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K4

Kirjava said:
BIPzj.png


Worst LL evar
Click to expand...


If I had that case I would probably start with the greens using : (x') U2 l2 U' L' U M2 U' L U r2 U2 (x)
Then the order in my method is locked so I have to do the blues after that : (y x') R U r U L' U' r' l' U L U' l U' R' (x)
Then it is the easiest parity case left for red and orange : (y) R' Lw2 U2 r' U2 l U2 l' U2 (x) U2 l' U2 Lw'

To setup that LL you can use:

Commutator : Lw' r' D2 r U2 r' D2 r U2 Lw
Checker parity: (y) r' U2 r2 U2 r U2 r' U2 r U2 r2 U2 r'
End it with a Z-PLL : M2 U' M E2 M E2 U' M2
 
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Did an example solve for someone

4x4 Scramble #1324: r d' U L R2 f2 u2 U2 l D' d F' r D f2 F U R r' U2 l D' U2 f' r D2 L2 l F' b r2 f l2 d2 b' l2 d2 U2 D l

13:54:00 <+Kirjava> first two centres
13:54:02 <+Kirjava> y'F2xRw'URw2ULw'ULw'U'LwyB'Uw2URwU2Rw'z
13:55:34 <+Kirjava> 1x3x4 block
13:55:36 <+Kirjava> xL'URwFRF'U2R'URU'R'ULUlU'L2U'Rw'Ux
13:58:52 <+Kirjava> last four centres
13:58:53 <+Kirjava> U' 3R2U2Rw'URw2l'U'RwU' 3R2URw2U2Rw2l'U'RwUlU'Rw'U2RwU'Rw'
13:59:12 <+Kirjava> 3R = (l'Rw)
13:59:46 <+Kirjava> finish off the first layer
13:59:48 <+Kirjava> U2R'URw'U'R2URwRU2z'
14:07:04 <+Kirjava> UR'Ur'U'RUrU'LwU'R'ULw'U'RU 3RU'L'U 3R'U'L2UrU'L'Ur'LU'Rw'UL'U'RwRU'rUR'U'r'U'RU'Lw'UR'U'Lw
14:07:08 <+Kirjava> F3L ^
14:07:20 <+Kirjava> CLL is just
14:07:51 <+Kirjava> U2R'U'RU 3LU'R'Ux
14:08:11 <+Kirjava> watch this
14:08:11 <+Kirjava> hehe
14:08:15 <+Kirjava> two look ELL I think
14:08:54 <+Kirjava> r'U'l'D2lUrU2r'U'l'D2lUrU2
14:08:56 <+Kirjava> then
14:09:34 <+Kirjava> Rw2U'lD2l'UlD2 3RRw
 
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Not sure if it's already mentioned somehwhere but does this lastlayer system give me the oportunity to avoid the so called "parity cases".

asked that I already know parity cases can't be avoided, but can I do something usefull while fixing parity?

When do you fix it? While doing the last 4 cubies?
 
oll+phase+sync said:
Not sure if it's already mentioned somehwhere but does this lastlayer system give me the oportunity to avoid the so called "parity cases".

asked that I already know parity cases can't be avoided, but can I do something usefull while fixing parity?

When do you fix it? While doing the last 4 cubies?
Click to expand...

exactly right. Parity in this method is not an extra step, it just is a different case for the last 2 edges.
 
Thanks to qqwref for rewriting all the LL algs in commutator notation. Hopefully it'll force people learning to actually understand what they're doing, which is important. ^_^

Also dan made a video tutorial for ELL some time ago that I should link here.
 
for some reason Chrome cant open it *cries*
will try Firefox... or Safari... nope nothing, not even my new iMac will open it =(
 
Kirjava said:
Thanks to qqwref for rewriting all the LL algs in commutator notation. Hopefully it'll force people learning to actually understand what they're doing, which is important. ^_^
Click to expand...
Hey Kirjava, do you want me to give you a very condensed outline for all (well, most if I can't break down some) of the odd permutation algorithms? That is, involving commutator notation, quarter turn, and conjugates?
 
Georgeanderre said:
for some reason Chrome cant open it *cries*
will try Firefox... or Safari... nope nothing, not even my new iMac will open it =(
Click to expand...

It's just HTML, nothing fancy - should be no reason why it doesn't work.

cmowla said:
Hey Kirjava, do you want me to give you a very condensed outline for all (well, most if I can't break down some) of the odd permutation algorithms? That is, involving commutator notation, quarter turn, and conjugates?
Click to expand...

Uh, possibly. It would have to be quite succinct to fit in the space provided.

Maybe that is better suited to another page. That's really more theory than you'd need to know as the parity algs are the ones you *don't* need to understand when you're learning.
 
Kirjava said:
Maybe your DNS is ****ed. Try switching to OpenDNS and see if it fixes it.
Click to expand...

now I can see the first page... but nothing else
only text, no links

this is the only webpage ive ever had problems viewing, even the old one took ages to load and then had to be loaded again and again until it worked.
this one just wont load at all
 
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found the problem, chrome is changing the link address from snk to skn for some stupid reason...
 
@Kirjava,

I have a few questions regarding solving the last layer wings.

First of all, I counted 54 cases on your ELL page, excluding the first section on "First/Second Edge Tricks," which may also include algorithms from the other 54 (I haven't compared the two to know).

[1] I recall that you said that you can solve any LL case in 3 algorithms, right?
[2] If [1] is true, are all of the algorithms on your ELL page sufficient to do this? I mean what IS there, not any algs for cases not on there which are mirrors, inverses, etc.
[3] If [2] is not true, then what's the minimum number of algorithms one needs to know to solve all LL cases in 3 or less algorithms?

The reason I ask is because I am currently putting together an alternate approach which requires the solver to know algorithms for 55 cases (55 algorithms total). That includes mirrors and inverses.

The Process:
[1] If one dedge is completely solved, skip this step. Otherwise, one dedge can be solved with one of 29 2 2-cycle algorithms (out of the 58 possible 2 2-cycle cases for the last layer).
[2] OLL. There are a total of 13 OLL cases (including mirros) when restricted to 3 unsolved dedges. Of those 13 OLLs, 11 of them can be done with 2 2-cycle algorithms from the selection of 29 used in step 1. The two new algorithms to be learned are to orient one dedge (a 3-cycle is probably best for this) and orient 3 dedges (a 3 2-cycle is probably best for this).
[3] PLL. There are a total of 32 PLL cases (including mirrors and inverses) when restricted to 3 unsolved dedges. Of those 32 PLLs, 8 of them can be solved with 8 of the 29 2 2-cycles used in step 1.

Hence,
Step 1 requires 29 algs
Step 2 requires 2 additional algorithms
Step 3 requires 24 additional algorithms

For a grand total of 55 algorithms. Again, that includes mirrors and inverses.

Another way to look at this is that
PLL (Step 2) requires 32 algorithms (including mirrors, inverses, etc.)
OLL (Step 3) requires 13 algorithms (including mirros, inverses, etc.)
and Step 1 therefore requires 10 additional 2 2-cycle algorithms to what can already be used in PLL and OLL.

Of those 55 algorithms, only 4 + 12 + 1 = 17 algorithms need to be odd permutations.
4 for the oriented 2-cycle cases, 12 for the oriented 1 2-cycle and 1 3-cycle cases, and 1 for the 3 dedge flip restricted to affecting the wings in those 3 dedges only.

I have algorithms for most of these cases, but I'll be trying to find the rest, as well as find faster ones for some of the longer ones that I have.

EDIT:
Well I guess only 16 odd permutation algorithms are required, as the 1 for the 3 dedge flip can be a specific 1 4-cycle and 1 2-cycle alg and possibly another cycle composition that I cannot think of right now. However, I think 3 2-cycle algorithms are more efficient anyway.
 
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cmowla said:
@Kirjava,

I have a few questions regarding solving the last layer wings.

First of all, I counted 54 cases on your ELL page, excluding the first section on "First/Second Edge Tricks," which may also include algorithms from the other 54 (I haven't compared the two to know).

[1] I recall that you said that you can solve any LL case in 3 algorithms, right?
[2] If [1] is true, are all of the algorithms on your ELL page sufficient to do this? I mean what IS there, not any algs for cases not on there which are mirrors, inverses, etc.
[3] If [3] is not true, then what's the minimum number of algorithms one needs to know to solve all LL cases in 3 or less algorithms?
Click to expand...

Kirjava's "algorithms" are hardly algs. I'd say I learned 2-3 algs (l2e parity cases) and can solve ELL in 3 looks or less. Its all intuitive, except for parity cases.
 
masterofthebass said:
Kirjava's "algorithms" are hardly algs. I'd say I learned 2-3 algs (l2e parity cases) and can solve ELL in 3 looks or less. It's all intuitive, except for parity cases.
Click to expand...
Regardless of how the 54 cases can be handled, they are still algorithms. I'm not talking about "intuition" or conjugation of other algs. In fact, many of the algorithms I have found for the 55 cases I mentioned can be handled much easier than conjugation (if that's what you mean by intuition).

Also, which 2-3 algs did you memorize that you can manage to solve all 8 wings for all 8! cases with 3 algs or less? I'm curious. I am not familiar with the "l2e parity cases" phrase to know what you are talking about. I mean, which cycle types do you use?

The total number of absolutely distinct algorithms required to handle all 55 cases is significantly reduced if we take into account inverses, mirrors, and minor modifications of other algorithms. I can't say exactly by how much as of yet (I haven't finished finding all 55 algs yet), but probably between about 1/4 and 1/3 of 55.
 
My point was that no one really has "learned" the ELL algorithms. Thom just put those up so he wouldn't have to explain commutators, but in reality, they aren't a set number of algorithms. It would be similar to comparing OLL algorithms to intuitive f2l. Your method probably requires a learned set vs. a group of intuitive commutators.


As for the algs I learned, they include (by colloquial names) OLL parity, Double Parity, opp 2cycle, PLL parity, and the 2 checkerboard 4cycles (named from bigcubes.com 5x5). I actually didn't need to learn any of these for ELL specifically, as I had known of them previously from 5x5 solving.
 
the l2e (technically l4e) parity cases are the cases when everything is solved minus 2 dedges and has a 2 or 4 cycle.
edit: the 2 or 4 cycles on kirjavas page are the l2e parity cases
 
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