Sierpinski (Triangle) for Other Polygons












5














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:



enter image description here



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?










share|cite|improve this question


















  • 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
















5














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:



enter image description here



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?










share|cite|improve this question


















  • 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














5












5








5







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:



enter image description here



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?










share|cite|improve this question













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:



enter image description here



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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










3 Answers
3






active

oldest

votes


















9














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:



enter image description here



Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.



enter image description here



A fun exercise is to find the general scaling factor and the resulting fractal dimension.






share|cite|improve this answer





















  • 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



















4














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...






share|cite|improve this answer





























    0














    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






    share|cite|improve this answer





















      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1237964%2fsierpinski-triangle-for-other-polygons%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      9














      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:



      enter image description here



      Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.



      enter image description here



      A fun exercise is to find the general scaling factor and the resulting fractal dimension.






      share|cite|improve this answer





















      • 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
















      9














      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:



      enter image description here



      Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.



      enter image description here



      A fun exercise is to find the general scaling factor and the resulting fractal dimension.






      share|cite|improve this answer





















      • 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














      9












      9








      9






      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:



      enter image description here



      Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.



      enter image description here



      A fun exercise is to find the general scaling factor and the resulting fractal dimension.






      share|cite|improve this answer












      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:



      enter image description here



      Here are the resulting fractals, if we iterate the procedure. I included the septa-version and octo-version as well.



      enter image description here



      A fun exercise is to find the general scaling factor and the resulting fractal dimension.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      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


















      • 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











      4














      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...






      share|cite|improve this answer


























        4














        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...






        share|cite|improve this answer
























          4












          4








          4






          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...






          share|cite|improve this answer












          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...







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 16 '15 at 20:24









          Zach466920

          6,18111142




          6,18111142























              0














              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






              share|cite|improve this answer


























                0














                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






                share|cite|improve this answer
























                  0












                  0








                  0






                  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






                  share|cite|improve this answer












                  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







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 11:51









                  ploosu2

                  4,5771023




                  4,5771023






























                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1237964%2fsierpinski-triangle-for-other-polygons%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Wiesbaden

                      Marschland

                      Dieringhausen