--- In [email protected], d_funny007 <[EMAIL PROTECTED] ..> wrote: > > Call your starting state A. Then what a program does is find an > optimal A^(-1). Say f(x) = x^(-1), but optimally so. > > So now say we have a goal state B. Then f(B^(-1)*A) should do the > trick. > > Thus I assert that all that needs to be done is to have a pre-parser > that takes A and B and passes B^(-1)*A into the program.
Let start state algorithm A = R' F R B' R' F' R B Let goal state algorithm B = F U R U' R' F' Then B^(-1)*A is: F R U R' U' F' - R' F R B' R' F' R B A solution the program (ignoring permutation of course, since Chris was interested in OLLs) might find this: F U R U' R' F' Now apply A to a solved cube and then the alg the program found, you will not reach the goal state. Where's the mistake? Btw, somebody once showed me this OLL alg: R U R B' R' B U' R' Do that starting with a solved cube. I immediately replied "hey I know a 6-turn alg for that case so if I use the inverse of that I'll have a 6-turner for the same case you showed me an 8-turner for". Cheers! Stefan ------------------------ 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/
