There's an apocryphal story about Richard Feynman about how he used to keep a dozen or so random problems in the back of his mind and made a little bit of progress on them every time he saw a connection, until finally he'd solve one and everyone would think he magically figured it out instantly. This was a bit similar except I'm not nearly at that level and I've only been able to do that for one problem instead of a dozen.
https://onlinetools.com/math/l-system-generator?draw=AB&skip...
edit: On second thought, this probably generates the other fractal, but I'm not sure.
https://onlinetools.com/math/l-system-generator?draw=ABCD&sk...
The previous one fills out the Koch island.
Edit: just noticed how you encoded a flip (AB <--> CD) between iterations like how the matrix flips the orientation of space. Super neat!
Exactly! There is also a less obvious relationship between A and B too: B is a A "backwards" (A rotated 180°, starting the curve from the opposite end).
The strategy was to put 5 lines on the plus sign on the sides of the 5 cells, with the idea that each line eventually fills out a neighboring cell in subsequent iterations. I found one such path that had a chance of working. Not sure if this makes sense.
Loved the 3d visualizations
It reminds me of this thing I built some time ago while playing with recursive decimation to generate effects similar to fractals from any image
You can play with it here: https://jsfiddle.net/nicobrenner/a1t869qf/
Just press Blursort 2x2 a couple of times to generate a few frames and then click Animate
You can also copy/paste images into it
There’s no backend, it all just runs on the browser
Don’t recommend it on mobile
Right now, roughly, the algorithm recursively divides the image by doing decimation (ie. picking every other pixel), and keeps the decimated pixels as a second image
Not sure how that algorithm would apply to a 3d data structure
Do you know how 3d objects/images are usually represented?
It would be cool to recursively decompose a 3d object into smaller versions of itself :)
https://chriskw.xyz/images/fractal/thumbnail.jpg
I think it would work perfectly as a mosaic eventually, but for the time being I'm perfectly content with the "rustic" 8x11 graph paper sized one taped to the wall. Currently planning to put up a slice of the orthotopeflower as a companion piece once I find matches for the colored pencils I used back then.
I expect 41+14 to be 12 (two right plus two up equals two right and two up).
Long addition in long form below uses:
'=' to show equivalent lines (reordering of terms (1+2=2+1), spliting numbers (41=40+1), adding single digits (1+4=22))
'->' for when the algorithm gives a digit
'<' for when we move over a column
41+14
= (40+1)+(10+4)
= 40 + 10 + (1+4)
= 40 + 10 + 22
-> 1s digit = 2
< 4 + 1 + 2
= 22 + 2
= 20 + 2 + 2
= 20 + 41
-> 10s digit = 1
< 2 + 4
= 0
-> done
== 12
It still works in the alternative system (14+21 should equal 12)
14+21
=10+20+42
->2
<1+2+4
=13+4
=10+3+4
=10+31
->1
<1+3
=0
==12https://raw.githubusercontent.com/cies/haskell-fractal/refs/... [17MB, sorry Github]
It contains the Haskell code that produced it: https://github.com/cies/haskell-fractal
Especially the `sharpen` function was interesting to come up with (I used some now-offline tool to do curve fitting for me): https://github.com/cies/haskell-fractal/blob/master/fractal....
Fun little project. :)
Question to the author: what would you recommend to hang on my kid’s wall today?
Relatable. Huge part of my decision on what degree to pursue was a list of problems (mostly linear algebra) I needed to solve, but didn't have the guidance (and internet connection) to.
Favorited—I'll be coming back to absorb more, as my aging semi-fluency in engineering physics and SQL doesn't help much with the notation I last saw in the 1980s.
If you start with the shape of iteration 3, it tessellates as a 5x5 square tile. Make an infinite grid of those tile shapes with one iteration 3 version in the center. Treat that center tile as the center square in the iteration 3 pattern and color the tiles around it according to how the 2nd and 3rd iterations were built of squares. This gives you the 4th and 5th iteration and you can continue to iterate on the coloring outwards to color the grid of tiles in the wallflower pattern.
I believe that there is a typo in the pattern formula (right after "Looking closely you might pick up on the pattern"): it should read
5**(n/2) instead of 5**n
5**((n-1)/2) instead of 5**(n-1)
Also, regarding Knuth's mistake: Youtube comments point out that his fractal is in fact correct; he just mistook the beginning point with the end point. Loosely speaking, the fractal is symmetrical about its middle turn, which is precisely the one Knuth believed to be incorrect. All in all, he still made a fractal-related mistake, so the conclusion holds.
taeric•1mo ago