Big Cube BLD Discussion

[xXzaKerXx]

Chris could you possibly make a complete tutorial on this subject and put it in your website and can you try to make the format like Joel Van Noort's website please? it's just an opinion i hope youthink about it thanks very much

 

[siva.shanmukh]

Chris could you possibly make a complete tutorial on this subject and put it in your website and can you try to make the format like Joel Van Noort's website please? it's just an opinion i hope youthink about it thanks very much

I can't find right now, but there is a thread in which Chris explainied a BLD method of his in this forum.

 

[siva.shanmukh]

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

Oh, just realized that you are not counting the cube rotations(makes sense)
That makes my pseudo cycle 7 moves long right?
Now to stop the other 3 cycles, I replaced the f and f' with U' f U and U' f' U respectively making it 11 move

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

Can't go ahead. I will think later about this.

 

[dbeyer]

Chris, that is the alg I found. Haha, yeah, great minds think alike I suppose.
All this algorithm does on a 3x3 is move the 6 centers of the core about the solved 3x3 cube. We are breaking the 3x3 core into 4 quadrants per face, and doing a few setups for the same effect on the 4x4.
Adding some extra moves too because we are trying to move certain pieces.

Okie doke gonna think on this. I have a competition tomorrow. Wish me luck Chris.

Later,
DB

 

[Chuck]

5x5x5 BLD Parity?

I need help really bad.

I just did a 5x5x5 BLD. This is the first time I do it again after my first successful solve about 2 months ago. On my first 5x5x5 BLD, I didn't got any odd target, so I hadn't encountered any 5x5x5 BLD parity.

But I had it on this solve, which was hand-scrambled by me.

I got the breakdowns:

1. Memo x centers: 03:04.10
2. Memo + centers: 03:42.40
3. Memo middle edges: 01:14.30
4. Memo wing edges: 02:21.40
5. Memo corners: 00:52.30
6. Refresh all: 04:36.30
7. Execution: 18:09.00
(too bad I didn't timed each execution steps)

TOTAL 33:59.90 (15:50.80).

It could've been my new PB by 38 seconds, but it's not :fp, because I got this:

[My BLD scheme is Red on Front & Yellow on Top, BOY]

FYI, I use M2 for middle edges, r2 for wing edges, and Classic Pochmann for corners. These are the detailed execution steps:

1. Solve x centers
2. Solve + centers
3. Solve middle edges > odd items/parity (fixed it with U'F2U M2 U'F2U)
4. Solve wing edges > even items/no parity
5. Solve corners > odd items/parity (no fixing)

So, where did I made mistake?
I'm very very sure I memorized all items correctly.

Please, I hope anyone would kindly help me.
Thank you.

 

[Zava]

probably when fixing middle layer edges parity, you forgot to "fix back" wings?

 

[Chuck]

I never do a sighted solve with 5x5x5...

How do we fix back wings?

 

[Zava]

umm, like: you do a J perm to fix corner-edge parity, but this way, you swap the 2-2 wings like on your latest attempt. in this case you just have to set them so they're diagonal (for example: if the blue is the R on the picture, then: RUR'), and do an 5x5-safe PLL parity, like:
(y, then) Rw2 Fw2 U2 r2 U2 Fw2 Rw2 or
(y, then) r2 U2 r U2 r2 U2 r2 U2 r U2 r2 U2
and after this, you just have to undo the setup.
I hope this solves your problem also I hope I understood the problem and could answer it correctly

 

[Mike Hughey]

Yes, I suspect that's the problem. Since I do it differently, it's hard for me to be specific to your case. When I do corner parity on 5x5x5, I always use a T-perm, swapping opposite edges with the corners. Then I do Bw2 Rw2 U2 b2 U2 Rw2 Bw2 to correct the edges.

 

[Zava]

my parity on the 5x5 is like: (corners: hungarian version of pochmann, freestyle, 3op, whichever is the easiest, when I get parity I always keep UFR and UBR swapped, and M2 for edges)
so, I end up with UB-DF swapped, and UFR-UBR. what I do is:
F2, then Rw2 U2 Rw U2 Rw2 U2 Rw2 U2 Rw U2 Rw2 (which does a pll parity on UF-UB, and as a side effect, does a H perm) F2, then a T perm. I find it quite handy

 

[dbeyer]

My parity fixes swap the URB corner and UR edge, which are my buffers with a random corner and a desigated edge.

I always swap the UR and UB if there is a parity. Then I can use an advanced Pochmann concept. Basically a bunch of PLL conjigations with cancellations.

It leads to a 1 turn setup max, and if I'm setting anything up you're limited to a single turn on the U, L, F faces. And this limitation isn't exactly a bad thing by no means.

Anyway, how do I always end swapping the UR and UB? Wouldn't you need to do extra setups? I'm confused.

Technically it's in the method of approach that I use. If there is parity. I am left with a lone edge that needs to be swapped with the UR. I do one extra 3x3 commutator on the edges. Buffer -> Parity edge -> UB. Leaving the Buffer corner and one lone corner, needing to be swapped. This is the perfect Pochmann scenario.

Now somebody asks, wouldn't it be easier to just do some setups, rather than an extra commutator? Look at all the extra moves. Well perhaps you might be right. This system that I use is engineered for success on big cubes. Nicely setting up the Edge and the Corner into a PLL case is not always that easy. Especially with a BH solve, where you have 462 2-cycles.
22 edges, choose 1, 21 corners, choose 1. 22x21 = 462.

I guess for an abbreviation, you could call this sub-step to my system
Permute Last Corner. PLC.

On 3x3 one thing must be done in conjuction with PLC to keep the cube solved. Likewise on 4x4 one different thing must be done with PLC to keep the cube solved. On a 5x5, since this contains both the 3x3 and 4x4, both steps must be done.

Okay. For all intensive purposes, I'll simplify this as a description of my method. I've created an algorithm set that swaps the URB with any of the other 21 corner cubie stickers. Tada!

All 21 algorithms swap the UR and UB. So to allow myself to just perform the algorithm and be done with the solve, I use a commutator to control the last two edges to be the UR and UB. Simple.

On the 4x4. All of the algorithms would swap the UR and UB wing sets. So I use a fast algorithm to swap them back. (3L)2 U [r2U2 xU2r2U2x' U2r2] U' (3L2)

Also, a little side note. The algorithm does affect 4x4 and 5x5 center orientation of the U layer. All of the PLC algorithms turn the U centers by one quarter turn.

 

[Chuck]

umm, like: you do a J perm to fix corner-edge parity, but this way, you swap the 2-2 wings like on your latest attempt. in this case you just have to set them so they're diagonal (for example: if the blue is the R on the picture, then: RUR'), and do an 5x5-safe PLL parity, like:
(y, then) Rw2 Fw2 U2 r2 U2 Fw2 Rw2 or
(y, then) r2 U2 r U2 r2 U2 r2 U2 r U2 r2 U2
and after this, you just have to undo the setup.
I hope this solves your problem also I hope I understood the problem and could answer it correctly

Wow, thank you so much, Balázs!
That is exactly the solution to my problem.

So from now on if I get middle edge parity, I will solve it with U'F2U M2 U'F2U and then BU'B' Rw2 Fw2 U2 r2 U2 Fw2 Rw2 BUB' at the end.

 

[dbeyer]

Chuck ...
This is my alg
(4L)2 U [r2 U2x U2r2U2 x'U2 r2] U' (4L)2
and for you
(4R)2 U' [r2 U2x U2r2U2 x'U2 r2] U (4R)2

 

[LarsN]

Yes, I suspect that's the problem. Since I do it differently, it's hard for me to be specific to your case. When I do corner parity on 5x5x5, I always use a T-perm, swapping opposite edges with the corners. Then I do Bw2 Rw2 U2 b2 U2 Rw2 Bw2 to correct the edges.

What?! Why havn't I seen that alg before? I always use T-perm aswell, but I've always used 2 coms to fix the wingedges. Even though the coms are very nice, this is so much better! Thanks

 

[Micael]

Here were my breakdowns for 4x4 BLD. I have been doing this two weeks now, so still a beginner.

Total memorization time: 8:54
Total solving time: 10:49
Total overall time: 19:43

1) 0:25 orient the cube
2) 2:13 Memorize centers
3) 3:52 Memorize edges
4) 0:43 Refresh
5) 1:41 Memorize corners
6) 1:52 Solve corners
7) 4:08 Solve centers
8) 4:49 Solve edges

 

[dbeyer]

By the way, if you take time to notice, Chris doesn't waste much time at all on orienting the cube. Realistically no orienation is better than any other.

Just you'd be just as randomly lucky to find the best orientation by tossing the cube up in the air with your eyes closed and then memorization begins when you catch the cube and open your eyes.

Later,
DB

 

[Mike Hughey]

I don't buy it, Daniel. I agree that it doesn't make as much difference as one tends to think it makes, but a few seconds choosing a good orientation can cut two or sometimes even three commutators off of your total solve. If you can choose a better orientation in a few seconds, that's bound to save a very significant amount of time. So I intend to continue trying to choose a best orientation.

However, that being said, if you spend too long orienting, you can't possibly come out ahead. I feel terrible if I go over 15 seconds orienting. (Although sometimes I do, because I'm stupid. )

I guess I should also mention that I use floating buffers on centers, which I believe gives me a significant advantage over Chris on some solves, since I have fewer commutators to do. It seems like proper orientation helps more if using floating buffers than it does when using a fixed buffer - with floating buffers you're more likely to cut an extra commutator off the total solve due to proper orientation.

 

[dbeyer]

You can't dwell or harp finding the best center orientation. Because that "perfect" choice later on down the yellow brick road could wind up with an ugly corner solution, and 1 large cycle on wings, and 2 other small cycles around the cube that you have to search for.

Expect to solve the centers in 8-9 commutators.
Wings in 12, and Corners in 4.
Therefore expect to solve the cube in an average of 25 commutators. The balance of the system comes into play, because you could get a better center solution, but wind up with 14 wing commutators and 5 corner commutators. Still you'd most likely then get 6-7 center commutators. Averaging 25 commutators.

Later,
DB

 

[Chuck]

Odd Cubes Edge Monoflip?

Dear all,

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

Obviously we can't use M'UM'UM'UM' U2 M'UM'UM'UM'.

Thank you

 

[yoruichi]

u cant?