4x4 Centres and 3x3BH Corners correlates

timelonade

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So I've been trying to learn 4BLD and 3x3BH Corners simultaneously.. I have managed to pretty much master the centres on the 4x4 using commutators but I am still having problems with even the pure comms when doing BH.

Is there any way that these two correlate to help me get a better understanding of BH? If so what are the similarities? I understand there are differences as centres are not corners, but there must be connections beside the use of commutators right?

Inb4 practice or read BYu's tut because I've done both, a lot :P!


~sorry if anyone thinks this belongs in the OAQ thread
 
Yes, the x-centers are often a lot like the BH corner cases. This is not always the case though, so be prepared.

Ubl -> Ufr -> Dfr can be done just the same way as UBL -> UFR -> DFR
Ubl -> Ful -> Rub can be done just the same way as UBL -> FUL -> RUB

Ubl -> Lub -> Ful is not a possible corner cycle, because it would involve two stickers on the same corner. This case is a strictly x-center type case called a "pseudo-slice" case. Cases like these you will have to learn to solve simply as x-center cases. (This case is solvable optimally in 8 moves STM by the way).

Corner cases like UBL -> FDL -> FDR have x-center equivalents, but it's relatively uncommon that you will actually shoot to a non-U layer face twice during the same cycle unless you're solving a supercube BLD or closing a cycle break.

To solve a corner 8 mover case you only need two things. Find 2 interchangeable corners. Interchangeable means that one can be moved to the spot of the other corner on the same side or slice. Then find a 3rd corner not on the interchanging slice that can be put into the interchanging slice with a 3 move insertion.

8 move commutators for BH are always of the form ABA'B' (or the inverse, BAB'A').

A = 3 move insertion
B = 1 turn interchange (either of a side or a slice)

Hope this helps.
 
Yes, the x-centers are often a lot like the BH corner cases. This is not always the case though, so be prepared.

Ubl -> Ufr -> Dfr can be done just the same way as UBL -> UFR -> DFR
Ubl -> Ful -> Rub can be done just the same way as UBL -> FUL -> RUB

Ubl -> Lub -> Ful is not a possible corner cycle, because it would involve two stickers on the same corner. This case is a strictly x-center type case called a "pseudo-slice" case. Cases like these you will have to learn to solve simply as x-center cases. (This case is solvable optimally in 8 moves STM by the way).

Corner cases like UBL -> FDL -> FDR have x-center equivalents, but it's relatively uncommon that you will actually shoot to a non-U layer face twice during the same cycle unless you're solving a supercube BLD or closing a cycle break.

To solve a corner 8 mover case you only need two things. Find 2 interchangeable corners. Interchangeable means that one can be moved to the spot of the other corner on the same side or slice. Then find a 3rd corner not on the interchanging slice that can be put into the interchanging slice with a 3 move insertion.

8 move commutators for BH are always of the form ABA'B' (or the inverse, BAB'A').

A = 3 move insertion
B = 1 turn interchange (either of a side or a slice)

Hope this helps.

Thanks a lot this is really helpful :).
 
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