Ignoring the solved case and counting mirrors together but inverses distinct...
First off, I looked at the number of totally distinct cases, that'd be 6!/2 = 360. Exploiting two planes of symmetry that gives 90 cases while exploiting 4 planes of symmetry gives 15 cases. But that was just to get a rough estimate so that I know that my answer is going to be between 14 and 79. Some cases are self-mirrors that shouldn't be counted twice and some cases have more or less symmetry than others... a bit of a mess trying to apply Burnside's Lemma on. So then I broke it up into categories: There are 8 cases involving a 5-edge cycle. There are 7 cases involving a 4-edge cycle and a 2-edge cycle. There are 6 cases involving two disjoint 3-edge cycles. There are 4 cases involving a single 3-edge cycle. There are 6 cases involving two disjoint 2-edge cycles. There is the solved case. Thus I conclude that there are 31 cases not counting the solved case. Also 3 of the algs needed are from PLL, so it's really only 28 algs to memorize, some of which maybe inverse of each other. That is a quick count, nothing rigorous... see if you can find any flaws in my reasoning. -Doug --- In zbmethod@yahoogroups.com, "Mike Bennett" <[EMAIL PROTECTED]> wrote: > Can someone help me work out the number of possible combinations for > when there are 1x3x4 blocks on the sides, all four U layer corners > correct, and the final 6 edges and centers oriented? I'd really like > to know, to see if this is a strategy worth pursuing. ------------------------ 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/zbmethod/ <*> 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/
