# Message #849

From: Roice Nelson <roice3@gmail.com>

Subject: Re: [MC4D] Re: Introducing "MagicTile"

Date: Sun, 07 Feb 2010 22:01:18 -0600

On Fri, Feb 5, 2010 at 12:29 AM, Melinda Green <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?

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…

seeya,

Roice