Managing Editor Carlos Pereira dos Santos, Center for Linear Structures and Combinatorics, University of Lisbon

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2 Editin Assciaçã Ludus, Museu Nacinal de História Natural e da Ciência, Rua da Escla Plitécnica Lisba, Prtugal rmm@ludus-puscula.rg URL: Managing Editr Carls Pereira ds Sants, cmfsants@fc.ul.pt Center fr Linear Structures and Cmbinatrics, University f Lisbn Editrial Bard Alda Carvalh, acarvalh@adm.isel.pt, ISEL & CEMAPRE Aviezri S. Fraenkel, Aviezri.Fraenkel@weizmann.ac.il, Weizmann Institute Carls Pereira ds Sants Clin Wright, juggler@slipsys.c.uk, Liverpl Mathematical Sciety David Singmaster, zingmast@gmail.cm, Lndn Suth Bank University David Wlfe, davidgameswlfe@gmail.cm, Gustavus Adlphus Cllege Edward Pegg, ed@mathpuzzle.cm, Wlfram Research Jã Pedr Net, jpn@di.fc.ul.pt, University f Lisbn Jrge Nun Silva, jnsilva@cal.berkeley.edu, University f Lisbn Keith Devlin, kdevlin@stanfrd.edu, Stanfrd University Richard Nwakwski, rjn@mathstat.dal.ca, Dalhusie University Rbin Wilsn, r.j.wilsn@pen.ac.uk, Open University Thane Plambeck, tplambeck@gmail.cm, Cunterwave, inc Infrmatins The Recreatinal Mathematics Magazine is electrnic and semiannual. The issues are published in the eact mments f the equin. The magazine has the fllwing sectins (nt mandatry in all issues): Articles Games and Puzzles Prblems MathMagic Mathematics and Arts Math and Fun with Algrithms Reviews News ISSN

3 Cntents Page Games and Puzzles: Gerge I. Bell Designing peg slitaire puzzles Articles: Maya Mhsin Ahmed Demystifying Benjamin Franklin s ther 8-square Games and Puzzles: Maurizi Palini Eplring the Rubik s magic universe Mathmagic: Ry Quinter On a mathematical mdel fr an ld card trick Games and Puzzles: Stephen B. Gregg, Britney Hpkins, Kristi Karber, Thmas Milligan, Jhnny Sharp Several bunds fr the k-twer f Hani puzzle Games and Puzzles: Urban Larssn, Israel Rcha Eternal Picaria

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5 Games and Puzzles Designing peg slitaire puzzles Gerge I. Bell Tech-X Crpratin Abstract: Peg slitaire is an ld puzzle with a 300 year histry. We cnsider tw ways a cmputer can be utilized t find interesting peg slitaire puzzles. It is cmmn fr a peg slitaire puzzle t begin frm a symmetric bard psitin, we have cmputed slvable symmetric bard psitins fr fur bard shapes. A new idea is t search fr bard psitins which have a unique starting jump leading t a slutin. We shw many challenging puzzles uncvered by this search technique. Clever slvers can take advantage f the uniqueness prperty t help slve these puzzles. Keywrds: Slitaire puzzles, game design. Intrductin Peg slitaire was invented in France in the late 17th century, where it started an early puzzle craze. Tday mst peple recgnize the puzzle, althugh its ppularity has declined. We will refer t a bard lcatin as a hle, which can either be empty r ccupied by a peg. Figure 1 shws three peg slitaire bards the first tw bards are based n a square lattice f hles, while the third is based n a triangular lattice. While the first tw bards are cmmn, the 37-hle heagn bard is nt. Pressman Ty Cmpany has manufactured this bard under the name Think N Jump, althugh it is nt identical since they remved sme f the uter jumps (ne can still play n this bard by allwing these jumps). The puzzle begins frm sme specified pattern f pegs, three eamples are shwn in Figure 1. A jump cnsists f ne peg jumping a neighbr int an empty hle, the jumped peg is remved. Jumps are allwed nly alng lattice lines, i.e. alng clumns and rws fr the first tw bards, and alng three lines n the heagn bard (as indicated by the arrws). The gal f the puzzle is t chse Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

6 6 designing peg slitaire puzzles jump directins Figure 1: Sample puzzles n the 33-hle English bard, the 37-hle French bard and the 37-hle heagn bard. a sequence f jumps which finish at a bard psitin with ne peg. The hle where the final peg ends at is called a finishing hle. A bard psitin where there eists a sequence f jumps ending with ne peg is said t be slvable. Figure 1 a shws the starting bard psitin f a special puzzle called the central game. The starting psitin has square symmetry, and the gal is t finish with ne peg in the center, a bard psitin which nt nly has square symmetry, but is als the cmplement f the starting psitin (where each peg is replaced by a hle, and vice-versa). If we take either f the 37-hle bards, and fill them with pegs, but remve the central peg, then we are in an unslvable bard psitin (we will prve this). Therefre, the analgus central game is unslvable n these tw bards, but many ther symmetric bard states are slvable. The cnfiguratin shwn in Figure 1b, fr eample, is slvable. This bard psitin is symmetric with respect t reflectin abut bth diagnals. Finally, the bard psitin in Figure 1c is slvable and is symmetric with respect t 60 rtatins. If yu slve this puzzle yu will discver that the finishing hle is nt the central hle. Fr the English and French bards, there are a ttal f seven symmetries pssible fr a cnfiguratin f pegs, summarized in Table 1. These crrespnd t subgrups f D 8, the dihedral grup f rder 8 (the symmetries f the square). We use the terminlgy that an rthgnal reflectin is ne that ccurs alng lattice lines, while a diagnal reflectin des nt. The English central game is interesting because the bard begins and ends at psitins with square symmetry. Jhn Beasley prved [1] that n matter hw the central game is slved, the bard cannt pass thrugh an intermediate psitin with square symmetry (type 1) r 90 degree rtatinal symmetry (type 2). We Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

7 Gerge I. Bell 7 Type Symmetry descriptin Order Eamples 1 square symmetry 8 Fig. 1a, 7a 2 90 rtatin 4 Fig. 5c, 7b 3 bth diagnal reflectins 4 Fig. 1b 4 bth rthgnal reflectins 4 Fig. 5a, 7c rtatin 2 Fig. 5d 6 ne diagnal reflectin 2 Fig. 7d 7 ne rthgnal reflectin 2 Fig. 5b Table 1: The seven pssible symmetries fr an English r French bard psitin. nte that this des nt mean that a square symmetric bard psitin cannt be reached starting with the centre vacant, nly that if a square symmetric psitin is reached, it is nt slvable. In what fllws we will determine what types f symmetric bard psitins can appear during a slutin t the central game. Psitin class thery Given a bard psitin, if we determine its psitin class we will knw which finishing hles are pssible. We begin with the English and French bards by labeling the hles diagnally with the numbers 0-2, and again with 3-5, as shwn in Figure Figure 2: Diagnal labeling f hles fr the English and French bards (left), and the heagn bard (right). Let N i be the number f pegs in the hles labeled i. We nw bserve what happens t N 0, N 1, N 2 after a peg slitaire jump is eecuted. One f the three increases by 1, while the ther tw decrease by 1. Therefre, if we add any tw f N 0, N 1, N 2, the parity f the sum can never change as the game is played. Fr eample, (N 1 + N 2 ) md 2 is an invariant f the game, its value can nly be 0 r 1. The same hlds fr N 3, N 4, N 5, s the binary 6-tuple N = (N 1 + N 2, N 0 + N 2, N 0 + N 1, N 4 + N 5, N 3 + N 5, N 3 + N 4 ) (1) is an invariant f the game (here each cmpnent f N is taken mdul 2). The siteen values f N separate all bard psitins int siteen equivalence classes [1], which we call psitin classes. A peg slitaire game is played Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

8 8 designing peg slitaire puzzles entirely in ne psitin class, s it is interesting t figure ut which psitin classes have representatives with ne peg. The psitin class f ne peg in the centre is an imprtant ne and we call it psitin class A, it crrespnds t N = (0, 1, 1, 0, 1, 1). The reader can check that all three bard psitins in Figure 1 are in psitin class A 1. A A A A B B B C C A B C C Figure 3: The three pssible patterns fr finishing hles n the English r French bards. Nte that N = (0, 1, 1, 0, 1, 1), (1, 1, 0, 1, 1, 0) and (0, 1, 1, 1, 0, 1), respectively. Figure 3 shws the pssible finishing hles fr any bard in psitin class A, as well as the ther ne-peg psitin classes B and C. Figure 3 shws that there are essentially nly three pssible patterns fr the finishing peg, up t rtatins and reflectins. It shuld be nted that the hles labeled A in Figure 3a are nly a necessary cnditin fr a ne-peg finish. If we are in psitin class A, the nly pssible finishing hles are thse marked by A s. Hwever, it may nt be pssible t finish with ne peg at all, r nly at sme A s. The symmetry f the three patterns in Figure 3 turns ut t be very imprtant fr what fllws. We nte that the pattern f A s has square symmetry (type 1), while the B s and C s have ne reflectin symmetry (types 7 and 6, respectively). One psitin class t be avided is the psitin class f the empty bard, N = (0, 0, 0, 0, 0, 0). The reasn t avid it is that this psitin class has n representatives with ne peg, s any bard in the psitin class f the empty bard is unslvable. Fr eample, if we take either f the 37-hle bards and fill the bard, then remve the central peg, we are in the psitin class f the empty bard and therefre in an unslvable bard psitin. The same idea can be applied t bards n a triangular lattice. On the 37-hle heagn bard we use the hle labeling f Figure 2c, the vectr N has nly three cmpnents, and there are nly fur psitin classes (fr details see [5]). Three f the fur psitin classes have representatives with ne peg, the eceptin being the psitin class f the empty bard. The three pssible patterns fr finishing hles are given in Figure 4. As befre, the psitin class f ne peg in the centre is psitin class A. We nte that patterns B and C are related by a reflectin abut the y-ais, s there are 1 Fr (a), ne shuld find N 0 = N 3 = 10, N 1 = N 2 = N 4 = N 5 = 11, fr (b), N 0 = 2, N 1 = N 2 = 5, N 3 = 6, N 4 = N 5 = 3, and (c), N 0 = 1, N 1 = N 2 = 6. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

9 Gerge I. Bell 9 A A A A A A A A A A A A A B B B B B B B B B B B B C C C C C C C C C C C C Figure 4: The three pssible patterns fr finishing hles n the heagn bard. essentially tw patterns fr the finishing hles. Symmetric bard psitins We dente a bard psitin by a lwer case letter, while sets f bard psitins will be dented by upper case letters. If b is a bard psitin then we dente the cmplement f b by b, and if B is a set f bard psitins, B is the set f cmplemented bard psitins (b B if and nly if b B). The English and French bards Suppse we begin frm an English r French bard psitin b which is square symmetric (type 1) and slvable. Where can the final peg be? A pwerful bservatin is that the set f finishing hles must have the same symmetry type as b itself. S n ne hand we knw the set f finishing hles must be square symmetric, and we als knw it can nly be a subset f ne f the three patterns in Figure 3. But nly finishing pattern A is square symmetric, s we must be in psitin class A, and we can nly finish in the hles marked A. Nt nly that, if the finishing hle is ne f thse alng the edge f the bard, then by reversing the directin f the last jump, we can always finish in the centre. This same argument wrks fr any symmetry f type 1-5, s we have prved: Therem 1. On the English and French bards, if a bard psitin is slvable and has symmetry type 1-5, then it lies in psitin class A and is slvable t the centre. It is critical in Therem 1 that the bard be slvable. If a bard psitin is nt slvable, then the set f finishing hles is empty, which is trivially a square symmetric pattern. An unslvable bard may be in the psitin class f the empty bard. In fact, if we take any bard psitin in psitin class A, and remve r add the centre peg (depending n whether it is present r nt), we are in the psitin class f the empty bard, and therefre nt slvable. What happens if the bard psitin has nly a single reflectin symmetry (type 6 r 7)? In that case it may be in psitin class A, B r C, and it may finish smewhere ther than the centre. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

10 10 designing peg slitaire puzzles Therem 1 tells us that all slvable symmetric psitins (types 1-5) are slvable t ne peg in the centre. This suggests that we can find them all by playing backward frm ne peg in the centre. Playing peg slitaire backward is equivalent t playing frward frm the cmplement bard psitin [2, 4], s here is a simple algrithm fr calculating them: Let b 1 be the bard psitin with a full bard with the centre peg missing (Figure 1a fr the English bard), and define B 1 = {b 1 }. Nw define B n+1 as the set f all bard psitins which can be reached frm any bard psitin in B n by eecuting any single jump. Nte that by design, B n is the set f n-peg bard psitins slvable t the centre. We nw search thrugh B n fr bard psitins with varius symmetries. The set f all n-peg slvable psitins with type j symmetry can be fund by searching B n fr bard psitins with type j symmetry. Fr details n hw these calculatins are dne, see [7]. We d nt stre duplicate cpies f bard psitins which are rtatins r reflectins f ne anther, each symmetric bard psitin has a single entry, determined by the mincde() (the minimum value f the bard cde ver all symmetry transfrmatins). A bard psitin is cnveniently (but nt efficiently) stred in a single, 64-bit integer. 21 pegs, type 4 17 pegs, type 7 12 pegs, type 2 13 pegs, type 5 Figure 5: Sample slvable bards with an assrtment f symmetry types. Psitin English French Square Type Symmetry descriptin Order class 33-hle 37-hle 36-hle 1 square symmetry 8 A rtatin 4 A bth diagnal reflectins 4 A bth rthgnal reflectins 4 A rtatin 2 A 2,238 7,051 9,148 6 ne diagnal reflectin 2 A 5,139 40,722 64,135 6 ne diagnal reflectin 2 C 15,187 n/c n/c 7 ne rthgnal reflectin 2 A 34, ,375 20,961 7 ne rthgnal reflectin 2 B 92,732 n/c n/c Ttal 150, ,576 94,658 Table 2: A cunt f slvable bard psitins fr the varius symmetry types. n/c means nt calculated. Table 2 shws the results f such a cmputatin, and fur sample psitins are Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

11 Gerge I. Bell 11 shwn in Figure 5. Fr the English bard, the ttals fr type 1 and 2 symmetries have been calculated by Beasley [1], and ur results agree with his. We nte that any bard psitin appearing in the English list is slvable n the French bard as well. We have remved these duplicates, s the 17 type 1 psitins n the French bard d nt include the 13 English bard psitins. The largest B i fr the English bard has size B 18 = 3, 626, 632 and fr the French bard, B 20 = 53, 371, 113. This is small enugh that a binary search tree f these sets fits int memry 2. By Therem 1, fr symmetry type 1-5 we need nly start in the centre and all bard psitins are in psitin class A. Fr symmetry type 6, we need a separate run fr psitin class C, and fr type 7, psitin class B. Nte that fr these etra runs, the starting set B 1 cntains mre than ne bard psitin. Fr eample, fr psitin class B we begin with a full bard with ne peg missing at each B in Figure 4b, althugh due t symmetry it suffices t use nly thse in bld. The reasn fr this is that we need t capture all pssible finishing hles. These runs are time cnsuming fr the French bard, and we have nt cmpleted them. Figure 5c is an interesting case, because this bard psitin fits n the English bard but is nt slvable there. The fur added hles are necessary in rder t slve it. Yu can try all type 1-4 puzzles n my Javascript web prgram fr the English bard [9] and the French Bard [10] (slutins can als be displayed). Yu can make these puzzles mre challenging by trying t slve them in the minimum number f mves (where a mve is ne r mre jumps by the same peg). Having cmputed these symmetrical bard psitins, we are in a psitin t demnstrate: Therem 2. A slutin t the central game n the English bard cannt pass thrugh an intermediate psitin with symmetry type 1-5. Prf: A bard psitin b can appear during a slutin t the central game if and nly if b is slvable t the centre and b is slvable t the centre [4]. If S j is the set f all slvable bard psitins with type j symmetry, then a bard psitin b with type j symmetry can appear during the central game if and nly if b S j and b S j. We can easily check each element f S j, and we find n matches amng types 1-5 (ecept, f curse, fr the initial and final bard states). Beasley [1] prves Therem 2 fr symmetry types 1 and 2. Therem 2 fr type 5 symmetry (180 rtatin). In [6] he prves Amng the 5, 139 type 6 bard psitins (in psitin class A), we find 198 which frm 99 cmplement pairs, all can appear during a slutin t the central game. Similarly, there are 912 type 7 bard psitins which frm 456 cmplement pairs. Martin Gardner gave a slutin he calls Jabberwcky [3] which passes 2 Run n a PC with a clck speed f 2.4 GHz and 8 GB f RAM. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

12 12 designing peg slitaire puzzles thrugh eleven intermediate psitins with reflectin symmetry abut the y-ais (type 7). On my web site [8] I shw a slutin t the central game which passes thrugh seven psitins with diagnal reflectin symmetry (type 6). The 36-hle square bard This bard is different because it des nt have a central hle. Nnetheless, it can be analyzed fr symmetrical bard psitins using the same technique. The smallest slvable bard psitin with square symmetry is fur pegs in the centre, shwn in Figure 7a, this bard psitin defines psitin class A. Figure 6 shws the three pssible patterns fr finishing hles, crucially their symmetry types are the same as thse in Figure 3. A A B B C C A A B B C C Figure 6: The three pssible patterns fr finishing hles n the square 6 6 bard. Therem 1 is valid n this bard (ecept fr the part abut being slvable t the centre ). Symmetrical bard psitins may be calculated by playing backwards, and Table 2 (right clumn) includes ttals fr each symmetry type. Fur eamples f symmetrical bard psitins n the 36-hle square bard are shwn in Figure 7. Unfrtunately, mst symmetrical bard psitins are easy t slve, because an bvius sequence f symmetrical jumps reduces them t fur pegs in the centre. Figure 7b-d shw three f the harder eamples where this is nt pssible. 4 pegs, type 1 16 pegs, type 2 16 pegs, type 4 15 pegs, type 6 Figure 7: Sample slvable 6 6 bards with varius symmetry types. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

13 Gerge I. Bell 13 The 37-hle heagn bard We nw repeat the analysis f symmetrical bard psitins fr the 37-hle heagn bard. We need t be aware f several imprtant differences. First, the symmetries are subgrups f D 12, the dihedral grup f rder 12 (the symmetries f the regular heagn). There are nine pssible symmetries, shwn in Table 3. Psitin Type Symmetry descriptin Order class Cunt Eamples 1 heagnal symmetry 12 A 20 Fig. 8a 2 60 rtatin 6 A 14 Fig. 1c rt. & rth. refl. (y-ais) 6 A 30 Fig. 8b rt. & diag. refl. (-ais) 6 A 87 Fig. 8c rt. & diag. refl. (-ais) 6 B 185 Fig. 8d 5 bth -ais and y-ais refl. 4 A 1,438 Fig. 9c rtatin 3 A rtatin 3 B 754 Fig. 9b rtatin 2 A 34,894 8 ne rthgnal refl. (y-ais) 2 A 219,295 Fig. 9d 9 ne diagnal refl. (-ais) 2 A 436,697 9 ne diagnal refl. (-ais) 2 B n/c Table 3: The nine pssible symmetries fr a bard psitin n the 37-hle heagn. n/c means nt calculated. We nw cnsider the symmetry f the tw pssible patterns fr finishing pegs n the bard, shwn in Figure 4. Psitin class A has heagnal symmetry, and psitin class B (and C) have type 4 symmetry (120 rtatin plus reflectin abut the -ais), as well as the sub-symmetries type 6 and 9. The same argument as befre leads t: Therem 3. On the 37-hle heagn bard, if a bard psitin is slvable and has symmetry type 1-3, 5, 7 r 8, then it lies in psitin class A. Anther difference is that when the bard is slvable and in psitin class A, it may nt be slvable t the centre (as in Figure 1c). Therefre, when we initialize the set B 1, we need t start with three bard psitins with ne peg missing at each f the bld A s in Figure 4a. The calculatin f all B n fr psitin class A is time cnsuming, taking abut a week f CPU time and 20 GB f disk space. The largest set B n in this case is B 19 = 364, 696, 466, and the binary search tree cntaining it is t large t fit int memry n my machine. The calculatin therefre has t be split int pieces, which increases the cmputatin time. Figures 8 and 9 shw seven bard psitins btained frm these calculatins, with bard cunts given in Table 3 and Javascript web prgram fr types 1-4 here [12]. Nte that Figure 8d and Figure 9b shw bard psitins in psitin class B. All ther bard psitins shwn in this dcument are in psitin class A. The bard psitins in Figure 9c and d were selected because they are particularly difficult t slve. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

14 14 designing peg slitaire puzzles 25 pegs, type 1 18 pegs, type 3 10 pegs, type 4 12 pegs, type 4 (B) Figure 8: Sample slvable heagn bards with an assrtment f symmetry types pegs, type 6 (B) 15 pegs, type 5 11 pegs, type 8 Figure 9: A template fr 120 symmetry, and mre sample slvable heagn bards. T cmplete all entries in Table 3 fr psitin class B wuld require a secnd run, even mre time cnsuming than the first. Fr psitin class B and symmetry types 4 and 6, we used a different technique. All bard psitins with 120 symmetry can be btained by mapping every 13-bit binary integer t the bard in Figure 9a, where a peg is present at lcatin i iff the i th bit is set. We can then ehaustively attempt t slve each bard, ne by ne, t derive a cmplete list f slvable bards by psitin class and symmetry type (1-4 r 6). This technique is reasnable when the ttal number f bards is under a few thusand, and it gives us a way t duble check ur results (at least fr types 1-4 and 6). Bard psitins with a unique winning jump Many f the symmetrical bard psitins fund in the previus sectin tend t be easy t slve by hand. The prblems shwn in Figures 5-9 are nt typical, they are sme f the harder prblems. Often, it is pssible t make a few symmetrical jumps which reduce the pattern t a smaller symmetrical pattern which has been slved previusly. At any initial bard psitin, a number f starting jumps are available. Often, any starting jump can be eecuted, after which the bard remains slvable. But suppse we search specifically fr initial psitins where nly ne f the starting jumps gives a slvable bard psitin. It is nt bvius that such bard psitins eist, because during the English central game (fr eample) mst bard psitins have many pssible jumps that can lead t a slutin. Frtunately, the sets B n calculated in the last sectin are eactly what we need t search fr these unique winning jump bard psitins. Cnsider a bard Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

15 Gerge I. Bell 15 psitin b B n. We are lking fr b s where nly a single jump ends at a slvable bard psitin. We can eecute jump k n b, prducing the bard psitin b k. b k is slvable if and nly if b k B n 1. In rder t find a puzzle which is difficult, the number f dead ends shuld be large, this suggests we want a large number f starting jumps. It wuld seem that the mst difficult n-peg initial psitins are thse which 1. Have a single winning jump, and 2. Have as many starting jumps as pssible. 6 pegs, 7 jumps 12 pegs, 13 jumps 17 pegs, 17 jumps 24 pegs, 12 jumps Figure 10: English puzzles with a unique winning jump. Under each diagram is the number f pegs and the ttal number f starting jumps. Playable n the web at [13]. 10 pegs, 13 jumps 12 pegs, 16 jumps 17 pegs, 19 jumps 22 pegs, 20 jumps Figure 11: French puzzles with a unique winning jump. Playable n the web at [14]. 7 pegs, 12 jumps 11 pegs, 18 jumps 16 pegs, 21 jumps 24 pegs, 12 jumps Figure 12: Heagnal puzzles with a unique winning jump. Playable n the web at [15]. Table 4 summarizes the results f these calculatins, and Figures shw eample bard psitins calculated using this strategy. All f these puzzles can be played n my Javascript prgrams [13, 14, 15] (the prgrams can als display Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

16 16 designing peg slitaire puzzles English bard French bard heagn bard n (pegs) ma jumps cunt ma jumps cunt ma jumps cunt Table 4: A summary f bard psitins with a unique winning jump by pegs and maimum starting jumps, fr each f the three bard types. : case appears in Figures slutins). We nte that fr a particular bard and number f pegs n, there is smetimes a unique bard psitin with as many jumps as pssible and ne winning jump. When an entry in Table 4 is blank, this indicates there are n n-peg bard psitins with a unique winning jump. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

17 Gerge I. Bell 17 These puzzles tend t be challenging t slve by hand, particularly as they becme larger. As an aid t the slver, we have identified the first peg t jump in red fr the larger bard psitins. If sme jump is a winner n a symmetrical bard psitin, then the symmetrical partners f this jump are als winners. Thus, bard psitins with a unique winning jump tend t have n symmetry. The nly eceptin wuld be a bard psitin with a single reflectin symmetry, it culd have a single winning jump alng the ais f reflectin. We have fund a few eamples like this, but nne with the maimum number f jumps. These puzzles have an entirely different character frm the symmetrical puzzles in the last sectin. The fact that there is a unique winning jump can be used t help slve these puzzles. After the first jump is eecuted, any jump which culd have been eecuted first must still be a dead end. The secnd jump can nly be a jump which was pened up by the first jump, and s n. Smetimes, if yu can identify the first jump, the rest f the slutin fllws mre easily. While the first jump f a slutin is unique, subsequent jumps can ften be eecuted in either rder, r the final jump can g in either directin, s the slutin is nt unique. Hwever, a few f these puzzles d have a unique slutin, which is quite rare in peg slitaire. An eample is Figure 12c there is nly ne sequence f jumps which slves this puzzle. When the number f pegs is relatively small (say, under 13), the bard may nt limit the jumping pssibilities. We can ften translate the pattern f pegs, and this gives a slutin which is cunted as different. This effect is respnsible fr the large cunts n the heagnal bard (26 slutins with 6 pegs and 10 jumps), this is nt 26 different slutins but a few slutins translated. These bard psitins with less than 13 pegs can be cnsidered as puzzles n an infinite bard. These puzzles retain the prperty that the number f starting jumps is large, and there is a unique winning jump. As the puzzles becme larger, n an infinite bard the unique winning jump prperty tends t be lst. Finally, we nte that all these unique winning jump puzzles were calculated using psitin class A. Since symmetry plays n rle here, there is n reasn why we culd nt use psitin class B r C. This wuld prduce a whle new set f prblems with a unique winning jump, and Table 4 wuld be different fr each psitin class. Summary We have presented tw different strategies fr creating peg slitaire puzzles. The first searches fr slvable symmetric psitins, while the secnd identifies slvable psitins with a unique winning jump. The tw strategies dn t seem t have anything in cmmn, but they can bth be calculated using the sets f bard psitins B n btained by playing the game backwards. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

18 18 designing peg slitaire puzzles The central game n the English bard is an attractive puzzle because it begins and ends at psitins with square symmetry, but in between symmetry is lst, and by Therem 2 symmetry is nt pssible (ecept fr reflectin symmetry). Fr a gd puzzle, it is desirable that symmetry is nt pssible in the middle, fr symmetric intermediate psitins ften indicate an easy slutin where jumps are repeated in a symmetrical fashin. We have identified all slvable symmetric bard psitins (f mst types), bth n a square a triangular grid. Many f these make nice puzzles t slve by hand. The unique winning jump puzzles have a cmpletely different feel they lack symmetry and are much harder t slve. The fact that they have a unique winning jump can be eplited by crafty slvers. Any slvable bard psitin presented abve is als slvable when cnsidered n an infinite bard. This means that in sme sense these puzzles eist independently f any particular bard. We saw in ging frm the English t French bard that additinal puzzles were fund that were slvable n the French bard but nt n the English bard (Figure 5c). Similarly, in ging frm the French bard t an infinite bard, we wuld epect additinal prblems slvable nly n a sufficiently large bard. Searching fr all n-peg symmetric r unique winning jump puzzles n an infinite bard is an interesting cmputatinal challenge. References [1] Beasley, J. The Ins and Outs f Peg Slitaire, Ofrd Univ. Press, [2] Berlekamp, E., Cnway, J., Guy, R. Purging pegs prperly, in Winning Ways fr Yur Mathematical Plays, 2nd ed., Vl. 4, Chap. 23: , A K Peters, [3] Gardner, M. Peg Slitaire, in Knts and Brrmean Rings, Rep-Tiles and Eight Queens, Cambridge Univ. Press, [4] Bell, G. Triangular peg slitaire unlimited, Games and Puzzles Jurnal, 36, [5] Bell, G. Slving triangular peg slitaire, Jurnal f Integer Sequences, 11, [6] Beasley, J. On 33-hle slitaire psitins with rtatinal symmetry, [7] Bell, G. Ntes n slving and playing peg slitaire n a cmputer, [8] Bell, G. Peg Slitaire web site, Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

19 Gerge I. Bell 19 [9] Bell, G. Symmetric English psitins, [10] Bell, G. Symmetric French psitins, [11] Bell, G. Symmetric 66 psitins, [12] Bell, G. Symmetric Heagn psitins, [13] Bell, G. Difficult English psitins, [14] Bell, G. Difficult French psitins, [15] Bell, G. Difficult Heagn psitins, Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

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21 Articles demystifying benjamin franklin s ther 8-square Maya Mhsin Ahmed Punjiri Web and Mbile Technlgies maya.ahmed@gmail.cm Abstract: In this article, we reveal hw Benjamin Franklin cnstructed his secnd 8 8 magic square. We als cnstruct tw new 8 8 Franklin squares. Keywrds: Magic squares, Benjamin Franklin s 8-squares. Intrductin The well-knwn squares in Figure 1 were cnstructed by Benjamin Franklin. The square F2 was intrduced separately and hence is generally knwn as the ther 8-square. The entries f the squares are frm the set {1, 2,..., n 2 }, where n = 8 r n = 16. Every integer in this set ccurs in the square eactly nce. Fr these squares, the entries f every rw and clumn add t a cmmn sum called the magic sum. The 8 8 squares have magic sum 260 and the square has magic sum In every half rw and half clumn the entries add t half the magic sum. The entries f the main bend diagnals and all the bend diagnals parallel t it add t the magic sum. In additin, bserve that every 2 2 sub-square in F1 and F2 adds t half the magic sum, and in F3 adds t ne-quarter the magic sum. The prperty f the 2 2 sub-squares adding t a cmmn sum and the prperty f bend diagnals adding t the magic sum are cntinuus prperties. By cntinuus prperty we mean that if we imagine the square is the surface f a trus; i.e. ppsite sides f the square are glued tgether, then the bend diagnals r the 2 2 sub-squares can be translated and still the crrespnding sums hld. In fact, these squares have innumerable fascinating prperties. See [1], [3], and [4] fr a detailed study f these squares. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

22 22 benjamin franklin s ther 8-square F F F3 Figure 1: Squares cnstructed by Benjamin Franklin N N N3 Table 1: Franklin squares cnstructed using Hilbert basis [1]. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

23 Maya Mhsin Ahmed B B2 Table 2: New Franklin squares. Frm nw n, rw sum, clumn sum, r bend diagnal sum, etc. mean that we are adding the entries f thse elements. Based n the descriptins f Benjamin Franklin, we define Franklin squares as fllws (see [2]). Definitin 1 (Franklin Square). Cnsider an integer, n = 2 r such that r 3. Let the magic sum be dented by M and N = n We define an n n Franklin square t be a n n matri with the fllwing prperties: 1. Every integer frm the set {1, 2,..., n 2 } ccurs eactly nce in the square. Cnsequently, M = n 2 N. 2. All the the half rws, half clumns add t ne-half the magic sum. Cnsequently, all the rws and clumns add t the magic sum. 3. All the bend diagnals add t the magic sum, cntinuusly. 4. All the 2 2 sub-squares add t 4M/n = 2N, cntinuusly. Cnsequently, all the 4 4 sub-squares add t 8N, and the fur crner numbers with the fur middle numbers add t 4N. We call permutatins f the entries f a Franklin square that preserve the defining prperties f the Franklin squares symmetry peratins, and tw squares are called ismrphic if we can get ne frm the ther by applying symmetry peratins. Rtatin and reflectin are clearly symmetry peratins. See [1] fr mre symmetry peratins like echanging specific rws r clumns. In [1], we shwed that F1 and F2 are nt ismrphic t each ther. Cnstructing Franklin squares are demanding and nly a handful such squares are knwn t date. We shwed hw t cnstruct F1, F2, and F3 using methds frm Algebra and Cmbinatrics in [1]. In the same article, fr the first time since Benjamin Franklin, we cnstructed new Franklin squares N1, N2, and N3, given in Table 1. We prved that these squares were nt ismrphic t each ther nr t F1 r F2. In ther wrds, these squares were really new. Thse methds being cmputatinally challenging are nt suitable fr higher Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

24 24 benjamin franklin s ther 8-square rder Franklin squares. Mrever, the cnstructins used cmputers and hence lacked the intrigue f Franklin s cnstructins. In [2], we fllwed Benjamin Franklin clsely, and used elementary techniques t cnstruct a Franklin square f every rder. With these techniques we are able t cnstruct F1 and F3, but nt F2. In Sectin, we mdify the methds in [2] t cnstruct F2. In Sectin, we create new Franklin squares B1 and B2, given in Table 2, but this time using elementary techniques, in keeping with the true spirit f Benjamin Franklin. Franklin s cnstructin f his ther 8-square In this sectin, we shw hw t cnstruct the Franklin square F N 13 N N 3 N N 12 N N 6 N N 10 N N 8 N N 15 N N 1 N 16 Figure 2: Cnstructin f the left half f the Franklin square F N 29 N N 19 N N 28 N N 22 N N 26 N N 24 N N 31 N N 17 N Figure 3: Cnstructin f the right half f the Franklin square F1. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

25 Maya Mhsin Ahmed 25 We present a quick recap f ur cnstructin f F1 in [2]. Nte that N = 65 thrughut this article. The 8 8 square was divided in t tw halves: the left half cnsisting f first fur clumns and the right half cnsisting f the last fur clumns. The left half is cnstructed by first placing the numbers 1, 2,..., 16 in a specific pattern as shwn in Figure 2. Then, the number N i is placed in the same rw that cntains i, where i = 1,..., 16, such that i and N i are always equidistant frm the middle f the left half. See Figure 2. Similarly, the secnd half is cnstructed by first placing numbers 17, 18,..., 32 in a specific pattern. Finally, the numbers N i is placed in the same rw that cntains i, where i = 17,..., 32, such that i and N i are always equidistant frm the middle f the right half. This prcedure is eplained in Figure 3. The algrithm fr generating the pattern f 1, 2,..., 32 in F1 is given in [2]. We als generalized this prcedure t cnstruct a Franklin square fr any given rder in [2]. Frm nw n, thrughut the article, when we say pattern f a square, we mean the pattern f the numbers 1, 2,..., 32 in the square Figure 4: Cnstructin f Franklin square S. In ur search fr new Franklin squares, it is natural t swap rws fr clumns in the abve methd. Thus, we divide the 8 8 square in t tw halves : the tp half cnsisting f the first fur rws and the bttm half cnsisting f the last fur rws. We place the numbers 1,..., 16 in the bttm half, and the numbers 17,..., 32 in the tp half in the pattern shwn in Figure 4. This time we place the numbers N i in the same clumns as i, equidistant frm the middle f each part. The Franklin square S we get frm this prcedure is ismrphic t F1: check that we can derive this square frm F1 by rtating it and then reflecting it. The square S is nt interesting because it is essentially the same as F1. But the pattern f entering the numbers 1, 2,..., 32 in S is algrithmic since it is just a rw versin f the pattern f F1. That is, instead f filling tw clumns at a time, we fill tw rws at a time. We mdify the pattern f S t get a pattern fr F2. After that, we place N i in the same clumn as i, fr i = 1,..., 32, equidistant frm the center f each half, as usual. See Figure 5 fr the cnstructin f F2. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

26 26 benjamin franklin s ther 8-square N N N N N N N N N N N N N N N N N N 3 4 N N N 1 2 N N 5 6 N N 2 14 N 13 5 N 6 10 N N 15 3 N 4 12 N 11 7 N 8 Figure 5: Cnstructing Franklin square F2. The cnstructin f F2 invlved mving clumns f in the 1,..., 32 pattern f the square S and then switching elements alng the diagnals. This pattern is nt easily generalized t higher rders. But in the net Sectin, we mdify the patterns f knwn squares t create new squares. Cnstructing new Franklin squares In this sectin we cnstruct the new Franklin squares B1 and B2 (see Table 2) N1 N2 N3 Figure 6: The patterns f Franklin squares N1, N2, and N3. T begin with, we lk at the patterns f the numbers 1, 2,..., 32 appearing in the Franklin squares N1, N2, and N3. These are shwn in Figure 6. We fund that even in these squares that were cnstructed using Hilbert bases (see [1]), the strategy f finding the pattern f the Franklin square, and then placing i and N i in the same rw r clumn, as the case may be, equidistant frm the center f relevant half f the square, wrked. Observe that the patterns f N1 and N2 are derived by mdifying the pattern f F2. S in their cnstructins, we will place i and N i in the same clumns. On the ther hand, since, N3 is a permutatin f the pattern f F1, we place i and N i in the same rws while cnstructing N3. Benjamin Franklin s patterns always restrict the entries Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

27 Maya Mhsin Ahmed 27 1, 2,..., 16 t ne half f the square. Observe that N2 is the nly square with this prperty. Guided by these bservatins, we mdified the pattern f F2 t build the new Franklin squares B1 and B2. See Figure 7 fr the patterns f B1 and B2. Since the patterns were derived frm F2, t cnstruct them, we will place i and N i in the same clumns. B1 and B2 are nt ismrphic t each ther r any ther knwn squares because the clumns were made different by cnstructin B1 B2 Figure 7: The patterns f the Franklin squares B1 and B2. Clearly, there are mre Franklin squares. T find them, we need t find mre patterns that will yield a Franklin square. S it is time t ask again Hw many squares are there Mr. Franklin? References [1] Ahmed, M. Hw many squares are there, Mr. Franklin?: Cnstructing and Enumerating Franklin Squares, Amer. Math. Mnthly, 111, , [2] Ahmed, M. Unraveling the secret f Benjamin Franklin: Cnstructing Franklin squares f higher rder, arxiv: [math.ho]. [3] Andrews, W. S. Magic Squares and Cubes, 2nd. ed., Dver, New Yrk, [4] Pasles, P. C. The lst squares f Dr. Franklin: Ben Franklin s missing squares and the secret f the magic circle, Amer. Math. Mnthly, 108, , [5] Pasles, P. C. Franklin s ther 8-square, J.Recreatinal Math., 31, , Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

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29 Games and Puzzles eplring the rubik s magic universe Maurizi Palini Dipartiment di Matematica e Fisica, Università Cattlica Sacr Cure, Brescia, Italy palini@dmf.unicatt.it Abstract: By using tw different invariants fr the Rubik s Magic puzzle, ne f metric type, the ther f tplgical type, we can dramatically reduce the universe f cnstructible cnfiguratins f the puzzle. Finding the set f actually cnstructible shapes remains hwever a challenging task, that we tackle by first reducing the target shapes t specific cnfiguratins: the ctminid 3D shapes, with all tiles parallel t ne crdinate plane; and the planar face-up shapes, with all tiles (cnsidered f infinitesimal width) lying in a cmmn plane and withut superpsed cnsecutive tiles. There are still plenty f interesting cnfiguratins that d nt belng t either f these tw cllectins. The set f cnstructible cnfiguratins (thse that can be btained by manipulatin f the undecrated puzzle frm the starting situatin) is a subset f the set f cnfiguratins with vanishing invariants. We were able t actually cnstruct all ctminid shapes with vanishing invariants and mst f the planar face-up cnfiguratins. Particularly imprtant is the tplgical invariant, f which we recently fund mentin in [7] by Tm Verheff. Keywrds: Rubik s Magic puzzle, ctminid 3D shapes, tplgical invariants. Intrductin The Rubik s Magic is anther creatin f Ernő Rubik, the brilliant hungarian inventr f the ubiquitus cube that is named after him. The Rubik s Magic puzzle is much less knwn and nt very widespread tday, hwever it is a really surprising bject that hides aspects which makes it quite an interesting subject fr a mathematical analysis n mre than ne level. We investigate here tw different invariants that can be used t prve the unreachability f many spatial cnfiguratins f the puzzle, ne f these invariants, f tplgical type, is t ur knwledge never been etensively studied befre, althugh it is presumably the same mentined in [7], and allws t Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

30 30 eplring the rubik s magic universe significantly reduce the number f theretically cnstructible shapes. Hwever even in the special case f planar face-up cnfiguratins (see Sectin 8) we dn t knw whether the cmbinatin f the tw invariants, tgether with basic cnstraints cming frm the mechanics f the puzzle, is cmplete, i.e. if it characterizes the set f cnstructible cnfiguratins. Indeed there are still a few planar face-up cnfiguratins having bth vanishing invariants, but that we are nt able t cnstruct. In this sense this Rubik s inventin remains an interesting subject f mathematical analysis. In Sectin 2 we describe the puzzle and discuss its mechanics, the lcal cnstraints are discussed in Sectin 3. The additin f a ribbn (Sectin 4) allws t intrduce the tw invariants, the metric and the tplgical invariants, described respectively in Sectins 5 and 6. The set f ctminid shapes (all tiles are parallel t a crdinate plane and n tw f them are superpsed) is described in Sectin 7. There are a ttal f 460 distinct ctminid cnfiguratins f the undecrated puzzle with vanishing invariants and all f them are actually cnstructible with the real puzzle [4, 3D ctminids] meaning that within this special class f shapes the tw invariants are cmplete. The special face-up planar cnfiguratins are defined in Sectin 8 and their invariants cmputed in Sectin 9. There are a ttal f 25 cnfiguratins with vanishing invariants, all f which we were able t actually cnstruct [4, planar face-up cnfiguratins] with nly 5 eceptins (Sectin 10). The tw basic cnfiguratins f Figures 1 and 2 are cntained in bth ctminid and planar face-up classes, fr a ttal f 485 cnfiguratins. Of curse there are still many cnfiguratins that are nt cntained in either f the tw classes (actually there are infinitely many f them), making the eplratin f the Rubik s Magic universe far frm cmplete. We cnclude the paper with a brief descriptin f the sftware cdes used t help in the analysis f the ctminid and f the planar face-up cnfiguratins (Sectin 11). The puzzle The Rubik s Magic puzzle (see Figure 1 left) cnsists f 8 decrated square tiles psitined t frm a 2 4 rectangle. They are ingeniusly cnnected t each ther by means f nyln strings lying in grves carved n the tiles and tilted 45 degrees [2]. The tiles are decrated in such a way that n ne side f the 2 4 riginal cnfiguratin we can see the picture f three uncnnected rings, whereas n the back side we find nn-matching drawings representing parts f rings with crssings amng them. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

31 Maurizi Palini 31 T 7 T 5 T 6 T 4 T 0 T 1 T 3 T 2 Figure 1: The riginal puzzle in its starting cnfiguratin (left). Orientatin scheme fr the tiles (right). Figure 2: The puzzle in its target cnfiguratin, the tiles are turned ver with respect t Figure 1. The declared aim is t manipulate the tiles in rder t crrectly place the decratins n the back, which can be dne nly by changing the glbal utline f the eight tiles. The slved puzzle is shwn in Figure 2 with the tiles psitined in a 3 3 square with a missing crner and verturned with respect t the riginal cnfiguratin f Figure 1. Detailed instructins n hw the puzzle can be slved and mre generally n hw t cnstruct interesting shapes can be cpiusly fund in the internet, we just pint t the Wikipedia entry [1] and t the web page [2]. The bklet [3] cntains a detailed descriptin f the puzzle and illustrated instructins n hw t btain particular cnfiguratins. The decratins can be used t distinguish a frnt and a back face f each tile and t rient them by suitably chsing an up directin. After dealing with the puzzle fr sme time it becmes apparent that a few lcal cnstraints are always satisfied. In particular the eight tiles remain always cnnected tw by tw in such a way t frm a cyclic sequence. T fi ideas let Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

32 32 eplring the rubik s magic universe us dente the eight tiles by T 0, T 1,..., T 7, with T 0 the lwer-left tile in Figure 1 and the thers numbered in cunterclckwise rder. Fr eample tile T 3 is the ne with the Rubik s signature in its lwer-right crner (see Figure 1). With this numbering tile T i is always cnnected thrugh ne f its sides t bth tiles T i+1 and T i 1. Here and in the rest f this paper we shall always assume the inde i in T i t be defined mdul 8, i.e. that fr eample T 8 is the same as T 0. We shall cnventinally rient the tiles such that in the initial cnfiguratin f Figure 1 all tiles are face up (i.e. with their frnt face visible), the 4 lwer tiles (T 0 t T 3 ) are straight (nt upside dwn), the 4 upper tiles (T 4 t T 7 ) are upside dwn (as a map with the nrth directin pinting dwn), see Figure 1 right. At a mre accurate eaminatin it turns ut that nly half f the grves are actually used (thse having the nyln threads in them). These allw us t attach t a crrectly riented tile (face up and straight) a priviledged directin: directin ( slash : Nrth-East t Suth-West) and directin ( backslash : Nrth-West t Suth-East). The used grves are shwn in Figure 1 right. Frm nw n we shall disregard cmpletely the unused grves. In the initial cnfiguratin tiles T i with even i are all tiles f type, whereas if i is dd we have a tile f type. The directin f the used grves in the back f a tile is ppsite (read rthgnal) t the directin f the used grves f the frnt face, but beware that when we revert (turn ver) a tile a grve becmes, s that the reversed tile remains f the same type ( r ). Frm the pint f view f an idealized physical mdelling a natural chice wuld be t assume that the tiles are made f a rigid material and have infinitesimal thickness, and that the nyln threads are perfectly fleible but inetensible (and f infinitesimal thickness). This allws fr tw r mre tiles t be jutapsed in space, hwever in such a case we still need t retain the infrmatin abut their relative psitin (which is abve which). Hwever in this mdel there are mves that can be perfrmed n the real puzzle (that entail a small amunt f elngatin n the nyln wires) but that are nt allwed in the ideal mdel (see e.g. the tw interesting shapes dented armchair and hard-t-reach planar shape linked frm [4]. Fr this reasn in the real cnstructins we shall allw fr mves that entail a small amunt f defrmatin f the tiles and elngatin f the wires. We shall nt be rigrus abut what is allwed and what is nt, the rule being that we shall generally allw fr mves that can actually be perfrmed n the real puzzle. On the cntrary the real puzzle has nn-infinitesimal tile thickness, which can lead t cnfiguratins that are allright fr the idealized physical mdel but that are difficult r impssible t achieve (because f the impsed stress n the nyln threads) with the real puzzle. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

33 Maurizi Palini 33 Undecrated puzzle We are here mainly interested in the study f the shapes in space that can be btained, s we shall neglect the decratins n the tiles and nly cnsider the directin f the grves cntaining the nyln threads. In ther wrds we nly mark ne diagnal n each face f the tiles, ne cnnecting tw ppsite vertices n the frnt face and the ther cnnecting the remaining tw vertices n the back face. Nw the tiles (marked with these tw diagnals) are indistinguishable; distinctin between and is nly pssible after we have riented a tile and in such a case rtatin f 90 degrees r a mirrr reflectin will echange with. Definitin 1 (rientatin). A tile can be riented by drawing n ne f the tw faces an arrw parallel t a side. We have thus eight different pssible rientatins. We say that tw adjacent tiles are cmpatibly riented if their arrws perfectly fit tgether (parallel and pinting t the same directin) when we ideally rtate ne tile arund the side n which they are hinged t make it jutapsed t the ther. There is eactly ne pssible rientatin f a tile that is cmpatible with the rientatin f an adjacent tile. A cnfiguratin f tiles is rientable if it is pssible t rient all tiles such that they are pairwise cmpatibly riented. Fr an rientable cnfiguratin we have eight different chices fr a cmpatible rientatin f the tiles. An eample f cmpatible rientatin f a cnfiguratin is shwn in Figure 1 right, which makes the initial 2 4 cnfiguratin rientable. Once we have a cmpatible rientatin fr a cnfiguratin, we can classify each tile as r accrding t the relatin between the rienting arrw and the marked diagnal: a tile is f type if the arrw alignes with the diagnal after a clckwise rtatin f 45 degrees, it is f type if the arrw alignes with the diagnal after a cunterclckwise rtatin f 45 degrees. Tw adjacent tiles are always f ppsite type. Definitin 2. A spatial cnfiguratin f the puzzle that is nt cngruent (als cnsidering the marked diagnals) after a rigid mtin with its mirrr image will be called chiral, therwise it will be called achiral. Nte that a cnfiguratin is achiral if and nly if it is mirrr symmetric with respect t sme plane. The initial 2 4 cnfiguratin is achiral since it is specularly symmetric with respect t a plane rthgnal t the tiles. Definitin 3. We say that an rientable spatial cnfiguratin f the puzzle (withut decratins) is cnstructible if it can be btained frm the initial 2 4 cnfiguratin thrugh a sequence f admissible mves f the puzzle. Once we have identified all the cnstructible spatial cnfiguratins, we als have all cnstructible cnfiguratins f the decrated puzzle. This is a cnsequence Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

34 34 eplring the rubik s magic universe f the fact that all pssible 2 4 cnfiguratins f the decrated puzzle are cmpletely classified (see fr eample [2] r [3]). We nte here that all 2 4 cnfiguratins f the undecrated puzzle are cngruent, hwever the presence f the marked diagnal might require a reversal f the whle cnfiguratin upside-dwn in rder t btain the cngruence. Fr chiral cnfiguratins (thse that cannt be superimpsed with their specular images) the fllwing result is useful. Therem 1. A spatial cnfiguratin f the undecrated puzzle is cnstructible if and nly if its mirrr image is cnstructible. Prf. If a cnfiguratin is cnstructible we can reach it by a sequence f mves f the puzzle starting frm the initial 2 4 cnfiguratin. Hwever the initial 2 4 cnfiguratin is specularly symmetric, hence we can perfrm the specular versin f that sequence f mves t reach the specular image f the cnfiguratin that we are cnsidering. Lcal cnstraints We nw cnsider a versin f the puzzle where in place f the usual decratin we draw arrws n the frnt face f the tiles as in Figure 1 right. The linking mechanism with the nyln threads is such that tw cnsecutive tiles T i and T i+1 are always hinged tgether thrugh ne f their sides. In particular, if we suitably rient T i with its frnt face visible and straight, i.e. with the arrw visible and pinting up) and we rtate tile T i+1 such that its center is as far away as pssible frm the center f T i (like an pen bk), then als T i+1 will have its arrw visible and ˆ pinting up if the tw tiles are hinged thrugh a vertical side (the right r left side f T i ); ˆ pinting dwn it the tw tiles are hinged thrugh a hrizntal side (the tp r bttm side f T i ). The surprising aspect f the puzzle is that when we clse the bk, i.e. we rtate T i+1 s that it becmes superimpsed with (stacked n r belw) T i, we than can repen the bk with respect t a different hinging side. The new hinging side is ne f the tw sides that are rthgnal t the riginal hinging side, which ne depending n the type f the invlved tiles (directin f the marked diagnals) and can be identified by the rule that the new side is nt separated frm the previus ne by the inner marked diagnals. Fr eample, if T i is f type (hence T i+1 is f type ) and they are hinged thrugh the right side f T i (as T 0 and T 1 f Figure 1 right) then after clsing the tiles by rtating T i+1 up arund its left side and placing it n tp f (stacked abve) T i, then we can repen the tiles with respect t the bttm side. On the cntrary, if we rtate dwn T i+1, s that it becmes stacked belw T i (and the invlved marked diagnal f T i is the ne n the back face), the new hinging side will be the upper side. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

35 Maurizi Palini 35 We remark that if a cnfiguratin des nt cntain superimpsed cnsecutive tiles, then the hinging side f any pair f cnsecutive tiles is uniquely determined. If the tiles are (cmpatibly) riented, than fr each tile T i we have a unique side (say East, Nrth, West r Suth, in shrt E, N, W r S) abut which it is hinged with the preceding tile T i 1 and a unique side (E, N, W r S) abut which it is hinged with T i+1. The tw sides can be the same. Definitin 4. Fr a given spatial riented cnfiguratin f the (undecrated) puzzle withut stacked cnsecutive tiles we say that a tile is straight if the tw hinging sides are ppsite; curving if the tw hinging sides are adjacent (but nt the same). In this case we can distinguish between tiles curving left and tiles curving right with the bvius meaning and taking int accunt the natural rdering f the tiles induced by the tile inde; a flap if it is hinged abut the same side with bth the previus and the fllwing tile. Flaps Flap tiles (thse that, fllwing Definitin 4, have a single hinging side with the tw adjacent tiles) require a specific analysis. The term flap is the same used in [3] and refers t the similarity with the flaps f an airplane, that can rtate abut a single side. Given an riented cnfiguratin with a flap T i, let us fi the attentin t the three cnsecutive tiles T i 1, T i, T i+1 and ignre all the thers. Place the cnfiguratin s that T i is hrizntal, with its frnt face up and the arrw pinting Nrth, then rtate T i 1 and T i+1 at maimum distance frm T i s that they becme reciprcally superimpsed. Nw all three tiles have their frnt face up and we can distinguish between tw situatins: Definitin 5. Tile T i is an ascending flap if tile T i 1 is belw tile T i+1 ; it is a descending flap in the ppsite case. Tile T i is a hrizntal ascending / descending flap if it is hinged at a vertical side (a side parallel t the arrw indicating the lcal rientatin f the flap tile), it is a vertical ascending / descending flap therwise. The ribbn trick In rder t intrduce the metric and the tplgical invariants we resrt t a simple epedient: we insert a ribbn in between the tiles that mre r less fllws the path f the nyln threads. The ribbn is clred red n ne side (frnt side) and blue in its back side and is riented with lngitudinal arrws printed alng its length that allws t fllw it in the psitive r negative directin. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

36 36 eplring the rubik s magic universe T 0 T 1 T 2 Figure 3: Ribbn path amng the tiles. Let the tiles have side f length 1, then the ribbn has width that des nt eceed 2 4 (the distance between tw nearby grves), s that it will nt interfere with the nyln threads. We insert the ribbn as shwn in Figure 3. Mre precisely take the 2 4 initial cnfiguratin f the puzzle and start with tile T 2. Psitin the ribbn such that it travels diagnally alng the frnt face f T 2 as shwn in Figure 3, then wrap the ribbn arund the tp side f T 2 and travel dwnwards alng the back f T 2 t reach the right side. At this pint we mve frm the back f T 2 t the frnt f T 3 (the ribbn nw has its blue face up) and cntinue dwnward until we reach the bttm side f T 3, wrap the ribbn n the back and s n. In general, every time that the ribbn reaches a side f a tile that is nt a hinge side with the fllwing tile, we wrap it arund the tile (frm the frnt face t the back face r frm the back face t the frnt face) as if it bunces against the side. Every time the ribbn reaches a hinging side f a tile with the fllwing tile it mves t the net tile and crsses frm the back (respectively frnt) side f ne tile t the frnt (respectively back) side f the ther and maintains its directin. In all cases the ribbn travels with sectins f length δ = 2 2 between tw cnsecutive tuchings f a side. It can stay adjacent t a given tile during ne, tw r three f such δ steps: ne r three if the tile is a curving tile (Definitin 4), tw if the tile is a straight tile. After having psitined the ribbn alng all tiles, it will clse n itself nicely (in a straight way and with the same rientatin) n the starting tile T 2, and we tape it with itself. In this way the ttal length f the ribbn is 16δ with an average f 2δ per tile, mrever if we remve the ribbn withut cutting it (by making the tiles disappear ), we discver that we can defrm it in space int the lateral surface f a large and shallw cylinder with height equal t the ribbn thickness and circumference 16δ. Direct inspectin als shws that the inserted ribbn des nt impact n the pssible puzzle mves, whereas its presence allws us t define the tw invariants f Sectins 5 and 6. We remark a few facts: Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

37 Maurizi Palini The ribbn is riented: it has arrws n it pinting in the directin in which we have inserted it, and while traversing the puzzle alng the ribbn the tiles are encuntered in the rder given by their inde. 2. Each time the ribbn bunces at the side f a tile (mving frm the frnt face t the back face r viceversa) its directin changes f 90 degrees and simultaneusly it turns ver. This des nt happen when the ribbn mves frm ne tile t the net, it des nt change directin and it des nt turn ver. 3. Each δ sectin f the ribbn cnnects a hrizntal side t a vertical side r viceversa; cnsequently the ribbn tuches alternatively hrizntal sides and vertical sides. 4. Each time the ribbn tuches a lateral side it ges frm ne side f the tiles t the ther (frm the frnt t the back r frm the back t the frnt). The abve pints 3 and 4 prve the fllwing Prpsitin 2. Fllwing the rientatin f the ribbn, when the ribbn tuches/crsses a vertical side, it emerges frm the back f the tiles t the frnt, whereas when it tuches/crsses a hrizntal side, it submerges frm the frnt t the back. Here vertical r hrizntal refers t the lcal rientatin assigned t the tiles. Behaviur f the ribbn at a flap tile. It is nt bvius hw the ribbn behaves at a flap tile (such tiles are nt present in the initial 2 4 cnfiguratin). We can recnstruct the ribbn psitin by imagining a mvement that transfrms a cnfiguratin withut flaps t anther with ne flap. It turns ut that there are tw different situatins. In ne case the ribbn cmpletely avids t tuch the flap tile T i and directly ges frm T i 1 t T i+1, this happens when in a neighburhd f the side where the flap tile is hinged the ribbn is n the frnt face f the upper tile and n the bttm face f the lwer tile (in the cnfiguratin where T i 1 and T i+1 are furthest away frm T i, hence superpsed), this situatin is illustrated in Figure 4 left. In the ther case the ribbn wraps arund T i with fur δ sectins alternating between the frnt face and the back face, this situatin is illustrated in Figure 4 right. The first f the tw cases arises at an ascending flap hinged at a vertical side (hrizntal ascending flap) r at a descending flap hinged at a hrizntal side (vertical descending flap); this is independent f the type r f the flap tile. The secnd f the tw cases arises at a vertical ascending flap r at a hrizntal descending flap. Metric invariant Whatever we d t the puzzle (with the ribbn inserted) there is n way t change the length f the ribbn! Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

38 38 eplring the rubik s magic universe Figure 4: Psitin f the ribbn in presence f a flap tile. The flap tiles are f type, The superpsed tiles are all f type. Left: ascending flap, the ribbn des nt even tuch the flap tile. Right: descending flap, the ribbn cmpletely wraps the flap tile with fur sectins, tw n the upper (frnt) face and tw n the back face. This allws t regard the length f the ribbn assciated t a given spatial cnfiguratin as an invariant, it cannt change under puzzle mves. The cmputatin f the ribbn length can be carried ut by fllwing a few simple rules, they can als be fund in [3]. The best way t prceed is t cmpute fr each tile T i hw many δ sequences f the ribbn wrap it and subtract the mean value 2. The resulting quantity will be called i and its value is: ˆ i = 0 if T i is a straight tile (Definitin 4); ˆ i = 1 if T i is f type and is curving left, r if it is f type and is curving right ; ˆ i = +1 if T i is f type and is curving right, r if it is f type and is curving left; ˆ i = 2 if T i is a hrizntal ascending flap (Definitin 5) r a vertical descending flap (see Figure 4 left); ˆ i = +2 if T i is a hrizntal descending flap r a vertical ascending flap (see Figure 4 right). The last tw cases ( i = 2) fllw frm the discussin in Sectin 4.1. We call = 7 i=0 i, the sum f all these quantities, then the ttal length f the ribbn will be 16δ + δ and hence is invariant under allwed mvements f the puzzle. Since in the initial cnfiguratin we wuld have = 0 it fllws that Therem 3. Any cnstructible cnfiguratin f the puzzle necessarily satisfies = 0. This invariant can als be fund in [3, page 19], thugh it is nt actually justified. A few cnfiguratins (e.g. the 3 3 shape withut the central square, called windw shape in [3]) can be ruled ut as nn-cnstructible by cmputing the invariant. The windw shape has a value = ±4, the sign depending n hw we rient the tiles. It is nn-cnstructible because 0. Figure 15 (left) shws a defrmed versin f this shape. Anther interesting cnfiguratin that can be ruled ut using this invariant is sequence (7), t be discussed in Sectin 9.1. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

39 Maurizi Palini 39 Tplgical invariant Sticking t the ribbn idea (Sectin 4) we seek a way t knw whether a given ribbn cnfiguratin (with the tiles and nyln threads remved) can be btained by defrmatins in space starting frm the cnfiguratin where the ribbn is the lateral surface f a cylinder. Tplgically the ribbn is a surface with a bundary, its bundary cnsists f tw clsed strings. One thing that we may cnsider is the center line f the ribbn: it is a single clsed string that can be cntinuusly defrmed in space and is nt allwed t crss itself. Mathematically we call this a knt, a whle branch f Mathematics is dedicated t the study f knts, ne f the tasks being finding ways t identify unknts, i.e. tangled clsed strings that can be unkntted t a perfect circle. This is precisely ur situatin: the center line f the ribbn must be an unknt, therwise the crrespnding cnfiguratin f the puzzle cannt be cnstructed. Hwever we are nt aware f puzzle cnfiguratins that can be ecluded fr this reasn. Anther (and mre useful) idea cnsists in cnsidering the tw strings frming the bundary f the ribbn. In Mathematics, a cnfiguratin cnsisting in pssibly mre than ne clsed string is called a link. Here we have a tw-cmpnents link that in the starting cnfiguratin can be defrmed int tw unlinked perfect circles. There is a tplgical invariant that can be easily cmputed, the linking number between tw clsed strings, that des nt change under cntinuus defrmatins f the link (again prhibiting selfintersectins f the tw strings r intersectins f ne string with the ther). In the riginal cnfiguratin f the puzzle, the tw strings brdering the ribbn have linking number zer: it then must be zer fr any cnstructible cnfiguratin. Cmputing the linking number In the field f knt thery a knt, r mre generally a link, is ften represented by its diagram. It cnsists f a drawing n a plane crrespnding t sme rthgnal prjectin f the link taken such that the nly pssible selfintersectins are transversal crssings where tw distinct pints f the link prject nt the same pint. We can always btain such a generic prjectin pssibly by changing a little bit the prjectin directin. We als need t add at all crssings the infrmatin f what strand f the link passes ver the ther. This is usually dne by inserting a small gap in the drawing f the strand that ges belw the ther, see Figure 5. In rder t define the linking number between tw clsed curves we need t select an rientatin (a traveling directin) fr the tw curves. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

40 40 eplring the rubik s magic universe Figure 5: Signature f a crssing fr the cmputatin f the linking number. + Figure 6: The ribbn bunces at the side f a tile. + + Figure 7: The ribbn passes ver/belw itself. In ur case the rientatin f the ribbn induces an rientatin f the tw brder strings by fllwing the same directin. The linking number changes sign if we revert the rientatin f ne f the tw curves, s that it becmes insensitive upn the chice f rientatin f the ribbn. Once we have an rientatin f the tw curves, we can assciate a signature t each crssing as shwn in Figure 5 and a crrespnding weight f value ± 1 2. Crssings f a cmpnent with itself are ignred in this cmputatin. The linking number is given by the sum f all these cntributins. Since the number f crssings in between the tw curves in the diagram is necessarily even, it fllws that the linking number is an integer and it can be prved that it des nt change under cntinuus defrmatins f the link in space. Tw far away rings have linking number zer, tw linked rings have linking number ±1. In ur case we shall investigate specifically the case where all tiles are hrizntal and face-up, in which case we have tw different situatins that prduce crssings between the tw bundary strings. We shall then write the linking number as the sum f a twist part (L t ) and a ribbn crssing part (L c ) L = L t + L c (1) Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

41 Maurizi Palini 41 where we distinguish the tw cases: 1. The ribbn wraps arund ne side f a tile (Figure 6). This entail ne crssing in the diagram, that we shall call twist crssing since it is actually prduced by a twist f the ribbn. A curving tile (as f Definitin 4) can cntain nly zer r tw f this type f crssings, and if there are tw, they are necessarily f ppsite sign. This means that curving tiles d nt cntribute t L t. 2. The ribbn crsses itself (Figure 7). Cnsequently there are fur crssing f the tw bundary strings, tw f them are selfcrssings f ne f the strings and d nt cunt, the ther tw cntribute with the same sign fr a ttal cntributin f ±1 t L c. The presence f this type f crssings is generally a cnsequence f the spatial dispsitin f the sequence f tiles and in the specific case f face-up planar cnfiguratins (t be cnsidered in Sectin 8) there can be crssings f this type when we have superpsed tiles, r in presence f flap tiles, hwever the cmputatin f L c must be carried ut case by case. Cntributin f the straight tiles t L t. The ribbn bunces eactly nce at each straight tile (Definitin 4), hence it cntributes t L t with a value δl t = ± 1 2. After analyzing the varius pssibilities we cnclude fr tile T i as fllws: ˆ δl t = if T i is a vertical tile (cnnected t the adjacent tiles thrugh its hrizntal sides) f type, r if it is a hrizntal tile f type ; ˆ δl t = 1 2 if T i is a hrizntal tile f type r a vertical tile f type. Cntributin f the flap tiles t L t. A flap tile can be cvered by the ribbn either with fur sectins (three bunces ) f nne at all. In this latter case there is still a bunce f the ribbn when it ges frm the previus tile t the net (superpsed) tile: the ribbn travels frm belw the lwer tile t abve the upper tile r viceversa. We need t keep track f this etra bunce. After analysing the pssibilities we cnclude fr tile T i as fllws: ˆ δl t = if T i is a vertical flap f type (cnnected t the adjacent tile thrugh a hrizntal side), r if it is a hrizntal flap f type ; ˆ δl t = 1 2 if T i is a hrizntal flap f type r a vertical flap f type. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

42 42 eplring the rubik s magic universe Linking number f cnstructible cnfiguratins. Therem 4. A cnstructible spatial cnfiguratin f the puzzle necessarily satisfies L = 0. Prf. The linking number L des nt change under legitimate mves f the puzzle, s that it is sufficient t cmpute it n the initial cnfiguratin f Figure 1. There are n superpsed tiles nr flaps, s that the ribbn des nt crss itself, hence L c = 0. The nly cntributin t L t cmes frm the fur straight tiles, and using the analysis f Sectin 6.2 it turns ut that their cntributin cancel ne anther s that als L t = 0 and we cnclude the prf. Eamples f cnfiguratins with nnzer linking number. Due t Therem 4 such cnfiguratins f the puzzle cannt be cnstructed. Figure 8: This cnfiguratin is nt cnstructible because it has linking number L 0. One such cnfiguratin is shwn in Figure 8 and wuld realize the maimal pssible diameter fr a cnfiguratin. The metric invariant f Sectin 5 is = 0 s that it is nt enugh t eclude this cnfiguratin, hwever we shall shw that in this case L 0 and cnclude that we have a nncnstructible cnfiguratin. It will be studied in Sectin 9. Anther interesting cnfiguratin that can be ecluded with the tplgical invariant and nt with the metric ne is a figure eight crrespnding t the sequence (6) f Sectin 9. Figure 15 (right) shw a 3D cnfiguratin that cannt be cnstructed because L 0. Octminid cnfiguratins A class f special cnfiguratins that can be studied using the tw invariants intrduced in Sectins 5 and 6 cnsists f the s-called ctminid cnfiguratins. These crrespnd t psitins f the eight tiles t frm a 3D shape with all tiles parallel t ne f the crdinate planes and n pair f superpsed tiles. The term ctminid was intrduced by Jürgen Köller in [2] and is suggested by the term plymines t dente planar shapes made f sme fied number f adjacent unit squares jined by their sides. The ttal number f distinct ctminids is the large number , mst f which can be immediately ecluded as pssible cnfiguratins f the Rubik s Magic because the eight squares cannt be cyclically cnnected by their sides; then a further reductin is btained by enfrcing the lcal cnstraints f Sectin 3. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

43 Maurizi Palini 43 The crrect way f cunting the set f feasible shapes (cnfiguratins f the undecrated puzzle that satisfy the lcal cnstraints) must take int accunt the cyclical rdering f the tiles tgether with the directin f their marked diagnals. It is thus pssible t btain the same 3D ctminid shape with different puzzle cnfiguratins: they cme ften in pairs, ne cnfiguratin btained frm the ther by reverting the directin f the diagnals, but there can be mre than tw cnfiguratins, r just ne. The ttal number f cnfiguratins in the shape f an ctminid that satisfy the lcal cnstraints turns ut t be 1291 realizing a ttal f 582 different ctminid shapes. These numbers are btained by using a sftware cde that can be dwnladed frm [4] and that will be briefly described in Sectin Any given ctminid cnfiguratin can be described by cnstructing the s-called magic cde. It cnsists f a sequence f characters (like RRmRRmRUmDUm) with eight capital letters taken frm the set RLUD (standing fr right, left, up, dwn) ptinally fllwed by the lwer letter m (muntain fld) r v (valley fld). They encde the relative adjacency infrmatin f each f the eight tiles with the net ne. The first tile (say T 0 ) is riented by drawing an arrw n ne f its faces (frnt) parallel t a side. The selected rientatin fr T 0 must be such that the nyln strings crss the frnt face in the directin suth-west t nrth-east ( directin). The first capital letter indicates which ne f the fur sides f T 0 is cnnected t T 1 (the subsequent tile), the presence f the lwercase m r v indicates that T 0 and T 1 frm a 90 degrees angle either with a muntain fld (letter m) r with a valley fld with respect t the frnt face f T 0. Otherwise T 1 is cplanar with T 0. The rientatin f T 1 is cmpatible with the selected rientatin f T 0 (as defined in Definitin 1), i.e. the drawn arrw n T 1 (in case f cplanarity) is eactly the mirrr image f the arrw n T 0 with respect t the hinging side (the side f T 0 adjacent t T 1 ). As a simple eample the starting 2 4 cnfiguratin f the puzzle can be encded as RRRURRRU where it shuld be nted that the fur tiles in the tp rw have dwnward arrws. Since we are interested in cnfiguratins f the undecrated puzzle, many distinct magic cdes describe the same cnfiguratin based n which tile we select as T 0, hw we rient it (such that T 0 becmes a tile) and in which directin we traverse the circular chain f eight tiles. The crrespnding magic cdes are all cnsidered equivalent. A cannical magic cde is then selected by taking all the disting equivalent magic cdes and selecting the first ne with respect t a suitable leicgraphic rdering. The leicgraphic cmparisn is defined such that the fur directins are rdered as R < U < L < D, and if the directin is the same, n fld is less than muntain fld which is less than a valley fld. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

44 44 eplring the rubik s magic universe The mirrr image f a cnstructible cnfiguratin can als be cnstructed by using the mirrr images f the sequence f cnstructin mves starting frm the riginal 2 4 cnfiguratin, s that we als include all the magic cdes f the mirrr images in the same equivalence class. b #10 b #1 b #8 Figure 9: Eamples f ctminid cnfiguratins. The label indicates the size f the smallest bunding b fllwed by +k where k is the number f tiles that lie cmpletely in the bundary f the bunding b. Finally #n is the sequential number in the table f the ctminid shapes in [4, 3D ctminid shapes]. Figure 9 shws three eamples f ctminid cnfiguratins, the shape n the left is encded by the magic cde RRmULmLLmURm, btained by selecting as T 0 the lwer-left frnt tile, riented with an upward arrw in the visible face and traversing the cnfiguratin cunterclckwise. Then tile T 1 is hinged at the right side f T 0 (hence the first R character in the magic cde) and is cplanar with T 0 (n m r v character fllwing the first R. Tile T 2 is als hinged at the right side f T 1 and tilted 90 degrees with a muntain fld (Rm). Tile T 3 is hinged at the upper side f T 2, nte that the rientatin f T 3 is cnsequently riented with a dwnward arrw. We have a number f equivalent magic cdes by changing the chice and rientatin f the starting tile, hwever the ne cming first in the leicgraphic rdering (the cannical magic cde) is the string described abve. Similarly, the cannical magic cde f the secnd image f Figure 9 is given by RRRLmDvUDvLm, btained by selecting as T 0 the leftmst tile. The third cnfiguratin f the figure is finally encded by RUUmLDvUvDmD by suitably chsing and rienting the tile T 0. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

45 Maurizi Palini 45 Figure 10: All 265 ctminid shapes with vanishing invariants. btained with PvRay. Syntetic images In the special class f ctminids, we were able t autmate the cmputatin f bth the metric and the tplgical invariants f the cnfiguratin having a given magic cde, thus allwing t quickly rule ut all cnfiguratins with Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

46 46 eplring the rubik s magic universe nnzer invariants. Enfrcing zer metric invariant reduces the cnfiguratins frm 1291 dwn t 737, then enfrcing als the vanishing f the linking number further reduces the number f cnfiguratins t 460 (265 distinct ctminid shapes). These are all listed in [4, 3D ctminids], rdered accrding t the dimensin f the smallest bunding b. Surprisingly, all f them culd be actually cnstructed with the real puzzle, and building istructins are available in [4, 3D ctminids]. A few f such shapes are nt present in the cmprehensive table f symmetric 3D shapes in [2] and might be pssibly btained by us fr the first time. Figure 10 shws all 265 such shapes in ne sht. The images where btained synthetically using the ray-tracing sftware PvRay, in cnjunctin with the utput f ur sftware cde t btain the list f admissible magic cdes. Planar face-up cnfiguratins We shall apply the results f the previus sectins t a particular chice f spatial cnfiguratins, we shall restrict t planar cnfiguratins (all tiles parallel t the hrizntal plane) with nn-verlapping cnsecutive tiles. Superpsed nncnsecutive tiles are allwed. They can be btained starting frm strings f cardinal directins in the fllwing way. The infinite string s : Z {E, N, W, S} is a typgraphical sequence with inde taking values in the integers Z where the fur symbls stand fr the fur cardinal directins East, Nrth, West, Suth. On s we require 1. Peridicity f perid 8: s n+8 = s n fr any n Z; 2. Zer mean value: in any subsequence f 8 cnsecutive characters (fr eample in {s 0,..., s 7 }) there is an equal number f characters N as f characters S and f characters E as f characters W. An admissible sequence is ne that satisfies the tw abve requirements. Peridicity allws us t describe an admissible sequence by listing 8 cnsecutive symbls, fr definiteness and simplicity we shall then describe an admissible sequence just by listing the symbls s 1 t s 8. The character s i f the string indicates the relative psitin between the tw cnsecutive tiles T i 1 and T i, that are hrizntal and face-up. The first tile T 0 can be f type r type, all the thers T i are f the same type as T 0 if i is even, f the ppsite type if i is dd. The lcal cnstraints allws t recver a spatial cnfiguratin f the puzzle frm an admissible sequence with tw caveats: Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

47 Maurizi Palini Fr at least ne f the tiles, say T 0, it is necessary t specify if it is f type r. We can add this infrmatin by inserting the symbl r between tw cnsecutive symbls, usually befre s 1 ; 2. In case f superpsed tiles (same physical psitin) it is necessary t clarify their relative psitin (which is abve which). We can add this infrmatin by inserting a psitive natural number between tw cnsecutive symbls that indicates the height f the crrespnding tile. In the real puzzle the tiles are nt f zer width, s that their height in space cannt be the same. In case f necessity we shall insert such numbers as an inde f the symbl at the left. Remark 1 (Cnfiguratins that can be assembled). Given an admissible sequence it is pssible t cmpute the number f superpsed tiles at any given psitin. An assemblage f a sequence entails a chice f the height f each f the superpsed tiles (if there is mre than ne). We d this by adding an inde between tw cnsecutive symbls. Hwever fr this assemblage t crrespnd t a pssible puzzle cnfiguratin we need t require a cnditin. We hence say that an assemblage is admissible if whenever tile T i is superpsed t tile T j, i j, and als T i±1 is superpsed t T j+1, then the relative psitin f the tiles in the tw pairs cannt be echanged. This means that if T i is at a higher height than T j, then T i±1 cannt be at a lwer height than T j+1. It is pssible that a given admissible sequence des nt allw fr any admissible assemblage r that it can allw fr mre than ne admissible assemblage. Observe that the mirrr image f an riented spatial cnfiguratin f the undecrated puzzle entails a change f type, tiles becme f type and viceversa. If the mirrr is hrizntal the reflected image is a different assemblage f the same admissible sequence with all tiles f changed type and an inverted relative psitin f the superpsed tiles. On the set f admissible sequences we intrduce an equivalence relatin defined by s t if ne f the fllwing prperties (r a cmbinatin f them) hlds: 1. (cyclicity) The tw sequences cincide up t a translatin f the inde: s n = t n+k fr all n and sme k Z; 2. (rder reversal) s n = t k n fr all n and sme k Z; 3. (rtatin) s can be btained frm t after substituting E N, N W, W S, S E; 4. (reflectin) s can be btained frm t after substituting E W e W E. Let us dente by S the set f equivalence classes. We develped a sftware cde capable f finding a cannical representative f each f these equivalence classes, they are 71 (cardinality f S). In Table 1 we summarize imprtant prperties f these cannical sequences, subdivided with respect t the number f flap tiles. It is wrth nting that sme f the 71 sequences admit mre than ne nnequivalent admissible assemblages in space due t the arbitrariness in chsing the type f tile T 0 and the rdering Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

48 48 eplring the rubik s magic universe f the superpsed tiles. A few f the 71 admissible sequences d nt admin any admissible assemblage, ne f these is the nly sequence with 8 flaps: EW EW EW EW. Since the cnstructability f a spatial cnfiguratin is invariant under specular reflectin (which entails a change f type f all tiles) we can fi the type f tile T 0, pssibly reverting the rder f the superpsed tiles. The cannical representative f an equivalence class in S is selected by intrducing a leicgraphic rdering in the finite sequence s 1,..., s 8 where the rdering f the fur cardinal directins is fied as E < N < W < S. Then the cannical representative is the smallest element f the class with respect t this rdering. The surce f the sftware cde can be dwnladed frm the web page [4]. Table 1: Sequences in S. number number f number f = 0 = L = 0 nt f flaps sequences assembl. assembl. assembl. classified nne ttal In Table 2 the sequences with tw and fur flaps are subdivided based n the distributin f the flaps in the sequence. Therem 5. All planar face-up cnfiguratins have zer twist cntributin t the tplgical invariant: L t = 0. Cnsequently we have L = L c and t cmpute the linking number it is sufficient t cmpute the cntributins cming frm the crssing f the ribbn with itself. Any planar face-up cnfiguratin with an dd number f selfintersectns f the ribbn with itself has L 0. Prf. 1 We dente with k 1,..., k s the number f symbls in cntiguus subsequence f E, W (hrizntal prtins) r f N, S (vertical prtins). Each prtin f k i symbls cntains k i 1 straight tiles r flaps, all hrizntal r vertical, hence each tile cntributes t L t with alternating sign due t the fact that the tiles are alternatively f type and. If k i 1 is even, then the cntributin f this prtin is zer, while if it is dd it will be equal t the cntributin f the first straight r flap tile f the prtin. It is nt restrictive t assume that the first prtin f k 1 symbls is hrizntal and the last (f k s symbls) is vertical. In this way if i is dd, then k i is the number f symbls in a hrizntal prtin whereas if i is even, then k i is the number f symbls in a vertical prtin. Up t a change f sign f L t we can als assume that the first 1 This prf is due t Givanni Palini, Scula Nrmale Superire f Pisa. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

49 Maurizi Palini 49 tile is f type. Finally we bserve that k i > 0 fr all i. Twice the cntributin t L t f the i-th prtin is given by ( 1) i 1 ( 1) k1+k2+...+ki 1 (1 + ( 1) ki ) (2) where the last factr in parentheses is zer if k i is dd and is 2 if k i is even; the sign changes n vertical prtins with respect t hrizntal prtins (factr ( 1) i 1 ) and changes when the type ( r ) f the first straight r flap til f the prtin changes (factr ( 1) k1+k2+...+ki 1 ). Summing up 2 n i and epanding we have 2L t = s ( 1) i 1 ( 1) k1+k2+...+ki 1 + i=1 = = s ( 1) i ( 1) k1+k2+...+ki 1 + i=1 s ( 1) i 1 ( 1) k1+k2+...+ki 1 ( 1) ki i=1 s ( 1) i+1 ( 1) k1+k2+...+ki i=1 s s+1 ( 1) i ( 1) k1+k2+...+ki 1 + ( 1) i ( 1) k1+k2+...+ki 1 i=1 = 1 + ( 1) s+1 ( 1) k1+k2+...+ks = 0 i=2 because s is even and k k s = 8, even. Cnfiguratins with vanishing invariants We shall identify admissible assemblages whenever they crrespnd t equivalent puzzle cnfiguratins, where we als allw fr specular images. In particular this allws us t assume the first tile t be f type. Assemblages crrespnding t nn-equivalent sequences cannt be equivalent, n the cntrary there can eist equivalent assemblages f the same sequence and this typically happens fr symmetric sequences. The tw invariants can change sign n equivalent sequences r equivalent assemblages, this is nt a prblem since we are interested in whether the invariants are zer r nnzer. In any case the cmputatins are always perfrmed n the cannical representative. The cntributin c f = c + f (cming frm the curving tiles) can be cmputed n the sequence (it des nt depend n the assemblage). On the cntrary the cntributin f cming frm the flap tiles depends n the actual assemblage. With the aid f the sftware cde we can partially analyze each cannical admissible sequence and each f the pssible admissible assemblages f a sequence. In particular the sftware is able t cmpute the metric invariant f an assemblage, s that we are left with the analysis f the tplgical invariant, and we shall perfrm such analysis nly n assemblages having = 0, since ur aim is t identify as best as we can the set f cnstructible cnfiguratins. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

50 50 eplring the rubik s magic universe Sequences with n flaps There are seven such sequences, three f them d nt have any superpsed tiles, s that they cver a regin f the plane crrespnding t 8 tiles (cnfiguratins f area 8). Fr these three sequences we nly have ne pssible assemblage (having fied the type f tile T 0 ). The sequence EEENW W W S (3) crrespnds t the initial cnfiguratin 2 4 f the puzzle. The sequence EENNW W SS (4) crrespnds t the windw shape, a 3 3 square withut the central tile. The sequence EENNW SW S (5) crrespnds t the target cnfiguratin f the puzzle (Figure 2). Tw sequences cver 7 squares f the plane (area 7), the sequence and the sequence EENW SSW N (6) ENENW SW S. (7) A sequence withut flaps and area 6 (tw pairs f superpsed tiles) is The last pssible sequence (with area 4) wuld be ENESW NW S. (8) ENW SENW S, (9) this hwever cannt be assembled in space since it cnsists f a clsed circuit f 4 tiles traveled twice (see Remark 1). The metric invariant is nnzer (hence the crrespnding assemblage is nt cnstructible) fr the tw sequences (4) and (7), the tplgical invariant L further reduces the number f pssibly cnstructible cnfiguratin by ecluding als the tw sequences (6) e (8). The remaining tw cnfiguratins, crrespnding t sequences (3) ed (5), are actually cnstructible (Figures 1 and 2). Sequences with ne flap We find seven (nnequivalent) sequences with eactly ne flap. Three f these have area 7: EENNW SSW (10) EENW NSW S (11) EEENW W SW (12) Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

51 Maurizi Palini 51 and fur have area 6: EENW SW SN (13) EENW SW NS (14) EENW SNW S (15) EEENW SW W. (16) In all cases it turns ut that there are tw nnequivalent assemblages f each f these sequences accrding t the flap tile being ascending r descending, and they have necessarily a different value f, s that at mst ne (it turns ut eactly ne) has = 0. We shall restrict the analysis f the tplgical invariant t thse having = 0. The tw sequences (10) and (12) have = 0 if the (hrizntal) flap tile is descending (Figure 4 right). The linking number reduces t L = L c (Therem 5). Since in bth cases we have eactly ne crssing f the ribbn with itself we cnclude that L 0 and the sequences are nt cnstructible. T have = 0 the vertical flap f the sequence (11) must be descending. Then there is ne crssing f the ribbn with itself, s that L 0 and the cnfiguratin is nt cnstructible. Sequences (13) and (14) have = 0 prvided their flap is ascending. We have nw tw crssings f the ribbn with itself and they turn ut t have ppsite sign in their cntributin t L c, s that L = 0 and the tw sequences might be cnstructible. Sequences (15) and (16) have = 0 prvided their flap is ascending. Sequence (15) is then nt cnstructible because there is eactly ne selfcrssing f the ribbn s that L = L c 0. On the cntrary, sequence (16) ehibits tw selfcrssings with ppsite sign and L = L c = 0. In cnclusin f the 7 different sequences with ne flap, fur are necessarily nn cnstructible because the tplgical invariant is nn-zer, the remaining three sequences: (13), (14), (16) are actually cnstructible (refer t [4, planar face-up] fr building instructins: first, secnd and third image f sectin One flap ), see als Figure 11, first three images. The tw sequences with tw adjacent flaps Adjacency f the tw flaps entails that bth are ascending r bth descending (Remark 1) and als they are bth hrizntal r bth vertical since they are hinged t each ther s that they cntribute t the metric invariant f = ±4 whereas c = 0. Hence the metric invariant is nnzer and the tw sequences are nn-cnstructible. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

52 52 eplring the rubik s magic universe The five sequences with tw flaps separated by ne tile Sequence EENW SEW W = 0 implies that the tw (hrizntal) flaps are ne ascending and ne descending. There are tw nn-equivalent admissible assemblages satisfying = 0, cmputatin f the tplgical invariant gives L = L c = ±4 fr ne f the tw assemblages whereas the ther has L = L c = 0 and is actually cnstructible (Figure 12, secnd image): E 3 E 2 NW S 2 E 1 W 1 W. (17) Fr building instructins, see [4, planar face-up], secnd image f sectin Tw flaps, area 5. Sequence ENEW SNW S = 0 implies that bth flaps (ne is hrizntal and ne vertical) are ascending r bth descending. The tw crrespnding distinct admissible assemblages have bth L = L c = 0. The tw assemblages are: E 2 N 3 EW 2 S 2 N 3 W S and E 1 N 1 EW 2 S 2 N 3 W S. (18) They are bth cnstructible (Figure 12, third and first image respectively). Fr building instructins: [4, planar face-up], third and first image f sectin Tw flaps, area 5. Sequence ENEW NSW S We can fi the first tile T 0 t be f type, then c = 4 and = 0 implies that the first flap (hrizntal) is ascending and the secnd (vertical) is descending. Cmputatin f the tplgical invariant gives L = L c = 0 and we have anther unclassified sequence: EN 1 EW 3 NS 2 W S. The tw lwest superpsed tiles can be echanged, hwever the resulting assemblage is equivalent due t the reflectin symmetry f the sequence f symbls. Sequence ENW ESNW S Impsing = 0 the tw flaps (ne is hrizntal and ne is vertical) must be bth ascending r bth descending. In bth cases we cmpute L = L c = 0. Actually the tw assemblages are equivalent by taking advantage f the symmetry f the sequence, ne f these is E 1 N 1 W 1 E 2 S 2 N 3 W 2 S (19) and is cnstructible (Figure 12, furth image); building instructins in [4, planar face-up], image in sectin Tw flaps, area 4. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

53 Maurizi Palini 53 Sequence EENW SSNW Impsing = 0 the tw flaps (ne hrizntal and ne vertical) must be bth ascending r bth descending. In bth cases we cmpute L = L c = 0. The tw assemblages are equivalent as in the previus case, ne f these (Figure 11, 8-th image) is E 1 ENW S 3 SN 2 W (20) and building instructins can be fund in [4, planar face-up], fifth image f sectin Tw flaps, area 6. The three sequences with tw flaps separated by tw tiles All three admissible sequences with tw flaps at distance 3 (separated by tw tiles) have c = 0. Tw f these sequences have bth hrizntal r bth vertical flaps, s that = 0 entails that ne flap is ascending and ne is descending. The third sequence has an hrizntal flap and a vertical flap s that = 0 entails that bth flaps are ascending r bth descending. In all cases we have tw selfcrssings f the ribbn with ppsite sign, hence L = L c = 0 and might be cnstructible. Each f the three sequences admit tw distinct assemblages bth with = L = 0: E 2 E 2 EW 1 NW S 1 W, E 1 E 1 EW 2 NW S 2 W (21) E 2 ENW 1 NS 2 S 1 W, E 1 ENW 2 NS 1 S 2 W (22) E 2 ENW 2 W E 1 S 1 W, E 1 ENW 1 W E 2 S 2 W (23) The first tw and the last tw are actually cnstructible (Figure 11, 4-th, 5-th, 6-th and 7-th image respectively); building instructins in [4, planar face-up], images 1 t 4 f sectin Tw flaps, area 6. The first f the secnd rw is als cnstructible (Figure 11, 9-th image) althugh with a cnsiderable amunt f strain n the nyln wires. Building instructins in [4, planar face-up], 6-th image f sectin Tw flaps, area 6. The twelve sequences with tw flaps in antipdal psitin Table 2: Sequences with tw flaps (left) and fur flaps (right) subdivided based n the relative psitin f the flaps. sequences distributin with 2 flaps f flaps 2 ff 5 ff 3 ff 12 ff sequences dist. f flaps 1 ffff 5 ffff 1 ffff 4 ffff 1 ffff 6 ffff Of the 12 sequences with tw flaps in ppsite (antipdal) psitin we first analyze thse (they are 10) in which the tiles fllw the same path frm ne flap t the ther and back. One f these is shwn in Figure 8. All have c = 0 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

54 54 eplring the rubik s magic universe Figure 11: The three cnstructible cnfiguratins with ne flap and thse with tw flaps and area 6. Fr building instructins we refer t [4], click n planar face-up. s that the cntributin f the tw flaps must have ppsite sign in rder t have = 0. If ne flap is hrizntal and the ther vertical, then they must be bth ascending r bth descending and we have n pssible admissible assemblage (Remark 1). We are then left with thse sequences having bth hrizntal r bth vertical flaps, ne ascending and ne descending. In this situatin we find that the ribbn has 3 selfcrssings, s that necessarily L = L c 0 and these sequences are als nt cnstructible. We remain with the tw sequences EENEW W SW and EENNSW SW that bth have a cntributin c = 4 (fiing T 0 f type ), s that the tw flaps must cntribute with a psitive sign t the metric invariant. The first sequence has bth hrizntal flaps, and they must be bth descending, this is nw pssible thanks t the different path between the tw flaps. The secnd sequence has ne hrizntal and ne vertical flap, s that the first must be ascending and the secnd descending. There are eactly tw selfcrssings f the ribbn in bth cases, hwever they have the same sign in the first case implying L = L c 0, hence nn cnstructible. They have ppsite sign in the secnd case and we have bth zer invariants. In cnclusin the nly ne f the 12 sequences that might be cnstructible is E 1 EN 1 NS 2 W S 2 W. (24) Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

55 Maurizi Palini 55 Figure 12: Cnstructible cnfiguratins with tw flaps, area 5 and 4 and cnfiguratins with three and fur flaps. Fr building instructins we refer t [4], click n planar face-up. Sequences with three flaps Of the 10 admissible sequences with three flaps there is nly ne with all adjacent flaps, having c = ±2. The three flaps being cnsecutive are all hrizntal r all vertical and all ascending r all descending, with a ttal f f = ±6 and the metric invariant cannt be zer. Fur sequences have tw adjacent flaps, and in all cases c = ±2. Impsing = 0 allws t identify a unique assemblage fr each sequence (with ne eceptin). In all cases a direct check allws t cmpute L = L c = 0. These sequences are: E 3 E 2 W 1,2 E 1 NW S 2,1 W, E 3 ENW 1 S 1 N 2 S 2 W (25) E 2 EN 2 W 2 E 1 W 1 S 1 W, E 2 E 1 N 1 S 2 N 2 W S 1 W. (26) The last tw are actually cnstructible (Figure 12, 5-th and 6-th images), building instructins in [4, planar face-up], first and secnd images f sectin Three flaps. One f the tw assemblages f the left sequence in (25) can be actually cnstructed (Figure 12, 7-th image), building instructins in [4, planar face-up], third image f sectin Three flaps. If the puzzle had sufficiently defrmable nyln threads and tiles we culd cnceivably defrm the first Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

56 56 eplring the rubik s magic universe assemblage int the secnd. We d nt knw at present if the right sequence in (25) is cnstructible (unclassified). The five remaining sequences all have c = ±2. Impsing = 0 leaves us with 10 different assemblages: the sequence with all three hrizntal flaps has three different assemblages with = 0, three f the remaining fur sequences (with tw flaps in ne directin and the third in the ther directin) have tw assemblages each, the remaining sequence has nly ne assemblage with = 0. In all cases a direct check quantifies in 3 r 5 (in any case an dd value) the number f selfcrssings f the ribbn, s that L = L c 0. Nne f these sequences is then cnstructible. Sequences with fur flaps There are 18 such sequences. Si f these have a series f at least three cnsecutive flaps and a cntributin c = 0. They are nt cnstructible because the cnsecutive flaps all cntribute with the same sign t f. The sequences ENW EW SNS and ENW EW SEW have c = 0 and tw pairs f adjacent flaps riented in different directins in the first case and in the same directin in the secnd case. T have = 0 they must cntribute with ppsite sign and hence must be all ascending r all descending in the first case whereas in the secnd case ne pair f flaps must be ascending and ne descending. Thanks t the symmetry f the sequences the tw pssible assemblages f each are actually equivalent. The linking number turns ut t be L = 0 and we have tw pssibly cnstructible cnfiguratins: EN 1 W 1 E 2 W 2 S 1 N 3 S 2 and E 1 N 1 W 1 E 2 W 2 S 2 E 2 W 1. (27) Bth turn ut t be cnstructible (Figure 12, 8-th and 9-th images), building instructins in [4, planar face-up], images f sectin Fur flaps. There are fur sequences with a single pair f adjacent flaps, the ther tw being islated, all with c = 0. The tw islated flaps must cntribute t the metric invariant with the same sign, ppsite t the cntributin that cmes frm the tw adjacent flaps. In three f the fur cases the tw islated flaps have the same directin and hence bth must be ascending r bth descending. It turns ut that there is n admissible assemblage with such characteristics. The pair f adjacent flaps f the remaining sequence (EENSW EW W ) are hrizntal. If they are ascending the remaining hrizntal flap must be descending whereas the vertical flap must be ascending (t have = 0). This situatin (r the ne with a descending pair f adjacent flaps) is assemblable and we can cmpute the linking number, which turns ut t be L = L c = ±2. Even this cnfiguratin is nt cnstructible. The remaining si sequences (flaps alternating with nn-flap tiles) all have the nn-flap tiles superpsed t each ther. An invlved reasning, r the use f the sftware cde, allws t shw that fr tw f this si sequences, having area 3, namely EEW W EEW W and ENSW ENSW, there is n pssible admissible assemblage with = 0. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

57 Maurizi Palini 57 Sequence ENSEW NSW This sequence has area 4, with tw f the fur flaps superpsed t each ther. Using the sftware cde we find tw different assemblages having = 0. Cmputatin f the linking number leads in bth cases t L = L c = ±2, hence this sequence is nt cnstructible. Sequence EEW NSEW W This sequence has als area 4, with tw f the fur flaps superpsed t each ther. Using the sftware cde we find nly ne assemblages having = 0. Cmputatin f the linking number leads in t L = L c = ±2, hence this sequence is als nt cnstructible. Sequence EEW NSSNW This sequence has area 5 with n superpsed flaps and cntributin c = 0, s that t have = 0 tw flaps cntribute psitively and tw cntribute negatively t the metric invariant. The sftware cde gives three different assemblages with = 0. An accurate analysis f the selfcrssings f the ribbn due t the flaps shws that flaps that cntribute psitively t the metric invariant als cntribute with an dd number f selfcrssings f the ribbn, besides there is ne selfcrssing due t the crssing straight tiles. In cnclusin we have an dd number f selfcrssings, hence the sequence is nt cnstructible. Sequence ENSEW SNW This sequence has als area 5 with n superpsed flaps, but nw the cntributin f the curving tiles t the metric invariant is c = 4, s that t have = 0 eactly ne f the flaps has psitive cntributin t the metric invariant. This flap will als cntribute with an dd number f selfcrssings f the ribbn. In this case there are n ther selfcrssings f the ribbn because there are n straight tiles, s that we again cnclude that the number f selfcrssings f the ribbn is dd and that L = L c 0. This sequence is als nt cnstructible. Sequences with five flaps Bth sequences have c = 2. The sftware cde quickly shws that the sequence EEW NSNSW des nt have any admissible assemblage with = 0. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

58 58 eplring the rubik s magic universe The ther sequence is EEW EW NSW and t have = 0 the three cnsecutive hrizntal flaps must be ascending, and als the vertical flap must be ascending. There is ne admissible assemblage satisfying these requirements, but the resulting number f selfcrssings f the ribbn is dd, and s als this cnfiguratin is nt cnstructible. Sequences with si flaps Nne f the fur sequences with si flaps is cnstructible. Indeed it turns ut that all have c = 0, s that in rder t have = 0 they must have three flaps with psitive cntributin and three with negative cntributin t f. Tw f the fur sequences have fur r mre flaps that are cnsecutive and hence all cntribute with the same sign t f, s that 0. The sequence EEW EW W EW must have three ascending cnsecutive flaps and three cnsecutive descending flaps (all flaps are hrizntal). Analyzing the ribbn cnfiguratin shws that there are an dd number f ribbn selfcrssings, hence L = L c 0. Finally, all the si flaps f sequence ENSNSW EW must be ascending in rder t have = 0, there is n admissible assemblage with this prperty. Sequences with seven flaps There is nne. Sequences with eight flaps The nly ne is EW EW EW EW, but there is n admissible assemblage f this sequence. Unclassified cnfiguratins Sequences with tw flaps There are 3 unclassified assemblages, crrespnding t 3 sequences. assemblage EN 1 EW 3 NS 2 W S The has tw flaps separated by ne tile. This assemblage f the sequence (22, right) has tw flaps separated by tw tiles: E 1 ENW 2 NS 1 S 2 W. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

59 Maurizi Palini 59 Finally there is ne unclassified assemblage with tw tiles separated by three tiles: E 1 EN 1 NS 2 W S 2 W. (28) A schematic visualizatin f these assemblages is depicted in Figure 13??? Figure 13: Schematic structure f the three unclassified sequences with tw flaps.?? Figure 14: Schematic structure f the tw unclassified sequences with three flaps. Sequences with three flaps There are 2 f them: E 3 E 2 W 2 E 1 NW S 1 W E 3 ENW 1 S 1 N 2 S 2 W. The first ne wuld cnceivably be cnstructible starting frm E 3 E 2 W 1 E 1 NW S 2 W by echanging the psitin f tw tiles, which is pssible nly with a very large amunt f stretching n the wires and defrmatin f the tiles, nt available n the real puzzle. A schematic visualizatin f these assemblages is depicted in Figure 14. The sftware cdes The sftware cde is cntained in the Subversin (svn) repsitry [5], tgether with a pvray mdule that can be used t prduce syntetized images f 3D cnfiguratins and scripts t prduce a printut with custmized decratins fr the puzzle. The cde can als be dwnladed frm [4] and shuld wrk n any cmputer with a C cmpiler. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

60 60 eplring the rubik s magic universe There are actually tw distinct sftware cdes, described in Sectin 11.1 (ctminid shapes) and Sectin 11.2 (planar face-up cnfiguratins). 3D ctminid shapes The name f the eecutable is rubiksmagic. If run withut arguments, it will search fr all cannical representatives f the set f ctminid cnfiguratins that satisfy the lcal cnstraints (feasible cnfiguratins), using the magic cde described in Sectin 7. This is part f its utput: $./rubiksmagic RRRURRRU b=024+8 plyminid= s f=0 delta=0 linking=0 symcunt=4 typeinv=yes RRRUmRRRUm b=114+8 plyminid= s f=0 delta=0 linking=0 symcunt=4 typeinv=n [...] RmUvRvLmUvUmUvLv b=222+0 plyminid= f=4 delta=4 linking=0 symcunt=1 typeinv=n RmUvUmLvRmUvUmLv b=222+0 plyminid= s f=4 delta=0 linking=0 symcunt=4 typeinv=yes Fund 1291 sequences $ The utput actually cnsists f a single (lng) line fr each cnfiguratin, here wrapped in tw lines fr cnvenience, and cntains the fllwing infrmatin. b The bunding b f the ctminid shape with synta yz+t where, y, z are the dimensins f the smallest b that cntains the shape (rtated such that y z. The special bes 024 and 033 crrespnd t the tw basic flat ctminid cnfiguratins, apart frm these, we have eactly eight distinct bes: 112 (clred purple in [2]), 113 (green), 114 (blue), 122 (red), 123 (grey), 133 (dark grey), 222 (light blue), 223 (yellw). The number fllwing the + sign 0 t 8 dentes the number f tiles f the cnfiguratin that lie at the bundary f the bunding b; plyminid Describes the actual 3D shape, there is a sequence f eight grups f triplets f digits. Each grup describes the 3D psitin f a tile as fllws: bserve that in each grup eactly ne f the three digits is even, after dividing all digits by tw we btain the 3D crdinates f the center f the tile. This uniquely determines the rientatin f the tile by bserving that the nly integral crdinate indicates the directin f the nrmal vectr t the tile; f The number f flaps in the cnfiguratin; Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

61 Maurizi Palini 61 delta This is the cmputed value f the metric invariant intrduced in Sectin 5; linking This is the cmputed linking number L (tplgical invariant) intrduced in Sectin 6; symcunt The rder f the grup f symmetries. Fr eample, a cnfiguratin that is mirrr symmetric will have symcunt at least tw (Figure 9, center, is an eample). The pssible values are 1, 2, 4, 8. The cnfiguratin f Figure 9 (right) is cmpletely unsymmetric, hence symcunt is 1. There are 4 cnfiguratins with the maimal value 8 fr symcunt, but nly ne f these: RRmRRmRRmRRm has bth vanishing invariants and has the shape f the lateral surface f a square prism f side 2 and height 1. Because f the marked diagnals f the tiles, this cnfiguratin is nt symmetric with respect t reflectin abut an intermediate plane parallel t the base f the prism, thus reducing the number f symmetries frm 16 (symmetries f a square prism) t 8; typeinv Usually, echanging the type f all the tiles frm t and viceversa prduces a nn-equivalent cnfiguratin (typeinv=n). Hwever, 141 f the 1291 feasible cnfiguratins prduce an equivalent cnfiguratin upn such echange (typeinv=yes), 50 f them have vanishing invariants. One f these is shwn in Figure 9 (left). It is clear frm the descriptin abve that the sftware cde is able t cmpute bth invariants (in cntrast with the cde fr the planar face-up cnfiguratins described in the net subsectin, fr which the cde is currently nt able t cmpute the linking number). Using ptin -M (./rubiksmagic -M) has the effect f filtering ut all cnfiguratins with nnvanishing metric invariant ( 0). Similarly ptin -T filters ut cnfiguratins with nnvanishing tplgical invariant (L 0). Optin -w allws t display the warp cde described by Verheff in [7]; a few ther ptins are described by./rubiksmagic --help. The rubiksmagic cmmand accepts an argument, in the frm f a magic cde, in which case the cmputatin is limited t the crrespnding cnfiguratin: the cde first cmputes the cannical (equivalent) magic cde and displays all the abve infrmatins fr that cnfiguratin. Finally, the cde is nt limited t the case f the puzzle with eight tiles; ptin -n n can be use t cmpute with the puzzle with n tiles (n must be even). As an eample, the cnfiguratin having magic cde RmRvUvDmLvLmDmUv (Figure 15, left) is analyzed as Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

62 62 eplring the rubik s magic universe $./rubiksmagic RmRvUvDmLvLmDmUv b=222+2 plyminid= s f=0 delta=-4 linking=0 symcunt=2 typeinv=yes $ where ptin -c inhibits the cannizatin f the magic cde. Since 0 such cnfiguratin cannt be cnstructed. The cnfiguratin with magic cde RRRmRmRRLmLm (Figure 15, right) is analyzed as $./rubiksmagic RRRmRmRRLmLm b=113+7 plyminid= s f=2 delta=0 linking=1 symcunt=2 typeinv=yes $ and is nt cnstructible since the tplgical invariant L = 1 des nt vanish. Figure 15: The cnfiguratin n the left has magic cde RmRvUvDmLvLmDmUv. The cnfiguratin n the right has magic cde RRRmRmRRLmLm. These cmputer generated images were btained frm the magic cdes f the cnfiguratins using the PvRay raytracing prgram and the include file prvided in [5]. Planar face-up shapes The name f the eecutable is rubiksmagic2d. If run withut arguments, it will search fr all cannical representatives f the set S f equivalent classes f sequences. This is part f its utput: Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

63 Maurizi Palini 63 $./rubiksmagic EEEEWWWW f=2 area=5 Dc=0 symcunt=8 assemblages=1 deltaiszer=1 EEENWWWS f=0 area=8 Dc=0 symcunt=4 assemblages=1 deltaiszer=1 EEENWWSW f=1 area=7 Dc=-2 symcunt=1 assemblages=2 deltaiszer=1 [...] ENSWENSW f=4 area=3 Dc=0 symcunt=8 assemblages=0 deltaiszer=0 EWEWEWEW f=8 area=2 Dc=0 symcunt=32 assemblages=0 deltaiszer=0 Fund 71 sequences $ It searches fr all admissible sequences that are the cannical representative f their equivalent class in S (it finds 71 equivalent classes), fr each ne it prints the sequence fllwed by sme infrmatin (t be eplained shrtly). The sftware allws fr puzzles with a different number f tiles, fr eample fr the large versin with 12 tiles f the puzzle it finds 4855 equivalence classes, with a cmmand like $./rubiksmagic EEEEEEWWWWWW f=2 area=7 Dc=0 symcunt=8 assemblages=1 deltaiszer=1 EEEEENWWWWWS f=0 area=12 Dc=0 symcunt=4 assemblages=1 deltaiszer=1 EEEEENWWWWSW f=1 area=11 Dc=-2 symcunt=1 assemblages=2 deltaiszer=1 [...] ENSNSWENSNSW f=8 area=3 Dc=0 symcunt=4 assemblages=0 deltaiszer=0 ENSNSWENSWEW f=8 area=3 Dc=0 symcunt=2 assemblages=0 deltaiszer=0 ENSWENSWENSW f=6 area=3 Dc=0 symcunt=12 assemblages=0 deltaiszer=0 EWEWEWEWEWEW f=12 area=2 Dc=0 symcunt=48 assemblages=0 deltaiszer=0 Fund 4855 sequences $ Hwever the cmputatinal cmpleity grws epnentially with the number f tiles. Anther use f the cde allws t ask fr specific prperties f a given sequence, we illustrate this with an eample: $./rubiksmagic EEWENWSW f=3 area=5 Dc=-2 symcunt=1 assemblages=6 deltaiszer=2 Assemblage with delta = 0: sla E3 E2 W2 E1 N1 W1 S1 W1 Assemblage with delta = 0: sla E3 E2 W1 E1 N1 W1 S2 W1 $ The first line f utput displays sme infrmatin abut the sequence given in the cmmand line, specifically we find Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

64 64 eplring the rubik s magic universe ˆ the sequence itself; ˆ the number f flaps (3 in this case); ˆ the area f the plane cvered (5); ˆ the cmputed cntributin c cming frm the curving tiles; ˆ the cardinality f the grup f symmetries f the sequence, this particular sequence des nt have any symmetry; ˆ the number f admissible assemblages f the sequence, cunting nly thse that start with T 0 f type and identifying assemblages that are equivalent under transfrmatins in the grup f symmetries f the sequence; ˆ the number f admissible assemblages with vanishing metric invariant ( = 0), we have tw in this case. Then we have ne line fr each f the pssible assemblages with = 0 with a printut f each assemblage, the numbers after each cardinal directin tells the level f the tile reached with that directin. It will be 1 fr tiles that are nt superpsed with ther tiles, therwise it is an integer between 1 and the number f superpsed tiles. The ptin -c n the cmmand line can be mitted in which case the sftware cmputes the cannical representative f the given sequence and prints all the infrmatins fr bth the riginal sequence and the cannical ne. Nte that the sign f the invariants is sensitive t equivalence transfrmatins. References [1] Rubik s Magic - Wikipedia, s_magic, retrieved Jan 15, [2] Köller, J.. Rubik s Magic, retrieved Jan 15, [3] Nurse, J.G. Simple Slutins t Rubik s Magic, New Yrk, [4] Palini, M. Rubik s Magic, retrieved Jan 15, [5] Palini, M. rubiksmagic prject, Subversin repsitry, 2015, [6] Scherphuis, J. Rubik s Magic Main Page, retrieved Jan 15, [7] Verheff, T. Magic and Is Nh Magic, Cubism Fr Fun, 15, 24 31, Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

65 Mathmagic On a mathematical mdel fr an ld card trick Ry Quinter Department f Mathematics, University f Iwa, USA ry-quinter@uiwa.edu Abstract: T he three-pile trick is a well-knwn card trick perfrmed with a deck f 27 cards which dates back t the early seventeenth century at least and its bjective is t uncver the card chsen by a vlunteer. The main purpse f this research is t give a mathematical generalizatin f the three-pile trick fr any deck f ab cards with a, b 2 any integers by means f a finite family f simple discrete functins. Then, it is prved each f these functins has just ne r tw stable fied pints. Based n this findings a list f 222 (three-pile trick)-type brand new card tricks was generated fr either a package f 52 playing cards r any apprpriate prtin f it with a number f piles between 3 and 7. It is wrth nting that all the card tricks n the list share the three main prperties that have characterized the three-pile trick: simplicity, self-perfrming and infallibility. Finally, a general perfrming prtcl, useful fr magicians, is given fr all the cases. All the emplyed math techniques invlve naive thery f discrete functins, basic prperties f the qutient and remainder f the divisin f integers and mdular arithmetic. Keywrds: Mdular arithmetic, fied pint thery f discrete functins, three-pile trick. Intrductin Amng ther reasns, the three-pile trick (TPT) is preferred by mathematicians because f its infallibility and simplicity. The rigin f this interesting card trick ges back arund fur centuries [1]. Accrding t Gardner [6] fr a pack f 27 cards it is ne f the ldest f mathematical tricks that invlve the rdering f cards, and ne f the mst intriguing. Prbably, the first persn t whm the TPT was tractable in a frmal way mathematically speaking was Jseph Diez Gergnne [7], a French gemeter frm the nineteenth century. He generalized the trick fr any deck f m m cards (m > 1 any integer) by applying sme cmbinatrial ideas. Additinal papers were written alng that century Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

66 66 n a mathematical mdel fr an ld card trick incrprating new results [5, 9], but in all f them the technique emplyed was similar t Gergnne s. During the past century the trick was included in bks n recreatinal mathematics, the main are by Ball and Ceter [2], Hugard [10], and Gardner [6]. It might sund strange saying that nly a few mathematical advances n the prblem and its generalizatin were achieved in the twentieth century. But lately the TPT has recvered its prminence as a math bject in varius cntets. Fr instance, it has been treated in an algrithmic manner [12], als as a numerical prcess in cnnectin with the psitinal ntatin system [3, 11], and as a discrete dynamical system [4, 8] t. In this nte the trick is seen as a simple functin frm a finite set f integers t itself. The TPT and its generalizatin Let us begin by intrducing sme basic ntatin. Given any pair f nnnegative integers j k the set f natural numbers {j,..., k} is dented by [j : k]. The symbls q(, a) and r(, a) represent the qutient and the remainder respectively when the integer is divided by the nnzer integer a. Given any functin D f D, the cmpsite functin f f (n times) is dented by f n and the set f its fied pints by F i(f), i.e., F i(f) = { D f() = }. As it was established in [11], given the c th card (cunting frm the tp) in a deck f 27 cards face dwn is said that the TPT is perfrmed n it, if the functin G 2 : [1 : 27] [1 : 27] defined by q(, 3) sign(r(, 3)) is cmpsed three times at = c. Befre cntinuing with ur presentatin, let us figure ut why the algebraic epressin q(, 3) sign(r(, 3)) encdes what is happening with the TPT. In fact, a simple eample will make it clear. Suppse that the chsen card is the 19th (cunting frm the tp). Well, q(19, 3) = 6 gives us the number f cards belw it after dealing the cards in three piles when the TPT is perfrmed. S, after cllecting the three piles the 19th card will ccupy the 16th psitin (remember that piles while cllecting are all flipped ver) which is equal t the sum f 6 (= q(19, 3)), 9 (the cards f the pile placed at the tp) and 1 (= sign(r(19, 3))). T precise mre the ntatin emplyed, the sub-inde 2 in G 2 indicates that after dealing the whle deck f 27 cards the pile cntaining the chsen card will be placed in the middle f the ther tw but this is equivalent t say that this particular pile is the secnd ne cunting frm the bttm after cllecting the three piles face dwn. Tw imprtant prperties f this discrete functin are: 1. Its nly fied pint is 14 (i.e., F i(g 2 ) = {14}). 2. The cmpsite functin G (i.e., G3 2 () = 14 fr every [1 : 27]). Remark 1. What makes the TPT an infallible card trick is prperty 2. Als Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

67 Ry Quinter 67 bserve ne has t apply the same prcedure three times and this is what makes the trick easy t perfrm and impressive in sme sense, hence its simplicity. Let us shw hw the TPT shuld be generalized in a natural way. T start just assume that ur deck has ab cards (face dwn). If we apply the trick with a piles, then each will have b cards. Of curse the trick is trivial when a, b = 1. Frm nw n, let us assume a, b 2. In [11], the general case under study was als cnsidered and the apprpriate functin fr a deck f ab cards is G p : [1 : ab] [1 : ab] defined by q(, a) + b(p 1) + sign(r(, a)) fr every p [1 : a]. In this general setting, the value p indicates that when the new trick is perfrmed the pile with the chsen card will be placed in the p-th psitin cunting frm the bttm nce the a piles are being recllected. Our first mathematical result is Lemma 2 which gives a list f relevant prperties f the functins G 1,..., G a. Lemma 2. Let a, b 2 be fied integers. The finite family f functins {G p } a p=1 satisfies the fllwing prperties: 1. The functin G p is nndecreasing fr every p. 2. Im(G p ) = [b(p 1) + 1 : bp] fr every p. 3. If p 1 < p 2, then G p1 < G p2. 4. The functin G 1 has 1 as its unique fied pint, i.e., F i(g 1 ) = {1}. 5. The functin G a has ab as its unique fied pint, i.e., F i(g a ) = {ab}. 6. Assume a 3. Fr each p [2 : a 1] hlds: (a) If a 1 b(p 1), then F i(g p ) = { p,1, p,1 + 1} where p,1 = (b) If a 1 b(p 1), then F i(g p ) = { p,2 } where p,2 = ab(p 1) a 1. (1) ab(p 1) + a j(p) a 1 being j(p) the unique integer in [2 : a 1] that is slutin f the mdular equatin X ab(p 1) + a md (a 1). Prf. 1. Let 2 p a. Frm the definitins f G 1 and G p fllws the relatinship G p () = G 1 () + b(p 1) fr every [1 : ab]. S, we nly need t prve that G 1 is nndecreasing. Set I k = [(k 1)a + 1 : ka] fr k = 1,..., b. Clearly, the finite class {I 1,..., I b } is a partitin f [1 : ab]. By bserving carefully the behavir f G 1 n each interval f the partitin we have Im(G 1 Ik ) = {k} fr k = 1,..., b, and frm this fllws that G 1 is certainly nndecreasing. (2) Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

68 68 n a mathematical mdel fr an ld card trick 2. By the latter cnditin G 1 maps [1 : ab] n [1 : b] and frm this we get the epected result. 3. It fllws immediately frm prperty Let us see that 1 is fied by G 1. In fact, G 1 (1) = q(1, a) + b(1 1) + sign(r(1, a)) = = 1. Suppse nw that [1 : ab] is any fied pint f G 1. Hence satisfies q(, a) + sign(r(, a)) =. Observe that r(, a) has t be nnzer. Otherwise, q(, a) = 0, and = 0 which is a cntradictin. Then, it hlds q(, a) = 1 r(,a) a 1, but q(, a) 0 and 1 < 1 r(,a) a 1 0. S, r(, a) = 1, and = q(, a)a + r(, a) = 1. Therefre, F i(g 1 ) = {1}. 5. First, ab is a fied pint f G a since G a (ab) = q(ab, a)+b(a 1)+sign(r(ab, a)) = b + b(a 1) + 0 = ab. Nw, suppse [1 : ab] is any fied pint f G a. Hence satisfies q(, a) + b(a 1) + sign(r(, a)) =. We claim that r(, a) = 0. Otherwise, by writing = q(, a)a + r(, a) in the latter equatin and slving fr b q(, a) fllws b q(, a) = r(,a) 1 a 1, but b q(, a) is a nnnegative integer and 0 r(,a) 1 a 1 < 1. S, r(, a) = 1, q(, a) = b, and = ab + 1 which is a cntradictin. Then, it hlds (b q(, a))(a 1) = 0. Thus, q(, a) = b, and = ab. Therefre, F i(g a ) = {ab}. 6. (a) By hypthesis, p,1 is well defined. Since 0 < ab(p 1)/(a 1) < ab, p,1 [1 : ab 1] and p,1 + 1 [2 : ab]. S, bth values are included in the dmain f G p. Nw bserve that and G p ( p,1 ) = q( p,1, a) + b(p 1) + sign(r( p,1, a)) = b(p 1) a 1 + b(p 1) + 0 = p,1 G p ( p,1 + 1) = q( p,1 + 1, a) + b(p 1) + sign(r( p,1 + 1, a)) = b(p 1) a 1 + b(p 1) + 1 = p, Thus, bth pints are fied by G p. Net, assume that [1 : ab] is any fied pint f G p. We claim that either r(, a) = 0 r r(, a) = 1. Otherwise, 2 r(, a) a 1 and satisfies q(, a) + b(p 1) + 1 =. Multiply the latter equatin by a and epress q(, a)a = r(, a) in rder t get ab(p 1)+a r(, a) = (a 1) which indicates that a 1 ab(p 1)+a r(, a), but a 1 ab(p 1). Then, a 1 a r(, a) an impssibility because 1 a r(, a) a 2. S, r(, a) {0, 1} as claimed. The first case makes = p,1 and the secnd = p, Therefre, F i(g p ) = { p,1, p,1 + 1}. This prves 6(a). (b) Suppse that [1 : ab] were a fied pint f G p. If r(, a) = 0 r r(, a) = 1, then shuld satisfy q(, a) + b(p 1) = q(, a)a. Frm which we get b(p 1) = q(, a)(a 1). In ther wrds, a 1 b(p 1) cntradicting the hypthesis. Therefre, r(, a) must belng t {2,..., a 1}, and must satisfy q(, a) + b(p 1) + 1 =. Multiply the latter equatin by a and replace q(, a)a Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

69 Ry Quinter 69 by r(, a) t get ab(p 1) + a r(, a) = (a 1). In ther wrds, r(, a) is cngruent t ab(p 1) + a mdul a 1, i.e., r(, a) is the unique slutin f equatin X ab(p 1) + a md (a 1). Let j(p) be this slutin. Thus, the value p,2 is well defined and = p,2. Mrever, j(p) = r( p,2, a) and G p ( p,2 ) = q( p,2, a) + b(p 1) + sign(r( p,2, a)) = 1 a [ p,2 + ab(p 1) + a j(p)] = 1 a [ p,2 + p,2 (a 1)] = p,2, as epected, and this cmpletes the prf f 6(b). An immediate cnsequence f Lemma 2 is the fllwing crllary. Crllary 3. Let a 3 and n 2 be fied integers. If b = a n 1, then F i(g p ) = { p,2 } fr every p [2, a 1]. Prf. It fllws frm the fact a 1 a n 1 (p 1) fr every p [2 : a 1]. In [4, Therem 3] the TPT was generalized fr a deck f pq cards where p, q 3 are any dd integers thrugh a functin dented by h and prved that pq+1 2 is a fied pint f h. Well, that result als fllws frm Lemma 2 after making sme easy symblic translatins. Crllary 4. Therem 3 given in [4] hlds. Prf. Befre starting the prf we need t translate ur ntatin t Champanerkar and Jani s. Well, a = q, b = p, and G q+1 = h. When p, q 3 2 are dd numbers then q 1 p( q+1 2 1). Otherwise p wuld be an even integer. By Lemma 2.6(b), we have t find the unique fied pint q+1 2,2 f h by slving first the mdular equatin X pq(q 1)/2 + q md (q 1) in [2 : q 1]. Since (pq(q 1)/2 + q) (q + 1)/2 = ((pq + 1)/2)(q 1) the slutin searched fr is j( q+1 2 ) = q+1 2. Then, pq + 1 q+1 2,2 = 2 as epected. When dealing the cards fr the first time prviding that a 1 b(p 1) (2 p a 1) the values p,1 and p,1 + 1 always fall in the a th pile and the 1 st pile respectively. By taking s = b(p 1) a 1 [1 : b], p,1 = sa and p,1 + 1 = sa + 1, bth are indicated inside a b in Table 1. Fied pints f type p,2 (2 p a 1) nly appear at intermediate clumns and can nt be lcated at the 1 st and b th rws. Let us see the eamples a = 4, b = 6 and a = 3, b = 5. Eample 5. Since 3 6 we need t apply equatin (1). S, 2,1 = 8 and 3,1 = 16. Hence F i(g 2 ) = {8, 9} and F i(g 3 ) = {16, 17} (Figure 1(a)). Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

70 70 n a mathematical mdel fr an ld card trick Table 1: First dealing pile 1 pile 2... pile j... pile a 1 pile a (b 1)a + 1 (b 1)a (b 1)a + j... (b 1)a + a 1 ba sa + 1 sa sa + j... sa + a 1 (s + 1)a (s 1)a + 1 (s 1)a (s 1)a + j... (s 1)a + a 1 sa a + 1 a a + j... a + a 1 2a j... a 1 a Eample 6. Since 2 5 we need t slve the mdular equatin X 18 md 2 whse unique slutin inside interval [2 : 2] is j(2) = 2. Then, by applying frmula (2), we get 2,2 = 8. Hence F i(g 2 ) = {8} (Figure 1(b)). Frm Eamples 5 and 6 ne might ask whether the finite class {F i(g p )} a 1 p=2 is always frmed by sets f ne element r by sets f tw elements. Net eample shws that it can be mied. Eample 7. Take a = 5, b = 2 (Figure 1(c)). Then F i(g 2 ) = {3}, F i(g 3 ) = {5, 6}, and F i(g 4 ) = {8}. G 4 G 3 G 2 G 1 G 3 G 2 G 1 (a) (b) (c) Figure 1: Eamples 5, 6, and 7. G 5 G 4 G 3 G 2 G 1 Main results In this sectin we prve sme imprtant facts f the family f functins {G p } a p=1 in terms f discrete dynamical systems. The first result generalizes the TPT because in this case b = 9 is a pwer f a = 3. Therem 8(b) shws that p,2 is an attractr and this is the cnditin that makes the TPT wrk. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

71 Ry Quinter 71 Let us fi mre ntatin. Let A, B, C and D dente the sets [2 : a 1], [1 : ab], [1 : p,1 ], and [ p,1 + 1 : ab] respectively. Set y n := an 1 a 1 fr every n 2. Therem 8. If p A and b = a n 1, then (a) p,2 = y n (p 1) + 1, and (b) G n p () = p,2 fr every B. Prf. (a) Observe that a n (p 1) + a p = (a n 1)(p 1) + (a 1) = (y n (p 1) + 1)(a 1). Thus, by Lemma 2.6(b), p,2 = y n (p 1) + 1. (b) By [11, Thérème II.2.4.] we have G n p () = (p 1)a n (p 1)a + (p 1) + 1, but the latter value cincides with y n (p 1) + 1. By taking a = 3 and b = 9 in Therem 8 we get 2,2 = y = 14 and G 3 2() = 14 fr every [1 : 27]. In ther wrds, the TPT is checked. The fllwing therem cnstitutes the mst imprtant result f this research. Therem 8(b) and Therem 9 guarantee that all the fied pints btained in Lemma 2 parts 4, 5 and 6 are attractrs. Based n this prperty we will be able t etend in the net sectin the TPT by listing mre than 200 new variants f the trick by varying the number f cards in the deck, the number f piles and the perfrmance prtcl. Therem 9. Let a 3, b 2 and n 2 be fied integers. If p [1 : a] and a n 2 < b < a n 1, then G n p () = 1 if (p, ) {1} B (3a) p,1 if (p, ) A C, a 1 b(p 1), b a n 2 +1 (3b) p,1 + 1 if (p, ) A D, a 1 b(p 1), b a n 2 +1 (3c) p,2 if (p, ) A B, a 1 b(p 1), b an 1 (3d) a 2 ab if (p, ) {a} B, (3e) { G n 1 p,1 if (p, ) A C, a 1 b(p 1), b = a n 2 +1 (4a) p () = p,1 + 1 if (p, ) A D, a 1 b(p 1), b = a n 2 +1, (4b) and fr a > 3 G n+1 p () = p,2 if (p, ) A B, a 1 b(p 1), b > an 1 a 2. (5) Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

72 72 n a mathematical mdel fr an ld card trick Prf. Equatins (3a) and (3e) were prved in [11, Thérème II.2.1.] and [11, Thérème II.2.2.] respectively. Fr the remaining si parts we use the same technique emplyed in [11]. Fr (3b), take p,1 arbitrary, say, = (s 1)a + j as is shwn in Table 1, then s b(p 1) a 1 = p,1 a. Frm [11, Thérème II.1.] it fllws that A(s) G n p () B(s), (6) where A(s) = 1 a n 1 ( p,1 a (an 1) + s ) and B(s) = A(s) a n 1. Our net gal is t demnstrate the fur inequalities p,1 1 < A(s) p,1 B(s) < p,1 + 1 (7) fr every s a 1 p,1. After ding many calculatins and simplificatins we get p,1 1 < A(s) p,1 < a n + sa (8a) A(s) p,1 p,1 sa (8b) p,1 B(s) p,1 a n + sa a (8c) B(s) < p,1 + 1 p,1 > sa a. (8d) (8e) Observe that the Right Hand Side (RHS) (8b) the RHS (8d) and the RHS (8c) the RHS (8a). But the RHS (8b) hlds fr the cnditin s a 1 p,1 and the RHS (8c) als hlds since p,1 ab a n a n + sa a. S, equatin (7) is true and this implies that G n p () = p,1 fr every C cmpleting the prf f (3b). The prf f (3c) is quite similar. Nw we take > p,1, which is equivalent t s a 1 p, This time we have t prve p,1 < A(s) p,1 + 1 B(s) < p,1 + 2 (9) fr every s a 1 p, But nw, we have the fllwing equivalences p,1 < A(s) p,1 < sa (10a) A(s) p,1 + 1 p,1 sa a n (10b) p,1 + 1 B(s) p,1 sa a (10c) B(s) < p,1 + 2 p,1 > sa a n a. (10d) (10e) In this ccasin, the RHS (10b) the RHS (10d) and the RHS (10c) the RHS (10a). But the RHS (10b) hlds since p,1 + a n a n ab sa and the RHS (10c) als hlds fr the cnditin s a 1 p, S, equatin (9) is true and this implies that G n p () = p,1 + 1 fr every D cmpleting the prf f (3c). Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

73 Ry Quinter 73 Fr (3d), take B arbitrary, say, = (s 1)a + j as is shwn in Table 1. Frm [11, Thérème II.1.] it fllws that C(s) G n p () D(s), (11) where C(s) = 1 a n 1 (by n + s) and D(s) = C(s) a n 1. Net, we need t demnstrate the fur inequalities p,2 1 < C(s) p,2 D(s) < p,2 + 1 (12) fr every s [1 : b]. And nw, the equivalences are p,2 1 < C(s) f(s) > a n 1 (1 j(p)) (13a) C(s) p,2 f(s) a n 1 (a j(p)) (13b) p,2 D(s) f(s) a n 1 (1 j(p)) + a 1 (13c) D(s) < p,2 + 1 f(s) < a n 1 (a j(p)) + a 1, (13d) where f(s) = s(a 1) b(p 1), but the RHS (13b) the RHS (13d) and the RHS (13c) the RHS (13a). Then, we have t shw the duble inequality a n 1 (1 j(p)) + a 1 f(s) a n 1 (a j(p)), fr every s [1 : b]. Fr this, it is enugh t prve these tw inequalities a n 1 (1 j(p)) + a 1 f(1) and f(b) a n 1 (a j(p)) and bth are true since b an 1 a 2. Thus, equatin (12) hlds and it implies that G n p () = p,2 fr every B cmpleting the prf f (3d). In rder t prve equatins (4a) and (4b), we nly need t prve that equatins (8c) and (10b) keep true if n is replaced by n 1. The first is true because p,1 (b 1)a = a n 1 a n 1 + sa a, and the secnd since p,1+an 1 a a+a n 1 a = 1 + a n 2 = b s. Finally, t prve equatin (5), everything dne fr prving equatin (3d) wrks well fr n + 1 instead f n. But, b satisfies the cnditin an 1 a 2 < b < an 1 (a > 3). Thus, b an a 2 and this cnditin guarantees the veracity f frmula (5), and the prf is dne. Remark 10. In [4, Therem 4] was prved that the pint pq+1 2 is a stable fied pint f h the generalized TPT fr a deck f pq cards where p, q 3 are dd integers. Well, frm Therem 8(b) and Therem 9 equatins (3d) and (5) fllws the same result. The minimum value f k fr which the pwer G k p is a fied pint is nt necessarily the ne given in Therem 9. T see this, just take a = 4, b 1 = 13 and b 2 = 14. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

74 74 n a mathematical mdel fr an ld card trick Fr p = 3 Therem 9 says that n = 3 and G fr b 1 and G fr b 2 since 4 < 13 < 14 < 4 2 and 14 > 13 > = 8. In this particular situatin the minimum value is 3 in the first case and 4 in the secnd. Fr a better understanding n the behavir f the family {G 1,..., G a }, based n Therem 9, we present a cuple f eamples graphically. Figure 2 shws the case a = 3, b = 4 fr which n = 3 and the fied pints are 1, 6, 7, and 12. Observe that fr p = 1, 3 three iteratins are needed, therwise just tw. G 3 G 2 G 1 G 2 3 G 2 2 G 2 1 G 3 3 G 3 2 G 3 1 Figure 2: a = 3, b = 4. Figures 3 and 4 shw the case a = 4, b = 14. Again n = 3, and the fied pints are 1, 19, 38, and 56. Figure 3 shws the graphics f G p and G 2 p, and Figure 4 the crrespnding nes f G 3 p and G4 p. Observe that fr p = 1, 4 three iteratins are needed, therwise fur. G 4 G 3 G 2 G 1 G 2 4 G 2 3 G 2 2 G 2 1 Figure 3: a = 4, b = 14 (G p and G 2 p). Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

75 Ry Quinter 75 G 3 4 G 3 3 G 3 2 G 3 1 G 4 4 G 4 3 G 4 2 G 4 1 Figure 4: a = 4, b = 14 (G 3 p and G 4 p). A bunch f brand new TPT-type card tricks Based n Therems 8 and 9 we are in a psitin f generating sme brand new TPT-type card tricks. Fr magicians sme reasnable deck s sizes and perfrmable with a prtin r the whle pack f regular playing cards are thse with ab [8 : 52] (3 a 7). The 49 pssible such decks and the 222 TPT-type variants f tricks are shwn at Table 2. Let us see the case a = 4, b = 13. Eample 11. (Perfrming prtcl) Previusly, the magician has t chse the value f p, say p = 3 (1 p a). Then, she lks at Table 2 t get n = 3 and the fied pint 35. Nw, by the cmment after Therem 9, she knws that G 3 3 () = 35 fr every [1 : 52]. Immediately, she has t perfrm the fllwing general prcedure: Start with the cards dwnwards. Ask smebdy frm the audience t remember ne f the cards withut revealing it. Deal the 52(= ab) cards face up in 4(= a) rws, making 4 piles each 13(= b) high. After dealing, ask in which pile the card is. Take up the 4 piles, placing that pile 3 rd (= p th ) frm the bttm. Turn the packet ver. D this twice mre and cunt ut the packet t the 35 th (= th p,2 ) card and turn it ver t yur audience s amazement and vatin. Suppse the magician chses p fr which G p has tw cnsecutive fied pints, i.e., p,1, p, Once she picks up the cards fr the last time and cunts ut the packet t the th p,1 card she des nt knw whether it is the spectatr s r net. T slve this situatin she may add sme patter at this mment and puts bth cards tgether back t back and hlds them with tw fingers withut shwing them. Then, she asks the spectatr: is this yur card? while shwing ne. If the answer is yes, everything is dne, therwise she turns her hand and says here is yur card and receives anyway her audience s applause. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

76 76 n a mathematical mdel fr an ld card trick Table 2: The 49 decks and the 222 TPT-type variants a b; n fied pints fr each p b; n fied pints fr each p 3 3; ;3 1-6, ; ;3 1-9, ; ;3 1-12, ; ;4 1-15, ; ;4 1-18, ; ;4 1-21, ; ;4 1-24, ; ; ;2 1-4,5-8, ; ; ; ; ; ;3 1-12,13-24, ; ; ;3 1-16,17-32, ; ; , ; ;2 1-5,6-10,11-15, ; ; , ; ;3 1-10,11-20,21-30, ; ; , ; ; ; ;2 1-6,7-12,13-18,19-24, ; ; ; ; , ; , , ; , ; ;2 1-7,8-14,15-21,22-28,29-35, ; Acknwledgements: I thank Dr. Kasturi Viswanath fr prpsing me a questin f interest frm which emerged this article. References [1] Bachet, C. G. Prblèmes plaisans et délectables, qui se fnt par les nmbres, partie recueillis de divers autheurs, et inventez de nuveau, avec leur démnstratin, par Claude Gaspar Bachet, Sr. de Méziriac. Très utiles pur tutes srtes de persnnes curieuses qui se servent d arithmétique (1612). Editin cnsulted: C. Bachet, Prblèmes plaisants et délectables qui se fnt par les nmbres, Gauthier-Villars, Paris, [2] Ball, W. W. R., Ceter, H. S. M. Mathematical Recreatins and Essays and Prblems f Past and Present Times, Macmillan, Lndn & New Yrk, [3] Blker, E. Gergnne s card trick, psitinal ntatin, and radi srt, Mathematics Magazine, 83, 46-49, Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

77 Ry Quinter 77 [4] Champanerkar, J., Jani, M. Stable Fied Pints f Card Trick Functins, arxiv: v1[math.ho], 1 10, [5] Dicksn, L. Gergnne s pile prblem, Bull. Amer. Math. Sc., 1, , [6] Gardner, M. Mathematics Magic and Mystery, Dver Publicatins Inc., Minela, N. Y., [7] Gergnne, J. D., Récréatins Mathématiques: Recherches sur un tur de cartes, Annales de Mathématiques Pures et Appliquées, 4, , [8] Harrisn, J., Brennan, T., Gapinski, S. The Gergnne p-pile prblem and the dynamics f the functin ( + r)/p, Discrete Applied Mathematics, 82, , [9] Hudsn, C. T. Slutin f questin 2594, Educatinal Times Reprints, 9, 89 91, [10] Hugard, J. Encyclpedia f card tricks, Faber and Faber, Lndn, [11] Quinter, R., Gerini, C. Le tur de cartes de Gergnne, Quadrature, 78, 8 17, [12] Quinter, R. Un algritm basad en el frmalism de aplicación-códig, 2d. Cngres Venezlan de Ciencia, Tecnlgía e Innvación, 7th-10th Nvember, Caracas, Venezuela, Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

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79 Games and Puzzles several bunds fr the k-twer f hani puzzle Stephen B. Gregg, Britney Hpkins, Kristi Karber, Thmas Milligan, Jhnny Sharp University f Central Oklahma TMilligan1@uc.edu Abstract: We cnsider special cases f a mdified versin f the Twer f Hani puzzle and demnstrate hw t find upper bunds n the minimum number f mves that it takes t cmplete these cases. Keywrds: Twer f Hani Puzzle, minimum number f mves. Intrductin The Twer f Hani puzzle is a game that is played with disks f graduated sizes n three pegs. When the game begins, the disks are arranged n the first peg accrding t their size. The largest disk is lcated at the bttm f the peg and abve it are all f the ther disks arranged in increasing size. The smallest disk is therefre at the tp f the peg. One must relcate all f the disks t the third peg, but there are strict rules abut hw the disks are mved. Mving ne disk at a time, and never placing a larger disk n tp f a smaller ne, the player slwly makes their way t the end f the puzzle, with the stack f disks back in their riginal rder n the third peg. The number f mves that it takes t cmplete the puzzle grws epnentially with the number f starting disks. It is fr this reasn, that when the priests at the temple f Benares were asked t cmplete the same task, but with sity-fur disks, the wrld wuld end at their cmpletin. The legend speaks f sity-fur disks f gld being placed n a brass plate with diamnd needles acting as the pegs. The priests were required t mve each disk, ne at a time, until the disks were back in their riginal rder n the third diamnd peg. Using the same cnditin that n gld disk ever be placed upn a smaller ne, their task wuld eceed 500 billin years befre finishing. Suppse we relaed the rule demanding the priests nt place a glden disk n tp f a smaller ne by nly requiring the bttm disk in any stack t be the largest. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

80 80 several bunds fr the k-twer f hani puzzle Then hw lng wuld it take the priests t finish their divine task? What abut if we nly required the bttm tw disks in any stack be the largest and allwed them t place any smaller disks n tp f thse? Furthermre, suppse we cntinue this and ask hw lng it will take t cmplete this puzzle if we allw a disk t be placed n tp f a smaller disk, prvided that the k disks n the bttm f the stack be the largest in the stack such that thse k disks are arranged frm largest n bttm, increasing in size t the k th disk? It is this questin that is f interest in this paper. The resulting puzzle will take lnger t slve fr larger values f k, and it can be slved relatively quickly fr smaller values f k. We will discuss several upper bunds n the ptimal slutins fr the minimum number f mves that it takes t cmplete these type f puzzles. In Sectin 2, we discuss the fundamentals f the apprach we will take t btain ur upper bunds. Net, in Sectin 3, we give upper bunds fr sme specific cases when the value k depends n the number f starting disks, n. Then in Sectin 4, we give an upper bund fr the case whenever k is three and the number f starting disks can be any value greater than r equal t k. Finally, in Sectin 5, we discuss ur future wrk t give a bund fr an arbitrary value f k and an arbitrary number f disks, n. Preliminary The structure f the prfs fr the upper bunds f the mdified Twer f Hani puzzles (called k-twer f Hani puzzles) is based n the cncept f building stacks f disks, frm the n starting disks. Because the gal f the puzzle is t relcate all disks t the third peg, at sme pint in the puzzle, the largest disk must be placed nt the empty third peg. This can nly be dne when the largest disk is the nly disk n the first peg and the remaining (n 1) disks are n the secnd peg. These remaining disks frm a stack. Nw befre we can have a stack f (n 1) disks, we must have a stack f (n 2) disks s that the secnd largest disk can be relcated, and s n. S, lking at the puzzle frm the frward directin, we must first frm a stack f ne, then a stack f tw, etc. until a stack f (n 1). We refer t this as the midpint. After this, we mve the largest disk nt the third peg. T cmplete the puzzle, we then reverse the prcess. We will lk at the puzzle frm the viewpint f building these stacks f disks, and we will lk at sme patterns cncerning hw these stacks are frmed. It shuld be nted that in rder t minimize the number f mves t cmplete the puzzle, we cannt simply minimize the number f mves that it takes t frm each stack, and then add thse tgether. We find that in minimizing the number f mves t frm sme stacks, we greatly increase the number f mves that it takes t frm the net stack. Hwever, there is a way t build successive stacks such that the number f mves f the tw tgether is kept t a minimum. We see that the value f k will determine hw we need t build these stacks. We have written a cmputer prgram t give us pssible mves frm ne stack t the net, with the minimum paths t each cnfiguratin stred (since we d nt knw which cnfiguratin we want t use in the verall minimum slutin). Then fr each f these minimum paths, the prgram repeats the Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

81 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 81 prcess, stpping at every cnfiguratin f the net stack. Once the midpint f the puzzle has been reached, the prgram takes the verall minimum path (which is the cncatenatin f the paths frm the first stack t the secnd stack, and then frm the secnd stack t the third stack, etc.) and uses that as the minimum slutin. Thus, what we are ding is minimizing the number f mves that it takes t build each stack, such that the disks are in sme desired cnfiguratin in each stack. When cmpleting any k-twer f Hani puzzle, ne always makes the first mve based n whether n is dd r even. If n is dd, then ne will mve the first disk t the third peg. If n is even, then ne will mve the first disk t the secnd peg. Because f this, and because ur prfs invlve an arbitrary n, when we speak f mving disks, we will use the terms surce peg, intermediate peg, and destinatin peg. Fr dd values f n, the intermediate peg will be the secnd peg, and the destinatin peg will be the third peg. Fr even values f n, the intermediate peg will be the third peg, and the destinatin peg will be secnd peg. Fr bth cases, the surce peg will be the first peg. Thus, when we build stacks f disks cntaining an even number f disks, they will always be n the intermediate peg, and stacks f disks cntaining an dd number f disks will always be n the destinatin peg. S, whenever we are making mves that depend n whether n is even r dd, we will use this terminlgy. Otherwise, if the peg is knwn, we will use the traditinal terms. With sme f these prfs, when frming stacks f disks, we can take advantage f the special rule invlving k. Fr this, we will use the cncept f flipping a grup f disks. A flip f a grup f disks is the prcess f mving disks, which are at the tp f a stack, in successin t sme ther peg s that their new arrangement is the inverted arrangement that they started in. Nte that a flip f disks takes mves. Als, the peg that the disks are being mved t must cntain at least k disks that are larger than each disk being flipped (s as t nt vilate the rule invlving k). Als, the mves that we can make that take advantage f the rule invlving k will be highlighted in the prfs. Such mves will be cnsidered saves as cmpared t mves made while cmpleting the classic Twer f Hani puzzle, as these are the mves that allw the puzzle t be cmpleted mre quickly. Whenever we make mves that fllw the traditinal Twer f Hani puzzle, the case where k = (n 0), we say that we fllw the classic rules. We ften leave these mves t the reader fr the sake f simplicity. Additinally, because ur prfs invlve many specific mves, we have included figures fr clarity. The figures use arrws cnnecting distinct steps f the puzzles and these arrws dente either a single mve r a grup f mves. Fr grups f mves, we will make them using the classic rules, using a flip, r by reversing mves that were previusly shwn. The type f mve will be listed beneath the arrw. Fr the case f a flip, which always results in sme number f mves being saved (as cmpared t hw the classic Twer f Hani puzzle is cmpleted), we typically indicate the number f saves by a negative number, which will be listed abve the arrw. Clr-cded disks are used t help identify the disks being mved. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

82 82 several bunds fr the k-twer f hani puzzle Rules fr the k-twer f Hani Puzzle Disks can nly be mved ne at a time. Only the tp disk n a stack can be mved frm that peg. A disk can be placed n tp f a smaller disk, prvided that the bttm k disks f that stack are the largest in the stack (where the largest disk is at the bttm, the net largest n tp f it, etc., up t the k th largest disk). Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = (n ) Here we cnsider the k-twer f Hani puzzle where k depends n the number f disks, n. We shw an upper bund T k (n) fr the cases f k = (n ) fr = 0, 1, 2, 3, 4, 5. 2 n 1 if = 0, 1, 2 2 T n (n) = n 3 if = 3 2 n 4n + 3 if = 4, n 7 2 n 20n + 57 if = 5, n 9 Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = (n 0) Since k = (n 0) = n, this puzzle is the classic Twer f Hani puzzle (since n disk can be placed n tp f a smaller ne), s its upper bund is 2 n 1. Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = (n 1) This puzzle requires the same number f mves as the k = (n 0) puzzle because a disk can be placed n tp f a smaller disk prvided the bttm k = (n 1) disks are the largest in that stack. If this cnditin is true, then there is nly ne disk that is nt in the stack f (n 1) disks. Thus, there des nt eist anther disk that can be placed n tp f sme smaller disk. Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = (n 2) Once again, this puzzle requires the same number f mves as the k = (n 0) puzzle because the nly time we can place disks ut f rder is whenever we have a stack f (n 2) disks with the tw remaining disks being smaller than each f the (n 2) disks. Obviusly the nly cnfiguratins that allw this are the beginning and ending cnfiguratins, but these cnfiguratins are fied. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

83 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 83 Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = (n 3) This puzzle has a bund f 2 n 3. Ntice that this is 2 fewer mves than using the classic rules. S, there are tw places in the k = (n 3) case where a mve is saved as cmpared t the k = (n 2) case. The saves ccur when we are building a stack f (n 1) disks. We cmplete this puzzle differently based upn whether n is even r dd. If n is even, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains the n th and (n 1) st largest, and the largest disks; the third peg cntains n disks; and the secnd peg cntains the remainder f the disks. If n is dd, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains largest disk; the third peg cntains the n th and (n 1) st largest disks; and the secnd peg cntains the remainder f the disks. Net, we flip the n th and (n 1) st largest disks nt the secnd peg, which saves 1 mve, getting us t the midpint. Then we mve the largest disk nt the third peg. Net, we reverse the steps prir t mving the largest disk, which saves 1 mre mve and cmpletes the puzzle. See Figure 1. Figure 1: Upper Bund fr k = (n 3). Whenever n = 4, the disk labelled n 2 cincides with the disk labelled 2. Thus, it takes mves t cmplete the puzzle. (2 n 1) 1 1 = 2 n 3 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

84 84 several bunds fr the k-twer f hani puzzle Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = (n 4) T btain this bund, we will divide it int the fllwing steps. Building a stack f 3 Building stacks f 4, 5,..., (n 3) Building a stack f (n 2) Building a stack f (n 1) Building a stack f n Nte that this bund assumes a starting number f disks n 7. Frm the starting cnfiguratin, we build a stack f 3 disks using the classic rules. Thus, it takes = 7 mves. See Figure 2. Figure 2: Building a Stack f 3 Disks Net, we build stacks f 4, 5,..., (n 3) using the classic rules until we reach a cnfiguratin where we can take advantage f the rule invlving k. This cnfiguratin ccurs, fr each f these stacks, right befre we are t place the 2 nd largest disk f the stack we are building nt the largest disk f the stack we are building. Building a stack f 4 disks illustrates this cncept immediately. We first mve the (n 3) rd largest disk (this is the largest disk in the stack f 4 that we are building) frm the surce peg nt the intermediate peg. Ntice that because k = (n 4) and because we have (n 4) disks n the surce peg, we can flip 2 disks frm the destinatin peg t the surce peg, which saves 1 mve. Net, we mve the (n 2) nd largest disk (this is the 2 nd largest disk in the stack f 4 that we are building) frm the destinatin peg nt the intermediate peg. We then flip 2 disks frm the surce peg nt the intermediate peg, which saves 1 mre mve. We have nw cmpleted the stack f 4 disks and we have saved 2 mves. See Figure 3. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

85 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 85 Figure 3: Building a Stack f 4 Disks. Whenever n = 7, the disk labelled n 4 cincides with the disk labelled 3. Fr the net stacks f 5, 6,..., (n 3) disks we repeat this prcess. Fr a stack f an even number f disks, we make mves using the classic rules until we reach a cnfiguratin where the destinatin peg cntains the n th and (n 1) st largest disks and the 2 nd largest disk f the stack we are building; the intermediate peg cntains nly the largest disk f the stack we are building; and the surce peg cntains the remainder f the disks. Fr a stack f an dd number f disks, we make mves using the classic rules until we reach a cnfiguratin where the destinatin peg cntains the n th and (n 1) st largest disks and the largest disk f the stack we are building; the intermediate peg cntains nly the 2 nd largest disk f the stack we are building; and the surce peg cntains the remainder f the disks. Net, we flip the n th and (n 1) st largest disks nt the surce peg, which saves 1 mve. Then we mve the 2 nd largest disk f the stack we are building nt the largest disk f the stack we are building. Net, we flip the n th and (n 1) st largest disks frm the surce peg nt the intermediate peg, which saves 1 mre mve. We then use the classic rules t finish building the stack. S, whenever we are building these stacks, we save 2 mves frm the tw times that we flip disks and the ther mves are dne accrding t the classic rules. See Figure 4. Thus, it takes n 4 ( )+( )+ +(1+2 n 4 1 2) = (2 i 2) = 1 8 (32+2n 16n) mves t build stacks f 4, 5,..., (n 3) disks. i=3 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

86 86 several bunds fr the k-twer f hani puzzle Figure 4: Overview f Building Stacks f 4, 5,, (n 3) Disks. Ntice that the pegs fr the final cnfiguratin are fied and are labelled accrdingly. Net, we wish t build a stack f (n 2) disks. If n is even, we begin by making mves using the classic rules until we reach a cnfiguratin where the secnd peg cntains the n th and (n 1) st largest disks n tp f the 4 th largest disk; the third peg cntains nly the 3 rd largest disk; and the first peg cntains the remainder f the disks. If n is dd, we begin by making mves using the classic rules until we reach a cnfiguratin where the secnd peg cntains nly the 4 th largest disk; the third peg cntains the n th and (n 1) st largest disks n tp f the 3 rd largest disk; and the first peg cntains the remainder f the disks. Then we flip the n th and (n 1) st largest disks nt the first peg, which saves 1 mve. Net, we mve the 4 th largest disk nt the third peg. If n is even, we flip the n th and (n 1) st largest disks nt the third peg. If n is dd, we flip the n th and (n 1) st largest disks nt the secnd peg. This flip saves 1 mre mve. If n is even, we make mves using the classic rules until we reach a cnfiguratin where the first peg cntains the n th and (n 1) st largest, the 2 nd largest, and the largest disks; the secnd peg cntains n disks; and the third peg cntains the remainder f the disks. If n is dd, we make mves using the classic rules until we reach a cnfiguratin where the first peg cntains the 2 nd largest and the largest disks; the secnd peg cntains the n th and (n 1) st largest disks; and the third peg cntains the remainder f the disks. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

87 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 87 Then we flip the n th and (n 1) st largest disks nt the third peg. This flip saves 1 mre mve and cmpletes the stack f (n 2). See Figure 5. Figure 5: Overview f Building a Stack f (n 2) Disks Thus, we save 3 mves and it takes mves t build this stack. 1 + (2 n 3 1) = 2 n 3 3 We nw frm a stack f (n 1) disks. We start by mving the 2 nd largest disk frm the first peg nt the secnd peg. If n is even, we flip the n th and (n 1) st largest disks nt the secnd peg. If n is dd, we flip the n th and (n 1) st largest disks nt the first peg. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

88 88 several bunds fr the k-twer f hani puzzle This flip saves 1 mve. If n is even, we fllw the classic rules until we reach a cnfiguratin where the secnd peg cntains the n th and (n 1) st largest, and the 2 nd largest disk; the third peg cntains the 3 rd largest disk; and the first peg cntains the remaining disks. If n is dd, we fllw the classic rules until we reach a cnfiguratin where the secnd peg cntains nly the 2 nd largest disk; the third peg cntains the n th and (n 1) st largest, and the 3 rd largest disk; and the first peg cntains the remaining disks. We then flip the n th and (n 1) st largest disks nt the first peg. This saves 1 mre mve. Net, we mve the 3 rd largest disk nt the secnd peg. If n is even, we flip the n th and (n 1) st largest disks nt the third peg. If n is dd, we flip the n th and (n 1) st largest disks nt the secnd peg. This saves us 1 mre mve. If n is even, we fllw the classic rules until we reach the cnfiguratin where the first peg cntains the n th, (n 1) st and (n 2) nd largest disks and the largest disk; the third peg cntains n disks; and the secnd peg cntains the remaining disks. If n is dd, we fllw the classic rules until we reach the cnfiguratin where the first peg cntains nly the largest disk; the third peg cntains the n th, (n 1) st and (n 2) nd largest disks; and the secnd peg cntains the remaining disks. We flip the n th, (n 1) st, and (n 2) nd largest disks nt the secnd peg. This saves 4 mre mves and cmpletes the stack f (n 1). See Figure 6. We have nw reached the midpint. S far we have made ( ) ( ) 2 8 (32 + 2n 16n) + (2 n 3 3) + (2 n 2 n 7) = 2 2n + 1 mves. Nw, we wish t frm a stack f n. We d this by mving the largest disk nt the third peg and then reversing the previus steps fr building stacks f 1, 2,..., (n 1). See Figure 7. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

89 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 89 Figure 6: Overview f Building a Stack f (n 1) Disks Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

90 90 several bunds fr the k-twer f hani puzzle Figure 7: Overview f Building a Stack f n Disks Thus, t build this stack it takes ( ) 2 n n + 1 mves. Therefre, t cmplete the puzzle, it takes ( ) ( ) 2 n 2 n 2 2n n + 1 = 2 n 4n + 3 mves. Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = (n 5) T btain this bund, we will divide it up int the fllwing steps. Building a stack f 3 Building a stack f 4 Building stacks f 5, 6,..., (n 4) Building a stack f (n 3) Building a stack f (n 2) Building a stack f (n 1) Building a stack f n Nte that this bund assumes a starting number f disks n 9. Frm the starting cnfiguratin, we build a stack f 3 disks using the classic rules. Thus, it takes = 7 mves. See Figure 8. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

91 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 91 Figure 8: Building a Stack f 3 Disks Net, we build a stack f 4 in the same manner that we built a stack f 4 in the bund fr k = (n 4). This takes 6 mves, as befre. See Figure 9. Figure 9: Building a Stack f 4 Disks We nw wish t build stacks f 5, 6,..., (n 4). These stacks are built in a similar manner that stacks f 4, 5,..., (n 3) fr k = (n 4) were built. Since k = (n 5), we nw nly need a stack f (n 5) in rder t take advantage f the special rule invlving k. As a result, we can flip 3 disks nt stacks f (n 5) instead f flipping 2 disks nt stacks f (n 4) (as we did whenever k = (n 4)). Fr a flip f 3 disks, we save 4 mves each time. S, we build stacks f 4, 5,..., (n 3) by using the classic rules until we reach a cnfiguratin where we can take advantage f the rule invlving k. This cnfiguratin ccurs, fr each f these stacks, right befre we are t place the 2 nd largest disk f the stack we are building nt the largest disk f the stack we are building. Building a stack f 5 illustrates this cncept immediately. We first relcate the (n 4) th largest disk (this is the largest disk in the stack f 5 that we are building) frm the surce peg nt the destinatin peg. Ntice that because k = (n 5) and because we have (n 5) disks n the surce peg, we can flip 3 disks frm the intermediate peg t the surce peg, which saves 4 mves. Net, we mve the (n 3) rd largest disk (this is the 2 nd largest disk in the stack f 5 that we are Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

92 92 several bunds fr the k-twer f hani puzzle building) frm the intermediate peg nt the destinatin peg. We then flip 3 disks frm the surce peg nt the destinatin peg, which saves 4 mre mves. We have nw cmpleted the stack f 5 disks and we have saved 8 mves. See Figure 10. Figure 10: Building a Stack f 5 Disks Fr the net stacks f 6, 7,..., (n 4) disks we repeat this prcess. Fr each f these stacks, we begin by making mves using the classic rules until we reach a cnfiguratin where the surce peg cntains every disk ecept: the largest disk f the stack we are building, the 2 nd largest disk f the stack we are building, and the n th, (n 1) st and (n 2) nd largest disks. Mrever, the n th, (n 1) st and (n 2) nd largest disks must be n a peg with nly the largest disk f the stack we are building r the 2 nd largest disk f the stack we are building. Nte that where these disks are depends n whether the stack we are building cntains an dd r an even number f disks. Fr eample, when building a stack f 5 disks, the n th, (n 1) st and (n 2) nd largest disks will be n the 2 nd largest disk f the stack we were building. When building a stack f 6, they will be n largest disk f the stack we are building. Nw, because we have (n 5) disks n the surce peg, we can take advantage f the special rule invlving k. S, flip the n th, (n 1) st and (n 2) nd largest disks nt the surce peg, which saves 4 mves. Mve the 2 nd largest disk f the stack we are building nt the largest disk f the stack we are building, and then flip the n th, (n 1) st and (n 2) nd largest disks frm the surce peg nt ne f the ther pegs (if we are building a stack f an even number f disks, we flip the disks nt the empty peg, therwise we flip the disks nt the peg cntaining disks), which saves 4 mre mves. We then use the classic rules t finish building the stack. S, whenever we are building these stacks, we save 4 mves frm the tw times that we flip disks and the ther mves are dne accrding t the classic rules. See Figure 11. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

93 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 93 Figure 11: Overview f Building Stacks f 5, 6,, (n 4) Disks. Whenever n = 9, the disk labelled n 6 cincides with the disk labelled 3. S, t build all f these stacks, it takes n 5 ( ) + + (1 + 2 n 5 1 8) = (2 i 8) = (2n 128n + 768) 16 mves. Nw we want t build a stack f (n 3) disks. We start by mving the 4 th largest disk nt the secnd peg. If n is even, we fllw the classic rules until we reach the cnfiguratin where the secnd peg cntains nly the 4 th largest disk; the third peg cntains the n th, (n 1) st and (n 2) nd largest disks and the 5 th largest disk; and the first peg cntains the remainder f the disks. If n is dd, we fllw the classic rules until we reach the cnfiguratin where the secnd peg cntains the n th, (n 1) st and (n 2) nd largest disks and 4 th largest disk; the third peg cntains nly the 5 th largest disk; and the first peg cntains the remainder f the disks. We then flip the n th, (n 1) st and (n 2) nd largest disks nt the first peg, which saves 4 mves. Net, we mve the 5 th largest disk nt the secnd peg. If n is even, we flip the n th, (n 1) st and (n 2) nd largest disks frm the first peg nt the secnd peg. If n is dd, we flip the n th, (n 1) st and (n 2) nd largest disks frm the first peg nt the third peg. This flip saves 4 mre mves. i=4 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

94 94 several bunds fr the k-twer f hani puzzle If n is even, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains the largest, 2 nd largest and 3 rd largest disks; the third peg cntains the n th and (n 1) st largest disks; and the secnd peg cntains the remainder f the disks. If n is dd, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains the n th and (n 1) st largest, 3 rd largest, 2 nd largest and largest disks; the third peg cntains n disks; and the secnd peg cntains the remainder f the disks. Then we flip the n th and (n 1) st largest disks nt the secnd peg, which saves 1 mre mve.see Figure 12. Figure 12: Overview f Building a Stack f (n 3) Disks. Whenever n = 9, the disk labelled n 6 cincides with the disk labelled 3. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

95 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 95 This cmpletes the stack f (n 3) disks and it takes 1 + (2 n 4 1) 9 mves. Nw we want t build a stack f (n 2) disks. We begin by mving the 3 rd largest disk nt the third peg. If n is even, we flip the n th and (n 1) st largest disks frm the secnd peg nt the first peg. If n is dd, we flip the n th and (n 1) st largest disks frm the secnd peg nt the third peg. This flip saves 1 mve. If n is even, we fllw the classic rules until we reach the cnfiguratin where the secnd peg cntains nly the 4 th largest disk; the third peg cntains the n th, (n 1) st and (n 2) nd largest disks and the 3 rd largest disk; and the first peg cntains the remainder f the disks. If n is dd, we fllw the classic rules until we reach the cnfiguratin where the secnd peg cntains the n th, (n 1) st and (n 2) nd largest disks and 4 th largest disk; the third peg cntains nly the 3 rd largest disk; and the first peg cntains the remainder f the disks. We then flip the n th, (n 1) st and (n 2) nd largest disks nt the first peg and save 4 mre mves. Nw, we mve the 4 th largest disk nt the third peg. If n is even, we flip the n th, (n 1) st and (n 2) nd largest disks frm the first peg nt the secnd peg. If n is dd, we flip the n th, (n 1) st and (n 2) nd largest disks frm the first peg nt the third peg. This flip saves 4 mre mves. If n is even, we fllw the classic rules until we reach the cnfiguratin where the first peg cntains the 2 nd largest and the 1 st largest disks; the secnd peg cntains the n th, (n 1) st and (n 2) nd largest disks; and the third peg cntains the remainder f the disks. If n is dd, we fllw the classic rules until we reach the cnfiguratin where the first peg cntains the n th, (n 1) st and (n 2) nd largest, the 2 nd largest and the 1 st largest disks; the secnd peg cntains n disks; and the third peg cntains the remainder f the disks. Net, we flip the n th, (n 1) st and (n 2) nd largest disks nt the third peg, which saves 4 mre mves. See Figure 13. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

96 96 several bunds fr the k-twer f hani puzzle Figure 13: Overview f Building a Stack f (n 2) Disks. Whenever n = 9, the disk labelled n 6 cincides with the disk labelled 3. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

97 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 97 This cmpletes the stack f (n 2) disks and it takes mves. 1 + (2 n 3 1) = 2 n 3 13 Net, t build a stack f (n 1) disks, we will divide this step up int tw parts. Fr the first part, we start by mving the 2 nd largest disk nt the secnd peg. If n is even, we flip the n th, (n 1) st and (n 2) nd largest disks frm the third peg nt the first peg. If n is dd, we flip the n th, (n 1) st and (n 2) nd largest disks frm the third peg nt the secnd peg. This flip saves 4 mves. If n is even, we fllw the classic rules until we reach the cnfiguratin where the first peg cntains nly the largest disk; the third peg cntains the n th and (n 1) st largest disks, the 4 th largest disk and the 3 rd largest disk; and the secnd peg cntains the remainder f the disks. If n is dd, we fllw the classic rules until we reach the cnfiguratin where the first peg cntains the n th and (n 1) st largest disks and the largest disk; the third peg cntains the 4 th largest and the 3 rd largest disks; and the secnd peg cntains the remainder f the disks. See Figure 14. We flip the n th and (n 1) nd largest disks nt the secnd peg, which saves 1 mre mve. Net, we mve the 4 th largest disk frm the third peg nt the first peg. If n is even, we flip the n th and (n 1) st largest disks frm the secnd peg nt the first peg. If n is dd, we flip the n th and (n 1) st largest disks frm the secnd peg nt the third peg. This flip saves 1 mre mve. If n is even, we fllw the classic rules until we reach the cnfiguratin where the secnd peg cntains nly the 2 nd largest disk; the third peg cntains the n th, (n 1) st and (n 2) nd largest disks and the 3 rd largest disk; and the first peg cntains the remainder f the disks. If n is dd, we fllw the classic rules until we reach the cnfiguratin where the secnd peg cntains the n th, (n 1) st and (n 2) nd largest disks and the 2 nd largest disk; the third peg cntains nly the 3 rd largest disk; and the first peg cntains the remainder f the disks. See Figure 15. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

98 98 several bunds fr the k-twer f hani puzzle Figure 14 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

99 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 99 Figure 15 We then flip the n th, (n 1) st and (n 2) nd largest disks nt the first peg and save 4 mre mves. Then we mve the 3 rd largest disk frm the third peg nt the secnd peg. If n is even, we flip the n th, (n 1) st and (n 2) nd largest disks frm the first peg nt the secnd peg. If n is dd, we flip the n th, (n 1) st and (n 2) nd largest disks frm the first peg nt the third peg. This flip saves 4 mre mves. See Figure 16. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

100 100 several bunds fr the k-twer f hani puzzle Figure 16 This cncludes the first part f this step. Nw we begin the secnd part f this step. If n is even, we make mves using the classic rules until we reach a cnfiguratin where the first peg cntains the 5 th largest, 4 th largest, and largest disks; the third peg cntains the n th largest and (n 1) st largest disks; and the secnd peg cntains the remainder f the disks. If n is dd, we make mves using the classic rules until we reach a cnfiguratin where the first peg cntains the n th largest, (n 1) st largest, 5 th largest, 4 th largest, and largest disks; the third peg cntains n disks; and the secnd peg cntains the remainder f the disks. Then we flip the n th largest and (n 1) st largest disks nt the secnd peg, which saves 1 mre mve. We then mve the 5 th largest disk nt the third peg. If n is even, we flip the n th largest and the (n 1) st largest nt the first peg. If n is dd, we flip the n th largest and the (n 1) st largest nt the third peg. This flip saves 1 mre mve. See Figure 17. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

101 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 101 Figure 17 Net, we make a similar sequence f mves. If n is even, we make mves using the classic rules until we reach a cnfiguratin where the first peg cntains nly the largest disk; the third peg cntains the n th largest, (n 1) st largest, 6 th largest and 5 th largest disks; and the secnd peg cntains the remainder f the disks. If n is dd, we make mves using the classic rules until we reach a cnfiguratin where the first peg cntains the n th largest, (n 1) st largest, and largest disks; the third peg cntains the 6 th largest and 5 th largest disks; and the secnd peg cntains the remainder f the disks. Then we flip the n th largest and (n 1) st largest disks nt the secnd peg, which saves 1 mre mve. We then mve the 6 th largest disk nt the first peg. If n is even, we flip the n th largest and the (n 1) st largest nt the first peg. If n is dd, we flip the n th largest and the (n 1) st largest nt the third peg. This flip saves 1 mre mve. See Figure 18. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

102 102 several bunds fr the k-twer f hani puzzle Figure 18 Ntice that in Figure 17, we tk advantage f the special rule invlving k whenever we had the 4 th largest and 5 th largest disks tgether n a peg separate frm a stack f (n 5) disks. Als, in Figure 18, we tk advantage f the special rule invlving k whenever we had the 5 th largest and 6 th largest disks tgether n a peg separate frm a stack f (n 5) disks. In general, we can take advantage f the special rule invlving k whenever we have a j th largest and (j + 1) st largest disk tgether n a peg separate frm a stack f (n 5) disks, where 4 j (n 4). That is, If n is even and j is even, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains the (j + 1) st largest, j th largest, and largest disks; the third peg cntains the n th largest and the (n 1) st largest disks; and the secnd peg cntains the remainder f the disks. If n is even and j is dd, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains nly the largest disk; the third peg cntains the n th largest, (n 1) st largest, (j + 1) st largest and j th largest disks; and the secnd peg cntains the remainder f the disks. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

103 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 103 Then we flip the n th largest and the (n 1) st largest disks nt the secnd peg, which saves 1 mre mve. If j is even, we mve the (j + 1) st largest disk nt the third peg. If j is dd, we mve the (j + 1) st largest disk nt the first peg. Net, we flip the n th largest and the (n 1) st largest disks nt the first peg, which saves 1 mre mve. See Figure 19. Figure 19: Mves made fr 4 j (n 4) whenever n is even. The disks n the secnd peg are uniquely determined by the disks depicted n the ther tw pegs. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

104 104 several bunds fr the k-twer f hani puzzle Otherwise, If n is dd and j is even, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains the n th largest, (n 1) st largest, (j + 1) st largest, j th largest, and largest disks; the third peg cntains n disks; and the secnd peg cntains the remainder f the disks. If n is dd and j is dd, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains the n th largest, (n 1) st largest, and the largest disk; the third peg cntains the (j + 1) st largest and j th largest disks; and the secnd peg cntains the remainder f the disks. Then we flip the n th largest and the (n 1) st largest disks nt the secnd peg, which saves 1 mre mve. If j is even, we mve the (j + 1) st largest disk nt the third peg. If j is dd, we mve the (j + 1) st largest disk nt the first peg. Net, we flip the n th largest and the (n 1) st largest disks nt the third peg, which saves 1 mre mve. See Figure 20. Ntice that because we use values f j = 4, 5, 6,..., (n 4), there are (n 7) distinct values f j. Thus, we will have (n 7) pairs f j and (j + 1) where we can take advantage f the special rule invlving k. Als, fr each f these pairs we save 2 mves. Thus, we will save 2(n 7) mves after we have made these sequences f mves, as depicted in Figure 21, we reach the fllwing cnfiguratin: Net, If n is even, we reach the cnfiguratin where the first peg cntains the n th largest, (n 1) st largest, (n 4) th largest and the largest disks; the third peg cntains nly the (n 3) rd largest disk; and the secnd peg cntains the remainder f the disks. If n is dd, we reach the cnfiguratin where the first peg cntains the (n 3) rd largest and the largest disk; the third peg cntains the n th, (n 1) st, and (n 4) th largest disks; and the secnd peg cntains the remainder f the disks. If n is even, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains nly the largest disk; the third peg cntains the n th largest, (n 1) st largest, (n 2) nd largest and (n 3) rd largest disks; and the secnd peg cntains the remainder f the disks. If n is dd, we make mves using the classic rules until we reach the cnfiguratin where the first peg cntains the n th largest, (n 1) st largest, (n 2) nd largest and (n 3) rd largest disks and the largest disk; the third peg cntains n disks; and the secnd peg cntains the remainder f the disks. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

105 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 105 Figure 20: Mves made fr 4 j (n 4) whenever n is dd. The disks n the secnd peg are uniquely determined by the disks depicted n the ther tw pegs. Net, we flip the n th, (n 1) st, (n 2) nd and (n 3) rd largest disks nt the secnd peg, which saves 11 mre mves. See Figure 22. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

106 106 several bunds fr the k-twer f hani puzzle Figure 21 Figure 22 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

107 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 107 This cmpletes the stack f (n 1) disks and it takes 1 + (2 n 2 1) (n 7) 11 = 2 n 2 2n 11 mves. Nw, we wish t frm a stack f n. We d this by mving the largest disk nt the third peg and then reversing the previus steps fr building stacks f 1, 2,..., (n 1). Thus, it takes 1+ ( (2n 128n + 768) + (2 n 4 9) + (2 n 3 13) + (2 n 2 2n 11) ) = 1 + ( 2 n 2 10n + 28) mves t build this stack. See Figure 23. Figure 23: Overview f Building a Stack f n Disks Therefre, it takes ( 2 n 10n ) mves t cmplete the puzzle. ( ) 2 n n + 28 = 2 n 20n Upper bund n the minimum number f mves t cmplete the k-twer f Hani Puzzle whenever k = 3 We nw lk at the k-twer f Hani puzzle whenever k = 3. We skip t k = 3 because upper bunds n the minimum number f mves required t cmplete the k-twer f Hani puzzle whenever k = 1 and k = 2 are knwn. An upper bund f n 2 n + 1 is shwn in [2] whenever k = 1, and an upper bund f 2n 2 4n + 1 whenever k = 2 is shwn in [1]. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

108 108 several bunds fr the k-twer f hani puzzle Rules fr the k-twer f Hani Puzzle Whenever k = 3 Disks can nly be mved ne at a time. Only the tp disk n a stack can be mved frm that peg. A disk can be mved n tp f a smaller ne prvided that the three disks n the bttm f that stack are the largest n that peg (with the largest disk n the bttm, the secnd largest disk n tp f that ne, and the third largest disk n tp f that ne). The upper bund, T 3 (n), fr the minimum number f mves required t cmplete the k-twer f Hani puzzle with n disks whenever k = 3 is T 3 (n) = 2 n 1 if n = 3, 4, 5 61 if n = if n = if n = 8 2n n 153 if n 9. Ntice that fr the cases f n = 3, 4, 5, 6, 7, we just refer back t the bunds frm Sectin 3 fr k = n fr = 0, 1, 2, 3, 4 respectively, t btain bunds f 7, 15, 31, 61 and 103 mves, respectively. We begin with the case f n 9. Therem 1. The ptimal slutin fr the k-twer f Hani puzzle fr n 9 disks whenever k = 3 has an upper bund f T 3 (n) = 2n n 153. Prf. T btain this bund, we will divide it up int the fllwing steps. Building a stack f 3 Building stacks f 4, 5,, (n 4) Building a stack f (n 3) Building a stack f (n 2) Building a stack f (n 1) Building a stack f n Nte that this bund assumes a starting number f disks n 9. Als, fr the previus bunds, we fcused n ttaling the number f saves at varius steps in cmpleting thse puzzles and then subtracting these saves frm the number f mves that wuld have been made when using nly the classic rules. Hwever, t btain the bund fr k = 3, we will simply ttal the number f mves that we make as we cmplete the puzzle. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

109 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 109 Building a Stack f 2 Disks: We build a stack f 3 disks n the destinatin peg using the classic rules. This takes 7 mves. Building Stacks f 4, 5,, (n 4) Disks: T build a stack f 4 disks, we mve the (n 3) rd largest disk frm the surce peg nt the intermediate peg. This takes 1 mve. Then we flip the n th and (n 1) st largest disks frm the destinatin peg nt the surce peg. This takes 2 mves. We then mve the (n 2) nd largest disk frm the destinatin peg nt the intermediate peg. This takes 1 mve. Net, we flip the n th and (n 1) st largest disks frm the surce peg nt the intermediate peg. This takes 2 mves. Thus, it takes = 2(3) mves t build a stack f 4 disks. See Figure 24. Figure 24: Overview f Building a Stack f 4 Disks We repeat this same prcess t build the net stack f 5 disks. We begin by mving the (n 4) th largest disk frm the surce peg nt the destinatin peg. We then flip all disks, ecept the (n 3) rd largest disk, frm the intermediate peg nt the surce peg. This takes 3 mves. We then mve the (n 3) rd largest disk nt the destinatin peg. This takes 1 mve. Net, we flip the same disks that were flipped nt the surce peg, nt the destinatin peg. This takes anther 3 mves, and gives us a stack f 5. Thus, it takes = 2(4) mves t build a stack f 5 disks. See Figure 25. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

110 110 several bunds fr the k-twer f hani puzzle Figure 25: Overview f Building a Stack f 5 Disks In general, t build a stack f disks, fr 4 (n 4), it takes 1 mve t relcate the largest disk f the stack we are building, ( 2) mves t flip disks, 1 mve t put the 2 nd largest disk f the stack we are building nt the largest disk f the stack we are building, and ( 2) mre mves t flip the remaining disks nt the stack we are building. That is, it takes 1 + ( 2) ( 2) = 2( 1) mves t build a stack f disks. See Figure 26. Figure 26: Overview f Building a Stack f Disks. Stacks f an even number f disks will be n the intermediate peg and stacks f an dd number f disks will be n the destinatin peg. The final stack f (n 4) disks will be n the third peg. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

111 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 111 S, it takes n 4 (2(i 1)) = n 2 9n + 14 i=4 mves t build stacks f 4, 5,..., (n 4). Building a Stack f (n 3) Disks: After building a stack f (n 4) disks, we reach the cnfiguratin where the first peg cntains the 4 th, 3 rd, 2 nd, and largest disks; the secnd peg cntains n disks; and the third peg cntains the remainder f the disks. We nw need t build a stack f (n 3) disks, and we start by mving the 4 th largest disk nt the secnd peg. This takes 1 mve. Net, we flip the tp (n 5) disks frm the third peg nt the first peg. This takes (n 5) mves. Then we mve the 5 th largest disk frm the third peg nt the secnd peg. This takes 1 mve. Net, we mve the 6 th largest disk frm the first peg nt the secnd peg. This takes 1 mve. Then we mve the 7 th largest disk nt the third peg. This takes 1 mve. We then flip (n 7) disks frm the first peg nt the secnd peg, which takes (n 7) mves. Lastly, we mve the 7 th largest disk frm the third peg nt the secnd peg, giving us a stack f (n 3) disks. See Figure 27. Figure 27: Overview f Building a Stack f (n 3) Disks. Ntice that we make a cunterintuitive mve by taking ut the 7 th largest disk. Ding this will save many mves in the lng term. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

112 112 several bunds fr the k-twer f hani puzzle Thus, it takes 1 + (n 5) (n 7) + 1 = 2n 7 mves t build a stack f (n 3) disks. Building a Stack f (n 2) Disks: We nw build a stack f (n 2) disks. First, we mve the 3 rd largest disk nt the third peg. This takes 1 mve. Net, mve the 7 th largest disk frm the secnd peg nt the first peg. This takes 1 mve. Flip (n 7) disks frm the secnd peg nt the first peg. This takes (n 7) mves. Mve the 6 th largest disk frm the secnd peg nt the third peg. This takes 1 mve. Flip (n 7) disks frm the first peg nt the secnd peg. This takes (n 7) mves. Mve the 7 th largest disk frm the first peg nt the third peg. This takes 1 mve. Flip (n 7) disks frm the secnd peg nt the third peg. This takes (n 7) mves. Mve the 5 th largest disk frm the secnd peg nt the first peg. This takes 1 mve. Mve the 8 th largest disk frm the third peg nt the secnd peg. This takes 1 mve. Flip (n 8) disks frm the third peg nt the first peg. This takes (n 8) mves. Mve the 8 th largest disk frm the secnd peg nt the first peg. This takes 1 mve. Flip the 7 th largest and 6 th largest disks frm the third peg nt the first peg. This takes 2 mves. Mve the 4 th largest disk frm the secnd peg nt the third peg. This takes 1 mve. Flip (n 5) disks frm the first peg nt the secnd peg. This takes (n 5) mves. Mve the 5 th largest disk frm first peg nt the third peg. This takes 1 mve. Flip (n 5) disks frm the secnd peg nt the third peg. This takes (n 5) mves and cmpletes the stack f (n 2) disks. See Figure 28. Thus, it takes 1+1+(n 7)+1+(n 7)+1+(n 7)+1+1+(n 8) (n 5)+1+(n 5) = 6n 28 mves t build stack f (n 2) disks. Building a Stack f (n 1) Disks: We nw build a stack f (n 1) disks. Due t the length f this step, we break the sequence f mves up int three sectins. We start by mving the 2 nd largest disk frm the first peg nt the secnd peg. This takes 1 mve. Then we flip the 6 th largest and 7 th largest disks frm the third peg t the first peg. This takes 2 mves. Mve the 8 th largest disk frm the third peg nt the secnd peg. This takes 1 mve. Flip (n 8) disks frm the third peg nt the first peg. This takes (n 8) mves. Mve the 8 th largest disk frm the secnd peg nt the first peg. This takes 1 mve. Mve the 5 th largest disk frm the third peg nt the secnd peg. This takes 1 mve. Flip (n 7) disks frm the first peg nt the secnd peg. This takes (n 7) mves. Mve the 7 th largest disk frm the first peg nt the third peg. This takes 1 mve. Flip (n 7) disks Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

113 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 113 Figure 28: Overview f Building a Stack f (n 2) Disks. Whenever n = 9, the disk labelled n cincides with the disk labelled 9. frm the secnd peg nt the third peg. This takes (n 7) mves. Mve the Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

114 114 several bunds fr the k-twer f hani puzzle 6 th largest disk frm the first peg nt the secnd peg. This takes 1 mve. Flip (n 7) disks frm the third peg nt the secnd peg. This takes (n 7) mves. Mve the 7 th largest disk frm the third peg nt the secnd peg. This takes 1 mve. Mve the 4 th largest disk frm the third peg nt the first peg. This takes 1 mve. Flip (n 6) disks frm the secnd peg nt the first peg. This takes (n 6) mves. See Figure 29. Mve the 6 th largest disk frm the secnd peg nt the third peg. This takes 1 mve. Figure 30 will begin after this mve is made. Figure 29 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

115 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 115 Flip (n 7) disks frm the first peg nt the secnd peg. This takes (n 7) mves. Mve the 7 th largest disk frm the first peg nt the third peg. This takes 1 mve. Flip (n 7) disks frm the secnd peg nt the third peg. This takes (n 7) mves. Mve the 5 th largest disk frm the secnd peg nt the first peg. This takes 1 mve. Flip (n 7) disks frm the third peg nt the first peg. This takes (n 7) mves. Flip the 7 th largest and 6 th largest disks frm the third peg nt the first peg. This takes 2 mve. Mve the 3 rd largest disk frm the third peg nt the secnd peg. This takes 1 mve. Mve the 6 th largest disk frm the first peg nt the secnd peg. This takes 1 mve. Mve the 7 th largest disk frm the first peg nt the third peg. This takes 1 mve. Flip (n 7) disks frm the first peg nt the secnd peg. This takes (n 7) mves. Mve the 7 th largest disk frm the third peg nt the secnd peg. This takes 1 mve. Mve the 5 th largest disk frm the first peg nt the third peg. This takes 1 mve. Mve the 7 th largest disk frm the secnd peg nt the first peg. This takes 1 mve. Mve the 8 th largest disk frm the secnd peg nt the third peg. This takes 1 mve. See Figure 30. Flip (n 8) disks frm the secnd peg nt the first peg. This takes (n 8) mves. Figure 31 will begin after this mve is made. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

116 116 several bunds fr the k-twer f hani puzzle Figure 30 Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm

117 S. B. Gregg, B. Hpkins, K. Karber, T. Milligan, J. Sharp 117 Mve the 8 th largest disk frm the third peg nt the first peg. This takes 1 mve. Mve the 6 th largest disk frm the secnd peg nt the third peg. This takes 1 mve. Flip (n 7) disks frm the first peg nt the secnd peg. This takes (n 7) mves. Mve the 7 th largest disk frm the first peg nt the third peg. This takes 1 mve. Flip (n 7) disks frm the secnd peg nt the third peg. This takes (n 7) mves. Mve the 4 th largest disk frm the first peg nt the secnd peg. This takes 1 mve. Flip (n 4) disks frm the third peg nt the secnd peg. This takes (n 4) mves. See Figure 31. Figure 31 Thus, it takes (n-8)+1+1+(n-7)+1+(n-7)+1+(n-7)+1+1+(n-6)+1+(n-7)+1+(n-7)+1 +(n-7) (n-7) (n-8)+1+1+(n-7)+1+(n-7)+1+(n-4)=13n-63 mves t build a stack f (n 1) disks. Recreatinal Mathematics Magazine, Number 7, pp DOI /rmm