Message #851

From: alexander.sage@jacks.sdstate.edu
Subject: RE: [MC4D] Re: Introducing "MagicTile"
Date: Mon, 08 Feb 2010 11:10:09 -0600

As far as klein’s quartic, what is happening is that if you solve a single face, then solve all layers bordering that face, and continue, then on a megaminx (or any other platonic solid besides a pyramid) you will eventually get to a single layer left to solve. if you try to start from one face on the {7,3}, and solve every face bordering it, then every face bordering what you have solved, you will end up with multiple unsolved faces. essentially, there seems to be more than one face that is the maximum distance from the first face.

I noticed something unfortunate. tthe random seed looks like it resets, so I always seem to get the same puzzle every time I open the program, unless I continue to shuffle it by using a couple extra five twists. Could you make a random seed that saves itself?

To: 4D_Cubing@yahoogroups.com
From: melinda@superliminal.com
Date: Sun, 7 Feb 2010 22:49:59 -0800
Subject: Re: [MC4D] Re: Introducing "MagicTile"

Roice Nelson wrote:
>
> On Fri, Feb 5, 2010 at 12:29 AM, Melinda Green
> <melinda@superliminal.com <mailto:melinda@superliminal.com>> wrote:
>
>
> Roice Nelson wrote:
>> About solving towards a single cell, I think the program can help
>> answer this by using the setting to only show the fundamental set
>> of tiles. I’m not perfectly confident, but my intuition says it
>> is possible to avoid multiple disjoint unsolved cells at the
>> end. My reasoning is that even if the topology of the object as
>> a whole (fundamental and orbit regions glued up) is not simply
>> connected, the fundamental region alone is simply connected (as
>> long as you never leave the boundary of it, so that it behaves
>> topologically as a disk). Looking at only the fundamental set of
>> Klein’s Quartic, it is easy to pick a solve order leaving only
>> one cell at the end, by using sort of a "solve wave" to push
>> unsolved cells from one end to the other. This would be my
>> strategy to "keep the edge a simple closed curve", as Melinda
>> nicely described things.
>
> Can you solve a puzzle on a torus this way? It may be different
> than on surfaces of higher higher genus but it’s a good first test
> of the idea.
>
>
>
> I went through a solve of the 12-color octagonal puzzle this weekend
> (fun!), and have to say I’ve been thoroughly confusing myself thinking
> about the group’s discussion on finishing with a single unsolved cell
> :) This makes me want to not make any claims about the torus puzzles!
>
> My thinking above was flawed due to the fact that pieces (2C edges and
> 3C corners) are connected across the border of the fundamental region.
> I was picturing solving the stickers on cells without regard to these
> basic connections.
>
> This leaves me wondering what other people have been claiming though.
> From a stickers-only perspective, it will always be impossible to
> have a single unsolved cell at the end, due to the simple fact that if
> there is a sticker unsolved on a cell, it had to come from another
> cell! So is it fair to say that what we are talking about is whether
> a single unsolved "layer" is possible, and how many of those there can
> be? (Sorry, this distinction may be obvious to everyone.)
>
> In that case, there is a much simpler demonstration that you can
> always end up with one unsolved layer at the end - just do a single
> twist of a single cell! This sounds a little trivial to say, but
> maybe that really does answer the question. Admittedly, this "final
> single layer" can be very weird on some puzzles, e.g. on the 4-colored
> torus, since a single twist changes stickers on all 4 cells.
>
> Continuing on the assumption that we’re talking about layers, I think
> it is easy to show that more than 3 disjoint unsolved layers can be
> had on Klein’s Quartic. Quickly experimenting, I seem to be able to
> make up to 5 disjoint twists.
>
> I hope I’m not being excessive in describing silly tautology-like
> things. Am I off track?

I think that you are right that we should use the term "layer" instead
of cell in this discussion, but I suspect that you may be off track when
it comes to the question at hand. Or rather the question that’s in my
mind, in case we’re not in sync. In my mind, the question is whether
it’s possible to solve KQ or other puzzles of genus > 0 by using a
simple layer-by-layer method. The key word here is "simple" as in
"simple closed curve" of the expanding edge. Ignoring that constraint,
I’m sure that you can end up with all sorts of non-simple curves
including ones that are disjoint. The question instead is whether you
necessarily have to go through such phases in a layer-by-layer solution.

>
> And here is a related question for you all. Can you checkerboard
> Klein’s Quartic? If so, does it have to be a 3-cycle checkerboard,
> due to there being "two bottoms" to every cell that are in-a-sense its
> opposites? As I started investigating just now, I was able to easily
> 2-cycle checkerboard both the 6-colored and 12-colored octagonal
> puzzles…

I haven’t the slightest idea about the checkerboard possibilities but
would love to see some. :-)
-Melinda