This puzzle has been solved. Please skip to the bottom of this question for the answer.

Introducing the mysterious foo encoding... based on the common computer programming template variables foo, bar, and baz. This encoding is truly unknown, as a simple sentence of characters returns an unreadable mess of foos and bars and bazes.

Let's say I input the following into my personal encoder: Noodle pie is delicious.

The program spits out the following, which is "foo-encoded":

baz-foo-baz-br- bar-baz-fo-bar- bar-baz-foo-foo- bar-foo-bar-baz- bar-foo-fo-baz- bar-fo-foo-baz- bar-foo-fo-baz- bar-foo-baz-fo- bar-fo-baz-foo- bar-fo-baz-fo- baz-fo-fo-bz- bar-baz-fo-bar- bar-foo-fo-baz- baz-fo-fo-bz- bar-fo-baz-foo- bar-foo-fo-baz- bar-baz-fo-fo- baz-fo-fo-bz- bar-fo-baz-foo- bar-foo-baz-fo- bar-fo-baz-fo- bar-foo-bar-baz- bar-foo-bar-baz- baz-fo-bar-br-

So how does this mysterious encoding work? That's for you to figure out!

EXTRA NOTE: The encoded messages may contain the following phrases: foo, bar, baz, fo, br, and bz. No, they are not mistakes, they are meant to be there.


Sirius. Two. There's your hint.

EXTRA CHALLENGE: Create an encoder/decoder for the foo encoding.


The newer versions of the encoding contain absolutely no dashes (so an example string would be barfobzbaz instead of bar-fo-bz-baz-. Also, the encoding has been renamed "FBZ" instead of the "foo" encoding.

View benji2240's solution here. I figured I would release some more resources, since the puzzle is solved:

  • The project on repl.it. Allows encoding a string and decodes the string that it developed. Written in Python.

  • A tutorial showing each step of the encoding process.

  • A website that encodes/decodes as you type, using an encoder/decoder written in JavaScript.

  • 1
    $\begingroup$ Welcome to Puzzling.SE! This looks like a fun puzzle. I see you already have a couple of badges, good for you. You can get another easy one by taking the tour. Is your fourth term br supposed to be a bar? same with 7th term fo should be a foo? $\endgroup$
    – Chowzen
    Oct 25, 2018 at 22:40
  • 1
    $\begingroup$ @Chowzen The program utilizes the keywords foo, bar, baz, fo, br, and bz. I will edit accordingly. $\endgroup$
    – moltarze
    Oct 25, 2018 at 22:42
  • $\begingroup$ @connectyourcharger Could you double-check the sixth group? It's not that I'm confident it's wrong or anything, but it surprises me a bit. $\endgroup$
    – Gareth McCaughan
    Oct 26, 2018 at 3:23
  • $\begingroup$ @GarethMcCaughan Nope, I've triple-checked the encoding and that final string is definitely correct. $\endgroup$
    – moltarze
    Oct 26, 2018 at 10:01
  • $\begingroup$ OK! Thanks, and sorry for doubting you. $\endgroup$
    – Gareth McCaughan
    Oct 26, 2018 at 11:18

2 Answers 2


Partial answer:

The string is reversed, and then each character (including punctuation) is translated to a 4-tuple of {foo, bar, baz, fo, br, bz}.

We know this because

The number of characters is the same as the number of 4-tuples. Frequency analysis shows that i is bar-foo-fo-baz. A little educated guessing on the other recurring characters shows the reversal pretty clearly, which confirms the 1:1 substitution.

Now that we know that,

We know the encoding for each letter:
. -> baz-foo-baz-br-
[space] -> baz-fo-fo-bz-
N -> baz-fo-bar-br-
c -> bar-fo-foo-baz-
d -> bar-fo-baz-fo-
e -> bar-fo-baz-foo-
i -> bar-foo-fo-baz-
l -> bar-foo-baz-fo-
o -> bar-foo-bar-baz-
p -> bar-baz-fo-fo-
s -> bar-baz-fo-bar-
u -> bar-baz-foo-foo-


This is not the obvious choice of a base-6 encoding of ASCII values. In fact, I doubt it's base 6 at all: We have the values for both o and u, which you would expect to be congruent mod 6 and therefore end in the same digit. However, o ends in baz- and u ends in foo-! If it is base 6, some additional transformation is being applied.

It is interesting that

None of that relates the clue, as far as I can tell. And the symbol bz is used only once (in the example string), for the space character.


It almost seems like, since the characters encode to 4-tuples, and 4 symbols are much more commonly used than the other 2, that we should translate each symbol into a bit pair. However, this looks like another dead end. For example, [00, 01, 10, 11] => [fo, bar, foo, baz] works for l and p, but is only "close" for the rest (and doesn't explain how br and bz might be used). I don't know where else to go, though, especially considering the hint (Sirius is a binary star).

Edit 2: Solved it! The key is

Different symbols can represent different numbers of bits:

[00, 01, 10, 11, 0, 1] => [fo, foo, br, bar, bz, baz]

This leads to a complication where a single bitstring may represented in multiple ways. One resolution to this is to apply substitutions in a priority order. For example, 11 could be baz-baz-, but bar- is preferred. The above is an example of an ordering that produces the same output as OP for the example string.

I have written an encoder and a decoder in Ruby.

  • 1
    $\begingroup$ That's exactly as far as I got and now just looking for some other kind of encoding that would translate. Nothing so far. $\endgroup$
    – MetaZen
    Oct 26, 2018 at 1:24
  • $\begingroup$ I have to say, I'm impressed, but unfortunately, you didn't get the correct answer. Keep trying, but don't over think it! It's actually quite simple once you figure it out. $\endgroup$
    – moltarze
    Oct 26, 2018 at 9:59
  • $\begingroup$ To add on, you are correct that it is not base 6. But if you really use some of the keywords in the hint, you may be able to pull out a computer a related word that can help you... $\endgroup$
    – moltarze
    Oct 26, 2018 at 10:04
  • $\begingroup$ Once you have found the encoding for each letter, isn't it over? What are you going for? $\endgroup$
    – Wais Kamal
    Oct 27, 2018 at 18:12
  • $\begingroup$ @WaisKamal, we have the encoding for each letter of the example, but this is tagged as a "computer-puzzle", so we are still looking for the underlying logic. Ideally, we'll be able to encode/decode any string. $\endgroup$
    – benj2240
    Oct 29, 2018 at 19:57

I also figured out the @benj2240 part (though I wouldn't have written it up so nicely!). Here's something else that I have observed. Again, not a complete answer. If you imagine that c,d,e are consecutively encoded in base-$n$, then we have the following:

c -> bar-fo-foo-baz-
d -> bar-fo-baz-fo-
e -> bar-fo-baz-foo-

Looking at the units column, baz would be $n-1$ (about to roll over), fo would be 1, foo would be 2. Looking at the $n$s column, baz is foo+1. So we have $n=3$, ie ternary numbers (and we ignore bar and bz and br for now).

Now, unfortunately, this doesn't quite work when you roll it out to all the lowercase letters, because there's a couple of ternary numbers that are repeated. But interestingly, it almost works, and the repeated numbers have bar in them. Here's how it works:

3 a bar-fo -foo-fo -
4 b bar-fo -foo-foo-
5 c bar-fo -foo-baz- correct
6 d bar-fo -baz-fo - correct
7 e bar-fo -baz-foo- correct
8 f bar-fo -baz-baz-
9 g bar-foo-fo -fo -
10 h bar-foo-fo -foo-
11 i bar-foo-fo -baz- correct
12 j bar-foo-foo-fo -
13 k bar-foo-foo-foo-
14 l bar-foo-foo-baz- one off
15 m bar-foo-baz-fo -
16 n bar-foo-baz-foo-
17 o bar-foo-baz-baz- correct except for bar<->baz
18 p bar-baz-fo -fo - correct
19 q bar-baz-fo -foo-
20 r bar-baz-fo -baz-
21 s bar-baz-foo-fo - one off and bar<->baz
22 t bar-baz-foo-foo-
23 u bar-baz-foo-baz- one off
24 v bar-baz-baz-fo -
25 w bar-baz-baz-foo-
26 x bar-baz-baz-baz-
27 y bar-fo -fo -fo -
28 z bar-fo -fo -foo-

Unfortunately, this coding is not correct. However, it is within one line of being correct in all cases, apart from a substitution of bar for baz.

So, somehow the bar is shifting things off by one every now and then.

Hopefully this helps someone else. I'm pretty stumped...

  • $\begingroup$ Wow, this is, again, extremely impressive. This work is amazing! But unfortunately, I think you're overthinking it a bit... think binary. $\endgroup$
    – moltarze
    Oct 27, 2018 at 0:12
  • $\begingroup$ I would say this is underthinking it. I do not think your method is going to be "simpler" in any meaningful sense. $\endgroup$
    – Dr Xorile
    Oct 29, 2018 at 16:27
  • $\begingroup$ Actually, I do think you are overthinking it, and I only say that because when I developed this, I didn't even come close to the level of complexity you are demonstrating here. I'm going to post the answer in a couple days, but I encourage you to keep trying to decipher the code. That being said, I want to commend you on your work so far, you've done a great job and you are starting to get close! $\endgroup$
    – moltarze
    Oct 29, 2018 at 21:19
  • $\begingroup$ Just so you are aware, this puzzle was solved by benji2240. $\endgroup$
    – moltarze
    Oct 30, 2018 at 11:02
  • $\begingroup$ Nice one. I got some ideas in that direction, but couldn't quite figure it out. $\endgroup$
    – Dr Xorile
    Oct 30, 2018 at 17:22

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