Big Cube BLD Discussion

[cmhardw]

Dear all,

How do you flip edge that flipped in position on odd cubes?

I use monoflip commutators:
R' e' R2 e2 R' U2 R e2 R2 e R U2

And if you use some foresight you can use cyclic shifts of this style commutator to cover other cases:
e' R2 e2 R' U2 R e2 R2 e R U2 R'

 

[yoruichi]

why cant u just do the M move one?

 

[cmhardw]

why cant u just do the M move one?

I flip these central edges first, so I have not yet solved centers. The MU alg is not supercube safe, so I opt for supercube safe commutators.

I think the time saved by memorizing these pieces visually, and solving first with longer commutators far outweighs the other option of memorizing them and having to retain the memory of them being flipped until all centers all solved, and then use a faster/shorter alg. My opinion of course, but I have tried both ways and I greatly prefer to flip first with longer commutators.

Chris

 

[cmhardw]

BH method update:

I know this topic is not a new topic, it has been covered by blah (Chester) on this forum, and if my memory serves me correctly it was first suggested by Stefan Pochmann around 2003-2004 on the Yahoo Group, but Daniel and I have also now looked very seriously into optimizing centers via a cube rotation on odd cubes. Basically, we now orient the 5x5x5 cube in such a way as to optimize solved centers, just like on an even cube like 4x4x4. This creates a situation where you need to rotate in such a way as to avoid center parity.

Anyway, we both know that this topic has been covered many times before, but Daniel has come up with an incredibly simple way to intuitively solve, and memorize, all 12 even center parity cycles, and I have come up with a way to quickly identify whether or not the centers are in an even or odd parity state seeing only two faces (U and F).

We both admit that it is certainly possible that both our recognition and solving/memory methods have already been discovered by others. However, we're both very excited about this and are now making it part of the BH method. Since the methods are so easy to apply, and so intuitive, we are going to implement them always in our 5x5x5 solves.

A writeup will soon follow on our methods. Again, I want to make it very clear that we both realize that others before us have already analyzed this and have already optimized this. Consider this only Daniel and my take on the same subject.

Chris

 

[dbeyer]

Chris and I have been talking about previous cube theories. Pochmann after his groundbreaking 5x5 blindfolded solve years ago, suggested this idea. Back then it was incredible to even solve a Rubik's cube blindfolded.

Earlier this week, I commented on taking too long to find that perfect orientation. I said it would be better to just toss it up in the air and start memorizing the 4x4 as it falls in yours hands. You are just as likely to get a good solve.

However, on the 5x5 we are binding ourselves to a simple rule. Solving the 6 centers before memorizing the cube. If you were to solve big center blocks on the 5x5 like you would on the 4x4, you would save time. Lots of time. Several images and algorithms on average in a x/+ center memo.

All that you have to do is avoid center parity. That is directly solving each piece ignoring the 3x3 centers, you will wind up with an odd number of permutations between the centrals and corners. Whether on a 3x3 or 5x5.

So on a 3x3 if you add up the edge permutation and corner permutations, you will always add up to an even number, generally around 20.
If you were to solve within your color scheme, but incorrectly about a set of centers, you will wind up with center parity.

There is a 50% chance you can solve your cube and not have center parity.
24 cube rotations, 12 of them are without parity, if you directly solved the cube using your color scheme, ignoring centers.

1 case is of course your preferenced cube orientation.
Such as myself I have Red on Top, and Blue to the Front,
Chris has Yellow on Top, and Green to the Front.
The other 11 cases without parity can be solved in 11 moves.
We want you to be able to use these other 11 rotations to help optimize your 5x5 memorization. Rectifying and solving the central-most centers can be done in as little as 4 moves.

We have systemized this method. It's profoundly easy, and can save dozens of moves, and several letter pair images on the 5x5 bld attempt. Here is an extention to the BH method ...

Coming soon ...

 

[aronpm]

Hi, I have a question about using r2 in 4x4x4 BLD. While I've been learning r2 I've been thinking of the edges like the stickers in 3x3x3 BLD using M2. So, BU would be the same as UBl. Also I use colours instead of positions. This has been working well for me, but I've been having trouble checking what piece would be next. Tracking the piece around the cube is really tedious, and I'm wondering if there is a general rule for pieces outside the r and l slices.

As an example, I have RB (FRd piece) and I know how to shoot there. My problem is what the next piece is. I know it is whatever is in RB, but I often get mixed up between whether that piece is FRd or FRu. My hypothesis is that if the piece I'm shooting starts with a front/up/back/down colour, the piece should be the most counter-clockwise piece in that slot. So in the FR edge, starting with FRu, it goes counter-clockwise to FRd, so that is the most counter-clockwise edge in the slot, and that is the next piece. So when I've shot to RB, my next piece is GY, so I'd apply my idea in reverse, because the first colour isn't one of the up/front/back/down colours. Shooting to GY (the LDb piece) which is the buffer, YR (FDr). This is assuming that using M2, the buffer is DF and not FD like some people use.

I think if you hold a slot in the front with either a F/U/B/D colour on top if it's a side edge or a F/U colour on top if it's a r/l slice edge, then the left piece is the top and the right piece is the front sticker. But for me it would be easier to check with the counter-clockwise idea, on the side edges at least.

I think my hypothesis works, but could someone else try it and tell me if it works for them? Sorry for the long post but it was hard for me to explain this.

PS: I won't be trying a 4x4x4 BLD solve for quite a while because I don't have enough memory routes set up, and I fail hard at the centres.

 

[Zava]

why cant u just do the M move one?

maybe because of what Chris said, or Chuck thought of the (UM')*4 (M'U)*4 as the MU flip

 

[Mike Hughey]

With regard to the new ideas by Chris and Daniel, here's the old thread by blah on the subject:

http://www.speedsolving.com/forum/showthread.php?p=61837

Also, somewhere in those posts is the link to Stefan's original post on the idea.

Also, deadalnix has an interesting idea there for how to very quickly and easily limit yourself to non-parity cases. I'm thinking that might be the way I would go with it.

 

[dbeyer]

Oh, yes we are basing our recognition of parity and non-parity cases on the orientation of the UFR corner. Looking at the U face, then the F face, you can determine if you have parity or not. If there is parity, you can do a simple cube rotation that preserves the major block that you solved. Such as preserving the block on the L face, you'd do a x cube rotation fixing parity.

There are 12 cube states that do not have parity. We tried counting them at first and then we laughed when we were missing a case.

2x2 Swaps on the M, E, and S slices.
3 Algs.
M'E2ME2, EM2E'M2, and xEM2E'M2x'
respectively. M'EM2E'M' might actually work for the S slice ...
I tried something earlier that I visualized incorrectly.

Then 3 cycles.
URB, UFR, ULF, and UBL, and the inverses of these.
Solving these 8 cases are very intuitive.
Finally, the solved cube state. -- the Missing case lol.

 

[dbeyer]

Basing our recognition off of the UFR corner, and looking at the U centres point, then the F centres point (courtesy of deadalnix's vocab).

You follow the UFR corner heirarchy. IF you have a U/D center on the U layer, THEN you must have an F/B center on the F face. If you have a F/B center on the U layer, THEN you must have an L/R center on the F face. Lastly IF you have an L/R center on the U layer, THEN you must have a U/D center on the F face.

 

[dbeyer]

Now, solving the 3 cycles. Its very intuitive, and can be solved using 4 moves, and no cube rotations. We are going to use M and E turns, to solve these other 8 cases.

Looking at the center point of the U layer.

Think U goes x. X goes to y.
Let x be F, R, B, or L. Let Y be one of the adjacent centers of the first choice.
IF x is F/B THEN y must be L/R.
IF x is L/R THEN y must be F/B.

Continuing.
2 centre points are mutually on the M slice, and E slice. The F and B centre points.

So we permute from the Buffer Layer, the U layer.
Buffer to x to y.
X may be F, B, L or R. When X is on the M slice, perform the M slice quarter turn that permutes the U centre point to the X.
F/B are on the M slice to begin with.
If x is L/R, then a quarter turn on the E slice must be made to bring the centre point to the M slice.
The L/R centre point must take place of the F/B centre point during permutation.

Such as U to F to R to U.
M inserts the U's centre point to the F face.
E' will bring the R centre point to the F face.
M'E will complete the commutator of centre points.

so there are 8 combiniations.
ABA'B'
Starting from any point in the AB commutator, you can go from left to right, or from right to left.
Leading to
ABA'B'
.BA'B'A
.A'B'AB
.B'ABA'
.B'A'BA
.A'BAB'
.BAB'A'
.AB'A'B

Exchange A and B for M and E. Here are the 8 cases.
If you have any further questions please ask.

Later,
DB

 

[Sakarie]

I think it is a good idea to solve the centers, that's great!

But how do you, dbeyer and cmhardw, think when you claim that "Cubeorientation is unimportant! You might as well get a hard/good solve anyway!"

Why don't you say exactly the same thing for 5x5? Me myself, I think that you raise the chancesof getting a better solve if you do a better cube orientation. But I can't see why you think it do on 5x5, but not on 4x4?

 

[trying-to-speedcube...]

I have been thinking about this, and at first it looked like a good idea, especially for bigger cubes, because the advantage you'd get would be bigger in terms of number of solved pieces.

On the other hand, on bigger cubes you would spend more time counting pieces and deciding on the best orientation. I don't know what takes/wins more time.

I have also been thinking about the "center parity". You're saying you have to avoid it, but you can also use an algorithm to solve the parity at the end of the solve. As I use Old Pochmann for corners, I would end up with the UB edge and the UL edge swapped. When you notice you have a center parity, to make the fix easier, swap the Lsu and the Bmu center pieces.

Now there are a few parity cases. You can have the centers an M/M'/S/S'/E/E' move off, or you could have a "Z-perm" in the M-slice, E-slice or the S-slice, with the other 2 centers being swapped.

With M S M' S' and similar trivial algorithms, and M' E2 M E2 and similar you can always reduce the parity case to the case where the centers are an M-slice off. Then you fix it with Uw Fw' Bw Lw Fw Bw' Uw2 Rw2 Dw Rw Lw' Fw' Rw2.

Although all this might seem as if it takes too much time, I think it's not that bad.

Example solve (on 5x5):
Scramble: U2 Bw F2 Uw2 R2 Dw2 Lw U' B2 Uw Dw' Bw' R' Fw2 U' Fw Dw' Lw' Bw2 Fw2 F' U' Rw Fw' Uw' L' F' Fw B2 Rw F Bw Fw' Rw2 F2 D2 R2 Lw' D R2 Lw' Dw2 Rw' F' U2 L Uw2 Lw' Fw B2 F' Dw L2 Lw F' Fw2 B' Bw' D2 B2

Hey, I see lots of D-colors on F, let's do x'.

Solve (with U2 for centers, M2/r2 for edges and Old Pochmann for corners)

D2 R (R U' R' U' R U R' F' R U R' U' R' F R) R' D2
D F' (R U' R' U' R U R' F' R U R' U' R' F R) F D'
F2 D (R U' R' U' R U R' F' R U R' U' R' F R) D' F2
F R' (R U' R' U' R U R' F' R U R' U' R' F R) R F'
D' (R U' R' U' R U R' F' R U R' U' R' F R) D
R' F (R U' R' U' R U R' F' R U R' U' R' F R) F' R
F' R' (R U' R' U' R U R' F' R U R' U' R' F R) R F
U' (to fix U-centers)
l F' l b2 l' F l b2 l2 U2
f u' f' U2 f u f'
f u2 f' U2 f u2 f'
U l u l' U2 l u' l' U'
D l B d2 B' l' U2 l B d2 B' l' D'
f' d f U2 f' d' f
U2
l' u' l U2 l' u l
l d l' U2 l d' l'
l' u2 l U2 l' u2 l
r2 f2 r' B' r f2 r' B r' U2
l d2 l' U2 l d2 l'
D2 l B d2 B' l' U2 l B d2 B' l' D2
d l d' l' U2 l d l' d'
U' r' d r U2 r' d' r U
f' d' f U2 f' d f
U2 r' u2 r U2 r' u2 r U2
U2
M u M' U M u' M' U M u M' U2 M u' M' U2 (don't know a better alg for this)
U S' d S U2 S' d' S U'
U' r' E' r U2 r' E r U
f' E f U2 f' E' f
U2
M d' M' U2 M d M'
M u M' U' M u' M' U' M u M' U2 M u' M' U2
f E2 f' U2 f E2 f'
f' E' f U2 f' E f
M d M' U2 M d' M'

I have now solved all B-centers, except Bmu. This center will be swapped with Lsu in the center parity fixing algorithm. That's why I'm going to shoot it to Lsu instead.

M' u M U2 M' u' M
U' r E r' U2 r E' r' U
d M d' M' U2 M d M' d'
f E f' U2 f E' f'
D M d2 M' U2 M d2 M' D'

Here is the actual Lsu piece, I'm going to shoot it to Bmu.

U S u S' U2 S u' S' U'
f E f' U2 f E' f'
U (to fix U-layer again)

B' R B M2 B' R' B
B L2 B' M2 B L2 B'
M2 F R U R' E R U' R' E' F'
M2
B L B' M2 B L' B'
U R2 U' M2 U R2 U'
B L' B' M2 B L B'
B' R2 B M2 B' R2 B
M2
D' L2 D M2 D' L2 D (U' F2 U M2 U' F2 U wouldn't work here, because it puts the Bmu center in Bmd)
B M' U M' U M' U2 M U M U M U2 B'
l B' R' B R' U R U' r2 U R' U' R B' R B l'
L B L' B' r2 B L B' L'
L U' L' U r2 U' L U L'
B L B' r2 B L' B'
R' U R U' r2 U R' U' R
F d R U R' d' R U' R' F' r2
B' R B r2 B' R' B
U R2 U' r2 U R2 U'
F d R U R' d' R U' R' F' r2
l' B' R' B R' U R U' r2 U R' U' R B' R B l
U' L U r2 U' L' U
r2
R B' R' B r2 B' R B R'
B L' B' r2 B L B'
B L2 B' r2 B L2 B'
l2 B' R' B R' U R U' r2 U R' U' R B' R B l2
U' L' U r2 U' L U
B' R' B r2 B' R B
U R' U' r2 U R U'
U R U' r2 U R' U'
B' R2 B r2 B' R2 B
U' L2 U r2 U' L2 U
U' L' U r2 U' L U
r2 D' L' F l' U2 l' U2 F2 l' F2 r U2 r' U2 l2 F' L D
M' E2 M E2
Uw Fw' Bw Lw Fw Bw' Uw2 Rw2 Dw Rw Lw' Fw' Rw2

 

[Mike Hughey]

With M S M' S' and similar trivial algorithms, and M' E2 M E2 and similar you can always reduce the parity case to the case where the centers are an M-slice off. Then you fix it with Uw Fw' Bw Lw Fw Bw' Uw2 Rw2 Dw Rw Lw' Fw' Rw2.

This is quite impressive. I knew there were algorithms that should be able to fix the parity cases, but I wasn't sure how to construct them easily. How did you come up with this? Since I always use a T-perm to solve my corner-centraledge parity cases, I would need a different one. In any event, I really suspect I'm slow enough that this sort of solution would almost never be worth the effort for me.

I think it is a good idea to solve the centers, that's great!

But how do you, dbeyer and cmhardw, think when you claim that "Cubeorientation is unimportant! You might as well get a hard/good solve anyway!"

Why don't you say exactly the same thing for 5x5? Me myself, I think that you raise the chancesof getting a better solve if you do a better cube orientation. But I can't see why you think it do on 5x5, but not on 4x4?

You are of course exactly right. Chris indicated to me in a PM that Daniel was really just trying to make a point with hyperbole that 4x4x4 is not worth spending very much time choosing an orientation - you shouldn't spend very much time choosing. (In other words, even though Daniel said this, I don't think he really meant it - he was exaggerating.) I believe this is equally true with 5x5x5. When you get to 6x6x6 and 7x7x7, you can probably spend a little longer choosing orientation, since the savings are greater. But with 4x4x4 and 5x5x5, you can't afford to spend very long, and it's about equally important on both, based on my experiments. (However, it would be interesting to see someone actually do the math and calculate how much it really helps.)

Here's how I think I'm going to try it the first time for 5x5x5 (either tonight or tomorrow):
1. Orient the cube in my normal orientation. Then count the number of center pieces that are solved. If the count is 10 or greater, then I will probably not even try to do this - I will just use my normal orientation. If that many centers are solved, I suspect it's not worth the effort. (I thought about 11, but with experimentation, it seems like 10 is probably a really good number.) This takes me about 10 seconds, and it takes about 2 of those to orient the cube in the normal orientation. If I get 10 or greater, I'll write this one off as a case where I lose by trying this - I will be 8 seconds slower because I wasted time counting centers. A shame, but not too bad.
2. Memorize centers and edges as normal for 5x5x5.
3. Memorize center fix. I've worked out a simple way to do this using an image, so I can use this comfortably for multis on big cubes. But I think I may not need it for a single 5x5x5, based on my experiments so far.
4. Memorize corners.
5. Pull on blindfold.
6. Solve corners.
7. Do center fix.
8. Solve the rest normally.

I've tried about 10 attempts doing steps 1 and 3-7 (skipping steps 2 and 8), and it takes me an average of about 1:40 to do all of that, of which about a minute is the corners. I suspect that's a win over solving the extra centers, so I'm going to give it a try.

Chris or Daniel, any suggestions on a better approach? I really think this could work okay for me. I'm still not sure it's going to save me much time on average, but it seems like it will probably save me a little, so it's worth trying it for a while.

I'm really looking forward to seeing how this works out on 7x7x7, though. It seems like it could help a lot there!

 

[trying-to-speedcube...]

With M S M' S' and similar trivial algorithms, and M' E2 M E2 and similar you can always reduce the parity case to the case where the centers are an M-slice off. Then you fix it with Uw Fw' Bw Lw Fw Bw' Uw2 Rw2 Dw Rw Lw' Fw' Rw2.

This is quite impressive. I knew there were algorithms that should be able to fix the parity cases, but I wasn't sure how to construct them easily. How did you come up with this? Since I always use a T-perm to solve my corner-centraledge parity cases, I would need a different one. In any event, I really suspect I'm slow enough that this sort of solution would almost never be worth the effort for me.

Cube explorer

With some more effort I could probably find a faster and easier to memorize algorithm, but this should do for now. For me the concept of being able to solve the parity at all is more important now, making it fast will come later

 

[dbeyer]

Indeed. I used to count centers solved, and look for solved wings. Now I just accept that this is what my expected value is.

I can attain the expected value of 8 solved x-centers by a simple method. Permute a 1x2 block or "/" diagonal of x-centers. Then I preserve this block, and rotate about that axis to solve another 1x2 block or "/" diagonal block.

There I've solved 4 cubies. Now I expect another 4+ cubies to solve themselves on the other 4 faces. Which is reasonable. I expect that the other 4 faces each contain one of their proper x-centers.

On a 4x4, there are 3 independent variables.
x-centers, wings, and corners.

You are able to reduce the image count, and solve time quite simply by orienting the non-super cube to solve large amounts of x-centers.

By orienting the cube state to solve x-centers before memorization, your expected value of wings and corner permutations remain the same.

using the same system as described before, and applying it to +centers you have the same expected value of solved pieces, ignoring other portions of the cube.

You can find an orientation for +centers that solves an average of 8+ cubies.
Likewise for x-centers. However, when you take these two variables, and center point permutations into account, and only being able to choose one orientation, you are do not have the net expected value of 16+ center cubies solved.

When you use the simple system for non-parity centre point permutations, you are able to increase your gains in solved cubies. Meanwhile, during these center optimizations, you are maintaining the basic structure of your amd expected value of solved cubies on the wings, centrals, an corners.

All 12 orientations without center parity, the solving structure will be different, but the other parity structures remain the same.

Anyway, good luck.
I just was overexagerating the random toss up approach. By solving blocks, you can know that you've increased your chances of meeting or exceeding the expected value of solved cubies.

Later,
DB

 

[dbeyer]

5x5x5 Blindfolded
1. Uw B' Bw L B D Uw' Rw' D' Rw' U F L2 Fw U' Rw2 R B2 R Uw U' Lw' B2 Rw B' Dw2 R' D' U2 Lw Rw Dw B' L2 B2 Dw' B2 L2 D U' Fw' L' Lw Rw' Bw2 Uw B2 Dw2 Fw Rw Dw' Uw' R Uw Lw R' Uw2 Rw2 Uw

Scrambling from the Official orientation. White U / Green F.
I solve with the color scheme Red U / Blue F. I chose orienting to Blue U / White F. (So x' cube rotation)

ME'M'E -- Solving Centers.

U' lul' U lu'l'; Urb -> Ubl -> Lub
b2 r'F2r b2 r'F2r; Urb -> Ful -> Dlb
f'U2f u f'U2f u'; Urb -> Ruf -> Bur
b UfU' b' Uf'U; Urb -> Lfu -> Rdb
U2lU' r2 Ul'U' r2 U'; Urb -> Dbr -> Blu
Bw r2 BwlBw' r2 Bwl'Bw2; Urb -> Fdr -> Brd
ldl' U2 ld'l' U2; Urb -> Lbd -> Ulf
r'd2r U r'd2r U'; Urb -> Bdl -> Ufr

x-centers complete, 8 commutators, 65 moves.

U' Mu'M' U MuM'; Ur -> Lu -> Ub
U f2 US'U' f2 USU2; Ur -> Ru -> Df
yM2 U'r'U M2 U'r(4D); Ur -> Lb -> Dl
r'Er U2 r'E'r U2; Ur -> Ul -> Lf
U M' uR'u' M uRUw'; Ur -> Bu -> Rb
r' ER2E' r2 ER2E r'; Ur -> Dr -> Fl
r BM'B' r' BMB'; Ur -> Br -> Uf
ER'E' r' ERE' r; Ur -> Rd -> Fr
D'r2D M D'r2D M'; Ur -> Db -> Fd
U M' UlU' M Ul'U2; Ur -> Bl -> Fd

+centers complete, 10 commutators, 83 moves

L2B'L b L'BL b' L; URb -> DBr -> LFu (9
RF'R' b' RFR' b; URb -> DFl -> RDb (8
r2Dr U rD'r U' r; URb -> LDf -> BUr (9
UF'U' b2 UFU' b2; URb -> DLb -> RFd (8
l'D2l U' l'D2l U; URb -> FDr -> UBl (8
f R2 fL2f' R2 fL2f2; Urb -> RUf -> LUb (9 brealomg into new cycle
L'b2L B2 L'b2L B2; URb -> BLu -> BRd (8
R'F2R b' R'F2R b; URb -> LUb -> FLd (8 breaking into a new cycle
U'lU R U'l'U R'; URb -> FRu -> BDl (8
r' F2 Db'D' F2 DbD' r; URb -> FUl -> UFr (10
y U' R U'l2U R' Ul2U2 y'; URb -> DRf -> RBu (8
d L'U2L d' L'U2L; URb -> LBd -> FLd (8

wings complete, 12 commutators, 101 moves.

y U MD'M' U' MDM' y'; UR -> BD -> UB
U MD2M' U' MD2M'; UR -> UF -> DF
E' FUF' E FU'F'; UR -> FR -> LF
R BMB' R' BM'B' ; UR -> BR -> LB
L E'RE L' E'R'E; UR -> LD -> LB

yM' U2MU2 M' U2M.y' R'E'R U2 R'ER (cancellations of U2.y' U2 = U4y' = y')
*Centers Solved -- Using non supercube safe alg.
R2 y M'UM'UM'U2 MUMUMU2 y' R2
Edges complete, 7 commutators, 55 moves. (5 commutators + 1 alg, 54 moves.)

F R2 FL2F' R2 FL2F2; URB -> RUF -> LUB
B R2 BL'B' R2 BLB2; URB -> FDR -> BRD
yR2 DL'D' R2 DL(4U)' y; URB -> DLB -> LFU
U' R' D2 R'UR D2 R'U'R2 U; URB -> FLD -> DRF

Corners complete, 4 commutators, 35 moves.

Every set of pieces averaged sub-9 turns per commutator.
8.27 moves/commutator. 41 commutators.
4 additional moves to solve centers.
343 moves. Cube complete.

This is my system. Chris would solve very similarly. Enjoy.
Later,
DB

 

[dbeyer]

Oh comparing methods.
M2,r2,U2 + 2-cycle corners vs BH.
Trying-to-speedcube's example. 728 moves.
My example. 343 moves.

I used half the moves ... rightfully so too based on the concepts of the methods.
I didn't have parities. I actually was fortunate to have 0 parities. However, still compare the move count. This isn't a hard system.

Ehh ... oh well
Later,
DB

 

[Jude]

Ok, I have a question about 4x4x4 centre commutators, I hope this is the right place to ask it.

I just solve 2 at a time doing 3 cycles, starting with the buffer Ubl.

My question is, what do I do if my first cycle goes something like Ubl --> Lub --> Ubl? Normally what I do is I make it Ubl --> Lub --> Uxy where Uxy is a piece that is already correct on the U face. For the first time ever today however, I had a scramble where the best place to hold it had 0 correct centres on the U face, so I wasn't sure what to do. In the end I did Ubr (which was a piece of the same colour as Ubl) --> Lub --> Ubl and then started my next cycle with the buffer piece Ubr. However, at the end of the solve I was just 2 centres off (one of which was Ubl) so I thought what I did might not work.

Hope I made that clear, thanks.

 

[peterbat]

Ok, I have a question about 4x4x4 centre commutators, I hope this is the right place to ask it.

I just solve 2 at a time doing 3 cycles, starting with the buffer Ubl.

My question is, what do I do if my first cycle goes something like Ubl --> Lub --> Ubl? Normally what I do is I make it Ubl --> Lub --> Uxy where Uxy is a piece that is already correct on the U face. For the first time ever today however, I had a scramble where the best place to hold it had 0 correct centres on the U face, so I wasn't sure what to do. In the end I did Ubr (which was a piece of the same colour as Ubl) --> Lub --> Ubl and then started my next cycle with the buffer piece Ubr. However, at the end of the solve I was just 2 centres off (one of which was Ubl) so I thought what I did might not work.

Hope I made that clear, thanks.

You can either start with the next piece over (like Ubr) as your buffer, and then use a three cycle at the end (your U-layer centers will be done at that point). Or you could not change buffers and do a cycle

Ubl --> Lub -->(some random center that needs to be solved).

That's breaking into a new cycle. You just have to remember to close off the cycle at the end with the same (some random center) you started with. You may have to do this more than once:

Ubl --> Lub --> (random center 1)
...
Ubl --> (random center 1 ) --> (random center 2)
...
Ubl --> (blah) --> (random center 2)

Or you may end up with

...
Ubl --> (random center)

at the very end. But if it's at the very end, all the other U face centers will be solved at that point, so you can just use one of those in your last cycle.

That make sense?