Message #3843

From: Marc Ringuette <ringuette@solarmirror.com>
Subject: Re: [MC4D] Melinda’s 2x2x2x2 solved
Date: Sun, 26 Nov 2017 19:15:48 -0800

Hi Bob,

Good stuff!   I’m glad to see that you put your puzzle together and made
progress on it.   Unless we unexpectedly identify a problem, I think
you’ve got the first valid solution recipe sketched out. Congrats!

I had used the identical approach – totally identical in all respects
except for how to achieve the basic 4-cycle of the top 2x2x2 upon which
it depends – and nicknamed it 4tega because it orients a pair of faces
first like the Ortega method does for the 2^3 cube.   However, I was
simply using parity-violating "quarter puzzle twists" to walk through
the solution, which was very unsatisfying and non-kosher, and prevented
me from saying "First!".   You have resurrected quarter puzzle twists in
a kosher way, using the minimalist move set.  Yay!   A kosher 4tega!

Since our approaches were so similar, I could recognize what you’re
doing right away, and I endorse it as a valid approach.

When I had worked through it before, I was worried that I would still
occasionally end up with the single-piece-double-twist problem at the
end.  I think I was hallucinating, since it seems clearly impossible to
be in that state with two faces oriented.  However, I would be reassured
if you had a look at the following.

Question:  if you take a single piece of your puzzle and give it a 180
degree twist along any axis, how would you solve the resulting puzzle
state?   I believe that’s a correct statement of the basic
single-piece-double-twist problem.   Maybe you’d like to give that a
try.   I had sketched out a tentative solution using quarter-puzzle
twists and 3 sequential puzzle reorientations, based on pure guesswork. 
I should dig it out and try to rehab it.   I’m still working out my
understanding of the 12 orientations of the pieces – I find that to be
the most fascinating part of this puzzle, theory wise – and I think a
good step to getting the hang of it will be to fully understand the
solution to the single-piece-double-twist. What’s the fewest number of
full puzzle reorientations required to undo a single-piece-double-twist
leaving the rest of the puzzle unchanged?  I think it might be three,
and that the reason will become clear to me after a bit.   But it might
be two, given how confused I still am about how the 12 orientations work.

It would be fun to see a video of you manipulating your puzzle. I’m
curious what feels natural for you.

Some time, a truly dedicated person should make a video of the
following:   generate a MC4D scramble of the 2^4.   Duplicate it on the
physical 2^4 by taking the puzzle apart.   Solve the physical 2^4 while
following along with MC4D with one click per physical puzzle move,
leaving both puzzles solved.

I’m also looking forward to jointly tackling some entries on the list of
open questions about the physical 2^4.  After writing down the list of
open questions, which I have not done.  :)   Do you have any entries for
it?  Directions you’d like to explore?

Cheers
Marc