Mixing
What is the average number of sides of a cell of a Voronoi pattern on a flat torus?
$begingroup$
Consider a random voronoi pattern with a uniform distribution on a large flat torus. What are the average number of sides of a cell. My guess is 6.
What about 3D or 4D?
My guess for 4D would be 24. For 3D I'm guessing somewhere between 12 and 13.
Is there a way to work this out?
P.S. I suppose this also depends on which "average" you are using. I'm thinking of the mean.
voronoi-diagram
asked Jan 25 at 5:28
zoobyzooby
1,022716
$endgroup$
|
show 1 more comment
$begingroup$
Consider a random voronoi pattern with a uniform distribution on a large flat torus. What are the average number of sides of a cell. My guess is 6.
What about 3D or 4D?
My guess for 4D would be 24. For 3D I'm guessing somewhere between 12 and 13.
Is there a way to work this out?
P.S. I suppose this also depends on which "average" you are using. I'm thinking of the mean.
voronoi-diagram
asked Jan 25 at 5:28
zoobyzooby
1,022716
$endgroup$
$begingroup$
Interesting question. Is it known for a square or a sphere? It's not obvious to me why it would not depend on the particular dimensions of the torus.
$endgroup$
– Jair Taylor
Jan 25 at 5:33
$begingroup$
@Jair I would consider the torus dimensions so big as to be essentially infinite.
$endgroup$
– zooby
Jan 25 at 5:38
$begingroup$
Then why would it differ from a plane? I guess it wouldn't.
$endgroup$
– Ivan Neretin
Jan 25 at 9:22
$begingroup$
(Except that you can't have a uniform distribution on a plane, of course, since it is infinite.)
$endgroup$
– Ivan Neretin
Jan 25 at 9:53
$begingroup$
@Ivan Probably wouldn't differ but one could do computer simulations with the first and not with the second.
$endgroup$
– zooby
Jan 25 at 16:15
|
show 1 more comment
$begingroup$
Consider a random voronoi pattern with a uniform distribution on a large flat torus. What are the average number of sides of a cell. My guess is 6.
What about 3D or 4D?
My guess for 4D would be 24. For 3D I'm guessing somewhere between 12 and 13.
Is there a way to work this out?
P.S. I suppose this also depends on which "average" you are using. I'm thinking of the mean.
voronoi-diagram
asked Jan 25 at 5:28
zoobyzooby
1,022716
$endgroup$
Consider a random voronoi pattern with a uniform distribution on a large flat torus. What are the average number of sides of a cell. My guess is 6.
What about 3D or 4D?
My guess for 4D would be 24. For 3D I'm guessing somewhere between 12 and 13.
Is there a way to work this out?
P.S. I suppose this also depends on which "average" you are using. I'm thinking of the mean.
voronoi-diagram
voronoi-diagram
asked Jan 25 at 5:28
zoobyzooby
1,022716
asked Jan 25 at 5:28
zoobyzooby
1,022716
edited Jan 25 at 5:37
zooby
asked Jan 25 at 5:28
zoobyzooby
1,022716
asked Jan 25 at 5:28
zoobyzooby
1,022716
asked Jan 25 at 5:28
zoobyzooby
1,022716
1,022716
$begingroup$
Interesting question. Is it known for a square or a sphere? It's not obvious to me why it would not depend on the particular dimensions of the torus.
$endgroup$
– Jair Taylor
Jan 25 at 5:33
$begingroup$
@Jair I would consider the torus dimensions so big as to be essentially infinite.
$endgroup$
– zooby
Jan 25 at 5:38
$begingroup$
Then why would it differ from a plane? I guess it wouldn't.
$endgroup$
– Ivan Neretin
Jan 25 at 9:22
$begingroup$
(Except that you can't have a uniform distribution on a plane, of course, since it is infinite.)
$endgroup$
– Ivan Neretin
Jan 25 at 9:53
$begingroup$
@Ivan Probably wouldn't differ but one could do computer simulations with the first and not with the second.
$endgroup$
– zooby
Jan 25 at 16:15
|
show 1 more comment
$begingroup$
Interesting question. Is it known for a square or a sphere? It's not obvious to me why it would not depend on the particular dimensions of the torus.
$endgroup$
– Jair Taylor
Jan 25 at 5:33
$begingroup$
@Jair I would consider the torus dimensions so big as to be essentially infinite.
$endgroup$
– zooby
Jan 25 at 5:38
$begingroup$
Then why would it differ from a plane? I guess it wouldn't.
$endgroup$
– Ivan Neretin
Jan 25 at 9:22
$begingroup$
(Except that you can't have a uniform distribution on a plane, of course, since it is infinite.)
$endgroup$
– Ivan Neretin
Jan 25 at 9:53
$begingroup$
@Ivan Probably wouldn't differ but one could do computer simulations with the first and not with the second.
$endgroup$
– zooby
Jan 25 at 16:15
$begingroup$
Interesting question. Is it known for a square or a sphere? It's not obvious to me why it would not depend on the particular dimensions of the torus.
$endgroup$
– Jair Taylor
Jan 25 at 5:33
$begingroup$
Interesting question. Is it known for a square or a sphere? It's not obvious to me why it would not depend on the particular dimensions of the torus.
$endgroup$
– Jair Taylor
Jan 25 at 5:33
$begingroup$
@Jair I would consider the torus dimensions so big as to be essentially infinite.
$endgroup$
– zooby
Jan 25 at 5:38
$begingroup$
@Jair I would consider the torus dimensions so big as to be essentially infinite.
$endgroup$
– zooby
Jan 25 at 5:38
$begingroup$
Then why would it differ from a plane? I guess it wouldn't.
$endgroup$
– Ivan Neretin
Jan 25 at 9:22
$begingroup$
Then why would it differ from a plane? I guess it wouldn't.
$endgroup$
– Ivan Neretin
Jan 25 at 9:22
$begingroup$
(Except that you can't have a uniform distribution on a plane, of course, since it is infinite.)
$endgroup$
– Ivan Neretin
Jan 25 at 9:53
$begingroup$
(Except that you can't have a uniform distribution on a plane, of course, since it is infinite.)
$endgroup$
– Ivan Neretin
Jan 25 at 9:53
$begingroup$
@Ivan Probably wouldn't differ but one could do computer simulations with the first and not with the second.
$endgroup$
– zooby
Jan 25 at 16:15
$begingroup$
@Ivan Probably wouldn't differ but one could do computer simulations with the first and not with the second.
$endgroup$
– zooby
Jan 25 at 16:15
|
show 1 more comment
1 Answer
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Poisson-Voronoi tessellation. Consider the point process on the $d$-dimensional torus, where each point is uniformly distributed given the total number of points. As the size of the torus goes to infinite while the expected number of uniform points per area converges to some positive number, the point processes converge in distribution to the homogeneous Poisson point process on $mathbb{R}^d$. (This is a simple generalization of the law of rare events.)
The case of $d = 2$. Now consider the planar case ($d=2$) and the Voronoi tessellation associated to this point process. Then with probability one, the planar graph obtained by this Voronoi tessellation is $3$-regular, and so, a typical Poisson-Voronoi cell has $6$ sides, see the addendum below. (Basically $6$ is the 'spatial average' of the number of edges of cells, but this also becomes average number of a typical Poisson-Voronoi cell by ergodicity.)
The case of $d=3$. It is known by Meijering (1953)[1] that the average number of vertices of a typical cell is $frac{96}{35}pi^2 approx 27.07$. Then the number of edges is $frac{3}{2}$-times of the number of vertices, hence is $frac{144}{35}pi^2 approx 40.61$. Finally, by Euler's formula, the average number of faces is $frac{48}{35}pi^2 + 2 approx 15.54$.
[1] Meijering, J. L. (1953). Interface area, edge length, and number of vertices in crystal aggregates with random nucleation. Philips Res. Rep. 8.
Addendum - a simple fact about $3$-regular planar graph. Consider any $3$-regular planar graph $G = (V, E)$ in the torus, where $V$ is the vertex-set and $E$ is the edge-set. If $F$ denotes the set of faces of $G$, then
By counting the number of all pairs $(v, e)$ in $Vtimes E$ for which $v$ is an end-point of $e$ in two ways, we have $2|E| = 3|V|$.
For each $f in F$, let $e_f$ denote the number of edges of $F$. As before, counting the number of pairs $(e, f)$ in $E times F$ for which $e$ is an edge of $f$, we obtain $sum_{fin F} e_f = 2|E|$.
Since the Euler characteristic of the torus is $0$, we have $|V| - |E| + |F| = 0$.
Combining altogether, we obtain
$$ sum_{fin F} e_f = 6|F|. $$
In other words, the average number of the edges of a face chosen uniformly at random is $6$.
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Poisson-Voronoi tessellation. Consider the point process on the $d$-dimensional torus, where each point is uniformly distributed given the total number of points. As the size of the torus goes to infinite while the expected number of uniform points per area converges to some positive number, the point processes converge in distribution to the homogeneous Poisson point process on $mathbb{R}^d$. (This is a simple generalization of the law of rare events.)
The case of $d = 2$. Now consider the planar case ($d=2$) and the Voronoi tessellation associated to this point process. Then with probability one, the planar graph obtained by this Voronoi tessellation is $3$-regular, and so, a typical Poisson-Voronoi cell has $6$ sides, see the addendum below. (Basically $6$ is the 'spatial average' of the number of edges of cells, but this also becomes average number of a typical Poisson-Voronoi cell by ergodicity.)
The case of $d=3$. It is known by Meijering (1953)[1] that the average number of vertices of a typical cell is $frac{96}{35}pi^2 approx 27.07$. Then the number of edges is $frac{3}{2}$-times of the number of vertices, hence is $frac{144}{35}pi^2 approx 40.61$. Finally, by Euler's formula, the average number of faces is $frac{48}{35}pi^2 + 2 approx 15.54$.
[1] Meijering, J. L. (1953). Interface area, edge length, and number of vertices in crystal aggregates with random nucleation. Philips Res. Rep. 8.
Addendum - a simple fact about $3$-regular planar graph. Consider any $3$-regular planar graph $G = (V, E)$ in the torus, where $V$ is the vertex-set and $E$ is the edge-set. If $F$ denotes the set of faces of $G$, then
By counting the number of all pairs $(v, e)$ in $Vtimes E$ for which $v$ is an end-point of $e$ in two ways, we have $2|E| = 3|V|$.
For each $f in F$, let $e_f$ denote the number of edges of $F$. As before, counting the number of pairs $(e, f)$ in $E times F$ for which $e$ is an edge of $f$, we obtain $sum_{fin F} e_f = 2|E|$.
Since the Euler characteristic of the torus is $0$, we have $|V| - |E| + |F| = 0$.
Combining altogether, we obtain
$$ sum_{fin F} e_f = 6|F|. $$
In other words, the average number of the edges of a face chosen uniformly at random is $6$.
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
$endgroup$
add a comment |
$begingroup$
Poisson-Voronoi tessellation. Consider the point process on the $d$-dimensional torus, where each point is uniformly distributed given the total number of points. As the size of the torus goes to infinite while the expected number of uniform points per area converges to some positive number, the point processes converge in distribution to the homogeneous Poisson point process on $mathbb{R}^d$. (This is a simple generalization of the law of rare events.)
The case of $d = 2$. Now consider the planar case ($d=2$) and the Voronoi tessellation associated to this point process. Then with probability one, the planar graph obtained by this Voronoi tessellation is $3$-regular, and so, a typical Poisson-Voronoi cell has $6$ sides, see the addendum below. (Basically $6$ is the 'spatial average' of the number of edges of cells, but this also becomes average number of a typical Poisson-Voronoi cell by ergodicity.)
The case of $d=3$. It is known by Meijering (1953)[1] that the average number of vertices of a typical cell is $frac{96}{35}pi^2 approx 27.07$. Then the number of edges is $frac{3}{2}$-times of the number of vertices, hence is $frac{144}{35}pi^2 approx 40.61$. Finally, by Euler's formula, the average number of faces is $frac{48}{35}pi^2 + 2 approx 15.54$.
[1] Meijering, J. L. (1953). Interface area, edge length, and number of vertices in crystal aggregates with random nucleation. Philips Res. Rep. 8.
Addendum - a simple fact about $3$-regular planar graph. Consider any $3$-regular planar graph $G = (V, E)$ in the torus, where $V$ is the vertex-set and $E$ is the edge-set. If $F$ denotes the set of faces of $G$, then
By counting the number of all pairs $(v, e)$ in $Vtimes E$ for which $v$ is an end-point of $e$ in two ways, we have $2|E| = 3|V|$.
For each $f in F$, let $e_f$ denote the number of edges of $F$. As before, counting the number of pairs $(e, f)$ in $E times F$ for which $e$ is an edge of $f$, we obtain $sum_{fin F} e_f = 2|E|$.
Since the Euler characteristic of the torus is $0$, we have $|V| - |E| + |F| = 0$.
Combining altogether, we obtain
$$ sum_{fin F} e_f = 6|F|. $$
In other words, the average number of the edges of a face chosen uniformly at random is $6$.
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
$endgroup$
add a comment |
$begingroup$
Poisson-Voronoi tessellation. Consider the point process on the $d$-dimensional torus, where each point is uniformly distributed given the total number of points. As the size of the torus goes to infinite while the expected number of uniform points per area converges to some positive number, the point processes converge in distribution to the homogeneous Poisson point process on $mathbb{R}^d$. (This is a simple generalization of the law of rare events.)
The case of $d = 2$. Now consider the planar case ($d=2$) and the Voronoi tessellation associated to this point process. Then with probability one, the planar graph obtained by this Voronoi tessellation is $3$-regular, and so, a typical Poisson-Voronoi cell has $6$ sides, see the addendum below. (Basically $6$ is the 'spatial average' of the number of edges of cells, but this also becomes average number of a typical Poisson-Voronoi cell by ergodicity.)
The case of $d=3$. It is known by Meijering (1953)[1] that the average number of vertices of a typical cell is $frac{96}{35}pi^2 approx 27.07$. Then the number of edges is $frac{3}{2}$-times of the number of vertices, hence is $frac{144}{35}pi^2 approx 40.61$. Finally, by Euler's formula, the average number of faces is $frac{48}{35}pi^2 + 2 approx 15.54$.
[1] Meijering, J. L. (1953). Interface area, edge length, and number of vertices in crystal aggregates with random nucleation. Philips Res. Rep. 8.
Addendum - a simple fact about $3$-regular planar graph. Consider any $3$-regular planar graph $G = (V, E)$ in the torus, where $V$ is the vertex-set and $E$ is the edge-set. If $F$ denotes the set of faces of $G$, then
By counting the number of all pairs $(v, e)$ in $Vtimes E$ for which $v$ is an end-point of $e$ in two ways, we have $2|E| = 3|V|$.
For each $f in F$, let $e_f$ denote the number of edges of $F$. As before, counting the number of pairs $(e, f)$ in $E times F$ for which $e$ is an edge of $f$, we obtain $sum_{fin F} e_f = 2|E|$.
Since the Euler characteristic of the torus is $0$, we have $|V| - |E| + |F| = 0$.
Combining altogether, we obtain
$$ sum_{fin F} e_f = 6|F|. $$
In other words, the average number of the edges of a face chosen uniformly at random is $6$.
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
$endgroup$
Poisson-Voronoi tessellation. Consider the point process on the $d$-dimensional torus, where each point is uniformly distributed given the total number of points. As the size of the torus goes to infinite while the expected number of uniform points per area converges to some positive number, the point processes converge in distribution to the homogeneous Poisson point process on $mathbb{R}^d$. (This is a simple generalization of the law of rare events.)
The case of $d = 2$. Now consider the planar case ($d=2$) and the Voronoi tessellation associated to this point process. Then with probability one, the planar graph obtained by this Voronoi tessellation is $3$-regular, and so, a typical Poisson-Voronoi cell has $6$ sides, see the addendum below. (Basically $6$ is the 'spatial average' of the number of edges of cells, but this also becomes average number of a typical Poisson-Voronoi cell by ergodicity.)
The case of $d=3$. It is known by Meijering (1953)[1] that the average number of vertices of a typical cell is $frac{96}{35}pi^2 approx 27.07$. Then the number of edges is $frac{3}{2}$-times of the number of vertices, hence is $frac{144}{35}pi^2 approx 40.61$. Finally, by Euler's formula, the average number of faces is $frac{48}{35}pi^2 + 2 approx 15.54$.
[1] Meijering, J. L. (1953). Interface area, edge length, and number of vertices in crystal aggregates with random nucleation. Philips Res. Rep. 8.
Addendum - a simple fact about $3$-regular planar graph. Consider any $3$-regular planar graph $G = (V, E)$ in the torus, where $V$ is the vertex-set and $E$ is the edge-set. If $F$ denotes the set of faces of $G$, then
By counting the number of all pairs $(v, e)$ in $Vtimes E$ for which $v$ is an end-point of $e$ in two ways, we have $2|E| = 3|V|$.
For each $f in F$, let $e_f$ denote the number of edges of $F$. As before, counting the number of pairs $(e, f)$ in $E times F$ for which $e$ is an edge of $f$, we obtain $sum_{fin F} e_f = 2|E|$.
Since the Euler characteristic of the torus is $0$, we have $|V| - |E| + |F| = 0$.
Combining altogether, we obtain
$$ sum_{fin F} e_f = 6|F|. $$
In other words, the average number of the edges of a face chosen uniformly at random is $6$.
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
edited Feb 25 at 6:44
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
answered Feb 25 at 5:31
Sangchul LeeSangchul Lee
96k12171280
96k12171280
add a comment |
add a comment |
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$begingroup$
Interesting question. Is it known for a square or a sphere? It's not obvious to me why it would not depend on the particular dimensions of the torus.
$endgroup$
– Jair Taylor
Jan 25 at 5:33
$begingroup$
@Jair I would consider the torus dimensions so big as to be essentially infinite.
$endgroup$
– zooby
Jan 25 at 5:38
$begingroup$
Then why would it differ from a plane? I guess it wouldn't.
$endgroup$
– Ivan Neretin
Jan 25 at 9:22
$begingroup$
(Except that you can't have a uniform distribution on a plane, of course, since it is infinite.)
$endgroup$
– Ivan Neretin
Jan 25 at 9:53
$begingroup$
@Ivan Probably wouldn't differ but one could do computer simulations with the first and not with the second.
$endgroup$
– zooby
Jan 25 at 16:15