PLL 21 cases or 22

is it considered a case?

  • yes a PLL skip is considered a case

    Votes: 35 61.4%
  • No it is not

    Votes: 22 38.6%

  • Total voters
    57
No. I'm pretty sure it's 72

most cases count as four cases (like A, or U)
because there is a possibility of 4 different corner or edge cycles.

However
H perm only count once, really, because the angle which you perform the alg is kinda arbitrary
and Z perm counts as only two
etc

Just found this from a while ago:
There are 72 PLL's if you consider all starting orientations:

- H, the two N's and the solved case have only one starting orientation since they're completely symmetric
- Z and E have two starting orientations
- the remaining 16 cases have 4 starting orientations

4*1+2*2+16*4=4+4+64=72

After the PLL you can have 4 AUF's (U/U'/U2/nothing).
 
No. I'm pretty sure it's 72

most cases count as four cases (like A, or U)
because there is a possibility of 4 different corner or edge cycles.

However
H perm only count once, really, because the angle which you perform the alg is kinda arbitrary
and Z perm counts as only two
etc

Just found this from a while ago:
There are 72 PLL's if you consider all starting orientations:

- H, the two N's and the solved case have only one starting orientation since they're completely symmetric
- Z and E have two starting orientations
- the remaining 16 cases have 4 starting orientations

4*1+2*2+16*4=4+4+64=72

After the PLL you can have 4 AUF's (U/U'/U2/nothing).

yes
 
Of course there are 22. The solved state is one of 22 attainable states on a scrambled last layer using legal moves. Therefore it's a case.
 
@statue: I got 288 (in the sense that d' changes which case it would be). First corner (4 spots), second corner (3 spots), third and fourth corner (2 positions), first edge (4 spots), second edge (3 spots), third and fourth corner (only 1 because of parity). 4*3*2*4*3 = 288
 
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