# Big Cube BLD Discussion

#### cmhardw

Hi everyone,

I've noticed that there are often very interesting posts related to big cube single BLD, and they are always spread out across various different threads. I'm posting this thread so that hopefully we can gather all of that interesting information into one easy to find place.

Consider this thread as a spring board thread for all things big cube BLD related. By Big Cube BLD I mean solving a cube larger than 3x3x3 blindfolded. It can be a "one answer question thread" for big cube BLD, or a place to discuss methodology or theory as well.

So, if you have something interesting to say relating to big cube BLD, post it here!

Chris
---------------------------------------------------

To start off, I've noticed that recently a lot of us are posting our time breakdowns for all steps for 4x4x4 BLD, but it's buried inside of the Blindfold Accomplishments thread. Well, to start off this thread let's all collect our time breakdowns here and maybe it will be easier to discuss!

4x4x4 BLD solve:

Scramble: L' F' U' F' Uw2 B2 D' B' Fw2 Uw' Rw Fw Uw2 F Rw' U2 L2 Rw D R' B U2 F' Uw' R' Fw2 Uw' B2 D Uw2 U2 Rw B' F' U' Rw Fw U2 B2 F'

Total memorization time: 2:38.34
Total solving time: 3:21.05
Total overall time: 5:59.39

1) 06.69 orient the cube
2) 1:02.03 Memorize centers
3) 1:17.98 Memorize edges
4) 11.64 Memorize corners
5) 21.45 Solve corners (with the exception of parity)
6) 1:23.86 Solve centers
7) 09.02 Fix corner parity
8) 1:26.72 Solve edges

#### Chuck

##### Member
Nice, Chris!

My latest breakdowns:

1. Orient: 00:08.00
2. Memo Centers: 02:24.40
3. Memo Edges: 01:25.20
4. Memo Corners: 00:28.00
5. Solve Centers: 02:24.40
6. Solve Edges: 02:13.00
7. Solve Corners: 00.44.00

TOTAL: 09:47.40 (04:26.00)

Memorizing and solving centers are new things for me, so they took longer than the edges.

I can see from your breakdowns that you're memorized very fast, was it without any reviewing?

Do you solve edges with commutators too?

#### dbeyer

##### Member
One thing that I've been looking into is x-center cases. Chris and I try to view the x-centers as extentions of the 3x3 corners. Likewise the t-centers are the extentions of the 3x3 edges.

This works well, until you get into the extra 128 cases that are not present on the 3x3 corners.

Cases such as Urb -> Fld -> Ldf
Urb -> Brd -> Rdb
Especially Urb -> Rbu -> Bur
84 cases are not present because no cycle involves two pieces from the buffer.
2 more cases are lost because you can never cycle all 3 pieces on the buffer cubie.
42 more cases are lost because two stickers from the same non-buffer corner can be in the same cycle.

The original 378 corner extention cases are directly proportional from the optimal 3x3 algorithms.
However, there are some tricks that we have already found for certain cases that we didn't optimize.

Such as the Urb -> Rdb -> Brd
r' (3U')l(3U) r (3U')l'(3U)

Lets look at this classification.
We have already discussed the AnI relationship in times past.

URB and DBR are adjacent non-interchangeable cubies.
However URB is interchangeable with BRD and RDB

So Urb is interchangeable with Brd and Rdb

View one of the x-centers that is paired on a corner extention as the interchangeable with the buffer. Now view the other as the lone cubie to be inserted.

Urb and Brd are interchangeable on the r slice. So we need to insert the Rdb into the Urb. Insertions and interchanging must be done on parallel planes as we've seen before. So lets try this neat trick, the deep turns will help you see what's going on.

r' interchanges the Urb and Brd.
now we need to insert on and L/R planes. Is that really possible? (3U') brings both the Urb and the Rdb to the l slice. Now you can insert, (3U')l(3U). Interchange again, and undo the insertion. 3-cycle of x-centers complete.

Urb -> Rbu -> Fru and its inverse
Urb -> Bur -> Lub and its inverse
these six cases can all be solved in similar manner.

I am really curious about the Urb -> Rbu -> Bur case. Is this case optimal at 10 moves? =/

Last edited:

#### Chuck

##### Member
Reading Mr. Beyer's post, suddenly I feel so noob...

#### Muesli

Reading Mr. Beyer's post, suddenly I feel so noob...
You feel like a noob?!

I've just started learning

#### LarsN

##### Member
Wow, this took me a while to decifer, but I think that's because of some mistypes. If I'm wrong then it's simply because I don't understand.

One thing that I've been looking into is x-center cases. Chris and I try to view the x-centers as extentions of the 3x3 corners. Likewise the t-centers are the extentions of the 3x3 edges.

This works well, until you get into the extra 128 cases that are not present on the 3x3 corners.

Cases such as Urb -> Fld -> Ldf
Urb -> Brd -> Rdb
Especially Urb -> Rbu -> Bur
84 cases are not present because no cycle involves two pieces from the buffer.
2 more cases are lost because you can never cycle all 3 pieces on the buffer cubie.
42 more cases are lost because two stickers from the same non-buffer corner can be in the same cycle.

The original 378 corner extention cases are directly proportional from the optimal 3x3 algorithms.
However, there are some tricks that we have already found for certain cases that we didn't optimize.

Such as the Urb -> Rdb -> Brd
r' (3U')l(3U) r (3U')l'(3U)

Lets look at this classification.
We have already discussed the AnI relationship in times past.

URB and DBR are adjacent non-interchangeable cubies.
However URB is interchangeable with BRD and RDB

So Urb is interchangeable with Brd and Rdb

View one of the x-centers that is paired on a corner extention as the interchangeable with the buffer. Now view the other as the lone cubie to be inserted.

Urb and Bdr are interchangeable on the r slice. So we need to insert the Rbd into the Urb. Insertions and interchanging must be done on parallel planes as we've seen before. So lets try this neat trick, the deep turns will help you see what's going on.

r' interchanges the Urb and Bdr.
now we need to insert on and L/R planes. Is that really possible? (3U') brings both the Urb and the Rbd to the l slice. Now you can insert, (3U')l(3U). Interchange again, and undo the insertion. 3-cycle of x-centers complete.

Urb -> Rbu -> Fru and its inverse
Urb -> Bur -> Lub and its inverse
these six cases can all be solved in similar manner.

I am really curious about the Urb -> Rbu -> Bur case. Is this case optimal at 10 moves? =/
And then the alg is for 4x4 and won't work for 5x5 which I used to test the case. On 5x5 it's:
r' (4U')l(4U) r (4U')l'(4U)

It's not because I tried to be rude. But if I'm correct then these corrections might help people understand. And I really like that alg

#### HASH-CUBE

##### Member
i know the concept of solving a big cube blindfolded, but i don't dare to, at least i can't memorize much :S

but i might in the future

cool idea that we all gather here to discuss this, specially if someone has questions

#### cmhardw

Mixing Visual Memory with Image Memory?

Ok, so this is something I have been thinking about a bit recently. What got me thinking was the fact that Ville is so incredibly fast with visual memo on big cubes, and also that Rafal admits that he sometimes uses visual mixed with memory methods for big cubes. I've been thinking about creating an image that, when it comes up, means that I have the next part memorized visually. I've noticed that sometimes I do see a neat pattern traced out by the cycles of the cube, or maybe for some reason the images I have to interact are all inanimate objects that for whatever reason don't fit well together and just become a list of three inanimate objects from left to right. Perhaps in this case I could memorize in my journey location some sort of marker that tells me "hey the next stuff is visual!" and then that could spark the fact that I have to remember what came visually after the last piece I memorized via images.

I wonder if Rafal or Ville would be willing to comment on how they use visual in their memorization. Seeing as how I tend to always solve slower when memorizing the whole cube visually (3x3x3), as compared to using images I would only do this occasionally. I think the "visual marker" image would be the best way to still memorize with images, but occasionally have a short burst of visual memory added to this.

Does anyone have any opinions on this? Of course I'm interested to hear from Ville and Rafal, but also how do you pure visual, or pure images, memorizers feel about mixing the two methods? I'm fairly positive on the idea, but I haven't actually tried to do it yet during a solve.

Chris

#### LarsN

##### Member
I can see what your getting at Chris. Only I see it from the opposite perspective. Sometimes when I use visual memo I get some cycles that doesn't have nice patterns. These I have to remember "brute force" which means repeatting them a lot, which obviously makes my memo time longer. At the moment I'm thinking of that as lack of practise, but maybe an image would help with these cases?

But then I would go back to learning a lot of images, which was the reason why I started using visual instead.

#### MatsBergsten

Does anyone have any opinions on this? Of course I'm interested to hear from Ville and Rafal, but also how do you pure visual, or pure images, memorizers feel about mixing the two methods? I'm fairly positive on the idea, but I haven't actually tried to do it yet during a solve.
Chris
Even if I'm not Ville or Rafal I am still a neighbour to both their countries

I mix memory methods sometimes, when, like you say, the normal method does not feel good or visual seems to be very easy. But I have not thought of the concept of a "marker" for switching method. I do not use images (yet) but just "silly sentences", but the same thing you note that for me the letters does not build sentences well. If there are pairs of consecutive opppsite cubies or some centers in a row on lower left corner or so I change method. But mostly just for the trailing cubies in a cycle, I do not swap back to sentences again.

#### mande

##### Member
I did a memo on Chris' scramble since I have a terrible cube, and was sure to DNF even if the cube was good.

Orient: 7s
Centre memo: 3:30
Edge memo: 6:40
Corner memo: 00:30
Total memo = 10:48

My solve time is generally about 20 minutes which I consider pretty hopeless, so I was hoping I could pick up a few tips here. I use pure commutators for centres and edges and do corners like on 3x3. The parity solve is the last thing I do (if applicable). For centres, I get two centres (with successive positions in the cycle) onto the U face, then perform a commutator aqa'q' where q is either U, U' or U2 (depending on the positions of the two centres on the U face). For edges, it is quite similar. I setup two edges (successive edges in the cycle) in positions (preferably on U face) such that I can move 1 to the other's position in just one move (its hard to explain in words). Then I do a commutator with the q move being the move required to do what I just said. For corners, I just use 3OP like I do for 3x3.
I memo using letter object combinations. I just make up stories on the flow and do not have fixed words for fixed letter pairs.
Any comments and suggestions would be greatly appreciated (if anyone understands what I have written ).

#### Mike Hughey

Staff member
Wow, Chris, that sounds like a totally cool idea! The only problem I can see with it is that I'm so conditioned now to do the images, I think it would be hard to swap. But I'm going to have to think about it a bit - it seems like it could be so powerful.

#### Mike Hughey

Staff member
I tried Chris's scramble. It was kind of a disaster for me time-wise, although I solved it. Memorization did not go well for me at all - it's apparently a bad night for me (bad news because I'm about to do the weekly multi ). (I was also treating this solve as my solve for the Time Machine Competition.) Anyway, here were my breakdowns for Chris's scramble in post #1:

Total memorization time: 5:34.22
Total solving time: 4:54.61
Total overall time: 10:28.83

1) 14.54 orient the cube
2) 1:56.02 memorize centers
3) 2:28.21 memorize edges
4) 33.92 review memorization
5) 21.53 memorize corners and pull on blindfold
6) 28.53 solve corners except parity
7) 1:45.56 solve centers
8) 14.45 fix corner parity
9) 2:26.07 solve edges

Since it was a bad solve for me, it's not necessarily a good measure of my relative times; it seems like these breakdowns are most useful on good solves, not average or bad solves. So it's probably not very useful data.

#### rachmaninovian

##### Member
dbeyer: to find the optimal solution I believe that you can set parameters and check on Clement Gallet's wonderful solver. However it only runs on mac/linux

cheers

#### dbeyer

##### Member
Would anybody be interested in looking up the optimal algorithm for
Urb -> Rbu -> Bur

I've found several nice 10 move solutions.
On a 4x4:
(3L) y [rU2r' u' rU2r' u] y' (3L)'
U x [uR2u' r' uR2u' r] x' U'

For clarificaiton, I am not giving the directional solution but rather the components fully written in SABA'B'S' form
S: U x
A: uR2u'
B: r'

A nice trick to look at for this case, [Urb, Fld, Ldf]

Fld and Ldf are in the same corner orbit around the DFL corner.
So lets view the Urb as the lone cubie.
r' and we have interchangeability with the Urb and Fld on the F plane.
r2 and we have interchangeability with the Urb and the Ldf on the f slice.

So we see our insertion and interchanging points for an A9 cancellation.
Obviously, they are on the F plane and f slice. Now lets make this a finger trick friendly case. Quite simply by doing an x cube rotation. Preserving r turns, and transforming F and f turns into U and u respectively.

x r' U2 r'u'r U2 rur2 x'

Enjoy,
Later,
DB

#### cmhardw

Would anybody be interested in looking up the optimal algorithm for
Urb -> Rbu -> Bur
I have a truly marvelous demonstration of a 9 mover for this case which this margin is too narrow to contain.

Actually I really did discover a way to do this cycle in 9 turns, but as a side effect it also does a 3 cycle of x-centers around the corner diagonally opposite through the cube.

I'll try to see if the idea I am using for my 9 cycle can be done in a such a way as to avoid the "side effect" cycle on the back of the cube.

Chris

#### V-te

##### Member
I'm not really a BLD cuber, but may I ask Mike and Chris, How do you memorize so fast? I assume it has to do with intelligence, or memo system righ?

#### cmhardw

I'm not really a BLD cuber, but may I ask Mike and Chris, How do you memorize so fast? I assume it has to do with intelligence, or memo system righ?
Having a memory system certainly helped me to speed up my memorization quite a bit, but there are people who are truly out of this world fast at memorization using either pure visual (pure rote memory) techniques, or a mix of this and more standard memory techniques.

Part of it also is practice. There is a certain "fudge factor" when memorizing that you have to learn to fight through. For example, when you memorize something and you know that you know it, but you can't recall it. Then you REALLY think hard and finally it comes to you. You get a lot of that, but you have to learn to fight through and recall it faster and faster with each practice solve. BLD is what I practice, I don't really speedsolve much anymore compared to what I used to. Like anything, the more you practice the better you get at it. Now if only I could figure out how Ville and Rafal achieve their level of craziness

If you're interested in journey/image memorization techniques here is a link to what I use.

Chris

#### siva.shanmukh

##### Member
Would anybody be interested in looking up the optimal algorithm for
Urb -> Rbu -> Bur
I have a truly marvelous demonstration of a 9 mover for this case which this margin is too narrow to contain.

Actually I really did discover a way to do this cycle in 9 turns, but as a side effect it also does a 3 cycle of x-centers around the corner diagonally opposite through the cube.

I'll try to see if the idea I am using for my 9 cycle can be done in a such a way as to avoid the "side effect" cycle on the back of the cube.

Chris
That will add some moves to it. Won't be 9 anymore. anyway is it something this:

y U' r' f r' f' r2 U y'

anyway I liked dbeyer's version : written in SABA'B'S' form
S: U x
A: uR2u'
B: r'

#### cmhardw

That will add some moves to it. Won't be 9 anymore. anyway is it something this:

y U' r' f r' f' r2 U y'
Wow, no your alg is much more efficient at the pseudo-cycle than mine. My pseudo cycle was:
U' r U2 f' U2 r' U2 f U'

anyway I liked dbeyer's version : written in SABA'B'S' form
S: U x
A: uR2u'
B: r'
Yes but it is a 10 mover Daniel and I were just talking to try and see which strategy would be more fruitful: try to prove there does not exist a 9 mover for this case somehow, or, hopefully, discover something amazing that only takes 9 moves to solve this case!

Chris