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Square-1 Yoyleberry Method

luizborges

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Joined
Sep 11, 2018
Messages
2
I used to use this method before I learnt Vandenbergh! The coolest part is probably that you can do 4b using 3-cycle commutators, so the whole solve requires zero algorithms. (Fun fact: when I did this, I used slice moves on both the left and the right, which looked pretty weird to a classmate who was an experienced squanner.)
How do you do commutators on a Square-1? How can you create an intersection of A and B of just one piece/block?
 

xyzzy

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Joined
Dec 24, 2015
Messages
1,759
How do you do commutators on a Square-1? How can you create an intersection of A and B of just one piece/block?
(Notation: R = /; L = y2 / y2; U = (2, 0); D = (0, 2).) [D' L D, R] is a 3-cycle. If you expand and rewrite it to use only normal slices, it becomes D' / U / U' / D /.

If you're asking about squan commutators in general (not in the context of the Yoyleberry method), (i) this is not the right thread to ask that question and (ii) that is a lot trickier because of the bandaged nature of the puzzle. Or if you mean comms once the puzzle is in cubeshape, well, just mess around with the puzzle for a while and you'll probably find something. M2 (which is 1,0/−1,−1/0,1) is a useful primitive since it does two 2-cycles on the edges.
 

luizborges

Member
Joined
Sep 11, 2018
Messages
2
@xyzzy I was talking about the commutator you used. I tried it out, I understood how it worked, but I'm still baffled by the Sq-1 UNintuitiveness...

It is so hard to track pieces in it, that even your simple alg D' / U / U' / D / looks confusing when trying to swap 2 pieces (instead of doing a 3 cicle)... o_O
 
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