Voxeling – js1k 2016 postmortem

CeAcHlAUYAAP-ua (1)

This year, as usual, I got the inspiration from somewhere 🙂
Checking twitter while commuting I saw a re-tweet with a voxel image, I digged a bit and, to be honest, I was really impressed by the amazing work of @Sir_carma:

More work of Sir_carma

More specifically in this voxel image:
mm2WNK7 copy

Source: https://twitter.com/Sir_carma/status/651500940822974464

So I did my best to recreate it in 1k and here is the result:
http://js1k.com/2016-elemental/demo/2497

or go to the js1k page for bigger preview: http://js1k.com/2016-elemental/demo/2497

and feel free to leave a message in the reddit thread:
https://www.reddit.com/r/js1k/comments/48ndh9/demo_2497_voxeling_by_raimon_r%C3%A0fols_canvas_1024/

and, of course, do not hesitate to check all the other amazing entries this year: http://js1k.com/2016-elemental/demos

Some details of the source code, scroll down to see the whole thing:

First I created a height map of the plane, but to optmize it a bit, I made it a bit smaller & symmetric:

//full plane
p = [0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
1, 1, 0, 0, 1, 2, 2, 2, 2, 2, 2, 1, 0,
8, 4, 3, 2, 2, 3, 3, 5, 5, 4, 3, 2, 1,
1, 1, 0, 0, 1, 2, 2, 2, 2, 2, 2, 1, 0,
1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0];

//reduced & symmetric plane
p = [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 1, 2, 2, 2, 2, 2, 2, 1, 0,
5, 2, 2, 2, 3, 3, 5, 5, 4, 3, 2, 1
]

I tried to optimize it by using a string and then splitting the array into numbers but after executing the code compressor, final size was bigger, so at the end I left it this way.

Terrain and clouds are a interpolated noise function. First we need to generate some random noise:

V = 50
H = 100
D = 1000
// ...
n = new Uint8Array(D)
w.crypto.getRandomValues(n)

V & H are the number of cells we will render (Vertical and Horizontal). We just reuse the variable D instead of calculating the right amount of noise values we need… so, we actually generate too many noise values, but memory is infinite.. right?

Then, to generate a smooth terrain, we interpolate values using a cosine interpolator. To generate a field of 50×1000 points, we need 5×100 and interpolate each single value for 10×10 new generated values.

To get the integer values of the coordinate we divide by 10 (B in our case) and we do a binary OR logic operation by 0 to get rid of the floating point part. It is one of the javascript tricks that save me few precious bytes.

Another approximation I used in the code below is to get rid of PI. As we have to calculate (1.f – Math.cos(value * Math.PI)) / 2.f and value has to be between 0 and 1 I originally did the following:

x = (j / 10) | 0
y = (i / 10) | 0
F = (i % 10) / 10
E = (j % 10) / 10
E = (1 - Math.cos(E * Math.PI)) / 2
F = (1 - Math.cos(F * Math.PI)) / 2

but, as turns out, both i % 10 and j % 10 values will be between 0 & and 9, so dividing them by 3, gives us a result between 0 and 3 which is “good enough” as an approximation of a value between 0..1 multiplied by PI, hence the division by 3 and no PI mentioned anywhere in the code:

B = 10
o = []
T = S = s = 0
i = V
while (i--) {
  j = D
  while (j--) {
    y = (i / B) | 0
    x = (j / B) | 0
    k = y * H + x
    F = (i % B) / 3
    E = (j % B) / 3
    E = (1 - Math.cos(E)) / 2
    u = 1 - E
    F = (1 - Math.cos(F)) / 2
    v = 1 - F
    v = n[k] * u * v + n[k + 1] * E * v + n[k + H] * u * F + n[k + H + 1] * E * F
    o[s++] = v < 180 ? 0 : (v - 180) / 7
  }
}

Last part is to “normalize” the value. If the interpolated value is smaller than 180 we assume is water or 0. Otherwise we scale it down dividing by 7 (7 felt good, not any big reason why it is 7 and not 8 or at least, not any I could remember right now, as far as I remember was trial & error)

Another optimization I had to do, although not really proud of this one is to get rid of one gradient color. Initially voxels had 3 colors (top, front, left) but at the end, for size constraints, front and left shared the same color value. At the time I did this optimization it felt acceptable enough, but definitely looks way nicer with 3 colors, you can judge yourself in the screenshot below (left – current version, right – old version with 3 colors)

screenshots_combined

To render the voxels and avoid any depth checks or z-fight between voxels, they are always drawn from the far end (top of screen) to the bottom of the screen and from right to left, so all overdrawn voxels are correct. In the initial version I introduced checks to avoid drawing left & front faces of the voxel if they were occluded, but I had to get rid of the optimizations because of size constraints.

Clouds are rendered vertically-centered, and small mountains are just a height function from 0.

clouds_terrain

Finally all the voxels are rendered (water, mountains, clouds & plane) in a single loop with the help of a precalc table:

A = [0, -6, 1,
9, -7, 1,
7, -11, 1,
-2, -10, 1,
0, 0, 0,
9, -1, 0,
9, -7, 1,
0, -6, 1,
0, 0, 0,
-2, -4, 0,
-2, -10, 1,
0, -6, 1
]

Each value indicates the X, Y and height multiplier of the rectangle to draw. There are 3 groups of 4 coordinates, rendering 3 distorted (perpective) rectangles for each voxel.

C = [
"#A87", "#754", //0-mountains
"#f96", "#C53", //3-plane
"#fff", "#ccc", //1-clouds
"#799" //2-water
]

// number of faces to draw (on water this will be set to 1)
r = 3

// h is the value of the height map for that voxel, if height multiplier is 1
// in the 3rd coordinate of the precalc table it will be added to the y
// coordinate of the rectangle
G = -h * 4

v = 0

// s contains the value of the clouds above only when rendering water &
// mountains, "shadows" are only an alpha function
// color is computed depending on which layer are we rendering and
// which face
while (r--) {
  c.fillStyle = C[f * 2 + (!v ? 0 : 1)]
  c.globalAlpha = 1 - s / 2
  c.beginPath()
  c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
  c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
  c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
  c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
  c.fill()
}

This code is executed three times, each with a different translate (see the while(–z) loop below) and some logic depending on which phase it is. Initially there were 3 different loops which was way more clear what was being drawn each time, but took too many space and the whole thing didn’t fit in 1k.

I had plans to actually move the plane using the keyboard and even fire bullets, but I was, maybe, a bit too optimistic 😉

Below you can find the whole source code (works using the js1k shim file). Code is quite unreadable due to being highly optimized for space. Compressed file under 1k is achieved by using: uglify and RegPack with 2, 1, 1 score weight values..

p = [0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 1, 2, 2, 2, 2, 2, 2, 1, 0,
5, 2, 2, 2, 3, 3, 5, 5, 4, 3, 2, 1
]

V = 50
H = 100
B = 10
D = 1000
w = window
n = new Uint8Array(D)
w.crypto.getRandomValues(n)
w = w.innerWidth

A = [0, -6, 1,
9, -7, 1,
7, -11, 1, -2, -10, 1,
0, 0, 0,
9, -1, 0,
9, -7, 1,
0, -6, 1,
0, 0, 0, -2, -4, 0, -2, -10, 1,
0, -6, 1
]

C = [
"#A87", "#754", //0-mountains
"#f96", "#C53", //3-plane
"#fff", "#ccc", //1-clouds
"#799" //2-water
]

o = []
T = S = s = 0
i = V
while (i--) {
  j = D
  while (j--) {
    y = (i / B) | 0
    x = (j / B) | 0
    k = y * H + x
    //F&E will be less than 3.333 - good approximation for PI
    F = (i % B) / 3
    E = (j % B) / 3
    E = (1 - Math.cos(E)) / 2
    u = 1 - E
    F = (1 - Math.cos(F)) / 2
    v = 1 - F
    v = n[k] * u * v + n[k + 1] * E * v + n[k + H] * u * F + n[k + H + 1] * E * F
    o[s++] = v < 180 ? 0 : (v - 180) / 7
  }
}

setInterval(function() {
  s = w / (D + V)
  t = T % 9
  a.width = w
  b.style.background = '#635'
  c.save()
  c.scale(s, s)
  c.translate(9, H * 3)

  c.fillStyle = "#568"
  c.beginPath()
  c.lineTo(0, -H * 2)
  c.lineTo(H * 9, -H * 3)
  c.lineTo(D, -H)
  c.lineTo(D, H)
  c.lineTo(0, 0)
  c.fill()

  c.save()
  c.translate(H * 9 - t, t / 9 - H)

  c.save()
  z = 4
  while (--z) {
    if (z == 2) c.translate(-t, t / 9)
    if (z == 1) c.translate(V + B, B)

    i = 0
    while (i++ < V) {
      j = z == 1 ? 12 : H
      while (j--) {
        k = i * D + j + (z == 3 ? S : S * 2)

        if (z == 2) k += H
        f = (z == 3 && o[k]) ? 0 : z
        h = z == 1 ? p[(i 1) s = 1
        v = G = E = F = 0

        if (f < 3) {
          r = 3
          G = -h * 4

          if (f) {
            E = 40
            F = -160 + h * 2
            s = 0
            if (!h) r = 0
          }
        }

        while (r--) {
          c.fillStyle = C[f * 2 + (!v ? 0 : 1)]
          c.globalAlpha = 1 - s / 2
          c.beginPath()
          c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
          c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
          c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
          c.lineTo(A[v++] + E, A[v++] + F + A[v++] * G)
          c.fill()
        }
        c.translate(-9, 1)
      }
      c.translate(z == 1 ? H + B : H * 9 + 2, z == 1 ? -9 : -H + 4)
    }
    c.restore()
  }

  T++
  S = (T / 9) | 0
  }, B)

JS1k 2014 entries

Last year I said I had a very ambitious idea for the js1k 2013 competition but I didn’t had the time and I thought to leave it for the following year. This year, once again, I could only work on my entry the day before the deadline, so I decided to do something else and start from scratch.

I saw this animated gif long time ago (which, by the way, I don’t know the original source or author)

So I thought it would be good to make a 1k version of it.

To be honest I’m not very happy with the result, I’m pretty sure it could be done better & nicer. Anyway I submitted it to the competition and here is the result:

http://js1k.com/2014-dragons/demo/1973

Find below the full source code or check the details page on js1k

//values are just trial & error
w=window
f=w.innerHeight/w.innerWidth
w=1343
h=w*f

_=Math
C=_.cos
S=_.sin
R=_.random

F=512
Z=2200
P=6.28

l=[]
A=[]
B=[]

L=54
p=P/L
Y=N=0

//generate lamp posts in a circle on both sides and pointing to the center
df='000100110010000001011010010110110010010011011010'.split('')
for(i=0;i=3;k--) {
    x=k*F*S(p*i)
    z=k*F*C(p*i)
    x2=(k+v)*F*S(p*i)
    z2=(k+v)*F*C(p*i)

    for(j=0;j9&&j&2){
        xv=x2
        zv=z2
      }
      A[N++]=xv+df[j*3  ]*10
      A[N++]=-df[j*3+1]*512
      A[N++]=zv+df[j*3+2]*8
    }
    i+=.5
    v=-v
  }
}

//store where lamp posts end
K=N

//generate random lines
k=0
for(i=L*8;i--;) {
  l[k++]=R()*P
  l[k++]=R()
  l[k++]=-R()*128-64
  l[k++]=R()*.2+.2;
}

N+=L*96;	//8*12

setInterval(function(){
  Y+=.004
  b.style.background='#000'

  //clear screen
  a.width=w
  a.height=h

  //generate trail lines & move them
  j = K
  k=i=0;
  for(;i<l*8;i++) {
    d=l[k++]
    p=l[k++]
    y=l[k++]
    e=l[k++]
    r=(3.1+p*.8)*F

    v=4*S(y+i)
    g=[0,v,v,0]
    for(v=0;v<4;v++) {
      x=r*S(d)
      z=r*C(d)

      A[j++]=x
      A[j++]=y+g[v]
      A[j++]=z
      if(v&1) d+=e
    }
    l[k-4]+=.02

    // // move lines in both directions
    // if(i= K - trail lines, half red #f22, half white #fff
  k=i=0
  for(;i=K) {
      if(i*12=2) {
        c.lineTo(B[k],B[k+1]);
      }
      k+=3
    }
    c.fill()

    //draw lamp post light after drawing the last top horizontal quad
    //lamp posts are formed of 4 quads:
    //0 & 1 vertical
    //2 & 3 top horizontal
    //avoid drawing lights on trail lights
    if(i % 4 == 3 && i*12<k) {
      c.fillStyle="#fff"
      c.beginPath()
      c.arc(B[k-3], B[k-2], (2-B[k-1])*13,0,P)
      c.fill()
    }
  }
}, 20)

Code is compressed under 1k thanks to Google Closure Compiler and RegPack v3

As I wasn’t really happy with the result I did another quick entry based on the old good daCube2 colors & idea:

Find below the result:

http://js1k.com/2014-dragons/demo/1969

and the source code or check the details page on js1k:

w=window
h=w.innerHeight
w=w.innerWidth

_=Math
mC=_.cos
mS=_.sin
R=_.random

F=512
Z=2048

N=L=X=Y=0

A=[]
B=[]
C=[]
D=[]

addCube=function(s) {
  k=N/3
  j='000100010110001101011111'.split('')
  for(i=0;i<24;i++) {
    A[N++]=(j[i]-.5)*2*s
  }

  l='011332200445514662577367'.split('')
  for(i=0;i<24;i++) {
    D[L++]=k+(l[i]|0)
  }
}

for(z=200;z--;){
  addCube(100+z*5)
}
anim=.004

setInterval(function(){
  Y+=anim*R()+anim
  X+=anim*R()+anim
  y=Y
  x=X
  //X+=.01
  b.style.background='#444'

  //clear screen
  a.width=w
  a.height=h

  k=0
  //dy rotation
  for(i=N;i--;) {
    if(k%24*3==0) y-=anim*2
    C[k+2]=A[k+2]*mC(y)-A[k]*mS(y)
    C[k  ]=A[k+2]*mS(y)+A[k]*mC(y)
    k+=3
  }

  k=0
  //dx rotation
  for(i=N;i--;) {
    if(k%24*3==0) x+=anim*2
    B[k+1]=A[k+1]*mC(x)-C[k+2]*mS(x)
    B[k+2]=A[k+1]*mS(x)+C[k+2]*mC(x)
    k+=3
  }

  //transform 3d-2d
  k=p=0
  for(i=N;i--;) {
    C[k++]=F*C[p  ]/(C[p+2]+Z)+w/2
    C[k++]=F*B[p+1]/(C[p+2]+Z)+h/2
    p+=3
  }


  p=0
  for(i=0;i<l/24;i++) {
    c.strokeStyle="#fff"
    c.globalAlpha=(i/(L/24))
    c.beginPath()
    for(k=24;k--;){
      j=D[p++]
      c.moveTo(C[j*2],C[j*2+1]);
      j=D[p++]
      c.lineTo(C[j*2],C[j*2+1]);
    }
    c.stroke()
  }
}, 20)

My js1k entry

I was thinking to take part into the javascript 1k competition (here) but as I didn’t had too much free time and never really did that much in javascript targeting at 1k, I discarded the idea. Until I got a very ambitious idea the last day before the deadline.. how appropriate.. So I started trying few things and, to be honest, they didn’t work as I expected.. so I saved my ambitious idea for the next competition 😉

At the end, as I was already programming in Javascript I did a waving catalan flag in roughly 1k without paying attention to optimization. Pretty simple but enough for being my first time on a 1k javascript competition. Anyway, after doing some basic optimizations I still got a couple hundred bytes left so I’ve decided to add some extras. Check the demo for more details 🙂

Click the image below for the demo:
Screen Shot 2013-04-01 at 1.11.46 AM

Just in case anyone is interested here is the fully uncompressed code with few comments.

k = 0
w=window
h=w.innerHeight
w=w.innerWidth
h2=h/2
W=H=40

setInterval(function(){
    // clear screen
    c.width=w
    c.height=h

    o = 0;
    p = []

    //Generate flag points. Before size-optimizing the code they were generated only once and stored in an external array.
    //To save one loop from the code they're generated every single frame.. oh well..
    for(j = 0; j < H; j++) {
        for(i = 0; i < W; i++) {
            x = 1.2 * i * w / W +8 * Math.sin((j + k) * 0.15)
            y = 1.2 * j * h / H + 8 * Math.sin((i + k) * 0.3)
            z = 512 + 32*Math.sin((i+j+k) * 0.2)

            p[o] = (512 * x) / z
            p[o+1] = (512 * y) / z
            p[o+2] = z;
            o+=3
        }
    }

    // draw the flag
    o = 0
    for(j = 0; j < H-4; j++) {
        for(i = 0; i < W-1; i++) {
            // compute cross product to compute light factor
            // this can be probably done in waaaay less bytes
            l = []
            l[0] = p[o+3] - p[o]
            l[1] = p[o+4] - p[o+1]
            l[2] = p[o+5] - p[o+2]

            m = []
            m[0] = p[o+W*3] - p[o]
            m[1] = p[o+W*3+1] - p[o+1]
            m[2] = p[o+W*3+2] - p[o+2]

            n =[]
            n[0] = l[1]*m[2] - l[2]*m[1]
            n[1] = l[2]*m[0] - l[0]*m[2]
            n[2] = l[0]*m[1] - l[1]*m[0]

            n = n[2] / Math.sqrt(n[0] * n[0] + n[1] * n[1] + n[2] * n[2]);

            // square it to make results more dramatic
            n *= n;

            // choose flag color (red/yellow) to render the different color stripes depending on the vertical coordinate
            g = 221
            b = 9
            if(((j / 4) % 2|0) == 0){
                r = 253
            } else {
                r = g
                g = b
            }

            // apply the lightning factor and convert it back to integer using "|0"
            r=r*n|0
            g=g*n|0
            b=b*n|0

            a.fillStyle="rgba("+r+","+g+","+b+",1)"
            a.strokeStyle=a.fillStyle

            // actual render of the quad. Fill & stroke to polish (antialiasing, subpixel,..)
            a.beginPath()
            a.lineTo(p[o], p[o + 1])
            a.lineTo(p[o+3], p[o + 4])
            a.lineTo(p[o + 3 + W*3], p[o + 4 + W*3])
            a.lineTo(p[o + W*3], p[o + 1 + W*3])
            a.fill()
            a.stroke()
            o+=3
        }
        o+=3
    }

    // draw the "nyan cat" like trail
    w4=4*w/6
    w1=w/10
    cols=["#5E96FF","#387DFF","#0F47AF","#0D3E99","#0B347F","#08265B"]
    for(j = 0; j < 32; j++) {
        for(i = 0; i< 6;i++) {
            a.fillStyle=cols[i]
            a.strokeStyle=a.fillStyle
            wb=w4 / 32
            hb=w1 / 6
            x=j * wb
            y=h2 + (i-3) *hb - hb*Math.sin(k*0.2 + j *0.6)
            a.fillRect(x,y,wb,hb)
            a.strokeRect(x,y,wb,hb)
        }
    }

    // draw the star
    a.fillStyle="#ffffff"
    a.strokeStyle=a.fillStyle
    a.beginPath()
    for(i = 0; i < 5; i++) {
        s = Math.sin(k * 0.2)*32
        l = w1 * Math.sin(i* 1.256) + s/4
        m = w1 * -Math.sin(i* 1.256 + 1.57) + s
        n = w/25 * Math.sin(i* 1.256 + 0.628)
        o = w/25 * -Math.sin(i* 1.256 + 2.19) + s
        a.lineTo(w4 + l, h2 + m)
        a.lineTo(w4 + n, h2 + o)
    }
    a.fill()
    a.stroke()
    k+=0.5
},20);

And the compressed result thanks to Google Closure Compiler and JSCrush:

k=wJdow;h&HV;w&Wid^;h2=h/2;H=W=4setInterval(function_{c.wid^;c.hV=h;o=p=[]j<hui<wx=~i*w/W+815*(j!y=~j*h/H+83*`!zK2+Z`+j!]Kx/z,1]Ky/z,2]=z,o=ji=cols[i,wb4/Z,hb1/6,x=j*wb,y=N`-3)*hb-hbk+6*j,;="#ffffff";;;5>is=Zkl1*O/4,m1*-+~57O,nY+628oY-+2.19Ow4+l,Nm)w4+n,No);_;_;k+=5},20);GJ(+X,a.lJeTo(a.strokefor`=~256*i[2]	[0][1]a.fillX=Rect(x,y,wb,hb)],++)3*W*;for(j=a.begJPa^_=w),p[o*n|0,0;2*F #;i]-o+=3!+k)$])&.Jner@*mC=GMa^.sJinK=51L+",Nh2+O)+sQ=[U;jVeightXStyleY/25*Z32^th_()`(i~1.0.';for(Y in $='~`_^ZYXVUQONLKJGC@&$!	')with(_.split($[Y]))_=join(pop());eval(_)