Re: [math-fun] Radial disk-dissection
I'm trying to guess what RWG meant without peeking at his drawings. In order to make Jim Propp's statement exact, I would have to make precise 1) what "dissect and reassemble" mean and 2) what "converges" to a 1-by-pi rectangle means. A typical meaning for 1): For subsets A, B of R^2, to dissect A and reassemble it to B means that there is a partition A = X_1 + ... + X_n of A as a finite disjoint union, such that there exist isometries f_1, ..., f_n of R^2 such that B = f_1(X_1) + ... + f_n(X_n) forms a partition of B as a finite disjoint union. * * * One meaning for 2) could be in the sense of Hausdorff distance between compact sets in the plane. The only problem I see here is that if strict partition are used in 1) as above, then the resulting rectangle B will not be compact, as it will not contain all of its boundary. I have complete faith that appropriate hand-waving will not incur the wrath of the math gods. —Dan ----- Jim Propp wrote:
If you dissect a unit disk radially into a large number of equal wedges, it’s well known that you can reassemble them to form a shape that in the limit converges to a 1-by-pi rectangle.
RWG wrote: ----- gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. ----- -----
I think (1) means that we have an infinite sequence of sets S_k where S_k is composed of k wedges (joined only along full edges), each with angle 2*pi/k; the limit is just the set of points p such that p is contained in all but finitely many S_k. You can definitely get an interesting collection of shapes this way. On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I'm trying to guess what RWG meant without peeking at his drawings.
In order to make Jim Propp's statement exact, I would have to make precise
1) what "dissect and reassemble" mean
and
2) what "converges" to a 1-by-pi rectangle means.
A typical meaning for 1): For subsets A, B of R^2, to dissect A and reassemble it to B means that there is a partition
A = X_1 + ... + X_n
of A as a finite disjoint union, such that there exist isometries
f_1, ..., f_n of R^2
such that
B = f_1(X_1) + ... + f_n(X_n)
forms a partition of B as a finite disjoint union.
* * *
One meaning for 2) could be in the sense of Hausdorff distance between compact sets in the plane. The only problem I see here is that if strict partition are used in 1) as above, then the resulting rectangle B will not be compact, as it will not contain all of its boundary. I have complete faith that appropriate hand-waving will not incur the wrath of the math gods.
—Dan
----- Jim Propp wrote:
If you dissect a unit disk radially into a large number of equal wedges, it’s well known that you can reassemble them to form a shape that in the limit converges to a 1-by-pi rectangle.
RWG wrote: ----- gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. ----- -----
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It seems to me that, in the limit, we have a behavior something like this: We have a unit line segment AB moving in the plane. Each of its endpoints is moving perpendicular to the line, toward the same side of the line, at speeds that add up to 1. Subject to that constraint, their speeds are an arbitrary function of time. Say the speed of point A is given by f(t); then point B is moving in the same direction at speed 1-f(t). Because the speeds of the endpoints can differ, the line can gradually change orientation; its angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at a constant speed of 1/2. The curvatures of the curves traced out by A and B are related by the equation 1/a + 1/b = 1. The whole process continues until t = 2pi, so the total area swept out is pi. On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins <mjcollins10@gmail.com> wrote:
I think (1) means that we have an infinite sequence of sets S_k where S_k is composed of k wedges (joined only along full edges), each with angle 2*pi/k; the limit is just the set of points p such that p is contained in all but finitely many S_k. You can definitely get an interesting collection of shapes this way.
On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I'm trying to guess what RWG meant without peeking at his drawings.
In order to make Jim Propp's statement exact, I would have to make precise
1) what "dissect and reassemble" mean
and
2) what "converges" to a 1-by-pi rectangle means.
A typical meaning for 1): For subsets A, B of R^2, to dissect A and reassemble it to B means that there is a partition
A = X_1 + ... + X_n
of A as a finite disjoint union, such that there exist isometries
f_1, ..., f_n of R^2
such that
B = f_1(X_1) + ... + f_n(X_n)
forms a partition of B as a finite disjoint union.
* * *
One meaning for 2) could be in the sense of Hausdorff distance between compact sets in the plane. The only problem I see here is that if strict partition are used in 1) as above, then the resulting rectangle B will not be compact, as it will not contain all of its boundary. I have complete faith that appropriate hand-waving will not incur the wrath of the math gods.
—Dan
----- Jim Propp wrote:
If you dissect a unit disk radially into a large number of equal wedges, it’s well known that you can reassemble them to form a shape that in the limit converges to a 1-by-pi rectangle.
RWG wrote: ----- gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. ----- -----
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Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But it requires a time-parametrization. Is there a way to characterize such shapes directly? Jim On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
It seems to me that, in the limit, we have a behavior something like this:
We have a unit line segment AB moving in the plane. Each of its endpoints is moving perpendicular to the line, toward the same side of the line, at speeds that add up to 1. Subject to that constraint, their speeds are an arbitrary function of time. Say the speed of point A is given by f(t); then point B is moving in the same direction at speed 1-f(t). Because the speeds of the endpoints can differ, the line can gradually change orientation; its angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at a constant speed of 1/2. The curvatures of the curves traced out by A and B are related by the equation 1/a + 1/b = 1. The whole process continues until t = 2pi, so the total area swept out is pi.
On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins <mjcollins10@gmail.com> wrote:
I think (1) means that we have an infinite sequence of sets S_k where S_k is composed of k wedges (joined only along full edges), each with angle 2*pi/k; the limit is just the set of points p such that p is contained in all but finitely many S_k. You can definitely get an interesting collection of shapes this way.
On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I'm trying to guess what RWG meant without peeking at his drawings.
In order to make Jim Propp's statement exact, I would have to make precise
1) what "dissect and reassemble" mean
and
2) what "converges" to a 1-by-pi rectangle means.
A typical meaning for 1): For subsets A, B of R^2, to dissect A and reassemble it to B means that there is a partition
A = X_1 + ... + X_n
of A as a finite disjoint union, such that there exist isometries
f_1, ..., f_n of R^2
such that
B = f_1(X_1) + ... + f_n(X_n)
forms a partition of B as a finite disjoint union.
* * *
One meaning for 2) could be in the sense of Hausdorff distance between compact sets in the plane. The only problem I see here is that if strict partition are used in 1) as above, then the resulting rectangle B will not be compact, as it will not contain all of its boundary. I have complete faith that appropriate hand-waving will not incur the wrath of the math gods.
—Dan
----- Jim Propp wrote:
If you dissect a unit disk radially into a large number of equal wedges, it’s well known that you can reassemble them to form a shape that in the limit converges to a 1-by-pi rectangle.
RWG wrote: ----- gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. ----- -----
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Maybe what you are looking for is this. The "ribbon" has two curvilinear edges. From any point A on one edge, draw a perpendicular line; it will turn out to be perpendicular to the other edge as well. (By "perpendicular" I mean "perpendicular to the tangent at that point".) On Fri, Apr 20, 2018 at 12:52 PM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But it requires a time-parametrization. Is there a way to characterize such shapes directly?
Jim
On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
It seems to me that, in the limit, we have a behavior something like this:
We have a unit line segment AB moving in the plane. Each of its endpoints is moving perpendicular to the line, toward the same side of the line, at speeds that add up to 1. Subject to that constraint, their speeds are an arbitrary function of time. Say the speed of point A is given by f(t); then point B is moving in the same direction at speed 1-f(t). Because the speeds of the endpoints can differ, the line can gradually change orientation; its angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at a constant speed of 1/2. The curvatures of the curves traced out by A and B are related by the equation 1/a + 1/b = 1. The whole process continues until t = 2pi, so the total area swept out is pi.
On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins <mjcollins10@gmail.com> wrote:
I think (1) means that we have an infinite sequence of sets S_k where S_k is composed of k wedges (joined only along full edges), each with angle 2*pi/k; the limit is just the set of points p such that p is contained in all but finitely many S_k. You can definitely get an interesting collection of shapes this way.
On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I'm trying to guess what RWG meant without peeking at his drawings.
In order to make Jim Propp's statement exact, I would have to make precise
1) what "dissect and reassemble" mean
and
2) what "converges" to a 1-by-pi rectangle means.
A typical meaning for 1): For subsets A, B of R^2, to dissect A and reassemble it to B means that there is a partition
A = X_1 + ... + X_n
of A as a finite disjoint union, such that there exist isometries
f_1, ..., f_n of R^2
such that
B = f_1(X_1) + ... + f_n(X_n)
forms a partition of B as a finite disjoint union.
* * *
One meaning for 2) could be in the sense of Hausdorff distance between compact sets in the plane. The only problem I see here is that if strict partition are used in 1) as above, then the resulting rectangle B will not be compact, as it will not contain all of its boundary. I have complete faith that appropriate hand-waving will not incur the wrath of the math gods.
—Dan
----- Jim Propp wrote:
If you dissect a unit disk radially into a large number of equal wedges, it’s well known that you can reassemble them to form a shape that in the limit converges to a 1-by-pi rectangle.
RWG wrote: ----- gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. ----- -----
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Perfect! Thanks. I guess the theorem here is that the area of such a ribbon is equal to half the sum of the lengths of the two non-straight sides. Come to think of it, this is just a consequence of what’s-his-name’s theorem about the area swept out by a line segment that moves perpendicular to itself. The only nonobvious step is relating the distance traveled by the midpoint of the segment to the distances traveled by the endpoints of the segment. Jim On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Maybe what you are looking for is this. The "ribbon" has two curvilinear edges. From any point A on one edge, draw a perpendicular line; it will turn out to be perpendicular to the other edge as well. (By "perpendicular" I mean "perpendicular to the tangent at that point".)
On Fri, Apr 20, 2018 at 12:52 PM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But it requires a time-parametrization. Is there a way to characterize such shapes directly?
Jim
On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
It seems to me that, in the limit, we have a behavior something like this:
We have a unit line segment AB moving in the plane. Each of its endpoints is moving perpendicular to the line, toward the same side of the line, at speeds that add up to 1. Subject to that constraint, their speeds are an arbitrary function of time. Say the speed of point A is given by f(t); then point B is moving in the same direction at speed 1-f(t). Because the speeds of the endpoints can differ, the line can gradually change orientation; its angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at a constant speed of 1/2. The curvatures of the curves traced out by A and B are related by the equation 1/a + 1/b = 1. The whole process continues until t = 2pi, so the total area swept out is pi.
On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < mjcollins10@gmail.com> wrote:
I think (1) means that we have an infinite sequence of sets S_k where S_k is composed of k wedges (joined only along full edges), each with angle 2*pi/k; the limit is just the set of points p such that p is contained in all but finitely many S_k. You can definitely get an interesting collection of shapes this way.
On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov <dasimov@earthlink.net> wrote:
I'm trying to guess what RWG meant without peeking at his drawings.
In order to make Jim Propp's statement exact, I would have to make precise
1) what "dissect and reassemble" mean
and
2) what "converges" to a 1-by-pi rectangle means.
A typical meaning for 1): For subsets A, B of R^2, to dissect A and reassemble it to B means that there is a partition
A = X_1 + ... + X_n
of A as a finite disjoint union, such that there exist isometries
f_1, ..., f_n of R^2
such that
B = f_1(X_1) + ... + f_n(X_n)
forms a partition of B as a finite disjoint union.
* * *
One meaning for 2) could be in the sense of Hausdorff distance between compact sets in the plane. The only problem I see here is that if strict partition are used in 1) as above, then the resulting rectangle B will not be compact, as it will not contain all of its boundary. I have complete faith that appropriate hand-waving will not incur the wrath of the math gods.
—Dan
----- Jim Propp wrote:
If you dissect a unit disk radially into a large number of equal wedges, it’s well known that you can reassemble them to form a shape that in the limit converges to a 1-by-pi rectangle.
RWG wrote: ----- gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. ----- -----
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Guys! Guys! Here is a question! You'll see, it will end with a question mark and everything. Can such a ribbon self-intersect? I'm guessing the answer is no, but I can't see a proof path. On Fri, Apr 20, 2018 at 1:28 PM, James Propp <jamespropp@gmail.com> wrote:
Perfect! Thanks.
I guess the theorem here is that the area of such a ribbon is equal to half the sum of the lengths of the two non-straight sides.
Come to think of it, this is just a consequence of what’s-his-name’s theorem about the area swept out by a line segment that moves perpendicular to itself. The only nonobvious step is relating the distance traveled by the midpoint of the segment to the distances traveled by the endpoints of the segment.
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Maybe what you are looking for is this. The "ribbon" has two curvilinear edges. From any point A on one edge, draw a perpendicular line; it will turn out to be perpendicular to the other edge as well. (By "perpendicular" I mean "perpendicular to the tangent at that point".)
On Fri, Apr 20, 2018 at 12:52 PM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But it requires a time-parametrization. Is there a way to characterize such shapes directly?
Jim
On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
It seems to me that, in the limit, we have a behavior something like this:
We have a unit line segment AB moving in the plane. Each of its endpoints is moving perpendicular to the line, toward the same side of the line, at speeds that add up to 1. Subject to that constraint, their speeds are an arbitrary function of time. Say the speed of point A is given by f(t); then point B is moving in the same direction at speed 1-f(t). Because the speeds of the endpoints can differ, the line can gradually change orientation; its angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at a constant speed of 1/2. The curvatures of the curves traced out by A and B are related by the equation 1/a + 1/b = 1. The whole process continues until t = 2pi, so the total area swept out is pi.
On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < mjcollins10@gmail.com> wrote:
I think (1) means that we have an infinite sequence of sets S_k where S_k is composed of k wedges (joined only along full edges), each with angle 2*pi/k; the limit is just the set of points p such that p is contained in all but finitely many S_k. You can definitely get an interesting collection of shapes this way.
On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < dasimov@earthlink.net> wrote:
I'm trying to guess what RWG meant without peeking at his drawings.
In order to make Jim Propp's statement exact, I would have to make precise
1) what "dissect and reassemble" mean
and
2) what "converges" to a 1-by-pi rectangle means.
A typical meaning for 1): For subsets A, B of R^2, to dissect A and reassemble it to B means that there is a partition
A = X_1 + ... + X_n
of A as a finite disjoint union, such that there exist isometries
f_1, ..., f_n of R^2
such that
B = f_1(X_1) + ... + f_n(X_n)
forms a partition of B as a finite disjoint union.
* * *
One meaning for 2) could be in the sense of Hausdorff distance between compact sets in the plane. The only problem I see here is that if strict partition are used in 1) as above, then the resulting rectangle B will not be compact, as it will not contain all of its boundary. I have complete faith that appropriate hand-waving will not incur the wrath of the math gods.
—Dan
----- Jim Propp wrote: > If you dissect a unit disk radially into a large number of equal wedges, > it’s well known that you can reassemble them to form a shape that in the > limit converges to a 1-by-pi rectangle. >
RWG wrote: ----- gosper.org/picfzoom.gif gosper.org/semizoom.gif --rwg I don't see how to get anything other than allowing unequal wedges. ----- -----
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The answer is yes. On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <acwacw@gmail.com> wrote:
Guys! Guys! Here is a question! You'll see, it will end with a question mark and everything.
Can such a ribbon self-intersect?
I'm guessing the answer is no, but I can't see a proof path.
On Fri, Apr 20, 2018 at 1:28 PM, James Propp <jamespropp@gmail.com> wrote:
Perfect! Thanks.
I guess the theorem here is that the area of such a ribbon is equal to half the sum of the lengths of the two non-straight sides.
Come to think of it, this is just a consequence of what’s-his-name’s theorem about the area swept out by a line segment that moves perpendicular to itself. The only nonobvious step is relating the distance traveled by the midpoint of the segment to the distances traveled by the endpoints of the segment.
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Maybe what you are looking for is this. The "ribbon" has two curvilinear edges. From any point A on one edge, draw a perpendicular line; it will turn out to be perpendicular to the other edge as well. (By "perpendicular" I mean "perpendicular to the tangent at that point".)
On Fri, Apr 20, 2018 at 12:52 PM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But it requires a time-parametrization. Is there a way to characterize such shapes directly?
Jim
On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
It seems to me that, in the limit, we have a behavior something like this:
We have a unit line segment AB moving in the plane. Each of its endpoints is moving perpendicular to the line, toward the same side of the line, at speeds that add up to 1. Subject to that constraint, their speeds are an arbitrary function of time. Say the speed of point A is given by f(t); then point B is moving in the same direction at speed 1-f(t). Because the speeds of the endpoints can differ, the line can gradually change orientation; its angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at a constant speed of 1/2. The curvatures of the curves traced out by A and B are related by the equation 1/a + 1/b = 1. The whole process continues until t = 2pi, so the total area swept out is pi.
On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < mjcollins10@gmail.com> wrote:
I think (1) means that we have an infinite sequence of sets S_k where S_k is composed of k wedges (joined only along full edges), each with angle 2*pi/k; the limit is just the set of points p such that p is contained in all but finitely many S_k. You can definitely get an interesting collection of shapes this way.
On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < dasimov@earthlink.net> wrote:
> I'm trying to guess what RWG meant without peeking at his drawings. > > In order to make Jim Propp's statement exact, I would have to make precise > > 1) what "dissect and reassemble" mean > > and > > 2) what "converges" to a 1-by-pi rectangle means. > > A typical meaning for 1): For subsets A, B of R^2, to dissect A and > reassemble it > to B means that there is a partition > > A = X_1 + ... + X_n > > of A as a finite disjoint union, such that there exist isometries > > f_1, ..., f_n of R^2 > > such that > > B = f_1(X_1) + ... + f_n(X_n) > > forms a partition of B as a finite disjoint union. > > * * * > > One meaning for 2) could be in the sense of Hausdorff distance between > compact sets > in the plane. The only problem I see here is that if strict partition are > used in > 1) as above, then the resulting rectangle B will not be compact, as it > will not contain > all of its boundary. I have complete faith that appropriate hand-waving > will not incur > the wrath of the math gods. > > —Dan > > > ----- > Jim Propp wrote: > > If you dissect a unit disk radially into a large number of equal wedges, > > it’s well known that you can reassemble them to form a shape that in the > > limit converges to a 1-by-pi rectangle. > > > > RWG wrote: > ----- > gosper.org/picfzoom.gif > gosper.org/semizoom.gif > --rwg > I don't see how to get anything other than allowing unequal wedges. > ----- > ----- > > _______________________________________________ > 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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-- -- http://cube20.org/ -- http://golly.sf.net/ --
Really? You only have 2pi time units to do it in. Also ... I want a clean intersection: the overlap has to have area. Pics or it didn't happen. On Fri, Apr 20, 2018 at 10:12 PM, Tomas Rokicki <rokicki@gmail.com> wrote:
The answer is yes.
On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <acwacw@gmail.com> wrote:
Guys! Guys! Here is a question! You'll see, it will end with a question mark and everything.
Can such a ribbon self-intersect?
I'm guessing the answer is no, but I can't see a proof path.
On Fri, Apr 20, 2018 at 1:28 PM, James Propp <jamespropp@gmail.com> wrote:
Perfect! Thanks.
I guess the theorem here is that the area of such a ribbon is equal to half the sum of the lengths of the two non-straight sides.
Come to think of it, this is just a consequence of what’s-his-name’s theorem about the area swept out by a line segment that moves perpendicular to itself. The only nonobvious step is relating the distance traveled by the midpoint of the segment to the distances traveled by the endpoints of the segment.
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Maybe what you are looking for is this. The "ribbon" has two curvilinear edges. From any point A on one edge, draw a perpendicular line; it will turn out to be perpendicular to the other edge as well. (By "perpendicular" I mean "perpendicular to the tangent at that point".)
On Fri, Apr 20, 2018 at 12:52 PM, James Propp <jamespropp@gmail.com> wrote:
Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But it requires a time-parametrization. Is there a way to characterize such shapes directly?
Jim
On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
It seems to me that, in the limit, we have a behavior something like this:
We have a unit line segment AB moving in the plane. Each of its endpoints is moving perpendicular to the line, toward the same side of the line, at speeds that add up to 1. Subject to that constraint, their speeds are an arbitrary function of time. Say the speed of point A is given by f(t); then point B is moving in the same direction at speed 1-f(t). Because the speeds of the endpoints can differ, the line can gradually change orientation; its angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at a constant speed of 1/2. The curvatures of the curves traced out by A and B are related by the equation 1/a + 1/b = 1. The whole process continues until t = 2pi, so the total area swept out is pi.
On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < mjcollins10@gmail.com> wrote:
> I think (1) means that we have an infinite sequence of sets S_k where S_k > is composed of k wedges (joined only along full edges), each with angle > 2*pi/k; the limit is just the set of points p such that p is contained in > all but finitely many S_k. You can definitely get an interesting collection > of shapes this way. > > On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < dasimov@earthlink.net> > wrote: > > > I'm trying to guess what RWG meant without peeking at his drawings. > > > > In order to make Jim Propp's statement exact, I would have to make > precise > > > > 1) what "dissect and reassemble" mean > > > > and > > > > 2) what "converges" to a 1-by-pi rectangle means. > > > > A typical meaning for 1): For subsets A, B of R^2, to dissect A and > > reassemble it > > to B means that there is a partition > > > > A = X_1 + ... + X_n > > > > of A as a finite disjoint union, such that there exist isometries > > > > f_1, ..., f_n of R^2 > > > > such that > > > > B = f_1(X_1) + ... + f_n(X_n) > > > > forms a partition of B as a finite disjoint union. > > > > * *
> > > > One meaning for 2) could be in the sense of Hausdorff distance between > > compact sets > > in the plane. The only problem I see here is that if strict partition are > > used in > > 1) as above, then the resulting rectangle B will not be compact, as it > > will not contain > > all of its boundary. I have complete faith that appropriate hand-waving > > will not incur > > the wrath of the math gods. > > > > —Dan > > > > > > ----- > > Jim Propp wrote: > > > If you dissect a unit disk radially into a large number of equal > wedges, > > > it’s well known that you can reassemble them to form a shape that in > the > > > limit converges to a 1-by-pi rectangle. > > > > > > > RWG wrote: > > ----- > > gosper.org/picfzoom.gif > > gosper.org/semizoom.gif > > --rwg > > I don't see how to get anything other than allowing unequal wedges. > > ----- > > ----- > > > > _______________________________________________ > > 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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Oh, wait. I see it. Tomas is right, of course. I wonder what the maximum self-overlap is. I can see how to get pi^2 / 64. On Fri, Apr 20, 2018 at 10:18 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Really? You only have 2pi time units to do it in. Also ... I want a clean intersection: the overlap has to have area.
Pics or it didn't happen.
On Fri, Apr 20, 2018 at 10:12 PM, Tomas Rokicki <rokicki@gmail.com> wrote:
The answer is yes.
On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <acwacw@gmail.com> wrote:
Guys! Guys! Here is a question! You'll see, it will end with a question mark and everything.
Can such a ribbon self-intersect?
I'm guessing the answer is no, but I can't see a proof path.
On Fri, Apr 20, 2018 at 1:28 PM, James Propp <jamespropp@gmail.com> wrote:
Perfect! Thanks.
I guess the theorem here is that the area of such a ribbon is equal to half the sum of the lengths of the two non-straight sides.
Come to think of it, this is just a consequence of what’s-his-name’s theorem about the area swept out by a line segment that moves perpendicular to itself. The only nonobvious step is relating the distance traveled by the midpoint of the segment to the distances traveled by the endpoints of the segment.
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Maybe what you are looking for is this. The "ribbon" has two curvilinear edges. From any point A on one edge, draw a perpendicular line; it will turn out to be perpendicular to the other edge as well. (By "perpendicular" I mean "perpendicular to the tangent at that point".)
On Fri, Apr 20, 2018 at 12:52 PM, James Propp <jamespropp@gmail.com
wrote:
Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But it requires a time-parametrization. Is there a way to characterize such shapes directly?
Jim
On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
> It seems to me that, in the limit, we have a behavior something like this: > > We have a unit line segment AB moving in the plane. Each of its endpoints > is moving perpendicular to the line, toward the same side of the line, at > speeds that add up to 1. Subject to that constraint, their speeds are an > arbitrary function of time. Say the speed of point A is given by f(t); then > point B is moving in the same direction at speed 1-f(t). Because the speeds > of the endpoints can differ, the line can gradually change orientation; its > angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area at > a constant speed of 1/2. The curvatures of the curves traced out by A and B > are related by the equation 1/a + 1/b = 1. The whole process continues > until t = 2pi, so the total area swept out is pi. > > On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < mjcollins10@gmail.com> > wrote: > > > I think (1) means that we have an infinite sequence of sets S_k where S_k > > is composed of k wedges (joined only along full edges), each with angle > > 2*pi/k; the limit is just the set of points p such that p is contained in > > all but finitely many S_k. You can definitely get an interesting > collection > > of shapes this way. > > > > On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < dasimov@earthlink.net> > > wrote: > > > > > I'm trying to guess what RWG meant without peeking at his drawings. > > > > > > In order to make Jim Propp's statement exact, I would have to make > > precise > > > > > > 1) what "dissect and reassemble" mean > > > > > > and > > > > > > 2) what "converges" to a 1-by-pi rectangle means. > > > > > > A typical meaning for 1): For subsets A, B of R^2, to dissect A and > > > reassemble it > > > to B means that there is a partition > > > > > > A = X_1 + ... + X_n > > > > > > of A as a finite disjoint union, such that there exist isometries > > > > > > f_1, ..., f_n of R^2 > > > > > > such that > > > > > > B = f_1(X_1) + ... + f_n(X_n) > > > > > > forms a partition of B as a finite disjoint union. > > > > > > * *
> > > > > > One meaning for 2) could be in the sense of Hausdorff distance between > > > compact sets > > > in the plane. The only problem I see here is that if strict partition > are > > > used in > > > 1) as above, then the resulting rectangle B will not be compact, as it > > > will not contain > > > all of its boundary. I have complete faith that appropriate hand-waving > > > will not incur > > > the wrath of the math gods. > > > > > > —Dan > > > > > > > > > ----- > > > Jim Propp wrote: > > > > If you dissect a unit disk radially into a large number of equal > > wedges, > > > > it’s well known that you can reassemble them to form a shape that in > > the > > > > limit converges to a 1-by-pi rectangle. > > > > > > > > > > RWG wrote: > > > ----- > > > gosper.org/picfzoom.gif > > > gosper.org/semizoom.gif > > > --rwg > > > I don't see how to get anything other than allowing unequal wedges. > > > ----- > > > ----- > > > > > > _______________________________________________ > > > math-fun mailing list > > > math-fun@mailman.xmission.com > > > https://mailman.xmission.com/c gi-bin/mailman/listinfo/math-fun > > > > > _______________________________________________ > > math-fun mailing list > > math-fun@mailman.xmission.com > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-f un > > > _______________________________________________ > 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 picture, please? Jim On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Oh, wait. I see it. Tomas is right, of course. I wonder what the maximum self-overlap is. I can see how to get pi^2 / 64.
On Fri, Apr 20, 2018 at 10:18 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Really? You only have 2pi time units to do it in. Also ... I want a clean intersection: the overlap has to have area.
Pics or it didn't happen.
On Fri, Apr 20, 2018 at 10:12 PM, Tomas Rokicki <rokicki@gmail.com> wrote:
The answer is yes.
On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <acwacw@gmail.com> wrote:
Guys! Guys! Here is a question! You'll see, it will end with a question mark and everything.
Can such a ribbon self-intersect?
I'm guessing the answer is no, but I can't see a proof path.
On Fri, Apr 20, 2018 at 1:28 PM, James Propp <jamespropp@gmail.com> wrote:
Perfect! Thanks.
I guess the theorem here is that the area of such a ribbon is equal to half the sum of the lengths of the two non-straight sides.
Come to think of it, this is just a consequence of what’s-his-name’s theorem about the area swept out by a line segment that moves perpendicular to itself. The only nonobvious step is relating the distance traveled by the midpoint of the segment to the distances traveled by the endpoints of the segment.
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Maybe what you are looking for is this. The "ribbon" has two curvilinear edges. From any point A on one edge, draw a perpendicular line; it will turn out to be perpendicular to the other edge as well. (By "perpendicular" I mean "perpendicular to the tangent at that point".)
On Fri, Apr 20, 2018 at 12:52 PM, James Propp < jamespropp@gmail.com
wrote:
> Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. But > it requires a time-parametrization. Is there a way to characterize such > shapes directly? > > Jim > > On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com> wrote: > > > It seems to me that, in the limit, we have a behavior something like > this: > > > > We have a unit line segment AB moving in the plane. Each of its endpoints > > is moving perpendicular to the line, toward the same side of the line, at > > speeds that add up to 1. Subject to that constraint, their speeds are an > > arbitrary function of time. Say the speed of point A is given by f(t); > then > > point B is moving in the same direction at speed 1-f(t). Because the > speeds > > of the endpoints can differ, the line can gradually change orientation; > its > > angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out area > at > > a constant speed of 1/2. The curvatures of the curves traced out by A > and B > > are related by the equation 1/a + 1/b = 1. The whole process continues > > until t = 2pi, so the total area swept out is pi. > > > > On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < mjcollins10@gmail.com> > > wrote: > > > > > I think (1) means that we have an infinite sequence of sets S_k where > S_k > > > is composed of k wedges (joined only along full edges), each with angle > > > 2*pi/k; the limit is just the set of points p such that p is contained > in > > > all but finitely many S_k. You can definitely get an interesting > > collection > > > of shapes this way. > > > > > > On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < dasimov@earthlink.net> > > > wrote: > > > > > > > I'm trying to guess what RWG meant without peeking at his drawings. > > > > > > > > In order to make Jim Propp's statement exact, I would have to make > > > precise > > > > > > > > 1) what "dissect and reassemble" mean > > > > > > > > and > > > > > > > > 2) what "converges" to a 1-by-pi rectangle means. > > > > > > > > A typical meaning for 1): For subsets A, B of R^2, to dissect A and > > > > reassemble it > > > > to B means that there is a partition > > > > > > > > A = X_1 + ... + X_n > > > > > > > > of A as a finite disjoint union, such that there exist isometries > > > > > > > > f_1, ..., f_n of R^2 > > > > > > > > such that > > > > > > > > B = f_1(X_1) + ... + f_n(X_n) > > > > > > > > forms a partition of B as a finite disjoint union. > > > > > > > > * *
> > > > > > > > One meaning for 2) could be in the sense of Hausdorff distance > between > > > > compact sets > > > > in the plane. The only problem I see here is that if strict partition > > are > > > > used in > > > > 1) as above, then the resulting rectangle B will not be compact, as > it > > > > will not contain > > > > all of its boundary. I have complete faith that appropriate > hand-waving > > > > will not incur > > > > the wrath of the math gods. > > > > > > > > —Dan > > > > > > > > > > > > ----- > > > > Jim Propp wrote: > > > > > If you dissect a unit disk radially into a large number of equal > > > wedges, > > > > > it’s well known that you can reassemble them to form a shape that > in > > > the > > > > > limit converges to a 1-by-pi rectangle. > > > > > > > > > > > > > RWG wrote: > > > > ----- > > > > gosper.org/picfzoom.gif > > > > gosper.org/semizoom.gif > > > > --rwg > > > > I don't see how to get anything other than allowing unequal wedges. > > > > ----- > > > > ----- > > > > > > > > _______________________________________________ > > > > math-fun mailing list > > > > math-fun@mailman.xmission.com > > > > https://mailman.xmission.com/c gi-bin/mailman/listinfo/math-fun > > > > > > > _______________________________________________ > > > math-fun mailing list > > > math-fun@mailman.xmission.com > > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math- f un > > > > > _______________________________________________ > > 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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Render three-quarters of a circle, using 3Pi/2 edges. Extend each flat edge straight out by Pi/8; this gives an overlap of Pi/8 x Pi/8. This can almost certainly be improved by extending by one eighth of a circle using the other point as a radius, rather than extending straight out, but I am too lazy to do the math to determine if this is actually an improvement. This is almost certainly not the best one can do and I am interested in Allan's question about what the maximum possible overlap might be. -tom On Sat, Apr 21, 2018 at 3:03 AM, James Propp <jamespropp@gmail.com> wrote:
A picture, please?
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Oh, wait. I see it. Tomas is right, of course. I wonder what the maximum self-overlap is. I can see how to get pi^2 / 64.
On Fri, Apr 20, 2018 at 10:18 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Really? You only have 2pi time units to do it in. Also ... I want a clean intersection: the overlap has to have area.
Pics or it didn't happen.
On Fri, Apr 20, 2018 at 10:12 PM, Tomas Rokicki <rokicki@gmail.com> wrote:
The answer is yes.
On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <acwacw@gmail.com> wrote:
Guys! Guys! Here is a question! You'll see, it will end with a question mark and everything.
Can such a ribbon self-intersect?
I'm guessing the answer is no, but I can't see a proof path.
On Fri, Apr 20, 2018 at 1:28 PM, James Propp <jamespropp@gmail.com> wrote:
Perfect! Thanks.
I guess the theorem here is that the area of such a ribbon is equal to half the sum of the lengths of the two non-straight sides.
Come to think of it, this is just a consequence of what’s-his-name’s theorem about the area swept out by a line segment that moves perpendicular to itself. The only nonobvious step is relating the distance traveled by the midpoint of the segment to the distances traveled by the endpoints of the segment.
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
> Maybe what you are looking for is this. The "ribbon" has two curvilinear > edges. From any point A on one edge, draw a perpendicular line; it will > turn out to be perpendicular to the other edge as well. (By "perpendicular" > I mean "perpendicular to the tangent at that point".) > > On Fri, Apr 20, 2018 at 12:52 PM, James Propp < jamespropp@gmail.com
> wrote: > > > Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I wanted. > But > > it requires a time-parametrization. Is there a way to characterize such > > shapes directly? > > > > Jim > > > > On Thursday, April 19, 2018, Allan Wechsler <acwacw@gmail.com
wrote:
> > > > > It seems to me that, in the limit, we have a behavior something like > > this: > > > > > > We have a unit line segment AB moving in the plane. Each of its > endpoints > > > is moving perpendicular to the line, toward the same side of the line, > at > > > speeds that add up to 1. Subject to that constraint, their speeds are > an > > > arbitrary function of time. Say the speed of point A is given by f(t); > > then > > > point B is moving in the same direction at speed 1-f(t). Because the > > speeds > > > of the endpoints can differ, the line can gradually change orientation; > > its > > > angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out > area > > at > > > a constant speed of 1/2. The curvatures of the curves traced out by A > > and B > > > are related by the equation 1/a + 1/b = 1. The whole process continues > > > until t = 2pi, so the total area swept out is pi. > > > > > > On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < > mjcollins10@gmail.com> > > > wrote: > > > > > > > I think (1) means that we have an infinite sequence of sets S_k where > > S_k > > > > is composed of k wedges (joined only along full edges), each with > angle > > > > 2*pi/k; the limit is just the set of points p such that p is > contained > > in > > > > all but finitely many S_k. You can definitely get an interesting > > > collection > > > > of shapes this way. > > > > > > > > On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < dasimov@earthlink.net> > > > > wrote: > > > > > > > > > I'm trying to guess what RWG meant without peeking at his drawings. > > > > > > > > > > In order to make Jim Propp's statement exact, I would have to make > > > > precise > > > > > > > > > > 1) what "dissect and reassemble" mean > > > > > > > > > > and > > > > > > > > > > 2) what "converges" to a 1-by-pi rectangle means. > > > > > > > > > > A typical meaning for 1): For subsets A, B of R^2, to dissect A and > > > > > reassemble it > > > > > to B means that there is a partition > > > > > > > > > > A = X_1 + ... + X_n > > > > > > > > > > of A as a finite disjoint union, such that there exist isometries > > > > > > > > > > f_1, ..., f_n of R^2 > > > > > > > > > > such that > > > > > > > > > > B = f_1(X_1) + ... + f_n(X_n) > > > > > > > > > > forms a partition of B as a finite disjoint union. > > > > > > > > > > * *
> > > > > > > > > > One meaning for 2) could be in the sense of Hausdorff distance > > between > > > > > compact sets > > > > > in the plane. The only problem I see here is that if strict > partition > > > are > > > > > used in > > > > > 1) as above, then the resulting rectangle B will not be compact, as > > it > > > > > will not contain > > > > > all of its boundary. I have complete faith that appropriate > > hand-waving > > > > > will not incur > > > > > the wrath of the math gods. > > > > > > > > > > —Dan > > > > > > > > > > > > > > > ----- > > > > > Jim Propp wrote: > > > > > > If you dissect a unit disk radially into a large number of equal > > > > wedges, > > > > > > it’s well known that you can reassemble them to form a shape that > > in > > > > the > > > > > > limit converges to a 1-by-pi rectangle. > > > > > > > > > > > > > > > > RWG wrote: > > > > > ----- > > > > > gosper.org/picfzoom.gif > > > > > gosper.org/semizoom.gif > > > > > --rwg > > > > > I don't see how to get anything other than allowing unequal wedges. > > > > > ----- > > > > > ----- > > > > > > > > > > _______________________________________________ > > > > > math-fun mailing list > > > > > math-fun@mailman.xmission.com > > > > > https://mailman.xmission.com/c gi-bin/mailman/listinfo/math-fun > > > > > > > > > _______________________________________________ > > > > math-fun mailing list > > > > math-fun@mailman.xmission.com > > > > https://mailman.xmission.com/ cgi-bin/mailman/listinfo/math- f un > > > > > > > _______________________________________________ > > > 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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There are two fun problems here. One is Allan’s question about the maximum overlap when a full disk is dissected. Another is, what is the critical angle theta such that, for all sectors exceeding theta, positive overlap can be achieved by some dissection/recomposition? Is it 3 Pi / 2? Jim Propp On Sunday, April 22, 2018, Tomas Rokicki <rokicki@gmail.com> wrote: > Render three-quarters of a circle, using 3Pi/2 edges. Extend each flat > edge > straight out by Pi/8; this gives an overlap of Pi/8 x Pi/8. > > This can almost certainly be improved by extending by one eighth of a > circle using the other point as a radius, rather than extending straight > out, > but I am too lazy to do the math to determine if this is actually an > improvement. > > This is almost certainly not the best one can do and I am interested in > Allan's question about what the maximum possible overlap might be. > > -tom > > On Sat, Apr 21, 2018 at 3:03 AM, James Propp <jamespropp@gmail.com> wrote: > > > A picture, please? > > > > Jim > > > > On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote: > > > > > Oh, wait. I see it. Tomas is right, of course. I wonder what the > maximum > > > self-overlap is. I can see how to get pi^2 / 64. > > > > > > On Fri, Apr 20, 2018 at 10:18 PM, Allan Wechsler <acwacw@gmail.com> > > wrote: > > > > > > > Really? You only have 2pi time units to do it in. Also ... I want a > > clean > > > > intersection: the overlap has to have area. > > > > > > > > Pics or it didn't happen. > > > > > > > > On Fri, Apr 20, 2018 at 10:12 PM, Tomas Rokicki <rokicki@gmail.com> > > > wrote: > > > > > > > >> The answer is yes. > > > >> > > > >> On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <acwacw@gmail.com> > > > wrote: > > > >> > > > >> > Guys! Guys! Here is a question! You'll see, it will end with a > > > question > > > >> > mark and everything. > > > >> > > > > >> > Can such a ribbon self-intersect? > > > >> > > > > >> > I'm guessing the answer is no, but I can't see a proof path. > > > >> > > > > >> > On Fri, Apr 20, 2018 at 1:28 PM, James Propp < > jamespropp@gmail.com> > > > >> wrote: > > > >> > > > > >> > > Perfect! Thanks. > > > >> > > > > > >> > > I guess the theorem here is that the area of such a ribbon is > > equal > > > to > > > >> > half > > > >> > > the sum of the lengths of the two non-straight sides. > > > >> > > > > > >> > > Come to think of it, this is just a consequence of > > what’s-his-name’s > > > >> > > theorem about the area swept out by a line segment that moves > > > >> > perpendicular > > > >> > > to itself. The only nonobvious step is relating the distance > > > traveled > > > >> by > > > >> > > the midpoint of the segment to the distances traveled by the > > > >> endpoints of > > > >> > > the segment. > > > >> > > > > > >> > > Jim > > > >> > > > > > >> > > On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> > > wrote: > > > >> > > > > > >> > > > Maybe what you are looking for is this. The "ribbon" has two > > > >> > curvilinear > > > >> > > > edges. From any point A on one edge, draw a perpendicular > line; > > it > > > >> will > > > >> > > > turn out to be perpendicular to the other edge as well. (By > > > >> > > "perpendicular" > > > >> > > > I mean "perpendicular to the tangent at that point".) > > > >> > > > > > > >> > > > On Fri, Apr 20, 2018 at 12:52 PM, James Propp < > > > jamespropp@gmail.com > > > >> > > > > >> > > > wrote: > > > >> > > > > > > >> > > > > Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to > what I > > > >> > wanted. > > > >> > > > But > > > >> > > > > it requires a time-parametrization. Is there a way to > > > characterize > > > >> > such > > > >> > > > > shapes directly? > > > >> > > > > > > > >> > > > > Jim > > > >> > > > > > > > >> > > > > On Thursday, April 19, 2018, Allan Wechsler < > acwacw@gmail.com > > > > > > >> > wrote: > > > >> > > > > > > > >> > > > > > It seems to me that, in the limit, we have a behavior > > > something > > > >> > like > > > >> > > > > this: > > > >> > > > > > > > > >> > > > > > We have a unit line segment AB moving in the plane. Each > of > > > its > > > >> > > > endpoints > > > >> > > > > > is moving perpendicular to the line, toward the same side > of > > > the > > > >> > > line, > > > >> > > > at > > > >> > > > > > speeds that add up to 1. Subject to that constraint, their > > > >> speeds > > > >> > are > > > >> > > > an > > > >> > > > > > arbitrary function of time. Say the speed of point A is > > given > > > by > > > >> > > f(t); > > > >> > > > > then > > > >> > > > > > point B is moving in the same direction at speed 1-f(t). > > > Because > > > >> > the > > > >> > > > > speeds > > > >> > > > > > of the endpoints can differ, the line can gradually change > > > >> > > orientation; > > > >> > > > > its > > > >> > > > > > angle (in radians) is changing at a speed 1/2 - f(t). It > > > sweeps > > > >> out > > > >> > > > area > > > >> > > > > at > > > >> > > > > > a constant speed of 1/2. The curvatures of the curves > traced > > > out > > > >> > by A > > > >> > > > > and B > > > >> > > > > > are related by the equation 1/a + 1/b = 1. The whole > process > > > >> > > continues > > > >> > > > > > until t = 2pi, so the total area swept out is pi. > > > >> > > > > > > > > >> > > > > > On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < > > > >> > > > mjcollins10@gmail.com> > > > >> > > > > > wrote: > > > >> > > > > > > > > >> > > > > > > I think (1) means that we have an infinite sequence of > > sets > > > >> S_k > > > >> > > where > > > >> > > > > S_k > > > >> > > > > > > is composed of k wedges (joined only along full edges), > > each > > > >> with > > > >> > > > angle > > > >> > > > > > > 2*pi/k; the limit is just the set of points p such that > p > > is > > > >> > > > contained > > > >> > > > > in > > > >> > > > > > > all but finitely many S_k. You can definitely get an > > > >> interesting > > > >> > > > > > collection > > > >> > > > > > > of shapes this way. > > > >> > > > > > > > > > >> > > > > > > On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < > > > >> > > dasimov@earthlink.net> > > > >> > > > > > > wrote: > > > >> > > > > > > > > > >> > > > > > > > I'm trying to guess what RWG meant without peeking at > > his > > > >> > > drawings. > > > >> > > > > > > > > > > >> > > > > > > > In order to make Jim Propp's statement exact, I would > > have > > > >> to > > > >> > > make > > > >> > > > > > > precise > > > >> > > > > > > > > > > >> > > > > > > > 1) what "dissect and reassemble" mean > > > >> > > > > > > > > > > >> > > > > > > > and > > > >> > > > > > > > > > > >> > > > > > > > 2) what "converges" to a 1-by-pi rectangle means. > > > >> > > > > > > > > > > >> > > > > > > > A typical meaning for 1): For subsets A, B of R^2, to > > > >> dissect A > > > >> > > and > > > >> > > > > > > > reassemble it > > > >> > > > > > > > to B means that there is a partition > > > >> > > > > > > > > > > >> > > > > > > > A = X_1 + ... + X_n > > > >> > > > > > > > > > > >> > > > > > > > of A as a finite disjoint union, such that there exist > > > >> > isometries > > > >> > > > > > > > > > > >> > > > > > > > f_1, ..., f_n of R^2 > > > >> > > > > > > > > > > >> > > > > > > > such that > > > >> > > > > > > > > > > >> > > > > > > > B = f_1(X_1) + ... + f_n(X_n) > > > >> > > > > > > > > > > >> > > > > > > > forms a partition of B as a finite disjoint union. > > > >> > > > > > > > > > > >> > > > > > > > * * > > > >> * > > > >> > > > > > > > > > > >> > > > > > > > One meaning for 2) could be in the sense of Hausdorff > > > >> distance > > > >> > > > > between > > > >> > > > > > > > compact sets > > > >> > > > > > > > in the plane. The only problem I see here is that if > > > strict > > > >> > > > partition > > > >> > > > > > are > > > >> > > > > > > > used in > > > >> > > > > > > > 1) as above, then the resulting rectangle B will not > be > > > >> > compact, > > > >> > > as > > > >> > > > > it > > > >> > > > > > > > will not contain > > > >> > > > > > > > all of its boundary. I have complete faith that > > > appropriate > > > >> > > > > hand-waving > > > >> > > > > > > > will not incur > > > >> > > > > > > > the wrath of the math gods. > > > >> > > > > > > > > > > >> > > > > > > > —Dan > > > >> > > > > > > > > > > >> > > > > > > > > > > >> > > > > > > > ----- > > > >> > > > > > > > Jim Propp wrote: > > > >> > > > > > > > > If you dissect a unit disk radially into a large > > number > > > of > > > >> > > equal > > > >> > > > > > > wedges, > > > >> > > > > > > > > it’s well known that you can reassemble them to > form a > > > >> shape > > > >> > > that > > > >> > > > > in > > > >> > > > > > > the > > > >> > > > > > > > > limit converges to a 1-by-pi rectangle. > > > >> > > > > > > > > > > > >> > > > > > > > > > > >> > > > > > > > RWG wrote: > > > >> > > > > > > > ----- > > > >> > > > > > > > gosper.org/picfzoom.gif > > > >> > > > > > > > gosper.org/semizoom.gif > > > >> > > > > > > > --rwg > > > >> > > > > > > > I don't see how to get anything other than allowing > > > unequal > > > >> > > wedges. > > > >> > > > > > > > ----- > > > >> > > > > > > > ----- > > > >> > > > > > > > > > > >> > > > > > > > _______________________________________________ > > > >> > > > > > > > math-fun mailing list > > > >> > > > > > > > math-fun@mailman.xmission.com > > > >> > > > > > > > https://mailman.xmission.com/c > > > >> gi-bin/mailman/listinfo/math-fun > > > >> > > > > > > > > > > >> > > > > > > _______________________________________________ > > > >> > > > > > > math-fun mailing list > > > >> > > > > > > math-fun@mailman.xmission.com > > > >> > > > > > > https://mailman.xmission.com/ > > cgi-bin/mailman/listinfo/math- > > > f > > > >> un > > > >> > > > > > > > > > >> > > > > > _______________________________________________ > > > >> > > > > > 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 > > > >> > > > > >> -- > > > >> -- http://cube20.org/ -- http://golly.sf.net/ -- > > > >> _______________________________________________ > > > >> 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 > > > > > > -- > -- http://cube20.org/ -- http://golly.sf.net/ -- > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
Replacing 3 Pi/2 in my argument with Pi+epsilon still gives overlap, so the bound is at most Pi+epsilon for arbitrarily small epsilon. I don't think Pi by itself is sufficient and I think this can be shown formally. On Sun, Apr 22, 2018 at 4:36 PM, James Propp <jamespropp@gmail.com> wrote:
There are two fun problems here. One is Allan’s question about the maximum overlap when a full disk is dissected. Another is, what is the critical angle theta such that, for all sectors exceeding theta, positive overlap can be achieved by some dissection/recomposition? Is it 3 Pi / 2?
Jim Propp
On Sunday, April 22, 2018, Tomas Rokicki <rokicki@gmail.com> wrote:
Render three-quarters of a circle, using 3Pi/2 edges. Extend each flat edge straight out by Pi/8; this gives an overlap of Pi/8 x Pi/8.
This can almost certainly be improved by extending by one eighth of a circle using the other point as a radius, rather than extending straight out, but I am too lazy to do the math to determine if this is actually an improvement.
This is almost certainly not the best one can do and I am interested in Allan's question about what the maximum possible overlap might be.
-tom
On Sat, Apr 21, 2018 at 3:03 AM, James Propp <jamespropp@gmail.com> wrote:
A picture, please?
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Oh, wait. I see it. Tomas is right, of course. I wonder what the maximum self-overlap is. I can see how to get pi^2 / 64.
On Fri, Apr 20, 2018 at 10:18 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Really? You only have 2pi time units to do it in. Also ... I want a clean intersection: the overlap has to have area.
Pics or it didn't happen.
On Fri, Apr 20, 2018 at 10:12 PM, Tomas Rokicki <rokicki@gmail.com
wrote:
The answer is yes.
On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <acwacw@gmail.com>
wrote:
> Guys! Guys! Here is a question! You'll see, it will end with a
question
> mark and everything. > > Can such a ribbon self-intersect? > > I'm guessing the answer is no, but I can't see a proof path. > > On Fri, Apr 20, 2018 at 1:28 PM, James Propp < jamespropp@gmail.com> wrote: > > > Perfect! Thanks. > > > > I guess the theorem here is that the area of such a ribbon is equal to > half > > the sum of the lengths of the two non-straight sides. > > > > Come to think of it, this is just a consequence of what’s-his-name’s > > theorem about the area swept out by a line segment that moves > perpendicular > > to itself. The only nonobvious step is relating the distance traveled by > > the midpoint of the segment to the distances traveled by the endpoints of > > the segment. > > > > Jim > > > > On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote: > > > > > Maybe what you are looking for is this. The "ribbon" has two > curvilinear > > > edges. From any point A on one edge, draw a perpendicular line; it will > > > turn out to be perpendicular to the other edge as well. (By > > "perpendicular" > > > I mean "perpendicular to the tangent at that point".) > > > > > > On Fri, Apr 20, 2018 at 12:52 PM, James Propp < jamespropp@gmail.com > > > > wrote: > > > > > > > Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I > wanted. > > > But > > > > it requires a time-parametrization. Is there a way to characterize > such > > > > shapes directly? > > > > > > > > Jim > > > > > > > > On Thursday, April 19, 2018, Allan Wechsler < acwacw@gmail.com
> wrote: > > > > > > > > > It seems to me that, in the limit, we have a behavior something > like > > > > this: > > > > > > > > > > We have a unit line segment AB moving in the plane. Each of its > > > endpoints > > > > > is moving perpendicular to the line, toward the same side of the > > line, > > > at > > > > > speeds that add up to 1. Subject to that constraint, their speeds > are > > > an > > > > > arbitrary function of time. Say the speed of point A is given by > > f(t); > > > > then > > > > > point B is moving in the same direction at speed 1-f(t). Because > the > > > > speeds > > > > > of the endpoints can differ, the line can gradually change > > orientation; > > > > its > > > > > angle (in radians) is changing at a speed 1/2 - f(t). It sweeps out > > > area > > > > at > > > > > a constant speed of 1/2. The curvatures of the curves traced out > by A > > > > and B > > > > > are related by the equation 1/a + 1/b = 1. The whole process > > continues > > > > > until t = 2pi, so the total area swept out is pi. > > > > > > > > > > On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < > > > mjcollins10@gmail.com> > > > > > wrote: > > > > > > > > > > > I think (1) means that we have an infinite sequence of sets S_k > > where > > > > S_k > > > > > > is composed of k wedges (joined only along full edges), each with > > > angle > > > > > > 2*pi/k; the limit is just the set of points p such that p is > > > contained > > > > in > > > > > > all but finitely many S_k. You can definitely get an interesting > > > > > collection > > > > > > of shapes this way. > > > > > > > > > > > > On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < > > dasimov@earthlink.net> > > > > > > wrote: > > > > > > > > > > > > > I'm trying to guess what RWG meant without peeking at his > > drawings. > > > > > > > > > > > > > > In order to make Jim Propp's statement exact, I would have to > > make > > > > > > precise > > > > > > > > > > > > > > 1) what "dissect and reassemble" mean > > > > > > > > > > > > > > and > > > > > > > > > > > > > > 2) what "converges" to a 1-by-pi rectangle means. > > > > > > > > > > > > > > A typical meaning for 1): For subsets A, B of R^2, to dissect A > > and > > > > > > > reassemble it > > > > > > > to B means that there is a partition > > > > > > > > > > > > > > A = X_1 + ... + X_n > > > > > > > > > > > > > > of A as a finite disjoint union, such that there exist > isometries > > > > > > > > > > > > > > f_1, ..., f_n of R^2 > > > > > > > > > > > > > > such that > > > > > > > > > > > > > > B = f_1(X_1) + ... + f_n(X_n) > > > > > > > > > > > > > > forms a partition of B as a finite disjoint union. > > > > > > > > > > > > > > * * * > > > > > > > > > > > > > > One meaning for 2) could be in the sense of Hausdorff distance > > > > between > > > > > > > compact sets > > > > > > > in the plane. The only problem I see here is that if strict > > > partition > > > > > are > > > > > > > used in > > > > > > > 1) as above, then the resulting rectangle B will not be > compact, > > as > > > > it > > > > > > > will not contain > > > > > > > all of its boundary. I have complete faith that appropriate > > > > hand-waving > > > > > > > will not incur > > > > > > > the wrath of the math gods. > > > > > > > > > > > > > > —Dan > > > > > > > > > > > > > > > > > > > > > ----- > > > > > > > Jim Propp wrote: > > > > > > > > If you dissect a unit disk radially into a large number of > > equal > > > > > > wedges, > > > > > > > > it’s well known that you can reassemble them to form a shape > > that > > > > in > > > > > > the > > > > > > > > limit converges to a 1-by-pi rectangle. > > > > > > > > > > > > > > > > > > > > > > RWG wrote: > > > > > > > ----- > > > > > > > gosper.org/picfzoom.gif > > > > > > > gosper.org/semizoom.gif > > > > > > > --rwg > > > > > > > I don't see how to get anything other than allowing unequal > > wedges. > > > > > > > ----- > > > > > > > ----- > > > > > > > > > > > > > > _______________________________________________ > > > > > > > math-fun mailing list > > > > > > > math-fun@mailman.xmission.com > > > > > > > https://mailman.xmission.com/c gi-bin/mailman/listinfo/math-fun > > > > > > > > > > > > > _______________________________________________ > > > > > > math-fun mailing list > > > > > > math-fun@mailman.xmission.com > > > > > > https://mailman.xmission.com/ cgi-bin/mailman/listinfo/math- f un > > > > > > > > > > > _______________________________________________ > > > > > 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 > -- -- http://cube20.org/ -- http://golly.sf.net/ -- _______________________________________________ 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 not sure I understand the question about the critical angle. Certainly you can get some overlap even if there are no contiguous sectors (hard right or hard left turn). But perhaps you are talking about total "swerve", essentially the difference in arc length between the two sides. My intuition is that unless the swerve exceeds pi, you can't get any overlap. On Sun, Apr 22, 2018 at 12:39 PM, Tomas Rokicki <rokicki@gmail.com> wrote:
Replacing 3 Pi/2 in my argument with Pi+epsilon still gives overlap, so the bound is at most Pi+epsilon for arbitrarily small epsilon.
I don't think Pi by itself is sufficient and I think this can be shown formally.
On Sun, Apr 22, 2018 at 4:36 PM, James Propp <jamespropp@gmail.com> wrote:
There are two fun problems here. One is Allan’s question about the maximum overlap when a full disk is dissected. Another is, what is the critical angle theta such that, for all sectors exceeding theta, positive overlap can be achieved by some dissection/recomposition? Is it 3 Pi / 2?
Jim Propp
On Sunday, April 22, 2018, Tomas Rokicki <rokicki@gmail.com> wrote:
Render three-quarters of a circle, using 3Pi/2 edges. Extend each flat edge straight out by Pi/8; this gives an overlap of Pi/8 x Pi/8.
This can almost certainly be improved by extending by one eighth of a circle using the other point as a radius, rather than extending straight out, but I am too lazy to do the math to determine if this is actually an improvement.
This is almost certainly not the best one can do and I am interested in Allan's question about what the maximum possible overlap might be.
-tom
On Sat, Apr 21, 2018 at 3:03 AM, James Propp <jamespropp@gmail.com> wrote:
A picture, please?
Jim
On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com> wrote:
Oh, wait. I see it. Tomas is right, of course. I wonder what the maximum self-overlap is. I can see how to get pi^2 / 64.
On Fri, Apr 20, 2018 at 10:18 PM, Allan Wechsler <acwacw@gmail.com
wrote:
Really? You only have 2pi time units to do it in. Also ... I
want a clean
intersection: the overlap has to have area.
Pics or it didn't happen.
On Fri, Apr 20, 2018 at 10:12 PM, Tomas Rokicki < rokicki@gmail.com
wrote:
> The answer is yes. > > On Fri, Apr 20, 2018 at 8:21 PM Allan Wechsler <
acwacw@gmail.com> wrote:
> > > Guys! Guys! Here is a question! You'll see, it will end with a question > > mark and everything. > > > > Can such a ribbon self-intersect? > > > > I'm guessing the answer is no, but I can't see a proof path. > > > > On Fri, Apr 20, 2018 at 1:28 PM, James Propp < jamespropp@gmail.com> > wrote: > > > > > Perfect! Thanks. > > > > > > I guess the theorem here is that the area of such a ribbon is equal to > > half > > > the sum of the lengths of the two non-straight sides. > > > > > > Come to think of it, this is just a consequence of what’s-his-name’s > > > theorem about the area swept out by a line segment that moves > > perpendicular > > > to itself. The only nonobvious step is relating the distance traveled > by > > > the midpoint of the segment to the distances traveled by the > endpoints of > > > the segment. > > > > > > Jim > > > > > > On Friday, April 20, 2018, Allan Wechsler <acwacw@gmail.com
wrote:
> > > > > > > Maybe what you are looking for is this. The "ribbon" has two > > curvilinear > > > > edges. From any point A on one edge, draw a perpendicular line; it > will > > > > turn out to be perpendicular to the other edge as well. (By > > > "perpendicular" > > > > I mean "perpendicular to the tangent at that point".) > > > > > > > > On Fri, Apr 20, 2018 at 12:52 PM, James Propp < jamespropp@gmail.com > > > > > > wrote: > > > > > > > > > Thanks, Allan! The relation 1/a(t) + 1/b(t) is close to what I > > wanted. > > > > But > > > > > it requires a time-parametrization. Is there a way to characterize > > such > > > > > shapes directly? > > > > > > > > > > Jim > > > > > > > > > > On Thursday, April 19, 2018, Allan Wechsler < acwacw@gmail.com
> > wrote: > > > > > > > > > > > It seems to me that, in the limit, we have a behavior something > > like > > > > > this: > > > > > > > > > > > > We have a unit line segment AB moving in the plane. Each of its > > > > endpoints > > > > > > is moving perpendicular to the line, toward the same side of the > > > line, > > > > at > > > > > > speeds that add up to 1. Subject to that constraint, their > speeds > > are > > > > an > > > > > > arbitrary function of time. Say the speed of point A is given by > > > f(t); > > > > > then > > > > > > point B is moving in the same direction at speed 1-f(t). Because > > the > > > > > speeds > > > > > > of the endpoints can differ, the line can gradually change > > > orientation; > > > > > its > > > > > > angle (in radians) is changing at a speed 1/2 - f(t). It sweeps > out > > > > area > > > > > at > > > > > > a constant speed of 1/2. The curvatures of the curves traced out > > by A > > > > > and B > > > > > > are related by the equation 1/a + 1/b = 1. The whole process > > > continues > > > > > > until t = 2pi, so the total area swept out is pi. > > > > > > > > > > > > On Thu, Apr 19, 2018 at 9:26 AM, Michael Collins < > > > > mjcollins10@gmail.com> > > > > > > wrote: > > > > > > > > > > > > > I think (1) means that we have an infinite sequence of sets > S_k > > > where > > > > > S_k > > > > > > > is composed of k wedges (joined only along full edges), each > with > > > > angle > > > > > > > 2*pi/k; the limit is just the set of points p such that p is > > > > contained > > > > > in > > > > > > > all but finitely many S_k. You can definitely get an > interesting > > > > > > collection > > > > > > > of shapes this way. > > > > > > > > > > > > > > On Wed, Apr 18, 2018 at 10:21 PM, Dan Asimov < > > > dasimov@earthlink.net> > > > > > > > wrote: > > > > > > > > > > > > > > > I'm trying to guess what RWG meant without peeking at his > > > drawings. > > > > > > > > > > > > > > > > In order to make Jim Propp's statement exact, I would have > to > > > make > > > > > > > precise > > > > > > > > > > > > > > > > 1) what "dissect and reassemble" mean > > > > > > > > > > > > > > > > and > > > > > > > > > > > > > > > > 2) what "converges" to a 1-by-pi rectangle means. > > > > > > > > > > > > > > > > A typical meaning for 1): For subsets A, B of R^2, to > dissect A > > > and > > > > > > > > reassemble it > > > > > > > > to B means that there is a partition > > > > > > > > > > > > > > > > A = X_1 + ... + X_n > > > > > > > > > > > > > > > > of A as a finite disjoint union, such that there exist > > isometries > > > > > > > > > > > > > > > > f_1, ..., f_n of R^2 > > > > > > > > > > > > > > > > such that > > > > > > > > > > > > > > > > B = f_1(X_1) + ... + f_n(X_n) > > > > > > > > > > > > > > > > forms a partition of B as a finite disjoint union. > > > > > > > > > > > > > > > > * * > * > > > > > > > > > > > > > > > > One meaning for 2) could be in the sense of Hausdorff > distance > > > > > between > > > > > > > > compact sets > > > > > > > > in the plane. The only problem I see here is that if strict > > > > partition > > > > > > are > > > > > > > > used in > > > > > > > > 1) as above, then the resulting rectangle B will not be > > compact, > > > as > > > > > it > > > > > > > > will not contain > > > > > > > > all of its boundary. I have complete faith that appropriate > > > > > hand-waving > > > > > > > > will not incur > > > > > > > > the wrath of the math gods. > > > > > > > > > > > > > > > > —Dan > > > > > > > > > > > > > > > > > > > > > > > > ----- > > > > > > > > Jim Propp wrote: > > > > > > > > > If you dissect a unit disk radially into a large number of > > > equal > > > > > > > wedges, > > > > > > > > > it’s well known that you can reassemble them to form a > shape > > > that > > > > > in > > > > > > > the > > > > > > > > > limit converges to a 1-by-pi rectangle. > > > > > > > > > > > > > > > > > > > > > > > > > RWG wrote: > > > > > > > > ----- > > > > > > > > gosper.org/picfzoom.gif > > > > > > > > gosper.org/semizoom.gif > > > > > > > > --rwg > > > > > > > > I don't see how to get anything other than allowing unequal > > > wedges. > > > > > > > > ----- > > > > > > > > ----- > > > > > > > > > > > > > > > > _______________________________________________ > > > > > > > > math-fun mailing list > > > > > > > > math-fun@mailman.xmission.com > > > > > > > > https://mailman.xmission.com/c > gi-bin/mailman/listinfo/math-fun > > > > > > > > > > > > > > > _______________________________________________ > > > > > > > math-fun mailing list > > > > > > > math-fun@mailman.xmission.com > > > > > > > https://mailman.xmission.com/ cgi-bin/mailman/listinfo/math- f > un > > > > > > > > > > > > > _______________________________________________ > > > > > > 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 > > > -- > -- http://cube20.org/ -- http://golly.sf.net/ -- > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >
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participants (5)
-
Allan Wechsler -
Dan Asimov -
James Propp -
Michael Collins -
Tomas Rokicki