Permutation of the last six edges of the Roux method

[Robert-Y]

Situation: You are on the last step of the roux method (The centres and U corners have been correctly permuted). All that needs to be done is to permute the remaining 6 edges (U edges, DF and DB (which have been oriented correctly)) and the cube will be solved.

Question: How many algorithms would you have to learn to permute all of the edges in one go? (Assuming you don't know any algorithms at the moment.)

 

[Tiw]

Are the 6 edges already oriented or not?

 

[Robert-Y]

Are the 6 edges already oriented or not?

Yes. (Just changed the situation slightly)

 

[Kirjava]

It's more about intuition than algorithms.

 

[Johannes91]

It's more about intuition than algorithms.

I agree, but I'd say it's less about intuition than Floppy Cube is, and the OP asked the same question about that here.

If you want a number as an answer, you need to give a clear definition for "algorithm".

 

[Robert-Y]

It's more about intuition than algorithms.

I agree, but I'd say it's less about intuition than Floppy Cube is, and the OP asked the same question about that here.

If you want a number as an answer, you need to give a clear definition for "algorithm".

Hmm I think perhaps I can give a similar question which will give the same answer that I want:

How many ways are there of arranging the last 6 edges of the Roux method, given that the edges have been correctly oriented and the rest of the cube is solved (including the centres)?

(I'm just talking about legal positions here)

 

[Johannes91]

How many ways are there of arranging the last 6 edges of the Roux method, given that the edges have been correctly oriented and the rest of the cube is solved (including the centres)?

6! / 2 = 360

That counts many practically identical positions as different, like M'U2MU2 and M'B2MB2.

 

[Robert-Y]

6! / 2 = 360

That counts many practically identical positions as different, like M'U2MU2 and M'B2MB2.

Thanks again Johannes, that's all I really wanted for now.

I was just wondering about how the Roux method can be improved.

 

[miniGOINGS]

(6*5*4*3*1*1 = 360)
If you permute DB, recognition would be a lot better and there would be only
(5*4*3*1*1) 60 cases to learn.

 

[Johannes91]

If you permute DB, recognition would be a lot better and there would be only
(5*4*3*1*1) 60 cases to learn.

It should be obvious that these calculations do *not* tell us how many cases need to be learned. Unless you absolutely refuse to use your brain, things like AUF, symmetries and understanding the step reduce it significantly.

 

[miniGOINGS]

It should be obvious that these calculations do *not* tell us how many cases need to be learned. Unless you absolutely refuse to use your brain, things like AUF, symmetries and understanding the step reduce it significantly.

Sorry, 60 cases exist.

 

[deadalnix]

We can fuse the DB edge solve in the orientation step.

 

[StachuK1992]

I actually like that idea.

At first, learn all of the orientation cases.
Next, learn all the cases where you orient all the edges, while permuting the DB edge.
After that, move on the the DF edge.

Anyone wanna figure out how many cases this is?

 

[deadalnix]

It depend on how much you want to use your brain

Theyre are 2^5 * 6 = 192 case for orientation and permutation of DB edge. But many of them are trivialy equivalents, redundants and/or very similars with others.