Message #3076

From: Melinda Green <>
Subject: Re: [MC4D] Introducing myself [1 Attachment]
Date: Sun, 08 Mar 2015 20:04:20 -0700

Hello Ryan, and welcome to our group!

I hadn’t realized that yours was a custom solution. That is very
impressive! It’s still impressive when people follow Roice’s solution,
but you earn a special sort of respect when you develop your own solution.

20 second 3x3 solutions are very fast! Are you old enough to remember
when the world record was approaching 1 minute? At my best I could do it
in around 5 minutes, so sub-one minute solves are magic to me, and
sub-10 second solves are miraculous. We’ve done some 4D speedsolving
competitions which were quite fun. If you get interested in that, we can
try to get enough people to do more.

Have you explored any of the graph theoretic aspects of twisty puzzles?
It seems right up your alley. I’m guessing that you are interested in
machine learning, is that right? I’m wondering what sort of things you
most want to work on.

Happy puzzling!

On 3/7/2015 4:58 PM, [4D_Cubing] wrote:
> [Attachment(s) <#TopText> from [4D_Cubing]
> included below]
> Hello everybody!
> My name is Ryan Echols. I’m 17 years old (turning 18 later this
> month), and currently a first-year student BYU in Provo, Utah,
> studying Math. For fun, I like to speedsolve the Rubik’s cube, roller
> blade, play accordion, do a few card tricks, program, and play Portal
> 2. My typical time on a standard Rubik’s cube is 20 seconds, and I
> mention Portal 2 because I was at once a co-world-record holder (the
> record has long since been broken, but I’m still top 100 in multiple
> places on the global leaderboards). For work, I’m a research assistant
> in the Math department with a group that’s currently looking at
> algorithms to make good Tree Decompositions in Spectral Graph Theory.
> As for solving the MC4D, I had quite a fun time. I downloaded it
> early on February 12th, and had bee working on it now and then in my
> free time for the last 3 weeks until yesterday, March 6th, when I
> finally finished it. I used no macros, but of course I did use
> algorithms I knew from the standard Rubik’s cube.
> So, here comes a long-winded explanation of how I solved it. Don’t
> feel obligated to read this if you don’t want to. I’ve attached my
> .log file if you’d like to follow along. anyway, Here we go:
> My overall approach was similar to F2L (but, more like "first two
> nested cubes"), with the last cube being solved a bit like Petrus. I
> started mostly on the red cube, building it in blocks like 2x2x2,
> then 2x2x3, and 3x3x2, but as I did so, I didn’t only solve the
> adjacent faces on adjacent cubes, I also solved the next layer out on
> the adjacent cubes. To solve the last ("upper") layer on the red cube
> (corresponding to the brown cube), I began thinking about it as
> starting to solve the brown cube. I built up the brown cube in an F2L
> fashion, which in turn solved the last layer of the red cube. The
> particulars of solving F2L of the brown cube were intriguing, though.
> I’d rotate the totally untouched "middle blue" or "mBlue" cube
> (opposite red) as to get the desired piece of the brown cube so that
> it had the brown colored cublet in the over-all brown cube. From that
> point, I’d one-by-one do 3 or 4 cube turns that each a mounted to a
> single face turn on the brown cube, as if the brown cube was a typical
> Rubik’s cube. I did this by rotating one of the cubes adjacent to
> brown so that the face of brown in question would slide up onto the
> untouched mBlue cube, then I’d rotate the mBlue cube (with the brown
> face on it) in the way that would rotate the brown face as desired,
> then I’d undo the first cube’s turn, putting the brown face back on
> the brown cube in its new orientation. In certain cases, reversing the
> turn done to the mBlue cube was necessary also.
> With the F2L of the brown cube solved, all that was left was the
> mBlue cube (and the attached adjacent faces of 6 other cubes). At this
> point, I had to take a step back and think about some theory. Each
> cube exists in 3 dimensions, of course. our space has 4. In our
> "flattened-to-3D" representation, let’s suppose our center cube exists
> in x,y,z. The adjacent cubes sharing the two x,y faces would then
> exist in x,y,w. The adjacent cubes sharing the two x,z faces would
> then exist in x,z,w. Likewise, the adjacent cubes sharing the two y,z
> faces would exist in y,z,w. The totally opposite cube is also in
> x,y,z. If we only look at the 6 cubes that do exist in z, but ignore
> z, they make up the faces of just a cube. I call this mindset a
> "dimension squash". In this case, we squashed z. So, if we dimension
> squash z, the mBlue cube becomes like the last, unsolved 3rd layer of
> a normal Rubik’s cube. Of course, any different pieces on the m Blue
> cube that are in the same x,y spot, but not z, will all act as one as
> we ignore z. So, I solved the mBlue cube in a Petrus fashion for the
> start, then just three or four LL algorithms for the last layer, using
> temporary setup moves and dimension-squashing the various 3 dimensions
> of the mBlue cube. The specifics of that were the hardest part to
> figure out.
> Anyway, thanks for welcoming me to the community!
> -Ryan Echols