## An interesting thought experiment about Rubik's hypercubes

For the discussion of math. Duh.

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### An interesting thought experiment about Rubik's hypercubes

Fig. 1:

Yay, a Rubik's cube! Actually, it's a lot more specific than that. We can say that this is a 3x3 Rubik's cube in 3 dimensions, and that therefore it has 6 colours and its stickers are 2-dimensional.

Fig. 2:

Yay, it's a...a what exactly? This is still 3x3, and it's still a Rubik's n-cube, but it is 4-dimensional. Each of those solid-coloured cubelets is in fact a sticker. You can't quite see the edge/corners/centres/whatever the hell you have in 4-D that they're on, of course, as those are 4-D, but it's worth showing that there are 8 sides each containing 27 stickers; you can see one in the centre, 6 bordering it, and one that you can't see but can switch out with another side by rotating the 4-cube in the fourth dimension. So rather than the 54 stickers of a 3x3 3-cube, we now have 216 to keep track of, and you can't see how they're bordering each other. That makes solving it rather more difficult (though still possible; you obviously can't manufacture this sort of Rubik's cube, but you can program a computer to emulate one that you can play around with.)

Then a thought occured to me: Those stickers in the centre are 3-cubes.

A normal Rubik's cube is a 3-cube.

So now let's imagine that each of those stickers that we see is in fact a 3-cube in miniature. Obviously you couldn't mark these mini-stickers they have on them with colour; you could mark them with texture, or shading, or something. It doesn't really matter, of course; from the standard of mathematics you would just assign each substicker two numbers: the first indicating what the colour of the big, 3-D sticker it's on is, from 1 to 8, and the second indicating what its texture is, from 1 to 6. In actuality it will be more complicated than this, because the part of the cube it's on- a corner, an edge, a centre- and the part of the 4-cube its mother sticker is on- must also be marked. In any case, then, there are 216*54= 11,664 2-stickers to keep track of.

And in fact you can extend this indefinitely! You can define all Rubik's cubes with just 3 numbers: n, q and k. The Rubik's cube is marked as n by n; in other words, each side has n by n by n (however many dimensions it is) stickers in it. Q is the cube's dimension. K is defined as the iteration. If the stickers of the cube are not further broken up, then k=0; it has not been iterated. For obvious reasons, all run-of-the-mill 3-cubes (you know, the ones you can actually manufacture and don't fuck your mind up too much) have a K of zero, because there are no 2-D Rubik's cubes strictu sensu (I don't think). A Rubik's tesseract can have a K of zero (stickers are not broken up) or a K of 1 (3-D stickers are treated as 3-cubes; cannot go further since that would involve treating a 2-D sticker as a 2-cube, which doesn't exist.) A Rubik's 5-cube (which is a penteract) has tesseractal stickers; thus it can have a K of zero (4-stickers not broken up, thus no iteration), a K of 1 (stickers treated as 4-cubes [iteration 1], stickers of resulting 4-cubes not further broken up), or a K of 2 (stickers treated as 4-cubes [iteration 1], stickers of 4-cubes as 3-cubes [iteration 2], stickers of resulting 3-cubes are 2-D and thus we cannot iterate further), and so on and so on; it is thus clear that n≥2, q≥3, k≤q-3. You could even imagine a fully fractal (q=∞) cube floating in a Hilbert space!

Actually, no you couldn't...that's nonsense. I think...Interesting thing to think about, though.

The first question is not really a question at all. It's really just a question of practicality- if we consider that any move on a lower-dimension sticker is counted as one move, just as any move on a higher-dimensional sticker or the whole n-cube is counted as one move- in other words, rotating the whole Rubik's penteract is counted as the same kind of move as rotating a 3-D substicker inside one of the penteract's tesseract stickers, are there defined algorithms to solve a scrambled cube?

(Alright, I'll make it a real question- are changing each of n, k or q a P problem, or an NP problem? In other words, can we still solve the n-cube in polynomial time relative to a changing number of stickers, a changing dimension, or a changing number of iterations? Which of these is P, and which of these is NP?)

The second question is, of course, completely impossible and just me having my head in the clouds- find a function f(n, k, q) which is equal to God's Number for a fully scrambled n by n Rubik's k-cube that's gone through q iterations. For example, f(3, 3, 0)- God's number for a normal Rubik's Cube- is equal to 20; f(2, 3, 0)- God's Number for a 2x2x2 cube, or what's sold under the name of the Pocket Cube- is equal to 11. For the purposes of God's number, half-turns or quarter-turns are counted as one turn; in other words, so long as you are moving only one face, you have only made one move.

Happy mathing?

dhokarena56

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### Re: An interesting thought experiment about Rubik's hypercub

I don't see how solving a (3,4,1) is any different to solving a (3,4,0) and 216 (3,3,0)s. In which case, I'm not sure if I understand the meaningfulness of the construction.

Talith
Proved the Goldbach Conjecture

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### Re: An interesting thought experiment about Rubik's hypercub

Talith wrote:I don't see how solving a (3,4,1) is any different to solving a (3,4,0) and 216 (3,3,0)s. In which case, I'm not sure if I understand the meaningfulness of the construction.

Well, it's not really...but it would still affect the P vs. NP question.

I suppose if you did want to make it harder, you could state that any n-D sticker could end up on any n+1-D sticker when you scrambled it, but then you'd have to define a realistically Rubik's-like algorithm for how you move a substicker between stickers.

dhokarena56

Posts: 179
Joined: Fri Mar 27, 2009 11:52 pm UTC

### Re: An interesting thought experiment about Rubik's hypercub

Well, if a problem can be solved by solving a finite number of polynomial time subproblems, then the initial problem can also be solved in polynomial time.

If you want the action of twisting an n-dimensional surface to act in some non-trivial way (ie twisting the surface of some k-dimensional sub/supercube, n=/=k) Then you'd have to define that action carefully. Once that definition has been made, you can then see if the new problem is solvable, and if it is, what complexity class the problem lies in.

Talith
Proved the Goldbach Conjecture

Posts: 848
Joined: Sat Nov 29, 2008 1:28 am UTC
Location: Manchester - UK