# Message #4175

From: Luna Harran <scarecrowfish@gmail.com>

Subject: Re: [MC4D] 5D parity ?

Date: Mon, 17 Dec 2018 23:34:54 +0000

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年12月17日(月) 20:07、henrikengebretsen@gmail.com [4D_Cubing] さん（

4D_Cubing@yahoogroups.com）のメッセージ:

>

>

> 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!

>

>