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ssakgul

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Oct 14, 2012
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WHICH IS EASY?

I am studying about a new aproach to solve the cube. Maybe somebody has used and named it before me, I don't know and I don't care.

1. EOLine
2. F2L - 1 Slot (LB is my choice)
3. LS + CP without O at the same time
4. CO + EP at the same time (Finishing the Cube)

For the 4th stage, there are two recognition alternatives.

1. Look the colors for URF corner and find the edges in the same colors.
2. Look the UF and UR edges, and determine the right position of them using the URF corner colors.

Both have pros and cons. You can try and notice them yourself. I want to ask you which was easier. Not logicaly but in practice.
The images have the same algorithm but different recognition techniques.

R' U2 R2 U R2 U R U' R U' R'

Thanx for your interest...
 

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xyzzy

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I am studying about a new aproach to solve the cube. Maybe somebody has used and named it before me, I don't know and I don't care.
You might not care, but for reference, this is basically full CPLS + 2GLL.

For the 4th stage, there are two recognition alternatives.

1. Look the colors for URF corner and find the edges in the same colors.
2. Look the UF and UR edges, and determine the right position of them using the URF corner colors.

Both have pros and cons. You can try and notice them yourself. I want to ask you which was easier. Not logicaly but in practice.
The images have the same algorithm but different recognition techniques.

R' U2 R2 U R2 U R U' R U' R'
#2 is absolutely, 100% the faster option if you can put time into practising it. It's essentially two-sided recognition, whereas the other one (finding two edges) is three-sided/four-sided.

However, if you're asking for "easier", looking for bars and other obvious patterns is the easiest, imo. The case you have in your pictures is easily identified by the lack of bars, as well as the H-perm-like pattern around the oriented corner.
 

ssakgul

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You might not care, but for reference, this is basically full CPLS + 2GLL.

CPLS+2GLL The aim is same, but all actions are different. I don't use RF slot as LS. I use LB. And "CPLS setup" is also different. I stack them in a different way an location. And my starting position needs 15 algs, not 26 ;) (CORRECTION: NOT 15! NEAR 90! GO BACK AGAIN TO THE CLASSICAL POSITION. OR SOLVE THE FL SLOT, GET THE LBD CORNER TO UFL AND MAKE AN L MOVE TO HIDE IT. THIS IS THE CPLS SETUP. AGAIN 26 ALGS. EXCUSE ME.) 2GLL is same but maybe some algorithms are different.

ZZ-Orbit solves the pair but does a kind of EPLL without inserting it. And hide them to the back before recognition. Not a benefit. Insert it and use COLL. Much easier.

The case you have in your pictures is easily identified by the lack of bars, as well as the H-perm-like pattern around the oriented corner.

Can you expand the explanation?

Edit 1: (WRONG! NO NEED TO READ IT.) Inspect my CPLS recognition position. R U' L U R' U L2 U2 L U' L2 U L U' L U' L U L. That needs 15 algs, because the corner is always at the ULB corner in 3 different positions and the edge can be only in 5 different positions. One of the algorithms is only L'. The solving alg of this scramble is R' U2 R U' L' U R' U2 R. Too easy. And the recognition code is LD. Because 1st sticker of the URF piece is Orange, and the other corner includes Orange is on the L side. Other Blue corner "is not" at B, So it must be Down. We look for only two corners. One is Front Left, second is Back Right, always. No need to look to Down...

Edit 2: And you can prepare a Special Case Algorithms Set. (SCAS, good name:) If these 3 corners (all U except BL) oriented correctly, we can use one of these algs to Skip the CO stage. That needs again only 15 algs, 30 in total. There is only easy EP. I can do all EP cases with AUF before, not after.

Edit 3: There is a different Special Case in which all the corners are at the right positions. In such a case, again a skip of CO is possible. But this requires 135 different algs.
 
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ssakgul

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However, if you're asking for "easier", looking for bars and other obvious patterns is the easiest, imo. The case you have in your pictures is easily identified by the lack of bars, as well as the H-perm-like pattern around the oriented corner.

Oh, yes. I understood. Weakness of my English... First I'd tried this but thought this one was more difficult. Because you must look for 3 edges again. It is the result of my second technique. So then I found these two. And yes, I've chosen #2 already. I'm memorizing the Sune cases now. First 6 is OK...
 

xyzzy

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Can you expand the explanation?
If you look at the case solved by R' U' R U R U R' U' R' U R U R U' R', it has two bars: one on the front face and one on the back face. The case solved by R U R' U R U2 R' has three bars. This case has no bars, but the edges around the oriented corner are adjacent colours rather than opposite colours.

You can make a flowchart to determine which case you have, based on matching/adjacent/opposite colour patterns and the number of bars you can see. It's not the fastest method of recognising 2GLL cases (or ZBLL in general), however.
 

PapaSmurf

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As we’re talking about LBL methods, here’s one which genuinely could be good:
1) layer (10)
2) edge (6)
3) 2 edges (10)
4) CFRLL - CLL but ignoring FR (not my original idea) (10.75)
5) L5E (12.5)

In total 49.25 moves. An improvement would be solve layer- corner then do L5C L5E.

That would make it more like this:
layer-corner (9)
Edge (6)
2 edges (8)
L5C (12)
L5E (12.5)
Which is 47.5. it saves 2 moves, and could make the third step less algorithmic.

Advantages: it's mostly algorithmic, as the first 2 steps are intuitive, then alg spam, so high tps. More efficient than cfop by a large amount.

Disadvantages: lots of algs (approx 370), and if you switch to the (probably) better L5C variant, it's around 951 algs. Which is a lot, although some of them are short (eg R U' R').
 

ssakgul

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If you look at the case solved by R' U' R U R U R' U' R' U R U R U' R', it has two bars: one on the front face and one on the back face. The case solved by R U R' U R U2 R' has three bars. This case has no bars, but the edges around the oriented corner are adjacent colours rather than opposite colours.

You can make a flowchart to determine which case you have, based on matching/adjacent/opposite colour patterns and the number of bars you can see. It's not the fastest method of recognising 2GLL cases (or ZBLL in general), however.

I did understand wrong. I understood as PLL positions in OLL, like Sune+Z or BowTie+E, etc. I'd used your explanation for PLL only. Yes, it is easier than others. But it requires a "head/cube circle" all the times, like my first recognition technique. In fact, my #1 rec. tec. need only -at most- 3 edges to look, not all four. Because if it is not in the first 3, then it must be at 4th. But you must count 2 or 3 bars. Maybe the 3rd is on the forth. So, you must look to all 4.

Despite everything, I'll try it before being too late...

Edit: Your long example algorithm is my Sune-LB alg for 2 techs both. Or Sune+Z for my wrong understanding ;) Is it 2-Bar-Sune for your tech?
 
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ssakgul

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It is 2-opposite-bar-sune.

R' U2 R . U R U' R' . U R U . R U R' U' . R U R is 2-neighbor-bar-sune. For me only Sune-FR for both techs. Naming your technique is difficult. But I think, it is easier for eyes without words...

Edit: And not only the number of bars, but the position of them is important for your rec. tec. I've found more than one 1-bar-sune.
 
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Hazel

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As we’re talking about LBL methods, here’s one which genuinely could be good:
1) layer (10)
2) edge (6)
3) 2 edges (10)
4) CFRLL - CLL but ignoring FR (not my original idea) (10.75)
5) L5E (12.5)

In total 49.25 moves. An improvement would be solve layer- corner then do L5C L5E.

That would make it more like this:
layer-corner (9)
Edge (6)
2 edges (8)
L5C (12)
L5E (12.5)
Which is 47.5. it saves 2 moves, and could make the third step less algorithmic.

Advantages: it's mostly algorithmic, as the first 2 steps are intuitive, then alg spam, so high tps. More efficient than cfop by a large amount.

Disadvantages: lots of algs (approx 370), and if you switch to the (probably) better L5C variant, it's around 951 algs. Which is a lot, although some of them are short (eg R U' R').
Isn't this 'improved' version just Keyhole followed by L5C/L5E? Solving the last F2L pair normally and following it with last layer is probably better, or maybe MGLS.
 

shadowslice e

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Another silly idea: using EOLS during f2l to orient other f2l edges so you can do rotationless f2l.
Maybe not using EOLS properly, but influencing F2L pairs is already a very common trick. Using full Eols wouldn't be very useful most of the time as inserting the pair and dealing with the next one would be better than rotationless
 

PapaSmurf

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Isn't this 'improved' version just Keyhole followed by L5C/L5E? Solving the last F2L pair normally and following it with last layer is probably better, or maybe MGLS.
It is just improved keyhole. It is definitely better to do L5C->L5E, as one less look and less moves by a long way. Also, L5E algs are out there and are good.
 
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