Megaminx LL (2x2 block)

Holger

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Nov 27, 2008
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WCA ID
2008BORU01
Hi People

I have been working on another way to solve the megaminx last layer for some time. I have discussed this approach with Henrik, who has been helping a lot.

The main idea is to reduce the last layer by building a 2x2 block on it before finishing the second last face, and then finish the last layer in 2 or 3 steps. There is several ways of doing this, I prefer to do the following steps, when I have reached the last two faces.

1.Solve bad edges. (Petrus style)
2.Build a 2x2 block on R and U
3.Solve 2 F2L pairs on either R or U
4.OLL
5.PLL

If you find bad edges difficult you can do this:

1.Build a 2x2 block on R
2.Solve bad edges
3.Build a 2x2 on U
4 .Solve 2 F2L pairs on either R or U
5.OLL
6.PLL

Another approach is to ignore bad edges and preserve the 2x2 block while doing EO, but I find it difficult to build the 2x2 since you can have 4 flipped U-edges.

The algorithms needed for this LL:
LL: http://www.cubing.dk/megaminx (28 PLL, 21 OLL-algs)
Last F2L: http://www.11time.dk/fspo/Mega/F2L.doc
Example: http://www.11time.dk/fspo/Mega/Example.doc


I am not very good at megaminx, so I find it hard to say whether this method has potential or not, so I would like to hear what you think about it. At least it can be used to force a skip in a normal solve if you find a 2x2 block solved before EO, CO or EP.

Please post if you have some better algs for the “F2L”, some of those I found is long, but most of the longer is 11-12 moves. There probably is some typos in my documents.

I hope you get the idea, but if I am unclear look at the example solves or ask.

/Holger
 
Last edited:
or: Heise style
-EO
-2x2x1
-2 pairs
-all but 3 corners (no odd parity since its megaminx)
-commutator

mvg
 
The algorithms needed for this LL:
LL: http://www.cubing.dk/megaminx (28 PLL, 21 OLL-algs)

It's only the two PDF's that cover this method. The first link are my incomplete collection of "normal" megaminx LL algs. (Yes, there are now arrows on the PLL's. When I begin practicing megaminx again I might complete my collection and update the images.)
 
I tried the first scramble that you did in 59 turns:

Scramble 1: R U2 R2' F2 R U' L' U L2 F2 L' U2' F2' U F2 U2' F U' F'
Two not solved sides on R and U
Block: R U2 R U2' R2' U' F R' F' R' 10
F2L#1: U' R' U' R U R' U2' R 9
F2L#2+ZBF2L: y' U2 R' U2 R U2' R' U R U2' R B' R' B 13
OLL: y' R' U' R U2' R' U2' R 7
EP: y R L U2' L' U R' L U' R U2 L' U2 R' U' 14
CP skip
53 moves

If 5 or 4 corners wrong at CP,then you can do it in 23 turns with commutators. The 3-cycle takes 14 moves to do. So lets say that my average CP is 19 moves (5 or 4 corners are probably more common). I didn't tried the other scrambles because the first part of this solve wasn't lucky.
So lets say that I turn average 72 (53 + 19) moves, without any luck (but there is some trick which you can do it less moves, e.g. using more EP-s so you can force corner 3-cycles).
Your average turns of the 4 solves was 64.5. So I do 7.5 turns more.
In fact I don't think that it's really worth it to learn more than 50 algorithms for 7 moves.
Naturally there are some really useful algorithms, that I have already learnt. :p
 
I tried the first scramble that you did in 59 turns:

Scramble 1: R U2 R2' F2 R U' L' U L2 F2 L' U2' F2' U F2 U2' F U' F'
Two not solved sides on R and U
Block: R U2 R U2' R2' U' F R' F' R' 10
F2L#1: U' R' U' R U R' U2' R 9
F2L#2+ZBF2L: y' U2 R' U2 R U2' R' U R U2' R B' R' B 13
OLL: y' R' U' R U2' R' U2' R 7
EP: y R L U2' L' U R' L U' R U2 L' U2 R' U' 14
CP skip
53 moves

If 5 or 4 corners wrong at CP,then you can do it in 23 turns with commutators. The 3-cycle takes 14 moves to do. So lets say that my average CP is 19 moves (5 or 4 corners are probably more common). I didn't tried the other scrambles because the first part of this solve wasn't lucky.
So lets say that I turn average 72 (53 + 19) moves, without any luck (but there is some trick which you can do it less moves, e.g. using more EP-s so you can force corner 3-cycles).
Your average turns of the 4 solves was 64.5. So I do 7.5 turns more.
In fact I don't think that it's really worth it to learn more than 50 algorithms for 7 moves.
Naturally there are some really useful algorithms, that I have already learnt. :p

R U2 R2' F2 R U' L' U L2 F2 L' U2' F2' U F2 U2' F U' F'
on R and U also (so y'?)
Block: R U2 R U2' R2' U' F R' F' R' 10
F2L#1: y' R' U' R U y R' U2 R 7
F2L#2+OLL (well, CO skip): y' R' U R U2' R B' R' B 8
EP: y2' U R2 U2' R2' U' R2 U2' R2' 8
CP: A perm, can be done in 9 turns, just hard to write down.
total: 42, with CO skip (don't know the average move count for that, but must be around 10-12) I think you should try another example solve ;)
 
I get solves in the 60's with both my normal and this approach, so I don't think there is a lot to save, but it would be a waste of time if I didn't post it :)

You don't really have to learn 50 new algorithms.
You probably know most of the OLL's allready, some are simply mirrored.

Of the PLLs:
7 is pure CP
2 is pure EP
2 is Y-perms
2 is J-perms
Then there is 15 PLLs and a few OLLs to learn, my biggest concern is actually the F2L cases.

Thank you for looking at this approach :)
 
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