[math-fun] Gaussian origami
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?
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
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
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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
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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
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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
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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
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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
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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
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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
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(Actually, I was just now able to find a construction for that set. I can now add any second point to (0,0). But the general procedure is still not obvious.) On Thu, Oct 22, 2020 at 12:07 AM 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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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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... a spiral is more efficient than raster rows ... On Wed, Oct 21, 2020 at 9:12 PM Tomas Rokicki <rokicki@gmail.com> wrote:
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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To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0). If I understand this correctly, the sequence is: 1. Start: {(0,0)} 2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)} This is a bottom-to-top unfold. 3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)} This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold. 4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)} This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold. This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others. In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below). Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed. To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point. Tom Tomas Rokicki writes:
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)} ?
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse. On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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Christian, this is fun to explore, thanks! Best, É.
Le 22/10/2020 12:47, Christian Lawson-Perfect <christianperfect@gmail.com> a écrit :
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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Thanks, Christian! Jim On Thu, Oct 22, 2020 at 6:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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This is lovely. I can do (0,0) (2,1) in two moves. Question: in folding mode, is there a way to specify which side of the line you are folding to? On Thu, Oct 22, 2020 at 8:42 AM James Propp <jamespropp@gmail.com> wrote:
Thanks, Christian!
Jim
On Thu, Oct 22, 2020 at 6:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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The side you fold on depends on which side of the origin your mouse is. A system involving more than one click might be more intuitive... On Thu, 22 Oct 2020, 16:39 Allan Wechsler, <acwacw@gmail.com> wrote:
This is lovely. I can do (0,0) (2,1) in two moves.
Question: in folding mode, is there a way to specify which side of the line you are folding to?
On Thu, Oct 22, 2020 at 8:42 AM James Propp <jamespropp@gmail.com> wrote:
Thanks, Christian!
Jim
On Thu, Oct 22, 2020 at 6:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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This is very nice. Thank you! Now we need some challenges---goal dots, as it were. And a leaderboard! And a world record page, and sponsors, and twitch feeds, and a competition rules committee. On Thu, Oct 22, 2020 at 3:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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And a dodgy eastern-european billionaire or two? WFL On 10/22/20, Tomas Rokicki <rokicki@gmail.com> wrote:
This is very nice. Thank you! Now we need some challenges---goal dots, as it were. And a leaderboard! And a world record page, and sponsors, and twitch feeds, and a competition rules committee.
On Thu, Oct 22, 2020 at 3:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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Okay, two things: a comment about notation, and a starting puzzle. I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, presented as a way to suggest an unambiguous and fairly terse notation. 1. Unfold 0 to 2+2i. 2. Fold 2+2i to 2+i. In each case the operation is performed so as to put a copy of the first point onto the second. This specifies the crease axis unambiguously. Some moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because there is no permissible crease that does that. The second point has to be a queen's move from the first. Now the puzzle, the simplest one I haven't been able to do yet:
From {0}, construct {0, 2, 2+i}.
On Thu, Oct 22, 2020 at 1:40 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
And a dodgy eastern-european billionaire or two? WFL
On 10/22/20, Tomas Rokicki <rokicki@gmail.com> wrote:
This is very nice. Thank you! Now we need some challenges---goal dots, as it were. And a leaderboard! And a world record page, and sponsors, and twitch feeds, and a competition rules committee.
On Thu, Oct 22, 2020 at 3:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ?
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Spoiler space. 3 1 4 1 5 9 2 6 5 3 5 8 Unfold (0,0) to (2,0) Unfold (2,0) to (3,0) (creates (5,0) as well) Fold (5,0) to (3,0) Fold (3,0) to (2,i). This is just my raster strategy, only using a spiral instead of a raster. -tom On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler <acwacw@gmail.com> wrote:
Okay, two things: a comment about notation, and a starting puzzle.
I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, presented as a way to suggest an unambiguous and fairly terse notation.
1. Unfold 0 to 2+2i. 2. Fold 2+2i to 2+i.
In each case the operation is performed so as to put a copy of the first point onto the second. This specifies the crease axis unambiguously. Some moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because there is no permissible crease that does that. The second point has to be a queen's move from the first.
Now the puzzle, the simplest one I haven't been able to do yet:
From {0}, construct {0, 2, 2+i}.
On Thu, Oct 22, 2020 at 1:40 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
And a dodgy eastern-european billionaire or two? WFL
On 10/22/20, Tomas Rokicki <rokicki@gmail.com> wrote:
This is very nice. Thank you! Now we need some challenges---goal dots, as it were. And a leaderboard! And a world record page, and sponsors, and twitch feeds, and a competition rules committee.
On Thu, Oct 22, 2020 at 3:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes:
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)} ? >
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Tom Rokicki builds 0, 2, 2+i from 0 in four steps. It's easy to prove that it takes at least three steps ... and I just realized that three steps can indeed be done. So there's a certain sort of "code golf" that can be played with this sort of puzzle. Answer below. 1 6 1 8 0 3 3 9 8 8 7 4 9 8 1. Unfold 0 to 2 2. Unfold 0 to 2+2i (2 is on the crease) 3. Fold 2+2i to 2+i. On Thu, Oct 22, 2020 at 3:12 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Spoiler space.
3
1
4
1
5
9
2
6
5
3
5
8
Unfold (0,0) to (2,0) Unfold (2,0) to (3,0) (creates (5,0) as well) Fold (5,0) to (3,0) Fold (3,0) to (2,i).
This is just my raster strategy, only using a spiral instead of a raster.
-tom
On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler <acwacw@gmail.com> wrote:
Okay, two things: a comment about notation, and a starting puzzle.
I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, presented as a way to suggest an unambiguous and fairly terse notation.
1. Unfold 0 to 2+2i. 2. Fold 2+2i to 2+i.
In each case the operation is performed so as to put a copy of the first point onto the second. This specifies the crease axis unambiguously. Some moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because there is no permissible crease that does that. The second point has to be a queen's move from the first.
Now the puzzle, the simplest one I haven't been able to do yet:
From {0}, construct {0, 2, 2+i}.
On Thu, Oct 22, 2020 at 1:40 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
And a dodgy eastern-european billionaire or two? WFL
On 10/22/20, Tomas Rokicki <rokicki@gmail.com> wrote:
This is very nice. Thank you! Now we need some challenges---goal dots, as it were. And a leaderboard! And a world record page, and sponsors, and twitch feeds, and a competition rules committee.
On Thu, Oct 22, 2020 at 3:49 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
I've arguably wasted my morning by making an interactive version of this, as Jim requested: https://somethingorotherwhatever.com/gaussian-origami/ While it sort of works on touchscreens, it's easiest with a mouse.
On Thu, 22 Oct 2020 at 10:56, Tom Karzes <karzes@sonic.net> wrote:
To fold (0,101) onto (51,50), the / fold needs to pass through (0,50) rather than (51,0).
If I understand this correctly, the sequence is:
1. Start: {(0,0)}
2. Unfold(-) through {(0,50.5)} to obtain: {(0,0), (0,101)}
This is a bottom-to-top unfold.
3. Fold(/) through (0,50) to obtain: {(0,0), (51,50)}
This is a nw-to-se fold, so (0,0) is unchanged. This is the first described fold.
4. Fold(\) through (51,1) to obtain: {(0,0), (2,1)}
This is a ne-to-sw fold, so (0,0) is unchanged. This is the second described fold.
This general approach seems to work well when it's possible to isolate a single point on one side of a fold, allowing that one point to be moved without changing any of the others.
In general, if you have a cluster of points, you can replicate it at a great distance away by unfolding about a distant crease line. You then need to eliminate those replicated points by folding inward, until only one replicated point is left. I believe this is always possible (see below).
Once you're down to a single replicated point, you position it at the desired position near the original cluster. The process can then be repeated until the desired set has been contructed.
To reduce a distant replicated cluster of points to a single point, I believe the following will work. Create the replicated cluster vertically above the original by folding upward about a distant horizontal crease line. Then perform a series of \ folds, folding ne-to-sw, until all of the replicated points lie on a single nw-se line. Then perform a series of / folds, folding nw-to-se, until you're down to a single replicated point.
Tom
Tomas Rokicki writes: > 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)} ? > >
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Nice. I have a different 3-fold solution: 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1. Unfold 0 to 2+2i 2. Fold 2+2i to 2+i 3. Unfold 2+i to 2 Tom Allan Wechsler writes:
Tom Rokicki builds 0, 2, 2+i from 0 in four steps. It's easy to prove that it takes at least three steps ... and I just realized that three steps can indeed be done. So there's a certain sort of "code golf" that can be played with this sort of puzzle. Answer below.
1
6
1
8
0
3
3
9
8
8
7
4
9
8
1. Unfold 0 to 2 2. Unfold 0 to 2+2i (2 is on the crease) 3. Fold 2+2i to 2+i.
On Thu, Oct 22, 2020 at 3:12 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Spoiler space.
3
1
4
1
5
9
2
6
5
3
5
8
Unfold (0,0) to (2,0) Unfold (2,0) to (3,0) (creates (5,0) as well) Fold (5,0) to (3,0) Fold (3,0) to (2,i).
This is just my raster strategy, only using a spiral instead of a raster.
-tom
On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler <acwacw@gmail.com> wrote:
Okay, two things: a comment about notation, and a starting puzzle.
I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, presented as a way to suggest an unambiguous and fairly terse notation.
1. Unfold 0 to 2+2i. 2. Fold 2+2i to 2+i.
In each case the operation is performed so as to put a copy of the first point onto the second. This specifies the crease axis unambiguously. Some moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because there is no permissible crease that does that. The second point has to be a queen's move from the first.
Now the puzzle, the simplest one I haven't been able to do yet:
From {0}, construct {0, 2, 2+i}.
I think Tom's 3-step solution is incorrect, and produces i as well as the desired 3 points. But this could be interpreted as a difference of opinion about what the "unfold" operation does. I imagined "unfold" as the union of the original set and its reflection around the crease. Christian's web app agrees with me. Christian: how big is the original Elm program? On Thu, Oct 22, 2020 at 4:16 PM Tom Karzes <karzes@sonic.net> wrote:
Nice. I have a different 3-fold solution:
1
2
1
3
1
2
1
4
1
2
1
3
1
2
1. Unfold 0 to 2+2i 2. Fold 2+2i to 2+i 3. Unfold 2+i to 2
Tom
Allan Wechsler writes:
Tom Rokicki builds 0, 2, 2+i from 0 in four steps. It's easy to prove that it takes at least three steps ... and I just realized that three steps can indeed be done. So there's a certain sort of "code golf" that can be played with this sort of puzzle. Answer below.
1
6
1
8
0
3
3
9
8
8
7
4
9
8
1. Unfold 0 to 2 2. Unfold 0 to 2+2i (2 is on the crease) 3. Fold 2+2i to 2+i.
On Thu, Oct 22, 2020 at 3:12 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Spoiler space.
3
1
4
1
5
9
2
6
5
3
5
8
Unfold (0,0) to (2,0) Unfold (2,0) to (3,0) (creates (5,0) as well) Fold (5,0) to (3,0) Fold (3,0) to (2,i).
This is just my raster strategy, only using a spiral instead of a raster.
-tom
On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler <acwacw@gmail.com> wrote:
Okay, two things: a comment about notation, and a starting puzzle.
I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, presented as a way to suggest an unambiguous and fairly terse notation.
1. Unfold 0 to 2+2i. 2. Fold 2+2i to 2+i.
In each case the operation is performed so as to put a copy of the first point onto the second. This specifies the crease axis unambiguously. Some moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because there is no permissible crease that does that. The second point has to be a queen's move from the first.
Now the puzzle, the simplest one I haven't been able to do yet:
From {0}, construct {0, 2, 2+i}.
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You are correct about my interpretation of "unfold" differing from yours. I assumed it was meant to be identical to "fold", except that no points were removed. In other words, "unfold" has a direction just as "fold" does, and only the points on one side of the crease get replicated. So when I said "unfold a to b", that was a different operation from "unfold b to a". That seemed most analogous to unfolding a physical piece of paper. If the intent was that "unfold" has no direction, and is symmetrical about the crease, with points on both sides of the crease being replicated on the opposite side, then my solution would indeed include i in the set, and would be incorrect. In that case, saying "reflect a with b" would be clearer than "unfold a to b", since it avoids arbitrarily introducing a direction into the operation. In the context of the original definition, the description of "unfold" followed the description of "fold", and it sounded to me like it was meant to be same as "fold" except that no points were removed, but it could be interpreted either way. If the definition is taken out of context, then it's much clearer that it should be symmetrical about the crease, and that it has no direction. The general procedure for constructing arbitrary sets of points should work with either definition, but when trying to find minimal solutions, the difference clearly matters. Tom Allan Wechsler writes:
I think Tom's 3-step solution is incorrect, and produces i as well as the desired 3 points. But this could be interpreted as a difference of opinion about what the "unfold" operation does. I imagined "unfold" as the union of the original set and its reflection around the crease. Christian's web app agrees with me.
Christian: how big is the original Elm program?
Christian: how big is the original Elm program?
The code is at https://github.com/christianp/gaussian-origami. It's just over 400 lines of elm at the moment. I really really should be doing my actual job, but I think it would help matters to be able to share links to sequences of moves, so I'll try to add that. On Thu, 22 Oct 2020, 21:39 Allan Wechsler, <acwacw@gmail.com> wrote:
I think Tom's 3-step solution is incorrect, and produces i as well as the desired 3 points. But this could be interpreted as a difference of opinion about what the "unfold" operation does. I imagined "unfold" as the union of the original set and its reflection around the crease. Christian's web app agrees with me.
Christian: how big is the original Elm program?
On Thu, Oct 22, 2020 at 4:16 PM Tom Karzes <karzes@sonic.net> wrote:
Nice. I have a different 3-fold solution:
1
2
1
3
1
2
1
4
1
2
1
3
1
2
1. Unfold 0 to 2+2i 2. Fold 2+2i to 2+i 3. Unfold 2+i to 2
Tom
Allan Wechsler writes:
Tom Rokicki builds 0, 2, 2+i from 0 in four steps. It's easy to prove that it takes at least three steps ... and I just realized that three steps can indeed be done. So there's a certain sort of "code golf" that can be played with this sort of puzzle. Answer below.
1
6
1
8
0
3
3
9
8
8
7
4
9
8
1. Unfold 0 to 2 2. Unfold 0 to 2+2i (2 is on the crease) 3. Fold 2+2i to 2+i.
On Thu, Oct 22, 2020 at 3:12 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Spoiler space.
3
1
4
1
5
9
2
6
5
3
5
8
Unfold (0,0) to (2,0) Unfold (2,0) to (3,0) (creates (5,0) as well) Fold (5,0) to (3,0) Fold (3,0) to (2,i).
This is just my raster strategy, only using a spiral instead of a raster.
-tom
On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler <acwacw@gmail.com> wrote:
Okay, two things: a comment about notation, and a starting puzzle.
I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, presented as a way to suggest an unambiguous and fairly terse notation.
1. Unfold 0 to 2+2i. 2. Fold 2+2i to 2+i.
In each case the operation is performed so as to put a copy of the first point onto the second. This specifies the crease axis unambiguously. Some moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because there is no permissible crease that does that. The second point has to be a queen's move from the first.
Now the puzzle, the simplest one I haven't been able to do yet:
From {0}, construct {0, 2, 2+i}.
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A little update: I've changed my interactive thing so that you have to click once to select a line, then again to select a side to fold onto. I've also made it keep track of moves and store them in the URL, so you can share solutions more easily. Finally, I changed "unfold" so it only folds the points on one side. Allan, is it OK if I share this outside math-fun, crediting you with the idea? On Fri, 23 Oct 2020 at 07:05, Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
Christian: how big is the original Elm program?
The code is at https://github.com/christianp/gaussian-origami. It's just over 400 lines of elm at the moment.
I really really should be doing my actual job, but I think it would help matters to be able to share links to sequences of moves, so I'll try to add that.
On Thu, 22 Oct 2020, 21:39 Allan Wechsler, <acwacw@gmail.com> wrote:
I think Tom's 3-step solution is incorrect, and produces i as well as the desired 3 points. But this could be interpreted as a difference of opinion about what the "unfold" operation does. I imagined "unfold" as the union of the original set and its reflection around the crease. Christian's web app agrees with me.
Christian: how big is the original Elm program?
On Thu, Oct 22, 2020 at 4:16 PM Tom Karzes <karzes@sonic.net> wrote:
Nice. I have a different 3-fold solution:
1
2
1
3
1
2
1
4
1
2
1
3
1
2
1. Unfold 0 to 2+2i 2. Fold 2+2i to 2+i 3. Unfold 2+i to 2
Tom
Allan Wechsler writes:
Tom Rokicki builds 0, 2, 2+i from 0 in four steps. It's easy to prove that it takes at least three steps ... and I just realized that three steps can indeed be done. So there's a certain sort of "code golf" that can be played with this sort of puzzle. Answer below.
1
6
1
8
0
3
3
9
8
8
7
4
9
8
1. Unfold 0 to 2 2. Unfold 0 to 2+2i (2 is on the crease) 3. Fold 2+2i to 2+i.
On Thu, Oct 22, 2020 at 3:12 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Spoiler space.
3
1
4
1
5
9
2
6
5
3
5
8
Unfold (0,0) to (2,0) Unfold (2,0) to (3,0) (creates (5,0) as well) Fold (5,0) to (3,0) Fold (3,0) to (2,i).
This is just my raster strategy, only using a spiral instead of a raster.
-tom
On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler <acwacw@gmail.com> wrote:
Okay, two things: a comment about notation, and a starting puzzle.
I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, presented as a way to suggest an unambiguous and fairly terse notation.
1. Unfold 0 to 2+2i. 2. Fold 2+2i to 2+i.
In each case the operation is performed so as to put a copy of the first point onto the second. This specifies the crease axis unambiguously. Some moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because there is no permissible crease that does that. The second point has to be a queen's move from the first.
Now the puzzle, the simplest one I haven't been able to do yet:
From {0}, construct {0, 2, 2+i}.
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This is great! I was thinking: is it possible to make "u" unfold and "f" fold, so we don't have to keep hitting the button to switch? Or if you prefer, make click fold and shift-click unfold? Or left click fold and right click unfold? Even though I mostly use vim these days, I'm not a big fan of modality . . . -tom On Sun, Oct 25, 2020 at 7:54 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
A little update: I've changed my interactive thing so that you have to click once to select a line, then again to select a side to fold onto. I've also made it keep track of moves and store them in the URL, so you can share solutions more easily. Finally, I changed "unfold" so it only folds the points on one side.
Allan, is it OK if I share this outside math-fun, crediting you with the idea?
On Fri, 23 Oct 2020 at 07:05, Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
Christian: how big is the original Elm program?
The code is at https://github.com/christianp/gaussian-origami. It's just over 400 lines of elm at the moment.
I really really should be doing my actual job, but I think it would help matters to be able to share links to sequences of moves, so I'll try to add that.
On Thu, 22 Oct 2020, 21:39 Allan Wechsler, <acwacw@gmail.com> wrote:
I think Tom's 3-step solution is incorrect, and produces i as well as the desired 3 points. But this could be interpreted as a difference of opinion about what the "unfold" operation does. I imagined "unfold" as the union of the original set and its reflection around the crease. Christian's web app agrees with me.
Christian: how big is the original Elm program?
On Thu, Oct 22, 2020 at 4:16 PM Tom Karzes <karzes@sonic.net> wrote:
Nice. I have a different 3-fold solution:
1
2
1
3
1
2
1
4
1
2
1
3
1
2
1. Unfold 0 to 2+2i 2. Fold 2+2i to 2+i 3. Unfold 2+i to 2
Tom
Allan Wechsler writes:
Tom Rokicki builds 0, 2, 2+i from 0 in four steps. It's easy to prove that it takes at least three steps ... and I just realized that three steps can indeed be done. So there's a certain sort of "code golf" that can be played with this sort of puzzle. Answer below.
1
6
1
8
0
3
3
9
8
8
7
4
9
8
1. Unfold 0 to 2 2. Unfold 0 to 2+2i (2 is on the crease) 3. Fold 2+2i to 2+i.
On Thu, Oct 22, 2020 at 3:12 PM Tomas Rokicki <rokicki@gmail.com> wrote:
Spoiler space.
3
1
4
1
5
9
2
6
5
3
5
8
Unfold (0,0) to (2,0) Unfold (2,0) to (3,0) (creates (5,0) as well) Fold (5,0) to (3,0) Fold (3,0) to (2,i).
This is just my raster strategy, only using a spiral instead of a raster.
-tom
On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler < acwacw@gmail.com> wrote:
> Okay, two things: a comment about notation, and a starting puzzle. > > I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my solution, > presented as a way to suggest an unambiguous and fairly terse notation. > > 1. Unfold 0 to 2+2i. > 2. Fold 2+2i to 2+i. > > In each case the operation is performed so as to put a copy of the first > point onto the second. This specifies the crease axis unambiguously. Some > moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", because > there is no permissible crease that does that. The second point has to be a > queen's move from the first. > > Now the puzzle, the simplest one I haven't been able to do yet: > > From {0}, construct {0, 2, 2+i}. > >
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My original conception (poorly expressed in my initial posts) was that the "unfold" operation didn't have a preferred side: it simply produced the union of the original set and its reflection. Tom Karzes had a different definition for "unfold": the union of the original set with the reflection of one half-plane. There are other possibilities: unfold could mean, unfold one side of the crease onto the other, deleting whatever was originally on the other side. Or it could only be legal to unfold when one half-plane was already empty. Each of these choices will lead to different puzzles being solvable or not, and will make the minimum number of moves different. I don't have any good reasons to prefer any definition over the other, but discussions should be clear about which definition they are using. What I would really like, from a puzzling perspective, is a set of rules that is "universal" in the sense that any set can be transformed into any other, but some puzzles are insanely hard, and the difficulty is not just correlated with the number of dots. There are endless possible basic operations (union with 90-degree rotation; xor with one-cell shift; and so on and on). But I would like to find a basic toolbox that spawns a nice set of puzzles. On Sun, Oct 25, 2020 at 11:09 AM Tomas Rokicki <rokicki@gmail.com> wrote:
This is great!
I was thinking: is it possible to make "u" unfold and "f" fold, so we don't have to keep hitting the button to switch?
Or if you prefer, make click fold and shift-click unfold?
Or left click fold and right click unfold?
Even though I mostly use vim these days, I'm not a big fan of modality . . .
-tom
On Sun, Oct 25, 2020 at 7:54 AM Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
A little update: I've changed my interactive thing so that you have to click once to select a line, then again to select a side to fold onto. I've also made it keep track of moves and store them in the URL, so you can share solutions more easily. Finally, I changed "unfold" so it only folds the points on one side.
Allan, is it OK if I share this outside math-fun, crediting you with the idea?
On Fri, 23 Oct 2020 at 07:05, Christian Lawson-Perfect < christianperfect@gmail.com> wrote:
Christian: how big is the original Elm program?
The code is at https://github.com/christianp/gaussian-origami. It's just over 400 lines of elm at the moment.
I really really should be doing my actual job, but I think it would help matters to be able to share links to sequences of moves, so I'll try to add that.
On Thu, 22 Oct 2020, 21:39 Allan Wechsler, <acwacw@gmail.com> wrote:
I think Tom's 3-step solution is incorrect, and produces i as well as the desired 3 points. But this could be interpreted as a difference of opinion about what the "unfold" operation does. I imagined "unfold" as the union of the original set and its reflection around the crease. Christian's web app agrees with me.
Christian: how big is the original Elm program?
On Thu, Oct 22, 2020 at 4:16 PM Tom Karzes <karzes@sonic.net> wrote:
Nice. I have a different 3-fold solution:
1
2
1
3
1
2
1
4
1
2
1
3
1
2
1. Unfold 0 to 2+2i 2. Fold 2+2i to 2+i 3. Unfold 2+i to 2
Tom
Allan Wechsler writes:
Tom Rokicki builds 0, 2, 2+i from 0 in four steps. It's easy to prove that it takes at least three steps ... and I just realized that three steps can indeed be done. So there's a certain sort of "code golf" that can be played with this sort of puzzle. Answer below.
1
6
1
8
0
3
3
9
8
8
7
4
9
8
1. Unfold 0 to 2 2. Unfold 0 to 2+2i (2 is on the crease) 3. Fold 2+2i to 2+i.
On Thu, Oct 22, 2020 at 3:12 PM Tomas Rokicki <rokicki@gmail.com
wrote:
> Spoiler space. > > 3 > > 1 > > 4 > > 1 > > 5 > > 9 > > 2 > > 6 > > 5 > > 3 > > 5 > > 8 > > Unfold (0,0) to (2,0) > Unfold (2,0) to (3,0) (creates (5,0) as well) > Fold (5,0) to (3,0) > Fold (3,0) to (2,i). > > This is just my raster strategy, only using a spiral instead > of a raster. > > -tom > > On Thu, Oct 22, 2020 at 10:50 AM Allan Wechsler <
acwacw@gmail.com> wrote:
> > > Okay, two things: a comment about notation, and a starting puzzle. > > > > I mentioned that I could do 0 -> 0, 2+i in two moves. Here is my > solution, > > presented as a way to suggest an unambiguous and fairly terse notation. > > > > 1. Unfold 0 to 2+2i. > > 2. Fold 2+2i to 2+i. > > > > In each case the operation is performed so as to put a copy of the first > > point onto the second. This specifies the crease axis unambiguously. Some > > moves are illegal, so it isn't acceptable to say "unfold 0 to 2+i", > because > > there is no permissible crease that does that. The second point has to > be a > > queen's move from the first. > > > > Now the puzzle, the simplest one I haven't been able to do yet: > > > > From {0}, construct {0, 2, 2+i}. > > > >
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participants (7)
-
Allan Wechsler -
Christian Lawson-Perfect -
Fred Lunnon -
James Propp -
Tom Karzes -
Tomas Rokicki -
Éric Angelini