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One-Answer Puzzle Theory Question Thread

qwr

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Actually, it´s (4^6)/2 = 2048. One quarter turn rotates the center piece by 90 degrees, but also makes corners and edges unsolvable while keeping that center piece rotated - so only 1/2 of combinations which can be reached by executing quarter turns are solvable.

What you are describing is a supercube, if you are only interested in centers, click here.
can it be proved just from EO?
 

bcube

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Doing two T Perms one after the other rotates the centre by 180 degrees for example

In that case "also makes corners and edges unsolvable while keeping that center piece rotated" wouldn´t be true anymore because corners and edges would be solvable again.

can it be proved just from EO?

I am sorry, I don´t understand the question. All I am saying is that each quarter turn of outer layer on a 3x3x3 cube toggles the permutation of corners (and the permutation of edges) from being odd to being even and vice versa. 3x3x3 cube (i.e. also Supercube) is solvable only if the permutation of corners (or the permutation of edges) is even. That´s why 1 rotated center by 90 degrees on 3x3x3 Supercube is unsolvable, while 1 rotated center by 180 degrees is solvable.
 

Gerry

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So with a supercube, The sides, and corners never move.or change direction, and only the centers spin, but still never move, right? I don't mean symmetry, just one position. Say white on top and blue front. Just thinking of a solve it the way you found it, (before scramble) variation of solving. Would be much harder I think.
 

Christopher Mowla

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So with a supercube, The sides, and corners never move.or change direction, and only the centers spin, but still never move, right? I don't mean symmetry, just one position. Say white on top and blue front. Just thinking of a solve it the way you found it, (before scramble) variation of solving. Would be much harder I think.
The cage portion of the supercube behaves the same as the cage portion of the regular (6-colored) cube. And the only reason that the centers on the supercube "behave" differently than the centers of the regular cube is because they are marked with a design of some kind which tells you which direction the center is turned (3x3x3) and/or which positions the centers are "solved" in (4x4x4 and up).

For example, there are this many number of "solved positions" that the regular 2x2x2 - 20x20x20 non-supercubes technically have. Similarly, there are this many number of "solved positions" that the regular minx^2 to minx^20 (where minx^3 = the megaminx, minx^5 = the gigaminx, etc.) can have.

(The number of "solved positions of the regular 6-colored nxnxn and 12-colored minx^n are the number of center positions which the supercube and superminx have, independently of the number of positions of their cages. That's not to say that the parity of the centers and the cage are independent, but this quantity shown in the formulas has already taken into account what we have to divide by to calculate the total number of positions for these supercube and superminx puzzles.)

You can see that the regular 6-colored 2x2x2 only has ONE solved position because the entire 2x2x2 is the cage. So there is no supercube of order 2.

The point being, you can technically say that the 3x3x3 supercube, for example, is 2048 times harder to solve than the regular 3x3x3 (by chance) provided that you make no concious effort to solve the centers in their correct directions as you solve the cage portion of that 3x3x3.
 
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DNF_Cuber

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Something I noticed made me wonder about how many ways there are to "solve" a 3x3. From the perspective of the logo there is at least 2 different solves. I noticed this while watching beginner tutorials that when the blue side is towards me and white side up, the logo is correctly oriented. Then I scrambled it and solved it. Now when the white side is up, and blue towards me, the logo is upside down. If I drew a picture on each side so you could still solve with the colors, would the images be scrambled? Or would just the centers of the images spin? Originally my guess was just the centres would spin, with 3 axis. For 64 different ways. (4x4x4) Now I'm thinking that the centers can spin separately from each other for 4x4x4x4x4x4 for 4096 different ways. Can anyone confirm or deny my guess?
they can "Spin" but they are not entirely independent from each other. If 5 centers are in one orientation, then the last one has only 2 possible positions, rather than 4, so there are 2048 combinations
 

Gerry

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The cage portion of the supercube behaves the same as the cage portion of the regular (6-colored) cube. And the only reason that the centers on the supercube "behave" differently than the centers of the regular cube is because they are marked with a design of some kind which tells you which direction the center is turned (3x3x3) and/or which positions the centers are "solved" in (4x4x4 and up).

For example, there are this many number of "solved positions" that the regular 2x2x2 - 20x20x20 non-supercubes technically have. Similarly, there are this many number of "solved positions" that the regular minx^2 to minx^20 (where minx^3 = the megaminx, minx^5 = the gigaminx, etc.) can have.

(The number of "solved positions of the regular 6-colored nxnxn and 12-colored minx^n are the number of center positions which the supercube and superminx have, independently of the number of positions of their cages. That's not to say that the parity of the centers and the cage are independent, but this quantity shown in the formulas has already taken into account what we have to divide by to calculate the total number of positions for these supercube and superminx puzzles.)

You can see that the regular 6-colored 2x2x2 only has ONE solved position because the entire 2x2x2 is the cage. So there is no supercube of order 2.

The point being, you can technically say that the 3x3x3 supercube, for example, is 2048 times harder to solve than the regular 3x3x3 (by chance) provided that you make no concious effort to solve the centers in their correct directions as you solve the cage portion of that 3x3x3.

Thank you so much! This is exactly the answer I was looking for. I appreciate the time and effort that was put into this answer.

I'm a little weird, and like the logos bottom to line up with the blue face. Not really much harder, I just do U until the bottom of the logo faces blue then start solving.
 
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for the megaminx, you can scramble it by just holding the top/left pieces and doing R++/R-- succeded by D++/D--. then a U/U' after certain number of moves
My question is:
can you scramble a 3x3 using the same method or a modification of it?
 
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Yes, just use wide R and wide D moves.
Using <rdU>, you can definitely scramble CP, CO, EO and EP, so yes.
thanks guys.
i made this:
 

WoowyBaby

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what is the god's number for one side on a 3x3 cube? i mean side, not layer

Let's get someone smart to figure this out!

I'll start with some basic bounds:

3x3 One Face God's Number:
Lower Bound - 8
Maximum moves required to solve a face already reduced into Domino move set.
Upper Bound - 19
God's Number minus one, because all positions one move away from solved have at least one face solved.

My guess is 13 moves.
 
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MethodNeutral

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Not sure if this deserves its own post, but my question is why can’t we change corner permutation with just <R,U>? Intuitively I can’t think of a reason, and it seems like the only way to discover this for yourself is to brute force it in some way. Is there an explanation as to why this is the case?

To further explain what I’m asking: any algorithm which swaps two corners (T-perm, J-perm, Niklas, etc.) is at least 3-gen. It could be RUD/RUF/RUL, but it can’t just be RU. Why is this the case?

So far my thinking is that we can obviously do a 4-cycle of corners in 2-gen, so the problem is that we can’t form commutators of corners 2-gen. But then, I don’t understand why this is the case either. Also this reasoning would imply that edges shouldn’t be able to be cycled either, since we can’t form a standard commutator of edges with just RU (but of course this is false, a counterexample is the RU U-perms).
 

Christopher Mowla

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Also this reasoning would imply that edges shouldn’t be able to be cycled either, since we can’t form a standard commutator of edges with just RU (but of course this is false, a counterexample is the RU U-perms).
Watch from 12:14 to 15:00 of a 4x4x4 2 gen video I made for parity algorithms. You will see that cycling 3 edges in 2 gen is not exactly a commutator. It's actually of the form: A B A' C. So instead of A B A' B' = [A, B], . . . that is, instead of B', we have a sequence C, where C undoes the permutation that A B A' did to the corners (and sequence C happens to just be one turn). So we have the conjugate A B A' and then a (one move) sequence C.

So that's how we can get away with a 3-cycle of edges in <R,U>. Although I proved that all even permutations on the 3x3x3 (and nxnxn) cube can be generated with a single commutator move sequence (which implies that all even permutations are technically commutators), that's the case assuming that you allow all turns. It's not necessarily true that a 3-cycle = a commutator if you restrict moves.

So far my thinking is that we can obviously do a 4-cycle of corners in 2-gen, so the problem is that we can’t form commutators of corners 2-gen. But then, I don’t understand why this is the case either.
To understand this, you can see my reasoning for why we can't perform a 3-cycle of corners in <R,U> in this post. (Lemma 4)

Intuitively I can’t think of a reason, and it seems like the only way to discover this for yourself is to brute force it in some way. Is there an explanation as to why this is the case?
Obviously the above linked post answers this question too. But you can see that Proof 1 kind of required semi-brute force observation that we can only do two of the six possible 4-cycles between those 6 corners. Proof 2 can be observed and may be more of a satisfactory reasoning which you may feel is more straightforward.

but my question is why can’t we change corner permutation with just <R,U>?
I guess by now you have reasoned that you can change corner permutation in <R,U> where I assume you mean change corner parity (as it's obvious that we can permute corners to some extent in <R,U>), because you acknowledged that we can do a 4-cycle of corners in <R,U>, which is an odd permutation like a 2-cycle is. (And you will probably see by now that this portion of your question has now become redundant since the core of your question has already been answered.)
 
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xyzzy

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Let's get someone smart to figure this out!
If nobody else has taken a crack at it by tomorrow, I guess I might as well.

(Btw, for your lower bound, did you account for a scenario where it takes n moves to solve when restricted to domino moves, but possibly less if non-domino moves are allowed?)

Not sure if this deserves its own post, but my question is why can’t we change corner permutation with just <R,U>? Intuitively I can’t think of a reason, and it seems like the only way to discover this for yourself is to brute force it in some way. Is there an explanation as to why this is the case?
An explanation I wrote for someone else who does know group theory. (It doesn't go into the nitty gritty details, but it provides enough information for the reader to do it themself.) Unfortunately, it'll probably go over your head if you're not versed in the terminology, sorry.

Jaap's Puzzle Page has a less dense and probably easier-to-understand proof.

To understand this, you can see my reasoning for why we can't perform a 3-cycle of corners in <R,U> in this post. (Lemma 4)

Obviously the above linked post answers this question too. But you can see that Proof 1 kind of required semi-brute force observation that we can only do two of the six possible 4-cycles between those 6 corners. Proof 2 can be observed and may be more of a satisfactory reasoning which you may feel is more straightforward.
Proof 2 is straight up not even a proof – it relies on the incorrect assumption that any 3-cycle can only be attained via a commutator, something that you yourself even said to not be true in your post above! ("It's not necessarily true that a 3-cycle = a commutator if you restrict moves.")

Proof 1 is… in a very weird position. Strictly speaking, it's also not a proof (it's missing too much detail), but it's also not wrong; with brute force, you can establish that only 2 of the 6 top-face 4-cycles can be reached in ⟨R,U⟩. The problem is that if you're already going to use brute force to show this, it's zero additional effort (!!!) to directly show that exactly 120 corner permutations are reachable in ⟨R,U⟩.

edit (2021-06-09): I originally wrote the below response in a private message to Christopher, because I was trying to not absolutely wreck his reputation, but since he seems more interested in publicly spiting me rather than actually correcting his mistakes, I'm reproducing the message below. (This took hours to write, by the way, and if you know what time zone I'm in, it's obvious I stayed up extremely late trying to explain exactly where his reasoning fell short.) I noticed that he edited his math.se answer to say that his proofs are "sketches", which addresses none of what I said.
No real need to argue this in public, methinks.

My critique of your "Proof 2" is not merely that it's only a sketch of a supposed proof; it's that it's just heading in the wrong direction entirely. Filling in all the details to make it a complete proof would essentially be throwing it out and writing your Proof 1 instead.

>it's "quite trivial" to isolate an edge in 2-gen by using the non-commutator 3-cycle itself as X. [R' U' R' U' R U R U R U', R] So this assumption I mentioned may not actually be true after all. (It in of itself requires proof and therefore you capitalizing on it as an argument against what I wrote was an error from your end.

This does not refute what I said, which was:
>incorrect assumption that any 3-cycle can only be attained via a commutator

What I mean: there exists a move sequence that causes a 3-cycle of pieces, where the move sequence is not a commutator. (We both know this is true, e.g. R U R U R U' R' U' R' U' is not a commutator within ⟨R,U⟩. There does exist a commutator that cycles the same three pieces, but that is irrelevant to what I'm talking about.)
What I think you interpreted it as: there exists a 3-cycle of pieces that cannot be solved with a commutator. (Almost certainly false, as you've noticed!)

Now that I'm thinking about this again, though, I might have made a mistake too, but on a meta level. If we're operating on the assumption that every 3-cycle in ⟨R,U⟩ can be attained via a commutator (#1), then showing that there is no 3-cycle alg is the same as showing that there is no 3-cycle commutator, in which case your refutation makes more sense.

But let's refer to Proof 2 again:

>Since we cannot isolate a corner in the U or R face using < R,U > turns only, we cannot make a 3-cycle corner commutator, as many simple commutators are based on isolating a single piece into a face with moves X and then turning that face for move Y for the commutator [X,Y].

This part of your argument doesn't cover commutators in general – it covers only commutators with one-move interchanges. Even if I grant you the condition that "we cannot isolate a corner in the U or R face" (#2), you have not ruled out the possibility of a 3-cycle commutator where the X and Y parts are both multiple moves long.

To wit, Proof 2 has:
- unproven assumption #1: every 3-cycle that can be solved, can be solved with a commutator
- unproven assumption #2: corners cannot be isolated in a layer
- conclusion that there is no 3-cycle commutator with a one-move interchange (follows from assumption #2)
- invalid conclusion that there is no 3-cycle commutator (true, but does not follow from above)
- invalid conclusion that there is no 3-cycle (follows from above invalid conclusion and assumption #1)

The two assumptions happen to be correct, but they're also very non-obvious. Is there a way to prove them that doesn't just amount to proving Lemma 4 (or Lemma 5) by a separate method first? I can't think of one.

(Assumption #1 does break if we change the context a bit. Here's an example of a permutation group that contains 3-cycles, but none of the 3-cycles are commutators: ⟨(1 4)(2 5)(3 6), (1 2 3)⟩. This is why it needs to be proven – it's not self-evident!)

Maybe I'm barking up the wrong tree and Assumption #1 isn't what you have in mind. If that's the case (and if you're still interested in defending yourself), then you'll have to state what exactly your assumptions are.
 
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WoowyBaby

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If nobody else has taken a crack at it by tomorrow, I guess I might as well.

(Btw, for your lower bound, did you account for a scenario where it takes n moves to solve when restricted to domino moves, but possibly less if non-domino moves are allowed?)

Oh no! You're correct. The lower bound of 8 is flawed, because indeed it's restricting to domino moves. I got this number from Hans Kloosterman's 42 move proof from 1992. Now it's fairly likely that even if you allow non-domino moves, the worst cases won't be able to be solved in fewer than 8 moves, but obviously I'm not totally sure, because for example, HTR state god's number is significantly lower when you allow quarter turns vs. only half turns, so we couldn't know. One face god's number is very likely higher than 8, too. I'm looking forward to your brute-force solution! If you decide to do it, maybe see what is it for just one specific face vs. any face.
 

MethodNeutral

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Thank you both for your input!

Watch from 12:14 to 15:00 of a 4x4x4 2 gen video I made for parity algorithms. You will see that cycling 3 edges in 2 gen is not exactly a commutator. It's actually of the form: A B A' C. So instead of A B A' B' = [A, B], . . . that is, instead of B', we have a sequence C, where C undoes the permutation that A B A' did to the corners (and sequence C happens to just be one turn). So we have the conjugate A B A' and then a (one move) sequence C.

Wow, I subscribed to that channel years ago when I was studying integer power sums and found your videos on them, I never thought I'd meet the channel creator! It's a small world I guess.

I did know of this edge 3-cycle, but this provides some formality to it that I didn't have before. Thank you.

Although I proved that all even permutations on the 3x3x3 (and nxnxn) cube can be generated with a single commutator move sequence (which implies that all even permutations are technically commutators)

This is really cool! Can I still send you some scrambles for you to find the commutator? Also how do you do this, is it by hand or do you have a program to do it for you?

To understand this, you can see my reasoning for why we can't perform a 3-cycle of corners in <R,U> in this post. (Lemma 4)

I didn't quite follow the response posted here, I'll have to look at it some more. Thank you for the resources.

An explanation I wrote for someone else who does know group theory. (It doesn't go into the nitty gritty details, but it provides enough information for the reader to do it themself.) Unfortunately, it'll probably go over your head if you're not versed in the terminology, sorry.

Yes, I have taken a couple of university courses in group theory, but some of the terminology still escaped me. No matter though, I appreciate your help regardless.


This helped a lot! I had already seen the pairs of pieces it describes from practicing the 2GR and yruRU methods, so this was a super straightforward proof (and honestly one I should have been able to come up with on my own if I had given this any thought). Thank you!

@xyzzy, do you have any link to a website explaining the 2GR method? I'd like to look at it some more, and it seems like my link isn't working anymore.
 

Christopher Mowla

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Proof 2 is straight up not even a proof –
Yeah, it's potentially a straight up start to a potential sketch of proof to an alternate proof than Jaap's diagram. If anything can come out of it, it would probably lead to another case-by-case proof, but maybe the number of cases to observe could be less. Don't know, don't care to look into that. (My 2-gen 4x4x4 parity quest ended my fascination with 2-gen.) Maybe someone else can in the future.

Although definitely not an excuse for failing to write Sketch of proof instead of Proof, I should have spent more than a couple of hours writing that entire stackexchange post (and coming up with all of that independently of Jaap in that time frame too), especially since I had zero knowledge of it before that day, but it is what it is. (I was in a hurry to write a response, because that was a Q/A board, not a website where I have all the time in the world to write up something and think it through. Still no excuse for me not editing it after the fact.)

I can't say I neglected the needed details for my commutator proof mentioned in my previous post. I unfortunately spent 2 years (rather than a few hours) on that to come up with two different actual proofs!

it relies on the incorrect assumption that any 3-cycle can only be attained via a commutator, something that you yourself even said to not be true in your post above! ("It's not necessarily true that a 3-cycle = a commutator if you restrict moves.")
I wrote the above post at 3:00 AM when I abruptly woke up from sleep. So I wasn't thinking too far ahead when I wrote:
So that's how we can get away with a 3-cycle of edges in <R,U>.

Because . . . now that I'm wide and awake . . . it's "quite trivial" to isolate an edge in 2-gen by using the non-commutator 3-cycle itself as X. [R' U' R' U' R U R U R U', R] So this assumption I mentioned may not actually be true after all. (It in of itself requires proof and therefore you capitalizing on it as an argument against what I wrote was an error from your end. Just returning the "favor" if that was your intent. I'm not quite sure what your intent actually was because again, you could have simply told me that my Proofs were sketchs of proofs or something along those lines. Saying the below didn't help me confirm your intent either, due to what I comment about what you wrote there . . . mentioning something that I already said as if I didn't mention it.)

but it's also not wrong; with brute force, you can establish that only 2 of the 6 top-face 4-cycles can be reached in ⟨R,U⟩.
Thanks for saying that. I wasn't aware of that at all . . . I just mentioned the following because it was a hunch.
But you can see that Proof 1 kind of required semi-brute force observation that we can only do two of the six possible 4-cycles between those 6 corners.



Wow, I subscribed to that channel years ago when I was studying integer power sums and found your videos on them, I never thought I'd meet the channel creator! It's a small world I guess.
It's hard to believe, but of the little subscribers that I have, this actually happened once before. (The other member was on the twistypuzzles forum I think.

I did know of this edge 3-cycle, but this provides some formality to it that I didn't have before. Thank you.
You will probably see this one (or cyclic shifts of it) around town, but it follows the exact same logic. Just conjugate the move R2 instead of R (and therefore finish with U2 instead of R').
[R' U' R' U' : R2] U2

This is really cool! Can I still send you some scrambles for you to find the commutator?
Why certainly. Preferably 2x2x2-4x4x4 only, as larger cubes than that are more time-consuming to do. For convenience (if you didn't happen to read that entire thread or my post about commutators on math.stackexchange!),
2x2x2 Example | 3x3x3 Example | 4x4x4 Example

Also how do you do this, is it by hand or do you have a program to do it for you?
I do it by hand, but it can be programmed. I use Cube Explorer to reduce the number of moves for X and Y in [X,Y] for the 2x2x2-3x3x3. I have to manually solve a 4x4x4 supercube by hand (using reduction) to reduce the moves of X and Y for the 4x4x4.
 

PapaSmurf

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@xyzzy, do you have any link to a website explaining the 2GR method? I'd like to look at it some more, and it seems like my link isn't working anymore.
Here's the link: https://john-ml.github.io/2GR/2GR.html

There is theory behind the website which I understand informally (ie. I can't "prove prove" it, but I do understand it and can teach it) which makes it arguably easier to learn. Give it a crack though but one word of advice - don't use it for serious speedsolving. I've taked to the creator of the method and he says that CP is never worth it and I would agree. Just learn ZBLL instead and use ZZ or something like that. Don't get me wrong, knowing the method is cool and it's very fun, but it's not fast.
 

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Christopher Mowla

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Could you do a 4x4 with just the UF dedge "flipped", plus a single r move? I think that would be a cool one, I can send you exactly what I'm talking about if that was unclear.
It is by no means short, but I did find this one a little while back.
[2R2 B2 D2 U2 F2 2R2 D2 F2 2L U2 2R2 F2 2L' U2 2R U2 2R', U2 B2 2R' B2 U2 2R U2 2R' B2 2R2 B2 U2 2R2] 2R

Let's get someone smart to figure this out!

I'll start with some basic bounds:

3x3 One Face God's Number:
Lower Bound - 8
Maximum moves required to solve a face already reduced into Domino move set.
Upper Bound - 19
God's Number minus one, because all positions one move away from solved have at least one face solved.

My guess is 13 moves.
I was just searching using Cube Explorer (manually). I know the question was just to solve the first face, not necessarily the first layer. But even with solving the entire face, I couldn't find a position (with about 50 tries) which required more than 11. You mentioned 13 moves. Do you know of a position which requires 12 or 13?
 
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