Comments on: Fermilab’s Strange Letter

By: Ralph

Ralph — Thu, 17 Jul 2008 04:48:02 +0000

I think we have to consider the possibility that the “Frank Shoemaker” decoding is a deliberately planted red herring. It’s way too easy, or to put it another way, the apparatus of the base-3 and base-2 coding is out of place, given the ridiculous simplicity of the cipher.

Furthermore, the implication is that something which passes for noise here is NOT in fact noise. For example, is “BASSE” truly just a typo? C’mon. If so, this isn’t really a very sophisticated or interesting exercise.

I’m guessing there is another message hidden beneath or within the portions that have already been decoded.

By: "It was beautiful, kind of like abstract art" | MetaFilter

"It was beautiful, kind of like abstract art" | MetaFilter — Thu, 17 Jul 2008 04:22:33 +0000

[...] (creator of the Homebrew Air Conditioner and employee of the Canadian Space Agency) noted that the first and last paragraphs appeared to be in base-3 and base-2, and got to work. By the 17th, he had experimented with possible mappings and reached a solution. [...]

By: partypants69

partypants69 — Tue, 20 May 2008 14:38:53 +0000

Oh, I’m stupid, “0″ is “i”, my bad.

By: partypants69

partypants69 — Tue, 20 May 2008 14:36:17 +0000

I also attempted to decode based on the number of times a symbol was defined, and also got gibberish. Although my method was different. “0″ is also undefined in the letter, could S simply be 0? If that’s the case, then the employee number is just 252. Anyone know who that is, if they exist?

By: mysteryman

mysteryman — Tue, 20 May 2008 10:31:21 +0000

the hex code seems to define everything one, two or three times but simply replacing each number with the number of times its defined and then decoding gave me gibberish (ddft/dzer, im known for my frequent rather stupid errors, so someone may want to double check this) but since the other two sections both rely on this key it may be more than cooincidence.

By: gmilburn.ca » Uncategorized » Fermilab’s Strange Letter - Progress

Sat, 17 May 2008 19:23:22 +0000

[...] now we’re getting somewhere with the strange letter Fermilab received. Looking at the top paragraph first, with I=1 II=2 and III=0 we [...]

By: neonsignal

neonsignal — Sat, 17 May 2008 09:28:53 +0000

Using a base 3 code, let
000=space
001=a
002=b
010=c
…
220=x
221=y
222=z

If |||=0, |=1, ||=2, then the first section can be rewritten as:
020 200 001 112 102 000 201 022 120 012 111 001 102 012 200 000 212 120 210 110 011 000 010 001 211 211 000 202 022 100 201 000 112 120 100 201 012
(errata: the line break at the end of line 6 has broken a symbol in two, and one of the symbols in line five is wrong)
Then the first section reads:
“Frank Shoemaker would call this noise”.

In the last section, the double bar is a gap between symbols as suggested above. Again, |||=0, |=1, and ||=2. The last section can be rewritten as:
012 111 121 110 120 221 012 012 000 112 210 111 002 012 200 000 002 001 201 201 012 000 201 100 220 202 012 012 112
(errata: the line break at the end of line 2 has broken a symbol in two, and there is a duplication of a whole triad in line 5)
Then the last section reads:
“employee number base sixteen”

Perhaps the three symbols before the last section is the employee number of the author (S252 in base 16). Or perhaps ’s’ encodes to either 1 (or less likely A), in which case the employee number would be 508. Or maybe just “S F C” for “Shoemaker, F C” as earlier suggested.

Not sure about the middle section (apart from it acting as a decoder for the employee number).

By: MidoriKid

MidoriKid — Sat, 17 May 2008 07:56:49 +0000

The first part can be decoded as “frank shoemaker would call this noise”.

Interpret ‘I’ as 1, ‘II’ as 2, ans ‘III’ as 0. The resulting trinary numbers are mapped to 000=’ ‘ (space, 001=’a', …etc This results in “frank shoemaker would camv ftvtcapsbc”. Some minor corrections to the input results in the final string.

It seems that Shoemaker is not a dead end.

By: lost

lost — Sat, 17 May 2008 04:56:54 +0000

Sneaky. The “indentation” is whitespace. Remove linebreaks and end up with this grouping.

323 233 331 112 132 … 1
333 231 322 123 312 … 2
111 331 132 312 233 … 3
333 212 123 213 113 … 4
311 333 313 331(111 … 5 113
2) … 6 (this is really 113 after bringing the leading II mark to join the I mark on previous line)
113 333 232 322(11 … 6 133
23) … 7 (this is really 133, I III III after bringing the II III to join previous line)
231 333 112 123 133 … 7
231 312 … 8

Make this base-3 and you get

212 122 220 001 021 222 120 211 012 201
000 220 021 201 122 222 101 012 102 002
200 222 202 220 002 002 222 121 211 022
120 222 001 012 022 120 201

where 000 = a, 001 = b, and so on …

Then do substitution cipher …

By: lost

lost — Fri, 16 May 2008 19:32:30 +0000

I disagree about the “binary” portion having five numbers. I see three numbers. They are rows 1/2, 3/4/5, and 6/7/8.