Big Cube BLD Discussion

[Ollie]

What program do you use? I need to learn to code in something other than VBA.

Top job!

 

[DrKorbin]

What program do you use? I need to learn to code in something other than VBA.

Top job!

A python script

 

[Mike Hughey]

So in my opinion, this trick is not worth. 1.5 solved centers are not worthy of awkward memorization of centers and additional alg for wings.

I always had a gut feel this was true - it's very nice of you to do the math to confirm it. Now I'm sure I'll never try - I agree that 1.5 solved centers aren't worth the effort.

To mention a similar-but-different idea, how do you feel about François' approach (I think it's François who does this?) of solving centers to arbitrary locations (so two centers that are supposed to be adjacent are opposite, or vice-versa), then fixing the centers afterwards? He's slower than you are - more my speed - but apparently he considers it to be worthwhile.

 

[DrKorbin]

To mention a similar-but-different idea, how do you feel about François' approach (I think it's François who does this?) of solving centers to arbitrary locations (so two centers that are supposed to be adjacent are opposite, or vice-versa), then fixing the centers afterwards?

Sounds interesting - could you please give some details?
For example, my front side is green, my right side is red. But I see that there are many red centers on front and green centers on right side, so I solve all red centers on F, green on R, and after that I swap all green centers with all red centers, am I right?
I'll try it.

 

[bobthegiraffemonkey]

To mention a similar-but-different idea, how do you feel about François' approach (I think it's François who does this?) of solving centers to arbitrary locations (so two centers that are supposed to be adjacent are opposite, or vice-versa), then fixing the centers afterwards? He's slower than you are - more my speed - but apparently he considers it to be worthwhile.

I can't think of a fast way to swap 2 adjacent centres, do you know a good alg? I might try this if the centre fixes aren't too horrible.

 

[Mike Hughey]

Sounds interesting - could you please give some details?
For example, my front side is green, my right side is red. But I see that there are many red centers on front and green centers on right side, so I solve all red centers on F, green on R, and after that I swap all green centers with all red centers, am I right?
I'll try it.

That's the idea as I understand it. I seem to recall it was him (TMOY) who was said to have done this, but I can't find a post that says it, and I don't know algorithms for it. He also memorizes where the center pieces are, rather than memorizing where they need to go. So he can memorize centers really fast, but it slows him down because he has to work out the cycles while solving. He talks about that part of it here.

I think he's one of the more innovative big BLD solvers out there, but he's not fast enough to get as much attention as perhaps he deserves.

Edit: I just realized that he goes ahead and describes the first part I talked about later in that same thread. So it really is him that did it.

 

[Mike Hughey]

I can't think of a fast way to swap 2 adjacent centres, do you know a good alg? I might try this if the centre fixes aren't too horrible.

I just found one!
M' f M F2 M' f' b' M F2 M' b M

(M means r' l)

It's not too bad. I wonder if he has a better one, though.

 

[TMOY]

The ones I'm using are l E' L2 E l' r' E' L2 E r for 4BLD, and l Ew' L2 Ew l' r' Ew' L2 Ew (M' r) U' l' r U M U' l r' U (Ew means all 3 inner slices, M means only the middle one) for 5BLD.
For bigger cubes, just use multiple slices instead of single ones.

 

[DrKorbin]

I always had a gut feel this was true - it's very nice of you to do the math to confirm it. Now I'm sure I'll never try - I agree that 1.5 solved centers aren't worth the effort.

To mention a similar-but-different idea, how do you feel about François' approach (I think it's François who does this?) of solving centers to arbitrary locations (so two centers that are supposed to be adjacent are opposite, or vice-versa), then fixing the centers afterwards? He's slower than you are - more my speed - but apparently he considers it to be worthwhile.

And finally my hands reached to do the math again.

1) 7.92 centers are solved in the best orientation (on the average)
2) 7.92 + 1.63 = 9.55 centers are solved using this approach.

I haven't tried this in practice, but I admit that sometimes you choose an orientation and still have a bunch of F centers on the R side, so maybe this can be worthy. I'll try it in the nearest future.

 

[211421]

4*4*4 BLD is interesting .

 

[Cubenovice]

6x6x6 obliques frustrate me so much!

I'm treating them as + centers on 5BLD and solve them the same way: commutators.
1st I solve the ones clockwise from the adjacent corner, then 2nd I handle their neigbours.

While I typically do not have any problems with + centers in 5BLD these obliques freak me out.
I'm having a very hard time "seeing it" and keep making mistakes in moving the correct "E" slice.

Apart from the obvious "practice" is there anything I can do to get better at these darn pieces?

 

[DrKorbin]

When solving first centers, imagine this picture:

     *
     *
      *
      *   ***
   *** ***
***   *
      *
       *
       *

In any cube rotation you can put this picture on the front or upper face, and the lines will point you your current obliques. When solving second centers, imagine symmetric picture.

 

[Cubenovice]

In any cube rotation you can put this picture on the front or upper face, and the lines will point you your current obliques. When solving second centers, imagine symmetric picture.

Thank you for your feedback
I can actually find the pieces on their faces quite well but it is performing the commutators that I keep messing up...

As ik won't be breaking any speed records anyway I'm considering using "M2" on obliques.
I just did a couple this way and it seems pretty straight-forward.

 

[cmhardw]

Thank you for your feedback
I can actually find the pieces on their faces quite well but it is performing the commutators that I keep messing up...

As ik won't be breaking any speed records anyway I'm considering using "M2" on obliques.
I just did a couple this way and it seems pretty straight-forward.

When executing the commutators I try to rememer when I am executing a "3" slice or a "2" slice on an axis.

Take the 4x4x4 center comm [r'u'r,U] which exanded is r' u' r U r' u r U'. On 6x6x6 you might have a cycle like:
[r3' u2' r3, U] and expanded that would be r3' u2' r3 U r3' u2 r3

Whenever you are turning on a slice parallel to the R and L planes you are always turning a "3" slice, or the 3rd layer from the outside. Whenever you are turning an inner layer slice parallel to the U and D layers you are always turning a "2" layer or a 2nd inner layer from the outside.

When executing this comm I would remember to myself something like "R3, U2" so that I know RL plane turns are "3" turns and UD plane turns are "2" turns.

Hope this helps, it helped me a ton when I started thinking this way!

 

[DrKorbin]

Whenever you are turning on a slice parallel to the R and L planes you are always turning a "3" slice, or the 3rd layer from the outside. Whenever you are turning an inner layer slice parallel to the U and D layers you are always turning a "2" layer or a 2nd inner layer from the outside.

:confused:
I think this is not true. Consider an example: [3r, U 2r' U'], when you turn both 2nd and 3rd layers.
If you mean commutators with B part parallel to the U/D planes, this is also wrong. Consider [3l' U2 3l, 2d2] and [2l' U' 2l, 3u2].

 

[cmhardw]

:confused:
I think this is not true. Consider an example: [3r, U 2r' U'], when you turn both 2nd and 3rd layers.
If you mean commutators with B part parallel to the U/D planes, this is also wrong. Consider [3l' U2 3l, 2d2] and [2l' U' 2l, 3u2].

Perhaps I don't understand the notation as well as I thought. Does 3r mean to turn three consecutive layers, or only the 3rd inner r layer counting from the outside?

I think it means to only turn the inner layer, and not 3 consecutive layers. If that's the case then I would say to myself that [3l' U2 3l, 2d2] is turning "3" turns when you turn the 3l'. I put the "3" in quotes because you're not turning three layers, you are turning the third inner layer. So in a sense "3" really means "third". "2" would really mean "second" by the same line of thinking.

Perhaps I wasn't clear with how I wrote my post, and I very likely am misunderstanding the bigger cube notation, but on bigger cubes BLD solves I try to say "3" or "2" turns whenever turning a "third" or "second" inner layer respectively.

 

[Cubenovice]

Hi Chris,

Thank you very much for your feedback
I understood your notation but I still think there is something not quite right (for my solving system) about:

Whenever you are turning on a slice parallel to the R and L planes you are always turning a "3" slice, or the 3rd layer from the outside. Whenever you are turning an inner layer slice parallel to the U and D layers you are always turning a "2" layer or a 2nd inner layer from the outside.

This is true if your oblique buffers are on the two middle slices.
But here’s the catch: I considered my buffers on the 2l slice…

This means that I can bring up pieces with both 2r and 3r so the “slice parallel to the R and L planes you are always turning a "3" slice” no longer holds true.

I copied this buffer position from my 5BLD position where I Use both M and r moves to bring pieces up.
This works great on 5BLD because it very right-hand oriented.
But now in 6BLD I find that this might become too confusing: too many options.

I’ll have a go with placing my oblique buffers at the two middle slices instead of on the 2l slice.
Then I am indeed confined to using the 2l and 2r slice.

 

[cmhardw]

Hi Chris,

Thank you very much for your feedback
I understood your notation but I still think there is something not quite right (for my solving system) about:

Ah, ok I understand why everyone is confused about what I wrote, because I wrote it poorly.

Let's take the commutator: [3r U 3r', 2u']
When executing this specific commutator, then anytime you are turning a RL layer turn you are turning a "3" turn (a "third" inner layer), and any time you are turning a UD layer turn you are turning a "2" turn (a "second" inner layer). The UD part only applies to the inner layer u turns, as I don't care about tracking the outer layer turns because they're comparatively easy to track.

Now let's take the commutator: [2r U 2r', 3u']
When executing this specific commutator, then anytime you are turning a RL layer turn you are turning a "2" turn (a "second" inner layer), and any time you are turning a UD layer turn you are turning a "3" turn (a "third" inner layer). The UD part only applies to the inner layer u turns, as I don't care about tracking the outer layer turns because they're comparatively easy to track.

The "status" so to speak of whether the inner layer only RL turns or UD turns or even FB turns are "2" or "3" turns changes from commutator to commutator. I should have made that more clear in my original post. For the part you quoted I was referring only to the commutator I had written up, namely [r3' u2' r3, U]. Not every commutator would have the specific properties I mentioned, but the commutator [r3' u2' r3, U] does have those properties. Again, you have to assume that for the UD layer I am ignoring tracking the outer layer U turns, because they are comparatively easy to track.

--edit--
Actually now that I think about it I don't really call the "3" turns "3" turns. I actually call them "inner" turns. The "2" turns I actually call "outer" turns.

"Inner" refers to "innermost inner layer", and "outer" refers to "outermost inner layer".

 

[Cubenovice]

The "status" so to speak of whether the inner layer only RL turns or UD turns or even FB turns are "2" or "3" turns changes from commutator to commutator. I should have made that more clear in my original post. For the part you quoted I was referring only to the commutator I had written up, namely [r3' u2' r3, U]. Not every commutator would have the specific properties I mentioned, but the commutator [r3' u2' r3, U] does have those properties.

Yes that's what I thought. My assumption was indeed that you meant for every commutator..

Just tried some cycles with buffers on the middle two slices and quickly realised that I was wrong in concluding I would be restricted to just 2l and 2r...
I found that I would still be using 2 and 3 lices to bring pieces up

But I may still keep the buffers there, somehow this feels better.

thx for your help,

Ralph

 

[Cubenovice]

Update: as with most thing cubing, practice did the job.
I can now say that I 'see' the obliques

I will stick to solving obliques (and some center-centers) from written memo for while to improve accurace. Only then I will invest the time in memoing for full solves.
With my 5BLD times in the 35 min range and the extra care needed for taking the correct slices in 6BLD I expect going over the hour...