# 3x3 last layer: number of positions

#### Die_d00pel_Null

##### Member
Hi

Well I just wanted to know how many different positions there are on the last layer of the 3x3 and how you can calculate this. I tried on my own with this:
4! * 4! * 3^4 2^4 /12 = 7 776
(corner perm)*(edge perm)*(corner orientation)*(edge orientation) / (parity)
But on two other sites it said 3916 or 1212 positions. So which number is correct?

#### Lord Voldemort

##### He-Who-Must-Not-Be-Named
I think 3916 accounts for inverses and reflections as a separate case.

Oh, I see.
You have extra cases for orientation.
There can only be an even number of unoriented edges, and for every corner that twisted clockwise, there has to be 2 more or 1 twisted counter clockwise.
SO there are less than 2^4 * 3^4 orientations.

I think I'm missing something though.

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#### Stefan

##### Member
4! * 4! * 3^4 2^4 /12 = 7 776
You might want to check that computation.

SO there are less than 2^4 * 3^4 orientations.
I think I'm missing something though.
Yeah. That he already dealt with the orientation issue you're telling him about.

#### Die_d00pel_Null

##### Member
4! * 4! * 3^4 2^4 /12 = 7 776
You might want to check that computation.
oh yes you are right. it should be 62 208. Well this number is quite huge. it probably contains all symetries, inversions and so on.. but is it correct or did I miss something?!

##### Member
I think there are 57 OLL (58 if there is OLL skip) and there are 21 PLL (22 if there is a PLL skip). The 1 Look LL is hence equal to 58 x 22 = 1276 cases.

Though on Bernard Helmsetter's page there are 1212 algorithms for 1 Look LL.

#### Stefan

##### Member
but is it correct or did I miss something?!
It's correct, that's the raw number of LL positions.The smaller numbers are indeed for considering different positions equivalent in certain ways.

#### Swordsman Kirby

##### Member
The 1 Look LL is hence equal to 58 x 22 = 1276 cases.
You can't count it like that.

anyway, there are in fact 62208 last layer positions, but there are far fewer algorithms to solve those positions. Note that we can do an AUF and we'd still use the same algorithm (with the AUF tacked onto the beginning or the end) to solve that. That already reduces the number of algorithms to 15552. Then, there are other positions that are reflectionally symmetric to each other, for example, the two R-perms. That would remove some cases. Furthermore, there are positions that are rotationally symmetric: there are 4 R-perm positions, so we may remove more cases. The 1212 number comes from also taking into account inverses as well.

However, we can't just immediately divide the number of positions by 2 or 4 because there are positions that are reflectionally or rotationally symmetric to itself or its own inverse (or even all three like the H-perm!).

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#### cuBerBruce

##### Member
I think there are 57 OLL (58 if there is OLL skip) and there are 21 PLL (22 if there is a PLL skip). The 1 Look LL is hence equal to 58 x 22 = 1276 cases.

Though on Bernard Helmsetter's page there are 1212 algorithms for 1 Look LL.
Using 58*22 is flawed since it doesn't take into account the fact that generally an OLL can be applied four different ways for each PLL (This doesn't apply to all combinations, so it's not exactly a factor of 4 overall). This factor of approximately 4 can be roughly "recovered" by not counting mirrors and inverses separately. This is why 58*22 and 1212 are roughly the same magnitude, but not the same value.

#### Die_d00pel_Null

##### Member
Hm I got another question why are there 21 PPL algorithms?
The permutation for the last layer should be (((4!)*(4!) /2) /4) . I considered AUF and parity but the result is 72 but it should be 22. So what did I miss?

#### dbax0999

##### Member
Hm I got another question why are there 21 PPL algorithms?
The permutation for the last layer should be (((4!)*(4!) /2) /4) . I considered AUF and parity but the result is 72 but it should be 22. So what did I miss?

Another example:
There are two Z-perms
One has Orange and Blue swapped and Green and Red swapped
The other has Orange and Green swapped and Blue and Red swapped
(On American color scheme with yellow on top)

They can both be solved with one algorithm however you counted them in your computation.

#### Mike Hughey

##### Super Moderator
Staff member
American color scheme
???
I still think American color scheme and Japanese color scheme are synonyms. Somewhere along the line, in the US they switched from the American color scheme (in the early 80's) to the European one. I think dbax0999 is talking about the European color scheme.

#### qqwref

##### Member
They've been selling the white-opp-yellow color scheme in America for a *long* time... maybe you remember when it was the other way, Mike, but you'll find that most forum members are too young

Personally I call that color scheme "Western" or "Standard" (since as far as I know it's the one officially sold in the US (where I live) and Europe), as opposed to "Japanese". My opinion is that anything that is understandable to a typical experienced cuber is fine.

#### dbax0999

##### Member
Oops. I meant European, Western, Standard or whatever its called. I think I've also heard addition color scheme or BOY color scheme.

#### royzabeast

##### Member
Hm I got another question why are there 21 PPL algorithms?
The permutation for the last layer should be (((4!)*(4!) /2) /4) . I considered AUF and parity but the result is 72 but it should be 22. So what did I miss?
I think the 22nd state would be a solved state, aka a PLL skip.

#### trying-to-speedcube...

##### Member
Hm I got another question why are there 21 PPL algorithms?
The permutation for the last layer should be (((4!)*(4!) /2) /4) . I considered AUF and parity but the result is 72 but it should be 22. So what did I miss?
I think the 22nd state would be a solved state, aka a PLL skip.
Well, in your calculation, the Z-perm green-orange and blue-red, is different from blue-orange and green red, while they can be solved in 1 algorithm.

It is good that you considered AUF, so you won't have awkward PLLs with 4-cycles and such. You still came to 72, just because you haven't calculated symmetries. There are 4 Ua-perms. There are 4 Ub-perms. There are 4 Acw-perms. etc. So go through each PLL and count them as 4 when you have no symmetries, 2 when you have 1, 1 when you have 2. And as you will see, you will come to 72.

Another nice fact is that from the number of symmetries also yields the chance of occurance. If you have no symmetries, like a U-perm, the chance of occuring is 4/72, because in the raw calculation, everything has the same chance of occuring. However, in your calculation, there are 4 Ua-perms, yet there is only 1 H-perm and 2 E-perms. That means that the chance that an H-perm occurs, is 1/72, and the chance that an E-perm occurs, is 2/72.

Pretty nice, eh?

#### Die_d00pel_Null

##### Member
Another nice fact is that from the number of symmetries also yields the chance of occurance. If you have no symmetries, like a U-perm, the chance of occuring is 4/72, because in the raw calculation, everything has the same chance of occuring. However, in your calculation, there are 4 Ua-perms, yet there is only 1 H-perm and 2 E-perms. That means that the chance that an H-perm occurs, is 1/72, and the chance that an E-perm occurs, is 2/72.

Pretty nice, eh?
That means that the chance for a PPL skip is the same as the chance for the H-perm?