I thought I'd see if I can calulate PI … by Jezza

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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Jezza

function u(t) {

c.width|=0;x.font='99px monospace';if(!t)p=3,a=2,o=0,q=2;p+=24/a/(a+1)/(a+3)/(a+=4);x.fillText((p+o).toFixed(15),9,99);if(p>q)p-=q,o+=q;q/=2

Jan 31 2018 10:32 PM UTC

}// 140/140

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I thought I'd see if I can calulate PI in 140 chars. Math.PI is 3.141592653589793, which is 15 decimal places. 15 decimal places is precise enough to calculate the circumference of the solar system (including Pluto) with an accuracy of a meter. The formula is Nilakantha's from the 15th century. You could also just print 245850922/78256779 but doing that doesn't seem much better than just printing Math.PI.

u/donbright

#pi #math


      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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