# Message #3101

From: Eduard Baumann <ed.baumann@bluewin.ch>

Subject: Re: [MC4D] New 3^4 shortest solution! (227 twists)

Date: Sat, 04 Apr 2015 14:38:11 +0200

This is not the shortest solution for a specific scrambling but for the worst case scrambling.

If ‘gods number for 3^3 got under 20 you try ‘under 200’ for 3^4 now.

Correct?

Ed

—– Original Message —–

From: damienturtle@hotmail.co.uk [4D_Cubing]

To: 4D_Cubing@yahoogroups.com

Sent: Saturday, April 04, 2015 2:10 PM

Subject: [MC4D] New 3^4 shortest solution! (227 twists)

Hi everyone,

For those newer members, I used to be fairly active here a few years back and set the 3^4 fewest moves record with 251 twists, amongst other things. I’m not actually returning to being active (I probably won’t be taking part in the speedsolving contest being talked about, much as I would like to), I just felt like tidying up this solve since I hadn’t finished optimising the last step before, but when I became busy I just submitted what I had so far. Essentially, I had ended with a parity situation which cost moves to fix, and I wasn’t happy with that. After going to a speedsolving competition recently (I had a free weekend and had fun even though I was very out of practice) and having good luck in FMC with a 28 twist solution (which should have been 25 twists, maybe if I had practiced at all I would have spotted the really obvious insertion that I missed), I decided to correct the ending of my 3^4 solution with something much better, thoug h I’m still busy these days so I could probably have found something even better if I was willing to spend longer on it.

I’ll try to explain roughly how my solution works, especially since the ending now isn’t very clear now that I’m better at fewest move solving. The first 182 twists of the solution are unchanged from before, since I didn’t really have the time to work through a full solve and because I always considered my previous submission as an unfinished solve which I wanted to finish eventually. It is simple blockbuilding mostly, followed by orienting the last layer. This means that what is left is essentially a 3^3 solve embedded into the 3^4 puzzle, and it is this which I revisited. As for the parity problem I had, try doing a U2 move on one 3^3 cell, leaving the rest of the puzzle solved, my best solution for this is 22 twists if I remember correctly.

Optimising the last step is weird. It’s based on optimising a 3^3 solve, but it’s not quite as simple as that. Let’s look at that first though, and the metric used (as a simplification since the ‘correct’ metric is too complicated) is quarter-slice turn metric. A quarter turn of any layer is one move, a half turn of any layer counts as 2 moves. For some orientation I happened to choose, I looked at what the 3x3 scramble was, and used CubeExplorer to find a short scramble to work with:

D L2 F D2 U’ F U’ F D’ U’ R’ B2 L’ B F’ U’ B U’

Now, due to the 3^4 context and my solution before this point, I needed to perform a net result of an L turn, which in practice means the net result of adding the twists of the solution (ignoring which faces are turned) must be a quarter turn clockwise (so L, R, U2 B’, L’ F’ D’, etc. would be valid solutions in this regard). Also, by being able to change the previous twist in the solution, I get a free turn of the front face in terms of movecount, which I tak e advantage of. My solution looks like this (knowledge of FMC techniques required, I can explain in more detail if anyone wants me to):

(F) R’ (U’ D) L F’ //222 block 4/4

U’ L2 U L D //223 block, 6/10

(B’ L D L D’ L’ U’ L U L2) //Performed on inverse, solves all but 4 corners + 3 edges, 11/21

Skeleton: (F) R’ (U’ D) L F’ U’ L2 * @ U L D ** L2 U’ L’ U L D L’ D’ L’ B //21 moves

Insertions:

* = (F B’) U’ L U L’ (B F’) //Solves 3 edges + 1 corner, no moves cancel, 6/27

@ = L U R’ U’ L’ U R U’ //Solves 3 corners, 2 moves cancel, 6/33

Solution: (F) R’ (U’ D) L F’ U’ L2 (FB’) U’ L U L’ (BF’) L U R’ U’ L’ U R L D L2 U’ L’ U L D L’ D’ L’ B //33 moves QSTM

Bonus: insert ** = D’ (LR’) B L2 B’ (RL’) D L2, to solve 3 edges, which cancels 7 moves. Never really checked the corner insertions on this, so it might have given a better result actually, but probably not.

Now, I just try to adapt this to the 3^4 solve as efficiently as possible. My first serious attempt managed 230 twists, then 229. After a while I found a 228 twist solution which I nearly kept, but some instinct told me that I could save one more twist, which I managed to do.

The idea is as follows. Say we need to do R U R’ U’ to solve the 3^3. The simple solution is to do R [F] R [F]’ R’ [F] R’ [F]’ (8), where [F] means to rotate the whole cube as in an F turn, and notice that if we ignore the rotations all the moves cancel, which is necessary. I think Ray put some info on this on the wiki for how this works, if you aren’t familiar with it, try doing these moves on the ‘inside’ face in the most obvious way, with one twist in MC4D for each twist and for each rotation. Instead, we could do R U [UR] U’ R’ [UR] (6), where [UR] rotates the cub e around the axis though the UR edge and the opposite edge DL, and again all moves cancel once rotations are ignored. With this idea, two moves are saved here, and for full solutions the same idea applies but it becomes far more difficult to optimise. With this, my solution turned out to be:

[F] F’ M’ [F]’ R B’ U’ R2 S R’ [UL] R D [DR] D’ S’ R U [DR] U’ L’ D’ [F]’ D L R U R [F]’ U [UR] U’ R’ U R D [UFL] D’ [RB] R’ U’ [DR] B

where S is a slice turn which follows F, M is a slice turn which follows L, and [UFL] is a clockwise cube rotation about the UFL corner.

I didn’t do this on paper, I worked in MC4D, then typed up what my final solution was. The strangest part for me was trying (R z’ U), which is essentially a double turn with a rotation in the middle, which actually saved a rotation overall.

I think I’m unlikely to try beating this, so if anyone wants to take the record, they are welcome to it. I just hope I’ve made it very challenging to do so :), but it is certainly possible. I reckon sub-200 twists is very possible, so maybe if this isn’t achieved for many years, I might try to do it myself …

These days, I’m doing a PhD is mathematics, working on network theory. I seen that someone new here recently is also in this area, so I might be able to have an interesting conversation with them :). I have one paper recently resubmitted to a journal which will hopefully be accepted soon, and another related paper nearly ready to be submitted. This is why I don’t have much time to spend on cubing, but that’s life.

I hope as many people take part in the speedsolving contest as possible, it was very fun the first time so I recommend having a go. Even though I won’t have time to practice for it, I’ll be interesting in the results so please make them exciting! Good luck everyone!

Happy hypercubing,

Matt