Cubelike Puzzles-What Are They and How Do You Solve Them?

Home , David Singmaster, Pyramid (geometry), Pyraminx

CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM?

J.A. EIDSWICK Departmentof Mathematics and Statistics,University of Nebraska, Lincoln, NE 68588 1. Rubik'scube, Meffert's pyramid, Halpern's dodecahedron, Alexander's great dodecahedron, and thelist goes on. Suchobjects are partitionedinto smaller pieces that get mixed up whenyou turntheir faces or layers,yet theydon't fall apart.All are groupsin disguise;solving them amountsto findingalgorithms for factoring arbitrary elements into products of certaingenerators. Justwhat constitutes a "cubelikepuzzle"? Is therea universalalgorithm for solving them? These are questionssuggested by Douglas Hofstadter'sScientific American article [12]. Concerninga puzzle he calls theIncrediBall, Hofstadter writes: "I foundthat when I loosenedmy conceptualgrip on the exact qualitiesof my hard-won operatorsfor the Cube and tookthem more metaphorically, I could transfer some of myexpertise fromCube to I-ball.Not everythingtransferred, needless to say.What pleased me mostwas when I discoveredthat my 'quarkscrew'and 'antiquarkscrew'were directly exportable. Of course,it took a whileto determinewhat such an exportwould consist of. What is theessence of a move? What aspects of it are provincialand shedable?How can one learn to tell easily?These are difficultquestions for which I do nothave theanswers. I graduallylearned my way around the IncrediBall by realizingthat a powerfulclass of moves consistsof turningonly two overlapping 'circles' in a commutatorpattern (xyx'y'). I therefore studiedsuch two-circlecommutators on paper untilI foundones thatfilled all my objectives. They includedquarkscrews, double swaps and 3-cycles,which form the basis of a complete solution.In doingso I cameup withjust barelyenough notation to covermy needs, but I did not developa completenotation for the IncrediBall.This, it seemsto me, wouldbe mostuseful: a standarduniversal notation, psychologically as wellas mathematicallysatisfying, for all cubelike puzzles. It is, however,a veryambitious project, given that you would have to anticipateall conceivablevariations on thisfertile theme, which is hardlya trivialundertaking."

Concerningalgorithms for solving the cube, David Singmaster[19, p. 12] writes:"... . we need to proceedin two directions.First, by examiningthe cube and its group,we discoverwhich patternsare possible and, second,we show that we can achieveall possiblepatterns." The unscramblingproblem, then, is directlyrelated to theproblem of determiningthe structure of the underlyinggroup. Later [19, pp. 58-9], Singmasterdescribes this group as a subgroupof index12 of a directproduct of wreathproducts. Singmasteralso writes:"... it is a remarkablephenomenon that everyone seems to finda differentcombination of processesand strategy."A surveyof themany "how-to" books on the subjectwill confirmSingmaster's assertion. But commonthreads will be foundrunning through all of thesealgorithms. These include the commutators and 3-cyclesmentioned by Hofstadter. This articleis an attemptto put some algebraicorder into thebusiness of solvingcubelike puzzles.Ideally, an algebraictheory would unfold that would, in an elementaryway, yield highly efficientalgorithms for all suchpuzzles. The workhere is a stepin thatdirection. Our strategy, like Singmaster's,will lead to a determinationof theunderlying group structures. A descriptionof thesestructures is givenin ?6. Our startingpoint will be to put the above-mentionedcommon threads together to forma commonstrategy. The strategywill thenbe applied to variouspuzzles includingthe general

J. A. Eidswick:I receivedmy Ph.D. in 1964 at PurdueUniversity under the direction of Louis de Branges.Since thenI have publishedmodestly in theareas of real analysisand topology.In 1981, I fellunder the spell of thecube and wrotethe booklet Rubik's Cube Made Easy. I also designedRubik's Cube EngagementCalendar 1982. The spell continues.My hobbies includebackpacking, gourmet cooking, and joggingenough to justifyeating.

157

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 158 J. A. EIDSWICK [March n x n x n cube.The efficiencyof thesealgorithms will not be a majorconsideration of thispaper, althoughsome attentionis givento this topic in ?7. It would be veryinteresting to obtain informationabout the length of theshortest possible algorithm. Group-theoretically, this amounts to calculatingthe least upper bound for the lengths of wordsrequired to expressall elementsof the underlyinggroup in termsof a certainset of generators.This is a difficultproblem about whichlittle is known.(See [11,p. 35],[19, pp. 52-3]; also [20] forrelated results.) In ?2, commonthreads of cube solutionsare summarized.In ?3, theconcept of wreathproduct is developedand examplesare giventhat illustrate the relationshipthat exists between wreath productsand cubelikepuzzles. In ?4, a generalstrategy is based on Propositions1-7. These resultsserve as keysfor solving most cubelike puzzles. Exceptions are the" twofaces puzzle" of [2, p. 768] (see Example4 below)and the"skewb" of [12,p. 20]. Applicationsin ?5 includethe cube puzzle, threedifferent partitions of the tetrahedron,two of theoctahedron, two of thedodeca- hedron,one of theicosahedron, and thegeneral n x n x n cube.The latterillustrates the essence of Proposition1. The only prerequisitefor readingthis articleis an elementaryknowledge of permutation groups.In particular,the reader does nothave to knowhow to "do thecube". For a discussionof permutations,see almostany introductoryalgebra text (e.g., [3], [10]). For a fairlycomplete treatmentof thesubject of permutationsand a glimpseat its evolution,see in order,[5], [6], [21], and [22]. For the generaltheory of groups,see, e.g., [9], [15], or [16]. A few "cube theory" referencesare includedat theend forthe interested reader. 2. Commonthreads. All intelligiblesolutions of thecube puzzleseem to have thesecommon features: (i) Two distinctsubproblems are recognized:the positioningproblem and the orientation problem.Mathematically, these relate to permutationgroups and wreathproducts, respectively. (ii) Two distinctorbits are recognized:the cornercube orbit and the edgecube orbit. (iii) A special parity-adjustingprocess is needed. (iv) Cubeletsare restoredone-by-one. (v) Processesfor restoring individual cubelets (which involve anywhere from zero to twenty quarter-turns)often involve conjugationsand commutators. A typicalsolution begins with operations that affect both orbits,but laterrestricts to those which affectonly one. Likewise,cubelets are positionedand orientedsimultaneously at the beginning,then separated later on. Thereare obviouslymany ways to do thisand therein,no doubt,lies theexplanation to Singmaster'sobserved "phenomenon". 3. Notation,wreath products. Throughout, 8 denotesa finiteset, G a permutationgroup actingon X, and F a subgroupof a wreathproduct. In ?5, 8 will be interpretedas a set of unorientedpuzzle pieces, G a group of permutationsof such pieces, and F a group of permutationsof orientedpuzzle pieces. Permutations will act on theright, other functions on the left.By an orbitof X is meanta setof theform

(9= ((x) = {xg: ge G}. If (9 is aa orbit,then GI( denotesthe restrictionof G to ( and, forg E G, gIO denotesthe restrictionof g to (. We willwrite Act(g) forthe action set { x E 8(: xg x }, Sym8 and Alt8, respectively,for the symmetricand alternatinggroups on 8, S, = Sym{1,..., n}, An = Alt{l, ... , n } and Zr forthe group of integers mod r. If S is a finiteset, ISI denotesits cardinality and S` theset of functionsfrom 8 to S. If g and h are elementsof a group,then [g, h] denotes the commutatorghg-lh-1. Wreathproducts are usually studied along with semidirect products and/or group extensions as, e.g.,in [13],[15], and [18].The definitionbelow (cf. [14, p. 32]) assumesno previousknowledge of thesecompanion ideas. We mention,though, the important theorem of Kaloujnineand Krasner

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 159

(see [18, p. 100] or [13,p. 49]): Everyextension of a groupA bya groupB can be embeddedin a wreathproduct.

DEFINITION. Let G and H be permutationgroups that act on .T= {1,. . , n } and MY= {1, ... , r}, respectively.Then the wreathproduct of H by G, writtenH I G, is the subgroupof Sym(T x C) generatedby permutationsof thefollowing two types: 7T(g) :(i, j) (ig, j) forg E G and a(hl, . .., hn) (i0 j) (1* , ihi) forhl,..., hn ( H. We mayvisualize the situation as follows:Suppose n decksof cardsoccupy positions 1,..., n and supposethat each deck contains r cardswhich occupy levels 1,. . . , r. Then7r(g) permutes the decks accordingto g, maintainingcard levels,and a(hj,.. ., hn) shufflesthe decks in positions 1,..., n accordingto hl,.. ., hn,respectively, maintaining deck positions. In the applicationsto be considered,H willbe cyclic.Accordingly, we will take H = Zr for some positiveinteger r and use additivenotation for this group. Notice thatfor such groups, shufflingsamount to whatcard players call "cuts". For g,p E G and hl,..., hn,kl,..., knE H, we have = TT(g)'7(p) r(gp)? a(hj,...,hn) a(kj,...,kn) = a(h + kl,..., hn + kn), and

ar(hl,..., hn)gT(g) = gT(g) a(hlg-l. .hng-1). It followsthat any elementa of H I G has a uniquerepresentation of the form T(a') a(a"), wherea' E G and a" e H. In particular,we have IH I GI = IG X Hfl = IGIIHIl'l. Henceforth,we willidentify elements a of H I G and correspondingpairs (a', a") of G x H. This will simplifynotation. For example,the identity1 in H I G will be written(1,0) or (1, (0,... , 0)) insteadof q(1)a(0,... , 0). The followingformulas may be routinelyestablished:

(1) (a/)' =?f'

(2) (a-1)' =a (3) (a13 )' = 'a'1 (4) [#,:] = [a',f']

(5) (a/3)"(i) = "(ii'1) + 13"(i)

(6) (x1)"(i) = -a"(ia') (7) (afla1)"(i) = a"(ia'/3') + /3"(ia) - a"(ia)

(8) [ a,:]"(i) = a,(if3'a'f'-1) + /a"(i'a') - "(i'a') - 3"(i: ). The followingthree examples show how wreath products relate to thepuzzles under considera- tion. They also illustratethe abovementionedtheorem of Kaloujnineand Krasner.The first exampleis a variationof " thethree coins game" of [4, p. 24]. EXAMPLE1. Threecoins are placed in a rowin positions1, 2,3 and thefollowing operations on

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 160 J. A. EIDSWICK [March

themare permitted. a: interchangethe coins in positions1 and 2. ,B: turnover the coin in position3, theninterchange coins in positions2 and 3. We assumethat the coins have distinguishable orientations "heads" and "tails" and (unlikein [4]) thatthe coins themselvesare distinguishable(say, penny, nickel, and dime).If we associate coin positionswith deck positions and "heads" and "tails",respectively, with levels 1 and 2, it is clear thata and ,Bmay be regardedas permuting-shufflingoperations on three2-card decks; i.e., as elementsof Z72 l S3. Explicitlywe have: a = ((12),(0,0,0)) and 3 ((23), (0, 1,0)). One may verifyby directcalculation that a and '3 generateall 48 elementsof ZL2 1 S3 (however,compare this with Examples 2 and 3 below). EXAMPLE 2. Six equilateraltriangles are placed in a rowin positions1,... , 6 and thefollowing operationson themare allowed. a: cycle the trianglesin positions 3, 4, 5, 6 in that order (i.e., 3 -> 4 -> 5 -> 6 -> 3). '3: rotatethe triangle in position1 clockwise1200, then cycle the triangles in positions1, 2, 3,4 in that order (i.e., 1 -- 2 -- 3 -- 4 -- 1). (It is to be understoodhere that when triangles are cycled,they are to maintaintheir orientations. Thus, verticespointing upward before cycling will remainpointed upward after cycling.) If we identifythe idea ofhaving i clockwiserotations and thatof occupyinglevel i (mod3), thena and '3 may be interpretedas permuting-shufflingoperations on six 3-carddecks. Hence, the group generatedby a and , maybe regardedas a subgroupF of Z73 l S6. Here, a ((3456), (0,0,0,0,0,0)) and 3 ((1234), (0,1,0,0,0,0)), and it maybe verifiedthat F has index6 in Z3 l S6. Indeed,an argumentsimilar to [19,pp. 55-7] will showthat F is isomorphicto 73 l S5. This exampleis relatedto the"Tricky Six Puzzle" of [2, p. 759] (see also [19,p. 57],[23], and Example3 below). EXAMPLE 3 (the 2-facesgroup). Six blocksare arrangedin an L-shapedstack as shownin Fig. l(a), and thefollowing operations are allowed(see Fig. 1(a)). 5: rotatethe front layer 900 clockwise. p: rotatethe right-hand layer 90? clockwise.

1 2 3 4 5 6 (a) (b)

FIG. 1

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 161

This exampleis closelyrelated to Example2. To see this,note that p and p takedarkened cornersto darkenedcorners (see Fig. 1(a)), and thatblocks can rotateabout thesecorners (e.g., qp: B -* Y -* P -- B). Now imagine the six blocks as spread out beforeus in some arrangement, say,as in Fig. l(b). We maythen view 0 and p as permuting-rotatingoperations as in Example2. Clearly,the representations of p and p willdepend on thearrangement, but thepoint is thatwe may regardthe group F generatedby 4 and p as a subgroupof Z3 1 S6. In termsof the arrangementshown, =((1234), (0, 0, 0, 0, 0, 0)) and p = ((3654), (0,0,1,1,0,1)).

It maybe verified(see [19,pp. 55-7]) thatF has index18 in Z3 l S6.

4. A commonstrategy. In thepuzzles we are consideringthere are variouspieces which can occupycertain positions with certain orientations. The basic movesof thepuzzle may be viewedas actingon both thepositions and theoriented positions. When we viewthese moves as actingon positions,we will obtainorbits 01,.r.., ON. Whenwe view themoves as actingon the oriented positionsrestricted to an orbit (,, the situationwill be analogousto Example3. Thus, these puzzlesmay be regardedas subdirectproducts of wreathproducts of theform Zr I S,i. Our strategywill be to proceed throughthe orbits (91,... 09N restoringthose pieces in a given orbitwithout disturbing the other pieces. To accomplishthis, we willfirst need to makea parity adjustment.Let a be theproduct of movesthat scrambled up our puzzle.We will producea productX of basic movessuch that a7r permutes the pieces evenlyon everyorbit. We willthen produceproducts ao I ... I aN ofbasic movessuch that a, actson (9i as theidentity for j * i, and a, acts on C, as theinverse of as7r.(In fact,we needonly require that a, be theidentity on 67Jfor j < i and that a, act on (9i as the inverseof aoal ... a, -1. This improvementto the strategywill be discussedin ?7.) Thus,the product a7al -1 *. aN willunscramble the puzzle. The followingresult shows that a parityadjustment of the typedescribed above is always possible.We definev(g, (9) to be 0 or 1 accordingas gJ( is evenor odd.

PROPOSITION 1. For any permutationgroup G thereexist orbits ,. . . , 0,, and elements hl,..., h,, e Gsuch that (i) v(h,,C(9) = 83J(i, j = 1,...,m) and (ii) for any orbit(9 thereexist Cl, . . , c,,mE {0, 1} such that

v(h,(9) = Z c,v(h,(9,) .==1 mod2 for everyh E G. Moreover, if (91,...,(m and h,...,h,,, are as above, then (iii) for any g e G, v(gp, (9) = 0 for everyorbit (3 wherep is theproduct of the h, 's, taken in order, for which v(g, (9,) = 1.

Proof.The followingargument, contributed by T. Shores,is "programmable".Let G = {g1,... ,gk}, let {f-,... ,XI} be the set of orbits of G, and consider the k x I matrix M = -- [v(g,, X,)]. Since the mappingg [v(g, _J)] is a homomorphismfrom G into Z,(), the row- reduced echelon matrix E of M has the form: E = [v(gr,_J)]. If E has m nonzero rows, let t1,... , t', be thecolumns corresponding to theleading 1's in theserows. Then (i) and (ii) follow with h, = gr and (9, = t,. Also forg and p as in (iii), we have v(gp,C9,) = Ofor i =1,...,m and hence v(gp, (9) = 0 for everyorbit (9 by (ii).- Since thestrategy outlined above deals witheach orbitseparately, we willconcentrate for the remainderof thissection on theaction of a wreathproduct on a singleorbit.

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 162 J. A. EIDSWICK [March

Recall thatfor a e H I G, a' and a" denotethe "permuting"and "shuffling"parts of a, respectively,i.e., a = (a', a") wherea' e G and a" e H`. We will also writeAct'(a) forthe projectionof Act(a) on S. Thus,

Act'(a) = { -iE: ia' * i or a"(i) * 0).

An elementa ECH I G willbe calleda k-cycleif a' is a k-cycle,Act'(a) = Act(a'), and a has orderk. Cubelikepuzzles are generatedby productsof such cycles.Examples: (i) p and p of Example 3 are 4-cycles.(ii) ,B of Example1 [Example2] is not a cyclebecause /t' is a 2-cycle [4-cycle]and theorder of ,Bis 4 [12]. By formula(5), k-1 (ak) "(i)=E "(ia'J) for i= n. j=O Hence,it followsthat n (9) a"( i) = O i=l holds forany cycle a,. Since property(9) is clearlypreserved under multiplication in a wreath product,Proposition 2 below follows.The result,almost trivial in thiscontext, translates into puzzle "conservationlaws" like "the numberof flippedcubies is always even" and "total clockwisetwist equals totalcounterclockwise twist mod 3". Elaboratearguments have been given to explainthese phenomena. See, e.g.,[11, p. 28],[19, p. 17],and [2, p. 762].

PROPOSITION 2. If F is a subgroupof H I G whichis generatedby cycles, then(9) holdsfor all a e F.

By a completeset of 3-cyclesfor a, b, c E X willbe meanta set { T,..., Tr-,} of 3-cyclessuch thatfor each j = 0,1,..., r - 1,

(i) T'= (abc) and (ii) Tr"(a) = Tr"'(a) +?j, T"( b) = o'( b), T"( c) = r'(c) -j. The followingresult shows how to produce3-cycles. PROPOSITION 3. If a, /3and a satisfy (i) Act'(a) n Act'(/3) = { b}, whereaa,' = b = c,6', and (ii) Act'(a) n { a, b, c} = { c}, whereca' = c and a"(c) = 1, thena completeset of 3-cyclesfor a, b, c is givenby

Tj = 'J[a]-J forj = 0, 1,., r-1. Proof.Let T = [a,,/]. By formulas(4) and (8) and condition(i),

T = [Ia',/'] = (abc), Tj'(a) = ,6"(b) - a"(b),

T"'( b) = a"( b), T"'( c) = - /3"( b), and T"'(i) = 0 for i + a, b, c.

Hence, T0 is a 3-cycle.Let T, = aToa-1. Thenby formulas(3) and (7) and conditions(i) and (ii), = a'(abc)a'' = (abc), T,'(a) = To'(a) + 1, T"(b) = To"'(b), <"'(c) = "'(c) 1, and ,'(i) = 0 for i * a,b,c. The resultfollows by induction.-

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 163

Elementsof theform ,(a, b) = (1,s), wheres(a) = 1, s(b) = -1, and s(i) = 0 otherwise, play a specialrole in cubelikepuzzles. For thecase of thecorner orbit of thecube puzzle,these elementsare popularlycalled mesonsbecause of an interestinglink with particle physics (see [8]). Since Trp-' = t(c,b), Proposition3 also showsus how to producethese generalized mesons. COROLLARY. Underthe assumptions of Proposition 3, we have

p(c,b) =[aja/flf. Typically,Proposition 3 willbe appliedwith /8 having one of thesetwo forms: / = y8y'- or 3 = [-y,83]. Propositions 4 and 5 belowshow how to deal withthese special situations. PROPOSITION4. If a, y,and 8 satisfy Act'(a) -y' n Act'(8) = { by'), thenAct'(a) A Act'(y3y') = { b}. Proof.By (3) and (7), i e Act'(3yfl/`) if and onlyif iff'G Act'(y) and theresult follows. U PROPOSITION5. If a, y, and 8 satisfy (i) Act'(a) n Act'(y) n Act'(S) = {b} (ii) (Act'(a) n Act'(y))y' n Act'(8) = 0, and (iii) (Act'(a) n Act'(3))a' n Act'(y) 0, thenAct'(a) A Act'([y,8]) = {b}. Proof. By hypothesis,by'8' = by' b3 ba=y'; hence, b e Act([y',8]). Conversely,if i 5 Act'(y), then i e Act'([y,8]) if and only if i' Ee Act(Qy).Therefore, if i E Act'(a) A Act'([y,8]), thenby (iii), we musthave i e Act'(y). Similarly,by (ii), i e Act'(8) and so i = b by (i). U For all of thepuzzles to be consideredthe group G willbe triplytransitive and thefollowing conditioneasy to verify. PROPOSITION6. Let r be a subgroupof H I G. If F containsa completeset of 3-cyclesfor three elementsa, b, c and iffor each x e 5W,x * a, b, c, thereexists a E I' suchthat Act(a) n{a,b,c} = {xa' }, thenr containsa completeset of 3-cyclesfor any three elements of X. Proof.From theproof of Proposition3, we see thatconjugation of a completeset of 3-cycles yieldsa completeset of 3-cycles.Hence, to provethe result we needonly express 3-cycles of X as conjugatesof (abc). For x E 5, x * a, b, c, assume Act(p) A{a,b,c} = {a) = {xp}. Then (xbc) =p(abc)p p ( axe) = (abc) (xbc)( abc), and (abx) = (abc)( xbc)( abcY) From thesewe getall 3-cyclesas follows:

(xyc) =(aby)'(xc)(aby), ( xby) = axe) 1( aby) axe), (axy) = (ybc)'(axc)(ybc),

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 164 J. A. EIDSWICK [March and

(xyz) = (zbc) '(xyc)(zbc). E

PROPOSITION 7. If F is a subgroupof H I G thatcontains a completeset of 3-cyclesfor any three elementsof J, then F containsall elementsa such thata' is evenand a" satisfies(9). For any such a, thereexist 3-cyclesT1, ... . TN E F such that aT1 TN = 1. Proof. The proofis by inductionon thenumber of elementsof Act'(a). If a' * 1, thereexist distinctelements a, b, and c of Act(a') withb = aa'. Let T be a 3-cyclesuch thatT' = (acb) and T"(a) = -a"(b). Then IAct'(aT)I < IAct'(a)I.If a' = 1, thereexist distinct elements a and b witha"(a) = 0 and a"(b) = 0. Then IAct'(att(a,b)a"(b))l < IAct'(a)I. m 5. Applications.This sectioncontains various applications. Discussion is limitedto puzzles based on partitionedregular polyhedra. It is assumedthat puzzle pieces are markedso thathome locationsand orientationsare uniquelydetermined. To keep the notationmanageable, in most cases onlythose generators and puzzlepieces that play direct roles in theillustrated moves will be givennames. The readermay find it helpfulat timesto expressgenerators as productsof k-cycles usinga morecomplete notational system. A completenotational system will be developedfor the n X n X n cube puzzle. Except for slightvariations, the devicesconsidered here actuallyexist. The n X n X n cube existsfor n = 2, 3, 4, and 5, and thereis evena transparent5 x 5 x 5 cubewhich is heldtogether by magnets. The icosahedronpuzzle. We startwith this example because it is simplestand displaysmany of thebasic ideas. The generatorsare thetwelve 720 vertexturns as indicatedin Fig. 2(a). Products of thesegenerators cause permutations of thetwenty triangular faces. They also cause rotationsas can be seen by studyingthe action of themove L2R2 (see Fig. 2(b)) on thevertices of theface markedc. By the discussionin ?3, we see thatthe combined permuting-rotating action can be regardedas takingplace in thewreath product Z3 l S20-

(a) (b) FIG. 2

Now let's see how to cyclethe pieces marked a, b, c in Fig. 2(b). To applyProposition 3 we need "disjoint" moves a and ,Bsuch that a': a -- b and /B': c -- b. That U-', RT'R2RI is such a pair can be seen fromFig. 2(b) and can also be verifiedby Proposition4. Since L2R2 causesa 1200 clockwiserotation of piece c and stabilizesa and b, it followsthat a completeset of 3-cycles fora, b, c is givenby:

(abc)1 = (L2 R2 )[U',R1R2R1](L2R2) (j = 0 ? 1). We can regardthis set as a vehiclefor moving piece b into locationc endingup withany

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 165 orientationwe please. Note thatthe move corresponding to j = 0 affectsthe orientation in the same way as the move RT'R-' and the move correspondingto j = -1 [j = 1] adds a 1200 clockwise[counterclockwise] rotation. Since thegenerators for this puzzle are 5-cycles,all patternscorrespond to evenpermutations and Proposition1 playsno role.Clearly, Proposition 6 (and itsproof) apply and we mayproceed as in Proposition7 to unscrambleany possible configuration.

(a) (b)

FIG. 3

The cubepuzzle. Thispuzzle can be viewedeither as beinggenerated by thesix 90? faceturns indicatedin Fig. 3(a) or by thenine 90? layerturns which include the three "slice turns"indicated in Fig. 3(b). Since a sliceturn is clearlyequivalent to a pair of parallelface turns followed by a 90? rotationof theentire cube, no new patternsarise by allowingthese moves. Group-theoretically,if Mf and Ml are the groupsgenerated by facesand layers,respectively, then Ml is an extensionof Mf by thegroup of rigidmotions of thecube. We willwork in thelarger group, allowing slice turns, because it willlead to neaterformulas. The pricefor this convenience will be theinconvenience of having to deal witha trivialorbit. The orbitsare: I, consistingof theeight corner positions, 9' consistingof thetwelve edge positions, and C, thetrivial orbit, consisting of thesix face centers. Corresponding to elementsof Y/,9, and C, respectively,there are 3, 2, and 4 possibleorientations. We note,however, that in themost popularversion of thispuzzle, the six facesare solidlycolored rendering center piece rotations indistinguishable. Let h, be any faceturn, h2 any sliceturn, and let (p, = Y/, (p2 = W. Then,for m = 2, (i) of PropositionI is clearlysatisfied. Moreover, if h is anygenerator, (ii) of PropositionI is satisfied by ?7= 6' and cl = C2 = '- It followsthat (ii) holds forall productsof generators,i.e., forall movesh. The conclusion(iii) of PropositionI saysthe obvious in thiscase: Anyelement can be made even on all orbitsby makingat most two quarter-turns.Note thatif face centersare correctlypositioned, at mostone quarter-turnis needed,and, if,in addition,face centersare correctlyoriented, then no adjustmentis needed. As with the icosahedronpuzzle, Propositions3-4 give us the followingset of 3-cycles, Proposition6 applies to both orbitsYl and 6' and we can proceedas in Proposition7. The notationrefers to Fig. 4.

(abc)j = (LD) J[U, R-1DR](LD) y ( j = 0, ?1), _J (123)j = (p-'F 2 )J[0-1,R-1FR](p-'F 2) ( j= 0, 1). Exceptfor a poss-ibleparity adjustment at thebeginning, the solution is a productof movesof theform -y'[a,7A]7-y which stabilizethe oriented centers. It followsthat if thecenter squares are

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 166 J. A. EIDSWICK [March

b~~~~

(a) (b)

FIG. 4

v/~ ~ ~vertex

piece

B FIG. 5

:~~~~~~~~~~~~~~~~~~~~~~~

(a) (b)

FIG. 6

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 167

correctlypositioned and orientedat thebeginning, they will end up thatway. Tetrahedronpuzzles. Two versionswill be discussed.When combined, a thirdversion (called theMaster Pyraminx in [12]) resultswhich can be solvedby combiningthe two solutions. The firstpuzzle is generatedby thefour 1200 vertexturns as indicatedin Fig. 5. Note thateach vertexconsists of a clusterof fourpieces which rotates as a unitindependent of theother vertex clusters.(For thepuzzle called the Popular Pyraminx in [12] thelittle tetrahedron tips can also be rotated.This complicationis trivialand is disregardedhere.) The vertexclusters play the same pivotalrole as thecenter squares for the cube puzzle without slices. As withthe cube puzzle, it will be seen thatthey can be fixedup at theoutset and neveragain worried about. Thus, there is only one orbitwhich consists of thesix edge pieces (see Fig. 5). As before,the following complete set of 3-cyclescan be obtainedvia Propositions3-4, Proposition6 applies,and we can proceedas in Proposition7. The notationrefers to Fig. 5:

(123)j = (RTBT-')J[T,B-1](RTBT-')-J (j =0,1). The secondversion is generatedby the180? edgeturns as indicatedin Fig. 6(a). Thereare four orbits:Yf consistingof thefour corner pieces and threeface orbits J1,2, 3 each consistingof fourface pieces. The orbitsare indicatedin Fig. 6(b) by differentshadings. Note thatthe six edgepieces remain fixed relative to one anotherand play thesame pivotal role as the centersquares in thecube puzzle and thevertex clusters in thepreceding example. Clearly,

V( g,1-) + v( g,2) + V( g9,') = 0 mod2 forany generatorg and hencefor all g, and Proposition1 is satisfiedwith m = 3. Note, though,that parities are takencare of automaticallyby correctlyorienting the edge pieces. Note also thatorientations of theremaining pieces are uniquelydetermined. By Propositions3-4, we obtain: (123) = [RT, LT BT LT] wherethe notation refers to Fig. 6 and V1V2denotes the turn with ends V,, V2. Similar 3-cycles can be foundfor the other two face orbits. It is easy to see thatthe hypothesis of Proposition5 is satisfiedwith a = BR, -y= BT, and 8 = BL, and, hence, (abc) = [BR, [ BT, BL]I. Proposition6 clearlyapplies to all orbitsand thesolution can be completedvia Proposition7. Octahedronpuzzles. Two versionswill be discussed.The firstis generatedby thesix 90? vertex turnsas indicatedin Fig. 7(a). As pointedout in [12] thereis a vertex/centerduality between this puzzleand thecube puzzle.The octahedronis easierbecause the eight corner cubelets of thecube puzzle correspondto merepoints on the octahedronpuzzle. Thus, a simplestrategy is to first orientthe vertex clusters, then use edgeprocesses developed for the cube puzzle(compare Figs. 4 and 7(b)):

(123)J = (p-1F2)J [1, R-1FR](p-IF2) J (j = 0, 1). The secondpuzzle is theface-turning puzzle described in [12]. Like thecube puzzle,we will allow sliceturns (see Fig. 9). The solutiongiven here provides for the addition of orientablepieces at the face centers,the idea being that such a centerpiece would rotate only when its correspondingface rotated.See Fig. 8. The puzzlewithout orientable center pieces is somewhat easier. Thereare fourorbits: f consistingof thesix corners, f consistingof thetwelve edges, and two face orbits.Sj and JF2 each consistingof twelvepieces. The orbitsare indicatedin Fig. 9 and

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 168 J. A. EIDSWICK [March

(a) (b)

FIG. 7

FIG. 8 againin Fig. 10, whichis a flattenedversion of the puzzle. In theflattened version, one can readily view the entireaction of the generators.Of course,these can all be expressedas productsof 3-cycles(five for face turns, four for slice turns) and thereader, in checkingthe formulas below, maywell find it advantageousto do so. Observethat edge and facepieces have onlyone possibleorientation and thatcorners have two.By Propositions3-4,

(ee2e3) =[F2 P and by Proposition5,

(fif2f)(111213) =[01,=211 [B-, R2] and ( (vLv2v3)3 = (L2F2)'[F1,[ R 'F'R,B1]]( L2F2- = 0,1). A similar3-cycle can be givenfor the other face orbit, and it is clearthat Proposition 6 applies to all orbits.The solutioncan thenbe completedvia Proposition7.

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 169

6 R,~~~~~~~~~~~~~~~B

U'00

FIG. 9

Dodecahedronpuzzles. Two versionswill be discussed.The generatorsfor the first one are the twelveface turns as indicatedin Fig. 11(a). Thereare twoorbits: J/ consisting of 20 cornerpieces and 5' consistingof 30 edgepieces. Clearly, each generatoris theproduct of two5-cycles and we can proceedin theusual way, obtaining the following moves which can be used to solvethe puzzle (see Fig. 11(b)): ( (abc), = (R2R1)'[L'1,U-'R1U](R2R1) = 0,?1) RJ1, FB F) (j = 1). (123)j = (LlL2F)J[ 1U-1BF-1] (LL2 J 0, The secondpuzzle, shown in Fig. 12(a), is a deepercut version of thefirst one, the extreme case beingthe Magic Crystalof [12]shown in Fig. 12(b). Notingthat the face pieces a and a' act as a unit(with two possible orientations), we can see thatthere are threeorbits Y/-, 0, and Y of sizes 20, 30, and 30, respectively. The followingbasic 3-cyclescan be obtainedvia Proposition3. Orientedvariations and other detailsare leftto thereader.

= (xyz) [L', RI],

(abc)-[ L-, L- R, 1+22R L

(123) = [L- 1, 2R2L 1RX-Ry1R1A72L2Ry1f1].

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 170 J. A. EIDSWICK [March

S ~~~~~~,.

FIG 11 ~

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 171

2~~~~~

(a) (b)

FIG. 12

l z 7 z 7

_~~~~~~IZ. ,I,f ,

t~~~~~~~. ZZZ.,

I Z nZ

FIG. 13

The n X n X n cubepuzzle. As theultimate cubelike puzzle, consider a cubepartitioned into n3 congruentsubcubes and suchthat each of the3n layerscan be rotatedas indicatedin Fig. 13. The algorithmto be describedis forunscrambling any generated configuration of surfacecubelets. It willbe seen thatthe algorithm leaves the (n - 2)3 internalcubelets invariant, given that they are fixedup in advance.Thus, if one wanted,in addition,to unscramblethe internal cubelets, this could be done by applyingthe algorithm successively to a nestedsequence of cubesstarting with eithera 1 X 1 x 1 centeror a 2 X 2 x 2 center(depending on the parityof n) and working outward.Each applicationof thealgorithm would fix up a new"shell" (the surrounding cubelets, of course,would just "come alongfor the ride").

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 172 J. A. EIDSWICK [March

We assumethat n is odd: n = 2m + 1, m > 2. The evencase is a corollaryand is leftto the reader.We set up notationin an xyz-coordinatesystem with the center of thecube at theorigin and theedges parallel to thecoordinate axes. A typicalcubelet is describedby theset

[ 2 'i+2 ]X[j 2 ' +2 ]X[k-2 , +2] or morebriefly by thecoordinates (i, j, k) of its center.The descriptionmay seeminadequate becauseit ignoresorientations. It is easyto see,though, that except for those cubelets corresponding to the3 x 3 x 3 case,only one orientationis possible. Each generatoris a productof m(m + 1) 4-cycles.Clockwise turns Xi, Yj, Zk (i,j,k = 0,+l +m) are givenby: m m

X = rI TI ((i, j, k)(i, k, -j)(i, -j, - k)(i, - k, j)), j~=O k=1 m m II ((i, j, k)( - k, j, i)(- i,j, - k)(k, j, - i)), yi-=iO k=1 m m Zk = II I ((isj, k)(j, - i, k)( - i, -j, k)( -j, i, k)). i=Oj=1 Here,"clockwise" is fromthe point of viewof lookingalong a positiveaxis towardthe origin.

(i, j, k)K=

FIG. 14

See Fig. 14. Noticethat the central slices X0, Y0, Z0 can be omittedfrom the list of generatorsas theycan be replacedby thecomplementary moves. Thereare m2 + 1 orbits(see blockedportion in Fig. 15): -/= 0(( m,m, m)) , & 0((m,m,0)),

6'k' = ((m, m,k)) (k= I,..,m - 1), and g5Ck((m,j,k)) = (j = l,...,m - 1; k = 0, +1 ,.. (j - l),j). Orbits Y/ and 4' correspondto the 3 x 3 x 3 case and consistof 8 and 12 cubelets, respectively.Each 4'k consistsof 24 edgecubelets and each ,k consistsof 24 facecubelets. The shadingsin Fig. 15 illustratethe four types of orbits.

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 173

_ (m, m, m)

akI i (E . g g ) ---I--I---. _ _ - : :~~~~ 1 ~ , : (r,rnO)

I ,

*m m

Fi- 15. Orbits in the face x m.

Now consideran arbitrarygenerated configuration g. Let .l...., LN be anysequence of layer turns* X0,Y0, Z0 suchthat g = LI-- N Let f((9) and s((9) denotethe number of L,'s which are, respectively,face and sliceturns such that P( L,, ?9)= 1. Then,clearly,

i(g _) f ( +) ?s((9) mod 2 from which we obtain the following:

V(grn JZt) = f( *J**)=( f V) = tgL d

V(9g ,'Vjk) =f(g) ? SOj( ) Sd(? )+ = = V(g,9') + v(g,JF]0) + P(g,0) for k j,

V(g9,) = S(d'k) = V(g9g) + V(g9, F)k and v(g,F) =1f() = p(g .

Thus condition (ii) of Proposition 1 is met with 6, = o. for j = 1,... , m - 1 and 6P)m= . Furthermore,it is clear that z(Y1,Oj) = S,j and, therefore,condition (1) of Proposition1 is satisfiedwith h, = Y. The surprisingconclusion is thatail paritiescan be made evenby making at most m quarter-turns. Noticethat if theface centers are correctlyoriented at thebeginning, then v(g, ) = 0 and the above adjustmentwill not affectthe face centers. Via Propositions3-4, we obtain the followingformula which gives 3-cyclesfor all orbits simultaneously: ((-j,m,k)(m j,k)(-k,m,-;))= [Zk,XmZ-j X ] for 0 < IkI < j < m, k * -j. The solution then proceeds as expected. To see thatthe internal cubelets remain invariant under the algorithm if theyare fixedup in

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 174 J. A. EIDSWICK [March

,J~~,

FIG. 16

advance,assume that v(g, 6(t)) = 0 forall internalcubelets t. Then,from Fig. 16,it is clearthat

= = = v( g, = V( g, 6(9s)) V(g,K9(t,)) 0 for j 1, ..., m-1. Hence, in this situation,parity adjustments which would affectthe internalcubelets are not needed.Therefore, since all othermoves are commutators,the internal cubelets are unalteredby the algorithm. 6. The underlyinggroups. If, in theabove examples, we ignorethe matter of orientingpivotal centers,the underlyinggroup P forsuch a puzzle can be describedas a subdirectproduct of groupsof theform H, l Gz, whereH, = Zr and Gi is eitherAlt -T or SymXi. By Propositions2 and 7, PIH, I G, has indexr, in H, I G,. Also,if f .., m} is a paritybasis in thesense of Proposition 1, then in forminga typicalelement g = g ...** . * I one has I J!choices for 1 each g,, i = 1,... , m, and JI-'! choicesfor each ofthe remaining g, 's. It followsthat the order of P is equal to

1 N 2N-m i=1 i

In particular,we see thatthe (2m + 1) X (2m + 1) X (2m + 1) cube contains

8! . 12! . 24!m2 . 211 . 37

2m2 + 1-m

distinctpatterns. For the(2 m) x (2 m) x (2 m) cube,the reader will find that there are m2 + 1 - m orbits,one of size 8 and all othersof size 24, and thata paritybasis has size m. Hence,taking intoaccount the fact that the 24 rigidmotions of thecube yield duplications, the (2m) X (2m) X (2 m) cube has

8! . 24!m(m-1) . 37

2(m-1)2 . 24

distinctpatterns. In case the faces are solidlycolored, the numberof distinguishablepatterns decreases.The followingtable summarizes our findings for the cases n = 2, 3, 4 and 5. Conflicting values forthe case n = 4 haveappeared in print.

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 1986] CUBELIKE PUZZLES-WHAT ARE THEY AND HOW DO YOU SOLVE THEM? 175

Coloredcube size Numberof distinguishablepatterns

8! -37 2 x 2X2 24

8! 12! - 21 . 37 3 x x3 3 2

4x4x4 8!.-24!2 -37 2 24 .(4!6/2)

5 x5x5 8! 12! 24!3. 211. 37 23 .(4 !6/2)2

7. Efficiency.Many of the algorithmsoutlined in ?5 can be shortenedconsiderably by replacing3-cycles by relative3-cycles. If O,... 0N is anyordering of theorbits, then the "pure" 3-cyclesused to unscramblethe orbit 0, can be replacedby 3-cyclesrelative to 91 U ... U O9. These,in general,will have shorterfactorizations, and can stillbe obtainedvia Proposition3. A measurementof theefficiency of anyparticular ordering is givenby

N ()m(21, ..., 9N) = Z liili i=l where 14denotes the shortestpossible 3-cyclein 69,relative to 691u ...* u6,. The followingtable is a summaryof our findingsof mostefficient orderings with the relevant additional3-cycles given in theright-hand column (cf. Figs. 4,6,9,11,12,14). The findingsare based on minimizing(*) overthe N! orderingswith the li's takenas lengthsof commutator 3-cycles.

Puzzle Most efficientordering Relative3-cycles 3x3x3cube (,f) [R,F]

tetrahedron# 2 GFI, -2-3, Y) [ RT, BT],[BT, LT],[LT, RT] octahedron#2 (,4, 'F,I F2) [Fl, F2 ], [ F1, B1] dodecahedron# 1 (,1") [ R1, R2] dodecahedron#2 (Y, g) [F1, F2], [L1, Rj1 R 2R1] ii X n X n cube (O(m, 1,0),..., O(9m,m, 0), [XI,? Y1],. ,[x"jq Yn,I thenany order)

References

1. ChristophBandelow, Inside Rubik's Cube and Beyond,Birkhauser, Boston, 1982. 2. E. R. Berlekamp,J. H. Conway, and R. K. Guy, WinningWays for Your MathematicalPlays, Vol. 2, Academic Press,New York, 1982. 3. GarrettBirkhoff and SaundersMac Lane, A Surveyof ModernAlgebra, Macmillan, New York, 1941. 4. F. J. Budden,The Fascinationof Groups,Cambridge University Press, London, 1972. 5. W. Burnside,Theory of Groups of FiniteOrder, Cambridge University Press, London, 1897. 6. Robert Carmichael,Introduction to theTheory of Groups of FiniteOrder, Ginn, Boston,1937. 7. A. H. Frey and D. Singmaster,Handbook of Cubik Math,Enslow Publishers,Hillside, NJ, 1982. 8. Solomon W. Golomb, Rubik's cube and a model of quark confinement,Amer. J. Phys.,49 (11), November 1981. 9. Marshall Hall, The Theoryof Groups,2nd ed., Chelsea,New York, 1976. 10. I. N. Herstein,Topics in Algebra,Blaisdell, New York, 1964.

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions 176 E. J. P. GEORG SCHMIDT [March

11. Douglas R. Hofstadter,Metamagical Themas: The Magic Cube's cubies are twiddledby cubistsand solved by cubemeisters,Scientific American, March 1981, pp. 20-39. 12. _ _, MetamagicalThemas: Beyond Rubik's Cube: Spheres,pyramids, dodecahedrons and God knows what else, ScientificAmerican, July 1982, pp. 16-31. 13. M. I. Kargapolov and Ju. I. Merzljakov,Fundamentals of the Theoryof Groups,2nd ed., Springer-Verlag Graduate Texts in Mathematics,New York, 1979. 14. Derek J. S. Robinson,A Course in theTheory of Groups,Springer-Verlag, New York, 1982. 15. JohnS. Rose, A Course on Group Theory,Cambridge University Press, London, 1978. 16. JosephJ. Rotman,The Theoryof Groups: An Introduction,Allyn and Bacon, Boston,1968. 17. Chris Rowley,The groupof theHungarian Magic Cube, AlgebraicStructures and Applications,Proceedings of the FirstWestern Australian Conference on Algebra,1982. 18. Eugene Schenkman,Group Theory,Van Nostrand,Princeton, NJ, 1965. 19. David Singmaster,Notes on Rubik's Magic Cube, Enslow Publishers,Hillside, NJ, 1981. 20. JohnC. Thompson,Rational functionsassociated to presentationsof finitegroups, J. Algebra,71 (1981) 481-489. 21. Helmut W. Wielandt,Finite Permutation Groups, Academic Press, New York, 1964. 22. _ _, PermutationGroups throughInvariant Relations and InvariantFunctions, Lecture Notes, Ohio State University,1969. 23. Richard M. Wilson, Graph puzzles, homotopyand the alternatinggroup, J. Combin. Theory Ser. B, 16 (1974) 86-96.

AN ALTERNATIVE APPROACH TO CANONICAL FORMS OF MATRICES

E. J.P. GEORG SCHMIDT* Departmentof Mathematics, McGill University,Montreal, Quebec, Canada, H3A 2K6 0. Introduction.The purposeof thispaper is to describea new approachto thederivation of themain structuretheorems for linear transformations and matrices.The approachis unconven- tionalin thatit reversesthe traditional order between the studyof variouscanonical forms and the structureof exponentialsof matrices(as treated,for example, in thebeautiful book [2]). It is efficientin thatit leads in a unifiedand directway to themain facts needed for the derivation of canonicalforms. This approachis accessibleto studentswho have masteredthe topics in linearalgebra which usuallyprecede the structuretheorems (as given,for example, in ChaptersI to IX of [3]). It is particularlyappropriate within the contextof a differentialequations course, where it is often necessaryto reviewor supplementa previoustreatment of linearalgebra. The expositionis given forreal and complexmatrices, but can easilybe adaptedto a moreabstract treatment. 1. The structureof explAti. Let A = [A,j] be an n X n real or complexmatrix and let the totalityof such matricesbe denotedby MJ(R) or MJ(C), respectively.It is customaryin the treatmentof linearsystems of differentialequations to define 00 Aktk exp[ At]- E k k=O

E. J. P. GeorgSchmidt: I was bornand raisedin SouthAfrica, where I completedmy undergraduate training. I receivedmy Ph.D. fromStanford University working under the supervisionof Ralph Phillips.After a year at the M.R.C. in Madison and two years in Aarhus, Denmark, I moved to Canada and McGill University.My mathematicalinterests are centeredaround partial differential equations. Otherwise I enjoycamping and hikingwith my family,and play tennisless frequentlyand less well thanI would like to. *This work was partiallysupported by a Grant of the Natural Sciences and EngineeringResearch Council of Canada.

This content downloaded from 65.206.22.38 on Mon, 19 May 2014 12:59:51 PM All use subject to JSTOR Terms and Conditions