# Message #3075

From: ryan@echolsphoto.com

Subject: Introducing myself

Date: Sat, 07 Mar 2015 17:58:57 -0700

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 amounted 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 mBlue 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