#breakout #tribute 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/13487 by u/KilledByAPixel

function u(t) {

for(i=t?c.width|=d=_=>u[j=q&-64|p>>6]=-!u[j]|1:(p=q=540,r=7,s=4);i--;)u[i]||x.fillRect(i?i%64*64:p+=r*=d(p+=r),i?i:q+=s*=d(q+=s),z=i?64:9,z)

Apr 27 2019 4:26 AM UTC

}// 140/140

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#breakout #tribute

Show more comments…
u/pavel

Had to make some visual sacrifices to get this #uncompressed
u/pavel

I discovered p+=r*=d(p+=r);q+=s*=d(q+=s) as part of messing around with this. Why does it work?
u/pavel

:)
u/KilledByAPixel

Oh wow, nice! That's the code that updates the position, velocity and applies collision for x and y. It needs to call the d function to check its colliding and also set the block to 1 if it does. It needs to check x and y separately so it knows which side to bounce off
u/pavel

Notice how d only takes one argument, and it ignores it.
u/KilledByAPixel

I see what you did there. So many great tricks in this one.
u/pavel

p+=r*=d(p+=r) the inner p+=r increments p and d ends runs with that value, however the outer p+= caches the old value of p so when d resolves the inner p+=r is silently ignored!
u/pavel

#physics


      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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