U2 for centers

[RobinBloehm]

Yeah, that works, but you have to make sure that have to do U' instead of U (for example) if you are at an odd position in the cycle, the same thing as in R2 or M2 or... bla

 

[alexc]

I actually was just thinking about an alternative method to centers besides commutators a couple days ago. I thought about using U2 to solve one center at a time, but I didn't know it would actually work. That's why I didn't mention it on this forum. Nice job Stefan!

 

[mrCage]

I actually was just thinking about an alternative method to centers besides commutators a couple days ago. I thought about using U2 to solve one center at a time, but I didn't know it would actually work. That's why I didn't mention it on this forum. Nice job Stefan!

Hi

I'd say the "U2" method is pretty much the same as doing commutator centers solving actually. Except leaving out the last turn in the commutator :rolleyes: The "U2" method would only work for centers first obviously - since it messes up edges and corners.

Btw, the method extends to higher order cubes right? And for that matter also "intends" to 5x5x5 and 4x4x4 would it not ?

- Per

 

[Stefan]

The "U2" method would only work for centers first obviously - since it messes up edges and corners.

What? It doesn't mess up edges/corners. And if it did, why would that work for centers first, and why *only* for that? I'm completely stumped by that statement.

But while you're here... do you have a faster way for 3-cycling two opposite U centers and one D center (and doing an additional U2)? Here's what we got so far and I always thought about asking you because of your cage method:

To shoot to Dbr I can use: (r F d2 F' r') U2 (r F d2 F' r')

 

[Swordsman Kirby]

Huh, now that I actually bothered to read this topic, I realized that it was very very similar to my center 3-cycle approach. Basically, a three cycle would look like the following (after setup moves):

(r U2 r') u (r U2 r') u'

It's possible to use any face for setups. For U2 2-cycle you would get, for the same case:

u r' u' r U2 r' u r u' r' u r U2 r' u' r

Mine is shorter by half. Of course, there are other cases where mine is shorter by a smaller degree, but I think it's all about personal preference. I don't think you (Stefan) like doing three-cycle.

 

[Lucas Garron]

Swordsman Kirby: Isn't that a way to just "freestyle" comms? I also use a few that way...

Anyhow, I thought of this a while ago: Would R2 work as well for this? If someone tries it, could you also try R2?

 

[ROOT]

howabout using ufL as a buffer and just adjusting the whole middle three layers so you can get UDFB. R will be a problem though. once your buffer is already solved, just do L/L'/L2 to adjust the buffer to a center that isnt solved. and once your done with that center do the same thing untill theyre all solved, then fix it with an L/L'/L2. i dunno

 

[Lucas Garron]

howabout using ufL as a buffer and just adjusting the whole middle three layers so you can get UDFB. R will be a problem though. once your buffer is already solved, just do L/L'/L2 to adjust the buffer to a center that isnt solved. and once your done with that center do the same thing untill theyre all solved, then fix it with an L/L'/L2. i dunno

Actually, that's a good idea. I'd prefer Rub and R2, though.
The only issue I see with this is that visualization seems a bit more confusing right now.
I'll have to try this, though...

 

[Stefan]

howabout using ufL as a buffer and just adjusting the whole middle three layers so you can get UDFB.

I don't understand.

 

[mrCage]

Hi

Sorry for ignoring this thread for such a long time. My forum search/contributions tend to be a bit random to put it mildly

How about : (Ff) - [r' u' r,U] - (Ff)' ??? Yes a conjugate, so in some sense not a pure commutator. But good enough for me ;-)
Different cutplane versions is of course possible, for even larger cubes ...

- Per

(edit)

Another direct approach - though longer - would be:
[r' B' d2 B r,U*] U* = U/U2/U'

Note similarity with the following corner cycle:
[R' B' D2 B R,U*]

(/edit)

 

[KConny]

Wait! I forgot why I forgot. Now I remember why I forgot: I didn't forget. Shooting from U to U is unnecessary. That's why I didn't cover it. I think. Maybe it is necessary. Or not? Once you have a U center in the buffer, you do just a U2 turn to break into the cycle starting at the target. Unless that one already contains a U center, too. In which case you can do a single U turn and start using the other two U center positions as buffer and target. I didn't fully think this through, though. Maybe that's the reason I didn't publish this method earlier. I'm confused now. Great.

I'm also confused, are you still confused? I've now failed about 6 attempts because of not really understanding how to handle shooting to U. Well, doing the actuall shooting is easy, but then reversing setup confuses me. Help? Anyone?

 

[KConny]

Seriously, is there no one here who uses U2 that can tell me how to handle shooting to U?

Shooting to UFR is either U or U' depending on if it's in an odd or even place in the cycle, U layer might be good or bad, just as in M2, right? And when you're done with your cycle and need to go back to your original buffer are you then supposed to do U or U'? This makes me so confused, I thought I had it thinking that if you need to reverse the setup moves at the same status (odd or even) as you did the shooting it's just to reverse the U turn. Is this correctt?

 

[shelley]

Starting new cycles from U centers always confused me the few times I tried U2 as well.

What if you just shoot from wherever the center is. E.g. you start a cycle from UFR, you shoot to UBL? It would eliminate a setup move at the beginning/end that I always seem to mess up.

 

[mazei]

Err this might sound a little noobish but if any of you don't mind helping me, how do I apply the U2 method in a 4x4 BLD solve. I have a rough idea of how it works but the way I can think of takes too much to remember. So care to explain how you guys apply the U2 method. Perhaps this should be on another thread. Maybe you can send me an e-mail on the explanation. [email protected].

[Edit]
Never mind. I got it now. But just one question. What if your U centers are all done before you finish other centers? Do you just start a new cycle by just switching centers with an unsolved center? or do you turn the cube and start a new cycle on a new face?

 

[KConny]

Start a new cycle by just switching centers with an unsolved center.

 

[KConny]

So I think I've got this figured out, shooting to U centers that is. I made a little tabel of all the cases.

SR | U | U' | U2
----------------
11 | U | U' | -
12 | U' | U | U2
21 | U | U' | -
22 | U' | U | U2

S=Shooting R=Reverse, U/U'/U2=setups
1=odd place in cycle, 2=even place in cycle.
So, if you shoot with U' at an even place in the cycle (2) and later need to reverse it at an odd place in the cycle (1) the reverse setup move should be U'. Yeah, hope you get it.

 

[siva.shanmukh]

It was really hard memorizing the entire 4x4. But, I was finally able to finish my first 4x4 bld solve using 3OP for corners, r2 for edges and U2 for centers. U2 is actually very easy and efficient. I find it much easier to execute than using free style commutators as it is very easy to adapt to. I am not sure about how good I am at explaining but I will try to explain in the next post about how I did the U face(only if you want to know that is)

 

[siva.shanmukh]

I dunno if this is the right thread to post this. But lemme jus do it and get corrected if I am wrong.

I tried to solve a 4x4 super-cube blindfolded once and got stuck with the parity case. No problems with edges. But the centers and corners were little problematic.

When I solved it, I was stuck at UBL and UBR swapped and ubl and ufr swapped. So I had to go ahead and solve it by doing the following from the current situation:

1. A UBR <-> UFL <-> UFR cycle
2. A ufr <-> ufl <-> ubr cycle
3. A ufr <-> ubl <-> ubr cycle
   And solved the edges by shooting them in the following order
4. UFL -> URF -> UBR -> ULB
5. UFR -> URB -> UBL -> ULF
6. And a final U move.

Any idea of an easier way?

 

[KConny]

Sure, do explain how you solve the U-centers.

 

[siva.shanmukh]

Sure, do explain how you solve the U-centers.

OK. I guess this is by default the same way anyone would solve it basing this concept. But anyway let me put it forward.

My buffer is ufr and I shoot it to ubl every time. My setups are pretty intutive and I don't think are worth posting it here.

I do all the thinking during memorization itself and there is only one action that I do that is shooting the ufr to the memorized location. I don't memorize it like breaking into a new cycle or finishing a cycle and stuff like that.

My thought process for memo broken into steps for easy unerstanding:

Step1:

I start memorizing the sequence of locations as to where the ufr should go to and then where the next one should go to and so on. If I come across a U center some where for the first time, then I memorize it as ubl(move: U2).

Step 2:

If I come across the second white center in the sequence, then I memorize it as a ufl (move: U') if this is the odd position of the sequence or I memorize it as a ubr (move: U) if this is the even position of the sequence.
Now what ever was in ufl initially will come to buffer and I repeat Step1.

Step 3:

If we come across all the 4 U centers, this is when i break into new cycle. But I won't memorize it as a breaking into new cycle. I just memorize it as shooting it to some unsolved center and continue step 1 till I reach the same center again.

Step 4:
I actually manage to easily find that odd position or even position of the sequence by memorizing it in pairs. So if I my last one is not a pair, then I memorize it as an extra ubl and then an extra ubr/ufl(the compliment of the one that I memorized in the step 2)

Execution:
It is very straight forward now:
Just use setups to bring the center in the sequence perform a U2 and undo the setup. If it is one of the cases mentioned in step 2 or step 4, then the moves are as I mentioned there.

If my explanation is not clear enough, I can write an example solve.

I guess this can be used for 5x5/6x6/7x7 centers also. There will be exceptional situation in the case of 5x5/7x7 outer layer parity situation.

I am still expecting a reply for my post about the super cube BLD