#tree #waitforit #uncompressed 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/18153 by u/Kirjava

function u(t) {

with(x)for(beginPath([X,Y=X,r=200,a=2]=u[T=-~T]||[999]),fillStyle=R(i=4,3e3/r);i--;fill())lineTo(...u[2*T+i]=[X-C(--a)*r,Y+S(a)*r,r*.8,a-9])

Apr 7 2020 3:38 AM UTC

}// 140/140

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#tree #waitforit #uncompressed

u/pavel

This is a breadth-first traversal of a binary tree, one frame at a time. The tree is maintained in a 1D array (2n+i notation) that I simply iterate left-to-right. The loop does double-duty for drawing the current node and inserting the children. The shape has four corners so the loop runs four times and inserts four children: 2n+0, 2n+1, 2n+2, 2n+3. My mind was blown when I realized no extra code was needed to prevent the adding of the extra children because the first two children of the next iteration will overwrite them!
u/joeytwiddle

#broccoli
u/katkip

interesting
u/pavel

https://i.imgur.com/FZ9Whj4.png


      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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