CheesecakeCuber
Veteran
I've also encountered some cases where a CLL alg messes up the blocks when used in place of good ol' CMLL
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scramble: F' R' F' L2 U2 L F' R2 D' U R B2 U2 L2 U L2 R' U R U' R2 L2 B' R F2
x2 y
B' U r F' (4)
R2 r U' r2 U2 R' U R U M' r U R' (13)
U F U R U' R2 F' R U2 R U2 R' (12)
U M' U' M' U' M2 U M' U2 M U M2 (12)
scramble: F' R' F' L2 U2 L F' R2 D' U R B2 U2 L2 U L2 R' U R U' R2 L2 B' R F2
y' x2 // inspection
U M' F' B U R U' B // (8)
M' U' R' U r' U' M' r U R' U r' U' r // (14)
U' x U R' U' L U2 R U2 R' // Sune C3 (10)
U M U M' U2 M U M U' M U2 M U M U2 M' U2 (17)
// Total: 49
Any advice where I can save some moves or where I'm doing something goofy would be greatly appreciated.
Hello Everybody, I, GG, am learning CMLL.
What does this mean?
Nobody knows....
anyway, what sites could you guys recommend?
http://www.speedsolving.com/wiki/index.php/CMLL
"Solving of corner orientation and permutation, without regard to the M-slice pieces."
also note that CLL = "corners of last layer"
Also:
http://wafflelikescubes.webs.com/cmllalgorithms.htm
http://rouxtorial.webs.com/cmll.htm
http://www.speedsolving.com/forum/showthread.php?20355-CMLL
http://www.speedsolving.com/wiki/index.php/CxLL
Question to the others:
Do you know any other great databases of CMLL sets?
scramble: F' R' F' L2 U2 L F' R2 D' U R B2 U2 L2 U L2 R' U R U' R2 L2 B' R F2
LSE: U' M' U M' U' M2 U' M' U M' U2 M' U2 M' U2 M2 U M' U2 M2 U2 M' (22 Ouch)
CMLL recognition: the APR and the 4SPR system
(APR is for Adjacent Pairs Recognition, 4SPR is for 4 Stickers Pattern Recognition. Yes, I made up the names.)
As my 1000th post, I show you my CMLL recognition system.
It has many similarities with Waffle's Bars and Slashes system. Furthermore, in many cases it is the same. But in my opinion mine is a bit less complicated and a little more explained; moreover, APR is quite easily understandable and learnable.
For the case naming, I use the same terminology (for example, E2 is where top-colors are BUL, BUR, UFL, UFR, and the first two need to be swapped). As for the color scheme, I use orange for the top colors; but this is just for the example images.
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For further notation, I will call a sticker pair (or color pair) same, if they are the same (e.g., green-green), opposite, if they belongs to opposite sides (green-blue), and neutral, if they belongs to adjacent sides (green-yellow).
From now on, I assume you can recognise the orientation case easily (A, B, ..., H), and only the matching permutation case (1, 2, ..., 6) means some trouble to you.
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Case A
First we have to analyze the A cases: A1 (the solved case), A6 (the diagonal swap), and A2/A3/A4/A5 (the adjacent swaps).
Take a look at these images:
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All the top-colors (orange) are on top, so obviously all of them are A cases. Now look around on the side stickers:
- In the first case, each side have same color pairs: green-green (back side), yellow-yellow (right side), blue-blue (front side), and white-white (left side). Therefore we know that this is an A1, no corners need to be swapped.
- In the second case, each side shows opposite colors, they are opposite pairs. Therefore we need a diagonal swap, A6.
- In the third case, things seems a bit complicated. On the back, we see a neutral pair (green-yellow). On the right, there is an opposite pair (blue-green). On the front, there is neutral pair again (yellow-blue), and on the left there is a same pair (white-white). In this case, the two right corners need to be swapped (A3).
And here is the main thing: If we have an A case, we immediately know which A case it is only by looking at two adjacent side pairs. So, let's say we are looking at the front and the right pairs:
- Same pair front, same pair right? A1.
- Opposite front, opposite right? A6.
- If one of the pairs is neutral, then we have to do an adjacent swap. The corners which should be swapped forms an opposite pair (in the example image: the right side has an opposite pair -> the right corners should be swapped). So:
- Neutral front, opposite right? We have to swap the right corners: A3.
- Neutral front, same right? The opposite pair is always on the far side (opposite side) of the same pair. In this case, if the right side shows a same pair, then the left side must have an opposite pair: A5.
- Same front, neutral right? Then the opposite pair must be the back: A2.
- Opposite front, neutral right? Swap the front corners: A4.
This is the basic concept about recognizing the cases, and we will use this in half the time. I will refer to this kind of recognition as Adjacent Pairs Recognition, APR for short.
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Case B
So let's jump to the B cases, how we can use APR:
![]()
First, let's see the left image. Orange shows the orientation (here, it is B). There is a sticker pair showed in purple, and there is a pair showed in brown. What happens, when we turn both the UFL corner and the UFR corner counter-clockwise? (So we force the B case into an A case. The UBR has no purpose here). Then the purple stickers become the left side pair, and the brown stickers become the front side pair! And as we saw, we can predict the permutation case just from two adjacent pairs. If both pairs are same pairs, it is a permutation 1 (in this case, B1), if both pairs are opposite pairs, it is a permutation 6 (B6). Finally, if one of them is neutral, the other must be same or opposite (because they are adjacent pairs), and we need to swap the the corners belonging to the opposite pair.
Take a look at the image on the right. The left pair (purple pair on the previous image) is white-blue, which forms a neutral pair. The front pair (brown pair on the previous image) is yellow-yellow, which forms a same pair. So, we need an adjacent swap on the side which has opposite pair. Where is it? It is the opposite side of the same pair, our yellow-yellow, which is front. So we need to swap the back, and our case is B2.
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Case C
Very similar to B.
![]()
If we rotate the UBL and the UBR corners clockwise, then we have an A case (UFR has no power here), then the purple stickers become the left pair, and the brown stickers become the back pair. With them we can predict the permutation case just like before. See the second image: left pair is opposite (green-blue), back pair is opposite (yellow-white), so this is an adjacent swap: C6.
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Case D
Again, the same.
![]()
Purple shows the right pair, brown shows the front pair (rotate the UFR corner clockwise). On the example, right pair is opposite (white-yellow), front pair is neutral (white-blue), so adjacent swap on the right side: D3.
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Case E
So far nothing new.
![]()
Purple shows the right pair, brown shows the front pair. On the example image, both pairs are same (blue-blue and white-white), no permutation is needed, E1.
However, we can use a different technique for the E cases too, which I call the Four Stickers Pattern Recognition (4SPR). Take a look at this image:
![]()
On the left side, you can see 4 purple stickers. This 4 stickers form some kind of pattern. In the pattern, look for same-opposite-neutral pairs. Depending on their relative position, you can recognise the permutation case for E.
On the bottom of the image, you can see the patterns belonging to the cases. Don't look here the colors itself; rather than which colors form same or opposite pairs, and how: in a row, or diagonally, etc. So the pattern of E3 is: " slash-diagonal is same, backslash-diagonal is opposite".
For example, take a look on the right image. The four colors are blue (UBL), green (UBR), green (FUL) and blue (FUR). We can say that the top row has opposite colors and the bottom row has opposite colors; ar alternatively, the same colors form an X pattern. This is the same as the E4 in the pattern list, so this is an E4.
4SPR seems a bit difficult to learn / get used to, but with practice it will be a piece of a cake (especially when you learn your CMLL algorithms together with the recognition system). For me, using 4SPR became the superior recognition method after APR.
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Case F
F is a bit problematic, because if we want to use APR, we need to look at the back side of the cube. We don't like to do that. Instead, we can use 4SPR, similarly to the E case. Beware, because the same pattern in E and F can mean different permutation case! (For example E2 has the same pattern as F1.)
![]()
On the right example, we can see that the top row forms an opposite, the bottom row forms a same pair, and this is equivalent to F2 according to the pattern list.
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Case G
Here, just like in E, we can use both recognition methods, either APR or 4SPR.
![]()
Purple is the right pair, brown is the front pair. On the example, green-green (sameon right), blue-yellow (neutral on front), so swap the left side: G5.
Or with 4SPR, the patterns are:
![]()
On the example, the four colors are white-blue-white-green. Left two forms a same, right two forms an opposite, which is G6.
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Case H
This very similar to case F. Not just we have trouble again with APR, but their 4SPR patterns are the same! This is not a coincidence, and I leave you to figure out the reason.
![]()
On the example, top two colors form an opposite pair, bottom two form a same pair, which means H2.
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There. My 1000th post. Yaaay!
Have fun!
I think F2B is suppose to take a little longer than CMLL and LSE combined. For me, F2B is about 6 seconds and CMLL+LSE is about 5 seconds.
My CMLL+LSE is around 12 to 15 seconds.
It seems that your F2B is pretty good now, and you have to work mostly on CMLL and LSE. As GuRoux said, F2B is a bit longer than the rest of the solve. For me, F2B is around 8 seconds and CMLL+LSE is around 6 seconds.
If your F2B is around 12 seconds, then try to reach 10 seconds for CMLL + LSE.