Noah's CP-Block Method

[Noahaha]

Just something I've been toying around with. Probably not very useful.

This is based on Petrus:

Step 1: create a 2x2x2 block with all D-layer corners placed. One of the two adjacent to the block must be oriented.

Step 2: Identify the corner swap in the U-layer and swap the corners using R U/U'/U2 R'.

Step 3: Place the last 2 edges of the 2x2x3 around the oriented D-layer corner while trying to make sure that you end up with 0 or 4 bad edges.

Step 3.5: If you ended up with 4 bad edges, use M' U M or M F M' to fix them. If you have 2 bad edges, which is undesirable, use R U R' U' M' U R U' r' or M U M' U2 M U M'.

Step 4: 2-gen F2L

Step 5: 2GLL.


Example solve with commentary:

U2 F' U' L' U2 F2 L2 U B' U F2 U2 F' R2 F L2 B' R2 F (scrambled with white on top and green in front)

CP-Block: (8)
D2 F2 B2

E2 ; instead of u2 since the blue-red-white corner is already placed.

R' U' B U ; It's a little unfortunate that the white-orange-blue corner is at DFR. Otherwise I would only have needed 2 moves here. For example, if it was at DBR, B R' would have sufficed.

CP: (4)
x' U ; places the UBL and UBR corners, so it is easy to see that UFL and UFR need to be swapped.

R U' R' ; accomplishes this.

2x2x3: (7)
z' y2 L U2 L' ; places the U F edge. I know that this insertion does not affect EO.

B' ; sets up an F' U' F insertion which flips the edges at RF and UR. I look at the edges at this point and I only see one bad edge at RD. So, I have to bite the bullet and accept 4 bad edges.

F' U' F

EO: (5)
x' y ; bad edges are at FL, FD, FR and UL
U F2 ; set up
M F M' ; kind of like arrow case in Roux.

F2L: (15)
y'
R U R2 U R2 U' R U R' U' R U R' U' R

2GLL: (14)
U2
R2 U' R U R U R' U2 R U R2 U R2

Total: 53


Conclusion: This might be good for OH sometimes. CP blocks often just happen, or are one or two moves extra. The example solve was probably about average movecount if not a little higher.

 

[monkeytherat]

Very cool! This might be useful.

 

[Rpotts]

Here's an alg.garron for that example.

 

[Noahaha]

Here's an alg.garron for that example.

Thanks!

 

[A Leman]

It's an interesting idea! the rotations and recognition seem toublesome for speedsolving though.

 

[Noahaha]

It's an interesting idea! the rotations and recognition seem toublesome for speedsolving though.

They're actually my normal rotations lol. Recognition for the CP step can't be any harder than CMLL recognition, so I think the only step that has a real disadvantage is the 2x2x3 expansion. There is some freedom though, as you can place one or both edges before CP as long as you can avoid the nasty 2 edge EO case.

 

[A Leman]

I am not that great at OH and would really struggle with the M F M' and z' y2 for OH. Rotations and slices moves are much easier 2H and there would be more freedom. The cp step would need an extension of cmll recognition since the bottom corners are not always oriented correctly for CmLL.
you COULD use 3-cycles for step three to solve the 2x2x3 and avoid the 2 bad edges.
for 2 bad edges the best i know are RUR'U'M'URU'r' AND MUM' U2 MUM'
I am having trouble understanding this method so please correct me If I am misunderstanding you

 

[Noahaha]

I am not that great at OH and would really struggle with the M F M' and z' y2 for OH. Rotations and slices moves are much easier 2H and there would be more freedom. The cp step would need an extension of cmll recognition since the bottom corners are not always oriented correctly for CmLL.
you COULD use 3-cycles for step three to solve the 2x2x3 and avoid the 2 bad edges.
for 2 bad edges the best i know are RUR'U'M'URU'r' AND MUM' U2 MUM'
I am having trouble understanding this method so please correct me If I am misunderstanding you

M =r' R
M' = r R'

The rotations in the example were really just the ones I thought would demonstrate the method most easily. You can do those moves from other angles in an actual solve.

Those algorithms you gave don't preserve CP which is the whole point. Normal 2-edge flips during EO like R U R' also don't preserve CP, which is why you either need to force an EO skip or force there to be 4 bad edge where you can do M F M'/ M' U M which does preserve CP.

Doing 3-cycles for step 3 sort of misses the point. 3 move insertions of the type I did in my example solve do preserve CP, so they are safe. The important part is understanding how they affect EO, so that you can always know you'll get a 0 or 4 bad edge case. So basically the goal is to be able to do EO without affecting CP.

I think you might be confused about the EO step of Petrus, which you can find an explanation of at lar5.com/cube. I think EJ (captaincrash44) has one on YouTube as well.

 

[A Leman]

I know what the M is, I just happen to find them difficult OH

It does preserve cp because the RUR'U' is the edge insertion before the interchange move M’ of the edge commutator and is undone with a cancellation in U R U’ r’.

I understand the Eo step of petrus and also know that the 3 move 2 flip creates corner parity with is why you would need to 3cycle(to flip 2 edges and create parity and then flip one of the already flipped edges with an unflipped edge to fix parity and solve 2 bad edges while maintaining correct cp)

since you would need to 3 cycle later on(if you had 2 bad edges), you could do it wll making the 2x2x3 and save the trouble later on.

 

[Noahaha]

I know what the M is, I just happen to find them difficult OH

It does preserve cp because the RUR'U' is the edge insertion before the interchange move M’ of the edge commutator and is undone with a cancellation in U R U’ r’.

I understand the Eo step of petrus and also know that the 3 move 2 flip creates corner parity with is why you would need to 3cycle(to flip 2 edges and create parity and then flip one of the already flipped edges with an unflipped edge to fix parity and solve 2 bad edges while maintaining correct cp)

since you would need to 3 cycle later on(if you had 2 bad edges), you could do it wll making the 2x2x3 and save the trouble later on.

I see. I should probably try algs on a cube before I judge them. That does seem like the best way to deal with 2-flips then.

Could you do an example solve where you use three cycles to finish the 2x2x3? I'm not sure exactly what you mean, or how you could use that to avoid 2 bad edge scenarios. I think that it would involve too many moves though, especially for OH. You say that a 3-cycle could avoid 2 bad edges, but they can be avoided just as easily by controlling which edges get flipped during your insertions. The one scenario where that would be useful is if you had no bad edges going into the last edge of the 2x2x3, which I think is what you're talking about.

EDIT: added those two 3-cycles for step 3.5 to the original post.

 

[applemobile]

Does CP stand for what I think it does?

 

[A Leman]

I see. I should probably try algs on a cube before I judge them. That does seem like the best way to deal with 2-flips then.

Could you do an example solve where you use three cycles to finish the 2x2x3? I'm not sure exactly what you mean, or how you could use that to avoid 2 bad edge scenarios. I think that it would involve too many moves though, especially for OH. You say that a 3-cycle could avoid 2 bad edges, but they can be avoided just as easily by controlling which edges get flipped during your insertions. The one scenario where that would be useful is if you had no bad edges going into the last edge of the 2x2x3, which I think is what you're talking about.

EDIT: added those two 3-cycles for step 3.5 to the original post.

Here's an example
Scramble
L' D L B' D' L2 U B' F U2 D2 B' R2 B' F' U' B2 L U' B R' D2 B R2 F'

Inspection x’
ULU’ R’ D2 2x2 block (5)
R2 F2 B’ R B R’ bottom corners and a 2x2x3 edge (11)
U R U R’ U’ F’ U’ F CP (19)
U z’ RU’RURURU’R’U’R2 z (The 2x2x2 block while making the bad edge count 0 instead of 2)(31)
y’ U R U’ R’ U R2 U’ R’ U’ R’ U’ R(F2L)(43)
U2 (RU2R’) U’ (RUR’) U’ (RU’R’) U2 (2GLL)(56)

This method averages high compared to petrus.Notice that since the block is in the back and I did a z’, the a-perm did affect the EO.The movecount could have been 52 but I used the non-optimal A-perm.

 

[mariano.aquino]

This is very similar to what I published on my page, sort of a method I use when I'm bored =) they differ many things, but the main idea is "similar."
You can see a video and reconstruction here:
http://sites.google.com/site/recursoscuberos/f2g
This version is oriented towards optimization of roux method, with the help of BRASS 2x2x2 method, however, in this example I finish the solve with this variant of freeFOP, which I personally find very finger-friendly (after reduction, blocks, f2l, oll and pll are done exclusively using r/R/u/U)

 

[Noahaha]

Here's an example
Scramble
L' D L B' D' L2 U B' F U2 D2 B' R2 B' F' U' B2 L U' B R' D2 B R2 F'

Inspection x’
ULU’ R’ D2 2x2 block (5)
R2 F2 B’ R B R’ bottom corners and a 2x2x3 edge (11)
U R U R’ U’ F’ U’ F CP (19)
U z’ RU’RURURU’R’U’R2 z (The 2x2x2 block while making the bad edge count 0 instead of 2)(31)
y’ U R U’ R’ U R2 U’ R’ U’ R’ U’ R(F2L)(43)
U2 (RU2R’) U’ (RUR’) U’ (RU’R’) U2 (2GLL)(56)

This method averages high compared to petrus.Notice that since the block is in the back and I did a z’, the a-perm did affect the EO.The movecount could have been 52 but I used the non-optimal A-perm.

For CP, U R U R' (15) would have sufficed
After which I would complete the solve with:
U2 F U'* B' R' B 21 so I have 4 bad edges which can be fixed with:
F U2 M' U M (26). Do you see how the A-perm is unnecessary? The reason I did the U' with the * is because otherwise you end up with two bad edges. Thats the whole point.

 

[Kirjava]

Doesn't placing D corners just to detect CP seem wasteful to you?

Why not do 2x2x3 -> EOCP -> F2L -> 2GLL?

 

[Noahaha]

Doesn't placing D corners just to detect CP seem wasteful to you?

Why not do 2x2x3 -> EOCP -> F2L -> 2GLL?

It's definitely wasteful but it can usually just be an extra move or two if you count placing the oriented corner as "useful".

I've tried an EOCP step before, but it was hard to find situations where the CP fit with the EO, which is actually why I switched to this once I figured out that you can complete the 2x2x3 without affecting CP.

 

[Kirjava]

It's definitely wasteful but it can usually just be an extra move or two if you count placing the oriented corner as "useful".

Movecount is not the only issue, thinking and looking time is more what I was getting at.

Also, completing the 2x2x3 by adding two edges seems silly and inefficient.

 

[Noahaha]

Movecount is not the only issue, thinking and looking time is more what I was getting at.

Also, completing the 2x2x3 by adding two edges seems silly and inefficient.

True. It is possible just to do one edge like that and have one be part of the CP-block like in Leman's example. I think it would be worth it if you could get an EO skip every time. Then it becomes:
1. 2x2x3 minus an edge
2. CP
3. Place edge + EO
4. 2GF2L
5. 2GLL

Which looks more efficient to me. Then the question is about whether there's a significant difference between placing the last edge after CP and just finishing the block. The problem is that the way I do it only half of all edge insertions affect EO, so I may need more tricks for step 3 if I want to do EO while only placing one edge.

 

[Kirjava]

eh, whatever.

I just really want someone to make a good/simple system for detecting CP after 2x2x3.

 

[Escher]

eh, whatever.

I just really want someone to make a good/simple system for detecting CP after 2x2x3.

Permuting 1 corner to UFL/UBL/DBR/DFR makes it really easy to trace the cycle and detect CP, just not that fast. Obviously actually recognising a specific case and solving CP is a different kettle of fish... I'm not sure it can be effectively done for speedsolving methods - if you can't trace x block -> CP in inspection.

It might be possible to do 2x2x2 + CP inspection, if you limit your movegroup while building the block it would be possible to calculate what type of case you'd end up with. I think this would be stupidly difficult to accomplish in <15s however.