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CUBIC CIRCULARA quarterly newsletter for Rubik Cube addicts

Issue 3 & 4 Spring & Summer 1982

CONTENTS Editor's Corner 2Songs 2Answer 2Cubic Quarterlies 3The First World Championship 4Cubic Cannabis 7New Cube Products 8A. Pryl - Fool! 13

Rubik's Revenge - The 43 14Winning Ways on the U Group 17

The 53 18The Magic Dodecahedron and Alexander's Star 19Magic Polyhedra 21Rubik Robots 25Pretty Patterns 26Orders of Elements 34The Magic Disc 36Comments on the U Group 36



Published by David Singmaster Ltd., London, England. ISSN NO 0261-8362

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EDITOR'S CORNER

When I started the Cubic Circular last year, I. intended to have 12 pages per issue and I wondered how Iwould fill them. As it turns out, material continues to arrive faster than I can deal with it or publish it, and twoother magazines have started. The first two Circulars were already 16 pages and I have about 40 pages oftype-script in front of me. Because of this and because the Spring issue is a bit late, we have decided to makethis a DOUBLE ISSUE for Spring and Summer, i.e. issues number 3 & 4. Depending on how the material fitsand on the printer, this issue will be 32 or 36 pages.

Hence this issue completes the first year of the Circular. The next issue will appear in the Autumn, probably inOctober or November. Because of the extra weight of larger issues, the need to employ technical assistanceand the fact that circulation is still confined to a small circle of cubologists, we find the current subscription of £2.50 per year is not really covering our costs. Hence we must raise the subscription to £4.00 for the nextyear (or $8.00). This may be revised later but we will accept renewals now at this rate.

Do it faster!

SONGS

Mister Rubik, by the Barron Knights, (C1-15) is now on their LP "Twisting the knights Away". An Englishschool group did a song of their own on TV while trying to restore their cubes at the same time. They didn'tfinish.

A Hungarian husband and wife, Bea Muszty and András Dobay, have made an English single called "Trick inthe Middle" about Rubik's cube (Start Records SP5 70537) -

"Try to harmonize the colours of your life Or the faces of the cube to be alike. You can twist, twiddle and fiddle, But the trick's in the middle of the cube."

Don Taylor has sent a photocopy of the cover of the Australian song. It's called "The Cube" by Mike Brady and



the Cubettes. It has been done as a video clip with dancers.

Claude Shannon has composed "A Rubric on Rubik Cubics" to the tune of 'Ta! Ra! Ra! Boom De Ay!'. It hassix verses of 8 lines with six 4 line choruses, and ten footnotes: It's a bit long to copy here, but perhaps I'll putit in a later Circular. [ See C7/8-36. - J ]

ANSWER

The answer to the film question on C2-4 is that the film was run backwards!

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CUBIC QUARTERLIES

The first issue of Rubik's Logic and Fantasy in Space has appeared (C2-3). It appears in several languagessimultaneously. The price is now listed at $2.00 per issue. The contents are as follows: (Some of theinformation will be excerpted elsewhere.)

Introduction by Rubik The Order of Disorder. This is an essay on entropy and the Cube. Mr. Cube - Ernö Rubik - a sketch. This is an interview with Rubik, one of the best and most informative that Ihave seen. Mister Rubik. Words of the song and a story about the Barron Knights (C1-15). A world you can twist around, by Gerzson Keri. An essay on patterns - achievable, constructible and evenunconstructible ones! Letters to the editor. From five thousand to fifteen millions. A report on Politechnika's production of the cube. Student Olympics. A report on the International Mathematical Olympiad. Geometrical Art - The Magic of Shapes. An essay on cubism and on cube images in art. Cube Fans - Competitions and clubs ; Our London correspondent reports Puzzles page Around the World - press cuttings. Reviews of "You can do the Cube" and of my Notes. Toy-Business; Varikon. An article on Konsumex products, especially the Varikon barrel puzzles. The Computer, my playmate. An essay on computer games. Corner - a page of cartoons.

The first issue of Rubik's Cube Newsletter (from PO Box 72, Hollis, New York, 11423, USA for $5.00)appeared in May and has just arrived. It's 8 sides, printed on a single piece of paper and folded, in red andblack. The contents are:

At the US National Championships The Editor's Cubie Upcoming Cube Clinics Sneak Preview: 2 new Rubik's Puzzles Cube Contest Kits Cube Club Calendar Membership Offer Letters to the Editor



Local Clubs Final Standings & Times 1981 Regional Cube-A-Thon Records and Results

The membership offer includes several new spinoffs: 'Keep on Cubin' tee shirts and badges, pencils, patches,a mobile!, neck ties and memo pads!

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THE FIRST WORLD CHAMPIONSHIP

The first Rubik's Cube World Championship was held in Budapest on 5 June 1982. This was jointly sponsoredby Politoys, Konsumex and Ideal. There were contestants from 19 countries. Minh Thai, the US champion, wonwith a time of 22.95 sec.

The competition was efficiently organized by Brian Cartmell, the public relations firm for Ideal (UK), so it hadthe same basic structure as the UK contest. The cubes were selected by Rubik from a special production runand competitors were given a sample to practice with. Competitors described them as pretty good. All had astandardized colour pattern with white opposite yellow, red opposite orange and blue opposite green (DavidSibley calls this the 'plus and minus yellow pattern' since opposite faces differ by yellow.) and with blue,orange, yellow clockwise at a corner (spelling BOY). Rubik worked with some mathematicians to developseveral patterns of similar complexity. (The contestants remarked that all the patterns seemed of equaldifficulty.) Four sets of cubes were put in four of these patterns by Rubik and sealed in separate briefcaseswhich were kept in a bank overnight and brought to the competition by the supervising lawyer. One case wasopened and placed by the competitor's entrance. Each contestant picked one at random as he came on stage.(The competitors remained offstage until their turn came so they could not watch the previous attempts.) Eachcontestant had three attempts and the best time was taken. The fourth cube was used as an extra cube incase a cube came apart. In this event, the contestant got one extra trial. This happened seven times. TheFinnish contestant's cube broke twice, so he lost one trial.

The contestant had fifteen seconds to examine the cube in his hand. It was then set down on a photoelectricbase. The timing was then based on the time from picking up the cube to setting it down again in completedform.

There was a jury consisting of Rubik, myself, Georgina Tamás (Commercial Director, Konsumex), PeterPeacock (Marketing Director, Ideal (UK)), Rainer Seitz (Product Manager of Arxon and founder of the Rubik'sCube Club). Basically the jury was to guarantee fairness in the event of unexpected phenomena. Indeed thepower for the TV and the display timer gave out in the middle of one trial. Fortunately the timing computersare battery powered and a valid time was obtained. One contestant started to do the cube during the studytime - he was given an extra try. Once the display timer wouldn't reset after the study time and there wassome delay, so the contestant was allowed a new examination period.

The order in the first round was determined by a drawing the previous night. On the following rounds, theslowest contestants went first. The logo of the contest was a cube with one corner replaced by a globe. Theprizes were three examples of these which were plated in gold, silver and bronze, together with a number ofelegant Hungarian handicrafts.



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The Contestants

I give a summary about each contestant: Country. Name, age, home city, occupation, personal best time,winning time in national championship (labelled: won in ...). Comments.

Austria. Josef Trajber, 25, Vienna, computer programmer, 29 sec. Author of two books on the Cube. Inventorof an octahedron (C1-7/8)

Belgium. Luc van Laethem, 25, Kraainem (near Brussels), student, 17 sec. Did 100 Cubes in 54 minutes (32.4sec average).

Bulgaria. Svilen Tenev, 18, Plovdiv, student of languages, 40 sec.

Canada. Duc Trinh, 14, Kitchener (Ontario), student, won in 26 sec. Originates from Viet Nam.

Czechoslovakia. Jiri Fridrich, 17, Ostrava-Poruba, student, won in 23.55 sec.

Finland. Jan Sandqvist, 21, Helsinki, advertising representative, 19 sec.

France. Jerome Jean-Charles, 26, Paris?, journalist, won in 25.60 sec.

Great Britain. Julian Chilvers, 15, Eaton (near Norwich), student, won British Isles Championship in 25.79 sec.He has done 44 cubes in 20 minutes (27.3 sec average!).

Holland. Guus Razoux Schultz, 17, ?, student, 18.50 sec.

Hungary. Zoltán Lábas, 26, Nyiregyhaza (near Budapest), student, 34.5 sec, won in 36.9 sec average.

Italy. Guiseppe Romeo, 16, Rome, student, won in 26 sec.

Japan. Ken'ichi Ueno, 18, Tomakomai (near Sapporo City), student, 22 sec, won in 42.30 sec.

Peru. Eduardo Valdiva Chacón, 21, Lima, student, 26.80 sec.

Poland. Piotr Serbenski, 17, Warsaw, student, 28.50 sec.

Portugal. Manuel Moura Galrinho, 24, Setubal, professor of philosophy, 25.80 sec.

Sweden. Lars Petrus, 21, Uppsala, student of technical physics at Uppsala University, won in 24 sec.

United States. Minh Thai, 16, Los Angeles, student, won in 26.06 sec. Seven days after buying a Cube inJune 1981, he could do it in under two minutes. Came to US 3 years ago from Viet Nam, speaking noEnglish.

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West Germany. Ronald Brinkmann, 17, Hamburg, student, won in 19 sec.

Yugoslavia. József Borsos, ?, ?, student, won in 28 sec.

The average age was about 20 years, which was older than I expected. Jean-Charles and Lábas were theoldest.

Results

Best times are underlined. B indicates breakage resulting in one extra trial.

Round Best Ave.Name (country) 1 2 3 time Rank timeBorsos (Yug) 36.75 35.33 30.02 30.02 13 34.03Brinkmann (BDR) 34.80 30.59 32.32B 30.59 14 32.57Chacón (Per) 34.91 29.62 30.01 29.62 11 31.51Chilvers (GB) 30.59 25.95 27.46 25.95 7 28.00Fridrich (Cze) 31.49 29.11 33.20 29.11 10 31.27Galrinho (Por) 40.74 48.67 37.11B 37.11 16 42.17Jean-Charles (Fra) 27.87 31.18 25.06 25.06 6 28.04Lábas (Hun) 24.49 27.58 28.21B 24.49 3 26.76van Laethem (Bel) 32.92 34.98 29.73 29.73 12 32.54Petrus (Swe) 35.42 33.11 24.57 24.57 4 31.03Razoux Schultz (Hol) 24.32 31.51 26.15 24.32 2 27.33Romeo (Ita) 34.23* 41.75 28.11 28.11 9 34.70Sandqvist (Fin) 31.17B B 31.56 31.17 15 31.37Serbenski (Pol) 44.40 37.50 40.86 37.50 17 40.92Tenev (Bul) 51.88 47.29 47.35 47.29 18 48.84Thai (USA) 27.16B 22.95 27.97 22.95 1 26.03Trajber (Aus) 50.16 54.93B 58.99 50.16 19 54.69Trinh (Can) 37.44 26.63 36.09 26.63 8 33.39Ueno (Jap) 27.56 27.90 24.91 24.91 5 26.79

* Romeo's first cube broke three times but he reinserted the piece each time and carried on. Had he stopped,he would have been given an extra trial.

I have computed the average time for each contestant out of curiosity. The averages do not give resultsgreatly different than the best time. Thai would have won on average as well as best time!

Championship Techniques

I asked all the contestants what algorithm they used. Borsos' algorithm was unclear to me. Brinkmann andChilvers had two possible algorithms. The following descriptions assume doing the D face first, but this doesn'tmean they hold it that way. I write DC for D corners, ME for middle edges, etc. The most popular method wasto do DC, UC, U & D E, ME (C2-3). This was used by Brinkmann, Chacón, Chilvers, Jean-Charles, Labas,Sandqvist, Thai, Trinh, van Laethem. Ueno did DC, UC, ME, 3UE, rest.



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The second most popular technique was Pierre Jullien's method of leaving a DC and the ME above it undoneas working or parking spaces. Galrinho, Romeo and Tenev used this method. Brinkmann used a method DC,3DE, 4UC, rest. Variants of the D, M, U layer method were used by Fridrich, Serbenski and Trajber. Chilversdid DC, DE, U, M. Razoux Schultz did DE, DC & ME together, U (see N32). Petrus does a 2 × 2 × 3 blockthen orients edges, then orients corners, then completes D & M and then U. (See N32-33.) He gets a 57move average.

OTHER COMPETITIONS

The Hungarian Championship was held on 27 September 1981. A local elimination round gave 300competitors and a second round on 20 September left 128, classified as under 15, 15 to 18, over 18,containing 58, 31 and 39 competitors. Three trials were made and the average time taken. The best five ineach class then formed a three trials, average time, final. Zoltán Lábas won the over 18 class with an averageof 37.1 sec and the final with 36.9 sec.

The Czech competition, held on 11 May, had three categories: children up to 10 years; women and girls; menand boys. In the semifinals they had seven trials and the middle five were averaged, but the finals were forbest time out of three trials. Jiri Fridrich, 17, of Ostrava-Poruba, was first in the semifinals with an average of32.24 seconds and in the finals with a best time of 23.55 seconds.

There have been a number of charity cube contests in the UK. These try to do as many cubes as possible ina given time, often 10 or 20 minutes. Dame Kathleen Ollerenshaw organized one in Manchester for the St.John Ambulance on 6 March. Mal Davies organized one in Birmingham for the English Schools AthleticsAssociation on 15 May. At the first, Anthony Willetts did 29 cubes in 20 minutes. At the second, John Duffyand Anthony Willetts did 33 cubes in 20 minutes (36.4 sec average). I saw John White do a cube completelybehind his back (136 sec). Richard Hodson improved his one-handed time to 77.50 sec., but Duncan Dicks, a2nd year student at Warwick University, did 59.32 and 53.29 sec. Hendon School (outer London) had a contestfor the Barnet Mayor's Appeal for research into Crohn's Disease.

Two London policemen, Tim Timoney and Barry Pullen, held a 24 hour cube marathon at Keddies in Romfordto raise money for a five year old girl suffering from an inoperable brain tumour. They restored 1000 cubes in22 hours 25 minutes and acquired severe cases of Cubist's Thumb. They raised over £2000 to send the girland her mother to Lourdes and Disney World. [ See also C5/6-2. - J ]

CUBIC CANNABIS

Rubik Cubes were banned at Wandsworth Prison, London, after one was found to be stuffed with cannabis!

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NEW CUBE PRODUCTS

In C2, I discussed a number of new sliding block puzzles. Here I will discuss new cube type puzzles. Severalof these are discussed elsewhere.



Ideal is bringing out a 'Rubik's Range' of cube products, apparently all selected and supervised by Rubik andsome of which originated with Rubik. The 43 is part of this range. There is also a 23 with Ishige's mechanismcalled Rubik's Pocket Cube, and a 33 globe with a rather schematic map called Rubik's World. The CalendarCube (C1-5) is also in the Range. This was devised by Marvin Silbermintz of Ideal in New York, and strikesme as one of the neatest variations. Because Wednesday is Mittwoch, it is possible to set two days at once onthe German version. (Exercise: find all such pairs. This depends on the year!) [ see also C7/8-4. - J ]

Rubik's Challenge is a 33 with holes for pegs in the facelets. Each player chooses a colour, puts a peg in afacelet of that colour and then gives a turn. The object to get three of your pegs in a row.

Rubik's Race is a 5 × 5 board with four pieces of 6 colours in it. A container of nine tiny cubes is shaken toproduce a 3 × 3 pattern. The race is to slide the pieces on the board so as to get the 3 × 3 pattern in themiddle.

Ideal is also bringing out the Popeye 23 (C1-6), made by Tsukuda (Japan) and Alexander's Star.

Konsumex, the Hungarian trading company which exports the Cube, has a number of other cube productsproduced by Politoys. Rubik's Mini Cube is a small 23 with only three colours and a mechanism which is a bitlike Ishige's, but without an inner sphere. See below for schematic cross sections of both mechanisms. In bothcases one corner is fixed to the centre.

Ishige's Mechanism Rubik's Mini cube

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There are several Taiwanese 23's with a different mechanism. I haven't taken one apart yet. [ see also C5/6-5. - J ] They come with two types of die pattern as well as Popeye and ordinary patterns. The Mini Cube has threepairs of adjacent faces with identical colours. It turns out that this allows two colour patterns, one of which isimpossible to achieve!

The Magic Domino has been re-engineered and now has a much smoother action. It is also being made in aversion with raised spots for the blind.

General Impex is the other Hungarian trading company which exports the Babylon Tower (C2-12), theHungarian Rings (C2-14) and the Magic Dodecahedron. They displayed a 23 Magic Globe attributed to a Mr.Csillam.

Mondadori, the Italian distributor of the cube, is doing two variants. The Fifteen Cubo is a



Magic Square Magic Cube with black numbers on gold labels. (See C1-5.) The MasterCubo has 12 colours which essentially colour the edges. Here the centre orientations areimportant but the edge orientations are not and there are 8!·38·12!·46/2·2·3 patterns,which is the same as on the ordinary cube! [ See also C7/8-5. - J ]

Toni Obermaier, of the German sports equipment firm Togu, has applied for patents on a23, a 43 and a 53. He has started production of a series of eight 23 with different colourpatterns. The simplest has only two colours with four solidly coloured cubelets of eachcolour. Though advertised as having 70 patterns, there are just 7 when symmetries of thecube are considered. Any two patterns are at most two moves apart! [ three moves actually. See also C7/8-5. - J ] Obermaier describes this as suitable for preschoolers. Enumerating the patterns, considering the symmetries,is quite a tricky task for the different colour patterns. The mechanism is like Ishige's.

When I was in New York over Christmas, I found examples of cubes with flags, die patterns, animals andnudes (not to be confused with the other Boob Cube which only turns on one axis).

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Ole Jacobsen sends an example of an idea of Jim Saxe and Dan Hoey. There are 6 cyclic arrangements offour colours. Take a blank cube and put four coloured spots in the corners of each facelet, so the 6arrangements are each on one face. I find this cube makes you colour blind! A Korean variation of the cubeuses thin (about 1 mm) plastic covers on the facelets rather than the PVC stickers. I saw these in the USbeing sold as the Deluxe Version of Rubik's Cube at $15.00.

Another variant is the Banana Puzzler which has six colours of bananas on white labelsand the box says "This puzzler will drive you bananas". Peter Redmill sends a differenttype of edge colouring. He has each face split in half. The halves can be viewed ascorresponding to edges. There is a way to colour the edges with four colours so the threecolours of each edge are along three orthogonal intersecting edges, as shown.

John White, of Warwick University, has invented a non-trivial Irish Cube (C1-5). It hastwo colours arranged as shown. He has made a large number of other colour patterns. [ see also C7/8-10. - J ]

Henry and Bruce Dan, of Atlanta, Georgia, have been making solid colour cubes bybuying lots of ordinary cubes and peeling the labels. (They could have bought white orblack cubes directly from Ideal!) The Taiwanese have sold a black cube as The ExecutivePolish Cube. "Its total lack of reason for being will keep you fascinated for severalminutes, or, perhaps, only a few seconds ... try to solve the puzzle with your eyes closed... work the puzzle in the dark."



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In New York, I found the small rhombi-cubo-octahedron (C1-9) with three colours instead of ten. Recently Ihave seen the shape obtained by intersecting three octagonal prisms at right angles. This is a bit like the smallrhombi-cube-octahedron, but the corner pieces are pyramids rather than being flat. I haven't yet discovered thename of this shape. It has twelve colours corresponding to the edges. [ Truncated rhombic dodecahedron, see also C7/8-5. - J ].

Trajber's Octahedron (C1-7/8) has been produced in Taiwan, in two forms. The other form has the vertexpieces slightly truncated. Tsukuda (Japan) advertise a 2 cube-octahedron.

The range of cube related products is also amazing. Perhaps most intriguing are cube earrings! These areabout 12.5 mm (1/2") on a side. One version doesn't turn at all. Another turns on one axis only. In New York, Isaw hair slides (or barrettes) and a cube jig-saw puzzle. [ see also C5/6-5. - J ]. In the UK are mugs ("It's a Mug's Game"!), a memo cube and a poster version of the solution. At least adozen cubic tee-shirts are available, mostly in the US. Kit-Kat and Olympic Holidays used the Cube in bill-beard advertising. Hamlet Cigars has a Cube being restored, but it turns out to have a blue centre in the whiteface, so the user throws it away and lights up a cigar. The cube rotation is by me.

I have seen a number of chimerical cubes (C1-9). The simplest are obtained by taking one, two or threecolumns from an octagonal prism and putting them into an ordinary cube. See below for an example. Actually,I have just realised that one need net make them parallel and a rather odd shape is obtained with threeorthogonal nonintersecting columns, as shown below.

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Tony Fisher, of Ipswich, has made what I call Siamese Cubes. Take two ordinary cubes each with one columnreplaced by a column from an octagonal prism. Glue the cubes together along these columns. The pieces donot move from one cube to the other, so each cube behaves like an ordinary cube with one column gluedtogether. Exercise: Consider any pattern on the cube, which has a column correctly lined up. Prove ordisprove that the cube can be restored without breaking up the column. [ see also C7/8-13, and my Siamese Cubes page - J ].



Fisher's Siamese Cubes

Fisher's Siamese Cubes

One could join two Cubes at a corner, using a piece from a truncated Cube, or crossways at an edge: I can'tsee any way to join two Cubes which allows mixing pieces.

Another chimera from Taiwan is a mixture of corners from a truncated Cube and edges from a cubo-octahedron.

Uwe Mèffert came to the London Toy Fair with about 30 new products, several of which will be appearingsoon. The Magic Dodecahedron is mentioned elsewhere. The Pyraminx Star is a stellated and sparkly versionof the Pyraminx. It is even easier to solve, but it makes a most handsome object.

The most interesting of Mèffert's products is the Pyraminx Cube, invented by Tony Durham, an Englishjournalist. It is a cube which is bisected by the four planes through the centre and perpendicular to thediagonals. It rotates on each of the planes, a movement which people have difficulty in discovering. There are6 faces and 8 corners. There is a dual structure which is an octahedron with 8 faces and 6 edges. I find4!·4!·6!·38/2·2·24 = 283 43520 patterns for the cube and 4!·4!·6!·26/2·2·2·4! = 1 38240 patterns for theoctahedral version. (We get 26 rather than 46 for the reasons explained under Magic Polyhedra.) Theoctahedral version was invented earlier by Rubik.

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Pyraminx Cube Magic Octahedron

It is possible to make both the 6 four-sided pieces and the 8 three-sided pieces have recognizable orientationeither by various colourings or by embedding the four axes into a less symmetric solid - Durham uses adodecahedral shape. I get 4!·4!·6!·38·26/2·2·2·24 = 9069 92640 patterns in this case. However, I have justread over Durham's work on the number of patterns and he finds two further conservation rules which arequite interesting. I haven't had time to fully assimilate them, but they give two twist conservation laws for the



cube and dodecahedron cases, yielding 1/9 of the results I have just given, namely, 31 49280 and 100776960. I will return to this one once the objects become available - no point in spoiling your fun completely. [ see also C7/8-15, and my Skewb and Skewb Diamond pages. - J ]

A. PRYL - FOOL!

The Prague newspaper, Mlada Fronta, ran a story on 1 April about a school giving courses on the cube. Theschool's director was A. Pryl whose best time for the cube was 69 hours, 37 minutes, 58 seconds. Theteachers were Frantisek Kubík, age 12, and Miloslav Kostečka, age 9. The Associated Press picked up thestory and The International Herald Tribune (IHT) ran it, but they lost some of the points by christening A. Prylas Antonin Pryl and omitting his time. When I read the IHT report it seemed quite genuine. Mlada Fronta ranthe IHT article in English with Czech commentary, "Believe or Not Believe - We Have Caught a Whale".Mojmír Sob, who has kindly sent the details of this epic, points out that Pryl, Kubík and Kostecka are commonCzech names, though the last two are also Czech words for a small cube.

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RUBIK'S REVENGE - THE 43

The 43 is appearing. I have received an early sample via Rainer Seitz and it works pretty well. Ideal isadvertising it as Rubik's Revenge in the US and UK and as Rubik's Master Cube in Germany, but has onlymade a few deliveries so far in the UK. It is 66 mm (about 2 5/8 ") on an edge and weighs about 240 gm(about 8 1/2 oz) which makes it about one-third denser than the ordinary cube (57 mm edge, weighing 120gm).

The mechanism is basically a grooved sphere, but there are several points where it varies from designs I haveseen. The centres are the only pieces hooked on to the sphere. The four centres of a face turn in a circularopening where two grooves meet. There is only one central groove in each direction for the slices to turnalong, but these grooves are asymmetric so that one slice will turn with respect to the sphere, but the parallelslice carries the sphere with it. No piece is attached to the central sphere. The 43 will come apart in much thesame way as the 33, but one must remove all the edges and corners of a face and spread out the centrepieces in order to disengage a centre piece. I don't know the source of this mechanism.

Three books on the 43 are about to appear from Minh Thai, Jerome Jean Charles and Tom Werneck. All ofthese have been written without a 43 in hand!

I received my 43 in Budapest and began to work on it on Saturday evening. Thinking that the methods of the33 would solve the edges and corners, I put the centres right first. This is not too difficult. I then got the edgepieces in pairs, but stopped when I needed a single 3-cycle of edges. The next morning, while flying back toLondon, I saw how to do this and then put the edges and corners correct. However, I made a boo-boo laterand then tried the same method and discovered I needed a single 2-cycle of edges! After applying a 3-cycle Iwanted a 4-cycle and I had to leave it like this. I returned to it a few days later and saw how to form a 4-cycle,but this moved centres. Finally, I saw an easy method to 3-cycle centres only. So, my strategy is to get theedges correct first, then centres and corners, which are easily done, but everything takes some time. A layerby layer method is feasible, but the edges of the last layer will need some tricks. If the last layer is a face, itwill be very awkward to get a 2-cycle of edges. Hence it may be best to do two opposite faces and then theslices between them.



Number of patterns

The number of patterns on the 43 has been considered by Herb Taylor, Solomon Golomb and David Kantor inLos Angeles.

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Taylor and I have located an error in the argument at C2-8. The factor of 24 in the denominator arises sincethe 6 sets of 4 identical centres can be permuted in 24 ways each and the cube has 24 indistinguishableorientations. However since the parity of centre turns is the same as the parity of corner turns, only half of the24 cases are possible. This is the same reasoning that gave the 2 in the denominator at C2-6 and was usedseveral times on C2-9/12. Thus the correct (I hope!) answer is 8!·24!·24!·38/(3·247) = 7 40119 68415 6490186987 40939 74498 57433 60000 00000 = 7.4 × 1045

Notation

Kurt Endl has proposed a notation using A, B, C, D, 1, 2, 3, 4; I, II, III, IV torepresent the levels in a 43. This has the advantage of being indefinitelyexpandable, but I prefer a notation which retains some mnemonic character likethe B, L, U, D, F, B notation for the 43 . I find the following seems to be themost convenient for me. Label the 'slice' layer adjacent to each face by thelover case letter of that face. Thus the four levels from left to right are denotedL, l, r, R as shown in the diagram. Similarly we use U, u, d, D and F, f, b, B.Because we do not have centre pieces, one must be careful physically toremember which directions are which, or one can keep a corner fixed, as withthe 23 (N53).

It is clear how the notation records moves, but we must adapt it a bit to labelpieces and positions. We now have two FU edges and four F centres. So weshall now label each piece with three symbols, corresponding to the three levelsit lies in. Each edge lies in two faces and a slice and each centre lies in a face and two slices. The pieces ofthe front face are shown.

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Obviously, this notation is awkward to implement on computers which have onlyupper case. In such situations, one could use the next letter of the alphabet torepresent a slice - e.g. L, N, S, P; U, V, E, D; F, G, C, B.



Some Sample Processes

I give a few sample processes to illustrate the notation and the way in which commutators give simpleprocesses. I will leave you to build a strategy from these.

[F, R] = (FLU, FUR)+ (RDF, RBD)- (FUl, FRu, DRf) (FUr, FRd, DRb)

[[F, R], L] = (FLU, ULB, RFU)

[[F, R], 1] = (FUl, UBl, DRf)

[U, r] = (ULb, UBr, FUr, DFr, RUb) (Ulb, Urb, Urf, Frd, Fru)

[[F, r], L] = (FLu, ULb, RFu)

[[F, r], l] = (Flu, Ulb, Drb)

[[F,R],l] [[F,r],L]

17

[[F, r], d'] = (Fru, Frd, Rbd)



[f,r] = (Luf, Urf, Frd) (Ruf, Drf, Brd)

[r,b] = (Fur, Ubr, Rbd) (Bur, Dbr, Lbd)

[[F,r],d'] Half of [r,b]

[[r, b], F] = (Fru, Flu, Rbd)

Let me fill in this space by remarking that the 43 exhibits the properties of both the23 and the 33, if you move halves or just faces. Thus one can obtain a largenumber of patterns by using known patterns on the 33, together with certainvariations from the 23. In particular, one can obtain the cube in a cube in a cubein a cube or quadruple cube pattern. It is not possible to get chessboards on allsix faces. I don't know if you can get stripes on all faces??

WINNING WAYS ON THE U GROUP

Winning Ways (= BCG) has appeared as noted in C2-4. The BCS Tables (pp.808-809) have been revised andany U position can be restored in at most 29 moves (N49). David Seal has shown that most of the results arebest possible. The hardest patterns are (A, B) (b, d) (in any orientation) and A+ C- a+ b+ c+ d+ requiring 13and 16 moves. [ see also C3/4-36, C5/6-2, and C7/8-8. - J ].

18

THE 53

Pierre Bouchard sends an article from Journal de Montréal about a 53 invented by Gaston Saint-Pierre, a 21-year old Quebec boy. The first prototype was in wood, 14 inches (about 35 cm) on a side. He has built a



prototype in aluminium, about 7 cm on a side, which took 176 hours to make and weighs 2 pounds (about 1kg). It has 99 pieces, presumably 98 (= 125 - 27) exterior pieces and one interior piece. He estimates that ithas 1090 positions and it takes him 1 1/4 hours to restore. Quite a number of 53 designs and patents haveappeared, but this is the first working one I have heard of.

Let us work out a first guess as to the number of patterns on the 53. There are 6kinds of pieces, indicated in the diagram: A = corners; B = edges; C = edge middles;D = inner corners; E = inner edges; F = centres. There are 8; 24; 12; 24; 24; 6 ofthese, but pieces B, C, E are asymmetric so their orientation is determined by theirposition. The constructible group thus has

8!·38·24!·12!·212·24!·24! = 123 96654 10985 74050 19981 53455 12384 8121240682 69122 60338 23429 49382 61930 84194 48832 00000 00000 00000 = 1.24 × 1092 patterns.

Just as with the 33, the corners and edges have their total twist and flip conserved so we will divide by 3 and2. There are two kinds of moves: faces and offset 'slices'. Each of these gives a number of 4-cycles of piecesas tabulated.

A B C D EFace 1 2 1 1 1Slice 0 1 0 2 1

Consequently the parity of the A, C and D movements must all be the same, which gives only 1/4 of thepossible permutations of the positions. Now we also have 6 sets of four D end E pieces which can bepermuted without changing the appearance (assuming the faces are uniformly coloured). However, the Dpieces have the same parity as the A pieces, so only half of the permutations of the D pieces are possible.Using commutators of the form [X,[Y,Z]], just as on the 43, we can get 3-cycles of each type of piece, so wecan get all the patterns permitted by the above conservations. Thus there are 8!·38·24!·12!·212·24!·24! / (3·2·4·246·246/2) = 28287 09422 77741 85653 61803 33107 15032 82931 2773198567 21347 21536 00000 00000 00000 = 2.83 × 1074 patterns. [ see also my 5×5×5 Cube page. - J ].

19

THE MAGIC DODECAHEDRON AND ALEXANDER'S STAR

The Magic Dodecahedron has been contemplated for some time. So far I have seen photos or models from:Ben Halpern (USA), Boris Horvat (Yugoslavia), Barry Lockwood (UK) and Miklós Kristóf (Hungary) whileKersten Meier (Germany) sent plans in early 1981. I have heard that Christoph Bandelow and Doctor Moll(Germany) have patents and that Mario Ouellette and Luc Robillard (Canada) have both found mechanisms.The Hungarian version 15 notable as being in production (being exported by General Impex) and as having theplanes closer to the centre so each face has a star pattern.



Face patterns of the Dodecahedron

Uwe Mèffert has bought the Halpern and Meier rights, which were both filed on the same day about a monthbefore Kristóf. However, there is an unresolved dispute over the extent of overlap in the designs.

The theory of the Magic Dodecahedron is discussed elsewhere. [ see also my Megaminx page. - J ].

A simpler version of the dodecahedron is due to Adam Alexander, a US mathematician. Theshape is a great dodecahedron which is an icosahedron with its faces pushed in so thepentagonal pyramid about each vertex becomes a five-pointed star sitting on a plane and itcan rotate on this plane (marked F). [ see also C7/8-25/26. - J ] One can view this as a Magic Dodecahedron without corners (or centres) [ Compare C5/6 - 10 - J ]. Thus there are 30!/2 · 230/2 = 7 12032 72378 98958 11841 60204 98546 68800 00000 = 7.1 × 1040 patterns.

20

The 'faces' of the star are the planes on which the rotations take place. They comprisefive triangular areas, marked F above. Now the 12 faces fall into 6 pairs of parallelfaces. Alexander makes the parallel faces have the same colour, giving just 6 colours.Hence there are two pieces of each of 15 types and the number of patterns must bedivided by 215/2. There are 60 rigid motions of a dodecahedron (reflections areexcluded) which form a group isomorphic to A5. (Exercise: Show you can inscribe 5cubes in a dodecahedron so a symmetry of the dodecahedron is a permutation ofthese cubes.) The rigid motions are rotations of order 5, 3, 2 and it is readily checkedthat each of these can be obtained as an achievable position of the edges whencentres are fixed. I.e. the spots group or the centres group (N46/47) is the whole group A5 of symmetries.Hence the effect of the centres being unseen is to divide the number of patterns by 60 and there are 30! ·214/60 = 72431 71425 27156 38411 62130 22720 00000 = 7.2 × 1029 patterns. The fact that the star hasidentical pieces means that one can produce an apparent 2-cycle! The arrangement of colours in the solvedstate is arbitrary, as can be seen by observing that two colours can be exchanged by 8 two-cycles and twoflips, so this is an achievable process. There are 6! = 720 ways to assign the 6 colours to the 6 axes, but weget to divide by 60 to obtain 12 distinct START colour patterns. Alternatively, if we look at a white face, we canrotate to put blue in a standard direction, and then the remaining 4 colours can be in any order, which seemsto give 24 START patterns. However looking at the opposite white face gives the mirror image arrangement ofcolours so there are only 12 START patterns. [ see also my Alexander's Star page. - J ].



21

MAGIC POLYHEDRA

For any polyhedron, or any planar graph, it is possible to imagine a magic version withpieces at the vertices and edges 50 that each face turns. I have reported the work ofAndrew Taylor (N29) and Uldis Celmins (N60) on this. Here I outline the general resultwhich subsumes both works. If we have vertices of different valencies, i.e. with differentnumbers of edges at them, then Andrew Taylor's idea (N29) is to imagine corner pieceswith the least common multiple (LCM) of the corner valencies. E.g. if we have valencies3 and 4, then corner pieces are 12-sided! See the diagram. One must then make someconvention regarding what effect face measurements have on corners. This is notcritical as long as the inverse of a move restores the pieces and the n-th power of ann-gonal face is the identity. For convenience, I assume that turning face A will carrysector 0 to sector 0, etc. That is, as we travel clockwise around the perimeter of A, the0 sector is the one on the right as we leave each vertex. Alternatively, we say that turning face A carries itsedges to its edges.

Now we have some conservation theorems.

Theorem 1. If every face has an odd number of vertices, then each face turn (and hence each achievablepattern) is an even permutation of corners and of edges. Otherwise each turn is an even permutation of thecorners and edges together.

Theorem 2. Only an even number of edge pieces can be flipped.

Chris Rowley has given me a simple proof.

Consider all S edges. Each of these has two sides or facelets. Each face turn is an even permutation of the2E facelets.

In order to deal with twists of corners, we must make some definitions following BCG and Neumann. In theinitial START position, assign a principal direction to each piece and the same principal direction to its initialposition. These directions correspond to some sector of the piece.

22

For a piece in any position, its twist is the difference between the principal direction of the piece and of theposition that it is in. The twist is measured in units of 360°/v where v is the LCM of all the corner valencies v0,and we work (mod v). There is freedom to choose the sign of the twist, but we consider the orientation of thepiece with respect to its position and measure clockwise to agree with the previous usage in the cyclerepresentation. E.g. on the cube, we can take our principal directions as U and D. Then if UHF is at BRU, wehave a twist of +1. The total twist of a position is the sum of all the twists (mod v). (Similar concepts hold for



the edges, where v = 2.) The total twist at START is 0.

Theorem 3. The total twist of corners is 0 (mod v) for any face turn (and hence any achievable pattern). (Asimilar theorem holds for edges.)

Proof. Consider any face F and let its vertices be V1, V2, ..., Ve in clockwise order. Label the sectors at eachVi by 0, 1, ..., v-1 so that F carries the sector at Vi into the same sector at Vi+1 - e.g. as described above.Suppose the principal directions of the piece at Vi and of the position Vi are sectors pi and qi in our labelling.Then the twist at Vi is pi-qi and the contribution of these pieces to the total twist is Sum(pi-qi) (mod v).

23

Now F carries the piece at Vi to Vi+1 making the principal direction of the piece at Vi+1 be pi while theprincipal direction of the position Vi+1 remains qi+1. Thus the twist at Vi+1 is pi-qi+1 and the contribution to thetotal twist is Sum(pi-qi+1) (mod v), which is the same as the previous contribution. Since all other pieces arefixed by F, the total twist has been preserved, hence it always remains 0 (mod v).

I think the above is the clearest proof of theorem 3. I gather that Andrew Taylor's approach was similar.However Taylor did not try to show all the permitted patterns were achievable and thereby missed the nextresults, which are rather complex.

Theorem 4. If all Vi are even, then no edges can be flipped.

Proof. Because of the assumption, it is possible to colour the faces with two colours so that adjacent faceshave opposite colours. (Exercise - verify this.) Now each face turn preserves the colouring, so that thecolouring is always preserved and this fixes the orientation of each edge piece.

The corresponding result for corners is a bit mere complex, but this argument works in simple cases, such asthe octahedron. If we build a magic octahedron and two-colour it, it would always look the same.



Let kC = v/vC where vC is the valence of vertex C

and v = LCM{vC}. Let C and D be two adjacent vertices and set tCD = kC + kD. Let d be the greatest commondivisor (GCD) of v with the set {tCD | C, D are adjacent vertices}

Theorem 5. A corner can only take on v/d of its possible positions.

The proof here is based on the same idea as Theorem 4. We cyclically colour the v sectors atvertex C with v/d copies of d colours. As we move the piece C about, we use it to define thecolouring at other vertices. Verifying the consistency of this colouring is trickier than inTheorem 4 and is based on the fact that if F and G are two adjacent faces at vertex C (asshown), then FG, which brings C back to C via D, imparts a twist of -kC-kD to C. Since this isdivisible by d, the colouring is preserved.

24

Example 1. The octahedron. Here all vC = 4, v = 4, all kC = 1 all tCD = 2, d = 2. The colouring is the sametwo-colouring mentioned above.

Example 2. The infinite baby-blocks pattern. Here vC is alternately 3 and 6, v = 6, k isalternately 2 and 1, all tCD = 3, d = 3.

The second half of the theory is to show that all the patterns permitted by theconservation theorems can be achieved. This requires a moderate amount ofcalculation, the exclusion of a number of degenerate cases and the separateconsideration of v >= 4, v = 3. The basic approach is to see that the Z and Ycommutators have the same structure as they have on the cube, although the twists onthe cycles are replaced by -kC-kD and kC+kD. For v > 4, we can use the commutatorsto build up pairs of corner 2-cycles and corner twists and then edge 3-cycles and pairsof edge flips. When v = 3, we combine the three Y commutators at a vertex as in Benson's process (N24&45)to get 3 twists and 2 flips, whence any twisting and flipping is obtained. Then we get 3-cycles of edges.Finally, we separately consider the tetrahedron and all other cases to see that even permutation of corners canbe achieved.

I believe the arguments work if we assume

. a the graph is connected;

. b two faces meet in at most one edge or at most one vertex;

. c all faces have at least three edges.

These conditions are certainly satisfied for a polyhedron.

Now we can write down the number of patterns for a graph of V vertices, E edges, v = LCM{vC}, d = GCD{v,tCD}, w= GCD{2, vC}, as N = V! E! / (2 or 4) · (v/d)V-1 · (2/w)E-1 where 2 or 4 is 4 if all faces have an odd number of edges, 2 otherwise.



25

For the regular polyhedra of V vertices, E edges, valence v and e edges on each face, we have kC = 1, tC =2, d = w = GCD(2, v).

V E v e 2 or 4 d=wTetrahedron 4 6 3 3 4 1Cube 8 12 3 4 2 1Octahedron 6 12 4 3 4 2Dodecahedron 20 30 3 5 4 1Icosahedron 12 30 5 3 4 1Tetrahedron N = (4! 6! / 4) 33 25 = 37 32480

Cube N = (8! 12! / 2) 37 211 = N (on N12)

Octahedron N = (6! 12! / 4) 25 = 275 90492 16000

Dodecahedron N = (20! 30! / 4) 319 229 = 1006 69616 55352 33471 22516 03231 3645505168 68811 64110 19768 62720 00000 00000 = 1.01 × 1068

Icosahedron N = (12! 30! / 4) 511 229 = 83 26777 68427 04628 34605 13497 17691 9244800000 00000 00000 00000 = 8.33 × 1056

These results include both Taylor's and Celmin's results. However, the magic makers are more masterful thanmere mathematicians. With different shapes it is easier to make magic with different styles of cutting. E.g. thePyraminx has no centre pieces. This factors N by the number of patterns in which only the centres move, orthe number of symmetries of the shape which are achievable patterns when the centres are ignored. Rubikand others have shapes where faces have no edge pieces and movement moves some centres as well asedges. Each of these gives rise to complications which will not be considered further here.

RUBIK ROBOTS

A group of students at the University of Illinois have built a robot called Robbie Rubik, to solve the cube. Thepattern must be entered into the computer however, - it does not have a colour sensor. It only turns the topface - flippers and pistons turn the whole cube to bring the appropriate face to the top. The computer isapparently programmed with my algorithm - one of the students is holding a copy of my book in the photo.

Claude Shannon (of information theory fame) sent me an outline for a Rubik robot about the beginning of thisyear.

26

PRETTY PATTERNS

I intended to include this material in C2, but there was too much other material. I have several hundred,



perhaps a thousand letters on my desk waiting to be read. Most of these contain patterns. However, I find thatthe same patterns are rediscovered often, so I am presenting what I have got so far in the hopes that it willanswer some of the letters.

Sources

Several books have appeared which present a fair number of patterns. I haven't yet collated all these, but letme list the books.

M. Razid Black & Herb Taylor. Unscrambling the Cube. Zephyr Engineering Design, PO Box 1668, Burbank,California, 91507, USA. 40pp, 1980 (printed in 1981). (This includes Black's earlier booklet: ConstructingPatterns on the Cube.)

Mal Davies (227 Belly Oak Road, Kings Norton, Birmingham, P30 1HP, UK). Pretty Patterns. 2pp, 1981.

Don Taylor & Leanne Rylands. Cube Games. Penguin (UK), Holt Rinehart & Winston (USA), Greenhouse(Australia). 54pp, 1981

French

A. Deledicq, J. C. Deledicq & J. B. Touchard. Le Cube - mode d'emploi. Editions CEDIC. 47pp, 1981.

André Warusfel. Réussir le Rubik's Cube. Editions Denoël. 190pp, 1981.

German

Christoph Bandelow. Einführung in die Cubologie. Vieweg. 144pp, 1981. [ See also C7/8-4. - J ]

Josef Trajber. Der Würfel fur Fortgeschrittene. Falken Verlag. 144pp, 1981.

Tom Werneck. Der Zauberwürfel fur Könner. Heyne Verlag. 160pp, 1982.

Generalities

I will adapt the presentation of patterns to give both the number of moves (as before) and the number ofquarter turns. E.g. 8,16q will mean 8 turns, but 16 quarter turns (i.e. 90° moves).

A. H. (Sandy) Frey has extended the notation to include moves of the whole cube, as follows. Consider makinga R face turn. If we let go of the rest of the cube, then the whole cube turns with the R face. We denote thismove by R. Don Taylor points out that this puts us into a larger group - the oriented group - which has 24N =10 38048 07858 77565 44000 patterns and that processes which only appear to be commutators when wechange coordinate systems (e.g. in the slice group) will now appear naturally as commutators. For example,the 6-spot RsFsUsRs is written RDR'D' = [R,D] in slice notation. This becomes [RsR', DsD'] in our newnotation. That is, a slice move R (in slice coordinates) is equal to RsR' in our new notation. Some authorsprefer to call RsR' the R slice move, viewing the slice as moved clockwise, rather than the R face. I will notuse the notation very often, but there are situations where it is very useful. Whole cube moves do not count asturns since they are part of the process of moving from one turn to the next and do not take any extra time toexecute.



27

Orbit names

Neil J Rubenking suggests names for the 12 orbits of the cube within the constructible group. I have changedSwitch to Swap. Start, Flip, Plus, Minus, Swap, Flip Plus, ... , Flip Swap Minus.

Diagonal Patterns

William J. Blewitt asked if there was any pattern with diagonals on all six faces. John Butler sent an elegantexample in a Twist orbit (i.e. in Plus or Minus). Consider alternate vertices of the cube. These are the verticesof a regular tetrahedron whose six edges are diagonals of the six cube faces. Imagine this tetrahedron rotatedby 120° about a main diagonal of the cube. This pattern is readily obtained by the corner 3-cycle on N25,followed by a 6-spot and then a corner twist.

I posed Blewitt's question in an article and John O. Kiltinen of Northern Michigan University hassent a solution. His argument is based entirely on the corners so it is most easily expressed interms of the 23 On the 23, a diagonal face has the shown pattern, which is also a chessboardpattern. Then Kiltinen's result is that there is no 6-diagonal pattern on the 23.

Proof. Take URF as fixed. Consider the UBL position, which would have to have a U face in its U direction,assuming that such a pattern exists. If UBR is at UBL, then the piece at DRB must have two R facelets.Similarly, UFL cannot be at UBL, so we must have UBL at UBL. Likewise, DRB must be at DRB and DLFmust be at DLF. So the four corners URF, UBL, DRB, DLF must be a correct tetrahedron of pieces.

By the same argument, the other four corners are a correct tetrahedron of pieces, but the two tetrahedra arenot correct with respect to each other. Taking the first tetrahedron as fixed, we need only consider the groupS4 of 24 permutations of the other corners, which correspond to the rigid motions (including reflections) of thesecond tetrahedron. The reflections can be excluded at once since they reverse the colours on the cornerpieces. So we have just the group A4 of 12 even permutations, which correspond to the rotations of thetetrahedron. One of these is the identity I. Eight of these are 3-cycles corresponding to rotations about a cubediagonal and involve a twist of a corner, so are impossible. The other three are pairs of 2-cyclescorresponding to rotations about a 'midline' of the tetrahedron, which is a face to face axis of the cube. Thisleaves two edges of the tetrahedron in place and hence leaves two faces of the cube of one colour. These arethe zig-zag patterns which have four diagonal faces.

(The beginning of the argument is Kiltinen's. I have brought in the consideration of S4 and A4 which allows usto deduce that the only 'generalised diagonal' patterns (i.e. 6-diagonal patterns where a = b can occur) areSTART and Zig-Zag. This also shows that one cannot get 6 chessboards on the 43.)

28

Two-face Patterns

Michael Holroyd points out that his 2-X on N53 is more efficient than the one on N47. This 2-X was alsodiscovered by G. Kéri and Charles Marks and other variations come from Kevin Lewis and Ross & KarlAnderson.

2-X F2L2F2L2R2F2L2B2 (M. Holroyd & WK - 8, 16q; N8,47,53)



R2F2sR2·B2Rs

2B2 (K. Lewis - 8, 16q; N8,47,53)

R2F2L2B2F2R2F2B2 (R. & K. Anderson - 8, 16q; N8,47,53)

The last is Crossbars (N48) followed by Fs2 end is a conjugate of Holroyd's method.

Four-face Patterns

There are two 4+ patterns. One rotates the corners 180° about a face axis, which I take as UD (N48), and theother rotates by 900 about the UD axis and I take the clockwise form.

4+ (180°) R2sU2R2

s·F2sD2Fs

2 (H. & K. Anderson - 10, 20q; N11,21,33,35,47,48)

UaFaUa2FaUaRa

2 (A. Treep - 12, 16q; N11,21,33,35,47,48)

4+ (90°) Rs2U2Fs

2D . Rs2D2Fs

2U' (V. Toth & L. Varga - 12, 22q; N48)

H T U

The H face pattern is the same as the I face pattern seen sideways. H faces are easily obtained fromprocesses like (UF,UB)(DF,DB), but these yield 4-H patterns corresponding to 180° rotations. We also get 4-Hpatterns from 4+ patterns by a slice move.

4-H (180°) = 4+ (180°)·Us2

Rs2U2Rs

2·Fs2D2Fs

2·Us2 (GK, DBS & WK - 12, 24q )

(This is a conjugate of two uses of the second process in 5.8-D on N45.)4-H (90°) = 4+ (90°)·U's

Rs2U2Fs

2DRs2D2Fs

2U2D (GK - 13,24q)

[ Slightly shorter is Rs2D2Fs

2DFs2D2Rs

2D' (12,22q) - J ]

29

There are several 4-T patterns and the two obvious ones can be converted to 4-U patterns by a U turn(though the Us are upside down).

4-U (180°) RaU2Ra'·FaU2Fa' (DBS - 10,12q; N49)

4-T (180°) RaU2Ra'·FaU2Fa'·U2 (E. Fuss - 11,14q)



Edgar Fuss calls the upside down 4-U a 'Temple' and it does look like a temple or table with doors in thecentre of each face. The 4-T can be viewed as a Temple with its pillars in the centre of each face rather thanat the corners. Fuss gives a friend's method for 90° versions of these and two shorter methods have beenreceived.4-U (90°) ([R,U]U)7 (? - 28, 28q)

FaU'Fa'URaURa'U'FaUFa'U'·U2 (H. Walker - 19,20q)

RUs'B'D'BUsR'DLFRUsF'L'B' (B. G. Jones - 18,18q)

[ Shortest is RF'L'U2B'D'RBU2LUF' (12,14q) - J ] David B. Shapiro sends a different 4-T arrangement which has the T's sideways with the crosspiecesadjacent. I have reduced this from 28 moves to 8 moves.4-T (sideways) L2FR2Us

2L2BR2 (D. B. Shapiro & DBS - 8,14q)

There is also a sideways version of 4-U.4-U (sideways) UsR2UsF2 (A. Treep - 6,8q)

The 4-X pattern is easily obtained from 4-bar followed by Us

2

4-X (R2F2L2)2Us2 (R. & K. Anderson, GK & WK - 8,16q; N47, 48)

RaUa2RaU2Ra

2U2 (A. Treep - 10,16q; N47,48)

Frank O'Hara sends a shorter 4-Z and K. P. Furness sends a 4-2L pattern which is the slice 4-flip (N31, 35,36, 45, 47) followed by Fa.

4-S FaRa'FaRaFa'Ra (F. O'Hara - 12,12q; N11,21,47)

4-2L Fs2LF2UsR2BRs

2F'L2Us'B2R'Fa (K. P. Furness - 18,24q)

[ Shorter is F2RD'L2BU'RBF'U'LB'D2RU' (15,18q) - J ]. Six-face Patterns One can get Hs or Is on all six faces in several ways. The most symmetric is to have the bars mutuallyorthogonal. Someone suggested this 6-H pattern be called the Irish cube (i.e. the H-block!).6-H RaFaRaFa

2RaFaRa (A. Treep - 14, 16q)

[ Using half turns only U2B2R2D2U2R2F2U2 (8,16q) - J ]. Several people have sent 6-T patterns. Interestingly, these all give the same asymmetrical arrangement.Exercise. Is there any other 6-T pattern?6-T (LsF2) 2 (D2B2) 3 (R. Ager - 12, 20q)

LsF2LsD2F2U2B2U2 (F. Schmidt - 10, 16q)

[ Shorter is LsU2B2LsB2U2 (8,12q) - J ].

Miroslav Goljan and Jiri Fridrich have analysed the 6-U patterns and found 10 types. I have their processesfor 6 types, one of which is Pretzel's, for which Richard Walker has a shorter version. Since these patterns are



easier to describe as though the centres move and the corners are fixed, I will do so. E.g. (F,U) means the Fand U centres are exchanged. Note that any pattern is even on centres and edges together (N46-47). Thisrepresentation must be treated carefully!

30

a)R'D'B'D2FaRaD2Ra'F'DRFs'DaRs

2Da'= (F B)(R,D)(U,L)LU+RD+(FD,BD) (22, 26q)

[ Shortest is FR2Fs2L'DFRs

2B'U'L'F' (13,18q) - J ].

b) R'B'RsF'R'U'RFRs'BLB2Rs'U' = (F,R)(L,B)(U,D)(UF,DF,UR)(UL,UB) (22,26q)

[ Shortest is D2LsB'L2UsB2R'U2BsL2 (13,18q) - J ].

c) R'B'RsF'R'DsBRBs'LFD2Bs'L2 = (F,L)(R,B)(U,D)(UF,UL)(UR,DB,UB) (18,20q)

[ Shortest is F2DFsL'F2L2UsB'LsU' (13,16q) - J ].

d) LDBsR'FL'F'DsRFR'D' = (F,U,R)(B,D,L)(UF,RU)+ (DB,LB)+ (14,14q)[ Shortest is RF'DsR'FsDLsBD' (12,12q) - J ].

e) DaLFD'FLFUFs'L'F'D'B'DF'L'U2FsRsDs= (F,D,R)(B,U,L)(FR,DR,DF)+(LB,LU,BU)+ (25,26q)

[ Shortest is U'B2L'DsFRaBDBUsR'U' (15,16q) - J ].f) Rs·U'F(UF)2(U'F')2·Rs'·RsFsUsRs

= Rs·U'F(UF)2U'F'U'B'UsRs = (F,R,U)(B,L,D)(FL,RD,UB)(R. Walker - 16,16q;

N49)[ Shortest is UsF'URsB'UFUsR' (12,12q) - J ].

The last three can be formed by moving the edges and then applying a 6-spot.

There is a possible 6-bar pattern obtained by applying Rs to the 4-bar, but the bars are not all orthogonal asdescribed at N28, 47.6-bar RaU2Ra

2U2R2 (A. Treep - 7,12q)

More interestingly, Ross end Karl Anderson suggest R2F2L2R2F2Us2L2Us

2 [ orB2D2Fs

2Rs2U2F2 (8,16q) - J ] which almost gives the impossible 6-bar and if you hold it with your thumb

and forefinger covering the F and B centres, then it looks like the impossible 6-bar.

Kate Fried has found a 6-flip.6-flip (LFL'U)5 = UL+UB+UR+UF+FR+FD+ (K. Fried - 20,20q)Conjugating this with F changes FD+ to FL+ and then the non-flipped edges are in the same configuration asthe flipped ones.

Mel Perry and Charles Marks send a new face pattern as shown. Marks calls this'pinwheel' and I called it 'spiral'. The pinwheel comes in two orientations. Let us call thedepicted one 'clockwise' and its reflection 'anticlockwise'. Perry's pattern has 4 pinwheels



in one direction and 2, on opposite faces, in the other direction, Marks has 5 pinwheels inone direction. Neither has a simple pattern-producing process. [ Try these sequences: 4-2 RaU'Fa'DsFa'Rs

2URa (14,16q) 5-1 FaU2L2F'D'R2L'F'D'RFUsF2BLD (18,22q) 3-3 R2BD2FaUaLFLsFsL'UL2FL2 (18,22q) 3-3 D2R2B2U2R2FD2F2LFaR'D'FR2F2DF'U' (19,28q) - J ]

31

Exercises. Can we get 6 pinwheels in the same direction or 3 in each direction? Are there patterns with a = c,b = d ? [ The answers are no, yes, no. -J ]

David Sibley provides a process for producing tri-coloured faces.

S = S(U,R,F) = R'FD2F'D·L2·D'FD2F'R = (FUR,FLU)(BRU,BUL)(UR,LU)(LF,LB)

This exchanges the UR and LU columns and also (LF,LB). Applying this twice he obtains a Tricolours wherethree faces use the same three colours in parallel columns. If we use the mono-column-flip (N31,48), C =C(R,F,D) = R2FDR2D'R, we can get the same pattern.

Tricolours(Sibley) = S(U,R,F)·D·S(D,R,B)·U2D (D. Sibley - 25,32q)

= C(R,D,B)U2C(R,D,B)-1 · C(F,U,L)D2C(F,U,L)-1 ·UFs

2U2Fs2U

(D. Benson & DBS -33,48q)

You may find these easier to carry out if I write them as (SDR2)2UD2 and RCB2C'·FR2CB2C'·FRs2F2Rs

2F

Joe Nemeth also found this pattern, but by a longer process.

[ Shorter is D'BLsBsL'D2Fa'UL2FsU2R'U' (17,20q) - J ]

Some Diagonal Patterns

Charles Marks send a number of patterns which led me to study ways of combining the processes of Ahrens(N29) and Walker (N48) which twist corner 'pyramids', i.e. a corner and its three adjacent edges considered asa unit. These give a number of diagonal face patterns which I name as follows.



Latin Square

Diagonal Tricolour Stairs Latin Square Quincolour

A = A(U,R,F) = URF+(FU,UR,RF)UFL- = [R',F][U,F'][U',R]

twists the URF pyramid clockwise and the adjacent corner UFL anticlockwise.

W = W(U,R,F) = URF+(FU,UR,RF)DFR- ULB+(BU,UL,LF)DBL- = RaURa'·Fa'UFa

twists the URF and ULB pyramids clockwise and the adjacent corners DFR and DBL anticlockwise.

Note that W = FR A R2 A RF'.

32

AU2F'LA'L' is a stage toward the Double-cube pattern (N28,29,48) and is a 6-stairs pattern. [ Shortest is FD2BRB'L'FD'L2F2RF'R'F2L'F' (16,20q) - J ]

AUR'A' gives one diagonal tricolour, four stairs and a plain face (C. Marks) [ or RF2L'UBU'F2U'B'RFULF'R2 (15,18q) - J ]. Applying RU'(RaFa) 3 to this gives 5-diagonal tricolours with one plain face (C. Marks - 36,36q) [ or DRF2U'FDB'RU2BR'DsL'BD' (16,18q) - J ]. This was the first case I saw of a 5:1 distribution of face patterns. Usually one gets 3:3, 4:2 or 2:2:2, althoughthe first stage has 4:1:1 and Marks' Pinwheel is also a 5:1 (but came later). Marks says that when he firstmade this, a friend picked it up and said "So you've got one face correct".

WFU'AUF' followed by DBL-DFR+ gives three diagonal tricolours with their diagonals radiating from UFL andthree stairs with their bulk at DRF (C. Marks).

[W, Fa'UFa UR2] = RaURa'Fa'U2FaU · Fa'D'FaRaD'Ra' · U'Fa'U'Fa (C. Marks - 27,28q)

[ or U'R2U2B2D'R2F2U'R'F'U2L'BsL2RFUa' (19,26q) - J ] gives 6-diagonal tricolours, a pattern which is mentioned in passing on N29,48.

While considering the above pattern, I discovered a new 6-2L pattern. In this one, all corners are in place,merely twisted, and no uprights adjacent.

6-2L WF2W (DBS - 20,20q)



[ or UaFaRa'Fa'RaUa (12,12q) - J ]

(Note that 2L patterns are Pinwheels with a = b, c = d. This pattern has 6 anticlockwise 2Ls while the earlier 6-2L had 4 anticlockwise 2Ls.)

Marks says he can get two Latin square faces and the other four close to it. Exercise. Is there any 6-Latinsquare pattern? [ The answer is no. - J ]

John O. Kiltinen sends a pattern consisting of 6-quincolours obtained by twisting the corners of the 4 pyramidswhich are twisted in 6-diagonal tricolours. This can be achieved by DWD'F2DW'D' (24,24q) [ or DB2R'B2RaU'L2BsL2DRa'F2RF2U' (18,24q) - J ]. If one twists the corners in the opposite ways, one gets the tricolour patterns which are quincolour witha = c = e.

The pattern given by (FLU,UBR,RDF)(FU,UR,RF)(FL,UB,RD)(FD,UL,RB) [ e.g. LUL2FUB2D'RsFaDR'B (14,16q) - J ] takes the three pyramids at FLU, UBR, RDF and rotates them as a unit 120° clockwise about the FUR-BLDaxis. This is a single pyramid twist away from Butler's impossible tetrahedron.

Miscellaneous Patterns

The Double-cube pattern (N28,29,48) can obviously be modified to a Triple-cube by twisting the corners at theends of the rotation axis. The faces can now all have two or three colours. I can produce either of these in 32moves by following Walker's process (N48) with a conjugate of the U corner process +0-0. [ or try these: F2D'R2D'L'U'RsBUsBLF2LU2 (16,20q) U'L2F2D'L'DU2RU'R'U2R2UF'L'UR' (17,22q) - J ].

David B. Shapiro sends a 'Flat meson' which consists of the two 23 at URF and ULB rotated 180° about theUD axis. Turned over, it looks like a Temple with two corner pillars or a two-legged table. It can be viewed as(UBR,UFL)(BR,FL) followed by D2, so it could be easily obtained from the U process (A,C)(a,c). [ F2LD2B2R2B2L'D2R2F2R'D2R2 (13,24q) - J ].

Anneke Treep sends an odd but attractive pattern which she calls Skein. Form UFR+DBL-UF+FR+RU+DB+BL+LD+ and then apply the 6-spot FsUsRsFs. (Picture on next page.) [ DR2U'B2UF2U'BsLFRsB2R2U'F2R2U' (19,26q) - J ].

Pretty Random Patterns

A number of people seem more interested in disorder than order. There are several different criteria that havebeen used to measure disorder. Let us consider the number of facelets of each colour on a face and writedown these numbers in decreasing order. We can have 900000, 810000, 720000, 711000, ...

33

We abbreviate these as 9 (= 905), 81, 72, 712 , .... The complete list of these 'partitions



of 9 into at most 6 parts' is: 9, 81, 72, 712, 63, 621, 613, 54, 531, 522, 5212, 514, 421,432, 4312, 4221, 4213, 415, 33, 3221, 3213, 323, 32212, 3214, 2413, 2313. Of these,only 415, 3214, 2313 have all six colours.

David C. Maxwell and K. P. Furness both send FaRaFaRa2 which gives six colours on

each face - four 2313and two 3213 - which is perhaps not the most random distribution.Moreover, no two facelets of the same colour have a common edge, so this is a propercolouring of the 54 facelets as a map. Brian Martin sends (RsUs)2 which is a proper mapcolouring with six 4221 faces.

Charles Marks obtains six colours on each face without moving the edges: twist all the U corners + and all theD corners - (e.g. by using the U corner process ++-- on the F and B faces); apply Zig-zag (RaFa)3; exchangeU and D edges with a 2X pattern. [ D'FRU2L2F2D2RBL'B2F'D2L2F'R'U' (17,24q) - J ].

Brian Martin gets six 2313 faces by: 12-flip; diagonal exchange of corners exchanging F & B (i.e.(FUR,BLD)(FLU,BDR)(FDL,BRU)(FRD,BUL)); then FLU-FRD-BRU+BLD+. [ RaU2R'DB'U'R2B2LUR2B2U'2LB'R'FU' (19, 24q) - J ]. This is also a proper map colouring. Indeed no two facelets of the same colour meet at an edge or a corner,so it is a 'strong' map colouring. In fact, any two cochromatic facelets on a face are separated by a knight'smove (16 times) or are diagonal corners (2 times). Frank Carey finds that (UL,LB,DL,RB)(UR,LF,DR,RF)UF+UB+DF+DB+ gives a strong map colouring.

[ See also C7/8-8. - J ]

Entropy of a Pattern

George Marx and his colleagues in Budapest have used the Cube as a model for the physical concept ofentropy, which is a measure of disorder and is the logarithm of the number of states of a given type. That is,the more common a type is, the greater its entropy, and the most common types are generally those with theleast structure, i.e. the maximum disorder.

A simple example is to consider each of the 9 facelets of a face as being equally likely to have each of the 6colours, giving 69= 100 77696 equally likely patterns. (This ignores the effects of the other faces.) If there areCi facelets of colour i, then there are 9!/C1! ... C6! ways to distribute the 9 facelets in this colour pattern.However, we shall just consider the 'partition' of 9 given by the Ci, so we want to count all the different waysthe Ci can be rearranged - i.e. we consider all the possible reassignments of colours. If there are Nj values ofCi which are equal to j, then there are 6! / N0!...N9! distributions of the Ci for a particular partition. Forexample, if we take the partition 712 , then the corresponding distributions (or ordered partitions) are 711000,710100, 710010, 710001, ..., 000117. Here N0 = 3, N1 = 2, ..., N7 = 1 and there are

6! / 3!2!0!0!0!1! = 60 distributions corresponding to the partition 712 (=71203). Each of these distributionsoccurs 9! / 7!1!l!0!0!0! = 72 times as a distribution of facelets, so we have 72.60 = 4320 patterns for thispartition. Straightforward calculations show that the most common partition is 32212 which has 27 21600patterns of the 69, which is about 27% of them.



34

For the 23, the partitions of one face are 4, 31, 22, 212, 14 and they have 6, 120, 90, 720, 360 patternsrespectively. On the 43 there are 136 partitions of 16 into at most 6 parts. Exercise. Calculate the number ofpatterns for each partition of a face on the 43.

Some Pretty Patterns in Other Orbits

The pattern of Walker's Snake (N49) followed by (FL,RB) has the appearance of the unique type of circuitalong edges through all the vertices of a cube. The pattern is reminiscent of the curve on a tennis ball or abaseball. Such circuits are called Hamiltonian and so I call this the Hamiltonian circuit or Tennis ball pattern. Itis in the Swap orbit. The Worm and the Snake are the only types of Hamiltonian circuits through the faces of acube, or, equivalently, through the vertices of an octahedron. Zig-zag can be viewed as such a circuit alongthe edges of a tetrahedron, and is the only type of such circuit.

Edgar Fuss gives a Swap Double-cube pattern. Imagine the ordinary Double-cube (about the FUR-BLD axis)and then rotate the two 23 180° about the LF-RB axis.

Ray Ager tried to construct patterns with (2+1)2 faces, with a = d, b = c. He can get it onfour faces, with two 81 faces, by U2R2U2R2U2B2R2U2. I have found a version with a =d on six faces arranged to show (2+1)3. The pattern is (FU,LU,BU)+ (RU,RB,RD)+(FR,FD,FL)+ (FLU,LBU)+ (UBR,BDR)+ (FRD,FDL)+ which is in Swap-Flip. Exercises. Isthere any achievable (2+1)3? Is there any constructible (2+1)3 with b = c (and a = d)?The name of this face pattern derives from its being a geometric representation of (2+1)2= 2·2 + 2·1 + 1·2 + 1·1.

ORDERS OF ELEMENTS

Several people have found that there are just 73 possible orders of elements (N50/50). Jesper C. Gerved andTorben Maack Bisgaard, of Copenhagen, have computed the number of elements of each order as shown intheir Table 4 on the adjoining page. I find that the average order is 122 and the median is 67.3. In ascendingfrequency of appearance, the orders are:

1, 11, 2, 3, 5, 7, 22, 4, 55 & 110, 80, 15, 33, 14, 21, 1260, 9, 10, 35, 280, 28, 315, 44, 99, 720, 112, 6, 16,495 & 990, 77, 154, 45, 20, 165, 330, 140, 105, 504, 840, 336, 63, 8, 70, 231 & 462, 630, 66; 240, 126, 18,360, 132, 56, 252, 144, 42, 198, 48, 420, 168, 40, 210, 72, 84, 90, 120 30, 12, 180, 36, 24, 60.

About 10.6% of the positions have order 60.

35

Order Number of elements Order Number of elements1 1 77 187 23810 94133 760002 17 09115 49183 80 13 34938 37266 94400



3 3389 45406 22394 84 1697 72581 86780 672004 4 34695 70301 44256 90 1925 06905 16173 831685 13352 81725 14624 99 104 36790 91359 744006 140 62105 92987 55526 105 232 82441 94233 548807 15324 55171 48800 110 4 85432 13551 616008 294 99863 89819 39200 112 128 72620 02216 960009 55 33375 23984 28896 120 1947 04401 14631 47520

10 65 25089 78363 52192 126 425 69635 22238 8736011 4 45906 94400 132 637 12967 78649 6000012 2330 23282 74555 54048 140 223 12541 37176 0640014 23 29837 43830 21440 144 714 19202 93781 5040015 14 38547 13332 09856 154 187 23810 94133 7600016 150 73188 62708 73600 165 213 59013 96271 1040018 520 62284 91581 24832 168 1050 26923 92665 0880020 198 73224 56649 27744 180 2453 88900 57383 1168021 39 33715 15593 33120 198 759 70129 20827 9040022 92708 51272 70400 210 1339 23273 20469 5040024 3293 93251 97962 44480 231 374 47621 88267 5200028 97 41976 09076 73600 240 407 15620 36641 7920030 2033 28420 89667 40224 252 689 87708 04473 8560033 15 87401 96622 33600 280 68 65397 34515 7120035 65 52621 89125 63200 315 99 30987 96525 1584036 3186 20259 79731 76320 330 213 59013 96271 1040040 1136 25838 02544 12800 336 257 45240 04433 9200042 737 19977 68310 97600 360 571 01988 89093 5296044 100 12037 79502 08000 420 961 15562 83219 9680045 197 32944 16597 27104 462 374 47621 88267 5200048 911 49764 74103 80800 495 174 75556 87858 1760055 4 85432 13551 61600 504 238 38185 22624 0000056 671 20530 68464 12800 630 395 38014 01624 1664060 4601 52469 28929 25952 720 120 14445 35402 4960063 264 37143 37053 08160 840 240 28890 70804 9920066 404 05117 52503 29600 990 174 75556 87858 1760070 353 49083 42737 30560 1260 51 49048 00886 7840072 1681 09266 03396 71040

Table 4. The number of elements of each possible order in the group Go



36

THE MAGIC DISC

In N51/52, I discussed planar versions of the Magic Cube. During my recent visitto Budapest, Tibor Szentiványi gave me a Magic Disc which is a different kind ofplanar magic cube. It comprises a circular disc divided into two halves (A and B)surrounded by a ring of 6 pieces (1 to 6). The outer ring rotates with respect tothe inner ring. Further, when aligned as shown, the right half of the disc can beturned over, exposing the reverse sides B', 1', 2', 3' as shown. The two sideshave pictures of dogs in blue and brown and they are not easy to keep track of. Ifwe hold it with A up, there are only 1152 distinct patterns and these readily reduceto 96. I find the diameter of the reduced Cayley graph is 7 and there are nineantipodes. Despite the simplicity of the problem, it is not trivial to produce analgorithm nor to determine just what patterns are possible. If the outer ring of thedisc has only 4 parts, then there are 48 patterns which reduce to 6, where theCayley graph is a 6-cycle. The case with 2n parts, for n≠3, is quite different to the6 part case.

[ The 6 segment magic disc is called Kép Korong. An 8 segment magic disc isstill available as the Saturn puzzle by Mag-Nif. A 12 segment magic disc is alsocommercially available and is called the Hockey Puck. These are indeed verydifferent to the 6 segment version. See also my Kép Korong, Saturn and HockeyPuck pages. - J ]

COMMENTS ON THE U GROUP

Guus Razoux Schultz has a method for the U group (N49) in at most 26 moves. Jerome Jean-Charlesconjectures its diameter is 19q. [ See also C3/4-17, C5/6-2 and C7/8-8. - J ]

© David Singmaster

Index Issue 1 Issue 2 Issue 3/4 Issue 5/6 Issue 7/8

The text and original drawings are copyright © David Singmaster, reproduced here with permission. Conversion to htmland additional commentary by Jaap Scherphuis.

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