Exploring The Einstein Problem
Thesis submitted at the University of Leicester in partial ful llment of the requirements for the degree of Master of Mathematics
by
Zoe C. Allwood Department of Mathematics University of Leicester
May 2014 Contents
Declaration iii
Abstract iv
Introduction 1
1 Describing the Tile 4
1.1 Lines and Flags ...... 4
1.2 Lines and Coloured Diagonals ...... 5
1.3 Other Representations ...... 6
2 Proof of Aperiodicity 7
3 The Substitution Tiling 12
3.1 Simplifying the Tile Set ...... 13
3.2 The Substitution ...... 17
3.3 The Substitution on the Simpli ed Tile Sets ...... 19
3.4 Amalgamation ...... 22
3.5 Proof of Aperiodicity ...... 22
4 A Better Einstein 23
4.1 Unconnected Tile ...... 23
4.2 3D Tile ...... 24
4.2.1 Construction ...... 24
4.2.2 Issues ...... 26
i 4.3 Schmitt, Conway and Danzer Tile ...... 26
4.4 Issues ...... 27
4.4.1 Height ...... 27
4.4.2 Tile Orientation ...... 27
4.4.3 Aperiodic Layers ...... 28
4.5 Danzer Implementation Attempt ...... 28
4.5.1 Rotation of 2π ...... 28 3 4.5.2 Irrational Rotation ...... 30
4.6 Aperiodic Stack ...... 31
4.6.1 Thue Morse exploration ...... 32
4.6.2 General explanation ...... 33
4.6.3 Two dimensional exploration ...... 33
5 Conclusions 34
5.1 Socolar and Taylor Tile ...... 34
5.2 Einsteins and Dimension ...... 35
Bibliography 36
ii Declaration
All sentences or passages quoted in this project dissertation from other people's work have been speci cally acknowledged by clear cross referencing to author, work and page(s). I understand that failure to do this amounts to plagiarism and will be considered grounds for failure in this module and the degree examination as a whole.
Name:
Signed:
Date:
iii Abstract
The Einstein problem is an attempt to nd an aperiodic prototile set that consists of a single tile. After some basic de nitions and a more detailed explanation of the problem we look at the tile that ts this de nition the best, created by Socolar and Taylor. The
rst chapter is an explanation of two of the ways this tile can be de ned. The second chapter is a proof of the non-periodicity of all tilings created by this prototile based on the hierarchical structure of them. The third chapter is looking at the tile set as a substitution tiling creating an easier way to make tilings in a computer and also show all tilings are non-periodic more easily. The fourth shows other constructions of the tile, including a three dimensional one, and explorations into trying to improve the tile by trying to remove some of the issues with the tile in its current forms.
iv Introduction
The Einstein problem is very simple to explain but very hard to nd a solution to, like
Fermat's last theorem. It is the problem of whether a single aperiodic prototile exists. The name comes from a play on words as Einstein in German can be translated roughly as "one stone". As Fermat hinted in the margin of a book regarding his problem, there is quite possibly a simple and rigorous way to nd solutions to this type of problem, but at the moment we are working with trial and error. If a solution, or groups of solutions, can be found it will have a profound eect on the understanding of certain crystalline structures which are aperiodic and can be understood better in terms of these tilings. The current solutions to this problem are not complete; there are many ways that the current tiles can be improved. In this work we explore a tile that is a good Einstein, found by Socolar and
Taylor [2, 3], show it is aperiodic in a way that is complementary to the original proof. We will then discuss the issues with it and try and look at ways to improve it.
Basic Notions
These are some de nitions that we are going to be using throughout this work. The general details of tilings are explored thoroughly by Grunbaum and Shephard[1].
n De nition 0.1. A tiling is a subdivision of R into pieces called tiles. These tiles only n intersect at their boundaries and their union is R .
Tiling is most commonly done in 2 dimensions, tiling the plane, but an equivalent is possible in any dimension. In one dimension line segments can be used to `tile' the line. A tile can be given a label, letting you distinguish between tiles of the same shape.
1 Thinking of these labels as colour would dierentiate between a simple square tiling and a checkerboard tiling.
De nition 0.2. Prototiles are the set of tiles that make up a tiling ignoring position.
For the checkerboard tiling the prototile set would be a black square and a white square. Another way to think of a tiling is as a collection of translates, translations and/ or rotations, of a prototile set so that each translate only intersects with others at its
n boundary and the union of the collection is R .
De nition 0.3. Matching rules are the guidelines for how tiles are allowed to be placed in regards to each other.
The rule that tiles must meet full edge to full edge is an example of a matching rule that is generally applied to tilings. The two ways to force a matching rule are by markings on the tile or by the shape of the tile. The relationship between markings on the tiles can be de ned in any way that creates the tilings wanted and most of these relationships can be de ned instead using shape, though this can create very complicated tiles when dealing with complex matching rules.
De nition 0.4. A patch is a nite section of a tiling.
If you have a nite set of protiles and a set of matching rules, for example having to meet full edge to full edge, then you can only create a nite number of patches. This number generally increases very quickly as you increase the possible size of the patch but at any nite size there are only a nite number of possible patches. This is called having
nite local complexity.
De nition 0.5. A tiling is called periodic if the whole tiling is created by regular repetition of a nite patch of the tiling. This means the placement of one version of this patch can completely de ne the position of all other tiles in the tiling. It is called non-periodic if this is not the case.
Using the example of the checkerboard tiling, you can determine what colour of tile will be a certain distance from a known tile in the tiling without having to work out all the tiles
2 in between because of the regularity. If you took a standard square tiling and randomly split each square into two triangles by adding in one of the diagonals there is no way to deduce which diagonal was added using already known ones as it is completely random.
This created tiling is therefore non-periodic. Though there is not any regularly repeating pattern, because the tiling has nite local complexity there are only nite patches of every size. Because of this there will be in nitely many copies of every patch, but they will not appear with any regularity.
There are many prototiles sets that can produce some periodic tilings and some non- periodic tilings as well. If you add the same diagonal to every square, instead of a random choice, you would get a periodic tiling from the same triangular prototile as the non-periodic tiling already de ned.
De nition 0.6. A prototile set is called aperiodic if every tiling it can create is non- periodic.
These are the prototile sets we are interested in. A famous example of these are the
Penrose tilings but with these there are always at least two dierent tiles in the prototile set, so they are not an Einstein.
Using the de nitions above an Einstein is an aperiodic prototile set consisting of a single tile. This means that using a single tile you can cover the entire plane without ever having a regularly repeating pattern. Because of the nite local complexity every patch of the tiling will repeat but this will never happen regularly. The tile we will describe below
ts this de nition.
3 Chapter 1
Describing the Tile
Though Socolar and Taylor have de ned a single tile, created from a hexagon, it can be useful to describe it in dierent ways. The matching rules cannot be de ned simply by shape, unless the tile becomes unconnected or you use a substitution with multiple tiles.
Also you need to use the fundamentally dierent mirror image to successfully tile the plane; you do not need this if you make the tile three dimensional. All these interpretations are explained in this work. Any pictures that do not have a reference were created for this work.
1.1 Lines and Flags
The prototile is a hexagon with markings to enforce local matching rules. There are black lines, two short lines around the top and bottom vertexes and a long one connecting the two remaining edges o center. The distance between the end of the short black line and nearby top or bottom vertex must be the same distance as the end of the long black line and the top vertex of the edge it is touching. At every vertex there is a ag. The top and bottom ags always point the same way and the other four ags point away from the long black line. As the long black line is o-center of the edges it touches this prototile is chiral, its mirror image is fundamentally dierent to itself. This mirror image will look the same except the ags at the top and bottom vertices will be pointing the other way. For this tiling to work we need both the tile and its mirror image. The local matching rules in
4 Figure 1.1: The tile, its mirror image and an allowed patch of tiling respect of these markings are as follows:
R1: All black lines must be continuous
R2: The ags at either end of a hexagonal edge must be pointing in the same
direction.
Though this interpretation is helpful in visualizing the rules on a small scale the ags quickly become hard to keep track of.
1.2 Lines and Coloured Diagonals
To make visualization easier the ags can be recoded as coloured lines along the diameter towards a vertex. Blue radii for ags pointing clockwise, red for anticlockwise. This will make the top-bottom diameter half of each colour, later on called the red blue diameter, and the other two all one colour, one red one blue. This does not aect the black lines or change the fact that the prototile is chiral but does change the phrasing of rule R2:
R1: All black lines must be continuous
R2: The diameters at either end on a hexagonal edge must be dierent colours
This is the main representation used throughout this work.
5 Figure 1.2: The tile, its mirror image and an allowed patch of tiling
1.3 Other Representations
Later on in this report we will look at other related representations. All of the tilings created by this tile can also be created by substituting a group of tiles. You need 28 half hexagons, 14 distinct with their mirror images, that are substituted for 4 every time you do the substitution. It is possible to simplify the substitution but we cover this in-depth in chapter 3. When trying to nd a better tile we base it o of a combination of an unconnected version of the tile and a three dimensional version. These are de ned and explored in chapter 4.
6 Chapter 2
Proof of Aperiodicity
The basic aim of this proof is to show there are rings of ever-increasing size created by the black lines on the tile. As there is no largest ring created there is not a single patch that can be repeated to tile the plane, showing that the tiling is always non-periodic. The proof will be broken down into smaller theorems that together will prove the aperiodicity of the prototile. We also need to prove that such a tiling of the plane does exist, otherwise it could just create a nite patch. The format of tile that we are using is the connected tile with short black lines around the top and bottom vertices and a long black line connecting the other two edges. Each diameter is also coloured (three red, three blue). The top and bottom halves of the vertical diameter are dierent colours; the remaining two diameters are all one colour. It is obvious that the regular hexagons without markings can tile the plane so we need to check that the markings can ll all tiles without a contradiction and without repeating.
Theorem 2.1. A small ring, created by three short black lines, occurs in every possible tiling.
Proof. If at a given vertex there are already two short black lines the only way to preserve continuity is to use a third short black line to complete a small ring. If there is a short black line connected to a long black line this forces the other side to also be connected to a long black line, to prevent a non-continuous spiral. Those two long black lines force two short black lines to be together on the next vertex, which must create a small ring. Since
7 Figure 2.1: The formation of a small ring, either at the vertex in question or round its neighbour
Figure 2.2: The existence of a ring at both ends of a tile and the uncontinuity of the black lines if we take the partially marked diameter to be red and blue the rst marked tile has a short black line on, one of these two situations must occur in the beginning stages of a tiling and so at least one ring will be created
Theorem 2.2. If there is a small ring at one end of the tile there is also one at the other.
Proof. Regardless of whether you are looking at the tile or its mirror image, once you have placed a short black line or a single coloured diameter on it you can automatically place both short black lines and both single coloured diameters as their position in comparison to each other is always identical. So if we have a small black ring, we choose one of the three tiles to check for a ring at the other end of. With the addition of the known diameters we can determine the colour of half of one of the diameters at the other end of the tile; this limits which diameter it can be. If it is the red blue diameter the short black line must go around this vertex, leading to discontinuous black lines. Therefore the diameter must be one colour, so both short black lines can be placed. Regardless of which side you look at there will now be two short black lines at the other end of the tile, forcing a third.
As there are three tiles creating a small ring, each of them must have a small ring at
8 Figure 2.3: A section of the L1 lattice the other end. Expanding this idea a lattice of small rings is created; a collection of single unmarked tiles surrounded by tiles with their short black lines decided. These partially marked tiles are labeled L1 tiles; the rest are currently called unmarked.
Theorem 2.3. Unmarked tiles must follow R1 and R2 amongst themselves.
Proof. Marking an unmarked tile, called X in this proof, forces a lot of the currently undecided marks on the nearby L1 tiles. Wherever a black line meets the edge of X it must be met by the long black line of the L1 tile, as the short ones have already been placed. The next unmarked tile must have a black line in line with X for the continuity to hold, preserving R1. If the half of a diameter of X at a vertex is blue, The L1 tile at the other end of the edge has its red blue diameter determined. If you look an edge away from the opposite vertex of this L1 tile you nd an unmarked tile, whose nearest vertex must be red to dier from the blue half of the red blue diameter. If you swap the colours this logic still holds and R2 is preserved among the unmarked tiles. See gure 2.4.
If you mark larger hexagons around each unmarked tile, taking half of each neighbouring
L1 tile, you get a new hexagonal tiling completely determined by these central unmarked tiles. As these follow R1 and R2 amongst themselves you can treat this new tiling the same way as the original. A small black ring amongst unmarked tiles translates to a black
9 Figure 2.4: Example of R1 and R2 amongst unmarked tiles. Note that a lot more is de ned but removed for clarity and the marking on the top right tile could be the long black line. ring formed of three short black lines with a long black line between each; these are called
L1 rings with the original rings being L0 rings.
These rings determine L2 tiles, partially marked tiles that create a lattice of L1 rings from identical logic of theorems 2.1 and 2.2 on the larger tiles. This leaves unmarked tiles surrounded by L2 tiles, themselves each surrounded by L1 tiles.
Theorem 2.4. These new unmarked tiles must follow R1 and R2 amongst themselves.
Proof. Marking an unmarked tile, Y , with a black line touching an edge forces the place- ment of the long black line of the L1 tile, the L2 tile and the other L1 tile extending out from that point where it will reach another unmarked tile, preserving R1. If half a diameter of Y is red, looking radially out, the next three tiles met will be L1, L2 and L1 respectively all with blue towards Y and red away, preserving R2 with the unmarked tile met next.
Using the fact that these rules still hold you can iterate this process inde nitely, creating larger and larger rings and lattices of tiles to in nitely large constructions. The issue is that at every level there are some unmarked tiles, so this does not show that a complete tiling can be created.
Theorem 2.5. A complete tiling of this form can be created.
Proof. By going back to the placement of L2 tiles it is clear to see that 3/4 of the unmarked tiles become partially marked leaving 1/4 still unmarked. So at every iteration the number
10 Figure 2.5: A section of the tiling when all iterations of the construction are done. The manual placement of the blank tile will determine all unknown lines and diameters in this diagram. In this case red and blue have been replaced with pink and purple.[2] of unmarked tiles quarters; theoretically this tiles the plane as there are in nite iterations.
If a tile's long black line is determined during the position of Ln tiles, so will its red-blue diameter. So every tile whose long black line is part of a nitely sized ring, created by the placement of Ln tiles, will be completely determined at a nite level. The only tiles we need to consider are ones that are part of a ring of in nite size. As we know there will always be a tile left unmarked, due to the form of the construction, we choose such a tile and designate it as the one that will not be marked as part of any lattice. At the end of the construction this unmarked tile will determine the rings of in nite size, namely lines, which we mark separately. All rings have then been determined and therefore all tiles, giving us a complete, non-periodic tiling.
There is an easier way to show that this tiling is aperiodic by looking at it as a substi- tution tiling.
11 Chapter 3
The Substitution Tiling
Though the nal aim of the Einstein is to have a single tile to be used, exploring the tilings as substitution tilings can be helpful in understanding them better. If you have a substitution tiling you start with a set of prototiles and a patch or full tiling created by those tiles. When you iterate the substitution each tile is scaled up in size and then split into sections each of which is one of the prototiles. As the patch you have gets larger at each iteration this method will eventually ll the plane. As long as the matching rules for the tiling de nitely still hold after the substitution then a valid tiling will be created.
First we will reduce the number of necessary tiles down to 6; however when we perform the substitution on these the next level of tiles will be split amongst the level below. To guarantee that the substitution is eective you have to start with half hexagons as they are the only tiles that substitute completely in themselves. Below is Socolar and Taylor's representation of the 28 half hex substitution tiles [2]
Here all black lines must be continuous and edges can only meet if they have the same letter on; all three letters must match up on the long edges. In this situation A¯ is the mirror image of A, along with all the other letters, making A¯L the mirror image of AR. The numbers around the rst pair are simply a way to numerate the sides. To make visualization easier I have also coded these letters as colours; if the line is dashed then it is N¯. First we will look at joining these tiles so they are easier to work with; we will then
12 Figure 3.1: Socolar and Taylor's tile set using letters and the same tile set using coloured and dashed lines look at the substitution itself and check that it preserves all necessary rules when used on the smaller set of tiles.
3.1 Simplifying the Tile Set
Theorem 3.1. The 28 half hexagons can be combined into 14 hexagonal tiles.
Proof. It is easy to see that each pair can be joined at the long edge; the issue is the additional 4 you get from a combination of C and E tiles
If you look at these and decide which of the other tiles can be put against sides 1 and
6 (the top two) in all cases you nd 4.
13 Figure 3.2: The 4 additional tiles. The mirror image is below each original tile.
Tile Side Possibilities Connecting Side
1 A A¯ B¯ G¯ 3 or 6 H 6 B B¯ F¯ G 5
1 A A¯ D G 5 I 6 D D¯ F G¯ 1 or 4
1 B B¯ F G¯ 5 H¯ 6 A A¯ B G 1 or 4
1 D D¯ F¯ G 3 or 6 I¯ 6 A A¯ D¯ G¯ 2
Regardless of which of those 4 you choose the style of black line from that side will be the same. Sides 5 and 2 have long black lines whereas 1,4,3 and 6 have short black lines.
For each tile you always get a non-continuous spiral so you can never have these tiles in any successful tiling. This means that although they are legitimate tiles we can disregard them from the set of useable tiles, leaving us with the obvious 14 tiles
These tiles are now labeled with the letters they had before without the need for the
L,R su xes.
Theorem 3.2. The 6 hexagonal tiles A,B, G¯, A,¯ B,G¯ only appear in a speci c 3-hex and
14 Figure 3.3: The 14 possible hexagonal tiles its mirror image throughout any possible tiling and must appear in each tiling.
Proof. First, at least one of these tiles must appear in every tiling. If you start with one of them it must be there. By taking every other tile and looking at what can connect with side 1 it has to be one of these 6 except for C. The only tiles with an appropriate red edge for the D tile are A,A,B¯ , and G; for the purple edge of E and F you can only use A, A,¯ B,¯ or G¯. For a C tile The tile at position 1 could be G or D but then as the connection is not on side 1 of the added tile a tile from the above six must again be on side one of the
D tile if G is not there. The mirror images of these tiles follow the same logic. Now we know at least one of them is in the tiling, we look at how they are arranged.
If you start with the B tile, the only tiles that can be connected to side 2 that preserves both line colour and black line continuity is C or C¯. Regardless of which it eventually is, the next edge round must be solid pink. This forces the next tile round to be an A tile.
Also on side 2 of tile A the only possibilities are C and C¯. This necessitates a G¯ tile to complete the small ring at the top of tile B. This argument works regardless of which of the three tiles you start with and works exactly the same way for the A,¯ B,G¯ 3-hex. Taking these two facts together shows that a 3-hex must be in every tiling and it is the only way these 6 tiles will appear.
Looking at the structure created in the above proof it is possible to see that the only options for the nal point in the triangle are C or C¯.
Theorem 3.3. If there is a 3-hex at one end of a C or C¯ tile there must be one at the
15 Figure 3.4: The formation of the 3-hex. The partially marked tiles are C or C¯ and are the only possible tiles in those positions other end.
Proof. The only possible tiles that can connect to sides 1,3,4 and 6 of C or C¯ tiles are the six in the 3-hexes unless it is in the center of a substitution. As we shall see later the position of the rst 3-hex contradicts the latter meaning it must be one of the six, which in turn must be part of a 3-hex.
So a 3-hex must be in every tiling, a C or C¯ tile must be in each dip of said 3-hex and there must be another 3-hex at the other end of each of those tiles. Extending this logic throughout the plane we create a lattice, with room for 7 tiles in each space.
Theorem 3.4. C or C¯ tiles form a regular triangular grid admitting only one tile between them.
Proof. To prove this we need to show that the central tile in each space of the lattice is a
C or C¯ tile. If the central tile was part of a 3-hex the other two hexagons in it would be touching the lattice; there is no way these tiles can be put next to a C or C¯ tile in this way as there are no green edges on them, this is the colour of all edges not used in the formation of the lattice. As shown at the start of theorem 3.2 every tile that is not C or C¯ or in a 3-hex has a tile that must be part of a 3-hex on side 1; meaning a 3-hex would have to be completed and overlap the lattice. This leaves C or C¯ as the only possibilities as the central tile.
16 Figure 3.5: The 7- hex
Theorem 3.5. The only way the tiles can ll the gaps in the lattice is with a speci c 7-hex and its mirror image.
Proof. You are now very limited in your choice of tile. There is no space for the 3-hex and
C or C¯ tiles cannot connect with each other, the black lines cannot be continuous, which leaves you with D,E, F , D¯, E¯, F¯ as possible tiles surrounding the central C or C¯ tile.
Of these 6 tiles only the D and D¯ tiles have a pink edge, so must be placed on side 1 of the C or C¯ tile. Once you have done this there is only one way to place the other tiles around the C and C¯ tile giving you the 7-hex and its mirror image.
This leaves you with 6 tiles; C, the 3-hex and the 7-hex each with its mirror image. we now need to look at the substitution itself
3.2 The Substitution
Theorem 3.4 shows that there is a regular grid of C or C¯ tiles with only one hexagon between them. You can therefore draw larger hexagons, with the C or C¯ tile in the center and half of each surrounding hexagon. When you go back to the half hexagon tiles you
nd that they will all t exactly into one of these larger hexagons and have a very obvious way to split them in half.
What you can see now is how the previous substitution step happened; each of these
17 larger areas was one tile in the tiling before the substitution. This is what we are aiming to arrive at. To do this we will look at all possible ways to arrange half hexes around C and C¯ half hex tiles and make these created groups the substitutions for our set.
These are the possible ways that you can arrange tiles around the CL and CR half hexes.
CL Side 6 Side 5 Side 4 Label CR Side 6 Side 5 Side 4 Label
1 GR DR BL - 1 A¯L D¯L AR AR
2 GR DR GL AL 2 GL F¯L AR BR
3 B¯R F R G¯L BL 3 GL E¯L A¯R F R
4 B¯R F R B¯L F L 4 GL E¯L G¯R ER
5 B¯R ER BL EL 5 DL EL D¯R CR
6 F¯R E¯R F L CL 6 AL F¯L AR GR
7 BR DR G¯L GL 7 AL E¯L A¯R DR
8 BR DR B¯L DL 8 AL E¯L G¯R -
If CL arrangement 1 was part of the tiling there would be no group of CR tilings that could connect with it, similarly with CR arrangement 8, so the two can be disregarded. The label assigned to each valid arrangement is the tile that that arrangement of tiles came from in terms of substitution. The letters of the added tiles correspond to the letters assigned to the edges in Socolar and Taylor's representations of the tiles.
So, to actually perform the substitution on a tile you perform the following actions:
1. Scale it up by a factor of 2 in every direction
2. Add lines such that each shorter edge is now the long edge of a half hex
3. Add the black lines to the half hex of the long edge in the same way as the original
tile and infer the black lines of the other three
4. Choose the label/ colour of each edge by relating each new half hex to the colour of
the side it used to represent
When looking at just the half hexes this is a proper substitution. when we simplify down
18 Figure 3.6: Two half hexes and what they look like after substitution, rescaled. There is one complete hexagon formed and as the two halves were de ned by the same edge they will have the same letter. to our smaller tile set this substitution either must contain parts of the tile split amongst the tiles of the new level, or be a substitution by amalgamation to counteract this.
Theorem 3.6. This substitution preserves the matching rules of the tiling.
Proof. Looking at just the black lines, the C half hex from the long edge will have the same orientation as the original tile. For the black lines to be continuous an extension of the original pattern must be sent over all 4 tiles. Looking at tiles from two neighbouring tile groups the black lines will be continuous as it was on the original set; because of this one must be L the other R. As they represent the same original edge, they must have the same letter. It is clear to see that the L and R of the same letter can match up long edge to long edge so the rules hold
3.3 The Substitution on the Simpli ed Tile Sets
Looking at a hexagonal tile, it is made of two half hexes. When the substitution is per- formed you will get 6 half hexes surrounding two C or C¯ half hexes. These two central halves must match, as the original two halves did, and will form a hexagonal C or C¯ tile. The other halves are on edges and must match with the neighbouring half from the next
19 tile group as in theorem 3.6. As there is a C or C¯ tile at the center of every substitution and one half from each set between them, the regular grid of C or C¯ tiles is still there.
Making the central C or C¯ half hexes one hexagonal tile and joining all other halves with the ones from their neighbours creates a valid set of hexagonal tiles that the substitution can be used on again.
Looking at the set of six tiles we know that most half hexes can only appear in one tile type; C or C¯ can appear by itself or in the 7-hex. We know the edge halves will match up but we must check the pattern of tiles can be recreated after the substitution.
The single C or C¯ tiles are surrounded by tiles that are only found in the 7-hex, so must be the central part of that tile.
With the 3-hex and 7-hex you can nd whole tiles within the substitution and half- hexes around the outside, which can be arranged into sections of tile sets. As the outside tiles are labeled according to the edge labels, they must match up.
So these tiles after substitution will still have the same form, so can be seen as a tile set.
20 21 Figure 3.7: The six tiles divided into their half hexes, them after substitution and then with thick black lines to show where the 6 tiles are within the substitutions. 3.4 Amalgamation
Another way to look at the tiles on this simpli ed set is to slightly alter the substitution so that whole tiles are substituted in. As the C tile is the center of a 7-hex and the other two tiles have sections of these round the edges that match up it is possible to redraw the limits of the substitution such that the C tile becomes a whole 7-hex. To allow for this the equivalent half hexes on the other tiles must be removed from the substitution. This change makes it obvious that we loose the simple ability to scale. There are also almost complete 3-hexes round the edge of the 7-hexes with corresponding sections on the points of the 3-hexes, so a second change of the limits is now needed.
As the scaling is now not exact the actual substitution has to be done collectively; that is replace the tile by its amalgamated substitution, making sure the orientation of the central tile or central vertex is the same as the original, altogether rather than section by section.
3.5 Proof of Aperiodicity
To reverse a substitution is to group all tiles together such that each group represents one tile after substitution, scale them all down and then replacing the group with that single tile. The uniqueness of the way that the substitution can be reversed guarantees that the tiling is non periodic. If there was a regular pattern of the same patch you could treat each of them as the patch that was de ned rst and have multiple starting points.
All of the tilings that can be made by our tile set are non-periodic due to the lack of choice when splitting our tiling into substitution groups. From theorems in this chapter we know that, at the hexagonal level, there is one tile between any neighbouring C or C¯ tiles and that there is one at the center of every tile group after substitution of a hexagon.
These two facts together mean that each tile that is not C or C¯ must be split in half to de ne the after substitution hexagons. As there is only one valid way to split each C or C¯ tile in half we have found the only possible way to reverse this substitution. Since every possible tilling created by our tile set is not periodic it is an aperiodic tile set.
22 Chapter 4
A Better Einstein
In all of the representations we have looked at above you need marks on the tiles to make the system work. If you make the tile unconnected this is completely de ned by shape but creates other issues. Another issue is you need to have at least two fundamentally dierent tiles. Though this is an Einstein as,with the lines and ags representation 1.1, the two tiles are mirror images and can therefore be counted as one; we would like to create one object that aperiodically tiles completely by itself. To do this we can make the two tiles opposite sides of a three dimensional tile. With the addition of this third dimension it is possible to make the unconnected tile become connected. How to created this single three dimensional, connected tile whose aperiodicity is only de ned by shape is explained in stages below. We will then try to improve the tile further by attempting to make the tile aperiodic in this third dimension.
4.1 Unconnected Tile
We are starting our construction from the representation with Lines and Flags1.1. Again we begin with a hexagonal tile. As the ags are relating between tiles that are not touching the shape coding this is not touching the bulk of the tile at all, it is radially out from the vertex it is representing. Using two rectangles of the same size that meet at one corner halfway between the two vertices that are being related and each have an edge exactly on the line radially out from the vertex. If we de ne the rectangle closer to the tile to be o
23 Figure 4.1: Unconnected tile with the Lines and Flags representation for comparison, highlighted sections are equivalent to the side that the ag is pointing the two rectangles from the other tile will ll in the two related gaps, creating one large rectangle with no overlaps. To make sure that this rectangle does not overlap with the two tiles creating the edge between the vertices we need to remove a central section of each edge to make space for these rectangles.
To represent the black lines that were on the tile we pull out a rectangle of the tile where the black line would meet the edge. If this point has been removed to make room for the ag replacements we can shift it closer to the nearest vertex or further away so that this rectangle is removed from the edge of the removed section. As you pull it out so that only the corner is still in contact the corresponding rectangle from the neighbouring tile will ll the gap created and have a gap for the rectangle to ll.
This unconnected tile has the same properties as the tiles we have been looking at but does not need markings. We can now translate this new tile into three dimensions.
4.2 3D Tile
4.2.1 Construction
We are now looking at a three dimensional tile that is a short hexagonal prism with an unconnected section radially out from each vertex and some kind of shape showing where
24 Figure 4.2: The Three dimensional tile. The two pictures have the same face up just rotated. The second picture shows how two tiles may connect. black lines meet the edge.
If we imagine putting the mirror image of the lines and coloured diameters tile 1.2 such that the two red-blue diameters were lined up each vertex has a red diameter on one side and a blue one on the other. As such instead of de ning two rectangles that meet at a corner we can de ne a block whose edge furthest from the tile is slanted. The end near the red diameter is always longer and the blue end shorter. This way when two diameters of opposite colours meet the two blocks will t together and if they are both the same colour they wont. This guarantees the ags or red and blue diameters condition holds. These diameter de ning sections are still not connected to the central section of the tile but as we have the third dimension we can put a connector in on the central third of the tile edge to make it work.
Because of the way we have arranged the faces of the tile the black lines on each side would be at the same point on the edge. To de ne the black lines therefore we can have two plugs on each edge, both that can only t together with themselves. All the points where a black line meets an edge has one type of plug and where there theoretically could be a black line there is another. These plugs guarantee the equivalent of continuous black lines. As the blocks coding the diameters are seperating the tiles that the black lines are coding between, holes must be put in these to allow interaction.
25 4.2.2 Issues
The central connector does create some problems as three of these connectors want to exist in the same space at every vertex. To make this tiling work we have to put tiles in a layer at three dierent heights. To make sure that the plugs for the black lines still make sense we have to put them at three heights on each edge so that regardless of the height dierence between two neighbouring tiles they still can t together, also three holes must be put in each small block.
If we therefore look at a single layer of these three dimensional tiles we get three latices of the tiles each at a dierent height, as any two tiles that have radial blocks meeting will have to be at the same height. We can now replicate this layer and stack it on top
3 of the layer we already have an in nite number of times to create a tiling of R . As the stacking does not have to change the position or orientation of the layer, it is possible just to stack the layer exactly on top of itself, we can create tilings that are periodic in this third dimension. For this tile to be considered a true Einstein periodicity should not be able to happen in any dimension. This periodicity is what we will try to eliminate.
4.3 Schmitt, Conway and Danzer Tile
A tile that was rst shown by Schmitt Conway and Danzer [3] is aperiodic in just one of the three dimensions it is in. It is a rhombic bi-prism, a polyhedron formed of 4 identical rhombi connected by triangles. The only way this tile can successfully tile 3-space is if each triangular face is connected with another creating ridges formed by the rhombic sides. If the smaller interior angle of each rhombus is not an integer multiple of π or π then the 3 4 ridges at the bottom of a layer and at the top are irrationally related, as such when a layer is stacked on top of another it has to be rotated by an irrational amount for the ridges to
t together. This is an idea we had as a way to improve the Socolar and Taylor tile, to try and nd a way to create an irregular rotation of each layer by altering the height of parts of the tiles hexagonal faces.
26 Figure 4.3: The Scmitt, Conway and Danzer Tile, alongside an example of the rotation created by the layers [3]
4.4 Issues
In trying to transfer the properties of this tile to the Socolar and Taylor tile there are fundamental dierences that makes it very di cult or impossible for this to be done.
Below each of the major dierences are explored and how this aects the way tiles could be altered. We believe that it is not possible to transfer this property across because the dierences combined leave no possibilities.
4.4.1 Height
One problem is that the Danzer tile has all the tiles in one layer at the same height. With the three dierent layers of Socolar and Taylor's tile this complicates matters a lot. If we were to rotate by some multiple of 2π it would lessen this issue but then it would be periodic 3 at least every 3 layers. This is because the three dierent heights in each layer create a regular lattice formed by the fact that two tiles whose protruding sections must be at the same height. A rotation of 2π , or some multiple, would keep these latices in alignment. 3 If we were to rotate by some other amount the height interactions would become very complicated. To nd a height change that successfully rotated a layer it would have to work if we made a dierent choice for the orientation of the tile we have freedom over at each level.
4.4.2 Tile Orientation
Also with the Danzer tile if you turn it upside-down it will look the same merely rotated; if you ipped the tile over you would have to orient it exactly the same way. Socolar and Taylor's tile have fundamentally dierent sides, so this may limit the way we can
27 Figure 4.4: A top down veiw of the three dimensional tile. When two layers are overlayed with an irrational rotation between them the di culties in height become obvious dierentiate the sides of the tile. So if it were possible to incorporate all height dierences we would also have to be able to have the interactions work for tiles with the same and opposite orientations, as tiles are not all the same way up in each layer.
4.4.3 Aperiodic Layers
The nal, and biggest, problem is that the Danzer tile forms periodic layers, creating regular ridges on both sides of the tile. As this periodicity is what we are trying to avoid this is the most fundamental issue to overcome; the periodicity of the Danzer tile is essential to its eectiveness.
4.5 Danzer Implementation Attempt
We will now try and nd a way to force an irregular rotation of each layer taking into account that we need to resolve all of the issues described above. As a stepping stone we will try to implement a rational one.
4.5.1 Rotation of 2π 3
By looking at a rotation of 2π we remove the issues of height dierences as tiles will be 3 rotated to positions where the tile was the same height. This means we can focus on resolving the other issues and then worry about rectifying for the change in heights. As
28 the rotation is periodic we know any tiling we create will be periodic but this may change with an irrational rotation.
Simply putting a ridge on each side with an angle between them will not work as there will be no regularity to the ridge shapes created. This is because they are created within an aperiodic structure by tiles that have to be rotated. The periodicity is what makes the
Schmitt, Conway and Danzer structure work. By dividing each side of the tile in half such that the two lines have an angle of 2π between them and then by lowering everything on 3 one side of the line and raising the other side the tile will have to be rotated to t together and will therefore work on a tile by tile basis.
As a side note this is not a rotation of the layer. The Schmitt, Conway and Danzer tile has no rotational symmetry when taken as a single tile. As the tile we are looking at is based o a hexagon it has 6 fold rotational symmetry except for the plugs and slanted edges forcing aperiodicity. This means in trying to rotate the layer what we actually create is a completely new tile layer as each tile has to rotate independently.
There are only 4 possible ways these tiles can be related in terms of layering. They both have their mirror face down, their mirror faces touching, both have their mirror face up or have their regular faces touching. Because of the addition of the line splitting height dierences on the hexagonal faces there is only one way to place any tile once you have chosen which face is up.
As an example we will make our divide on the blue diameter, the same on both sides.
When looking at each side they will look the same but because of the ip involved in the creation of the tiles these two lines have an angle of 2π between them and force a rotation 3 of 2π regardless of which two faces you put together. Regardless of the orientation of either 3 tile the rotation is the same so we know all combinations lead to the same alterations of the black lines. The theorems we have proven in regards to the black lines cannot hold for this altered arrangement. By looking at a small black ring with the long black lines closer to it, which must appear, the rotated black lines reach a contradiction. The rst thing we proved was that if you have a short black line there is a small black ring on that vertex or on a neighbouring one 2.1. Each vertex where two tiles meet cannot be a small black
29 Figure 4.5: Both faces of the tile and how any face will look when stacked on top of a correctly oriented face. Also three tiles that show the contradiction. ring and has only one neighbour where it could be. The next theorem we proved was that if there is a short black line at one end of a tile there is one at the other 2.2. The vertex we just made a small black ring cannot have one at the other end, the vertices where two meet, and we reach a contradiction.
There is a way to change this tile that may lead to a successful second layer. That is to swap which side of the line is higher, to swap both would be equivalent. With a swap in heights there is either a rotation of π in one direction or 2π in the other, see gure 4.6. 3 3 The short black lines are in the same position as when the heights were the same. When looking at the same three tiles the position of the long black lines are forced to be the same way and so we will reach the same contradiction.
4.5.2 Irrational Rotation
Even if we found a choice of line placement and height change that worked it would still create a periodic rotation. If the rotation was not a multiple of π the hexagonal faces of 3 each tile would not line up where layers met and there would be a lot of height interactions that cannot at this point be accounted for. The aperiodicity of the tile layer dictates that as well as each non hexagonal edge meeting somewhere, each of these combinations will interact in every possible height combination in the layer somewhere. As this is the case at least two dierent height relations must be dealt with in the same way, the chances of all
30 Figure 4.6: Dierentiating the heights on the faces leads to four dierent tiles above. The tile below the one shown is the same as on the far right of the row of these interactions being successfully resolved with a single tile is very improbable. So we believe that there is no way to force an irrational rotation without reaching a contradiction in the dierent height interactions.
4.6 Aperiodic Stack
As the issues described above rule out the possibility of rotation to force non-periodicity another way we could form this complete non-periodicity is by forcing each layer to be a translate of the one below or a fundamentally dierent tiling when considered in two dimensions. By forcing the tile above a given one to be in a dierent orientation that layer of tiling has to be dierent in some way. We now have to see if there is a way we can alter the hexagonal faces of the tiles so that they still t together and force non periodic layering. To do this we can just focus on nding a non-periodic stack of tiles and then working out if it is possible to create non-periodic layers from them.
The rst way we may be able to do this is to translate a one dimensional tiling into a stack of tiles. As we want the stack to be aperiodic it makes sense to use a non-periodic tiling as the basis. Though there are lots of possibilities we will use the example of the
Thue-Morse tiling to demonstrate why this will not work.
31 Figure 4.7: By rotating the tile you get a regular system of divots on each side as both must t with either
4.6.1 Thue Morse exploration
This tiling or sequence contains two elements, a and b, and has a very simple substitution, a → ab, b → ba, that creates a non-periodic sequence. If we relate these two letters to dierent orientations of the corresponding tiles the created stack would be aperiodic. We could then extend out from this stack to create appropriate layers; just taking a layer and creating another by this substitution cannot work as the changes in orientation will lead to contradictions in the layer's two rules. The problem arises in translating the substitution into a shape that can work.
If you try to do this by raising a section of the hexagonal face and add divots to allow for this you very quickly see that this is not helpful if you do not introduce a rotation. The necessary alterations needed are the same on both sides and therefore do not distinguish the sides and in turn do not force the orientation.
If you add a rotation in so the divot is in a dierent position on the face you again get a situation where even if the faces are not exactly the same you cannot distinguish the two.
This makes sense as you still have to be able to let an a tile be followed by both another a tile and on other occasions a b tile. By making holes in the tile and adding sections that can ll them and interact with tiles they would not normally touch you can force the Thue
Morse structure on a section. If say for example one tile was in contact with four others, its neighbours and the next tiles out, these tiles could have the right structure. However this group would need to be able to interact with the same group rotated both ways and you reach the same issue with groups of the tile or reach a contradiction.
32 4.6.2 General explanation
Looking at any one dimensional tiling there is no way to force interaction by shape; any two tiles meet at a point and there is no way you can distinguish between two. As shown for the Thue-Morse above even when implemented in higher dimensions you cannot dis- tinguish between the dierent tile faces involved. This is not because the tiling is one dimensional but because it is aperiodic. As all interactions must be possible any aperiodic one-dimensional tilings cannont be de ned by shape.
4.6.3 Two dimensional exploration
As we know there are some aperiodic prototile sets completely de ned by shape it makes sense to consider the fact that there may be a way to combine these tiles with our existing tile to create a single tile aperiodic in all three dimensions.
Unfortunately a very similar contradiction as with one dimensional situation occurs.
The six edges that are not hexagonal as they are before alteration are indistiguishable, up to rotation. On one side there are three plugs denoting the position of a black line, on the other three that are denoting the lack of a black line. As the layer is aperiodic we know that at some point in each layer each of these six edges will meet; they can only meet in one combination of orientations, such that the black lines will remain continuous, but they will all meet. As such these six edges must stay indistinguishable. In combination with the regularity of the position of tiles themselves this means that that the tiling must be periodic in this dimension. If you have periodicity in one dimension there exists a tiling with the tiles used that is periodic in both dimensions, negating the aperiodicity of tiling we are attempting to integrate.
33 Chapter 5
Conclusions
From what we have explored here are some conclusions we may come to. We realise that some of these ideas are merely speculation and may easily be refuted with exploration in other areas of mathematics but they are still ideas worth noting.
5.1 Socolar and Taylor Tile
The Socolar and Taylor tile can not be improved from its current forms as far as we can tell. Depending on the circumstances dierent representations may be useful but there is no better way than the ways explored here and posed by Socolar and Taylor. The Lines and Flags representation 1.1 makes the most sense if you want the tile to be connected, tile every dimension it lls aperiodically and only contains only the tile and its true mirror image. The Lines and Diameters representation 1.2 may make the matching rules easier to visually create though the two versions of the tile used are not actually mirror images of each other. The unconnected representation 4.1 is merely de ned by shape. Though each tile is split amongst multiple shapes this representation could model complex molecular structures or other areas where shape de nition is vital. To truly make this Einstein one tile you must move into three dimensions, but as far as we know there is no way to make the tile aperiodic in this extra dimension by shape or markings. Substitution can work but is a very dierent form of construction to what we have been dealing with.
34 5.2 Einsteins and Dimension
Looking at the explorations above it may be the case that there is no better Einstein in two dimensions; there may be a simpler one or one that is easier to understand but it will still have the constraints of Socolar and Taylor's tile. The argument being that there are not enough degrees of freedom in just two dimensions to allow for all the restrictions we wish to impose on the tile. There are only a few shapes that can tile the plane on their own, all periodically, so the tile must be an alteration of one of them. A possible way around this would be to take a tiling containing two tiles and changing the tiles until they are the same, the two tiles being an aperiodic prototile set. As the markings on the Socolar and
Taylor tiles are based on the Penrose tiles this may not be enough.
Another idea is that it is not possible to have a three dimensional Einstein that is aperiodic in every dimension it occupies. Though this third dimension allows the possibility of a single prototile by combining the tile and mirror image as one, this extra dimension also interferes with the established aperiodicity of the other two dimensions. It may be the case that the aperiodicity in one of the three dimensions is dependent on the other two being periodic, like with the tile exhibited by Schmitt, Conway and Danzer [3]. If there is not a periodic plane then only planes can be aperiodic. As there is not an even number of dimensions this will always leave one dimension that will be periodic.
If these conjectures were to be right this would mean that an Einstein could exist in four dimensions. There would be enough degrees of freedom to combine the tile and its mirror image, or possibly incorporate completely dierent two dimensional tiles into one.
Also if two dimensions formed an aperiodic plane there would be another pair that could create another. Each two dimensions together would have to create an aperiodic plane.
The fact that this is now in higher dimensions the concepts may be very dierent but it is an idea to consider.
35 Bibliography
[1] Grunbaum, B. and Shephard, G. C., Tilings And Patterns, W. H. Freeman, New york,
1987
[2] Joshua E.S. Socolar, Joan M. Taylor, An aperiodic hexagonal tile, Journal of Combi-
natorial Theory Series A, 2011, 118, 22072231
[3] Joshua E.S. Socolar, Joan M. Taylor, Forcing Nonperiodicity with a Single Tile, The
Mathematical Intelligencer, 2012, 34 (1), 18-28
[4] Online http://makezine.com/2010/03/26/worlds- rst-aperiodic-tiling-with/
36