L2L4 - The Lost Methode ? Help Find it!

[FatBoyXPC]

I've done too much thought about this with other people, I can't remember what I've said here or in various conversations. I was giving the number of cases without mentioning the solved states. In the event I got to a solved case, I'd continue on to the next step, all this means is doing one step twice (but it lowers alg count).

 

[mr. giggums]

For the case numbers I got 185 = 59+16+51+59 so how did you get 90 Thom? Although I'm pretty sure I messed up somewhere I'll double check this later when I have the time.

 

[FatBoyXPC]

I asked him how he got it and he said it was an estimation, no math at all was involved.

 

[nickvu2]

Let me just think out loud here. Here's how I might do a solve. Blue forward, yellow up. I’m trying to do this without cube turns; not sure if it’s the most efficient.

1. First layer: intuitive

2. CO: I’ll always shoot to BL since it’s out of the way. AUF and locate green/orange edge. If the edge is solved, you still need CO. If CO is solved, you still need to place the edge. 117 cases.

T: 8 (edge positions) * 2 (orientations) = 16 cases
U: 8 (edge positions) * 2 (orientations) = 16 cases
L: 8 (edge positions) * 2 (orientations) = 16 cases
Pi: 8 (edge positions) * 2 (orientations)= 16 cases
Sune: 8 (edge positions) * 2 (orientations)= 16 cases
Anti-sune: 8 (edge positions) * 2 (orientations)= 16 cases
H: 6 (edge positions w/ symmetry) * 2 (orientations) = 12 cases
Solved: 5 (edge positions w/ symmetry) * 2 (orientations) – 1 (CO skip)= 9 cases​

3. CP: Shoot to BR. AUF and locate green/red edge. 31 cases.

T: 7 (edge positions) * 2 (orientations) = 14 cases
Y: 5 (edge positions w/ symmetry) * 2 (orientations) = 10 cases
Solved: 4 (edge positions w/ symmetry) * 2 (orientations) – 1 (CP skip)= 7 cases​

4. EO: Shoot to FL. AUF and locate blue/orange edge. Recognition might be difficult since LL edges might be in E-slice. 31 cases.

No edges: 3 (edge positions w/ symmetry) * 2 (orientations) = 6 cases
Opposite edges: 4 (edge positions w/ symmetry) * 2 (orientations) = 8 cases
Adjacent edges: 6 (edge positions) * 2 (orientations) = 12 cases
Solved: 3 (edge positions w/ symmetry) * 2 (orientations) – 1 (EO skip)= 5 cases​

5. EP: Shoot to FR. AUF and locate blue/red edge. Also hard to recognize case with missing LL edge. 21 cases.

U(a): 5 (edge positions) = 5 cases
U(b): 5 (edge positions) = 5 cases
Z: 5 (edge positions) = 5 cases
H: 5 (edge positions) = 5 cases
Solved: 2 (edge positions w/ symmetry) – 1 (EP skip) = 1 cases​

117 (CO) + 31 (CP) + 31 (EO) + 21 (EP) = 200 cases

Now we don’t necessarily need a separate alg for each case. But we would have to not be using a fixed target and also have d/d2/d’ setup moves. For example, jms_gears1 gave us Ru2R'u'Ru'R' which sends 3 edges in 3 different directions. Sure that’s 2 algs less you have to learn, but the trade off is you have to keep track of what algs send what edges where, keep track of changing orientations on the fly (simple ZZ orientation recognition no longer works), and keep track of the bottom 2 layers for setup. I’m not so sure the memorization load would be reduced at all. But for the sake of argument, if each algs we come up with has an average of 2 edge movements, that would put us down to 97 algs. And if we somehow standardize these edge movements, for example always shooting to BL and BR in the EO step, the tradeoffs I listed above would be a lot more manageable.

Surely I’ve made mistakes in these calculations and analysis. What have I got wrong?

 

[FatBoyXPC]

Have you followed the thread closely? If so, do you thoroughly understand the content? When you use jms_gears1's odd sune, 2 edges from U layer go into E, which could potentially be a bad thing (I would prefer to keep all E edges in U). How are you determining the amount of edge positions? The number of edges in U + the number of available positions in E? This counteracts using the d/d' moves for setup.

 

[Kenneth]

It was up at this page in 2005

http://mysite.wanadoo-members.co.uk/cubertscubicle/L2L4.htm

But gone now...

Edit: discussed in Swedish : http://www.svekub.se/forum.html?func=view&id=2237&catid=1#2237

"Administrator" is Gustav Fredell who generated optimal algs for the last 4 steps and sent them to Duncan who published them at the site. I hoped to find the algs at SveKub but it seems they never got into there.

 

[mr. giggums]

3. CP: Shoot to BR. AUF and locate green/red edge. 35 cases.

T: 7 (edge positions) * 2 (orientations) = 14 cases
Y: 7 (edge positions) * 2 (orientations) = 14 cases
Solved: 4 (edge positions w/ symmetry) * 2 (orientations) – 1 (CP skip)= 7 cases​

Surely I’ve made mistakes in these calculations and analysis. What have I got wrong?

Y has symmetry to so it should be 5 (edge positions) * 2 (orientations) = 10 cases

EDIT: and this too.

5. EP: Shoot to FR. AUF and locate blue/red edge. Also hard to recognize case with missing LL edge. 11 cases.

U(a): 5 (edge positions) = 5 cases
U(b): 5 (edge positions) = 5 cases
Z: 5 (edge positions) = 5 cases
H: 5 (edge positions) = 5 cases
Solved: 2 (edge positions w/ symmetry) – 1 (EP skip) = 1 cases​

Surely I’ve made mistakes in these calculations and analysis. What have I got wrong?

5+5+5+5+1 = 21 not 11

 

[qqwref]

5. EP: Shoot to FR. AUF and locate blue/red edge. Also hard to recognize case with missing LL edge. 11 cases.

U(a): 5 (edge positions) = 5 cases
U(b): 5 (edge positions) = 5 cases
Z: 5 (edge positions) = 5 cases
H: 5 (edge positions) = 5 cases
Solved: 2 (edge positions w/ symmetry) – 1 (EP skip) = 1 cases​

Oops, careful, you can't do this. If you swap an edge in U with an F2L edge, you do a 2-cycle, and then the U cases are not necessarily PLL algs (in fact they can't be PLL algs because there is parity).
Cases for EP step:
- If the F2L edge is in place, there are 4 cases (Ua Ub H Z) plus solved (which we won't count).
- If it's not in place, let's say we AUF so that the edge in FR must shoot to UF. If that's part of a 2-cycle, we need another 2-cycle (3 possibilities). If that's part of a 3-cycle, there are 3 possibilities for that, but only one possibility for the remaining two pieces. It can't be part of a 4-cycle. If it's part of a 5-cycle, we can choose the next 3 pieces in any order, so there are 6 possibilities for that.
Hence the EP step has 4+(3+3*1+6) = 16 possibilities in total. Plus solved.

Didn't look at the rest, no clue if there are problems there. I'm not hugely fond of doing casework.

 

[mr. giggums]

Oops, careful, you can't do this. If you swap an edge in U with an F2L edge, you do a 2-cycle, and then the U cases are not necessarily PLL algs (in fact they can't be PLL algs because there is parity).
Cases for EP step:
- If the F2L edge is in place, there are 4 cases (Ua Ub H Z) plus solved (which we won't count).
- If it's not in place, let's say we AUF so that the edge in FR must shoot to UF. If that's part of a 2-cycle, we need another 2-cycle (3 possibilities). If that's part of a 3-cycle, there are 3 possibilities for that, but only one possibility for the remaining two pieces. It can't be part of a 4-cycle. If it's part of a 5-cycle, we can choose the next 3 pieces in any order, so there are 6 possibilities for that.
Hence the EP step has 4+(3+3*1+6) = 16 possibilities in total. Plus solved.

Didn't look at the rest, no clue if there are problems there. I'm not hugely fond of doing casework.

For EP I got (5!/2) - 1 = (60) - 1 = 59 + solved

 

[FatBoyXPC]

qqwref you forgot about the other 16 algs for the other orientation (such as if you want RF into UF), making this a total of 32 total. The math I did was 4 EP cases times 8 (two orientations for each U edge location).

 

[mr. giggums]

qqwref you forgot about the other 16 algs for the other orientation (such as if you want RF into UF), making this a total of 32 total. The math I did was 4 EP cases times 8 (two orientations for each U edge location).

But everything is all ready oriented I thought?

 

[qqwref]

For EP I got (5!/2) - 1 = (60) - 1 = 59 + solved

Mhm. But you're completely ignoring that two cycles like (FR UR UB) and (FR UF UR) - with respect to a solved cube with the U layer corners aligned - can both be solved with the same alg, as long as you AUF before and after as needed. My number only looks at the number of actually different algs you have to know, just like a PLL count wouldn't consider T-perm and U T-perm U' as different.

qqwref you forgot about the other 16 algs for the other orientation (such as if you want RF into UF), making this a total of 32 total. The math I did was 4 EP cases times 8 (two orientations for each U edge location).

"Other orientation"? I suppose you could consider two orientations if you want - having the 5 edges oriented with respect to <R,U> or <F,U> - but if you solve EO in the same way each time you don't have this problem. And you do NOT add 16 more, you add 12, because the EPLL cases stay the same whether the U layer is oriented to R or F. So 32 is wrong; if you are counting both orientations it should (with my calculations) be 28 + solved.

And again, I'm very skeptical of using EP cases - how exactly are you coming up with 4 EPs for the edge not in place? I don't think you can just do that.

 

[FatBoyXPC]

I guess my school of thought was when you swap out UF for FR, you'll end up with one of the 4 EP cases since the CO/CP/EO are all complete. But there was a very good point about EO being already done, so it'd put it at 4 (U edges) * 4 EP cases, which is still 16.

Is there any reason why we couldn't use a sort of EO-Line to preorient bad edges prior to starting the 4 LL steps? It would cut down a fair bit of algorithms needed (because we could skip a step). The only "problem" with this is that you'd have to do one of the 3 steps twice to make up for the loss.

 

[qqwref]

I guess my school of thought was when you swap out UF for FR, you'll end up with one of the 4 EP cases since the CO/CP/EO are all complete.

But you won't, you'll end up with one of the "parity EP" cases, because doing a 2-swap from a valid cube position leaves you in an *invalid* cube position. So your possibilities are now adjacent swap, opposite swap, O-perm (clockwise or counterclockwise), and W-perm. But one of those has two-way symmetry and two have 4-way symmetry, so you can't just multiply by 4... and you're still not counting the cases with FR in place, anyway.

Be careful - it's not 16. It's 16 *plus solved*. If you're getting a total of 16 cases, something in your calculations is off.

Is there any reason why we couldn't use a sort of EO-Line to preorient bad edges prior to starting the 4 LL steps? It would cut down a fair bit of algorithms needed (because we could skip a step). The only "problem" with this is that you'd have to do one of the 3 steps twice to make up for the loss.

Have fun efficiently making a layer once you've EO'd. Also have fun remembering to put in one of the middle edges (it'd take too many moves/piece to insert it later if you're not also affecting LL). You'd probably be better off just doing ZZ.

 

[mr. giggums]

Mhm. But you're completely ignoring that two cycles like (FR UR UB) and (FR UF UR) - with respect to a solved cube with the U layer corners aligned - can both be solved with the same alg, as long as you AUF before and after as needed. My number only looks at the number of actually different algs you have to know, just like a PLL count wouldn't consider T-perm and U T-perm U' as different.

:fp I don't know why I didn't think of that.
Anyway I would think that it's 14 now.
4 for FR being solved (Ua,Ub,Z,H).
Now if you AUF FR to UR if on the top layer pretend you did a 2 swap of FR and UR. Then you would have a last layer of odd swaps. You would have 4 different adjcent swaps, 2 different opposite swaps, 4 different 4-cycles.
4+2+4+4=14

 

[qqwref]

There are 6 possible 4-cycles (4xW, counterclockwise, clockwise).

 

[nickvu2]

Have you followed the thread closely? If so, do you thoroughly understand the content? When you use jms_gears1's odd sune, 2 edges from U layer go into E, which could potentially be a bad thing (I would prefer to keep all E edges in U). How are you determining the amount of edge positions? The number of edges in U + the number of available positions in E? This counteracts using the d/d' moves for setup.

Yes, I've heard you mention keeping E edges in U but I have no understanding of how you intend to do that. If there are misplaced E edges in the E slice, which most of the time there will be, how are you bringing them up to the top? Is this an extra step after 1st layer? And then you plan on using only "pure" algs that don't touch any of the E edges? If so, these will be longer/slower alg, but perhaps worth it with the control you gain.

 

[FatBoyXPC]

Misplaced edges will be brought into the U layer as you place edges in the correct slots.

qqwref: Let me do more thinking about the recent posts before I look myself look like an idiot (or possibly bigger idiot than I've already made myself look ) You said I'm assuming FR is not in place, did you also mean that all other 3 E edges are in placed correctly (including orientation) as well?

 

[qqwref]

They should be if you're solving EP Doesn't really make sense to not do EP last.

 

[FatBoyXPC]

I wanted to make sure you meant that before I ended up missing something by making an assumption. I really need to play with ACube a bit, or cube explorer and test this out.

I guess I'm just confused about something. If we swap UF/FR (and it doesn't matter what happens to UL/UB/UR permutation wise), we'll just end up with a standard EPLL (Ua/Ub/H/Z). It would be the same if we swap UL with FR, UB with FR, UR with FR. Why can't I take 4 (U edge locations) * 4 (EP cases)?