> I can't prove it, but I am almost 100% convinced that counting the > full cycles is the only way.
Hi, I think it is not necessary counting the full cycles with the 24 cubies, but we can calculate a combination between the cycle for the 12 pairs and the orientation for each 12 pairs separatly. Here's the method: Definition : Current Pair : Any combinations of two edges cubies. FirstEdge and SecondEdge are two edges cubies whith the two same colors. MyCounter values 0. I take one pair. It's my Current Pair. I add 1 to MyCounter. The left part of my Current Pair is my First Edge. The pair where my Second Edge is becomes the Current Pair. If the Second Edge is good oriented (according blindfold), I add 1 to my counter, else I add nothing. The left part of my Current Pair is my First Edge. The pair where my Second Edge is becomes the current pair. If the Second Edge is good oriented (according blindfold), I add 1 to my counter, else I add nothing. I do this until my Second Edge is on the first pair I choose at the beginning. The first pair must be twice the current pair. When I finish one cycle, I choose for current pair one pair i have not seen yet.ETC..... Until the 12 pairs have been seen. One example: On a boy color sheme up/yellow face/blue. I do an "f" and a b2 move and I flip the UL and UR edges( as we do when we have 2 edges flipped in a blindfold resolution) . The parity/signature will be odd. I take the UR pair. the left part is yellow/orange. I add one to my counter , it values 1. The other yellow/orange is on the UL pair and good oriented . I add 1 to my counter = 2. The other UL edge is White/red. The other white/red is on the DL pair and not good oriented . I add nothing and my counter = 2. The other DL edge is Yellow/red. The other yellow/red is on the DR pair and not good oriented . I add nothing and my counter = 2. The other DR edge is Yellow/orange. The other yellow/orange is on the UR pair and good oriented . I add 1 and my counter = 3. My counter = 3 , so the parity is odd. Do this with all cycles if they are more cycle. Don't forget the pair already "good paired" and add 1 if it is not good oriented. Let's suppose the cube solved with only the UL pair flipped. I take this pair UL and add 1. The left part is yellow/red. the other yellow/red is on UL pair and not good oriented. I add nothing. The cycle is finish. the parity is odd. Have fun, Frédérick. ------------------------ Yahoo! Groups Sponsor --------------------~--> Get fast access to your favorite Yahoo! Groups. Make Yahoo! your home page http://us.click.yahoo.com/dpRU5A/wUILAA/yQLSAA/MXMplB/TM --------------------------------------------------------------------~-> Yahoo! Groups Links <*> To visit your group on the web, go to: http://groups.yahoo.com/group/speedsolvingrubikscube/ <*> To unsubscribe from this group, send an email to: [EMAIL PROTECTED] <*> Your use of Yahoo! Groups is subject to: http://docs.yahoo.com/info/terms/
