ZBLL for megaminx

[CxCxC]

Has anyone thought on how many algs would constitute a ZBLL-like method for a megaminx?
I'm just curious.

 

[irontwig]

Something less than 16*5*152=12080. The same reasoning gives at most 7*4*22=616 for the 3x3, which is certainly a over count, but in the right ball park.

 

[xyzzy]

There are 11672 mega-ZBLL cases if we treat mirrors and inverses as different cases, one of which is the solved case, so you'd need 11671 algs.

Even if you could mirror and invert algs on the fly, that'd still be 3236 cases.

 

[Chree]

There are 11672 mega-ZBLL cases if we treat mirrors and inverses as different cases, one of which is the solved case, so you'd need 11671 algs.

Even if you could mirror and invert algs on the fly, that'd still be 3236 cases.

Here... let me read that for you: "Nope!"

 

[CxCxC]

Something less than 16*5*152=12080. The same reasoning gives at most 7*4*22=616 for the 3x3, which is certainly a over count, but in the right ball park.

Why have you multiplied those numbers? can you explain it a bit more?

 

[AlphaSheep]

Why have you multiplied those numbers? can you explain it a bit more?

You can solve every ZBLL case by doing an OCLL then AUF then PLL, so he's just multiplied the number of OCLL cases by the number of AUFs and the number of PLLs, including a PLL skip. It will always be greater than the actual number of cases but gets closer as the proportion of cases that are rotationally symmetric reduces.

 

[CxCxC]

Thanks!!!
But, why to include the number of cases wherein you get a PLL skip, can't we exclude that?

 

[AlphaSheep]

Thanks!!!
But, why to include the number of cases wherein you get a PLL skip, can't we exclude that?

No, because it covers the cases where the ZBLL alg is just your normal OCLL alg.