# Unscrambling the cube

Discussion in 'General Science & Technology' started by noodler, Nov 11, 2009.

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1. ### noodlerBannedBanned

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Mathematics is not a theory the way economics is a working model. The whole idea of there being a single element, is a mathematical theory - in reality there is no really constantly unitary element. There are a lot between 0 cubes and 1 that functions, in the simple way you describe.

Who told you there were any real integers or numbers? Why do you think mathematics or any other -ics isn't a theory?
(ye gods)

3. ### alephnullyou can count on meRegistered Senior Member

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OK we're just talking semantics now.

Mathematics is the study of structure, space and change. It might be that whilst studying these things, one uncovers theories or conjectures, but mathematics is just the study of these things.

Why have you gone on to the philosophy of whether numbers exist or not?

5. ### noodlerBannedBanned

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751
Ah, so the study of mathematics isn't theoretical either.
For my answer on the philosophy question, is study a philosophy and if not, what do you mean by "the study of structure..."?

And does a number exist or not goes to does a cube exist or not if it's disassembled partly, or otherwise? You must know, if you've studied lambda calculus that you can't prove numbers exist.
If you assume they do, like you assume you can put a cube's parts together, you can prove that arithmetic exists, primes and all that. Numbers must be a function of "logic" or whatever.

7. ### alephnullyou can count on meRegistered Senior Member

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This has nothing to do with your original post's topic.

It doesn't surprise or interest me in the slightest that lambda calculus cannot prove numbers exist.

8. ### AlphaNumericFully ionizedRegistered Senior Member

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The bits of the cube transform as a group but the cube itself is not a cube. Objects are not mathematical groups. They can have structures assigned to them which can have a group structure but objects are not mathematical groups.

If a cube is a group, tell me the identity element. Demonstrate how it is associative.

You just throw out buzzwords. You mentioned holonomy and when I stated I know about it, you changed topic. You haven't done any group theory about the Rubik's cube, you went onto thermodynamics equations. In this thread you throw around Lorentz transforms and entropy but you illustrate you don't understand what they mean. You mistake the fact the orthochronos Lorentz group is continuous, as I'm sure you read on Wikipedia, with the notion of a 'continuous Lorentz transformation', since you don't get the difference between a parametrised quantity (ie a function) and a parametrised curve through a group.

Why do you bother? Every time you try to BS your way past someone on a topic they know and you don't you just get exposed as a hack.

9. ### noodlerBannedBanned

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This thread is about a packing problem, which like all computational problems is in a problem domain.
The objects, apart from the ones you need to construct a Rubik's cube, are threads and "statements" in them.

Now entropy, as we all know, is parameterized by the capacity of a communication channel. All physical channels have a limit, but we know we can construct arbitrarily accurate channels, that will faithfully transmit signals. Entropy in a permutation puzzle is the amount of repackaging you have to do, to restore the "color symmetry".

This is far from being a continuous curve through a group; the entropy is a function of the channel's width and depth (I mean, doh!). Shannon entropy is therefore a function of any packing or repacking class of problem, in the full group of packing problems that pertain to n-cubes...

Does it surprise or interest you that lambda calculus exists? Can you prove it does, by not using a lambda calculus?

Last edited: Jan 12, 2010
10. ### AlphaNumericFully ionizedRegistered Senior Member

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No, since the bits which make up a cube always make up a cube, no matter how you twist and turn them. A packing problem is about getting seperate objects into an optimal arrangement. In a Rubik's cube the arrangement is predetermined to be a cube. Rocket science, I know!

Clearly you just hit Google or Wikipedia again to find a new buzzword.

Entropy is defined in terms of the likelihood of a particular macroscope state in terms of microscope states. The fewer microscopic state configurations which give a particular macroscopic state the lower the entropy. Look up 'partition function' in thermodynamics when you're next Wiki'ing for a buzzword.

No, the entropy is the measure of the likelihood of a given colour configuration. The 'solved cube' has, up to global rotations, one microscopic state and thus very low entropy compared to a cube configuration which has all colours on all faces, ie 'random', which occurs in a great many configurations and thus is of high entropy.

In my last post and the thread I linked to I made it clear I was talking about paths through the Lorentz group. Clearly your lack of knowledge with regards to anything I've said leads you to have no clue what I'm talking about. You make sly comments "(I mean, doh!)" as if you're familiar with all this and its all easy to you but you haven't managed to say anything relevant or coherent in your entire post.

I specifically used the phrase 'mathematical group' to make it clear I was referring to a group and not a set (for which 'group' is a layperson's synonym). You failed to understand the meaning of 'group' even when I spell it out. And thus you used 'group' in two different ways in the same sentence :

"This is far from being a continuous curve through a group....... in the full group of packing problems that ..."

Yet more evidence you really don't have a clue. And the fact you completely ignored my previous post and retorted nothing I'd said shows you know you're doing it. It was just nice of you to give yet another example to illustrate my point.

11. ### noodlerBannedBanned

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I'll just do a rewrite here and present the argument, one more time
An agin:

"...the bits which make up a cube always make up a cube, no matter how you twist and turn them."

This is why you have to untwist and unturn them when you get a constructor kit. You have to do some figuring as well, too bad if you're color blind, at this point of the build. The kit is also something that always makes up a cube, as long as there is always a way to build it.

"A packing problem is about getting seperate objec4s into an optimal arrangement. In a Rubik's cube the arrangement is predetermined."

As the picture shows for anyone's benefit, you have to get the separate objects into an arrangement, optimal or otherwise, and so doing is a rearrangement, again in the same general class of packing problems directly related to the group of problems in the domain class.

Cheating, and getting someone else to do it isn't predetermined, and there aren't any repair shops, you have to learn how to restore the optimality. After all, removing the stickers and replacing them "blindly" is a perfectly valid construction, a random permutation generated without rotating any [other] cube elements [only the stickers].

"the entropy is the measure of the likelihood of a given colour configuration. "
And the proof is complete.

Last edited: Jan 12, 2010
12. ### AlphaNumericFully ionizedRegistered Senior Member

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Packing problems are the issue of putting a set of objects into the smallest region possible.

For instance, packing circles into the shape which can be surrounded by the shortest amount of string. For 6 or less the best arrangement is in a line. For 7 you make them into a hexagon with a circle in the middle. The arrangement of Rubik's cube pieces always has the same surface area, volume etc. There is no packing problem.

Well done on responding to nothing I actually said.

13. ### noodlerBannedBanned

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751
(and little fishes)

"The arrangement of Rubik's cube pieces always has the same surface area, volume etc."

Go to a toy store and purchase an optimized color-symmetric Rubik's cube. Also purchase a kit for the same object (and remain objective, use "objectification" at maximum amplitude at this step).
Note that: the kit pieces aren't stickered, but have "almost" the same volume - you can prove this by assembling the black plastic pieces and not the stickers.

But (i) to construct an "arrangeable" cube - that has cubic symmetry and optimization rules/relations, you do have a packing problem, and (ii) when you "encounter" a scrambled cube, you also have a packing problem. The solution to (i) is demonstrably an optimization and a packing problem - the algorithms involve selecting "best fit" and discarding "worst fit" configurations, as in (ii). You simply are not thinking, old boy.

And to close the argument, the pieces always have the same volume, but since you can sticker, desticker and resticker (that's two functions, one has an inverse and reflections are excluded, but only by convention, sticker rotations are "symmetric" as 1-facets only rotate dynamically, translations are "asymmetric" as 2-facets and 3-facets have this dynamic).
This remains true for stickers when the surface area is distributed, optimized or deoptimized. Optimization is a computational class, so is parallelization (and just look at the way they're slouching!).

"There is no packing problem." Not after you've solved the packing problem, no, that's correct. Who told you?

14. ### AlphaNumericFully ionizedRegistered Senior Member

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The packing problem is finding an optimal shape. Putting a Rubik's cube together is about making a pre-determined shape. A jigsaw is not a packing problem in the same way fitting oranges into a cuboid box is. An assembled cube in a configuration which isn't 'the solution' is not a packing problem either, it's a permutation problem.

You continue to illustrate you don't grasp the basic terminology and none of the buzzword mutterings you say like "that's two functions, one has an inverse and reflections are excluded, but only by convention, sticker rotations are "symmetric" as 1-facets only rotate dynamically, translations are "asymmetric" as 2-facets and 3-facets have this dynamic)." go any way to you addressing any of the problems relating to a Rubik's cube this thread is supposedly about. Just more of you putting your foot in it.

15. ### noodlerBannedBanned

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I disagree that assembling a Rubik's cube "from scratch" is not a packing problem. Assembling anything physical at all is a packing problem, unpacking something is a packing problem.

The stickering function, which acts once on centers, twice on edges, and thrice on corners is two functions, one that places and unplaces stickers with rotation, and another that translates them; there is a map that goes between stickering in a "pre-determined" or framed configuration (i.e. the "store-bought" model) and the "elements in the neighborhood" unpacked model.

Again, this map of rotations, inversions and translations (but not reflections, because the three operations use up the available dimensions), is made of the same stuff as the pre-packed model.
That is to say, "someone" has to iterate a construction formula that takes the cube's elements to the optimal shape or configuration. The function "space" is the set of rotations, translations and inversions of all of the elements, and the full permutation group is larger than for a cube that is stickered "optimally".

Then sorting optimally stickered cubes (cells) from the larger group is a function of stickering, a decision problem, or "is the cube stickered properly"? Design an algorithm that can decide, in a minimum number of steps, say O(n), that a given cube can be "symmetrized" so all faces have individual color. Initially assume that the stickers can be rearranged, so you can either swap stickers two at a time, or some even number, then assume that the inner uncolored elements have to "carry" the stickers around the surface instead. The sticker swap algorithm will restore the symmetry of the faces, in color; the inner-element swap algorithm will not. At some point if the algorithms are run in parallel, there will be a conflict (resolved by swapping a pair of stickers). Thus, the pairing of colors, over elements is a pivoting function - the centers are physical pivots, but 'meh' centers are redundant.

Last edited: Jan 13, 2010
16. ### noodlerBannedBanned

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751
Closed, but no banana:

Something I read somewhere (it might make sense):

A mechanism is a general term, used to denote that a "theory of mechanisms" is being applied. The theory is based on references to physical "matter and motion", a set of observable phenomena. In the beginning of the 19th century the notion of "mechanical philosophy" was revived, a Hellenistic influence here was the idea of atomism. The science of mechanics the 19th century required, also demanded a quantitative set of "laws" that fit the philosophical paradigm of the age.

Descartes wrote: "Give me extension and motion and I will recreate the world", a reference to Archimedes. To the scientists and thinkers of the time, the universe was both mechanical and also a machine, in which "everything happens by figure and motion".

Boyle's philosophy was expressed as explanations, of physical phenomena, by means of elements which were "little bodies variously figured and moved". Among the important problems of the time, was finding a distinction between primary and secondary qualities, made by both Galileo and Locke, which the Cartesian dualism of mind and body introduces to the paradigm. In dualism, the body is purely mechanical (i.e. "colorless") and the mind is either also a body which is part of the mechanical one, or is a "design problem" with tautological implications.

Later, Kant took the view that physicochemical modes of operation, while being necessary, were ultimately inadequate explanations, since, no "disembodied" mind is exposed or explained. 20th century thinking saw two paradigmatic extremes: one that holds all biological forms are physicochemical (biochemical) phenomena, the other that each living thing possesses a directive and vital principle, an entelechy, which explains emergence.

17. ### noodlerBannedBanned

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751
So, on to Einstein's puzzle, the one he wrote before he wrote the other thing.
In which there are various characters, nationalities with houses of different color.

Which employs the "color", function -> function of color application. I have a solution written by someone else in Prolog, that uses the same "apply" type of logic. The program is a theory, that a machine proves there is a solution for. It's the application.

I could post or publish this, but there is any number of solutions to the problem. The important part is that, it's a location problem - the colors locate the houses and the clues or relations are conjunctives like "next to", "left of", and the associated set of properties (the data in the object, or list) with associatives "owns", "drinks", etc used to order the relations between the houses (the colored ones).

In the listing, colors are just words or symbols, Red, Yellow, etc.

18. ### noodlerBannedBanned

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Is the kitset form of a puzzle in the cube group (the full group) a jigsaw?

Let's see: assemble the black plastic elements into a cubic solid form (you may figure out that solving this problem doesn't require a frame, then that it doesn't require any of the edge pieces, or elements either).
Using a single color set of stickers, white say, begin applying colored squares to the elements (so, you know, they "fit"). After the first sticker is stuck, how many permutations exist?, after two stickers, then two colors, etc? Does the number of permutations depend on which elements are colored the same?

Does reassembling a jigsaw puzzle depend on which elements are colored the same, apart from being able to "stick" them in (the shape man, the shape)?

Last edited: Jan 16, 2010
19. ### noodlerBannedBanned

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751
Golomb's conjecture:

"Mesons are made with a quark and an anti-quark, and baryons
with three quarks or three anti-quarks; well, an element of the
Subgroup [of the cube] can create a twist plus an anti-twist, or
three twists or anti-twists, but never an isolated twist."

John Baez:

"One other little observation... Z_3 is deeply related to SU(3),
because the center of SU(3), consisting of the diagonal matrices
exp(2 pi i n/3), is isomorphic to Z_3. More generally, the center
of SU(n) is Z_n in the same way. For SU(2) it's Z_2, and when
we mod out by this Z_2 we get good old SO(3).

From this one would suspect that Z_3 representation theory is
important in SU(3) representation theory. My guess: Each irrep
of SU(3) is obviously (or should I say Schurly) an irrep of Z_3,
of which there are 3, and to tell which one, just count the number
of boxes in your Young diagram mod 3.

The corresponding Z_2 invariant of SU(2) irreps is just the
'integer/half-integer spin' distinction!"

20. ### noodlerBannedBanned

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751
So after all, the initial "excitation energy" generated by the cube's appearance as a permutable, symmetric object (it's a verbal noun too like the word color, or colour) appears to have a deep connection to the symmetries of space and time at much smaller distances.

This is something I would like my niece to appreciate (she already has a good grip on the solving a scrambled configuration problem), so that is sorta what this is about.

Because it's easy to explain that color is a function, it's a function of itself (multiply the color by its identity), and how to compose the function, bypassing all the conventional equations that describe frequency, intensity, and so on.

If white is a function that is true for all colors, then we can build a list of functions of color which are in the set of "true or false", and then interpret what that means - we get 6 functions for free in a kitset. But the 7th is the absence of color, which is false for white.

It's easy to show that at least one color (which we can choose to be the white one) is redundant, and in fact only four adjacent sides need colors on the sections, as long as the other missing color is from the opposite face. Reducing the colors to 4 means the black faces are in the center of the 2-cube.

And if the connection to Einstein exists puzzle-wise, my conjecture is that he didn't design the puzzle as an IQ test or for other's benefit necessarily. I've heard it was published while he was a student, but of course he went to what passes for "high school" as well, I'm not sure how old he was, or how many years to the solution to another puzzle. The "clue" if it exists, in as insight into M, natch, and how it works, or why he designed the puzzle or problem - what sort of device is it really?

Recall that he once claimed he rarely thought in words. What did he mean and is there a device in the problem (a Prolog list) that he employed later on?

Last edited: Jan 18, 2010
21. ### noodlerBannedBanned

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751
Now to reflect something, using what I generally characterize as the "GTFOH channel", aka the Eddie Murphy hotline.

...entropy, as we all know, is parameterized by the capacity of a communication channel.

"Entropy is defined in terms of the likelihood of a particular [message] in terms of [bits]. The fewer [bit] configurations which give a particular [message] the lower the entropy. Look up [Shannon entropy and Von Neumann entropy in IS] when you're next Wiki'ing [with] a [binary word]."

Entropy in a permutation puzzle is the amount of repackaging you have to do, to restore the "color symmetry". [This entropy is the length of the word that corresponds to the number of steps.]

"No, the entropy is the measure of the likelihood of a given [word]. The 'solved cube' has, up to global rotations, one [bit the identity,] and thus very low entropy compared to a cube configuration which [is a word with a large length], ie 'random', which occurs in a great many configurations and thus is of high entropy."

:shrug:

22. ### noodlerBannedBanned

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751
Now to the structural character, literally, of the cube group.
This is a multiplex:

The formula for a computable number and hence any code that can be written by a machine (like Enigma) is

$(2i -1)n + \sum_{r=1}^{\infty} (2c_r -1)(\frac {2}{3})^r$

Suppose a recursive number $N$, and an algebraic character $r\; =\ 2 \sqrt {\frac {3} {2}}$, not the same as r in the above formula (hence a correspondence is needed).

$N$ has a commutative relation to k, the length of any algorithm which can be written with a general structure $[(AB)^n + (A'B')^n]$ and the formula
$N_k = [(3+r)(6+3r)^n + (3-r)(6-3r)^n]/4$, where $N$ is any character (in a standard basis of the sliced cube) generated of length k.

N the slice number is the 'computational surface' over which $N_k$ is iterated.

Suppose also the sequence $s_0, s_1, ..., s_n$ is the formula: $s_n\; = 12.s_{n-1} + 18.s_{n-2}$, in respect of n. There are two numbers $\alpha, \beta$ that start equal at n = 1. The recurrence formula in n for these is:

$\alpha_1= \beta_1 = 9,\; \alpha_{n+1} = 6.\alpha_n + 9.\beta_n,\; \beta_{n+1} = 6.\alpha_n + 6.\beta_n$

Note the 3/2 nature of the character's symmetry. (hope i got this right)

Last edited: Jan 21, 2010
23. ### AlphaNumericFully ionizedRegistered Senior Member

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You think incoherent nonsense could be 'right'?