--- In [email protected], Gilles van den Peereboom <[EMAIL PROTECTED]> wrote: > > Thank you, > That's I asked ACube to do but your solution is better. > However it's still very very long and you can't resonably finish with > such an algorithm on the 3x3x3. > Another possibility is to fix the centers at the beginning. > But then it just goes back to a normal Fridrich Solve. > > So I guess I won't look much more into that possibility, maybe I'll > from time to time do the cross on one side and fix the center but it > needs to be a very lucky case then. > > Thanks :-) > Gilles.
You could always use cases when the cross is already solved on the opposite side, and finish the solve with M'E2ME2. If there are any nice blocks or pairs, this may work better than F2R2B2L2. -Mike ------------------------ 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/
