Performativizing the Polygonic

Maths is a mountain: you can start climbing in different places and reach the same destination. There are many ways of proving the irrationality of √2 or the infinitude of the primes, for example. But you can also arrive at the same destination by accident. I’ve found that when I use different methods of creating fractals. The same fractals appear, because apparently different algorithms are actually the same underneath.

But different methods can create unique fractals too. I’ve found some new ones by using what might be called point-to-point recursion. For example, there are ten ways to select three vertices from the five vertices of a pentagon: (1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 3, 4), (1, 3, 5), (1, 4, 5), (2, 3, 4), (2, 3, 5), (2, 4, 5), (3, 4, 5). Find the midpoint of the first three-point set, (1, 2, 3). Then select two vertices to go with this midpoint, creating a new three-point set, and find the midpoint again. And so on. The process looks like this, with the midpoints shown for all the three-point sets found at each stage:

v5_p3_stage1

vertices = 5, choose sets of 3 points, find mid-point of each

v5_p3_stage2

v5_p3_stage3


At stage 5, the fractal looks like this:

v5_p3_static

v = 5, p = 3


Note that when pixels are used again, the colour changes. That’s another interesting thing about maths: limits can sometimes produce deeper results. If these fractals were drawn at very high resolution, pixels would only be used once and the colour would never change. As it is, low resolution means that pixels are used again and again. But some are used more than others, which is why interesting colour effects appear.

If the formation of the fractal is animated, it looks like this (with close-ups of even deeper stages):
v5_p3


Here are some more examples:

v4c_p2_static

v = 4 + central point, p = 2 (cf. Fingering the Frigit)

v4c_p2

v = 4c, p = 2 (animated)


v4_p3_static

v = 4, p = 3

v4_p3


v5_p4_static

v = 5, p = 4

v5_p4


v5c_p3_static

v = 5 + central point, p = 3

v5c_p3


v5c_p4

v = 5c, p = 4


v5c_p5

v = 5c, p = 5


v6_1_p6

v = 6 + 1 point between each pair of vertices, p = 6


v6_p2

v = 6, p = 2


v6_p3_static

v = 6, p = 3

v6_p3


v6_p4

v = 6, p = 4


v6c_p2_static

v = 6c, p = 2 (cf. Fingering the Frigit)

v6c_p2


v6c_p3_static

v = 6c, p = 3

v6c_p3


v6c_p4

v = 6c, p = 4


v7_p3

v = 7, p = 3


v7_p4_static

v = 7, p = 4

v7_p4


v7_p5_static

v = ,7 p = 5

v7_p5


v7_p4

v = 7c, p = 4


v3_1_p2

v = 3+1, p = 2


v3_1_p3

v = 3+1, p = 3


v3_1_p4

v = 3+1, p = 4


v3_2_p5

v = 3+2, p = 5


v3c_1_p2

v = 3c+1, p = 2


v3c_1_p4

v = 3c+1, p = 4


v3c_p2

v = 3c, p = 2


v3c_p3

v = 3c, p = 3


v4_1_p3

v = 4+1, p = 3


v4_1_p4

v = 4+1, p = 4


v4_1_p5

v = 4+1, p = 6


v4_1_p6

v = 4+1, p = 2


v4c_1_p4

v = 4c+1, p = 4


v4c_p3_static

v = 4c, p = 3

v4c_p3


v5_1_p4_va

v = 5+1, p = 4 (and more)


v5_p2

v = 5, p = 2


4 thoughts on “Performativizing the Polygonic

  1. The triangle rule v = 3c, p = 2 is very similar, if not equal to the Sierpiński triangle and Rule 90 celullar automatton.

    Is this creation process any way similar with those?

    i’m really enjoying your posts, congratulations.

    • The triangle rule v = 3c, p = 2 is very similar, if not equal to the Sierpiński triangle and Rule 90 celullar automatton. Is this creation process any way similar with those?

      Yes, I assume they’re algorithmically equivalent, in one or another sense. (v = 3, p = 2) produces the actual Sierpiński triangle, but I didn’t include that because it wasn’t new.

      i’m really enjoying your posts, congratulations.

      Thank you.

      Btw, if you speak Portuguese, could you tell me if “gadanha da prata” is the right translation for the English “silver scythe / scythe (made) of silver”?

      • Gadanha de prata.

        The preposition ‘da’ is the contraction with a definite feminine article: ‘de + a’. It gives the impression that belong to or come from ‘prata’. To clearly mean ‘made of’, just use the simple ‘de’.

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