Getting Sub-5 min

[Carson]

http://www.youcandothecube.com/secret-unlocked/solution-stage-one.aspx

Lol thats the one I'm using

This method is a great segway into CFOP, which is by-far the most common method used. (Don't wanna start an argument about whether it is the BEST method, but it's pretty obviously the most common)

 

[robindeun]

learn full zbll

 

[AngeL]

Well guess what? I LEARNED THE REAMAINING 3 ALGS IN 1 HOUR! WOOT

Well then congrats! I hope you enjoy. Learning algs sucks, but it's a one time payoff for infinite future benefit. Just push through and you'll find it rewarding.

 

[Joker]

I'm afraid I have to throw in with Rpotts. 12 algs isn't very many, I don't encourage that you quit, but I do recommend that you commit and learn the small handful beginner's method requires.

I always say that this is the best beginner's method out there:
http://www.youtube.com/watch?v=609nhVzg-5Q

Agreed. But, if you don't like learning algs, you probably won't get fast.

 

[Joker]

learn full zbll

LMAO

 

[comady25]

learn full zbll

Woah I think 177 algorithms is a bit much for a begginer.

 

[Rpotts]

its 493 if you count mirrors/inverses individually. And when you're a beginner they all feel individual, LOL.

 

[comady25]

its 493 if you count mirrors/inverses individually. And when you're a beginner they all feel individual, LOL.

Case in Point

 

[Joker]

learn full zbll

Woah I think 177 algorithms is a bit much for a begginer.

Theres way more than 177 for ZBLL
And if you think thats alot, try learning all 1197 for 1LLL (that is, if the way to calculate all possible algs is 57*21 ) :tu

 

[comady25]

learn full zbll

Woah I think 177 algorithms is a bit much for a begginer.

Theres way more than 177 for ZBLL
And if you think thats alot, try learning all 1197 for 1LLL (that is, if the way to calculate all possible algs is 57*21 ) :tu

AHHHHHHHHHH!!!!!!!!!!!!!!!!!!!!!

 

[Cride5]

And if you think thats alot, try learning all 1197 for 1LLL (that is, if the way to calculate all possible algs is 57*21 ) :tu

Yeah, just like the number of PLLs is 2*4 = 8 algs :rolleyes:

The number of cases for pure 1LLL is 15,552 ... but the number of algs required to solve them will be less than this, because many of the cases are rotationally asymmetric (meaning they have equivalents solvable by applying the same alg from different angles), and also because mirrors and inverses can be used to solve different cases. Establishing exactly which cases are rotationally asymmetric, or mirrors/inverses of each other requires a lot more than a simple multiplication. Related info here:
http://www.speedsolving.com/forum/showthread.php?t=12815

EDIT: The [wiki]1LLL[/wiki] page now explains how the number of 1LLL cases/algs is derived.


The example below shows how the number of PLL cases is derived:

Spoiler

The total number of cases is:
number of possible corner permutations * the number of possible edge permutations / 2

Numerically this is 4! * 4! / 2 = 288.

The product is divided by 2 because the 3x3x3 cube cannot enter an odd permutation state, which means only half of the permutations are reachable.


Assuming [wiki]AUF[/wiki]'d cases are the same the number of cases is: 288/4 = 72. For example:

==

Any rotationally asymmetric case can be formed in 4 possible ways. For example there are 4 clockwise U-perms. One preserving the UF-edge, one preserving UB, UR, and UL. Visually:

==

Treating these cases as the same the total for PLL is 22 (or 21 excluding solved). If you solve PLL using mirrors (for example, solving anticlockwise U-perm, by mirroring the clockwise U-perm alg) ie:

==

...then the number of cases is reduced to: 22 - 7 = 15 (including the solved case)

Also including inverses, the G-perm can be solved with a single alg and the number of cases becomes 14, which makes PLL solvable with knowledge of just 13 algorithms.

Apologies for going off-topic, but I felt this needs to be explained.