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function u(t) {
} //
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u(t) is called 60 times per second. t: Elapsed time in seconds. S: Shorthand for Math.sin. C: Shorthand for Math.cos. T: Shorthand for Math.tan. R: Function that generates rgba-strings, usage ex.: R(255, 255, 255, 0.5) c: A 1920x1080 canvas. x: A 2D context for that canvas.
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  • u/ivylab
    "That does not look like anything to me" :)

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remix of d/17271 by u/tomxor

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  • yote

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remix of d/10534 by u/tomxor

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  • u/tomxor
    A replacement for 0|Math.random()*2 (floored) I meant to say.
  • u/rodrigo.siqueira
    I was trying to understand what is the intuitive meaning of this value "0|Math.random()*2" in the equation. I am still not sure (and I am probably wrong) but It seems that adding this "R" value to variable "a" (X position), it will displace the image to the right or left and create a similar image. It seems that it also works as a radius (zoom factor).
  • u/tomxor
    Well my intuition from my limited understand here is that it's kind of like a fork at each point in the path that [a,b] is taking, as long as R is uniform enough the bias in the function sort of emerges and makes [a,b] follow certain paths more than others... I imagine it like a stochastic marble rolling into certain grooves in a landscape, just goto roll enough marbles to start to see the grooves.
  • u/tomxor
    Sounds a bit like 2d gradient descent O_o

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remix of d/17223 by u/tomxor

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  • u/katkip
    Amazing!!!
  • u/DataMeta
    WOW!
  • u/magna
    wowza
  • u/aleamb
    yeah!
  • u/ivylab
    Looks like a giant godlike creature, head and eyes on the left. No idea how you created this awesomeness
  • u/tomxor
    hah, yeah i didn't see that before, like some sort of weird ornate crustacean.

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remix of d/17189 by u/yonatan

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  • u/katkip
    awesome!
  • u/Coalshork
    on old systems this would be called handball and be a seperate game lol

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  • u/pavel
    If you take every point on this picture, turn it into z=x+i*y and feed it through (sqrt(3)+z)/(1+sqrt(3)*z) then decode the resulting complex number as (x', y') and put a point there, you will get d/17063.
  • u/pavel
    Of course you can't take infinitely many points so you will get a pixelated version. To get a proper version you need to solve the equation for the new, mapped, circle given the old circle so that you can use x.arc..

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