A friend told me to try to decrypt this as a fun puzzle but won’t tell me the answer:

bmf kgf dc yufy ckd yb crysn

He gave me this (which I think is the key or at least related to it):

  • shark
  • C
  • beta
  • 256
  • orioncaa
  • lu ca
  • include

He also told me to use google search where I could input things to decrypt the message, but I am stuck with no clue.

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    $\begingroup$ I think you're looking for cryptii.com $\endgroup$ Mar 21 at 6:21
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    $\begingroup$ When you go to choose the "MODEL" on cryptii.com one of the options is "Enigma M4 Shark", is that correct? $\endgroup$ Mar 21 at 7:26
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    $\begingroup$ @theooofsof2023 I've managed to figure out some of the clues, hopefully this helps. On cryptii.com, you need to decode the cipher text using the Enigma Machine, with model Enigma M4 "Shark", reflector UKW C thin, Rotor 1 Beta, and from lu ca and include, it can be determined that you need to use a plugboard lu ca and include foreign characters. Although that's all I've figured out so far. Edit: nvm,you also need Rotor 2 set to II, Rotor 3 set to V, and Rotor 4 set to VI (that's how 256 comes into play) $\endgroup$
    – CrSb0001
    Mar 21 at 13:25
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    $\begingroup$ Might I suggest putting the cypher and clues in a code block (using ` symbols) for formatting purposes? The edits were too small for me to make. $\endgroup$ Mar 21 at 13:51
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    $\begingroup$ @CrSb0001 orioncaa might stand for the position and ring. When choosing the position and ring for each rotor, there's a letter next to the number. The amount of letters in orioncaa corresponds with the amount positions + rings (4 each). Edit: nvm, didn't work. It might also be that you need to use the letters in a different order, than I used them $\endgroup$
    – WOWOW
    Mar 21 at 15:03

1 Answer 1


It's most likely undecodable (due to undercluing), but I would like to give the benefit of the doubt to OP and say that they transcribed the hints and cipher correctly. (so not OP's fault)

Here's how:

We know that from the hints, the following are correct:

1. Cipher: Enigma Machine
2. Reflector: UKW C Thin
3. Model: Enigma M4 "Shark"
4. Rotor 1: Beta
5. Rotor 2-4: II, V, VI
6. Plugboard: lu ca

Note that it doesn't matter whether foreign characters are included or not, since that would only matter if there were characters outside of the Ascii value range of 97-122. Also, characters with an Ascii value of 65-90 would only get changed to their lowercase variants, so if the solution has to do with capitalized letters, then that is very bad.

Now, here is the only reason one is able to tell why rotors 2-4 were II, V and VI in the first place, which was something that was really bugging us in chat:

Well, what if instead three letters from ORIONCAA had to be the rotors? To see why we couldn't do that, let's convert the letters to Roman Numerals:


There's a problem here: If 256 was supposed to clue to either the positions or rings, then what's the point of the ORIONCAA? We can't have ORI->XV XVIII IX be the rotors, since their Roman Numeral values fall outside the allowed rotor selection range of I through VIII.

Then there is the problem of what the solution was even supposed to look like, which led to a probably unintended red herring in the puzzle. As @WOWOW mentioned in chat:

...although we quickly ran into a problem with that, as we quickly found out. So now we knew that yes, deciphering using the Enigma Machine is the way to go.

However, assuming that everything is correct and the only problem is with the positions/rings, why isn't it solvable? Here's why:

Note: Position/Ring are from top -> bottom

Ciphertext: bmf kgf dc yufy ckd yb crsyn

|Position|Ring|      Output (Decoding)     |
|  ORIO  |NCAA|xsy ren ft gqdx mpz qc zoxus|
|  OINC  |ROCA|tzb zsg sq osql bpa an hcbdq|

|Position|Ring|      Output (Encoding)     |
|  ORIO  |NCAA|xsy ren ft gqdx mpz qc zoxus|
|  OINC  |ROCA|tzb zsg sq osql bpa an hcbdq|

So as we can see, both encoding and decoding the ciphertext with our current setting gives the same result.

Now, here's some quick math:

Assuming everything but the positions/rings are correct, then there is a $\frac1{26^8}$ or a 208827064576 to 1 chance of randomly guessing all 8 positions/rings (4 positions and 4 rings) correctly first try.

Assuming we know that ORIONCAA is correct, but not necessarily in that order, that brings it down to searching through all 10080 possible combinations (or $\frac{8!}{2!2!}=\frac{8!}{(2!)^2}$) of arrangements of the letters in ORIONCAA.

If it is to be fully together, then we have a total of 8 possible combinations to use.

However, I would like to assume best intentions and say that either the ciphertext is incorrect or that it was underclued, therefore making the cipher impossible to decode.


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