I have argued in my writing that the meaning of the word platonist has changed in the past thirty years in the context of set theory, since formerly it was often taken to indicate a singularist view, giving a definite value for CH etc., but I argued that platonism should instead be about the realism of the objects. By now, this usage seems to have been more widely adopted (and indeed this is how you seem to be using the word). So yes, I am platonist, but pluralist, even about arithmetic, and I think it is an illusion that we have a singular conception of ℕ.

I've written careful accounts of my views in my various articles on the topic and also in my book, Lectures on the Philosophy of Mathematics. Regarding arithmetic, I think we are mistaken to think there is an absolute conception of the finite, and I have argued that all arguments that have been given that there is an absolutely standard model are ultimately circular.

@JuanAtacama You can read my paper on the set-theoretic multiverse, which is linked in your SEP link. I am a pluralist realist, about arithmetic as well as set theory.

There is a definable global choice function iff there is a $\Delta_2$ definable global choice function, so it is expressible.

@Lucenaposition It can talk about definable classes, as I explain.

@JackEdwardTisdell Another issue is that you've described the set-up in terms of an order, and perhaps that is how we imagine it geometrically, but it seems to me that the order doesn't matter. You have a set $L$ and another copy of it $-L$. Is this correct?

Every tile is part of a 2×2×2 Rubik's subcube, and indeed any finitely many tiles are part of a finite Rubik's subcube. So can we solve the countable case by solving and then preserving increasingly large finite subcubes? I guess the problem is that we have to temporarily upset the solved part to solve larger parts.

@JackEdwardTisdell About the 24, yes, I had come to the same conclusion myself, and that is what I had meant when saying it isn't possible without repeating twists. Meanwhile, I guess every infinite scramble that uses every twist only finitely often will be convergent, and my question is whether these are exactly the convergent scrambles.

No, I guess it isn't clear how to move a tile to infinity that way without repeating twists. Is there a simple (or any) convergent scramble that uses some twist infinitely many times, but is not possible with only finitely many of each twist?

Do you know if you can unscramble the operation of doing every twist exactly once in some sort of organized $\omega$-enumeration? It seems that this will be convergent, since any tile location is affected by only three twists, but it seems that we can in effect move a tile to infinity in the limit, making it in effect disappear, which would suggest you can't invert the whole operation. For example, this phenomenon also happens with infinite convergent compositions of permutations on an infinite set. See mathoverflow.net/a/17655/1946

I'm still confused about your rotations, since under your description I think you should mean $(\hat 0,-\infty,\alpha)$ then. That is, there should be only one $-$ sign, not two. In 2D, for example, a quarter turn rotates $(x,y)$ counterclockwise to $(-y,x)$ or clockwise to $(y,-x)$, not $(-y,-x)$. And yes, about the group, I was referring to the question whether the infinite convergent operations form a group.

You've presented the question as a game, but as I see it, the core question is simply whether infinite convergent scrambles can be unscrambled. Essentially, is this a group? Is that right?

Very nice question. But have you defined the quarter turn properly? I would expect $(\hat 0,\infty,-\alpha)$.

@JuanAtacama Yes, thank you, I am quite familiar with Balaguer's view, and I have written at length on it. My work on the set-theoretic multiverse is cited in the SEP link you had provided in the section on Plenitudinous platonism.

Of course you should require that the theory is consistent.

I think the OP is asking what is the complexity of verifying that a given solution does in fact solve a given decision problem in P. And my answer is that this is actually very difficult, not even computably decidable. If this is indeed what the OP intends, then I suggest leaving it as is, but if not, then I agree the question is not clear enough and should be closed.