Thread: "3^4 parity?"

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Hi.

I think someone must have popped your puzzle apart, put it back together in
a different orbit, and is playing a trick on you! But seriously, as far as
I understand, you can't have two swapped edge (3-colored) pieces on the 3^4
in isolation. See this message
639>
from
the archives (and the entire thread really, for tons of discussion about
it).

If you had only two unsolved 3C pieces, they would necessarily be in the
correct position, but wrong orientation. If that's the case, you can
leverage the 3rd sequence on this page
to correct
the orientations. That sequence twirls a 3C and two 4C pieces. If you
build a commutator with it and a middle slice move that permutes that 3C
piece but not the 4C pieces, you'll end up twirling two 3C pieces, that is
do:

3rd sequence
move
reverse of 3rd sequence
reverse of move

Or maybe your situation is different than I've inferred here (say two
*flipped* edge pieces like here
). Let us
know more specifics if so. Hope this helps.

Roice


On Mon, Jun 27, 2016 at 6:51 PM, legomany3448@gmail.com [4D_Cubing] <
4D_Cubing@yahoogroups.com> wrote:

>
>
> My first four solves, I got the better side of a coin flip and didn't get
> any parity. But ever since I had completed my second solve, it occurred t=
o
> me that I probably should be encountering some sort of parity issues; may=
be
> flipped/swapped 3c/4c pieces ... and now on my fifth solve I have two
> swapped edge pieces. I'm not really sure what to do now. I've been thinki=
ng
> about this for most of the day, and I understand 100% *why* this happens,
> but I don't entirely know how to fix it. I have a vague idea to rotate on=
e
> face 90=C2=B0 and "get away with it" (i.e. reposition the 1c/2c pieces of=
the
> middle slice of that face back to where they started, while keeping an ev=
en
> number of 90=C2=B0 rotations of the center) but I'm not really sure how t=
o
> accomplish that, because I can't do a four-cycle of 2c pieces.
>
>
> I am not looking for a solution to this parity; I am looking for an
> explanation of how to go about creating one. If an example solution is
> necessary to do that, that's fine. Thanks. :)
>
>
>=20
>

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Sorry, I forgot to mention my method. I solve layer-by-layer: 1c pieces of
the first layer, then 2c, then 3c, then the 2c pieces of the second layer,
then the 3c pieces of the second layer. The last layer requires six steps,
consisting of orientation and permutation of the three types of pieces.
Usually I orient all pieces first, then permute the 1c, 2c, and 3c pieces
respectively, though it's possible to solve 3c before 2c. In this case, I
had oriented the entire last layer, and was almost done permuting the 2c
pieces, while the 3c pieces were still unsolved (oriented, though). If
you're curious I can supply a screenshot or log file.

--

I'd love to change the world, but they haven't released the source code yet.

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That is an interesting result. Does your 1344 number exclude the=20
symmetries of the whole puzzle or other symmetries? Have you tried=20
visualizing the resulting graph with GraphVis or other tool?

-Melinda

On 6/29/2016 6:59 AM, phamthihoa4444@gmail.com [4D_Cubing] wrote:
>
>
> I want to calculate number of permutation of half-turn-only 2^4 cube.
>
>
> First, I split the cubies into two groups, called "even" and "odd"=20
> ones. Fix one piece in odd group. The permutation in each group is=20
> even. Thus an upper bound is 8!/2 * 7!/2.
>
> Intuitively I think single 3-cycle in each group is impossible, so=20
> there is much less permutation. But I can't think how to prove it.
>
> Using computer brute force, there are 1344 (surprisingly small)=20
> permutations, if the program has no bug.
>
>
>
>=20


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">


That is an interesting result. Does your 1344 number exclude the
symmetries of the whole puzzle or other symmetries? Have you tried
visualizing the resulting graph with GraphVis or other tool?



-Melinda

I want to calculate number of permutation of
half-turn-only 2^4 cube.

First, I split the cubies into two groups, called
"even" and "odd" ones. Fix one piece in odd group. The
permutation in each group=C2=A0is even. Thus an upper bound is 8!=
/2
* 7!/2.

Intuitively I think single 3-cycle in each group is
impossible, so there is much less permutation. But I can't
think how to prove it.

Using computer brute force, there are 1344 (surprisingly
small) permutations, if the program has no bug.

=20=20=20=20=20=20