Thread: "5D parity ?"

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On Mon, Dec 17, 2018, 4:40 PM Alvin Yang wrote:

> Hi Henri!
>
> I believe I have a clue about what your issue is, but I can't be too sure=
.
> Next time, maybe post a screenshot of your issue. Based on what you've
> described, the problem seems to be a swapping permutation of pieces
> belonging to the same cell (well, hypercell anyway). In the screenshot
> below, on the 4D Cube, I've tried to create something that seems to match
> what you're describing, where two pieces are swapped in terms of
> permutation on the same cell. If that's similar to the issue you're deali=
ng
> with, I have also attached a video showing how I resolved my particular 2=
c
> swapping issue. Hope this helps any bit.
>
> Alvin
> [image: image.png]
>
> On Mon, Dec 17, 2018, 3:07 PM henrikengebretsen@gmail.com [4D_Cubing] <
> 4D_Cubing@yahoogroups.com> wrote:
>
>>
>>
>> Hello all!
>>
>>
>> It's been a couple years since I first solved both the 3^4 and the 5^4,
>> and just recently I thought I should take on the daunting challenge that=
is
>> the 3^5.
>>
>> Holy mother, my brain. It hurts.
>>
>> Anyway, I have a problem that I can't for the life of me figure out.
>>
>>
>> Basically, I have managed to solve all the 2-sticker pieces, apart from =
2.
>>
>> I.e. the yellow-red needs to swap with the yellow-blue.
>>
>> I know this is impossible on the 3^3, and to the best of my knowledge
>> it's not possible on the 3^4 either (though I might be wrong on that one=
).
>>
>>
>> My question to you all is how would I go about fixing this? Is this a
>> parity case?
>>
>> I feel like there's a very trivial solution, but my brain is fried..
>>
>>
>> Thanks!
>>
>>=20
>>
>

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If it's what I think it is, you've got two pieces adjacent that need to
swap right? If you just do one twist of that cell to solve one, you'll then
have three unsolved. Then it's just a 3-cycle.

For example, if IF and IR need to swap, do the equivalent of an IU twist,
and that will solve IF, then you have IL-IR-IB as a 3-cycle. Ignore that if
it doesn't make any sense.

~Luna

2018=E5=B9=B412=E6=9C=8817=E6=97=A5(=E6=9C=88) 20:07=E3=80=81henrikengebret=
sen@gmail.com [4D_Cubing] =E3=81=95=E3=82=93=EF=BC=88
4D_Cubing@yahoogroups.com=EF=BC=89=E3=81=AE=E3=83=A1=E3=83=83=E3=82=BB=E3=
=83=BC=E3=82=B8:

>
>
> Hello all!
>
>
> It's been a couple years since I first solved both the 3^4 and the 5^4,
> and just recently I thought I should take on the daunting challenge that =
is
> the 3^5.
>
> Holy mother, my brain. It hurts.
>
> Anyway, I have a problem that I can't for the life of me figure out.
>
>
> Basically, I have managed to solve all the 2-sticker pieces, apart from 2=
.
>
> I.e. the yellow-red needs to swap with the yellow-blue.
>
> I know this is impossible on the 3^3, and to the best of my knowledge it'=
s
> not possible on the 3^4 either (though I might be wrong on that one).
>
>
> My question to you all is how would I go about fixing this? Is this a
> parity case?
>
> I feel like there's a very trivial solution, but my brain is fried..
>
>
> Thanks!
>=20
>

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