How to generate a Penrose tessellation around a given tile?
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Given a starting Penrose tile, I need to build a "spiraling" tessellation around it.
The following picture illustrates the request:
In this example, the starting tile is a "thin rhombus" (the pink one).
I need to write an algorithm which is able to generate the $n$ tiles (and whose output is, for instance a, SVG file), starting from any given tile, and with the possibility of coloring the tiles according to a given sequence of $n$ colors.
Thanks for your help!
NOTE: This post is related to this one.
geometry discrete-mathematics math-software hamiltonian-path tiling
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
 |Â
show 3 more comments
up vote
3
down vote
favorite
Given a starting Penrose tile, I need to build a "spiraling" tessellation around it.
The following picture illustrates the request:
In this example, the starting tile is a "thin rhombus" (the pink one).
I need to write an algorithm which is able to generate the $n$ tiles (and whose output is, for instance a, SVG file), starting from any given tile, and with the possibility of coloring the tiles according to a given sequence of $n$ colors.
Thanks for your help!
NOTE: This post is related to this one.
geometry discrete-mathematics math-software hamiltonian-path tiling
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
1
From a graph-point-of-view, you are traversing the dual-graph of the tiling, with the convention of taking the right-most turn each time.
– Alex R.
Sep 10 at 19:07
@AlexR. The dual graph is the network of the centers of the tiles, right? In this case, the problem becomes how to predict the $n$-th center, right?
– Andrea Prunotto
Sep 10 at 19:11
How is your penrose tiling stored?
– Alex R.
Sep 10 at 20:46
So far I made a Xfig file (mcj.sourceforge.net), which can be easily transformed into a SVG file and viceversa, perhaps more common. In practice, for each tile I have 4 points. But I'm not sure it is the best way to store it.
– Andrea Prunotto
Sep 10 at 21:10
Say, I would prefer not to store the underlying Penrose tiling, but to generate the next tile from first principles, a bit as i tried to depict in the figure above.
– Andrea Prunotto
Sep 10 at 21:16
 |Â
show 3 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Given a starting Penrose tile, I need to build a "spiraling" tessellation around it.
The following picture illustrates the request:
In this example, the starting tile is a "thin rhombus" (the pink one).
I need to write an algorithm which is able to generate the $n$ tiles (and whose output is, for instance a, SVG file), starting from any given tile, and with the possibility of coloring the tiles according to a given sequence of $n$ colors.
Thanks for your help!
NOTE: This post is related to this one.
geometry discrete-mathematics math-software hamiltonian-path tiling
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
Given a starting Penrose tile, I need to build a "spiraling" tessellation around it.
The following picture illustrates the request:
In this example, the starting tile is a "thin rhombus" (the pink one).
I need to write an algorithm which is able to generate the $n$ tiles (and whose output is, for instance a, SVG file), starting from any given tile, and with the possibility of coloring the tiles according to a given sequence of $n$ colors.
Thanks for your help!
NOTE: This post is related to this one.
geometry discrete-mathematics math-software hamiltonian-path tiling
geometry discrete-mathematics math-software hamiltonian-path tiling
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
edited 3 hours ago
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
asked Sep 10 at 18:43
Andrea Prunotto
1,401726
1,401726
1
From a graph-point-of-view, you are traversing the dual-graph of the tiling, with the convention of taking the right-most turn each time.
– Alex R.
Sep 10 at 19:07
@AlexR. The dual graph is the network of the centers of the tiles, right? In this case, the problem becomes how to predict the $n$-th center, right?
– Andrea Prunotto
Sep 10 at 19:11
How is your penrose tiling stored?
– Alex R.
Sep 10 at 20:46
So far I made a Xfig file (mcj.sourceforge.net), which can be easily transformed into a SVG file and viceversa, perhaps more common. In practice, for each tile I have 4 points. But I'm not sure it is the best way to store it.
– Andrea Prunotto
Sep 10 at 21:10
Say, I would prefer not to store the underlying Penrose tiling, but to generate the next tile from first principles, a bit as i tried to depict in the figure above.
– Andrea Prunotto
Sep 10 at 21:16
 |Â
show 3 more comments
1
From a graph-point-of-view, you are traversing the dual-graph of the tiling, with the convention of taking the right-most turn each time.
– Alex R.
Sep 10 at 19:07
@AlexR. The dual graph is the network of the centers of the tiles, right? In this case, the problem becomes how to predict the $n$-th center, right?
– Andrea Prunotto
Sep 10 at 19:11
How is your penrose tiling stored?
– Alex R.
Sep 10 at 20:46
So far I made a Xfig file (mcj.sourceforge.net), which can be easily transformed into a SVG file and viceversa, perhaps more common. In practice, for each tile I have 4 points. But I'm not sure it is the best way to store it.
– Andrea Prunotto
Sep 10 at 21:10
Say, I would prefer not to store the underlying Penrose tiling, but to generate the next tile from first principles, a bit as i tried to depict in the figure above.
– Andrea Prunotto
Sep 10 at 21:16
1
1
From a graph-point-of-view, you are traversing the dual-graph of the tiling, with the convention of taking the right-most turn each time.
– Alex R.
Sep 10 at 19:07
From a graph-point-of-view, you are traversing the dual-graph of the tiling, with the convention of taking the right-most turn each time.
– Alex R.
Sep 10 at 19:07
@AlexR. The dual graph is the network of the centers of the tiles, right? In this case, the problem becomes how to predict the $n$-th center, right?
– Andrea Prunotto
Sep 10 at 19:11
@AlexR. The dual graph is the network of the centers of the tiles, right? In this case, the problem becomes how to predict the $n$-th center, right?
– Andrea Prunotto
Sep 10 at 19:11
How is your penrose tiling stored?
– Alex R.
Sep 10 at 20:46
How is your penrose tiling stored?
– Alex R.
Sep 10 at 20:46
So far I made a Xfig file (mcj.sourceforge.net), which can be easily transformed into a SVG file and viceversa, perhaps more common. In practice, for each tile I have 4 points. But I'm not sure it is the best way to store it.
– Andrea Prunotto
Sep 10 at 21:10
So far I made a Xfig file (mcj.sourceforge.net), which can be easily transformed into a SVG file and viceversa, perhaps more common. In practice, for each tile I have 4 points. But I'm not sure it is the best way to store it.
– Andrea Prunotto
Sep 10 at 21:10
Say, I would prefer not to store the underlying Penrose tiling, but to generate the next tile from first principles, a bit as i tried to depict in the figure above.
– Andrea Prunotto
Sep 10 at 21:16
Say, I would prefer not to store the underlying Penrose tiling, but to generate the next tile from first principles, a bit as i tried to depict in the figure above.
– Andrea Prunotto
Sep 10 at 21:16
 |Â
show 3 more comments
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1
From a graph-point-of-view, you are traversing the dual-graph of the tiling, with the convention of taking the right-most turn each time.
– Alex R.
Sep 10 at 19:07
@AlexR. The dual graph is the network of the centers of the tiles, right? In this case, the problem becomes how to predict the $n$-th center, right?
– Andrea Prunotto
Sep 10 at 19:11
How is your penrose tiling stored?
– Alex R.
Sep 10 at 20:46
So far I made a Xfig file (mcj.sourceforge.net), which can be easily transformed into a SVG file and viceversa, perhaps more common. In practice, for each tile I have 4 points. But I'm not sure it is the best way to store it.
– Andrea Prunotto
Sep 10 at 21:10
Say, I would prefer not to store the underlying Penrose tiling, but to generate the next tile from first principles, a bit as i tried to depict in the figure above.
– Andrea Prunotto
Sep 10 at 21:16