did anyone else find the codes manually? I did. I quickly realised that it'd break off the code as soon as it couldn't lead to a match, greatly reducing the search space (which was only 256 options to begin with).

As soon as I noticed that, I worked out all the codes by hand in only 15 minutes (including playing with the experiments I found).

I just iterated through the search space (starting at all zeroes, counting up in binary) and eliminated whole sets of numbers as soon as it broke off the input. e.g. starting with all zeroes it breaks off at the third zero, meaning anything under 00100000 was impossible (immediately eliminating 32 of those 256 options in my search).

Did anyone else do that too? Perhaps more interesting: did anyone have a completely different approach?

pong: 10000001

simone: 11010011

eightbit: 01010011

song: 11011011

synth: 10001000

ascii: 01111111

bowling: 01110101

rocket: 01000101

burger: 00111001

cat: 11100111

bacon: 10010000

This was done a while ago.

    ww.mode.register("home", ww.mode.HomeMode, null);
    ww.mode.register("cat", ww.mode.CatMode, 231, 8);
    var isAndroid = navigator.userAgent.match(/Android/);
    isAndroid || ww.mode.register("space", ww.mode.SpaceMode, 42, 8);
    ww.mode.register("pong", ww.mode.PongMode, 129, 8);
    ww.mode.register("bacon", ww.mode.BaconMode, 144, 8);
    ww.mode.register("simone", ww.mode.SimoneMode, 211, 8);
    ww.mode.register("eightbit", ww.mode.EightBitMode, 83, 8);
    ww.util.getAudioContextConstructor() && (ww.mode.register("song", ww.mode.SongMode, 219, 8), ww.mode.register("synth", ww.mode.SynthMode, 136, 8));
    ww.mode.register("ascii", ww.mode.AsciiMode, 127, 8);
    ww.mode.register("bowling", ww.mode.BowlingMode, 117, 8);
    ww.mode.register("rocket", ww.mode.RocketMode, 69, 8);
    ww.mode.register("burger", ww.mode.BurgerMode, 57, 8);

Also assuming that everyone has the skill to dig through javascript files.

    Number(x).toString(2)

  (x).toString(2)

128, 64, 32, 16, 8, 4, 2, 1

----

Let's decompose 231.

----

231 >= 128; That's a 1 in first position. 1

231 - 128 = 103 >= 64; We have a one in the second position. 11.

103 - 64 = 39 >= 32; One in the third position. 111.

39 - 32 = 7 < 16; Zero in the fourth position. 1110.

7 < 8; Zero in the fifth place. 11100

7 >= 4; 111001

7 - 4 = 3 >= 2; 1110011

3 - 2 = 1 >= 1; Final result: 11100111. Enjoy your cats :)

All you have to do is integer division by two and keep the remainder. You don't even need to know the powers of two.

Here's an example with 137.

  137 % 2 = 1
  68  % 2 = 0
  34  % 2 = 0
  17  % 2 = 1
  8   % 2 = 0
  4   % 2 = 0
  2   % 2 = 0
  1   % 2 = 1

Final result: 10001001

You can use a similar method to convert the fractional part of a float, but with multiplication.

And 10010000 took me to https://developers.google.com/events/io/experiment-bacon

Time to see if I can find any more! :)