Have you seen this fractal?

[ChristopherJones]

Hello, everybody! My name is Christopher Jones, new to this forum.

I am wondering if anyone recognizes this. I've been scouring the internet and can't find it anywhere:







I made that in MSPaint. It's rather unusual in that you can divide it into three equal parts and it keeps the same shape. For example:




That's kind of unusual. Actually it makes a mirror image, slightly rotated. You have to split it up twice to get the orientation back.

Self-similarity, outside in:




And inside-out:






All of the above were done in MSPaint using a tiny seed and cut/paste. I've been playing with this shape for years and I keep staring at it because of the odd way it retains its shape when divded.

Has anyone seen this before? Does it have a name?

[lkmitch]

Interesting!  I haven't seen it before, to my knowledge.  Can you explain how you did it?  The MS Paint part.

[claude]

http://classes.yale.edu/fractals/Labs/FracTileLab/FracTile.html seems related?

[ChristopherJones]

Interesting!  I haven't seen it before, to my knowledge.  Can you explain how you did it?  The MS Paint part.

I will. It's hexagonal. When I get off of work I'll post a diagram.

[kram1032]

ah I see what you are doing: You start with a hexagon, then put three copies of it around it to form what looks like a T made from four hexagons. From there it's rinse-repeat: Use the new shape, stack it around the original hexagon (I think - or are you just using three copies directly without a central hexagon?) and keep going. The new shape is stacked together to form the old.

As such it appears to be closely related to something like the Koch Snowflake, stacking more and more Triangles onto each other - although in that case, you keep stacking triangles, not new shapes.
Another related one appears to be the dragon curve. This one does stack new copies of itself.
Both the Koch Snowflake and the dragon curve also feature similar tiling patterns to what your shape has.

Really nice find.

[ChristopherJones]

ah I see what you are doing: You start with a hexagon, then put three copies of it around it to form what looks like a T made from four hexagons. From there it's rinse-repeat: Use the new shape, stack it around the original hexagon (I think - or are you just using three copies directly without a central hexagon?) and keep going. The new shape is stacked together to form the old.

As such it appears to be closely related to something like the Koch Snowflake, stacking more and more Triangles onto each other - although in that case, you keep stacking triangles, not new shapes.
Another related one appears to be the dragon curve. This one does stack new copies of itself.
Both the Koch Snowflake and the dragon curve also feature similar tiling patterns to what your shape has.

Really nice find.

@kram1032: You got it! ... and thank you!


Ok, here's the building strategy:










Each increase in size makes a mirror image and a slight rotation.







Look for the "big triangle" to help see whether it's a right hand or left hand shape.

I have to tilt my head so the triangle is pointing "down", and then I can see it.







Every other iteration, the shape regains its orientation.






I used a four-pixel "T" tetramino to make the first pictures I posted.




The tetramino gives more detail faster. In the image below, the top is 3x zoom and the bottom is 10x.






Finally, look at the recursion. This creates external 'structures', or at least the possibility of making them.






@Claude, yes definitely related. That was helpful. I took a closer look at the perimeter measure after reading that, and have a couple of questions.

With each iteration you you 'lose' two sides of the 'hex' to the inside. So with each iteration you triple the area, but only double the perimeter.  (If you think of the area as constant, then the perimeter increases by 3/2 with each iteration.) So the perimeter is infinite, right? 

Is that enough information to calculate the dimension? (It' been twenty years since I studied formal math... help.)

[Minor edits]

[claude]

With each iteration you you 'lose' two sides of the 'hex' to the inside. So with each iteration you triple the area, but only double the perimeter.  (If you think of the area as constant, then the perimeter increases by 3/2 with each iteration.) So the perimeter is infinite, right?

Is that enough information to calculate the dimension?

Yes, the perimeter is infinite, and here's the dimension calculation according to this page on the site I mentioned ( http://classes.yale.edu/fractals/Labs/FracTileLab/FracTileAP.html )

  2 = (P2/P1) = (A2/A1)^(d/2) = 3 ^ (d/2)
so
  2 = 3^(d/2)
so
  log 2 = (d/2) log 3
so
  d = (2 log 2) / (log 3) = log 4 / log 3
approximately
  d = 1.2618595071429148...

I hope I got that right!

[kram1032]

The way this is build also is vaguely related to diffusion aggregation fractals.

Indeed now I wonder whether anything interesting happens if you change the typical algorithm:

- start with a point at the center
- add another point somewhere nearby
- let the second point Brownian walk until
-- either the particles collide (in which case they stick together)
-- or the walking particle leaves a given (perhaps adaptively expandable) area
- repeat

to add another step:
- rescale the new system to the original size of one patricle (keep the shape)
- change particle shape primitive to this rescaled system

So more and more complex particles are going to form (similarly to how they do here)

Additional possibility: assign random initial orientation and rotation speed to spawned particles. Grid based methods probably won't work well in this case)

[ChristopherJones]

This fractal has one unique property that I can see, in that it can be divided into three equal parts and retain its shape.



Is this the only shape that can do this? I am thinking it has to be. The iteration is turning one hex into three, then treating the group of three as a single hex and repeating.

1) Can anyone think of any other shape (at all) that can be divided into three parts and retain its shape? (Including reflected/rotated versions.) Really, I can't think of a single one. Can you?

2) If so, is it a significant find in any way. Is this something that should be published somewhere?

Or is it just a doodle, like the drawings I made with it?

Honest feedback please, I want to know one way or the other.

[kram1032]

I already told you that the dragon curve admits various tilings similar to this http://en.wikipedia.org/wiki/Dragon_curve
And so does the Kock Snowflake

And by that token, so do the Sierpinsky Triangle and the Sierpinsky Carpet and the Menger Sponge and the...
there are an infinitude of examples.

[ChristopherJones]

The tilings are not unique, I'll readily admit.

However, if you divide a Koch snowflake or pretty much any shape into thirds, it doesn't keep the same shape. This one does. I'm wondering if a) it's unique in this property and b) if that's significant.

[kram1032]

Sorry to rewake an age-old thread but I thought up an answer for you:
It's really easy to contruct a very simple shape with this property:

So you want a shape where, if you split it into thirds, the individual pieces all are similar to the original shape.
Let's try a rectangle:
It has two side lengths and .

All we are asking is

This can be reformed into
or, expressed as proportionality relation,
So there you have an infinite family of solutions to this problem. You can pick one out of an infinite number of arbitrary extra constraints. For instance, what happens if or ? As long as you get a solution that is real and positive for both and , the resulting rectangle's proportions will properly allow for splitting it into 3 similar copies of itself.

That same argument can be used for any desired proportionaliy: If you want copies, just go for propotions of or or  for - for instance, if you have a 4:8-rectangle and split it into 4 pieces, you'll end up with 4 2:4-rectangles.

And doing this for halving is actually what gave rise to the DIN-A4 standard. A0 has an area of 1m². All higher ones are repeatedly halved versions of A0 and are designed to have the same side-length-ratios as A0. (With the caveat that actual formats are rounded to the nearest millimeter).

Finding more complex examples like the one you found here will be more difficult, but no doubt there are infinitely many of those around.

[ChristopherJones]

Thank you so much for your reply! I'm glad I checked back in.

[Lee Oliver]

My initial reaction to your shape really reminded my of a Gosper Island!  I will come back to look at yours closer later, but I just thought it might be worth mentioning because I didn't see anyone else mention it.
Here is a wiki link to it https://en.wikipedia.org/wiki/Gosper_curve