How many algorithms would you need to memorize to finish the cube after f2l?

[toronto754]

Out of curiousity I'd like to know how many possible positions there are after you've finished F2L and if it would be possible for somebody to memorize all of them so they can do OLL and PLL with one algorithm...Probably not, but I figured it's worth asking.

 

[Ingo]

The Speedsolving Wiki is your friend

Just click on "Speedsolving wiki" in the forum navigation, search for 1LLL and you'll be there.

https://www.speedsolving.com/wiki/index.php/1LLL

 

[obelisk477]

Your best bet to brute force LL is learn all of OLLCP and then the remaining EPLL algs.

 

[Kirjava]

1211, ZBLL, use the wiki, etc

 

[SpeedSkewber]

Theoretically 1LLL would be the faster than doing an OLL and then a PLL. The problem is the amount of time it would take to recognize the case and do the algorithm would take longer than just doing an OLL and a PLL. The L111 algorithms are used for fewest move solving, because they don't finger trick well.

 

[SenorJuan]

As Mr. Toronto is obviously a new cuber, the answer is somewhat more than 1211, as there are mirror cases, which to a newbie means a different algorithm. I'm curious myself as to the actual number, but the only info I've ever seen on the subject is Bernards' work.

 

[GuRoux]

As Mr. Toronto is obviously a new cuber, the answer is somewhat more than 1211, as there are mirror cases, which to a newbie means a different algorithm. I'm curious myself as to the actual number, but the only info I've ever seen on the subject is Bernards' work.

(3^3)(2^3)(4!)(4!)/2 = 62208 cases.

divide that by 4 to remove AUF at the end and it is 15552 cases.

 

[SenorJuan]

Hmmm....if you needed a mirror, inverse, or (both mirror AND inverse) of EVERY one of Bernards 1211 algorithms, you would still have only 4 x 1211 as your total?
I've never tried actually using the Helmstetter tables for completing a LL (for fun), even though I downloaded it many years ago. But surely it has a better 'hit rate' than about 1 in 13?

Edit: A search of the forum turned up this thread:
https://www.speedsolving.com/forum/showthread.php?12815-3x3-last-layer-number-of-positions