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estimate

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I ran a simulation and painted a raster with the tree 200*26 by 200*18, simulating an area of 6.5*4.5 up to a minimum square size of side length of .001, squares may have been rendered a little large, the area I got (#pixels/(200*200*16))=13.88431875 68.144.80.168 (talk) 20:08, 1 November 2008 (UTC)[reply]

No square is rendered smaller than 1 pixel. When I ran the simulation with a minimum square size .0001 area=14.8469515625. When raster size set to 150*26 by 200*18 area=14.219888888888889. Clearly a much better simulation is needed to establish a proper result. 68.144.80.168 (talk) 20:26, 1 November 2008 (UTC)[reply]
Raster=400*26*400*18, min square size=1 pixel, area=13.5057578125, min square size=1/2 pix, area=14.078665234375. These numbers lead me to believe the answer could be 14.

68.144.80.168 (talk) 20:40, 1 November 2008 (UTC)[reply]

Raster 800*..., squ size (1,1/2), area=(13.79561943359375,14.2233669921875), same except resolution 1000*..., (13.4443739375,13.898576625) 68.144.80.168 (talk) 21:21, 1 November 2008 (UTC)[reply]
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Exact area

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I believe I have proved the area of the Pythagoras tree to be exactly the rational number 12823413011547414368862997525616691741041579688920794331363953564934456759066858494476606822552437442098640979 / 877512406035620068631903180662851572553488753575243048137500508983979170248733422547196905684808937723408093

This is roughly 14.6133694787067. I have shown that the Pythagoras Tree can be described as a (fairly large) set of tiles, each of which is composed of 4 other tiles in that set, giving rise to a system of linear equations that allow calculating the area of each tile. My proof, along with code to construct and solve the system of linear equations can be found at https://penteract.github.io/pythagTree.html . I understand that I cannot add this to the article myself as it is original research. Tesseract16 (talk) 17:33, 21 July 2022 (UTC)[reply]

@Tesseract16 I used a very similar strategy to recreate your result, and I am confident that you are correct. I would be interested in further discussing the solution with you, If you'd like to talk privately you can send me a message via this tool. Dchew89 (talk) 21:12, 14 March 2024 (UTC)[reply]
I'm very happy to talk more about it, although I haven't made enough edits to be able to email you using the linked tool. Hopefully my settings now let you email me; I can also talk here or on a user talk page. Tesseract16 (talk) 05:21, 5 April 2024 (UTC)[reply]
@Dchew89 I should have pinged you in the reply above. Tesseract16 (talk) 18:27, 5 April 2024 (UTC)[reply]
Oops looks like I can't email you either lol, we can just talk here. I was wondering if you were planning on or interested in writing a paper on this topic, possibly considering other partitions of this fractal as well as applications to other fractals (like you mentioned briefly on your website). I personally believe that a significantly more efficient partitioning exists but I am still working on finding it. @Tesseract16 Dchew89 (talk) 22:17, 5 April 2024 (UTC)[reply]
I've been sporadically working on a paper about this (area.tex in https://github.com/penteract/pythagorasTree) and I'd be keen to work together on it if you're interested. The repository also contains some further results, including the fractal dimension of the boundary, and a similar calculation for a probabilistic fractal involving decomposition of equilateral triangles rather than squares.
What do you mean by a more efficient partitioning? Showing some of the squares are equivalent to each other, or using a different base shape? Something close to the decomposition into half-square triangles used in https://www.hindawi.com/journals/ijmms/1997/850947/ might work, since it's analysing a fractal closely tied to the Pythagoras Tree.
I think I've found you on Discord and sent a message. @Dchew89 Tesseract16 (talk) 03:10, 6 April 2024 (UTC)[reply]

[sentence moved to talk page]

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"It would be interesting to know if there's an algorithmic relationship between the value of the base half-angle and the iteration at which the squares first overlap each other."

This doesn't belong on a full wiki page, but it should be ok here. 172.59.191.66 (talk) 02:59, 13 April 2023 (UTC)[reply]