Talk:Rubik's Cube

Reworked mathematics section

I have a few concerns about the reworked mathematics section, largely the work of User:Josephbrophy. The explanation seems to have become far longer and more complex than necessary and I would challenge its accuracy in places. This statement in particular,

the corner, edge, and center pieces represent separate solution domains. They can be solved independently of each other, and consequently, their permutations are determined independently of each other.

is fundamentally flawed. It leads the author to the conclusion that permutations of corner piece positions are 8!, but that permutations of edge piece positions are 12!/2. The more usual way of stating this is that only even combinations are achievable and therefore the permutations of corners and edges together as a group is (12!8!)/2. Any skilled cube solver can easily disprove by counter-example the premise that odd permutations of edges are impossible by swapping a block consisting of one edge and one corner with another similar block. This is an odd permutation of edges and an odd permutation of corners but an even permutation overall. This position should be impossible by Joe's explanation but makes perfect sense in the conventional explanantion. Conversely, Joe's explanation would allow odd permutations of corners, something that is universally accepted as impossible. So while Joe may have arrived at the correct size of the group, the members of the group his explanation predicts is radically different.

I also have a problem with this;

Here is yet another twist. If the Center Faces have a picture rather than synchronized color markings, then the entire cube does not have a fixed orientation in space, and can be rotated freely to any starting position. Thus one set of orientations is free or effectively eliminated, reducing the potential configurations to 4⁵ = 1096 permutations. Because of parity restrictions the total number of solvable configurations would be reduced to 4⁵/2 = 512 permutations.

I am struggling to work out what is meant here. A picture indicates orientation of the centres just as much as colour markings so it should make no difference. If what is meant is that the centres are effectively unmarked then the standard Rubik's cube group is reduced in size by a factor of 6x4=24, not 512. I am also struggling with what that binomial expansion is maant to represent or prove. There are quite a few other things, but I'll stop there and see what other editors think. SpinningSpark 22:43, 22 January 2009 (UTC)

What is said about the centre faces seems to be at least partly gibberish; certainly a Cube having coloured markings on the centre giving the colours of the adjacent faces is self-evidently isomorphic with a Cube having an asymmetrical picture on each face, thus giving the centres a definite orientation in either case; so the claim that these cases are somehow distinct is nonsense. Certainly what is true and verifiable (I think it's in Singmaster's Notes) about either case is that, provided of course that all six centres have just one definite "correct" orientation, within any orbit the total twist on the centres cannot differ from the start position by an odd number of quarter-twists. This means of course that the picture-on-each-face colouring has 24 orbits, as compared to the 12 of the standard solid-colour-on-each-face colouring. -- 92.40.224.25 (talk) 14:19, 24 January 2009 (UTC)

I've spotted another problem with the wording, namely the passage "The centre pieces can be rotated without disturbing the corners or edges but it requres the rotation of two centre faces at a time [emphasis mine]. The simplest algorithms are slow and tedious requiring 105 turns." This is wrong on two counts; firstly, correcting the centres does not necessarily involve an even number of quarter-turns, it could equivalently require an odd number of half-turns; and although I've unfortunately lost my copy of Singmaster's Notes, ISTR that Singmaster's algorithm to give a half-twist to a single centre is considerably less than 105 turns, in fact only about 15 or 20. -- 92.40.224.25 (talk) 14:37, 24 January 2009 (UTC)

My initial inclination was to restore the original section in its entirety. I did not do this because the editor has clearly put a lot of work into this and it seemed wrong to wipe it all out without discussion. You seem to be agreeing with me, though, that we should restore to previous version. SpinningSpark 15:15, 24 January 2009 (UTC)

response to spinningspark

it will take me a bit to respond to this discussion.joe (talk) 16:43, 29 January 2009 (UTC)joe

regarding the statement: the corner, edge, and center pieces represent separate solution domains. They can be solved independently of each other, and consequently, their permutations are determined independently of each other.

the cube consists of 3 independent puzzles. each (corner, edge, center) can be solved independently of each other. In your footnote 21, Phillip Marshall demonstates elegantly that edge pieces can be solved independently of corner pieces. he has two simple algorithms that demonstrate this phenonmenon and it is the key to his simple method of solving the cube.

another footnote, which has now been deleted, talked to the independence of center faces. You can still find it in wikibook "How to solve the Rubik's Cube" under the section "Rotating the Center Faces." It refers to the tedious 105 move algorithm that you questioned.

more to come. joe (talk) 16:50, 29 January 2009 (UTC)joe

you state: It leads the author to the conclusion that permutations of corner piece positions are 8!, but that permutations of edge piece positions are 12!/2

i am not concluding that. FROM A DISASSEMLED CUBE, the corners can be arranged in 8! ways and the edges in 12!, not all such arrangements are solvable.

you state, and i believe you contradict yourself when you say: "Any skilled cube solver can easily disprove by counter-example the premise that odd permutations of edges are impossible by swapping a block consisting of one edge and one corner with another similar block"

It has been proven mathematically that a single edge piece in the wrong orientation cannot be corrected. Your example: points to swapping two pairs of corner/edge pieces. Well you are implying the existence of 2 edge pieces. And of course, two incorrect edge pieces can be corrected. Therefore, there is a parity requirement for edge pieces, and not for corner pieces. joe (talk) 17:22, 29 January 2009 (UTC)joe

more to come. joe (talk) 17:23, 29 January 2009 (UTC)joe

the statement that the total permutations is 8! x 12! /2 is accurate but misleading. when i first read the original writeup, i reacted to that representation. the divisor of 2 only applies to the 12! because of its parity constrains.

corners do not have a parity constraint. they do have an orientation constraint in that the total number of changes in orientations must be modulo3.

Edge pieces have both a parity constraint and an orientation constraint. A single edge piece cannot cannot be manipulated by itself; it must involve another edge piece. therefore 12! must be divided by 2. Edge pieces also have a orientation constraint. edge pieces have to be disoriented in pairs to produce a solvable cube. this requires another divisor of 2.

there is plenty of literature on this subject.

more to comejoe (talk) 17:49, 29 January 2009 (UTC)joe

i am repeating myself a bit here to make a point: the statement above "as a group is (12!8!)/2" is mathematically correct. But as a mathematician and a teacher i must say that it is misleading and will not help the student understand the real dynamics of the cube. a more appropriate way of representing the total permutations is "as a group is (8!)(12!/2). since each set of permutations is independent of the other. more to come joe (talk) 18:27, 29 January 2009 (UTC)joe

Spinningspark: Please do not take anything I say as a rude remark. You said "This is an odd permutation of edges and an odd permutation of corners but an even permutation overall. This position should be impossible by Joe's explanation but makes perfect sense in the conventional explanantion. Conversely, Joe's explanation would allow odd permutations of corners, something that is universally accepted as impossible. " more to comejoe (talk) 20:35, 29 January 2009 (UTC)

If you have an even number of corners, you need to have an even number of permutations. I do not what leads you to believe that I am permitting an odd number. none of the calcuations shown in my article produce an odd number. I must, in all respect, say that i am a speed cubit - 45 seconds; i have taken every one of the dozens of cubes i own, apart numerous times. I am also recognized in my field as a world class mathematician and actuary.

please take my comments as respectfully submitted. i need to write some more explanation joe (talk) 20:41, 29 January 2009 (UTC)joe

Spinningspark said: "I am struggling to work out what is meant here. A picture indicates orientation of the centres just as much as colour markings so it should make no difference."

if the center faces are marked with color on the edges that match the adjacent center faces, then their orientation is absolutely tied to the color scheme of the cube. But if the center faces are simply faces (of people for example), there is no restriction where the top of the head points to. if we start with say the blue face, the picture on the blue face can point to any four center faces, but not to the green face which is opposite. when you start to solve the puzzle you have the latitude therefore to pick any of four adjacent faces. therefore one set of permuations is lost, or is an indistinguishable freebe. joe (talk) 21:48, 29 January 2009 (UTC)joe

I apologize for repeating myself, but i find better thoughts to answer the questions raised by Spiningspark. He said: "A picture indicates orientation of the centres just as much as colour markings so it should make no difference." This is not correct; there is a big difference; the color markings are linked to the adjacent faces. A picture face is not linked to any of the adjacent faces. Its orientation can start in any direction, and therefore produces few permutations.

in summary about center faces: if there are no markings on the center face, then there is only one distinguishable permutation. If each center face has the numbers 1,2,3,4 in its four corners, then there are 4^6 power distinguishable permutations. It the center faces are color coordinated with their adjacent faces, then there are only (4^6)/2 permutations because each face is linked to an opposite face. if the center faces contain a people picture, then any given center face can have its person's head point in a north, south, east, or west position by simply rotating the total cube to a desired orientation. there the number of permutations is given by (4^5)/2. joe (talk) 23:15, 29 January 2009 (UTC)joe

spinningspark said: "I am also struggling with what that binomial expansion is maant to represent or prove. There are quite a few other things, but I'll stop there and see what other editors think."

One of the criticisms of the article is that it contains little real mathematics. Most of the articles reference by this article point to websites containing statements about he cube based on set theory. and most of those demonstrations are computer derived solutions, by simply counting all the possible configurations. that is not math, that is counting with a fast machine.

the EXAPNSION OF THE binomial theorem is a very elegant way of describing every possible edge configuration including those that are solvable and not solvable. "Even E" represents correct orientation. "Odd O" represents incorrect orientation. a cube with incorrect orienations can be solved in those number cases where the exponent is even. The number of cases is determined by the coefficients of the expansion.

young children learning the rubics cube, who are studying albegra in grammar or highschool, will be able to relate to this binomial theorm, and it would produce quite a revelation in a class room discussion with an algebra math teacher.

in summary; set theory or computer computation can be used to prove that one half of the total permutations of the edge pieces. But the binomial expansion does the same thing in a very elegant way. joe (talk) 23:31, 29 January 2009 (UTC)joe

I think the bottom line is that you have completely turned upside down a mature section of an article without discussion or providing the source of your theories. I really am not going to answer all this in detail. Just one point, you claim that the solution of corners and edges are entirely separate domains and quote Philip Marshall's method as proof of this. Marshall's method is to solve all the edges first; quoting from Marshall at an intermediate stage of solving the edges; some corner pieces have also moved but that is immaterial. In other words his method requires disruption of the corner domain during solution of the edge domain and so proves no such thing. The move U = {urf, ufl, ulb, ubr}{ur, uf, ul, ub}. This is a 4-cycle of corners and a 4-cycle of edges. A 4-cycle is an odd permutation, by which I mean an odd number of swaps must be carried out to achieve it. Clearly odd permutations of corners and odd permutations of edges are allowed. It is odd permutations overall that are not allowed, which is why it is (8!12!)/2, not (8!/2)12!. That the domains cannot (always) be solved independantly can be seen from the simple U move above: the corners cannot be solved without disrupting the edge formation. Likewise and equivalently the edges cannot be solved without disrupting the corner formation. SpinningSpark 00:30, 30 January 2009 (UTC)

spinningspark: i apologize for not following discussion protocol. i am at a disadvantage to debate with you. so i will withdraw without further argument. maybe i will return later with some simple examples. i must add that in my article i did provide links to sites supporting my arguments. Furthmore, the 4 cycle has an odd permutation of swaps; but the changes in orientation of the corner pieces are modulo 3, and the changes in orientation of the edges occurs in pairs, so one edge piece does a double flip. if what you say were to be true, then it violates a fundamental theory of the cube, and that is: an odd number of incorrectly orientated edges cannot be solved. this occurs when the cube is being assembled and an odd number of edges are placed in the cube frame in an incorrect orientation. I must tell you that I thoroughly understand the theory of the cube and its math. joe (talk) 02:34, 30 January 2009 (UTC)joe

Then you will understand that rotations and flips are independant solution domains from the permutations domain (at least if you are not interested in minimal solutions) and have nothing to do with the point I raised above. I fail to see why my claim above violates a "fundamental law of the cube". Let me restate my claim in easy to understand terms. I claim that the edges of the position {urf, ufl, ulb, ubr}{ur, uf, ul, ub} cannot be solved and simultaneously leave all the corners in place. This cube is only very lightly scrambled and if you are really the "world class mathematician" you claim above it should be a piece of cake for you to provide me with the written solution if I am incorrect. SpinningSpark 11:10, 30 January 2009 (UTC)

Or here is an even simpler position {ub, ur} which we both agree, I think, is unsolvable. So on a scrambled cube wherever {ub, ur} occurs it cannot be solved independantly of the corners (or the rest of the cube in general). The simplest scramble containing this group is {ulb, ubr}{ub, ur} and this is solvable. A solution is;
LdL^-1ULd^-1L^-1ULdL^-1U²Ld^-1L^-1U = {ulb, ubr}{ub, ur}

SpinningSpark 12:21, 30 January 2009 (UTC)

Spinningspark - i am 80 years old, so nothing is easy any more. let me go back to the drawing board, and brush up on group theory and make sure i am not missing something. i will return with some proof, or i will concede to your superior intellect. joe (talk) 16:37, 30 January 2009 (UTC)joe

Ok Joe, sorry if I have come across as hostile, I don't mean to be, I am just very intense at times. I don't think you are the oldest Wikipedian I have been giving a hard time recently though. SpinningSpark 19:49, 30 January 2009 (UTC)

Spinningspark: no need to back off; you are doing your job, and i appreciate that. someone has got to keep wikipedia honest. here is my progress: (1) i tried to disprove your case with group theory last night and i did not succeed. I am rusty. my specialty is differential equations which i used, (hey, i am bragging) as a rocket science to prove the feasibility of the polaris weapon system for the Admiral Raeburn in 1957. (2) on the other hand, i think i can solve it my way with the cube, and i plan to demonstate that to you also. (3) I plan to get some help from local university to get up to speed on subject group theory. (4) i cannot find my Singmaster Book, so I ordered a new one on Amazon a few minutes ago. (5) the reason I remain persistent is that i believe we are not in disagreement, but a failing on my part not to communicate effectively. (6) I just dug out my old copies of Hofstadter Metamagical Themas on the cube that were published in Scientific America (I can tell what issue). On page 22, second paragraph, right most column, he talks about permutations. As I read it, he is confirming what I am saying; so i feel good about that. For example, his explanation would support my arguement that the permutations are = (8!) x (12!/2). Please do not get mad at me for disagreeing with you. Let me come back and prove it to you.joe (talk) 21:37, 31 January 2009 (UTC)joe

Not able to read that online Joe, but here's a book that explains it quite nicely. Also, the reference that is actually in the article, which is what we are meant to be writing from, states whenever we exchange a pair of corners we must also exchange a pair of edges. SpinningSpark 23:49, 31 January 2009 (UTC)

Spinningspark:

I have been trying to crash learn my group theory as it relates to the cube, and it is a bear. I mentioned in the article that i posted - (inappropriately without discussion, and i do apologize, i guess i was too confident of my work) - that the edge permutations were independent of the corner permutations. Further i mentioned that a single edge could not be flipped because it violated cube theory.

i still have not been able to write a proof. But i can solve the edges and corners independently of each other. but how do i show you? I just came across a link to an article written by Ann Scott that makes the case for the points i was trying to make: (by written, I mean she proved the theorems)

for example:

1. the number of twisted corners must remain intergral, i.e. mod3 (found on page 181)

2. the number of flipped edges must remain even, i.e. mod2 (found on page 181)

3. there is a move that twists a corner, and preserves the orientation and postions of all other subcubes (found in ponderable 11.2.1, page 183)

4. there is a move that flips two edges and preserves the orientation and positions of all other subcubes (found in ponderable 11.2.1, page 183)

these references are found in Adventures in Group Theory;Adventures in Group Theory: Rubik's Cube, Merlin's Machine, and Other Mathematical Toys

By David Joyner, Edition: illustrated, Published by JHU Press, 2002, ISBN 0801869471, 9780801869471, 262 pages

at the following link: http://books.google.com/books?::::id=VePv84nsfWIC&pg=PA182&lpg=PA182&dq=anne+scott+rubik's+cube&source=web&ots=kjHamZ5QU6&sig=B5XBBkc3KmRSR02Lv4enE5Pi0JA&hl=en&sa=X&oi=book_result&resnum=1&ct=result

i used the following argument in google: ann scott runk's cube and took the first pick.

i will check out the reference you mentioned. joe (talk) 03:28, 1 February 2009 (UTC)joe

Joe, you seem to be quoting back to me the same book I gave you a link to, although your link doesn't work. There is no argument from me about the above properties (except that you have made a mistake in number 3, it should read twists two corners). I think your basic mistake is that you are conflating arguments about twists and flips with arguments about position permutations. They are different things, but you have confused them a number of times in this discussion. Theorem 11.2.1 could not make it clearer (paraphrasing slightly to avoid defining all the maths terms):

A position on the Rubik's cube is reachable if and only if there is;

(a) equal parity of corner and edge permutations

(b) conservation of total twists of corners (mod 3)

(c) conservation of total flips of edges (mod 2)

I have been talking about (a), whereas you commonly reply with a statement about (b) or (c). I agree that the article would benefit from making some of these points, but your version just had too many errors. You especially do not seem to have grasped the meaning of (a) and want to prove it with the requirement (c). Hope that makes what I am saying clear. SpinningSpark 11:03, 1 February 2009 (UTC)

You are still claiming to be able to solve corner and edge permutations independently. You can solve edge flips (c) and corner twists (b) independently. You cannot solve edge and corner permutations independently, that is what (a) is saying. If you could there would exist an algorithm operator X such that X{ulb, ubr}{ub, ur}={ulb, ubr}. You can prove me wrong by counter-example with a value of X, but none such exist. SpinningSpark 11:57, 1 February 2009 (UTC)

Spinningspark: thank you for your patience. you are correct, "twist two corners", of course. i am making typos all over the place: to get to the link, i made a typo. The link refernce should be: go to google and use the argument: ann scott rubik's cube

here is what i did this morning to prove to myself what i think i am talking about.

(1) using jeffrey varasano's method in "Jeff conquers the cube in 45 seconds" isbn 0-8128-7097-2, i solved the edge pieces independently of the corner pieces. I took my camera, took a picture of a scrambled cube, and then solved the edge pieces without disturbing the position or orientation of any of the corner pieces.

(2) using phillip marshalls method for corners, described in your link to his site. i took a picture of a scrambled cube, and solved the corner pieces without disturbing the positions or orientations of the edge pieces. in his paper he solves the edge pieces first. i didn't. i simply took the scrambled cube and solved the corner pieces independly of the edge pieces.

I know that you can't solve corners or edges without "temporarily" moving other pieces, because they are in the way. And that is what Hofstadter is implying in his paper on permutations. I will make an OCR copy of his remarkes and post them asap.

thank you for your patience. and i will read the reference that you provided. joe (talk) 16:56, 1 February 2009 (UTC)joe

Try the same exercise using one of the specific examples I gave you rather than a random scramble. You will then soon see why you are wrong. SpinningSpark 18:40, 1 February 2009 (UTC)

you may have trumped me! i have to take a time out and watch the super bowl. i have to do some deep thinking. thank you. joe (talk) 20:51, 1 February 2009 (UTC)joe

Spinningspark: I see my error. You are an outstanding editor, and wikipedia is fortunate to have someone of your caliber. If one sets all the corner pieces, then half the otherwise legal positions of the edge pieces cannot be reached. i got the right answers for the wrong reason because the associative law of mathematics is forgiving. I have to check out for a month, and get the cube out of my head. but i will be back. thanks for valuable help and again i apologize for not discussing my writeup before i placed it. joe (talk) 03:10, 4 February 2009 (UTC)joe

Deathtrap image

My edit summary was accidentally cut off. Per WP:NFCC we don't include nonfree images for tangential topics, for example for a move poster that just happens to include a Rubik's cube. — Carl (CBM · talk) 21:47, 26 January 2009 (UTC)

I disagree. A major point of the section is to demonstrate how the Rubik's Cube became (at least for a little while) a shorthand icon for cleverness and/or complexity. The poster demonstrates that in a way that the text does not. - Richfife (talk) 23:26, 26 January 2009 (UTC)

There is an argument for using an image in that section. There is no argument at all for using a non-free image. For one thing, the subject is now doubly tangential since the bulk of the material is now in its own article. For another, the non-free argument is only valid if it is impossible to obtain a free image. I very much doubt that it is impossible to find or create a free image demonstrating that Rubik's cube is a cultural icon for cleverness. SpinningSpark 09:40, 29 January 2009 (UTC)

Is the Visaria method solution nobable at all ??

I would guess not, but this is mere to alert editors to the proposed deletion (PROD) - please reply at that talk page Power.corrupts (talk) 21:13, 4 February 2009 (UTC)

Onyx cube?

The following uncited text has been removed from the article:

There is a rubik's cube called the "Onyx edition" where all the sides are a different side of grey, with one side being black.

It looks like a good faith edit, so I'm keeping it here. It should probably be merged into the variants section.—Tetracube (talk) 20:22, 16 March 2009 (UTC)

It's called the Rubik's Icon cube 69.136.72.16 (talk) 01:50, 31 August 2010 (UTC)

Larry Nichols and Frank Fox

In March 1970, Larry Nichols invented a 2×2×2 "Puzzle with Pieces Rotatable in Groups" and filed a Canadian patent application for it. Nichols's cube was held together with magnets. Nichols was granted U.S. patent 3,655,201 on April 11, 1972, two years before Rubik invented his improved cube. On April 9, 1970, Frank Fox applied to patent his "Spherical 3×3×3". He received his UK patent (1344259) on January 16, 1974.

I have removed these two from the beginning of "conception and development", as they were, to the best of my knowledge, not involved with the original conception of what we've come to know as the Rubik's Cube. They are unsourced, and I'm not quite sure where we should mention them, if at all since this page is specific to Rubik's conception. Should we move them to the variations section? ←Spidern→ 12:51, 23 March 2009 (UTC)

Nichols fought a well publicised legal battle with Ideal which he won in respect to the 2x2 (pocket) cube, but not the 3x3. Apparently, he would have won the 3x3 as well if he had bothered to put a 3x3 sketch in his patent. I tend to agree with you that the Nichols cube bears no relationship to Rubik's invention, but the courts said otherwise and it would be our own OR not to have this in. Citations for this are easily found, I have been planning an article on Magic polyhedra mechanisms for some time (though I rarely work on it) and you will find some refs in my draft article. Don't really have a take on where the Frank Fox patent fits in. SpinningSpark 19:45, 23 March 2009 (UTC)

Anytime invarient?

Does for any N>=2, in rubik's cube NxNxN, the orientation of one corner is depending by the orientations of the other corners? —Preceding unsigned comment added by 77.124.153.109 (talk) 07:21, 7 April 2009 (UTC)

Yes. The corner piece's behavior is not effected by the number of layers of the cube and the orientation of the last corner is controlled by the first 7. - Richfife (talk) 19:40, 7 April 2009 (UTC)

Refactorization

In factoring the number of possibilities of the Rubik's cube, there is an alternate factorization: $43252003274489856000=9!12!12^{5}$ would this be relevant? —Preceding unsigned comment added by AJRobbins (talk • contribs) 04:14, 27 April 2009 (UTC)

That's ok if you are just want mathematical brevity. However, the factorisation given is directly related to the derivation of the permutations and thus makes the explanation more understandable. SpinningSpark 20:47, 1 May 2009 (UTC)

GA Reassessment

This discussion is transcluded from Talk:Rubik's Cube/GA1. The edit link for this section can be used to add comments to the reassessment.

I am performing a reassessment of this article as part of the GA Sweeps process. Here are some problems I see with this article:

• The lead is too short. It should summarize all the main points of the article.
• Quite a few unsourced paragraphs and entire sections ("Mechanics").
• A cleanup tag in "Move notation".
• A few citation needed tags.
• A cleanup tag in "Custom-built puzzles".
• Un-formatted citations in the reference section.

I'll give the editors of this article a weeks to fix these initial problems, and then'll I give the article a more thorough look-through. If the problems are not fixed, I will delist the article. Thanks. Nikki♥311 00:36, 13 May 2009 (UTC)

I am delisting the article due to lack of improvement. Nikki♥311 01:28, 21 May 2009 (UTC)

Parity Errors in some versions of 3x3x3, illustrated/mapped style solutions

I found that there is a way for some versions of the 3x3x3 to have parity errors in such cases as the "Hollow Cube" or "Void Cube", and was wondering if anyone else knew of other versions of the 3x3x3 that also had this type thing. It seems only the fixed centers of the original version of the Rubik's Cube kept the errors from occurring in the standard cube, but it's likely that other 3x3x3 versions might not have the fixed centers. On the Rubik's Cube I know the orientation of the fixed centers can require additional moves to orientate them into the correct solution, if they are marked with shapes or images rather then colors. I think that mentioning that parity errors could occur would make an interesting addition to the "Solutions" section. It'd be interesting to find out if anyone knows which company first built and made the Void/Hollow Cube, the version I got was from a company called LanLan Toys, the only reference I know of that mentions the parity errors and shows its inner rails and how it works is this link from youtube http://www.youtube.com/watch?v=RGrPS1ez3Lg but I don't think we can use that as a reference. I also think it'd be good to include mentioning the solution that didn't use the U+, B', type solution that David Singmaster used, something more in the line of this http://www.youtube.com/watch?v=EnZ2Zu2iZW8 where the cube is being solved by illustrations rather then letters. Those two styles seem to work with both 'corners first/edges first' and 'from top downwards' solutions. I think it might be of interest and help to others to mention the mapped/arrow type solutions, and the void cube and it's 3x3x3 mechanics, but there are probably other Wikipedians out there that know more about them, and know of better references that could be used. I hope someone will help take these ideas and try incorporate them into the article, I think it'd help build up the article and give a better view of the possibilities of the 3x3x3 cubes and how they are solved. (Floppydog66 (talk) 02:49, 14 May 2009 (UTC))

ryanheise.com has a good page on Rubik's Cube theory which discuss the parity of normal cubes. I think void cubes should probably be discussed in their own article, finding reliable sources could be a problem.--Salix (talk): 06:40, 14 May 2009 (UTC)

I apologize if I'm doing this unconventionally; I've never added to a disscussion before. It just so happens that I know where the Void Cube came from. Katsuhiko Okamoto, a sheet metal worker from Japan first invented the Void Cube. He then worked out an arrangement with Gentosha Toys to mass produce them. Check these links for more info: http://puzzle3d.hp.infoseek.co.jp/index.html (Void Cube prototype can be found under Products 4 near the bottom) http://www.gentosha-edu.co.jp/products/div-cl.html Unfortunately I don't read Japanese, but I know Okamoto invented the Void Cube and perhaps someone else can translate and confirm? Feel free to incorporate or delete my contribution :) —Preceding unsigned comment added by 205.196.178.27 (talk) 17:39, 22 June 2009 (UTC)

Notation Error for y axis

(y) is a rotation of the cube on the Y axis, clockwise. As of today, it is listed as rotation counterclockwise, which is incorrect. —Preceding unsigned comment added by Skanda swamy (talk • contribs) 10:19, 18 July 2009 (UTC)

Done. SpinningSpark 18:35, 18 July 2009 (UTC)

Permutations

Maybe it is just my problem but I don't undrestand one thing in the part "Permutations". That is "... giving 2¹¹ (2,048) possibilities" and in the equation next there is only 2¹⁰?
Iffcool (talk) 20:09, 10 September 2009 (UTC)

It's because the equation is properly;

${{\frac {8!\times 12!}{2}}\times {\frac {3^{8}}{3}}\times {\frac {2^{12}}{2}}}$

the divisors being due to unreachable domains in each class of operation. However, if you write it like that in the article it will not be long before someone comes along and divides out the factors again. SpinningSpark 20:54, 10 September 2009 (UTC)

"Odd permutations" not defined

Int the permutations section the term "odd permutations" is used, but not defined, and is not immediately evident.

There are 12!/2 (239,500,800) ways to arrange the edges, since an odd permutation of the corners implies an odd permutation of the edges as well.

I cannot edit it myself, because I don't understand it :D — Preceding unsigned comment added by Tjips (talk • contribs)

It simply refers to the number of pairs of cubies that have swapped positions. All scrambled positions of the cube, however complicated, can be described as a number of swapped pairs. This number (obviously) must be either even or odd, often called its parity. Only even parity positions can be achieved without disassembling the cube. The statement above means odd parity of the edges implies odd parity of the corners since the parity is required to be even overall (odd+odd=even). SpinningSpark 07:46, 27 September 2009 (UTC)

Covering the Earth with cubes.

TomZ has a great point in this post: http://twistypuzzles.com/forum/viewtopic.php?p=186906

He says that:

R(n) = 6371000+0.057n (radius of n-th layer, zero being the earth's surface) A(n) = pi*(6371000+0.057n)^2 (surface area of n-th layer) C(n) = pi*(6371000+0.057n)^2/0.057^2 (cubes on n-th layer, calculated by dividing surface area of sphere by that of a cube)

Then we simply need to sum the cubes on every layer... 1102 layers would be required.

Apparently whoever wrote the original wiki article probably mistook the Earth's diameter for its radius.

Please correct it.

Thanks. —Preceding unsigned comment added by 206.248.157.60 (talk) 00:13, 15 November 2009 (UTC)

I have removed this sentence: "Alternatively, if laid out on the ground, this is enough to cover the earth with 273 layers of cubes, recognizing the fact that the radius of the earth sphere increases by 57 mm with each layer of cubes." It is uncited, disputed here, and to be honest considering that there are 261 light years of combination, it seems highly unintuitive that there should be only a few hundred layers of cubes if laid out on the Earth's surface! I don't think any such claim should be made until we have a reliable source. The forum linked to above is interesting but I don't think it is a reliable source. This whole section is uncited, so I am going to add a citations needed tag to the section too. Unnachamois (talk) 16:36, 31 January 2010 (UTC)

Brute force algorithm?

Do you think it would be impracticable to solve the Rubik's cube on a standard 1.x Ghz machine by a brute force recursive algorithm? See User:Tisane/Rubik's cube saga Tisane (talk) 12:14, 21 February 2010 (UTC)

Two basic problems combine to make this impractical. First, the number of possible positions is 43,252,003,274,489,856,000. Assuming, say, 200 cycles per position (and that's assuming a VERY efficient program, way more that PHP could ever produce), you can do 5,000,000 positions per second. So you're looking at 8650400654897.9712 seconds to look at every possible combination. That's over 274,000 years.

But wait, it gets worse. As anyone can tell you, it's very easy to repeat yourself when solving a cube. Wildly different series of moves can lead to the same result. So, to avoid this, you need to maintain a table of all the positions you've already looked at so you don't loop back on yourself. Assuming one bit per position, that's 5406500409311232000 bytes. That's just under 5035195881 Gigabytes of RAM. It has to be RAM, because you don't have time to check the hard drive for each position as you go . What was your budget for this project?

These two issues just scratch the surface of the problem, by the way. The path solving issues by themselves are extremely complex. - Richfife (talk) 18:26, 21 February 2010 (UTC)

11x11 Cubes

I just want to weigh in on this because the edits are rolling in.

1) If you pay attention to the demonstration videos of the 11x11 cube, you will note a curious thing: They only turn the cube on a single axis. The guy flips it around a lot and makes a lot of quick moves, but all of the twists are along the red/orange axis. It's not at all clear that what we're seeing isn't 11 plates around a single axle dressed up to look like a functional cube.

2) (Not really important from a Wikipedia standpoint, but still...) The 11x11 cube violates V-Cubes patents.

Thanks! - Richfife (talk) 20:00, 1 March 2010 (UTC)

I'm pretty sure the 11x11x11 is fully functional. But, in addition, a 12x12x12 is available too... not infringing any copyrights... http://www.bedardpuzzles.com/index.php?puzzle/49 173.168.177.217 (talk) 18:40, 20 March 2010 (UTC)

number of permutations

I'm probably completely wrong here, but since there are only six colours on the rubik's cube, isn't the number of combinations actually much lower than 43 quintillion, since many squares of the same colour can be in different locations and still produce the same combination, for example it isn't a different combination if the cube has two combinations that are identical, but have different squares of the same colour in the same place. Sorry if I'm explaining it really badly, but could anyone explain to me how this is taken into account. —Preceding unsigned comment added by Js1056 (talk • contribs) 09:36, 22 May 2010 (UTC)

You are, as you say yourself, completely wrong. The explanation of how this number is arrived at is in the article immdediately above the 43 quintillion figure. The mistake you are making is that you are considering the elements of the puzzle to be the squares (ie the coloured stickers) when you really need to consider the solid pieces, or "cubies" as the solvers call them. For instance an edge cubie might have one red sticker and one blue sticker. The red sticker of this cubie in a solved cube cannot go just anywhere on the red face. In fact there is one, and only one, position it can go. That position is the edge between the red face and the blue face since the blue sticker has to be on the blue face as well as the red sticker on the red face. What is more, the solver has to place it there in the correct orientation - it is no good having it flipped over, with the red sticker on the blue face and the blue sticker on the red face. Hope that helps. SpinningSpark 10:34, 22 May 2010 (UTC)

Conjugacy classes

How many conjugacy classes has the Rubik's cube group? --84.61.131.18 (talk) 07:56, 15 July 2010 (UTC)

1x1x1 cube

Shouldn't this article also explain that there exists a novelty 1x1x1 Rubik's Cube? [1][2][3]

76.66.193.119 (talk) 11:27, 4 August 2010 (UTC)

If I were you, I'd add it to the politics section of Rubik's Cube in popular culture and rename the section if that floats your boat. - Richfife (talk) 19:22, 4 August 2010 (UTC)

Jargon in the Solutions section

I am a Rubik's noob, and I don't understand many of the terms in the Solutions section. None of these terms are defined anywhere in the article. Many of you probably know what these mean, but a regular Rubik's noob like me certainly doesn't. Undefined terms include:

• Middle layer, as in: "ƒ (Front two layers): the side facing you and the corresponding middle layer"
• Orienting, as in: "This is then followed by orienting the last layer then permuting the last layer..."
• Permuting, as seen above
• The cross, as in: "The cross is done first followed by first-layer corners and second layer edges simultaneously"
• First-layer corners, as seen above
• Second-layer edges, as seen above

SnottyWong ^converse 23:29, 6 August 2010 (UTC)

I can explain the middle layer and the cross but the others I can not explain very well. When holding the cube the middle layer is the layer in the middle. The cross is when you have the first layer edges solved and nothing else.Sumsum2010 · Talk · Contributions 19:23, 7 August 2010 (UTC)