Sure. Let's see, 2,1 is odd so set a point at (0,101). Flip that bad boy up to (51,50) with a / mirror at (51,0), and then flip it to (2,1) with a \ mirror at (51,1). On Wed, Oct 21, 2020 at 9:08 PM Allan Wechsler <acwacw@gmail.com> wrote:
I'm afraid I'm not following the explanation. Can you illustrate by showing a construction for {(0,0), (2,1)} ?
On Thu, Oct 22, 2020 at 12:02 AM Tomas Rokicki <rokicki@gmail.com> wrote:
Just make a set of raster lines on Z_1. Then using pairs of folds, fold those raster lines on top of each other, in a serpentine pattern.
This should let you build anything in Z_2.
-tom
On Wed, Oct 21, 2020 at 8:58 PM Allan Wechsler <acwacw@gmail.com> wrote:
But I have no strong feeling about Z^2. I guess if I had to bet, I'd bet that it will turn out that all sets are also constructable in Z^2, but in two dimensions I don't see any obvious analogue to the toolkit Tomas Rokicki used to defeat Z^1.
On Wed, Oct 21, 2020 at 11:25 PM James Propp <jamespropp@gmail.com> wrote:
I now believe that you can build any finite subset of Z from {0}, but finding the quickest way to build it might be tricky.
Jim
On Wed, Oct 21, 2020 at 10:43 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Yeah, it seems clear any finite pattern can be built from {0} in Z_1.
You can always set a point any number of cells to the right of the current point (and a bunch of other points further to the right).
You can always clear a single rightmost point (except when there are no points left.
These two operations are all you need to build any pattern.
-tom
On Wed, Oct 21, 2020 at 7:19 PM Allan Wechsler <acwacw@gmail.com> wrote:
It's not? Then that's obviously something to work on first.
It's still true, even in Z^1, that any finite set can be transformed to {0}, at the worst by alternating folding toward 0 around 1/2 and -1/2. (The minimum number of steps it takes is clearly something to do with ln_2 of the most distant element.)
My intuition is that, at least in Z^1, any finite set can be transformed into any other. If this is not the case, then I would be *very* interested to see a set that can't be constructed from {0}. (To build {0,1,3}, first unfold around 1/2, then unfold around 2. Now we have {0,1,3,4}, and folding toward 0 at 7/2 merges the 4 with the 3.)
On Wed, Oct 21, 2020 at 7:43 PM James Propp < jamespropp@gmail.com> wrote:
> Even 1D isn’t trivial. > > Jim > > On Wed, Oct 21, 2020 at 4:04 PM Allan Wechsler < acwacw@gmail.com
wrote:
> > > I thought of that! > > > > > > On Wed, Oct 21, 2020 at 3:47 PM James Propp < jamespropp@gmail.com> > wrote: > > > > > Puzzles based on these moves would be fun to play on a laptop or phone! > > > > > > Jim Propp > > > > > > On Wed, Oct 21, 2020 at 12:47 PM Allan Wechsler < acwacw@gmail.com> > > wrote: > > > > > > > Two different Math-Fun threads have cross-fertilized in my brain, > > leading > > > > to the following rumination. Thanks to Jim Propp, and indirectly to > my > > > > wife, for the origami theme; and thanks to Tom Karzes for his > > > > Gaussian-integer puzzle (to which I think the answer is mostly > > "Gaussian > > > > GCD"). > > > > > > > > I propose two operations, folding and unfolding, on any set of > Gaussian > > > > integers. > > > > > > > > For both operations you start by selecting a crease-line, which must > be > > > one > > > > of the reflection lines of the Gaussian integers -- that is, a > > horizontal > > > > or vertical line with integer or half-integer intercept, or a > 45-degree > > > > diagonal with integer intercepts. > > > > > > > > To "fold" a set, you remove all the points on one side of the crease, > > > while > > > > adding their mirror images to the other side. (The constraint on the > > > crease > > > > line ensures that the mirror images will also be Gaussian integers). > > > > > > > > To "unfold" a set, just add all mirror images, with no erasing. > > > > > > > > The central question of the "Theory of Gaussian origami" is, what > sets > > > are > > > > achievable from what starting sets? In particular, what sets are > > > achievable > > > > starting from a single point? I know that by repeated folding I can > > get a > > > > single point from any finite starting set. But, for example, starting > > > with > > > > {0}, can I get {0, 2+i} by any number of folding and unfolding > > > operations? > > > > _______________________________________________ > > > > math-fun mailing list > > > > math-fun@mailman.xmission.com > > > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > > > _______________________________________________ > > > math-fun mailing list > > > math-fun@mailman.xmission.com > > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > _______________________________________________ > > math-fun mailing list > > math-fun@mailman.xmission.com > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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