Just an idea -> 2-cycles

[clement]

Hi !

Maybe someone speak about this before :
I was thinking, it IS possible to do a 2-cycle of edges... if you don't look at the centers !
For exemple :
M'U'MU'M'U'MU'M'UM'UMU'MU'M
It swap UF and UB
So we could build up a blindfold method using this, and there is no parity !
The only problem is that the alg is quite long.
I tried with optimal solver to find a better one, and I came up with this one, still very bad :
R U r' U M U' M' L F' l' U M U'

 

[joey]

UR -> UB I used this when solving the void cube.
M' U M' U' M U' M U2 r U R' U' M U R U' R' U'

Messy, but it works!

 

[Stefan]

Came up a little over two years ago here:

Thread start:
http://games.groups.yahoo.com/group/blindfoldsolving-rubiks-cube/message/461

And the first post with the idea of using a centers side effect of order 2:
http://games.groups.yahoo.com/group/blindfoldsolving-rubiks-cube/message/464

 

[cuBerBruce]

Technically speaking, clement's algs will trade an odd edge permutation for an odd center permutation, so I would say "no parity" isn't exactly accurate. But this does seem to be a way to "mindlessly" deal with parity.

I'm looking at this from the point of view of orient-first 3-cycle solving. In that solving style, a parity solve is often resolved by using a PLL alg that swaps two corners and two edges, but this can result in not-so-simple setups. But if you have an alg to swap two edges while permuting centers, and another alg that swaps two corners while doing the inverse permutation of the centers of the edge-swap alg, then I think you should be able to fix parity using two algs with easy setup moves. Of course, the two algs could use the same self-inverse permutation of centers, such as mentioned in the thread Stefan referenced.

Another possibility: fix edge parity using y and corner parity using a corner swapping alg that has same effect on centers as y'.

 

[philkt731]

with this, if you use the algorithm given to add to an odd edge permutation so that you can then solve the last two corners and UF and UB with a PLL alg, you also have to be careful of having it look like this: E2 M' E2 M, by seeing if it was a 1 mod 4 edge permutation or a 3 mod 4 edge permutation...

 

[Stefan]

Hmm, I never thought of it like Bruce suggests, using this just for the parity fix. I thought about using this to solve all edges and all corners. That's why I wanted the center permutation to be self-inverse so I wouldn't have to think mod 4.

 

[cuBerBruce]

I'm unclear what philkt731 s referring to. It seems he may be referring to my 2nd paragraph of my previous message, but may have misunderstood it, so I'll try to explain what I mean in more detail. (It appears Stefan understood. And yes, Stefan, I understood why you wanted self-inverse algs for your purposes.) My suggestion is to *avoid* having to set up a PLL alg by using two algs instead, one that swaps two edges and affects centers, and one that swaps two corners and undoes what the other does to the centers.

An odd permutation is a permutation that contains an odd number of even-length cycles. In 3-cycle solving, odd-length cycles are easily solved. Pairs of even-length cycles can be solved with 3 cycles by "bridging" the two cycles (no hyphen) using an extra 3-cycle. So any odd permutation can be solved with only 3-cycle algs except for a single swap (2-cycle) needing to be done.

So for a given solve, you only execute the edge swap alg once, and you only execute the corner swap once (and then only if you have odd parity for corners and for edges). Since you choose algs that have inverse effects on the centers, you do not have to worry about ending up with the cube in the "E2 M' E2 M" (or similar) state.

The 3rd paragraph of my previous post is an alternate approach something similar to what Kenneth talked about in the "One turn parity fix." thread. Instead of using U (Kenneth's suggestion) as a parity fix, y is used instead, but it fixes only edge parity. You then only need (as far as fixing parity is concerned) a corner swap alg that also does a center 4-cycle in the opposite direction. (For example, the alg U' F L D R' D' F2 D2 B' U R D2 L' F D' B' y' works.) Like Kenneth's approach, it requires memo after determining parity, or some technique for working around that issue. One thing I now realize is that y affects EO, so you have to be careful. It seems to me you simply do the y after determining you have parity, but before memo (WCA rules *allow* you to do y before putting the blindfold on!), and pretend the centers are the colors they would be if you hadn't done y. You keep the the cube oriented that way when putting the blindfold on and start the solve.