UNDECIDABILITY AND APERIODICITY OF TILINGS OF THE PLANE

Transcription

1 UNDECIDABILITY AND APERIODICITY OF TILINGS OF THE PLANE A Thesis to be submitted to the University of Leicester in partial fulllment of the requirements for the degree of Master of Mathematics. by Hendy Wickham-Pusey Department of Mathematics University of Leicester May 2014

2 Contents Declaration Abstract iii iv Introduction 1 1 An Aperiodic Tile Set Robinson's Aperiodic Tile Set Determining All Robinson Tilings Tiling by Extension Substitution Tilings Non-Periodicity of Substitution Tilings The Robinson Tiles as a Partial-Substitution The Robinson Tiles as a Substitution Two Classical Decision Problems The Halting Problem Counter Machines The Immortality Problem The Origin-Constrained Problem Turing Machines as Tilings The Turing Tile Sets for the Origin-Constrained Problem The Undecidability of the Origin-Constrained Problem i

3 5 Reducing the Domino Problem to the Halting Problem Alterations to the Robinson Tiles Constructing the New Tile Set Reducing to the Halting Problem The Periodic Problem Undecidability of the Periodic Problem Reducing the Domino Problem to the Immortality Problem Introduction to Ane Functions and Immortal Points Reducing the Domino Problem to the Immortality Problem for Ane Functions Kari's Aperiodic Tile Set Reducing the Immortality Problem for Ane Functions to the Immortality Problem for Turing machines The Row-Constrained Problem Undecidability of the Row-Constrained Problem ii

4 Declaration All sentences or passages quoted in this project dissertation from other people's work have been specically 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: Hendy Wickham-Pusey Signed: Date: May 2014 iii

5 Abstract As a whole this paper seeks to answer the Domino Problem, or more precisely, to prove its undecidability. The rst chapter establishes whether or not periodicity is an appropriate way to decide whether or not a given tile set tiles the plane. In fact this is not the case and the construction of an aperiodic tile set (namely the Robinson tiles) is used to prove this. Given that an aperiodic tile set exists it is not possible to determine whether or not a tile set tiles the plane based purely on whether or not it can do so periodically, as this would imply that aperiodic tile sets cannot tile the plane. The Robinson tiles are then considered in terms of substitution. First a partialsubstitution is shown which follows from the construction of the tilings in the previous chapter. This allows for another proof of their aperiodicity. In fact a proof is given that any substitution with a unique composition leads to a non-periodic tiling. The Robinson tiles are then used to construct a tile set with a formal substitution tiling, but that still replicates the Robinson tiles. This allows (among other things) for the identication of the minimal subspace of the tiles. The classical decision problems of the Halting Problem and Immortality Problem are then considered and proved to be undecidable. The Halting Problem has a simple enough proof by construction of a Turing machine which contradicts itself. The Immortality Problem is also reduced to the Halting Problem, by way of counter machines, and this proves the undecidability of the former. Two proofs of the undecidability of the Domino Problem have been given. The rst is by Robinson and requires the combination of his tile set with Turing tiles to show that the iv

6 Domino Problem reduces to the Origin-Constrained Problem. It is therefore necessary to explore the undecidability of this problem too, and its proof is to show that it reduces to the Halting Problem. This is followed by the Periodic Problem as the construction of the tile set used for this leads directly on from that which Robinson described, and goes onto prove that this too is an undecidable problem. The second proof of the undecidability of the Domino Problem is more recent (2008) and is the work of Kari. This converts the tile set into a system of ane maps, and thereby a Turing Machine. This eectively reduces the Domino Problem to the Immortality Problem, and so the second proof is complete. It is natural to follow this with a consideration of the Row-Constrained Problem. The method reduction is the same, but because of the extra constraint it is the Halting Problem not the Immortality Problem is a reduction of the Row-Constrained Problem. The gures throughout this report are of my own construction with the exception of those in Chapter 7, namely Figure 7.2.1, Figure 7.3.1, Figure 7.2.3, which have been taken from Kari's On the Undecidability of the Tiling Problem [7]. The gures of Chapter 1 have been created from descriptions and diagrams of Robinson's original construction in [10]. And similarly the gures of Chapter 2 have been created from descriptions and diagrams in Gähler's paper [3]. The concepts considered are either my own or have been replicated from others papers and appropriately referenced. Particularly Sections 2.2 and 8.1 being entirely my own work (although chapter 8 draws heavily on the work of Kari in the previous chapter). Similarly the complete proof of the Immortality Problem is my own, although individual theorems involved in it have been used elsewhere. v

7 Introduction This thesis aims to explore the question of whether or not it is possible to tile the plane with a given nite set of tiles, that is the Domino problem. Problem. The Domino Problem: Given a nite set of tiles, does there exist a valid tiling of the plane using this set? To begin we rst need a mathematical denition for both a tile and a tiling, and more specically Wang tiles. Denition. A tile set is a set of n-dimensional shapes that may possess colours or labels on their edges and matching rules placing a restriction on how the edges meet in a tiling. A tiling is a division of R n into tiles that intersect only on their boundaries. A patch is a tiled subsection of R n. Here tiles are only considered as two dimensional objects with a tiling covering the plane. A matching rule is simply some condition placed on a tile edge such that only a particular subset of tiles from the set may be placed next to that edge. A matching rule may be given as a label or colour on the edge or it may be determined by the geometric shape of the tile. A matching rule may not be part of the tile but just an instruction with the tile set that limits the way that the tiles may be placed. Denition. A Wang tile is a unit square tile that may be placed in a unit square on the plane, it is understood that innitely many copies of tiles from a tile set may be used to tile the plane. The tiles can not be rotated or reected, that is they may only be translated. Additionally the edges of Wang tiles are coloured (or labelled in some other way) and tiles 1

8 can only be placed next to each other if the intersecting edges match. In any given set of Wang tiles there is only a nite number of colours, and therefore only a nite number of tiles. Wang tiles may also be represented geometrically. Let the left-hand edges be altered to have indentations corresponding to the colours. Right-hand edges are similarly altered with the colours represented by appropriately shaped protrusions to match the same colour indentations. Top and bottom edges of the tiles can be altered in the same manner. Colours on the vertical edges are considered distinct from those on the horizontal edges, so they are associated with dierently shaped indentations and protrusions. In this way rotations and reections of the tiles will necessarily admit a tiling unless all tiles are rotated or reected in the same way. In this case patches and tilings are reections and rotations of those achieved by the original tile set. So Wang tiles, including the matching conditions can be completely represented geometrically. Denition. A tile set is said to have finite local complexity if there are only a nite number of arrangements of tiles around any given vertex. Denition. A tiling is said to be periodic if there exists some vector v such that translating the tiling by v does not aect the tiling. A tiling which is not periodic is said to be non periodic. If a tile set does not admit any valid periodic tilings then it is said to be aperiodic. Theorem 0.1. Let a tile set admit a tiling of the plane that is periodic in one direction, and let the tiling have nite local complexity, then there exists a tiling that is also periodic in a linearly independent direction. Proof. Let a tiling be periodic in direction v (call this direction horizontal) and have nite local complexity, where translating by v leaves the tiling invariant. Since the tiling has nite local complexity there are only a nite number of ways that tiles can be arranged in cross-section of a vertical strip, width v. Say for instance that v is measured from the centre of tiles, although it could be measured from the far left point, or any other point so long as it is consistent. Given that a tiling exists at least one of these strips must be repeated 2

9 vertically within some horizontal interval of length v. So the top and bottom edges of some interval in the vertical tiling are given by the same strip, as are the tiles on each side due to the horizontal periodicity. So between these strips there exists a square that could be repeated across the plane. That is to say the tiling is also periodic vertically. Note that although squares are discussed here, the argument works with tiles of any shape. In fact the square isn't a strict square at all, but the a shape in which the east and west sides are equidistant apart and the north and south sides are equidistant apart. The sides of the square follow the boundaries of the tiles on the edge, but the shape may be approximated by a square, with the corners at the centre of tiles (or whichever point was used previously to measure distances from). Many problems in this paper are undecidable, so let this term be dened here. Denition. A problem, generally of deciding whether a particular statement holds, is decidable if there exists an algorithm (that is a nite set of steps) that will either prove or disprove the statement made in the problem. An undecidable problem is one that is not decidable. To say a problem is undecidable is not to say that the problem cannot be solved, but rather that there is no consistent way of doing so following a single problem. Denition. A T uring machine is a machine that manipulates symbols on a strip of tape according to a table of rules, it may only perform one action at a time and cannot store information other than the state that it is in and what is written on the strip. An arbitrary Turing Machine, T, has a nite number of states s 1,..., s n and a nite number of tape characters α 1,..., α m. The Turing machine is able to follow only a single set of instructions. Each instruction reads If in state s i and reading symbol a j then change a j to a k and move left/right and into state s l. with left or right being chosen as applicable. It is often simplest to order these instructions into a table with the states marking the rows and the characters marking the columns, each entry being the action the machine should take when reading the given character whilst in the given state. 3

10 An alternative way to understand the concept of undecidability is with regards to Turing machines. Consider an even relatively basic Turing machine that performs dierent arithmetic tasks based on the current value or string, then there are three possible results when running the machine for a given input. Either the programme halts by reaching some pre-set halting condition, or the programme loops, repeatedly producing the same set of outputs, or nally the programme could continue to run forever but without entering into a loop. Given sucient time it is possible to prove if the programme will halt or loop for a given input, but the converse, proving that the machine does not stop is impossible for some problems. These problems are said to be undecidable. In order to prove the undecidability of many problems discussed here they are shown to reduce to known undecidable problems. Denition. A problem A is said to reduce to problem B if the decidability of B follows from the decidability of A. That is to say that if B is undecidable, then A must also be undecidable. Reducibility is in general not a reversible quality. To say that problem A reduces to problem B is not the same as saying problem B reduces to problem A. For instance a problem (with regards to a given set) is undecidable if the problem is undecidable for some subset. And if a problem is decidable for a set then it is decidable for every subset. However a problem may be undecidable over a given set, and yet be decidable for some subsets. Note here the specic case of if a subset has only one element, then then problem can simply be tested on this element (although in some cases this still won't be decidable). 4

11 Chapter 1 An Aperiodic Tile Set In 1961, Hao Wang [12] conjectured the following: Conjecture 1.1. If a nite tile set can tile the plane then it can do so periodically. If this conjecture were to be proved true then there would be a way of deciding whether or not a given tile set can tile the plane, and hence the Domino Problem would be decidable. Namely if a tile set tiles the plane then there exists a nite patch of tiles that can be repeated, and this patch could be constructed in a nite number of steps, simply by extending the tiling from the edge of a single tile until such a patch was obtained. Given a nite set of tiles that are known to tile the plane there are three possibilities for how they do so. Either all tilings of the plane are periodic, some of the tilings are periodic and some are non-periodic, or all of the tilings are non-periodic (that is the tile set is aperiodic). If the conjecture were to be true then the third option would not be possible, there would not exist a non-periodic tile set. So to disprove the the conjecture it is enough to prove the existence of a nite aperiodic set of tiles. Theorem 1.2. There exist nite aperiodic tile sets. It was Wang's student Robert Berger who rst disproved the conjecture [1] and shortly after produced a set of 20,426 Wang tiles that that tiled the plane only non-periodically. Since then many more (and often notably smaller) aperiodic tile sets have been found, this includes Robinson's set of 28 tiles [10]. 5

12 1.1 Robinson's Aperiodic Tile Set This section will show the construction of an aperiodic tile set. Doing so will mean that not all tile sets that can tile the plane do so periodically. Thus nding a periodic tiling is not a method which can be used to see if the tile set tiles the plane. The tiles discussed here were originally discovered by Raphael Robinson in 1971 and given in Undecidability and Nonperiodicity for Tilings of the Plane [10]. They are based on square tiles with notches and projections to restrict the tiling and are shown in Figure below. Note that these tiles can be reected (in the horizontal or vertical axis) or rotated as necessary, so the tile set in fact consists of 28 tiles. Figure This tile set can be represented by the one below (Figure 1.1.2) and will be used from this point forth for simplicity. Here the notches and projections are represented by the tails and heads of arrows respectively, those which are symmetric are represented by green arrows and those which are asymmetric are represented by o-centre red arrows. The cornered tile (labelled A,(a)) is given blue corners to account for the tiles all being the same basic shape. Figure Firstly, consider a tiling that has no matching rules for the arrows, that is there are two tile types: a cornered tile (a) and an uncornered tile (~a). At each intersection of four tiles there must be exactly one cornered tile. This leads to two possibilities illustrated below in Figure 1.1.3, but in either case there is a row of tiles...(a)(~a)(a)(~a)(a)... so this will be the start point for constructing the tiling. 6

13 Figure Clearly, if two adjacent (a) tiles were to face red arrow to green arrow there is no tile in the set that is able to connect the two, so this possibility is eliminated and it is concluded that adjacent (a) tiles face either red arrow to red arrow or green arrow to green arrow (and will necessarily take each case alternately). Similarly if two adjacent (a) tiles were to face red arrow to red arrow, but with the second red arrows facing diering directions (i.e. one pointing upwards and one will pointing downwards) then the facing red arrows on these two (a) tiles will not be aligned and there is no tile in the set to join them. So it is necessary to conclude that there exists within the tiling the arrangement shown in Figure of two (a) tiles. Figure There are only two options for ling this gap, a tile of type (c) or (d), regardless of which is chosen the topmost edge of the tile will be marked with the tail end of an arrow. The tile above the right (a) tile must be of the form (c), (d), or (e), regardless of which is chosen the right hand edge of the tile will be marked by the tail end of an arrow. Therefore the central tile on this second row must have the heads of arrows on two adjacent edges and, since it can not be a tile of type (a), must be a tile of type (b). Select arbitrarily the orientation of tile (b), this will then force the selection of tiles for the previous two positions discussed. 7

14 Figure Since the (b) tile has an arrow head on its topmost edge (and would have regardless of which orientation was chosen) an (a) tile cannot be placed above it. Therefore the (a) tiles in the third row must be placed directly above the initial (a) tiles. As before these tiles must face those below red arrow to red arrow with the arrows in alignment, forcing the arrangement shown to the left below. This then forces the types of tile to be placed in the nal two positions, namely type (c) and (d), to give a complete 3 3 square (to the right below). Figure Note that the choice of orientation for the central tile (the only choice that was made) would change the resulting 3 3 square only by rotating it by some multiple of π 2, and since the tiles are permitted to be rotated this would not change whether or not the tile set admits a tiling starting from this arrangement. This 3 3 square patch must now be extended into a 7 7 square patch, in fact the construction of this extension occurs in the same way as before. Let the 3 3 square above be the bottom left corner of the extended square. Clearly there must be a single vertical row of tiles to the right of the given 3 3 square followed by another 3 3 square of some orientation in order to continue the pattern of (a) tiles. Consider the rules established 8

15 for selecting the orientation of (a) tiles, these same rules will apply to (b) tiles with the exception that they may be placed more than one square apart (since their location is not restricted by corners). So each corner of the 7 7 square patch will contain a 3 3 square patch with the orientation of the central tile of each patch corresponding to the the orientation of the corresponding (a) tile in the 3 3 square, this is made clearer looking at Figure 1.18 below. Consider the currently blank cross in this 7 7 tile, the tiles in the middle of each arm must be of type (c) or (d) so the central tile must necessarily be of type (b). As before the orientation of this (b) tile is chosen arbitrarily and the remaining tiles in the square patch are then forced. Figure The tiling can be arbitrarily extended in this way creating successive squares of size 2 n 1 so the extension theorem, discussed in greater detail in the following section, can be applied. Intuitively this means that there exists successive squares of size 2 n 1 for 9

16 n N and since n tends to innity so does the tiling. So a tiling of the plane exists, and all tilings created from the tile set are of this form. As mentioned before the only choice when constructing the tiling is the orientation of the centre tile of each consecutively larger square. Since there are innitely many centre tiles within the tiling it follows that there are innitely many possible dierent tilings. Consider the case where each centre tile is chosen to have red arrows pointing north and east, then the tiling would extend only in the north-east direction and hence the tiling obtained would be a tiling of only the north east quadrant. Say the centre tiles are chosen to have red arrows pointing north and east or red arrows pointing north and west with equal probability, then the tiling will extend north and both east and west leading to a tiling of the upper half-plane. Similarly, if the centre tiles are chosen arbitrarily (that is each orientation is chosen with equal probability) then the tiling extends to cover the entire plane. From this, it is seen that if the probability of choosing a particular orientation of the centre tile is altered then the tiling will dier, however the form of the tiling will remain the same as that given above. By extending the tiling in this way there is always, to either the left or right and to either the top or bottom, a row of green arrows pointing outwards. By extending the tiling in such a way as to maintain this edge in the same position allows for a row of either (e) or (f) tiles to be placed. The other side of this row may be matched to some tiling of the other half plane which does not necessarily line up with the half that has been constructed That is tiles of type (a) and (b) may not occur with the same spacing or orientation on either side of this row. Such a row is called a fault line and only one horizontal and one vertical fault line may occur in a tiling. This is because to each side of such a line there exists on any patch an outward-facing red arrow preventing a further fault line. In any case it will be shown later (in section 1.2) that an innite partial tiling leads to the existence of a tiling of the plane, so it is only necessary to consider at this point tilings which to not contain fault lines. So it is left to show that the tiling is non-periodic and, since the tiling is forced, that the tile set is aperiodic. Consider only the red arrows on the tiles, then a patch of the 10

17 tiling will look like the image below. Figure Assume there exists a patch that tiles the plane periodically. Then there exists within this patch some red square of greatest size. However, at the centre of this square exists a (b) tile meaning that it is the corner of some larger red square. So tiling the plane with this patch would mean failing to complete this red square and hence the tiling would not be valid. Thus there does not exist a periodic tiling by the Robinson tile set, and the tile set is aperiodic. An alternative way to look at is to again assume that there exists a patch that tiles the plane periodically. Such a patch will either be a square of size 2 n 1 as described above or some patch contained within such a square. If the patch is a square of size 2 n 1 then it will have a centre tile, with the arrow pointing to the left being a dierent colour to the arrow pointing to the right. Since this coloured arrow is carried to the edge of the square, the square cannot in fact be periodic. If the periodic patch is smaller than the square of size 2 n 1 but larger than a square of size 2 n 1 1 then the same argument applies (except the centre tile in this case does not lie in the centre). If the patch is smaller than 2 n 1 1 then the argument can be applied to this square instead. So again it is concluded that the Robinson tile set is aperiodic. The method used to prove the aperiodicity of this tile set can be generalised to prove that certain tilings constructed by substitution are non-periodic as explained in Chapter 2. In the above case since the tiling was forced by the tile set, this led to the aperiodicity 11

18 of the tile set Determining All Robinson Tilings In Robinson's original paper [10] 2-adic numbers were used to determine each tiling by the Robinson Tiles. Let the tiles be placed on the plane such that their centres lie at integer points. So for an expanding sequence of squares (of size 2 n 1), if they are to tile the rst quadrant, the centre (b) tiles will lie at positions (2 n 1, 2 n 1 ) and each will have red arrows pointing upwards and to the right. If a (b) tile lies at a point with coordinates (a, b) then it must be the centre of some 2 n 1 square, so a = 2 n 1 + s2 n and b = 2 n 1 + t2 n. Similarly if a tile lies at such a point then it will necessarily be an (a) or (b) tile. In fact if the point is (1 + 2s, 1 + 2t) then the corresponding tile will be of type (a), else it will be of type (b). A 2-adic integer has the form A = a 0 +a 1 2+a a with a i {0, 1}. It can be viewed as an innitely long binary number, traditionally written left to right as opposed to right to left. Those 2-adic integers which eventually end in a string of all 0s is a binary representation of some natural number, and those 2-adic integers which eventually end in a string of all 1s is a binary representation of some negative integer. Consider some square of size 2 n 1 then let the vertical row of tiles to the left and right of the square satisfy A + x 0 (mod2 n ) and the horizontal row of tiles above and below the square satisfy B + y 0 (mod2 n ). If the tiles are placed such that they tile the rst quadrant then the tiles to the left and below any 2 n 1 square will have x and y coordinates of the form s2 n and tiles to the right and above will have a coordinate of (s+1)2 n so this reduces to A 0(mod2 n ) and B +y 0 (mod2 n )so A = B = 0. Otherwise A and B will represent the distance that the tiling needs to be translated by for this to be the case. The nth digits of A and B represent the horizontal and vertical distances needed for the centre of a 2 n 1 to lie at the point (2 n 1, 2 n 1 ). If A or B are integers then there will be a horizontal or vertical boundary respectively, that is to say the tiling does not cover the whole plane. If say A is a positive integer then at some point no further translations to the right are required for the tiling to tile only 12

19 the rst quadrant, meaning that the tiling could not originally have tiled the whole plane. Similarly if A is a negative integer then for some k there exists no squares of size 2 n 1 > k to the right of the y-axis as a translation is required for such a square to exist, meaning that the tiling could not originally have tiled the whole plane. The argument follows for B in the vertical direction. So if either A or B (but not both) are integers then the tiling is one of the half plane. If A and B are integers then the tiling is one of the quarter plane. And if neither A not B are integers then the tiling is one of the entire plane. If A is an integer then there may be a tiling of the remaining half plane, the two halves separated by a fault line. In the case that no fault line exists the pair A, B uniquely denes the tiling and vice versa. In the case that a fault line exists then the tiling could be determined by multiple pairs of the 2-adic numbers. But for multiple pairs to determine a tiling the condition must be added that for the integer part of the pair (say A) the corresponding part of the other pair must be it's negative, that is (A + 1), the +1 being to account for the fault line. This ensures that each half of the tiling meets the other. 1.2 Tiling by Extension It is natural to start the construction of a tiling with a few tiles, and then extending that patch outwards until the plane is covered. In some circumstances, such as with the Robinson tiles, extending a tiling in this way doesn't necessarily produce a tiling of the plane. It may only produce a partial tiling, which still extends innitely in some direction (or in the two-dimensional space here, two linearly independent directions). The theorem below is given in Tilings and Patterns[4] and is reproduced in a similar manner here. Theorem 1.3. If a tile set correctly tiles some innitely large patch, then there also exists a valid tiling of the plane. Proof. Construct a lattice covering the plane and label each point of the lattice by the natural numbers, cycling outwards from the origin as shown below. Since the tile set 13

20 correctly tiles some innitely large patch there can be constructed increasingly large discs that can be correctly tiled. Let the lattice be superimposed upon the tiling such that the origin lies at the centre of a disc. So there exists a section of the tiling that contains increasingly many points of the lattice. Consider a sequence of discs of increasing size that are correctly tiled. Any nite patch of the tiling will be repeated innitely often throughout the tiling. Create a subsequence of the disks such that each contains some given patch at its centre. In fact let each disc contain the previous disc at its centre. Then there is a sequence of disks of increasing size each of which extend a patch of the tiling further. Since the size of discs may be arbitrarily large, and the discs are strictly increasing in size, the size of the discs tends to innity. So there is a patch of tiling that can be extended (by each element of this sequence) to be innitely large, thus tiling the plane. Figure

21 Chapter 2 Substitution Tilings Denition. A substitution tiling is a tiling such that is extended by stretching each tile by a given scale factor and which is then subdivided into tiles (from the tile set). It can be written more formally by the pair (Q, σ) where Q is the scale factor and σ is the substitution. A substitution matrix M contains information about the proportions of dierent tiles that exist within the substitution tiling. M i,j gives the number of ith-tiles in a substituted jth-tile. So in a patch generated by σ n (J), where J is the jth tile, there will be (M n ) ij tiles of the ith type. Denition. A primitive substitution is a substitution by which for any two tiles a, b the tile b lies in the patch σ n (a) for some natural number n. For the substitution matrix of a primitive substitution there exists a natural number n such that every entry of M n is positive. 2.1 Non-Periodicity of Substitution Tilings Consider a substitution tiling such that there is only one way to compose super-tiles, that is to say that following an extension the subdivided tiles can only be recomposed into the original tiling, then the composition of the substitution tiling is said to be unique. 15

22 Theorem 2.1. If the composition of the substitution tiling is unique then the tiling is non-periodic. Proof. Assume that the tiling is periodic and the composition of the substitution tiling is unique. Since the tiling is periodic there must exist a vector v such that translating by v does not change the tiling. Consider a ball of radius v then the tiles within this ball can be uniquely recomposed into a new tiling such that the tiles are larger than the ball v. If the original tiling was periodic then this would not be possible since a periodic super-tile would not have a unique recomposition. As the tiling is periodic, the periodic patch could be joined to the periodic patch on either side of it. In either recomposition the super-tiles will be identical, so the recomposition is not unique. Thus the tiling cannot be periodic, a contradiction to the original assumption of a periodic tiling. Therefore the theorem is proven and if the composition of the substitution tiling is unique then the tiling is non-periodic. 2.2 The Robinson Tiles as a Partial-Substitution As mentioned at the end of the previous chapter the tiling generated by the Robinson Tiles can intuitively be represented in a similar way to a substitution tiling where the rules in the table on page 18 (in Figure 2.2.1) apply. This is not a true substitution as, among other things, the scale factor each tile is stretched by is not uniform across the tile set. Additionally there are conditions on the tiles that determine which substitution rule is applied, as such this shall be called a partial-substitution. To prove the aperiodicity of the tile set using the theorem from the previous section it must be show that the composition of the partial-substitution is unique. Since each tiling by the tile set is forced to have the same form as is generated by the partial-substitution, the non-periodicity of this tiling will then prove that the tile set is aperiodic. Clearly any 3 3 square that shows a red square will be replaced by an (a) tile in the appropriate orientation, and (b) tiles remain unchanged. Consider the remaining tiles, if a (d) tile occurs then it does so either singularly or between a pair of (f) tiles. If the second is the case and in particular if a row (f)(d)(f) exists alongside a 3 3 square that shows a 16

23 red square then it is replaced by a single (d) tile, else (d) tiles will be considered singular and remain the same. This argument can also be followed for tiles of type (c). If an (e) tile appears alone then it remains unchanged, if it appears in a row of three or more then it could be replaced with a single or multiple (e) tiles. If a row of three runs alongside a 3 3 square that shows a red square then it will be replaced by a single (e) tile else tiles will be considered singular and remain the same. The same argument applies to (f) tiles. If these rules are not followed for recomposition of the partial-substitution then the new tiling will not correctly tile the plane, there will be gaps or tiles placed together when there edges do not match. Therefore the composition of the partial-substitution is unique, and as such the tiling must be non-periodic. Since the tiling is forced, as explained in Section 1.1, this means that the tiling is in fact aperiodic. To generate a tiling by the partial-substitution, the partial-substitution must be recursively applied to tile (a). To do so with any other will either not extend the patch or only do so in one direction. If such a tiling is generated (with the (a) tile in the orientation shown over the page) then this will only lead to a tiling of the north-east quadrant as it has been determined that the centre tile will always have red arrows pointing north and east. But in fact it can be shown that if a tile set tiles a quadrant then the tile set will also tile the entire plane, as was explained in Section

24 18

25 2.3 The Robinson Tiles as a Substitution The Robinson tilings can also be represented by a tile set which admits a formal substitution. The following describes a tile set that is explored in Combinatrorics and Topology of the Robinson Tiling [3]. In section 2.2 the partial-substitution required conditions on the tiles to determine the amount by which they were stretched and sub-divided. In order to remove this problem the new tiles will be square tiles which represent the intersections of the Robinson tiles within a tiling. Since there are only nitely many Robinson tiles and they may only meet full-edge to full-edge there are only nitely many possible tiles in the intersection tile set. In the following explanation of the construction of the intersection tile set for Robinson tiles the following representation will be used for each of the Robinson tiles. Figure Since the orientation of the Robinson tiles aects the identity of the intersection tiles, it is important to note the direction each triangle faces. When constructing the Robinson tilings each intersection had to include exactly one blue corner, so each of the intersection tiles will contain a blue triangle. As with the Robinson tiles, each intersection tile will be shown in only one orientation, although may be rotated or reected in order to construct a tiling. So each intersection tile (only showing the representations of (a) tiles) will look like one of the following. Figure Each of the (a) tiles in the Robinson tiling laid at the corner of some 3 3 square, at the centre of which was a (b) tile. Including representations of (b) tiles in the intersection 19

26 tiles gives the following set. Figure The matching rule that is imposed upon these tiles is that a tile may only be placed next to a tile if the joining edge displays the same colour triangle facing the same direction. Using the intersection tiles, with their rotations and reections, in this state (that is with blank sections) to form a tiling will lead to super-tiles of the form shown in gure 2.3.4, and the tiling will be comprised only of these super-tiles. The has been used to represent either a purple triangle or a blank. These super-tiles will then be used to tile the plane completely. Figure However if the construction of the Robinson tiling is recalled then the only choices made were the orientation of centre tiles. Due to this fact, and noting that the above patch corresponds to a 3 3 square of Robinson tiles, the the remaining interior symbols 20

27 can be added. Some elements of the external symbols will also be known. Here double bars have been used to signify that the colour and orientation of the triangle is unknown, in terms of the Robinson tiles the colour arrow with two tails is known but no information is known about the arrow with a head. A single bar signies that the orientation of the triangle is partially known, in terms of the Robinson tiles the colour of the arrow with two tails is known and the direction of the arrow with one head is known, but the colour of this arrow and it's location (central, to the left, or to the right) is not. The is again either a purple triangle or some red/green triangle combination. Knowing the location and orientation of the purple triangles in the corners will then determine the remaining sections of the intersection tiles. Figure Since the tiling is then completely determined by the blue and purple triangles on the tiles, although the red/green triangle combinations are needed to ensure the matching rules occur on the edges of the super-tiles, the substitution associated with these tiles need only specify the locations and orientations of the blue and purple triangles. In essence the patch above represents a 3 3 patch of Robinson tiles, and logically it would make sense if this patch were an extension of the patch corresponding to a central Robinson tile (in a similar manner to the rst line of the partial-substitution table in Section 2.2 of this chapter). So in the case of these intersection tiles each tile will be 21

28 stretched by a scale factor of 2 and substituted by four other tiles. The diagram below shows this substitution graphically. Figure To give a complete set of tiles and their substitutions the red/green triangle combinations must also be added. For those with purple triangles this is fairly straight-forward as each of the remaining blank section is forced. However for those with more blank sections all possibilities must be considered. The way to nd the markings on all tiles, as was done in [3] is to ll the substitutions for those with purple triangles and then generate the substitutions for each new tile that is fully marked, thus generating the entire tiles set and a substitution rule for each. This complete set of tiles and their substitutions is shown below. It is important to note again at this point that all tiles (and their corresponding substitutions) may be reected or rotated as required. 22

29 23

30 This substitution is a primitive substitution, that is to say that applying the substitution repeatedly to any tile a sucient number of times will lead to a patch that contains at least one of every other tile. This can be seen from the fact that each substitution eectively places an (a) tile in the centre of the corresponding Robinson tile patch. And it is known that extending from a single (a) tile will generate a forced tiling which necessarily must contain every type of intersection. So a patch generated from any of these intersection tiles will eventually form a patch containing every other tile. From this it can also be seen that the substitution will therefore lead to an innite order super-tile. Providing that the initial seed tile is placed at the origin this will lead to the tiling of the plane as the substitution will extend the tiling in all directions. If the seed tile is placed elsewhere then the substitution will lead to a tiling of either the half plane or quarter plane. The intersection tiles may also be used to replicate Robinson tilings with fault lines. However, in this case the substitution has not been used to generate the tiling since the substitution is used to generate larger and larger 2 n 1 squares in the corresponding Robinson tiling. Each square will contain tiles of type (a) or (b) in ever row and every column, so the substitution cannot generate a fault line. Rather, tilings of the half or quarter plane generated by the substitution could be combined to form an appropriate tiling of the plane. Denition. A minimal subspace of a tiling space is a set that is closed and translation invariant and contains no proper closed subsets that are translation invariant. Loosely, this is the set of limits of translations of all tilings in the tiling space. Each tiling that has a fault line, if translated innitely far in an appropriate direction, will tend towards a tiling of the plane by one super-tile. This is similar to the extension theorem in that the section of the tiling to the other side of the fault line could be considered to be empty in this case. So it can be concluded that the minimal subspace of the Robinson tilings is represented by the the set of substitution tilings which only generate tilings of this form (that is without fault lines). Although a proof is not provided here it is also of interest to note that knowing that the substitution is primitive is enough to conclude that the set of substitutions tilings is 24

31 the minimal subspace. 25

32 Chapter 3 Two Classical Decision Problems Consider the Domino Problem, a decision problem for deciding if a tile set admits a valid tiling of the plane. Problem. The Domino Problem: Given a nite set of Wang tiles, does there exist a valid tiling of the plane? It is entirely possible that there is no answer to this problem (or any other given decision problem). For a given tile set it may be perfectly clear whether or not a tiling of the plane exists, think plain square or circular tiles. Clearly the rst will admit a tiling but the second will not. Yet an arbitrary tile set will likely be more complex than this, and an answer to the question of whether it tiles the plane is less clear. Denition. A problem, generally of deciding whether a particular statement holds, is decidable if there exists an algorithm, that is a nite set of steps, that will either prove or disprove the statement made in the problem. An undecidable problem is a problem for which there is no single method that will universally determine whether or not the statement is true. Denition. A problem A is said to reduce to problem B if the decidability of B follows from the decidability of A. That is to say that if B is undecidable, then A must also be undecidable. 26

33 It is of importance to recognise that saying A reduces to B does not necessarily mean that A reduces to B. If a decision problem applies to some set A then recognising that some subset A results in the problem being undecidable will mean that A is also undecidable, hence A reduces to A. However, A is more than likely to have some subset A for which the decision problem is decidable (even if it is only a trivial case of the problem), so it is not true that A reduces to A. But without knowing the decidability of A there is no reason to assume that A A, so A does not reduce to A. If the Domino Problem could be reduced to a known undecidable problem then the Domino Problem would also be undecidable. Both the Halting Problem and the Immortality Problem are such undecidable problems. The proof of the undecidability of the Immortality Problem has been included here, but a more traditional version may be seen in the papers of Hooper [5] and Jeandel [6] among others. Problem 3.1. The Halting Problem: Does an arbitrary Turing machine halt for a given input? Problem 3.2. The Immortality Problem: For a given a Turing machine, does there exist an immortal point? That is to say does a given Turing machine halt when started on an arbitrary tape? 3.1 The Halting Problem Problem. The Halting Problem: Does an arbitrary Turing machine halt for a given input? Theorem 3.3. The Halting Problem is undecidable. Proof. Assume there exists some Turing machine M that can correctly decide whether or not a Turing machine halts for a given input, it has input < X, x > where X is a Turing machine and x is the input for X. Say that the M returns 'yes' if the machine halts and 'no' if the machine does not halt. Construct a second Turing machine T such that it takes as input some machine X, copies it and the inputs the pair < X, X > into M. If M returns 27

34 'yes' then T re-inputs < X, X > into M, thereby entering an innite loop. If M returns 'no' then T immediately halts. It is possible to use X as input to X as any Turing machine X can be written as a set of instructions, and therefore a string of symbols. Any Turing machine can accept any string of symbols as input. However if it reaches a symbol that is not recognised then it immediately enters the halting state, in fact it is possible for the starting state to be the halting state. Consider the case where T is an input to T, then < T, T >is the input to M. There are two cases, either M returns 'yes' or M returns 'no'. If M returns 'yes' then it is because T will halt, but by returning 'yes' T must enter into an innite loop and cannot halt. If M returns 'no' then it is because T will not halt, but by returning 'no' T halts immediately. So it is not possible for such a machine M, that correctly decides whether or not a given Turing machine will halt, exists, and as such the halting problem is undecidable. 3.2 Counter Machines Before considering the Immortality Problem further it is necessary to introduce the concept of counting machines and how they are able to carry out basic operations. For a more detailed introduction to counter machines refer to A. J. Myers Compiling Scheme to Nano-Computers A counter machine is a nite set of counters, each able to hold a single natural number. A counter machine can compute only three things; increment (inc), decrement (dec), if zero (if). The function inc1 increases counter C1 by 1. The function dec2 decreases counter C2 by 1. The function if3 I4 I5 checks to see if counter C3 is zero, if it is then the machine will follow instruction I4 otherwise it will follow instruction I5. An instruction is a string of functions that the machine will follow in order. Assuming the counter machine is formed from multiple counters, basic operations can be computed. Suppose the value of C2 should be added to the value of C1. Dene instruction Iadd(1,2) to be as follows: inc1 dec2 if2 H I1. Then the instruction Iadd increases C1 as C2 is 28

35 decreased, and when C2 is zero halts. Suppose the value of C2 should be subtracted from the value of C1. Dene instruction Isub(1,2) to be as follows: dec1 dec2 if2 H Isub. Then the instruction Isub decreases C1 and C2 simultaneously, and when C2 is zero halts. Suppose the value of C1 should multiplied by the value of C2. Dene the instruction Itimes(1,2) to be the system (I1,I2,I3,I4) as follows: I1: if1 I4 dec1 I2 and I2: dec2 inc3 inc4 if2 I3 I2 and I3: dec4 inc2 if4 I1 I3 and I4: dec3 inc1 if3 H I4. Then Itimes(1,2) increases C3 the amount that is in C2 for each unit in C1 before copying C3 back to C1 and halting. Suppose the value of C1 should be divided by the value of C2 (this assumes that C1 is divisible by C2). Dene the instruction Idiv(1,2) to be as follows: I1: dec1 dec2 inc3 if2 I2 I1 and I2: inc4 if1 I5 I3 and I3: dec3 inc2 if3 I1 I3 and I5: dec4 inc1 if4 H I5. Then Idiv(1,2) simultaneously decreases C1 and C2 and copies C2 to C3, when C2 is empty it is returned to it's original value and C4 is increased by 1. The process is repeated until C1 is empty then the value in C4 is copied to C1 and the machine halts. So the counter machine can perform the basic operations of addition, subtraction, multiplication, and division. In fact it can use these to perform any of the actions of a Turing machine. 3.3 The Immortality Problem Problem. The Immortality Problem: For a given a Turing machine, does there exist an immortal point? That is to say does a given Turing machine halt when started on an arbitrary tape? To show that the Immortality Problem for Turing machines is undecidable a series of reductions will be proven until eventually the Immortality Problem reduces to the Halting Problem. The idea behind such a reduction comes from both Meyer [9] and [8], the proof of the second theorem occurring on page 14 of [8]. Theorem 3.4. The Halting Problem for 2-CM reduces to the Halting Problem for Turing 29

36 machines. Proof. The Halting Problem for Turing machines is undecidable as shown in section 3.1. For every Turing machine it is possible to construct a system of counting machines (CM) that simulate the Turing machine started on a blank tape. The tape of the Turing machine at any given instant will be nite in length, where the length is taken to be the distance between the rst and last non-blank character, since there will have been a nite number of steps, each extending the length by at most 1. Without loss of generality assume that the Turing machine works on a tape innite in only one direction. So let the character sequence on the tape a 1, a 2,..., a n be represented by the number p a 1 1 pa pan n where p i is the ith prime number. The sequence is unique to this number as the number is unique to the sequence. Let the number representing the sequence be the number on the rst counter C1, so C1 initially starts at 0 since the tape is initially blank. Let the second counter C2 represent the (numerical representation of the) state that the machine is in, so C2 will initially start at 2. The third counter C3 will represent the position of the head on the tape (that is the distance the head is from the origin). C4 will represent the character on the tape at a given point. An additional 2 counters will be used to aid in computations and will be called K1 and K2. For an arbitrary Turing machine instruction T MI = (s i c i c i+j Ls i+l ) where the machine head is in position k the CM will follow a set of instructions. First the character is changed, that is to say C1 is multiplied by p j k. This is done by initially copying the value of C1 to K1. At each step C1 is reduced by 1 and K1 and K2 are both increased by 1. When C1 is empty at each step K2 is reduced by 1 and C1 is increased by 1. When K2 is empty at each step K1 is decreased by 1 and C1 is increased by p k. Then the position of the head is changed, so C3 is reduced by 1. And the state is changed, so C2 is increased by l. If the halting state of the Turing machine has been reached then there is no numerical representation of this and so the CM will also halt. Assuming that the CM has not halted the next symbol on the tape must be read. The machine (now in position k 1) divides C1 by p k 1. At each step C1 is reduced by p k 1 30

37 and K1 and K2 are increased by 1. When C1 reaches 0, C1 is returned to it's original value by reversing the process and reducing K2 back to 0. Counter K1 is divided by p k 1 and the result held on K2, the process is repeated dividing counter K2 by p k 1 and holding the result on K2. Each time either K1 or K2 (or C1) reaches 0 counter C4 is increased by 1. If a counter cannot be reduced by p k 1 (that is the counter reaches 0 before the reduction can be completed) both K1 and K2 are emptied. The values on counters C3 and C4 determine the next instruction set to be followed. This CM instruction set will correspond to the next Turning machine instruction. This has shown the Halting Problem for Turing machines to be a reduction of the Halting Problem for 6-CM. But any counter machine can be reduced to a 2-CM in a similar manner. Let the counters of the 2-CM be represented by N1 and N2. Then the rst counter will represent the contents of each counter of the original system of counters. That is N1 = 2 C1 3 C2 5 C3 7 C4 11 K1 13 K2 and N2 is used to perform computations. As before to increment/decrement a counter from the original system the value of N1 is multiplied/divided by the appropriate prime, to check if a counter is zero check if there is a remainder when dividing by the appropriate prime. The halting state of the Turing machine has been replaced by a halting instruction, but the instruction will occur within the series of instructions corresponding to the Turing machine action, and this is the only time that the halting instruction will occur. So the 2-CM will halt if and only if the Turing machine halts. Theorem 3.5. A 4-CM can be constructed from a 2-CM such that the 4-CM has an immortal conguration only if the 2-CM does not halt. This leads to the conclusion that the Halting Problem for 2-CM is a reduction of the Immortality Problem for 4-CM. Proof. Let an arbitrary 2-CM be called M and the corresponding 4-CM be called M*. Then M* performs (bounded) simulations of M, that is it runs the rst n steps of M and then restarts the simulation. Counters C1 and C2 of M* are identical to counters C1 and C2 of M. Counters C3 and C4 of M* are used to bound the length of the simulation. For each step of the 2-CM counter C3 is decremented and C4 is incremented. When C3 is empty 31

38 C1 and C2 are set to zero (to restart the simulation of M) and C4 is copied back to C3 which is then also incremented by 1, as such the next simulation of M will be increased by one step. Clearly if M does not halt then M* must have at least one immortal point. If M does halt then M* will eventually halt either because the input to M* is not valid for a simulation of M (in which case M* will halt as it cannot continue an impossible simulation) or because it will lead to a restart of the simulation of M (in which case it will halt when the simulation length is suciently large). So M does not halt if and only if M* has an immortal point. And so the Halting Problem for 2-CM is a reduction of the Immortality Problem for 4-CM. Theorem 3.6. Any 4-CM can be written by a Turing Machine. Proof. Assume the counters have their number written in binary. Let the Turing machine run on a semi-innite tape. Each character on the tape will be a four-symbol string. That is each character represents the units from each counter. So to increment a counter is to return to the origin, and change the relevant 0 or 1 to the other respectively. If a 1 is changed to 0 then the head must move to the right and repeat. A counter is decremented in the same manner. To check if a counter is zero is to move to the far right, to the point at which the tape is blank rather than containing characters (such a place exists since a counter may only hold a nite number). Moving left from this point, check that each character has a 0 in the appropriate position. If a 1 is found then the head moves into a state appropriate to the counter not being zero, otherwise if the origin is reached the head moves into a state appropriate to the counter being zero. So a Turing machine can hold all the information of a 4-CM and simulate each operation of a 4-CM. If the 4-CM reaches a halting instruction then the Turing machine will halt as it will not have any other action to perform. The Turing machine therefore accepts as input a conguration corresponding to the initial values of the 4-CM counters. And for a given input the Turing machine will halt if and only if the 4-CM also halts as it exactly simulates it. So the Immortality Problem for 4-CM is a reduction of the Immortality Problem for Turing machines. 32

39 On the basis of the above three theorems it is seen that the Immorality Problem for Turing machines reduces to the Immorality Problem for 4-CM, which reduces to the Halting Problem for 2-CM, which reduces to the Halting Problem for Turing Machines. So the Immortality Problem for Turing machines reduces to the Halting Problem for Turing machines. And since the Halting Problem for Turing machines was proved to be undecidable in Section 3.1 of this chapter it has been prove, that the Immortality Problem for Turing Machines is also undecidable. A more natural approach is to attempt to construct for each Turing machine T a Turing machine M that has an immortal point if and only if T does not halt. Indeed this was the direction taken by Hooper [5]. However, the diculty which is run into here is preventing an unbounded search. In order to bound the simulation of the machine T, the simulation must end on reaching some boundary marker. But the input tape and input state is arbitrary so there is no guarantee that such a marker exists. In which case M may have an immortal point despite the fact that T halts when started on a blank tape, because the simulation of T is never restarted. 33

40 Chapter 4 The Origin-Constrained Problem The Origin-Constrained Domino Problem is similar to the Domino problem but with a further restriction. Problem 4.1. The Origin-Constrained Problem: Given a set of Wang tiles and a given tile which must be included in any tiling, does the tile set admit a valid tiling of the remaining plane. Before examining this problem, rst consider Turing Machines and how they may be represented as tilings. 4.1 Turing Machines as Tilings Recall that a Turing machine is a machine that manipulates symbols on a bi-innite tape according to a table of rules, it may only perform one action at a time and cannot store information other than the state that it is in and what is written on the strip. In this way such a Turing machine is entirely deterministic, regardless of how many times the machine is run the process and result will always remain the same for a given starting tape. Consider an arbitrary Turing Machine, T, with a nite number of states s 1,..., s n and a nite number of tape characters α 1,..., α m. Then each action of the Turing machine can be uniquely represented by the string s i α i α j L/Rs j meaning that if T is in state s i reading symbol α i then it changes α i to α j and moves left or right and into state s j. 34

41 Let the tape after each action be copied onto the plane such that each tape symbol lies exactly on the upper edge of a unit square. The tiles of the tile set will then be the unit squares, so each tile has a colour on it's upper and lower edge that represents the tape symbol at that point. If the tape symbol does not change then the tile is called an alphabet tile. But the tiles must also represent the actions of T. The position of the head at a given instant will be represented on the tile by the state it is in. So the lower edge of the tile is coloured to represent both the tape symbol and the head state. Similarly the state that T moves into is represented by a colour on the left or right edge of the tile (depending on which direction the head moves in). These tiles will be called action tiles and those shown below represent the actions s i α i α j Rs j and s i α i α j Ls j respectively. In order for a tile to be placed next to an action tile (in the position the head has moved to) the tile must have a corresponding state colour on the relevant edge, and the upper edge is also coloured to represent both the tape symbol and the state of the head. Tiles such as these are called merging tiles and exist for each character and each state except the halting state. The tiles shown below illustrate those discussed above. Tiles of this form for an arbitrary Turing machine will be called a Turing tile set or Turing tiles from this point forth. Figure Providing any row of the tiling contains only one action tile and one merging tile, the tiling will simulate the Turing machine. Note also that when constructing a tiling using these tiles arrow heads must be placed next to arrow tails, and vice versa. 4.2 The Turing Tile Sets for the Origin-Constrained Problem In order to specify a tile that must be included in the tiling the starting tape must be specied, in fact let the Turing machine always start on a blank tape. The tile set then 35

42 includes a set of starting tiles given as below and a blank tile which will be used to tile the lower half-plane. Figure It is the central starting tile (i.e. the starting tile including the state colour) that is the tile that is specied must be included in the tiling. Without loss of generality it can be placed at the origin. Given that this tile must exist in the tiling, and it is placed at the origin, the only tiles that can be placed along the x-axis are the other two starting tiles. Notice that the direction of the arrows mean that the central starting tile cannot be used again. The only tile that can tile the lower half-plane is the blank tile, and the blank tile cannot be placed in the upper half-plane due to the arrows on the starting tiles. So the upper half-plane will stimulate the Turing machine being run on a blank tape. Since the Turing machine is deterministic and the tiles exactly simulate it, there is only one possible tiling that the tiles can form given that the starting tile is included. 4.3 The Undecidability of the Origin-Constrained Problem To prove that the Origin-Constrained Problem is undecidable it is enough to show that there exists a sub-set of tile sets for which the Origin-Constrained Problem is undecidable. Consider the tile set discussed above, including both the starting tiles and the Turing tiles. For each Turing machine there exists a corresponding tile set of this form. Theorem 4.2. A Turing machine halts if and only if the corresponding tile set doesn't admit a valid tiling of the plane. Proof. Any Turing machine either halts or does not halt. Assume the Turing machine halts, then it will have reached the halting state. Recall that there does not exist a merging tile for the halting state. So if the Turing machine 36

43 reaches the halting state by a given action tile then there is no merging tile that can be placed to the left (or right, respectively) side of it. Since no tile can be placed in this position the tiling cannot be completed and a valid tiling of the plane does not exist. Assume the Turing machine does not halt, then the halting state is never reached. For any other state that is reached there exists a corresponding merging tile that may be placed next to it so that row of the tiling can be completed. Since the halting state isn't reached each row of the tiling can be completed, and hence a valid tiling of the plane exists. So a valid tiling of the plane by the corresponding tile set exists if and only if the Turing machine doesn't halt. The Origin-Constrained Problem is therefore reduced to the Halting Problem and as such is undecidable. 37

44 Chapter 5 Reducing the Domino Problem to the Halting Problem This chapter relies on an understanding of both Chapter 1 and Chapter 2. Both Robinson's tile set and the Turing tile sets are combined to create a new subset of tile sets that have the aperiodic structure of the Robinson tiles and forces the Origin-Constrained Problem to occur within the tiling. Such a reduction is included in Robinson's paper [10] 5.1 Alterations to the Robinson Tiles Remember the Robinson tiles are labelled with red and green arrows. Change the red arrows to either pink or orange arrows according to the following rules. The one end of an arrow is orange then the other end is the same colour. For tiles of type (a) or (b) the red arrows either become both pink or both orange. If two red arrows intersect, that is a tile of type (c), then one becomes orange and one becomes pink. So the tile set becomes the set that follows. 38

45 Constructing a tiling using these new tiles, and then only considering the pink lines, gives a system of squares, none of which intersect. The following diagram shows what the tiling will look like when only the pink lines are considered. 39

46 To run a Turing machine as a set of tiles, each tile represents a vertical unit of time and a horizontal unit of space. So simulating a Turing machine in a section of a tiling will require a 'square' amount of space. In order to achieve this sort of space a sections of the tiling must be prevented from simulating the machine. Therefore a rule is set that within any pink square, those tiles that lie in the same row and column smaller pink squares will not contain any Turing tiles, but may transfer the information held on adjacent tiles. In order to visualise this the above diagram shows the area in which a Turing machine may be run as black, and the areas only permitted to transfer information as grey. As such each pink square contains a smaller black space the tiles in which may have Turing tiles imposed upon them. 40

47 5.2 Constructing the New Tile Set The Turing tiles are superimposed onto the Robinson tiles in a manner explained below. Turing tiles are used in order to allow the corresponding Turing machine to run within the tiling, and the starting tiles are only placed on the lower edge of pink squares. Tiles bearing a are used to represent those areas in which that the Turing machine may not run, whilst also bearing state and symbol markings to transfer the information of the Turing tiles. A U symbol marks the internal edges of pink squares, except the lower edge, and this allows for the Turing patch to legally end at the edge of the pink square. Unlike before, tiles of dierent orientations (including reections) are considered to be distinct from each other. If a tile is a pink corner then the two edges that the corner does not touch are marked with a. If a tile has a horizontal pink headless arrow on it and the other arrow (which must have a head) is pointing downwards then the lower edge is marked with a and the upper edge marked with the starting symbol and state. If the second arrow points upwards then the upper edge is marked with a and the lower edge with a U. For the remaining tiles with horizontal pink lines then two tiles are created. The rst with a above and a U below, the other with a starting state symbol above and a U below. The exception to this is if the second arrow has a head, in which case the appears only on the side with the head. Similarly, if the pink line is vertical two tiles are created each with a on one side and a U on the other. Again the exception being if the second arrow is a head, in which case as before the appears only on the side with the head. For the remaining tiles (that is to say those without pink lines) each must be replicated at least three times. The rst tile will be marked with a on its upper and lower edges, and the left and right sides will be marked with a state symbol (the same on each side), a copy of each tile must be made for each state of the Turing machine. Similarly, the second tile will be marked with a on the left and right sides and a symbol or state/symbol pair on the top and bottom edges; again there must be a copy of each tile for each symbol and state/symbol pair. The third and nal tile is marked as each of the Turing tiles, with the exception of the Turing tiles. So it must be copied for each action, merging, and alphabet tile that exists in the Turing tile set. 41

48 A visual explanation of this construction is given on the next page (Figure 5.2.1). The Robinson tiles on the left have each of the Turing tiles on the right superimposed upon them. This has been followed by an example of the Robinson tiles superimposed with a 2-state 2-symbol Turing machine (Figure 5.2.2). It is seen that even adding a very simple Turing machine to the tile set quickly produces a very large tile set. This is due in part to the fact that all rotations of the tiles must be included; nonetheless increasing the Turing Machine by even a single state or symbol increases the size of the tile set very quickly. [Please note that this example covers more than one page.] Strictly these tiles should include the arrows of the Turing tiles; however, for visual clarity, they have been omitted. This omission, whilst oering more possibilities for tiling the plane, does not aect whether or not the tile set will tile the plane. This new tile set follows the same matching rules as both the Robinson tiles and the Turing tiles. Arrows must line up head-to-tail of the same colour, with a blue square at every intersection. Similarly adjacent tiles must have matching states and symbols. This means that in addition to the regular matching rules of Turing tiles any edge with a must be placed adjacent to another edge with a. An exception to these matching rules is tiles with edges marked U. The U represents a universal matching symbol, meaning that it may legally be placed next to any state or symbol (but not a ) or a blank edge. 42

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54 5.3 Reducing to the Halting Problem Theorem 5.1. The tile set described in Section 5.2 tiles the plane if and only if the corresponding Turing machine does not halt. Proof. A tile set of this form may be constructed to correspond to any Turing machine. As the pink squares become increasingly large as does the number successive congurations of the Turing machine that are simulated. If the Turing machine were to halt after n congurations then there exists a suciently large pink square that will simulate more than n congurations. This halting state (not represented on any of the tiles) would therefore prevent a valid tiling of the plane past this point, and therefore a valid tiling of the plane would not exist. If the Turing machine does not halt then each pink square may be correctly tiled, and since the Robinson tiles tile the plane, so will this new tile set. Since this theorem is an 'if and only if' statement, in this instance it has also been shown that the Halting Problem reduces to the Domino Problem. So had the decidability of the Halting Problem not been known, this could be given by the undecidability of the Domino Problem. 48

55 Chapter 6 The Periodic Problem The Periodic Problem is related to the Domino Problem. It asks not whether or not the tile set tiles the plane, but whether or not there exists a periodic tiling by the tile set. Problem 6.1. The Periodic Problem: Is it possible, with a given tile set, to tile the plane periodically? The following idea was covered at the start of Chapter 1 but is repeated here to expand on the given problem. Given a nite set of tiles they will either tile the plane (option 1) or they will not. Knowing that they tile the plane either all tilings of the plane are periodic (option 2), some of the tilings are periodic and some are non-periodic (option 3), or the tile set is aperiodic (option 4). The construction of an aperiodic tile set (option 4) has proved that not all tile sets that tile the plane do so periodically. But it says nothing of the inverse, of whether or not it can be decided that a tile set tiles the plane periodically. Is it perhaps the case that distinguishing between option 1 and options 2-4 is undecidable, but it is possible to decide if a tile set falls into options 1,4 or options 2,3? As it turns out this is an undecidable problem too. 49

56 6.1 Undecidability of the Periodic Problem The proof of the undecidability of this problem follows on from the proof of the undecidability of the Domino Problem given in Chapter 5. It therefore requires the construction of a tile set that is an extension of that used is the previous chapter. The proof given here is given in the same way as in The Classical Decision Problem [1]. Theorem 6.2. The Periodic Problem is undecidable. Proof. Take as a starting point the tile set constructed in Chapter 5. The idea is to make sucient alterations such that if the Turing machine halts, the tile set tiles the plane periodically rather than not tiling it at all. The rst change is to include the halting state H in the list of states to be put on tiles. In fact the halting state is treated as any other state and on reaching the halting state the state/symbol pair is transferred upwards as a symbol would on an alphabet tile. The matching rule which allows all states and symbols to be placed next to a universal symbol U does not apply to to the halting state H. Instead, all tiles with pink lines are copied and recoloured with purple to replace the pink. Tiles that lie along the base of the now purple square will maintain their starting symbols (or starting state/symbol), but markings below the purple line will be erased. This includes both markings from the Turing tiles and from the Robinson tiles. Similarly, the remaining tiles with purple lines (including the corners) will have any edge marked by a star erased, the opposite edge will retain the U and any Robinson markings. An extra arrow will be added to these tiles parallel to the purple one, and the tiles are copied to accommodate an arrow in either direction. As usual these arrows must meet head to tail and the lower corner tiles need only depict the tail end of a single arrow. Another copy of these tiles is made in which the extra arrow is double-headed and the U symbol is changed to H. The requirement of the extra arrow is to ensure that purple squares may not exist unless the halting state is reached. As a result of these changes any Turing machine that does not halt within the given space must be simulated within a pink square as before. However, if the Turing machine does halt at some point, then any square that simulates the machine past this point must 50

57 necessarily be purple. Remember that purple squares have no matching rules on their outer edges (they are completely blank). So if the Turing machine halts, and therefore necessitates a purple square, that particular square patch may be repeated across the plane to produce a periodic tiling. Similarly if the Turing machine does not halt then no purple squares will exist in a valid tiling and the tile set remains aperiodic. So the Periodic Problem has been reduced to the Halting Problem, thus proving its undecidability. 51

58 Chapter 7 Reducing the Domino Problem to the Immortality Problem 7.1 Introduction to Ane Functions and Immortal Points This section introduces some basic concepts key to understanding the Immortality Problem and how to reduce the problem to it. An ane transformation is a map which preserves points and straight lines (and planes when working in higher dimensions). Parallel lines remain parallel after an ane transformation but the transformation does not necessarily preserve angles between lines or distances between points, although ratios between points on a line remain constant. For the purposes of this paper an ane space (that is a space in which an ane map may be applied to a point) can be thought of as an Euclidean space, although an ane space has a more general denition. Ane transformations are similar to linear transformations but the zero point is not necessarily preserved. Given a system of ane maps it is possible to select a point as the origin, then any ane map is equivalent to a linear transformation followed by a translation. That is every ane map f : X Y is of the form x Mx + b where M is a linear transformation and b is a translation in Y. Consider an ane map f and it's domain D where f(x) = f i (x) for some system of ane maps f i that are applied to domains U i (that is i U i = D). The orbit of x D is 52

59 the set of f n (x) for n N < N such that for all n < N the point f n (x) lies in D but f N (x) / D, that is to say each iteration of f such that f(x) lies within D. If there exists some N N for some x D such that f N (x) / D then the system is said to halt. The system is said to be mortal if the system halts regardless of the value x D and immortal if there exists a point x D such that the system does not halt, that is to say the system is immortal if for some x D the value of f n (x) lies with D for all n N. Note that if f : D D then each point in D is immortal. 7.2 Reducing the Domino Problem to the Immortality Problem for Ane Functions This reduction to prove the undecidability of the Domino Problem was rst introduced by Jarkko Kari in 2008 [7]. Let a set of Wang tiles be labelled in such a way that they may be associated with a system of ane functions. Consider the following labeling of a tile set. Take a set of Wang tiles T and let the colours on their edges be replaced with pairs of real numbers. In fact let the rst coordinate represent the colour of the edge, and let the second represent the function that the tile is able to compute on the vertical edges and be simply 0 on the horizontal edges. Note that tiles have a north, south, east, and west side which will be shortened to n, s, e, w. A tile is said to compute an ane function f : R 2 R 2 if f(n) + w = s + e, that is if the numbers corresponding to each side fulll the equation. In order for the tile to compute a function f the function must not change the second coordinate of n. Let all tiles that compute the same function f i belong to the subset T i T. If a tile computes more than one function the it is split into multiple tiles each labelled as computing one of the functions. So for each i j the intersection of the respective subsets is the empty set, that is T i T j =. Take a horizontal strip of length m then it follows that f( n) + w m = s + e m where n, s are the averages of the north and south colours respectively and w, e are the colours of the respective end tiles. This can be seen by adding the equations for each individual 53

60 tile, remembering that e i = w i+1 since the connecting edges of adjacent edges must match and that f(n 1 + n 2 ) = f(n 1 ) + f(n 2 ) by the properties of ane functions. As the strip increases in length (that is as it tends towards crossing the entire plane) the value of 1 m tends towards 0 meaning w m and e m f( n) = s. are negligible and so the equation tends towards being As with the reduction to the Halting Problem since these theorems are proved in both directions it can also be concluded that the undecidability of the Domino Problem would prove the undecidability of the Immortality Problem. Theorem 7.1. A nite tile set T admits a valid tiling of the plane if and only if the corresponding system of ane maps f has an immortal point. Proof. First assume a valid tiling of the plane by the tile set T exists. Then an immortal point would be the average of the north colours n 0 on a horizontal strip. The northern average of subsequent strips is therefore an iteration f n ( n) of the ane function. Let the tiles be labelled as before then an appropriate system of ane functions would be f : D D where f(x) = {f i (x) x U i and f i : U i D (that is U i is the domain of f i and D = i U i ). The function f i is given by f i ( n) = s where n lies in U i and the tiles in the strip all computed the function f i. If it is not possible to calculate the average of northern colours, then the limit of averages of northern colours on nite strips of increasing length can be used. Since there are only nitely many tiles in the set, there need only be nitely many functions f i that the tiles compute, hence the system of ane functions is also nite. If the tile set tiles the plane then n D for all n since each row computes some f i by construction. If f n ( n 0 ) / D then n n / D, but n D for all n so f n ( n 1 ) / D. So if the nite tile set T admits a valid tiling of the plane then the corresponding system of ane maps has as immortal point. So it is left to show that if a system of ane maps has an immortal point then the corresponding tile set admits a valid tiling of the plane. For any α R 2 let α be represented by the bi-innite sequence B(α) where the kth digit of B k (α) is given by B k (α) = kα (k 1)α for k Z. The oor function is applied to each coordinate separately, that is (α 1, α 2 ) = ( α 1, α 2 ). Let the tile set corresponding to a system of nitely many 54

61 rational ane maps f i and their domains U i is given by the tiles below, where α is taken to be any value within the domain U i for a given f i. As usual the system is represented by f : D D where f(x) = {f i (x) x U i ). Figure By this construction for xed α and successive k the east/west sides of consecutive tiles match, so the tile set is able to tile a bi-innite horizontal strip, and the top and bottom edges of the strip read B(α) and B(f(α)) respectively so the tile computes some f i of the system. Despite there existing innitely many values of k and α there are only nitely many tiles in the set due to the use of the oor function in the colours. If an immortal point exists then the bottom edge of the strip, reading B(f(α)), will always lie within D, that is within some domain U i of the system, and so the next row of the tiling is possible. If however an immortal point did not exist then there would exist some n such that B(f n (α)) does not lie within D, that is within some U i of the system. Each possible row is associated with some α D, and if the representation of the lower edge of the previous row does not lie within D then the representation of the upper edge of the new row would also not lie within D (since the colours, and hence the representations, must match). So the new row of the tiling cannot be formed from tiles in the tile, else it could be represented by α D. Such a construction leads to a tiling of the lower half plane, but a similar construction could be used to tile the upper half plane. Regardless a tiling of the half plane can only exist if there is a tiling of the plane. So if a system of nitely many pairs (f i, U i ) has an immortal point then a nite tile set can be constructed which admits a valid tiling of the plane. Thus A nite tile set T admits a valid tiling of the plane if and only if the corresponding system of ane maps (f i, U i ) has an immortal point. 55

62 So the Domino Problem has been reduced to the Immortality Problem for ane functions. Therefore the Immortality Problem for ane functions is undecidable if and only if the Domino Problem is undecidable. 7.3 Kari's Aperiodic Tile Set Along with his reduction from the Domino Problem to the Immortality Problem, in 2008 Kari also produced an aperiodic tile set of 14 tiles [7] as shown below. Both the construction of the tiles set and the proof of it's aperiodicity use methods described in the previous section. So the tile set will be described here. Figure A tile is said to multiply by q R if qn + w = s + e. The rst four tiles all multiply by q 1 = 2, let these belong to T 2 T. The remaining ten tiles all multiply by q 2 = 2 3, let these belong to T 2 3 T. The colours of the vertical edges of these two subsets have been dened such that tiles can only form a horizontal row with other tiles from the same subset. Theorem 7.2. Kari's tile set is indeed aperiodic. Proof. Now that the tile set has been dened assume that it tiles the plane periodically. So there exists some rectangle R (of height k) that repeats periodically across the plane. For each row i = 1,..., k (counting downwards) of R let n i denote the sum of the northern 56