Message #149

From: Roice Nelson <roice@gravitation3d.com>
Subject: RE: [MC4D] Cubes, beer and Beethoven
Date: Sun, 08 May 2005 23:07:09 -0500

Hi Guy,

I had a few comments…

> I will check the sites you refer me to and
> let you know if they answer my question.
> I suspect I was not very clear in my original
> posting, but by ‘3D version’ I meant the 4^3,
> not the 3^3 and it was from this that I made
> guesses about the 4^4. Each of us when solving
> probably creates his own lingo but I should be
> more careful when communicating: by ‘centres’
> I meant the 8 one-colour pieces on the interior
> of each cubic face of the 4^4. Since you cannot
> distinguish orientations and positions of these
> I wondered if their arrangement influenced the
> subsequent placing of the two-colour pieces,
> such that if it was wrong these last could
> not be aligned.

My memory had failed me as usual, but I dug up some emails from when I
had worked on the 4^4. I did have exactly what you are describing
here happen to me (a parity issue such that I could not line up all
the 2 color pieces after placing the centers, forcing me to undo work
on the centers).

I had the parity issue you had as well, forcing me to break and
rearrange 2 color pieces to fully align the 3 color ones. Very
thankfully, I did not have the issues with the corners! After working
through the other 2 problems, I distinctly remember hoping hard for
that :) Maybe some of other 4^4 solvers have had a corner parity
issue, allowing us to empirical verify the existence of all of these
using the whole group.

I think there could be something to your question about the solution
method affecting parity problems arising in the 4^4 (centers-out vs.
layer by layer). Say for instance there are multiple situations that
can lead to parity problems for 4 color corners, and that these depend
not just on the 3 color edges (if an even/odd number of twists was
used to solve them or whatever), but on both the 3 color and 2 color
pieces. In the centers-out approach, certain situations leading to 4
color problems might never show up since some work to get 2 color and
3 color pieces "in parity" has already been done, and so maybe such
issues could only arise in a layer-by-layer approach. This is
hypothetical, but my point is that the solution algorithm itself
places additional constraints on the state of the puzzle and may limit
what is possible.

In fact, I wonder if the fully unconstrained 4^4 has more than 3
contributing situations that can lead to parity problems (maybe 3! or
something). I am not familiar enough with group theory to back this
up mathematically, but others might be able to. As David suggested,
maybe the information could be gleaned from Eric’s page on the number
of permutations. It would be cool to have a specific explanation of
the number of these problems and their occurrence percentage.

Take Care,

Roice

P.S. I had put marks in my 4^4 solution at the time, so you can check
out the parity problems I encountered on it if you like. Here are
descriptions for my marks…

a Centers Placed
b First parity problem
c Done Combining Face Pieces
d Done Combining Edge Pieces
e Second parity problem
f Fixed Parity Problem
g Faces Placed
h Edges Placed
i Finished