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I'm thinking of a possible CFOP subset for the cross. It should be a one-look cross but I would have to generate too many algs, but it would be good I guess?
Three are over 190,000 cross cases for each colour. Symmetries still won't reduce this to the point where humans could learn it. And as Shadowslice said: would we need this? It's not usually difficult to see a near-optimal cross in inspection.
 
I was thinking about using the the Lin method for squan (I'm sub-25 with noob EP), and I was wondering if it would be even better if you used 2 algs for each CP + Edge insert, with one solving parity and the other not, then you could eliminate W-perm and adj. swap for the last step.
 
If you would use this method, you can easily look ahead to the F2L pairs, thus improving your times. I plan to only make this for people with a fixed color cross (mainly white) and people can easily adapt it to other color crosses.
 
I was thinking about using the the Lin method for squan (I'm sub-25 with noob EP), and I was wondering if it would be even better if you used 2 algs for each CP + Edge insert, with one solving parity and the other not, then you could eliminate W-perm and adj. swap for the last step.
uhh this is 3x3
 
If you would use this method, you can easily look ahead to the F2L pairs, thus improving your times. I plan to only make this for people with a fixed color cross (mainly white) and people can easily adapt it to other color crosses.

Can you though? If you just have random algorithms memorized and you don't understand why they work, that doesn't make it easier to look ahead to F2L. Intuition is always better for lookahead, because you can understand where the pieces are going and why they're going there. Think about how hard it is to predict PLL before you do OLL. Besides, if you're willing to put the time into learning the impossible amount of algs (190K+, according to TDM; equivalent to learning ZBLL over 385 times) that this method would require, your time would be much better spent working on lookahead.

Also, like others have mentioned, it's not that hard to get an optimal/near optimal intuitive cross with practice. I just pulled up a random scramble on csTimer: F2 R D L' U2 B' D F R F2 D2 L' B2 L' U2 L2 D2 L2 D'
I solved cross on green intuitively like this: x' y R' F2 D F D F2
csTimer's "Solve Cross" feature gave this: x' B' R2 D R D R2
Which is exactly the same thing; I just added a rotation to make it more fingertrick friendly. No algs necessary. :)
 
Recently on vacation, I thought of a pretty cool LLOB-ZZ hybrid that went like this (I call it "cantelope" because I was eating a cantelope when I thought of it).

1. FB (1×2×3 block on left, ~7 moves)
2. SB (1×2×2 on back right, ~9 moves)
2.9. Use an M/ M' to orient the FR edge (~0.5 moves)
3. Solve the FR edge into its slot while orienting the corners (this is just TSLE in ZZ-CT but you don't need to preserve EO, so probably ~7 moves)
4. EOFB (Orient the edges while solving the DB and DF edge, same step as most LLOB methods, ~8 moves)
5. TTLL (~12 moves)

This method seems pretty decent, it could very good for someone switching from ZZ to Roux (or any casual solver) and probably averages in the ballpark of 44-46 moves (with AUFs). Anybody think this is worth genning algs for?
 
Your movecount estimates are on the low side:
7 FB
8 SBsq
0.5 EO
10 CO+FR (EO does not matter as much as you think)
8 EOFB
14 TTLL + 1.5 AUF

= 49
 
If you would use this method, you can easily look ahead to the F2L pairs, thus improving your times. I plan to only make this for people with a fixed color cross (mainly white) and people can easily adapt it to other color crosses.

uhh this is 3x3


this thread is not for 3x3 only

Well that didn't go as I expected, It was a question about square-1 substep for the Lin method and generating the CP and one edge insert + parity for it.
 
Your movecount estimates are on the low side:
7 FB
8 SBsq
0.5 EO
10 CO+FR (EO does not matter as much as you think)
8 EOFB
14 TTLL + 1.5 AUF
Whoops looks like I mixed up the average PLL movecount (12) with TTLL (14). I stil think the method would be around 45 moves, as I doubt the TSLE step would be 10 moves on average since a lot of cases are <5 moves and the original TSLE is about 9-10 moves on average.
 
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Can you though? If you just have random algorithms memorized and you don't understand why they work, that doesn't make it easier to look ahead to F2L. Intuition is always better for lookahead, because you can understand where the pieces are going and why they're going there. Think about how hard it is to predict PLL before you do OLL. Besides, if you're willing to put the time into learning the impossible amount of algs (190K+, according to TDM; equivalent to learning ZBLL over 385 times) that this method would require, your time would be much better spent working on lookahead.

Also, like others have mentioned, it's not that hard to get an optimal/near optimal intuitive cross with practice. I just pulled up a random scramble on csTimer: F2 R D L' U2 B' D F R F2 D2 L' B2 L' U2 L2 D2 L2 D'
I solved cross on green intuitively like this: x' y R' F2 D F D F2
csTimer's "Solve Cross" feature gave this: x' B' R2 D R D R2
Which is exactly the same thing; I just added a rotation to make it more fingertrick friendly. No algs necessary. :)

i need the possible combinations for one cross color lol
 
Whoops looks like I mixed up the average PLL movecount (12) with TTLL (14). I stil think the method would be around 45 moves, as I doubt the TSLE step would be 10 moves on average since a lot of cases are <5 moves and the original TSLE is about 9-10 moves on average.
As someone that genned some preliminary algs for this exact substep: yes, it is 10 moves on average. TSLE is not 9-10 moves, it is 10 moves (and that is without AUF even)
 
i need the possible combinations for one cross color lol
12 * 11 * 10 * 9 EP * 2 * 2 * 2 * 2 EO = 190,080 cases, just as @TDM said.
You can reduce by LR, FB, and (if you really want) UD reflections and mirror across FR/RU/UF diagonals to make that into ~2970 cases, but even then (and mind you I would love to see someone try to mirror across FR on the fly) the case count is really high
 
12 * 11 * 10 * 9 EP * 2 * 2 * 2 * 2 EO = 190,080 cases, just as @TDM said.
You can reduce by LR, FB, and (if you really want) UD reflections and mirror across FR/RU/UF diagonals to make that into ~2970 cases, but even then (and mind you I would love to see someone try to mirror across FR on the fly) the case count is really high
Maybe I could make this like only one edge inserted or I do 999-look last layer
 
Well that didn't go as I expected, It was a question about square-1 substep for the Lin method and generating the CP and one edge insert + parity for it.
Haha RIP. I for one think it's a nice idea. Honestly the best judge would be to test it out.
 
Yes there are two AUFs

Edit: Actually, there should even be a third AUF before CO+FR. So really it's 49.75 total
 
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