cmhardw
Premium Member
Okay, I've given the parity issue more thought, and have determined that I was right to doubt my first impression. Parity excludes some of the scenarios I was thinking were possible. If we assume that the observer is clever enough to calculate the corner parity of a scrambled cube in his head, then the result increases to 6/7, or about 85.714%.
Do you mean permutation parity? The corners can have either even or odd permutation in a legal state. Perhaps I don't follow?
The third and final scenario is that the two pieces share exactly two colors, i.e. they belong adjacent to each other. Let us call the two colors they have in common color A and color B. As before, there are three parity cases. In one case, an orientation exists that reveals stickers A and B, but not one that reveals stickers B and A. In another parity case, an orientation exists that reveals stickers B and A, but not one that reveals stickers A and B. (This bit is what tripped me up the first time 'round.) It seems like you would be able to confuse them, but if you have perfect understanding of parity, it should be possible to determine which corner is which just by seeing A and B or B and A along with all the other stickers that are in view.
I don't follow this. If the two corners in DBL and DBR are ones that belong in UFL and UFR when solved, then I could see a F sticker color at LDB and a U sticker color at RDB in two possible ways, both of which are allowed since I can have even or odd permutation parity in the corners.
Am I not following your argument? Am I making a mistake in my logic somewhere?