# Edge Flip and Corner twist Proof?



## PalashD (Oct 6, 2010)

Does anyone know a proof for the fact that on a 3x3x3 cube always an even number of edges are flipped?[Defining flipped as the edges cannot be put to their right position in the right orientation with the <U,D,F2,B2,R,L> subgroup]

Also is there a proof for the fact that if a cw rotation of a corner is put as 1 and a ccw rotation is put as 2 and a correct orientation as 0. Then the summation of the numbers after any moves made on the 3x3x3 is always divisible by 3?


----------



## cmhardw (Oct 6, 2010)

PalashD said:


> Does anyone know a proof for the fact that on a 3x3x3 cube always an even number of edges are flipped?[Defining flipped as the edges cannot be put to their right position in the right orientation with the <U,D,F2,B2,R,L> subgroup]
> 
> Also is there a proof for the fact that if a cw rotation of a corner is put as 1 and a ccw rotation is put as 2 and a correct orientation as 0. Then the summation of the numbers after any moves made on the 3x3x3 is always divisible by 3?


 
There are three different types of quarter turns considering your subgroup. Of interest we have R, F, and U. Any other turn can be considered as either a combination of one of those, reflection or cube rotation, or both, of one of the three mentioned.

U does not change the orientation of any corner or edge, based on your defined orientation scheme. R does not change the orientation of any edge, but it does rotate some corners. After the turn R has been completed, then the corner now at UBR has been rotated clockwise. The corner at DBR has been rotated counter-clockwise. The corner now at DFR has been rotated clockwise, and the corner at UFR has been rotated counter-clockwise. The sum of all clockwise twists is congruent to 0 (mod 3). 

Considering the turn F it changes the orientation of all four edges on the F slice, giving 4 flipped edges based on your orientation scheme. This is congruent to 0 (mod 2). For the corners, after the F turn, UFR is rotated clockwise, DFR rotated counter-clockwise, DFL rotated clockwise, UFL rotated counter-clockwise. The sum of all clockwise twists is congruent to 0 (mod 3).

Every quarter turn either flips an even number, or no, edges. Thus it's not possible to flip an overall odd number of edges on the cube. All quarter turns that rotate corners will perform a number of clockwise twists that is congruent to 0 (mod 3). Thus it is not possible to perform an overall number of clockwise twists that is congruent to either 1 or 2 (mod 3).

Not a precise proof, but this is the basic idea.

Chris


----------



## qqwref (Oct 6, 2010)

Simpler proof for edges: Consider the stickers on the edges, and label each one uniquely. Now, a quarter turn of any layer on any axis does two 4-cycles of edge stickers, which is an even permutation. Every move sequence can be written in terms of quarter turns, so every move sequence performs an even permutation on the edge stickers. But flipping one edge does one 2-cycle of edge stickers, which is an ODD permutation. So no move sequence can have the effect of flipping a single edge (or an odd number of edges, since an odd number of odd permutations is odd).

Incidentally this proof works perfectly (well, with slight variations in some of the numbers I wrote) on similar puzzles, such as the Pyraminx and Megaminx.


----------



## PalashD (Oct 6, 2010)

cmhardw said:


> There are three different types of quarter turns considering your subgroup. Of interest we have R, F, and U. Any other turn can be considered as either a combination of one of those, reflection or cube rotation, or both, of one of the three mentioned.
> 
> U does not change the orientation of any corner or edge, based on your defined orientation scheme. R does not change the orientation of any edge, but it does rotate some corners. After the turn R has been completed, then the corner now at UBR has been rotated clockwise. The corner at DBR has been rotated counter-clockwise. The corner now at DFR has been rotated clockwise, and the corner at UFR has been rotated counter-clockwise. The sum of all clockwise twists is congruent to 0 (mod 3).
> 
> ...


 
Nice!! Thanks!!!


----------



## PalashD (Oct 6, 2010)

qqwref said:


> Simpler proof for edges: Consider the stickers on the edges, and label each one uniquely. Now, a quarter turn of any layer on any axis does two 4-cycles of edge stickers, which is an even permutation. Every move sequence can be written in terms of quarter turns, so every move sequence performs an even permutation on the edge stickers. But flipping one edge does one 2-cycle of edge stickers, which is an ODD permutation. So no move sequence can have the effect of flipping a single edge (or an odd number of edges, since an odd number of odd permutations is odd).
> 
> Incidentally this proof works perfectly (well, with slight variations in some of the numbers I wrote) on similar puzzles, such as the Pyraminx and Megaminx.


nice proof!

can we extend it to corners also?
I mean a quarter turn does 3 4-cycles of edge stickers. And a corner twist can be represented as a 3-cycle of corner stickers. Now since every move sequence can be written in terms of quarter turns. It means that any permutation of the cube can be written as 3x 4-cycles of corner stickers(x being some number). And every n-cycle can be written as n-1 2-cycles. So, a quarter turn gives us 9 2-cycles of stickers(3 disjoint sets of 3 2-cycles). And a corner twist is 2 2-cycles. Can we go down that road and show something?

Also is there a proof that if there is an odd number of even-cycle of edges then there will be an odd number of even-cycle of corners also?


----------



## Stefan (Oct 6, 2010)

cmhardw said:


> There are three different types of quarter turns considering your subgroup. Of interest we have R, F, and U.



I'd say there are just two types. Because U and R are equivalent.



cmhardw said:


> U does not change the orientation of any corner or edge, based on your defined orientation scheme.



He didn't provide a scheme for corners! At least not a complete one. He only talked about cw and ccw *rotations* of corners but failed to consider corners moving to other places! I suggest to complete his definition by saying a corner is (mis)oriented judging by where its F or B sticker is relative to the F or B side (on it, or cw/ccw off). Don't know why you made your life hard by apparently (cause you didn't specify it either) using U/D instead.

And as always: Ryan's Theory pages are helpful.


----------



## qqwref (Oct 6, 2010)

PalashD said:


> nice proof!
> 
> can we extend it to corners also?


Unfortunately, no. A 3-cycle of stickers is even, and if you can reach other even permutations (and other odd ones, since 3 4-cycles is odd) you can't prove that you can't reach this particular one with that technique.

I've been trying to find a good corner orientation proof that will work for many different things (cubes, megaminx, pentultimate...) but I haven't found any luck yet. The best one I know is pretty much what cmhardw said, the proof where you setup a way of measuring orientation and prove that you can't change the sum-of-twists-mod-3, but as you might imagine that is pretty cumbersome to expand to other puzzles.


----------



## cmhardw (Oct 6, 2010)

Stefan said:


> I'd say there are just two types. Because U and R are equivalent.
> 
> He didn't provide a scheme for corners! At least not a complete one. He only talked about cw and ccw *rotations* of corners but failed to consider corners moving to other places! I suggest to complete his definition by saying a corner is (mis)oriented judging by where its F or B sticker is relative to the F or B side (on it, or cw/ccw off). Don't know why you made your life hard by apparently (cause you didn't specify it either) using U/D instead.
> 
> And as always: Ryan's Theory pages are helpful.


 
Wow, Stefan I never thought of using F/B as well for corner orientation. It makes much more sense for proofs like this, and I agree it makes it easier. The reason I used U/D, and should have mentioned it yes, is that back when I used 3OP for blindfolded this is how I thought of it. I oriented edges to L/R and corners to U/D. This mental separation obviously must have become habit. I do like how oriented both corners and edges to FB makes this proof have only 2 possible turn types, we could say U and F.

Chris


----------



## Stefan (Oct 7, 2010)

qqwref said:


> I've been trying to find a good corner orientation proof that will work for many different things (cubes, megaminx, pentultimate...) but I haven't found any luck yet. The best one I know is pretty much what cmhardw said, the proof where you setup a way of measuring orientation and prove that you can't change the sum-of-twists-mod-3, but as you might imagine that is pretty cumbersome to expand to other puzzles.



If you temporarily switch to a nicer orientation scheme for the duration of the turn, the proof generalizes rather easily.

In the picture below, the top-right cube shows the usual reference orientations. The red arrows are the reference orientations for the *places*, these stay fixed. The blue circles are the reference orientations for the *pieces*, these get moved around when turning. The green numbers tell us how far the pieces are (mis)oriented, i.e., how far the blue circles are turned from the red arrows. That's a solved cube and we want to call a solved cube fully correctly oriented, so the arrows and circles all match, the numbers are all zero, and the orientation sum is 0. The goal of the proof is to show that it *always* is 0 (modulo 3).

But let's not use this standard orientation scheme. Let's instead use the one on the top left cube, with just UFR having a different reference orientation. Why? Cause I'm a rebel. And, more importantly, because it illustrates the whole thing better.

Then do an F turn. This moves blue circles around and the green numbers change accordingly. This step is just to get to an unsolved cube, again just to illustrate things better. But note that while the green orientation numbers change, the orientation sum modulo 3 (os%3) doesn't change.

Now... instead of doing an R turn just as quickly, let us look at it differently. For the duration of the turn, let us temporarily switch to a different orientation scheme, where the red reference orientation arrows on the right layer point to the right. This of course also changes the green numbers, as the red arrows move relative to the blue circles. And the os%3 changes accordingly (+2 here). However, and this is important, these *changes* in numbers only depend on how the two orientation scheme differ, how their red arrow directions differ. It is independent of the actual cube state, i.e., where the blue circles are and what the green numbers are.

In the new orientation scheme, the R turn's effect on the orientation numbers is easy to analyze. The green numbers are just moved around, they move with the pieces but don't change, they just end up at other places. What was 0122 before is 2012 now. Of course, the os%3 doesn't change either, then.

After the turn, we switch back to the main orientation scheme. As before, this changes the orientation numbers and the sum. Also as before, this does not depend on the actual state, the *changes* only depend on the difference of the two orientation schemes. Thus, the original os%3 of +2 is reversed, it gets -2 now.

So... overall, instead of directly analyzing the R turn, we did three substeps. The first and third each changed the os%3, but cancel. And the second doesn't change it. So overall, the os%3 didn't change. I hope it is clear that this applies not just to my slightly non-standard orientation scheme (with just an unusual URF) but also to arbitrary others. And also applies to other puzzles, e.g., megaminx. And also applies to edges, not just corners. Just think of using any arbitrary main orientation scheme to define orientations, and for each turn temporarily switching to another scheme where the red arrows all point in the same direction of the layer so that the actual turn just moves the green orientation numbers around without changing them.


----------



## PalashD (Oct 7, 2010)

Stefan said:


> So... overall, instead of directly analyzing the R turn, we did three substeps. The first and third each changed the os%3, but cancel. And the second doesn't change it. So overall, the os%3 didn't change. I hope it is clear that this applies not just to my slightly non-standard orientation scheme (with just an unusual URF) but also to arbitrary others. And also applies to other puzzles, e.g., megaminx. And also applies to edges, not just corners. Just think of using any arbitrary main orientation scheme to define orientations, and for each turn temporarily switching to another scheme where the red arrows all point in the same direction of the layer so that the actual turn just moves the green orientation numbers around without changing them.


 
I did not understand why you switched to a different orientation after the F turn to analyse the R turn? You could have done without it. The objective of the proof was to show that os%3 remains constant. If it is so for one orientation now if you change the orientation(reference) automatically the os%3 will change. But, now if you made a move it will remain constant as it remains constant for the normal(unchanged) reference. I mean why do you want to change the reference scheme here. Is it just because it generalizes in some way to other puzzles?

Also in a 5x5x5 why is it possible to have a single two cycle of edges?
I mean in a 4x4x4 the cube's edges and corners can be represented as


```
C1 E1 E2 C2
E3       E5
E4       E6
C3 E7 E8 C4
```
Now a quarter turn transforms this to

```
C3 E4 E3 C1
E7       E1
E8       E2
C4 E6 E5 C2
```
which is two 4-cylces of edges and a 4-cycle of corners. This is equivalent to
(E1 E5)(E1 E8)(E1 E4)
(E2 E6)(E2 E7)(E2 E3)
(C1 C2)(C1 C4)(C1 C3)

So an odd number of swaps is performed thus odd number of 2-cycles can occur

But I do not understand how this happens in a 5x5x5
There are 3 4-cycles of edges and a 4-cycle of corners.
Thus having 12 2-cycles which should not break down to a single 2-cycle but that happens in the parity case.


----------



## Stefan (Oct 7, 2010)

PalashD said:


> But, now if you made a move it will remain constant as *it remains constant for the normal(unchanged) reference*.



And how do you know that? (the bolded part)



PalashD said:


> But I do not understand how this happens in a 5x5x5



r


----------



## FMC (Oct 7, 2010)

PalashD said:


> But I do not understand how this happens in a 5x5x5
> There are 3 4-cycles of edges and a 4-cycle of corners.
> Thus having 12 2-cycles which should not break down to a single 2-cycle but that happens in the parity case.


you should also consider inner slice turns!


----------



## PalashD (Oct 7, 2010)

Stefan said:


> And how do you know that? (the bolded part)


you can do a R turn(on the normal reference) and see that. No need to change the reference for that. 



Stefan said:


> r


:fp


----------



## Stefan (Oct 7, 2010)

PalashD said:


> you can do a R turn(on the normal reference) and see that. No need to change the reference for that.



And then I've only shown that os%3 miraculously remains zero when doing an R turn from that particular state. I have not shown it for any other turns or for any other states, and more importantly it is not clear why it should be true for them. There'd be no explanation. But the detour through the modified orientation scheme does offer an explanation and shows why/that os%3 remains zero for all turns on all sides for all states. And it shows that it doesn't depend on the orientation scheme (e.g. UD vs FB), also explains the edges, and also explains megaminx and other puzzles. All without analyzing different cases individually (like Chris's six cases, two piece types times three turns) but by showing a single insightful idea.



PalashD said:


> :fp



Inner slice turns are the real reason for it on the 4x4x4 as well, btw. You're right that an outer layer turn is an odd permutation of edge and corner pieces, but it's an even permutation of edges. So that's not how you get the single edges swap there.


----------



## PalashD (Oct 7, 2010)

Stefan said:


> And then I've only shown that os%3 miraculously remains zero when doing an R turn from that particular state. I have not shown it for any other turns or for any other states, and more importantly it is not clear why it should be true for them. There'd be no explanation. But the detour through the modified orientation scheme does offer an explanation and shows why/that os%3 remains zero for all turns on all sides for all states.


but as chris said "Of interest we have R, F, and U. Any other turn can be considered as either a combination of one of those, reflection or cube rotation, or both, of one of the three mentioned." and as you said "U and R are equivalent." and an R or an F on a solved cube does not change os%3. Assuming n R and F moves (allowing reflections rotations in between[since they also do not change os%3]) do not change os%3. Then, if we apply another R or F then again os%3 does not change. So by Induction we can easily say os%3 is conserved. But your proof is much more elegant though it requires switching orientation.



Stefan said:


> Inner slice turns are the real reason for it on the 4x4x4 as well, btw. You're right that an outer layer turn is an odd permutation of edge and corner pieces, but it's an even permutation of edges. So that's not how you get the single edges swap there.



Since it is an even perm of edges and a odd perm of corners. Can we have 2 2-cycles of edges with the outer layers turns only[The perm parity]? I mean can I do something like (E1 E4)(E2 E3) with outer layer turns only. (E1 E2)(E3 E4) is obviously possible[Considering the pervious notation I have used].


----------



## Stefan (Oct 7, 2010)

PalashD said:


> as you said "U and R are equivalent."



In your original orientation scheme, properly completed. Not in mine. And again: I didn't show that F turns *always* preserve os%3, as I only observed the change (of green numbers and os%3) for the specific cube state. I did not, like Chris, analyze why it changed like that (how the red arrows are arranged and how *any* cube state is affected by an F turn). My observation of that F turn wouldn't allow any generalization, and neither would a the likewise observation of the R turn.



PalashD said:


> Since it is an even perm of edges and a odd perm of corners. Can we have 2 2-cycles of edges with the outer layers turns only[The perm parity]? I mean can I do something like (E1 E4)(E2 E3) with outer layer turns only.


 
No.


----------



## TMOY (Oct 8, 2010)

FMC said:


> you should also consider inner slice turns!


 
Actually, on the 5^3 an inner slice turn is an even permutation. It performs one 4-cycle of wings, two 4-cycles of x-centers and one 4-cycle of +-centers. Hence if you want to swap 2 wings you have to swap 2 +-centers as well. But since you can always apply the swap to +-centers of the same colour, ithe apparent result is just a 2-wing swap.


----------



## pyraminx (Oct 8, 2010)

slightly off-topic,sorry:
Is there any proof showing that a 90 degree turn of centre facelet on a 3x3x3 super cube is impossible?
can we have a general result which tells the impossible centre facelet turns an nxnxn super cube?


----------



## Stefan (Oct 8, 2010)

pyraminx said:


> Is there any proof showing that a 90 degree turn of centre facelet on a 3x3x3 super cube is impossible?


 
Center orientation parity always matches corner permutation parity.


----------



## pyraminx (Oct 8, 2010)

Stefan said:


> Center orientation parity always matches corner permutation parity.


 
ok..if i understand well,a more general way to state this is
center orientation parity matches with corner permutation or edge permutation(correct me if i am wrong)
but how can we extend to NxNxN super cube?
first of all we should have a definition of orientation of center as the centers 
also permute


----------



## Stefan (Oct 8, 2010)

pyraminx said:


> ok..if i understand well,a more general way to state this is
> center orientation parity matches with corner permutation or edge permutation(correct me if i am wrong)



And. Not or. Bot corner permutation parity and edge permutation parity match center orientation parity. But one is enough for a proof. I chose corners because there are fewer and they're in a simpler shape.



pyraminx said:


> but how can we extend to NxNxN super cube?
> first of all we should have a definition of orientation of center as the centers
> also permute



Since they can't rotate in place, it's enough to consider only their permutation.


----------



## qqwref (Oct 8, 2010)

Stefan said:


> If you temporarily switch to a nicer orientation scheme for the duration of the turn, the proof generalizes rather easily.
> 
> [...details...]
> 
> So... overall, instead of directly analyzing the R turn, we did three substeps. The first and third each changed the os%3, but cancel. And the second doesn't change it. So overall, the os%3 didn't change.



It took me a while to work through your post, but I think basically you're saying that:
- changing the orientation scheme changes co%3 by a constant (the "difference" between the two schemes)
- while in an orientation scheme which is rotationally symmetric on one layer, turning that layer does not affect co%3
- so, we can prove that each turn does not affect co%3 by changing to an orientation scheme symmetric about that turn's layer, doing the turn, and changing back, for a net co%3 affect of 0.

Interesting idea, for sure. And the really cool thing is that it works for any puzzle where a rotationally symmetric orientation scheme is possible, and it should work for EO as well. So this proof even works for Pentultimate and Helicopter Cube; you can only not find such a scheme on a corner turning puzzle, which is in my experience the only puzzle type which allows a single corner twist (bar Skewb type puzzles, which are a notable exception). And the EO version works for everything which isn't edge turning (although mine does too), which again is the only puzzle type which allows a single edge twist.


----------



## Stefan (Oct 8, 2010)

Yeah, that's it. I like the term "rotationally symmetric", good way to put it. I know it was long, but I hope it was worth reading . And yeah, I think I originally came up with this for megaminx corner orientation a while back and was very pleased that it is so general and applies to many things. I really wasn't keen on defining certain arbitrary orientation schemes for every single puzzle and analyzing them one turn at a time, like Chris did for 3x3x3 early in this thread. I mean, it's not *that* bad, but it's still ugly repetitive work.



qqwref said:


> And the EO version works for everything which isn't edge turning (although *mine* does too)


 
What are you referring to there?


----------



## RussianWhiteBoi (Sep 4, 2011)

I just had to bump this thread because this post is genius!



qqwref said:


> Simpler proof for edges: Consider the stickers on the edges, and label each one uniquely. Now, a quarter turn of any layer on any axis does two 4-cycles of edge stickers, which is an even permutation. Every move sequence can be written in terms of quarter turns, so every move sequence performs an even permutation on the edge stickers. But flipping one edge does one 2-cycle of edge stickers, which is an ODD permutation. So no move sequence can have the effect of flipping a single edge (or an odd number of edges, since an odd number of odd permutations is odd).
> 
> Incidentally this proof works perfectly (well, with slight variations in some of the numbers I wrote) on similar puzzles, such as the Pyraminx and Megaminx.


----------



## Godmil (Sep 4, 2011)

Well I'm glad for the bump. Really interesting thread.


----------



## Christopher Mowla (Nov 18, 2011)

I am not saying this is correct, but this might be a corner twist proof similar to qqwref's edge flip proof.

Axiom 1:
The identity (solved cube) is an even permutation of fixed centers (or X-center pieces on big even cubes) by the properties of a supercube.

Axiom 2:
If the fixed centers (or X-center pieces on big even cubes) are in an odd permutation, then the minimum number of corners which can be unsolved is 2, and the minimum number of edges which can be unsolved is 2. (Think of a J-Perm, T-Perm, etc.)

Axiom 3:
When the centers are in an odd permutation, one can twist the corners with algorithms just as one can twist the corners when the cube is in an even permutation (the ability to twist corners does not depend on whether the permutation of the centers is even or odd).


The the sum of clockwise (or anti clockwise, but not both) twists of corners is always evenly divisible by 3.
*Proof*
Consider the stickers on the corners and label each one uniquely. Each corner has 3 stickers.
By axiom 2, there is a minimum of 2(3) = 6 corner stickers which must be unsolved if the fixed centers (or X-center pieces on even cubes) are in an odd permutation.

By axiom 1, when there is an even permutation of centers, a minimum of 0 corners and 0 edges can be unsolved (despite that the edges and corners are independent from each other when the centers have an even permutation).

Note that 3 corner stickers is neither the minimum amount to be unsolved for even or odd permutations.

Now, to take care of the last case, that is, to determine the second to least minimum number of corner stickers which can be unsolved for even permutations of centers,

When the centers are in an odd permutation, since the minimum number of corner stickers to be unsolved is 6, then it is not possible to have only 3 corner stickers unsolved. This implies that one corner cannot be twisted once (in either direction) when the centers are in an odd permutation.
By axiom 3, the previous statement implies that 3 corner stickers cannot be unsolved when the centers are in an even permutation.

To take care of all other cases for which the sum of the clockwise (or anti clockwise, but not both) twists of corners is not evenly divisible by 3, it follows that since one cannot twist a single corner (in either direction), one cannot generate any other of the possible situations in which the sum of clockwise (or anti clockwise, but not both) twists of corners is not evenly divisible by 3.

Lastly, notice that even if this proof were true (it might be, but...), it does not fully justify the 2x2x2 case. However, note that a quarter turn (say, R) does a 3 4-cycle of corner stickers (an odd permutation), and R2 does a 6 2-cycle of corner stickers (an even permutation). Hence the centers in an odd permutation if and only if the corner stickers are in an odd permutation (so you can just substitute "centers in an odd permutation" with "corner stickers in an odd permutation" for the 2x2x2 case to generate both the axioms and the proof). I didn't do this so that it's easier to see.

Feedback anyone?


----------



## Lucas Garron (Nov 18, 2011)

cmowla said:


> Feedback anyone?



I think your axioms are more like lemmas. An axiom is something you want to be true but can't prove (so you assume it).

(Also, I think a fixed-corner proof would be simplest.)


----------



## Stefan (Nov 18, 2011)

You seem to forget the case where two corners are swapped and one of them is misoriented.

edit: lolfact: While I still had this thread open, I played with my cube and somehow managed to misorient one corner.


----------



## Christopher Mowla (Sep 6, 2014)

qqwref said:


> Simpler proof for edges: Consider the stickers on the edges, and label each one uniquely. Now, a quarter turn of any layer on any axis does two 4-cycles of edge stickers, which is an even permutation. Every move sequence can be written in terms of quarter turns, so every move sequence performs an even permutation on the edge stickers. But flipping one edge does one 2-cycle of edge stickers, which is an ODD permutation. So no move sequence can have the effect of flipping a single edge (or an odd number of edges, since an odd number of odd permutations is odd).


qqwref, I believe your edge orientation proof proves corner orientations as well.

For those who cannot visualize what a 2 4-cycle of edges stickers is, here is an image. (The image below illustrates what happens to the edge stickers in slice R when we do the move R to a solved 3x3x3 cube.)






When you mentioned


qqwref said:


> But flipping one edge does one 2-cycle of edge stickers, which is an ODD permutation






, we cannot say that twisting a corner is a 3-cycle of corner stickers as we could say that flipping an edge is a 2-cycle because there are two different 3-cycles in 3 objects (where as there is only one unique 2-cycle in 2 objects--for the edge).

To keep a corner "intact", we cannot move one sticker without removing the remaining two. (We could keep an edge "intact" by choosing to switch its stickers as you illustrated in your proof.)

Therefore there is only one version of a "3-cycle" of corner stickers in a corner, and thus we can abstract any twist of a single corner as the 2-cycle (sticker A joined with sticker B)↔(sticker C).


----------



## elrog (Sep 17, 2014)

Interesting read!

cmwola, as far as I can tell your proof is correct.


----------



## Herbert Kociemba (Sep 17, 2014)

cmowla said:


> Therefore there is only one version of a "3-cycle" of corner stickers in a corner, and thus we can abstract any twist of a single corner as the 2-cycle (sticker A joined with sticker B)↔(sticker C).



Hm, I am not convinced. What is the difference between a clockwise twist and an anticlockwise twist using your 2-cycle?
And what happens, if you do three times the same twist? This should be equivalent to no twist - but three 2-cycles cannot cancel.


----------



## Christopher Mowla (Sep 17, 2014)

Herbert Kociemba said:


> Hm, I am not convinced. What is the difference between a clockwise twist and an anticlockwise twist using your 2-cycle?
> And what happens, if you do three times the same twist? This should be equivalent to no twist - but three 2-cycles cannot cancel.


Although I did not claim it to be a proof, below is an image I made which illustrates what went through my mind when I posted, at least.




As far as the = (rotate) sign, let's look at the counterclockwise rule (in the image above), for example.

Let 1 = the white sticker, 2 = the red sticker, and 3 = the green sticker on the solved cube to the left. We have the list {1,2,3} when the corner is solved.

For the counterclockwise rule, we say we have {1,{2,3}}. 

*Interpretation 1*
When we want to swap either 2 or 3 with 1, we have {{2,3},1} = {2,3,1}, which is a 3-cycle.

*Interpretation 2*
When we want to swap either 2 or 3 with 1, we have {{2,3},1} = {1,{3,2}} = {1,3,2}, which is a 2-cycle. (When you swap two objects, there isn't a rule that says that you cannot change the orientation of one or both of the objects: as long as you exchange their locations, the swap is valid.)

That is, as you can see after the = (rotate) sign in the counterclockwise example, we can either view us swapping the green and red stickers (flip the 1x2 bar of the green and red sticker horizontally) and then swapping it with the white "rectangle"/square (flip the two bars vertically) (2-cycle),
or
we can view it at rotating both rectangles 180 degrees (3-cycle).

I'm not disputing your mention of the 2-cycle being repeated 3 times is not the identity permutation, but we cannot rule out "interpretation 2" just because it "behaves" like a 3-cycle since when either the clockwise or counterclockwise rule is applied three times, it returns the stickers to their original state. (If this was easy to see, I'm sure someone would have posted it earlier.)

Lastly, when I mentioned that we cannot view a corner twist as a 3-cycle (which is incorrect from one interpretation), I had in mind that since a 3-cycle is a composition of two overlapping 2-cycles, then if we can do a 3-cycle, we should also be able to do just a 2-cycle. However, I just realized that when qqwref mentioned that a quarter face turn does a 2 4-cycle of edge stickers, we cannot do one 4-cycle, as this destroys edge stickering. (In short, disregard that statement.)


----------



## qqwref (Sep 17, 2014)

I'm not sure if this helps because you have to "redefine" the corners after every move. That is, when you move a corner along two different axes, you now have to choose the pair differently to consider it as a swap. So I don't think you can apply a sequence of moves and meaningfully ask whether the total orientation of corners is an odd permutation (and thus whether one corner is twisted).


----------



## Christopher Mowla (Sep 18, 2014)

qqwref said:


> I'm not sure if this helps because you have to "redefine" the corners after every move. That is, when you move a corner along two different axes, you now have to choose the pair differently to consider it as a swap. So I don't think you can apply a sequence of moves and meaningfully ask whether the total orientation of corners is an odd permutation (and thus whether one corner is twisted).


I'm confused (I'm not calling you wrong, I just really don't understand) as to why we need to see how the moves R U, for example, affect the orientation of the ufr corner in terms of 2 to 1 corner sticker pairing when we can observe that the move R and the move U do a 2 4-cycle of corner stickers independently.

Again, if my interpretation about a corner twist being a 2-cycle is incorrect, then I understand completely, and we can put this one approach for finding another proof for corner orientations behind us. However, if I successfully justified my 2-cycle interpretation, then why can't we treat corner orientations just like how you treated middle edge orientations using just one face quarter turn?


----------



## Herbert Kociemba (Sep 20, 2014)

I think I can prove a quite general theorem about the orientations of a permutation puzzle under the following conditions:

1. Applying a move to the puzzle does not change its shape.
2. If a piece p has k possible orientations in a place P, p has a k-fold rotational symmetry axis which defines the k possible orientations.

For a move M we call a place P M-unambiguous if it is not possible to apply only M several times such that piece p moves to P again, but in a different orientation.

Then for an arbitrary orientation scheme for the pieces we have:

1. If a place P is M-unambiguous and p a piece in P with k possible orientations, the sum of the orientations of all pieces which are in the same orbit (with respect to M) as p does not change modulo k if M is applied.

Simple but important conclusions are

2. If a place P is unambiguous for several moves M1, M2,... Mn and p a piece in P with k possible orientations, then the sum of the orientations of the pieces in the union of the orbits does not change modulo k if any move M1,.. Mn is applied.

and

3. If place P is unambiguous for all moves of a puzzle it is impossible to rotate a piece in P in place without changing the orientation of other pieces.

For nxnxn Cubes all conceivable moves M are M-unambiguous for all places P for example (except for odd n the centerplace in the middle of a face).

Edit: I now put the proof here.
Edit2: And without using the "unambiguous" term we have:

Let us have a rotational permutation puzzle which does not change its shape and where the moves are spacial rotations of some of its pieces. Let a certain kind of piece have a k-fold rotational symmetry which allows it to take k different orientations within a place.
Then if these pieces do not intersect with the axes of the spacial rotations of the moves the sum of the orientations of these pieces modulo k is an invariant.


----------

