Barb.lat.bar

The main library of the Vatican owns an encrypted letter from 1628. Can a reader break this cryptogram?

Everybody interested in codebreaking should know MysteryTwister C3 (MTC3), the largest crypto puzzle website in the world.

On MTC3, you can try your luck on hundreds of encryption challenges in four different classes of difficulty: simple (1), medium (2), hard (3), and unsolved (X). Some of the challenges were designed by me. I mainly contributed level 3 and level X challenges – i.e., some of the tough ones. Beginners should start with the level 1 challenges, of course. You will see that most of them are easy to break, even for a layman.

 

The Vatican Challenge #1

Prof. Bernhard Esslinger, MTC3 steering group member and reader of this blog, has informed me about a challenge that was published on MTC3 in January this year. It’s named Vatican Challenge part 1 (there’s also a second part, which I will cover in  a later blog post). The Vatican Challenge part 1 was authored by linguist and crypto history expert Prof. Beáta Megyesi from Sweden. It’s a level X challenge, which means that the solution is not known.

The Vatican owns a large collection of historical writings, including manuscripts, letters, and printed books. Many of these writings are currently digitized to be made available to the public. Some of these old documents contain ciphertexts that have not been deciphered yet. The following three double pages (they probably represent a letter) come from the Biblioteca Apostolica Vaticana, the Main Library of the Vatican:

Barb.lat.6956_0004_m

Barb.lat.6956_0005_m

Barb.lat.6956_0006_m

According to Beáta, these pages were labled with the following information:

Germania: Giovanni Battista Pallotto, Nunzio all’Omp eratore. Cifre del 14 maggio al 29 agosto 1628.

This means that this cryptogram was created in 1628. Giovanni Battista Maria Pallotto (1594-1668) was an Italian cardinal.

 

What kind of encryption was used?

Beáta Megyesi has transcribed the whole encrypted text. The complete transcription is available for download here. Here’s the transcription of the first page:

# 002r.jpg

<IT De Inenunchi? 14 Maggio 1628.>
6 2 3 9 6 7 5 0 1 7 3 7 8 2 3 3 2 3 6 5 0 2 3 4 3 0 5 1 8 2 2 0 0 4 6 2 3? 9 0 1 5 6 7 3 8 6 3 0 2 1 4 4 6 3 3 7 2 0 8 5 2 1 8 7 4 1 3 6 7 0

3 4 2 4 0 9 8 1 5 7 1 5 0 4 1 9 0 8 6 3 1 2 3 9 0 3 6 0 6 5 0 1 8? 5 7 1 2 2 7 0 2 5 9 0 3 6 3 0 8 2 3 0 9 2 9 2 6 0 2 1 6 7 0 0 8 4 3 7 0 2 3

2 2 0 0 2 5 3 5 7 0 1 6 2 3 9 6 7 3? 6 3 0 8 0 2 1 2 3 0 2 2 2 0 9 3 6 0 9 1 7 7 4 8 4 7 1 5 5 0 2 6 0 9 0 2 3 4 3 0 9 8 3 6 5 0 2? 2?

7 0 4 6 2 5 2 0 3 5 0 7 ? 1 7 6 8 7 9 3 1 5 4 1 8 2 0 0 1 9 0 8 4 5 4 8 3 6 0 8 4 7 1 6 3 8 2 2 0 0 3 6 3 0 6 0 2 0 1 6 9 1 5 3 1 2 2 7 0

3 2 0 3 8 1 7 7 0 3 7 7 9 0 2 9 9 3 1 5 7 8? 7 3 1 6 3 0 2 1 5 4 5 1 8 6 2 3 9 6 7 5 0 1 0 0 8 7 7 5 1 0 0 4 4 3 0 2 5 3 5? 7 0 1 7? 6 8 1 8

4 4 0 9 3 6 9 0 2 5 3 5 5 0 1 8 6 3 7 0 1 4 4 0 2 6 3 6 6 5 0 2 0 3 6 0 9 8 5 2 8 5 6 9 1 4 7 8? 2 3 7 0 9 3 1 2 2 3 6 9 0 2 5 9 0 9 2

7 0 1 6 9 8? 3 2 7 0 3 9 2 2 6 6 5 0 7 7 9 0 2 9 9 3 1 8 6 5 1 2 0 ? ? 3? 4 7 0 3 2 2 0 2 3 6 4 1 5 2 8 1 7 9 3 1 7 3 7 1 6 4 9 9

6 0 5 0 8 9 0 8 7 6 6 8 1 7 6 1 3 3 9 0 2 5 9 3 1 3 2 0 2 6 3 9 6 1 5 7 1 5 0 4 8 9 0 1 2 2 5 0 6 0 1 3 4 3 6 9 0 6 0 3 0 5? 1 8

6 9 1 7 7 4 8 4 7 9 4 1 4 4 9 0 5 2 5 0 8 0 0 1 8 6 9 3 6 5 0 2 5 6 0 4 6 2 0 8? 1 5 7 8 6 0 3 0 4 6 7 0 3 9 0 2 1 6 6 1 0 0 2 3 2 2

9 0 2 5 3 5 2 6 5 0 1 8 4 3 0 7 3 5 2 5 0 7 1 3 2 0 7 6 6 0 3 3 6 0 9 1 6 9 1 2 0 2 3 2 2 3 0 2 5 7? 5 0 0 7? 6 9 0 8 7 3 1 4 4

0 2 6 3 3 7 2 0 1 0 2 1 3 6 7 0 6 0 3 0 4 6 0 9 8 1 6 6 0 4 7 0? 9 9 0 7 1 3 3 7 0 3 3 0 9 3 2 5 0 1 7 7 5 6 3 7 0 2 8 2 2 4 6

0 5 6 2 1 6 9 8 8 0 4 6 0 5 6 0 2 0 3 6 6 5 8 1 0 0 1 2 5 3 6 9 0 2 9 3 1 7 6 8 7 9 3 8 1 6 3 1 5 2 7 7 6 0 2 9 5 0 3 2 7 0 1 7 3 1

2 3 6 0 4 3 2 5 0 0 2 9 2 6 2 0 1 6 9 8 7 7 4 1 4 7 8 4 5 5 0 2 1 2 2 0 2 4 3 2 3 2 0 2 5 7 0 8 5 2 1 6 6 5 0 1 2 3 6 0 3 3 9 0

2 5 7 3? 0 5 8 9 0 1 7 7 4 1 4 7 1 8 5 5 2 0 6 0 3 0 6 0 2 0 1 8 5 2 1 4 9 7 8 5 6 8 2 2 0 0 2 9 3 3 0 7 6 6 4 4 5 0 3 9 0 7 8 1

5 6 6 6 5 0 1 2 7 2 0 3 5 2 5 3 6 7 0 1 9 0 8 3 2 5 0 2 9 2 7 9 0 1 6 9 8 1 6 2 9 9 5 0 7 0 1 8 8 2 3 9 0 2 2 9 0 1 6 7 5 1 8

6 6 1 0 0 3 6 4 6 0 5 0 6 7 0 1 6 9 1 7 7 5 8 2 2 0 2 3 6 2 5 0 9 1 6 9 1 0? 6 3 0 2 1 4 4 0 2 7 7 5 0 3 7 0 0 8 9 0 1 5 7 8 1

6 0 2 9 1 2 3 0 4 3 9 1 9 9 0 9 2 9 2 5 0 5 6 6 7 0 7 7 0 3 8 3 2 0 2 7 7 9 0 3 6 5 0 9 0 3 6 7 0 1 5 6 7 3 1 3 2 5 5

5 0 2 0 6 0 3 0 8 3 3 9 3 4 6 0 0 3 5 0 0 8 6 2 3 9 3 3 7 0 1 0 0 3 3 1 5 0 2 9 6 0 4 0 2 5 0 0 3 6 6 6 0 2 1 9 0 8 4 4

0 2 3 6 8 1 5 0 2 9 9 2 0 0 3 9 2 6 0 2 1 4 7 8 4 4 6 0 2 3 4 3 0 9 3 6 7 0 1 0 0 3 3 1 8 2 0 6 6 0 7 9 9 0 5 0 2 3 6

9 0 8 9 7 1 2 2 0 2 6 6 2 0 2 6 0 3 1 4 4 6 0 8 3 6 0 5 4 2 0 6 4 3 2 0 3 5 0 7 8 6 6 9 7 6 0 4 0 9 3 3? 5 2 1 3 4 6 0

7 0 2 9 8 1 2 5 9 0 3 6 6 3 9 5 1 9 0 8 2 2 0 7 0 5 1 7 6 8 4 4 9 0 3 6 0 9 2 3? 3 5 4 0 8 5 2 1 5 6 7 1 6 9 5? 1

0 4 0 5 4 3 5 0 2 5 0 2 3 6 9 0 8 5 0 6 6 1 2 9 3 8 6 6 0 9 7 0 2 9 0 2 3 6 3 3 7 0 1 0 0 3 6 5 0 6 0 2 0 3 4 9 0

9 5 8 2 3 6 0 7 0 1 0 0 6 3 3 2 7 0 1 9 0 8 3 2 7 0 2 9 0 7 2 3 4 2 5 0 6 0 9 3 1 5 9 8 3 2 7 0 6 6 7 4 9 9 5 0

7 0? 1 7 7 0 7 2 9 2 5 0 0 6 6 9 3 1 0 2 8 3 4 0 3 6 6 0 7 9 9 2 9 0 2 8 3 3 0 5 4 9 ? 0 3 6 7 0 8 4 7 1 5 3? 0 0 6 0

9 0 6 0 2 0 1 3 6 5 0 3 2 6 0 4 3 2 0 2 5 7 0 8 6 6 1 5 0 2 9 6 0 4 0 9 3 8 1 5 7 8 6 4 3 6 0 4 0 5 3 6 9 0 1 2 0 8 6 7 5

1 8 7 7 0 9 3 3 0 9 6 5 6 2 8 9 2 0 0 2 9 2 5 0 9 1 6 6 1 5 6 7 8 9 0 1 7 4 8? 3 2 5 0 4 6 4 2 0 7 9 2 0 0 2 9 2 6 3 0 1

5 6 7 3 1 6 3 0 2 8 0 0 2 9 6? 7 2 5 2 0 8 8 6 7 1 7 7 5 1 7 7 0 9 3 3 9 0 6 5 7 7 0 3 1 2 9 3 8 6 3 1 3 6 5 0 4 4 0 9? 4 6 0 5

1 7 6 8 4 4 0 9 3 6 9 0 2 5 3 5 4 0 1 7 4 7 8 2 3 0 2 2 2 6 0 9 3 1 5 2 8 7 7 5 1 5 6 7 1 4 7 8 6 9 1 4 2 2 8 1 2 0

2 3 2 2 0? 9 3 5 2 5 2 0 6 0 2 0 2 9 7 0 8 3 5 3 6 2 0 1 3 4 3 6 0 9 6 0 9 0 1 3 5 0 9 7 7? 2 2 0 5? 1 6 0 9 0 2 9

As can be seen, the ciphertext consists of a long sequence of digits. There are no separations. This means that the digits need to be read in groups of equal length, probably in pairs. Here’s a similar cryptogram from about the same time.

The encryption of Pallotto’s letter is probably based on an encryption table that assigns a letter or a word or a syllable to each digit group. If the table contains only letters, we deal with a letter substitution (probably a homophonic one, as there are certainly more different digit groups than letters in the alphabet). If the table also lists words or syllables, the cipher used is a nomenclator.

It is known that the Vatican used nomenclators already in the 14th century. So, it would come as no surprise to encounter such a method in the 17th century. It is difficult to break a well-designed nomenclator. However, many nomenclators were not well-designed. For instance, many used low numbers for letters and high numbers for words. In many nomenclators the entries are sorted, like A=11, B=12, C=13, which makes the cipher easier to use but is less secure. The following picture shows a nomenclator (taken from the paper Briefe durch Feindesland published by Gerhard Kay Birkner in the proceedings of the conference Geheime Post in 2015; it’s, of course, not the one used here):

Nomenclator-Post

If you have a clue about these encrypted texts, please leave a comment. However, please don’t publish the solution here. If you know the solution, please register at MTC3 and hand it in. You will receive many points for solving a level X challenge.


Further reading: An unsolved message sent to Robert E. Lee

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Kommentare (19)

  1. #1 Thomas
    8. März 2018

    Presumably in 1628 the Roman Curia still used former cipher secretary Matteo Argenti’s ‘Trattato familiare di cifre’ from1610, in which he had devised several number cipher systems (https://archive.org/stream/diegeheimschrift00meis#page/70)

  2. #2 Klaus Schmeh
    8. März 2018

    @Thomas:
    Thanks for the hint. I knew that there is this book about Vatican ciphers by Aloys Meister. I should have known that it makes sense to check whether Pallotto’s cipher is mentioned there.

  3. #3 Thomas
    8. März 2018

    Pallotto’s cipher isn’t mentioned there, because Meister’s ‘Geheimschrift’ doesn’t cover the time after Argenti’s ‘Trattato’. But I think Argenti’s successor as cipher secretary (probably Antonio Ferragallo according to Meister) still used his methods to devise new number ciphers.

  4. #4 Thomas
    8. März 2018

    There is a book containing Pallotto’s letters from 1628, esp. to cardinal Barberini (“Nuntiaturberichte aus Deutschland, 4. Abteilung, 1. Band: Hans Kiewning, Die Nuntiatur des Pallotto, 1628” (1895), content: http://d-nb.info/366957422/04). This website https://faccion.hypotheses.org/materiales/documentos provides a link to archive.org, but on my device it doesn’t lead to a legible google-copy of the book. But Kiewning’s book is also available in many libraries.

  5. #5 Thomas
    8. März 2018

    If I’m not wrong, the manuscript Barb.lat. 6596 contains not only the ciphered letters, but also the deciphered texts: The pages with numbers are preceded by pages with plaintext, beginning with “Recifrato di…” (deciphered by …). For example the first letter and the first enciphered text – as shown in the images above – were both written in Innsbruck (“Di Ispruck”) on May 14 (1628): https://digi.vatlib.it/view/MSS_Barb.lat.6956. The Curia in Rome received Pallotto’s letters and deciphered them. (In 1628 Pallotto was the nuntius to the Emperor in Vienna and sent a couple of letters from Austria). I suppose every letter from May 14 – Aug. 29 was enciphered with the same key. Since the decipherer’s handwriting is difficult to read, it would be easier to have Kiewning’s printed edition at hand.

  6. #6 Thomas
    8. März 2018

    This is Kiewning´s book which I have found. https://archive.org/stream/nuntiaturberich07romgoog#page/n25/mode/2up/.Here he describes how the letters to the curia were deciphered in Rome by the recipient. After “Recifrato di” the date is given when the letter was deciphered. Unfortunately the book doesn´t contain the letters from Ms. Barb. lat. 6956, Kiewning states that the Barberini Archive wasn´t accessible for him at that time.

    But apparently also the letters in Ms. Barb. lat. 6956 have been deciphered in Rome. You can see from the Vat.Lib. scan (link #5) that the plaintext of the letter is given on the previous page: In image 3 of the scan the plaintext starts with “M…(illegible for me) come sorrisi”. The same words have been written – presumably by the decipherer to mark the cipher letter – above the 1. line of the number cipher in image 4. That means image 3 shows the deciphered plaintext of the cipher letter from May 14 in image 4.

  7. #7 Thomas
    8. März 2018

    Who can read the Italian handwriting in image 3 of the scan to compare it to the cipher letter: “M… come scrissi p… della mia famiglia con…”?

  8. #8 Norbert
    8. März 2018

    Mandai come scrissi parte della mia famiglia con le robbe per Mantova.

    robbe = goods (cf. https://books.google.de/books?id=qdAtp13yp_AC&hl=de&pg=PA665#v=onepage&q&f=false)

  9. #9 Norbert
    8. März 2018

    Let me propose that each new level X challenge on MTC3 should be counterchecked by Thomas before being published – if the solution is already out there, hidden in the furthest corner of the internet, he will reveal it within a few hours :-)

  10. #10 Thomas
    8. März 2018

    @Norbert
    Thanks a lot, “Mandai”, that’s it! Have you already compared “Mandai come scrissi parte della mia famiglia con le robbe per Mantova.” to Beata’s transcription in order to reveal the cipher key?

  11. #11 Thomas
    8. März 2018

    The recipient of Pallotto’s letter was cardinal Barberini: Kiewning’s edition has a letter from Barberini to Pallotto from June 3, 1628, in which he refers to Pallotto’s letter from May 14: “la sua cifera d’Ispruch de14. di maggio” and Pallotto’s report on Mantua (Mantova) in this letter (https://archive.org/stream/nuntiaturberich07romgoog#page/n181, No.24). Kiewning’s commentary in footnote No. 3. “Not existant, maybe an error.” No, Barberini was right, Kiewning didn’t know Ms. Barb. lat. 6956 containing Pallotto’s letter to Barberini.

  12. #12 BE
    10. März 2018

    Thanks to Thomas, this MTC3 challenge became a known-plaintext challenge instead of a ciphertext-only challenge. So Norbert’s proposal to check each new level X challenge on MTC3 by Thomas before publishing might be very helpful :-)

    However, level X challenges require not only to reveal the plaintext but also to explain the whole method including its key (and probably its homophons here).

    In this special case, where a ciphertext is embedded in a cleartext context, further things have to be clarified:
    1. To decide whether a cleartext in the context of ciphertext is the plaintext or not.
    2. If it is the plaintext, create the key given a partly interpretable plaintext because of unclear handwriting – worth to develop methods for it in general in CrypTool as well. This info can be very useful for nomenclatures.
    3. If it is a cleartext (and not the plaintext), use some kind of topic analysis to reduce the search space during the decryption.

    So this challenge is still considered as unsolved by its author, and we keep the challenge online.

  13. #13 Thomas Ernst
    Latrobe
    11. März 2018

    Have been following this exchange with great pleasure …

  14. #14 Thomas
    11. März 2018

    Meanwhile I have figured out the key and posted it on MTC3. Principally, it is based on Argenti’s method No. 5 (as numbered by Meister, link #1). Two digits for the alphabet with homophones, two one-digit nulls as word separators, but additionally with three digit entries for certain persons and cities (nomenclator).

  15. #15 BE
    11. März 2018

    Hello Thomas.

    Thanks a lot. Your solution was received at MTC3. As usual for level X challenges, it’s now up to the author to check the solution and to accept or reject the proposal.

  16. #16 Thomas
    12. März 2018

    Prof.Beata Megyesi has confirmed my solution. But I was asked by MTC3 not to post the key here, at least for now.

  17. #17 Norbert
    12. März 2018

    @Thomas: Congratulations! :-)

  18. #18 Klaus Schmeh
    12. März 2018

    @Thomas: Congratulations!! I’m proud that a reader of this blog has solved another crypto mystery.

  19. #19 Marc
    12. März 2018

    Congratulations, Thomas !