Yes theres 3...the 2 3 Cycles and then the one that switches the 2 sets of the adjacent corners...unless you are counting the 3 cycles are 1 cuz one is just the inverse...
Craig P.S. - Stefan you were the 25000th post...intense --- In [email protected], "Stefan Pochmann" <[EMAIL PROTECTED]> wrote: > > Oh Craig... :-) > > Less than 3. > > Cheers! > Stefan > > P.S. Yes I got your mail but I'm too busy to chat now, sorry... > > --- In [email protected], "Craig Bouchard" > <[EMAIL PROTECTED]> wrote: > > > > yah...its only 3 i'm pretty sure...did you get my e-mail??? > > > > Craig > > > > --- In [email protected], "Stefan Pochmann" > > <[EMAIL PROTECTED]> wrote: > > > > > > Oh Craig... :-) > > > > > > Less than 4. > > > > > > Cheers! > > > Stefan > > > > > > --- In [email protected], "Craig Bouchard" > > > <[EMAIL PROTECTED]> wrote: > > > > > > > > More??? meh...at least 4...the 3 cycles...the adjacents...and... > > > maybe > > > > 3...w/e > > > > > > > > Craig > > > > > > > > --- In [email protected], "Stefan Pochmann" > > > > <[EMAIL PROTECTED]> wrote: > > > > > > > > > > --- In [email protected], "Craig > Bouchard" > > > > > <[EMAIL PROTECTED]> wrote: > > > > > > > > > > > > >>>>>>>>Someone tossed out some algorithms...I think if you > want > > > a > > > > > > 4LLL(every time) or less with skips...then you need to > learn: 3 > > > for > > > > > > (as you call it) the cross on top, 4 for permuting corners.. > . > > > > > > > > > > 4 ? > > > > > > > > > > Cheers! > > > > > Stefan > > > > > > > > > > > > > > > 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/
