# One Look Last Layer



## Zarxrax (Feb 21, 2009)

I'm still very much a noob, so perhaps this is a silly idea. Friedrich 2-LLL is quite easy to learn due to the small number of algorithms, but if you wanted to do it all in just 1 step, then it turns into 1212 algorithms, which is simply absurd for most people to bother with.

But I was thinking, what if you only learn SOME of those 1212 algorithms? On Badmephisto's website the PLL page has some interesting statistics. It turns out that some PLL cases are much more likely to occur than others. Many cases have a 1/18 chance of occurring, while some have a 1/36 chance, or even just 1/72.

Now, what if we take the 1212 algorithms required for a 1-LLL, and determine the chances of each one occurring? Perhaps some of the cases are much more common than others? What if you could get a 1-LLL 50% of the time by learning only 25% of the algorithms? 

Has anyone explored this path in detail? I think it may have some merit.


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## EmersonHerrmann (Feb 21, 2009)

wow...this is such an old idea called ZBF2L and ZBLL. With ZBF2L you orient all of the edges on the last slot, then with ZBLL you solve the entire last layer in one alg.

P.S. - this is a really really old idea...the Zborowski-Bruchem method. Have fun


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## Jhong253 (Feb 21, 2009)

But you are assuming that the scrambled cubes will always follow those chances. There are too many different chances that play upon what LL algorithm you will need -- depends on the scramble and how the person solves the F2L. I don't know how the chances of each algorithm out of the 1212 occurring works, but what comes out of that may not be the actual probability for everyone. It still might work to some degree though.

Emerson -- I think what he's saying is doing the Fridrich F2L for 4 slots and then doing the entire last layer in 1 alg.


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## EmersonHerrmann (Feb 21, 2009)

jhong253 said:


> Emerson -- I think what he's saying is doing the Fridrich F2L for 4 slots and then doing the entire last layer in 1 alg.



I know what he's talking about...but his idea is more insane than ZB...his idea would take many more AGES for recognition of an alg during a solve than ZB AND it would take a lot more learning than ZB.


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## Zarxrax (Feb 21, 2009)

Hmm, yea I guess ZB does sound kind of like what I was talking about, but it still kinda does it in two steps.


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## EmersonHerrmann (Feb 21, 2009)

Zarxrax said:


> Hmm, yea I guess ZB does sound kind of like what I was talking about, but it still kinda does it in two steps.



There's less algs and a bit better recognition.


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## Escher (Feb 21, 2009)

you could always learn the one look last layer algs for the OLLs that you are slowest at... as long as you are only slow at two or three OLLs this would probably be feasible


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## cuBerBruce (Feb 21, 2009)

Among the 1212 cases, there are:

801 cases with probability 1/972
287 cases with probability 1/1944
98 cases with probability 1/3888
22 cases with probability 1/7776
4 cases with probability 1/15552 (including the solved case)

The probabilities assume you ignore the LL pieces while solving the first two layers.

You would have to learn 486 of those 1/972 cases to reach 50% probability of 1-look. If you count mirrors and inverses as separate algs, that's 486*4 = 1944 algs.


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## EmersonHerrmann (Feb 21, 2009)

cuBerBruce said:


> Among the 1212 cases, there are:
> 
> 801 cases with probability 1/972
> 287 cases with probability 1/1944
> ...



But then 50% of the time you will look to see if you know the alg...and then realize that you don't know it and there goes a few seconds


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## jcuber (Feb 21, 2009)

If we create a test-tube child and never teach them to talk, just cube, then perhaps when they are in their teenage years, they may have learned those 1944 algs... 

Remember, many speedcubers probably know hundreds of algs if they know: multiple 2x2 methods + alg-based fridrich + other 3x3 methods + parity algs for bigcubes + last centers for cubes 6x6 and larger + square-1 + praminx + megaminx +...

You see what I mean? If you are willing to spend a few years, it might be theoretically possible.


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## ChromisElda (Feb 23, 2009)

jcuber said:


> If we create a test-tube child and never teach them to talk, just cube, then perhaps when they are in their teenage years, they may have learned those 1944 algs...
> 
> Remember, many speedcubers probably know hundreds of algs if they know: multiple 2x2 methods + alg-based fridrich + other 3x3 methods + parity algs for bigcubes + last centers for cubes 6x6 and larger + square-1 + praminx + megaminx +...
> 
> You see what I mean? If you are willing to spend a few years, it might be theoretically possible.



Haha better start now before others find out that it's possible


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## AvGalen (Feb 23, 2009)

jcuber said:


> If we create a test-tube child and never teach them to talk, just cube, then perhaps when they are in their teenage years, they may have learned those 1944 algs...
> 
> Remember, many speedcubers probably know hundreds of algs if they know: multiple 2x2 methods + alg-based fridrich + other 3x3 methods + parity algs for bigcubes + last centers for cubes 6x6 and larger + square-1 + praminx + megaminx +...
> 
> You see what I mean? If you are willing to spend a few years, it might be theoretically possible.


3x3x3: about 30 (including beginner algs, blindfold algs, fmc algs, etc)

2x2x2: 3 extra
parity algs for bigcubes: 3 (2 for 4x4x4, 1 for 5x5x5)
last centers for 6x6x6 and larger: 0
square-1: 3 (parity fix, and 2 semi-algs)
pyraminx: 1 (2 edge flip)
megaminx: 0 (nothing new to learn after 3x3x3)

Only 10 extra algs in total. I am not a world class speedcuber, but you don't need many extra algorithms after you know a basic 3x3x3 system

There is nobody that knows full ZB and that is mostly because of the amount of algs. Learning less algs, but being able to execute them quickly is much more important than knowing many algs

If you disagree, please prove it by learning > 400 algs


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## daniel0731ex (Feb 24, 2009)

AvGalen said:


> jcuber said:
> 
> 
> > If we create a test-tube child and never teach them to talk, just cube, then perhaps when they are in their teenage years, they may have learned those 1944 algs...
> ...





world class: 1 minute


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## nitrocan (Feb 24, 2009)

But...

What about the recognition. Fridrich is fast because it's so easy tor recognize.


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## qqwref (Feb 25, 2009)

I think as time goes on a lot of Fridrich users slowly learn more and more 1LLL algs, or at least alternate OLLs which control the PLL (such as COLL) and thus make a 1LLL more likely. In general 1LLL is really hard to recognize, so people often pick up the easier-to-recognize ones first - algs to flip/twist two or three pieces, three-cycles, setups to standard PLL algs, that kind of stuff. Even if you only gain maybe a second or so on a 1LLL solve (minus like a second for recognition but plus two for not having PLL), it feels way cooler. I think I had a solve today where the LL was just two flipped edges.


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## dChan (Feb 25, 2009)

jcuber said:


> If we create a test-tube child and never teach them to talk, just cube, then perhaps when they are in their teenage years, they may have learned those 1944 algs...
> 
> Remember, many speedcubers probably know hundreds of algs if they know: multiple 2x2 methods + alg-based fridrich + other 3x3 methods + parity algs for bigcubes + last centers for cubes 6x6 and larger + square-1 + praminx + megaminx +...
> 
> You see what I mean? If you are willing to spend a few years, it might be theoretically possible.



What a creepy idea...


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