Smooth, approximately space-filling curves in high dimensions The Next CEO of Stack OverflowSpace-filling sineHow are asymptotes actually defined in rigorous mathematics?Dense simple smooth immersed curves in manifolds.Space filling curves: initial definitionsLine integrals along space filling curvesFormulas for space curvesIs the restriction (to lower dimensions) of a smooth function still smooth?Space filling curve's intersections with closed jordan curvesSpace filling curves: Hilbert vs PeanoSmooth curves and velocity
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Smooth, approximately space-filling curves in high dimensions
The Next CEO of Stack OverflowSpace-filling sineHow are asymptotes actually defined in rigorous mathematics?Dense simple smooth immersed curves in manifolds.Space filling curves: initial definitionsLine integrals along space filling curvesFormulas for space curvesIs the restriction (to lower dimensions) of a smooth function still smooth?Space filling curve's intersections with closed jordan curvesSpace filling curves: Hilbert vs PeanoSmooth curves and velocity
$begingroup$
I'm looking for smooth (infinitely differentiable everywhere) functions (curves) $mathbbRrightarrowmathbbR^d$ that are approximately space-filling, i.e. scaling allows the curve to eventually get arbitrarily close to all points in $mathbbR^d$.
An intuitive example for $mathbbRrightarrowmathbbR^2$ would be the Archimedean Spiral, e.g.:
Example function:
$$
mathrmf(t)=
rhocdot
beginpmatrix
cos(t) cdot t \
sin(t) cdot t
endpmatrix
$$
As $rho$ approaches zero, the spiral will eventually get arbitrarily close to every point in $mathbbR^2$.
It would also be great if the computational complexity of calculating such a function only increases linearly with the dimension $d$.
differential-geometry analytic-geometry curves smooth-functions
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
$endgroup$
add a comment |
$begingroup$
I'm looking for smooth (infinitely differentiable everywhere) functions (curves) $mathbbRrightarrowmathbbR^d$ that are approximately space-filling, i.e. scaling allows the curve to eventually get arbitrarily close to all points in $mathbbR^d$.
An intuitive example for $mathbbRrightarrowmathbbR^2$ would be the Archimedean Spiral, e.g.:
Example function:
$$
mathrmf(t)=
rhocdot
beginpmatrix
cos(t) cdot t \
sin(t) cdot t
endpmatrix
$$
As $rho$ approaches zero, the spiral will eventually get arbitrarily close to every point in $mathbbR^2$.
It would also be great if the computational complexity of calculating such a function only increases linearly with the dimension $d$.
differential-geometry analytic-geometry curves smooth-functions
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
$endgroup$
$begingroup$
I have examples that are continuous but not smooth, based on covering size-$2n-1$ hypercubes such that the end point is on the hypercube surface, and then stepping outward and covering the surface of the size-$2n+1$ hypercube. The smoothness requirement makes this a very intriguing question.
$endgroup$
– Mark Fischler
Mar 18 at 21:57
$begingroup$
Nice question. Is self-intersection allowed? Also do you require the convergence to be "uniform"?
$endgroup$
– Rob Arthan
Mar 18 at 22:02
$begingroup$
@RobArthan Though self-intersection would not pose a technical problem for my application, it would seem preferable if there were no redundancies in the way the curve explores the space.
$endgroup$
– Markus Mottl
Mar 18 at 22:30
$begingroup$
I see. How about the convergence? I don't think the Spiral of Archimedes gives uniform convergence.
$endgroup$
– Rob Arthan
Mar 19 at 0:01
$begingroup$
@RobArthan I don't make any assumptions about limits of $mathrmf$. E.g. a point in $mathbbR^d$ may be equally distant to different parts of the curve, or a limit for $t$ that gives you the closest point along the curve may not exist as you increase the density of the curve. All I need is that the curve can at least in principle be made to wiggle through space so densely that at least one point anywhere along the curve will come arbitrarily close to any point in space.
$endgroup$
– Markus Mottl
Mar 19 at 2:01
add a comment |
$begingroup$
I'm looking for smooth (infinitely differentiable everywhere) functions (curves) $mathbbRrightarrowmathbbR^d$ that are approximately space-filling, i.e. scaling allows the curve to eventually get arbitrarily close to all points in $mathbbR^d$.
An intuitive example for $mathbbRrightarrowmathbbR^2$ would be the Archimedean Spiral, e.g.:
Example function:
$$
mathrmf(t)=
rhocdot
beginpmatrix
cos(t) cdot t \
sin(t) cdot t
endpmatrix
$$
As $rho$ approaches zero, the spiral will eventually get arbitrarily close to every point in $mathbbR^2$.
It would also be great if the computational complexity of calculating such a function only increases linearly with the dimension $d$.
differential-geometry analytic-geometry curves smooth-functions
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
$endgroup$
I'm looking for smooth (infinitely differentiable everywhere) functions (curves) $mathbbRrightarrowmathbbR^d$ that are approximately space-filling, i.e. scaling allows the curve to eventually get arbitrarily close to all points in $mathbbR^d$.
An intuitive example for $mathbbRrightarrowmathbbR^2$ would be the Archimedean Spiral, e.g.:
Example function:
$$
mathrmf(t)=
rhocdot
beginpmatrix
cos(t) cdot t \
sin(t) cdot t
endpmatrix
$$
As $rho$ approaches zero, the spiral will eventually get arbitrarily close to every point in $mathbbR^2$.
It would also be great if the computational complexity of calculating such a function only increases linearly with the dimension $d$.
differential-geometry analytic-geometry curves smooth-functions
differential-geometry analytic-geometry curves smooth-functions
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
edited Mar 19 at 16:28
Markus Mottl
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
asked Mar 18 at 21:46
Markus MottlMarkus Mottl
1213
1213
$begingroup$
I have examples that are continuous but not smooth, based on covering size-$2n-1$ hypercubes such that the end point is on the hypercube surface, and then stepping outward and covering the surface of the size-$2n+1$ hypercube. The smoothness requirement makes this a very intriguing question.
$endgroup$
– Mark Fischler
Mar 18 at 21:57
$begingroup$
Nice question. Is self-intersection allowed? Also do you require the convergence to be "uniform"?
$endgroup$
– Rob Arthan
Mar 18 at 22:02
$begingroup$
@RobArthan Though self-intersection would not pose a technical problem for my application, it would seem preferable if there were no redundancies in the way the curve explores the space.
$endgroup$
– Markus Mottl
Mar 18 at 22:30
$begingroup$
I see. How about the convergence? I don't think the Spiral of Archimedes gives uniform convergence.
$endgroup$
– Rob Arthan
Mar 19 at 0:01
$begingroup$
@RobArthan I don't make any assumptions about limits of $mathrmf$. E.g. a point in $mathbbR^d$ may be equally distant to different parts of the curve, or a limit for $t$ that gives you the closest point along the curve may not exist as you increase the density of the curve. All I need is that the curve can at least in principle be made to wiggle through space so densely that at least one point anywhere along the curve will come arbitrarily close to any point in space.
$endgroup$
– Markus Mottl
Mar 19 at 2:01
add a comment |
$begingroup$
I have examples that are continuous but not smooth, based on covering size-$2n-1$ hypercubes such that the end point is on the hypercube surface, and then stepping outward and covering the surface of the size-$2n+1$ hypercube. The smoothness requirement makes this a very intriguing question.
$endgroup$
– Mark Fischler
Mar 18 at 21:57
$begingroup$
Nice question. Is self-intersection allowed? Also do you require the convergence to be "uniform"?
$endgroup$
– Rob Arthan
Mar 18 at 22:02
$begingroup$
@RobArthan Though self-intersection would not pose a technical problem for my application, it would seem preferable if there were no redundancies in the way the curve explores the space.
$endgroup$
– Markus Mottl
Mar 18 at 22:30
$begingroup$
I see. How about the convergence? I don't think the Spiral of Archimedes gives uniform convergence.
$endgroup$
– Rob Arthan
Mar 19 at 0:01
$begingroup$
@RobArthan I don't make any assumptions about limits of $mathrmf$. E.g. a point in $mathbbR^d$ may be equally distant to different parts of the curve, or a limit for $t$ that gives you the closest point along the curve may not exist as you increase the density of the curve. All I need is that the curve can at least in principle be made to wiggle through space so densely that at least one point anywhere along the curve will come arbitrarily close to any point in space.
$endgroup$
– Markus Mottl
Mar 19 at 2:01
$begingroup$
I have examples that are continuous but not smooth, based on covering size-$2n-1$ hypercubes such that the end point is on the hypercube surface, and then stepping outward and covering the surface of the size-$2n+1$ hypercube. The smoothness requirement makes this a very intriguing question.
$endgroup$
– Mark Fischler
Mar 18 at 21:57
$begingroup$
I have examples that are continuous but not smooth, based on covering size-$2n-1$ hypercubes such that the end point is on the hypercube surface, and then stepping outward and covering the surface of the size-$2n+1$ hypercube. The smoothness requirement makes this a very intriguing question.
$endgroup$
– Mark Fischler
Mar 18 at 21:57
$begingroup$
Nice question. Is self-intersection allowed? Also do you require the convergence to be "uniform"?
$endgroup$
– Rob Arthan
Mar 18 at 22:02
$begingroup$
Nice question. Is self-intersection allowed? Also do you require the convergence to be "uniform"?
$endgroup$
– Rob Arthan
Mar 18 at 22:02
$begingroup$
@RobArthan Though self-intersection would not pose a technical problem for my application, it would seem preferable if there were no redundancies in the way the curve explores the space.
$endgroup$
– Markus Mottl
Mar 18 at 22:30
$begingroup$
@RobArthan Though self-intersection would not pose a technical problem for my application, it would seem preferable if there were no redundancies in the way the curve explores the space.
$endgroup$
– Markus Mottl
Mar 18 at 22:30
$begingroup$
I see. How about the convergence? I don't think the Spiral of Archimedes gives uniform convergence.
$endgroup$
– Rob Arthan
Mar 19 at 0:01
$begingroup$
I see. How about the convergence? I don't think the Spiral of Archimedes gives uniform convergence.
$endgroup$
– Rob Arthan
Mar 19 at 0:01
$begingroup$
@RobArthan I don't make any assumptions about limits of $mathrmf$. E.g. a point in $mathbbR^d$ may be equally distant to different parts of the curve, or a limit for $t$ that gives you the closest point along the curve may not exist as you increase the density of the curve. All I need is that the curve can at least in principle be made to wiggle through space so densely that at least one point anywhere along the curve will come arbitrarily close to any point in space.
$endgroup$
– Markus Mottl
Mar 19 at 2:01
$begingroup$
@RobArthan I don't make any assumptions about limits of $mathrmf$. E.g. a point in $mathbbR^d$ may be equally distant to different parts of the curve, or a limit for $t$ that gives you the closest point along the curve may not exist as you increase the density of the curve. All I need is that the curve can at least in principle be made to wiggle through space so densely that at least one point anywhere along the curve will come arbitrarily close to any point in space.
$endgroup$
– Markus Mottl
Mar 19 at 2:01
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I have it!
The idea is to multiply a parameter $t^frac1d+1$ by various periodic functions of arguments that depend non-linearly on $t$ and contain coefficients which are not rationally related. Then eventually any spot in space gets approached arbitrarily closely, yet in more than 2 dimensions the curve is non-intersecting.
The example I have in mind is something like:
$$
x = sqrt[4]t sin left( t + fract^2sqrt2 right) \
y = sqrt[4]t cos left( sqrt3t + t^2 right) \
z = sqrt[4]t sin left( pi t^2 right)
$$
The low-radius area gets filled fairly thoroughly because the curve keeps zipping through it when all the periodic functions coincide near zero. And that statement appears to be scale independent.
The reason that $t^frac1d+1$ is used is so that the (hyper)volume being traversed grows more slowly than the length of curve within that volume. That may or may not be a necessity.
I would have difficulty with a rigorous proof for any given fixed $epsilon$ and any point $vecx$ there exists some $delta(epsilon,vecx)$ such that the curve intersects an $epsilon$-ball about $vecx$ for some $t < delta$ but for generic (non-special) choices of the coefficients in the periodic functions I would be shocked to learn that "holes" exist.
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
$endgroup$
$begingroup$
Thanks, this proposal seems interesting. Visualizing it, the curve definitely seems to cover 3D space well, even without multiplying by $t^frac1d+1$. That factor would not work well for me anyway, because $d$ can be extremely large in my application (possibly millions). It would be nice if randomly chosen hypervolumes of the same size had the same expected length of curve going through it as seems to be the case for Archimedes in 2D. I also wished there were a principled way of constructing the function (choosing appropriate coefficients).
$endgroup$
– Markus Mottl
Mar 19 at 17:29
add a comment |
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1 Answer
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$begingroup$
I have it!
The idea is to multiply a parameter $t^frac1d+1$ by various periodic functions of arguments that depend non-linearly on $t$ and contain coefficients which are not rationally related. Then eventually any spot in space gets approached arbitrarily closely, yet in more than 2 dimensions the curve is non-intersecting.
The example I have in mind is something like:
$$
x = sqrt[4]t sin left( t + fract^2sqrt2 right) \
y = sqrt[4]t cos left( sqrt3t + t^2 right) \
z = sqrt[4]t sin left( pi t^2 right)
$$
The low-radius area gets filled fairly thoroughly because the curve keeps zipping through it when all the periodic functions coincide near zero. And that statement appears to be scale independent.
The reason that $t^frac1d+1$ is used is so that the (hyper)volume being traversed grows more slowly than the length of curve within that volume. That may or may not be a necessity.
I would have difficulty with a rigorous proof for any given fixed $epsilon$ and any point $vecx$ there exists some $delta(epsilon,vecx)$ such that the curve intersects an $epsilon$-ball about $vecx$ for some $t < delta$ but for generic (non-special) choices of the coefficients in the periodic functions I would be shocked to learn that "holes" exist.
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
$endgroup$
$begingroup$
Thanks, this proposal seems interesting. Visualizing it, the curve definitely seems to cover 3D space well, even without multiplying by $t^frac1d+1$. That factor would not work well for me anyway, because $d$ can be extremely large in my application (possibly millions). It would be nice if randomly chosen hypervolumes of the same size had the same expected length of curve going through it as seems to be the case for Archimedes in 2D. I also wished there were a principled way of constructing the function (choosing appropriate coefficients).
$endgroup$
– Markus Mottl
Mar 19 at 17:29
add a comment |
$begingroup$
I have it!
The idea is to multiply a parameter $t^frac1d+1$ by various periodic functions of arguments that depend non-linearly on $t$ and contain coefficients which are not rationally related. Then eventually any spot in space gets approached arbitrarily closely, yet in more than 2 dimensions the curve is non-intersecting.
The example I have in mind is something like:
$$
x = sqrt[4]t sin left( t + fract^2sqrt2 right) \
y = sqrt[4]t cos left( sqrt3t + t^2 right) \
z = sqrt[4]t sin left( pi t^2 right)
$$
The low-radius area gets filled fairly thoroughly because the curve keeps zipping through it when all the periodic functions coincide near zero. And that statement appears to be scale independent.
The reason that $t^frac1d+1$ is used is so that the (hyper)volume being traversed grows more slowly than the length of curve within that volume. That may or may not be a necessity.
I would have difficulty with a rigorous proof for any given fixed $epsilon$ and any point $vecx$ there exists some $delta(epsilon,vecx)$ such that the curve intersects an $epsilon$-ball about $vecx$ for some $t < delta$ but for generic (non-special) choices of the coefficients in the periodic functions I would be shocked to learn that "holes" exist.
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
$endgroup$
$begingroup$
Thanks, this proposal seems interesting. Visualizing it, the curve definitely seems to cover 3D space well, even without multiplying by $t^frac1d+1$. That factor would not work well for me anyway, because $d$ can be extremely large in my application (possibly millions). It would be nice if randomly chosen hypervolumes of the same size had the same expected length of curve going through it as seems to be the case for Archimedes in 2D. I also wished there were a principled way of constructing the function (choosing appropriate coefficients).
$endgroup$
– Markus Mottl
Mar 19 at 17:29
add a comment |
$begingroup$
I have it!
The idea is to multiply a parameter $t^frac1d+1$ by various periodic functions of arguments that depend non-linearly on $t$ and contain coefficients which are not rationally related. Then eventually any spot in space gets approached arbitrarily closely, yet in more than 2 dimensions the curve is non-intersecting.
The example I have in mind is something like:
$$
x = sqrt[4]t sin left( t + fract^2sqrt2 right) \
y = sqrt[4]t cos left( sqrt3t + t^2 right) \
z = sqrt[4]t sin left( pi t^2 right)
$$
The low-radius area gets filled fairly thoroughly because the curve keeps zipping through it when all the periodic functions coincide near zero. And that statement appears to be scale independent.
The reason that $t^frac1d+1$ is used is so that the (hyper)volume being traversed grows more slowly than the length of curve within that volume. That may or may not be a necessity.
I would have difficulty with a rigorous proof for any given fixed $epsilon$ and any point $vecx$ there exists some $delta(epsilon,vecx)$ such that the curve intersects an $epsilon$-ball about $vecx$ for some $t < delta$ but for generic (non-special) choices of the coefficients in the periodic functions I would be shocked to learn that "holes" exist.
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
$endgroup$
I have it!
The idea is to multiply a parameter $t^frac1d+1$ by various periodic functions of arguments that depend non-linearly on $t$ and contain coefficients which are not rationally related. Then eventually any spot in space gets approached arbitrarily closely, yet in more than 2 dimensions the curve is non-intersecting.
The example I have in mind is something like:
$$
x = sqrt[4]t sin left( t + fract^2sqrt2 right) \
y = sqrt[4]t cos left( sqrt3t + t^2 right) \
z = sqrt[4]t sin left( pi t^2 right)
$$
The low-radius area gets filled fairly thoroughly because the curve keeps zipping through it when all the periodic functions coincide near zero. And that statement appears to be scale independent.
The reason that $t^frac1d+1$ is used is so that the (hyper)volume being traversed grows more slowly than the length of curve within that volume. That may or may not be a necessity.
I would have difficulty with a rigorous proof for any given fixed $epsilon$ and any point $vecx$ there exists some $delta(epsilon,vecx)$ such that the curve intersects an $epsilon$-ball about $vecx$ for some $t < delta$ but for generic (non-special) choices of the coefficients in the periodic functions I would be shocked to learn that "holes" exist.
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
edited Mar 18 at 22:24
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
answered Mar 18 at 22:19
Mark FischlerMark Fischler
33.8k12552
33.8k12552
$begingroup$
Thanks, this proposal seems interesting. Visualizing it, the curve definitely seems to cover 3D space well, even without multiplying by $t^frac1d+1$. That factor would not work well for me anyway, because $d$ can be extremely large in my application (possibly millions). It would be nice if randomly chosen hypervolumes of the same size had the same expected length of curve going through it as seems to be the case for Archimedes in 2D. I also wished there were a principled way of constructing the function (choosing appropriate coefficients).
$endgroup$
– Markus Mottl
Mar 19 at 17:29
add a comment |
$begingroup$
Thanks, this proposal seems interesting. Visualizing it, the curve definitely seems to cover 3D space well, even without multiplying by $t^frac1d+1$. That factor would not work well for me anyway, because $d$ can be extremely large in my application (possibly millions). It would be nice if randomly chosen hypervolumes of the same size had the same expected length of curve going through it as seems to be the case for Archimedes in 2D. I also wished there were a principled way of constructing the function (choosing appropriate coefficients).
$endgroup$
– Markus Mottl
Mar 19 at 17:29
$begingroup$
Thanks, this proposal seems interesting. Visualizing it, the curve definitely seems to cover 3D space well, even without multiplying by $t^frac1d+1$. That factor would not work well for me anyway, because $d$ can be extremely large in my application (possibly millions). It would be nice if randomly chosen hypervolumes of the same size had the same expected length of curve going through it as seems to be the case for Archimedes in 2D. I also wished there were a principled way of constructing the function (choosing appropriate coefficients).
$endgroup$
– Markus Mottl
Mar 19 at 17:29
$begingroup$
Thanks, this proposal seems interesting. Visualizing it, the curve definitely seems to cover 3D space well, even without multiplying by $t^frac1d+1$. That factor would not work well for me anyway, because $d$ can be extremely large in my application (possibly millions). It would be nice if randomly chosen hypervolumes of the same size had the same expected length of curve going through it as seems to be the case for Archimedes in 2D. I also wished there were a principled way of constructing the function (choosing appropriate coefficients).
$endgroup$
– Markus Mottl
Mar 19 at 17:29
add a comment |
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$begingroup$
I have examples that are continuous but not smooth, based on covering size-$2n-1$ hypercubes such that the end point is on the hypercube surface, and then stepping outward and covering the surface of the size-$2n+1$ hypercube. The smoothness requirement makes this a very intriguing question.
$endgroup$
– Mark Fischler
Mar 18 at 21:57
$begingroup$
Nice question. Is self-intersection allowed? Also do you require the convergence to be "uniform"?
$endgroup$
– Rob Arthan
Mar 18 at 22:02
$begingroup$
@RobArthan Though self-intersection would not pose a technical problem for my application, it would seem preferable if there were no redundancies in the way the curve explores the space.
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– Markus Mottl
Mar 18 at 22:30
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I see. How about the convergence? I don't think the Spiral of Archimedes gives uniform convergence.
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– Rob Arthan
Mar 19 at 0:01
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@RobArthan I don't make any assumptions about limits of $mathrmf$. E.g. a point in $mathbbR^d$ may be equally distant to different parts of the curve, or a limit for $t$ that gives you the closest point along the curve may not exist as you increase the density of the curve. All I need is that the curve can at least in principle be made to wiggle through space so densely that at least one point anywhere along the curve will come arbitrarily close to any point in space.
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– Markus Mottl
Mar 19 at 2:01