• Welcome to the Speedsolving.com, home of the web's largest puzzle community!
    You are currently viewing our forum as a guest which gives you limited access to join discussions and access our other features.

    Registration is fast, simple and absolutely free so please, join our community of 30,000+ people from around the world today!

    If you are already a member, simply login to hide this message and begin participating in the community!

3x3 last layer: number of positions

Joined
Feb 28, 2009
Messages
20
Likes
1
Thread starter #1
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
Joined
Nov 24, 2008
Messages
492
Likes
2
WCA
2009PRAS02
#2
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.
 
Last edited:
Joined
Oct 11, 2008
Messages
886
Likes
2
Location
New Brunswick
WCA
2009KADU01
YouTube
kadukarahul
#5
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.
 
Joined
Apr 29, 2006
Messages
1,802
Likes
3
#7
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!).
 
Last edited:
Joined
Oct 8, 2006
Messages
914
Likes
15
Location
Malden, MA, USA
WCA
2006NORS01
YouTube
cuBerBruce
#8
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.
 
Joined
Feb 28, 2009
Messages
20
Likes
1
Thread starter #9
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?
 
Joined
Jul 4, 2009
Messages
380
Likes
1
WCA
2009ADAM01
YouTube
davidadams123456
#10
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?
Read Swordsman Kirby's post above.

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
Joined
Jun 7, 2007
Messages
9,652
Likes
1,613
Location
Indianapolis
WCA
2007HUGH01
YouTube
MikeHughey1
#12
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. :p
 
Joined
Dec 18, 2007
Messages
7,830
Likes
33
Location
a <script> tag near you
WCA
2006GOTT01
YouTube
qqwref2
#13
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.
 
Joined
Apr 29, 2008
Messages
1,680
Likes
7
Location
Almelo, Holland
WCA
2008SMIT04
YouTube
RubiksAddict95
#16
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?
 
Joined
Feb 28, 2009
Messages
20
Likes
1
Thread starter #17
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?
 
Top