# How I Solve Centers on 4x4x4 BLD



## Mike Hughey (Nov 14, 2007)

I mentioned on another thread that I might try to describe how I go about doing centers on big cubes BLD. Since I'm really just a beginner at this point, my perspective might be useful because I don't take as much for granted as some of the experts might - at least, that's my hope. Anyway, I sent this to Dan Cohen earlier, and he seemed to think it was useful, so here it is. I hope it's helpful.

This will probably be a bit more rambling than a typical tutorial, since my goal here is not to give the perfect working system. Instead, I’m just trying to show you how I do them, to give you another perspective on how they can be done. I’m not as good at this as the other people who have done tutorials. I’ve probably attempted less than 200 big cubes BLD in my life, and I’m very slow, so I’m far from being an expert. But I do have about 50% accuracy on the 4x4x4, and about 25% accuracy on the 5x5x5, so at least this method can work (even though I’m pretty slow).

For now, I'll just focus on the 4x4x4 centers. The 5x5x5 x-centers work exactly the same as the 4x4x4, and the 5x5x5 +-centers are actually easier to see, in my opinion. So you should first try to learn the 4x4x4 centers, and then the 5x5x5 centers should follow fairly easily, probably without my help. But I might post something about the 5x5x5 +-centers later, if I get the chance.

Anyway, for the 4x4x4 centers, I just use straight commutators, much as Chris Hardwick has described before. I must admit I never really got much benefit from Daniel Beyer's approach to centers, although I more or less understand it, and I can see how it might help some other people. If you find my description confusing, you might want to go back and try his approach; it might make more sense to you. It can be found at http://www.speedsolving.com/showthread.php?t=697.

I really think the whole big cubes BLD thing is all about finding a method that works well for each individual - everybody sees them a little differently.

I'm going to go ahead and do something that Chris and Daniel have expressly avoided, if only because I think it will make it easier to describe. I'm going to reveal my lettering scheme for the centers. Sorry if this seems like the wrong thing to do for some people. Please just use this scheme to follow my description, and then consider changing it to something that feels comfortable for you. I only want to make it easy to follow my examples. For each face, starting with the top face, I assign 4 letters, beginning in the upper left of the face, and going clockwise. So, the faces look as follows:

Top (viewed from the top, with the front face towards you):

A B
D C

Front (viewed from the front with the top face above):

E F
H G

Right (viewed from the right with the top face above):

I J
L K

Back (viewed from the back with the top face above):

M N
P O

Left (viewed from the left with the top face above):

Q R
T S

Down (viewed from the bottom with the front face away from you):

U V
X W

This lettering scheme might not be very logical from a mathematical perspective, but it works for me. I just figured it might be helpful for people to see exactly how I implemented things, in case they’re having trouble working it out for themselves.

Memorization
-----------------

I realize that many people reading this will have already worked out memorization, so sorry if it’s a waste to describe this, but I probably need to do it so you can follow what I do with execution. So here is how I memorize the centers.

As all previous tutorials have mentioned, the first thing to do when solving the 4x4x4 is to pick the orientation for your cube. By picking an orientation with more centers already solved, you can potentially get out of solving some centers. Most people seem to favor pre-solved Down face centers, since they’re potentially harder to see the commutators for. I think I find Back centers even harder than Down centers, but I’ve always tried to encourage myself to get good at seeing the commutators anywhere on the cube, so I’ve avoided this approach. Instead, I just go for the orientation of the cube with the most solved centers, even if they’re all on top. I really don’t think it makes a lot of difference where they are, once you get used to freestyling the commutators anywhere on the cube.

Okay, so once you’ve picked an orientation, the next step is obviously to memorize the cube. For the centers, you just start at an arbitrary point, and keep track of what you’ve done so far as you go, so you can make sure you solve each center piece just once. I always start at A, unless it’s already solved, in which case I go to the first unsolved letter. If it’s unsolved, it will need to go to a different face. I find the first piece alphabetically on that face that still has not been solved, and add it to my memorization list. I have taken to using Chris Hardwick’s neat cube-holding trick for keeping track of what has already been done. He suggested placing your fingers on the most recently-used cubie on each face, to keep track. Since I always choose the next piece alphabetically, by having one finger placed on each face, I always know by the position of my fingers what the next piece should be. For instance, say I’ve memorized the sequence A-I-M-Q-N-R, and now the piece at R is the color of my Back face. Since I will have a finger on the N piece, I know that the piece at R should go to O, so I’ll add O to my sequence: A-I-M-Q-N-R-O.

Like Chris, after my first piece in a cycle, I pair up the letters as I go and memorize a word or small phrase to go with that pair of letters. Unlike Chris, I have not memorized a whole list of words for all possible letter pairs. Instead, I just try to make them up as I hit them. It’s not as efficient, but it’s a lot less overwhelming for a beginner – doing it this way, you can start trying solves as soon as you understand your method! Anyway, what happens is that after a while, you start to find that words or phrases “stick” for pairs you’ve seen several times before, so eventually you wind up memorizing a list of words anyway, without even trying. I’m guessing I probably already have standard values for probably 25% of the possible pairs already, even though I’m sure I’ve still done less than 100 successful big cubes BLD. But one nice thing about making up the words or phrases as you go is that, if you’re putting them together in a story, you can make up a different one that goes better with your story if it occurs to you quickly.

When I get back to the face I’ve started with (the Top face to begin with, unless it was already solved), if I’m in the middle of a pair of letters for my memorization, I’ll just keep going with my cycle. But if the piece that needs to go to the starting face is the second letter in a pair, I will just start a new cycle with the next piece. It’s really not important to do that, but that’s how I do it.

I don’t mind having many cycles – my memorization method (what little there is of it) works just as well with many cycles as with few cycles. I have a list of persons for each letter that I use at beginnings of cycles. I also try to use them at the ends of cycles, but I’m not strict about that. I have a different categorized list of memorized persons for each type of piece I’m solving, so I use people I know for x-centers, and singers for +-centers on the 5x5x5. (I use cartoon characters for wing edges and 4x4x4 edges.) So when I get to a person, I know it’s the start of a new cycle, and the type of person keeps me from mixing up the center cycles with the edge cycles, for instance.

I’ll do an example, to hopefully make all this a little clearer. Here’s a scramble (the first 4x4x4 BLD scramble from last week’s competition):

u2 r2 B2 f F2 L' r' R' f2 L' r' R' B2 u' U B2 f' F u' B' L2 r' D2 u' U' F D U B f2 u2 U' r2 B' f' F' R2 D2 u2 U'

(Remember that for scrambles, including this one, lower case letters mean turning both slices, but in the discussions below, I’m going to switch to using lower case letters to refer to individual slice moves.)

For this scramble, I chose to do z y’ to orient. That has 9 centers already solved – a somewhat lucky scramble from the perspective of the centers. Starting at A, my first cycle looks like this:

A->M->I

Now piece I can simply go back to A, so I can treat that as my whole first cycle. I would now have fingers placed on A, M, and I, to keep track of what I’ve already done. Now the second cycle starts with D, since B and C are already done:

D->O->Q->G->S->U->P->K->T->V

V needs to go to the top face, and D is already used, so that cycle is done. At this point I would have fingers on D, G, K, P, T, and V. That lets me know the next one that needs to be done is H, so I start the third cycle with H:

H->W

Now I have fingers on D, H, K, P, T, and W, and I can see that L and X are both already solved, so that means I’ve done all 6 faces – I’m finished memorizing. To double-check, I can count to see that I have 15 pieces in my memorization, and there were 9 already solved, so that adds up to 24, proving I’m done. Three cycles.

Execution
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Now I just start doing 8-move commutators to cycle 3 pieces at a time to solve the cycles. Much has been written on commutators, and it is most important that you get good at commutators to be successful at the big cubes BLD. But I’ll try to give an explanation that seems simple to me here, in the hopes that maybe another perspective on them will be helpful. I’m not going to bother with the math of it – the goal here is to give something that works. Please read the other articles on commutators if you want to understand why they work.

I think of each piece in a three-cycle as having a “target”. So in my example from the memorization section above, A has a target of M, M has a target of I, and I has a target of A. I say “A needs to go to M”, or “M needs to go to I”, or “I needs to go to A”. My way of looking at commutators is that I want to get any two of the three pieces that are going to cycle in a position where a single slice turn will replace one of those pieces with another piece. The two pieces are said to be interchangeable – a single slice move interchanges the two pieces. Once I manage that, I need to put the third piece in a place where performing the interchanging slice move will not move that third piece, and where a move that’s out of the plane of the interchanging slice move can make it possible for the third piece to go to its target with a move that’s parallel to the original interchangeability slice turn. If I can do that, I’m set up for my commutator.

All of that is really hard to understand as just a concept. I think the best way to understand it is not a description like this, but is rather an example. So let’s solve my example solve, and I’ll show you some of my “tricks” along the way. This stuff is probably going to be terribly confusing to just read, so I’d recommend the following as you’re reading this: look at the three-cycle I’m trying to solve, understand where each piece needs to go, and then find the place where I do the actual moves on a separate line and execute them to see where everything goes. Then, once you’ve tried it and seen how it works, then read the actual paragraph I’ve written and see if you can make sense of it. It’s probably going to be confusing anyway, but maybe that way it won’t be quite as confusing.

Don’t forget I did z y’ to orient.

First, we have the cycle A->M->I. This is set up nicely – no setup moves will be necessary. In general, I try to go for zero or one setup moves, but will settle for two in a couple of very specific cases where one doesn’t seem to me to be possible. In this case, we don’t need any, because M and I are on the u slice (remember I’m not scrambling now, so lower case letters means just a single slice turn for this whole execution section). If you make the turn u, you can say that “M goes to I”. So now I need a way for my third piece, A in this case, to be able to go to its target, which is M. So I look at this and say, “A needs to go to M.” I can see that if I perform r’, I’ve put M in a position where U2 will make A go to M, which is what I need to do. So I do r’ U2, then I do r to put it back, then I replace M with I by doing the interchanging move u’ (which now puts I at the place M was originally), then do r’ which will allow I to go to A (which is where it needs to go), and then U2 to put it there, and then r to put it back, then u to restore the interchanging move. The whole thing is:

(r’ U2 r) u’ (r’ U2 r) u

which as you can see is a standard commutator. I never bother to figure out what A and B in the commutator are – I just think of where each piece needs to go, and it all works out. When I’m doing the commutator, I always count to 8, to keep track of where in the commutator I am. I find that it really helps me avoid mistakes.

Now we can go to the second cycle. The first three in it are D->O->Q. You can look at these three pieces and quickly see that no two of them are interchangeable. Since none of the pieces are on opposite faces from each other, it’s bound to be possible to use only one setup move to make this work – that’s a rule of thumb that can be helpful. If you have three faces all next to each other like this, you know you need no more than one setup move. Also, since none of them are interchangeable, you’ll need at least one setup move, so you know you’re looking for exactly one setup move. We have a bunch to choose from – you can try most anything that will line up two pieces to be interchangeable. But remember that we will need to take the third piece’s target and move it out of the plane of the interchangeable slice move so the third piece can “go to” its target. From a practical standpoint, that means we want to keep the third piece from being on the same slice with its target. For this example, I can see that Q needs to go to D, and Q and D aren’t on the same slice with each other. Therefore, a good option is to see if I can move the other piece, O, to be interchangeable with D. So, I do B’ as a setup move, which makes D and O interchangeable. Now Q needs to go to D, so I do f’ to make that possible, then I do L2 to make Q go to where I moved D, then I do f to put it back, then I do the move which replaces D with O, which is l, then I move O over to the left face so I can move it to Q by doing f’, then I move O to Q with L2, then f to put it back, and finally l’ to undo the interchanging move. Now I need to undo my setup move, so B finishes it. The whole thing is:

B’ (f’ L2 f) l (f’ L2 f) l’ B

Now since Q went to D, D now needs to go to G, so our new three-cycle is D->G->S. Here G and S are interchangeable already with the slice move d’, so no setup move is necessary unless the third piece is also interchangeable with both of the other pieces, which it’s not here. So all I need to do is take my third piece D, and move its target out of the plane of the interchangeable slice move up to be replaced, by doing the move r. Now I make D go to G with U’, then put it back with r’, then I replace G with S by doing d, then I move S up in position to move it to D by doing r, then I do U to move S to D, then I do r’ to put it back, then finally d’ to undo the interchanging move. This whole thing is:

(r U’ r’) d (r U r’) d’

Now S went to D, so D now needs to go to U; our new three-cycle is D->U->P. This is a case that comes up commonly. When two of the pieces that need cycling are on opposite faces, and all 3 are on different faces, the easiest thing is to make it so the two pieces on opposite faces are interchangeable by turning a slice that’s parallel to the face of the third piece. In this case, D and U are on opposite faces, and they can be made interchangeable on the f slice by doing the move D. Then, we need P to go to D (P is the non-interchangeable piece), so we do l’ to allow P to go to D, B2 to move P to D, l to put it back, f2 to interchange D and U, l’ to allow U to go to P, B2 to move U to P, l to put it back, and f2 to undo the interchanging move. Then we undo our setup move with D’. The whole thing is:

D (l’ B2 l) f2 (l’ B2 l) f2 D’

I have noticed that when both of the middle moves are double turns like this, it’s easier, since you’re less likely to forget which direction the turn needs to go. So often I actually go a little out of my way to try to setup for double turns like this. I don’t know if anyone else agrees, but it works for me.

Next three-cycle is D->K->T. Again, we have 2 pieces that are on opposite faces: K and T. We can make them interchangeable parallel to the U face by doing the setup move L’, so they’re interchangeable on the d slice. Then we want D to go to K, which we do with b U2 b’, interchange with d2, move K to T with b U2 b’, and then undo the interchange move with d2. Then undo the setup move with L. The whole thing is:

L’ (b U2 b’) d2 (b U2 b’) d2 L

Now we have D->V left. We need that to be a three-cycle, so we can make the third piece be any piece on the same face as D which is already solved. One cool trick here is that if both of the pieces on the same face are not on the same slice as the piece on the opposite slice, we can do this in a single setup move. So I pick D->V->A as my three-cycle. Then, I can do (R r) as my setup move, and now the two interchangeable pieces are D and A (they’re both on the U face, so turning the U face interchanges them). So now I can make it possible for my third piece V to go to A by doing b’, then move V to A with u’, then put it back with b, then replace A with D by doing U, then put D in position to go to V with b’, then move D to V with u, then put it back with b, then undo the interchanging move with U’. Then I undo my setup move with (R’ r’). The whole thing is:

(R r) (b’ u’ b) U (b’ u b) U’ (R’ r’)

Now all that is left is H->W. We need to pick a third piece somewhere in H’s face for W to go to. F is a really good choice because by doing u’, W can move to F without affecting the front face. When doing these pairs, you have to be careful to make sure your middle commutator move is parallel to the face of the first and third piece – otherwise it won’t work. This whole thing is:

(u’ b u) F2 (u’ b’ u) F2

And we’re done with centers! Everything else should be left untouched (assuming I didn’t mess this up). One nice thing about centers is that there’s never any parity left over to worry about. By cycling with one of the solved pieces, you avoid this entirely.

I hope that was more helpful than confusing to at least someone out there. Good luck with your big cubes BLD!


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## masterofthebass (Nov 14, 2007)

Ok... Mike gave me this post ahead of time to make sure that people were able to understand it. I was never able to figure out commutators before, and after just one or two times through this tutorial I'm able to do the centers on an cube (if I have the cycle written out in front of me). This gives a great explanation of freestyling commutators for those who have never really understood it. 

Thanks again Mike!


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## tim (Nov 14, 2007)

Wow, great work, Mike. I'll read it completely later when i have more time, but it looks quite helpful, thanks .


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## AvGalen (Nov 14, 2007)

That is our faith Mike. We can explain things very well to others so they can do it faster than we can 

I will try this next year, after I get another succesfull multi-blind


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## Mike Hughey (Nov 14, 2007)

AvGalen said:


> That is our faith Mike. We can explain things very well to others so they can do it faster than we can


How very true. I figure Dan will probably beat my best time with his first successful solve. (Just like Lucas Garron and Tim Habermaas.)



AvGalen said:


> I will try this next year, after I get another succesfull multi-blind


I'm looking forward to it! Then at least I might not be the slowest big cubes BLD solver out there. Just the second slowest.


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## AvGalen (Nov 14, 2007)

> Then at least I might not be the slowest big cubes BLD solver out there. Just the second slowest.


I don't think that is a good reason for me to start doing it 

You should look on your big-blind-skills a little differently: You are not the slowest out there at all. Some people have a 100% DNF-rate (Erik) and "6.543.210" other people on this planet are even worse than people like Erik.

(this kind of logic helped me during my 100% unsuccesfull 3x3x3 blind-streak: "I am the number 7 of the world in multi-blind" (me) and "you are the fastest 2/2 multi-blinder in the world" (Gilles vdP))


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## tim (Nov 14, 2007)

Mike Hughey said:


> AvGalen said:
> 
> 
> > That is our faith Mike. We can explain things very well to others so they can do it faster than we can
> ...


I actually wasn't faster than you at my first attempt. My first successful attempt was about 22-25 minutes.



Mike Hughey said:


> AvGalen said:
> 
> 
> > I will try this next year, after I get another succesfull multi-blind
> ...



lol, that's mean


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## Mike Hughey (Nov 14, 2007)

Of course, I'm probably wrong about my being the second slowest after Arnaud tries it. His skills seem particularly well-matched to big cubes BLD. I have this sneaking suspicion he might be faster at 4x4x4 BLD than at 3x3x3 BLD. In which case he'll blow me away too. (And sorry Arnaud, I didn't intend to be mean.)

And I guess it's true that Lucas's first successful BLD solve was also slower than my (current) best. But probably not his second one. (And probably not yours either, Tim.)


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## masterofthebass (Nov 14, 2007)

Yeah... this week isn't really good for really learning anything because it's the week before our Thanksgiving break, but next week, I have off the entire week and don't need to do anything but BLD! Another thing, I think that someone should find a better parity fix for r2. With Erik's, you end up with a couple of U and D centers swapped, and it would be nice to just be able to use a k4 algorithm. My idea would be doing an F2 which puts the 2 swapped edges at FUl and BUr. Then you could do an algorithm that switches some U centers and swaps those 2 edges. I would try to find an algorithm, but I can't figure it out. If anyone else does something different for r2 parity, it would be great to know.


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## Mike Hughey (Nov 14, 2007)

masterofthebass said:


> Another thing, I think that someone should find a better parity fix for r2. With Erik's, you end up with a couple of U and D centers swapped, and it would be nice to just be able to use a k4 algorithm. My idea would be doing an F2 which puts the 2 swapped edges at FUl and BUr. Then you could do an algorithm that switches some U centers and swaps those 2 edges. I would try to find an algorithm, but I can't figure it out. If anyone else does something different for r2 parity, it would be great to know.



I'm confused about this. When you're talking about r2 parity, are you just talking about the situation where, at the end of the last cycle, if you do r2, you wind up with BUr and FDr swapped? If you do r2, aren't the centers okay, and then all you have to do is swap BUr and FDr? I've only tried 11 r2 solves on the 4x4x4 (all last weekend - and I got 2 of them right, but I don't remember if they had parity or not), so I wonder if I'm missing something. It seems like it's as easy as doing U2 (R r) as a setup and then doing the simple single edge flip algorithm to swap them. (r2 B2 U2 l U2 r' U2 r U2 F2 r F2 l' B2 R2) Is there something I'm missing that makes it harder than this? There probably is, but I just don't have enough experience with r2 to see what it is.

Actually, this was one of the things I liked about r2 over commutators for 4x4x4 edges - the parity fix was always done in the same place, so you don't have to think about it after you've done it a few times. But maybe my method doesn't really work (in all cases)?

Edit: Actually, I just realized that I typed in the setup wrong. I really meant U2 (R r) (B b f'), and this only works because I solve the centers first. So I guess the issue is the order you solve it in. I do corners, then centers, then edges, and then I fix any parities after that.


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## tim (Nov 14, 2007)

Mike Hughey said:


> masterofthebass said:
> 
> 
> > Another thing, I think that someone should find a better parity fix for r2. With Erik's, you end up with a couple of U and D centers swapped, and it would be nice to just be able to use a k4 algorithm. My idea would be doing an F2 which puts the 2 swapped edges at FUl and BUr. Then you could do an algorithm that switches some U centers and swaps those 2 edges. I would try to find an algorithm, but I can't figure it out. If anyone else does something different for r2 parity, it would be great to know.
> ...



Your method works in all cases. And Erik's algorithm isn't really a problem, if you solve centers first (except for supercubes).


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## Pedro (Nov 15, 2007)

nice one, Mike

looks like I think a lot like you on center commutators 

but...I still have some problems with ones like this:

Ubr -> Dfl -> Ufl

how am I going to solve that? 

I remember that at Gingko Hostel, Budapest, Chris showed Stefan something similar to this...but I can't remember what he did 

I tried [D l2 D', U2], but that messes up the edges...


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## Mike Hughey (Nov 15, 2007)

Pedro said:


> but...I still have some problems with ones like this:
> 
> Ubr -> Dfl -> Ufl
> 
> ...



That's one of the worst cases. I still don't know how to do that with zero or one setup moves. So I just do 2 setup moves for that case, I'm afraid. I think that's the only case for centers where I have to settle for 2 setup moves (at least, it's the only case I can think of right now). It happens when you have 2 centers directly opposite each other and the third center is in the same face as one of the other two.

For that particular case, I do (L' l') B' as my setup moves. Then I can do l' U' l u2 l' U l u2 (or [l' U' l, u2] in commutator form), and then undo the setup moves. I've gotten used to doing this case this way. I always use my second setup move to make it so that I can do the commutator in the forward rather than the backward direction, which is why I chose B' instead of F here. It gives me slightly less to think about.

However, if this was a case where I just had to exchange Ubr and Dfl, for instance (those are B and U, in my lettering scheme), instead of having to 3-cycle, I would definitely make this easier by picking a different piece from Ufl (D) to cycle with. By picking Ufr (C) instead (making the cycle Ubr -> Dfl -> Ufr, or B -> U -> C), I could make it so much easier by just doing (L' l') as my setup move. Then I can do r u2 r' U r u2 r' U' (or [r u2 r, U]) to fix it.


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## masterofthebass (Nov 15, 2007)

I actually should've checked chris's thread on 4x4 BLD. He adds an algorithm to the parity that switches the centers back to the way they were. You can just do U2 (Rr) and then do his algorithm for the parity fix. It works no matter when you fix it, because it only switches the 2 edges. I probably should've looked somewhere else.


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