Difference between revisions of "CPLS"

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~Statue will get to this. Just testing things in previews for now before releasing :D~
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{{Method Infobox
 +
|name=CPLS
 +
|image=
 +
|proposers=[[Baian Liu]], ?, ?
 +
|year=2009?
 +
|steps=1
 +
|algs=26
 +
|moves=~10
 +
|purpose=<sup></sup>
 +
* [[Speedsolving]]
 +
* [[One-Handed_Solving]]
 +
}}
 +
 
 +
''CPLS'' (short for ''Corner Permutation and Last Slow'') is used namely when one has completed the cube apart from [[EPLL]], [[CPLL]], [[COLL]], and the last corner of the final F2L slot.
 +
''CPLS'' is intended to both finish the last F2L corner, usually the DFR corner, while permuting the last-layer corners.  This sets up the last layer to be able to be solve with a 2-gen system, either with one or two steps.
 +
Although seemingly hard to recognize, this method can be very beneficial, especially for one-handed solvers with fast <R,U> or <L,U> -gen algs, depending on hand preference.
 +
 
 +
If one is to simply do a two-step process after reducing the [[LL]] to a 2-gen system, they would need only 11 algorithms - 7 for [[OCLL]], which in turn of course solves [[CLL]] entirely, and 4 for [[EPLL]], in that order.
 +
If they instead decide to learn a 1-step system, then [[2GLL]] is really the only choice.  When looking for this, be aware that it is also referred to as ZZ-d.  2GLL algorithms can be solved entirely with the use of only two faces, and is composed of 84 algorithms.
 +
 
 +
===Learning Approach===
 +
 
 +
Like almost any other set of algorithms, it's best to divide ''CPLS'' into subsets.
 +
 
 +
These subsets are the same ones that can be found in [[CLS]], which are -, +, O, I, Im, and C.  The first three sets have six algorithms each, while the next two have three each, and the final set has only two.
 +
 
 +
Perhaps a good order to learn these sets in would be C, O, I, Im, -, +.  This is recommended due to ease of learning and ease of recognizing.
 +
 
 +
===Recognition===
 +
 
 +
Recognition is definitely the most prominent downfall to this method, at least in terms of personally finding an intuitive way to do so.
 +
However, a system has been found that works very well.
 +
 
 +
Before you start, it must be emphasized that you *must* know your color scheme fairly well.
 +
 
 +
 
 +
Let's set up a case. Do x R' U R U' R' U R U' R' U R U' x' onto a scrambled cube with yellow on top and red on front.
 +
 
 +
Seeing this, I know that this is an O case, and that it's already positioned correctly, in URF, to be recognized.
 +
 
 +
Next, I need to look at the remaining three U-layer pieces.
 +
 
 +
Let's keep a system to *always* start with UFL.
 +
 
 +
Once reaching any of the three corners, in this order, the first goal is to locate the U-layer sticker, in my case yellow. This should be on the top for this case.
 +
 
 +
Next, look at the sticker *clockwise* from this sticker. In other words, the sticker on the F face, or the FLU sticker.  This should be red. I note this, and that my red FACE is currently on B.  I now pretend that the entire F face is orange, because when rotated, that sticker lies on that face.
 +
 
 +
Following that, do the same for the LUB and BRU corners.
 +
 
 +
For the L face, you should come out with green, since the sticker at LUB is green.
 +
 
 +
For the B face, you should come out with orange, since the sticker at BRU is orange.
 +
 
 +
One could write this case down as an [O BLF], being as though B is on F, L is on L, and F is on B.  Depending on initial AUF, however, this may change
 +
 
 +
So, from the above, we can visualize that F = red, L = green, and B = orange. 
 +
Knowing that yellow is our U-layer sticker, we know that the above can't be right - those colors aren't allowed to be like that! 
 +
So what we must do is switch those "face." To do this, we must switch the corners that these pieces are represented by. For example, the UFL piece represents the Front face. 
 +
In this case, we must switch the UFL and UBR corners.  To understand what to do now, we go over to https://sites.google.com/site/devastatingspeed/3x3x3/cpls and try to find the right case. 
 +
Knowing that we have a diagonal switch (switching UFL and UBR has essentially the same effect as switching UFR and UBL), we look for the O case with the diagonal swap, which is in the 3rd row, second column, yielding us the algorithm x (U R' U' R)*3 x'
 +
 +
After applying this, the [[F2L]] should be finished, and the [[LL]] should be reduced to a [[2GLL]] case, to be done in however many steps. More than two steps makes CPLS very unnecesarry and arguably a waste of time, though.
 +
 
 +
===Example Solves===
 +
 
 +
The best way to describe this recognition technique is simply with an example solve, so three shall be presented here.
 +
For convenience of most, a standard color scheme shall again be assumed here.  Also, please scramble with the standard white on top (U) and green on front (F).
 +
 
 +
 
 +
Scramble 1: D' B' D2 F2 R F' L' R B L' F' U' R' B F2 U R2 D L2 D2 U' B R' F2 D'
 +
 
 +
A Petrus approach:
 +
 
 +
2x2x2: x2 D B' R' D2 L2 (5/5)
 +
 
 +
2x2x3: x' y D R D' U' L' U L (7/12) -note that this sets up for an Im case later on.
 +
 
 +
EO: y U M' U M (4/16)
 +
 
 +
F2L: x y U2 R' U' R U R' U' R2 U2 R' (10/26) [ewww]
 +
 
 +
CPLS (Im FBL): y2 U R' F2 R D' L' U2 L' U' L2 D (11/37)
 +
 
 +
2GLL (Pi Ua1): y (y) R U2 R2 U2 R U R2 U R2 U' R2 U2 R' U2 R(16/53)
 +
 
 +
AUF: U2 (1/54)
 +
 
 +
 
 +
Scramble 2: L2 F' L2 F L' D B D2 U2 L2 D L' R B' L2 U B2 R2 D2 R B2 D U2 B' F'
 +
 
 +
A RH OH ZZ approach:
 +
 
 +
EOCross: L B' R' U D F R D R' D R' (11/11)
 +
 
 +
BL slot: U2 R U R' L U L' (7/18)
 +
 
 +
FL slot: L2 U2 L U L' U L2 (7/25)
 +
 
 +
BR slot: U R' U' R U' R' U R (8/33)
 +
 
 +
FR edge: R U R' (3/36) [gah, it kills me not to do CLS for this case :(]
 +
 
 +
CPLS setup: U2 (1/37)
 +
 
 +
CPLS (O): y U2 z' U L' D2 L U' L' D2 z (8/45)
 +
 
 +
2GLL (T Ub2): z' U L2 U' L' U L' U' L U L U' L U L2 U' (15/60)
 +
 
 +
AUF: L2 (1/61, or 57 with cancellations)
 +
 
 +
 
 +
Scramble 3: L' R' B F U B2 D2 F2 L F2 D U' B' U D2 F2 B L2 R2 B U2 B' D2 U2 R
 +
 
 +
A CFOP approach:
 +
 
 +
Cross: x2 F D L2 U2 L F' (6/6)
 +
 
 +
BL slot: F' U2 F L U' L' (6/12)
 +
 
 +
FR slot: U R U2 R' U R' F R F' (9/21)
 +
 
 +
BR slot: U' R' U2 R2 B R B' (7/28)
 +
 
 +
ELS setup: y U (1/29)
 +
 
 +
ELS: R U' R' F' U2 F (6/35)
 +
 
 +
CPLS (+): U' U' y R' U L' U' R U L (9/44)
 +
 
 +
2GLL (Pi Ub3): y' (y) R U2 R2 U' R' U R U' R' U' R' U' R' U2 R U2 (15/59, or 56 with cancellations)
 +
 
 +
== See Also ==
 +
 
 +
* [[2GLL]]
 +
* [[CLS]]
 +
* [[ZZ-d]]
 +
* [[ELS]]
 +
* [[One-Handed_Solving]]
 +
 
 +
== External Links ==
 +
* [https://sites.google.com/site/devastatingspeed/3x3x3/cpls Baian Liu's Original site]
 +
* [http://www.speedsolving.com/forum/showthread.php?p=454801#post454801 SpeedSolving CPLS+2GLL thread]
 +
* [http://db.tt/KIRzYvL Stachu Korick's printable OH algs]
 +
 
 +
[[Category:Methods]]
 +
[[Category:Advanced Methods]]
 +
[[Category:Experimental Methods]]
 +
[[Category:Cubing Terminology]]
 +
[[Category:Abbreviations and Acronyms]]
 +
[[Category:Sub Steps]]

Revision as of 13:16, 15 September 2010

CPLS method
Information about the method
Proposer(s): Baian Liu, ?, ?
Proposed: 2009?
Alt Names: none
Variants: none
No. Steps: 1
No. Algs: 26
Avg Moves: ~10
Purpose(s):


CPLS (short for Corner Permutation and Last Slow) is used namely when one has completed the cube apart from EPLL, CPLL, COLL, and the last corner of the final F2L slot. CPLS is intended to both finish the last F2L corner, usually the DFR corner, while permuting the last-layer corners. This sets up the last layer to be able to be solve with a 2-gen system, either with one or two steps. Although seemingly hard to recognize, this method can be very beneficial, especially for one-handed solvers with fast <R,U> or <L,U> -gen algs, depending on hand preference.

If one is to simply do a two-step process after reducing the LL to a 2-gen system, they would need only 11 algorithms - 7 for OCLL, which in turn of course solves CLL entirely, and 4 for EPLL, in that order. If they instead decide to learn a 1-step system, then 2GLL is really the only choice. When looking for this, be aware that it is also referred to as ZZ-d. 2GLL algorithms can be solved entirely with the use of only two faces, and is composed of 84 algorithms.

Learning Approach

Like almost any other set of algorithms, it's best to divide CPLS into subsets.

These subsets are the same ones that can be found in CLS, which are -, +, O, I, Im, and C. The first three sets have six algorithms each, while the next two have three each, and the final set has only two.

Perhaps a good order to learn these sets in would be C, O, I, Im, -, +. This is recommended due to ease of learning and ease of recognizing.

Recognition

Recognition is definitely the most prominent downfall to this method, at least in terms of personally finding an intuitive way to do so. However, a system has been found that works very well.

Before you start, it must be emphasized that you *must* know your color scheme fairly well.


Let's set up a case. Do x R' U R U' R' U R U' R' U R U' x' onto a scrambled cube with yellow on top and red on front.

Seeing this, I know that this is an O case, and that it's already positioned correctly, in URF, to be recognized.

Next, I need to look at the remaining three U-layer pieces.

Let's keep a system to *always* start with UFL.

Once reaching any of the three corners, in this order, the first goal is to locate the U-layer sticker, in my case yellow. This should be on the top for this case.

Next, look at the sticker *clockwise* from this sticker. In other words, the sticker on the F face, or the FLU sticker. This should be red. I note this, and that my red FACE is currently on B. I now pretend that the entire F face is orange, because when rotated, that sticker lies on that face.

Following that, do the same for the LUB and BRU corners.

For the L face, you should come out with green, since the sticker at LUB is green.

For the B face, you should come out with orange, since the sticker at BRU is orange.

One could write this case down as an [O BLF], being as though B is on F, L is on L, and F is on B. Depending on initial AUF, however, this may change

So, from the above, we can visualize that F = red, L = green, and B = orange. Knowing that yellow is our U-layer sticker, we know that the above can't be right - those colors aren't allowed to be like that! So what we must do is switch those "face." To do this, we must switch the corners that these pieces are represented by. For example, the UFL piece represents the Front face. In this case, we must switch the UFL and UBR corners. To understand what to do now, we go over to https://sites.google.com/site/devastatingspeed/3x3x3/cpls and try to find the right case. Knowing that we have a diagonal switch (switching UFL and UBR has essentially the same effect as switching UFR and UBL), we look for the O case with the diagonal swap, which is in the 3rd row, second column, yielding us the algorithm x (U R' U' R)*3 x'

After applying this, the F2L should be finished, and the LL should be reduced to a 2GLL case, to be done in however many steps. More than two steps makes CPLS very unnecesarry and arguably a waste of time, though.

Example Solves

The best way to describe this recognition technique is simply with an example solve, so three shall be presented here. For convenience of most, a standard color scheme shall again be assumed here. Also, please scramble with the standard white on top (U) and green on front (F).


Scramble 1: D' B' D2 F2 R F' L' R B L' F' U' R' B F2 U R2 D L2 D2 U' B R' F2 D'

A Petrus approach:

2x2x2: x2 D B' R' D2 L2 (5/5)

2x2x3: x' y D R D' U' L' U L (7/12) -note that this sets up for an Im case later on.

EO: y U M' U M (4/16)

F2L: x y U2 R' U' R U R' U' R2 U2 R' (10/26) [ewww]

CPLS (Im FBL): y2 U R' F2 R D' L' U2 L' U' L2 D (11/37)

2GLL (Pi Ua1): y (y) R U2 R2 U2 R U R2 U R2 U' R2 U2 R' U2 R(16/53)

AUF: U2 (1/54)


Scramble 2: L2 F' L2 F L' D B D2 U2 L2 D L' R B' L2 U B2 R2 D2 R B2 D U2 B' F'

A RH OH ZZ approach:

EOCross: L B' R' U D F R D R' D R' (11/11)

BL slot: U2 R U R' L U L' (7/18)

FL slot: L2 U2 L U L' U L2 (7/25)

BR slot: U R' U' R U' R' U R (8/33)

FR edge: R U R' (3/36) [gah, it kills me not to do CLS for this case :(]

CPLS setup: U2 (1/37)

CPLS (O): y U2 z' U L' D2 L U' L' D2 z (8/45)

2GLL (T Ub2): z' U L2 U' L' U L' U' L U L U' L U L2 U' (15/60)

AUF: L2 (1/61, or 57 with cancellations)


Scramble 3: L' R' B F U B2 D2 F2 L F2 D U' B' U D2 F2 B L2 R2 B U2 B' D2 U2 R

A CFOP approach:

Cross: x2 F D L2 U2 L F' (6/6)

BL slot: F' U2 F L U' L' (6/12)

FR slot: U R U2 R' U R' F R F' (9/21)

BR slot: U' R' U2 R2 B R B' (7/28)

ELS setup: y U (1/29)

ELS: R U' R' F' U2 F (6/35)

CPLS (+): U' U' y R' U L' U' R U L (9/44)

2GLL (Pi Ub3): y' (y) R U2 R2 U' R' U R U' R' U' R' U' R' U2 R U2 (15/59, or 56 with cancellations)

See Also

External Links