Hey Ron,

One last thing, I didn't address your one question.  I think some sort
of cycle-ish strategy will be required, since you can switch two edge
pieces literally without affecting anything else, not even centers.

That seems like it would be very hard to detect in any type of
orientation method for finding parity.  Two edges can be switched that
were both oriented that would still appear to be oriented.

Just the sheer complexity of it seems to have to require at least the
12 cycle counting method.  Again I have no idea how to prove that, but
it seems like some sort of cycling has to be done.

Just my hunch though :-(

Chris

--- In [email protected], "Ron van Bruchem"
<[EMAIL PROTECTED]> wrote:
>
> Hi friends,
> 
> Could Chris/Per/Stefan please explain why they think there cannot be
a method other than counting cycles?
> I would love to see a proof! Or better, I would love to see a proof
of a better method. :-)
> 
> Yesterday I found a counter example of my factor 4 hypothesis: just
swap the edge cubies UrF and URb. You have 0 flipped edge pieces, but
you do have parity.
> When I started this solve the total number of flipped edge pieces
was 14 (=parity, which was the indeed the case), so somewhere in the
solve it changed to a factor of 4.
> 
> CLL=>ELL is indeed an interesting approach. :-)
> 
> Orientation parity during F2L was an idea I already had, and I asked
Jaap to find an algorithm for it, using his program. At the moment he
is still on tour in Florida. :-)
> 
> Stefan, I don't understand what you mean with your idea about dedges.
> 
> Doing both parities in one is interesting. At the moment I use two
algorithms (the normal one by Chris, and the double parity one by
Frédérick), but both twist corners because I use multislice moves. It
would be great to have a fast algorithm that does not twist corners...
It would be great to have a fast algorithm for the orientation parity
anyhow. Yuki can do the parity algorithm amazingly fast. I need to
train for that. :-)
> 
> Is it true that once you solved 3 centers, you cannot change parity
anymore when you keep these centers intact? Because then the center
positions are fixed?
> In that case you only have to count the parity during the first
three centers. Which is not a big task.
> 
> Thanks and have fun,
> 
> Ron
> 
> ----- Original Message ----- 
>   From: Stefan Pochmann 
>   To: [email protected] 
>   Sent: Sunday, November 13, 2005 4:08 AM
>   Subject: [Speed cubing group] Re: 4x4x4 parity without counting cycles
> 
> 
>   --- In [email protected], "Ron" <[EMAIL PROTECTED]> 
> wrote:
>   >
>   > One of the subjects was avoiding the orientation parity, as already 
>   > investigated by Chris Hardwick.
>   > I really think there must be an easier way to check this parity, 
>   > than to count the cycles. Counting cycles takes quite a lot of
time, 
>   > because you have to follow all pieces around the cube.
> 
>   I don't think there's an easy way out and like Per I don't think it 
>   would be worth it anyway unless maybe you can really see it in 5-10 
>   seconds.
> 
>   Some in my opinion more realistic time-savers dealing with the 
>   parities in a centers>edges>3x3 method:
> 
>   - Do CLL->ELL, solving both parities as part of ELL. It's an easy 
>   2-look LL.
> 
>   - Recognize and fix orientation parity before solving the last F2L 
>   pair. Chris should already be able to recognize it since he's doing 
>   ZBF2L and it's not hard to learn anyway. Now we just need a fast alg 
>   that exploits the unsolved slot. Should be shorter/faster than
what we 
>   have so far.
> 
>   - While pairing up dedges, don't flip built dedges anymore (optimally 
>   don't change them at all) and count flipped dedges. Then you could
fix 
>   the parity before starting the 3x3 step. An even shorter/faster alg 
>   should exist.
> 
>   Cheers!
>   Stefan
> 
> 
> 
> 
> 
> 
> 
>
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