I wrote a new code as my last one seem to fail so badly. I'm doing this by the way, as it's to do with my Mathematics degree.

Here's the new code:

141813 61618118, 161415 115 1820181142618 31411262443

25 15114618 19 261815 1714318

614143 158171078

Each line is a sentence. The numbers for example if we had 123456, 1 could be a letter or 12, 123, 1234, 12345, 123456. Or it could be split up so 1 is a letter, then 23456 is another letter, the amount of numbers doesn't mean that's the amount of letters in the word. Each collection of numbers does mean it's a word but not necessarily the amount of letters in that word. Hope this one proves to be more challenging. Also I would like to know how people would think of going about decrypting it, either using ciphers or mod or anything else. Also special characters like "!" and "?" are encrypted, but "," and "." aren't. Hope this is more challenging.

All blocks of numbers make up a word/letter, and they do make grammatical sense as they are all english sentences, so it's not just a jumble of letters.

  • $\begingroup$ Can a letter contain another one ? if "12" is a letter, can "1234" be another letter ? $\endgroup$
    – Fabich
    Feb 24, 2018 at 13:04
  • $\begingroup$ Yeah, so 12 could be a letter, and 123 or 1234 could be another letter. $\endgroup$ Feb 24, 2018 at 21:49

1 Answer 1


(EDIT: the answer is now complete, there's a full decryption in the last spoiler block.)

Partial answer, since OP wants to know about the methodology:

The strings contain a huge number of 1s, and there are very few occurrences of several consecutive large digits. This is symptomatic of

strings consisting of concatenated small numbers. The ”teens” account for the abundance of 1s, and the lack of big two-digit numbers would cause the shortage of consecutive large numbers.

So, the first idea would be to split the ’words’ into

small integers. Indeed, accepting only numbers in the range 1-26, the few ”exceptions” requiring consecutive single digits occur mostly at the ends of the lines, where we might expect some punctuation.

So the first cipher to check would be

a substitution cipher, where each letter is replaced by its position in the alphabet. If it doesn’t work, check for caesar encryption, or do a letter (number) frequency analysis.

Going to check for that in the near future, but please do feel free to continue from here if you like.

EDIT: Here's my attempt at a decryption, after a bit of letter frequency analysis and guesswork.)

 14-18-13   6-16-18-1-18, 16-14-15   11-5   18-20-18-1-14-26-18   3-14-11-26-24-43 
  H  e  y   t h  e  r  e,  h  o w     i s    e  v  e r  o  n  e   d o  i  n  g  ?
 25   15-1-14-6-18 19 26-18-15 17-14-3-18
  I    w r  o t e  a   n e  w  c  o  d e
 6-14-14-3 15-8-17-10-78
 G  o  o d  l u  c  k !

Looks like there might a mistake in the ciphertext, or a typo in the original plaintext: the fifth number should probably be 182018113142618. Also, if my guesses are correct, the capital letter encodings overlap with the lower-case letters, which is not usually considered kosher in a cipher. I may of course have guessed wrong too.

  • $\begingroup$ Ok I see what you've gone for here, but I doubt very much any cipher or frequency analysis will work on this code. But see if you can decrypt it if possible.. $\endgroup$ Feb 24, 2018 at 21:44
  • $\begingroup$ That's amazing that you've come up with that tbh, as i thought frequency analysis wouldn't work, but i'm gonna try and readjust my code and see if i can get it to the point that frequency analysis and guesswork won't work. But thank you so much for the working out and showing how easy the code was to decrypt $\endgroup$ Feb 24, 2018 at 21:48
  • $\begingroup$ The method I used was to split the strings into biggest possible numbers not exceeding 26, and then 18 stood out as the most common number, so it was probably "e". There were two 18s in the second word, which provided a nice starting point for solving the rest of the text. The message length was probably exactly what was needed for decryption; a "regular" substitution cipher can theoretically be solved from 28 characters of ciphertext, but in this case there's quite a lot of ambiguity involved. (Wouldn't recommend this cipher for transferring missile strike coordinates, for example.) $\endgroup$
    – Bass
    Feb 24, 2018 at 23:43
  • $\begingroup$ Ok thank you so much for this... I'm going to upload an improved version... The English will be different but I'll upload it in a bit and would really like your feedback on whether it's solvable and how you'd go about doing it. I'll comment once I've posted it. $\endgroup$ Feb 24, 2018 at 23:59
  • $\begingroup$ If the cipher gets much harder than this, it's probably going to be out of scope for this community: already this one required using cryptanalysis methods that I wouldn't necessarily count as "normal puzzle solving". I'm not familiar with the crypto SE folks, but in case this community doesn't like a pure cryptanalysis puzzle, you might try your luck there. $\endgroup$
    – Bass
    Feb 25, 2018 at 0:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.