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Skewbed

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In all of my testing UL and UR are automatically solved. Can you provide an example solve were ZZLL is needed?
Nvm I found a case.
Reminds me of another post where someone thought they never got parity with their solving order for 3BLD.
If you can't figure out if it will be a skip or not intuitively, then please do some more testing before posting.
 

brododragon

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Reminds me of another post where someone thought they never got parity with their solving order for 3BLD.
If you can't figure out if it will be a skip or not intuitively, then please do some more testing before posting.
I'll try to do more testing, it was 12:00.
 

zimlit

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Steps:
1. 2x2x3 + eo
2. right cross edge and back right f2l pair
3. Remaining edges
4. last 5 corners
 
Last edited:

RedstoneTim

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Steps:
1. 2x2x3 + eo
2. right cross edge and back right f2l pair
3. Remaining edges
$. last 6 corners
This seems good for a 2 look LSLL approach. However there's an issue with alg count (for comparison, L5CO which only solves five corners is already 614 algorithms).
Also as I've said on the Discussion page of the wiki article, the name ZZ-EF is already taken so this should be renamed and possibly just be listed as a variant in the ZZ wiki article.
 

zimlit

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This seems good for a 2 look LSLL approach. However there's an issue with alg count (for comparison, L5CO which only solves five corners is already 614 algorithms).
Also as I've said on the Discussion page of the wiki article, the name ZZ-EF is already taken so this should be renamed and possibly just be listed as a variant in the ZZ wiki article.
the alg count should be the same as L5CO since I accidentally wrote the wrong thing it just needs to preserve edges. as for the name that is true all though I'm not sure to call it. also it only has those 614 algs since everything else is intuitive and you could reduce the alg count by using heise techniques to solve two more corners leaving you with a heise finish.
 
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The idea is to orient the corners and turn them right with only 2 algorithms. That works everytime nomatter how the LL look like after F2L.
You apply a trigger algorithm right after F2L and you either have an X or no correct color on the edges. That means you only have 3 OLL pattern left if X and slightly more 12 OLL if no color.
Isn't this basically just CLP + ELL?
 

RedstoneTim

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the alg count should be the same as L5CO since I accidentally wrote the wrong thing it just needs to preserve edges. as for the name that is true all though I'm not sure to call it. also it only has those 614 algs since everything else is intuitive and you could reduce the alg count by using heise techniques to solve two more corners leaving you with a heise finish.
If we just concentrate on the 2LLSLL after F2L-1 + EO (since the steps before are just Petrus), this would have to be more efficient than already existing approaches to LSLL in two looks (like ZZ-a and ZZ-Zipper) to be viable. (Maybe someone here can calculate the average movecount?).
Also to reduce algorithms, conjugated COLL (the idea is taken from 42) might be possible (although you have to consider that reducing alg count also means raising movecount).
 

zimlit

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If we just concentrate on the 2LLSLL after F2L-1 + EO (since the steps before are just Petrus), this would have to be more efficient than already existing approaches to LSLL in two looks (like ZZ-a and ZZ-Zipper) to be viable. (Maybe someone here can calculate the average movecount?).
Also to reduce algorithms, conjugated COLL (the idea is taken from 42) might be possible (although you have to consider that reducing alg count also means raising movecount).
conjugated coll seams like an interesting Idea it also probably would not take to many moves
 

Skewbed

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If we just concentrate on the 2LLSLL after F2L-1 + EO (since the steps before are just Petrus), this would have to be more efficient than already existing approaches to LSLL in two looks (like ZZ-a and ZZ-Zipper) to be viable. (Maybe someone here can calculate the average movecount?).
Also to reduce algorithms, conjugated COLL (the idea is taken from 42) might be possible (although you have to consider that reducing alg count also means raising movecount).
Conjugated COLL should work. Then, you would just have L5EP. Although this would probably be better with ZZ than Petrus.

1. EOCross (probably better than EOLine)
2. Left pairs/block
3. BR slot
4. Conjugated COLL (42 algs)
5. L5EP (16 algs, 12 without EPLLs)

Seems pretty good, especially for it's alg count.
 
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Conjugated COLL should work. Then, you would just have L5EP. Although this would probably be better with ZZ than Petrus.

1. EOCross (probably better than EOLine)
2. Left pairs/block
3. BR slot
4. Conjugated COLL (42 algs)
5. L5EP (16 algs, 12 without EPLLs)

Seems pretty good, especially for it's alg count.
When is the DFR corner solved?
 
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