# Solving 4x4 on your own



## Zane_C (Oct 31, 2009)

Hey,
I have a 4x4x4 cube and haven't been spending much time on it. I haven't bothered to learn a method and I play around with it every now and then. So far I can intuitively solve the first 3 layers, all corners, but havn't quite figured out how to orient and/or permute individual last layer edges.
The closest I've got to it being solved was all but 2 pieces solved, they were oriented correctly but needed to switch places.

I was wondering who here managed to solve the 4x4 all by themselves?
If so, how long did it take you?
And how fast can you solve it now? (perhaps even with your own method)


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## Tim Major (Oct 31, 2009)

Google, "4x4 Redux method." I don't know of people solving it like this. It is very different to a 3x3, but you end up solving it as a 3x3.


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## mcciff2112 (Oct 31, 2009)

What you have is a parity. And there's nothing you can do without an algorithm (unless, of course, you could fix it intuitively, which is close to impossible to do). If you wanna solve it on your own, I would rescramble it and hope for the best that you don't get a parity case.


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## Lucas (Oct 31, 2009)

> (unless, of course, you could fix it intuitively, which is close to impossible to do)



What you are calling impossible is what I do with every puzzle I solve on my own "intuitively" (for example the parity of square-1). It just having knowledge, understanding some "secrets" of twisty puzzles.

I would use my combination of a parity fix move + commutators to solve the 4x4

I remember that when I first solved 4x4 on my own I did the reduction about 4 times (messed up a bit the centers) until I got no parities; it took me about 30 minutes. But now I would have done it as I said above.


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## mcciff2112 (Oct 31, 2009)

Lucas said:


> > (unless, of course, you could fix it intuitively, which is *close to* impossible to do)
> 
> 
> 
> What you are calling impossible is what I do with every puzzle I solve on my own "intuitively" (for example the parity of square-1). It just having knowledge, understanding some "secrets" of twisty puzzles.



I never said it was impossible. I know that for a beginner it would be VERY difficult to successfully fix a parity case nusing pure knowledge of a cube. I couldn't even do it myself with my knowledge of cubes. I know there are people like you who are able to apply a more advanced knowledge to allow them to get past any challenge, so please don't take my message that way. You jumped to conclusions way too quickly there.


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## Lucas (Oct 31, 2009)

mcciff2112 said:


> Lucas said:
> 
> 
> > > (unless, of course, you could fix it intuitively, which is *close to* impossible to do)
> ...



Sorry, I didn't want to mean that... I meant that it was not SO hard. Perhaps I can do it easily because I know about cycles and commutators because of learning blindfolded.


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## cmhardw (Oct 31, 2009)

If you understand what parity is, then it is quite easy to solve. Having to switch two wing edges means that the wing edges are in an odd permutation. What this means is that it requires an *odd number* of inner slice quarter turns to place them into an even permutation (and thus, easily solvable with almost any intuitive or regular speedsolve method). The "difficult" part is that at this point in your solve your center pieces are usually completely solved. So I propose to you a challenge that will fix the parity of the wing permutation to be even (and thus you won't have any strange problems when trying to solve them).

*Can you scramble and resolve centers using an odd number of inner slice quarter turns?* It's ok if this messes up some other things on the cube, as long as centers end up solved again.

Hint: It only takes 9 overall turns to do this

Chris


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## Zane_C (Oct 31, 2009)

thanks, you all seem to know what your talking about, I will get to it now and figure out ways of fixing the parity.


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## AvGalen (Oct 31, 2009)

Zane_C said:


> Hey,
> I have a 4x4x4 cube and haven't been spending much time on it. I haven't bothered to learn a method and I play around with it every now and then. So far I can intuitively solve the first 3 layers, all corners, but havn't quite figured out how to orient and/or permute individual last layer edges.
> The closest I've got to it being solved was all but 2 pieces solved, they were oriented correctly but needed to switch places.
> 
> ...


Great that you actually tried to solve this all on your own. It is called a PUZZLE for a reason. Getting to all but 2 pieces solved is very impressive. You have solved all, but the hardest sub-puzzle of any sized cube (parity)
This parity case occurs 50% of the time. There is another one that might give you problems as well that also occurs 50% of the time. All this means that you could just rescramble and hope for the best (at least 25% of the time this will work!). Chris tip about the odd number of quarter-turn-slice-turns is indeed the missing link.
I tried to solve 4x4x4 on my own using layer by layer about 4 years ago and could also do it untill the parities. I first did the rescramble-and-hope-for-the-best thing.
Then I decided to first do centers layer by layer, then create "big edges" and then solve layer-by-layer (found out this is Pochmann-centers and reduction later). This still gave me problems with the last edges, but eventually I found a way to solve this problem


Spoiler



Do a D-move, then restore the middle 2 layer ceners with (dR2d')(u'R2u) (d2B2d2)(u2B2u2) (dF2d')(u'F2u)


I never figured out how to do OLL-Parity myself which is a really good thing because while looking for that on the internet I discovered the whole speedcubing community. It wasn't untill years later that I understood what parity actually is.



ZB_FTW!!! said:


> Google, "4x4 Redux method." I don't know of people solving it like this. It is very different to a 3x3, but you end up solving it as a 3x3.


Please don't tell people that ask for experiences with a puzzle-method to look up a speedsolving method.



Lucas said:


> ...Sorry, I didn't want to mean that... I meant that it was not SO hard. Perhaps I can do it easily because I know about cycles and commutators because of learning blindfolded.


Yes, knowing about blindfolded (cycles+setupmoves) is bascially the key to solving every puzzle. If you can find a couple of sequences that only effect a couple of pieces on the puzzle solving it becomes trivial


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## Zane_C (Oct 31, 2009)

Thanks your've given me hope, I will just scramble it until parity is fixed, then I will be happy! I won't look up a method until I've solved it on my own.


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## Cyrus C. (Oct 31, 2009)

I solved it intuitively, takes waaaaayyy to long. I am now sub-2 minutes.


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## Lucas (Oct 31, 2009)

cmhardw said:


> If you understand what parity is, then it is quite easy to solve. Having to switch two wing edges means that the wing edges are in an odd permutation. What this means is that it requires an *odd number* of inner slice quarter turns to place them into an even permutation (and thus, easily solvable with almost any intuitive or regular speedsolve method). The "difficult" part is that at this point in your solve your center pieces are usually completely solved. So I propose to you a challenge that will fix the parity of the wing permutation to be even (and thus you won't have any strange problems when trying to solve them).
> 
> *Can you scramble and resolve centers using an odd number of inner slice quarter turns?* It's ok if this messes up some other things on the cube, as long as centers end up solved again.
> 
> ...



I think that would be the easier intuitive way to fix the parity and rebuild centers. And I just needed to do one edges pairing after that.


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## Cride5 (Oct 31, 2009)

cmhardw said:


> If you understand what parity is, then it is quite easy to solve. Having to switch two wing edges means that the wing edges are in an odd permutation. What this means is that it requires an *odd number* of inner slice quarter turns to place them into an even permutation (and thus, easily solvable with almost any intuitive or regular speedsolve method). The "difficult" part is that at this point in your solve your center pieces are usually completely solved. So I propose to you a challenge that will fix the parity of the wing permutation to be even (and thus you won't have any strange problems when trying to solve them).
> 
> *Can you scramble and resolve centers using an odd number of inner slice quarter turns?* It's ok if this messes up some other things on the cube, as long as centers end up solved again.
> 
> ...



Despite quite a thorough search, I haven't been able to fine a proper explanation for OLL parity. As far as I can see (by looking at the cube), OLL parity is when only two dedges are swapped. The reason this maifests its self as an orientation problem is that when a pair of dedges are swapped in place, then they must also be flipped too because the internal mechanics of the pieces wouldn't allow otherwise.

My question is: How can only two dedges be swapped, and nothing else. Its counter to my intuition on how twisty puzzles operate. Does OLL parity possibly mean that two centre pieces are also swapped? If anyone could shed light on this I'd be very interested. I'd like to be able to solve OLL parity intuitively, but I've no idea how to at the moment


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## trying-to-speedcube... (Oct 31, 2009)

Well, if you do r, you have done an odd permutation off edges (a 4-cycle), but an even permutation of centers (2 4-cycles). That means that edge parity is not related to the center parity. The corner parity, however, *is* related to the center parity; if you do R, you have done an even permutation of edges (2 4-cycles), an odd permutation off corners (a 4-cycle) and an odd permutation of edges (a 4-cycle). That's why the OLL parity can be pure (on a 4x4).


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## Cride5 (Oct 31, 2009)

trying-to-speedcube... said:


> Well, if you do r, you have done an odd permutation off edges (a 4-cycle), but an even permutation of centers (2 4-cycles). That means that edge parity is not related to the center parity. The corner parity, however, *is* related to the center parity; if you do R, you have done an even permutation of edges (2 4-cycles), an odd permutation off corners (a 4-cycle) and an odd permutation of edges (a 4-cycle). That's why the OLL parity can be pure (on a 4x4).



Ah right, that makes sense. I checked the centres and found that OLL parity indeed doesn't affect them. So by what you were saying, does that mean any scramble + solve that involves an odd number of outer-layer turns (QTM) will result in OLL parity?


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## trying-to-speedcube... (Oct 31, 2009)

No, any scramble + solve that involves and odd number of inner-layer turns (QTM) will result in OLL parity.


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## rubixfreak (Oct 31, 2009)

http://web.comhem.se/solgrop/kub/4x4x4_PLL.htm will solve your problem.

its a site about how to solve the top edges of the 4x4x4 cube


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## cmhardw (Oct 31, 2009)

To expand on what Maarten said I have included a proof about the 4x4x4 cube below. Also, Maarten :fp I did not see your post before I started typing my own, so I apologize for the repeat parts.



Cride5 said:


> Despite quite a thorough search, I haven't been able to fine a proper explanation for OLL parity. As far as I can see (by looking at the cube), OLL parity is when only two dedges are swapped. The reason this maifests its self as an orientation problem is that when a pair of dedges are swapped in place, then they must also be flipped too because the internal mechanics of the pieces wouldn't allow otherwise.



Mostly correct. "Parity" is the state of a permutation of pieces, however we misuse this term to describe the state when the parity is odd (i.e. the permutation is in an odd permutation state). The proper term we should use is "wings have odd parity" or something like this. However, we incorrectly say "wings have parity", or "the cube has parity". In fact, each group of pieces on the cube *always* have parity, because pieces are always in either an even or odd permutation.

Your assertion that "odd parity" in the wings asserts itself as two wings swapped is the only part I truly, and completely, disagree with. I'll give you another example of odd parity. From a solved cube do the turn *r*. You cube now has OLL parity, by the true definition of what OLL parity is. So as not to be a complete stickler for terminology, I will add to your definition of OLL parity the additional: "When the wing edges are in an odd parity state when the cube is being solved with the reduction method, and centers are solved and edge groups are built, then this manifests itself as an impossible edge orientation case, considering the rules of a 3x3x3 cube."



> My question is: How can only two dedges be swapped, and nothing else. Its counter to my intuition on how twisty puzzles operate. Does OLL parity possibly mean that two centre pieces are also swapped? If anyone could shed light on this I'd be very interested. I'd like to be able to solve OLL parity intuitively, but I've no idea how to at the moment



This is my favorite question in cubing theory, and it's the reason why I love the 4x4x4 more than any other n x n x n cube. Let's for a second not only consider the 4x4x4 cube, but the 4x4x4 supercube. On the 4x4x4 supercube it is indeed possible to switch any two wing edges, and *affect no other part of the cube at all!* Not only is that true, but the 4x4x4 supercube is the only n x n x n supercube where you can swap any two wing edges in the same orbit without affecting any other piece on the cube. For all other cube sizes that have edge pieces, swapping only two of those edges *must have an effect on some other pieces of the cube!*

*Here is the proof that it is possible to swap two wings on a 4x4x4 and not affect any other pieces on the cube:*
When turning an inner layer of a 4x4x4 supercube by one quarter turn you are affecting four wing edges, and you are 4-cycling them (toggling the permutation parity, and yes I mean parity in the correct sense of the word here). You are also creating two 4-cycles of centers that are in the same orbit. This net effect is an even permutation, so the parity of the centers remains the same. From here you can use commutators and 3 cycles to return all the centers to their correct location (all even parity positions are achievable using 3 cycles) and you can do 1 commutator on the wing edges to solve 2 of the now unsolved pieces back to their original location. This leaves 2 wing edges swapped, and everything else in its original location.

*Proof for why this does not work on any other cube size:*
When turning an inner layer slice by one quarter turn on a cube larger than 4x4x4 there are now two types of center pieces. There are the "x" centers, and the "oblique centers" as I call them (you can rename them how you like, but I define my terms in the next sentence). Each inner slice contains exactly 8 "x-centers" from the same orbit, and only from one orbit of "x-centers". Each inner slice contains exactly 4, and no more, "oblique centers" from the same orbit, for 1 or more different orbits. I could provide you with the formula for how many orbits exactly, but I just want you to get the basic idea.

Here's how this affects the pieces when doing a single quarter turn of an inner layer. When turning the inner layer the wing edges are 4-cycled (toggling the parity). The "x-center" orbit that is affected has had two 4-cycles of pieces created which is a net even permutation. Parity of this orbit is not affected. Now, all the remaining center pieces on this slice are "oblique" centers, and each orbit contained on this slice has only 4 of its pieces on that slice. By doing the inner slice quarter turn you create a 4-cycle of pieces in every single one of these orbits, thus toggling the parity of the permutation in every single one of those orbits.

Therefore the 4x4x4 supercube is the only supercube where the parity of the wing edge permutation is completely independent of the parity of all other orbits on the cube.

Hope this helps, and yes I agree that the 4x4x4 is the most beautiful cube from a theory standpoint for exactly this same reason.

Chris


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## cmhardw (Oct 31, 2009)

trying-to-speedcube... said:


> No, any scramble + solve that involves and odd number of inner-layer turns (QTM) will result in OLL parity.




This is actually the idea behind my method for always avoiding the OLL parity case (and yes I am back to misusing that term colloquially haha). If you count the cycles of the wings during inspection in a BLD sense you can determine if the cube has odd or even wing edge parity. You then solve in such a manner as to ensure even wing parity, thus avoiding OLL parity.

However, I found that it is incredibly difficult to do this during the inspection time. After much practice I am down to maybe 25 seconds record for a single solve. However, there is a shorter way to do this than counting cycles of the 24 pieces, and more like counting the 12 cycles of a 3x3x3. I can explain this method if anyone is interested. Based on how fast the truly world class BLD people are at memorzing an entire 3x3x3, perhaps it is not so far fetched to count cycles using my shortcut pseudo-3x3x3 edge method to count cycles on a 4x4x4 and always avoid OLL parity.

Chris


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## Cride5 (Oct 31, 2009)

cmhardw said:


> To expand on what Maarten said I have included a proof about the 4x4x4 cube below. Also, Maarten :fp I did not see your post before I started typing my own, so I apologize for the repeat parts.
> 
> 
> 
> ...


#

Chris, you have really shed a lot more light on this, many thanks for the post! I'm reading this at a Halloween party, and I've had too many to really take in the stuff about permutation of the centre pieces (I'll have another read when i'm sober).

One further question, when you say this doesn't apply to larger dimensions of cube, are you also including the 6x6? My understanding was that the inner-slice edges of a 6x6 behave in the same way as those of a 4x4..

Thanks again for the explanation, you have a gift for teaching


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## Zane_C (Nov 1, 2009)

Ok Chris, I've turned the inner layers like you said, and after some thinkin' and some time, I realised I paired up all the edges!
I then started to solve all the centers and corners. With my adrenilan pumping, before I knew it, all centers, edges and corners were solved!....

Well, that's what I first thought, until I rotated the cube.
I now have another parity, 2 ADJACENT corners now need to be SWAPPED!!!!!!!!! 
This is driving me INZANE!


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## Zane_C (Nov 1, 2009)

If I scramble it like a 3x3x3 (no inner slices), then solve it like a 3x3x3, is there a chance it will fix problem?


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## masterofthebass (Nov 1, 2009)

no, you have what is known as "PLL parity". It actually is just 2 2-cycles of edges. You can solve it by solving everything but 2 edges, and then use intuitive commutators to place the edges correctly (while obviously breaking something else up).


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## Zane_C (Nov 1, 2009)

Maby I would do that, but my understanding of commutators is very limited


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## masterofthebass (Nov 1, 2009)

I highly doubt you were able to get this far without doing commutators. Just try and solve single edge pieces.


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## cmhardw (Nov 1, 2009)

To be honest there is not often discussion on an intuitive way to solve this case. I figured out an algorithm intuitively when I first started solving the 4x4x4 that I think is a good one for this case to understand what you are doing. I won't give you the algorithm itself, but rather the first step of the thought process I had. If you want to find this algorithm online, it is quite easy to do so, but if you want the puzzle challenge then read below.

*Cubing observation:*
On a 3x3x3 doing the algorithm R2 U2 R2 U2 R2 U2 achieves an interesting effect by swapping two pairs of edges.

r2 U2 r2 U2 r2 U2 (turn inner slices only on the lower case *r* turns) has a somewhat nice effect on a 4x4x4 that seems to fix this "PLL parity" or PLL error, but it messes up centers.

Also try R2 Uw2 R2 Uw2 R2 Uw2 (Uw2 means to turn the double layer, or both the outer and inner U layers) which is even a bit better than the algorithm above. This swaps the two edge pairs, but again messes up some centers.

I leave it to you to either restore centers, or find a way to change the application of the R2 U2 R2 U2 R2 U2 concept such that it *does not* destroy centers. Either of those options would result in an intuitive way to fix this case.

Chris


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## bigbee99 (Nov 1, 2009)

This is off topic, but can anyone tell me how to post a new thread?


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## phases (Nov 1, 2009)

bigbee99 said:


> This is off topic, but can anyone tell me how to post a new thread?




At the very top left, hit Home. On the left (also very top left, where Home was that you clicked, that changes to "Forum" too), hit "Forum". Pick the appropriate forum and click into it. Then top left click New Thread.


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## Zane_C (Nov 2, 2009)

Blast! I tried the first alg but while fixing up the centers in the 2nd layer it swapped the last layer edges.
I disturb the last layer edges when:
- I solve the d layer centers
- I solve the u layer edges.

Thanks very much anyway, what centers does the other algorithm disturb?


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## Zane_C (Nov 2, 2009)

Now I'm back with 2 edges needing to be swapped


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## Zane_C (Nov 2, 2009)

next time I will use the first algorithm.


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## Cride5 (Nov 2, 2009)

cmhardw said:


> *Proof for why this does not work on any other cube size:*
> When turning an inner layer slice by one quarter turn on a cube larger than 4x4x4 there are now two types of center pieces. There are the "x" centers, and the "oblique centers" as I call them (you can rename them how you like, but I define my terms in the next sentence). Each inner slice contains exactly 8 "x-centers" from the same orbit, and only from one orbit of "x-centers". Each inner slice contains exactly 4, and no more, "oblique centers" from the same orbit, for 1 or more different orbits. I could provide you with the formula for how many orbits exactly, but I just want you to get the basic idea.
> 
> Here's how this affects the pieces when doing a single quarter turn of an inner layer. When turning the inner layer the wing edges are 4-cycled (toggling the parity). The "x-center" orbit that is affected has had two 4-cycles of pieces created which is a net even permutation. Parity of this orbit is not affected. Now, all the remaining center pieces on this slice are "oblique" centers, and each orbit contained on this slice has only 4 of its pieces on that slice. By doing the inner slice quarter turn you create a 4-cycle of pieces in every single one of these orbits, thus toggling the parity of the permutation in every single one of those orbits.



Is this what you mean when relating the parity of centre pieces and edge pieces?







So once the centres are solved on an odd-dimensional cube, we can guarantee that the edges must be in an even parity state, because the centres must be. If the centres are in an odd parity state then they cannot be solved. On an even-dimensional cube, inner-slice turns do not affect the parity state of the centre pieces, while the parity state of the edges flips from odd to even for every inner-slice quarter turn. This means there is a 50% chance that the edges are in an odd parity state after solving the centres. After edge pairing, the odd parity state will manifest its self as a single pair of swapped edges (which appears as a single flipped edge pair after reduction).

So I hope thats about right for OLL parity, but I'm still not certain about PLL parity. As Dan pointed out, its 2x edge 2-cycles. If this is the case, is PLL parity fixed at the end of edge pairing? Ie, would it be possible to go back to the edge pairing phase to fix PLL parity? Again, PLL parity doesn't happen on odd-dimensional cubes. Is this because the permutation of the centre edge pieces determines the overall permutation of the edges after reduction (ie, its not possible to swap only two centre-slice edges)?


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## trying-to-speedcube... (Nov 2, 2009)

Your conclusion only applies for supercubes.


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## cmhardw (Nov 2, 2009)

Cride5 said:


> Is this what you mean when relating the parity of centre pieces and edge pieces?



The parts about the centers are not correct for the 2nd and 3rd pictures, and only partially correct for the 1st picture.

You have to consider what kinds of centers are cycled, meaning from which orbits. Remember that from a theory standpoint there are two kinds of centers on a supercube (and yes Maarten is right that any conclusion we're drawing here is really only valid on a supercube). When you turn an inner slice on a 4x4x4 cube you have done two 4-cycles of pieces, but you have to consider which center orbit those 4 cycles occur in.

On the 5x5x5 picture you are correct that you have done three center 4-cycles, but without determining how many cycles happen in each kind of center orbit, we can't make any conclusion about how parity of those orbits are affected. Same thing for the 6x6x6 picture.



> So once the centres are solved on an odd-dimensional cube, we can guarantee that the edges must be in an even parity state, because the centres must be.



If we're considering regular cubes (and not supercubes), then no - and I'll give you a counterexample as to why not. Say your scrambling algorithm on a 5x5x5 uses an odd number of inner slice quarter turns, and you completely solve the centers using an even number of inner slice quarter turns. The centers are now solved, but what is the parity of the wing edge orbit?

If you're considering supercubes, and you solve all centers to an even parity state, then yes all wing edge orbits must be in an even permutation.



> If the centres are in an odd parity state then they cannot be solved.



Sure they can. Keep in mind you are talking supercube centers right now. The way to toggle parity in a supercube "oblique" center orbit is by either turning an inner slice that contains this orbit, or by turning an outer face of the cube. To toggle the parity of a supercube "x-center" orbit you can only turn an outer face of the cube.

You have to look at all the center orbits to see which ones are odd parity and which are even parity. From this you can deduce which combination of inner slices quarter turns and/or outer face turns you must use to reduce all center orbits to even parity. Or, you can simply figure out which inner edge orbits are in an odd parity state, and the parity of the corners, and do the same thing. If all wing edge orbits are in an even permutation, and if corners are also in an even permutation, then every center orbit must be in an even permutation (on a supercube). This is a fundamental property of supercube centers.

Remember that supercubes are a little bit different from regular cubes, especially when it concerns centers.



> *On an even-dimensional cube, inner-slice turns do not affect the parity state of the centre pieces*, while the parity state of the edges flips from odd to even for every inner-slice quarter turn. This means there is a 50% chance that the edges are in an odd parity state after solving the centres. After edge pairing, the odd parity state will manifest its self as a single pair of swapped edges (which appears as a single flipped edge pair after reduction).



Inner slices certainly change the parity of center orbits on even cubes, just not on the 4x4x4 supercube  That was the whole point of my proof above. Remember, you have to consider not how many 4-cycles occur because of an inner slice quarter turn, but how many 4-cycles *from which center orbits* occur because of an inner slice quarter turn.



> So I hope thats about right for OLL parity, but I'm still not certain about PLL parity. As Dan pointed out, its 2x edge 2-cycles. If this is the case, is PLL parity fixed at the end of edge pairing? Ie, would it be possible to go back to the edge pairing phase to fix PLL parity? Again, PLL parity doesn't happen on odd-dimensional cubes. Is this because the permutation of the centre edge pieces determines the overall permutation of the edges after reduction (ie, its not possible to swap only two centre-slice edges)?



PLL parity happens during edge pairing. When pairing the edges into groups you are also creating an edge cycle out of those groups. You have some control over where these edge groups end up, but we never think about this we just think about simply pairing them up. A good example of one case that is easy to see on 4x4. You have to pair the red-white edge group and the red-yellow edge group. Those pieces are in FR and FL on the cube. You can choose to pair them up using *d R F' U R' F d'* or *u' R F' U R' F u*. These two different algorithms will switch which pair ends up on which side of the cube (either FR or FL). During a speedsolve you have not figured out the parity of the corner orbit, or the resulting parity of the edge orbit based on where you can place the last two edge groups. However, if you were doing 4x4x4 FMC you could do this and figure out which alg to use such that the edge permutation you create after making those edge groups will match the corner permutation parity. This will avoid the PLL error or the PLL parity error.

Chris


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## Cride5 (Nov 2, 2009)

Wow! OK, This goes a bit deeper than I thought. I had a wee look here, and I think I now understand the terminology.

So when we're looking at supercubes and referring to parities of the centres, we need to look at each orbit individually? If I've understood correctly, a 4x4 supercube only has x-centres, which are all within the same orbit since they can all be permuted with each other. An inner slice turn creates 2x 4-cycles of centres all within the same orbit (thus maintaining even parity). When doing a *2r* inner-slice move an a 6x6x6 supercube, the same applies to the inner-most centre pieces (also x centres), but the 8 oblique centres affected belong to two separate orbits, creating 2x odd permutations. As a result, the parity of a 6x6x6 centre pieces and edge pieces are related.

So why does this not apply to regular cubes? Is it because the lack of distinction between centre pieces in the same face means that odd and even parity are impossible to distinguish?


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## trying-to-speedcube... (Nov 2, 2009)

Yes. 2 centers can be swapped without anybody noticing. (Simply said )


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## cmhardw (Nov 2, 2009)

trying-to-speedcube... said:


> Yes. 2 centers can be swapped without anybody noticing. (Simply said )



Technically this is not true. We can appear to swap two centers of different colors, but this is actually a 3-cycle of physical pieces on the cube. On any sized supercube if you really truly do swap only two centers in the same orbit, you have changed the parity of the centers in that orbit. This must require a toggling of the parity on either the corners of the cube, or the wing edges on the same slice, depending on if you swapped two x-centers or two oblique centers (and even a bit on your approach to how you swapped them if they were oblique centers).

Even on a regular cube, if you truly did pick two centers of the same color to "swap" then this must still be the case. To swap those two centers, assuming they are in the same orbit and can be swapped, is effectively toggling the parity of that orbit in the supercube sense, so there would be some other effect on the other pieces of the cube.

If someone wants me to write up a very detailed explanation of how the parities of center orbits relate not only to eachother, but also to the rest of the cube, I can easily do so. I came up with a very simple way to do this using matrices when I came up with my explicit formula for finding the number of possible states on the super-supercube on this page.

Chris


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## Cride5 (Nov 2, 2009)

cmhardw said:


> If someone wants me to write up a very detailed explanation of how the parities of center orbits relate not only to eachother, but also to the rest of the cube, I can easily do so.


If you can spare the time that would be excellent, thanks


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## trying-to-speedcube... (Nov 2, 2009)

cmhardw said:


> trying-to-speedcube... said:
> 
> 
> > Yes. 2 centers can be swapped without anybody noticing. (Simply said )
> ...


I know, I was just giving an exaggeratedly simple example why it's impossible to distinguish between odd and even parity on cubes with 2 or more identical pieces.


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## KwS Pall (Nov 2, 2009)

After reading Your posts Chris i came up with an Idea :
count w's in the scramble = a (either do not count w2's or count them as 2  )
count w's while solving centers = b
if (a + b) is divisible by 2 then go with pairing
else do parity and go on with with pairing

imposible on official competition but great if you can solve while counting as fast as without counting  always no parity can give you some sub x0s avgs

( by w's I mean quater turn of a inner slice (see WCA Scrambler))


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## rjohnson_8ball (Nov 2, 2009)

I solved the 4x4 on my own back in the 80's. My first approach was to try layer by layer, but that gave me trouble. Then I realized myself that I could do centers and edges and solve it like a 3x3. That worked pretty well. Then I ran into a couple weird parity snags that I eventually figured out on my own.


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## Zane_C (Nov 3, 2009)

Well, I paired up all the edges and was inserting the u layer centers. I was expecting to have to use the corner parity fixing algorithm that Chris gave me. 
On the last layer I realised I just needed to cycle 3 edge pairs.

THEN IT WAS SOLVED!!!!!

I've never solved a 3x3 intuitively let alone 4x4 intuitively, actually, I did once experience an OLL and PLL skip.

Thanks alot everyone, and the next time I run into a corner permutation parity, I will remember what you told me Chris. 
R2 Uw2 R2 Uw2 R2 Uw2.


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## hawkmp4 (Nov 4, 2009)

Congrats! I have to say I have not yet solved a twisty puzzle on my own the first time. I'd like to. I'm trying to find a Rubik's UFO around here somewhere...
Anyway, I'm glad you did that and I think you will be too- once you start doing things like that you get a much better understanding of the cube and puzzles in general and all kinds of possibilities open up for you.


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## siva.shanmukh (Nov 12, 2009)

cmhardw said:


> *Can you scramble and resolve centers using an odd number of inner slice quarter turns?* It's ok if this messes up some other things on the cube, as long as centers end up solved again.
> 
> Hint: It only takes 9 overall turns to do this
> 
> Chris



I tried my best. Couldn't come up with a way to do it in 9 quarter turns. Can you tell how.



cmhardw said:


> However, I found that it is incredibly difficult to do this during the inspection time. After much practice I am down to maybe 25 seconds record for a single solve. However, there is a shorter way to do this than counting cycles of the 24 pieces, and more like counting the 12 cycles of a 3x3x3. I can explain this method if anyone is interested. Based on how fast the truly world class BLD people are at memorzing an entire 3x3x3, perhaps it is not so far fetched to count cycles using my shortcut pseudo-3x3x3 edge method to count cycles on a 4x4x4 and always avoid OLL parity.
> 
> Chris



Really interested in knowing how you find out by counting just 12 cycles instead of 24. Can you please explain.



cmhardw; said:


> You have to consider what kinds of centers are cycled, meaning from which orbits. Remember that from a theory standpoint there are two kinds of centers on a supercube (and yes Maarten is right that any conclusion we're drawing here is really only valid on a supercube). When you turn an inner slice on a 4x4x4 cube you have done two 4-cycles of pieces, but you have to consider which center orbit those 4 cycles occur in.



Sorry for my limited understanding of what an orbit is. When you said there are 2 kinds of centers, did you mean

Kind1: Centers that lie on the middle slice
Kind2: Centers that are not of Kind1

or

Kind1: Centers that lie on the geometric diagonal of the cube's face
Kind2: Centers that are not of Kind1

or

Kind1: The center-most center
Kind2: Centers that are not of Kind1

And if you don't mind can you tell what an orbit actually refers to.

Thanks a ton.


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## cmhardw (Nov 12, 2009)

siva.shanmukh said:


> I tried my best. Couldn't come up with a way to do it in 9 quarter turns. Can you tell how.



I don't think it can be done in 9 quarter turns. You can do this using 9 turns in face turn metric, but make sure that you use an odd number of turns in quarter turn metric.



> Really interested in knowing how you find out by counting just 12 cycles instead of 24. Can you please explain.



What I do is sort of like memorizing a 3x3x3 cube blindfolded, so it will require counting only 12 "cycle" rather than the full 24. Basically you don't remember how many transpositions it takes to reduce the edges to the solved state, but rather to a state where edge groups are paired together, and oriented correctly in a 3x3x3 blindfolded orient first sense. For example:

Scramble your 4x4x4 cube with white on top and green on front with the following scramble:
Rw B2 D R2 B Uw2 B2 R B U' L' Uw2 Rw D' F2 D R2 Uw' B' Fw2 D' F2 Rw' R2 D2 F2 L' U2 L D B L2 R U R U L' B' F2 D2

If I was using my cycle counting method on this scramble I would do exactly the following.

1) Orient the cube using the cube rotation *x*, or turn the entire cube as if you were doing the turn *R*. This will place the paired yellow-red piece such that the yellow stickers are on the U layer (which is where yellow should go for me in order for an edge to be "oriented" in a blindfold solving sense).

2) I will look at the piece in UBl and see which piece it is. It is the orange yellow wing edge. I now find the other yellow orange wing, which is at RBd. The yellow is on the B face, which is where yellow should be in my 3x3x3 edge blindfold orientation scheme to be oriented "correctly". So I will imagine that I am sending the piece at UBl to RBu, which will pair it up with the other yellow orange and make an oriented 3x3x3 edge pair. Now I don't memorize this, nor do I count how many cycles there are. All I do is think that I just did one cycle, so I toggle the parity of the permutation I am analyzing. When beginning I have imagined that I have done no cycles at all, and I now imagine that I will swap my UBl edge with the one at RBu, which is one transposition. So now the parity is odd, and I verbally say "odd"

3) The piece that is in RBu is the yellow green edge, and it's pair mate is at DFr with the yellow on the D layer. This is already oriented, so I send RBu next to it to DFl, one transposition which toggles parity. I now say "even" meaning the permutation I am creating has even parity.

4) The piece that was at DFl is green red, and it's mate is at DBl. This piece is "flipped" in a blindfolded sense for 3x3x3 edge orientation because the green is on the B face. So now I will do two transpositions. I imagine that I am sending the piece at DFl to DBl, and the piece at DBl goes to DBr. Since I have done two swaps in a row, I have not toggled parity. So I don't say anything, and the permutation I am creating is still "even" as it was at the end of the last step. Now, when actually doing this, if edges are flipped I simply don't do anything and just move on, but the reason is because I am actually doing two transpositions.

5) The piece at DBr goes to UBl to pair with it's mate (which is where I started). This piece was already sent to UBl when I did the swap earlier. So this tells me that I completed a partial cycle and I must start at a new piece somewhere else. The parity of the permutation is still "even"

6) I now need to start from a new location, because I completed a cycle. I will now start at URb. This piece is white orange, and pairs with the other one at DRb. The DRb piece is flipped in a 3x3x3 sense, so I will do two transpositions, or I will not toggle parity. Therefore I simply don't say or do anything and just skip along to the next piece. My permutation parity is still "even" as it was at the end of the last step.

7) DRf pairs with UFr, which is flipped. Again do nothing and skip to the next piece.

8) UFl goes to LBd. I say "odd"

9) LBd goes to LFu. I say "even"

10) LFu goes to LRu goes to LRd. I say nothing and move on.

11) LRd goes to DLb goes to DLf. I say nothing and move on.

12) DLf is the piece that pairs with URf (the place adjacent to where I started). So I know the parity of the wing edge permutation is the last word I said, so the parity of the wing permutation is even. This means that the wing edges on the cube are in an even permutation, and I must solve the centers using an *even* number of quarter turns to restore the parity of the wing edges to even. This will effectively skip the OLL parity case, and I will know that it is impossible to have the OLL parity during the solve, so I save time.

13-optional) You may wonder how I would know that I am done at step 12. Back when I was trying to do this seriously I used a base 5 counting system with my feet to count how many edge pairs I had built during my analysis. The reason I used base 5 was because it allows you to count to 24 on two feet, and this way I could also use it when counting the full 24 cycles (which I did for a while before coming up with the idea I just described).

To verify that the wing edges actually are in an even permutation go ahead and just count the cycles in the usual sense. After the initial cube rotation, here is the wing permutation using my blindfolded lettering scheme.

(QFJBIG)(NPHOUXATCLV)(RK)(DSW) and two wing edges are solved. If you analyze the actual wing edge permutation you will find that it is indeed in an even permutation.



> Sorry for my limited understanding of what an orbit is. When you said there are 2 kinds of centers, did you mean



I meant this option that you wrote:



> Kind1: Centers that lie on the geometric diagonal of the cube's face
> Kind2: Centers that are not of Kind1



Basically every inner slice contains 8 x-centers from the same orbit, and all the rest of the centers are in groups of 4 oblique centers from the same orbits.



> And if you don't mind can you tell what an orbit actually refers to.
> 
> Thanks a ton.



I wonder if we are using this in the correct sense, as I have heard some people say that cubers misuse this term in a pure mathematical sense. By orbit, I mean all possible locations on the cube that a certain piece may occupy. So for example the corner orbit consists of 8 locations that the corners may occupy. The wing edge orbit on a 4x4x4 consists of the 24 locations that the wing edges may occupy.

Chris


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## Ton (Nov 12, 2009)

Zane_C said:


> Hey,
> I have a 4x4x4 cube and haven't been spending much time on it. I haven't bothered to learn a method and I play around with it every now and then. So far I can intuitively solve the first 3 layers, all corners, but havn't quite figured out how to orient and/or permute individual last layer edges.
> The closest I've got to it being solved was all but 2 pieces solved, they were oriented correctly but needed to switch places.
> 
> ...



Took me one week to figure out the parities that was in the 80's, after I thought about it -I had some knowledge about permutation- the solution was just one turn away

I solve the 4x4 now in about 1:18


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## Thomas09 (Nov 12, 2009)

I did it all on my own. I was lucky though, I had no parity errors.


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## siva.shanmukh (Nov 12, 2009)

cmhardw said:


> siva.shanmukh said:
> 
> 
> > I tried my best. Couldn't come up with a way to do it in 9 quarter turns. Can you tell how.
> ...



u R2 u R2 u R2 u R2 u ?
I completely randomly got that. 
Its actually nice, only 3 swaps of edges induced. Or to put it in other words one 3 cycle commutator away from parity case.

And nice way of finding the odd/even parity of dedges. I didn't really get how you planned to use toes to count base 5 numbers. Is it something like left toes for 5's place and right for 1s place? That sounds as a good idea actually. I will give it a try.

I feel like sharing my way of memoing. Tell me if I should change it.
Your (QFJBIG)(NPHOUXATCLV)(RK)(DSW)
would be FJBI GNPH OUXA TCLV NRKR DSWD for me with Q as my buffer.

1) My memo seq doesn't have a letter for buffer.
2) Once the first cycles close, the rest of the cycles will be starting and ending with the same letter.
3) I memo them in groups of four. (I find it better than 2 as I map these four letter words to some word with some vowels added. If the last group is a 1 or a 3 long one, then it is odd parity)
4) I find this easy because all I need to do is pick 2 letters, and cycle them with buffer. I don't have to think of things like when I need to break into a new cycle.

Shanmukh


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## mrCage (Nov 15, 2009)

Zane, if you really understand what you have done so far you should realise that your second problem is almost like the first. A few setup moves is all you need to transform between them. Me like others would like to express my doubt as to how you get get so far and not to be able overcome the final small hurdle ??

Per


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## siva.shanmukh (Nov 20, 2009)

siva.shanmukh said:


> cmhardw said:
> 
> 
> > siva.shanmukh said:
> ...



Following is an observation when I did r U2 r U2 r U2 r U2 r
All this while I was thinking that the centers were intact and somehow I was not comfortable with the idea. On careful observation I found this:

Considering the four centers in each face as 2 pairs. An offset of one quarter turn is induced in the r slice pairs and There is an odd permutation in the pairs. The pairs on the U face end up swapped at the end of the algo.

So I don't know if we can say that it actually re-solves the centers. If you are okay with the odd permutation in centers then thats the best I can come up with. Let me know if there is some other algo is in your mind.

Shanmukh


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## Stefan (Nov 20, 2009)

siva.shanmukh said:


> So I don't know if we can say that it actually re-solves the centers.


Do they not looked solved afterwards?


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## siva.shanmukh (Nov 20, 2009)

My mistake for having put it that way. I was just considering using this for BLD odd dedge parity, which is when it bothered me because I usually solve dedges before centers.

In the current context, it answers the question. Thanks and apologies.


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## cmhardw (Nov 21, 2009)

siva.shanmukh said:


> I was just considering using this for BLD odd dedge parity, which is when it bothered me because I usually solve dedges before centers.



As long as you only use this on a 4x4x4 you could use an alg like (R' U' R U')*5 to rotate the center back to it's original state. On a 5x5x5 you will end up also swapping some t-centers (and on a supercube you would have rotated the centralmost center 180  ). For 4x4x4 though those swapped rows are equivalent to rotating the center 180 degrees. This way you can still solve dedges before centers.

Chris


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## siva.shanmukh (Nov 22, 2009)

cmhardw said:


> As long as you only use this on a 4x4x4 you could use an alg like (R' U' R U')*5 to rotate the center back to it's original state.



Thanks for this algo. I always use the commutator [RLdw2R'L' U] and a U2



cmhardw said:


> On a 5x5x5 you will end up also swapping some t-centers (and on a supercube you would have rotated the centralmost center 180  ). For 4x4x4 though those swapped rows are equivalent to rotating the center 180 degrees. This way you can still solve dedges before centers.
> 
> Chris



So to use the same method in 5x5 super bld:
Goal: swap UBl <-> UFl and swap USl <-> USr (I meant the tedges)

1) edge + t-center parity fix:
r U2 r U2 r U2 r U2 r

2) cyle (DBr DFr UFl) depending on the situation

3) correct the other centers:
[rld2r'l' U] U2

I don't have 5x5 supercube and currently my windows is not working so can't check it on gabbasoft.
Correct me if I am wrong.

anyway this method would fail for 6x6 or higher due to the introduction of oblique centers(or whatever they are called)
I can't think of a good alg for the case of cubes higher than 5x5


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## nigtv (Jan 7, 2010)

This thread inspired me a while ago to get a 4x4x4, and to do it a good number of times on my own until I got to a decent enough speed (maybe somewhere around 2-8 minutes?). Just got my 4x4x4 in the mail the other day, and scrambled it up (by hand, I still don't really know the notation), and got it in 6 hours. It's very strange though, I've had it take me 16 hours to do it and one solve in about 9-10 minutes, maybe I'll start getting a little more precision in a few weeks.

I'm still having trouble with a few key things, mainly figuring out whether or not I need to fix parity or not (I'm a little lost reading through the heavily technical portions of this thread that seem to explain what I'm missing). Oh well, I'll probably figure it out sooner or later, and that may even be some of the fun (?). 

Zane, did you end up getting better with your own intuitive solves or did you start to collect some methods and algs, or fully pick up a method or what? I'm curious!

Oh, and if anyone cares, I've been doing my best to follow the loose steps of firstly fixing centers, then pairing edges, then making some blocks and getting a F2L minus a slot style thing, trying my damndest to detect if I need to fix parity, trying to fix parity, and then checking if I actually needed to fix parity, at which point I just sort of push through to the end. Sloppy, right?


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## Zane_C (Jan 7, 2010)

nigtv said:


> This thread inspired me a while ago to get a 4x4x4, and to do it a good number of times on my own until I got to a decent enough speed (maybe somewhere around 2-8 minutes?). Just got my 4x4x4 in the mail the other day, and scrambled it up (by hand, I still don't really know the notation), and got it in 6 hours. It's very strange though, I've had it take me 16 hours to do it and one solve in about 9-10 minutes, maybe I'll start getting a little more precision in a few weeks.
> 
> I'm still having trouble with a few key things, mainly figuring out whether or not I need to fix parity or not (I'm a little lost reading through the heavily technical portions of this thread that seem to explain what I'm missing). Oh well, I'll probably figure it out sooner or later, and that may even be some of the fun (?).
> 
> ...



I hope I don't dissapoint you, but I havn't really touched it since I've solved it. Some moves I used to pair the last layer edges was just a simple r U r' U r U2 r'. 
I solved the effected centers without disturbing the edges or corners at all using intuitive moves such as, b l' b' U' b U l U' b'.
I don't literally perform it as (b l' b' U' b U l U' b'), because it is intuitive I use cube rotations to change the 'b' into say, an 'r' or an 'l' 

I also experimented with some Fridrich PLL algorithms, by applying some inner slice set-up moves and a PLL alg then undoing the set-up moves, I was able to minipulate individual edge peices, but that didn't work out to well.

I'm glad this thread inspired you 

Next time I order some more puzzles I might get a better 4x4x4 cube and learn reduction or something


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## nigtv (Jan 7, 2010)

I had to skim your post, like most of the posts in the last few pages of this thread, just for the sake of avoiding influence (stealing algorithms or something without knowing it), hope that doesn't make me a bad person...

And if I knew more algs, I'd probably use them, I only really know niklas and sune, and their inverses. No biggie that you haven't touched it since you solved it though (I've done the same with sq1). I've been trying a sort of 2 gen type thing (not sure if it counts as a 2 gen, but using only u/U r/R, as in leaving a 2x2x4 block), that might keep me interested if I do end up figuring it out.


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## vgbjason (Jan 24, 2010)

i found it out on my own the first time ^_^
then i had parity every other time. . .grr. . .


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## SebCube (Jan 24, 2010)

mcciff2112 said:


> What you have is a parity. And there's nothing you can do without an algorithm (unless, of course, you could fix it intuitively, which is close to impossible to do). If you wanna solve it on your own, I would rescramble it and hope for the best that you don't get a parity case.


its not impossible to fix it intuitively all you need is a brain.


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## Zane_C (Jan 24, 2010)

Why do old threads suddenly get opened again?


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## SebCube (Jan 24, 2010)

I was bored


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## Zane_C (Jan 24, 2010)

Fair enough.


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