UNIVERSITY OF STRATHCLYDE

DEPARTMENT OF

MATHEMATICS AND STATISTICS

The Inscribed Square Problem and Other

Theories Relating to the Number 4

by

Rebecca Alston

201210039

BSc Maths, Statistics and Finance

2015

Statement of work in project

The work contained in this project is that of the author and

where material from other sources has been incorporated

full acknowledgement is made.

Signed .........................................................

Print Name .........................................................

Date .........................................................

Professor Ernesto Estrada

2

Contents

1 Introduction 4

2 The Inscribed Square Problem 5

2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Stromquist’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2.1 Proof of Theorem 1.1 . . . . . . . . . . . . . . . . . . . . . 6

2.2.2 Proof of Theorem 1.2 . . . . . . . . . . . . . . . . . . . . . 9

2.3 Surrounding Results . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3.1 Symmetric Simple Closed Curves in the Plane . . . . . . . 12

2.3.2 Other Inscribed Shapes . . . . . . . . . . . . . . . . . . . . 13

3 Ducci Sequence 19

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2 The 4-number Game . . . . . . . . . . . . . . . . . . . . . . . . . 19

4 The Four Color Theorem 22

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

4.2 The History of the Four Color Theorem . . . . . . . . . . . . . . . 23

4.3 Kempe Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.4 The Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5 Rearranging Four Squares 26

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

5.2 The Wallace-Bolyai-Gerwein Theorem . . . . . . . . . . . . . . . . 26

5.3 Four Piece Puzzle . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

6 Conclusion 28

3

1 Introduction

Inspired by ”Single Digits. In Praise of Small Numbers” by Marc Chamberland

[3], this report is based on the chapter ”The Mathematics of the Number 4”. The

number 4 is often seen as a balanced number. It is the number of players in a

game of bridge, the number of legs on a table, the number of people involved in

a double date and the number of paws on a cat. It is the only number, in the

English Dictionary, that has the same number of characters as it’s value. Math-

ematically, it is the smallest composite number, the second square number and

the first positive non-Fibonacci number.

This report will study the mathematical importance of the number 4 in vari-

ous theorems, puzzles and problems. The main body will focus on an unsolved

problem in 2-dimensional geometry - The Inscribed Square Problem. We will then

introduce and prove the Ducci Sequence, a phenomenon attributed to Edda Ducci

first discovered in 1937 [3]. Moving on from this we will briefly discuss the con-

troversial Four Color Theorem, demonstrating a few important techniques and

discussing the use of Computer Aided Proofs in mathematics. To end, we will

look at The Four Piece puzzle, attributed to Henry Dudeney, which demonstrates

how to rearrange an equilateral triangle into a square.

4

2 The Inscribed Square Problem

2.1 Introduction

The Inscribed Square Problem, also known as the Toeplitz Problem, was pro-

posed by Otto Toeplitz in 1911 making it one of the oldest unsolved problems

in 2-dimensional geometry. Early studies produced some positive and interesting

results but the general case is still to be proven. The most significant of these

studies was carried out by Walter Stromquist in 1989 [12] and will be discussed

later. The claim is that on any simple closed curve there exists four points that

are the vertices of a square. A simple closed curve is defined as the image in the

R2 plane of the following continuous function [12]:

f : [0, 1] −→ R2.

The function is one-to-one except that f(0) = f(1). A simple closed curve is also

known as a Jordan curve. For a quadrilateral to be inscribed in a Jordan curve,

J , all four vertices must lie on J . The sides of the quadrilateral do not need to

lie in the interior of J [18]. Many partial and related results have evolved around

the Inscribed Square Problem but a general proof has been elusive.

To understand the problem more we will start by examining the simplest Jordan

curve, the circle. A circle has many inscribed squares, some of these are shown in

Figure 1: A Circle showing some of it’s Inscribed Squares.

Figure 1. As curves become more complicated it can be more difficult to imagine

their inscribed squares. Some more complicated Jordan Curves have many in-

scribed squares whereas some only have one. Are there some Jordan curves that

don’t have any?

5

Figure 2: Examples of Inscribed Squares in Jordan Curves [12]

2.2 Stromquist’s Theorem

The Inscribed Square Problem has sparked many interesting surrounding results.

The sharpest of these results are the following two theorem’s from Stromquist

[18].

Theorem 1.1: Each smooth simple closed curve in Rd admits an inscribed

quadrilateral with equal sides and diagonals

Theorem 1.2: Each locally monotone Jordan curve in R2 admits an inscribed

square.

The proofs involve complex homology and because of this only a brief review will

be provided. The full proofs of these theorems can be found in ’Inscribed Squares

and Square-Like Quadrilaterals in Closed Curves’ by Walter Stromquist [18].

2.2.1 Proof of Theorem 1.1

A curve, J , is smooth if it has a continuously non-vanishing derivative. In other

words, if J has a continuously turning tangent. We are mainly interested in

curves in R2 where the inscribed quadrilateral is a square.

To understand the proof of Theorem 1.1 we must introduce a few tools. The

simplex, Q, will represent the set of quadrilaterals inscribed in J . Q1, Q2, Q3, Q4

are four subsets which cover Q and whose intersection corresponds to the set of

inscribed rhombuses.

We will denote Q by the set:

Q = {(x1, x2, x3, x4) ∈ R4|0 ≤ x1 ≤ x2 ≤ x3 ≤ x4 ≤ 1} (1)

This represents a 4-simplex with vertices

6

v0 = (1, 1, 1, 1)

v1 = (0, 1, 1, 1)

v2 = (0, 0, 1, 1)

v3 = (0, 0, 0, 1)

v4 = (0, 0, 0, 0).

The faces F0, ..., F4 are numbered so that each Fi is opposite vi. The simplex is

illustrated in Figure 3. Figure 3 requires the following compromises as it is an

illustration of a 4-dimensional object: the right edge of the figure, drawn as a line

segment, is actually the 2-simplex with 3 vertices, F0 ∩ F4. The top and bottom

faces are F4 and F0 respectively and are both 3-simplex.

Figure 3: The Simplex Q of Quadrilaterals.[18]

Given a fixed curve, J , we associate an inscribed quadrilateral with each point x ∈

Q. The vertices of this quadrilateral are J(x1), J(x2), J(x3), J(x4). The points of

F4 and F0 represent the same quadrilaterals , therefore the correspondence is not

quite one-to-one. We can define h : F0 → F4 by

h(0, x, y, z) = (x, y, z, 1),

then h(x) represents the same quadrilateral as x with differently numbered ver-

tices. We refer to the points of Q as quadrilaterals and can identify these points

with their corresponding geometric figures despite this ambiguity. Now we know

that every point of Q corresponds to a quadrilateral whose vertices have the same

cyclic order in the quadrilateral as J . Some of these quadrilaterals may have sides

of zero lengths and some may have all sides with zero length. These are described

as degenerate and one-point quadrilaterals respectively.

7

For each i = 1, 2, 3, 4 we will define si(x) = ||J(xi+1)− J(xi)||. Here, si(x) refers

to the length of the ith side of a quadrilateral corresponding to x and each si is

a continuous function on Q.

Now for each i, we will define Qi to be the closure of the set

{x ∈ Q0|si(x) = maxj

sj(x)}. (2)

Here, Q0 denotes the interior of Q. Now, Qi is the set of quadrilaterals whose

ith side is the longest side. The purpose of this device is to prevent one-point

quadrilaterals from automatically being elements of every Qi. It is only successful

if the curve, J , is sufficiently smooth. Lemma 1 follows:

Lemma 1: If J is a smooth curve, then each one-point quadrilateral is contained

in only one set Qi. In particular, vi ∈ Q, for i = 1, 2, 3 and x ∈ Q4 for each x

on the edge connecting v0 and v4.

For what follows, assume that J is any curve for which the result of Lemma 1

is valid. Let R =⋂iQi and call a point, x ∈ R, a rhombus. A square-like

quadrilateral is a rhombus which satisfies d13(x) = d24(x) where,

d13 = ||J(x3)− J(x1)||

d24 = ||J(x4)− J(x2||

A square-like quadrilateral is just an inscribed quadrilateral with equal sides and

equal diagonals. This is a square in R2. If d13>d24 then a thin rhombus is cre-

ated. A fat rhombus satisfies d13<d24. RTHIN and RFAT are subsets of the set

of rhombuses, R, containing thin and fat rhombuses respectively. The set R, is

comprised of RTHIN and RFAT such that R = RTHIN ∪RFAT .

Stromquist then studies the set of rhombuses in Q showing that there must be an

odd number of rhombuses in F0. He finds that if the number of thin rhombuses

in F0 is even then the number of fat rhombuses in F0 must be odd and vice versa

creating an overall odd number of rhombuses. The correspondence based on the

function h shows that whatever happens on F0 will be reversed on F4. The next

part of Stromquist’s proof uses machinery from the homology theory to create

8

three important Lemmas. This can be found in Stromquist(1999) under the sec-

tion “The degree of a set of rhombuses” [18].

To understand how these lemmas are derived, we will introduce a few terms.

K will be the set of rhombuses and A will be an n-simplex. The degree is our

way of counting the rhombuses mod2. In a way the degree of K tells us whether

to consider K to be even or odd. Let γ be the n-chain consisting of the n-simplices

which touch K and L =⋂Ai−K. The boundary of γ then represents a homology

class υ ∈ Hn−1(A−⋂Ai) which is the unique homology class surrounding all of

K but none of L. This is denoted by υK . The degree of Ai around K is defined

as the image F∗(υK) in Hn−1(∂A) where f is any reversing map for {Ai}.

Using the above information and the homology theorem Stromquist provides 3

Lemmas:

Lemma 2: deg∅ = 0.

Lemma 3: If K = K1 ∪K2, then degK = degK1 + degK2.

Lemma 4: deg(⋂Ai) = 1.

Using these lemmas and some manipulation using homology theorem Stromquist

finds an equivalence between the degree of RTHIN ∩F0 and the degree of RTHIN ∩

F4. This equivalence contradicts his earlier findings and thus establishes Theorem

1.1.

2.2.2 Proof of Theorem 1.2

In this section all curves are in R2. The smoothness condition required now is

“local monotonicity” which we will prove also guarantees an inscribed square.

Smooth curves, convex curves, polygons and most piecewise C1 curves all satisfy

this condition [18]. We will first introduce some definitions.

Definition 1: A segment of a curve J , corresponding to an interval (a, b) is

the restriction of the function to that interval, J |(a, b).

Definition 2: The length of the segment is described as (b − a). This is mea-

sured in parameter space not R2 [18].

Definition 3: The segment is monotone in the direction u (where u is a non-

9

zero vector in R2) if the dot product J(x).u is a strictly increasing function of x

for x ∈ (a, b).

Definition 4.1: The curve, J is locally monotone if, for every y ∈ R, there

is an interval (y − µ, y + µ) and a direction u(y) such that J |(y − µ, y + µ) is

monotone in the direction of u(y) [18] .

If the restriction of local monotonicity holds for a curve, J , then the periodicity

of J and the compactness of [0, 1] allow us to choose the number µ to be constant

and independent of y. In this case, every segment of J with maximum length 2µ

is monotone in some direction. Therefore, J is locally monotone with constant µ.

The term local monotonicity depends only on the image of J and not on the

parametrization and can therefore be described as purely geometric. A new def-

inition of Locally Monotone emphasises the geometric nature:

Definition 4.2: A curve J is locally monotone if for every point J(y) of the

curve, there is a neighbourhood, U(y), in R2 and a direction n(y) such that no

chord of the curve is contained in U(y) and parallel to n(y). The direction n(y)

is normal to u(y) and can be thought of as a normal vector [18].

To see that smooth curves are locally monotone, let u(y) = J ′(y) or take n(y) to

be a normal vector. To see that convex curves are locally monotone take n(y) to

be a vector from J(y) toward any interior point.

What C1 curves are locally monotone?

A curve is piecewise C1 if there exist numbers x0, ..., xk satisfying 0 = x0 < ... <

xk = 1 such that J is smooth on each interval [xi−1, xi], including the one-sided

derivatives at the endpoints. We denote the one-sided derivatives at xi by J ′−(xi)

and J ′+(xi). The curve has a cusp at xi if these two vectors point in opposite

directions, i.e if J ′+(xi) is a negative multiple of J ′−(xi). A cusp is a point where

two branches of a curve meet such that the tangents of each branch are equal.

A piecewise C1 curve without cusps is locally monotone. With this we can now

state Theorem 1.2, as before.

Theorem 1.2: Each locally monotone Jordan curve in R2 admits an inscribed

square.

Assume J is locally monotone with constant µ. Our strategy in this proof will be

10

to approximate J using smooth curves Je, which must contain an inscribed square

by Theorem 1.1. As ε→ 0, a subsequence of these inscribed squares converge to

a square inscribed in J . The main difficulty arises when proving that the limiting

square does not have side zero. To show this, we must establish a lower bound

on the size of the square we find in J by showing that each Je is itself locally

monotone with a constant of value at least 12u.

First, we will define the size of ||x|| of an element x = (x1, x2, x3, x4) ∈ Q to be the

smallest segment of these four numbers: (x4−x1), ((1 +x4)−x4), ((1 +x2)−x3),

and ((1 + x1)− x2). Thus ||x|| is the same as the length of the smallest segment

of the curve that can contain all of the points J(x1), J(x2), J(x3), J(x4). The

only quadrilaterals with size zero are the one-point quadrilaterals. Since ||x|| is

continuous on Q, any sequence of quadrilaterals with sizes who have a positive

lower bound cannot converge to a one-point quadrilateral.

Let ε > 0. Choose δ > 0 such that |x − y|<δ implies that ||J(x) − J(y)|| < ε .

Now choose δ<12µ and define Jε : R→ R2 by

Jε(x) =1

δ

∫ δ

t=0

J(x+ t)dt. (3)

Now, Jε satisfies ||Jε(x)− J(x)||<ε for all x. We then conclude that J is smooth

with a continuous non-vanishing derivative given by

J ′ε(x) =1

δ(J(x+ δ)− J(x)). (4)

Using the local monotonicity of J and a sufficiently small ε, we can show that Jε

is actually a simple closed curve. We may therefore apply Theorem 1.1 to find

a square Sε inscribed in Jε, whose vertices are in the same cyclic order in the

square as in the curve.

We can then show that Jε is locally monotone with constant 12µ. Let y ∈ R and

u(y) be chosen such that J |(y − µ, y + µ) is monotone in the direction u(y). Let

x1, x2 be contained in (y − 12µ, y + 1

2µ), with x1<x2. Then

(Jε(x2)− Jε(x1)) · u(y) = 1δ

∫ δt=0

(J(x2 + t)− J(x1 + t))dt>0,

The monotonicity of J forces the integrand to be strictly positive. Therefore the

chord from Jε(x1) to Jε(x2) has a positive component in the direction of u(y), as

11

required.

It follows that the inscribed square, Sε cannot have size less than µ unless its

vertices are contained in some interval of length µ, in which Jε would have to

be monotone. We can repeat this construction for a sequence of values of ε ap-

proaching zero to find some subsequence of the squares Sε that converges to a

square, S. This S must have size at least µ and must be inscribed in J completing

the proof of Theorem 1.2.

A few prominent mathematicians believe that Stromquist’s theorem is good enough

to prove the full unsolved statement, ”All simple closed curves have inscribed

squares” as morally true. This is because any curve that can be drawn by a

human with a pencil on a sheet of paper satisfies Stromquist’s conditions. The

formal question is far from solved as most simple closed curves have fractal-like

behaviour that does not meet these conditions and therefore, from a topologists

point of view, The Inscribed Square Problem remains unsolved.

2.3 Surrounding Results

This section includes an exploration into studies carried out by numerous mathe-

maticians providing interesting results surrounding the Inscribed Square Problem.

2.3.1 Symmetric Simple Closed Curves in the Plane

Results considered to be the strongest solutions to the problem usually involve

some sort of implicit tameness on the curve. An example of this, first proved by

Mark Nielson [12], imposes that the curve, J must be symmetrical.

Theorem 1.3: Every closed curve that is symmetric about the origin has an

inscribed square [3].

We can prove this easily. Let J be a simple closed curve in the R2 plane which

is symmetric about the origin. Therefore for every point P on J there is a point

−P , also lying on J , with coordinates that are the negative values of those of P .

Let f : R2 −→ R2 be the function that rotates each point on the plane 90 degrees

about the origin as shown in Figure 4. To prove that J has an inscribed square

12

we must first prove that J and f(J) have a point in common. We will call this

point, f(P ).

Figure 4: J , f(J) and their Inscribed Square.[12]

The four points P , f(P ), −P and −f(P ) form the vertices of a square. To com-

plete this proof we need only to prove what we previously assumed, that J and

f(J) have a point in common.

Let Pnear be the point of J that is the minimum distance to the origin and Pfar be

the point that is the maximum. The function f does not change a point’s distance

to the origin and therefore f(Pnear) and f(Pfar) are as close or far, respectively,

from the origin as any point of J . We can then deduce that f(Pnear) is inside or

on J and f(Pfar) is outside or on J . This proof is complete if either of these lie

on J as that would be the common point of f(J) and J . If they do not lie on J ,

f(J) must connect a point inside J to a point outside J which can only happen

if it crosses J . Therefore we can conclude that there is a common point between

J and f(J) and furthermore that every curve that is symmetric about the origin

has an inscribed square [12].

2.3.2 Other Inscribed Shapes

Whilst proving that a simple closed curve has an inscribed square has been im-

possible so far, many people have used this idea to prove the same thing for other

quadrilaterals, triangles and polygons. In this section we will look at proofs and

examples for each of these shapes.

Theorem 1.4: Every Simple closed curve has lots of inscribed parallelograms

13

and lots of inscribed rhombuses [13].

Two climbers starting at sea-level at opposite ends of a mountain range can

coordinate their movements so that they are always at the same elevation and

therefore reach the summit at the same time. In order to do this, one of the

climbers will often have to reverse her course to allow the other to proceed up

or down a valley or peak making sure they stay at the same height before they

both continue onwards again. This is often referred to as the ’Mountain Climbing

Problem’ and was proved by Tatsuo Homma in 1952 [13].

Using the Mountain Climbing Problem, we can prove that any simple closed

curve, J , has an inscribed rhombus with two lines parallel to any line, L. For

simplicity, we will assume that L is the x-axis. The base of the mountain, P , will

have the minimum y-coordinate. Taking this as 0 allows P to lie on the x-axis,

L. The point with the maximum y-coordinate, Q, symbolizes the summit of the

mountain.

If we have two climbers starting at P , climbing upwards in a way that satisfies

the Mountain Climbing Problem, and two climbers starting at Q, repelling in a

similar manner, then we have two parallel lines that are also parallel to L. We

then require the climbers to coordinate their efforts so that the distance between

the ascending pair matches the distance between the descending pair. This will

require one of the pairs to move backwards sometimes, once again proved by

Homma [13]. If all four climbers stay in coordination with each other, the lines

between them will always form a parallelogram. This will be long and skinny at

Figure 5: Parallelogram Inscribed in J [12].

14

the start becoming increasingly shorter and fatter as the climbers become closer

together. At some point the shape created will be a rhombus. This rhombus will

have two lines parallel to L and can be found in any simple closed curve.

Theorem 1.5: Every simple closed curve has at least one inscribed rectangle.

In 1977, H.Vaughan proved, using basic topology, that there are four points on

any Jordan Curve that are the vertices of a rectangle.

Two objects are thought to be the same if there is a homeomorphism between

them. A homeomorphism is an instance of topological equivalence and therefore

requires a one-to-one connection.

Let J be a simple closed curve in the plane. J is homeomorphic to the standard

unit circle, C, and therefore the Cartesian product J × J is topologically similar

to a torus, C × C. A torus is commonly thought of as the square [0, 1] × [0, 1],

illustrated in Figure 6.

Figure 6: Points on a Torus [12]

Let X represent the set of all pairs of points on the curve J . A pair of points

on J would therefore be a single point in X. Topologically we can think of X

as what we get when we ’fold’ our torus over onto itself matching a point,(s, t),

with its reflection,(t, s), as shown in Figure 6. We now use a cutting and pasting

process to show that X is a Mobius strip, see Figure 7. The Mobius strip is a

non-orientable surface with only one side and one boundary [12].

Now, Imagine J lying on the plane z = 0 inside the x,y,z-space, R3.

Define a function f : X → R3 by the following rule:

• f(P,Q) is the point in R3 lying directly over the midpoint of the segment

PQ but with z-coordinate equal to the distance from P to Q.

15

Figure 7: Creating Mobius Strip [12]

The distance from P to Q becomes very small when the points, (P,Q) of our

Mobius strip, X, get close to the boundary of Z. The z-coordinate of f(P,Q)

becomes very close to zero. Using this we can see that f : X → R3 sews the

boundary of the Mobius strip, X, to the curve J and takes it into the region

above the x,y-plane.

Figure 8: J lying on R3 plane.[12]

An example of a compact non-orientable 2-dimensional manifold is the real pro-

jective plane. It can’t be embedded in standard 3-dimensional space without

intersecting itself [11]. The space that is formed when sewing a disk to a Mobius

strip is a projective plane and therefore can not be embedded without intersecting

itself. Therefore to obtain f(P,Q) = f(P ′, Q′) there must be two pairs of points,

P,Q and P ′, Q′. We now know that PP ′QQ′ is a rectangle as the segments PQ

and P ′Q′ meet at their midpoint and have the same length.

Thus, using basic topology, we have proved that every simple closed curve must

have at least one inscribed rectangle.

Theorem 1.6: Every simple closed curve has an inscribed equilateral triangle

[14].

This is a very simple proof constructed by Mark D. Meyerson in 1980 [9]. We

start with a simple closed curve, J , and a small circle in the interior that meets

J at the point x. y0 and z0 are other points on C, placed such that x, y0 and

16

z0 form an equilateral triangle. y and z are variable points which move about

the plane under the restraint that x,y and z will always be the vertices’s of an

equilateral triangle.

Let y = y0 and z = z0. Keeping xy = xz, let y and z move radially outward from

x through y0 and z0 until y or z meet J . If either of these points are already on

J then there is no need for the rotation. We can now set y1 = y and z1 = z.

Assume y1 now lies on J along with x. y2 is a point on J maximally distant from

x. Now let y move continuously along J until it reaches y2 whilst maintaining

the condition, xy = xz. As y moves from y1 to y2, z describes a continuous

curve from z1 to some point, z2. The arc created by z is a 60◦ rotation of the arc

described by y. Since xy2 = xz2, z2 must lie on or outside J . Since z1 lies inside

J , the arc must have crossed J at some point. At this point, x, y and z all lie on

J and are the vertices of an equilateral triangle. This process can be referred to

as The Goldilocks Effect as the first triangle is too small, the second too big and

the final fits perfectly. See Figure 9.

Figure 9: Goldilocks Effect with Inscribed Triangle. [3]

Theorem 1.7: If J is any simple closed curve and T is any triangle then J has

an inscribed triangle similar to T . It actually has so many similar to T that the

vertices of all these triangles are “dense” in the curve J [3].

The first part of Theorem 1.7 is proved using the same method as Theorem 1.6,

the only difference being that there is no need for the triangle to be equilateral.

This only proves that there must exist an inscribed triangle somewhere on a

Jordan curve but does not provide any information on where the triangle will

occur [12]. The second part of Theorem 2.4 shows that every point on the curve,

J , is close to a vertex of one an inscribed triangles.

Let J be any simple closed curve, T be any triangle and V be the set of points on

17

J that are vertex to a triangle, similar to T , that is inscribed in J [12]. Since some

points of V are arbitrarily close to any given point of J , the set V is described as

dense.

Let A,B and C be the vertices of a triangle. Let P be a point on J and define a

function f : R2 −→ R2 by the following rule:

• f(P ) = P ,

• If the point Q, Q 6= P , lies on the curve J in such a way that the triangle

PQf(Q) is directly similar then P will correspond to A, Q will correspond

to B and f(Q) will correspond to C.

Hence, f amounts to a rotation about P followed by either a stretch or shrink

(dilation) with center P .

The image f(J) of the curve J will also be a curve passing through the point,

P . If J , and hence f(J), are relatively well-behaved at P these curves will cross

each other at P . If f(J) crosses J at P , then part of f(J) must lie inside of J

whilst the other part lies outside. Therefore f(J) must meet J at another point.

This means that there is a point, f(R) which appears on both curves. Therefore

the points P,R and f(R) all lie on J and form a triangle similar to T ,as shown

in Figure 10. Hence, any point P will be a vertex to an inscribed triangle similar

to T as long as it is “nice enough” to make J and f(J) cross. Mark Meyerson

Figure 10: Triangles Inscribed in J.[12]

[9] proved, using some complex topological arguments, that “nice”, well behaved

points are so common in the curve J that they are dense and therefore we can

find one as close as we want to any other point on J . [12]

18

3 Ducci Sequence

3.1 Introduction

The Ducci sequence, also known as the n-number game, is a simple iterative

problem attributed to E.Ducci. It explores the limiting behaviour of finite strings.

The result below was first proved by Ciamberlini and Marengoni [4].

Let f : Rn −→ Rn be defined as:

f(x1, ..., xn) = (|x1 − x2|, |x2 − x3|, ..., |xn − x1|). (5)

The zero-string is the string whose terms are all zero.

Theorem 2.1: Every integer string of length n will iterate to the zero-string in

a finite number of steps if and only if n = 2m for some positive integer m.

3.2 The 4-number Game

Figure 11: Ducci Sequence Visualisation. [17]

In this report we will focus on the 4-number game. To start we will demonstrate

the ‘game’. Choose four numbers and place them at the corners of a square. At

the mid point of each edge, write the difference between the two adjacent numbers.

Always subtract the smaller one from the larger to avoid negative numbers. The

four new numbers now become the corners of a smaller, rotated square. The

Ducci Sequence describes what happens when this process is repeated. Figure 11

illustrates the first four steps of an example using the numbers 1,5,3,2.

The results can be seen more compactly using a table. By completing the first

example and trying a few more, shown in Figure 12, we can see the expected

result.

19

1 5 3 2

→ 4 2 1 1

→ 2 1 0 3

→ 1 1 3 1

→ 0 2 2 0

→ 2 0 2 0

→ 2 2 2 2

→ 0 0 0 0

3 2 6 11

→ 1 4 5 8

→ 3 1 3 7

→ 2 2 4 4

→ 0 2 0 2

→ 2 2 2 2

→ 0 0 0 0

0 653 1854 4063

→ 653 1201 2209 4063

→ 548 1008 1854 3410

→ ... ... ... ...

→ 0 0 128 128

→ 0 128 0 128

→ 128 128 128 128

→ 0 0 0 0

Figure 12: Examples of Ducci Sequence

Are there any examples that don’t end with a row of zeros?

Lists of examples never suffice as a full proof. One way of proving that the

4-number Ducci Sequence will always end in the zero-string is shown below:

If Q = (a, b, c, d) then Q′ is derived by taking the modulus of the differences

between each component. i.e The first entry of Q′ is the absolute value |a − b|.

Using this terminology we can see the following result:

Q = (a, b, c, d)

→ Q′ = (|a− b|, |b− c|, |c− d|, |d− a|)

Observation: If Q 6= (0, 0, 0, 0) then Q′ seems to be smaller than Q.

Since entries are always non-negative, for this observation to be true, our game

must end in the zero-string because a decreasing sequence of non-negative inte-

gers cannot do anything else. First we must define what it means for one row

to be smaller than another. The row size will not change so we cannot use that.

The sum of all the components does not work either. To decide the ‘size’ of a

row we will consider the maximum entry. When Q = (a, b, c, d) , let max(Q) be

the largest of the four numbers in Q. The Ducci Sequence uses only subtraction

of non-negative numbers, therefore max(Q) cannot increase throughout the iter-

ations. In most cases the maximum values of two iterations will be equal at some

point. For example, when Q = (3, 0, 0, 0) max(Q) = 3 and max(Q′) = 3. We can

then state; max(Q′) 6 max(Q).

Since there is a possibility of equality in maximal entries, we must add another

20

step to our proof using the following claim:

Claim: For any row R, at least one of the first four iterated rows must have all

even entries.

Let “e” and “o” describe an even and an odd number, respectively. Using the

fact that e − e = e, o − o = e, o − e = e and e − o = o, we can iterate every

combination of odd and even numbers to prove that a row with all even entries

will appear in first four steps.

for example,

Q = (e, e, o, o)

→ Q′ = (e, o, e, o)

→ Q′′ = (o, o, o, o)

→ Q′′′ = (e, e, e, e)

This result appears for all combinations of odd and even numbers. Now we

can prove that the game must stop at the zero-string for any initial row Q.

First we must run the game until a derived row, S, is all even. We can write

this row as; S = (2w, 2x, 2y, 2z) for some whole numbers w, x, y and z. Let

T = 12S, i.e T = (w, x, y, z). The next step is to replace S by T noting that

max(T ) < max(S). Continue the game, taking out a factor of 2 every time a

completely even row appears. This forms a decreasing list of positive integers

and therefore has to end in (0,0,0,0) at some point.

We have proved that the Ducci Sequence will end in the zero string for any four

numbers but we do not know how many iterations a sequence will take to achieve

this. A code, created by myself using the statistical package,R , calculates how

many iterations a 4-number chain will take to get to the zero-string. This code is

shown in Figure 13. By inputting any 4-number vector, x, into my code, it will

be reduced to (0, 0, 0, 0) using the method above and the number of iterations

taken to get there will be produced.

We can use this code to find that in Figure 12 the fourth example would have

taken 23 iterations to reach the zero string.

21

counter<− 0

while(any(c(x[1]>0, x[2]>0, x[3]>0, x[4]>0))){

x<− c(abs(x[1]-x[2]), abs(x[2]-x[3]), abs(x[3]-x[4]), abs(x[4]-x[1]))

counter<− counter + 1;

print(counter);

print(x)

}

Figure 13: Ducci Code created in R

What happens when irrationals are used?

From Figure 14 it appears that the Ducci Sequence works for all irrational

4-number String Iterations

0, 653, 1854, 4063 23

2, 77, 89, 3959 5

12, 1

5, 3

5, 4

54

0.846, 2.4874, 0.3467 , 0.912 7

π, π2, π

3, π

26

2√

15, 4√3

7, π ,

√19 9

Figure 14: Iterations for Sequences using R Code

numbers too. Although this may be the case, our proof that the game will end

in zero’s does not work as the assumption that a fully even row will appear is

not true for irrational numbers. Surprisingly, no one has created a proof using

irrational numbers and therefore we can not say with any confidence that there

are no cases when the four-number game is infinite.

4 The Four Color Theorem

4.1 Introduction

What is the least number of colors needed to color a map so that no

two adjacent regions share the same color? [12]

22

The Four Color Theorem proves that the answer to this question is 4.

The oxford dictionary provides a very generalized description of a map: “A di-

agram or collection of data showing the spatial arrangement or distribution of

something over an area”. This definition can describe geographical maps, astro-

nomical maps and maps used in various areas of science, such as electron density

maps [16]. In this section we will focus on maps representing Countries, Counties

and states. Maps display a huge amount of information and can be easily under-

stood and interpreted. One factor that contributes to this understanding is the

use of different colors to readily distinguish between neighboring regions. The

Four Color Theorem states that every map can be colored using only 4 colors.

According to this theorem if two regions touch at only one point, they are able

to be colored the same. A map of the United States Of America is an example

where no fewer than four colors can be used.

Figure 15: Map of States using Four Colors [1]

4.2 The History of the Four Color Theorem

The Four Color Problem began in the early 1850’s and intrigued many mathe-

maticians and scientists for 120 years before a proof was found. The theorem is

very well known as it’s solutions are so complex that it launched the study of chro-

matics and graph theory along with fueling a heated controversy over computer

generated proofs [2].

Francis Guthrie developed the problem whilst studying in 1950 after discovering

that it only took four colors to distinguish between counties in England. He then

passed on his ideas to many influential mathematicians in search of a proof. The

23

most notable of these was Augustus De Morgan, the inventor of Mathematical

Induction, who is now accredited for the popularity of the problem. The next

major step towards finding a proof was made by Alfred Kemp in 1879 who used

a method called Kempe Chains. His proof was perceived as correct for 11 years

before a flaw was found by Percy Haewood in 1890 [2]. Along with finding

the flaw, Haewood modified Kempe’s study and proved that any map is five

colorable. In the next few decades notable advancements were made from Errera,

Tait, Petersen and Berkhoff creating concepts used in the final proof such as

unavoidable sets and a trivalent vertex. In 1950 Heesh followed Kempe’s train of

thought deciding that the only way The Four Color Theorem could be proven

was by finding an unavoidable set of reducible configurations. To carry this

out in practise would have involved approximately ten thousand configurations

[2]. Heesch used technology and started a proof leading to the development of

discharging, an integral concept of Appel and Haken’s full proof. The Four Color

Theorem was successfully solved in 1976 by Appel and Haken. This was not

accepted by many mathematicians due to it’s reliance on technology. Simpler

proofs by Robertson, Sanders, Seymour and Thomas have been created since

then but, despite much controversy, they still rely on computer technology and

Theorem Proving Software.

4.3 Kempe Chains

During Kemp’s research he created a theory called Kempe Chains, a method of

rearranging colors on a map by switching adjacent colors. Many consider this

to be the foundation of the Four Color Theorem. Kempe Chains are created as

Figure 16: Kempe’s chain Layout [2]

AC and BD chains which cannot cross each other, as shown in Figure 16. The

manipulator can choose between using the AC chain or the BD chain. Using

24

Figure 16 as a Kempe Chain model, an AC chain can be utilized by switching

the C to an A and using the C in the blank middle block. The BD chain can

be used in the same way. This is the basic idea behind Kempe’s failed proof of

the Four Color Theorem. He believed that if a simple map, similar to Figure 16

can be colored with only four colors then a more complex map could be reduced

to a simple map and thus four colors would also suffice. After finding a flaw in

Kempe’s work, Haewood used the theory behind Kempe’s chains to formulate a

proof using 5 colors.[2]

4.4 The Solution

Mitchem [10] proposed that there are three main parts in proving The Four Color

Theorem. The first of these is to find an unavoidable set. Unavoidable means

that every map must be made of countries surrounded by only two, three, four or

five countries. A degree is the number of edges originating from each vertex. A

group of two, three,four or five countries that all border one country are considered

to be a set of configurations of the map. Furthermore, unavoidable sets are of

degree 5 and must contain one member of the set of configurations. Secondly,

Mitchem suggests that we must show that each element of S is reducible.

A configuration is thought of as reducible if every part of the graph can not be

dis-proven with a proven counterexample using known configurations. This is the

step that Kempe’s proof missed out. We must then introduce Discharging.

The basic idea behind discharging is to give every vertex in a map charge. Apple

and Haken underwent a labelling process where each vertex was given a charge

using the algorithm 6− k, where k is the degree.Using this all vertices of degree

less than or equal to 5 will have positive charges and all vertices of degree greater

than 5 will receive negative charges. For example 6− degv describes the vertex v

with charge 6. Euler’s theorem states that the overall charge of a graph must sum

to 12. Discharging moves around the charges on the graph whilst keeping the sum

at 12. The discharging procedure proves that no element of the set S can be in

the graph and therefore proves that S is unavoidable. Discharging moves charges

around so that all vertices have a positive charge and the resulting distribution

25

must be in a reducible configuration [2].

Appel and Haken created an algorithm that used reducible sets consisting of

less than two thousand configurations in order to solve the Four Color Problem.

Although techniques and discharging procedures were checked by hand, most of

the work was done using a computer [10]. Other mathematicians have since found

ways to reduce the number of configurations but no-one has found a way to prove,

what we now know is true, without the use of technology.

5 Rearranging Four Squares

5.1 Introduction

The Wallace-Bolyai-Gerwein Theorem: Any two Polygons of equal area are

equidecomposable. [12]

In other words, one of the polygons can be decomposed into a finite number of

pieces and rearranged to form a different polygon. In this section we will show

each of the steps needed to prove the Wallace-Bolyai-Gerwein Theorem and then

explore the puzzle attributed to Henry Dudeney involving only four pieces.

5.2 The Wallace-Bolyai-Gerwein Theorem

Bolyai first formulated the question in the early 19th century and it was proved

independently by Bolyai and Gerwein in 1833 and 1835 respectively. The Wallace-

Bolyai-Gerwein Theorem can be formulated in many different ways. The concept

of “equidecomposability” of polygons is the most common. If two polygons can

be split into finitely many triangles that only differ by some isometry they are

said to be equidecomposable. The Wallace-Bolyai-Gerwein Theorem states that

polygons must have the same area to be equidecomposable. [7]

The theorem can be understood and proved in very few steps. Every polygon

can be cut into triangles. This is easiest for convex polygons as we can cut off

each vertex in turn. This is not possible for concave polygons and thus another

method is required. We must first choose a line, L, not parallel to any of the sides

of the polygon and then draw a line parallel to L through each of the vertices

26

of the polygon. These lines will divide the polygon into triangles and trapezoids

which can then be converted into triangles. Now that we have have only triangles

we must transform each into a right angled triangle and subsequently a rectangle

with side of length 1. This can be achieved for any 2 polygons and therefore any

two polygons of equal area are equidecomposable.

Figure 17: Wallace-Bolyai-Gerwein Puzzle [6]

5.3 Four Piece Puzzle

This Puzzle is a more general version of the Wallace-Bolyai-Gerwein Theorem,

asking how to decompose an equilateral triangle into four pieces that can be re-

arranged to form a square, as shown in Figure 18. Figure 18 illustrates how the

Figure 18: Reversible 4-piece String [6]

pieces form a chain which forms a triangle when it is closed one way and a square

when closed the other. This puzzle has a surprising practical use in the building

industry where it has helped design brass hinges.

Henry Dudeney specialised in logic puzzles and mathematical games. Many in-

teresting puzzles, including the Four Piece Puzzle can be found in his book “The

Canterbury Puzzles And Other Curious Problems” [5].

27

6 Conclusion

Reviewing the first section, “The Inscribed Square Problem”, we see that many

shapes are inscribed in all Jordan Curves by use of topology and geometry. So

far the only way to prove that a square too, must be inscribed is to imply some

implicit tameness on the Jordan Curves such as symmetry or Local Monotonicity.

The Inscribed Square Problem may or may not have a full solution. No mathe-

matician has found a way to prove it thus far.

In section 3 we studied “The Ducci Sequence”, particularly the 4-number string.

We can conclude that, using the Ducci Sequence Process, any 4 rational num-

bers will result in the zero-string. More complex results of the Ducci Sequence

look into n-number strings and can be found in “Unbounded Ducci Sequences”

[4] amongst other research papers.

“The Four Color Theorem” was explored in section 4. Many believe that the

Four Color Problem will remain unsolved until someone finds a solution without

a computer. In a world relying more and more on technology, should computer

aided proofs be accepted? While this is a very contraversial issue in mathematics,

it can at least be noted that for now without computers the Four Color Theorem

would still be the Four Color Problem.

Lastly, we introduced the Wallace-Bolyai-Gerwein Theorem, specifically looking

at the special result involving 4 pieces. This puzzle has inspired many puzzles and

games, including Rush Hour ; a game where the player is expected to rearrange

cars on a square board in order to escape a traffic jam.

To conclude, the number 4’s importance in Maths, among other things, is infi-

nite. We have seen it shine throughout 4 different topics and there are countless

more to explore. Is the number 4 more important than other numbers? I believe

that is a matter of opinion. My answer is, probably not. If interested, I strongly

recommend reading “Single Digits: In Praise of Small Numbers” by Marc Cham-

berland [3] in order to appreciate the use and importance of all small numbers in

some very interesting mathematical processes.

28

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color.htm. Last accessed 20th Nov 2015.

[2] Calton, K. (2009) Four Color Theorem. Unpublished thesis (Msc.) The Uni-

versity of Texas.

[3] Chamberland, M. (2015). The Number 4. In: Single Digits: In Praise of

Small Numbers. Princeton: Princeton University Press. 111-131.

[4] Chamerland, M. (2003). Unbounded Ducci Sequences. Journal of Difference

Equations and Applications 10 (10), 887-895.

[5] Dudeney, H (1958). The Canterbury Puzzles and Other Curious Problems.

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[16] Oxford English Dictionary (2007). Vol.2. 6th ed. Oxford: Oxford University

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