JLarsen
Premium Member
- Joined
- Sep 30, 2008
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- Dover, New Hampshire, USA
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- 2009LARS03
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@Sn3kyPandaMan, there is no such cycle as (1 2 4 5 6 7 5 6 7 5 6 7 ...).
Your recognition is faulty if you loop back into the middle of the cycle. Why? In the given example, if the piece in position 4 belongs in position 5, then how can the piece in position 7 also belong in position 5? Look more carefully. Maybe you didn't distinguish white from yellow on a piece? For example, my position 4 is just above position 8; if I confuse top/bottom colors, I might go to the wrong one. The sample memos given at cubefreak.net might help. If I didn't answer your question, can you rephrase it?
Oh I figured it all out by myself at about 12:30 in my room. I'm still amazed at the fact that none of you could answer that, because now that I know, allow me to explain it. If you at any time encounter a number already contained within your cycle, you now have the buffer piece, begin a new cycle. If at any time you get your buffer piece, being a new cycle. And as far as parity goes, you need to swap the 2 edges that you swap during corner perm, and make sure that the 2 corners you swap during edge perm are contained within that alg, for me, its the Ra perm because my buffer piece is UBL, and my buffer piece is UR. This is only neccesary when you have an odd number of corner swaps, because you're edges are still left swapped, and never had a chance to return to their positions. Now see if what I was asking makes sense. Perhaps my explanation was off, and no, everything i just said was not asked in my question, because i figured it out myself prior to that. But nobody could explain piece by piece corner cycles, and that, is my summary of them.