[The Math Game] Get the left side of th… by MrM@_S

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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MrM@_S

remix of d/2573 by u/MrM@_S

function u(t) {

c.width=200;t||(z=X=2,s=-1);onkeyup=a=>z*=a.keyCode&15||1;x.fillText(X+"="+z+"? Score:"+s,2,9);z==X&&(X*=s++%4*~~t%4+1,z=1);z>X&&(frame=0)

May 5 2017 12:51 PM UTC

}// 138/140

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[The Math Game] Get the left side of the equation equal right side using the numbers 1-9. It is easier if you know how to multiply.

u/sigveseb

Cool game! Only problem is when the goal number's prime decomposition contains a prime number with more than one digit (e.g. 11 and up). Then it is impossible to enter the correct answer!
u/MrM@_S

"3+3+1=7" so X*=s++%4*~~t%4+1 would never have a prime composition of 11 etc. I am pretty sure that my algorithm prevents such causes.....
u/MrM@_S

wait it is "(0-3)*(0 to 3)+1"
u/MrM@_S

(0 to 3)
u/MrM@_S

that leaves a 1 through 10 possibility


      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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