The Domino Problem of the Hyperbolic Plane Is Undecidable
aa r X i v : . [ c s . C G ] J un The Domino Problem of the Hyperbolic Plane IsUndecidable
Maurice Margenstern,Universit´e Paul Verlaine − Metz,LITA, EA 3097, IUT de Metz,ˆIle du Saulcy,57045 METZ C´edex, FRANCE, e-mail: [email protected]
October 25, 2018
Abstract
In this paper, we prove that the general tiling problem of the hyper-bolic plane is undecidable by proving a slightly stronger version using onlya regular polygon as the basic shape of the tiles. The problem was raisedby a paper of Raphael Robinson in 1971, in his famous simplified proofthat the general tiling problem is undecidable for the Euclidean plane,initially proved by Robert Berger in 1966.
The question, whether it is possible to tile the plane with copies of a fixedset of tiles was raised by Wang, [24] in the late 50’s of the previous century.Wang solved the origin-constrained problem which consists in fixing an initialtile in the above finite set of tiles. Indeed, fixing one tile is enough to entailthe undecidability of the problem. The general case, later called the generaltiling problem in this paper,
GT P in short, without condition, in particularwith no fixed initial tile, was proved undecidable by Berger in 1966, [1]. BothWang’s and Berger’s proofs deal with the problem in the Euclidean plane. In1971, Robinson found an alternative, simpler proof of the undecidability of thegeneral problem in the Euclidean plane, see [22]. In this 1971 paper, he raisesthe question of the general problem for the hyperbolic plane. Seven years later,in 1978, he proved that in the hyperbolic plane, the origin-constrained problemis undecidable, see [23]. Up to now,
GT P remained open.In this paper, we give a synthetic presentation of the techniques containedon several technical papers deposited on arXiv , see [13, 18], and on the web siteof the author, in particular [15]. 1n the second section, we remind a few features of all proofs of this problem.Then, we turn to the hyperbolic case, assuming that the reader is familiarwith hyperbolic geometry, at least with its popular models: Poincar´e’s disc orhalf-plane. We refer the reader to [20] and to [10] for preliminaries and otherbibliographical references.In the third section, we sketchilly present the frame of the construction.In the fourth section, we present a needed interlude, a parnthesis on brackets,which is a basic ingredient of the proof. In the fifth section, we lift up this lineconstruction to a planar Euclidean one. In the sixth section, we show how toimplement the Euclidean construction in the hyperbolic plane, using the specificproperties indicated in the third section. In the seventh section, we completethe proof of the main result:
Theorem 1
The domino problem of the hyperbolic plane is undecidable.
From theorem 1, we immediately conclude that
GT P is undecidable in thehyperbolic plane.In the eighth section, we give several corollaries of the construction andwe conclude in two directions. The first one wonders whether it is possibleto simplify the construction. The second direction tries to see what might belearned from the construction leading to theorem 1.Before turning to section 2, let us remark that an alternative proof of
GT P is claimed by Jarkko Kari, see [6, 7]. His proof is completely different of thisone. It is completely combinatoric and it makes use of a non-effective argument.Here, we have no room to discuss this latter point.
In the proofs of
GT P in the Euclidean plane by Berger and Robinson, there isan assumption which is implicit and was, most probably, considered as obviousat that time.Consider a finite set S of prototiles . We call solution of the tiling of theplane by S a partition P such that the closure of any element of P is a copy ofan element of S . We notice that this definition contains the traditonal conditionon matching signs in the case when the elements of S possess signs.Note that GT P can be formalized as follows: ∀ S ( ∃ P sol ( P , S )) ∨ ¬ ( ∃ P sol ( P , S )) , where ∨ is interpreted in a constructive way: there is an algorithm which,applied to S provides us with ’yes’ if there is a solution and ’no’ if there is none.The origin-constrained problem can be formalized in a similar way by: ∀ ( S, a ) ( ∃ P sol ( P , S, a )) ∨ ¬ ( ∃ P sol ( P , S, a )) , where a ∈ S , with the same algorithmic interpretation of ∨ .Now, note that if we have a solution of GT P , wa also have a solution ofthe origin-constrained problem, with the facility that we may choose the firsttile. However, to prove that
GT P has no solution, we have to prove that,2hatever the initial tile, the corresponding origin-constrained problem also hasno solution.However, Berger’s and Robinson’s proofs consider that we start the con-struction with a special tile, called the origin . There is no contradiction withwhat we have just said, because they force the tiling to have a dense subset oforigins. In the construction, the origins start the simulation of the space-timediagram of the computation of a Turing machine M . All origins compute thesame machine M which can be assumed to start from an empty tape. Theorigins define infinitely many domains of computation of infinitely many sizes.If the machine does not halt, starting from an origin, it is possible to tile theplane. If the machine halts, whatever the initial tile, we nearby find an originand, from this one, we shall eventually fall into a domain which contains thehalting of the machine: at this point, it is easy to prevent the tiling to go on.The present construction aims at the same goal. { , } and itsmantilla Our construction takes place in a particular tiling of the hyperbolic plane: thetessellation { , } which we call the ternary heptagrid , simply heptagrid ,for short, see [2, 11, 20]. It is generated by the regular heptagon with vertexangle 2 π a b Figure 1
On the left: Robinson’s basic tiles for the undecidability of the tiling problemin the Euclidean case. On the right: the tiles a and b are a ’literal’ translation ofRobinson’s basic tiles to the situation of the ternary heptagrid. We consider a special tiling based on the tessellation { , } which is moti-vated b the following consideration. In figure 1, the left-hand side indicates thebasic shapes of the tiles devised by Robinson in the construction of the tilingused in his proof of the undecidability of GT P . The right-hand side of the figuregives the ’literal’ translation of these tiles for the heptagrid. It is not difficultto see that it is not possible to tile the hyperbolic plane with the tiles a and b .However, a slight modification of the tile b , see the tile c of figure 2, leads to thesolution. 3 c α β Figure 2
On the left: change in the tiles `a la Robinson. On the right: their transla-tion in pure Wang tiles.
On the right-hand side of figure 2, we have the translation of the tiles a and c into genuine `a la Wang tiles. The pattern α corresonds to a and the pattern β corresponds to c . In the ternary heptagrid, a ball of radius n around a tile T is the set oftiles which are within distance n from T which we call the centre of the ball.The distance of a tile T to another T is the number of tiles constituting theshortest path of adjacent tiles between T and T . We call flower a ball ofradius 1.Now, the tiles α and β of figure 2 can be assembled in flowers only: a tile α ,which we call centre , require to be surrounded by tiles β only, which we call petals . This is obtained by numbering the edges of the tiles α from 1 to 7, see[14, 15]. Now, a petal belongs to three flowers at the same time by the verydefinition of the implementation. From this, there is a partial merging of theflowers. By definition, the resulting tiling is the mantilla .
15 6
104 105 106
272 273 274 275 276 277 278 279
107 108 109
280 281 282 283284285286287
42 4344
A B C F
15 6
104 105 106
272 273 274 275 276 277 278 279
107 108 109
280 281 282 283284285286287 G
15 6
104 105 106
272 273 274 275 276 277 278 279
107 108 109
280 281 282 283284285286287
42 4344
A BC
Splitting of the sectors defined by the flowers. From left to right: an F -sector, a G -sector and an -sector. It is not difficult to see tha there can be several types of flowers, consideringthe number of red vertices for which the other end of an edge is a vertex of acentre. We refer the reader to [14] for the corresponding properties. Here, wesimply take into consideration that we have three basic patterns of flowers, whichwe call F -, G - and -flowers respectively. They are represented by figure 3.4he figure also represents the way which allows to algorithmically constructthe mantilla. It consists in splitting the sectors generated by each kind offlowers in sub-sectors of the same kind and only them, which we call the sons of the flowers. From this, we easily devise a way to recursively define a tiling.The construction is deterministic below the flower, and it is non-deterministicwhen we proceed upwards. We do not make the notion of top and bottom moreprecise: it will be done later. The exact description of the splitting can be foundin [14]. We simply remark that such a splitting is an application of the generalmethod described in [20, 10], for instance.Based on these considerations, we have the following result which is thor-oughly proved in [14]. Lemma 1
There is a set of tiles of type α and tiles of type β which allowsto tile the hyperbolic plane as a matnilla. Moreover, there is an algorithm toperform such a construction. Note that the leftmost flower of figure 3, which represents an F -sector, also in-dicates a region delimited by continuous lines, yellow in coloured figures. Theselines are mid-point lines, which pass through the mid-points of consecutiveedges of heptagons of the heptagrid. As shown in [2, 11], they delimit a Fi-bonacci tree. Let us remind that a Fibonacci tree has two kinds of nodes: blackones and white ones. A black node has two sons, a black and a white one. Awhite node has three sons, a black and two white ones. In both cases, the blackson is the leftmost son. The root of a Fibonacci tree is a white node. The tilesinside the tree which are cut by these mid-point rays are called the borders ofthe tree, while the set of tiles spanned by the Fibonacci tree is called the area of the tree.Say that an F -son of a G -flower is a seed and the tree, rooted at a seedis called a tree of the mantilla . As the seeds are the candidates for theconstruction of a computing region, they play an important rˆole. From figure 3,,we can easily define the border of a sector which is a ray crossing -centres.See [14] for exact definitions. Lemma 2
The borders of a tree of the mantilla never meet the border of asector.
From lemma 2, as shown in [14], we easily obtain:
Lemma 3
Consider two trees of the mantilla. Their borders never meet. Eithertheir areas are disjoint or the area of one of them contains the area of the other.
From this, we can order the trees of the mantilla by inclusion of their areas.It is clear that it is only a partial order. We are interested by the maximalelements of this order. We call them threads , see [14] for an exact definition.Threads are indexed by IN or ZZ . We call ultra-threads the threads which are5ndexed by ZZ . When there are ultra-threads, two of them coincide, startingfrom a certain index. Note that the union of the areas of the trees which belongto an ultra-thread is the hyperbolic plane. There can be realizations of themantilla with or without ultra-threads. In [15], we have a new ingredient. We define the status of a tile as black or white , defining them by the usual rules of such nodes in a Fibonacci tree.Then, we have the following property. Lemma 4
It is possible to require that -centres are always black tiles. Whenthis is the case, a seed is always a black tile.
15 6
104 105 106
272 273 274 275 276 277 278 279
107 108 109
280 281 282 283284285286287
42 4344
A B C
15 6
104 105 106
272 273 274 275 276 277 278 279
107 108 109
280 281 282 283284285286287
42 4344
A B
Figure 4
The black tile property and the levels:On the left-hand side, a black F -centre; on the right-hand side, a black G ℓ -centre. Wecan see the case of an -centre on both figures. As shown in [15], we can define arcs as follows: in a white tile, the arc joinsthe mid-point of the sides which have a common vertex with the side shared bythe father. In a black tile, the arc joins the mid-point of the sides shared bythe father and the side shared by the uncle, which is on the left-hand side ofthe father. Joining arcs, we get paths. The maximal paths are called isoclines .They are illustrated on figure 4. An isocline is infinite and it splits the hyperbolicplane into two infinite parts. The isoclines from the different trees match, evenwhen the areas are disjoint.
Lemma 5
Let the root of a tree of the mantilla T be on the isocline . Then,there is a seed in the area of T on the isocline . If an -centre A is on theisocline , starting from the isocline , there are seeds on all the levels. Fromthe isocline there are seeds at a distance at most from A . We number the isocline from 0 to 19 and repeat this, periodically. Thisallows to give sense to upwards and downwards in the hyperbolic plane.6
A parenthesis on brackets
We refer the reader to [15] for an exact definition. However, figure 5, below,illustrates the construction which now, we sketchily describe.The generation 0 consists of points on a line which are regularly spaced. Thepoints are labelled R , M , B , M , in this order, and the labelling is periodicallyrepeated. An interval defined by an R and the next B , on its right-hand side, iscalled active and an interval defined by a B and the next R on its right-handside is called silent . The generation 0 is said to be blue . R BB RB MM M M MBR M M BR MM M BR M M BR M M BRMR B M MR B M MR M MR B M MR B M MR B M MR B M MR BX YMMM MMM
Figure 5
The silent and active intervals with respect to mid-point lines. The lightgreen vertical signals send the mid-point of the concerned interval to the next gener-ation. The colours are chosen to be easily replaced by red or blue inan opposite way.The ends X and Y indicate that the figure can be used to study both active and silentintervals. Blue and red are said opposite . Assume that the generation n is defined.For the generation n +1, the points which we take into consideration are thepoints which are still labelled M when the generation n is completed. Then,we take at random an M which is the mid-point of an active interval of thegeneration n , and we label it, either R or B . Next, we define the active and silentintervals in the same way as for the generation 0. The active and silent intervalsof the generation n +1 have a colour, opposite to that of the generation n .When the process is achieved, we get an infinite model .Deep results on the space of all these realizations are given by an acurateanalysis to be found in [8]. The interested reader should have a look at thispaper.For our purpose, infinte models have interesting properties, see [15]. Wecannot mention all of them here. We postpone some of them to the Euclideanimplementation with triangles.Cut an infinite model at some letter and remove all active intervals whichcontain this letter. What remains on the right-hand side of the letter is calleda semi-infinite model .It can be proved that in a semi-infinite model, any letter y is contained inat most finitely many active intervals, see [15].7 The interwoven triangles
Now, we lift up the active intervals as triangles in the Euclidean plane. Thetriangles are isoceles and their heights are supported by the same line, calledthe axis , see figure 6.We also lift up silent intervals of the infinite model up to again isocelestriangles with their heights on the axis. To distinguish them from the others,we call them phantoms . We shall speak of trilaterals for properties sharedby both triangles and phantoms.
Figure 6
An illustration of the interwoven triangles.
We have very interesting properties for our purpose.
Lemma 6
Triangles of the same colour do not meet no overlap: they are dis-joint or embedded. Phantoms can be split into towers of embedded phantomswith the same mid-point and with alternating colours. Trilaterals can meet by abasis cutting the halves of the legs which contains the vertex.
From the construction of the abstract brackets, we get a simple algorithmto construct the inteerwoven triangles.The generation 0 is fixed by alternating triangles and phantoms. The mid-distance lines of the phantoms of generation 0 grow a horizontal green signalwhich crosses the legs of the phantom which they meet.When the generation n is completed, a vertex of a triangle or a phantomis put on the intersection of the axis with the mid-distance line of a triangle of generation n . The legs grow until they meet a green signal, traveling on ahorizontal line. If the trilateral is a tirangle, the legs stop the green signal. Ifthe trilateral is a phantom, the green signal crosses the legs. In both cases, thelegs go on, until thy meet a basis of their colour. They stop it and constitutea trilateral of the generation n +1. Now, the intersection of the basis of the8rilateral with the axis requires a vertex: of a triangle, if the basis belongs toa phantom, of a phantom, if the basis belongs to a triangle. The process isrepeated endlessly.The detection of the green signal and of the correct basis are facilitated bythe following mechanism: the legs of red triangles emit horizontal signals whichhave a laterality. A right-hand side signal emits a right-hand side signal and aleft-hand side leg emits a left-hand side signal. We do not provide a tile for themeeting of such signals inside a triangle. This allows to detect what correspondsto the free letters and which we call the free rows of a red triangle. Lemma 7
The interwoven triangles can be obtained by a tiling of the Euclideanplane which can be forced by a set of tiles.
In [15], we display the corresponding tiles which are in a square format,and we also describe them with the help of formulas taking into account theproperties of lemma 6.
The idea of the implementation in the hyperbolic plane is based on the followingobservation.From lemma 5, we define the isoclines 0, 5, 10 and 15 to play the rˆole of therows in the Euclidean implementation. The trilaterals will be constructed ontrees of the mantilla. A vertex will be a seed, and the legs are supported by theborders of the tree. The basis is defined by an isocline which cuts both bordersof the tree.As there are 6 seeds on the isocline 5 inside the tree defined by a seed at theisocline 0, there are 6 trilaterals of the generation 1 raised by a triangle of thegeneration 0. And so, contrarily to what happens in the Euclidean construction,we have several trilaterals of the same generation for the same set of isoclinescrossed by the legs of these trilaterals.Call latitude of a trilateral the set of isoclines which are crossed by its legs,vertex and basis being included.It is not difficult to see that there will be infinitely many trilaterals within agiven latitude. This requires to synchronize the choice of triangle or phantomwhen turing from one generation to the next one. Also, we can imagine thathorizontal signals coming from different triangles of the same latitude and withdifferent lateralities will meet. This will require to tune many details in orderto maintain the guidelines of the algorithm, described in section 5.The idea will be to synchronize the bases, the vertices and the signals on themid-distance lines. We shall also have to see how the axis is implemented.From what we have already seen, there is no moe one axis, but a lot ofthem. In fact, what we called an axis can be materialized by a thread. AS most9hreads are indexed by IN only, we have always the implementation of a semi-infinite model. Now, we shall manage the implementation in such a way that thesemi-infinite models are simply different cuts of the same infinite model. Thepossibility of the realization of the infinite model in the case of an ultra-threadbrings in no harm: it can be viewed as a cut at infinity. By definition, we decide that all seeds which are on an isocline 0 are active ,which means that they actually grow legs of a triangle of the generation 0. Thisis enough to guarantee that the set of active seeds is dense in the hyperbolicplane. Next, an active seed diffuses a scent inside its trilateral until the fifthisocline, starting from this seed, is reached. Seeds which receive the scent, andonly them, become active. An active seed triggers the green signal when itreaches an isocline 5 or an isocline 15. By construction, the generation 0 is notdetermined by the meeting of a green signal. But the others are.We can see that the scent process constructs a tree. The branches of thetree materilize the thread which implements the considered semi-infinite models.Note that the above synchronizatoin mechanism fixes things for spaces betweentriangles but also inside them.
Due to the occurrene of several trilaterals within a given latitude, now we haveto require that all triangles, both red and blue and blue 0, emit a signal alongtheir legs. The signal will be called upper when it is emitted by the legs, butthere is an exception: the vertices do not emit any horizontal signal. There isanother exception: the corners of a phantom, as well as those of a triangle, alsoemit an upper signal. The colour and the laterality of this signal are those ofthe corner.Now, in between two contiguous triangles of the same latitude, horizontalsignals of the same colour but with a different laterality will meet. We have toallow such a meeting, which will be performed by an appropriate tile which wecall a join tile. There is a join-tile for red and blue signals, as well as for theorange signals. The join-tile, see the pattern represented by figure 7, illustratessuch a junction. On the left-hand side, we have the right-hand side signal and,on the right-hand side, we have the left-hand side one. We also require that anupper horizontal signal of a given laterality may cross a leg of a trialteral of thesame colour, only if it has the same laterality as the leg. Note that the oppositejunction cannot be obtained, as turing tiles is impossible in our setting. Thisimpossibility is guaranteed by the existence of the isoclines and thir numbering.
Lemma 8
An upper horizontal signal with a laterality cannot join two legs ofthe same tilateral.
Proof: easy corollary of the rule about the meeting of legs with an upper hori-zonal signal. 10ccordingly, the legs of a trilateral may only be joined by a horizontal signalwhich has no laterality.
Figure 7
The pattern of a join-tile.
The first mechanism which we introduce to force the synchronization of differentconstriutions is that all bases on a given isocline merge. This changes the tilefor the coner, but this does not affect the algorithm of section 5.The first principle forces the presence of various signals on the same isocline.We have to look at the consequences of such simultaneous presence in order toavoid contradictions which would ruin the construction.As a first point, note that the upper signals allow to differentiate the variousparts of a basis. If the upper horizontal signal is of the same colour as the basis,we are outside any trilateral whose basis lies on the same isocline. If not, weare inside. We say that the basis is covered if it is accompanied by an upperhorizontal signal of its colour. Otherwise, we say that the basis is open .This distinction is important. When a leg meets a basis: if it is the first halfof a leg, i.e. between the vertex and the mid-point of the leg, it meets the basiswithouth changing it. When the second half of a leg meets a basis, it crosses itif the colour is different. If it is the same colour, it crosses it only if the basisis covered. If it is open, the leg has met the basis with which it forms a corner.Indeed, from lemma 8, an upper horizontal signal cannot go from one leg of atrilateral to the other: there must be a triangle of the same colour in between.And so, an open basis does not cross the second half of a leg of a trilateral ofthe same colour: and so, for the second half of a leg, when the leg meets such asignal, this means that the expected basis is found.Now, we have to look at the consequences of the synchronization on othersignals: on the green signals and on the horizontal blue and red signals.The problem is the following. Consider two triangles A and B of the samelatitude. They belong to the same generation n and they have the same colour.We may assume that A is on the left-hand side of B . Let ι be the isocline ofthe mid-distance line of A and B . From the study of the abstract brackets, weknow that there is a tower of phantoms inside A and inside B , with also ι as themid-distance line of the phantoms of the tower. We also know that the sametower is repeated along ι between A and B , possibly several times. Accordingly,there is a green signal on ι inside A and inside B , and also between A and B .11 similar problem occurs with the horizontal signals which are emitted bythe right-hand side leg of A , the vertex being excepted. Symmetrically, theleft-hand side leg of B emits a horizontal signal of the same colour and onthe same isoclines of the latitude of A . The join-tile of figure 7, replicated inappropriate colours, allows to connect together horizontal signals of oppositelateralities travelling between A and B in the appropriate directions and on thesame isocline. From lemma 8, if a horizontal signal enters a phantom of itscolour, we get into trouble for the meeting with the other leg of the phantom.Both problems can be solved in a similar way.The idea is to avoid the phantoms as it is not possible to cross them. Theadvantage is that the deviated signal does not distrub the construction insidethe phantom. Also, if during the detour, the signal keeps its laterality, it behavesas if the phantom were not present. In particular, the join-tile can be used forthe connection of the signal with the opposite one coming in front of it, fromthe other triangle.Technically, the solution is the following. The legs of a triangle stop the greensignal which meet them on the mid-distance line, as required by the algorithmof section 5. Now, at the mid-point of the leg, on the corresponding isocline ι ,the leg triggers an orange signal σ of its laterality, outside the triangle, say A .When σ meets the first phantom P on its way, it does not cross it: on the ohterside of the leg of P , there is a green signal. Instead of this, σ climbs up over theleft-hand side leg of P until it reaches the vertex. Then, it goes down along theright-hand side leg of P until it reaches the isocline σ . This is easy to recognize:it is the single tile such that there is a green signal on the other side of the leg.Also, during this travel on the first half of the legs of P , σ does not change itslaterality. From this, lemma 8 also applies to σ , which rules out the occurrenceof an orange signal on ι inside P . And so, σ does not disturb the constructioninside P . By lemma 8, we can see that σ matches its opposite signal on ι withthe join-tile.It is not difficult to see that the solution applies to a horizontal blue signal.Indeed, the structure of inner trilaterals inside a phantom is the same as insidea triangle of the same generation. Accordingly, the notion of a free row alsoapplies to phantoms. Now, for a blue-0 or blue phantom, there is a single freerow: the mid-distane line. Accordingly, a horizontal blue signal meeting the legof a blue-0 or blue phantom P on an isocline which is not the mid-distance lineof P also meets the opposite signal coming from a blue-0 or blue triangle insidethe phantom. The join-tile solves the problem if needed. For the blue signalwhich travels on the isocline ι of the mid-distance line of P , its behaviour isexactly that of an orange signal, which solves the problem.We remain with the case of a red triangle. This time, ι still denotes themid-distance line of the triangles A and B , and the right-hand side leg of A emits right-hand side horizontal red signals on all the isoclines of the latitudeof A , the line of the vertex of A being excepted. We have the same situation aswith a blue signal if the signal arrives on an isocline which does not correspondto a free row of the red phantom P which is first met on the way. If the signalarrives on an isocline of a free row of P , the idea is to collect all the red signals12ravelling within the latitude of P on the isocline of a free row in a single signal,as there are a lot of free rows in a red trilateral. This single signal has the form ofa red signal with a laterality: that of all the red signals arriving to P . It climbsup over the legs of P , this time from the vertex to the basis and conversely.Also, the laterality of the signal is not changed and, when it goes down alongthe other leg of P , the signal sends a copy of itself, outside P , on each isoclineof a free row. In this way, the red signals which arrive to P and which departfrom it on an isocline of a free row of P constitue a comb: on one side, thecomb gathers the signals, and on the other side, it dispatches them. In thisway, the avoiding is obtained without perturbing what happens inside P , andalso without disturbing what must be outside P . The signal passes as if P werenot present. We just remark that as the laterality is not changed, lemma 8 alsoapplies here. As a consequene, the same phenomenon may happen inside P ,but only within the latitudes of the trianles which P contains. As the mid-distance line of these triangles are different from that of P , the just describedphenomenon occurs for the inner phantoms inside P whose mid-distance line isthat of P . This is conformal with the requirement that what happens inside P must not be disturbed.Now, we can conclude that the tiling forces the construction of trilateralsgeneration after generation, as indicated by the algorithm os section 5. The active seeds were defined in sub-section 6.1. They allow to define thetrilaterals in the hyperbolic plane.Now, we ignore the blue-0 and the blue triangles, the phantoms of any colouras well as the parts of the bases of red triangles which are covered. Accordingly,we focus our attention on the red triangles only.We have already mentioned that lemma 8 applied to horizontal red signalsallows us to detect the free rows inside the red triangles. We shall agree thatta special signal, the yellow one, without laterality, will mark the free rowsof the red triangles. In fact, the set of tiles which implement the interwoventriangles in the Euclidean plane, see [13, 15], also implement this detection andthe installation of the yellow signal on the free rows.The free rows of the red triangles constitute the horizontals of the grid whichwe construct in order to simulate the space-time diagram of a Turing machine.Now, we have to define the verticals of the grid to complete the simulation.The verticals consist of rays which cross -centres. Figure 8 illustrates howthey are connected to the different possible cases of contact of the isocline offree row with the border of the tree.The computing signal starts from a the seed. It travels on the free rows.Each time a vertical is met, which contains a symbol of the tape, the requiredinstruction is performed. If the direction is not changed and the corresponding13order is not met, the signal goes on, on the same row. Otherwise, it goesdown along the vertical until it meets the next free row. There, it looks at theexpected vertical, going in the appropriate direction. Further details ared dealtwith in [15].
15 6
104 105 106
272 273 274 275 276 277 278 279
107 108 109
280 281 282 283284285286287
42 4344
A B C
15 6
104 105 106
272 273 274 275 276 277 278 279
107 108 109
280 281 282 283284285286287
42 4344
Figure 8
The perpendicular starting from a point of the border of a triangle whichrepresents a square of the Turing tape.On the left-hand side: the case of the vertex. On the right-hand side, the threeother cases for the right-hand side border are displayed on the same figure.
Note that in the case of the butterfly model, see [13, 15], the mechanism ofthe orange signal forces the green signal to run over the whole isocline which isthe mid-point of the latitude which contains no triangle. Indeed, the lateralityconstraints of the tiles for the crossing legs of trilaterals prevent an orange signalto run at infinity.
Within the frame of this paper, it is not possible to exhaustively describe thetiles needed for the construction which we described. In the reports [15] andin [13], we give a precise account of the tiles.Here, we just indicate how to construct the tiles, at the same time giving away to describe all the tiles and to count them.The finite set of tiles which we need to prove theorem 1, consists of two parts.First, we have the set of prototiles which forces the construction of the mantillaas well as the isoclines, the activation of seeds through the isoclines 0 and thesreading of the scent, and then the construction of the interwoven triangles,including the detection of the free rows in the red triangles.The second set consists of meta-tiles which, in fact, are variables for tiles,as the meta-tiles convey the signals directly connected with the simulation ofa given Turing machine. In the actual construction, the meta-tiles replace apart of the prototiles: they replace all the prototiles which are placed on anelement of the computation: either the tiles which convey the computing signal,or those which convey the evolution of each square of the Turing tape. But thereplacement is not systematic: depeding on the simulated machine, the samefree row may hold a computing signal in a precise interval and no computingsignal outside this interval. 14n each set, the tiles are constituted of a tile ( α ) or ( β ) on which we superposeseveral signals. We distinguish between horizontal and vertical signals. Thehorizontal signals can be viewed as channels which run along the path of theisocline inside the tile. We know that each tile has a top part and a bottom one.By definition, the top is determined by the side which is shared by the father ofthe tile. We define a local numbering by numbering the sides of a tile from 1up to 7, with the number 1 given to the side shared with the father, and theother numbers in increasing order while counter-clockwise turing around thetile. In the local numbering, the isocline goes from the side 2 to the side 7 ina white tile and it goes from the side 3 to the side 7 in a black one. For uppersignals;the channel is above the isoclines. A basis runs below the isocline. Thegreen and orange signals run on the same channel which is above the isoclineand below the channel for a horizontal red or blue signal.For vertical signals, we have two main kinds of verticals: the legs of a tri-lateral and the verticals for the computing grid. The signals of the legs alwayscross black tiles and they are split into two categories: right-hand side signalswhich go from the side 2 to the side 6 and left-hand side ones which go fromthe side 1 to the side 4. From this, we notice that a tile of a leg always knowsits laterality and, also, which of its sides are inside the trilateral and which areoutside it. The vertical signals for the grid go through -centres. In an -centre,the signal goes from the side 2 to the side 5 in terms of the local numbering. Interms of the numbering of the mantilla, the signal goes from the side 7 to theside 4. Next, in the petal 147 ◦ , it goes from the side 1 to the side 4 in terms ofthe local numbering. And then, in the petal 2 ◦
77, it goes from the side 1 tothe side 6, still with respect to the local numbering.We just remark that we have three colours for the trilaterals: blue-0 for thegeneration 0, blue for the even generations and red for the odd ones. Legs of tri-angkes are represented by a thick signal while legs of phantoms are representedby a thin one. First and second halves of legs are also distinguished. In blue-0and blue trilaterals, the first half is dark and the second half is light. In redtrilaterals, we use the opposite convention: the first half is light and the secondone is dark. The tile of figure 7 defines the general patterns to indicate thelaterality of a signal. These patterns are given the colour of the correspondingsignal.Then, the most intricate task comes: the superposition of all these kinds ofsignals. This can be precisely described and counted, as performed in [15, 13].With this, we completed the proof of theorem 1.
The construction leading to the proof of theorem 1 allows to get a few resultsin the same line of problems.As indicated in [3, 4], there is a connection between
GT P and the
Heeschnumber of a tiling. This number is defined as the maximum number of coronas of a disc which can be formed with the tiles of a given set of tiles, see [9] for15ore information. As indicated in [4], and as our construction fits in the caseof domino tilings, we have the following corollary of theorem 1.
Theorem 2
There is no computable function which bounds the Heesch numberfor the tilings of the hyperbolic plane.
The construction of [12, 14] gives the following result, see [16, 19].
Theorem 3
The finite tiling problem is undecidable for the hyperbolic plane.
Combined with the construction proving theorem 3 and a result of [20], theconstruction of the present paper allows us to establish the following result,see [17].
Theorem 4
The periodic tiling problem is undecidable for the hyperbolic plane,also in its domino version.
In this statement, periodic means that there is a shift which leaves thetiling globally invariant.At last, in another direction, we may apply the argments of Hanf and Myers,see [5, 21], and prove the following result, see [15].
Theorem 5
There is a finite set of tiles such that it generates only non-recursivetilings of the hyperbolic plane.
The first consequence is that, according to the estimations of [15, 18], we need ahuge number of tiles. Taking into account the changes introduced in [18], a newcounting indicates that we need 23,323 tiles for the prototiles and 6,541 addi-tional ones for the meta-tiles, this will precisely be presented in a forthcomingpaper.Of course, we may wonder whether the number of tiles can be reduced. Thismight be possible by a small tuning of the present signals. As an example,we could forbid yellow rows on the mid-distance line of a red triangle. Butthe advantage would not be important enough. Now, an attentive look at thetables, suggested in sub-section 7.2, indicates that the reason of the big numberof tiles lies in lemma 5. A consequence of the lemma is the important numberof passive tiles connected with the isoclines which do not bear the constructionsignals. And so, to get a significant reduction of the number of tiles means tofind a new, simpler setting. Professor Goodman-Strauss advised me to do so.In fact, recently, I realized that this is possible. I have not the room, here,to explain the idea. It is interesting to notice that this simplification allowsto implement a similar implementation of the interwoven triangles in both theternary heptagrid and in the pentagrid.The second consequence which could be derived from the construcion lies ona more abstract level. Let us look at the lifting of the abstract brackets to the16nterwoven triangles. At first glace, this seems to be a Euclidean construction.In the very paper, a whole section is devoted to the Euclidean implementationof the interwoven triangles. And in the next section, we still transfer this con-struction to the hyperbolic plane. It seems to me that the fact that this transferis possible has an important meaning. From my humble point of view, it meansthat a construction which looks like purely Euclidean has indeed a purely com-binatoric character. It belongs to absolute geometry and it mainly requiresArchimedes’ axiom. Note that absolute geometry itself has no pure model. Arealization is necessarily either Euclidean or hyperbolic. We suggest to concludethat, probably, the extent of aboslute geometry is somehow under-estimated.
Acknowledgement
I am very pleased to acknowledge the interest of several colleagues and friends tothe main result of this paper. Let me especially thank Andr´e Barb´e, Jean-PaulDelahaye, Chaim Goodman-Strauss, Serge Grigorieff, Tero Harju, Oscar Ibarra,Hermann Maurer, Gheorghe P˘aun, Grzegorz Rozenberg and Klaus Sutner.
References [1] Berger R., The undecidability of the domino problem,
Memoirs of the Amer-ican Mathematical Society , , (1966), 1-72.[2] Chelghoum K., Margenstern M., Martin B., Pecci I., Cellular automatain the hyperbolic plane: proposal for a new environment, Lecture Notesin Computer Sciences , , (2004), 678-687, proceedings of ACRI’2004,Amsterdam, October, 25-27, 2004.[3] Goodman-Strauss, Ch., A strongly aperiodic set of tiles in the hyperbolicplane, Inventiones Mathematicae , (1), (2005), 119-132.[4] Goodman-Strauss, Ch., Growth, aperiodicity and undecidability, invited ad-dress at the AMS meeting at Davidson, NC , March, 3-4, (2007).[5] Hanf W., Nonrecursive tilings of the plane. I. Journal of Symbolic Logic , ,(1974), 283-285.[6] Kari J., A new undecidability proof of the tiling problem: The tiling problemis undecidable both in the Euclidean and in the hyperbolic plane, AMS meet-ing at Davidson, NC, Special Session on Computational and CombinatorialAspects of Tilings and Substitutions , March, 3-4, (2007).[7] Kari J., The Tiling Problem Revisited,
Lecture Notes in Computer Science ,to appear in the volume , (2007).[8] Levin L.A., Aperiodic Tilings: Breaking Translational Symmetry,
The Com-puter Journal , (6), (2005), 642-645.179] Mann C., Heesch’s tiling problem, American Mathematical Monthly , (6),(2004), 509-517.[10] Margenstern M., Cellular Automata and Combinatoric Tilings in Hyper-bolic Spaces, a survey, Lecture Notes in Computer Sciences , , (2003),48-72.[11] Margenstern M., A new way to implement cellular automata on the penta-and heptagrids, Journal of Cellular Automata , N ◦
1, (2006), 1-24.[12] Margenstern M., About the domino problem in the hyperbolic plane froman algorithmic point of view, iarXiv:cs.CG/ v
1, (2006), 11p.[13] Margenstern M., About the domino problem in the hyperbolic plane, a newsolution, arXiv:cs.CG/
Technical report, - , LITA, Universit´ePaul Verlaine − Metz
Technical report, - , LITA, Universit´e Paul Verlaine − Metz arxiv : cs.CG/ v
1, (2007), March, 8p.[17] Margenstern M., The periodic domino problem is undecidable in the hy-perbolic plane, arxiv : cs.CG/ v
1, (2007), March, 12p.[18] Margenstern M., About the domino problem in the hyperbolic plane, a newsolution: complement, arxiv :0705 . v
4, (2007), May, 20p.[19] Margenstern M., The Fininte Tiling Problem Is Undecidable in the Hyper-bolic Plane,
Workshop on Reachability Problems , RP07 , July 2007, Turku,Finland, accepted.[20] Margenstern M., Cellular Automata in Hyperbolic Spaces, Volume 1, The-ory,
OCP , Philadelphia, to appear (2007).[21] Myers D., Nonrecursive tilings of the plane. II.
Journal of Symbolic Logic , , (1974), 286-294.[22] Robinson R.M. Undecidability and nonperiodicity for tilings of the plane, Inventiones Mathematicae , , (1971), 177-209.[23] Robinson R.M. Undecidable tiling problems in the hyperbolic plane. In-ventiones Mathematicae , , (1978), 259-264.[24] Wang H. Proving theorems by pattern recognition, Bell System Tech. J.vol.40