Message #3104

From: Melinda Green <>
Subject: Re: [MC4D] New 3^4 shortest solution! (227 twists)
Date: Tue, 07 Apr 2015 20:02:54 -0700

I’ve updated the HOF with Matthew’s new shortest solution and link to
his log file <>. I
like how it took 5 years to shave 24 twists! That brings it to under the
250 twist milestone and within sight of sub-200.

Nice going, Matthew!

On 4/4/2015 5:59 AM, [4D_Cubing] wrote:
> This was just for a random scramble generated by MC4D, I never tried
> to get an unusually good or bad scramble, I took the first one
> generated. 200 is just a nice milestone, in my own opinion, so there
> is no real significance to the 200 moves I mentioned as far as I know.
> I sent the solution to Melinda last weekend (it’s taken me this long
> to type up the post) but it doesn’t seem to have been updated in the
> HOF yet so I realise the log file isn’t available to look at.
> Matt
> —In, <ed.baumann@…> wrote :
> 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@… [4D_Cubing]
> <mailto:damienturtle@…%20[4D_Cubing]>
> *To:* <>
> *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</ p>