Ollie
Member
What program do you use? I need to learn to code in something other than VBA.
Top job!
Top job!
What program do you use? I need to learn to code in something other than VBA.
Top job!
So in my opinion, this trick is not worth. 1.5 solved centers are not worthy of awkward memorization of centers and additional alg for wings.
To mention a similar-but-different idea, how do you feel about François' approach (I think it's François who does this?) of solving centers to arbitrary locations (so two centers that are supposed to be adjacent are opposite, or vice-versa), then fixing the centers afterwards?
To mention a similar-but-different idea, how do you feel about François' approach (I think it's François who does this?) of solving centers to arbitrary locations (so two centers that are supposed to be adjacent are opposite, or vice-versa), then fixing the centers afterwards? He's slower than you are - more my speed - but apparently he considers it to be worthwhile.
Sounds interesting - could you please give some details?
For example, my front side is green, my right side is red. But I see that there are many red centers on front and green centers on right side, so I solve all red centers on F, green on R, and after that I swap all green centers with all red centers, am I right?
I'll try it.
I can't think of a fast way to swap 2 adjacent centres, do you know a good alg? I might try this if the centre fixes aren't too horrible.
I always had a gut feel this was true - it's very nice of you to do the math to confirm it. Now I'm sure I'll never try - I agree that 1.5 solved centers aren't worth the effort.
To mention a similar-but-different idea, how do you feel about François' approach (I think it's François who does this?) of solving centers to arbitrary locations (so two centers that are supposed to be adjacent are opposite, or vice-versa), then fixing the centers afterwards? He's slower than you are - more my speed - but apparently he considers it to be worthwhile.
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In any cube rotation you can put this picture on the front or upper face, and the lines will point you your current obliques. When solving second centers, imagine symmetric picture.
Thank you for your feedback
I can actually find the pieces on their faces quite well but it is performing the commutators that I keep messing up...
As ik won't be breaking any speed records anyway I'm considering using "M2" on obliques.
I just did a couple this way and it seems pretty straight-forward.
Whenever you are turning on a slice parallel to the R and L planes you are always turning a "3" slice, or the 3rd layer from the outside. Whenever you are turning an inner layer slice parallel to the U and D layers you are always turning a "2" layer or a 2nd inner layer from the outside.
:confused:
I think this is not true. Consider an example: [3r, U 2r' U'], when you turn both 2nd and 3rd layers.
If you mean commutators with B part parallel to the U/D planes, this is also wrong. Consider [3l' U2 3l, 2d2] and [2l' U' 2l, 3u2].
Whenever you are turning on a slice parallel to the R and L planes you are always turning a "3" slice, or the 3rd layer from the outside. Whenever you are turning an inner layer slice parallel to the U and D layers you are always turning a "2" layer or a 2nd inner layer from the outside.
Hi Chris,
Thank you very much for your feedback
I understood your notation but I still think there is something not quite right (for my solving system) about:
The "status" so to speak of whether the inner layer only RL turns or UD turns or even FB turns are "2" or "3" turns changes from commutator to commutator. I should have made that more clear in my original post. For the part you quoted I was referring only to the commutator I had written up, namely [r3' u2' r3, U]. Not every commutator would have the specific properties I mentioned, but the commutator [r3' u2' r3, U] does have those properties.
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