Metatron's Cube by pavel

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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pavel

remix of d/20898 by u/cantelope

function u(t) {

for(c.width=i=640;i--;T=!T&&!x.beginPath(x.stroke()))j=i/16-T*i&30,r=(j+4)/12<<6,x.arc(320+r*C(a=j/1.91+t),180+r*S(a),i&30?0:32,0,7)

Dec 4 2020 3:01 AM UTC

}// 132/140

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Metatron's Cube

Show more comments…
u/pavel

Golfed this one a little bit :) #symmetry
u/pavel

i=[4 bits for start circle index][4 bits for end circle index offset from start circle][1 bit to decide start vs end]
u/pavel

That's 2**9=512 which is too small for power-of-two circle sizes but start clicle index wraps around so larger i is only redundant.
u/pavel

end-circle-offset == 0 is easy to check and means start and end are same circle, so draw that circle.
u/pavel

Otherwise draw the current circle with radius 0 which is same as lineTo.
u/pavel

T=!T&&!x.beginPath(x.stroke()) starts/strokes the path every two circles.
u/pavel

Here is a version before all the bit hacking for(c.width=640,i=16;i--;)for(j=16;j--;x.beginPath(x.stroke()))for(k=2;k--;)q=(i+j*k)%16,r=((q+2)/6|0)*64,x.arc(320+r*C(a=Math.PI/3*q+t),180+r*S(a),!j*32,0,7)


      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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