3x3 Permutations?

[ChickenWrap]

I know other threads have been made on this topic, but none of them quite answered my question. Today, I wanted to try and determine the possible permutations for a 3x3, and got this far:

- 3 orientations for each corner, so 3^7 (since the last corner’s position is determined by the preceding 7)
- 8 edges can be permuted in one of 7 remaining places, so 8^7
- 12 edges can be flipped in 2 different orientations, giving 12^2
- 12 edges can be solved in one of 12 spots, giving 2^11 (since the 12th edge will be solved)

So my calculation was 3^7 * 8^7 * 12^2 * 2^11. Obviously, this is wrong....did I make a small error, or am I an idiot? Please explain where I went wrong, if possible!

 

[TheNextFeliks]

Let's see:
Corners: 3 orientations 3^7 [last depends on others]
8 permutation positions 8!/2 (no parity)
Edges: 2 orientations 2^11 (again final is dependent)
12 permutation positions 12!/2 (no parity)

This is definitely on the forums and even on Wikipedia.

 

[ChickenWrap]

Thanks, I understand this except....why do you divide by 2 for the permutations?

 

[kcl]

Thanks, I understand this except....why do you divide by 2 for the permutations?

You can't swap an even number of pieces.

 

[SpicyOranges]

You can't swap an even number of pieces.

T perm? Y perm? J perm? L perm? V perm? Ra perm? Rb perm? all swap 4 pieces. 4 is an even number.

 

[kcl]

T perm? Y perm? J perm? L perm? V perm? Ra perm? Rb perm? all swap 4 pieces. 4 is an even number.

You smartass

Ok, you can't swap an odd number of pairs.

 

[IcyBlade]

U-perm, A-perm.

You divide by 2 because there can't be parity. There is only 1 place for the last two edges.

 

[Kirjava]

U-perm, A-perm.

They are both an even number of 2-swaps.

 

[Stefan]

You smartass

Ok, you can't swap an odd number of pairs.

U A-perm.

Swaps an odd number of pairs (and does a four-cycle).

 

[Renslay]

http://www.speedsolving.com/wiki/index.php/Laws_of_the_cube
"Only an even number of cubie swaps is possible"

With different words, the signature of the edge-permutation must equal to the signature od the corner-permutation.
See http://en.wikipedia.org/wiki/Parity_of_a_permutation

And yes, U-perm and A-perm are both possible only with even number of swaps, for example:
The permutation UL -> UB -> UR -> UL (an U perm) equals to UL-UB swap plus UL-UR swap (which is two swaps).

 

[rj]

You smartass

Ok, you can't swap an odd number of pairs.

Sometimes I wonder how the [censored word] you're so fast. You don't even know PLL!

 

[kcl]

Sometimes I wonder how the [censored word] you're so fast. You don't even know PLL!

lol I know PLL. I was ridiculously tired when I wrote that and what I meant to say came out wrong XD

 

[CompanionCube]

I know other threads have been made on this topic, but none of them quite answered my question. Today, I wanted to try and determine the possible permutations for a 3x3, and got this far:

- 3 orientations for each corner, so 3^7 (since the last corner’s position is determined by the preceding 7)
- 8 edges can be permuted in one of 7 remaining places, so 8^7
- 12 edges can be flipped in 2 different orientations, giving 12^2
- 12 edges can be solved in one of 12 spots, giving 2^11 (since the 12th edge will be solved)

So my calculation was 3^7 * 8^7 * 12^2 * 2^11. Obviously, this is wrong....did I make a small error, or am I an idiot? Please explain where I went wrong, if possible!

you are right but your calculations are wrong


3 orientations for each corner, so 3^7 (since the last corner’s position is determined by the preceding 7)

8 edges can be permuted in one of 7 remaining places, so 8^7

8 edges permutation is 8!(this is just a different way of showing 8*7*6*5*4*3*2*1) divided by 2 (you divide it in two to avoid parity)

12 edges can be flipped in 2 different orientations, giving 12^2

the orientation for 12 edges should be 2^11

12 edges can be solved in one of 12 spots, giving 2^11 (since the 12th edge will be solved)

the permutation of edges should be 12! divided by 2