Neuro
Veteran
- Joined
- Dec 23, 2016
- Messages
- 622
VHLS is incredibly easy to perform intuitively, even on Mega
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I was literally writing out that exact idea as you posted it... that works pretty well, although I slightly prefer to just do EO without connecting the pair, inserting the edge while solving EP and then doing CO+CP with the last slot corner unsolved. Your idea is way less algs but mine is slightly more efficient.One thing I've experimented with is doing EO and building the pair, inserting the pair and permuting edge, and solving CO then CP. It works pretty well.
317 algs? I don't really get your calculation, I was thinking if you move the LS corner to the UFR spot the number of cases would just be the 17 CPLL cases+5!, which gives you 127 cases, less than half of what you got. I'm not an expert on this sort of math or anything, and I could easily be wrong, but I feel like maybe you mixed in EP or something...That'd make sense. There'd be a lot of opportunities for partial corner control and such, I think. EO and EP will probably be intuitive, CO would be around 3 times as many algs as OCLL, so 51, and L6CP would be CPLL plus 5 possible corner out of slot cases times 5!/2, so 17+5*60, 317, making ~368 algs. Maybe look at ways to reduce alg count?
I mean... this is basically just L4E but you solve one of the u-layer edges to reduce it to L3E. I actually used something like this for a while while I was learning L4E, the main difference is that I would just solve all three centers and an edge through blockbuilding and then as I inserted my second edge I would try to force one u-layer edge to be solved. So, yeah, this is a good stepping stone toward L4E that's often more efficient than LBL with only 4 "algorithms", but I would just stick with blockbuilding instead of solving a "hook".I don't know if this method exists or not, but here goes! it is a pyraminx method that I call "Hook".
Step 1, Hook: Create a "hook", by solving 2 centers and 2 edges on the bottom layer. (One solved edge must be adjacent to both solved centers)
Step 2, C+E: Without destroying the hook, simultaneously solve the remaining center on the bottom and place one of the edges that goes on the top so that its colors match up with the top center (the top center doesn't need to be solved.)
Step 3, L3E: Solve the last three edges, and auf.
Example solve:
*Scramble: l r' u' U' R' L' R' U' B R' L U R' U' B U L B' U R B' U B' L' B
Hook: l' R B U' B'
C+E: r u B' R' B
L3E: U L' U' L R' L R L'
Pros:
Hook is easy to plan out during inspection.
L3E requires very few algorithms.
Cons:
C+E can be tricky to do, and can take time to get used to.
* Sorry the scramble was so long, I got it from the WCA's website.
I mean... this is basically just L4E but you solve one of the u-layer edges to reduce it to L3E. I actually used something like this for a while while I was learning L4E, the main difference is that I would just solve all three centers and an edge through blockbuilding and then as I inserted my second edge I would try to force one u-layer edge to be solved. So, yeah, this is a good stepping stone toward L4E that's often more efficient than LBL with only 4 "algorithms", but I would just stick with blockbuilding instead of solving a "hook".
Your idea is a viable method but I'm not really understanding why someone would use it if they knew L4E... L4E is basically the same thing but you just don't have to solve any u-layer edges while solving your bottom, making it easier to blockbuild. I think this idea is better than both keyhole and LBL since the L3E algs are like 5 moves on average as opposed to some of the 7-8 move algs in LBL, but it's maybe a bit confusing for newer cubers.Thanks for your feedback! while I see your point, I think that this is more of a variant of L4E than a stepping stone. I plan to use both this and L4E as my main methods, and use whichever one most fits the scramble.
Also, I think your 1-flip esque method could be used, but only if it could be one-looked easily. Maybe you could call it 1-look 1-flip, or 1L1F. That would be cool.
Yeah, I think I overcalculated quite a bit. I didn't account for rotational symmetry. However, the good news is, you did too. You didn't account for parity, so it would only be 77, and even then the number would be cut down by mirrors. As for the pyraminx idea, I don't really think it's worth it, but as you said, the best way to find out would be to gen some algs.317 algs? I don't really get your calculation, I was thinking if you move the LS corner to the UFR spot the number of cases would just be the 17 CPLL cases+5!, which gives you 127 cases, less than half of what you got. I'm not an expert on this sort of math or anything, and I could easily be wrong, but I feel like maybe you mixed in EP or something...
Your idea is a viable method but I'm not really understanding why someone would use it if they knew L4E... L4E is basically the same thing but you just don't have to solve any u-layer edges while solving your bottom, making it easier to blockbuild. I think this idea is better than both keyhole and LBL since the L3E algs are like 5 moves on average as opposed to some of the 7-8 move algs in LBL, but it's maybe a bit confusing for newer cubers.
I would still like a bit more feedback on "1L1F" (I think a better name needs to be made for it) before I commit myself to making 100 pyraminx algs with ksolve, but it might be worth it. It's possible solving the centers and edges at the same time makes it a little easier 1-look a solve but doesn't really save any moves, in which case I really don't think the algs are worth generating.
Actually now that I think about it the only way to figure out is probably to start making the algorithms and seeing how they turn out... so I'll update people if I decide to continue it and eventually release it.
Yay! I'll make those algs as soon as I figure out how ksolve works. I always forget I'm making algs for megaminx and not 2×2, that's why I never account for parityYeah, I think I overcalculated quite a bit. I didn't account for rotational symmetry. However, the good news is, you did too. You didn't account for parity, so it would only be 77, and even then the number would be cut down by mirrors. As for the pyraminx idea, I don't really think it's worth it, but as you said, the best way to find out would be to gen some algs.
The "hook" variation you're thinking of could be good, but I recommend you do some tests with it to see how many moves on average it actually saves from doing regular L4E. It might not be worth genning ~125 algs if they only save 1-2 moves. As for 1L1F, I think I've come up with a better idea that just expands L4E and adds ~30 algs... So basically when you solve your centers, instead of trying to blockbuild two edges on the bottom layer to form a "V", you solve ANY 2 edges on the pyraminx. Most of the time you'll end up with two solved edges on the same layer, in which case you rotate the pyraminx so that that side is on the d-layer and you do L4E, but sometimes you'll have two edges solved that are on opposite sides of each other and end up leaving two u-layer edges and two d-layer edges unsolved. This could be a subest of L4E called "HLE" (the unsolved edges sort of form an H shape, plus H is the first initial of my last name, Higgs) or something like that, and would only be 20-30 algs. As a bottom-first solver myself I think this opens up a lot more options for blockbuilding during solves and is much more useful than learning 1L1F. I'll generate these algorithms as soon as I can, it does seem like a great subset to add on to L4E.I just realized that I could also make my Hook method into a 1-look method as well. It would just require ~125 algorithms to solve the other bottom center and l4e at once. I wish you luck with 1L1F!
Huh, that's pretty alright! and onlyApproximate movecount: 9 CPFB + 8 sq&FREO + 8 EODFDB + 5 pair + 13 LSLL = 43
How would one go about solving CPFB with that few moves, though? What's the best/fastest way to solve CP?
Here's a cool CP roux variant, potentially sub-40 moves and less algs than @Aerma's:Approximate movecount: 9 CPFB + 8 sq&FREO + 8 EODFDB + 5 pair + 13 LSLL = 43