PLL 21 cases or 22

[Jaysammey777]

everybody always says there are 21 PLL cases:
Aa Ab E F Ga Gb Gc Gd H Ja Jb Na Nb Ra Rb T Ua Ub V Y and Z

well is a PLL skip considered a case?
an F2L skip is considered a case in the speed solving wiki (case 37)
If so is OLL 58 cases?
And should F2L be 41?

Tell me your thoughts

 

[Cyrus C.]

There are 21 PLL cases that need algorithms to solve. 22 that you can get.

 

[Jaysammey777]

There are 21 PLL cases that need algorithms to solve. 22 that you can get.

that means there are 22 cases and that it should be included.
Just like in F2L

 

[Chapuunka]

I can't imagine why this would matter to anyone.

 

[Forte]

It's not like it matters too much. People usually say 21 because it's the number of algs you need to learn.

 

[Cyrus C.]

There are 21 PLL cases that need algorithms to solve. 22 that you can get.

that means there are 22 cases and that it should be included.
Just like in F2L

There are 22 cases. Without AUF. Why would they list a case that only requires an AUF?

 

[Ashmnafa]

an F2L skip is considered a case in the speed solving wiki

I think it would be very hard to get an accidental quad-x-cross.

 

[~Phoenix Death~]

an F2L skip is considered a case in the speed solving wiki

I think it would be very hard to get an accidental quad-x-cross.

This bastard got lucky

 

[nlCuber22]

Uhh, no.
LOL WHAT I GOT F2L OLL PLL SKIP

 

[TeddyKGB]

If that's the case shouldn't their be 58 OLL cases?

 

[Owen]

an F2L skip is considered a case in the speed solving wiki

I think it would be very hard to get an accidental quad-x-cross.

This bastard got lucky

Odds anyone?

 

[Jaysammey777]

If that's the case shouldn't their be 58 OLL cases?

yep

 

[Mitch15]

the odds that it's not a real scramble and he faked it all? i'd say pretty high. look at the cube at the start, it starts off practically solved.

Guess what? I can solve really fast if the scramble is U'

 

[riffz]

Of course it's a case. But no one cares and 21 refers to how many algorithms you need to learn.

 

[Jaysammey777]

I can't imagine why this would matter to anyone.

its been boggling around in my head for a while.

 

[ariasamie]

When you say PLL, you should know that it stands for permuting the last layer.
do you permute the last layer when the cube is already solved?
so there are 21 PLL cases.

 

[Kenneth]

44 cases if you include the paritys that may occure on for exampel a [wiki]4x4x4 cube[/wiki] (reduced), [wiki]Domino[/wiki] or a [wiki]Void Cube[/wiki].

As I see it a case is the situation you may get, in some cases it is a skip.

 

[Forte]

When you say PLL, you should know that it stands for permuting the last layer.
do you permute the last layer when the cube is already solved?
so there are 21 PLL cases.

Well technically it counts as an identity permutation :rolleyes:

 

[miniGOINGS]

Only 22 cases then? I would think 288.

 

[shelley]

When you say PLL, you should know that it stands for permuting the last layer.
do you permute the last layer when the cube is already solved?
so there are 21 PLL cases.

Or it stands for Permutation of the Last Layer, and the solved state is certainly a permutation.

video of kid doing two sunes in 7 seconds

lulz