1 Look LL?

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57x21 algs. 1197.
I don't think anyone will ever bother to learn them all.

Actually, not including mirrors:
41x14 = 574

Brings it way down.
Sigh, you guys should really pay atttention in math. Multiplying the number of oll algs by the number of pll algs will result in the amount of cases you can apply 1 look ll. However, you guys want the number of algs you need to learn, and to get to such it's: The number of oll algs plus the number of pll algs which will equal the number of total algs you need to learn. I really don't get why you'd multiply.
 
Most of my non-bld solves are one look as well :P

Frank Morris can solve the cube in -1 looks.
Edit:
y3k9 said:
Sigh, you guys should really pay atttention in math. Multiplying the number of oll algs by the number of pll algs will result in the amount of cases you can apply 1 look ll. However, you guys want the number of algs you need to learn, and to get to such it's: The number of oll algs plus the number of pll algs which will equal the number of total algs you need to learn. I really don't get why you'd multiply.
What? You WOULD have to multiply. You ADD for 2LLL. MULTIPLY for 1LLL.
But guys...I'm not sure if I am right on this, but wouldn't there be alot more than ~1100 cases? Because if you need to do something like this:
Sune OLL, U2, T Perm to solve a 2LLL case,
Sune OLL, T Perm would not work for a 1LLL case, because you miss the U2 inbetween, ending up not solving the cube.
So this means for 1 OLL and 1 PLL, you would need to learn 4 algs.
For example,
1. Sune OLL+T Perm
2. Sune OLL+U+T Perm
3. Sune OLL+U2+T Perm
4. Sune OLL+U'+T Perm
This would quadrouple the amount of algs needed to learn, making it ~4400 algs.
Please correct me if I am wrong.
Edit 2:
I just remembered some PLLs like H Perm can be solved without AUFing, because the alg still works (E Perm also can solve 2/4 states, and other PLLs I'm too lazy to think of).
So, there would not be 4400 algs, but there would still be alot more than 1100 algs.
 
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Which would take a too many moves and be hard to recog, and usually edge control is done in very few algs.

ZBF2L uses 125 algorithms. Also, I'm not going to claim I know what I'm talking about. I don't use LBL methods at all. You're probably right that recognition would be terrible, but at least this way you would only have to look at the OC and PC.
 
ZBF2L uses 125 algorithms. Also, I'm not going to claim I know what I'm talking about. I don't use LBL methods at all. You're probably right that recognition would be terrible, but at least this way you would only have to look at the OC and PC.
It definitely could be done, but it would take a lot of work to make it decent, and it would probably take longer than just OLL/PLL. And also, i believe it would be less than 42 corner algs because not all of the COLL cases are possible with edges solved.
 
It definitely could be done, but it would take a lot of work to make it decent, and it would probably take longer than just OLL/PLL. And also, i believe it would be less than 42 corner algs because not all of the COLL cases are possible with edges solved.

Well I was asking if that would be a reasonable option, not stating that it was. However, I've found the 42 cases: http://www.ai.univ-paris8.fr/~bh/cube/solutions_c2.html

Also, ZB takes longer than just OLL/PLL, so I'm not surprised that this would.
 
1293? LL algorithms

I am curious how the number could be this low. There are 42 CLL cases. Giving the CLL case a fixed orientation, there are 8 cases for edge orientation, and 12 cases for edge permutation. 42 * 8 * 12 is a lot more than 1293. Obviously mirrors reduce the number quite a bit, but I still don't see how the number could be as low as 1300.

EDIT: Oh, unless inverses are counted as the same alg as well? Then I guess that number actually sounds reasonable
 
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