issue in mathematics. Today combinatorics and Graph Theory are the most active research areas in Mathematics.

() The Beauty of Combinatorics November 1, 2012 3 / 19

“Elementary problems, a sample”

Not too long ago Graph Theory was dubbed as the The Slums of Topology combinatorics was treated as a curious side issue in mathematics. Today combinatorics and Graph Theory are the most active research areas in Mathematics. In this talk I will attempt to do justice to combinatorics.

() The Beauty of Combinatorics November 1, 2012 3 / 19

“Elementary problems, a sample”

Not too long ago Graph Theory was dubbed as the The Slums of Topology combinatorics was treated as a curious side issue in mathematics. Today combinatorics and Graph Theory are the most active research areas in Mathematics. In this talk I will attempt to do justice to combinatorics. Many combinatorial problems have an “elementary” flavor. They are easily understood even by high school students. I will discuss a sample of such problems, solutions and open problems.

() The Beauty of Combinatorics November 1, 2012 3 / 19

“Elementary problems, a sample”

Not too long ago Graph Theory was dubbed as the The Slums of Topology combinatorics was treated as a curious side issue in mathematics. Today combinatorics and Graph Theory are the most active research areas in Mathematics. In this talk I will attempt to do justice to combinatorics. Many combinatorial problems have an “elementary” flavor. They are easily understood even by high school students. I will discuss a sample of such problems, solutions and open problems. I will also try to demonstrate how combinatorics today actively interacts with many “classical” areas of

• mathematics. Let me start with a “topological” classic:

() The Beauty of Combinatorics November 1, 2012 3 / 19

Brouwer’s fixed point theorem

Theorem (Sperner) Let T1, T2, . . . Tk be a triangulation of a triangle T. Assume that the vertices of T are colored red, blue and green. Further assume that on the green-red edge of T all points of the triangulation are colored green or red, on the green-blue edge all points are colored blue or green and on the red-blue edge all points are colored red or blue. Then there is a “rainbow” tiriangle Ti.

() The Beauty of Combinatorics November 1, 2012 4 / 19

Brouwer’s fixed point theorem

Theorem (Sperner) Let T1, T2, . . . Tk be a triangulation of a triangle T. Assume that the vertices of T are colored red, blue and green. Further assume that on the green-red edge of T all points of the triangulation are colored green or red, on the green-blue edge all points are colored blue or green and on the red-blue edge all points are colored red or blue. Then there is a “rainbow” tiriangle Ti. Proof.

() The Beauty of Combinatorics November 1, 2012 4 / 19

Brouwer’s fixed point theorem

Theorem (Sperner) Let T1, T2, . . . Tk be a triangulation of a triangle T. Assume that the vertices of T are colored red, blue and green. Further assume that on the green-red edge of T all points of the triangulation are colored green or red, on the green-blue edge all points are colored blue or green and on the red-blue edge all points are colored red or blue. Then there is a “rainbow” tiriangle Ti. Proof.

1

Count the number of rainbow edges in each triangle Ti modulo 2.

() The Beauty of Combinatorics November 1, 2012 4 / 19

Brouwer’s fixed point theorem

Theorem (Sperner) Let T1, T2, . . . Tk be a triangulation of a triangle T. Assume that the vertices of T are colored red, blue and green. Further assume that on the green-red edge of T all points of the triangulation are colored green or red, on the green-blue edge all points are colored blue or green and on the red-blue edge all points are colored red or blue. Then there is a “rainbow” tiriangle Ti. Proof.

1

Count the number of rainbow edges in each triangle Ti modulo 2.

2

Deduce that there is an odd number of Rainbow triangles.

() The Beauty of Combinatorics November 1, 2012 4 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point.

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

1

As the topologists tell us, we may assume that D = {(x1, x2, x3) | x1 + x2 + x3 = 1, 1 ≥ xi ≥ 0.}

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

1

As the topologists tell us, we may assume that D = {(x1, x2, x3) | x1 + x2 + x3 = 1, 1 ≥ xi ≥ 0.}

2

Color (0, 0, 1) 1, (0, 1, 0) 2 and (1, 0, 0) 3.

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

1

As the topologists tell us, we may assume that D = {(x1, x2, x3) | x1 + x2 + x3 = 1, 1 ≥ xi ≥ 0.}

2

Color (0, 0, 1) 1, (0, 1, 0) 2 and (1, 0, 0) 3.

3

For every point u ∈ D, let φ(u) = i where i is the smallest index

• f u − f(u) which is > 0.

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

1

As the topologists tell us, we may assume that D = {(x1, x2, x3) | x1 + x2 + x3 = 1, 1 ≥ xi ≥ 0.}

2

Color (0, 0, 1) 1, (0, 1, 0) 2 and (1, 0, 0) 3.

3

For every point u ∈ D, let φ(u) = i where i is the smallest index

• f u − f(u) which is > 0.

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

1

As the topologists tell us, we may assume that D = {(x1, x2, x3) | x1 + x2 + x3 = 1, 1 ≥ xi ≥ 0.}

2

Color (0, 0, 1) 1, (0, 1, 0) 2 and (1, 0, 0) 3.

3

For every point u ∈ D, let φ(u) = i where i is the smallest index

• f u − f(u) which is > 0.

4

This is a “Sperner” coloring.

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

1

As the topologists tell us, we may assume that D = {(x1, x2, x3) | x1 + x2 + x3 = 1, 1 ≥ xi ≥ 0.}

2

Color (0, 0, 1) 1, (0, 1, 0) 2 and (1, 0, 0) 3.

3

For every point u ∈ D, let φ(u) = i where i is the smallest index

• f u − f(u) which is > 0.

4

This is a “Sperner” coloring.

5

Keep triangulating the “rainbow” triangles to get a sequence ui − → w0.

() The Beauty of Combinatorics November 1, 2012 5 / 19

Brouwer fixed point theorem

Theorem Let f : D − → D be a continuous function on the disc D. Then f has a fixed point. Proof.

1

As the topologists tell us, we may assume that D = {(x1, x2, x3) | x1 + x2 + x3 = 1, 1 ≥ xi ≥ 0.}

2

Color (0, 0, 1) 1, (0, 1, 0) 2 and (1, 0, 0) 3.

3

For every point u ∈ D, let φ(u) = i where i is the smallest index

• f u − f(u) which is > 0.

4

This is a “Sperner” coloring.

5

Keep triangulating the “rainbow” triangles to get a sequence ui − → w0.

6

Use continuity to coclude that f(w0) = w0.

() The Beauty of Combinatorics November 1, 2012 5 / 19

This is a nice example of using a pure combinatorial approach to prove a classical result in analysis.

() The Beauty of Combinatorics November 1, 2012 6 / 19

This is a nice example of using a pure combinatorial approach to prove a classical result in analysis. In the next problem we’ll see how we can use analysis (calculus) to prove a purely combinatorial result.

() The Beauty of Combinatorics November 1, 2012 6 / 19

Dissecting rectangles and squares.

1 Dissecting rectangles into rectangles with an

integral side.

() The Beauty of Combinatorics November 1, 2012 7 / 19

Dissecting rectangles and squares.

1 Dissecting rectangles into rectangles with an

integral side.

2 Can a rectangle be dissected into n triangles of

equal area?

() The Beauty of Combinatorics November 1, 2012 7 / 19

Dissecting rectangles and squares.

1 Dissecting rectangles into rectangles with an

integral side.

2 Can a rectangle be dissected into n triangles of

equal area?

3 How can we dissect the unit square into n

rectangles so that the largest perimeter is minimized?

() The Beauty of Combinatorics November 1, 2012 7 / 19

Rectangels with integral sides

A rectangle is tiled by m smaller rectangles. One of the sides of each smaller rectangle has integral length. Prove that one of the sides of the tiled rectangle has integral length.

() The Beauty of Combinatorics November 1, 2012 8 / 19

A proof

1 This is a simply stated elementary problem. () The Beauty of Combinatorics November 1, 2012 9 / 19

A proof

1 This is a simply stated elementary problem. 2 As such, it attracted many lovers of mathematics. () The Beauty of Combinatorics November 1, 2012 9 / 19

A proof

1 This is a simply stated elementary problem. 2 As such, it attracted many lovers of mathematics. 3 At least 14 diiferent proofs were published. () The Beauty of Combinatorics November 1, 2012 9 / 19

A proof

1 This is a simply stated elementary problem. 2 As such, it attracted many lovers of mathematics. 3 At least 14 diiferent proofs were published. 4 Here is one using classical analysis. () The Beauty of Combinatorics November 1, 2012 9 / 19

A proof

1 This is a simply stated elementary problem. 2 As such, it attracted many lovers of mathematics. 3 At least 14 diiferent proofs were published. 4 Here is one using classical analysis. () The Beauty of Combinatorics November 1, 2012 9 / 19

A proof

1 This is a simply stated elementary problem. 2 As such, it attracted many lovers of mathematics. 3 At least 14 diiferent proofs were published. 4 Here is one using classical analysis.

Assume that the sides of the “big” triangle are at (0, 0), (0, a), (a, b), (b, 0).

() The Beauty of Combinatorics November 1, 2012 9 / 19

A proof

1 This is a simply stated elementary problem. 2 As such, it attracted many lovers of mathematics. 3 At least 14 diiferent proofs were published. 4 Here is one using classical analysis.

Assume that the sides of the “big” triangle are at (0, 0), (0, a), (a, b), (b, 0). Claim: a b e2πi(x+y)dxdy = 0.

() The Beauty of Combinatorics November 1, 2012 9 / 19

proof (continued)

b a e2πi(x+y)dxdy = b e2πixdx a e2πiydy = 0 ⇒ b e2πixdx = 0

• r

a e2πiydy = 0

() The Beauty of Combinatorics November 1, 2012 10 / 19

proof (continued)

b a e2πi(x+y)dxdy = b e2πixdx a e2πiydy = 0 ⇒ b e2πixdx = 0

• r

a e2πiydy = 0 But b(a) e2πixdx = 0 ⇒ b (or a) is an integer. b a e2πi(x+y)dxdy =

m

• k=1

Tk

e2πi(x+y)dxdy

() The Beauty of Combinatorics November 1, 2012 10 / 19

proof (continued)

b a e2πi(x+y)dxdy = b e2πixdx a e2πiydy = 0 ⇒ b e2πixdx = 0

• r

a e2πiydy = 0 But b(a) e2πixdx = 0 ⇒ b (or a) is an integer. b a e2πi(x+y)dxdy =

m

• k=1

Tk

e2πi(x+y)dxdy Since each small rectangle has an integral side each integral inside the sum = 0.

() The Beauty of Combinatorics November 1, 2012 10 / 19

proof (continued)

b a e2πi(x+y)dxdy = b e2πixdx a e2πiydy = 0 ⇒ b e2πixdx = 0

• r

a e2πiydy = 0 But b(a) e2πixdx = 0 ⇒ b (or a) is an integer. b a e2πi(x+y)dxdy =

m

• k=1

Tk

e2πi(x+y)dxdy Since each small rectangle has an integral side each integral inside the sum = 0. This nice proof is due to the Dutch mathematician De Bruijn.

() The Beauty of Combinatorics November 1, 2012 10 / 19

More dissection problems Question Can a rectangle be dissected into n triangles of equal area?

() The Beauty of Combinatorics November 1, 2012 11 / 19

More dissection problems Question Can a rectangle be dissected into n triangles of equal area? Answer When n = 2k the answer is YES. And it is very easy to do. For intsance you can dissect the rectangle into k rectangles of equal area and split each such rectangle by its diagonal into two triangles of equal area.

() The Beauty of Combinatorics November 1, 2012 11 / 19

More dissection problems Question Can a rectangle be dissected into n triangles of equal area? Answer When n = 2k the answer is YES. And it is very easy to do. For intsance you can dissect the rectangle into k rectangles of equal area and split each such rectangle by its diagonal into two triangles of equal area.

() The Beauty of Combinatorics November 1, 2012 11 / 19

More dissection problems Question Can a rectangle be dissected into n triangles of equal area? Answer When n = 2k the answer is YES. And it is very easy to do. For intsance you can dissect the rectangle into k rectangles of equal area and split each such rectangle by its diagonal into two triangles of equal area. Surprisingly it is not possible to do it for n = 2k + 1 and the proof is not easy.

() The Beauty of Combinatorics November 1, 2012 11 / 19

An open elementary dissection problem

Question For a given dissection S of the unit square into k recatngles let β(S, k) be the largest perimeter among the k rectangles. What is min(β(S, k)) among all possible dissections?

() The Beauty of Combinatorics November 1, 2012 12 / 19

An open elementary dissection problem

Question For a given dissection S of the unit square into k recatngles let β(S, k) be the largest perimeter among the k rectangles. What is min(β(S, k)) among all possible dissections? Answer

() The Beauty of Combinatorics November 1, 2012 12 / 19

An open elementary dissection problem

Question For a given dissection S of the unit square into k recatngles let β(S, k) be the largest perimeter among the k rectangles. What is min(β(S, k)) among all possible dissections? Answer

1

The answer is easy when k = m2, dissect the unit square into m2

1 m × 1 m

squares.

() The Beauty of Combinatorics November 1, 2012 12 / 19

An open elementary dissection problem

Question For a given dissection S of the unit square into k recatngles let β(S, k) be the largest perimeter among the k rectangles. What is min(β(S, k)) among all possible dissections? Answer

1

The answer is easy when k = m2, dissect the unit square into m2

1 m × 1 m

squares.

2

Is it true that the minimum is obtained when all “little” squares have the same area?

() The Beauty of Combinatorics November 1, 2012 12 / 19

An open elementary dissection problem

Question For a given dissection S of the unit square into k recatngles let β(S, k) be the largest perimeter among the k rectangles. What is min(β(S, k)) among all possible dissections? Answer

1

The answer is easy when k = m2, dissect the unit square into m2

1 m × 1 m

squares.

2

Is it true that the minimum is obtained when all “little” squares have the same area?

3

The answer in general is negative!

() The Beauty of Combinatorics November 1, 2012 12 / 19

An open elementary dissection problem

Question For a given dissection S of the unit square into k recatngles let β(S, k) be the largest perimeter among the k rectangles. What is min(β(S, k)) among all possible dissections? Answer

1

The answer is easy when k = m2, dissect the unit square into m2

1 m × 1 m

squares.

2

Is it true that the minimum is obtained when all “little” squares have the same area?

3

The answer in general is negative!

4

In any partition of the unit square into k rectangles of equal area there must be rectangles with a side equal to ⌊ 1

√ k ⌋ and some

rectabgles with a side equal to ⌈ 1

√ k ⌉

() The Beauty of Combinatorics November 1, 2012 12 / 19

An optimal partitioning of the unit square into 76 rectangles of equal area

() The Beauty of Combinatorics November 1, 2012 13 / 19

Comments

1 The largest perimeter is 2(1

8 + 8 76).

() The Beauty of Combinatorics November 1, 2012 14 / 19

Comments

1 The largest perimeter is 2(1

8 + 8 76).

2 Clearly there are partitions into 76 rectangles with

smaller perimeter.

() The Beauty of Combinatorics November 1, 2012 14 / 19

Comments

1 The largest perimeter is 2(1

8 + 8 76).

2 Clearly there are partitions into 76 rectangles with

smaller perimeter.

3 The cases for which we know the optimal

solutions are: k = m2, m(m + 1), m2 ± 1.

() The Beauty of Combinatorics November 1, 2012 14 / 19

Comments

1 The largest perimeter is 2(1

8 + 8 76).

2 Clearly there are partitions into 76 rectangles with

smaller perimeter.

3 The cases for which we know the optimal

solutions are: k = m2, m(m + 1), m2 ± 1.

4 In all other cases the problem is open. () The Beauty of Combinatorics November 1, 2012 14 / 19

Comments

1 The largest perimeter is 2(1

8 + 8 76).

2 Clearly there are partitions into 76 rectangles with

smaller perimeter.

3 The cases for which we know the optimal

solutions are: k = m2, m(m + 1), m2 ± 1.

4 In all other cases the problem is open. 5 This problem arose in data allocation for parallel

algorithms.

() The Beauty of Combinatorics November 1, 2012 14 / 19

Distances among points in R2. This was a favorite topic of Paul Erd˝

• s. In 1946 he

proved that if you have an infinite set of points in the plane such that the distance between any two points is an integer then they must lie on a straight line. This led to many elementary problems, some solved and some are still waiting for a clever idea. Here is a list of some of these problems and their status.

() The Beauty of Combinatorics November 1, 2012 15 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

3

How big is the smallest circle?

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

3

How big is the smallest circle?

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

3

How big is the smallest circle? OPEN

4

How many points in general position (no 3 on a line, no 4 on a circle) with integral distances exist?

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

3

How big is the smallest circle? OPEN

4

How many points in general position (no 3 on a line, no 4 on a circle) with integral distances exist?

5

Best known is only 9 points.

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

3

How big is the smallest circle? OPEN

4

How many points in general position (no 3 on a line, no 4 on a circle) with integral distances exist?

5

Best known is only 9 points.

6

How many times can the distance 1 appear among n points in the plane?

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

3

How big is the smallest circle? OPEN

4

How many points in general position (no 3 on a line, no 4 on a circle) with integral distances exist?

5

Best known is only 9 points.

6

How many times can the distance 1 appear among n points in the plane?

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems”

1

How many points exist in the plane no three on a line such that the distance between any two is an integer?

2

Ans: arbitrarily many, all on one circle.

3

How big is the smallest circle? OPEN

4

How many points in general position (no 3 on a line, no 4 on a circle) with integral distances exist?

5

Best known is only 9 points.

6

How many times can the distance 1 appear among n points in the plane? OPEN

7

Among n points in the plane, how many time can the largest distance occur?

() The Beauty of Combinatorics November 1, 2012 16 / 19

“Elementary problems” and their status.

1 The smaller angle among any two diagonals of the

four diagonals of the regular cube centered at the

• rigin is the same: arccos(1

3).

() The Beauty of Combinatorics November 1, 2012 17 / 19

“Elementary problems” and their status.

1 The smaller angle among any two diagonals of the

four diagonals of the regular cube centered at the

• rigin is the same: arccos(1

3).

2 The smaller angle among any two diagonals of the

regular icosahedron centered at the origin is also the same: arccos( 1

√ 5).

() The Beauty of Combinatorics November 1, 2012 17 / 19

“Elementary problems” and their status.

1 The smaller angle among any two diagonals of the

four diagonals of the regular cube centered at the

• rigin is the same: arccos(1

3).

2 The smaller angle among any two diagonals of the

regular icosahedron centered at the origin is also the same: arccos( 1

√ 5).

3 In how many ways can you form four equiangular

lines through the origin in R3?

() The Beauty of Combinatorics November 1, 2012 17 / 19

“Elementary problems” and their status.

1 The smaller angle among any two diagonals of the

four diagonals of the regular cube centered at the

• rigin is the same: arccos(1

3).

2 The smaller angle among any two diagonals of the

regular icosahedron centered at the origin is also the same: arccos( 1

√ 5).

3 In how many ways can you form four equiangular

lines through the origin in R3?

4 Given n lines through the origin in R3 how large

can the smallest angle determined by any pair of lines can be?

() The Beauty of Combinatorics November 1, 2012 17 / 19

Equiangular lines

1

What are the possible angles among d + 1 equiangular lines in Rd?

() The Beauty of Combinatorics November 1, 2012 18 / 19

Equiangular lines

1

What are the possible angles among d + 1 equiangular lines in Rd?

2

For which angles α there are infinitely many dimensions d in which one can find d + 1 equiangular lines in Rd?

() The Beauty of Combinatorics November 1, 2012 18 / 19

Equiangular lines

1

What are the possible angles among d + 1 equiangular lines in Rd?

2

For which angles α there are infinitely many dimensions d in which one can find d + 1 equiangular lines in Rd?

3

There are 4d + 1 equiangular lines with angle arccos( 1

4) in

R4d for d = 1, 2, . . . 13.

() The Beauty of Combinatorics November 1, 2012 18 / 19

Equiangular lines

1

What are the possible angles among d + 1 equiangular lines in Rd?

2

For which angles α there are infinitely many dimensions d in which one can find d + 1 equiangular lines in Rd?

3

There are 4d + 1 equiangular lines with angle arccos( 1

4) in

R4d for d = 1, 2, . . . 13.

4

Are there 4d + 1 equiangular lines with angle arccos( 1

4) in

R4d ∀d?

() The Beauty of Combinatorics November 1, 2012 18 / 19

Equiangular lines

1

What are the possible angles among d + 1 equiangular lines in Rd?

2

For which angles α there are infinitely many dimensions d in which one can find d + 1 equiangular lines in Rd?

3

There are 4d + 1 equiangular lines with angle arccos( 1

4) in

R4d for d = 1, 2, . . . 13.

4

Are there 4d + 1 equiangular lines with angle arccos( 1

4) in

R4d ∀d?

5

For a give dimension d, what is the largest number of equiangular lines in Rd?

() The Beauty of Combinatorics November 1, 2012 18 / 19

Equiangular lines

1

What are the possible angles among d + 1 equiangular lines in Rd?

2

For which angles α there are infinitely many dimensions d in which one can find d + 1 equiangular lines in Rd?

3

There are 4d + 1 equiangular lines with angle arccos( 1

4) in

R4d for d = 1, 2, . . . 13.

4

Are there 4d + 1 equiangular lines with angle arccos( 1

4) in

R4d ∀d?

5

For a give dimension d, what is the largest number of equiangular lines in Rd?

6

An upper bound is d+1

2

• (Blokhuis).

() The Beauty of Combinatorics November 1, 2012 18 / 19

Cocluding remarks

My students are currently working on some of these problems.

() The Beauty of Combinatorics November 1, 2012 19 / 19

Cocluding remarks

My students are currently working on some of these problems. Mathematicians are persons with problems.

() The Beauty of Combinatorics November 1, 2012 19 / 19

Cocluding remarks

My students are currently working on some of these problems. Mathematicians are persons with problems. Thank you

() The Beauty of Combinatorics November 1, 2012 19 / 19

Cocluding remarks

My students are currently working on some of these problems. Mathematicians are persons with problems. Thank you Questions?

() The Beauty of Combinatorics November 1, 2012 19 / 19

Cocluding remarks

My students are currently working on some of these problems. Mathematicians are persons with problems. Thank you Questions? Answers?

() The Beauty of Combinatorics November 1, 2012 19 / 19