Smoler Isometric Divisions by tomxor

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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tomxor

remix of d/25593 by u/KilledByAPixel

function u(t) {

for(i=1e3;i--;b=a)a=u[j=i%461]||[960,540],x.fillRect(a[0]+=0|3*S(r=j*1.0472),a[1]+=0|3*C(r),u[a]?u[j]=[...b]:u[a]=2,3)

Jun 8 2022 3:56 PM UTC

}// 118/140

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Smoler Isometric Divisions

Show more comments…
u/tomxor

Seems to be easier to find good results if j=i%461 is restricted to primes.
u/pavel

-4 for(i=1e3;i--;x.fillRect(a[0]+=3*S(r=j/.955)|0,a[1]+=3*C(r)|0,u[a]?u[j]=[...b]:2,u[b=a]=3))a=u[j=i%461]||[960,540]
u/KilledByAPixel

nice!
u/pavel

Weirdly the numeric error from approximating pi/3 seems to matter a lot.
u/tomxor

Yeah, I think it's because it causes a deviation in the sequence at j=0. Although the output of 0|3*C(r) generally gives the expected low period of +1,+2,+1,-1,-2,-1 as j decrements, it ends on a 3, but only when using lower precision PI/3. full precision PI/3 pushes all of the 2s into 3s, which can be corrected by lowering the coefficient to 2.9, but this also makes j=0 output 2 which then seems to destroy the complexity.
u/pavel

I tried getting rid of the sin/cos by using hex coordinates but I was unable to make a complex picture. Usually just one of the six slices drew-in.


      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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