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Can you solve this challenge based on an 1877 cipher device?

Von / 29. November 2020 / 9 Kommentare
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In 1877, US engineer Frank S. Baldwin received a patent for a cipher disk. Blog reader Matthias Brüstle has encrypted two plaintexts with this method. Can a reader break these?

Deutsche Version des Artikels (Beta)

Among the books in my shelf I consult the most often is Jack Levine’s United States Cryptographic Patents, 1861-1989. This work lists thousands of patents related to encryption technology, all of which are today available on Google Patents. Among the patents from Levine’s book I have covered on this blog is Marie Vahjen’s device for secret writing.

There is still plenty to discover, as only small parts of the crypto patent history have been researched so far.

 

Frank S. Baldwin’s cipher

A patent in Levine’s book I didn’t know until recently is a cipher disk invented by a certain Frank S. Baldwin from St. Louis, Missouri. When I searched for this name on the internet, I was quite surprised. Apparently, Frank Stephen Baldwin (1838-1925) …

Source: Wikimedia Commons

… was a notable US engineer who invented a pinwheel calculator in 1875. According to Wikipedia, he started the design of an improved machine in 1905 and was able to finalize it with the help of Jay R. Monroe who eventually bought the exclusive rights to the machine and started the Monroe Calculating Machine Company to manufacture it.

Baldwin filed his cipher disk patent in 1877. Blog reader Matthias Brüstle thankfully made me aware of it.

Source: Patent

Apparently, this invention wasn’t successful and didn’t play a major role in Baldwin’s career.

Badwin’s disk consists of three character rings. On the two outer ones (ring B and ring C), the alphabet from A to Z is inscribed; they are not movable. The inner ring bears the numbers from 0 to 25; it can be turned. The pointer in the middle is not relevant. Note that there is a hook connected to ring C between 0 and 25.

The following excerpt from the patent explains how Baldwin’s cipher works. The explanation is based on the key H 1 3 7 9 14 and the plaintext READ THIS:

Source: Patent

If the plaintext is longer than the key, the latter is repeated.

The method implemented by Baldwin’s device is similar to an autokey Vigenère cipher. Each encryption step is dependent on the previously encrypted plaintext. The key consists of a letter used as a starting vector as well as a series of numbers used as offsets.

A Perl script Matthias Brüstle created is available here.

Matthias provided me the following example encryption:

Key: C 4 13 2

Plaintext: NOW AND THEN AUGUSTA THOUGHT OF HER PRIZEFIGHTER

Ciphertext: OJOIROWEUJSLXFURGAKZGSNORWHIRGCHDWRBBWXKT

 

Challenge

Matthias Brüstle has created two challenges based on Baldwin’s cipher disk. The plaintext language is English. The key was generated at random.

Challenge 1 (the key consists of a letter and six numbers):

ILVGPMTSSYTRFXFPGWYVDCLVEKWCUTQUYWDYRGTIPWPPCMQ
OBXLJDZBLQWVUDMAKZXRSEZWAXRNMAYLDPTZBYEQFICMWLI
QUYSFZNBLARCAXCUQPQFRJLQEJWZJSTMMYLSJFLEANLZYFJMZ
VVCFQLVIIKPZQVOERLUYPZJRPSJGUETEDDSIVDPHSYGPTYZRBFE
HOZYESOENZWHDFPSA

Challenge 2 (the key consists of a letter and up to ten numbers):

UEEVPCHHHRNVUDCEXIEMJHWGQYYTJWLNSAXBPJNFXYNVDVK
ZKWAOCOMFVXSBJNDUGPGBWGCOSOUEIXVDKYCBWTMNIGVA
KRRKYNVCHLIKSVHPLGXHRQSWEOIEMRKYZLRTAPNJWOXTRAQP
KMQSCVLPNDHDHTLJZSCMRHMUUKWFKPABVOETLUCJNBPDBR
QBVKOSEXQEPTBDXZXIDVRZKJEHXXZONIDUSRCYDRWETRC

Can a reader break these ciphertexts?

I want to thank Matthias Brüstle for informing me about this patent and for providing me these great challenges.


Further reading: Can a reader help to correctly decrypt this WW2 message?

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Kommentare (9)

  1. #1 Narga
    30. November 2020

    I think the key used to encrypt the example above the challenge is “H 4 13 2” instead of “C 4 13 2”. But I’m not 100% sure.

  2. #2 ShadowWolf
    Stepping is Key Dependent
    30. November 2020

    First, this is just a stepping version of the Beaufort cipher with an offset. The plaintext has no effect on the key sequence as per an autokey. The offset 6 is built into the wheel and the steps are the variable. This makes it a little more complex than a normal Vigenere or Beaufort which are identical except for the arrangement of the tables.

    If it goes unsolved to long, I do have a solution. It is easier to explain than to actually do on paper unless I figure out a better way.

  3. #3 Armin
    30. November 2020

    Challenge 1:
    There is an error in the description: the key consists of a letter and five numbers, not 6.

    Key: H 9 19 2 4 25

    Plaintext: HERE IS A YOUNG MARRIED WOMAN IN OUR VILLAGE WHO IS GENERALLY REGARDED AS A THOROUGHLY IMMORAL PERSON IN FACT AS THE REPREHENSIBLE INFLUENCE OF THE TOWN IT IS NOT THAT SHE SMOKES OR DRINGS COCKTAILS EVEN IN RURAL NEW ENGLAND WE HAVE LONG AGO GOT USED TO THAT

    Challenge 2:

    Key: H 19 13 23 5 2 19 17 5 3

    Plaintext: HER DEPRAVITY CONSISTS IN THE FACT THAT SEVERAL OLDER WOMEN HAVE DROPPED IN TO SEE HER AT ELEVEN O CLOCK IN THE MORNING AND FOUND HER SEATED IN AN ARMCHAIR IN THE GARDEN READING THE DAILY PAPER THE INFERENCE IS OBVIOUS THAT SHE MUST BE NEGLECTING HER DUTIES AS A HOUSEWIFE AND MOTHER

    I solved the challenges by first chosing a plaintext trigram for the first 3 letters of the ciphertext. That determines all plaintext trigrams at positions 0 (modulo key length). Then I searched for the maximum trigram score sum that a chosen first trigram would generate. Sliding the trigram window over the ciphertext then revealed the entire plaintext.

    @ShadowWolf: Did you solve it with pen and paper only?

  4. #4 Norbert
    30. November 2020

    I agree with Narga: Matthias’ example encryption has a key starting with H, not C.

    And imho, Klaus description is not correct: ring C is the only movable one. The inner ring D with the numbers and the attached hook is fixed, as is ring B (note the screws depicted in Fig. 1). Ring C is moved by inserting a “stile” (=stylus) in the hole opposed to the appropriate number and turning it counter-clockwise until the stylus is stopped by the hook.

    The index serves as a memory aid, as it points to the last used number of the key.

  5. #5 Klaus Schmeh
    30. November 2020

    @all: Sorry for the mistakes.
    @Armin: Your solutions are correct! Congratulations!

  6. #6 Klaus Schmeh
    30. November 2020

    Here’s solution 1 (according to Matthias):
    N 9 19 2 4 25
    thereisayoungmarriedwomaninourvillage
    whoisgenerallyregardedasathoroughlyim
    moralpersoninfactasthereprehensibleinfl
    uenceofthetownitisnotthatshesmokesord
    ringscocktailseveninruralnewenglandwe
    havelongagogotusedtothat

  7. #7 Klaus Schmeh
    30. November 2020

    Here’s solution 2 (according to Matthias):
    T 19 13 23 5 2 19 17 5 3
    herdepravityconsistsinthefactthatse
    veralolderwomenhavedroppedintos
    eeheratelevenoclockinthemorninga
    ndfoundherseatedinanarmchairinth
    egardenreadingthedailypapertheinf
    erenceisobviousthatshemustbeneg
    lectingherdutiesasahousewifeandm
    other

  8. #8 ShadowWolf
    Solving Method
    2. Dezember 2020

    As already noted, I simplified the cipher to just using a Beaufort alphabet table. As most here should know, the Beaufort is periodic cipher that is functionally identical to the Vigenere. That was the first clue. Next, note how the key steps. Each time you encrypt a letter, they key advances modulo 26. There is a really cool mathematical trick you can use. Once you know the period, you can then search for the stepping distance for each key repeat. This is different from the period. Ignoring the first key use, add the stepping distance mod 26 to each key repeat and increase the stepping distance mod 26 for each repeat. This turns the cipher into a simple Beaufort (Vigenere) and you can solve it just as easily.

    Say you have the numeric key 4,6,11,15, 3. The stepping distance is 17. This is what the key sequence looks like without actually encrypting anything. If I encrypted a message, you wouldn’t normally see this sequence. You might want a mono-spaced font to look at this part.

    L n n n n
    4 6 11 15 3 6 11 15 3 6 11 15 3 6 11 15 3
    0 1 2 3 0 1 2 3 0 1 2 3 0 1 2 3
    4 10 21 10 13 19 4 19 22 2 13 2 5 11
    17 8 25
    4 10 21 10 4 10 21 10 4 10 21 10 4 10 21 10

    The first line of numbers is they actual key sequence the user would perform on the wheel. The second line is just a key reference. The third line of numbers is the actual key sequence created. The fourth line is the stepping distance. The last line is the new Beaufort cipher key after applying the stepping distance changes. This makes it no harder to solve than a Vigenere.

    Worst case on finding the period, you can search and see what results you get. It is little more than a common Vigenere solver program with an added stepping correction search modification (0-25) and a slight mathematical change for the Beaufort alphabet table. You could do it with pencil and paper this way, but it would be a lot of work.

  9. #9 Klaus Schmeh
    2. Dezember 2020

    @ShadowWolf: Thanks for describing this method. Very interesting!