martijn_cube
Member
BU: (0292718) 1-1- (7281019)
Kind Regards
Martin
What do you mean with that? i do not know what those numbers mean.
BU: (0292718) 1-1- (7281019)
Kind Regards
Martin
BU: (0292718) 1-1- (7281019)
Kind Regards
Martin
What do you mean with that? i do not know what those numbers mean.
The algorithm I use for UL-UR is (U M' U2 M U), the inverse will of course cycle the opposite way, UR-UL (U' M' U2 M U'). My buffer is DF. If you get these specific cases you'll be saving 13 moves (best scenario), but with setup moves they can also be applied to any edges oriented like that. So if I get BL-FR for example, I do: L R (U M' U2 M U) R' L'. This shows that the maximum number of moves for similar R/L edges is 9, that is if neither of the stickers to be cycled are R/L colors.
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Sounds like TuRBo with DF as buffer. could be nice. the Algs seem to be shorter. Turbo is with UF as buffer, and then cycle UR/ UL.
Here's something I've been messing around with the last few days. I'm probably not the first to use it, but I haven't seen it mentioned. It's a quicker way to cycle similarly oriented edges that are on opposite L and R layers. Of course you don't need to use it all the time, but if you recognize an easy case early on it can save some time.
The algorithm I use for UL-UR is (U M' U2 M U), the inverse will of course cycle the opposite way, UR-UL (U' M' U2 M U'). My buffer is DF. If you get these specific cases you'll be saving 13 moves (best scenario), but with setup moves they can also be applied to any edges oriented like that. So if I get BL-FR for example, I do: L R (U M' U2 M U) R' L'. This shows that the maximum number of moves for similar R/L edges is 9, that is if neither of the stickers to be cycled are R/L colors.
That same case with normal M2 would look like this: (U' L U M2 U' L' U U R U' M2 U R' U'), which is 14 moves with the cancellation. Even with the improved setup moves as shown by joel and stefan it would be 11 moves (L U M2 U R U' M2 U R' U' U' L').
The same idea applies when you have two edges on opposite layers with both R/L stickers stickers to cycle. For that, I use (M U' M U2 M' U' M') which cycles LU-RU, and its inverse (M U M U2 M' U M'). With a maximum of 2 setup moves this means that the most moves for a pair of edges like this is 11. This can be used even in the nasty cases when the improved setup moves wont apply. Example: RB-LB, we can't cancel any setup moves in this case because doing B will move the LB edge. So for both regular M2 and M2 with improved setup moves, the total moves here is 18. It becomes 11 if you use the above algorithm: R' L (M U M U2 M' U M') L' R
If you really wanted to, you could use these algorithms for similar edges on the same layer. An example is RU-RF, which the setup would be (U2 R). I don't know if I really care to use something like that though, I think I'd rather keep it to opposite layers for now. Maybe I'll change my mind with more practice.
anyway, sorry if this is too far off of the M2 topic or if it's already been discussed... it seems like this is something that might be used in another edge method (maybe it is in one that I haven't learned), but I will definitely be using these two algorithms and their inverses in my M2 solves.
Hi,
I work on some improvement in M2 for many mounth now. Just see the dates of my firsts posts on this topic to get an idea.
I've made so many modifications that I think we can consider my method as a new one. You can solve edge by 2 with it and with a maximum of 11 STM move for 2 edges. I haven't made the calculation including probability for an average, but I'm pretty sure it's under 10 STM. The only slice turn you need is M. The number of moves and regrips is reduced compared to M2.
As some of you have noticed, BH seems to be great and I wonder if a kind of « super M2 » method could interest some of you ?
I'm not a native english speaker, so making the explaination in english is a big task for me, so I want to be sure that some of you are intersted.
I'm interested as well. I'd suggest you just briefly outline the method and give an example solve. If we need more details to understand it, we can ask.I'm not a native english speaker, so making the explaination in english is a big task for me, so I want to be sure that some of you are intersted.
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