BigFloat in 140 chars! This is the Long… by rodrigo.siqueira

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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rodrigo.siqueira

function u(t) {

D=(a,b)=>{L=a.length,i=r=0,R='';while(i<53){d=i<L?~~(a[i]):0,R+=(i==L?".":"")+~~((K=d+(r*10))/b),r=K%b,i++}throw(a+"/"+b+"="+R)};D("1",9998)

May 30 2019 5:23 PM UTC

}// 140/140

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BigFloat in 140 chars! This is the Long Division Algorithm used here to calculate 1/9998. Note that this simple division results in Powers of Two: 0001, 0002, 0004, 0008, 0016, 0032, 0064.... Try to change 9998 for 9997 and you will see Powers of Three (1, 3, 9, 27, 81...). Guess what 9996 does? D is the division function. (a,b) are (numerator, denominator). 53 is precision (number of digits). Add more "9" to fit larger power numbers. E.g. To see powers up to 2**40 (that is 1099511627776), change denominator from 9998 to 9999999999998 and total number of digits from 53 to 534.

u/rodrigo.siqueira

I have discovered a truly remarkable demonstration of this result, but this comment is too small to contain.
u/joeytwiddle

#math #division
u/marc8

quoting Fermat... LOL


      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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