Jaap's Puzzle Page

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A quarterly newsletter for Rubik Cube addicts

Issue 5 & 6
Autumn & Winter 1982






Editor's Corner2
More on the U Group2
The n-Dimensional Cube Solved3
New Cube Products5
Tsukuda's Square7
More Clubs8
Generalized Hungarian Rings9
The Incredi-Ball (or Impossiball)10
Going Ape over the Cube10
Some Geometric Puzzles
   A. Eight Cube Rings11
   B. Twelve Tetrahedra Rings13
   C. Cubic Snakes14
Some Notes on the 4315
Cubism and Religion22
Prehistory of the Cube23
Local Maxima24
The Trials of the Cube25
A Czech Check Problem25
God's Algorithms26
First Day Cover28
Block of 4 Stamps28


Published by David Singmaster, 87 Rodenhurst Road, London, SW4 8AF, UK
ISBN No. 0261-8362. All material © 1982 by David Singmaster.




IMPORTANT NOTICE: Due to retrenchment at David Singmaster Ltd., I am now running the Cubic Circular personally. All future correspondence should be addressed to CUBIC CIRCULAR, 87 RODENHURST ROAD, LONDON, SW4 8AF, UK. Subscription cheques should be made out to David Singmaster. Subscribe now - give a gift subscription for Christmas: Subscription rate for the second year is £4 or $8. Back issues of Volume 1 available for £3 or $6.

From July, Peter Light has been assisting me in dealing with the mass of accumulated correspondence. By now, almost everyone will have had some reply to their letters - some of which are eighteen months old! We have not had time to digest everything and there are a number of letters (e.g. in other languages) which he has not been able to deal with. We hope everything will be cleared away by the new year:

Let me urge correspondents to send stamped addressed envelopes (or at least a mailing label and a reply coupon): Also, please make sure your name and address and the date is on the contents as well as the envelope - envelopes do get separated from their contents and contents sometimes get filed in different ways.

Peter has also been testing the various computer programs which have been sent in.

Some of this Circular has been typed with 1 1/2 spacing instead of single spacing. This gives about 25% fever lines per page but is more readable. What do you think?

I would be delighted to receive contributions in the form of short articles for the Circular.

On page 3, I mention the Rubik's Cube stamps from Hungary. I have arranged to print these on the back page and the following has been transferred from there to make room for it.


MORE ON THE U GROUP (N49, C3/4-17,36)

Nicolas Hammond has a method for the U group in at most 21 moves.
[ See also C7/8-8. - J ]

J.F. Jarvis (96 Forrest Ave., Fair Haven, New Jersey, 07701, USA) has an inventory of U processes and would like to hear from anyone interested in compiling a directory with him. (Mr. Jarvis, meet Mr. Hammond - Mr, Hammond, meet Mr. Jarvis!)


In the 24 hour Cubathon of PCs Timoney and Pullen (C3/4-7), they completed 1043 Cubes in 24 hours. That's 21.73 cubes per man-hour, or an average of 2.76 min or 2 min 45.7 sec per cube.

Cardiff Students Rag 1982 had a Cubathon in which two students solved 2000 Cubes in 36 hours. That's 27.78 cubes per man-hour or an average of 2.16 min or 2 min 9.6 sec per cube.




Joe Buhler, Brad (3D) Jackson and Dave Sibley have drafted their work on the n-dimensional Rubik Cube. Like Keane and Kamack (C2-8), they allow orientation preserving movements of the n-1 dimensional facets of the n cube. The constructible group is a direct product of the appropriate wreath products

where Nd = 2n-d n! / d!(n-d)! is the number of d-dimensional faces of an n-cube, but when d = 0, we take An-d instead of Sn-d. They find several conversation laws which yield

Where [An:An'] = 3 for n=3, 4 and is otherwise 1. It is not clear how the expression works for n=2.




Several readers have been confused by expressions involving commutators. Recall that (N30,41) [A,B] = ABA'B'. Hence
[A,B]-1 = (ABA'B')-1 = BAB'A' = [B,A]!! Thus we have
[A,[B,C]] = A·[B,C]·A'·[B,C]' = A·[B,C]·A'·[C,B] = A·BCB'C'·A'·CBC'B', and
[A,[B,C]]' = [[B,C],A] = [B,C]·A·[C,B]·A' = BCB'C'·ACBC'B'·A'.

I forgot to mention that the First World Championship in Budapest was commemorated by a special 2-forint stamp, with First Day Covers, etc. If space and reproduction permit, I'll try to include a picture on the back page.

Several readers assert that WD40 and other aerosol spray lubricants can dissolve the central spindle!

The handbook of Cubik Math, by A. H. (Sandy) Frey and myself, is now available in the UK and Europe from Lutterworth Press. The UK list price is £5.95. I will be happy to dispatch copies to the UK and Europe for $6.50, including post and packing. The book has been well reviewed.
"A book that will delight enthusiasts"
"The solution in Chapter 3 is ... intelligent and breathtaking in its simplicity."

An article in Science 82 reports a 19 year old autistic child who doesn't speak but solved his first cube in 40 seconds!

An Auckland (New Zealand) paper reported a solution time of 5.51 seconds!

A French engineer, Alex Renault, at the firm AKR, has built a 'Rubik Robot with video colour detector which takes about 15 seconds per turn.

A cubical question was set on a UK entrance exam: "Exam Paper for Awards and Entrances in the College of the University of Cambridge - Paper 871, Physics & Maths, 23 Nov 1981". Question 12 asks how many ways one can reassemble a cube when cut into 27 cubies and one wants the colours outside, and what is the probability of getting the U face all blue.

Arthur P. Cracknell reports that the Cube proved an exception to a well known scientific law. The law is that no-one comes to lectures on the last afternoon of a conference. The exception was a lecture by Hans Zassenhaus at a colloquium on group theoretic methods in physics at the University of Kent, which was the last and most attended lecture of the colloquium.

At C2-4, I reported that 3 year old Glenn Quackenbush could do the Cube. His father writes that he can 'only' do several patterns - still that's not bad for a 3 year old.
[ See also C7/8-46. - J ]




I have been kindly sent cubes with five fruits and Chex (a US cereal box top offer) and with Pac Man symbols. Several full-face patterns have appeared: animals, butterflies, copulating couples: In Italy, I found several interesting Ideas: Football Cubo - with full-face colours of six leading football (= soccer) clubs; Caricatur Cubo - with full-face caricatures of six Italian political party leaders; Italian Political Cube - with symbols of the six major parties (It can't get any more confused than Italian politics. Perhaps it's used for predicting Italian election results!); Souvenir di Venezia - with five full Venetian scenes and the motto on the sixth face. (I saw a ample of a scenic cube in Nuremberg, but this is the first I've seen on sale.) Neil C. Rubenking sends a cube with four colours forming pyramids about four alternate corners. He also sends a slightly melted cube, transfixed by a large brass screw, mounted on a wood base with a brass plate: "SCREW IT". In Holland, the all black cube (C3/4-10) is called the Belgian Cube.

The unnamed shape mentioned on C3/4-11 is a truncated rhombic dodecahedron. I have examined a Taiwanese 23 (C3/4-9) and its mechanism is similar to Rubik's 23 patent. That is, it is a 33 with its corners concealing the edges and centres. It differs from Rubik's method in details, but it is still a very complex design.

Jorg Brown, age 13, of Fort Collins, Colorado, has designed (but not built) a 53.

Edward Hordern has made a 33 into a 43 by gluing extra pieces on 3 sides. This causes the 42 faces to rotate eccentrically about one of the centre pieces. It is quite difficult to get back into shape!
[ Idea due to Nob Yoshigahara, see C7/8-9. - J ]

Cube Mates (Cinderella Co., P C Box 265, Sykesville, Maryland, 21784, USA) is a set of 54 lettered stickers to put on your cube to play word games. Every letter and * (a wild card) occurs at least once and the following more often - 2 times: D, G, H, P, S; 3 times: B, E, R, T; 4 times: I, L, O; 6 times: A. The instructions describe several games.

Ideal has produced a Puzzle Poster in the US showing about 2,000 cubes - one of which is wrong: They are also marketing a Jigsaw with all the pieces the same shape, but which assemble into a pattern in just one way: (cf. C3/4-11)
[ See also my Zig-Zaw page. - J ]
The second issue of their Rubik's Cube Newsletter has appeared. I haven't yet seen the second issue of Rubik's (the Hungarian magazine), but I have just heard of it.

Cubical shoelaces have come from the USA. Solomon W. Golomb sports seeing cubical duvet covers in Amsterdam.

Although I have not yet seen examples, Mèffert has offered to send a catalogue of his products and to supply them by mail order. His address is Pricewell (Far East) Ltd., P.O. Box 31008, Causeway Bay, Hong Kong. The October issue of Omni shows a Magic Dodecahedron, called Megaminx (C3/4-19), a Pyraminx Cube (C3/4 - 12/13), and the Impossi-Ball (also called Incredi-Ball and described elsewhere in this issue). Some other products are mentioned. The offer to supply was made to Omni readers.



An article in La Recherche last December is partly by Raoul Raba inventor of the Rotascope (C2-14/15). Several variants are described. The flattened cube is three rings which behave like three faces of a cube - but each face has six corners and six edges. (A two-ring version of this was devised by Doug Engel of Colorado).
[ See also my Turnstile puzzle page - J ]
Two spherical versions are shown, along with cubical equivalents. These look very like Durham's Pyraminx Cube (C3/4-12/13), but it's not clear just how they move. Raba also shows that the Rotascope can be viewed as a cube with some pieces glued together as shown.
[ See also my Bandaged Cube page - J ]

Ruth Roufberg has tracked down the earlier version of these puzzles (C2-14/15). It was a 1971 Parker Brothers game called Orion with a 52 array of rotating discs, each containing lenses (as on C2-14)

Linear Aesthetic Systems, P O Box 23, West Cornwall, Connecticut, 06796, USA, have programs to display and simulate cubes up through 73.

Diane Turnbull reports seeing a 'Cubies' chair at Heal's (Tottenham Court Road, London). It has a plastic frame which holds small coloured cubes which you can take out and rearrange.

Van C. Wedgwood sends his "The Idiot's solution to the Rubiks Cube" which consists of a set of coloured labels. It costs $2.00 from PO Box 9261, Denver, Colorado, 80221, USA. (This reminds me of what someone called the 'chemical solution' to the Cube - use some solvent to dissolve the glue of the sticky labels and then reglue them!)

Early reports from Russia and China regarded the Cube as a symbol of Western decadence. Izvestia reporter Melor Sturua said: "Some assert that the Rubik Cube reflects the philosophy of the Reagan Administration - to build and destroy aimlessly in a futile search for a solution to the world situation. Others see Americans' efforts to solve the puzzle as the desire to escape from a disordered life." A report from China criticised be Cube as a time-waster and said that a boy had allowed a batch of buns to burn in the oven while he was distracted by a Cube.

However, a more recent report in Trud reveals that the Russians have succumbed to Cubitis. They will produce their own products -the Muldavian Pyramid and the Nikitin Cube - early next year!




I mentioned this briefly at C2-9. I have had a chance to study Edward Hordern's example and I can describe it in more detail.

My previous description was incorrect due to having seen it only briefly and in conjunction with a Crossover.

The puzzle is a 42 board with pieces numbered 1 to 16. Each horizontal row can be slid one place to the right, bringing in an empty space on the left and concealing the piece on the right as In Figure 1 which shows the effect of sliding row 2. The three columns on the right can be slid down together by one place as shown in Figure 2.




All the slides are spring loaded to return to the original positions. Let us denote the right slide of row n by n, its left slide by n', the down slide of columns 2, 3, 4 by V (for vertical and its up slide by V'. When V has been made, row 1 is obstructed and slides 1 and 1' cannot be made.



As with other sliding piece puzzles, we have two levels of group structure. If we include 8 extra pieces along the bottom and right edges, we can view each move as a 5-cycle on this set of 24 pieces but the set of patterns is not a group. We get a group structure on the 16 numbered pieces by considering the patterns in which each extra piece is restored, i.e. each slide is returned to its initial position. Since this requires an even number of moves for each slide, each such pattern is an even permutation of the 16 pieces. A sequence of moves is in the group if and only if:

  1. it has as many 1's as 1s, ..., V's as Vs;
  2. the 1s and 1's, ..., alternate, in this order, in the sequence;
  3. 1 and 1' do not occur when there is an uncancelled V.

Commutators of moves are obvious processes with these properties, but they are messy. To investigate them, I wrote a simulation program on my Sinclair ZX81. This yields the following:

[2,V] = 2V2'V' = (2, 5, 6, 7, 8, 4, 3)

[23,V] = 23V2'3'V' = (2, 5, 9, 10, 11, 12, 8, 4, 3) etc. (Note that 2'3' = 3'2'.) and

[1,[2,V]] = (1, 2) (4,8)

The last result was not quite expected since 1 and [2,V] overlap in 3 places. But 1 sends

1 → 2 → 3 → 4 →

and [2,V] sends

→ 8 → 4 → 3 → 2 → 5

so some of the interaction cancels in the middle, leaving only the 2 2-cycles at the ends of the overlap. In fact, this always happens when we take a commutator of two cycles that overlap in a sequence of positions.

From (1,2)(4,8), it is easy to obtain (4, 8, 12) and (3, 4, 8). It is easy to put whatever pieces are desired into locations 1 and 2. Then it is easy to get any piece in rows 2, 3, 4 into position 8 without affecting row 1. Thus we can achieve all the 3-cycles (3, 4, n) , n > 4, and so we can get any even permutation of the 16 pieces.

Added in Press. This puzzle is being distributed in the UK as "It".
[ See also my 'It' puzzle page - J ]


MORE CLUBS (Cont. from C1-12)

The Nederlanse Kubisten Club (NKC) has been established by Anneke Treep and others (address in C2-3).
[ The NKC (Dutch Cubists Club) and their quarterly newsletter Cubism For Fun (CFF) are still going strong after 20 years. See their homepage for contact details. - J ]

There is a Hungarian Rubik Cube Club with rooms in the Kossuth Club, Budapest.

Georges Helm has founded the Rubik's Cube Club Luxemburg at 76 Bomecht, L-4936 Bascharage, Luxemburg.




The Hungarian Rings, described at C2-14, have a nice generalisation. Suppose we have left and right rings with m and n balls with the intersection being a and b balls apart on these rings. Then there are N = m + n - 2 balls all together and we consider them as all distinct. Let G(m, n, a, b) be the group of achievable positions. Let L, R denote clockwise rotations of the left (right) ring. Philippe Paclet [Jeux et Stratégie 16 (Aug-Sep 82) 30-32] asserted that G(m, n, a, b) always contains AN the group of all even permutations of the N balls.

I have shown that this is not true. There is an infinite class of exceptions and one special case.

I. G(4, 4, 1, 1) is the same as the cube group <F,R> acting on the 6 corners it affects. In N55/57, this group is discussed and shown to be isomorphic to S5 - the group of permutations of 5 objects.

II. G(2a, 2b, a, b) is the antipodal case. Here the balls fall into antipodal pairs which cannot be separated. There are N/2 = a + b - 1 antipodal pairs and we can view the rings as an a-cycle and a b-cycle of these pairs. Thus it is easy to get any even permutation of pairs. Pairs may also be inverted, e.g. [La, R] inverts two pairs.

We can view G as a group of actions on the set of a + b - 1 oriented pairs, i.e. as a subgroup of the wreath product Z2 wr Sa+b-1

  1. If a and b are odd, we get only even permutations of pairs. But La inverts an odd number of pairs, so any number can be inverted. So G = Z2 wr Aa+b-1.
  2. If a and b are of opposite parity, we get G = Z2 wr Sa+b-1.
  3. If a and b are both even, we can only invert an even number pairs. The easiest method is to use Rowley's proof (N3/4-21). Both L and R are odd permutations of pairs and are odd permutations of the individual pieces. So if all pairs are in place, only an even number of them can be inverted. Hence G is the subgroup of Z2 wr Sa+b-1 in which the parity of orientation changes is the same as the parity of the permutation of pairs

Cases A and C give 2a+b-1 (a+b-1)!/2 patterns. Case B gives 2a+b-1 (a+b-1)!. In all cases other than mentioned in I and II, does contain AN and = SN if m or n is even; = AN otherwise.

In case you haven't yet solved your Hungarian Rings, here' s hint. Consider [La,Rb], which is [L5, R5] in the commercial version. This is equivalent to [F, R] on corners of a cube.

[ See also C7/8-19 and my Rubik's Rings puzzle page - J .]



The argument used in Case II-C above shows that I was wrong in C2-15 in asserting that each antipodal pair on the Equator could be inverted. Thus there are 15! 215/2 = 21 42493 68453 12000 permutations of the pieces, 15! 215 415/2 = 2 30048 50767 37011 27290 88000 patterns when all four orientations of each pair are distinct. (Pentangle is now importing a version with a map of the world on it which has this many patterns.) The number of patterns on the Equator is half the number at C2-15, line -6, namely 1560 11629 01048 52480.


THE INCREDI-BALL (or ImpossiBall)

This is described in Scientific American (July 1982) and Omni (October 1982). It consists of 20 spherical triangles (i.e. a spherical icosahedron) with a springy mechanism which allows a 'face' of five triangles about a vertex to rotate. There are coloured circles about each vertex. A little thought shows the triangles are equivalent to the corners of a Magic Dodecahedron, so there are
20!/2 320/3 /60 = 2 35639 02142 42189 66794 24000 distinct positions where the denominator of 60 corresponds to the dodecahedron having no centres. (Compare with C3/4-20.)

One version of the Incredi-Ball uses only six colours, like Alexander's Star. The fact that each triangle has three colours completely fixes the relationship of the colours, so the last discussion on C3/4-20 does not apply. But the ten pairs of antipodal pieces can be evenly exchanged in 210/2 ways, giving
(20!/2)·(320/3) / (210/2 · 60) = 460 23246 37191 77669 52000 distinguishable patterns.

[ Incorrect: Antipodal pieces are mirror images, so not identical. Hence there are (20!/2)·(320/3) / 60 = 2 35639 02142 42189 66794 24000 positions. - J ]

It is also possible to remove one piece and have an icosahedral sliding piece puzzle.

[ See also my ImpossiBall page - J .]



UK TV viewers will be familiar with the Brooke Bond tea advertisements featuring chimpanzees who pour tea, ride bicycles, move pianos, etc. Researchers into behaviour wondered if there was any instinctual component in playing with the cube and so gave some cubes to the chimps. Five year old Louis, the bicycle rider, mixed up his cube and then twiddled aimlessly and bemusedly. Four year old Jill was fascinated by the red squares, but then got angry with the cube and tried to eat it. The baby chimp, Charlotte took hers apart. The chimps' owner, Molly Badnam, reported "The chimps were clearly enthralled, playing with the cubes for long periods each day. However, I regret to say that they showed no promise and only became more and more exasperated. For the sake of their health, I was forced to abandon the experiment". Sounds terribly human!





Over the years there have been a number of 'fiddlies' based on hinging together 8 cubes along edges so that they will fold into a 23. The best known of these has its 8 cubes joined in a ring as shown in the diagram. The cubes are shown from above. Some of the vertical faces where cubes meet have hinges joining their edges as indicated by the letters. When the front and back pairs of cubes are folded up, we have a 23. The folding can be continued through three more steps to give another 23. But then all the cubes are inverted so their previous outsides are inside and vice versa This fiddly has been known long enough that origamists can make the whole thing from a single strip of paper. (There are also a number of other hingings. If the R and L are changed to L and R and the back D is changed to B and the front D is changed to F, the result can be folded into a cube in four ways with a tricky move required to get between two sets of two ways.)

This year, a variation of the first idea above has been a popular seller 'under the name of The Shinsei Mystery (or Twin Comet) in England. Each of the eight cubes is split into two congruent halves, with the halves meeting along the hinged edges. Each half is separately hinged. The result is two rings of 8 half-cubes. When joined the two rings fold together and lock like an 8-cube ring But the two rings can be separated and each ring folds in the same way. Each ring can be folded into an interior form and an exterior form, both with cubical symmetry. The exterior form is a cube with an interior hole congruent to the interior form. (Added later: It is called Starburst in the US.)

Manufacturers assert that the idea is due to Naoki Yoshimoto, a Japanese artist. I have found it depicted in two Japanese books, attributed to Yoshimoto and giving a date of 1972. However the idea is



shown in eight forms in a 1973 article "Flexicubes - Reversible Cube Shapes" by the Dutch puzzle inventor Jan Slothouber (J. Recreational Math. 6:l (Winter 1973) 39-46). He gives several examples where the two halves are not congruent, giving rise to some interesting shapes.

The halving used by Yoshimoto is also given by Slothouber. Consider the cube divided into 6 square pyramids, based on the faces with the apices at the cube centre. Three of these meeting at a corner are a half-cube congruent to the other half. The interior form is a star-like solid which rejoices in the name of the stellated rhombic dodecahedron. Each point of the star is a rhombic pyramid on a face of the rhombic dodecahedron. The shape is fairly well known because a popular assembly puzzle forms this shape from 6 pieces. (There are two versions - one with different pieces, including a key piece and the other with 6 identical pieces which is more interesting.)

Slothouber gives two other divisions into congruent halves.

1. Imagine each cube again cut into a 23 and take 4 of the small cubes comprising a 'pyramid' at a corner. 2 Cut each cube in half by a plane perpendicular to a main diagonal. The interior forms of these are shown in the figure. The second is a truncated octahedron - though Slothouber calls it a cuboctahedron.

An amusing sidelight is that an early Taiwanese copier of the Shinsei Mystery clearly misunderstood the idea and produced a version with only one of the of the rings of half-cubes, thereby losing 50% of the interest of the object. (The truncated octahedral form was independently devised in New York last year. I've unfortunately mislaid my notes, but I believe that It was done by two people, one of whom is Alton Bader. The first form is in production by the Japanese Asahi Corp. (and several Taiwanese firms).




If one divides one of the six face-based pyramids of a cube in half by a diagonal plane through the apex and two diagonal corners of the base, then we get two tetrahedra which are mirror images. These can be hinged in various ways to make a ring of 12 tetrahedra which flexes on various ways.

Bart Phillips makes an elegant wooden version called "Flexicube", which allows you to invert the six pyramids, making an exterior form which is a rhombic dodecahedron with a cubical hole inside, the size of the original cube. There is no way to put another cube in the interior and make them flex together.
[ see C7/8-5 for a picture of the rhombic dodecahedron. - J ]

One can view the cuts of the six pyramids as made by three diagonal planes of the cube which meet at a corner. The result is symmetric about the main diagonal of the cube through this corner. Six of the hinges are on this diagonal and six are along cube edges, forming a skew hexagon.

A second version of this fiddly, called Kaleido Puzzle or Kaleidoform, was sent from Taiwan. It is also available as a Corgi toy and via other importers. It has the same pieces in the same arrangement, but the hinges are all along the cube edges. It is trickier to flex and it took me some tome to realise that it could be inverted into a rhombic dodecahedron. It makes a lot of patterns including half a solid rhombic dodecahedron. I have only one copy of the ring but one version comes with two rings. One can combine two so they will flex together but only in a limited way and not like the Shinsei Mystery. I don't know if any hinging will allow the two rings to flex together through an inversion.

A third variation of this idea, called the Shinsei Miracle, starts by cutting each of the 6 face-based pyramids in quarters. This is the same as cutting the cube by all six diagonal planes as shown. These are divided into two congruent groups of twelve which are hinged alternately along cube edges and along midlines (i.e. lines from a face centre through the cube centre). Each half makes some nice shapes and there are some ways to make the two rings flex together but they don't have the interaction of the Mystery. It is possible to use both rings to make a rhombic dodecahedron with a cubical hole.




About two years ago, Diane Turnbull brought me an example of a 'folding cube'. It consists of 27 cubes strung together with elastic. The cubes have holes which go from a face centre to a face centre, either directly opposite a straight piece) or adjacent (a bend piece) as shown. The object is to fold it into a cube. For a particular arrangement of straight and bend pieces it takes about half an hour to check all possibilities. The example brought to me has only one solution as a 33 and no solution as a 3 × 9 rectangle. It struck me that it might be possible to have all bend pieces. However, I have shown that one must have at least two and at most eleven straights among the 25 interior pieces. (The end pieces cannot be classified as straight or bend since the string doesn't continue.)

This led me to consider making other versions of the puzzle. I've recently thought that a good name for these would be "cubic snakes". One way to make your own is to use Multilink blocks which are plastic cubes with one circular projection and five circular holes on the faces. These snap together and rotate, but do not hold together well. At Nuremberg, in February, I saw a better version of the idea developed by Antal Kelle for General Impex. This had a more sophisticated snap mechanism with two states: loose (for turning) and firm. I don't know if it is being produced.

This year, I have also seen three elastic versions made for Jeux Descartes, the French puzzle distributors. These have 36, 64 and 125 pieces, the first two being formed in a ring. All pieces are bends. Forming the 36 into a 3 × 3 × 4 is a bit tricky. I haven't yet made the 125 into a 5 - it looks like it will take some time!

Exercise. Find all the circuits through the 27 cells of a 33. Classify them by the sequence of straight and bend pieces. Are there any sequences which form a cube in many different ways?
[ See also my Snake Cubes page. - J ]




A Strategy for the 43

  1. Do two opposite face centres, say L and R.
  2. Pair up and place the F, D, B edge pairs on the L and R faces.
  3. Put D corners in place. (Perhaps combine steps 2 and 3.)
  4. Pair up FD edges and store in UR, say. Pair up BD edges. Put FD and BD edge pairs in place.
  5. Check the parity of the U corners. If in odd parity, apply U.
  6. Use the moves like [[R,U], l'] = (URb, UBl, FUl) to 3-cycle U edges to get UR and UL edge pairs correct.
  7. Examine (carefully!) the parity of the four edge pieces at UF and UB. If in odd parity, apply r2D2l'D2r2, which 4-cycles these edges (and moves some F, U, B centres). (Since the D centres are not affected by this or later moves, one can try to get some (or all) D centres correct at step 4.)
  8. Use the moves of step 6 to get all edge pieces together.
  9. If you've done the parity at step 7 correctly, you may have to flip an even number of edge pairs, which can be done as on the 3 × 3 × 3. If you find you have to flip an odd number of pairs, you didn't examine the parity closely enough. Flipping a single edge pair requires exchanging them, which is an odd permutation. Return to step 7 if you need to do this.
  10. Place the U corners by use of 3 × 3 × 3 moves.
  11. Use moves like [[r, b], U] =(Fur, Ubr, Ubl) to 3-cycle centres into place.

[ See also my 4×4×4 Cube page. - J ]

Useful moves

The trickiest step in restoring the 43 is getting the edges into correct parity, as done in step 7 (cf. step 9). Because this requires making a slice move, I put off getting the other centres correct until after this step.

It is possible to exchange a single pair of edges as follows:

r2D2l'D2r2 · r2D'R2 · [[R, U], l'] · R2Dr2
= r2D2l'DR'UR'U'l'URU'lRDr2
= (UBl,UBr)(Ful,Ubl,Bdl,Ufr)(Fdl,Ufl,Bul,Ubr)

Jeffrey Adams' book gives

l2D2[u, F'] D2l2·LUL'u'LU'L' = (UFl,FUr) times a 6-cycle and a 4-cycle of centres on four faces, The centres are not as easy to correct as in my version.

Applying r2U2r (u2r2l2)2 r'U2r2 will correct the F and B centres in my process, leaving a 180° rotation of U centres, which can be corrected by a standard 33 supergroup move.

You may find it convenient to have a 2-cycle of edge pairs.

r2D2l2D2r2 = (UFl, UBr) (UFr, UBl) (Ufl, Ubr) (Ufr, Ubl) (Ful, Bdl) (Fdl, Bul).

Applying r2U2r (l2u2d2)2 r'U2r2 corrects all the centres.

Other strategies

After writing the above, I decided it was time to study other methods. I mentioned Jeffrey Adams' book, called "How to Solve Rubik's Revenge" which I have only seen briefly. He uses the same notation that I use - describing it as a natural extension of my 33 notation. I didn't have time to work through his method, but he does all corners, then U edges, M edges, D edges and then centres.



Jerome Jean-Charles' "Solving Rubik's Revenge" aims to be suitable even for those who haven't yet done the 33. The notation is minimal and the directions of moves are not consistent. But he presents everything in pictures, so the notation is not important.

Jean-Charles' strategy is as follows.

  1. Put all corners in place.
  2. Use standard edge moves to place 6 U edges in 3 U edge pairs and then all D edges. (The missing U edges serve as 'parking places'.
  3. Place last U edges.
    (Here he uses F'u'Fu'F'u2F = (FRu, Ufr, BLu, RBu, LFu) + movement of centres. This is actually a 33 move. If necessary he flips the pair after getting them in place by Fu'dF'u'dF'u'dF, though many other 33 moves could be used.)
  4. Place u edges. (This uses fairly easy moves.)
  5. Place d edges. (Here his 'magic move' is
    R'dR'F2Rd'R'F2R2d' = (FRd, Lfd) (Fur, Bdr, Rfd, Fdl) (Ful, Bdl, Rbd, Frd).
  6. Place centres. (As in my Step 11.)

The key to the simplicity of Jean-Charles' strategy is that step 1 solves the corner parity and step 4 makes the edge parity correct (at the end). Leaving all centres till last gives a lot of freedom to move the edges. I suspect one could adapt the strategy to place some centres at the beginning. Michael Mrowka and Wolfgang J. Weber do the whole D face, then U corners, U edges, M edges, centres. (More strategies in a later issue.)


Peter Lees has pointed out a duality which I have studied in the different context of Noughts and Crosses (= Tic Tac Toe) on the 44 board. In our notation, this has a very simple form: exchange upper and lower case letters throughout an expression. Thus the dual of the corner URF is an invisible central piece urf and the dual of the edge URf is the centre Fur. This mapping arises in Noughts and Crosses because it reserves all the lines of four in a row.) It turns the 43 inside out - hence the name 'evisceration'. The dual of the move R is r.

Then we have a Principle of Duality: The dual of a sequence of moves has the same effect on the dual pieces.

This allows us to transfer knowledge from 33 moves to 43 moves as illustrated by the following. On the 33, (F2R2)3 = (FU, FD) (RU, RD). On the 43, (F2R2)3 = (FUl, FDr) (FUr, FDl) (RUf, RDb) (RUb, RDf) (Ful, Fdr) (Fur, Fdl) (Ruf, Rdf) (Rub, Rdf)

Let a 'column' be an intersection of two planes or layers, such as the Fl column. Then (F2R2)3 is a pair of exchanges of columns: (Fl, Fr) (Rf, Rb) where the columns get inverted. Thus
(f2r2)3 = (fuL, fdR) (fuR, fdL) (ruF, rdB) (ruB, rdF) (fUL, fDR) (fUR, fDL) (rUF, rDB) (rUB, rDF) which is a pair of exchanges of columns: (fL, fR) (rF, rB).



Processes which affect corner columns on the 33 lead to dual processes which affect central columns on the 43. E.g. U2R2L2U2R2L2 exchanges corner columns (UR, UL) (DR, DL) on the 33, so u2r212u2r2l2 exchanges central columns (ur, ul) (dr, dl), which gives a 2-spot pattern! The presence of the 2-spot means we can create many of the impossible 33 patterns on the 43.

Another useful process shown to me by Trevor Hutton, is based on the 33 process (F2R2)2 which is a 3-cycle of corner columns: (FL, FR, BR), so the 43 process (f2r2)2 is a 3-cycle of central columns (fl, fr, br). Two of these yield the 2-spot by (r2u2)2 (b2r2)2.

Note that the part of a 33 process which affects corners vanishes in the 43 dual process. We can also use duality in just one or two directions, especially if we are looking at square moves. With non-square moves, it doesn't work! - compare [F,R] with [F,r].

Special Subgroups

Many patterns have already come in the post, but here I will only mention some special subgroups with patterns.

If we only make moves like Rr (=R2 say?), we obviously are simulating a 23. Less obviously, moves like Rl' (=Ri for interleaved?) also simulate 23, but in 8 copies! The squared interleaved moves generate a number of elegant patterns of stripes and bars, notably Ri2Fi2Ui2Ri2 which has 6 orthogonal 4-stripe patterns.

If we only make face moves, we are simulating a 33 with fat slices. But since we can produce a single exchange of centres and of edge pairs and we can invert a single edge pair, we can construct all the impossible 33 patterns except those involving a twist of a single corner. Also, we can obtain any permutation of the centres, though only 24 of these have the colours in the correct relationship.

God's Algorithm on the 43

There are 44 = 256 possible moves along an axis, e.g. LalbrcRd where 0 <= a, b, c, d <= 4. These include four cases where a = b = c = d. If we hold one layer fixed, say L, then we have 64 moves, including I. It is not yet clear just many 'moves' a move takes. For example, r really requires two hand moves - Rr followed by R' - while Rr and even Rrl' could be counted as a single move. The number of positions achievable after n axial moves is at most

Tn = 1 + 192 + l92·128 + ... + 192·128n-1 = (192·128n - 65) / 127.

This exceeds the N on C3/4-15, namely 7.4 × 1045, when n = 22. Guessing about two hand turns or three quarter turns per axial move, this gives about 44 hand moves or 66 quarter turns as a minimum for God's Algorithm.
[ See also C7/8-9. - J ]




Several writers have sent letters on the Pyraminx. Most have pointed out that the method in Werneck's book is not very efficient and that the English edition did not include the pictures of Mèffert and the production line mentioned in C1-11, nor the material on other puzzles in the German.

Notation. Consider the Pyraminx on a table with a face toward you. The four corners are denoted L, R, B, U and the four faces are denoted F, R, L, D as shown in the top view. Vertex pieces (V in the diagram) are denoted l, r, b, u and corner pieces (C in the diagram) are denoted L, R, B, U. Edge pieces (E in the diagram) are denoted by pairs of faces, thus: FR, RL, LF, FD, RD, LD. Moves will be considered as rotations of a corner with its associated vertex and edges. E.g. the move U is shown in the diagram. Its cycle representation will be written (FR, LF, RL) Uu where U means a 120° clockwise twist of the U corner and U' will denote its opposite, etc. (Kersten Meier uses F, R, L, U for his corners - i.e. he looks at an edge rather than at a face.)

Solutions. For me, the simplest solution is to restore the vertices and corners and then use:
(FR,RD,DL) = RBR'B' = [R,B];
(FR,RD BR) = RB'R'B = [R,B'];
FR+RB+ = [R,B'] [U' ,B];
to restore and flip edges.
[ See also my Pyraminx page. - J ]

Several solutions and many patterns have been sent in. Thanks to James Allwright, Angus Lavery, Lawrence Cuthbert, Bill McKeeman, Martin Milner, and Tino and Daniel Delbourgo (ages 10 and 12). David Peter has an easy algorithm in at most 37 moves and a fast algorithm in at most 28 moves, available from him at 50p at Wentworth College, University of York, York YO1 5DD, UK. Another 28 move solution is given by Benjamin L. Schwartz in Journal of Recreational Mathematics 15:1 (1982-83) 31-38.



The Delbourgos' "How To Tackle the Tetrahedron" is available from them at 15 Willowdene Avenue, Sandy Bay, Hobart, 7005, Tasmania, Australia. Nicolas Hammond has sent some of his computer results which he asserts solve all those edge only processes which use at most 9 moves.

There is no method to just flip a pair of edges in less than 8 moves, but Werneck and Meier give:

The Groups of the Pyraminx. The constructible group (where we allow just the edges to be removed) has 34·34·6!·26 = 3023 30880 patterns. The conservation rules are that each move is an even permutation of edges and of edge faces. Hence we can only get an even permutation of pieces and an even number of flips. All such patterns are readily achieved, so the Pyraminx group has 34·34·6!·26 / 2·2 = 755 82720 patterns.

Since the corners and vertices can be moved independently of the edges, the constructible group is the direct product (N59) Z34 × Z34 × (Z2 wr S6). The Pyraminx group is the subgroup of index 4 (N39) obtained by restricting Z2 wr S6 to even permutations of position where the 'total flip' is 0 (mod 2).

God's Algorithm. The Pyraminx has 8 corner moves. After turning a corner, we have 6 turns available at other corners. So the number of patterns after n moves is at most Tn = 1 + 8 + 8·6 + 8·62 + ... + 8·6n-1 = (8·6n - 3)/5. These are the patterns where the vertices are not turned, so we are interested to know when Tn exceeds 34·6!·26 / 2·2 = 933120. The values of Tn are as follows.

n 0 1 2 3 4 5 6 7 8
Tn 1 9 57 345 2073 12441 74649 447897 2687385

Hence God's algorithm must be at least 8 moves Long. Hammond's work (C1-11) shows it is at most 13 moves long. (Add four moves if we consider vertices as well.)
[ See also C7/8-19. - J ]

Pyraminx Patterns. Werneck's book gives a selection of patterns, but most of these can be done in fewer moves. Ronald Turner-Smith "The Amazing Pyraminx" gives more patterns than any other book. Though he does not always give the best solution, he usually states what the least moves known is. Some patterns have better solutions in the Pyramaths appendix.



We abbreviate sources as follows.
TDD - Tino & Daniel Delbourgo "How to Tackle the Tetrahedron"
NJH - Nicolas J Hammond
KM - Kersten Meier "The Small Magic Tetrahedron"
DP - David Peter
DBS - Myself
T-S - Ronald Turner-Smith "The Amazing Pyraminx"
WK - Well known
TW - Tom Werneck "Mastering the Magic pyramid"

Jewel (T-S & TW) or Twist
Uu = (URUR')2 (T-S & WK - 8)
Joseph's Coat (T-S) or Till's Windmill (TW) or 4-Jewels or 4-Twist
RrLlBbUu = (R'L')5 (U'B')5 (T-S - 20)
(T-S asserts 16 moves is possible.) Each half of this is a 2-twist.
[ In 10 moves: UBR'U'BRLU'R'L - J ]
Queen's Jewel Box (T-S) or Fabian's Windmill (TW) or 4-twist
is the above followed by r'l'b'u' to give RLBU.
Doors (T-S) or Holy Temple Gates (TW) or 3-V (TDD)
(FD, RD, LD) = RL'R'L'B'L'B (T-S & WK - 7)
Cat's Paws (T-S) or Fire-Red Cat's Paw (TW) or Adjacent 2-flip
FR+FL+ = [R, U'] [R', L] (T-S & WK - 8)
Gates at Gizeh (T-S) or 4-V (TDD) or Opposite 2-flip
LR+FD+ = L'U'LU'RUR'U (NJH & KM - 8)
This puts a door or V pattern on all four faces. There is another such pattern mentioned without solution in T-S:
(FD, FL, RD) = R [U,L']R' (DBS - 6)
The square of this is another such pattern.
FR+RL+LF+FD+ = L'U'LU'RBUB'UR' (NJH & KM - 10)
The configuration of edges FL, FD, RD, LR is a central square and its symmetries give interesting patterns.
Sun and Moon (T-S) or Square 4-flip
[ In 10 moves: U'L'RUL'RB'L'RB - J ]
King's Treasure Chest (T-S) or Tutankamen's Treasure Chest or 6-flip
FR+RL+LF+FD+RD+LD+ = (UR'L)3 (T-S - 9)
Cleopatra's Needle (T-S & TW)
(FD, LD, RD) Uu = U'RU'R'LU'L'BU'B'RU'R' (T-S & TW - 13)
[ In 9 moves: BR'UB'R'LR'L' - J ]
Chariot (T-S) or Wagon Wheel (TW) or 3-wheeler
(FR, RL, LF)(FD, RD, LD) = RUR'URUL'R'L'B'L'B (DP - 12)
This rotates the edges 120° anticlockwise about the vertical axis.
[ In 9 moves: RL'RBUB'RLU' - J ]
Chariots of Egypt (T-S) or 4-wheeler
FR+DL+(FL, RD)(FD, RL) = LBR'U'·BLU'R'·LBR'U' (DBS - 12)
This rotates the edges as a unit 180° about the midline joining FR and DL. If one views the moves as moves of faces, rather than corners, this is [R, L]3 which exchanges two pairs of corners!
[ In 9 moves: UBUL'RB'ULR' - J ]



Priest's Pattern (T-S) or Whirlwind (TW). Let us call this face pattern a clockwise spiral and its mirror image an anticlockwise spiral. Four anticlockwise spiral faces correspond to a 6-flip and 4-clockwise-jewels:

FR+ FL+ FD+ LR+ LD+ RD+ Uu Rr Ll Bb = LRBL'RL'R'L'UB'LBL' (DP - 13)
[ In 11 moves: BU'L'B'R'U'BLRLR - J ]

TDD give a long pattern with 3 clockwise and one anticlockwise spiral which I have simplified to:

FR+RL+LF+(FD, DR, LD)+ U'u'R'r'B'b'L'l' = B'R'BL'B'U' (TDD & DBS - 6)

The face pattern shown is variously called spade (as in cards), palm (tree) or fish.


The next pattern is called Doubles (T-S). We can get 4-Doubles in two ways.

(FR, FD, LF, RL, DR) = ? (T-S asserts it can be done in 7)
[ In 7 moves: BU'B'U'L'U'L - J ]
(FR, DL)(FL, DR) = [R, B][L, U][R, B] (DBS - 12)
[ In 7 moves: BRB'RU'RU' - J ]

The latter is 180° rotation of the central square about the axis LR-FD.

2-Doubles or 2 plain & 2 cat's eyes (T-S)
(FR, LR)(FD, LD) = R'BR'BRBR'BRBR'BR' (T-S & DBS - 13)
[ In 9 moves: RBRB'R'B'RBR - J ]

This is a reflection of the central square in a bisecting plane through the UL edge of the tetrahedron.

T-S calls this a (left) Nile Temple. I think it's clearly a (left) Sphinx. He gives a method for 3 right Sphinxes and one left one. Can you get 4 rights?



T-S calls this a 3-Sceptre face. He believes it is impossible to get four 3-Sceptres??

There are a number of other patterns in T-S and TW but they aren't as interesting as the above.



CUBISM AND RELIGION (Cont. from C1-16)

Neil J. Rubenking writes from a Zen Center in California, and he sounds like a serious Buddhist. He has produced a set of stickers which converts a cube into a "Buddha's Cube". Each corner has a single noun three times and each edge has a single verb twice. The resulting sentences across and down express truths of various schools of Buddhism. The corner nouns are: Buddha, delusion, emptiness, enlightenment, existence, Nirvana, the self, suffering. The edge verbs are: abides in, conceals, contains, deceives, destroys, hinders, is, is not, knows, liberates, produces, reveals. This gives a total of 672 possible sentences and any configuration of the cube has 12 at a time. Problem for the masochistic - is there a set of 56 cubes on which the 672 sentences occur once each. The centre stickers are "OM MANI PADME HUM", which makes it a genuine Prayer Wheel. He has made a more complex version for a Zen teacher with 24 words on the corner facelets and 24 words on the edge facelets.

Corner triples:
Buddhaignorancethe selfconcentration
Dharmadelusionsentient beingswisdom
Sanghawrong viewsexistencemindfulness
compassionsufferingthe waytrue suchness

Edge pairs:
is notrevealsproduceshinders
deceivesjoinsabides insaves
knowsparts fromcontainsignores


Ian Williams sends an article from Aquarian Arrow by Lionel Snell entitled "Notes towards the deployment of the Hungarian Magic Tube as a system of divination".

"It is a portable permutable set ... smaller than the Rider Traveller's Tarot." He allocates the following meanings to the three axes:



"Up, down = WILL (cf. the upright rod or phallic symbol of will)

Left, right = THINKING (cf. the "weighing up" motion of the balance)

Front, back = FEELING (cf. a reaching forward and a shrinking back towards yourself)"

He finds the cube more harmonious to Libran aesthetic sensibilities when he recoloured his cube so

F = green, B = red, R = yellow, L = blue, U = white, D = brown (presumably a dark orange?).

He finds planetary and other associations with the colours - e.g. F = green, Venus, attraction ... !



As a result of the enormous publicity generated by the Cube, a number of independent inventions of the idea have surfaced.

The first French book on the cube (Le Cube Hongrois by André Deledicq and Jean-Baptiste Touchard) reports: "In fact, Inspector General Semah says he played with a similar cube (but in wood) in 1920 in Istanbul, then about 1935 in Marseilles (with a cube of five white faces and one green face)." Touchard says he went to see Semah, who is now an old man, but who seemed to recognise the cube as an old friend. In view of the Middle Eastern tradition of puzzle rings and the Far Eastern tradition of puzzle boxes and other wooden puzzles, I think there is some probability that a wooden cube might have been made by some craftsman, The cost of wooden cubes would have been prohibitive and only a few (one?) would have been made. I would be delighted if anyone can find more evidence of such a cube.

In 1960, a mathematics teacher named Gustafson, in Fresno, California, invented the 23 Magic Sphere, but only conceptually at first. One of his students, Dick Walton, reports that his class discussed the mathematics and possible mechanisms at length. Gustafson eventually proposed having a central sphere with grooves. I have been told that Gustafson obtained a patent on this, but I have not yet tracked it down.
[ See C7/8-47. - J ]

Larry D. Nichols, of Arlington, Massachusetts, filed for a US patent for "Pattern forming puzzle and method with pieces rotatable in groups" on 4 March 1970. This was granted as Patent no. 3,655,201 on 11 April 1972. This is for a 23 with a magnetic mechanism The concept of a magic cube is clear. However the magnetic mechanism works by separating the cube in half, then turning and putting the halves back together. If you try to rotate a half, magnetic repulsion pushes the halves apart. This is achieved by putting three small bar magnets on the inner faces of each cubelet as shown. So the mechanism is not at all like Rubik's. In particular, one can cheat too easily by disassembling the cube.



However, Nichols refers to other shapes and sizes, and "engagement by mechanical rather than magnetic means, as for example by using a pop-in snap linkage, or a tongue-in-groove arrangement". Nichols' company, which holds the patent rights, has written to Ideal claiming prior invention. Ideal feels that the cube does not infringe Nichols' patent. Nichols company has recently filed suit for $60,000,000!

The New York Times reports that Nichols' firm, Moleculon Research Corp., did try to market Nichols' idea in 1969 but wouldn't divulge the secrets until the patent came through. It was rejected by all the firms approached, including Ideal, generally because of the secrecy.

I hesitate to predict the outcome. Though Nichols has the concept, his claim (quoted above) must be stretched rather far to cover Rubik's mechanism and Gustafson's patent would seem to pre-empt Nichols. However, a large US law firm has taken the Nichols case on a contingency fee basis - i.e. they only get paid if they win! Gustafson has apparently stated that a cubical version did not occur to him.

Frank Fox of Bletchley, Bucks, filed for a UK patent on an "Amusement Device" on 9 April 1970 and this was published as UK patent 1,344,259 on l6 January 1974. This was for a 33 in spherical shape, using tongue and groove linking to hold the 26 exterior pieces together, leaving the centre hollow. Fox thought to use this as a kind of board for noughts and crosses (= tic tac toe), though he mentions use of 26 colours and the use of letters and numbers.



Consider the Cayley graph of the Cube group. We say a position P is a local maximum if it is further away from I than any of its 12 neighbours - i.e. PR, PR', PL, PL',... The 12-flip is a local maximum. Let P be any process which produces the 12-flip. Consider the last move of this process, say it is R. Since the pattern is symmetric with respect to the rigid motions of the cube and a mirror image of the process will give the same effect, we can transform P into processes which end with R', L, L', ... Any neighbour of P is thus obtained from some transform of P with the last move omitted, hence is obtained in one move less than P itself. There is no other pattern which is symmetric under all the rigid motions of the cube, but there may be a pattern with less symmetry which can be obtained by several processes of the same length such that we can obtain the pattern with any move at the end. Such a pattern will also be a local maximum.
[ See also C7/8-8. - J ]




Ideal Toys and Politoys (the new name of Politechnika commenced legal action against the Taiwanese 'knock off' cubes early in 1981. The first case to reach court was against Dallas Print Transfers and Dallas Marketing, two London distribution firms run by Alec Yuen. After many postponements, this case began on 25 January and lasted twelve days! There were two grounds - infringement of copyright and 'passing off'. UK copyright law is nearly unique in extending to working or production drawings. Copying of a three dimensional object made from the drawings infringes on the copyright in them. 'Passing off' means selling an imitation product which is so similar to an established original as to mislead the public. The defendants first claimed that the drawings produced were not the basis of the copied cube. This was due to the fact that Politoys first sent the drawings for the pre-1980 cube rather than those for the later production which was the basis for the copied cubes. Once the correct drawings were produced and the draughtsman, Miss H. Czák, made it clear that she had made both sets of drawings, then the infringement of copyright was established and the judgment upheld this. I thought 'passing off' was also clear, but Ideal was held not to have established that they had created a brand image, due to the pre-1980 sales and the lack of image promotion. Damages have not yet been assessed and Dallas is appealing the judgment. However, I have recently heard that the whole Dallas operation has gone bankrupt.

Cases have taken place in Holland and the US. In Holland the case went through several hearings and I am unclear as to the final outcome. In the US, Ideal has obtained several injunctions and is attempting to block the importation of pirate cubes.

I have heard from a student doing a law thesis on the cube. She has offered to write a summary for the Circular.



The Czech weekly, Mladý Svet (Young World), poses as an exercise the problem of having each face with 8 facelets of one colour. The other facelet is a second colour but can be at the centre or an edge or a corner. If all are at the centre, we have 6-spot; if all are at edges, we can get it easily as (UL, FD, RD); if all are at corners, we can get it easily as URF+BLD-. However, there may be other ways of getting these effects and one might have different places for the odd coloured facelets - or can one?




In C2-8, I discussed an improved lower bound for God's algorithm. I have since realised that the argument can be refined a bit. Recall that Sn is the number of sequences of n 90° moves with the obvious redundancies excluded. Previously I totaled Tn = S0 + S1 + ... + Sn until Tn > N. But each 90° move is an odd permutation of the corners and of the edges, so U2m = U2m+1 = S0 + S2 + ... + S2m is the number of sequences which give even permutations on corners and on edges while V2m+1 = V2m+2 = S1 + S3 + ... + S2m+1 is the number of sequences giving odd permutations on corners and edges. Both these sums must be >= N/2 = 2.16 × 1019. It turns out that

S20 = 3.56 × 1019, T20 = 3.98 × 1019, U20 = 3.60 × 1019,

S21 = 3.33 × 1020, T21 = 3.73 × 1020, V21 = 3.37 × 1020,

so that the refinement gives no gain, although I thought at first it would.

However, if we consider the supergroup, we want Tn, Un and Vn >= 2048·N/2 = 4.43 × 1022. Here T24 = 3.07 × 1023, U24 = 2.78 × 1023, V24 = 2.96 × 1022, so we do need to go one step further to get U25 = 2.60 × 1024 and God's algorithm for the supergroup has length > 25, instead of 24.

Herb Taylor has passed on some further results.

The 23 square group was observed to have 24 patterns, diameter 4 and 5 antipodes at N53/54. Taylor points out that the Cayley graph can be drawn on a torus with 12 hexagonal faces. The number Nd of positions at distance d from a position is given by


According to Joe Buhler, via Herb Taylor, three different programs have found the diameter of the 23 is 11. One of these was circulated as a UNIX news item. It took over 51 hours of computer time. Don Knuth, the well-known Stanford computer scientist, gave the problem to a graduate class, but wanted R2 to be counted as two turns, etc. Buhler recollects that this gives diameter 14 and says that this was done by Joan Feigenbaum or Richard Anderson. (Several people have suggested that we write 14q when we mean quarter or 90° turns. This seems useful.) The UNIX news item gives the following table for Nd.



72 27536
88 70072
918 87748
106 23800

Using simple counting, as at N26 and 60, one sees that some positions require at least 9 moves. There are no obvious redundancies to account for (we are keeping one corner fixed), so I expected a diameter less than 15. An antipode of START is


[ See also my 2×2×2 Cube page. - J ]

One of the groups who did the 23 went on to the 33 square group. They get diameter 15 which confirms Thistlethwaite's conjecture (N39) and hence reduces his algorithm to at most 50 moves. The numbers Nd are as follows

91 13892
101 78495
111 79196

The fact that these diameters are small leads me to believe that the length of God's algorithm will be about 22. However, it seems unlikely that there will be a unique antipode. The UNIX news item concludes "I just thought of a lovely way to get the answer for the 3 ×3 × 3 cube, but the margins on this news item are a little bit too small for me to include it."

The UNIX news item is a rather faint copy, but further examination shows that it comes from duke!chico!esquire!

Frank O'Hara has found the diameter of the antislice group is 8 antislice moves. (N54)



© David Singmaster



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