--- In [email protected], "Craig Bouchard" <[EMAIL PROTECTED]> wrote: > > I once heard a good joke...not gunna repeat it...but anyways... > > I like your thinking Stefan, but I was aiming for a 1 algorithm > solution to most cases in a 4LLL...the concept you described is > actually quite wonderful...i would try it...but I'm lazy :p as you > know...but if you are a beginner who wants something easy, but also > wants to learn a more advanced solution I would suggest trying what I > suggested, as what you suggested would take more moves on average... > > Craig
For simplicity, let's assume all algs are 12 moves. If you permute corners and edges *separately* you'll need (5/6)*12 + (11/12)*12 = 21 moves. If you *combine* permuting corners and edges using 5 algs, then it looks like this: (3*2*1)*(4*3) = 72 cases after solving one corner with a U turn. We need: 1 * 0 algs 21 * 1 alg 50 * 2 algs On average that's 20.17 moves, ha! :-) (I admit being a bit sloppy but I hope the general idea is clear) Furthermore, you can easily reduce twisting work for free with separate permutations. For permuting corners you have one alg to swap adjacent corners and one to swap diagonally opposite corners. Consider the latter case and let the algorithm be this one: http://www.cubewhiz.com/images/pll/pll21.gif Now: 1) Rotate U so that the FL corner is solved (BR is solved, too, and the other two need to be swapped). 2) Look where the edge at F belongs. - If it belongs at F, simply apply the alg. - If it belongs at R, do y2 and apply the alg. - If it belongs at B, do U'd and apply the alg. - If it belongs at L, do y2 and go to step 1) unless you've done that before, in which case you just apply the alg. This is fairly simple and guarantees that at least one edge is solved, and then you don't need the Z and H perms anymore, you're guaranteed a 3-cycle or even a edge permutation skip with 1/3 chance (hey, you don't even have to declare it lucky then :-). If two adjacent corners need to be swapped, it's not so easy. For swapping two diagonally opposite corners it doesn't matter which two opposite corners you swap, i.e. you can apply the alg from all four angles and the corners will be solved (at least relative to each other). But for adjacent corners, you just have one angle. So you can't influence the edges if you only know one alg, and you can't prevent getting e.g. Z or H. You need separate algs for that. But using J and its mirror is enough. And a rule to decide which to apply is also very easy. Summary: Using Y, J, mirror(J), U and mirror(U) is enough for easy 2-step PLL. No need learning H and Z. 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/
