mark49152
Premium Member
I have been experimenting with solving parity (M2/OP edges first) by memoing corners first and if parity exists, exchanging UB and UL during edge memo, so I shoot them to opposite places in anticipation of an odd number of corner swaps fixing that. This avoids a parity swap step. It also means there is always an even number of edge swaps, whereas if I had not swapped those targets, there would obviously be an odd number. So there is one more or one less edge swap.
My question is this: Will there always be one more edge swap, always one less, or 50/50 depending on cycle breaks? Statistically, if we count a dedicated parity fix as equal to any other swap, is there a swap count advantage to fixing parity by exchanging targets?
(My guess is that it's 50/50, so in half of parity solves I'd be replacing parity fix with an extra swap, and in the other half I'd be saving two steps, the fix and a swap.)
My question is this: Will there always be one more edge swap, always one less, or 50/50 depending on cycle breaks? Statistically, if we count a dedicated parity fix as equal to any other swap, is there a swap count advantage to fixing parity by exchanging targets?
(My guess is that it's 50/50, so in half of parity solves I'd be replacing parity fix with an extra swap, and in the other half I'd be saving two steps, the fix and a swap.)