Smiles
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Parity means that the states with an odd number of swaps are unreachable, eg if you just swapped 2 edges, then that is one swap and the case is unreachable (without disassembly)
Sorry i deleted my comment because I found out what parity was LOL.
I kinda meant parity in context to Roux, which I think isn't the same thing. I know what parity is in general though, like on 4x4, 6x6, void cube, etc.
According to google it's the E2 M E2 M' case.
The 2 is parity. The 6 is for circular permutations. Basic explanation: UF->UR->UL is the same as UR->UL->UF. So it is really (n-1)! or n!/n
are you sure the circular permutation argument can eliminate cases?
because here's what i'm thinking:
6! means first edge can go anywhere (6 possibilities), next edge can go in anywhere of the 5 remaining spots, etc.
now that has nothing to do with swaps (it's not counting swaps), just the position of each edge. UF -> UR -> UL is one of the 6! possible permutations. UR->UL->UF is the same permutation, it never got counted multiple times anyway so i don't see why you can divide to get rid of it.
anyway i might be wrong, can someone care to explain this?
my math is just L6E cases = 6!/2, or LnE = n!/2
and then there's the solved case, so number of algs is just subtract 1.
intuitively for like L3E there would be 1 solved case + 2 3-cycles, which makes 3 cases or 2 algs.
3!/2 = 3 cases, and subtract 1 is 2 algs.
and then obviously for L2E there would be 1 solved case and 0 algs.
2!/2 = 1 case, and subtract 1 is 0 algs.
shouldn't the same thing work for L6E?
edit: i meant permutation only, like PL6E rather than L6E
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