Annals of Mathematics, 0 (1991), 1{???The Pinwheel Tilings ofthe PlaneBy Charles Radin*IntroductionThe subject of forced tilings (also called aperiodic tilings) was createdby the philosopher Hao Wang in 1961 [9-12] as a tool in the study of certaindecidability problems in the propositional calculus. It is concerned with thepatterns generated by \tiles" as they are used to tile space; the formalism isthe following. One de�nes a �xed, �nite number of shapes (henceforth called\prototiles") in Euclidean n-space, En, for n � 2. The prototiles are usuallyrequired to be rather nice topologically, at least homeomorphs of the closedunit ball. One then makes arbitrarily many congruent copies, called \tiles", ofthese prototiles, and considers all ways (called \tilings") that such tiles mayprovide a simultaneous covering and packing of En; a tiling is thus an unorderedcollection of tiles for which the union is all of En, but such that the interiorsof each pair of tiles do not intersect.Wang's original problem was to determine if it was possible to have a �niteset of prototiles, with associated tilings of space, for which all the tilings werenonperiodic. (A tiling is \nonperiodic" if it is not invariant under any singlesimultaneous translation of its tiles.) This was settled in the a�rmative in thethesis of his student Robert Berger [1].Slowly over the years, due to the e�orts of Raphael Robinson [8], RogerPenrose [2], Shahar Mozes [4] and many others (see [3,6,7] for other references),Wang's problem has been generalized to ask whether it was possible for a �xed,�nite set of prototiles to generate tilings of space but only very complicatedtilings, where \very complicated" is interpreted as appropriate, but alwaysimplying nonperiodicity. In some interpretations, particularly the �rst onesassociated with physical models of quasicrystals, \very complicated" means\without crystallographic symmetry". In later work, \very complicated" often*Research was supported in part by NSF Grant DMS-9001475 and Texas ARP Grant 003658-113. I would also like to acknowledge discussions with John H. Conway from whom I learned, inJanuary, 1991, of the tesselation used in this paper, and many long hours of discussions with DaniBerend.

2 CHARLES RADINmeans \disordered" in the probabilistic sense used to study patterns in nonlin-ear dynamics. Within the �eld of logic, there was a brief line of developmentin which \very complicated" means \nonrecursive".All published examples, of �nite sets of prototiles which can only tile theplane nonperiodically, have the feature that in every tiling each prototile onlyappears in �nitely many orientations. Therefore, for these examples one couldadd the requirement that copies of a prototile be not just congruent to theprototile, but congruent by a translation; to recover the tilings of these previ-ous examples one might then need to increase the sets of prototiles to largerbut still �nite sets, e�ectively declaring certain rotated or re ected prototilesas new distinct prototiles. In other words, the use of congruence in the pro-duction of copies of the prototiles was e�ectively replaceable by translation inall these examples, at the expense of increasing the number of prototiles to alarger �nite number. This is not true of the example of this paper. In thisexample there is a �nite set of prototiles, and in every associated tiling of theplane tiles appear in in�nitely many orientations; all the connected part of theEuclidean group is needed to analyze the tilings by this set of prototiles, notjust its translation subgroup. This constitutes a major advance in the �eld,which began with models only requiring discrete translations (using \Wangdominoes"), and expanded to continuous translations with models such asthose of Penrose; this introduction of rotations adds a distinctly new elementto the subject, of particular interest to certain applications such as the physicsof quasicrystals. There are other features of this example which are novel, butwe postpone discussion of them to the end of the paper.1. The substitution tesselation of the plane;the tesselation and its LevelsWe begin with a certain hierarchical structure, a tesselation of the planewhich motivated our tiling example; the tesselation is due to John H. Conway(unpublished).We de�ne the \unmarked prototile" as the right triangle with the followingvertices in the Cartesian plane: (0,0), (2,0), (2,1). (To be precise, it is theclosed convex hull of these points.) The interior angles at the vertices of thistriangle (and any similar triangle) are of three sizes, small, medium and large,so the vertices are denoted S, M and L. Similarly the edges of this triangle (andany similar triangle) are of three sizes and are denoted S, M and L. See Figure1. We now de�ne the (substitution) tesselation � based on this unmarkedprototile. It consists of isometric copies of the unmarked prototile, obtainedby the following iterative procedure. De�ne \the �rst type C Triangle of Level0" to be the unmarked prototile. Consider the map which takes this Triangle

PINWHEEL TILINGS 3LMS M SLFigure 1into a similar Triangle (which we call \the �rst type C Triangle of Level 1"),with vertices (�2; 1), (2;�1) and (3; 1), and \composed of" �ve isometric copiesof the �rst type C Triangle of Level 0, with each of the �ve labeled with a \type"A, B, C, D, or E, as follows: a Triangle of type A with vertices (�2; 1), (0; 1)and (0; 0); a Triangle of type B with vertices (0; 1), (0; 0) and (2; 1); a Triangleof type C with vertices (0; 0), (2; 0) and (2; 1) (coinciding with the �rst type CTriangle of Level 0); a Triangle of type D with vertices (0; 0), (2; 0) and (2; 1);and a Triangle of type E with vertices (2; 1), (3; 1) and (2;�1). In summary,the �rst type C Triangle of Level 1 is a certain collection of �ve \component"Triangles. See Figure 2. EDCBAFigure 2Next we repeat this process, mapping the �rst type C Triangle of Level 1 intothe �rst type C Triangle of Level 2, with vertices (�5; 5); (1;�3) and (5; 0); itis similar to the former and composed of �ve Triangles (each composed of �vesubtriangles) isometric to the former, which are in the same geometric relationto the former as are the �ve created at the �rst step, with the �rst type CTriangle of Level 1 as the C in this Level 2 Triangle. See Figure 3.

4 CHARLES RADINED CBAEDCBA EDCBA ABCDEEDCBA

Figure 3Continuing this process leads to the desired tesselation � of the plane. SeeFigure 4. We de�ne Triangles of Level n as those Triangles, created above,which are isometric to the �rst type C Triangle of Level n; each has a \type"(A � � � E) de�ned by its relative position in the unique Level n + 1 Trianglecontaining it. We also de�ne a \class" (A1, B1,� � �, E1, A2, B2,� � �, E2) for eachTriangle; the �rst component in the symbol represents the type, and the secondcomponent, either 1 or 2, distinguishes between Triangles which are re ectionsof one another. See Figure 5. (This pattern of Triangles of all Levels is thekey feature of the tesselation � on which we will focus our attention. Also,we emphasize that by de�nition the above Triangles are each �xed sets in theplane, and not movable in any sense.) It is of prime importance to this paperthat the Triangles of each Level appear at in�nitely many orientations in �,as follows from the following simple lemma (Theorem 6.15 in [5]) applied tothe small angle S = tan�1(1=2).Lemma. For any rational number r other than 0 and �1, tan�1(r) isirrational with respect to �.

PINWHEEL TILINGS 5

Figure 42. Marks in the tesselation �Our main objective is to de�ne a �nite set of prototiles which can tile theplane but only with tilings that have some of the geometric relationships ofConway's tesselation �; in particular, the hierarchical relationships betweenconsecutive Levels of Triangles. This will be done in section 3. As preparation,we begin by adding \marks" to the (smallest) Triangles in �. Each Triangleof Level 0 in the tesselation � will be called a Tile, and will have a mark

6 CHARLES RADINassociated with (but not necessarily located at) each of its vertices, �S; �M; �L,where the subscript refers to the angle of the vertex. (We will do what we canto explain the signi�cance of the various marks, though this may only be fullyclari�ed by our proofs below, and the Remarks afterward. The basic idea willbe to record enough information in the marks of the Tiles so that similarlyde�ned but movable \tiles" will only have tilings with a hierarchical structureanalogous to that of the Triangles. There will be a brief summary of the marksat the end of this section.)A1 B1 C1D1 E1A1C1 B1D1E1A2B2 C2D2E2 A2 B2 C2D2 E2A2B2 C2D2E2Figure 5Each of the three vertex marks of a Tile will contain a variety of informationabout the edges of those Triangles in the tesselation � which are related insome sense to the vertex. In particular, with each vertex V of each Tile Tthere will be recorded information concerning each edge of each Tile whicheither has a coinciding vertex, or which has an edge with V at its center. Since(by de�nition) abutting Tiles have edges which coincide in an interval, thelines representing Tile edges will be separated into double lines, as follows.At vertex V of Tile T, the information at V is given in linear order. Thereare two edges of T meeting at V. Measuring angles as positive when referringto clockwise rotation, we call that edge of T at V the \�rst" one which is a

PINWHEEL TILINGS 7positive angle from the other, the other being called the \last" edge of T atV. We will record the angle between edges at V. The �rst angle will be thatof Tile T at V, and we use the abbreviations S, M and L noted above. Theremust be some Tile T0 abutting T at the �rst edge of T at V. Since this edge ofT0 at V and the �rst edge of T at V coincide in an interval, the angle betweenthem is zero, and does not need to be recorded. Therefore, we include in theinformation at V: the angle of T (called A1), then some information aboutthe �rst edge of T (called E11), then information about the abutting edge ofT0 (called E21), then the angle of T0 (called A2), information about the otheredge of T0 (called E12), : : :, and �nally, information about the last edge of T(called E2e ; e is used to denote the last two edges). So the information at Vis a sequence of triples, each beginning with an angle and then informationabout two edges which coincide in an interval; the kth such triple is referredto as Ak, E1k and E2k. Edge-marks Ejk will be said to \refer to" or \lie on"the appropriate edges of a Tile or Triangle (and are denoted by thick intervalsin some �gures). Now we need to specify what the information is in each Ejk,where j = 1; 2 and 1 � k � e.We are concerned with the edges not just of Tiles, but of Triangles ofall Levels. We call a \complete" edge only the following: the small edges ofTriangles of type A, B, C, and D; the medium edges of Triangles of type C, Dand E; and the large edges of Triangles of type B and C. Note that an edgeof a Triangle is complete if and only if it is not part or all of an edge of ahigher Level Triangle. For each of the three edges of each Tile, we will recordinformation about the complete edge of highest level of which it is a part;we record the \size" of that edge (S, M or L), what type Triangle (A � � �E)it is an edge of, and with what type Triangle (A � � �E) this last Triangle is\associated"; within each Ejk we use variable names J , N and P for the threepieces of information just described { J 2 fS;M;Lg for the size of the completeedge; N 2 fA; : : : ;Eg for the type of Triangle of which the complete edge is apart; and P 2 fA; : : : ;Eg for the type of Triangle of which that of N is one ofthe �ve components.As an example, the vertex mark �L of the Tile of type B in Figure 6has the following structure: �L = fA1; E11; E21 ; A2; E12 ; E22; Ae; E1e ; E2eg, whereEjk = (J;N; P; � � �) and so:�L = fL; (S;B;C; � � �); (S;A;C; � � �);L; (L;C; � � �); (L;B; � � �);�; (L;B; � � �); (L;C; � � �)gThere are three other pieces of information recorded in each Ejk, described byvariables F , G andH. Using the notation Ejk[�] for the edge-mark Ejm at angle� beyond that of Ejk, we assign F 2 fS;M;L;Z;Rg the value Z in E1k (resp.E2k) if there is an edge E2k[+�] (resp. E1k[+�]) at V, referring to a di�erent Tile

8 CHARLES RADINthan does E1k (resp. E2k), which is part of the same complete edge as that ofE1k (resp. E2k). F = S (resp. M, resp. L) if the complete edge of Ejk ends atV at an angle of S (resp. M, resp. L) with the rest of its Triangle. F = Rin Ejk if a di�erent value has not been speci�ed above. (As one sees in theabove example, it sometimes happens that in a vertex of a Tile T informationmust be given about an abutting Tile T0 which does not share this vertex, asillustrated in Figure 6; in such a case we use the value R for F .) To expandon the above example then, we have: Ejk = (J;N; P; F; � � �) and�L = fL; (S;B;C;L; � � �); (S;A;C;L; � � �);L; (L;C; �;Z; � � �);(L;B; �; R; � � �); �; (L;B; �; R; � � �); (L;C; �; Z; � � �)gB �L of B

Figure 6In order to discuss variables G and H we need the following convention.For each Triangle T of Level n � 1 we de�ne three Tiles in the Triangle,TS, TM and TL, by: TS shares its small vertex VSS � VS with that of theTriangle, another vertex VSM of TS lies on the large edge of the Triangle, andthe third vertex VSL of TS lies on the medium edge of the Triangle; TM sharesits medium vertex VMM � VM with that of the Triangle, another vertex VMSof TM lies on the large edge of the Triangle, and the third vertex VML of TMlies on the small edge of the Triangle; TL shares its small vertex VLL � VSwith the large vertex of the Triangle, and another vertex VLM of TL lies on the

PINWHEEL TILINGS 9small edge of the Triangle. For Level 0 Triangles, which are Tiles, we de�neTS, TM and TL as the Tile itself, with VSS as VS, VMM as VM, and VLL asVL. Note that this generalizes the concept of VS, VM and VL of Tiles to VSS,VMM and VLL of Triangles.The variable G has the following structure:G = (G�1MA ; G�1SE ; G�1SB ; G�1SD ; G�1MD ; G�2MA ; G�2SE ; G�2SB ; G�2SD ; G�2MD ; G�1MA ;G�1SE ; G�1SB ; G�1SD ; G�1MD ; G�2MA ; G�2SE ; G�2SB ; G�2SD ; G�2MD ), where eachGjknm takes val-ues in f+1;�1g. We will now de�ne these variables on each complete edge,assigning a value at the edge-marks at each end of the edge and prescribingan algorithm for assigning values at the other edge-marks on the edge. (Thegeometric signi�cance of variables G and H is less simple than that of theprevious marks. This will be discussed at the end of this section.)

V2 V1V2V1

1 32BB DFigure 7(M1) [GjkMX ]: Let X = A or D. Consider any Tile which is part of a TriangleT, with vertices V1 and V2 of the Tile lying on an edge of T, and E11 (resp.E2e ) of V2 and E2e (resp. E11) of V1 referring to that edge of T. If F = Z in E11(resp. E2e) of V2, then GjkMX has the same value in E2e (resp. E11) of V1 as it hasin E21 [��] (resp. E1e [+�]) of V2, unless V2 has an edge-mark E21[��=2] (resp.

10 CHARLES RADINE1e [+�=2]) which contains the values (J;N; F; P ) = (S;A;L;A) or (S;A;L;D),in which case the two G values have opposite sign. (In Figure 7, numbers 1and 2 each label a pair of edge-marks; for each pair there is the E2e [��] of VLand the E2e of VS of a Tile of type B. For pair 1, GjknD has di�erent valuesin the two edge-marks, and for pair 2, GjknD has the same values in the twoedge-marks.) If F 6= Z in E11 (resp. E2e) of V2, and we do not have: k = 2(resp. k = 1) and (J;N; F ) = (S;X;L) in E11 (resp. E2e) of V2, then G�kMX = +1in that edge-mark; k = 1 (resp. k = 2) and (J;N; F ) = (S;A;L) or (S;D;L)in E11 (resp. E2e) of V2, then G�kMX = +1 in that edge-mark. (In Figure 7,label 3 indicates an edge-mark E11 of vertex VM of a Tile of type D, in whichG�kMX = G�kMX = +1 from above.)G�2MXG�2MX

G�1MX G�1MXG�1SXG�1SX

X2Y2 X2 M5

M3M2

X1 Y1X1

G�2SXG�2MX G�2MX

G�1MX G�1MXG�2SXFigure 8a

PINWHEEL TILINGS 11(M2) [G�kMX inX]. Let T be a Triangle of class X1 = A1 or D1 (resp.X2 = A2or D2). In edge-mark E2e (resp. E11) of vertex VML, G�2MX (resp. G�1MX ) has thesame value as that of G�2MX (resp. G�1MX ) in the edge-mark E11 (resp. E2e) ofvertex VMS. See Figure 8A.(M3) [G�kMX in Y ]. Let T be a Triangle of class Y 1 = A1 or D1 (resp. Y 2 = A2or D2), and let X = A or D. In edge-mark E2e (resp. E11) of vertex VML, G�1MX(resp. G�2MX ) has the negative of the value of G�1MX (resp. G�2MX ) in the edge-mark E11 (resp. E2e) of vertex VMS. See Figure 8A.

G�2SBG�2SBG�2SX G�2SXG�1SX G�1SXY1B2Y1M9

M7M6

Y2B1Y2G�2SX G�2SX

G�1SXG�1SXG�1SB G�1SB Figure 8b(M4) [GjkSX ]: Let X = D or E. Consider any Tile which is part of a TriangleT, with vertices V1 and V2 of the Tile lying on an edge of T, and E11 (resp.E2e ) of V2 and E2e (resp. E11) of V1 referring to the edge of T. If F = Z in E11(resp. E2e) of V2, then GjkSX has the same value in E2e (resp. E11) of V1 as it has

12 CHARLES RADINin E21[��] (resp. E1e [+�]) of V2, unless: V2 has an edge-mark E21 [��=2] (resp.E1e [+�=2]) which contains the values (J;N; F; P ) = (S;A;L;E) or (S;A;L;D);or V2 has an edge-mark E2e (resp. E11) which contains the values (J;N; F; P ) =(S;A;M;B), in which case the two G values have opposite sign. If F 6= Z inE11 (resp. E2e ) of V2, and we do not have: k = 2 (resp. k = 1) and (J;N; F ) =(M;X;L) in E11 (resp. E2e ) of V2, then G�kSX = +1 in that edge-mark; k = 1(resp. k = 2) and (J;N; F ) = (M;E;L) or (M;D;L) or (L;B;M) in E11 (resp.E2e) of V2, then G�kSX = +1 in that edge-mark.(M5) [G�kSX in X]. Let T be a Triangle of class X1 = E1 or D1 (resp. X2 = E2or D2). In edge-mark E11 (resp. E2e ) of vertex VSL, G�2SX (resp. G�1SX ) has thesame value as that of G�2SX (resp. G�1SX ) in the edge-mark E2e (resp. E11) ofvertex VSM. See Figure 8A.(M6) [G�kSX in Y ]. Let T be a Triangle of class Y 1 = E1 or D1 (resp. Y 2 = E2or D2), and let X = E or D. In edge-mark E11 (resp. E2e) of vertex VSL, G�1SX(resp. G�2SX ) has the negative of the value of G�1SX (resp. G�2SX ) in the edge-markE2e (resp. E11) of vertex VSM. See Figure 8B.(M7) [G�kSX in B]. Let T be a Triangle of class B1 (resp. B2) and let X = Eor D. In edge-mark E11 (resp. E2e ) of vertex VSM, G�1SX (resp. G�2SX ) has thenegative of the value of G�1SX (resp. G�2SX ) in the edge-mark E2e (resp. E11) ofvertex VSL. See Figure 8B.(M8) [GjkSB ]. Consider any Tile which is part of a Triangle T, with vertices V1and V2 of the Tile lying on an edge of T, and E11 (resp. E2e ) of V2 and E2e (resp.E11) of V1 referring to the edge of T. If F = Z in E11 (resp. E2e) of V2, then GjkSBhas the same value in E2e (resp. E11) of V1 as it has in E21[��] (resp. E1e [+�]) ofV2, unless: V2 has an edge-mark E21 [+�=2] (resp. E1e [��=2]) which containsthe values (J;N; F; P ) = (S;A;L;E) or (S;A;L;D); or V2 has an edge-markE2e (resp. E11) which contains the values (J;N; F; P ) = (S;A;M;B), in whichcase the two G values have opposite sign. If F 6= Z in E11 (resp. E2e ) of V2,and we do not have: k = 2 (resp. k = 1) and (J;N; F ) = (M;E;L) or (M;D;L)or (L;B;M) in E11 (resp. E2e ) of V2, then G�kSX = +1 in E11 (resp. E2e) of V2;k = 1 (resp. k = 2) and (J;N; F ) = (L;B;M) in E11 (resp. E2e) of V2, thenG�kSX = +1 in E11 (resp. E2e) of V2.(M9) [G�kSB in Y ]. Let T be a Triangle of class Y 1 = E1 or D1 (resp. Y 2 = E2or D2). In edge-mark E11 (resp. E2e) of vertex VSL, G�2SB (resp. G�1SB ) has thenegative of the value of G�2SB (resp. G�1SB ) in the edge-mark E2e (resp. E11) ofvertex VSM. See Figure 8B.(M10) [G�kSB in B]. Let T be a Triangle of class B1 (resp. B2). In edge-mark E11(resp. E2e) of vertex VSM, G�2SB (resp. G�1SB ) has the negative of the value ofG�2SB (resp. G�1SB ) in the edge-mark E2e (resp. E11) of vertex VSL. See Figure8C.

PINWHEEL TILINGS 13(M11) [G�kSB in B]. Let T be a Triangle of class B1 (resp. B2). In edge-markE11 (resp. E2e ) of vertex VSM, G�1SB (resp. G�2SB ) has the same value as that ofG�1SB (resp. G�2SB ) in the edge-mark E2e (resp. E11) of vertex VSL. See Figure8C. (Note: the above values of G are complicated but well de�ned. To seethat they are well de�ned just note that even though one value is sometimesde�ned in terms of another, this process always ends since it always refers to avariable on a higher Level edge; in the tesselation � every Triangle is containedin in�nitely many Triangles of type C, on the edges of which all G variablesare explicitly de�ned, and the process cannot take us outside any Triangle oftype C.)G�1SX G�1SX

G�1SX G�1SXB1B1 M11

M10B2B2G�2SXG�2SX

G�2SXG�2SXFigure 8cH 2 f+1;�1g basically maintains a \parity" between two speci�c pointsin each Triangle of Level n � 1 , marked V1 and V5 in Figure 12A, and V6 andV10 in Figure 12B. The usefulness of this parity must remain obscure until theproof of Theorem 5.4.The variable H has the following structure:H = (H1MA , H1SE ;H1SB ;H1SD ;H1MD ;H2MA ;H2SE ;H2SB ;H2SD ;H2MD ), where each Hknmtakes values in f+1;�1g, as follows.

14 CHARLES RADIN(M12) Consider any Tile with E11 (resp. E2e) of vertex V2 and E2e (resp. E11)of vertex V1 referring to the same edge of the Tile, and assume F = Z inE11 (resp. E2e) of V2. Then the ratio of Hknm and G�knm has the same value inE21[+�] (resp. E1e [+�]) of V2 as in E2e (resp. E11) of V1. Also, Hknm = G�knm inall edge-marks except as prescribed below, in M13, M14 or M15, for certainedge-marks in small edges of Triangles of type A and D, and medium edges oftype D.

D2D1 M13M14M15

M13 M14 M15Figure 9(M13) Consider any vertex V of a prototile with edge-mark E11 (resp. E2e ) inwhich (J;N; F ) = (S;D;M); in particular, the edge-mark is lying in the small

PINWHEEL TILINGS 15edge of a Triangle of class D1 (resp. D2). See Figure 9. Then the ratio of H1Mmand G�1Mm (resp. H2Mm and G�2Mm ) in E11 (resp. E2e) equals the value of G�1Mm(resp. G�2Mm ) in E12 (resp. E2e�1). Also, the ratio of H1Sm and G�1Sm (resp. H2Smand G�2Sm ) in E11 (resp. E2e ) equals the value of G�1Sm (resp. G�2Sm ) in E2e (resp.E11).(M14) Consider any Tile with E11 (resp. E2e) of vertex V2 and E2e (resp. E11)of vertex V1 referring to the same edge of the Tile, and assume (J;N; F ) =(M;D;L) in E11 (resp. E2e ) of V2. See Figure 9. Then the ratio of H2nm andG�2nm (resp. H1nm and G�1nm ) is equal in E2e (resp. E11) of V1 with its value inE21 [��=2] (resp. E1e [+�=2]) of V2.(M15) Consider any Tile with E11 (resp. E2e) of vertex V2 and E2e (resp. E11)of vertex V1 referring to the same edge of the Tile, and assume (J;N; F ) =(M;D;S) in E11 (resp. E2e) of V2. See Figure 9. Then the ratio of H2nm andG�2nm (resp. H1nm and G�1nm ) is equal in E2e (resp. E11) of V1 with its value inE21 [+�=2] (resp. E1e [��=2]) of V2.We now summarize the above addition of marks to the Tiles in the tesse-lation �. The marks all concern the hierarchical structure of the Triangles in�, and more speci�cally they record certain facts about their complete edges.At each vertex in � it is recorded, on all the Tiles sharing that vertex, whichtypes of complete edges are associated with that vertex, what type Triangleeach such edge belongs to, and what type Triangle is the parent of the latterTriangle. We also record at what angles these edges meet, and whether or notthe edge is ending at that vertex. This is all recorded in the variables A; J;N; Pand F , and is rather easy to understand visually. The variables G and H referto less obvious features of the tesselation �. Both these (families of) variablestake values �1, and do not have a natural absolute meaning; they should bethought of as phases, and convey information through the agreement or dis-agreement of the values of variables on pairs of edges meeting at a vertex. In �the relative values of these variables are �xed at certain such intersections, asillustrated in Figures 8A,B,C and 9, and the values then oscillate as one movespast vertices on a given complete edge, with certain speci�ed exceptions, asillustrated in Figure 7; these exceptions each have simple geometric meaningin terms of edges of certain Triangles of lower Level appearing at the vertex.The reason for the de�nitions is that certain pairs of variables meet in phase ata vertex in � only under a unique circumstance; for example variables G�1MAand G�1MA meet in phase in the geometric relation shown in Figure 8A if andonly if they are meeting at the medium vertex of a Triangle of class A1. Thisis shown in the proof of Theorem 5.1 below. To repeat then, the actual valueof any of the variables in the G or H families has no geometric meaning, itis only the relative values of pairs of variables meeting at a vertex which mayhave a (simple geometric) meaning in terms of the hierarchy of Triangles.

16 CHARLES RADIN3. Tiling the plane; the prototiles and their matching rulesWe now consider the problem of trying to reproduce the structure of thesubstitution tesselation � by the tilings of a �nite set of prototiles. Beforewe begin, we need to introduce a simple technique used almost universally inforced tiling.As stated above, tiling consists of simultaneous covering and packing; thepacking condition intuitively requires (given the covering condition) that thetiles must �t together like a jig-saw puzzle, the boundary of one tile nestledinto the boundaries of its neighbors. Consider for example the system in theleft half of Figure 10 of two prototiles in the plane which are both roughlysquare with edges alligned (but with jagged edges). (They can only tile theplane in checkerboard fashion.) 4 1 23 20 3010 40Figure 10Now consider in the right half of Figure 10 the set of two perfect squares,with edges alligned, which have \colors" (1; 2; 3; 4; 10; 20; 30; 40) assigned to theiredges; the colors come in \complementary" pairs: j and j0 are complements.If we consider this a pair of prototiles, and allow translated copies (calledtiles) to be made, preserving the assignment of colors to edges, and add the\matching rule" that in a tiling tiles must abut full edge to full edge and withcomplementary colors meeting, then it is clear that we will reproduce, in anyreasonable sense, the previous example. In other words, one can often simplifyan example of prototiles by using simpler shapes, but with added \colors"and \matching rules" to make up for the simpli�ed boundary. Reversing thisprocess, if one is given a set of prototiles with colors and matching rules, itis straightforward to replace it by a set of prototiles with more complicatedboundaries and no colors or matching rules, and this is what we will do.(We sketch here a justi�cation for this claim. Assume we have a �nite setof polygonal prototiles, the set of all edges of which is called Q, and assumewe are given rules governing, for each edge, which edges it may abut in atiling, this information being summarized in the symmetric set K � Q�Q. We

PINWHEEL TILINGS 17assume that edges may only abut along their full length in a tiling. In someexamples not satisfying this condition, noted below, one can add \vertices" onthe original edges to produce smaller edges which then satisfy the condition.We now show that the e�ect of such rules K can be reproduced by suitablymodifying the edges of the polygons, and requiring that the new tiles tile theplane in the usual geometric sense, if and only if a certain mild condition Cis satis�ed by K, namely that if (a; b); (a; c) and (d; b) are in K then so is(d; c). The necessity of this condition for the desired conclusion is obvious.We sketch the su�ciency, using [4]. De�ne two edges a and b to be equivalentif there exists an edge c such that (a; c) and (b; c) are both in K. It followsfrom condition C that this is an equivalence relation. Next we de�ne for eachequivalence class a unique complementary class as follows. For class E thecomplement is that class E0 such that (a; b) is in K for some a in E and b inE0. It follows from condition C that this de�nes a unique class E0 independentof choice of representatives. This is su�cient to de�ne a unique family of zig-zag curves to modify the edges, one for each pair of complementary classes, sothat only edges from complementary classes will �t together; one could takethe same number of bumps for all edges { all congruent for a given curve { andvary the height of the bumps from class to class to guarantee the uniqueness.)We will have in fact only one basic shape (and its re ection), the unmarkedprototile of the tesselation �, but we will de�ne a set of colors, which we willcall \marks", with a notion of complementary pairs, and a set of \matchingrules", and we will analyze those tilings of the plane by tiles which satisfy thematching rules, the object being to show that these tilings all exhibit the basicstructure of Triangles of all Levels which we see in �. We begin by introducingthe marks.The marks were already introduced for the Tiles of �. Intuitively, wewould like to de�ne our prototile set as all the distinct versions of the unmarkedprototile and its re ection (re ecting about its small edge say) obtained byadding to them those marks which appear on the Tiles which are obtainedfrom these two unmarked prototiles by orientation preserving isometries. Thiscould be done, but it is somewhat unsatisfying in that it would then be di�cultto determine precisely which of the conceivable mark combinations are used.So we instead de�ne our prototiles as those obtained by using all possiblemarks, but satisfying an explicit list of \restrictions".

18 CHARLES RADIN

C2C1B2B1

(S,A,L)(S,A,M) A1A2(S,C,L)(L,C,M)(L,B,M)(S,B,L)(L,B,S)

(L,C,S)(L,B,S)(L,C,S)Figure 11aA \(marked) prototile" consists of the unmarked prototile, or its re ectionabout its small edge (which we will now call another unmarked prototile),together with three (vertex) marks �S; �M; �L, each of the three of the form(A1; E11 ; E21; A2; E12; E22 ; : : : ; Ae; E1e ; E2e), where e varies between 3 and 8 (estands for \end") and where:(a) Aj 2 fS;M;L; �g(b) each Ekj is of the form (J;N; P; F;G;H), where:(c) J 2 fS;M;Lg(d) N 2 fA;B;C;D;Eg(e) P 2 fA;B;C;D;Eg(f) F 2 fS;M;L;R;Zg

PINWHEEL TILINGS 19(g) G = (G�1MA ; G�1SE ; G�1SB ; G�1SD ; G�1MD ; G�2MA ; G�2SE ; G�2SB ; G�2SD ; G�2MD ; G�1MA ,G�1SE ; G�1SB ; G�1SD ; G�1MD ; G�2MA ; G�2SE ; G�2SB ; G�2SD ; G�2MD ), where each Gjknm takesvalues in f+1;�1g.(h) H = (H1MA ;H1SE ; H1SB ;H1SD ;H1MD ;H2MA ;H2SE ;H2SB ;H2SD ;H2MD ), where eachHknm takes values in f+1;�1g.(We note that although these are called \vertex marks", they could easilybe implemented by encoding their information in bumps and dents in the edgesof the tiles, away from the vertices.) As mentioned above, we do not allow ourprototiles to have all possible values of the above marks; we allow preciselythose combinations satisfying the following set of restrictions.Every prototile must contain marks of one of the following ten \classes",A1 � � � E2, (and thus �ve \types", A � � �E, where the type of class Xj is denotedX), the \original " ones, with vertices at (0,0), (2,0), and (2,1), satisfying oneof A2, B1, C1, D2, E2, and the \re ected" ones, with vertices at (2,0), (4,0)and (2,1), satisfying one of A1, B2, C2, D1, E1, as shown in Figures 11A,B.(It will follow from further restrictions, namely V5 and V10, that these tenclasses are mutually exclusive.) (M,D,S)(M,D,S)(M,E,S)(M,E,S) (M,E,L) E1E2

D2 D1(S,D,M)(S,D,L)Figure 11b(A1) In VL, E2e contains (J;N; F ) = (S;A;L), and in VM, E11 contains(J;N; F ) = (S;A;M).

20 CHARLES RADIN(B1) In VL, E11 contains (J;N; F ) = (S;B;L); in VM, E11 contains (J;N; F ) =(L;B;M); and in VS, E2e contains (J;N; F ) = (L;B;S).(C1) In VS, E2e contains (J;N; F ) = (M;C;S); in VM, E11 contains (J;N; F ) =(L;C;M); and in VL, E11 contains (J;N; F ) = (S;C;L).(D1) In VS, E2e contains (J;N; F ) = (M;D;S); in VL, E2e contains (J;N; F ) =(M;D;L); and in VM, E11 contains (J;N; F ) = (S;D;M).(E1) In VS, E2e contains (J;N; F ) = (M;E;S), and in VL, E11 contains(J;N; F ) = (M;E;L).(A2) In VL, E11 contains (J;N; F ) = (S;A;L), and in VM, E2e contains(J;N; F ) = (S;A;M).(B2) In VL, E2e contains (J;N; F ) = (S;B;L); in VM, E2e contains (J;N; F ) =(S;B;M); and in VS, E11 contains (J;N; F ) = (L;B;S).(C2) In VS, E11 contains (J;N; F ) = (L;C;S); in VM, E2e contains (J;N; F ) =(S;C;M); and in VL, E2e contains (J;N; F ) = (M;C;L).(D2) In VS, E11 contains (J;N; F ) = (M;D;S); in VL, E11 contains (J;N; F ) =(S;D;L); and in VM, E2e contains (J;N; F ) = (S;D;M).(E2) In VL, E2e contains (J;N; F ) = (M;E;L), and in VS, E11 contains(J;N; F ) = (M;E;S).

V2V1

(M,D,S)(M,C,S)(L,C,S)(L,B,M)(S,B,M)(S,A,M)F=Z

F=ZA B CD EV1

V2V3

V4V5F=Z F=Z(S,A,L) (S,B,L) Figure 12a

PINWHEEL TILINGS 21A B CD EV1

V2V3

V4 V5F=Z F=Z(L,B,S)(L,C,M) (S,C,M) (M,E,L)V3 (M,C,L)(M,D,L) (S,D,L)

(S,C,L) (M,E,R)(M,E,R)V4

Figure 12bThe following restrictions can be understood from Figures 12A-12F; werecord in the restrictions V1 � � � V10 almost everything that can be easilydeduced as occuring in the tesselation � at such vertices. The references tovariables H and G will need special analysis.(V1) Consider a prototile T, with vertices V1, V2 and V3, such that E11 of V2and E2e of V1 refer to the same edge of T. (For vertices V1, V2 and V3, of atile { as opposed to a Tile { the phrase \E11 of V2 and E2e of V1 refer to thesame edge" means by de�nition that V1, V2 and V3 are in clockwise order inthe plane.) If in V1 either of the following conditions are satis�ed: E2e contains(J;N; F ) = (S;A;L), or E1e contains (J;N; F ) = (S;B;L), then so is the other,and also: F = Z in E1e [+�=2] and E2e [��=2]; P has the same value in E1e andE2e ; the quotient of H1Mm by G�1Mm in E11 of V2 equals G�1Mm in E2e [��=2] of V1

22 CHARLES RADINif P = m in E2e of V1; and the quotient of H1Sm by G�1Sm in E11 of V2 equals G�1Smin E1e [+�=2] of V1 if P = m in E2e . And if in V1 we have, in E2e , that P hasvalue: B, then (J;N) = (L;B) in E1e [+�=2], and (J;N) = (L;B) in E2e [��=2];C, then (J;N) = (L;C) in E1e [+�=2] and in E2e [��=2].

V5A B CD EV1

V2V3

V4 V5(S,D,M) (M,E,S)Figure 12c(V2) If a vertex of a prototile satis�es any of the following conditions: E1kcontains (J;N; F ) = (S;A;M); E2k contains (J;N; F ) = (S;B;M); E1k+1 con-tains (J;N; F ) = (L;B;M); E2k+1 contains (J;N; F ) = (L;C;S); E1k+2 contains(J;N; F ) = (M;C;S); E2k+2 contains (J;N; F ) = (M;D;S), then it satis�es theothers, and also: F = Z in E2k�1 and in E1k+3; P has the same value in E1k, E2k ,E1k+1, E2k+1, E1k+2 and E1k+2.

PINWHEEL TILINGS 23(V3) If a vertex of a prototile satis�es any of the following conditions: E2k con-tains (J;N; F ) = (L;B;S); E1k contains (J;N; F ) = (L;C;M); E2k�1 contains(J;N; F ) = (S;C;M); E1k�1 contains (J;N; F ) = (M;E;L), then it satis�es theothers, and also: F = Z in E2k�1[��=2] and in E1k+1 ; P has the same value inE2k , E1k, E2k�1 and E1k�1.(V4) If a vertex of a prototile satis�es any of the following conditions: E1k con-tains (J;N; F ) = (S;C;L); E2k[��=2] contains (J;N; F ) = (M;C;L); E1k[��=2]contains (J;N; F ) = (M;D;L); E2k[��] contains (J;N; F ) = (S;D;L), thenit satis�es the others, and also: E2k contains (J;N; F ) = (M;E;R); E1k [��]contains (J;N; F ) = (M;E;R); P has the same value in E2k , E1k, E2k[��=2],E1k [��=2], E2k [��] and E1k [��].

(S,B,L) (S,A,L)F=Z F=ZV10V9

V8V7V6E DC B A

F=ZF=Z

(S,A,M)(S,B,M)(L,B,M)(L,C,S)(M,C,S)(M,D,S)V6

V7 Figure 12d(V5) If a vertex of a prototile satis�es either of the following conditions: E11contains (J;N; F ) = (S;D;M), or E21 contains (J;N; F ) = (M;E;S), then itsatis�es the other and: P has the same value in E11 and E21 ; the quotient ofH1Mmby G�1Mm in E11 equals G�1Mm in E12 if P = m in E11 ; the quotient of H1Sm by G�1Smin E11 equals G�1Sm in E2e if P = m in E11. Also, if, in E11, P has value: B, then

24 CHARLES RADINF = Z in E2e , and (J;N; F ) = (S;B;L) in E12; C, then (J;N; F ) = (M;C;L) inE2e and (J;N; F ) = (S;C;L) in E12 ; A, then (J;N; F ) = (S;A;L) in E12 , andF = Z in E2e ; D, then (J;N; F ) = (M;D;L) in E2e , and (J;N; F ) = (S;D;L) inE12; E, then (J;N; F ) = (M;E;L) in E2e , and F = Z in E12 and in E2e [+�].V4

(M,E,R)(M,E,R) (S,C,L)

(S,D,L) (M,D,L)(M,C,L) V3(M,E,L) (S,C,M) (L,C,M)(L,B,S)F=Z F=ZV5 V4

V3V2

V1E DC B A

Figure 12e(V6) Consider a prototile T, with vertices V1, V2 and V3, such that E11 ofV1 and E2e of V2 refer to the same edge of T. If in V1 either of the followingconditions are satis�ed: E11 contains (J;N; F ) = (S;A;L), or E21 contains(J;N; F ) = (S;B;L), then so is the other, and also: F = Z in E11[+�=2] andE21[��=2]; P has the same value in E11 and E21 ; the quotient of H�2Mm by G�2Mmin E2e of V2 equals G�2Mm in E11 [+�=2] of V1 if P = m in E11 of V1; and thequotient of H�2Sm by G�2Sm in E2e of V2 equals G�2Sm in E21 [��=2] of V1 if P = min E11 . And if in V1 we have, in E11, that P has value: B, then (J;N) = (L;B) inE21[��=2] and E11 [+�=2]; C, then (J;N) = (L;C) in E21[��=2] and E11 [+�=2].(V7) If a vertex of a prototile satis�es any of the following conditions: E2kcontains (J;N; F ) = (S;A;M); E1k contains (J;N; F ) = (S;B;M); E2k�1 con-

PINWHEEL TILINGS 25tains (J;N; F ) = (L;B;M); E1k�1 contains (J;N; F ) = (L;C;S); E2k�2 contains(J;N; F ) = (M;C;S); E1k�2 contains (J;N; F ) = (M;D;S), then it satis�es theothers, and also: F = Z in E1k+1 and E2k�3; P has the same value in E1k , E2k,E1k+1, E2k+1, E1k+2 and E1k+2.

(M,E,S) (S,D,M)V5 V4V3

V2V1E DC B A

V5 Figure 12f(V8) If a vertex of a prototile satis�es any of the following conditions: E1k con-tains (J;N; F ) = (L;B;S); E2k contains (J;N; F ) = (L;C;M); E1k+1 contains(J;N; F ) = (S;C;M); E2k+1 contains (J;N; F ) = (M;E;L), then it satis�es theothers, and also: F = Z in E1k+1[+�=2] and E2k�1; P has the same value in E1k,E2k , E1k+1 and E2k+1.(V9) If a vertex of a prototile satis�es any of the following conditions: E2k con-tains (J;N; F ) = (S;C;L); E1k[+�=2] contains (J;N; F ) = (M;C;L); E2k[+�=2]

26 CHARLES RADINcontains (J;N; F ) = (M;D;L); E1k [+�] contains (J;N; F ) = (S;D;L), thenit satis�es the others, and also: E1k contains (J;N; F ) = (M;E;R); E2k[+�]contains (J;N; F ) = (M;E;R); P has the same value in E2k, E1k , E2k [+�=2],E1k[+�=2], E2k [+�] and E1k[+�].(V10) If a vertex of a prototile satis�es either of the following conditions: E2econtains (J;N; F ) = (S;D;M), or E1e contains (J;N; F ) = (M;E;S); then itsatis�es the other, and: P has the same value in E2e and E1e ; the quotient ofH2Mm by G�2Mm in E2e equals G�2Mm in E2e�1 if P = m in E2e ; and the quotientof H2Sm by G�2Sm in E2e equals G�2Sm in E11 if P = m in E2e . Also if, in E2e ,P has value: A, then (J;N; F ) = (S;A;L) in E2e�1, and F = Z in E11 ; D,then (J;N; F ) = (M;D;L) in E11 and (J;N; F ) = (S;D;L) in E2e�1; B, then(J;N; F ) = (S;B;L) in E2e�1, and F = Z in E11; C, then (J;N; F ) = (M;C;L)in E11 and (J;N; F ) = (S;C;L) in E2e�1; E, then (J;N; F ) = (M;E;L) in E11 ,and F = Z in E2e�1 and in E1e�1[+�].Finally we make some restrictions which follow easily from examiningFigures 3, 13, 14 and 15; R9 and R10 are less obvious, and we will analyzethem carefully.(R1) In any Ejk, (J;N) 6= (M;A); (L;A); (M;B); (L;D); (S;E); (L;E), and F 6=J .(R2) In any Vk, Pej=1 Size(Aj) = 2�.(R3) A1 must have the value of the vertex of the � component to which itbelongs.(R4) If Emn and Epq belong to two di�erent vertices and refer to the same edgeof a prototile, then J (resp. N , resp. P ) must have the same value in Emn asin Epq .(R5) In any vertex of a prototile, if one of a pair E1k ; E2k of edge-marks has(J;N) values: (S;A), then for the other (J;N) = (S;B), and then one hasF = L (resp. F = Z or R , resp. F = M) if and only if the other has F = L(resp. F = Z or R, resp. F = M); (S;B), then for the other (J;N) = (S;A);(L;B), then for the other (J;N) = (L;C), and then one has F = S (resp.F = Z or R) if and only if the other has F = M (resp. F = Z or R); (L;C),then for the other (J;N) = (L;B); (M;C), then for the other (J;N) = (M;D),and then one has F = L (resp. F = S, resp. F = Z or R) if and only if theother has F = L (resp. F = S, resp. F = Z or R); (M;D), then for the other(J;N) = (M;C); (S;C), then for the other (J;N) = (M;E), and then theformer has F = M if and only if the other has F = L, and if the former hasF = L (resp. F = Z or R) then the other has F = Z or R (resp. F = Z or R);(S;D), then for the other (J;N) = (M;E), and then the former has F = M ifand only if the other has F = S, and if the former has F = L (resp. F = Zor R) then the other has F = Z or R (resp. F = Z or R); (M;E), then for

PINWHEEL TILINGS 27the other (J;N) = (S;C) or (S;D), and if the former has F = Z or R then theother has F = Z or R or L.(R6) If F = S (resp. M) in E1k then Ak = S (resp. M). If F = S (resp. M) inE2k then Ak+1 = S (resp. M).(R7) Assume for an edge-mark E1k (resp. E2k) of any Vm that F = R. Thenthere is an edge-mark E2k [+�] (resp. E1k[+�]) of Vm, and F = R in that edgetoo. The value F = R may not appear in any edge-mark numbered E11 or E2e .Also, the values of J are the same in E1k and E2k [+�] (resp. E2k and E1k[+�]),the values of N are the same in E1k and E2k[+�] (resp. E2k and E1k[+�]), and thevalues of P are the same in E1k and E2k[+�] (resp. E2k and E1k[+�]). Finally, itcannot happen that F = R for another edge-mark Epq of Vm.

V1V1V1

V2V2V2V3V3

V3R10R9

G�2Mm G�2MmG�2SmG�2SmSV2 V1V3G�1Sm G�1Sm

MG�1Mm G�1MmSMFigure 13(R8) Assume for an edge-mark E1k (resp. E2k) of any Vm that F = Z. Thenthere is an edge-mark E2k [+�] (resp. E1k[+�]) of Vm, and F = Z in that edgealso. Also, the values of J are the same in E1k and E2k [+�] (resp. E2k andE1k [+�]), the values of N are the same in E1k and E2k [+�] (resp. E2k andE1k [+�]), and the values of P are the same in E1k and E2k[+�] (resp. E2k and

28 CHARLES RADINE1k[+�]). Finally, it can only happen that F = Z for another edge-mark Epq ofVm if Epq has angle zero with one of the above pair.(R9) Let V1, V2 and V3 be the three vertices of a prototile, with E2e (resp. E11)of V1 and E11 (resp. E2e ) of V2 both referring to the same tile edge, and A1 = Sin �2. See Figure 13. If G�1Sm in E2e (resp. G�2Sm in E11) of V1, equals G�1Sm inE11 (resp. G�2Sm in E2e) of V3, then in V2: if m = D, and (J;N) = (M;D) in E11(resp. E2e), then F = S in E11 (resp. E2e ); if m = E, and (J;N) = (M;E) in E11(resp. E2e ), then F = S in E11 (resp. E2e); if m = B, and (J;N) = (L;B) in E2e(resp. E11), then F = S in E2e (resp. E11).

V1V2 R13V1V2 V3 V1V1V1V1

V2V2V2V2

V3V3V3

V3V3

SS (L,C)(M,C)(L,C)(M,C)MM(S,B) (L,B) (S,B) (L,B)R11

M M(S,C) (L,C) (S,C) (L,C)R12Figure 14

PINWHEEL TILINGS 29(R10) Let V1, V2 and V3 be the three vertices of a prototile, with E2e (resp.E11) of V1 and E11 (resp. E2e) of V2 both referring to the same physical edge,and A1 = M in �2. See Figure 13. If G�1Mm in E2e (resp. G�2Mm in E11) ofV1, equals G�1Mm in E11 (resp. G�2Mm in E2e ) of V3, then in V2: if m = D, and(J;N) = (S;D) in E2e (resp. E2e), then F = M in E2e (resp. E2e ); if m = A, and(J;N) = (S;A) in E2e (resp. E2e), then F = M in E2e (resp. E2e ).(R11) Let V1, V2 and V3 be the three vertices of a prototile, with E2e (resp.E11) of V1 and E11 (resp. E2e) of V2 both referring to the same physical edge,and A1 = M in �2. See Figure 14. If (J;N) = (L;B) in E2e (resp. E11) of V1,and (J;N) = (S;B) in E11 (resp. E2e ) of V3, then in V2 we have F = M in bothE11 and E2e .(R12) Let V1, V2 and V3 be the three vertices of a prototile, with E2e (resp.E11) of V1 and E11 (resp. E2e) of V2 both referring to the same physical edge,and A1 = M in �2. See Figure 14. If (J;N) = (L;C) in E2e (resp. E11) of V1,and (J;N) = (S;C) in E11 (resp. E2e) of V3, then in V2 we have F = M in bothE11 and E2e .(R13) Let V1, V2 and V3 be the three vertices of a prototile, with E2e (resp.E11) of V1 and E11 (resp. E2e) of V2 both referring to the same physical edge,and A1 = M in �2. See Figure 14. If (J;N) = (L;C) in E2e (resp. E11) of V1,and (J;N) = (M;C) in E11 (resp. E2e ) of V3, then in V2 we have F = S in bothE11 and E2e .(R14) If in two of the vertices of a prototile there are F values indicating thatthe edge between the vertices ends at the vertices, then the corresponding Nvalue must agree with the class of prototile that it is.V1V2 V3V3 V1V2 F=Z F=ZR19Figure 15(R15) Let X = A or D. Assume we have a prototile T, with vertices V1, V2and V3, and that E11 (resp. E2e ) of V2 and E2e (resp. E11) of V1 refer to thesame edge of T. Then if F = Z in E11 (resp. E2e ) of V2, GjkMX has the samevalue in E2e (resp. E11) of V1 as it has in E21[��] (resp. E1e [+�]) of V2, unless

30 CHARLES RADINV2 has an edge-mark E21[��=2] (resp. E1e [+�=2]) which contains the values(J;N; F; P ) = (S;A;L;A) or (S;A;L;D), in which case the two G values haveopposite sign. If F 6= Z in E11 (resp. E2e ) of V2, and we do not have: k = 2(resp. k = 1) and (J;N; F ) = (S;X;L) in E11 (resp. E2e) of V2, then G�kMX = +1in that edge-mark; k = 1 (resp. k = 2) and (J;N; F ) = (S;A;L) or (S;D;L) inE11 (resp. E2e ) of V2, then G�kMX = +1 in that edge-mark. See M1.(R16) Let X = D or E. Assume we have a prototile T, with vertices V1, V2and V3, and that E11 (resp. E2e) of V2 and E2e (resp. E11) of V1 refer to thesame edge of T. If F = Z in E11 (resp. E2e ) of V2, then GjkSX has the samevalue in E2e (resp. E11) of V1 as it has in E21[��] (resp. E1e [+�]) of V2, unless:V2 has an edge-mark E21[+�=2] (resp. E1e [��=2]) which contains the values(J;N; F; P ) = (S;A;L;E) or (S;A;L;D); or V2 has an edge-mark E2e (resp. E11)which contains the values (J;N; F; P ) = (S;A;M;B), in which case the two Gvalues have opposite sign. If F 6= Z in E11 (resp. E2e ) of V2, and we do not have:k = 2 (resp. k = 1) and (J;N; F ) = (M;X;L) in E11 (resp. E2e) of V2, thenG�kSX = +1 in that edge-mark; k = 1 (resp. k = 2) and (J;N; F ) = (M;E;L)or (M;D;L) in E11 (resp. E2e) of V2, then G�kSX = +1 in that edge-mark. IfF = Z in E11 (resp. E2e ) of V2 and we do not have k = 1 (resp. k = 2) and(J;N; F ) = (L;B;S) in E2e (resp. E11) of V1, then G�kSX = +1 in E11 (resp. E2e)of V2. See M4.(R17) Assume we have a prototile T, with vertices V1, V2 and V3, and thatE11 (resp. E2e) of V2 and E2e (resp. E11) of V1 refer to the same edge of T. IfF = Z in E11 (resp. E2e ) of V2, then GjkSB has the same value in E2e (resp.E11) of V1 as it has in E21[��] (resp. E1e [+�]) of V2, unless: V2 has an edge-mark E21[��=2] (resp. E1e [+�=2]) which contains the values (J;N; F; P ) =(S;A;L;E) or (S;A;L;D); or V2 has an edge-markE2e (resp.E11) which containsthe values (J;N; F; P ) = (S;A;M;B), in which case the two G values haveopposite sign. If k = 2 (resp. k = 1) and we do not have (J;N; F ) = (M;E;L)or (M;D;L) in E11 (resp. E2e) of V2, then G�kSB = +1 in that edge-mark. Ifk = 1 (resp. k = 2) and we do not have (J;N; F ) = (L;B;S) in E2e (resp. E11)of V1, then G�kSB = G�kSB = +1 in E11 (resp. E2e) of V2. See M8.(R18) Hknm = G�knm in an edge-mark if in that edge-mark we have (J;N) 6=(S;A); (S;D), (M;D), or if (J;N; F ) = (S;A;L); (S;D;L) or (M;D;S).(R19) Assume we have a prototile T, with vertices V1, V2 and V3, and thatE11 (resp. E2e) of V2 and E2e (resp. E11) of V1 refer to the same edge of T. IfF = Z in E11 (resp. E2e ) of V2, then the ratio of Hknm and G�knm is the same inE21[+�] (resp. E1e [+�]) of V2 as it is E2e (resp. E11) of V1. See Figure 15.(R20) Consider any prototile with E11 (resp. E2e) of V2 and E2e (resp. E11) ofV1 referring to the same edge. If F 6= Z in E11 (resp. E2e) of V2, and it isnot the case that: k = 2 (resp. k = 1) and (J;N; F ) = (S;X;L) in E11 (resp.E2e) of V2, then G�kMX = +1 in that edge-mark; k = 1 (resp. k = 2) and

PINWHEEL TILINGS 31(J;N; F ) = (S;A;L) or (S;D;L) in E11 (resp. E2e) of V2, then G�kMX = +1in that edge-mark (see M1); k = 2 (resp. k = 1) and (J;N; F ) = (M;X;L)in E11 (resp. E2e) of V2, then G�kSX = +1 in that edge-mark; k = 1 (resp.k = 2) and (J;N; F ) = (M;E;L) or (M;D;L) or (L;B;M) in E11 (resp. E2e )of V2, then G�kSX = +1 in that edge-mark (see M4); k = 2 (resp. k = 1) and(J;N; F ) = (M;E;L) or (M;D;L) or (L;B;M) in E11 (resp. E2e) of V2, thenG�kSX = +1 in E11 (resp.E2e ) of V2; k = 1 (resp. k = 2) and (J;N; F ) = (L;B;M)in E11 (resp. E2e) of V2, then G�kSX = +1 in E11 (resp. E2e ) of V2 (see M8).(R21) Assume we have a prototile T, with vertices V1, V2 and V3, and thatE11 (resp. E2e ) of V2 and E2e (resp. E11) of V1 refer to the same edge of T. If(J;N; F ) = (M;D;L) in E2e (resp. E11) of V1, then the ratio of H1nm and G�1nmin E11 (resp. the ratio of H2nm and G�2nm in E2e ) of V2 equals the ratio of H1nmand G�1nm in E2e [+�=2] (resp. the ratio of H2nm and G�2nm in E11 [��=2] of V1.ED CBAEDCBA EDCBA ABCDEEDCBA

Figure 16(R22) Assume we have a prototile T, with vertices V1, V2 and V3, and thatE11 (resp. E2e ) of V2 and E2e (resp. E11) of V1 refer to the same edge of T. If(J;N; F ) = (M;D;S) in E2e (resp. E11) of V1, then the ratio of H1nm and G�1nm

32 CHARLES RADINin E11 (resp. the ratio of H2nm and G�2nm in E2e) of V2 equals the ratio of H1nmand G�1nm in E1e [��=2] (resp. the ratio of H2nm and G�2nm in E21[+�=2] of V1. Wemust now de�ne the tiles of our system, and their \matching rules".We de�ne a \tile" as an (orientation preserving) isometric image of oneof the two unmarked prototiles, together with its marks. (We could de�ne thetiles of class X2 to be re ections { with appropriate changes of marks { of thoseof class X1 if it was desired to minimize the number of prototiles.) In order tode�ne \matching rules" for these tiles, we �rst de�ne what it means for twotiles to be \neighbors". Two tiles are neighbors if their intersection consistsof an edge of each. Also, a tile of type E will be called the neighbor of a tileof type C if they intersect precisely in the small edge of the C and half themedium edge of the E, with the medium vertex of the C and the large vertexof the E coinciding. A tile of type E is a neighbor of a tile of type D if theyintersect precisely in the small edge of the D and half the medium edge of theE, with the medium vertex of the D and the small vertex of the E coinciding.A tile of type E is a neighbor of a tile of type A if they intersect precisely inthe small edge of the E and half the medium edge of the A, with the smallvertex of the A and the medium vertex of the E coinciding. A tile of type Ais a neighbor of a tile of type B if they intersect precisely in half the mediumedges of each, with the small vertex of the B in the middle of the edge of theA, and the large vertex of the A in the middle of the edge of the B. And �nally,a tile of type B is a neighbor of a tile of type B if they intersect precisely inhalf the medium edges of each, with the large vertex of each B in the middleof the edge of the other B. See Figure 16. Using the above, the matching rulesfor tiles are: in a tiling two tiles Tm and Tn may abut if they are neighborsand whenever a vertex Vm of Tm coincides with a vertex Vn of Tn, and Tnis in the positive direction from Tm about the common vertex, then Ejk of Vnequals Ejk+1 of Vm, and Ak of Vn equals Ak+1 of Vm, where e+ 1 � 1:4. Hierarchical structure in the tilingsWe de�ne a triangle of level 0 (note the distinction from \Triangle" and\Level") as a tile; its class and type are those of the tile. We inductively de�nea triangle of level n � 1 as an \appropriate collection" (de�ned below) of �vetriangles of level n� 1 together with certain conditions on values of F , N andJ as noted below.An appropriate collection of �ve triangles of level n is the image, by asingle isometry of the plane, of: a triangle of type A with vertices 5n=2(�2; 1),5n=2(0; 1) and (0; 0); a triangle of type B with vertices 5n=2(0; 1), (0; 0) and5n=2(2; 1); a triangle of type C with vertices (0; 0), 5n=2(2; 0) and 5n=2(2; 1) ; atriangle of type D with vertices (0; 0), 5n=2(2; 0) and 5n=2(2; 1); and a triangle

PINWHEEL TILINGS 33of type E with vertices 5n=2(2; 1), 5n=2(3; 1) and 5n=2(2;�1). For any suchcollection we carry over the de�nition we used for Triangles of tiles TS, TMand TL, and vertices VSS, VSM etc.To be a triangle of level n � 1 we require moreover, F = Z in all edge-marks referring to tile edges which coincide in an interval with one of thethree edges of the collection, except possibly for edge-marks in those verticescoinciding with the three vertices of the collection, which however must satisfythe following. The large triangle must belong to one of these \classes" (seeFigures 17A,B), which from R8 are mutually exclusive:

B2B1(S,A,L)(S,A,M) A1A2(S,C,L)(L,C,M)(L,B,M)(S,B,L)(L,B,S)

(L,C,S)(L,B,S)(L,C,S)C1 C2

(F=Z)(F=Z)(F=Z)(S,B,M)F=Z F=Z

(M,C,S) (M,C,S)(S,C,M)(M,C,L)Figure 17a

34 CHARLES RADIN

F=Z(M,E,L)F=ZF=Z (S,D,L)D2 D1(M,D,S) (M,D,S)(S,D,M)(M,D,L) (M,E,S)(M,E,S) E1E2 Figure 17b(A1) (resp. A2) in VMM, E11 (resp. E2e) contains (J;N; F ) = (S;A;M), andE2e (resp. E11) contains F = Z, and in VSS, E2e (resp. E11) contains (J;N; F ) =(S;A;L), and E12 (resp. E1e�1) contains F = Z;(B1) (resp. B2) in VMM, E2e (resp. E11) contains (J;N; F ) = (S;B;M), and E11(resp. E2e) contains (J;N; F ) = (L;B;M), and in VSS, E11 (resp. E2e ) contains(J;N; F ) = (S;B;L), and E2e�1 (resp. E12) contains F = Z, and in VSS, E2e(resp. E11) contains (J;N; F ) = (L;B;S), and E11 (resp. E2e ) contains F = Z;(C1) (resp. C2) in VMM, E2e (resp. E11) contains (J;N; F ) = (S;C;M), and E11(resp. E2e) contains (J;N; F ) = (L;C;M), and in VSS, E11 (resp. E2e ) contains(J;N; F ) = (S;C;L), and E2e�1 (resp. E12) contains (J;N; F ) = (M;C;L), andin VSS, E2e (resp.E11) contains (J;N; F ) = (L;C;S), and E11 (resp. E2e) contains(J;N; F ) = (M;C;S);(D1) (resp. D2) in VMM, E11 (resp. E2e ) contains (J;N; F ) = (S;D;M), and inVSS, E2e (resp. E11) contains (J;N; F ) = (S;D;L), and E12 (resp. E2e�1) contains(J;N; F ) = (M;D;L), and in VSS, E11 (resp. E2e) contains F = Z, and E2e (resp.E11) contains (J;N; F ) = (M;D;S);(E1) (resp. E2) in VSS, E2e (resp. E11) contains F = Z, and E12 (resp. E2e�1)contains (J;N; F ) = (M;E;L), and in VSS, E2e (resp. E11) contains (J;N; F ) =

PINWHEEL TILINGS 35(M;E;S). A triangle of class Xj and level n � 1, de�ned above, is said to beof type X.In summary, a triangle of level n � 1 is de�ned inductively as a collec-tion of �ve triangles of level n � 1 in one of two possible speci�ed geometricrelationships, satisfying one of ten (mutually exclusive) sets of conditions onvalues of F , N and J values in those edge-marks referring to the edges ofthe collection; F = Z at all interior vertices, and (F;N; J) exhibit one of tenspeci�ed, mutually exclusive, patterns at the three vertices of the collection.Note (using V1 � � � V10) that triangles of level 0 (namely tiles) satisfy all theproperties of triangles of level n, for n = 0.5. ResultsTheorem 5.1. If in the substitution tesselation � of the plane weconsider the Tiles as tiles, we obtain a tiling of the plane with all matchingrules satis�ed.Proof. We need to prove that the marks on the Tiles satisfy the restrictionsfor marks of tiles, A1 � � � E2, V1 � � � V10, and R1 � � � R22, and that the neighborrules are satis�ed.For the most part, the restrictions on tiles and the de�nition of \neighbor"are constructed so as to obviously conform with the tesselation �. We �rstshow that abutting Tiles occur in � only in ways allowed by the neighbor rules.Our claim is obviously true for Level 1 Triangles. New types of neighborcombinations may appear as a result of some Triangle of a certain Level nbeing divided in �ve parts according to our procedure; two abutting Tiles ina Level n Triangle, when subdivided, may create along their common edgesomething new at Level n + 1. Consequently, we just have to verify that anytwo Tiles, neighbors according to the rules, give rise only to such neighboringTiles after being divided. We now examine four cases.If the Tiles match (full) edge-to-edge, with equal angles at coincidingvertices, then they produce only edge-to-edge matchings after the subdivision.If the Tiles match edge-to-edge, with unequal angles at coinciding vertices(this happens in the tesselation � only for L edges, but this is irrelevant toour argument), then they produce edge-to-edge matchings unless they borderalong L edges. In this case they produce the A-to-A and A-to-B matchings onhalf the medium edge and the A-to-E small edge to half medium edge matching(see Figure 3).All matchings of an S edge to half an M edge give rise to edge to edgematchings after a subdivision (see Figure 3).

36 CHARLES RADINAll half medium edge to half medium edge matchings give rise to edge toedge matchings after a subdivision (see Figure 3). This completes our argumentconcerning the neighbor rules. D2G�1SDG�1SD V3V2 V1Figure 18The only restrictions we consider are R9, R10, and those parts of V1 andV6 which refer to the H variables.Consider a typical case of R9: assume (J;N) = (M;D) in E11 of V2, andG�1SD in E2e of V1 equals G�1SD in E11 of V3. We will show that this only occursif the Tile is as shown in Figure 18, namely the Tile TS of a Triangle of classD2 (which is of course consistent with the conclusion of R9). We know fromthe hypothesis there is a Triangle T of type D, with the medium edge of Tcontaining vertices V1 and V2 of the Tile. Our proof will be by contradiction.G�1SDG�1SDV3 V2E1 V1

Figure 19

PINWHEEL TILINGS 37So assume that F 6= S in E11 of V2. It follows that V2 and V3 lie on a completeedge which ends at V2, and from the geometry this edge must be either: themedium edge of a type D or E Triangle, or the large edge of a type B Triangle.The three cases are similar, and we illustrate the argument with the case ofthe medium edge of a type E Triangle, T'. The geometry must then be as inFigure 19. But this cannot be our situation because of M6, with Y 1 = E1 andX = D, which completes our analysis of R9.We now consider R10. More speci�cally assume G�1MA in E2e of V1 equalsG�1MA in E11 of V3, and that (J;N) = (S;A) in E2e of V2; we must show thatF = M in E2e of V2 as in Figure 20.A2 G�1MAG�1MA V3V2V1 Figure 20 G�1MAG�1MA V3V2 D1V1 Figure 21We know from the hypothesis there is a Triangle T of type A, with the shortedge of T containing vertices V3 and V2 of the Tile. Our proof will be bycontradiction. So assume F 6= M in E2e of V2. It follows that V2 and V1 lie ona complete edge which ends at V2, and from the geometry this edge must bethe short edge of a type D or A Triangle. The two cases are similar, and we

38 CHARLES RADINillustrate the argument with the case of a type D Triangle, T'. The geometrymust then be as in Figure 21. But this cannot be our situation because of M3with Y 1 = D1 and X = A, and this completes our discussion of R10.For V1 and V6 we will need two lemmas.Lemma 5.2. a) Let V1 and V2 be vertices of a Tile, both lying on themedium edge (not necessarily a complete edge) of a Triangle of Level n � 1,with V1 at one end of the edge and E11 (resp. E2e ) of V2 and E2e (resp. E11) ofV1 referring to the edge. Assume V4 is a vertex of a Tile lying at the otherend of this medium edge, with E11 (resp. E2e) of V4 referring to the edge. ThenGjkMn has the same value in E11 (resp. E2e) of V2 as it has in E11 (resp. E2e) ofV4. See Figure 22A.b) Let V1 and V2 be a vertices of a Tile, both lying on the medium edge (notnecessarily a complete edge) of a Triangle of Level n � 1, with V1 at the endof the edge meeting the small angle, and E11 (resp. E2e ) of V2 and E2e (resp.E11) of V1 referring to the edge. Assume V4 is a vertex of a Tile lying at theother end of this medium edge, with E11 (resp. E2e) of V4 referring to the edge.Then GjkSn has the same value in E11 (resp. E2e ) of V2 as it has in E11 (resp.E2e) of V4. See Figure 22B.

Figure 22a

PINWHEEL TILINGS 39Figure 22bProof of Lemma 5.2. a) The case of Level n = 1 follows from M1. Thehigher Level cases then follow from the A-D symmetry of M1, since the mediumedge is composed of the large edges of an A and a D Triangle, and any changesGjkMn undergoes in the A is reproduced in the D.b) The case of Levels n = 1; 2 follow from M4 and M8. And now the sameargument used in part a) can be applied, which completes our proof of thelemma.Lemma 5.3. Consider a Triangle of type E and Level n � 0, with E11(resp. E2e) of VMM and E2e (resp. E11) of VMS referring to the large edge. ThenGjkMn has the same value in E2e (resp. E11) of VSM as in E2e (resp. E11) of VMSif and only if it has the same value in E11 (resp. E2e ) of VLL as in E11 (resp.E2e ) of VML. See Figure 22C.

Figure 22c

40 CHARLES RADINProof of Lemma 5.3. First note from M1 that cases n = 0; 1 hold. Thecases n � 2 then follow by induction, using Lemma 5.2 to reduce to the com-ponent E Triangle of one lower Level. This ends our proof of the lemma.The nonobvious portions of V1 and V6 concern the values of the Hknm .However the case for variables HkSm follows immediately from R10 and Lemma5.2b, and for variables HkMm from R9, Lemma 5.2a and Lemma 5.3. Thiscompletes our proof of Theorem 5.1.Theorem 5.4. In any tiling of the plane satisfying the matching rules,there is a unique decomposition into triangles of level n � 0.Proof. We begin with an outline of the proof. To prove by induction theexistence of unique triangles of all levels one needs to show that each triangle oflevel N is contained in a unique \appropriate collection" at level N +1, whichfurthermore has all the distinguishing characteristics of some class, so that onecan unambiguously form the level N + 2 \appropriate collection". Some ofthis is easy to arrange using the \accessible" parts of the components of thecollection, in particular the F values at the large vertex of the collection, whichis accessible from the E component. (Accessible means known directly from thede�nition of the class of the component.) The harder step is to arrange for theneeded characteristics of the collection at inaccessible parts of the components,in particular the F values at small and medium vertices of the collection. Thisis accomplished by encoding the information in the G variables, which canthen carry it from accessible to inaccessible parts along pathways de�ned bythe F and H variables, where R9 or R10 is applied.To begin the proof we �rst note that for n = 0 the conclusion is satis�edby the assumption of a tiling. So to continue the induction, assume we havea tiling of the plane (with all matching rules for tiles satis�ed of course), andthat the tiling is uniquely decomposable into triangles of level n for 0 � n � N .We must show that the tiling is uniquely decomposable into triangles of levelN + 1.It is routine to check using R5 and V1 � � � V10 that given any triangleof level N in the tiling there is a unique way to extend it to an \appropriatecollection" of �ve such triangles, and that there is a unique value of P in thenine complete edges of the �ve level N triangles, in particular the small edgeof the A and the small and medium edges of the D. It then follows from R4,R8, V1, V5, V6 and V10 that this P value is converted into the N values in allthose edges of the collection appropriate for the value of P , where the N valuesare meant to help de�ne a type, and thus class, for the collection, making it anN +1 triangle. It remains to show that the needed F values are automaticallycorrect, thereby justifying these N values.

PINWHEEL TILINGS 41The needed F values (namely all Z) in edge-marks interior to the edgesof the collection are automatically correct by properties of the �ve componenttriangles. The only possible di�culty is therefore at the three vertices of thecollection. There are ten cases, depending on the ten possible classes, A1 � � �E2, of the collection. The proof is similar for all cases. Using the P valueused to determine the class of the triangle of level N + 1, we get the neededinformation as follows, the general method being illustrated in the interests ofclarity for the cases of class B1 and A2.We consider �rst the case where P = B in the edge-marks in, say, thesmall edge of the component triangle of type D of the collection. The neededinformation in VLL of the collection follows from V5 using this value of P in thesmall edge of the triangle of type D, which shares its medium vertex with thesmall vertex of TL. The needed information in VMM of the collection followsfrom R11, applied to tile TM, again using the above P value, but now using itwith R4 and R8.The needed information in VSS of the collection follows from R9 appliedto tile TS as follows. We know from V5 together with R19, R21 and R22 thatG�1SB in E1e [+�=2] of vertex TLL of the component A triangle equals G�1SB inE21 [��=2] of vertex TLL of the collection. To carry this information to vertexVSS of the collection, in order to use R9 and V3, we use Lemma 5.2, which, likeLemma 5.3, holds for \appropriate collections" of triangles as well as Trianglessince it depends only on the variation of the G and H variables, which are thesame for Tiles and tiles.We now consider another case of the theorem, class A2, and only theneeded information at the medium vertex of the collection (the information atthe large vertex following as above.)The needed information in VMM of the collection follows from R10 appliedto tile TM, as follows. We know from V5 together with R19, R21 and R22 thatG�1MA in E1e [��=2] of vertex TLL of the component A triangle equals G�1MA inE11 of vertex TLL of the collection. To carry this information to vertex VMMof the collection, in order to use R10, we use V2, Lemmas 5.2 and 5.3.This completes our proof of Theorem 5.4.6. Concluding RemarksIn the above we have constructed the �rst example of a �nite set of shapes(prototiles) which can tile the Euclidean plane, but only using tiles in in�nitelymany orientations (and thus of course only nonperiodically.) This requires,essentially for the �rst time, the use of rotations in the study of forced tilings.We make the following supplementary remarks.

42 CHARLES RADIN� If our restrictions for tile marks are replaced by the stronger restrictionsthat the only marks allowed are those which appear on Tiles in the tesselation�, then of course Theorem 5.1 and Theorem 5.4 still hold. (The reason wedid not do this is that it would then be more di�cult to determine if a givenmarked prototile satis�ed the restrictions.)Figure 23�� It is common in the tiling literature to restrict attention to tilings ofpolygonal tiles which are \edge-to-edge", that is, in which the vertex of a tilecan only intersect another tile at a vertex, as distinct from the way our tilesof type E meet tiles of type C for example. To accomodate this (mistaken)prejudice, we note that our results can be recaptured with this additionalproperty by the doubling of the number of prototiles; each of our prototileslisted above could be decomposed into two by cutting on a straight line fromthe center of the medium edge to the medium vertex, each half retaining ofcourse the information on the two vertices it inherits: see Figure 23. One wouldthen add matching rules along these fresh edges, forcing only the original pairsto be able to meet along such an edge.Figure 24

PINWHEEL TILINGS 43� � � It is convenient to insert a distinguished point within each triangle of typeC, where the bisector of the large angle meets a straight line perpendicular tothe center of the small edge, as in Figure 24. It is easy to check that fora triangle of type C of level n � 1 this point coincides with that associatedwith the triangle of type C of one lower level (and of course continuing downto level 0). These points are of interest because the triangles appear rotatedabout these points, in opposite directions for triangles of the same type butdi�erent class. This is the origin of the name we have given to this tilingsystem, with reference to the pinwheel toy which has blades, attached to astick, which rotate in the wind.University of Texas, Austin, TX References[1] R. Berger, The undecidability of the domino problem, Mem. Amer. Math. Soc. no. 66,(1966).[2] M. Gardner, Extraordinary nonperiodic tiling that enriches the theory of tiles, Sci. Amer.January 1977, 116-119.[3] B. Gr�unbaum and G.C. Shephard, Tilings and Patterns, Freeman, New York, 1986.[4] S. Mozes, Tilings, substitution systems and dynamical systems generated by them, J.d'Analyse Math. 53 (1989), 139-186.[5] I. Niven and H. Zuckerman, An Introduction to the Theory of Numbers, 3rd ed., Wiley,New York, 1972[6] C. Radin, Global order from local sources, Bull. Amer. Math. Soc. 25 (1991), 335{364.[7] C. Radin and M. Wolff, Space tilings and local isomorphism, Geometriae Dedicata 42(1992), 355-360.[8] R.M. Robinson, Undecidability and nonperiodicity for tilings of the plane, Invent. Math.12 (1971), 177-209.[9] H. Wang, Dominoes and the AEA case of the decision problem, Mathematical Theory ofAutomata, ed. Jerome Fox, Polytechnic Press, New York, 1963, 23-56.[10] H. Wang, Proving theorems by pattern recognition II, Bell Systs. Tech. J. 40 (1961),1-41.[11] H. Wang, Notes on a class of tiling problems, Fund. Math. 82 (1975), 295-305.[12] H. Wang, Computation, Logic, Philosophy: A Collection of Essays, Kluwer AcademicPub., Dodrecht, 1990. Received ??.Revised ??.