# Message #4100

From: Andrew Farkas <ajfarkas12@gmail.com>

Subject: Re: [MC4D] 2x2x2x2: List of useful algorithms (please add yours)

Date: Thu, 02 Aug 2018 01:17:55 -0400

Gah, I mistyped

Thus by the time I’m encountering the apparent corner twist parity I don’t

> need to worry about the *z2* rotation since the *I*/*O* sticker shouldn’t

> be on the frame anyway.

That should read *x2*, not *z2*.

On Thu, Aug 2, 2018 at 1:14 AM Andrew Farkas <ajfarkas12@gmail.com> wrote:

> Oh goodness.

>

> You’ve brought up a lot of things to unravel here! I’ll go in order.

>

> But referring to them as "clockwise" and "counterclockwise" relative to

>> I/O didn’t help me. Aren’t we going to need to be able to recognize 3

>> distinct cases? I know I needed 3 cases for my sequences for "RUFI by

>> x2/y2/z2 while keeping the rest of In/Out fully solved".

>

>

> I think this is a result of a difference in our solving approaches. I find

> it easier (and faster) to first consolidate all stickers of the first color

> pair in any position except the frame, and only then form faces from them.

> Thus by the time I’m encountering the apparent corner twist parity I don’t

> need to worry about the *z2* rotation since the *I*/*O* sticker shouldn’t

> be on the frame anyway.

>

> Instead of performing a net 180 degree flip on the piece, you give it a

>> net 120 degree twist on a different axis, while exchanging some twists with

>> other pieces. So, for instance, applying either sequence 2 times to the

>> solved state does not lead to an aligned state like mine does. This

>> baffled me for a few minutes there. It takes applying it 3 times.

>

>

> Again taking a speedsolver’s approach here, I focused solely on achieving

> the desired effect at the given stage, without regard for the true nature

> of the algorithm. I lazily called them "double twist parity algorithms"

> simply because they solved an issue which others have called "double twist

> parity." I appreciate your analysis of what these algorithms actually

> accomplish, though I think I need some more hands-on testing of my own to

> fully grasp what you’re saying.

>

> I’ll call your two algs TTA and TTB (for Triple Twist A and B).

>

>

> Sounds great, if not a bit arbitrary. Generally when mirror cases of an

> algorithm are followed by "a" or "b," no one remembers which is which (e.g.

> A, G, J, N, R, and U PLL algorithms

> <https://www.speedsolving.com/wiki/index.php/PLL>), so if there’s a

> better way to distinguish the two, I’d be more satisfied. That said, the

> only solution I have (CW/CCW) is based on a very limited view of their

> effect.

>

> …

>> and note the colors of the piece that sits at RUFO (Red R, White U, Green

>> F, Pink O).

>

>

> Interesting that we both chose this as the "default" orientation. I

> suppose the red/white/green stems from the standard WCA scramble

> orientation, and pink just falls into place (unless one of our puzzles were

> rotated through a real fourth dimension!). Even pentaquark394’s scrambler

> (and thus my own) use red/orange on the frame with white on the "top"

> (really *O*) and green in front, making it easy to reach from the

> WCA-inspired horizontal orientation. (And the unfolded view is equivalent

> to removing the top half from our standard horizontal orientation and

> placing it to the left of the bottom half with a *z2*.) Anyway, back to

> the cube theory!

>

> There are two choices of CW and CCW in this alg, in step 1 and step 2, and

>> I think we’ll find that we need to use 3 of those 4 combinations in our 3

>> cases. At least, that’s what ended up happening when I created my similar

>> sequences. It looks like the 3rd case can be handled by applying the

>> inverse of the 1st alg.

>

>

> You know, as I was developing these I noticed that the *I*/*O* sticker

> landed on the frame (*R*/*L* face) if I rotated *I* in the opposite

> direction or applied the wrong algorithm for either case; I dismissed this

> at the time as an undesired result, but in retrospect it could certainly be

> useful. In my solution video, I accidentally left an *I*/*O* sticker on

> the frame, and had to spend quite some time resolving it when clearly a

> single short algorithm would have sufficed.

>

> The next step is to see how these conjugates with *I[…]* can be used to

> efficiently orient several pieces at once! I’m not sure whether this would

> be any faster than simple 3D moves, but perhaps even some existing 3D

> algorithms could take advantage of the extra dimension that the 2^4 has to

> offer.

>

> *TTA*: Twist *OFRU* counterclockwise (relative to its Front

>> tetrahedron):

>> *[ R[ U’ R’ U2 ]: Iy Ix ]**TTB*: Twist *OBRU* clockwise (relative to

>> its Back tetrahedron): *[ R[ U R U2 ]: Iy’ Ix’ ]*

>

>

> In my 3D mindset at this stage, I prefer to think of these as clockwise

> and counterclockwise respectively around the *R* hypersticker, hence my

> original naming.

>

> Recognition: put misaligned piece on *OFRU*. If the I/O color is on

>> the Front face, perform *Rx* and then *TTB*. Otherwise, the piece can

>> be aligned via a twist of the Front tetrahedron. Apply *TTA* (if a

>> counterclockwise twist is needed, i.e. the I/O color is on U) or *TTA’* (if

>> a clockwise twist is needed, i.e. the I/O color is on the R corner).

>

>

> I prefer to combine this into one, slightly more complicated step: Hold

> left and right subcubes such that all oriented pieces are on the *I* and

> *O* faces, and that the misaligned piece is in *ROFU* or *ROBU* with the

> *I*/*O* sticker facing *U*. If the piece is in *ROFU*, apply TTA; if it

> is in *ROBU*, apply TTB.

>

> The same result is achieved either way.

>

> Thanks for being such a fun co-conspirator.

>

>

> Right back at ya. 🙃

>

> Theorem: Every combination of three corner twists is equal to one of the

>> eight possible single corner twists (clockwise and anticlockwise around any

>> of the 4 colors) or the identity. Every combination of two corner twists

>> is equal to one of the three monoflips or the identity. (OK, OK, this is

>> still just a hypothesis until I enumerate the damn things or otherwise

>> prove it more thoroughly than I have done in my head so far.)

>

>

> Well, so much for sleeping tonight. 😛 I have a strong feeling that both

> of those are true, but of course a proof is necessary. Enumeration is

> pretty trivial at this scale – there’s probably only a dozen or so cases

> after removing mirrors and the like – but of course a rational argument is

> much more appealing. I’ll give it a shot.

>

> Random idea: at the beginning of a solve, if we notice that there’s a

>> color pair with exactly 1 piece on the corners, we should just probably

>> just go ahead and align the other 15 pieces of that color pair, then apply

>> one of these algs. Now that it’s so easy to fix this kind of misalignment,

>> futzing around with additional gyros doesn’t seem worth it if we’re only 1

>> piece off from having a color pair off of the corners.

>

>

> Certainly! It might even be worth it for two, if we can account for double

> tw– er, corner twist (?) parity along the way. I would still like to

> develop a general intuitive strategy and/or algorithm set for this stage; I

> think it’s the least consistent part of Fourtega and thus the one that

> could use the most improvement.

>

> Thank you very much for continued analysis and discussion! It’s fun to be

> exploring new territory.

>

> - Andy

>

> P.S. Counterexample to the first hypothesis: (execute on *ROFU*) CW

> around *R *+ CW around *U* + CW around *R* results in *x2*. The second

> hypothesis contradicts corner twist parity: two corner twists in the same

> direction violates corner twist parity, while monoflips and the identity do

> not.

>

> On Wed, Aug 1, 2018 at 9:57 PM Marc Ringuette ringuette@solarmirror.com

> [4D_Cubing] <4D_Cubing@yahoogroups.com> wrote:

>

>>

>>

>> Hey, Andy,

>>

>> I love your maybe-the-shortest-possible monoflip aligners. But referring

>> to them as "clockwise" and "counterclockwise" relative to I/O didn’t help

>> me. Aren’t we going to need to be able to recognize 3 distinct cases? I

>> know I needed 3 cases for my sequences for "RUFI by x2/y2/z2 while keeping

>> the rest of In/Out fully solved".

>>

>> Your algs are a bit more confusing for me to think about than mine were,

>> because they do three distinct corner twists on the misaligned piece,

>> whereas mine do two. Instead of performing a net 180 degree flip on the

>> piece, you give it a net 120 degree twist on a different axis, while

>> exchanging some twists with other pieces. So, for instance, applying

>> either sequence 2 times to the solved state does not lead to an aligned

>> state like mine does. This baffled me for a few minutes there. It takes

>> applying it 3 times.

>>

>> I’ll call your two algs TTA and TTB (for Triple Twist A and B).

>>

>> In tracing through your first alg, TTA, I found it useful to start from

>> my standard solved state and note the colors of the piece that sits at RUFO

>> (Red R, White U, Green F, Pink O).

>>

>> Step 1. R [ U’ R’ U2 ] – twists RUFO CCW around the Right center

>> (the red corner of the piece) and then places the piece on RUBI with R[ U2 ]

>> Step 2. Iy Ix – twists RUBI CW around the In

>> center (the anti-green corner of the piece) and does not permute it

>> Step 3. R [ U2 R U ] – places the RUBI piece on RUFO with R [

>> U2 ] and then twists it CW around the Right center (the pink corner of the

>> piece)

>>

>> There are two choices of CW and CCW in this alg, in step 1 and step 2,

>> and I think we’ll find that we need to use 3 of those 4 combinations in our

>> 3 cases. At least, that’s what ended up happening when I created my

>> similar sequences. It looks like the 3rd case can be handled by applying

>> the inverse of the 1st alg.

>>

>> Note that in TTA three different colors on the piece get twists applied

>> (Red CCW, Green CCW, Pink CW). The net result is Green CCW (!), the color

>> that was originally Front and still remains Front.

>>

>> Tracing similarly, TTB twists the Back tetrahedron of OBRU (Blue in this

>> case) CW.

>>

>> So I guess here’s how I’d have described your algs and the recognition:

>>

>> * TTA*: Twist *OFRU* counterclockwise (relative to its Front

>> tetrahedron): *[ R[ U’ R’ U2 ]: Iy Ix ]*

>> *TTB*: Twist *OBRU* clockwise (relative to its Back tetrahedron): *[ R[

>> U R U2 ]: Iy’ Ix’ ]*

>>

>> Recognition: put misaligned piece on *OFRU*. If the I/O color is on

>> the Front face, perform *Rx* and then *TTB*. Otherwise, the piece can

>> be aligned via a twist of the Front tetrahedron. Apply *TTA* (if a

>> counterclockwise twist is needed, i.e. the I/O color is on U) or *TTA’*

>> (if a clockwise twist is needed, i.e. the I/O color is on the R

>> corner).

>>

>> What do you think?

>>

>> (The three cases above could also be recognized as the ones where a y2

>> flip, z2 flip, and x2 flip are needed, respectively; although we do not

>> actually perform that flip, so it would seem a bit odd to do recognition by

>> figuring out what 180 degree flip we "could" use, and then not using it.

>> I might do it that way anyway..)

>>

>>

>>

>> I absolutely love this part of the puzzle-figuring-out process, because

>> I’m starting to get the hang of the 12 orientations, and how they divide up

>> into 4’s and 3’s, and how corner twists can combine into monoflips, etc.

>> Your triple twister algorithms are reminding me that I don’t fully grok it

>> yet, but I feel like I’m making good progress. Thanks for being such a fun

>> co-conspirator.

>>

>>

>> Theorem: Every combination of three corner twists is equal to one of the

>> eight possible single corner twists (clockwise and anticlockwise around any

>> of the 4 colors) or the identity. Every combination of two corner twists

>> is equal to one of the three monoflips or the identity. (OK, OK, this is

>> still just a hypothesis until I enumerate the damn things or otherwise

>> prove it more thoroughly than I have done in my head so far.)

>>

>>

>> Random idea: at the beginning of a solve, if we notice that there’s a

>> color pair with exactly 1 piece on the corners, we should just probably

>> just go ahead and align the other 15 pieces of that color pair, then apply

>> one of these algs. Now that it’s so easy to fix this kind of misalignment,

>> futzing around with additional gyros doesn’t seem worth it if we’re only 1

>> piece off from having a color pair off of the corners.

>>

>>

>> Cheers

>> Marc

>>

>>

>> On 7/31/2018 9:59 PM, Andy F legomany3448@gmail.com [4D_Cubing] wrote:

>>

>> I’ll include my "double twist" algorithms here. The rest are trivial or

>> simply 4D use of 3D methods. These algorithms preserve I/O orientation for

>> the other seven pieces, but do not preserve orientation on other axes or

>> permutation at all.

>>

>> Twist *OFRU* clockwise (relative to I/O): *[ R[ U’ R’ U2 ]: Iy Ix ]*

>> Twist *OBRU* counterclockwise (relative to I/O): *[ R[ U R U2 ]: Iy’ Ix’

>> ]*

>>

>> The *Iy Ix* and *Iy’ Ix’* moves can be executed by moving the right and

>> left endcaps around the inner face, as can be seen in my solution video

>> <https://youtu.be/Fd9NUaO5AYo?t=5m58s>.

>>

>>

>> On 7/28/2018 2:46 PM, Marc Ringuette ringuette@solarmirror.com

>> [4D_Cubing] wrote:

>>

>> Monoflip, solving In+Out faces only: (12 moves physical using ROIL

>> Zero, 3 cases)

>> RUFI by x2: Rzy I [ U F U’ F U F2 U’ ] Iy Lx2 Iy’ Rz’

>> RUFI by y2: Ry’z’ I [ U F U’ F U F2 U’ ] Iy Lx2 Iy’ Ry

>> RUFI by z2: Rzy Iy Lx2 Iy’ I [ U F2 U’ F’ U F’ U’ ] Rz’

>> (those are sideways Sune, Sune, and Antisune, inside the brackets)

>>

>>

>>

>

>

> –

>

> "Machines take me by surprise with great frequency." - Alan Turing

>

"Machines take me by surprise with great frequency." - Alan Turing