The New Method / Substep / Concept Idea Thread

[Axiom]

Yeah, I counted them like this. You have 2dfr positions, in Dfr or above it. So you will have 4 eo cases if it is in place and 8 if it's not, because you are cancelling simmetries.
Then you have 6 CP of the 4 resting corners, this makes you have 4•6+8•6 EOCP cases.
For JBLL you have 12 EP cases, and 14 co cases, you can check the ribbon method algcount. It has 116 tols cases instead of 57 because of having one corner more to orient. This means that there are 14 ocll cases. So it's just 14•12 that makes 168 algs.

Please tell me if you didn't understand

Oh, okay, I understand. EOCP also inserts DFR into it’s slot…
Yeah that makes sense, that also proves your 168 figure and it might explain EOCP? Haven’t checked. The reason I was mistaken is because I believed the DFR corner can be in the U layer too (I thought it was cool), so that made the alg count way higher.
Nice thinking, love the creativity!

 

[J-B]

Yeah, EOCP inserts it in the sense it goes at his place, because of Corner Permutation. Then the parity law reduces a lot the co situations.