Sierpinski (Triangle) for Other Polygons
The Sierpinski triangle can be "generated" by the algorihm where you start in the triangle, pick a vertex at random, then move half the distant towards it, draw a dot and then repeat this.
I wasn't able to find anything on whether doing this for other polygons produces similar sort of structures, so I tried it out and this is what I got:
After these there just seems to be a small hole in the middle. Well the square isn't interesting. But what about the holes in the pentagon and the looming star structure in the hexagon? Are these of any interest?
geometry fractals
add a comment |
The Sierpinski triangle can be "generated" by the algorihm where you start in the triangle, pick a vertex at random, then move half the distant towards it, draw a dot and then repeat this.
I wasn't able to find anything on whether doing this for other polygons produces similar sort of structures, so I tried it out and this is what I got:
After these there just seems to be a small hole in the middle. Well the square isn't interesting. But what about the holes in the pentagon and the looming star structure in the hexagon? Are these of any interest?
geometry fractals
1
There are other ways of defining the Sierpinski triangle, and some of them generalise in more interesting ways. Look up "Sierpinski triangle" and "Sierpinski carpet" on Wikipedia.
– Henrik
Apr 16 '15 at 19:59
1
Just wondering, what program/method did you use to generate these shapes? They look pretty cool.
– Asker
Apr 16 '15 at 20:55
1
@Asker Thanks, it's just a small Python code drawing dots according to the algorithm I mentioned.
– ploosu2
Apr 17 '15 at 0:03
add a comment |
The Sierpinski triangle can be "generated" by the algorihm where you start in the triangle, pick a vertex at random, then move half the distant towards it, draw a dot and then repeat this.
I wasn't able to find anything on whether doing this for other polygons produces similar sort of structures, so I tried it out and this is what I got:
After these there just seems to be a small hole in the middle. Well the square isn't interesting. But what about the holes in the pentagon and the looming star structure in the hexagon? Are these of any interest?
geometry fractals
The Sierpinski triangle can be "generated" by the algorihm where you start in the triangle, pick a vertex at random, then move half the distant towards it, draw a dot and then repeat this.
I wasn't able to find anything on whether doing this for other polygons produces similar sort of structures, so I tried it out and this is what I got:
After these there just seems to be a small hole in the middle. Well the square isn't interesting. But what about the holes in the pentagon and the looming star structure in the hexagon? Are these of any interest?
geometry fractals
geometry fractals
asked Apr 16 '15 at 19:34
ploosu2
4,5771023
4,5771023
1
There are other ways of defining the Sierpinski triangle, and some of them generalise in more interesting ways. Look up "Sierpinski triangle" and "Sierpinski carpet" on Wikipedia.
– Henrik
Apr 16 '15 at 19:59
1
Just wondering, what program/method did you use to generate these shapes? They look pretty cool.
– Asker
Apr 16 '15 at 20:55
1
@Asker Thanks, it's just a small Python code drawing dots according to the algorithm I mentioned.
– ploosu2
Apr 17 '15 at 0:03
add a comment |
1
There are other ways of defining the Sierpinski triangle, and some of them generalise in more interesting ways. Look up "Sierpinski triangle" and "Sierpinski carpet" on Wikipedia.
– Henrik
Apr 16 '15 at 19:59
1
Just wondering, what program/method did you use to generate these shapes? They look pretty cool.
– Asker
Apr 16 '15 at 20:55
1
@Asker Thanks, it's just a small Python code drawing dots according to the algorithm I mentioned.
– ploosu2
Apr 17 '15 at 0:03
1
1
There are other ways of defining the Sierpinski triangle, and some of them generalise in more interesting ways. Look up "Sierpinski triangle" and "Sierpinski carpet" on Wikipedia.
– Henrik
Apr 16 '15 at 19:59
There are other ways of defining the Sierpinski triangle, and some of them generalise in more interesting ways. Look up "Sierpinski triangle" and "Sierpinski carpet" on Wikipedia.
– Henrik
Apr 16 '15 at 19:59
1
1
Just wondering, what program/method did you use to generate these shapes? They look pretty cool.
– Asker
Apr 16 '15 at 20:55
Just wondering, what program/method did you use to generate these shapes? They look pretty cool.
– Asker
Apr 16 '15 at 20:55
1
1
@Asker Thanks, it's just a small Python code drawing dots according to the algorithm I mentioned.
– ploosu2
Apr 17 '15 at 0:03
@Asker Thanks, it's just a small Python code drawing dots according to the algorithm I mentioned.
– ploosu2
Apr 17 '15 at 0:03
add a comment |
3 Answers
3
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As Zach points out, your pictures are images of self-similar sets. For the Sierpinski triangle and the square, the scaling factor of $1/2$ is not arbitrary; rather, it's chosen so that the individual pieces just touch one another without overlap. While self-similar sets can have overlap, they tend to be more visually interesting and are definitely easier to analyze when there is no overlap. To arrange the "just touching" scenario for other polygons, we need a smaller scaling factor. Here's what I mean for a regular pentagon and a regular hexagon:
Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.
A fun exercise is to find the general scaling factor and the resulting fractal dimension.
Yes finding the general scaling factor certainly was a nice calculation. I got $x=frac{1-sqrt{1-beta_n}}{2beta_n}$ of the sidelength $d$, where $beta_n = sin^2(pi/n)$. Is this correct? But I don't seem to be able to find the amount to move in the "turtle-algorithm". It should be $d-x$, right? But that doesn't seem to work.
– ploosu2
Apr 18 '15 at 12:50
add a comment |
The Sierpinski triangle is related to itself, that's what makes it a fractal. However, lets mathematically define this "similarity" transformation, a transformation that maps the triangle into itself.
The Sierpinski triangle can be visually broken up into three smaller triangles, each with side length half the original. Its also very similar in form to an equilateral triangle. Expressing this mathematically...
$$A=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 0 \ 0 end{pmatrix}$$
$$B=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/2 \ 0 end{pmatrix}$$
$$C=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/4 \ sqrt 3/4 end{pmatrix}$$
Assuming you understand matrix theory, it should be clear that the union of the A,B, and C transformations of $S$ (the triangle) which has both $S_x$ (x components) and $S_y$ (y components) is S again. Now look at the equations in more detail. Notice that the parts that just add to the total, are the vertexes of the triangle halfed. That's why your algorithm works. I'd suggest you investigate from here to find out whether or not this rule generalizes to other shapes...
add a comment |
By also varying how we choose the next vertex to move towards to, we get different fractal patterns (for example: can't choose same twice in a row, must choose adjacent, etc.).
See: Chaos Game
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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As Zach points out, your pictures are images of self-similar sets. For the Sierpinski triangle and the square, the scaling factor of $1/2$ is not arbitrary; rather, it's chosen so that the individual pieces just touch one another without overlap. While self-similar sets can have overlap, they tend to be more visually interesting and are definitely easier to analyze when there is no overlap. To arrange the "just touching" scenario for other polygons, we need a smaller scaling factor. Here's what I mean for a regular pentagon and a regular hexagon:
Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.
A fun exercise is to find the general scaling factor and the resulting fractal dimension.
Yes finding the general scaling factor certainly was a nice calculation. I got $x=frac{1-sqrt{1-beta_n}}{2beta_n}$ of the sidelength $d$, where $beta_n = sin^2(pi/n)$. Is this correct? But I don't seem to be able to find the amount to move in the "turtle-algorithm". It should be $d-x$, right? But that doesn't seem to work.
– ploosu2
Apr 18 '15 at 12:50
add a comment |
As Zach points out, your pictures are images of self-similar sets. For the Sierpinski triangle and the square, the scaling factor of $1/2$ is not arbitrary; rather, it's chosen so that the individual pieces just touch one another without overlap. While self-similar sets can have overlap, they tend to be more visually interesting and are definitely easier to analyze when there is no overlap. To arrange the "just touching" scenario for other polygons, we need a smaller scaling factor. Here's what I mean for a regular pentagon and a regular hexagon:
Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.
A fun exercise is to find the general scaling factor and the resulting fractal dimension.
Yes finding the general scaling factor certainly was a nice calculation. I got $x=frac{1-sqrt{1-beta_n}}{2beta_n}$ of the sidelength $d$, where $beta_n = sin^2(pi/n)$. Is this correct? But I don't seem to be able to find the amount to move in the "turtle-algorithm". It should be $d-x$, right? But that doesn't seem to work.
– ploosu2
Apr 18 '15 at 12:50
add a comment |
As Zach points out, your pictures are images of self-similar sets. For the Sierpinski triangle and the square, the scaling factor of $1/2$ is not arbitrary; rather, it's chosen so that the individual pieces just touch one another without overlap. While self-similar sets can have overlap, they tend to be more visually interesting and are definitely easier to analyze when there is no overlap. To arrange the "just touching" scenario for other polygons, we need a smaller scaling factor. Here's what I mean for a regular pentagon and a regular hexagon:
Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.
A fun exercise is to find the general scaling factor and the resulting fractal dimension.
As Zach points out, your pictures are images of self-similar sets. For the Sierpinski triangle and the square, the scaling factor of $1/2$ is not arbitrary; rather, it's chosen so that the individual pieces just touch one another without overlap. While self-similar sets can have overlap, they tend to be more visually interesting and are definitely easier to analyze when there is no overlap. To arrange the "just touching" scenario for other polygons, we need a smaller scaling factor. Here's what I mean for a regular pentagon and a regular hexagon:
Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.
A fun exercise is to find the general scaling factor and the resulting fractal dimension.
answered Apr 16 '15 at 22:58
Mark McClure
23.4k34471
23.4k34471
Yes finding the general scaling factor certainly was a nice calculation. I got $x=frac{1-sqrt{1-beta_n}}{2beta_n}$ of the sidelength $d$, where $beta_n = sin^2(pi/n)$. Is this correct? But I don't seem to be able to find the amount to move in the "turtle-algorithm". It should be $d-x$, right? But that doesn't seem to work.
– ploosu2
Apr 18 '15 at 12:50
add a comment |
Yes finding the general scaling factor certainly was a nice calculation. I got $x=frac{1-sqrt{1-beta_n}}{2beta_n}$ of the sidelength $d$, where $beta_n = sin^2(pi/n)$. Is this correct? But I don't seem to be able to find the amount to move in the "turtle-algorithm". It should be $d-x$, right? But that doesn't seem to work.
– ploosu2
Apr 18 '15 at 12:50
Yes finding the general scaling factor certainly was a nice calculation. I got $x=frac{1-sqrt{1-beta_n}}{2beta_n}$ of the sidelength $d$, where $beta_n = sin^2(pi/n)$. Is this correct? But I don't seem to be able to find the amount to move in the "turtle-algorithm". It should be $d-x$, right? But that doesn't seem to work.
– ploosu2
Apr 18 '15 at 12:50
Yes finding the general scaling factor certainly was a nice calculation. I got $x=frac{1-sqrt{1-beta_n}}{2beta_n}$ of the sidelength $d$, where $beta_n = sin^2(pi/n)$. Is this correct? But I don't seem to be able to find the amount to move in the "turtle-algorithm". It should be $d-x$, right? But that doesn't seem to work.
– ploosu2
Apr 18 '15 at 12:50
add a comment |
The Sierpinski triangle is related to itself, that's what makes it a fractal. However, lets mathematically define this "similarity" transformation, a transformation that maps the triangle into itself.
The Sierpinski triangle can be visually broken up into three smaller triangles, each with side length half the original. Its also very similar in form to an equilateral triangle. Expressing this mathematically...
$$A=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 0 \ 0 end{pmatrix}$$
$$B=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/2 \ 0 end{pmatrix}$$
$$C=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/4 \ sqrt 3/4 end{pmatrix}$$
Assuming you understand matrix theory, it should be clear that the union of the A,B, and C transformations of $S$ (the triangle) which has both $S_x$ (x components) and $S_y$ (y components) is S again. Now look at the equations in more detail. Notice that the parts that just add to the total, are the vertexes of the triangle halfed. That's why your algorithm works. I'd suggest you investigate from here to find out whether or not this rule generalizes to other shapes...
add a comment |
The Sierpinski triangle is related to itself, that's what makes it a fractal. However, lets mathematically define this "similarity" transformation, a transformation that maps the triangle into itself.
The Sierpinski triangle can be visually broken up into three smaller triangles, each with side length half the original. Its also very similar in form to an equilateral triangle. Expressing this mathematically...
$$A=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 0 \ 0 end{pmatrix}$$
$$B=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/2 \ 0 end{pmatrix}$$
$$C=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/4 \ sqrt 3/4 end{pmatrix}$$
Assuming you understand matrix theory, it should be clear that the union of the A,B, and C transformations of $S$ (the triangle) which has both $S_x$ (x components) and $S_y$ (y components) is S again. Now look at the equations in more detail. Notice that the parts that just add to the total, are the vertexes of the triangle halfed. That's why your algorithm works. I'd suggest you investigate from here to find out whether or not this rule generalizes to other shapes...
add a comment |
The Sierpinski triangle is related to itself, that's what makes it a fractal. However, lets mathematically define this "similarity" transformation, a transformation that maps the triangle into itself.
The Sierpinski triangle can be visually broken up into three smaller triangles, each with side length half the original. Its also very similar in form to an equilateral triangle. Expressing this mathematically...
$$A=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 0 \ 0 end{pmatrix}$$
$$B=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/2 \ 0 end{pmatrix}$$
$$C=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/4 \ sqrt 3/4 end{pmatrix}$$
Assuming you understand matrix theory, it should be clear that the union of the A,B, and C transformations of $S$ (the triangle) which has both $S_x$ (x components) and $S_y$ (y components) is S again. Now look at the equations in more detail. Notice that the parts that just add to the total, are the vertexes of the triangle halfed. That's why your algorithm works. I'd suggest you investigate from here to find out whether or not this rule generalizes to other shapes...
The Sierpinski triangle is related to itself, that's what makes it a fractal. However, lets mathematically define this "similarity" transformation, a transformation that maps the triangle into itself.
The Sierpinski triangle can be visually broken up into three smaller triangles, each with side length half the original. Its also very similar in form to an equilateral triangle. Expressing this mathematically...
$$A=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 0 \ 0 end{pmatrix}$$
$$B=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/2 \ 0 end{pmatrix}$$
$$C=begin{pmatrix} 1/2 & 0 \ 0 & 1/2 \ end{pmatrix} cdot begin{pmatrix} S_x \ S_y end{pmatrix}+begin{pmatrix} 1/4 \ sqrt 3/4 end{pmatrix}$$
Assuming you understand matrix theory, it should be clear that the union of the A,B, and C transformations of $S$ (the triangle) which has both $S_x$ (x components) and $S_y$ (y components) is S again. Now look at the equations in more detail. Notice that the parts that just add to the total, are the vertexes of the triangle halfed. That's why your algorithm works. I'd suggest you investigate from here to find out whether or not this rule generalizes to other shapes...
answered Apr 16 '15 at 20:24
Zach466920
6,18111142
6,18111142
add a comment |
add a comment |
By also varying how we choose the next vertex to move towards to, we get different fractal patterns (for example: can't choose same twice in a row, must choose adjacent, etc.).
See: Chaos Game
add a comment |
By also varying how we choose the next vertex to move towards to, we get different fractal patterns (for example: can't choose same twice in a row, must choose adjacent, etc.).
See: Chaos Game
add a comment |
By also varying how we choose the next vertex to move towards to, we get different fractal patterns (for example: can't choose same twice in a row, must choose adjacent, etc.).
See: Chaos Game
By also varying how we choose the next vertex to move towards to, we get different fractal patterns (for example: can't choose same twice in a row, must choose adjacent, etc.).
See: Chaos Game
answered Dec 2 '18 at 11:51
ploosu2
4,5771023
4,5771023
add a comment |
add a comment |
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1
There are other ways of defining the Sierpinski triangle, and some of them generalise in more interesting ways. Look up "Sierpinski triangle" and "Sierpinski carpet" on Wikipedia.
– Henrik
Apr 16 '15 at 19:59
1
Just wondering, what program/method did you use to generate these shapes? They look pretty cool.
– Asker
Apr 16 '15 at 20:55
1
@Asker Thanks, it's just a small Python code drawing dots according to the algorithm I mentioned.
– ploosu2
Apr 17 '15 at 0:03