4

A friend of mine just wrote one thing on a board : TEJDAO QZ OCUQJ UVANBVCR

The only indication was "Enigma Puzzle".

If it is really an Enigma like code how could I have a chance to solve it ? Shall I try every possibility until I found a solution (I guess there is calculator for it) but knowing I'm not even sure of the language used...

Any ideas ?

Edit :

My friend remembered this indication :

Attiser les braises de la connaissance (French)

Which could be translated :

Stir up the embers of the knowledge

Baldráni
  • 141
  • 1
  • 6

3 Answers3

6

A mysterious stranger has told me that the default settings on the Cryptii V2 Enigma machine will solve this cipher, producing this plaintext:

ALICIA JE TAIME TOUJOURS

Or in English, "Alicia, I still love you"

The default settings are:

Model:           Enigma M3
Reflector:       UKW B
Rotors:          I II III
Rotor positions: M C K
Rings:           A A A
Plugboard:       (empty)
codewarrior0
  • 6,694
  • 1
  • 33
  • 57
6

Unfortunately, if this really is enciphered using an Enigma Machine, there is virtually zero chance that you could ever decrypt it without more information. An Enigma Machine uses a very complicated set of encryption criteria, and without knowledge of what settings were used for encryption, it is virtually impossible to decrypt. (Access to large amounts of encrypted messages and some known plaintext was the only way the British ever managed to crack any Enigma messages.)

I suspect that the word "Enigma" here doesn't refer to an enigma machine. Most likely the cipher is encrypted using some more basic cipher. But the ciphertext is very short, and the number of unknowns is too great, so without more information, finding a credible solution is unlikely.

Running the ciphertext through quipqiup gives a number of amusing potential solutions (my favourite is MUSCLE OF ETHOS HILARITY), but nothing that seems credible.

Similarly, trying to solve using a French dictionary at rumkin.com yields nothing credible.

My suspicion is that unless your friend is willing to be more forthcoming and supply at least a couple of the following, this cipher will likely remain unsolved indefinitely.

  • The cipher used
  • A (much) longer ciphertext
  • The language of the plaintext
  • Some/all of the keys/settings required for the given encryption method
GentlePurpleRain
  • 25,965
  • 6
  • 93
  • 155
0

I know the Ring settings (TEJ) and the ENCRYPTED ground settings (DAO) by looking at the first 6 letters if it is Enigma, so that helps massively. It, unfortunately, would be too hard to decrypt without him telling us what method he used.

garr890354839
  • 373
  • 1
  • 10
  • Wait, how can you know Ring settings from TEJ? Can that encode any 3 first letters of the message? Or is that some sort of standard? (If it is, I'd have to say it's a pretty silly one, shipping (part of) the key with every message...) – mr23ceec Oct 22 '16 at 13:10
  • @mr23ceec That is a standard in Enigma Messages, although it usually is denoted in Equals Signs, (so it would be: some junk, the day number (refer to a code sheet) = TEJ DAO = QZOCU QJUVA NBVCR). The german wehrmact knew what they were doing, the allies had NO IDEA, and it still remains the order to do decoding with Enigma. – garr890354839 Oct 23 '16 at 17:27
  • Re: "the allies had no idea" wasn't that how the Enigma was broken? (There was something about that in Imitation Game, but I didn't quite get it at the time.) – mr23ceec Oct 25 '16 at 13:51
  • The allies use the fact that a letter NEVER ENCODED into itself to break the Enigma. It was way harder to break than the Lorenz, but that was another machine. Say that you have some Enigma-encoded message – garr890354839 Oct 26 '16 at 14:58
  • (this is a continuation of my previous comment) ...in this case of this message: QZOCU QJUVA NBVCR. I know that there can't have a q in the first letter's position, a z in the 2nd, an o in the third, and so on. So therefore the message CANNOT be QZOCU QJUVA NBVCR. This, unfortunately does not inform us what the actual rings were. There are 8 wheels, and 3 reflector choices (A, B and C), along with 20 of 26 letters swapped, or "steckered" with each other, and after 1944 there was another security increase, and that just blew, because, of the 40 settings, 10 of them were just another combination – garr890354839 Oct 26 '16 at 15:07
  • ...of plugs, so with ALL of those combinations, it is pretty much impractical to try them ALL by hand. Therefore you need a machine to crack these messages, and you need a rather powerful computer to do so, because there are trillions of combinations, but only one of them is correct. Therefore, it is hard to break by hand. If this is a one time pad, however, then there are $26^17+26^16+26^15+26^14+...+676+26$ different choices, literally quadrillions of choices. Let's hope it is just the Enigma, and not a one time pad. – garr890354839 Oct 26 '16 at 15:16
  • It's interesting that you immediately jump to one-time pad, when it's much more likely (by puzzle design) that it's a Vernam cipher (or similar) with the "indication" being used as key. The only trouble is, it's way too long for that, by my reckon. – mr23ceec Oct 27 '16 at 09:24
  • This message is uncrackable by any means, and even going through all $876(either (22) or 3)*40$ and all 40 UHR settings is impossible, so therefore, it is impossible – garr890354839 Oct 27 '16 at 15:12
  • There are only 24 Uhren. ;) More to the point, I though we decided this was more likely Vernam then Enigma? – mr23ceec Oct 28 '16 at 08:51
  • but then decryption is nigh impossible. instead of that many wheels, we now have exactly $32^21$, or, worst-case scenario, $65536^21$ DIFFERENT KEYS to encode that. A vigenere is most likely. – garr890354839 Oct 28 '16 at 16:33
  • (stupid 5-min edit) but then decryption is nigh impossible. instead of that many wheels, we now have exactly $32^(21)$, or, worst-case scenario, $65536^(21)$ DIFFERENT KEYS to encode that. And besides, we would be using 12 letters, so that key would be TEJDAOQZOCUQ, not TEJ DAO. – garr890354839 Oct 28 '16 at 16:43