Peter Nüchterlein, the writer-in-residence in Wernigerode, Germany, has asked me for help. In a Vienna archive he has found over 100 encrypted documents from the Thirty Years’ War. Today, I am going to introduce five of them. Can a reader decipher them?

Last week, I received an email from Peter Nüchterlein, the writer-in-residence in Wernigerode, a town in the German state of Saxony-Anhalt. He wrote: “In the course of my research on the Thirty Years’ War, I have recently viewed and partly photographed several thousand archive pages in the Vienna Court Archive. There are a dozen or so pages with a numerical encryption code, nobody in the archive could decipher. It is a mixed text of German sentences and numbers.”

It  goes without saying that I suggested to publish these findings on my blog. Mr. Nüchterlein agreed. Meanwhile, he has even found over 100 encrypted pages (mainly letters) in the material he copied. These letters were written by ministers to the Austrian emperor. Mr. Nüchterlein believes that at least four different encryption methods have been used. Today, I am going to introduce excerpts from five documents that are encrypted in one of these methods.

Some readers will remember that last year blog reader Thomas Ernst solved a number of encrypted letters from the Thirty Years’ War. Two years ago, blog readers Christoph Tenzer and Thomas Bosbach broke a cryptogram created by Albrecht von Wallenstein in the same time period. This proves that some of my readers are quite experienced in cryptanalyzing cryptograms from the 17th century. I’m optimistic that the solution of the ciphertexts Mr. Nüchterlein provided me will be found, too.

 

Five encrypted passages

Here are the five excerpts:

#1

Wien-RKA-K1-17.09-506

#2

Wien-RKA-K1-17.09-525

#3

Wien-RKA-K1-17.09-527

#4

Wien-RKA-K1-17.09-528

#5

Wien-RKA-K1-17.09-529

 

Solution approaches

As can be seen, every encrypted word consists of an even number of digits, like 332014 or 2234. This means that we probably deal with an alphabet consisting of two-digit numbers. In other words, 332014 needs to be read as 33/20/14, while 2234 is actually 22/34. The Wallenstein cryptogram mentioned above is written in the same style. The cryptogram also contains a number of commas and colons.

As far as I can see, none of the two-digit numbers appearing in the cryptogram is greater than 60. This means that we probably deal with a letter substitution that includes a number of homophones or nulls or phrases.

Can a reader break this encryption? I am sure, Peter Nüchterlein will be very thankful for a solution.


Further reading: How Edgar Allan Poe broke about 100 ciphertexts

Linkedin: https://www.linkedin.com/groups/13501820
Facebook: https://www.facebook.com/groups/763282653806483/

Subscribe to Blog via Email

Enter your email address to subscribe to this blog and receive notifications of new posts by email.

Kommentare (21)

  1. #1 Breaker
    16. Oktober 2018

    This appears to be a round of sorts in the encoding of an alphabet that is matched to a certain starting position of the alphabet.

    My intuition comes from #1 where the number 65 is the only pair that acts as a single consonant or vowel….this more than likely being the “A”

    Starting at this numerical assignment A=65, the alphabet is then encoded multiple times over the array of numbers, with them sometimes repeating, but with the double bigrams covered up in each word by allowing the cryptographer to select different numerals for the same letters.

    The lowest and highest numbers in the first cipher are the keys to finding the placement as well. 12 is the lowest and 65 the highest.

    Start at A=65 and then count round to the B = 12, continuing the alphabets until the Z, and add the comma, then carry over the alphabet to the next Z and add a period. They occupy every space perfectly using two alphabets, and then offsetting them.

    That’s why this will not produce a logical alphabet from the normal methods

    It uses the layout of the alphabets to create the appearance of a repeating pattern and breaking the alphabets with two punctuation marks it offsets the counts even further…..simple and mathematical process of fitting two alphabets together into one numerical code.

    https://i63.tinypic.com/i1wx12.jpg

    Since the numbers appear out of synch directly with the alphabets, it naturally would divert attention AWAY FROM this technique.

    There is also a spliced section where there appears to be some form of coding mixing letters and numbers. This may operate a second layer of a wheel cipher as well.

  2. #2 Breaker
    16. Oktober 2018

    Note the timing of the date of the ciphers occurring during the post Shakespeare Era of Francis Bacon’s Studies, as well as the hints in the photo on twitter of the first numbers being ascribed to appear like a 5+3

    As in the page of the Folios, #53, that says “I know a trick with two”

    This appears to hold a dual meaning, or that a secondary complication is added to the results of these alphabets to create the final message…..

    Numerical assignments in the first as shown in the link, but then possibly a ROT or another guide will be found that reveals the second message within the ciphers.

    Note that one of the number seets almost appears to take the shape of a Y as well

    Perhaps this is the second shift position of the letter Y on the Alphabet from the first layout.

    What a cool system to use and very easy to understand in its complications through the use of the stacked letters that act as descriptors.

    Love it Klaus…..you made my century.

  3. #3 Thomas Ernst
    Atlantis
    16. Oktober 2018

    A variant of my „zifra picolominea”, for crying out loud. Yes. more material! Can’t figure out the kinks in it, yet.

  4. #4 Thomas
    16. Oktober 2018

    Solved II and V (no word separation due to wordpress)

    II
    D I E GE =
    F A S S E T E L U N E =
    B U R G U N D B R U N S =
    W E I G I S C H E R E =
    S O L U T I O N A L L E
    D I N G E O H N E
    U O R B E W U S T T E S
    H E R N E R T Z B I =
    S C H O F S Z U B R E
    M E N N I C H T E R
    F O L G T

    V
    F A U O R A B I L I A P R O =
    P O N I R T U N D S O W O I
    Z U R I N T E R P O S I T I O N
    A L S A S S I T E N T Z GE =
    M A C H E T W E R D E N D O R =
    F T E E S M O C H T E N
    A B E R D A H I N I E G E N
    W I E I C H S U O N W E I =
    T E N U E R M E R K T H A N 39
    U N D A N D E R E P O S T N =
    L A T A GE S C H E H E N
    W I E I C H D A N D U R C K
    D I I D R I T T E H A N D
    GE F R A G E T W O R D E N O B A U C H
    60 C O N S E N T I R E N M O C H T E N
    D A S 65 D I E S T A D T R O S T O K
    U O N D E M H E R Z O GE N I N U 02
    G E GE N D I E S U M M A F E L
    D E S D A M I T S E L B I G E R H E R
    Z O G D E M K O N I G U N D D E S =
    S E N

  5. #5 Thomas
    16. Oktober 2018

    II.
    DIE GEFASSETE LUNEBURG UND BRUNSWEIGISCHE RESOLUTION ALLE DINGE OHNE UORBEWUSST TES HERN ERTZBISCHOFS ZU BREMEN NICHT ERFOLGT

    V:
    FAUVORABILIA PROPONIRT UND SOWOI ZUR INTERPOSITION ALS ASSITENZ GEMACHET WERDEN DORFTE ES MOCHTEN ABER DAHINIEGEN WIE ICHS UON WEITEN UERMERKT HAND UND ANDERE POSTNLATA GESCHEHEN WIE ICH DAN DURCK DII DRITTE HAND GEFRAGET WORDEN OB AUCH 60 CONSETIREN MOCHTEN DAS 65 DIE STADT ROSTOK UON DEM HERZOGEN IN U 2 GEGEN DIE SUMMA FELDES DAMIT SELBIGER HERZOG DEM KONIG UND DESSEN

    60 and 65 are nomenclators for names

  6. #6 Thomas
    16. Oktober 2018

    The key; may some rare letters are still missing:

    8 W
    9 Z
    12 U
    13 T
    14 S
    15 R
    17 P
    18 O
    19 A
    20 A
    21 B
    22 C
    23 D
    24 E
    25 F
    26 G
    27 H
    28 I
    29 K
    30 L
    33 D
    34 E
    35 F
    36 G
    37 H
    38 I
    39 K
    40 M
    41 L
    42 M
    43 N
    44 O
    45 P
    47 R
    48 S
    49 T
    50 N
    51 U
    55 W

    60, 65: persons
    122 (in my solution above: “U 2”: city/town (“Herzog in 122”)

  7. #7 Thomas
    16. Oktober 2018

    I:
    (this yields more keys: 31 = B, 32 = C, 54 = Z )

    (only the enciphered parts of the text capitalized; up to now I could read only a few words of the “plaintext” parts in old German script)

    SOLCHE ALLERDINGE OHNE 65 UORBEWUST NICHT MUSTE GESCHEHEN SEIN ….DIESE TRACTATEN UOM 65 ALSO URGIRET WORDNN,…
    DIESE STETE BEWEGLICHST GESCHRIEBEN SIE ZUR BESTENDIGKEIT ANGEMANET, mitt UERTROSTUNG …. DER COMMERTION …. IM R. REICH …. GNADE ZU ERWEIEN, damit solche SCHREIBEN von dem MAGISTRAT der BURGERSCHAFT UORGEZEI

  8. #8 Thomas
    16. Oktober 2018

    III.

    DAS SICH SOLCHE DER ZEIT UND UND UMMSTENDEN NACH ACCOMODIREN WERDE

  9. #9 Thomas
    16. Oktober 2018

    IV.

    HABE AUCH AUS DEN MIT HERN PENTZEN GEPFLOGENEN DISCURSEN ENTNEMEN KONNEN DAS

  10. #10 Thomas
    16. Oktober 2018

    As to “Herrn Pentz”: https://www.30jaehrigerkrieg.de/pentz-pentz-bentz-penz-cuno-ulrich-von-kuno-ulrik/

    I suppose that “122” stands for “Friedland”, since in 1628 the emperor had given Mecklenburg including Rostock to Wallenstein, the “Herzog von Friedland”, who kept it until the Swedes captured it.

    It remains to find out who was “60” and 65″.

  11. #11 Narga
    16. Oktober 2018

    Fantastic job, Thomas! Congratulations!

  12. #12 Dario
    Italy
    16. Oktober 2018

    Yes Thomas, that was quick indeed! Kudos!

  13. #13 Dampier
    16. Oktober 2018

    That’s exciting.

    over 100 encrypted pages

    I hope, they will all be translated now – and I hope we can read about it somewhere : ]

  14. #14 Marc
    16. Oktober 2018

    Klasse Thomas, wie hast du das so schnell hinbekommen ?

  15. #15 Thomas
    16. Oktober 2018

    V. had some typos, and I overlooked the last lines. Corrected version:

    V.
    FAUVORABILIA PROPONIRT UND SOWOI ZUR INTERPOSITION ALS ASSITENTZ GEMACHET WERDEN DORFTE. ES MOCHTEN ABER DAHINIEGEN WIE ICHS UON WEITEN UERMERKT HAND UND ANDERE POSTULATA GESCHEHEN. WIE ICH DAN DURCH DIE DRITTE HAND GEFRAGET WORDEN OB AUCH 60 CONSENTIREN MOCHTEN DAS 65 DIE STADT ROSTOK UON DEM HERZOGEN IN 112 GEGEN DIE SUMMA GELDES DAMIT SELBIGER HERZOG DEM KONIG UND DESSEN SCHWESTER UERHAFTET; ERB: UND EIGENTHUMLICH EINGEREUMBT, UND GEGEN DIE 122 UIER EMBTER, WELCHE GEGEN DIESE SCHULD UERUNTERPFANDET WURDE: WORBEI(?) 65 FERNER

    Now I doubt that 112 means “Friedland”. Any ideas? what 60, 65 and 122 could mean?

    @ Marc:

    1. transscript, 2. frequency count and guesses concerning the most frequent letters in German 3. trial and error with the “finden” and “ersetzen” functions in Word

  16. #16 Thomas
    16. Oktober 2018

    The link in # 10 was misleading: “Herr Pentz” was Christian von Pentz, son-in-law and advisor to Christian IV., king of Denmark. Christian of Denmark was the archbishop of Bremen (1635 – 1648) who is mentioned in excerpt II. (Erzbischof von Bremen). Pentz was also King Christian’s ambassador to the court in Vienna.
    Although the excerpts don’t name the sender, I suppose it was Graf von Tattenbach-Reinstein, the Habsburg ambassador to Braunschweig who informed the emperor in Vienna about the events and negotiations in Northern Germany in 1644 (https://www.30jaehrigerkrieg.de/tattenbach-reinstein-wilhelm-leopold-graf-von-2/). But maybe Herr Nüchterlein can provide hints as to the sender?

  17. #17 Breaker
    16. Oktober 2018

    Thomas did you have an explanation for the numerous misspelled words in the translation of your alphabet.

    I tried to run it through piece by piece and there were a ton of words that were not recognized by the translators

    Possibly a series of nulls, but just saying isn’t that a lot of mistakes for a cipher that has been made in a professional manner?

    I would think it would reveal a perfect speaking plaintext that didn’t have the mistakes added?

    Was there a reason for the misspellings?

  18. #18 Thomas
    16. Oktober 2018

    @Breaker
    400 years ago the German language had a nowadays outdated vocabulary and a different spelling due to which translation software can’t recognize a lot of words.

  19. #19 Klaus Schmeh
    16. Oktober 2018

    @Thomas: Congratulations, really great job! I’m absolutely impressed! I’m sure, Peter Nüchterlein will be overwhelmed, too.

  20. #20 Thomas Ernst
    Atlantis
    17. Oktober 2018

    DIE Lösung einer historischen Chiffre des Jahres 2018! In Anbetracht solch einer Leistung kann man eigentlich nur tot umfallen … Herzliche Glückwünsche!

  21. #21 Peter Nüchterlein
    Wernigerode
    18. Oktober 2018

    Also dem Übersetzer erst einmal meinen Respekt und Dank, ich habe den Zahlencode mal in eigene Anwendung gebracht und siehe da, es lassen sich auch andere Code´s übersetzen. Demnächst gibt es dann über den Moderator noch eine andere Chiffrierung aus diesem Aktenconvolut.