Also ville t http://m-journl.eu ISSN 1855-3966 (printe en.), ISSN 1855-3974 (eletroni en.) ARS MATHEMATICA CONTEMPORANEA 10 (2015) 135 167 Unilterl n equitrnsitive tilings y squres of four sizes Csey Mnn University of Wshington Bothell, Shool of STEM, 18115 Cmpus Wy N.E., Bothell, WA 98011, USA Joseph DiNtle Armstrong Atlnti Stte University, USA Emily Peire Bylor University, USA Ellen Viterik Columi University, USA Reeive 3 April 2014, epte 15 July 2014, pulishe online 18 August 2015 Astrt D. Shttshneier prove tht there re extly eight unilterl n equitrnsitive tilings of the plne y squres of three istint sizes. This rtile extens Shttshneier s methos to etermine lssifition of ll suh tilings y squres of four ifferent sizes. It is etermine tht there re extly 39 unilterl n equitrnsitive tilings y squres of four ifferent sizes. Keywors: Tilings, equitrnsitive, unilterl, squres. Mth. Suj. Clss.: 05B45, 52C20, 54E15, 05C15 1 Introution A two-imensionl tiling, T, is ountle olletion of lose topologil isks {T i }, lle tiles, suh tht the interiors of the T i re pirwise isjoint n the union of the T i is the Eulien plne. A symmetry of T is ny plnr isometry tht mps every tile of T onto tile of T. Two tiles T 1 n T 2 re equivlent if there exists symmetry of T tht E-mil resses: mnn@uw.eu (Csey Mnn), j4732@stu.rmstrong.eu (Joseph DiNtle), EmilyPeire@ylor.eu (Emily Peire), emv2126@olumi.eu (Ellen Viterik) This work is liense uner http://retiveommons.org/lienses/y/3.0/
136 Ars Mth. Contemp. 10 (2015) 135 167 mps T 1 onto T 2. The olletion of ll tiles of T tht re equivlent to T 1 is lle the trnsitivity lss of T 1. T is equitrnsitive if eh set of mutully ongruent tiles forms one trnsitivity lss. This rtile will onern only tilings of the plne y squres of few ifferent sizes. A onnete segment forme y the intersetion of two squres of T will e lle n ege of T, n the enpoints of the eges re lle verties of T. T is unilterl if eh ege of the tiling is sie of t most one tile, mening tht if two ongruent tiles meet long n ege, they re never inient long the full length of tht ege. The ronym UETn will refer to unilterl n equitrnsitive tiling y squres of n istint sizes. A lssifition of ll UET3 tilings is given in [3]. There re only eight UET3 tilings, shown in Figure 1. Beuse the lssifition of UET4 tilings is se on the methology of [3], it will e helpful to outline those methos here. First, some nottion n terminology must e introue. () (...,...,...) () (...,...,...) () (...,...,...) () (...,...,...) (e) (...,...,...) (f) (...,...,...) (g) (...,...,...) (h) (...,...,...) Figure 1: The eight UET3 tilings lssifie in [3]. Let T e UET4 tiling of squres with sie lengths < < <. The skeleton of T is the union of ll of the eges of the tiling T. A vortex is tile T T for whih eh ege of the tile is extenle within the skeleton of T in extly one iretion, given n orienttion of T, s in Figure 2. Figure 2: A vortex tile A oron of tile T in tiling T onsists of ll tiles in T whose intersetion with T is nonempty. The oron signture of T is n orere list of the sizes of the tiles in T s oron. In UET4 tiling the orons of ny two ongruent opies of T must e ongruent ue to equitrnsitivity, so the oron signture of T unmiguously esries the oron of ny tile in T tht is ongruent to T. A smple oron (i.e. oron of tile) n its orresponing oron signture re given in Figure 3. The oron signtures of the eight
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 137 UET3 tilings shown in Figure 1 re given s triplet ( oron signture, oron signture, oron signture). Cyli permuttions of signture, s well s yli permuttions of signture re in reverse, re onsiere equivlent to the originl signture. Figure 3: This oron hs signture... 1.1 Shttshneier s Metho The metho of lssifition use in [3] to fin ll UET3 tilings n e roughly esrie s follows: 1. Determine ll extenle,, n orons. 2. Determine whih 3-tuples of extenle,, n orons re omptile in terms of their oron signtures. 3. Determine whih 3-tuples of omptile,, n oron signtures give rise to tilings (n how mny). A similr proess will e followe in this rtile (with the ition of fining ll extenle oron signtures). It is to e expete tht the sope of the UET4 lssifition prolem is roer in size thn the UET3 prolem; s result, the prolem is solve through two mjor ses. These re the ses where: 1. n re jent. 2. n re not jent. Two tiles re jent if their intersetion is n ege of the tiling. Setions 2-4 onern the se when tiles n tiles re jent. While the se where tiles n tiles re not jent employs themes estlishe these setions, the ifferenes etween these ses re suffiient to require seprte nlysis; this is one in Setion 5. 2 UET4 tilings in whih tiles n tiles re jent The ulk of the work one in lssifying ll UET4 tilings is enumerting ll possile,,, n orons. This jo is me mngele y first estlishing some neessry equtions relting,,, n. These equtions re estlishe in Susetion 2.1. After estlishing finite set of possile equtions relting,,, n, the orons orresponing to these equtions re foun; this is esrie in Setion 3. Finlly, one set of orons
138 Ars Mth. Contemp. 10 (2015) 135 167 orresponing to n eqution or equtions is estlishe, the proess for onstruting the possile tilings is esrie in Setion 4. One ft tht will e use throughout the rtile omes from [2], n n e stte s follows. Lemm 2.1. Let T e UET4 tiling of squres with sie lengths < < <. Then ll n tiles of T re vorties. 2.1 Equtions relting,,, n Lemm 2.2. Let T e UET4 tiling in whih n re jent. Then + = or + =. Proof. Begin y exmining n oron. Beuse n tiles re jent vorties, these two tiles must meet t orner s shown in Figure 4. The she lines epit the neessry skeletl extension in T require y the vortex onition on the n tiles. It is ler tht tile or group of tiles must fill the length inite in Figure 4 extly in orer for these vortex onitions to hol. () () () () (e) Figure 4 Suppose + n +. Neither tile nor tile n fill the spe inite extly, so some omintion of n tiles must e use inste. In ft, the vortex onition requires tht extly two suh tiles e use, n unilterlity implies tht extly one tile n one tile must e use. This yiels the rrngement shown in Figure 4. The length inite in 4 rings up the sme issue, n following the sme logi it is seen tht the rrngement in Figure 4 is the only vli rrngement for this spe. The sme is true for the length inite in Figure 4, yieling the full oron foun in Figure 4. Hving now omplete n oron, equitrnsitivity tells us tht every oron in T must e ientil to this, whih genertes the pth seen in Figure 4e. The only possile unilterl n equitrnsitive tiling tht this pth mits ontins only n tiles n is therefore UET2. This gives rise to two suses within the se of n eing jent, nmely tht where + = n tht where + =. These two ses re onsiere in turn. 2.1.1 + = The following two suses of tht when n re jent n the eqution + = is stisfie re onsiere seprtely:
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 139 1. The tile is not vortex. 2. The is vortex. The tile is not vortex: Figure 5: The extension of non-vortex tile. If is not vortex, then is must hve pir of prllel eges tht exten into the skeleton of T s in Figure 5. There must e some omintion of,, n tiles whih fit perfetly etween the she lines in Figure 5. Sine n re vorties, they must shre orner with the tile n therefore one of their eges must e ontine in she line. There re extly five possile wys to fill the region, ll of whih re shown in Figure 6. Figure 6: All possile wys of filling the region etween the she lines. This then gives us extly five possile reltionships for if it is not vortex: 1. = + + = 2 + 2 2. = + = + 2 3. = + = 2 + 4. = 2 + = 3 + 5. = 2 + = 3 + The tile is vortex: If is vortex, then the oron must ontin either n or, s expline elow; furthermore, there must e n or tile tht shres orner with the tile to stisfy the pertinent vortex onitions, s epite in Figure 7. () () Figure 7
140 Ars Mth. Contemp. 10 (2015) 135 167 If this ws not the se, then the tiling woul e UET2. To see this, notie tht sine is vortex, there must e tiles whih line up extly with the ote lines in Figure 8. If there re no or tiles in the oron of the tile, these tiles must e tiles, s in Figure 8. Finlly, the rest of the oron must e me up y tiles, s in Figure 8. This pth n only e extene to UET2 tiling. Therefore, there must e t lest one or tile whih shres orner with the tile. () () () Figure 8: A UET2 oron. Sine,, n re ll vorties in this suse, there must e omintion of tiles tht fits perfetly etween the she eges initing ege extentions into the skeleton of the tiling in Figure 7. No more thn three tiles my fit in this spe, for if there were four or more, then the two or more tiles snwihe in the mile woul hve to e nonvorties. However, only n e non-vortex, n two tiles nnot meet ege-to-ege y the unilterl onition. Only speifi onfigurtions of tiles n fit etween the she lines. By exmining these onfigurtions n using simple lger, it is esy to enumerte the possile reltionships etween n the smller squre sizes tht woul llow for tiling. This nlysis is summrize in Tle 1. Of ourse, it is possile for tile to shre orner with or tile even if is not vortex, whih is why some reltions re repete from the se where tiles were not vorties. In this se, however, only those reltionships for whih the tile is neessrily vortex re onsiere, so ny repete reltionships re isregre, resulting in the following omplete list of reltionships when = + : 1. = + + = 2 + 2 2. = + = + 2 3. = + = 2 + 4. = 2 + = 3 + 5. = 2 + = 3 + 6. = 2 7. = 3 8. = 3 2.1.2 Asie: : sie length proportions When + = n n re jent, there re ertin tile onfigurtions tht re possile only when the size of is speifilly relte to the size of. By exmining
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 141 All possile omintions of two or three tiles Coul they fit etween the she lines in Figure 7? If yes, wht reltions re require for the onfigurtion to fit perfetly? Coul they fit etween the she lines in Figure 7? If yes, wht reltions re require for the onfigurtion to fit perfetly? + No No + No No + Yes + = + No No new informtion + = Yes + = + 2 No 2 + = 2 + No Yes + = + No new informtion + No No + + = 3 + 2 No Yes + = 3 + = 3 + + = 2 + 2 Yes + = 2 + 2 = + 2 Yes + = 2 + 2 = 2 + + + No No + + = Yes + = + 3 No + 3 = 3 + + No No Tle 1 these speifi onfigurtions for eh se, ertin : rtios re etermine tht must e onsiere; suh rtios re etermine when set of tiles must fit perfetly etween the extene eges of two vorties. These speifi tile onfigurtions re shown in Figure 9 n Figure 10. Tles showing the rithmeti use to fin the : rtios re provie s well. In the first row of oth tles, the eight reltions previously generte for this se ( + = n n re jent) re onsiere. In the leftmost olumn, the onfigurtions s well s the generl proportions they neessitte re liste. For exmple, in Figure 9, tile n tile fit perfetly ove two s n. For this onfigurtion to our, + = 2 +. Therefore, = 2. It shoul e note tht the suses within the se where + =, eh of whih is tehnilly given y ifferent sie length, re not onsiere within the tul oron onstrution proess s seprte ses. Inste, the reer shoul er them in min s oron onstrution egins within the pproprite speifie suse. It shoul itionlly e note tht these : sie reltions re only pertinent when + =. They re not onsiere in the se where + =, whih follows. 2.1.3 + = Lemm 2.3. If + =, then must e vortex. Proof. Suppose the tile is not vortex. Then two eges of ny tile must exten into the
142 Ars Mth. Contemp. 10 (2015) 135 167 Tile Configurtions n Assoite Generl Proportions Figure 9: 2 + = + = 2 Figure 9: 2+ = 3+ = + = 3 Figure 9: + + = 2 + 2 = + = + 2 Figure 9: + + = 2 + 2 = + = 2 + Figure 9e: 2+ = +3 = + 3 = Figure 9f: 2 + = + 2 = Figure 9g: 2+ = ++ 2 = + Figure 18: + = + 2 = + 2 = = 2 + = + 2 = 3 + = 2 + 2 = + 3 = 2 = 3 = 3 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 3 = 2 + not possile 3 = + 2 not possile 3 = 3 + not possile 3 = 2+2 not possile 3 = + 3 not possile 3 = 2 not possile 3 = 3 not possile 3 = 3 not possile + 2 = 2 + not possile 2 + = 2 + possile 3 = 2 + possile +2 = + 2 not possile 2+ = + 2 possile 3 = + 2 possile + 2 = 3 + = 2 2 + = 3 + possile + 2 = 2 + 2 not possile 2 + = 2 + 2 possile 3 = 3 + 3 = 2+2 = 3 2 = 2 +2 = + 3 not possile 2+ = + 3 possile 3 = + 3 possile + 2 = 2 not possile 2 + = 2 = 2 3 = 2 possile + 2 = 2 not possile 2 + = 3 possile 3 = 3 possile + 2 = 3 not possile 2 + = 3 possile 3 = 3 possile 2 = 2 + = 2 2 = 3 + = 3 2 = + 2 possile 2 = 2+2 possile = 2 2 = + 2 possile 2 = 3 + = 3 2 = 4 + = 4 2 = 2+2 possile 2 = 3+2 possile = 3 2 = 2+2 possile 2 = + 3 possile 2 = 2+3 possile 2 = + 3 possile 2 = 2 2 = 3 possile possile 2 = + 2 2 = + 3 possile possile 2 = 2 2 = 3 possile 2 = 3 = 3 2 2 = 4 = 2 = 3 2 Tle 2
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 143 Tile Configurtions n Assoite Generl Proportions Figure 10: 2 + = + = + 2 + = 2 Figure 10: 2 + = + + = + 3 + = 3 Figure 10: 2 + = + = 2 Figure 20 + + = + + = No new informtion Figure 9e: 2+ = ++ 2 = 2 + = 2 Figure 10f: 2+ = 2++ 2 = 3 + = 3 = 2 + = + 2 = 3 + = 2 + 2 = + 3 = 2 = 3 = 3 2 = 3 + = 3 2 = 2+2 possile 2 = 4 + = 4 2 = 3+2 possile 2 = 2+3 possile 2 = + 2 possile 2 = + 3 possile 2 = 4 = 2 3 = 3 + 3 = 2+2 = 3 2 = 2 3 = 4 + = 2 3 = 3+2 = 3 3 = 2+3 possile 3 = + 2 possile 3 = + 3 possile 3 = 4 = 4 3 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 2 = 3 = 3 = 3 = 3 = 3 = 3 = 3 = 3 Tle 3
144 Ars Mth. Contemp. 10 (2015) 135 167 () () () () (e) (f) (g) (h) Figure 9 skeleton of the UET4 tiling T s in Figure 11: It is pprent then tht some tile or omintion of tiles must fit extly etween the she lines shown ove. A tile is lerly to lrge to fit etween these lines, n tile nnot e ple there y unilterlity. Sine < <, then some omintion of tiles n must fit etween these lines, n sine oth n must e vorties, then in ft only two tiles my fit etween these she lines. Beuse T is unilterl, it is seen tht one tile n one tile must fit extly etween the she lines in Figure 11. However, this implies tht = + = <, ontrition. Therefore, there is no tile or omintion of tiles tht n fit extly etween the she lines ove, so must e vortex. Next, enumerte possile wys to express in terms of,, n within the + = se. Note tht < = +, n oserve tht when the tile is surroune tiles (s in Figure 12 elow), no further speifitions s to vlues of n e me. Setting this speil se sie momentrily, ontinue, using the ft tht must e vortex, to fin ll possile reltionships for se on implitions tht rise through eh of the three ses foun in Figure 13: Note tht in the ses illustrte in 13 n 13, the skeleton of the tiling T must exten long the she lines y virtue of,, n ll eing vorties. Begin with sttement implying the impossiility of the existene of the prtil oron in Figure 13 in UET4 tiling. Lemm 2.4. If + = in UET4 tiling, then eh oron will not ontin n. Proof. Let T e UET4 tiling suh tht + = n suppose tht the tile s oron ontins t lest one tile. Beuse oth n re vorties, they must meet t one of their orners s shown in Figure 13 ove; the she lines show the neessry skeleton extension of T lso require y this vortex onition. Next, etermine whih tile or omintion of tiles n fit extly etween the she lines tht exten towr the left from the union of the left eges of n. Were one tile to fill this spe, it woul hve to e tile, whih woul imply tht = + > +, ontrition. Hene more thn one tile must fill this spe. Were three tiles to fill this spe, then the mile tile in the
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 145 () () () () (e) (f) Figure 10 Figure 11: Neessry onfigurtion when is not vortex group must e non-vortex tile; euse is the only non-vortex tile in this se, one of the three tiles must e. Then regrless of the other two tiles hosen, the sum of their sie lengths will lwys exee the length + etween the she lines. Hene three tiles nnot fill this spe extly; it is ovious tht four or more tiles similrly nnot fill the spe ppropritely. This leves the se where two tiles extly fill the spe etween these she lines. Then ll possile omintions of two istint tiles re liste s follows: n ; n ; n ; n ; n ; n. Of these omintions, the only one tht overs the length + extly is the omintion n. Therefore these two tiles must e ple long the left eges of n from Figure 13, n this rrngement is shown elow in Figure 14 long with the neessry skeleton extension require y the Figure 12: surroune y s
146 Ars Mth. Contemp. 10 (2015) 135 167 () () Figure 13 () vortex onitions of n. The sme issue rises gin: tile or group of tiles must fit etween the she lines tht exten ownwr from the union of the ottom eges of tiles 1 n, n y the rgument ove, only tiles n n fill this spe extly. This logi is repete gin in Figure 14 to rrive t the prtil oron in Figure 14, n it is ler y our ssumptions tht only tile n e ple long the top ege of tile 3 to omplete the oron, whih is shown in Figure 14. By equitrnsitivity, the tiling tht this pth genertes is in ft UET2. 1 1 1 2 1 2 1 3 1 2 2 3 1 4 3 1 2 2 3 () () () () Figure 14: Ajent n tiles results in UET2 tiling Hving onlue tht nnot e in the neighorhoo of, next onsier the speifi sie lengths for when + = tht rise from the onfigurtion in Figure 13 se on the knowlege tht the vortex onitions require tht tile or tiles fit extly etween the she lines tht exten towr the left of the union of the left eges of n in this piture. Note tht one tile is too smll to fill this spe ompletely, four or more tiles re too lrge to fill this spe ompletely, n in the se where three tiles extly fill this spe, non-vortex tile must e in the mile of the group, foring one of the three tiles to e s it is our only non-vortex. Using these fts, the following tle enumertes ll possiilities where they exist. Therefore, when hs tile in its oron, the sie lengths tht must e onsiere re = 2 n = 3. The finl se to e onsiere is tht in Figure 13. Now, euse no speifi reltionships for the sie length of n e estlishe when it is surroune y only s, onsier rrngement in Figure 13 where s oron ontins tile other thn only. Without loss of generlity, suppose tht this non- tile n e foun long the right ege of in Figure 13. It is ler tht this tile nnot e tile euse the resulting tiling woul not e
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 147 All possile omintions of two or three tiles Cn this omintion fill the length in question extly? no If yes, wht oes this imply out? n n no n yes + = + = 2 + = 2 n yes + = + no new info n no n no n n yes + = 2 + = 3 + = 3 n n no n n no n n no n n no n n no Tle 4 unilterl. Lemm 2.4 implies tht this tile nnot e tile. Then the non- tile tht must e foun in s oron is. Knowing tht must e vortex, there re two rrngements tht n result from this, shown in Figure 15: () () Figure 15 The rrngement in Figure 15 proues ontrition euse it implies = + > + =, so the rrngement in Figure 15 is the only vile UET4 possiility. This rrngement implies + = + = 2 +, so = 2. Note tht this sie length ws lrey foun in Tle 4. Therefore, the only sie lengths tht must e onsiere when + = re 1. = 2 2. = 3 long with the se in whih s oron ontins only tiles, in whih se the only restrition is given y < < +. Below is summry of the sie lengths onsiere when n re jent.
148 Ars Mth. Contemp. 10 (2015) 135 167 = + is not vortex is vortex = + 1. = + + = 2 + 2 2. = + = +2 3. = + = 2+ 4. = 2 + = 3 + 5. = 2+ = 3+ 1. = 2 2. = 3 3. = 3 1. = 2 2. = 3 3 Coron Constrution With the generl UET4 prolem hving een effetively ivie into suses within whih the prolem n e ppropritely exmine, exhustive lists of ll possile orons for the four sizes of tiles when n re jent n now e rete. The proess of onstruting ll possile orons for given se is egun y reting squres of the speifie imensions. Before eginning onstrution, it shoul e note tht, for ny given tile, there exists t lest one ege tht extens into the skeleton of the tiling in no more thn one iretion, mening it is omptile with the following figure: * Figure 16 As neighorhoos re onstrute, it is ssume tht this neessry ege is the top ege of the tile in question. The illustrtion of prtil exmple of oron onstrution is now presente so s to fmilirize the reer with the generl proess use y exmining speifi suse. In orer to illustrte the proess use to rete ll,,, n orons for given set of sie length proportions, prtil exmple is now outline. Consier the se where n re jent, = +, n = + 3. The proess is illustrte here y onstruting ll possile orons for this se, s these re the most omplite orons to onstrut; it shoul e note tht,, n orons woul lso nee to e onstrute for this se. It is known tht the rrngement in Figure 16 must pper in ny oron, so the tiles tht oul e ple in the mrke orner in tht figure re first onsiere. An,, or tile oul e ple there, reting three rnhes shown in Figure 17 tht will eh e onsiere in turn.
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 149 () () () Figure 17: Three rnhes to onsier for ll possile orons Tke first the rrngement in Figure 17; the oron is onstrute y pling tiles roun this enter in lokwise iretion. As the remining length long the top ege of is 3, pling n or next woul violte vortex onitions. Hene or n ome next in the oron, reting two new rnhes shown elow. () () Figure 18: Two rnhes from Figure 17 Consier the rrngement in Figure 18. The length remining long the top ege of is 2, so neither n nor n ome next ue to vortex onitions. A is lso not llowe y unilterlity. A tile must ome next, shown in Figure 19. Now moving to the right ege of the enter tile, it is evient tht tile nnot ome next y unilterlity; gin, vortex onitions sy tht nor n e in this next spe either. A tile must ome next, shown in Figure 19. There is now istne of 2 long the remining right ege of the enter tile, so it is gin onlue tht only tile oul ome next, seen in Figure 19. Similr logi is employe to onlue tht, ontinuing to move lokwise roun the enter tile, the remining sies re overe y, then, then, then. This rrngement is shown in Figure 19. However, this ontrits the vortex onition on the tile, so this rrngement is invli. Hving exhuste ll possiilities tht oul rise from the rrngement in Figure 18, it is onlue tht no vile orons ome from this rnh, n ttention is next given to the rrngement in Figure 18. In orer to minimize the retion of orons tht re ientil up to yli permuttions of their signture, the rrngement in Figure 18 is heneforth never onsiere long ny ege of the enter tile.
150 Ars Mth. Contemp. 10 (2015) 135 167 () () () () Figure 19: Suessively filling the prtil oron in Figure 19 Coron onstrution proeeing from the rnh in Figure 18 ontinues long the left ege of the lower tile, moving lokwise s lwys. Given this rrngement, it is possile to stisfy vortex onitions if n or were ple next in the oron (note tht the sie length = 2 is require if this next tile is ; n is require long the remining ege of this in orer to mke it vortex). A tile oul ome next s well, ut tile nnot y unilterlity. This les to three itionl rnhes, shown in Figure 20. () () () Figure 20: Three rnhes from Figure 18 Consier now the rrngement in Figure 20. Next, vortex onitions eliminte n or, so there re two options for the next tile in the oron, nmely or s shown in Figure 21. Rell tht the rrngement in Figure 21 ontins on its right sie prtil oron for whih ll possiilities were previously exhuste; this rnh oes not nee to e reonsiere. () () Figure 21: Two rnhes from Figure 20
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 151 Consier the rrngement in Figure 21. A tile is the only one tht oul not ome next in the sequene, y unilterlity. The three rnhes in Figure 22 result (note tht the onition = 2 is gin invoke in Figure 22). () () () Figure 22: Three rnhes from Figure 21 Consier the rrngement in Figure 22. In hoosing the next tile, vortex onitions eliminte or, so or oul ome next in the oron, ut s pling in this position woul le to yli permuttion of n rrngement previously exhuste, only the se where omes next is onsiere. Following this (seen in Figure 23), tile,, or oul e ple ( eing isllowe y unilterlity). These three rnhes re shown in Figures 23, 23, n 23. () () () () Figure 23: Prtil oron... n three resulting rnhes As pling next in the prtil oron of Figure 23 hs een shown to e n impossile rrngement (see the rnh from Figure 18), only oul omplete this oron ppropritely. This gives the omplete oron signture... Next, the vortex onition on the first tile in Figure 23 woul require tht omintion of tiles fit extly long the remining istne 2 + 2 long the left ege. Tile lengths require tht this inlue t lest one tile, whih woul violte the vortex onitions. Hene the rrngement in Figure 23 oes not le to vile oron. Finlly, the prtil oron of Figure 23 n only e omplete with, gin violting the vortex onition on ; no vile orons result from this rrngement. Now, ll rrngements rnhing from tht in Figure 23 hve een exhuste. Figure 22 is the rnh tht shoul e returne to next. The prtil exmple outline in this setion is illustrte y tree igrm in Figure 24; this shows rnhes of ll possiilities onsiere long with whih rnhes proue
152 Ars Mth. Contemp. 10 (2015) 135 167 vile UET4 orons n whih re unfruitful. The tree igrm shows ll rnhes tht rise from pling n tile in the steriske position in Figure 16; those rising from pling or in tht position re not inlue in the igrm for revity s ske. Figure 24: A tree igrm to ompny the illustrtion of oron onstrution foun in Setion 4. All oron possiilities illustrte in Setion 4 re piture here; those omitte re not feture here either. Corons in lk retngles re vile UET4 orons; those in re retngles (full or prtil) o not le to vile UET4 orons. The otte lines mrk where step or series of steps (for eh of whih only one possile tile oul pper next) in the onstrution proess hve een omitte for revity s ske. 4 Constrution of the UET4 Tilings from Vile Corons Hving ompile list of ll possile,,, n orons for eh of the ses when n re jent, it remins to etermine whih omintions of these orons n e omine to generte UET4 tiling. The extension of Shttshneier s metho of fining orons tht orrespon to tiling is illustrte with n exmple. Consier the se where n re jent, = +, n = 3. Using the proess expline erlier, the orresponing set of vile orons re:
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 153 orons orons orons orons....................................................................................................................................................................................................................................................................................................................................... Figure 25: An oron with signture... Choose the oron..., s illustrte in Figure 25. Oserve tht euse of the tiles neighoring the tile in Figure 25, n euse T is equitrnsitive, the oron for
154 Ars Mth. Contemp. 10 (2015) 135 167 T must ontin the prtil oron... Likewise, the oron for T must ontin the prtil orons.. n.. n the oron for T must ontin the prtil oron... Fining the prtil orons tht re omptile with our hoie of oron is esily utomte. For exmple, to fin the first prtil oron, fin the first instne of in the oron,... A prtil oron signture for woul then e the letter ylilly preeeing this instne of, followe y (sine this prtil signture is tken from n oron signture), followe y the letter ylilly following this instne of, yieling the prtil oron signture... This proess n e repete for eh ourrene of n for n s well. Note tht prtil oron signtures tht re yli permuttions of eh other re onsiere equivlent, n so re reverse orerings. Performing suh serh for prtil,, n oron signtures orresponing to the initil hoie of oron signture,..., yiels the following. A B C D oron Prtil orons Prtil orons Prtil orons......... Serh the list of full orons for ny whih ontin the prtil oron... In this exmple, fter the serh is performe, the five mthing orons inlue...,...,...,... n... Use these orons to rete five orresponing 2-tuples of omptile n oron signtures (e.g. (...,...), (...,...), et). For eh new 2-tuple (x, y), the oron s orresponing prtil, n orons to their respetive olumns s in Tle 5. A B C D E (x, y) prtil prtil prtil prtil Tuple orons in (x, y) orons in (x, y) orons in (x, y) orons in (x, y) 1 (...,...........).. 2 (...,...)................ 3 (...,...) 4 (...,...) 5 (...,...).............................. Tle 5.......... While eh of the orons in (x, y) ontin the neessry prtil oron signture.., some my ontin other prtil signtures tht re not omptile with the originl hoie of signture in (x, y). For exmple, the oron of (x, y) in line 2,...,
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 155 hs the prtil oron.., whih is not ontine in the originl oron,... So Line 2 is eliminte from further onsiertion. Next, for eh surviving tuple (x, y) in Tle 5, serh the list of full orons for those tht ontin the tuple s orresponing prtil orons. For exmple, the tuple in line 4 hs prtil orons..,..,.. n... All of these prtil orons re ontine only in full oron... For eh full oron tht is omptile with our 2-tuple (x, y), rete new 3-tuple (x, y, z) y ppening omptile oron signture, s in olumn A of Tle 6. Also, list the prtil,,, n orons tht re ontine in (x, y, z) (olumns B - E). There re now nineteen vile (x, y, z) tuples. For eh of these 3-tuples, hek if ll of the prtil n orons re ontine in the full n orons in the tuple. If not, remove tht tuple. For exmple, onsier the 3-tuple (...,...,...) of Line 4... is liste s prtil oron, ut.. is not ontine in this tuple s full oron, whih is... Therefore, elete line 4. After performing this hek for ll of the tuples, there re only three tuples whih pss the test, shown in Tle 7. Finlly, serh the list of full orons tht ontin ll of the prtil orons for eh (x, y, z) 3-tuple to rete new list of 4-tuples (x, y, z, w) where w is oron tht is omptile with the 3-tuple (x, y, z). For this exmple, this is Column A of Tle 8. A to this tle olumns ontining the prtil,,, n orons ontine in (x, y, z, w) whih will e use to hek the viility of (x, y, z, w) s efore. For eh tuple, hek if ll of the prtil, n orons re ontine in the full, n orons in the tuple. For the tuple in line 1, this is not the se. Prtil oron.. is not ontine in full oron... The tuple in line 2 psses the test. At this point, from our originl onitions n hoie of oron signture, there remins only one omintion of,,, n orons tht my result in tiling or tilings. When this proess is utomte n performe for ll possile ses (e.g. 2 + = )) n hoies of oron signtures, the following list of 4-tuples (x, y, z, w) is generte. The proess of onstruting tiling is emonstrte using 4-tuple from Tle 9. Consier the 4-tuple (...,...,...,...), for whih = 2, = +, n = 2 +. It is known tht = 2 euse it is neessry onition for tile to hve the oron... The tiles n their orons re isplye in Figure 26............. Figure 26: (...,...,...,...), for whih = 2, = + n = 2 + There re mny wys to onstrut tiling. The en gol is to rete pth tht will tile the plne. For exmple, strt with the tile n its oron. Complete the orons of the tiles whih surroun the using refletions n rottions of the orons in Figure 26. This
156 Ars Mth. Contemp. 10 (2015) 135 167 A B C D E (x, y, z) Tuple prtil orons in prtil orons in prtil orons in prtil orons in (x, y, z) (x, y, z) (x, y, z) (x, y, z) 1 (...,...,...)...................... 2 (...,...,...) 3 (...,...,...) 4 (...,...,...) 5 (...,...,...) 6 (...,...,...) 7 (...,...,...) 8 (...,...,...) 9 (...,...,...) 10 (...,...,...) 11 (...,...,...) 12 (...,...,...) 13 (...,...,...) 14 (...,...,...) 15 (...,...,...) 16 (...,...,...) 17 (...,...,...) 18 (...,...,...) 19 (...,...,...).......................................................................................................................................................................................... Tle 6..............................................................................................................................................................................................
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 157 A B C D E (x, y, z) Tuple prtil orons in (x, y, z) prtil orons in (x, y, z) prtil orons in (x, y, z) prtil orons in (x, y, z).... 1 (...,...,...) 2 (...,...,...) 3 (...,...,...)........................ Tle 7.................................... A B C D E (,,, ) Tuple Tuple s prtil Tuple s prtil Tuple s prtil Tuple s prtil orons orons orons orons.............. 1 (...,...,...,...) 2 (...,...,...,...)............................ Tle 8.............................. results in the pth illustrte in Figure 27.
158 Ars Mth. Contemp. 10 (2015) 135 167 4-tuple (...,...,...,...)* (...,...,...,...)* (...,...,...,...)* (...,...,...,...)* (...,...,...,...)* (...,...,...,...)* (...,...,...,...)* (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...) (...,...,...,...)** (...,...,...,...)** (...,...,...,...)** (...,...,...,...)** (...,...,...,...)** Proportions = + = + = + = + = + = + = 2; = + = 2 + 2 = 2 + 2 = 3; = 2 + = 3 = + 2 = + 2 = + 2 = + 2 = + 2 = + 2 = + 2 = 2; = + 2 = + 2 = + 2 = 2 + = 2; = 2 + = 2 + = 2; = 2 + = 2 + = 2; = 2 + = 2 + = 2; = 2 + = 2 + = 2 + = 2 + = 2 + = 2 + = 2 + = 2 + = 2 + = 2 + = 2 + = 2 + Tle 9: All 4-tuples generte when + =. Note tht ll mrke with (*) were lrey foun in Setion 2.1. All mrke with (**) nnot e extene to rete tiling of the plne.
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 159 Figure 27: A seon lyer of the oron for (...,...,...,...). By ing tiles in similr mnner, suh tht the orons re refletions n rottions of those in Figure 26, s well s eleting tiles where neessry, one will esily fin pth whih n tile the plne unilterlly n equitrnsitively using trnsltions. Suh pth is illustrte in Figure 28. Figure 28: A pth whih will tile the plne unilterlly n equitrnsitively y wy of trnsltions. However, for some 4-tuples, it will soon eome ler tht no tiling is possile. For exmple, onsier the 4-tuple (...,...,...,...), for whih = + n = + 2. If one ttempts to expn on the oron in similr mnner s ove - y ompleting inomplete orons while hering to the orering presrie y the 4-tuple - one will enounter the pth in Figure 29. * * * * Figure 29: A pth whih nnot e extene for 4-tuple (...,...,...,...), for whih = + n = + 2.
160 Ars Mth. Contemp. 10 (2015) 135 167 The (*) represent prolem res. To here to equitrnsitivity, one must ple tile in these spots. Oviously, this is impossile. For eh of the five tilings mrke with (**) in Tle 9, pth whih oul not e extene ws inevitle. 5 n re not jent In this se, there re six possile orons. These re illustrte in Figure 30. By repling the tiles with tiles in Figure 30, it is ler tht there re lso extly six orons when n re not jent. Lemm 5.1 elimintes one of these six suses of orons n one of these six suses of orons from onsiertion in UET4 tilings. 4 1 1 1 1 2 1 1 1 2 2 2 3 3 1 () () () 2 1 1 3 1 4 1 1 1 2 2 1 2 1 3 () (e) (f) Figure 30: All possile orons when n re not jent Lemm 5.1. Let T e UET4 tiling in whih n re not jent. Then orons n orons nnot ontin only tiles. Proof. Suppose tht the oron n ontin only tiles s shown in Figure 31 elow. At lest one orner forme y two tiles must ontin non- tile; otherwise the resulting tiling woul e UET2. Without loss of generlity, suppose tht this require orner is tht mrke y the sterisk in Figure 31; it will e etermine whih tiles n e ple in the orner mrke y the sterisk. Were tile to e ple here, then this tile woul overhng pst the right ege of 1. Sine every tile is vortex, the length of this overhng must e overe extly y tile (or tiles), n the only tile tht n over this length while mintining the pproprite reltive sie lengths of,,, n is n tile. However, this woul ontrit n not eing jent. Therefore tile must fill this spe s shown in Figure 31. Next, it is etermine whih tiles oul e ple in the orner mrke y the sterisk in FIgure 31. Were n tile to e ple here, there woul e tile in its oron n the tiling woul ese to e equitrnsitive y the ssumption tht s oron ontine only tiles. A tile nnot fill the steriske orner y unilterlity, so the two ses shown in Figures 31 n 31 must e onsiere, strting with Figure 31.
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 161 2 3 * 1 1 4 2 3 1 * 1 2 1 1 2 4 3 1 1 4 () () () * 1 1 5 1 2 2 1 1 3 1 2 3 1 1 4 1 1 3 2 4 1 1 4 () (e) (f) Figure 31: Progressively uiling ll possile orons The vertil istne remining long the left ege of tile 2 in this figure is of length ; hene the only tile tht oul ppropritely fill this remining ege length is n tile. However, the oron of this new tile woul ontin tile, ontrry to hypothesis. Therefore the rrngement of tiles in Figure 31 oes not give rise to UET4 tiling. Next onsier se illustrte in Figure 31. Note tht, s piture, the top ege of tile 1 must line up with the top ege of 1. Were this not the se, either 1 woul ese to e vortex or woul e fore to hve n tile in its neighorhoo. Thus, n tile nnot e ple in the orner mrke y the sterisk euse n nnot e jent; neither n tile e ple there y the unilterlity onition. This leves the two ses shown in Figure 31e n 31f. In oth of these ses, the remining vertil istne long the left ege of 2 is of length, so the only tile tht oul fill this spe is n tile. However, the istne tht the ottom ege of 5 in Figure 31e n the ottom ege of 2 in Figure 31f hng over the left ege of 2 is, in oth ses, stritly greter thn the length. Hene n tile ple long the remining left ege of 2 woul not e vortex. Therefore the rrngement of tiles foun in Figure 31 oes not give rise to ny UET4 tilings. A nerly ientil rgument shows tht tile nnot e surroune y only tiles. Beuse neither n tile nor tile n hve oron ontining only tiles, then eh of their orons must ontin tile. An immeite orollry to this is tht eh oron must ontin t lest one tile n t lest one tile. Lemm 5.1 implies tht there re five possile orons n five possile orons when n re not jent. Lemm 5.1 illustrtes the nlysis of only one possile oron, ut there re 5 more orons to onsier, 6 more orons, n severl possile n orons to onsier. Beuse eh possile oron involves exhustive exmintion, it woul e imprtil to present suh n nlysis in short rtile. However, the following exmple illustrtes the methoology use to eie if given oron is vile. Consier the oron of Figure 30. This oron is reprinte in Figure 32. Proee y onstruting ll possile orons tht rise from this rrngement y pling tiles long the eges of the tile 2 in Figure 32 moving in lokwise iretion. There re three suses here to onsier:
162 Ars Mth. Contemp. 10 (2015) 135 167 > +, = +, n < +. In this exmple, only the suse = + is emonstrte. To egin enumertion of ll possile orons tht n rise from the rrngement in Figure 32, first etermine whih tiles oul e ple in the orner mrke with n sterisk. An tile nnot e ple there euse the vortex restrition on tiles woul imply tht =. A tile nnot e ple there y unilterlity. This gives us two options to onsier: tile 1 n e ple there or tile 3 n e ple there. These two options re shown in Figures 32 n 32. Note tht for the se shown in Figure 32, the vortex onition on tiles n the ft tht tiles n tiles nnot e jent requires tht + = +. 1 1 1 1 2 2 2 2 1 1 1 1 1 1 1 1 2 * 2 1 * 2 * 3 * 2 3 3 () () () () Figure 32 In full nlysis, oth of the rrngements in 32 n 32 ove nee to e onsiere, ut for the purposes of this exmple, onsier only how to fill in the steriske orner in Figure 32. A tile nnot e ple there y unilterlity; however, n,, or oul e ple there uner the pproprite onitions. Eh of these options nees to e onsiere. Exmine the rrngement, shown in Figure 32, where tile 3 fills the steriske orner. Agin, it must e etermine whih tiles n e ple in the steriske orner in Figure 32. Were tile ple there, the onition = + implies tht the top ege of will hve overhng pst the left ege of 2 ; then in orer for s vortex onition to e stisfie, tile woul hve to e ple long the remining top ege of. This ontrits n not eing jent. A tile nnot e ple in the steriske orner y unilterlity. This leves two options to onsier. The se n tile in this position is seen in Figure 33, n the se where tile in this position is seen in Figure 34. 1 1 1 2 2 2 * 1 2 2 3 1 3 1 * 2 4 2 3 1 3 5 4 1 2 2 3 1 3 () () () Figure 33 Consier first the rrngement shown in Figure 33. By equitrnsitivity of, tile 4 must e ple in the steriske orner so tht its oron will mth tht of 1. This
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 163 is shown in Figure 33. The steriske orner Figure 33 n only e fille y tile 4, shown in Figure 33; n re not llowe there y the vortex onitions, n is not llowe y unilterlity. Then Figure 33 shows omplete oron. However, sine this oron ontins no tile, this will not result in UET4 tiling s result of orollry to Lemm 5.1 n is hene not vile oron. 1 1 1 * 2 1 2 1 3 4 * 2 1 2 1 3 5 4 2 1 2 1 3 3 3 3 4 4 4 () () () Figure 34 Next, onsiering the rrngement shown in Figure 34, it must e etermine whih tiles n e ple in the orner mrke y n sterisk. Vortex onitions prohiit n or from eing ple there, n tile is lso not llowe y unilterlity. Then the only option is to ple tile 4 in this position, shown in Figure 34. The steriske orner in this figure n only e fille y tile 5, s shown in Figure 34; n re not possile y vortex onitions n is not possile y unilterlity. Figure 34 shows omplete oron. However, euse this oron oes not ontin tile, it is not omptile with UET4 tiling. The next step woul e to onsier the possiilities when n tile or tile re ple in the position oupie y 3 in Figure 32. The metho ontinues in this fshion, enumerting ll possile tiles tht n e ple in lotion, moving roun tile, reting rnhing list of ll orons n weeing out orons tht re known to e impossile uner our onstrints. This is one for the three suses > +, = +, n < + for eh of the five possile orons in Figures 30-30f, n it seen tht = + is the only se tht yiels vile orons. It shoul lso e note tht in the se where n tile is surroune y three tiles n one tile, s shown in Figure 30, the sme metho use to uil roun tile inste of tile. It shoul lso e note tht, t times, it is neessry to speify ertin sie lengths for n tiles in terms of sie lengths of smller tiles in orer for ertin rrngements to e vile. This llows for further flexiility in oron onstrution n ensures tht ll possile potentilly vile orons re foun. Summrizing, the riteri use throughout this metho re s follows: 1. Equitrnsitivity of the tiling T. When n tile is ple in the oron of lrger tile, it is possile to ontinue uiling roun the lrger tile using the knowlege tht every neighorhoo must e ientil to tht of the lrey estlishe 1 oron. It is possile tht t times the only option is to ple tile within n tile s neighorhoo tht mkes it inomptile with the originl 1 oron. In this se, the metho n e ene on this rnh, s it will not yiel ny vile or orons.
164 Ars Mth. Contemp. 10 (2015) 135 167 2. Unilterlity of the tiling T. 3. Vortex onitions on n tiles. 4. The requirement tht tiles n tiles re not jent. 5. Reltive sizes of tile sie lengths: < < <. 6. Eh oron must ontin t lest one tile n t lest one tile. This metho essentilly egins with one of the five vile orons shown in Figure 30 n then employs the oron onstrution lgorithm outline previously in Setion 3, uiling roun tile in the oron (speifilly, roun tile in Figures 30, 30, n 30e or roun tile in Figures 30 n 30) until either ontrition is rehe for prtiulr rnh or full oron is rehe. One the onstrution proess hs een omplete uiling roun the hosen tile in the oron of the originl tile, one is left with n exhustive list of ll orons (or orons, epening on the oron from whih onstrution egn) tht re omptile with the originl oron. Then, using equitrnsitivity, the orons of ll new n tiles (or tiles, gin epening on the se) n e omplete, expning the pth until one n either estlish tht no UET4 tiling n result (ue to filure of equitrnsitivity, overlpping or gps etween tiles, ontrition of vortex onitions, et.) or until full n orons (or n orons) re foun. Note tht multiple tilings my result from the sme set of n orons (or n orons), s there my e multiple wys to tile the originl pth using these orons. Tle 10 lists the eleven UET4 tilings foun using this metho n the neessry sie length proprtions require for the tiling to e generte. ( oron, oron, oron, oron) Sie Reltions (...,...,...,...) = + ; = + (...,...,...,...) = + ; = + (...,...,...,...) = + ; = + (...,...,...,...) = 2; = + (...,...,...,...) = + ; = + (...,...,...,...) = + ; = + (...,...,...,...) = 2; = + (...,...,...,...) = + ; = + (...,...,...,...) = 2; = + ; = + (...,...,...,...) = 2; = 2; = + (...,...,...,...) = 2; = + Tle 10 Illustrtions of the eleven tilings when n re not jent n e seen in the finl setion. This onlues the se where tiles n tiles re not llowe to e jent.
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 6 165 The 39 UET4 Tilings (...,...,...,...) =+ =+ (...,...,...,...) = 2 =+ (...,...,...,...) =+ =+ (...,...,...,...) =+ =+ (...,...,...,...) =+ =+ (...,...,...,...) =+ =+ (...,...,...,...) = 2 =+ =+ (...,...,...,...) = 2 =+ (...,...,...,...) = 2 = 2 =+ (...,...,...,...) = 2 =+ (...,...,...,...) =+ =+ Below re the 28 tilings when n re jent. For ll, = +. = + oes not generte ny tilings. (..., (..., (..., (...,...,...,...,...,...,...,...,...,...)...)...)...) = 3 = 2 + 2 = 3 = 2 + 2 = 2 + (..., (...,...,...,...,...,......).) = + 2 = + 2
166 Ars Mth. Contemp. 10 (2015) 135 167 (..., (...,...,...,...,...,......).) = + 2 = + 2 (...,...,...,...) = + 2 (...,...,...,...) = + 2 (...,...,...,...) = + 2 (...,...,...,....) = 2 = + 2 (...,...,...,....) = + 2 (...,...,...,...) = 2 = 2 + (...,...,...,...) = 2 = 2 + (...,...,...,...) = 2 + (...,...,...,...) = 2 = 2 + (..., (...,...,...,...,...,...)...) = 2 = 2 + = 2 + (...,...,...,...) = 2 + (...,...,...,...) = 2 = 2 + (...,...,...,...) = 2 + (...,...,...,...) = 2 = 2 + (...,...,...,...) = + 2 (...,...,...,...) = 2 + (...,...,...,...) = 2 + (...,...,...,...) = 2 + (...,...,...,...) = 2 = 2 +
C. Mnn et l.: Unilterl n equitrnsitive tilings y squres of four sizes 167 Referenes [1] G. Grünum n G. C. Shephr, Tilings n Ptterns, Freemn, New York, 1987. [2] H. Mrtini, E. Mki n V. Soltn, Unilterl tilings of the plne with squres of three sizes, Beiträge Alger Geom. 39 (1998), 481 495. [3] D. Shttshneier, Unilterl n Equitrnsitive Tilings y Squres, Disrete Comput. Geom. 24 (2000), 519 526.