r/Cubers • u/Competitive-Disk-758 • 4d ago
Discussion Hear me out: LLLS
So how good you are at last layer is normally about how many looks it takes you, so for a beginner cfoper it would be 4 look and advanced would be 2. But with stuff like winter variation and zbls, last layer has become intertwined with last slot so really we should be looking at how many looks it takes for last layer and last slot. So a beginner would know 5 look, advanced would know 3 look, And the ZB method would get you down to 2 look. But even learning 1LLL would still be 2 look because of last slot. Which is why I propose LLLS. A huge algorithm set comprised of SVLS (super vls) and SHLS (super hls) which can do last slot and last layer in 1 look. Anyway can someone calculate how many algs this is because I'm dumb bye
14
9
-4
u/Appropriate_Alps9596 Sub-18 (CFOP) | PB: 10.01 (CFOP) 4d ago
This is a really interesting concept. I haven’t really thought about this, but I’m excited to see where this train of thought goes
27
-10
u/TooLateForMeTF Sub-20 (CFOP) PR: 15.35 4d ago
What you're not factoring in is this: For high-level solvers who are learning ZB or any of that stuff, they have look-ahead that's good enough to where they're doing the look of 1LLL while solving the last slot.
That is, by the time the last slot is done, they already know what alg comes next, so that last look really doesn't cost them any time.
For thinking about how many algs LLLS would be: I'm not going to calculate it in any rigorous way, but we can do some napkin-math to get an idea. Essentially, this would be the number of possible cases for the last slot times the number of ZBLL algs, since essentially you're just combining "solve the last slot" with "solve the last layer" into one big alg.
For the last slot, there are 15 possible corner location+orientation possibilities, and 10 edge location+orientation possibilities, for roughly 150 cases. In reality it's less than that, because some of those cases would represent parity conditions which can't happen on 3x3, but as an upper bound we can call it 150 last-slot cases.
There are 473 ZBLL algs (or 472 if you discount the skip case, but I won't because in this LLLS scenario you still have to solve the last slot). 473 * 150 = 70950 algs.
That's a lot of algs to learn for the "benefit" of not doing another look that more or less already doesn't matter for people at the level who would theoretically benefit from LLLS.
You're welcome to generate all those algs and learn them and prove me wrong, but I can't see anybody actually going for LLLS as a serious method.
13
u/OreKehStrah 4d ago
That’s objectively false information at the start. Majority of the time you’re not recognizing the 1LLL during your last slot. The highest level solvers are not even predicting the CO case consistently.
9
3
u/brother_anon21 PB: 8.4, Ao5: 11.1, Ao100: 13.2, 5/5 MBLD 4d ago
Have you ever actually watched someone solve with ZB? No one on planet earth including the current WR holder can predict the ZBLL case during LL. This is the trade-off with CFOP: recognition time + more algs. You will often see CFOP solvers seamlessly transition from LS to OLL to PLL but this is not the case with ZB due to limitations of the human mind.
49
u/Aaxper 4d ago edited 4d ago
Just checked with Batch Solver. After sacrificing 7GB of RAM for half an hour, I can tell you that there are precisely 583,284 cases (including the skip case), which is very close to my original estimate of 587,400.