Golfed my Langton's ant enough to fit i… by nabbydude

function u(t) {

c.width=1920 // clear the canvas
for(i=0;i<9;i++)
x.fillRect(400+i*100+S(t)*300,400,50,200) // draw 50x200 rects

} // 122/140

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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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nabbydude

remix of d/8811 by u/nabbydude

function u(t) {

x.fillStyle=R(a=w*!x.getImageData(...P).data[0])
d+=!a||3
x.fillRect(...P)
P[j=d%2]+=(d%4<2||h*w-1)*5
P[j]%=(j?h:w)*2}{P=[d=w=960,h=540,5,5]

Jul 5 2018 4:27 PM UTC

}// 140/140

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Golfed my Langton's ant enough to fit in some scaling

Show more comments…
u/nabbydude

(and fix the aspect ratio)
u/aleamb

#cellular #automaton
u/alien

Hi there what is this part doing a=w*!x. ? :)
u/BalintCsala

@alien I think that gets the image data at the current position and if it's red, makes the fillStyle black and vice-versa
u/nabbydude

@alien @BalintCsala is right, x.blahblah is 0 if the pixel is black and 255 if its red, inverting it makes 0 into 1 and 255 into 0, and then we multiply by w because its a big number (>255) to make the color more visible
u/nabbydude

@alien @BalintCsala is right, x.blahblah is 0 if the pixel is black and 255 if its red, inverting it makes 0 into 1 and 255 into 0, and then we multiply by w because its a big number (>255) to make the color more visible


      u(t) is called 60 times per second.
      t: elapsed time in seconds.
      c: A 1920x1080 canvas.
      x: A 2D context for that canvas.
      S: Math.sin
      C: Math.cos
      T: Math.tan
      R: Generates rgba-strings, ex.: R(255, 255, 255, 0.5)

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