constructing penrose-like tilings from a single prototile ...steinh/jeong1.pdf · known structural...

PHYSICAL REVIEW B 1 FEBRUARY 1997-IIVOLUME 55, NUMBER 6

Constructing Penrose-like tilings from a single prototile and the implications for quasicrystals

Hyeong-Chai JeongInstitute for Physical Science and Technology, University of Maryland, College Park, Maryland 20742

Paul J. SteinhardtDepartment of Physics and Astronomy, University of Pennsylvania, Philadelphia, Pennsylvania 19104

~Received 3 September 1996!

We present two sets of rules for constructing quasiperiodic tilings that suggest a simpler structural model ofquasicrystals and a more plausible explanation of why quasicrystals form. First, we show that quasiperiodictilings can be constructed from a single prototile with matching rules which constrain the way that neighborscan overlap. Second, we show that maximizing the density of a certain cluster of fat and thin tiles can force aPenrose tiling without imposing the usual Penrose matching rules.@S0163-1829~97!02706-9#

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I. INTRODUCTION

Quasicrystals are solids with quasiperiodic translatioorder and crystallographically disallowed rotational symmtry. A leading model for describing their structure and proerties has been the Penrose tiling picture,1 based on the two-dimensional aperiodic pattern invented by Roger Penrose2 in1974. The Penrose pattern is composed of fat andrhombi with matching rules which constrain the way neigboring tiles can join together edge-to-edge. Penrose shothat the only plane-filling tiling consistent with those matcing rules is uniquely the Penrose pattern. The generalizato three-dimensional structures with icosahedral symmcomposed of rhombohedral bricks with matching rulesbeen found. In the Penrose tiling picture, one imaginesthe tiles represent two distinct clusters of atoms andmatching rules represent atomic interactions.

The Penrose tiling picture successfully explains all ofknown structural and physical properties of quasicrystals,cluding the microscopic arrangement of atoms as viewwith scanning tunneling electron microscopy.3,4 Neverthe-less, serious theoretical doubts about its validity havemained. The physical conditions required to emulate a Prose tiling appear to be much more complex than whaneeded to form periodic crystals. For example, the Pentiling picture suggests that the atoms must organize intotwodistinct clusters which act as the building blocks of the qsicrystalline structure, whereas crystals require only a sinbuilding block. The condition for crystals seems intuitivesimple: it is easy to imagine a single building block arisias a low-energy atomic cluster of the given elements.quasicrystals, the energetics must be delicately balanceallow two distinct clusters to intermix with a specific ratiodensities. Furthermore, the atomic interactions must resclusters so that they join only according to the matchrules. Concerns about these conditions have led to a preration of alternative models for quasicrystals, includingicosahedral glass, random tiling, and entropic pictures.5 Thealternatives all predict imperfect quasiperiodic structuwith differing degrees and types of imperfection and thalso predict that quasicrystals are metastable. Yet, rechighly perfect quasicrystals have been discovered—as

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fect as periodic crystals made of similar elements—and soappear to be thermodynamically stable. The peculiar sittion has developed that the Penrose tiling picture has bincreasingly supported observationally, while it has remainquestionable theoretically.

In this paper, we discuss two schemes for constructquasiperiodic tilings which address the theoretical criticisof the Penrose tiling picture. Although the work presenthere is, for the most part, an abstract study of the properttilings, the results may profoundly influence our intuitioabout quasicrystals. First, in Sec. II, we present a simproof of the claim6 that a quasiperiodic tiling can be forceusing only a single type of tile~plus matching rule!. Second,in Sec. III, we show that matching rules can be discardaltogether. Instead, maximizing the density of a chosen cter of tiles suffices. The results are surprising from a maematical standpoint and suggest an explanation of why qsicrystals form. The basic conclusions were discussedshort paper,7 and here we produce the detailed proofs whestablish the results.

In particular, the second approach suggests a simple tmodynamic mechanism for quasicrystal formation. If oimagines that the chosen cluster of tiles represents someergetically preferred atomic cluster, then minimizing the frenergy would naturally maximize the cluster density anthereby, force quasicrystallinity. We first show that tground state is the perfect~Penrose-like! quasicrystal state ifthe clusters are assigned the same energy independelocal environment, as discussed in our short paper. Incase, maximizing density and minimizing energy are equilent. In this paper, we also show the same ground-statesults if different energies are assigned to the cluster depeing on its local environment. This is a significant extensiof our earlier results since some difference in cluster enerfor different environments seems a natural idea. Conversit shows that the same maximum-density configuration isground state for a range of energetic assignments. This cter model of quasicrystals had been conjectured previouby the authors,8 and now it is proven to be mathematicalpossible. In our concluding remarks in Sec. IV, we discuthe implications of our mathematical result for future theretical and experimental research in quasicrystals.

3520 © 1997 The American Physical Society

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55 3521CONSTRUCTING PENROSE-LIKE TILINGS FROM A SINGLE . . .

II. CONSTRUCTION SCHEME I: SINGLE TILEWITH A OVERLAPPING RULE

We first show that a quasiperiodic tiling can be forceusing a single type of tile combined with an overlapping ruleBecause neighboring tiles overlap, the result is an unconvetional tiling, perhaps better termed a ‘‘covering.’’

As an analogy to a real atomic structure for quasicrystathe overlaps should be construed as the sharing of atobetween neighboring clusters, rather than interpenetrationtwo complete clusters. This paper suggests that the sharof atoms by clusters may be an important structural mothat may play an important role in the formation and stabilitof quasicrystals.

The construction was originally proposed by Gummelwho presented an elaborate proof.6 Our contribution is a verysimple, alternative proof which makes clear the relationPenrose tilings and leads us to a second scheme.

For our example, the single tile is chosen to be the decgon shown in Fig. 1~a!. Unlike Penrose tiles, the decagonare permitted to overlap, but only in certain discrete wayTwo decagons may overlap only if similarly shaded regionoverlap and the overlap area is greater than or equal tohexagonal overlap area shown inA. Counting the relativeorientations of the neighboring decagons, one finds that toverlap rule permits five types of nearest-neighbor configrations: four A-type overlaps and oneB-type overlap asshown in Fig. 1~b!. TheA andB overlaps produce two dif-ferent separations between the centers of neighboring degons. Roughly speaking, these two distances, which havrelatively irrational ratio, replace the two tile types in theusual Penrose tilings.

Our proof that the overlap rules force uniquely a quasperiodic tiling analogous to a Penrose tiling is based on iscribing each decagon with a Penrose fat rhombus t

FIG. 1. A quasiperiodic tiling can be forced using marked decgons shown in~a! with overlapping rules. Overlapping rules de-mand that two decagons may overlap only if shaded regions overand if the overlap area is greater than or equal to the overlap regin A. This permits five types of pairwise overlaps, four types ofAoverlaps and one type ofB-type overlap as shown in~b!. If eachdecagon is inscribed with a fat rhombus, as shown in~c!, the in-scribed rhombi from two overlapping decagons share at least overtex ~d! and as much as a complete edge.

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marked with single and double arrows, as illustrated in F1~c!. We remind the reader that the original Penrose tilingconstructed from fat and thin rhombi. Two edges may jotogether only if the type and direction of arrows match.9,10

For any two overlapping decagons, the ‘‘inscribed’’ rhomshare at least one vertex and sometimes share an eWhere the rhombi join at a vertex only, there is an opangle formed by the edges which are the location and shwhere rhombi can be fit according to the Penrose matchrules as shown in Fig. 1~d!. For these ‘‘implied’’ rhombi, thearrows are fixed by the arrows on adjoining inscribrhombi.

We will show that the overlap rules force precisely tsame nearest-neighbor configurations of inscribed andplied rhombi as do the Penrose edge-matching rules. Gmelt showed that there are exactly 20 different ways a gidecagon can be surrounded by neighbors, where surrouna decagon means the edges of the decagon is covered binterior of other decagons byA or B overlaps.6 Figures 2~a!–2~c! show all 20 ways of surrounding a decagon with corsponding rhombus tile configurations. These 20 allowdecagon configurations have been broken into three gro

The first nine decagon configurations@Fig. 2~a!# are inone-to-one correspondence with the nine ways of surrouing the fat rhombus inscribed in the central decagonneighboring fat and thin rhombi according to the Penromatching rules. By ‘‘surrounding’’ a rhombus, we mean thall four of its edges are joined to neighboring rhombi wimatching arrows; furthermore, three of its vertices mustsurrounded by corners of neighboring rhombi. The neighbing rhombi may be inscribed in neighboring decagonsimplied. The fourth vertex~at which the two double-arrowedges of the central rhombus meet! is not completely sur-rounded in some of the nine configurations. Later, we dischow this last vertex is also forced to be a Penrose-like cfiguration when one applies the overlap rules beyondcentral decagon to neighboring decagons.

The next eight decagon configurations are shown in F2~b!. The inscribed rhombi of the central decagon in thgroup are only partially surrounded. However, when tdecagon overlap rules are applied to neighbors of the cendecagon, the result is one of the nine ways of completsurrounding the central rhombus, as discussed in thegrouping. Hence, these eight configurations do not addnew constraints beyond what is already imposed by thenine configurations.

The remaining three configurations are shown in Fig. 2~c!.These configurations of inscribed and implied rhombi dplay no matching rule violation. However, they never occin a perfect Penrose tiling. This is because a mismatchinevitable if one continues to add rhombi. Similarly, ththree configurations of decagons do not occur in a perdecagon tiling because, if one continues to add decagondecagon overlap-rule violation is inevitably forced; thatsome neighboring decagons cannot be surrounded by onthe 20 configurations. Hence, the three configurations pequivalent roles in decagon tilings and Penrose rhombusings.

We have shown that all decagons in a perfect decatiling are surrounded by one of the first nine configuratioin Gummelt’s list. In three of these configurations~Configu-

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3522 55HYEONG-CHAI JEONG AND PAUL J. STEINHARDT

FIG. 2. Twenty ways of surrounding a decagon and correspoing rhombus tile configurations.~a! First group; central fat rhombare surrounded by fat and thin rhombi according to the Penmatching rules. ~b! Second group; center decagons are surrounby other decagons but the central rhombi are only partially srounded. Each configuration is a subconfiguration of one of the nconfigurations in~a!, and so imposes no new constraints.~c! Thirdgroup; any way of adding decagons to these three configuratforces a mismatch. Similarly, any way of adding rhombi to trhombus configuration forces a mismatch~e.g., some open angleformed by the edges of fat rhombi cannot be filled by arrmatched rhombi!. Hence, these configurations never appear inperfect tiling.

ration 7, 8, and 9!, the central inscribed rhombus is surounded on all edges and all four vertices by neighborrhombi. So these map into a unique, Penrose rhombusfiguration. In the remaining six configurations, the fourvertex is incomplete. For four of the six cases~Configuration3, 4, 5, and 6!, simple experimentation shows that there isunique way to complete the fourth by applying the saoverlap rules to one of the neighboring decagons. Forremaining two configurations, Configuration 1 and 2, theare two distinct allowed ways of adding overlapping decgons so that the inscribed rhombi complete the vertex. Ttwo configurations of inscribed decagons are also distinbut each corresponds to an allowed Penrose configuraCounting all of these, the nine decagon configurations minto 11 ways of completely surrounding a central fat rhobus by neighboring rhombi, precisely the number and tyallowed by the Penrose arrow rules.@One of the 11 rhombusconfigurations, does not occur in a perfect Penrose tilingdoes the analogous decagon configuration occur in a pedecagonal tiling. Although the configuration itself obeys tmatching rules~overlap rules!, it is impossible to add theconfiguration without ultimately producing a matching ru~overlap rule! violation.# That is, restricting the surroundings of every fat rhombus to these 11 types is equivalenenforcing the Penrose arrow rules for fat rhombi; in a Prose tiling, every thin tile adjoins fat tiles, so fixing the surounding of every fat rhombus automatically fixes the enronment of every thin rhombus. Hence, the decagon overules are equivalent to Penrose arrow rules, and the proocompleted.

A significant corollary is that the two-tile Penrose tilincan be reinterpreted in terms of a single, repeating motif. Tanalog for quasicrystals is that the atomic structure, whhas been described in terms of two or more cluster unitsthe past, can be reinterpreted in terms of a single, repeaatomic cluster. This reinterpretation of the structure itself isimplification which is valid independent of whether overlenergetics is an important factor in quasicrystal formationnot.

III. CONSTRUCTION SCHEME II:MAXIMIZING CLUSTER DENSITY

Here we show the Penrose matching rules can be replaby a different tiling principle and still obtain the same tilingNamely, if one allows any possible configuration of fat athin rhombi, a perfect Penrose tiling can be picked outmaximizing the density of a particular tile cluster. Asmodel for quasicrystals, the notion is that the chosen clurepresents some low-energy, microscopic cluster of atoMinimizing the energy naturally maximizes the cluster desity and forces quasiperiodicity. Our result suggests thatenergetic conditions required for quasicrystal formationonly that a single chosen cluster be low energy, whichsimilar to the conditions needed to form periodic crystals asimpler to envisage than the conditions needed to formdifferent tiles plus matching rules.

For this paper, we will focus on the tile cluster,C, definedas

Definition.A C cluster is composed of five fat and twthin rhombi plus two side hexagons composed of two fat a

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one thin rhombus each, as shown in Fig. 3. Penrose tiliexperts will recognize the central seven tiles as being a Jaconfiguration and the central decagon as being‘‘cartwheel.’’ 11 It should be emphasized that this is onlyparticular example of a tile cluster that suffices; other choicare possible.

Our central proposition isProposition. The Penrose tiling uniquely has the maxi

mum density ofC clusters among tilings constructed fromPenrose rhombi. The density is defined as the number ofCclusters per unit area. Two tilings are considered to be dtinct only if they differ from one another by patches whosdensity has nonzero measure.

Our choice for theC clusters was motivated by the overlapping decagons described in the previous section. TheCclusters in a Penrose tiling and the decagons in a decagtiling ~Scheme I! are in one-to-one correspondence. That iif decagons are placed so that they circumscribe the censeven tiles of theC cluster, the decagons form a perfecdecagon tiling obeying the overlap rules described in thprevious section. Although the decagon andC cluster havedifferent exterior shapes, their key similarity is that twoneighboringC clusters can share tiles in two ways isomorphic to theA andB overlaps of decagons; see Fig. 4.

Figures 3~c! and 3~d! show the relation between a markeddecagon and aC cluster. Consider the central seven tiles otheC cluster. Shade a dart-shaped region in each fat rhobus and the entire interior of each thin rhombus as shownFig. 3~c!. For a particular choice of orientations for the farhombi, the resulting shadings within aC cluster join to-gether in a pattern that is equivalent to that of an overlaimarked decagon that circumscribes the central seven tilesshown in Fig. 3~d!. ~We have added a third hexagon anticipating our result that two side hexagons guarantee the th

FIG. 3. ~a! A C cluster consists of five fat and two thin rhombwith two side hexagons composed of two fat and one thin rhombeach. There are two possible configurations for filling each sihexagon. This gives rise to three types,Ca , Cb , andCc , dependingon the orientation of the tiles within the two side hexagons~b!.Shade a dart-shaped region in each fat rhombus and the entireterior of each thin rhombus as shown in~c!. For a particular choiceof orientations for the fat rhombi, the resulting shadings within aCcluster join together in a pattern that is equivalent to that of aoverlaid, marked decagon that circumscribes the central seven tias shown in~d!.

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hexagon in a tiling with maximum density ofC clusters; seeSec. III C!

In broad outline, our formal proof of the propositionbased on the properties of tilings under ‘‘deflation.’’ Defltion corresponds to replacing each completeC cluster by alarger, ‘‘deflated’’ fat rhombus using a slight generalizatioof the self-similar substitution rules introduced by PenroThe deflated rhombi form a new tiling~with holes, in gen-eral! which can be deflated again. We show that, if thewere a tiling with a greater density ofC clusters than Penrose tiling, then deflating the tiling increases the density fther. Repeated deflation leads to an unbounded denwhich is impossible. Hence, there can be no tiling wgreater density ofC clusters than Penrose tiling. Then wshow that the Penrose tiling is the unique one which hasmaximum density ofC clusters.

A. Terminology and notation

For the details of the proof, it is useful to introduce somspecialized terminology and notation.

~i! C(a,b,c) : By definition, a tile cluster is identified as aCcluster independent of the orientation of its side hexagoHowever, for the purpose of proof, it is useful to distinguithree types ofC clusters depending on which way the sidhexagons are flipped:Ca , Cb , andCc as shown in Fig. 3~b!.The reflection ofCb around the vertical axis is also considered to be typeCb .

~ii ! r andR: For any given configurationX, we userX torepresent the number per unit area where an acute rhomhas area equal to unity; and we useRX be the number perCcluster. For example,rC is the number ofC clusters per unitarea. Note thatRC51 by definition. To refer to the values or andR in a perfect Penrose tiling, we add a superscript:r0

andR0. For example,

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FIG. 4. ~a! Shows aC cluster with a decagon circumscribing thcentral seven rhombi. The center of theC cluster is defined to bethe center of the decagon.~b! Shows the two kinds of overlapbetweenC clusters which bring the centers closest together.~TheAtypes have the same separation between centers.! If decagons arecircumscribed about the central seven rhombi of eachC cluster, theA- andB-type overlaps betweenC clusters transform into preciseltheA- andB-type overlaps between decagons described in thevious section.

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wheret5~11A5!/2 is the golden ratio and subscriptB re-fers to pairs of neighboringC clusters withB overlaps.~Boverlaps play an important role in the proof.!

~iii ! Area per configuration,A~a!: A~a! represents thearea of a configurationa. aøb is the union of configurationsa andb. For example,

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~iv! C-kites (KC) andH-kites (KH): Because theC clus-ters can overlap, it is important for the proof to have a reable scheme for assigning, or at least bounding the areacupied by a givenC cluster. A useful trick is to decorateachC cluster with the kite-shape shadings shown in Fig.A C-kite, KC , is the shaded subregion of aC cluster shownin Fig. 5~a!. For any twoC clusters which do not have aBoverlap, the associatedC-kites do not overlap. Hence, ifgiven C cluster has noB overlaps whatsoever, it can bassigned at least the area ofKC . Conversely, the only way aC cluster can be assigned less area thanKC is if it has Boverlaps.

An H-kite, KH , refers to the shaded subregion of a hexgon belonging to aC cluster as shown in Fig. 5~b!. We applythe termH-kite only to hexagons which belong to someCcluster. Note that

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A~KH!5t2.

~v! Deflation: ‘‘Deflation,’’ as used in this paper, istransformation in which eachC cluster is replaced by a largfat rhombus, as shown in Fig. 6~a!. The large rhombus willbe referred to as the ‘‘deflated rhombus.’’ Note that theflated rhombus replacing oneC cluster is unable to overlapthe deflated rhombus replacing any otherC cluster no matterhow the two are juxtaposed or overlapped.

This definition of deflation is closely related to the sesimilar inflation and deflation operations introduced by Prose. The Penrose operations produce both deflated fatthin rhombi, whereas here we only introduce the fat rhomThe Penrose deflation operation acting on a perfect Pentiling produces a scaled-up version of a perfect Penrose ticomposed of deflated fat and thin rhombi, demonstratingself-similar structure. Penrose’s operation is defined onlyPenrose tilings and subconfigurations. Our deflation option acting on a perfect Penrose tiling reproduces

FIG. 5. ~a! C-kite; ~b! H-kite.

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scaled-up fat rhombi, but does not explicitly construscaled-up thin rhombi. Instead, holes are left betweendeflated fat rhombi where the thin rhombi fit. It is a mindifference if one is only considering perfect Penrose tilinFor more general tilings, as must be considered in this prthe difference is more significant. Our deflation is definedPenrose and non-Penrose. In either case, one obtains arangement of deflated rhombi separated by spaces or hbut, the holes do not have to be shaped so that scaled-uprhombi fit, in general.

~vi! DeflatedC cluster,DC: The deflation operation canbe repeated. Identify all configurations of deflated rhomwhich form a scaled-up version of theC cluster, as shown inFig. 6, and replace with a yet larger, fat rhombus: a doudeflated rhombus. We call the group of nineC clusters a‘‘deflatedC cluster,’’ DC. The positions of the first fiveCclusters in theDC relative to one another is fixed, but thpairs on either side can be flipped independently, just ashexagons can be independently flipped in aC cluster. Thisresults in three kinds ofDC clusters,DCa , DCb , andDCc .Repeated deflations are defined analogously.

B. Proving the maximum density proposition

The first goal is to prove thatrC<r C0 for any definite

tiling. Then, we show that the Penrose tiling is the uniqtiling with rC5r C

0 .

1. An overview

To obtain a lower bound on the average area occupiedC clusters, we sum theC-kite subareas and divide by thnumber ofC’s, correcting for cases whereC-kites overlap.TheC-kite is a subregion of aC cluster with area 3t12; wecan consider it to be a core area of theC cluster whichis only overlapped by a neighboringC-kite if there is aB

FIG. 6. Deflation replaces eachC cluster with a large, fat rhom-bus, as shown in~a!. A deflatedC clusterDC is composed of thenine deflated rhombi obtained by replacing each of nineC clustersnumbered from 1 to 9 with a fat rhombus, as shown in~b!. TheDCcluster can be deflated again by replacing it with the very larhombus~doubly deflated tile! shown in~b!.

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overlap. More precisely, althoughC clusters can overlap tosome degree, the only possibilities for close overlap arAoverlaps, in which the correspondingC-kites meet along anedge; orB overlaps, in which a specific overlap ofC-kitesoccurs~recall Fig. 4!. In a Penrose tiling, theC-kites fill theentire plane without holes. If theC-kite is not overlapped byany neighboring one, the correspondingC cluster can beassigned the entire area of aC-kite ~at least that!; for thesecases, theC clusters occupy area>3t12, so they decreasthe density relative to the Penrose valuerC51/~3t11!. TwoC clusters withB overlaps are assigned area less than 3t11due to the overlap of theirC-kites. Hence, we reach an important conclusion:B overlaps are the only mechanism fexceeding Penrose density.

To exceed the Penrose density, a tiling must have morBoverlaps than in a Penrose tiling. However, this conditionnot sufficient. In Penrose tiling, everyB overlap of twoCclusters is surrounded by aDC cluster. In a non-Penrostiling, a fraction ofB overlaps may not be part of aDCcluster~i.e., one or more of the seven otherC clusters thatcompose aDC cluster is absent!. In these cases, it is necesary to show by explicit constructions that one can alwidentify ‘‘extra’’ area ‘‘P’’ nearby the associatedB overlapwhich does not belong to theC-kite of anyC cluster and isnot associated with any otherB overlap. This important parof the argument is detailed in the next subsection. The ‘‘tra,’’ unassigned area occupies at least as much area as sby theB overlap. Hence, aB overlap which is not part of aDC cluster does not contribute to decreasing the densityC clusters below the Penrose value.

Suppose there were a tiling with a density ofC clustersgreater than the Penrose value. Then, it must have a hidensity ofDC clusters than in a Penrose tiling,RDC.t22,whereRDC is the number ofDC clusters divided by thenumber ofC clusters. Under deflation and rescaling the aby t2, eachDC cluster becomes aC cluster of the deflatedtiling whose density ist2RDCrC . SinceRDC.t22, the de-flated tiling has a density ofC clusters that is strictly greatethan the original tiling. Repeating the deflationad infinitumwould lead to a tiling with an unbounded density ofC clus-ters. This is impossible~sinceC clusters occupy finite area!,and so there can be no tiling with higherC-cluster density. Acorollary is that, if theC-cluster density equals the Penrovalue, thenRDC5t22 ~the Penrose tiling value! and theC-cluster density in thedeflatedtiling must also equal thePenrose value. This corollary is then used to establishthe Penrose tiling is the unique configuration with the mamal density value.

2. Formal argument

Let xm be a hypothetical tiling which has therC.r C0 . We

set out to prove:Theorem 1. In xm , RDC.t22. If Theorem 1 holds, then

the density of DC clusters in tiling xm isrDC5RDCrC.t22rC . Under deflation, eachDC cluster be-comes aC cluster composed of deflated tiles. If we rescathe deflated rhombi byt2 so that they have the same areathe original rhombi, the resulting tiling has a density ofCclusterst2rDC which is strictly greater thanrC . Repeateddeflation leads to a sequence of tilings with increasingCcluster density without bound, which is impossible. Hen

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Theorem 1 is the key to proving thatxm is impossible and alltilings must haverC<r C

0 . We build to Theorem 1 through aseries of subsidiary theorems.

By checking all hexagons which may share a tile withCclusters, it is straightforward to see that:

Lemma 1: For a givenH-kite, there are only two waysthat it can overlap with aC-kite, shown in Fig. 7.

Corollary 1.1 An H-kite is completely overlapped by aC-kite or it has no overlaps withC-kites at all.

Definition.Configurationn is the union ofC clusters from1 to n as enumerated in Fig. 6~b! ~n51, . . . ,9!. The numberorder in the figure is significant.Rn is theR value of Con-figurationn. By definitionR151. R2 meansRB , the pairs ofC clusters~belonging to aDC cluster! with B overlap.

Theorem 2. In xm , R2.t22.Proof: The overlap between theC-kites of any twoC

clusters with aB overlap corresponds to anH-kite with areat2. All other C-kites do not overlap. Hence, the fractionaarea occupied byC-kites overall,a1[A(øKC)/A(xm), isgiven by

a15rC~t42R2t2!.

~That is, areat4 for eachC-kite minus areat2 for eachBoverlap.! Sincea1<1 by definition andrC.r C

0 by assump-tion, R2t

2 must be greater than 1 @note thatrC051/~3t11!51/~t421!#. Hence,R2.t22.This proves that the number ofB overlaps perC cluster,

R2, must exceedt22 in order for tilingxm to haverC.r C

0 .This is necessary because onlyB overlaps bringC clustersclose enough together to exceed the Penrose density oCclusters locally. But,R2.t22 is not sufficient to prove thatrC.r C

0 : the area saved by a givenB overlap may be com-pensated by other areas which do not belong to anyC clus-ter.

In fact, we claim that exactly this happens if aB overlapis not surrounded by the seven otherC clusters needed toform aDC cluster about it. We first consider the case ofBoverlaps in Configuration 2~B overlapped! clusters forwhichC cluster number 3 in Fig. 6 is missing. We show thaeach suchB overlap is compensated by ‘‘extra area’’P2adjoining the Configuration-2 cluster which is not assigneto any otherC cluster. We then repeat a similar argument focases of Configuration-3 clusters whereC clusters 1 through3 are present, but 4 is missing, etc. We identify ‘‘extra areaP3, P4, etc., and use this to showR3, R4, etc., all mustexceedt22. Each case in the sequence requires its own dtailed analysis, although the basic structure of the argumeis the same. We stop once we have shown thRDC[Rg.t22. Then, our deflation argument described ithe previous subsection can be applied.

FIG. 7. An H-kite ~dark grey! is either a subarea of aC-kite~light grey!—the two ways shown in 6~a! and 6~b! above—or it hasno overlap with aC-kite at all.

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The technically complicated part of the proof is checkithat ‘‘extra area’’ is not double counted; that is,Pi must notoverlap any otherPj with j< i . If it does, we must identifyan alternative choice forPi . This is done on a case by casbasis below. In some cases, we may state that it is ‘‘impsible’’ for Pi andPj to overlap. By impossible, we mean onof two conditions:~a! it is geometrically impossible to juxtapose a Configuration-i and Configuration-j cluster so thatPi andPj overlap; or,~b! it is geometrically possible to jointhe Configuration-i and Configuration-j clusters, but then anadditionalC cluster is forced in between.Pi is defined onlyfor Configuration i which are not part of a Configuratioi11; but joining a Configurationi onto a Configurationjtransforms the Configuration-i cluster to a Configurationi11 cluster. Lemma 2 is an explicit example of~b!. AfterLemma 2, we simply state when an overlap is impossiwithout specifying whether the reason is~a! or ~b!.

Theorem 3. In xm , R3.t22.Proof: Figure 8 shows a Configuration-2 cluster which

not included in a Configuration-3 cluster. The rhombi wsolid lines belong to the Configuration 2. Dashed-line rhomwould also be there if this cluster were included in a Cofiguration 3. Since it is not, one or more of the dashed-lrhombi must be missing. Then the dark, shaded region‘‘extra area’’ P2 which is not assigned to anyC cluster. Theextra area, with magnitudet2, compensates for the aresaved by theB overlap betweenC1 andC2.

ThenA(P2)5t2 andRP25R22R3 since any twoP2’s do

not overlap. Leta2 be the fractional area occupied by thunion of all KC’s and P2’s. From Theorem 2, we knowR2.t22. Now we have

FIG. 8. The assignment of ‘‘extra area’’P2 as described inproof of Theorem 3.

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a25rC@t42R2t21~R22R3!t

2#5rC~t42R3t2!,

Sincea2<1, we haveR3.t22.Theorem 4. In xm , R4.t22.Proof: Figure 9 shows a Configuration 3 which is n

included in a Configuration 4 in part~a!. The rhombi withsolid lines belong to Configuration 3. Dashed-line rhomwould also be there if this were a subconfiguration of a Cfiguration 4. Since the Configuration 3 is not included inConfiguration 4, one or more of the dashed rhombi mustmissing. Then, the dark, shaded region is ‘‘extra area’’P3provided that this area does not overlap any otherP2 or P3assigned to another configuration. This must be checked.possible for the region shaded in~a! to overlap aP2 from aneighboring Configuration 2, as illustrated in~b!. But, then,it is possible to make a different, choice ofP3 which is notoverlapped, as shown by the dark shading in~b!. Anotherpossibility is that the shaded region in~a! overlaps aP3belonging to a neighboring Configuration 3, as shown in~c!.Once again, a different, choice ofP3 that is not overlappedcan be made, as shown by the dark shading. In any cthere is always a regionP3 of areat2 which can be assignedto any Configuration 3 that does not belong to a Configution 4.

The R value for P3 is R32R4 . From Theorem 3, weknowR3.t22. Let a3 be the area density occupied byKC’s,P2’s, andP3’s. Then, we have


2#5rC~t42R4t2!.

Therefore,R4.t22.Theorem 5. In xm , R5.t22.Proof: Figure 10~a! shows a Configuration 4 which is no

included in a Configuration 5. The rhombi with solid linebelong to Configuration 4. Dashed-line rhombi would alsothere if this were a subconfiguration of a ConfigurationSince it is not, one or more of the dashed rhombi mustmissing. Then, the dark, shaded region is ‘‘extra area’’P4provided that this area does not overlap any otherP2, P3, orP4 assigned to a neighboring configuration. Lemma 2 shothat it is impossible for aP4 to overlap aP3, leaving twocases. Figure 10~b! shows the case where the shaded regin ~a! overlaps aP2 belonging to a neighboring Configuration 2. In this case, a different, choice ofP4 can be made, asshown by the dark shading in~b!. Similarly, Fig. 10~c! showsthe case where the shaded region in~a! overlaps aP4 belong-ing to a neighboring Configuration 4. Then a different, uoverlapped choice ofP4 can be made, as shown by the dashading in~c!. In any case, there is always a regionP4 of

FIG. 9. The assignment of ‘‘extra area’’P3 asdescribed in proof of Theorem 4.


FIG. 10. The assignment of ‘‘extra area’’P4as described in proof of Theorem 5.

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areat2 which can be assigned to any configuration 4 tdoes not belong to a configuration 5.

Since any twoP4’s do not overlap,RP45R32R4. From

Theorem 4, we knowR4.t22. Let a4 be the area densitoccupied byKC’s, P2’s, P3’s, andP4’s. Then, we have


2#5rC~t42R5t22!.

Therefore,R5.t22.Lemma 2. A P4 cannot overlap anyP3.Proof: Figure 11 illustrates what happens if one were

try to bring together a Configuration 3 with aP3 and a Con-figuration 4 with aP4 so thatP3 and P4 overlap ~in theshaded region!. Different cases must be analyzed becaufor example,P4 may be anH-kite, as in Fig. 10~a!, or P4may have a different shape, as in Figs. 10~b! and10~c!. @Note: P4 in Fig. 10~b! has the same geometricshape as in Fig. 10~a!, but it is not anH-kite because it doesnot belong to a hexagon which is part of aC cluster. Simi-larly, only in Fig. 9~a! doesP3 correspond to anH-kite.#

If P4 andP3 are bothH-kites, an overlap is not possiblbecause it forces a tile conflict atX. If one is anH-kite shapebut not the other, a similar conflict results.

The remaining case to consider is if theP4 of a Configu-ration 4 and theP3 of a Configuration 3 do not correspondanH-kite, as in Fig. 9~b! or Fig. 10~b!. @Overlaps are notpossible ifP3 corresponds to the case shown in Fig. 9~c! orP4 corresponds to the case in Fig. 10~c!.# This case occurswhere there is also the overlap of aP2 belonging to a third,Configuration-2 cluster. Here, it is geometrically possiblejoin the Configuration-3 and Configuration-4 clusters, bthen an additionalC cluster is forced in between, as showin Fig. 11~b!. The Configuration 4 is transformed into a Cofiguration 5, and so there is noP4 region to identify. ~Re-call, P4 is only defined for Configuration-4 clusters that anot part of a Configuration 5.! Thus, there is never a caswhereP4 andP3, as defined in the previous theorems, coverlap.

Theorem 6. In xm , R6.t22.

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Proof: Figures 12 and 13 illustrate a Configurationwhich is not included in a Configuration 6, showingP5 pro-vided that this area does not overlap any otherP2, P3, P4, orP5 assigned to another configuration. The markings in~a! foreach figure are defined as in previous figures. Figure 12 cers the case where theC1 cluster in the Configuration 5 istypeCa ; Fig. 13 covers the case where it is typeCb .

Let us first consider Fig. 12: If the shaded region in~a!overlaps aP2 belonging to a neighboring Configuration 2, ashown in~b!, then a different, unoverlapped choice ofP5 canbe made, as shown by the dark shading. If the shaded rein ~a! overlaps aP4, there are two possibilities depending owhether theP4 corresponds to case~a! or ~b! in Fig. 10; ineither case, a different unoverlapped choice ofP5 can bemade as shown in the dark shading in~c! or ~d!, respectively.If the P4 corresponds to case~c! in Fig. 10, it cannot alsooverlap aP5. If the shaded region in~a! overlaps anotherP5there is only the possibility shown in~e! and the choice ofP5for one of the Configuration-5 clusters can be made diffently, as shown by the dark shading.

Next, let us consider Fig. 13: The only geometric posbilities for overlap are if the shaded region in~a! overlaps aP4 of a neighboring cluster of the type shown in Fig. 10~a! ora P5 of a neighboring cluster, similar to Fig. 12~e!. For eachcase, a different, unoverlapped choice ofP5 can be made, asshown by the dark shading in~b! and ~c!. In any case, aP5does not overlap with anyPi for i,5 and any twoP5’s donot overlap each other. SinceRP5

5R52R6, and, from Theo-rem 5, we knowR5.t22, we have

rC@t42R5t21~R52R6!t

2#<1,

and we haveR6.t22.Theorem 7. In xm , R7.t22.Proof: Figure 14 shows a Configuration 6 which is n

included in a Configuration 7 for the case where theC1 clus-ter is typeCa in part~a! and for the case where theC1 clusteris typeCb in part ~b!. In both cases, the ‘‘extra area’’P6 isshown for cases in which this area does not overlap any o

2.

s-

FIG. 11. Two cases considered in LemmaOverlap betweenP4 andP3 is ruled out becausethere is a geometrical conflict between tile@markedX in ~a!#, or because juxtaposing configurations force newC clusters where theP4 andP5 were supposed to overlap@shaded region in~b!#. See proof of Lemma 2 for discussion.


FIG. 12. The assignment of ‘‘extra area’’P5if the C1 cluster is typeCa , as described in proofof Theorem 6.

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P2 throughP6 assigned to another configuration. If theC1cluster is typeCa , there is no possible overlap with anothPi . If the C1 cluster is typeCb , the only possible overlap iswith theP3 of a neighboring Configuration 3, in which casa different, choice ofP6 can be made, as shown by the dashading in~c!.

SinceP6 can always be defined such that it does not ovlap otherPi , i<6, theR value forP6 is RP6

5R62R7. FromTheorem 6, we knowR6.t22.

rC@t42R6t21~R62R7!t

2#<1,

and we haveR7.t22.Theorem 8. In xm , R8.t22.Proof: From Fig. 6, one sees thatC6 andC8 are mirror

images with respect to the vertical line bisecting theB over-lap. Recall thatP5 ~or P7! is extra area defined whenC6 ~orC8! is missing. So the analysis of cases whereP7 overlapsPifor i52,3,4 is similar to those forP5. ~To be precise, thecase of overlap betweenP7 andP2 is equivalent toP5 and

r-

P2; betweenP7 and P3 is equivalent toP5 and P4; andbetweenP7 andP4 is equivalent toP5 andP3.!

The new checks are for the case whereP7 ~correspondingto anH-kite! overlaps with aP5. In this case, as shown inFig. 15~a!, we can find the dark region which does not ovelap with anyPi for i<6. Also, it is possible for twoP7’s tooverlap; we have not illustrated the resolution of this cabecause it is similar geometrically to the case of twoP5’soverlapping~see Theorem 6!. By similar analysis to theprevious theorems, we obtainR8.t22.

Finally, we are prepared to prove Theorem 1:Theorem 1. In xm , RDC[R9.t22.Proof: C9 is in a position that is a mirror toC7 in the

deflatedC cluster, so the analysis forP8 is similar toP6 inTheorem 7, so far as cases of overlap with anyPi for i,6.The only case of overlap ofP8 with Pi for i>6 is P8 ~cor-responding to anH-kite! overlapping withP6. In this case,as shown in Fig. 15~b!, we can identify the shaded regiowhich does not overlap with anyPi for i<7. It is not geo-metrically possible to arrange twoP8’s to overlap.

FIG. 13. The assignment of ‘‘extra area’’P5if the C1 cluster is typeCb , as described in proofof Theorem 6.

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Following the same analysis as in previous theorems,obtainR9.t22. Hence, deflatingxm and rescaling the deflated tiles byt2 produces a tiling withrC8 5t2RDCrC whichis strictly greater thanrC . How much so? Suppose we prametrize the fractional excess ofrC over rC

0 as

rC5rC0 ~11a!. ~1!

Then, since

rC~t42Rgt2!<1,

FIG. 14. The assignment of ‘‘extra area’’P6, as described inproof of Theorem 7.

FIG. 15. Assignment ofP7 in ~a! andP8 in ~b! for special casesdescribed in proofs of Theorems 8 and 1.

e

we have

Rg>S 11t4a

11a D Y t2; ~2!

consequently, the increase inC clusters under deflation isrC8 5t2RDCrC5rC

0 (11t4a). Repeated deflations woullead to an unbounded density@r C

0 (11t4na).rC asn→`#,which is impossible.

Now let us show that the Penrose tiling is the uniqtiling with rC51/~3t11!. If a tiling hasC-cluster densityrC5r C

0 ~the Penrose value!, then RDC5t22 and theC-cluster density in the deflated tiling must equal the Prose value from the same arguments given in Theor1, . . . ,8,Suppose there were a non-Penrose tiling withsame density. We have shown that the only local configutions which can increase the density above the Penrose vareDC clusters, and that the increase in density is due toB overlap ofC-kites, which is the same perDC cluster.Now, the hypothetical tiling has the same density ofDCclusters and, hence, the same density ofB overlaps sur-rounded byDC clusters as Penrose tiling. However, by denition, the non-Penrose tiling must also have patches wnonzero area measure which violate the Penrose matcrules, and so cannot belong to theC-kite of anyC cluster.Since theDC-cluster density is the same but there are thpatches, the average area perC cluster must be less than thPenrose density. A conceivable exception is if there hapto be additionalB overlaps which do not belong toDC clus-ters whose overlap area exactly compensates the area opatches. Even this possibility can be eliminated becausecorollary states thatRDC51/t2, which means that the densitof C clusters remains unchanged under deflation and resing. Yet, the patches grow: a patch excluded from aC clustermust also be excluded from aDC cluster, but, also, someCclusters that border the patches cannot be part of aDC clus-ter and add to the patch area. Since the number ofC clustersremains fixed but the patches grow, theC-cluster density inthe deflated tiling must be less than the Penrose value.contradicts the corollary; hence, uniqueness is establishe

C. Is the C cluster unique?

The particular choice ofC cluster considered in this papeis not the unique cluster whose density is maximal in a Prose tiling. ~It was chosen primarily because it makes fthe simplest proof.! As a simple example, consider thclusterC8 which consists ofC plus an additional side hexagon, as shown in the right most cluster of Fig. 3~d!. In aperfect Penrose tiling, everyC cluster lies within aC8 clus-ter. Due to the one-to-one correspondence between thetypes of clusters, the density ofC clusters andC8 clusters arethe same in Penrose tiling. On the other hand, the densitC8 clusters cannot be greater than that ofC cluster in anynon-Penrose tiling because everyC8 cluster guarantees aCcluster. The consequence of this observation is that Pentiling has the maximum density ofC8 clusters as well asCcluster. Or, equivalently, maximizing the density ofC8 clus-ters also forces a Penrose tiling.

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D. Maximum density and energetics

We have proven that maximizing the density ofC clustersforces a perfect Penrose tiling. The notion is that the clurepresents a low-energy atomic cluster and that minimizthe energy naturally maximizes the density. What kindHamiltonian is needed so that the minimum energy confiration is the one that maximizes theC-cluster density?

A simple Hamiltonian with this property is one that asigns energy2e~e.0! to everyC cluster and zero energy tall other clusters. It is obvious that the ground state of tHamiltonian has the maximum density ofC clusters. How-ever, this choice is not unique.

As an example, we illustrate a Hamiltonian which assigC clusters different energies depending on the orientationtheir side hexagons~which is the same as assigning differeenergies according to their local environments!. In Fig. 16,we show each of the three types ofC clusters and introducea way of assigning core areas, using a different shape foCathroughCc . This differs from the assignment in the previosections which assigns the sameC-kite core area to eachCcluster. Now, consider a Hamiltonian which assigea52gt3 to aCa , eb52g~t31t4!/2 to aCb andec52gt4 toaCc cluster and zero for all other clusters. These choiceei are proportional to the magnitudes of the respective careas perC cluster for each given type,ai . If Ni , i5a, b, orc is the number ofC clusters of typei , the total energy of aconfiguration is2SNie i}2SNiai , which is the total areaoccupied by the union of all core areas. The energy densiminimized by the configuration whose core areas have

FIG. 16. A core area assignment that differs from theC-kitesconsidered in the proof. EachC cluster in a tiling can be assignedrhombus-, trapezoid-, or kite-shaped core area dependingwhether it is of typeC(a,b,c) . In a perfect Penrose tiling, these joto form a plane-filling tiling without holes or overlaps, as shownthe figure. This assignment of core area differs from the consttion used in the proof which assigns the sameC-kite core area toeachC cluster. Different assignments are used to explore howferent ways of assigning energy per cluster can lead to the sfinal structure.

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greatest fractional coverage of the plane. The Penrose tis the unique tiling in which the core areas cover 100% oftotal plane. Consequently, the Penrose tiling must beground state of the Hamiltonian.

A second, similar example is shown in Fig. 17. In thcase, the core area assignment is a little more complicaFirst, as can be seen from~a!, parts of the ‘‘core area’’ lieoutside theC cluster itself. This technical difference does naffect our argument, though. Second, there are certain cfigurations, shown in~b!, for which the core areas overlapunlike the case of Fig. 16. In these cases, though, an altetive assignment is possible, as shown in~b!, such that theassigned core area areaa52t13 for Ca , ab55/2t13/2 forCb andac53t11 forCc for anyC cluster. For the particularcase of a perfect Penrose tiling, the core areas as assign~a! never overlap@so the subsidiary rule in~b! does not haveto be invoked#; instead, the core areas join to form a plafilling tiling without holes. Consequently, technical diffeences aside, the situation is the same as in Fig. 16 excepthe area assignments are different forC(a,b,c) . This enablesus to choose a different energy assignment,ea52d~2t13!,eb52d$~5t13!/2%, andec52d~3t11!, such that the Penrosis, again, the unique ground state of the Hamiltonian.

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FIG. 17. A third core area assignment forC clusters leading toa different prescription for assigning energy per cluster. The carea depends on whether a givenC cluster is of typeC(a,b,c) . Forcertain arrangements, shown in~b!, the core areas can overlap. Fthese configurations, though, it is possible to assign the overlapto one of theC-cluster pair and a different region of the same ar~black! to the other cluster. With this rule, the assigned core areaaa52t13 for Ca , ab55/2t13/2 for Cb andac53t11 for Cc foranyC cluster in a Penrose or non-Penrose arrangement. In a pePenrose tiling, the special cases shown in~b! do not occur; that is,the core areas as assigned in~a! join to form a plane-filling tilingwithout holes or overlaps as shown in~c!.

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Since the Penrose tiling is the ground state of threeferent Hamiltonians~one which assigns2e to all C clustersand two linearly independent choices which assign differenergies toC clusters in different local environments!, thePenrose tiling is also the ground state of any linear comnation of the three Hamiltonians~provided e,g and d arepositive. So, in the three-dimensional parameter space~ea ,eb , ec!, there is a three-dimensional region of nonzero msure for which the Hamiltonian maximizesC-cluster densityand selects the Penrose tiling for the ground state.

We note that the proofs that the two other Hamiltoniahave a Penrose-tiling ground state stand independently ocentral proof that Penrose tiling has the maximum densityC clusters. We have approached our investigation frompoint-of-view that maximizing the density is a simple anatural criterion, and so we have made this result our focHowever, others may consider the corresponding energeassignment in which allC clusters have the same energybe artificial since differentC clusters have different locaenvironments. Here, we have shown that the two pointsview are not contradictory—the Penrose tiling emergesther way.

We speculate that the Penrose ground state is stable ueven a wider class of Hamiltonians in which nonzero eneis assigned to other types of tile clusters provided theirsigned energy,ki satisfiesuki u!e. Suppose one tried to introduce a small, nonzerokX to increase the density of a certacluster X ~not C! with respect to its density in a perfecPenrose tiling. The density would be increased only by cating Penrose mismatches, which destroy one or morCclusters each. That is, for every fewX’s gained, one or moreC clusters would be destroyed. An energy proportional tokiwould be gained for each addedX, but this would be accompanied by a loss of energy proportional toe for eachCcluster that is destroyed. It seems that there must be a fiband of ukui!e for which the Penrose tiling is preservedthe ground-state configuration.

The last speculation, along with the rigorous resuabove, strongly suggest that there is a robust range of Hatonians which maximizeC-cluster density and pick out thPenrose tiling as the ground state. The principal requiremfor the Hamiltonian is thatC clusters have low energy,condition similar to the condition that determines crysstructure. It seems that the ground state does not depsensitively on other details, such as whether other clushave the same or different energies or whether allC clustershave the same energy or different energies. This is a sigcant improvement over the old Penrose tiling picture baon two types of tiles and matching rules, which seemedrequire delicate tuning of Hamiltonian parameters.

We note that the models we have been considering insection begin with two rhombus tiles~without matchingrules! as fundamental building blocks. We were seekingground state among all possible configurations of these tThis approach is convenient for comparing with the Penrmatching rule model and the random tiling model, bothwhich are constructed from rhombus tiles. An open issuwhether the presumption of two building blocks, rhombusotherwise, is necessary.

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IV. CONCLUDING REMARKS

Although the results of this paper are mathematicalnature, they have several important implications for tphysics of quasicrystals. First, the decagon construcshows that the atomic structure of a quasicrystal can be cacterized in terms of the decoration of a single cluster, ratthan two clusters as the Penrose tiles would suggest. Furmore, the single cluster has to have the property that itshare atoms with neighbors in accordance with overlaThis greatly simplifies the problem of searching for strutural models for quasicrystals. It immediately appliesthree-dimensional decagonal quasicrystals, where decagprisms would replace the decagons. The analog of the dgon for icosahedral quasicrystals is the triacontrahedron.

Second, the constructions imply a closer tie between qsicrystals and crystals. Now one can say that both candescribed in terms of the packing of a single cluster. Incrystal, the cluster is called the unit cell, and it packs edto-edge with its neighbors. In this picture, quasicrystals crespond to a generalization in which the ‘‘unit cells’’ ovelap. In both cases, the formation of the particular structmay be described in terms of a low-energy atomic clusHence, perhaps one heads towards a more unified pictuordered solids.

Third, the constructions imply a simpler explanationwhy quasicrystals form, shedding, new light on an old mtery. They make it plausible that quasicrystals can be undstood by considering the energetics of microscopic clusand that cluster overlap is an important structural elemeThe atomic structures of known quasicrystals include atclusters which can share atoms using geometries analoto those considered here.5 This has motivated several closerelated models of quasicrystals based on clusters.8,12–15

Our concept can be tested by studying theoreticallyenergetics of atom clusters found in real quasicrystals. Ocan use total-energy calculations to estimate the energeticthe clusters and to test if overlap is energetically preferrPerhaps the modeling studies will lead to suggestionsnew quasicrystals. The second construction scheme, in ation to posing an explanation of why quasicrystals form, asuggests a kinematic mechanism to explain how they fonamely, by local atomic rearrangement that increases thecal density of some certain atom cluster which then forquasicrystallinity. From future structural and kinematicstudies of known quasicrystals, these principles may betablished, perhaps enabling the reliable prediction of nquasicrystals.

ACKNOWLEDGMENTS

We thank F. Ga¨hler for bringing P. Gummelt’s work toour attention and for many helpful comments and criticisWe also thank P. Gummelt for sharing her results priorpublication and for suggestions regarding this paper. J.colar has also provided important suggestions. This resewas supported by the Department of Energy Grant No. DOEY-76-C-02-3071 at Penn~P.J.S.! and by the National Sci-ence Foundation Grant No. NSF-DMR-96-32521 at Maland ~H.C.J.!.

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9P. Steinhardt, Am. Sci.74, 586 ~1986!.10N. G. de Bruijn, Nederl. Akad. Wetensch. Proc. A84, 1 ~1981!.11M. Gardner, Sci. Am.236, 110 ~1977!.12J. L. Aragon, D. Romeu, and A. Gomez, Phys. Rev. Lett.67, 614

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constructing penrose-like tilings from a single prototile ...steinh/jeong1.pdf · known structural...

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