Cube Math: How Do You Figure Out the Number of U and D Edge Permutations...?

[dChan]

Could someone well-versed in cube math/theory walk me through how to find, for example, the number of permutations, assuming all pieces are oriented, of U and D-layer edges there are? I know a very primitive way to figure this out but what would the proper formula be for this kind of problem? Just to clarify, for my example problem, I want to figure out how many different configurations there are of wrongly permuted U-face and D-face edges. I know the number for one face of edges is 4 possible permutations(including inverses) so how would I find out the number for two opposite faces(besides using the primitive counting method)?

Thanks,
-dChan

 

[dChan]

I should mention that the U and D edges are in their respective layers so there are no edges that need to be placed from the U to the D layer or vice versa.

 

[watermelon]

In that case, I believe the solution would be like this.

There are 5 possible EP states for a single layer:
-solved
-3-cycle clockwise (U-Permutation)
-3-cycle counterclockwise (U-Permutation)
-2 adjacent 2-swaps (Z-Permutation)
-2 opposite 2-swaps (H-Permutation)

Each layer's edge permutation is independent of the other, so the total number of cases would be 5*5 = 25 cases, one of which is the solved/solved case. Of course, if one of the layers is solved, you can simply do a normal PLL, so the number of cases is really 4*5 = 20 cases.

 

[tim]

(Permutation of U edges)^2

 

[Ville Seppänen]

In that case, I believe the solution would be like this.

There are 5 possible EP states for a single layer:
-solved
-3-cycle clockwise (U-Permutation)
-3-cycle counterclockwise (U-Permutation)
-2 adjacent 2-swaps (Z-Permutation)
-2 opposite 2-swaps (H-Permutation)

Each layer's edge permutation is independent of the other, so the total number of cases would be 5*5 = 25 cases, one of which is the solved/solved case.

I think there are many more cases when both layers edges need to be permuted. For example one case could be solved by doing M2U2M2U2.

 

[dChan]

Yes, I think Ville is right. But how would you do the math for the number of parity cases(I assume all that is left is parity cases)?

EDIT: Okay, so there are 36 cases but how did you find 16 for the formula?

 

[watermelon]

When there is parity, the 5 cases are:
-adjacent 2-swap
-opposite 2-swap
-clockwise 4-cycle
-counterclockwise 4-cycle
-zigzag 4-cycle

The number of parity cases is therefore 5*5 = 25. Adding that to the original number, we get a total of 20 + 25 = 45 cases.

 

[dChan]

lol, I am a bit confused but I think I understood how that all worked out. Thanks. I think I will let that stew in my brain for a little while longer.

 

[Lofty]

I'm assuming we are ignoring the corners and allowing for U adjustments? Cause when I was finding ZBLL/2GLL there was 12 edge possibilities.

 

[dChan]

You can do U adjustments but we are assuming everything is solved except for the U and D layer edge permutation. I figured this would be simple enough to understand in terms of math which is why I chose it.

I believe that if we can do U or D adjustments the number of cases probaly decreases, right?

 

[Lucas Garron]

4 possible permutations(including inverses)

3, with inverses.

I'll, calculate it for you, but could you specify your counting scheme?
There's:

Standard orientation, no AUF, number of cases counted.
AUF/ADF allowed
Mirrors not counted
Inverses not counted.
U and D may be switched.

I think most practical is AUF/ADF, and not counting mirrors duplicate, but not inverses or cube flips.

If you only allow AUF/ADF, it should be equal to the number of even Square-1 EP algs, so half: 49
(Right?)


(How about AuL ?)

 

[dChan]

4 possible permutations(including inverses)

3, with inverses.

I'll, calculate it for you, but could you specify your counting scheme?
There's:

Standard orientation, no AUF, number of cases counted.
AUF/ADF allowed
Mirrors not counted
Inverses not counted.
U and D may be switched.

I think most practical is AUF/ADF, and not counting mirrors duplicate, but not inverses or cube flips.

If you only allow AUF/ADF, it should be equal to the number of even Square-1 EP algs, so half: 49
(Right?)


(How about AuL ?)

For practical use, don't count cases that are the same as a previous case with the exception of a U or D turn(e.g. UF and UB need to be switched with DF and DR - this is the 'same' as UF and Ub needing to be switched with DB and DR).

lol, Lucas, I am confused by what "A" means in "AUF" or "AUL." Can you explain? I am super lost(mostly because of my small brain).

 

[Lofty]

AUF= Adjust U face.
And if thats how you are counting then the numbers given to you already are correct.
Basically what Lucas is talking about is that you can't just cycle the edges to the right spot they also have to be right relative to the corners. So not allowing for AUF/ADF yields a much higher alg count but you would possibly save on moves for cutting out the AUF and probably recognition too.

 

[dChan]

Wouldn't they still be right if you adjusted the U face? If UF and UL need to be swapped while DF and DB need to be swapped you can do U' and apply the algorithm for swapping DF with DB and UF with UR and they will be correct right(except you need you do U at the end)?

 

[tim]

Wouldn't they still be right if you adjusted the U face? If UF and UL need to be swapped while DF and DB need to be swapped you can do U' and apply the algorithm for swapping DF with DB and UF with UR and they will be correct right(except you need you do U at the end)?

Of course, but these are different cube states and different permutations.

 

[Kenneth]

First U layer edge can be in 4 positions, the second in 3 and the third in 2 and the last one can go to one place. 4! = 4 * 3 * 2 * 1 = 24 (4 * 3 = 12 if only one layer is affected, then the paritys can not occur) and for two layers that will be 24 * 24 = 576 combinations but many are the same.

Number of uniqe combinations is much harder to calculate doe.

Paritys newer occur in only one layer. You will actully have two situations, parity or not. The non parity cases are the five normal ones (U, U', H, Z PLL and solved). Two layers makes 5 * 5 = 25 cases and the parity cases are the same number so in total you will have 25 non parity and 25 parity cases (one of the non paritys is solved so I got to the same number as Lucas =)

Parity cases are two opposite, two adjacent, circle, circle' and zig-zag.

You got to love this one: M' S' U' S U2 M