Timeline for Who wins the Scrambler-Solver game for infinitary Rubik's cubes?
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22 events
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| Sep 1 at 13:46 | comment | added | Jack Edward Tisdell | @JoelDavidHamkins I believe you are correct, the order may not matter. So long as we distinguish the "center" cell between $L$ and $-L$ (or the center cut in the "even" version), the definitions still work. However, there is a subtlety regarding whether or not $L$ has an "outermost" element (greatest in the order formulation); I stipulated in the OP that it does not for if $L$ has a distinguished "outermost" element, then it seems we should want edge and corner cubies which carry more than one tile. Intuitively, "$\mathbb Z^3$" has no such cubies while "$(\mathbb Q\cap[-1,1])^3$" does. | |
| Aug 31 at 18:53 | comment | added | Joel David Hamkins | @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? | |
| Aug 31 at 16:06 | comment | added | Jack Edward Tisdell | @JoelDavidHamkins I think I have a convergent scramble only attainable using some twist infinitely often if we allow half-turns $T^2$ as single operations, although a Rubik's cube expert might correct me. Take $L=\omega$. Apply half-turn twists in $x = \pm n$ then do a quarter turn face twist in the $\pm y,\pm z$ faces, call this group element $s_n$. Note $s_n^2$ results in a kind of windowpane pattern in the $z,y$ faces. $s_0^2,s_1^2,...$ yields an asterisk pattern in the $z,y$ faces but leaves the $x$ faces pristine. I don't think this is doable without infinitely many $y,z$ face twists. | |
| Aug 31 at 14:57 | comment | added | Joel David Hamkins | 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. | |
| Aug 31 at 14:47 | comment | added | Jack Edward Tisdell | @Gro-Tsen, I appreciate the feedback, this is for sure a useful change in perspective and I will think more along these lines. In defence of the peculiar set-up, as I see it, the underlying group structure is motivated by the Rubik's cube operations and there's a sense in which the "sanity" of this completion process (if it turns out to be non-trivial) is justified by the Rubik's cube interpretation. But certainly I agree that's it's worth thinking about this completion process in a more general context. | |
| Aug 31 at 14:20 | comment | added | Joel David Hamkins | @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. | |
| Aug 31 at 14:13 | comment | added | Gro-Tsen | [contd.] If my understanding is correct, I suspect that very little about this depends on the details of what the cube is like: it's probably more of a question of what kinds of conditions should be imposed for the “completion” process to yield something sane (and, hopefully, a group, or at least something symmetric). Couldn't you find a simpler and/or more general situation than an infinite Rubik's cube in which this “completion” makes sense, and start by studying this? | |
| Aug 31 at 14:08 | comment | added | Gro-Tsen | I think this question has great potential, but as it is currently phrased it is very confusing (although I admit it is rigorously defined). IIUC, as per @JoelDavidHamkins's comment, this isn't at all about games but about accessibility in a certain graph. Basically you have a group of finite operations which you “complete” into something that might not be a group but is still, at least, an oriented graph (with edges being “accessible in $\leq\omega$ moves”) and you are asking whether any out-neighbor of the origin is also an in-neighbor, or something of the sort. [contd.] | |
| Aug 31 at 13:45 | comment | added | Jack Edward Tisdell | @JoelDavidHamkins that's a good question, I will think about the every twist once operation. I don't think tiles can escape to infinity, though, unless I've misunderstood your meaning. Like an ordinary Rubik's cube, each tile can only ever occupy up to 24 distinct cells (corresponding four on each face). (This is why, e.g., illegal configurations are persistent.) | |
| Aug 31 at 13:40 | comment | added | Joel David Hamkins | 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? | |
| Aug 31 at 13:29 | comment | added | Joel David Hamkins | 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 | |
| Aug 31 at 13:14 | comment | added | Jack Edward Tisdell | @JoelDavidHamkins Ah yes of course, thanks for that correction, I've fixed that. I think I agree with your group formulation of the question regarding infinite operations. But also, even for finite scrambles, obviously, they are invertible, but I'd still like to know in the case that $L$ is uncountable, given only the scrambled configuration, can the Solver always find the inverse just by inspection? Or can/must she use $\ge \omega$ many twists to search for it? (I think the game perspective invites these kinds of questions.) | |
| Aug 31 at 13:06 | history | edited | Jack Edward Tisdell | CC BY-SA 4.0 |
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| Aug 31 at 13:05 | comment | added | Joel David Hamkins | 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. | |
| Aug 31 at 12:53 | comment | added | Jack Edward Tisdell | @JoelDavidHamkins I agree that the core question is indeed whether infinite convergent scrambles can be unscrambled. Definitely, the twists form a group (which acts on the configuration space) but it seems to me that the group structure alone cannot suffice to answer questions about infinite sequences. Four twists are the identity but a the action of a constant sequence of twists need not converge. Moreover, whether or not (the action of) a given sequence of twists converges depends on the initial configuration, not just the sequence of twists. | |
| Aug 31 at 12:42 | comment | added | Jack Edward Tisdell | @JoelDavidHamkins I think the quarter turn definition is simply a matter of which rotation direction one takes as "clockwise". In my example, I was imagining the $x,y,z$ coordinates with $+x$ to the right, $+y$ upward, and $+z$ towards myself and "clockwise" according to the right-hand-rule. So in an $x$ twist $T_{x,\hat 0}$, the "front" cell $(\hat 0,\alpha,+\infty)$ is twisted to the "bottom" cell $(\hat 0, -\infty, -\alpha)$. But your suggestion works just as well so long as the rotations are defined consistently. | |
| Aug 31 at 12:12 | comment | added | Joel David Hamkins | 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? | |
| Aug 31 at 11:15 | comment | added | Joel David Hamkins | Very nice question. But have you defined the quarter turn properly? I would expect $(\hat 0,\infty,-\alpha)$. | |
| Aug 30 at 22:15 | history | edited | Jack Edward Tisdell | CC BY-SA 4.0 |
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| Aug 30 at 22:12 | history | edited | Jack Edward Tisdell | CC BY-SA 4.0 |
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| S Aug 30 at 22:10 | review | First questions | |||
| Aug 31 at 1:51 | |||||
| S Aug 30 at 22:10 | history | asked | Jack Edward Tisdell | CC BY-SA 4.0 |