Unscrambling the cube

Polynomial: a formula of numbers with terms, coefficients and degree.
Moebius polynomial: a formula which reconfigures the pivot group, this is really on the centers group but the re-labeling of vertices generates pseudo-slices in the cube's elements.

The edges group is not included in the transform, and the centers vanish, so that this polynomial reconfigures the cube as a set of octants only.
The shape of the elements is "only" a convenience - you can handle the octants when they project "corners" into 3-space, or have hapticity

 

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So how exactly does a 'moebius polynomial' reconfigure a group?

Did you name it the moebius polynomial or did you reference it from somewhere? The only reference I could find (in three seperate papers) is this:

"...The Moebius Polynomial, \( \mu \left(\Sigma,I \right) \), of the graph \( \left( \Sigma, I \right) \) is defined by

\(\mu \left( \Sigma, I \right) = 1 + \sum_{u \in \mathfrak{C}} (-1)^{|u|}y^{|u|} \) ..."


EDIT: Also, what is a pivots group, or a centers group? When you say, 'the edges group is not included in the transform' , what transform are you referring to? What does it mean to reconfigure a group? I have never come across this term before.

 

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Hmm. It reconfigures the group of vertices. You choose 1 of 8 of these to be "at zero", so choosing a first vertex is step 1, etc and divides the cube. There is a step from vertex to vertex that passes through the long axis - the authors do this at 1 -> 2, but this isn't a convention, it could be inf -> 0 just as easily.

See if this: http://www.permutationpuzzles.org/rubik/crossgp.html says anything. When you get to: "We will label the vertices of the cube..." see if you think projecting the vertices is reconfiguring the cube. If you think it doesn't can you also say why?

 

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I don't really see how this answers my questions, I'm still rather confused sorry.

The cube, or any section of the cube, is not a group. You can split the different parts of a cube into sets, such as the set of pairs of opposite faces, or the vertices. You can then act on this set with a group, such as the group of symmetries of a cube. Where a (left) group action, of a group \(G\), acting on a set \(\Omega\) is a binary function \(\pi\) where

\( \pi : G \times \Omega \rightarrow \Omega \)

\( \pi : (g,\omega) \rightarrow g \cdot \omega \)

 

You can remove the vertices in two ways: disassemble the cube and reconstruct a smaller version using only 8 corner elements, or expand the cube into a sphere, making the cube into an object with 8 octants of a sphere. Each vertex of "the cube" is now in the center of a smooth surface, and each octant has a color on/for each vertex it has left.

The face group (4 octants on a slice) is in the same group as a square with colored edges, which has a rotation group of order 1. The squares group is a permutation group of faces (i.e. of 4 octants at a time).

Now, I have a question to ask (anyone at all). Is coloring the faces an operation, and if so what does it operate on, and what with? Is color injective or surjective, is it even a function?
If it is, what sort of function is stickering a cube, with fixed size, fixed color stickers (here, elements are fairly obvious).

 

I'm sorry but this just doesn't make any sense.

What is 'the face group' ? Do you mean the word group in the mathematical sense?

What do you mean by: the group of a square with coloured edges has a rotation group of order 1?

Do you mean there is a subgroup of the group of a square with coloured edges (whatever that means) which is isomorphic to SO(1)?

SO(1,F) to be more precise, but over what field?

 

Noodler is a known hack whose been banned before for just spouting nonsense about things he doesn't understand. I would recommend you don't bother to engage him in actual discussion, just point out his nonsense.

 

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The square puzzle has a reflection group of operations - these do not map to the rotation groups of permutation puzzles, and in any case require an extra dimension. That is, reflections in 2d are possible only because there is a 3rd dimension - the one the computer is in that projects the "puzzle".

As any idiot can see (including me) the 1-polytope is a 2-cube with 4 colored edges. The face groups are therefore 2-cubes that are 2-polytopes in the 3-cube.
Now rotating a single group of face elements, will not get any idiot far in solving or understanding the permutation dynamics. Nonetheless, even I can tell that you have to use each group of face elements one at a time - you have to invoke the single-step erm. automorphism (unless you design say, an automatic automorphism).

Hey, I know, why don't we have a collective attempt to do an all-out OO analysis - after collectively deciding a notational convention for it ..?? Who knows about Haskell and currying?

(note to self: don't get excited, or anything)

 

Turns out there is no mathematics in this thread so I am going to have to take my leave.

EDIT: I know a fair bit about Haskell. What has currying got to do with anything? Haskell isn't OO, it's a functional language.

What do you propose we analyse? It's incredibly unclear from your progression of posts what you are actually trying to do.

 

Another math student who didn't want to become a hack(er)... Ah well.

 

Did you even read what I said?

 

If you are a big enough fool to have looked at OO, and in particular the first step, you know that the analysis is independent of any implementation.
Therefore the analysis (OOA) is independent also of the language used for the analysis. This kind of recursion is (like a Moebius transformation) a fundamental property of computational "objects' which are subject to such rules of analysis.

All objects in OO are in a problem domain which is bounded by the problem class, and by the classes of machine available to encode the implementation. Encoding the Moebius transformation is an implementation of an OO analysis - which is encoded in a language which is freely chosen (does this sound familiar?).

So far the score is: one determined critic with nothing meaningful to say, a masked man from Los Alamos, and a math undergrad who went looking for math, and couldn't find it!

Here's the thing though - the cube is always mathematical, you can disagree with it, but it only has ONE solution, all the same. There is no math in a cube either, for the mathematically confused. This is located in a neural network between your ears.

 

alephnull: I posted a response, does it not make any sense either? You are still confused?
Is it because you are thinking I strung two words together that, to you, mean something else altogether?

Is it not that case that the 3-cube is a group of 2-cubes?

 

You are very strange.

You want to encode the Mobius Transformation?

mob :: C -> C -> C -> C -> C -> C
mob z a b c d =
| ad - bc == 0 = error "Undefined"
| c /= 0 && c*z == -d = a/c
| c == 0 && c*z == -f = infinity
| otherwise = (a*z + b)/(cz + d)


Why did you ask about currying?

 

Do you know what a group is?

http://mathworld.wolfram.com/Group.html

Now in what sense is a '3 cube a group of 2 cubes' ? Your questions are not well formed.

 

Do you know what this is?

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How would you describe it, in terms of group theory bearing in mind it is partially assembled or disassembled..?

 

OK you've dodged my questions, this is my last post.

That is not a well formed question either, group theory doesn't describe the cube itself, it describes what you can do to the cube. The cube is not a group, that doesn't make sense.

The cube, or any part of the cube for that matter, is not a group.

The "Rubik's Cube Group", is the group \( (R, \circ) \) . The group of all operations that can be applied to the cube (twists) , under functional composition.

 

What has OO (im assuming you man object oriented) languages to do with this. The solution is a very simple rule based algorithm. If position Z has colour X to move to position Y then execute moves A,B,C..... Any language can solve a cube.

 

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I disagree. Operations in the set of "natural" are a group the "operator" descends into readily.
The operator operating, is also part of the group. That is to say the theory of operation, that requires a theory of an operator of the cube, is a recursion that transforms the group (of recursive operators).

Application is again, in the group of operations -> operator. Mathematics is an operational theory, etc.


As to Haskell and currying - well let's start with "the parts of the cube are not a group". Prove that any part of the cube is not a group first then prove the stronger argument.
And with the OO: start with the problem domain, and objects:

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Let S(m,n) be the set of m rotations, n translations in the problem domain, for the elements or objects so that a cube symmetry is restored, starting at the initial state depicted. Hang on, I have a solution to hand already...

 

Who was I kidding.

That makes no sense. None whatsoever. You clearly don't know anything about group theory, so why keep up the charade? It's clearly impossible for me to debate group theory with someone that doesn't understand group theory.

As has been pointed out numerous times, the solution is a very simple rule based algorithm. Since that was pointed out you have begun to butcher so many mathematical terms and definitions in order to make the problem seem more complicated than it is.

Again, nonsense.

The cube is not a group, get over it. You can act on subsets of the cube with a group (say, of symmetries)

Groups acting on sets, read into it.

Mathematics is not a theory, let alone an 'operational theory'