At first sight, an encrypted postcard from 1909 looks like many others I have introduced on this blog. However, the encryption method used is unusual.

Readers of this blog undoubtly know Tobias Schrödel, …

Schroedel

… who is known as the world’s leading comedy hacker, an outstanding crypto book expert, a successful book author, a crypto items collector, and a TV journalist (Stern TV). Tobias, who is also a friend of mine, has now provided me his latest acquision: a postcard encrypted in a number code. Here’s a scan of the text side:

Postcard-Newport-tex

As can be seen, this postcard is dated 1909. The recepient is (like in most cases of this kind) an unmarried woman, a certain Laura Kauscher (?) living in Elgin, Illinois, a town near Chicago. I bet that the sender was Laura’s lover (most encrypted postcards were sent by young men to their spouse).

The card was stamped in Newport, Rhode Island, on the US East Coast. The picture side depicts a ship:

Postcard-Newport-pic

Most encrypted postcards I am aware of are MASC-encrypted, i.e., a simple letter substition was used. This is apparently not the case here. Instead, an unusual number code has been applied. Only the numbers 1 and 2, as well as “+” and “.” are used.

My first guess was that the dot separates letters, while the plus-sign separates words. However, this would mean that every letter is encoded with at most four 1’s or 2’s, which makes an alphabet consisting of 16 letters only. This doesn’t sound plausible.

As another possibility, the plus-sign together with the 1 and the 2 might be used for a ternary code (e.g., “12+”=A, “11+”=B, “2+1″=C, …). However, there are only four plus signs in the cryptogram, which would be unusual.

All in all, I have no idea what kind of encryption the sender of this postcard used. Can a reader find out more? If so, please leave a comment.


Further reading: Who can decipher this encrypted postcard sent from San Francisco to Paris?

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Kommentare (16)

  1. #1 Richard SantaColoma
    https://proto57.wordpress.com/
    28. November 2018

    I tried the Biliteral Cipher, both with the 2’s as A, the 1’s as B, and the other way around. I didn’t get anything… but perhaps it is some variation of that (using 4 character letters? 6?)…. considering there are only two characters.

  2. #2 TWO
    28. November 2018

    `Perhaps the + sign indicates a sentence break?

    I have a hunch the end decodes to : kiss kiss

    The method could be the adressee has a list of words with corresponding numbers.

  3. #3 Gerd
    28. November 2018

    >However, this would mean that every letter is encoded
    >with at most four 1’s or 2’s, which makes an alphabet
    >consisting of 16 letters only. This doesn’t sound
    >plausible.

    As also letters consisting of less than four symbols are possible, there are enough letters for a full alphabet. Look at morse code: the whole alphabet is encoded with up to four dits or dahs.
    However I found nothing useful here when translating 1 to dit and 2 to dah or vice versa.
    Gerd

  4. #4 TWO
    28. November 2018

    If you convert it to Morse it starts with GNJ,

    2 means a dash
    1 means dot

    If I did this correct even

  5. #5 Norbert
    29. November 2018

    Just a suggestion:

    LOVE AND A GREAT BIG HUG AND ???
    WILL

    Maybe at the end of line 4, a dot is missing in the middle of “1212”. “12.12” would yield a plaintext “ee”. Then “2121.12.12” could be something like “see” …

  6. #6 Norbert
    29. November 2018

    Too bad “… AND KEEP WELL” doesn’t work – it would make the most sense to me.

  7. #7 Norbert
    29. November 2018

    Or maybe like this?

    LOVE AND A GREAT BIG HUG AND MY
    X (for kisses)
    WILL

  8. #8 Norbert
    29. November 2018

    I am pretty convinced of my last proposal. Note that the short number sequences, consisting of one or two digits, are used for the frequent letters AEINOT. (Aside from that, I cannot make out a “system” the cipher is built upon, though.)

  9. #9 häh?
    29. November 2018

    @Norbert: Great work! Yes, I think, your last solution is correct, too.

  10. #10 Thomas
    29. November 2018

    Norbert:
    I like your solution. My guess was also “love and a…”, but “hug” at the end as the last 3-letter-word before the sender´s name. Thus I got stuck looking for an 8-letter adverb or adjective + a 6-letter adjective or noun (compound with “hug”) ending with “…and”, because there are no word separating crosses at the end of lines 2 and 3. Since there seem to be no matching 8- and 6-letter words, I think your assumption that the sender omitted the crosses at the end of lines 2 and 3 is correct.

    BTW: Familysearch has two entries on “Laura Kaucher” living in Elgin. If the sender was her soon-to-be husband (unfortunately no marriage is mentioned), their relationsship didn`t last long: In 1910 she was single, in 1920 her status was: “divorced”. Perhaps she split up with him because of marital communication problems due to messages she didn´t understand?:-)

  11. #11 Mike Schroeder
    Code ID
    29. November 2018

    The code is the General Service Code used by the US Military for Wig Wag telegraphy from 1862-1912. The message reads “Love and a great big hug and kiss. Will”.

    I recognized the code because I delivered a talk on Wig Wag in the US Civil War about 8 years ago.

    The post mark of Newport, RI suggests Will may have been a Signalman in the US Navy.

  12. #12 Gerd
    29. November 2018

    Anyone tried if there is a systematic MASC between the morse-code interpretation and the cleartext found by Norbert?
    Gerd

  13. #13 Tobias Schrödel
    München
    29. November 2018

    @Norbert – Bravo! Great work, really. Actually, I haven‘t speculated on a quick solution on this card.

  14. #14 Gerry and Andrea
    29. November 2018

    @Norbert: The cipher “system” is a somehow twisted frequency distribution:
    1 and 2 symbols: IT NEOA instead of ET AOIN
    3 symbols (with letters S and C missing): .U.HRXLD instead of SHRDLCUM
    4 symbols for the “mixed” rest

  15. #15 Klaus Schmeh
    29. November 2018

    @Norbert: Thanks and congratulations.
    @Thomas, Mike: Thanks for the background information.

  16. #16 Neal Weems Best
    Great Falls
    9. September 2021

    Solution is simple substitution.
    Pick out o Vekoslav devout why. Rapp