I had another idea for a 3x3 method:
1. EOCross. This can be tricky. If you do EOLine followed by the last two edges, this averages just under 10 moves (9.916 to be exact). Done more efficiently you should be able to get 8-9 moves.
2. Cross extension. Solve the edges of the U layer intuitively. I can do this in about 8 moves with very little practice.
3. Edge permutation and corner layers. Solve the remaining four edges and get all of the corners in their correct layers. I haven’t finished the algorithms but I believe that there are 78 of them.
4. Corner orientation. Orient all of the corners. I believe this is around 180 algorithms, which is a lot to swallow, but manageable.
5. Corner permutation. Permute the corners, but you only have to permute them within their individual layers. I think this is 39 algorithms.

Any suggestions are welcome. I would particularly like a way to reduce the algorithm count. I’m not too attached to anything after the first 2 steps.
 
Was in the FMC discord this weekend and happened to be playing around with some ideas. Just wanted to post a rough outline for a potential FMC method. The method is a mix of HSC, CEOR, and 2GB!
Here it goes:

(1) EO+Psuedo CP-line:
The goal here is the same as yRUru in solving CP and EO. Ideally you'd want to solve these simultaneously although you can do it in any order. The psuedo part is that the 2 corners in DL don't need to be permuted directly - they can be a mix of 2 opposite colours since CP would be solved.

(2) Psuedo 2x2x3:
Now the goal is to solve a 2x2x3 made of opposite colours. 2x2x3 will be built in dl. You'll put 2 E-Slice edeges in BR and FR and opposite edges in DF, DB, DL. If implementing HSC you can leave the 2 remain E-Slice edges in DF + DB or possibly in DL (Need to play around with this).

(3) DR:
The next goal is to reduce the cube into a Domino Reductive state <R2 L2 F2 B2 U D>. Because CP is already solved you will get a HTR skip for corners. General approach is to set up a psuedo-WV pairs. Things like L6CO and HSC-CO can also be used.

(4) HTR:
At this stage you should be left with partial HTR - that is the 2x2x3 + CP + E-slice will be made up of all opposite colours. This will leave us with UF UR UL UB DR being unsorted (not in a proper HTR state). You can now use 3/4/5- Cycles (Ua Ub H and Z perm setups/ L5EP stuff) to separate (or solve) the remaining 5 edges. This actually can be set up during DR/Solving corners.

(5) Endgame:
The end game is pretty straight forward - solve the cube in a HTR state. This is obviously done in <F2 U2 D2 L2 R2 B2> - go crazy, you HTR nuts are freaks.

So you might be asking the question, " why would I use this method?". Thats a great question. The main goal of this method is to implement it as an alternative/simpler path to an HTR.

The numbers would something like this

EO+CP: 9 HTM (4.5 for each)
Pusedo 2x2x3: 8 HTM (Need to do some example solves to come up with a more concrete number)
DR: 7.25 (WV) + 3.25 HTM (WV setup)
HTR: 15 HTM
Total: 42.5 HTM

These numbers will change based on edge inserts/cancelations (Since this is meant for FMC)
Notes: EO and/or CP can be done before, during, or after the 2x2x3.

If you wanted to speedsolve this it would just be CEOR (EO+CP+2x2x3) with a BR + FR + L6CO + L6CP (done simulatneously) into L5EP finish.

The main draw to this method is the application of Psuedo-Block Building, EO, CP, 2-Gen DR, and HTR. It's also cool that this method requires no algs.
The main insipiration for this method was to come up with an "easier" way to reach HTR. Some limitations is that 2-Gen doesn't work well for edge insertions. One way around this could be to use non-2-gen DR algs that preserve CP - will have to play around with this.

If I fudged the numbers, lmk. I'm sure a few people have come up with similar ideas - just wanted to document it here.

Cheers!
 
I had another idea for a 3x3 method:
1. EOCross. This can be tricky. If you do EOLine followed by the last two edges, this averages just under 10 moves (9.916 to be exact). Done more efficiently you should be able to get 8-9 moves.
2. Cross extension. Solve the edges of the U layer intuitively. I can do this in about 8 moves with very little practice.
3. Edge permutation and corner layers. Solve the remaining four edges and get all of the corners in their correct layers. I haven’t finished the algorithms but I believe that there are 78 of them.
4. Corner orientation. Orient all of the corners. I believe this is around 180 algorithms, which is a lot to swallow, but manageable.
5. Corner permutation. Permute the corners, but you only have to permute them within their individual layers. I think this is 39 algorithms.

Any suggestions are welcome. I would particularly like a way to reduce the algorithm count. I’m not too attached to anything after the first 2 steps.
Why not just use ZZ if you are starting with EOcross?
 
I have this stupid idea for the squan that I just wanted to throw here.

Basically I was scrambling 2nd block (including misaligning D by 1 but that doesn’t affect corner permutation), and I should have known before but every time I resolved it so were the corners which reminded me that cp principles that apply to 3x3 also apply to squan. This would be bad for speedsolving due to cp recog, but what if you could plan it in inspection like obl? Just saying I don’t know how to recognize cp and my knowledge of these methods involving cp first is bad. My guess is that it may make 1st block less efficient but it can reduce PLL+1 to L5E or PLL to EPLL. I’m not trying to push this as a good method, more just an interesting concept. It also gives room for new Alg sets such as LSE with the edge in the back and left solved and the DF and DR edges unsolved
 
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I have this stupid idea for the squan that I just wanted to throw here.

Basically I was scrambling 2nd block (including misaligning D by 1 but that doesn’t affect corner permutation), and I should have known before but every time I resolved it so were the corners which reminded me that cp principles that apply to 3x3 also apply to squan. This would be bad for speedsolving due to cp recog, but what if you could plan it in inspection like obl? Just saying I don’t know how to recognize cp and my knowledge of these methods involving cp first is bad. My guess is that it may make 1st block less efficient but it can reduce PLL+1 to L5E or PLL to EPLL. I’m not trying to push this as a good method, more just an interesting concept. It also gives room for new Alg sets such as LSE with the edge in the back and left solved and the DF and DR edges unsolved

Give me a bit to look through some of my notes. I believe Yau-1 does this however it doesn't utilize early cp. It solves L5C then L5E.
 
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I have this stupid idea for the squan that I just wanted to throw here.

Basically I was scrambling 2nd block (including misaligning D by 1 but that doesn’t affect corner permutation), and I should have known before but every time I resolved it so were the corners which reminded me that cp principles that apply to 3x3 also apply to squan. This would be bad for speedsolving due to cp recog, but what if you could plan it in inspection like obl? Just saying I don’t know how to recognize cp and my knowledge of these methods involving cp first is bad. My guess is that it may make 1st block less efficient but it can reduce PLL+1 to L5E or PLL to EPLL. I’m not trying to push this as a good method, more just an interesting concept. It also gives room for new Alg sets such as LSE with the edge in the back and left solved and the DF and DR edges unsolved
I thought about this many years ago (5 and a half now) and it's possible to do 2gr on squan, but there are two problems that need to be solved (that are exactly the same as for 3x3) for it to become a speedsolving viable technique. Firstly, how are you going to be able to do it in inspection every time in such a way that doesn't take away from the rest of the solve, then secodly will it actually benefit times more than the best methods (CSP OBL PBL)? It's a cool idea though.
 
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idea i had for a ls+ll zz variant:

when you get to last slot, you would do an algorithm that inserts the FR edge while permuting corners (ignoring orientation). after that you can solve the rest of the cube in one algorithm. because the corners are already permuted and the edges are already oriented, the last step would be R U gen.

this actually has a surprisingly low number of algs (~300)

i didn't realise this when i first thought of the method but this is pretty much just zz-ct and zz-d at the same time

edit: i'm calling this ZZ-CM bc those are my initials

first step is CMLE – calum melrose last edge

second is 2GLLLC – 2 gen last layer + last corner
 
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idea i had for a ls+ll zz variant:

when you get to last slot, you would do an algorithm that inserts the FR edge while permuting corners (ignoring orientation). after that you can solve the rest of the cube in one algorithm. because the corners are already permuted and the edges are already oriented, the last step would be R U gen.

this actually has a surprisingly low number of algs (~300)

i didn't realise this when i first thought of the method but this is pretty much just zz-ct and zz-d at the same time
those algs would probably be quite bad because it requires you to twist a bunch of corners in place, which typically creates longer algs.
 
those algs would probably be quite bad because it requires you to twist a bunch of corners in place, which typically creates longer algs.
i've already genned algs with trangiums batch solver. the avg mcc for the second step isn't actually that bad – just under 15. the main thing i'm worried about for the viability of this is cp recog.

i think this could be pretty good for one handed though. RU gen in 2GLLLC is usually amazing for OH and CMLE could be RUL or RUD gen, which is also really good for OH
 
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ZZ-D should be strictly better since the difference in speed between CPLS and CP-last-edge is significantly smaller than the difference in speed between 2GLL and 2GLL+DFR
yeah that's probably true

doesn't mean i won't develop the method further though. who cares if a solution is inferior to the standard way of solving a problem when you created the solution yourself?

while i know i just said that you are probably correct i am incredibly tempted to argue that you are not correct. cmle is about as fast as ocll and 2glllc is about as fast as pll (this is a statistic you shouldn't trust as it is just based off of random guessing and the avg mcc of my algs). i don't know if that's better than cpls + 2gll, but it's not like it matters; no-ones gonna learn 300 algs to test out an experimental method.
 
could anyone help me figure out some average movecounts using software?

I would like to know average movecounts for these steps:
  • Full LSE (I know it's not human feasible but I'm just curious)
  • Only Arrow LSE
  • Only 2adj/2 LSE
 
I am 100% sure that I am not the first to come up with this but it is something I invented independently, and there is a chance that it is not found of the forums.

The method I propose is a 2x2 method called Corners of Both Layers (CBL). If you look at CLL, you will find it is basically a way to do OLL and PLL in one alg. CBL is more advanced, where you do OBL and PBL in one alg.

To do the method, first you need to separate two opposite colors. For example, if all the yellow pieces are on the top layer and the white pieces are one the bottom, you have completed the first step. Separating the pieces is comparable to solving level one of the Rubik's Apprentice.

It takes at most 2 moves to do the first step, and it is very easy to one look into an alg. Once you separate the pieces into separate layers, you just need 1 more alg to solve the cube.

The number of cases for CBL is 5546. How this number is found is that you are doing CLL on both layers at that same time. So you would square the number of CLL cases. (43 when you include the solved layer as a case) This gets you 1849. However, because a corner can be twisted, you would have to multiply this number by 3, for each twist the corner is in. CBL requires you knowing TCLL recongnition of shapes like Hammer, Stollery, Pinwheel, etc. When you multiply 1849 by 3, you get 5547. Minus one from this for the case where you have a solved layer on both sides and you get 5546 cases.

While 5000+ algs is a lot, the number of algs that must be learned can be significantly reduced. If after separating colors, you do a z2 rotation, you could end up with a potentially better CBL case. While rotating in solve might be bad, if you one-look and do the z2 rotation in inspection, you can effectively almost half the number of algs you need to know.

I think that this method could be the future for 2x2. It takes at most 2 moves to get the first step, and there are many times I have found scrambles with the first step done already. Sometimes, I find scrambles where the first step is done on 2 axis. (Like you seperated White/Yellow and Green/Blue) While CBL might have a lot of algs, there is a very good chance you already know some of them. EG is a subset of CBL, and so is TCLL and TEG.
Zayn Khanani mentioned this method in his 0.92 recons.
 
Zayn Khanani mentioned this method in his 0.92 recons.
This is known as RoFL. It was proposed and developed by Stefan Pochmann around 2009. The 18 sets where the bottom front two corners are misoriented and or swapped were developed. However, all of the other possibilities are also part of RoFL. This is also where TCLL came from and, interestingly, is credited to Yau and Olson although Pochmann had already generated the algs.

 
I am 100% sure that I am not the first to come up with this but it is something I invented independently, and there is a chance that it is not found of the forums.

The method I propose is a 2x2 method called Corners of Both Layers (CBL). If you look at CLL, you will find it is basically a way to do OLL and PLL in one alg. CBL is more advanced, where you do OBL and PBL in one alg.

To do the method, first you need to separate two opposite colors. For example, if all the yellow pieces are on the top layer and the white pieces are one the bottom, you have completed the first step. Separating the pieces is comparable to solving level one of the Rubik's Apprentice.

It takes at most 2 moves to do the first step, and it is very easy to one look into an alg. Once you separate the pieces into separate layers, you just need 1 more alg to solve the cube.

The number of cases for CBL is 5546. How this number is found is that you are doing CLL on both layers at that same time. So you would square the number of CLL cases. (43 when you include the solved layer as a case) This gets you 1849. However, because a corner can be twisted, you would have to multiply this number by 3, for each twist the corner is in. CBL requires you knowing TCLL recongnition of shapes like Hammer, Stollery, Pinwheel, etc. When you multiply 1849 by 3, you get 5547. Minus one from this for the case where you have a solved layer on both sides and you get 5546 cases.

While 5000+ algs is a lot, the number of algs that must be learned can be significantly reduced. If after separating colors, you do a z2 rotation, you could end up with a potentially better CBL case. While rotating in solve might be bad, if you one-look and do the z2 rotation in inspection, you can effectively almost half the number of algs you need to know.

I think that this method could be the future for 2x2. It takes at most 2 moves to get the first step, and there are many times I have found scrambles with the first step done already. Sometimes, I find scrambles where the first step is done on 2 axis. (Like you seperated White/Yellow and Green/Blue) While CBL might have a lot of algs, there is a very good chance you already know some of them. EG is a subset of CBL, and so is TCLL and TEG.
imagine learning 5k algs to solve a meme cube
 
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