ESI The Erwin S hr�odinger International Boltzmanngasse 9Institute for Mathemati al Physi s A-1090 Wien, AustriaNeighbours of Self{AÆne Tiles in Latti e TilingsKlaus S hei herJ�org M. Thuswaldner

Vienna, Preprint ESI 1173 (2002) June 12, 2002Supported by the Austrian Federal Ministry of Edu ation, S ien e and CultureAvailable via http://www.esi.a .at

Neighbours of self-aÆne tiles in latti e tilingsKlaus S hei her and J�org M. ThuswaldnerAbstra t. Let T be a tile of a self-aÆne latti e tiling. We give an algorithmthat allows to determine all neighbours of T in the tiling. This an be used to hara terize the sets VL of points, where T meets L other tiles. Our algorithmgeneralizes an algorithm of the authors whi h was appli able only to a spe ial lass of self-aÆne latti e tilings. This new algorithm an also be applied to lasses ontaining in�nitely many tilings at on e. Together with the resultsin re ent papers by Bandt and Wang as well as Akiyama and Thuswaldnerit allows to hara terize lasses of plane tilings whi h are homeomorphi to adis . Furthermore, it sheds some light on the relations between di�erent kindsof hara terizations of the boundary of T .1. Introdu tionIn this paper we are on erned with some properties of tilings of the eu lideanspa e. As des ribed in Gr�unbaum-Shephard [11℄ the art of tiling the plane goesba k to an ient times, where oors and walls were overed with beautiful patterns.We mention here the mosai s of the an ient Rome as well as the beautiful tilingswhi h an be found for instan e in the Alhambra at Granada (we refer to [11,Chapter 1℄ for many beautiful pi tures).In the last entury, greatly in uen ed by the resear h of rystallographers,who have to deal with tilings of high symmetry ( f. Klemm [18℄), mathemati iansgained interest in the theory of tilings. We point out that the 18th problem [13℄in Hilbert's famous problem list of the beginning of the 20th entury is related totilings.Here we are mainly on erned with topologi al properties of tilings. Topo-logi al properties were often used in order to lassify tilings in di�erent types ( f.for instan e Hees h [12℄ and Gr�unbaum-Shephard [11, Chapter 4℄ or more re entlyBandt-Gelbri h [4℄, Gelbri h [8℄, Bandt-Wang [5℄ and Kirat-Lau [17℄). Before wegive a more detailed a ount on the topi s dis ussed in the present paper we wantto de�ne the lass of tilings we are on erned with ( f. for instan e Wang [31℄).Re eived by the editors July 3, 2002.1991 Mathemati s Subje t Classi� ation. 37B50.Key words and phrases. latti e tiling.The �rst author was supported by the Austrian S ien e Foundation Proje t S8305.The se ond author was supported by the Austrian S ien e Foundation Proje t P-14200-MAT.

2 Klaus S hei her and J�org M. ThuswaldnerLet A be an expanding matrix in Mm(R), that is a real m �m matrix withall eigenvalues greater than 1. Suppose that j det(A)j = q > 1 is an integer andlet D � Rn with jDj = q. By a result of Hut hinson [14℄ there exists a uniquenon-empty ompa t set T := T (A;D) su h thatAT = [v2D(T + v):For most pairs (A;D) the set T has Lebesgue measure zero. If it has positiveLebesgue measure we all T a self-aÆne tile.A self-aÆne tile is alled latti e self-aÆne tile ( f. Lagarias-Wang [19℄) if thedi�eren e set D �D is ontained in a latti e � that is A-invariant, i.e. A� � �.In the present paper we are on erned with a lass of latti e self-aÆne tiles, so- alled integral self-aÆne tiles with standard digit set. These are tiles T (A;D) withinteger matrix A whose digit set D �Zm is a omplete set of oset representativesofZm=AZm. Moreover, we shall assume that T (A;D) tiles by the latti e Zm, i.e.T +Zm = Rmwhere �m((T + 1) \ (T + 2)) = 0 for 1 6= 2 ( 1; 2 2Zm):Here �m denotes the m-dimensional Lebesgue measure. Following Bandt-Wang [5℄we shall all su h a tile a Zm-tile for short.Furthermore, we all a tiling whi h tiles the m-dimensional real spa e Rm bya latti e a latti e tiling. Hen e, ea h Zm-tile indu es a latti e tiling of Rm.There are standard methods for he king whether T forms a tiling or not.We refer for instan e to Vin e [30, Theorem 4.2℄ where a list of tiling riteria isgiven. Furthermore, we note that a ording to Lagarias-Wang [19, Lemma 2.1℄ea h latti e self-aÆne tile is aÆnely equivalent to an integral self-aÆne tile, su hthat the tiling properties of latti e self-aÆne tiles are the same as those of theasso iated integral self-aÆne tiles.There exists a vast literature on self-aÆne latti e tilings. In Lagarias-Wang [19, 20, 21℄ there are proved fundamental results for this kind of tiling.Among many other things there are given general onditions under whi h T formsa tiling or a latti e tiling. The Hausdor� dimension of the boundary of self-aÆnetiles was al ulated in the ase where A is a similarity in Duvall et al. [6℄ andKenyon et al. [16℄. If A is not a similarity the exa t value of the Hausdor� dimen-sion is unknown even for spe ial examples. Estimates where given by Veerman [28℄.We refer also to the re ent papers Wang [31℄ and Vin e [30℄, where a survey aswell as a large list of re ent literature on tilings is provided.An interesting lass of self-aÆne tiles are tiles whi h are homeomorphi to adis or dis like, for short. Bandt-Gelbri h [4℄ showed, that for a given positive inte-ger k there exist essentially only �nitely many dis like self-aÆne latti e tilings withjDj = k. Gelbri h [8℄ obtained similar results for the more general ase of rystal-lographi tilings. It seems to be a diÆ ult problem to hara terize all dis like tilesby means of their de�ning data (A;D). Some progress in this dire tion was made

Neighbours of self-aÆne tiles in latti e tilings 3re ently in Bandt-Wang [5℄, where a riterion for the dis likeness of plane tiles wasgiven. Unfortunately, in order to apply this riterion one needs information whi his not so easy to obtain. Namely, one has to know the set S of \neighbours" of T( f. Se tion 2 for an exa t de�nition of S). In fa t, there exists an algorithm whi hallows to determine S for a given tiling ( f. Stri hartz-Wang [26, Appendix℄). Butthis algorithm does not seem to be suitable to hara terize S for larger lasses oftilings. In the paper Song-Kang [25℄ dis like tilings related to polyominoes havinga small number of digits are hara terized.Another dire tion was pursued in Akiyama-Thuswaldner [1℄ (see also [2℄).Firstly, without any knowledge on the set of neighbours a large lass of dis liketiles stemming from ertain number systems in the plane is hara terized. On theother hand, using an algorithm of S hei her-Thuswaldner [24℄ the set of neighboursis determined for a lass of tilings related to these number systems. The algorithmin [24℄ was developed only for this lass of tilings. One purpose of the present paperis to generalize it in order to make it appli able for ea hZm-tile. In fa t, we will �rstestablish an algorithm whi h determines all neighbours of a tile T in the tiling.This algorithm has two advantages ompared to the known ones by Stri hartz-Wang [26℄ and Indlekofer et al. [15℄. It is faster and an be applied to a lass oftiles at on e ( f. Akiyama-Thuswaldner [1℄). With help of the set of neighbours we an even obtain the stru ture of the set VL of points, where T oin ides with Lother tiles (so- alled L-verti es; f. Se tion 6 for an exa t de�nition of VL). Thiswill be done in Se tion 6.-2 -1 0 1 2

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Figure 1. A non-dis like tiling having many verti esWe want to illustrate the importan e of the set V2 (whi h is alled the setof verti es of T ) for the topology of T with help of two pi tures ( f. also the

4 Klaus S hei her and J�org M. Thuswaldnerpi tures in Bandt-Wang [5℄). In Figure 1 we see that the bla k tile T in the middleis perforated by the tiles below and beyond it. At ea h \perforation point" theinterior of T is splitted in two parts. On the other hand, in some perforationpoints T meets two other tiles, whi h ontributes a vertex.Figure 2 ontains a tiling without perforation points whi h is formed by theso- alled tame-twindragon ( f. for instan e Ngai et al. [23℄). In this ase we ansee (and prove) that T has only 6 verti es. Furthermore, one is lead to think thatthe interior of T is onne ted and T is dis like. This is indeed the ase. For proofswe refer to Akiyama-Thuswaldner [1, 2℄.-2 -1.5 -1 -0.5 0 0.5 1 1.5

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Figure 2. A dis like tiling having 6 verti esBefore we state and prove the new algorithm we need ertain preliminaries.In Se tion 2 we will de�ne the onta t matrix in the sense of Gr�o henig-Haas [10℄.This matrix has been used in di�erent variants in literature in order to de�nethe boundary of T . We will dis uss these di�erent variants as well as the relationbetween the onta t matrix and the boundary �T of a tile T . Our neighbour�nding algorithm will then shed some light on how these variants are related toea h other. The algorithm together with some important de�nitions is presentedin Se tion 3. Se tion 4 and Se tion 5 are devoted to proving that it terminates andyields the desired output. Se tion 6 is devoted to the sets VL of L-verti es of T . In

Neighbours of self-aÆne tiles in latti e tilings 5Se tion 7 we will �nally dis uss some appli ations of the algorithm for obtainingresults of the topology of lasses of tilings.2. Graphs and matri es related to the boundary of a tileIn this se tion we want to present two graphs whi h are helpful for the hara teri-zation of the boundary �T of aZm-tile T in Rm. The adja en y matri es of bothof these graphs o ur frequently in literature. Ea h of them is alled \ onta t ma-trix". We will deal with both of them. In fa t, the algorithm proposed in Se tion 3will larify their relation to ea h other ( f. Se tion 4).The following lemma shows that Zm-tiles have ertain natural properties.Lemma 2.1. ( f. Bandt [3℄ and Wang [31, Theorem 2.2℄) Let T = T (A;D) bea Zm-tile. Then T has nonempty interior, T is the losure of its interior, and�m(�T ) = 0.Now we start with the de�nition of the �rst graph. It was impli itly de�nedin Wang [31℄ and used in spe ial ases in M�uller et al. [22℄ and Thuswaldner [27℄.In the remaining part of the paper let T be a Zm-tile. For s 2Zm let Bs :=T \ (T + s) and de�ne the set of neighbours of T byS := fs 2Zm n f0g jBs 6= ;g:Note that S is a �nite set be ause T is obviously ompa t. By Lemma 2.1 we anwrite �T = [s2SBs: (1)It turns out that the boundary of T forms a graph-dire ted system. Indeed, bythe de�nition of the sets Bs we getBs = T \ (T + s)= A�1(T +D) \A�1(T + D + As) (2)= A�1 [d;d02DBAs+d0�d + d:Now label the elements of S as S = fs1; : : : ; sJg and de�ne the graph G(S) =(V;E) with set of states1 V := S in the following way. Let Ei;j be the set of edgesleading from si to sj . ThenEi;j := nsi d�! sj jAsi + d0 = sj + d for some d0 2 Do :This yields together with (1) and (2) the following result.1Note that we adopt the notion \state" instead of the more ommon notion \vertex" in order toavoid onfusions with the \vertex" of a tile.

6 Klaus S hei her and J�org M. ThuswaldnerProposition 2.2. �T is a graph-dire ted system dire ted by the graph G(S). Inparti ular, �T = [s2SBswhere Bs = [d2D; s02Ss d�!s0 A�1(Bs0 + d):The union is extended over all d; s0 su h that s d�! s0 is an edge in the graph G(S).So the boundary an be determined by the graph G(S). Let CS denote itsadja en y matrix.This approa h has the disadvantage that it involves the graph G(S) whi h isnot so easy to determine. On the other hand, G(S) is interesting be ause knowingthis graph implies the knowledge of all neighbours of T . As mentioned in theintrodu tion this is of great interest for de iding whether T is dis like or not.Furthermore, we need this graph in order to explore the stru ture of the sets VLmentioned before (see also Se tion 6).There is also a se ond graph G(R) whi h determines the boundary of T . Itwill turn out later that this graph is a subgraph of G(S). G(R) has the advantagethat it is easier to determine than G(S), but there is no way to read the set S ofneighbours of T dire tly from this graph. However, we will show in Se tion 3 thatG(R) an be used as a starting point in order to onstru t G(S).We need some de�nitions. Let Q be a parallelogram spanned by a basis ofthe latti e Zm and setTn =[(A�nQ+ x �����x 2 nXi=1 A�iD) (n 2 N):It is lear that limn!1 Tn = T in Hausdor� metri . Note also that Tn +Zm tilesRm for ea h n 2 N. Furthermore, letBa;n := Tn \ Tn + a:Let fe1; : : : ; emg be a basis of the latti e Zm, set R0 := f0;�e1; : : : ;�emg andde�ne Rn indu tively byRn := fk 2Zm j (Ak +D) \ (l + D) 6= ; for l 2 Rn�1g:If R = Sn�0Rn then R is a �nite set, in parti ular we have Rn�1 = Rn for n largeenough ( f. Gr�o henig-Haas [10, Se tion 4℄).Now we de�ne the graph G(R) = (V;E) with set of states R in the followingway. There exists an edge r d�! r0 from r to r0 labelled by d if Ar + d0 = r0 + dholds for a d0 2 D. The adja en y matrix CR = ( kl)k;l2R is de�ned by kl = j(Ak + D \ (l +D)j:

Neighbours of self-aÆne tiles in latti e tilings 7Following [10, Se tion 4℄ let Tn := AnTn and Ba;n := AnBa;n. Tn is a union ofm-dimensional ubes. Two ubes Q+k; Q+ l �k; l 2Pn�1i=0 AiD� have an (m�1)-dimensional fa e in ommon if and only if k � l 2 R0. �Tn onsists of those fa esof the ubes Q+ k �k 2Pn�1i=0 AiD� that are not ommon to two su h ubes. Let�(�Tn) be the number of su h fa es on the boundary of Tn.The following lemma is an immediate onsequen e of the above de�nitions.Lemma 2.3. For n � 0 the impli ation�(Ba;n) > 0 ) a 2 Rn (3)holds. If n is large enough su h that Rn = R then�(Ba;n) > 0 ) a 2 R:Proof. It suÆ es to prove the �rst assertion. The se ond one is an immediate onsequen e of it. We will use indu tion on n. For n = 0 the impli ation in (3) isobviously true. Now assume that (3) holds for n� 1 instead of n. We see that�(Ba;n) = �(AnBa;n) = �(An(Tn \ (Tn + a)))= Xk;k02D �(An�1((Tn�1 + k) \ (Tn�1 +Aa+ k0)))= Xk;k02D �(BAa+k0�k;n�1+ k):By the indu tion assumption the last sum is greater than zero only if there existsa pair k; k0 2 D with Aa + k0 � k = l 2 Rn�1. But the existen e of su h a pairimplies a 2 Rn and we are done.We are now in a position to prove the following representation of �T .Proposition 2.4. �T an be de�ned via the graph G(R). In parti ular,�T = limn!1 [r2Rnf0gBr;nwhere the limit is taken with respe t to Hausdor� metri andBr;n = [d2D; r02Rr d�!r0 A�1(Br0;n�1 + d): (4)The starting values Br;0 are de�ned by Br;0 := Q\ (Q+ r). The union is extendedover all d; r0 su h that r d�! r0 is an edge in the graph G(R).

8 Klaus S hei her and J�org M. ThuswaldnerProof. Assume that n is large enough su h that Rn = R holds. Sin e Tn =Sd2D A�1(Tn�1 + d) we haveBr;n = Tn \ (Tn + r)= A�1 [d;d02D(Tn�1 + d) \ (Tn�1 + d0 + Ar)= A�1 [d;d02DBAr+d0�d;n�1 + d:By the de�nition of the graph G(R) this implies together with Lemma 2.3Br;n = [d2D; r02Rr d�!r0 A�1(Br0;n�1 + d):We now need to onsider a union of the Br;n. Sin e �Tn = Sr2Zmnf0gBr;nLemma 2.3 implies that �Tn = Sr2Rnf0gBr;n. Furthermore, sin e T forms a tiling,Vin e [29, Theorem 3℄ yields limn!1 �Tn = �T . Thus�T = limn!1 [r2Rnf0gBr;n:Remark 2.5. If G(R) is a primitive graph (i.e. G(R) is strongly onne ted and thegreatest ommon divisor of its y le lengths is 1) the set �T is a graph-dire tedself-aÆne set in the sense of Fal oner [7, Chapter 3℄. In parti ular, there exists aunique family of non-empty ompa t sets f ~Brgr2R su h that�T = [r2Rnf0g ~Brwhere ~Br = [d2D; r02Rr d�!r0 A�1( ~Br0 + d):Remark 2.6. Note that both matri es CR and CS were used in order to determinethe Hausdor� dimension of the boundary �T in the ase where A is a similarity.The en ountered formul� readdimH (�T ) = log�Rlog� = log�Slog�(for the CR version we refer to Duvall et al. [6℄, the CS version was used forinstan e in Wang [31℄). Here �R and �S denote the largest eigenvalue of the matrix~CR whi h emerges from CR by deleting the row and the olumn orresponding tothe state 0 in G(R) and CS , respe tively. � is the modulus of the eigenvalues ofA. This implies that the largest eigenvalues of ~CR and CS oin ide in this ase.

Neighbours of self-aÆne tiles in latti e tilings 9Similar formul� were obtained for the box ounting dimension of a lass of self-aÆne tilings for whi h A is not a similarity ( f. S hei her-Thuswaldner [24℄).One ould think that Rnf0g = S or even ~CR = CS. In general, this is not the ase. In Akiyama-Thuswaldner [1℄ it was shown that for ea h N 2 N there existsa plane tiling for whi h jR n f0gj = 6 and jSj > N:R and S were expli itely hara terized for a lass of plane tilings in this paper. Asmentioned before the onne tion between R and S as well as between G(R) andG(S) will be explored in Se tion 4.3. The Neighbour Finding AlgorithmIn the previous se tion we pointed out that the advantage of the graph G(S) isthat its set of states onsists exa tly of the neighbours of T . Thus if we knowG(S), we also know the set of neighbours of T . On the other hand, as we haveseen in Se tion 2 the graph G(R) an be onstru ted easily by applying a simplealgorithm starting with the set f0;�e1; : : : ;�emg (here fe1; : : : ; emg is a basisof the latti s Zm). In the present se tion we want to state the algorithm whi hallows the onstru tion of G(S) starting from G(R). This algorithm generalizes[24, Theorem 3.5℄ to arbitrary Zm-tiles. Compared with the known algorithms ofIndlekofer et al. [15℄ and Stri hartz-Wang [26℄ this algorithm is faster and an beapplied in order to hara terize the neighbours of large lasses of tilings ( f. [1℄).Remark 3.1. Note that in S hei her-Thuswaldner [24℄ the graphs G(R) and G(S)are de�ned in a slightly di�erent way. Namely, their edges are dire ted in theopposite dire tion. This is due to the fa t that G(R) is interpreted as ountingautomaton in that paper. Sin e we are not on erned with this interpretation wedire t the edges in the same way as in M�uller et al. [22℄. So their adja en y matri es oin ide with the onta t matri es used in Duvall et al. [6℄, Gr�o henig-Haas [10℄,Wang [31℄ and many other papers.Before we state the algorithm, we give some de�nitions whi h will be usedthroughout the remaining part of the paper.De�nition 3.2. Let G(Zm) be a labelled dire ted graph with set of statesZm and setof labels D � D whose elements are written as `j`0. The labelled edges onne tingtwo states s1 and s2 are de�ned bys1 `j`0��! s2 if and only if As1 � s2 = ` � `0 (s1; s2 2Zm; `; `0 2 D):Sin e `0 is uniquely determined by s1, s2 and a we sometimes omit it and writejust s1 �! s2 for the edges in G(Zm) and its subgraphs. A subgraph of G(Zm) withset of states V will be denoted by G(V ).

10 Klaus S hei her and J�org M. ThuswaldnerLet G be a subgraph of G(Zm). For abbreviation we will write s1 `j`0��! s2 2E(G) if s1 `j`0��! s2 is an edge in G.Note that G(R) and G(S) are subgraphs of G(Zm). Thus the notation of thegraphs in the previous se tion agrees with the above de�nition.De�nition 3.3. Let M �Zm. We say that a subgraph G(M ) of G(Zm) has property(C) if(C) for ea h pair (s2; `) 2M � D there exists a unique pair (s1; `0) 2 M � Dsu h that s1 `j`0��! s2 is an edge of G(M ).De�nition 3.4. Let G be a graph. We denote by Red(G) the graph that emergesfrom G if all states of G, whi h are not the starting point of a walk of in�nitelength, are removed.De�nition 3.5. Let G1 and G01 be subgraphs of G(Zm). The produ t G2 := G1G01is de�ned in the following way. Let r1; s1 be states of G1 and r01; s01 be states ofG01. Furthermore, let `1; `01; `2 2 D.� r2 is a state of G2 if r2 = r1 + r01� There exists an edge r2 `1 j`2���! s2 in G2 if there existr1 `1j`01���! s1 2 E(G1) and r01 `01j`2���! s01 2 E(G01)with r1 + r01 = r2 and s1 + s01 = s2 or there existr1 `01j`2���! s1 2 E(G1) and r01 `1j`01���! s01 2 E(G01)with r1 + r01 = r2 and s1 + s01 = s2.Furthermore, if G is a subgraph of G(Zm) we use the abbreviationGp := pj=1G := G � � � G| {z }p times :The algoritm now reads as follows.Algorithm 3.6. The graph G(S), and with it the set S, an be determined by thefollowing algorithm starting from the graph G(R).p := 1A[1℄ := Red(G(R))repeatp := p+ 1A[p℄ := Red(A[p� 1℄ A[1℄)until A[p℄ = A[p� 1℄G(S) := A[p℄ n f0gThis algorithm always terminates after �nitely many steps.For proving that this algorithm yields G(S) after �nitely many steps we needa list of preparatory results.

Neighbours of self-aÆne tiles in latti e tilings 114. On the relation between G(R) and G(S)In this se tion we explore the relations between the graphs G(R) and G(S). Inparti ular, we will \bound" G(S) from below and from above in terms of G(R).We start with a very easy result on G(R).Lemma 4.1. Ea h state r 2 R0 of the graph G(R0) := Red(G(R)) has in�nitelymany prede essors and in�nitely many su essors. Thus G(R0) is a union of y lesof G(Zm) and of walks onne ting two of these y les. Furthermore, G(R0) hasproperty (C).Proof. The fa t that ea h r 2 R0 has in�nitely many su essors follows from thede�nition of Red(�).Con erning the prede essors we only have to show that ea h r 2 R0 has aprede essor. The result then follows indu tively. By the de�nition of G(R0) wehave to �nd an r0 2Zm withr + d = Ar0 + d0 (d; d0 2 D): (5)Indeed, the de�nition of R implies that su h an r0 is even ontained in R and thatit is a prede essor of r in G(R). But sin e r 2 R0, r has in�nitely many su essorsin G(R). Sin e r0 is a prede essor of r in G(R) the same is true for r0. This impliesthat even r0 2 R0 and r0 is a prede essor of r in G(R0).We will now show that there exists an r0 2 Zm with (5). The fa t that Dis a set of omplete oset representatives of Zm=AZm implies that for ea h pair(r; d) 2 R0 � D 9! d0 2 D : r + d � d0 (AZm) and, hen e,9! r0 2Zm : r + d = Ar0 + d0:Thus r0 2 R0 is a prede essor of r. The assertion on erning the stru ture of G(R0)follows immediately sin e R0 is a �nite set. Furthermore, be ause r 2 R0 and d 2 Dwere arbitrary, we on lude that G(R0) has property (C).The next result is dedi ated to a similar property of the graph G(S [ f0g).Lemma 4.2. The graph G(S [ f0g) is the union of all y les of G(Zm) and allwalks onne ting two of these y les.Proof. The result is ontained in M�uller et al. [22, Remark 3.4℄ for a spe ial ase.Sin e the general ase is similar to prove we only want to give a sket h here. Wedivide the proof into three parts.First wee show that G(S [ f0g) ontains all y les of G(Zm). By Stri hartz-Wang [26℄ s 2 S [f0g if and only if it has a representation s =Pj�1A�j(dj�d0j)with dj; d0j 2 D. By [22, Proposition 3.3. (v)℄ ea h s 2Zm whi h is ontained in a y le s d1jd01���! s1 d2jd02���! s2 d3jd03���! � � � d`�1 jd0�1������! s` d`jd0���! s

12 Klaus S hei her and J�org M. Thuswaldnerof G(Zm) admits a representationXj�0B�`j Xi=1 B�i(di � d0i)! :Thus G(S [ f0g) ontains all y les of G(Zm).In the se ond step we show that G(S [ f0g) ontains all walks onne tingtwo y les of G(Zm). Suppose thats1 d1jd01���! � � � d`�1 jd0�1������! s` (6)is a walk onne ting two y les of G(Zm). Then s` 2 S[f0g, be ause it belongs toa y le of G(Zm). By [22, Proposition 3.3. (ii)℄ the existen e of s djd0��! s0 in G(Zm)with s0 2 S [ f0g implies s 2 S [ f0g. This yields by indu tion that ea h statein (6) is ontained in S [ f0g. Thus G(S [ f0g) ontains all walks onne ting two y les of G(Zm).In the last step we show that no other states are ontained in G(S [ f0g) Tothis matter note that ea h state of G(S [ f0g) has in�nitely many prede essorsand in�nitely many su essors ( f. [22, Proposition 3.3. (iii) and (iv)℄). Thus bythe �niteness of S ea h s 2 S [ f0g is ontained in a y le of G(Zm) or in a walk onne ting two of them. This �nishes the proof.The above lemmas have the following onsequen e.Corollary 4.3. Red(G(R)) � G(S [ f0g):We want to onstru t G(S) starting from G(R0) := Red(G(R)). To this mat-ter we have to be sure that the set R0 ontains a basis of the latti e Zm.Lemma 4.4. Let G(R0) := Red(G(R)). Then the set R0 ontains a basis fe01; : : : ;e0mg of the latti eZm. By symmetry and be ause 0 0�! 0 is a y le in G(R) it even ontains a set of the shape f0;�e01; : : : ;�e0mg.Proof. Let n0 be large enough su h that Rn0 = R. Furthermore, let n00 be largeenough su h that �(An00Br;n00 ) = 0 for r 62 R0 (su h a hoi e is possible byLemma 2.3 and (4)). Let n � max(n0; n00). Then�Tn = [r2R0nf0gBr;n and Tn +Zm = Rm:Let fe1; : : : ; emg be a basis of the latti eZm. We will show that ea h of the elementsof this basis an be represented as a Z-linear ombination of elements of R0. This learly proves the result.Fix j 2 f1; : : :mg, x0 2 Tn and xj 2 Tn + ej . Then draw a line L onne tingx0 and xj . If ne essary we slightly deform L in order to avoid that it passes verti esof the parallelepipeds A�nQ + k. By the ompa tness of Tn this deformed line is ontained in a �nite number of Zm-translates of Tn. Sin e the boundaries of the

Neighbours of self-aÆne tiles in latti e tilings 13Tn are polygons, there is a �nite hain of translates ti 2 Zm su h that L passesthrough the tiles Tn; Tn+ t1; : : : ; Tn + t`; Tn + ejin the indi ated order (note that the tj do not have to be pairwise disjoint). Sin e�(AnBr;n) > 0 only for r 2 R0 we on lude thatt1; t2 � t1; : : : ; t` � t`�1; ej � t` 2 R0:Thus ej an be written as aZ-linear ombination of elements of R0 in the followingway: ej = t1 + Xi=2(ti � ti�1) + (ej � t`):Sin e j was arbitrary, this on ludes the proof.Remark 4.5. Note that by this lemma in the onstru tion of the graph G(R)in Se tion 2 we an always sele t the base fe1; : : : ; emg in a way su h thatRed(G(R)) = G(R), i.e. R0 = R.Lemma 4.6. Let G1 and G2 be subgraphs of G(Zm) having property (C). ThenH := Red(G1 G2) has property (C) and there exists an edgem1 djd0��! m2in G1 G2 if and only if Am1 + d0 = m2 + d.Proof. We prove that H has property (C). The other laims an be proved in thesame way as [1, Lemma 8.8℄. Sin e during the redu tion pro ess only su essorsare removed the impli ationG1 G2 has property (C) =) H has property (C)holds. Thus it suÆ es to show that G1 G2 has property (C). Let m0 be a stateof G1G2 and d 2 D. We have to �nd a state m of G1G2 and d0 2 D su h thatm djd0��! m0 (7)is an edge in G1G2. Note that the de�nition of implies the existen e of statesm0i of Gi (i 2 f1; 2g) su h that m0 = m01 +m02. Sin e G1 has property (C) thereexists an edge m1 djd00���! m01 (8)in G1. Furthermore, sin e G2 has property (C) there exists an edgem2 d00jd0���! m02 (9)in G2. By the de�nition of the existen e of the edges in (8) and (9) impliy theexisten e of the edge in (7).The pre eeding lemmas have the following result as a onsequen e. It om-plements Corollary 4.3.

14 Klaus S hei her and J�org M. ThuswaldnerCorollary 4.7. There exists a positive integer p0 su h thatRed(G(R)p0) � G(S [ f0g):Furthermore, Red(G(R)p0) has property (C).Proof. By Lemma 4.4 R ontains a set of the shape f0;�e1; : : : ;�emg, wherefe1; : : : ; emg is a base ofZm. By the de�nition of this impliesG(R)p �8<: 2Zm ������ = mXj=1 jej ; j 2 f�p; : : : ; pg9=;for ea h positive integer p. Sin e S [ f0g is �nite there exists a p0 2 N su h thatG(R)p0 � G(S [ f0g). But sin e ea h state in G(S [ f0g) has in�nitely manysu essors it is even ontained in Red(G(R)p0). Thus the �rst laim follows. These ond laim follows immediately from Lemma 4.6.Corrolary 4.3 and Corrolary 4.7 bound G(S) from below and from above.They will play a prominent role in the proof of Algorithm 3.6.5. Con lusion of the proofIn this se tion we will show that Algorithm 3.6 indeed terminates after �nitelymany steps and yields the graph G(S) as output.The following lemmas are auxiliary results whi h are ne essary for provingProposition 5.3.Lemma 5.1. Let G1 and G2 be subgraphs of G(Zm). Suppose that there exist theedge r1 d1jd3���! r2 (10)in G1 and the edges s1 d1 jd2���! s2 and s3 d2jd03���! s4 (11)in G2. If r2 = s2 + s4 then r1 = s1 + s3 and d3 = d03.Proof. The existen e of the edge (10) implies thatAr1 + d3 = r2 + d1: (12)Similar we get from (11) that As1 + d2 = s2 + d1 and As3 + d03 = s4 + d2. Addingthese two equations leads toA(s1 + s3) = s2 + s4 + d1 � d03: (13)Subtra ting (13) from (12) yields A(r1 � s1 � s3) = d03 � d3. This implies d3 �d03 (AZm). Sin e d3; d03 2 D we even have d3 = d03 and, hen e, r1 = s1 + s3 whi hproves the result.

Neighbours of self-aÆne tiles in latti e tilings 15Lemma 5.2. Let p 2 N and suppose that for some N 2 N there exists a walkrp;1 `1j`01���! rp;2 `2 j`02���! � � �rp;N�1 `N�1 j`0N�1�������! rp;Nin G(R0)p. Then for suitable `001 ; : : : ; `00N�1 2 D there existrp�1;1 `1j`001���! rp�1;2 `2 j`002���! � � � rp�1;N�1 `N�1 j`00N�1�������! rp�1;Nin G(R0)p�1 and r1;1 `001 j`01���! r1;2 `002 j`02���! � � � r1;N�1 `00N�1 j`0N�1�������! r1;Nin G(R0) su h that rp;j = rp�1;j + r1;j for all j 2 f1; : : : ; Ng.Proof. By the de�nition of there exist a state rp�1;N of G(R0)p�1 and a stater1;N of G(R0) with rp;N = rp�1;N + r1;N . Lemma 4.6 implies that there exist theedges rp�1;N�1 `N�1 j`00N�1�������! rp�1;N 2 E(G(R0)p�1)r1;N�1 `00N�1 jhN�1�������! r1;N 2 E(G(R0))be ause it ensures that G(R0)p�1 (as well as G(R0)) has property (C). ByLemma 5.1 we have hN�1 = `0N�1 and rp;N�1 = rp�1;N�1 + r1;N�1. The resultnow follows by iterating this argument N � 1 times.Before we an �nish the proof we need a result on the redu tion of graphs.Namely, we will show that forming produ ts and redu ing an be ex anged in a ertain way.Proposition 5.3. Let G(R0) := Red(G(R)) and p 2 N. Then the identityRed(G(R0)p) = Red(: : :Red(Red(Red(G(R0)) G(R0)) G(R0)) : : :G(R0))| {z }p fold iteration (14)holds.Proof. We prove this assertion by indu tion on p. For p = 1 there is nothingto prove. Thus we assume, that (14) is true for p � 1 instead of p. Using thisassumption, (14) be omesRed(G(R0)p) = Red(Red(G(R0)p�1)G(R0)): (15)The lemma will be proved, if we an show that (15) holds. Sin eRed(G(R0)p) = Red((G(R0)p�1) G(R0)) (16)we on lude that Red(G(R0)p) � Red(Red(G(R0)p�1) G(R0)). It remains toestablish the opposite in lusion in order to prove (15). Suppose that rp1 d1 jd2���!rp2 2 E(Red(G(R0)p)). By (16) there existrp�1;1 d1jd01���! rp�1;2 2 E(G(R0)p�1) and r11 d01jd2���! r12 2 E(G(R0)) (17)

16 Klaus S hei her and J�org M. Thuswaldnerwith r11 + rp�1;1 = rp1 and r12 + rp�1;2 = rp2 or there existrp�1;1 d01jd2���! rp�1;2 2 E(G(R0)p�1) and r11 d1jd01���! r12 2 E(G(R0))with r11+rm�1;1 = rm1 and r12+rm�1;2 = rm2. W.l.o.g. we suppose that the pairof edges in (17) exists. By Lemma 4.1 r12 has in�nitely many su essors. Now wehave to distinguish two ases.Case 1: rp�1;2 has in�nitely many su essors. Thus rp�1;2 2 Red(G(R0)p�1)and, hen e, rp�1;2 + r12 2 Red((G(R0)p�1) G(R0). Sin e rp�1;2 + r12 = rp2 hasin�nitely many su essors in G(R0)p we on lude that rp1 d1jd2���! rp2 is ontainedin Red(Red(G(R0)p�1) G(R0)) and we are done.Case 2: rp�1;2 has only �nitely many su essors. Sin e rp2 has in�nitely manysu essors by the de�nition of there exist a state sp�1;2 of G(R0)p�1 and a states12 of G(R0) with rp2 = sp�1;2+s12 su h that sp�1;2 has in�nitely many su essors.To proof this sele t N 2 N su h that N � 2 is equal to the number of states ofG(R0)p�1. Then, sin e rp;2 has in�nitely many su essors there exists a walkrp;2 `2j`02���! rp;3 `3 j`03���! � � �rp;N�1 `N�1 j`0N�1�������! rp;Nin G(R0)p. By Lemma 5.2 this implies the existen e of the walkssp�1;2 `2j`002���! sp�1;3 `3 j`003���! � � � sp�1;N�1 `N�1 j`00N�1�������! sp�1;N (18)in G(R0)p�1 and s1;2 `002 j`02���! s1;3 `003 j`03���! � � � s1;N�1 `00N�1 j`0N�1�������! s1;Nin G(R0) su h that rp;2 = sp�1;2 + s1;2. Thus sp�1;2 has N � 2 su essors. But bythe sele tion of N the walk (18) ontains a y le. Hen e, sp�1;2 has even in�nitelymany su essors. This implies that rp1 d1 jd2���! rp2 2 Red(Red(G(R0)p�1) G(R0))and we are done.After these preparations we are �nally in the position to show that Algo-rithm 3.6 yields G(S) after �nitely many steps.Proof. We �rst show that the algorithm terminates. By Corollary 4.7 there exists ap0 2 N su h that G(S [ f0g) � Red(G(R)p0). Furthermore, Proposition 5.3 yieldsRed(G(R)p0) = A[p0℄. Thus G(S [ f0g) � A[p0℄: (19)But (19) implies together with Lemma 4.2 that A[p0℄ ontains ea h redu ed �nitesubgraph of G(Zm) having the property that ea h of its states has a prede essor.In parti ular, A[p0+1℄ � A[p0℄, and sin e the opposite in lusion is trivial we evenhave A[p0 + 1℄ = A[p0℄. Thus the algorithm terminates for a p1 � p0 + 1.

Neighbours of self-aÆne tiles in latti e tilings 17Sin e by Lemma 4.2 G(S [ f0g) ontains ea h redu ed �nite subgraph ofG(Zm) having the property that ea h of its states has a prede essor we on ludethat G(S [ f0g) � A[p0℄ (20)(note that A[p0℄ has property (C) by Lemma 4.6; thus ea h of its states has aprede essor). It is easy to see, that A[p℄ = A[p + 1℄ implies A[p℄ = A[p + p0℄ forea h p0 2 N. In parti ular, we have A[p1℄ = A[p0℄. Together with (19) and (20)this shows that G(S) = A[p1℄ n f0g and thus the algorithm yields G(S).6. On points, where more than two tiles meetFor the sake of ompleteness we now give a short a ount on points where morethan two tiles of the tiling indu ed by the Zm-tile T oin ide (for a detaileddis ussion we refer to Stri hartz-Wang [26, Appendix℄ or Akiyama-Thuswaldner [1,Se tion 8℄). To this matter we need the following de�nitions. For pairwise disjoints1; : : : ; sL 2 S we setVL(s1; : : : ; sL) := 8<:x 2 Rm ������x 2 T \ L\j=1(T + sj)9=; :The set of L-verti es of T is then de�ned byVL = [fs1;:::;sLg�S VL(s1; : : : ; sL)where the union is extended over all subsets of S ontaining L elements. As men-tioned in the introdu tion a 2-vertex is sometimes simply alled vertex.The sets VL an be hara terized by taking a ertain produ t of the graphG(S) with itself. This produ t is de�ned in the following way.De�nition 6.1. Let G be a subgraph of G(Zm). The L-fold power GL := �Lj=1G isde�ned as the redu tion Red(G0L) of the following graph G0L:� The states of G0L are the sets fs1; : : : ; sLg onsisting of pairwise di�erentstates si of G.� There exists an edgefs11; : : : ; s1Lg �! fs21; : : : ; s2Lgin G0L if there exist the edgess1i `j`i��! s2i (1 � i � L)in G for ertain `1; : : :`L 2 D.The next proposition now gives the way how to hara terize VL with help ofthis produ t.

18 Klaus S hei her and J�org M. ThuswaldnerProposition 6.2. Let L � 1 and let s01; : : : ; s0L 2 S be pairwise di�erent. Then thefollowing three assertions are equivalent.(i) x =Xj�1A�jdj 2 VL(s01; : : : ; s0L):(ii) There exists an in�nite walk of the shapefs01; : : : ; s0Lg d1�! fs11; : : : ; s1Lg d2�! fs21; : : : ; s2Lg d3�! � � �in �Lr=1G(S).(iii) There exist the L in�nite walkss0i d1�! s1i d2�! s2i d3�! � � � (1 � i � L)in G(S).For the proof of this result we refer to the papers mentioned at the beginningof the present se tion.7. Appli ations and on luding remarksIn this se tion we want to dis uss some possible appli ations of Algorithm 3.6.Consider �rst a lass of \fra tal stair ases", i.e.Z2-tiles de�ned by T := T (A;D),with A := �k 00 k� (k 2 N) andD := f(aj + i; j) j 1 � j � k; 0 � i � k � 1g (21)where aj 2 N (1 � j � k) ful�l a1 = 0 and �k < aj � aj�1 < k. Examples of su hsets an be found in Figure 3 and Figure 4. They are spe ial ases of tiles indu edby polyominoes (for a de�nition of polyomino f. Golomb [9℄). Tiles indu ed bypolyominoes were studied for instan e in Song-Kang [25℄.With help of our algorithm it is possible to hara terize the set S of neigh-bours of su h stair ases. With help of the results in Akiyama-Thuswaldner [1℄ orBandt-Wang [5℄ from this we an de ide if they are dis like or not.As mentioned before the advantage of our algorithm ompared with theknown ones is that it is faster and an be applied to large lasses of tiles at on e ( f.for instan e [1, Theorem 9.1℄). Thus it an be used for instan e to give riteria forthe dis likeness of a large lass of polyominoe tiles. Sin e the pro edure of gettingsu h hara terizations is rather long ( f. for instan e [1, Se tion 9℄) we will on�neourselves with giving some simple examples here. More detailed hara terizationsof lasses of dis like tiles will appear in forth oming papers.The �rst example is a tile T3 whi h is de�ned by (21) with k = 3, a2 = 2; a3 =1 and b1 = b2 = b3 = 3. Applying Algorithm 3.6 yields #R = 7 and #S = 6. Bythe shape of the set of digits it follows for instan e from [5, Theorem 2.1℄ that itis dis like.

Neighbours of self-aÆne tiles in latti e tilings 190 0.5 1 1.5 2

0

0.2

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Figure 3. A dis like stair ase0 0.5 1 1.5 2 2.5 3

00.20.40.60.8

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Figure 4. A stair ase with a ut pointFigure 4 shows the pi ture of a stair ase T4 whi h is given by (21) with k = 4,a2 = 3; a3 = 6; a4 = 3 and b1 = b2 = b3 = b4 = 4. One expe ts from the pi turethat it is not dis like and has a ut point. In fa t, this ut point appears as anelement of the set V2 of this tile. The non-dis likeness an easily be on�rmed by omputing R and S via Algorithm 3.6. They are given byR := �(0; 0); (1; 0); (�1; 0); (0; 1); (0;�1); (1;1); (�1;�1) andS := �(1; 0); (�1; 0); (0; 1); (0;�1); (1; 1); (�1;�1); (2;0); (�2;0); (2;1); (�2;1):Then an appli ation of Bandt-Gelbri h [4, Lemma 5.1℄ yields that T4 an not bedis like. On the other hand, note that T4 is onne ted as an be seen for instan eby applying the riterion proved in Kirat-Lau [17, Theorem 1.2℄.Next we give in Figure 5 an example of a tile whi h has 14 neighbors andun ountably many verti es. I.e. the bla k tile in the middle \tou hes" all the othertiles drawn in the �gure. In this ase the boundary has a very strong reentrantstru ture so that some tiles seem to be overlapped by others. Of ourse, these\overlapps" an o ur only at boundary points.

20 Klaus S hei her and J�org M. Thuswaldner-15 -10 -5 0 5 10

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Figure 5. A tile with many neighbours-2 -1.5 -1 -0.5 0 0.5 1

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4

Figure 6. A dis like tileWe �nish with a more beautiful example of a dis like tile with 6 verti eswhi h looks even more \tame" than the tame-twindragon (Figure 6).Referen es[1℄ S. Akiyama and J. M. Thuswaldner. Topologi al stru ture of fra tal tilings generatedby quadrati number systems. submitted.

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22 Klaus S hei her and J�org M. Thuswaldner[25℄ H. J. Song and B. S. Kang. Dis like lati e reptiles indu ed by exa t polyominoes.Fra tals, 7:9{22, 1998.[26℄ R. Stri hartz and Y. Wang. Geometry of self-aÆne tiles I. Indiana Univ. Math. J.,48:1{23, 1999.[27℄ J. M. Thuswaldner. Fra tals and number systems in real quadrati number �elds.A ta Math. Hungar., 90:253{269, 2001.[28℄ J. J. P. Veerman. Hausdor� dimension of boundaries of self-aÆne tiles in Rn. Bol.Mex. Mat., 3(4):1{24, 1998.[29℄ A. Vin e. Self-repli ating tiles and their boundary. Dis rete Comput. Geom., 21:463{476, 1999.[30℄ A. Vin e. Digit tiling of eu lidean spa e. In Dire tions in Mathemati al Quasi rystals,pages 329{370, Providen e, RI, 2000. Amer. Math. So .[31℄ Y. Wang. Self-aÆne tiles. In K. S. Lau, editor, Advan es in Wavelet, pages 261{285.Springer, 1998.Institut f�ur Analysis und Numerik, Abteilung f�ur Finanzmathematik, JohannesKepler Universit�at Linz, A-4020 Linz, AUSTRIAInstitut f�ur Mathematik und Angewandte Geometrie, Abteilung f�ur Mathematikund Statistik, Montanuniversit�at Leoben, Franz-Josef-Stra�e 18, A-8700 Leoben,AUSTRIA